Date: 12 JUL 1980 1342-EDT
From: ALAN at MIT-MC (Alan Bawden)
To: CUBE-LOVERS at MIT-MC

Welcome to the cube-lovers mailing list.  I don't know what we will be
talking about, but another mailing list cannot hurt (too much).

Mail to cube-lovers@mc or cube-hackers@mc whatever strikes your fancy.

Date: 15 July 1980 01:41-EDT
From: Jef Poskanzer <AQE at MIT-MC>
Subject: Complaints about :CUBE program.
To: CUBE-HACKERS at MIT-MC


This is a real-neat program, BUT:

In solving the cube, the description it gives for the action it is
about to perform can be ambiguous, I think.  For example, can't "Turn
TOP center to BOTTOM" be performed in two non-equivalent ways?

I don't know what the display format is on color LISP machines, but on
my terminal is sucks.  How about something like this:

		+--------+
	       / YL YL YL \
	      /		   \
	     /  YL  YL  YL  \
	    /		     \
	   /   YL   YL	 YL   \
	 ,-+------------------+-.
      ,-'  |		      |  `-.
   ,-'  GR |   RD   RD	 RD   | BL  `-.
+-'  GR    |		      |    BL  `-+----------+
| GR	   |		      |       BL | OR OR OR |
|	   |		      |		 |	    |
| GR GR GR |   RD   RD	 RD   | BL BL BL | OR OR OR |
|	   |		      |		 |	    |
| GR	   |		      |       BL | OR OR OR |
+-.  GR    |		      |    BL  ,-+----------+
   `-.  GR |   RD   RD	 RD   | BL  ,-'
      `-.  |		      |  ,-'
	 `-+------------------+-'
	   \   WH   WH	 WH   /
	    \		     /
	     \  WH  WH  WH  /
	      \		   /
	       \ WH WH WH /
		+--------+

Maybe add FRONT, BACK, LEFT, RIGHT, TOP, BOTTOM labels; maybe compress
it vertically so it fits on 24-line screens; maybe three-char abbrevs
instead of two; maybe the back face should be shown reversed (i.e.
from the inside of the cube looking out) to facilitate mental
manipulations; the basic format is better, though, don't you agree?

---
Jef

Date: 15 JUL 1980 1413-EDT
From: ALAN at MIT-MC (Alan Bawden)
To: CUBE-HACKERS at MIT-MC

Since we have this mailing list I am tempted to put on record an
interesting property of the cube that some of you might not be aware
of:

The cube has a degree of freedom that is rarely considered.  The six
center faces which are the pivots about which rotations are performed
do not always return to their original orentation.  In other words, if
you were to paint little arrows on the center square of each face of a
solved cube, randomized the cube, and then solved it, you might well
find that the arrows no longer pointed in the same directions as they
did before.  (Now you have to get them back.  And you thought you had
already solved the damn thing!)

This extended cube problem has been solved independently twice to my
knowledge (not by me, I cheated and learned someone else's solution).
The property was first shown to me by Spencer Love.  Does anybody know
if anybody else ever discovered it independently?

Without considering this property we already knew that there were
43252003274489856000 = (8! * 3^8 * 12! * 2^12)/12 different
rearrangements of the cube (all those numbers are obvious except the
12 in the denominator).  Considering this property raises that number
to 88580102706155225088000 = (8! * 3^8 * 12! * 2^12 * 4^6)/24 (the 24
being just as hard to explain as the 12 was).


CMB@MIT-ML 07/15/80 15:06:58
To: cube-lovers at MIT-MC
I had known about the center faces being turnable for a while, and have a cube
that I have marked up so I could work on solving the center face problem.  In
general, I solve the cube except for the center faces (since that was what I
already knew how to do), and then have three transforms:  one that will turn
any center face 180 degrees and leave everything else unchanged, one that will
turn one center face 90 in one direction, and an adjacent center face 90 in
the other direction, and the last turns two adjacent center faces 90 in the
same direction.   My transforms are rather long and rep(it)*ous.


Date: 15 JUL 1980 2158-EDT
From: ALAN at MIT-MC (Alan Bawden)
To: CMB at MIT-ML
CC: CUBE-HACKERS at MIT-MC

The last two transforms you describe sound similar to the two I
learned.  Mine are also rather repititous.  Perhaps it is the case
that the two configurations are very distant using the obvious metric:
smallest number of twists from one to the other.

I wonder if anyone knows very much about the nature of that metric
anyway?  I understand that it is known that no two points are more
than 94 (or is it 93?) twists apart (disregarding the extended
problem).  I don't know if that number is actually attained, or if it
is only the currently known upper bound based on the best algorithm.
(Or perhaps there isn't an algorithm that good yet, just a proof of
the fact.)

I believe that you and ACW and I once did the math to show that
whatever that longest distance is, it has to be greater than something
around 30, and for the extended cube problem it must be even bigger
than that, so since I can do the transformations you speek of in about
28 (I think) moves, those must not be most distant points.

Date:	  15 July 1980 2245-edt
From:	  Bernard S. Greenberg	     <Greenberg at MIT-Multics>
Subject:  Cube minima
To:	  CUBE-LOVERS at MIT-MC

I believe the 94 number comes from Singmaster's book. UNFORTUNATELY
Singmaster doesnt seem to know what the word "canonical" means,
and 180-degree twists count as single moves "too" in his religion.
This makes his number kind of worthless.

ED@MIT-ML 07/15/80 23:21:07 Re: Singmeister who?
To: cube-lovers at MIT-MC
I've never heard of Singmaster's Book.  Is it in the Old Testament
or the New?  Is it prophetic, historical or lackadaisical?

Date:  16 July 1980 09:35 edt
From:  Greenberg.Multics at MIT-Multics
Subject:  Singmaster
To:  cube-lovers at MIT-MC

David Singmaster is a prof at an English university whose
name escapes me, who publishes a 40-page pamphlet on cubing
for about $5.00.  It took me months to get; I have it for xeroxing
in my posession.  It contains many transforms, including many
wierd solve-the-edge-cubes-first solutions.  As I have
pointed out, his noncanonical move counting is a problem.

ACW@MIT-AI 07/16/80 15:08:09 Re: Cubespeak
To: CUBE-HACKERS at MIT-MC
I would like to see some talk about a good language for
describing cube manipulations.  I know Bernie has one that
he swears by, but I would rather not see the discussion
START with everybody giving their favorite cubespraak...
this can get confusing, dogmatic, counterproductive, nobody
listens, etc.  So let's keep the "My language is better than
your language" to a minimum at first, and see what desiderata
people consider as basic.

My first contribution is:  I think that turning the cube over,
rotating it, etc., without performing any twists -- that is,
all the things that you could do just as well to a solid block of
wood -- SHOULD be considered manipulations in their own right.
This includes performing a mirror reversal.  Why?  So that
manipulations that only differ in starting orientation will not
have representations that look completely different.

   ---Wechsler


Date: 16 JUL 1980 2051-EDT
From: RP at MIT-MC (Richard Pavelle)
To: CUBE-LOVERS at MIT-MC

A simple transformation with a pretty resulting design:
Let CS = center slice which faces you and is
parallel to the axis of your body.
Then CS up 90, rotate cube 90 in either direction so
that the CS turns in the direcion of rotation.
Performing these two operations 3 more times gives
the desired result.

Date: 16 Jul 1980 2130-PDT (Wednesday)
From: Lauren at UCLA-SECURITY (Lauren Weinstein)
Subject: confusion
To: CUBE-HACKERS at MC

I am relatively new to the whole idea of cubing, in fact I do not yet
even own a cube.  I am a littel confused as to what the state of the
art is in solving an actual randomized Rubik's Cube.  Are these
algorithms so complex that they are only usable with computer aid, or at
least pencil and paper?  Or are they such that, handed a random cube,
you can just sit there and (given sufficient time) put it right?

--Lauren--
-------


Date: 17 July 1980 01:22-EDT
From: Alan Bawden <ALAN at MIT-MC>
Subject: confusion
To: Lauren at UCLA-SECURITY
cc: CUBE-HACKERS at MIT-MC

Given a randomized cube it can take about 5 to 10 minutes to set it
straight with no aid whatsoever.  Pencil and paper or a computer can
be a great deal of help when one is first learning to solve the cube
(I used both), but I know of no one who uses such aids once they have
learned how.  A method of solution that required some computational
aid to perform (some hairy calculation based on the current
configuration of the cube, resulting in a single 259 twist sequence
that brings it immediately back to the solved state) is not
inconceivable, but most people have solutions composed of short, easily
comprehended steps.

Can anyone tell me who this Rubik character is?  His name appears to
be attached to the new American version of what some of us once knew
as the "Hungarian Cube".  Is Rubik the Hungarian who invented it?  Has
he done anything else?  I heard this rumor that there was a 4x4x4 cube
out there somewhere, anybody else heard about it?

Date: 17 JUL 1980 0846-EDT
From: JURGEN at MIT-MC (Jon David Callas)
To: CUBE-LOVERS at MIT-MC

Yeah, sounds great, I always loved linear Alg.
I'm not really COMPLETELY sure what you're doing, but
count me in. It sounds like I might learn something.
			Thanx,
				Jurgen@mc


Date:  17 July 1980 09:40 edt
From:  Greenberg.Multics at MIT-Multics
Subject:  Re: confusion
To:  Alan Bawden <ALAN at MIT-MC>
cc:  Lauren at UCLA-Security, CUBE-HACKERS at MIT-MC
In-Reply-To:  Message of 17 July 1980 01:23 edt from Alan Bawden

For the record, Erno^H" Rubik is the hungarian teacher of
architecture and design at some Budapest equivalent_high_school
, who invented the Cube. Apparently he has a solution,
but it is not a particularly good one.  I would also like to add to
ALAN's note that given my method or ALAN's method, I have never
seen either take longer than 5 minutes (I promise clocked
solutions in under 4 and have taken much less) and Singmaster
has heard of solutions in 2 minutes, but I find this difficult to believe.

I think I can also assert that most of the certified Cubemeisters
did NOT use a computer in solving it, although it can be of great help.

For the record for the purposes of this list, my algorithm
is implemented in the :CUBE program that runs on all ITS's.

Date: 17 JUL 1980 1042-EDT
From: RP at MIT-MC (Richard Pavelle)
Subject: confusion but simplicity
To: Lauren at UCLA-SECURITY
CC: CUBE-HACKERS at MIT-MC

I do not use a computer but I found it useful in keeping track of
moves and testing various transformations. I can fix the cube
now in 5 minutes or less. In fact, I have won cubes at
stores by challenging the owner to a
freeby if I can do it under 10 or 15 minutes. My transformations are
very easy to remember because after one face is complete I need only 4
to fix the cube. The transformations are positioning the corners, toppling
them, fixing edges and then toppling edges. These are not designed for
speed but for simplicity. I have taught some people to solve the cube this
way with a few hours of practice. To do it in 5 minutes, however, requires
a few more transformations.

Date: 17 JUL 1980 1114-EDT
From: ACW at MIT-MC (Allan C. Wechsler)
Subject:  Short Introductory Speech
To: CUBE-LOVERS at MIT-MC


Nobody else has made this kind of flame yet, so I
guess I will.

Welcome to CUBE-LOVERS.  We are devotees of a certain
mathematical puzzle variously called the Hungarian
Cube, the Magic Cube, and Rubik's Cube.  It is a hard
puzzle.  Very intelligent people often take weeks to
learn to solve it.  Once they have learned, though, they
can solve it in a few minutes.

The puzzle embodies mathematical sophistication and
mechanical ingenuity in a pleasing and intriguing
synthesis.  I have forgotten the Hungarian inventor's
name, but we should learn it: this person deserves
our profound respect.

For those who have not yet become Cube Solvers: you
can only solve the Cube for the first time ONCE.  After
that, although there are a lot of problems to think
about, the initial challenge is gone.  So, in the words
of Mr. Duffey:  SPOILER WARNING!  SPOILER WARNING!
Messages to this list will often deal with particular
solution techniques.  If you haven't solved the cube yet,
and want to do it on your own, reading these messages
may ruin your fun.

If there is any demand, I am willing to hack up a short
introduction to Group Theory for Cubans.  Group Theory
gives us a mathematical language for talking about the
cube.

Also, if anyone out there knows Hungarian, there are some
pamphlets we need to translate.  Cubans should inform
everyone on the list of any written material they know of,
so that we can compile a bibliography.

Happy cubing,

  ---Wechsler

Date: 17 JUL 1980 1420-EDT
From: RP at MIT-MC (Richard Pavelle)
Subject: the cross design
To: CUBE-LOVERS at MIT-MC

Does anyone know a transformation (or series of) which will produce
the following pattern on each face where X and O are two different colors.

	|X O X|
	|O O O|
	|X O X|

I think this may not be possible which brings me the question which
is how one can say whether a particular arrangement is or is
not possible.

Date:  17 July 1980 14:38 edt
From:  Greenberg.Multics at MIT-Multics
Subject:  Re: the cross design
To:  RP at MIT-MC (Richard Pavelle)
cc:  CUBE-LOVERS at MIT-MC
In-Reply-To:  Message of 17 July 1980 14:20 edt from Richard Pavelle

There are two such sets of patterns known, one with three
sets of paired colors (the Christman cross) and one
with two triplets of colors (the Plumer cross).
The Plummer cross is achieved by two orthogonal applications
of the transform to the Christman cross.  The transforms are
fairly long and hairy, and I hesitate a little before attempting
to describe them, but will if people want.

Date: 17 JUL 1980 1635-EDT
From: ALAN at MIT-MC (Alan Bawden)
Subject: the cross design
To: RP at MIT-MC
CC: CUBE-HACKERS at MIT-MC

Yes, the cross design is possible.  I will let BSG try and describe it
if people want.  It is fairly straightfoward to tell if a position is
reachable.  I am thinking of onlining the proof that there are 12
equivalence classes of cubes (I think I have a fairly simplified
version), and if I do it should contain an algorithm to tell if a
position is reachable.

Yekta@MC (Sent by ___117) 07/17/80 16:45:00 Re: Checker board pattern...
To: cube-lovers at MIT-MC
Is the pattern in which evry face of the cube is a checkerboard
of two colors possible??  (  I have not played with the cube at all,
so my apologies if it is too trivial to get... )


Date:  17 July 1980 17:33 edt
From:  Greenberg.Multics at MIT-Multics
Subject:  Re: Checker board pattern...
To:  Yekta at MIT-MC (Sent by ___117)
cc:  cube-lovers at MIT-MC
In-Reply-To:  Message of 17 July 1980 16:45 edt from Sent by ___117

Yes. This is absolutely trivial.  Rotate each face 180. Si x total
onehunred-eighty degree twists.

Date: 17 Jul 1980 1358-PDT (Thursday)
From: Mike at UCLA-SECURITY (Michael Urban)
Subject: Confusion
To: lauren
CC: cube-hackers at mit-mc

   Actually, dealing with the cube is a learn-as-you-go experience.
The appeal of the cube, which makes it superior to Soma, or "Instant
Insanity", etc, is that you actually have to analyze what's going on
in order to approach the solution.  For example, I began by learning
how to put one face right; this required certain simple transformations
that were useful later.  As you go, you develop your own heuristics
for moving the faces you need around without messing up what you've
done so far.  I can do the whole thing from an arbitrary starting
position in around 20 minutes; I'm still not very adept at moving
corners around.
   Even after you have solved it, there are still things to do
with the cube, including improving your personal algorithm, as well
as creating nifty patterns, etc.  A Worthy Toy.
   According to "Games & Puzzles" magazine, Rubik is the Hungarian
fellow that devised this evil little time-stealer.  The
article also impies a 2x2 version is in the works, which is even harder
to understand mechanically than the 3x3 version.
   How the heck IS the thing put together, anyhow?
	Mike
-------


Date:  17 July 1980 17:38 edt
From:  Greenberg.Multics at MIT-Multics
Subject:  Re: the cross design
To:  ALAN at MIT-MC (Alan Bawden)
cc:  RP at MIT-MC, CUBE-HACKERS at MIT-MC
In-Reply-To:  Message of 17 July 1980 16:35 edt from Alan Bawden

There is a wonderful alogorithm for ANY pattern, which constitutes
an existence proof.  Given that you know how to "solve cubes", you
can achieve a given
pattern if it is achievable simply by "solving" to that position,
which may in fact be faster than some set of arcane transforms.
Before the "neat" algorithms for the Cruces Plummeri et Christmani
were discovered, they were achieved by cubemeisters only by
"solving" to that position, lambda-binding the target state
(lessee, this guy wants to go here, etc.) to the desired pattern.
I will burn some neurocomputrons tonight to describe the
algorithms   for the Crosses.

Incidentallly, if you try to solve for some pattern and come to
a roadblock of the form "this cant come here because its here"
(we need two of these cubies ( a local jargon
for the little cubes)), or a parity/trinity argument, you have
proven that you can't achieve it.

Date:  17 July 1980 17:33 edt
From:  Greenberg.Multics at MIT-Multics
Subject:  Re: Checker board pattern...
To:  Yekta at MIT-MC (Sent by ___117)
cc:  cube-lovers at MIT-MC
In-Reply-To:  Message of 17 July 1980 16:45 edt from Sent by ___117

Yes. This is absolutely trivial.  Rotate each face 180. Si x total
onehunred-eighty degree twists.

Date:  17 July 1980 21:18 edt
From:  Greenberg.Multics at MIT-Multics
Subject:  Re: Confusion
To:  Mike at UCLA-Security (Michael Urban)
cc:  lauren at UCLA-Security, cube-hackers at MIT-MC
In-Reply-To:  Message of 17 July 1980 18:12 edt from Michael Urban

The easiest way to see how the thing is put together is to take it apart.
If you turn the "top" plane 45 degrees, you can pry out the
top edge cube (any of the 4) fairly easily, and it becomes clear
how the devilish thing is put together.

In three lines or less, all the center-of-face pieces are on a
6-armed, solid, rigid cross. They can spin, but that's it.
The EDGE pieces (i.e., not the corneres) have a rectangular
projecction from their non-visible edge which gets them stuck
behind whichever two center-of-face pieces they're currently
visiting.  While not rotating, it's stuck behind two. while
rotating it (a edge piece) is stuc behind the one
on its rotating face, and a groove that all edge pieces
have on their sides, in this case on the edge pieces of the
next plane back.  The corner pieces have little cube-like
projections on their invisible corners that basically
wedge in behind the edge pieces, which are stuck as desribed
above. No magnets, wires, universal
joints or rubber bands.

IF youshould decide to take one apart, be SURE to put it
together SOLVED to ensure solvability.

Date:	  17 July 1980 2228-edt
From:	  Bernard S. Greenberg	     <Greenberg at MIT-Multics>
Subject:  The Higher Crosses
To:	  CUBE-LOVERS at MIT-MC

For the amusement of the experienced cubemeister, and the
education of those desirous of learning the Art, I have here
produced a Description of the Methods, as I have used,
for the production of the Cruces Plummeri & Christmani.
These are the elegant configigurations of lid Crosses
of which we spoke earlier today.

**********************************************************************
I herein describe the algorithms for construction of
the higher (Plummer, Christman) crosses from a solved position.
From an unsolved position, it is faster to "solve" directly to
the desired configuration; by following the steps below,
the eager cubist may learn exactly what these configurations are.

Of the words and phrases I use:

I call the faces front, back, right, left, top, bottom. A face
has 9 cubies, viz., 4 corner cubies, 4 edge cubies, and its
center cubies.  Separating 2 opposite faces, is  a "center
slice", being of 4 center cubies and 4 edge cubies. As I
hold the cube, I call three center slices: floor-parallel,
body-parallel, body-slicing. For instance, the body-parallel
and body-slicing centerslices meet in the front face.

I name "double-swap" the transform which is performed as follows:
    Double-swap (front, back)		      ;parameter-faces
      Turn body-slicing centerslice 180.
      Turn bottom face 180.
      Turn body-slicing centerslice 180.
      Turn bottom face 180.

Observe well what it has done, viz. swapped the two cubies of the
turned centerslice on the front with those of the back.  You
will use it as needed during the following shenanigans:

----------------------------------------------------------------------
To achieve Christman's (DPC at MIT-MC) Cross, the simpler of the two:

Rotate the body-slicing centerslice 180.  Rotate the floor-parallel
centerslice 90 either way (your choice).
Stare hard at what you have.  The CORNER CUBIES and CENTER CUBIES
are in their final position for the Crux Christmani; all further
hacking will be simply to move the EDGE CUBIES, IN PAIRS, into
place.  To achieve ANY Crux Plummeri or Crux Christmani configuration,
learn how to do the initial rotations (see below for the CP)
so that you get the center cubies to corners you want, and hack from
there.

I will now describe the edge-cubie moves for the CC given that
the centerslices have been aligned to orient the center cubies
as needed:

  Among the six faces you now have, find one of the two that
have a solid stripe between two sides of the same color, i.e.,
     x y x
     x y x
     x y x
and align it like so, so that the stripe is vertical, and this
face is the front.

Note that the edge  cubies of the y y y stripe want to be exchanged
with the two x-showing edge cubies, i.e.,
     x x x
     y y y
     x x x
(Remember that the goal is x y x/y y y/x y x)
You can tell that they want to be inthe horiz. positions by their
non-showing faces, which you will observe match the center-cubies
on the right and left sides.
To do this:
   1.	   Perform doubleswap on front-back.
   2.	   Rotate the FRONT so that when you do (3), the two cubies
	   we just moved to the back will come to such place so that
	   when we undo this step (see 4), they will be in the right
	   place. This will be either 90 deg. left or right.
   3.	   Perform doubleswap on front-back.
   4.	   Undo step 2, i.e., turn FRONT 90 deg the "other" way.
Whehter you blew (2) or not, you will now find you have
 (x x x/y y y/x x x) on front. If you understood 2 and DIDNT
blow it, you will have the sides of the y y edge cubies matching
the side centers (if you blew it, doubleswaps on the side faces
can fix you up).

You will see the floor-parallel centerslice begin to form a band.
We will now finish that band.  The two appropriate cubies (to go
in the two rear positions of the floor-parallel centerslice are
now on the front plane, the x x cubies of the last step. Note
that a simple doubleswap on front-back would move them to the
back face, but the WRONG two places on the back face. Easy.
So, turn the back face 90 degrees and do the doubleswap,
and unturn the back.  Choose which 90 such that these two
cubies wind up in the right place.

You will now find you have solid bands and solid crosses
galore. The front and back should have solid crosses, and the
floor-parallel slice should now be a solid band.
Look at the top of the cube.  Make it the front.
Orient it so that it is (a b a/c c c/a b a). Do a front-back
doubleswap, and now look at the remaining face pair we
havent been thinking about. Do the appropriate doubleswap on
them to get solid crosses, and then you should have the Crux Christmani.
Study well what you have: three pairs of alternated crosses.

----------------------------------------------------------------------


The Crux Plummeri (after DCP at MIT-MC who first came up with it,
altho by solving-to) is exactly equal to doing the entire
above transformation twice, at 90 degrees.  The following, however,
is a direct route from solved that is more intuitive.

Take the cube, turn the body-slicing centerslice up 90 deg.
Turn the floor-parallel centerslice 90 deg clockwise as seen from
the top. Note well the configuration of corners versus centers;
it is the final one. Note that you will have two triplets
of trebly-interleaved colors: that is the characteristic of the CP.

Look at the TOP or BOTTOM. Let's say the TOP. Make it the front.
Orient it so that you see
    x y x
    x y x
    x y x

Only the top or bottom look like this; this is what you have to
remember to look for after aligning centers to taste.
We're gonna rotate the y y y band into the horiz position.
Do this exactly as for the CC above, producing
   (x x x/ y y y/x x x)
Next goal is again to complete the solid band of the floor-parallel
centerslice by doubleswapping front/back so that
the x x edgecubies,w hich would complete that band, go to
the back. Of course, we must temporarily rotate the
back during this doubleswap, so that they go to the side
positions ofthe back when swapped. Do so, completing the
solid color-band of the floor-parallel slice.

Now consider the top and bottom. You note that exactly one appropriate
doubleswap between top and bottom would give us solid
crosses on both. Do it.


Take what had been the top just now, and call that the front.
Note that there are solid crosses on front and back, and the
body-parallel plane is correct and complete.
Think about the front: it looks like
   a b a
   b b b
   a b a

Although it looks right fromt the front, the two vertical b-edge
cubies want to be the two horizontal b-edge cubies, as a cursory
inspectionof the top bottom and sides of the cube will show.
This is true of the back, as well.

Tofix up the FRONT do this:

     1. Doubleswap front/back
     2. Rotate the FRONT (temporarily) 90 degrees sothat the two
	vertical b-edge cubies are gonna come to the right place,
     3. And doubleswap front/back
     4. Undo 2.
     5. Doubleswap front/back.

Now you see all is right save the back. It wants the same
thing done to it.  Do it for it;  Do this same thing
just doNe in the last 5 steps for the back (viewing it as the
temporary front).

It is done. Consider it.

An exquisite variant ont he CP is obtained by taking on of the
trebly-bound sides and rotating the centers via the well-known
center-cubie rotating algorithm. As the centers are rotated
left or right, either a sextuple checkerboard or a stunning
triply-rotated canon of centers , edges, and corners appears.
The checkerboard is amusing insofar as it appears to a
novice cubist to be the Pons Asinorum 6tuple checkerboard
made by 6 twists (described earlier today), but cannot be
fixed (solved, or produced) without the consummate hair
of the CP that only true cubemeisters can execute.

The application of the Pons Asinorum checkerboard transform
to the CP (as well as the CC) produces interesting and
suprising results.

----------------------------------------------------------------------
The Higher Crosses are fascinating insofar as they appear to be
very simple edge-cube hacks, but are in fact quite "far"
from home; the CP being exactly twice as "hairy" (far)
as the CC (discovered by ALAN) is in itself a source of
wonderment.

Date: 17 JUL 1980 2245-EDT
From: ALAN at MIT-MC (Alan Bawden)
Subject: The Higher Crosses
To: Greenberg at MIT-MULTICS
CC: CUBE-HACKERS at MIT-MC

    Date:     17 July 1980 2228-edt
    From:     Bernard S. Greenberg	 <Greenberg at MIT-Multics>

    ...

    The Higher Crosses are fascinating insofar as they appear to be
    very simple edge-cube hacks, but are in fact quite "far"
    from home; the CP being exactly twice as "hairy" (far)
    as the CC (discovered by ALAN) is in itself a source of
    wonderment.

I wish I could really say something was "exactly twice as far" from
home as something else.  Unfortunately, as I complained before, the
metric by which one measures the distance of one configuration from
another is not well enough understood to be able to make claims like
this.  It might well be that the CP is less than twice as far as the
CC, all we can really be sure is that it is not any further than that.

Date:  17 July 1980 22:52 edt
From:  Greenberg.Multics at MIT-Multics
Subject:  Postscript to above
To:  cube-hackers at MIT-MC

I should have noted in the above flamage
(btw, will all of those of us who are paid by our respective employers
to hack this lunacy please let me know at once)
that there is room for opitimization and lookahead in the final
doubleswaps in correcting the top/bottom, and the
doubleswap preceding it, but I do not do this, so that
I might better keep track of what I am doing.

Date:  17 July 1980 22:54 edt
From:  Greenberg.Multics at MIT-Multics
Subject:  Bug in above
To:  cube-hackers at MIT-MC

In the terminOlogy section, note that the
body-slcing and FLOOR-PARALLEL centerslices meet in the front face,
not the body-parallel and body-slicing as given.

ED@MIT-AI 07/18/80 00:12:53 Re: Patterns, designs &c.
To: cube-lovers at MIT-MC
To jump the gun slightly on the group-theoretic explanation,
any sequence of rotations of any number of faces can be
thought of as an atomic "transformation" for the purposes of
group theory.  One of the precepts of this theory is that any
such transformation, repeated often enough, will return the
cube to the original state.  For instance, given the transform
"rotate top ccw 90", 4 iterations suffice to return the cube
to the original state.
Mike Speciner, a fellow Camexian, claims that no transformation
can be created that requires more than 216 (=6^3) iterations to
return to the virgin state.  (He doesn't yet have a cube, but
has been stealing his daughter's blocks and modeling the cube
with them.)
Where does this number come from, and is it true?
I have been playing with various transforms, and have found
at least one reasonably trivial one that requires the 216
iterations:  rotate a face 90, then turn the cube 90 and repeat.
The transform here is 4 twists in a band around the axis of
cube rotation.  The patterns generated in the process are interesting,
too, though none of them are as unique as the cruciform or center-face patterns.

Date: 18 JUL 1980 0205-EDT
From: ALAN at MIT-MC (Alan Bawden)
Subject: Patterns, designs &c.
To: ED at MIT-AI
CC: CUBE-HACKERS at MIT-MC

Er, perhaps I don't understand the move you describe, but in any case,
by my calculations it would take 1260 repetitions to come back home
doing that one (1260=2^2*3^2*5*11).  It is certainly the case that
1260 > 216 so that number must be wrong (I could be miscalculating,
but I don't see how).  1260 is rather similar in appearance to 216,
perhaps you spazzed somehow?

Date: 18 JUL 1980 0351-EDT
From: ALAN at MIT-MC (Alan Bawden)
Subject: 1260
To: ED at MIT-MC
CC: CUBE-HACKERS at MIT-MC

I now have a proof that no element of the cube group can be of order
greater than 1260.  Since you have so thoughfully provided me with an
element of order 1260, I must conclude that this element is indeed the
maximum, as you claimed (but where did the number 216=6^3 come from?).

My proof contains no nice derivation of the number 1260, you will be
dissapointed to see where it comes from, it is just all that is left
after a number of cases have been eliminated.

Perhaps someone can devise a "nice" proof of this fact.  Is this fact
in the literature?  (Bernie?)

Date: 18 JUL 1980 0932-EDT
From: RP at MIT-MC (Richard Pavelle)
To: CUBE-LOVERS at MIT-MC
CC: ZIM at MIT-MC

The file which contains all cube-lovers mail is ALAN;CUBE MAIL on MC.

ACW@MIT-AI 07/19/80 01:42:56 Re:   Where to get them in the Boston Area, Cube Language.
To: CUBE-LOVERS at MIT-MC

"Games People Play" on Mass Ave, near the fork where Mt. Auburn
St. branches off, carred them the last I knew.  According to
rumor, the domestic product ("Rubik's Cube"), selling for
about $7, is mechanically inferior to the Hungarian import,
costing $15.  I don't know how everyone feels about money,
but to me, not having to fight with the thing would be worth
the extra $8.

Bernie's explanation of how to achieve the Plummer and Christman
Crosses is a prime example of why we need a cube language.

Since no one has proposed anything, I will jump into the fray.

Center the cube at the origin of a 3-d coordinate system.  The
axes of the coordinate system protrude from the centers of the faces.

Make a hitch-hiker's gesture with your right hand and point the thumb
along the X axis.  Imagine rotating the WHOLE CUBE one quarter turn
in the direction your curled fingers are pointing.  I call this operation
"I".  (The X axis is the horizontal axis that does not skewer you.)

If the cube was lying on a table, "I" would roll it toward you.

Now point up, along the Y axis, with your thumb.  A quarter-rotation
in the direction your curled fingers point is the operation I call "J".

The Z axis goes right through your belly.  A quarter turn around it
I call "K".  Actually, K=IIIJI.

To simplify things a little, we define I'=III, J'=JJJ, and K'=KKK.
In general, M' is M done backwards.  If we call the do-nothing
manipulation "1", then MM'=M'M=1.

For my own nefarious reasons I define "H" as the operation (unachievable
in real life) of REFLECTING the cube right-to-left through the YZ plane.
We note H'=H.

Twisting the front (Z=1) face 90 degrees counter-clockwise I call
"T".

One more piece of notation: For any manipulations M and N, I write
M'NM as N[M], reading this as if N were a function: "N of M".

Note M[1]=M.

Examples:

T[I] means "Twist the top face"
T[II]=T[JJ] means "Twist the back face"
T[I'] means "Twist the bottom face"
T'[J] means "Twist the left face CLOCKWISE"
T[I] T'[I'] J' means "Rotate the floor-parallel center-slice a
	quarter turn counter-clockwise as seen from above."

Note that (MN)'=N'M', and (N[M])'=N'[M].

Also notice that (MN)[P] = M[P] N[P].

To do the Pons Asinorum checkerboard:

Set Q= (TT)[J] (TT)[ZJ] "Half turn body-slicing center-slice."

Then the Pons is Q Q[J] Q[K].

Does anybody see what I'm getting at or am I a lone, mad
genius?

Set Q= T[J] T[J'].

Then (Q Q[J])^3 is quite pretty.

  ---Wechsler


Date: 19 JUL 1980 0249-EDT
From: ALAN at MIT-MC (Alan Bawden)
Subject: 1260 vs. 2520
To: ED at MIT-MC, CUBE-HACKERS at MIT-MC

Sorry folks, there was a slight error in my last message about the
maximum order of any element in the cube group.  To understand why,
let us examine the transformation that ED described.  I like to label
the cube this way:

	  A0 AE A1
	  AB AX AD
	  A3 AC A2

B0 BA B3  C3 CA C2  D2 DA D1  E1 EA E0
BE BX BC  CB CX CD  DC DX DE  ED EX EB
B7 BF B4  C4 CF C5  D5 DF D6  E6 EF E7

	  F4 FC F5
	  FB FX FD
	  F7 FE F6

(I won't bother to describe the features of this labeling, and I will
presume that the correspondence between this unfolded labeling and an
actual 3D cube is obvious.)

If I understand ED's directions this is the way the cube looks after
applying his transformation:

	  A3 AB A0
	  AC AX AE
	  D1 DA D6

C3 CA E1  A1 AD F6  E6 EA E0  B0 BA B3
CB CX BF  FB DX DE  ED EX EB  BE BX BC
C2 EF E7  B7 CF B4  C4 DC C5  D5 DF D2

	  F7 FC F4
	  FE FX CD
	  A2 FD F5

Note that the center faces BX, CX, DX and EX have all moved.  My proof
that the maximum order of any element is 1260 assumes that these
center faces remain fixed.  There is nothing wrong with that
assumption, it simply relects the intuition that picking a cube up and
putting it down again on a different face doesn't change the
configuration at all.  If we decide that the orentation of the cube
DOES matter, then we get transformations like ED's here.  The group
we are dealing with is also 24 times larger than before.  And my proof
now shows that no element has an order greater than 2520 (twice as big).

Could it be that ED's transformation is not actually a maximal one?
Could there be one with higher order?  Or can my proof be tightened up
some, so that even in the larger group the maximum order remains the
same?

To answer these questions (hopefully) I present a transformation
that I claim has order 2520:

	  D2 AB A0
	  DC AX AE
	  D5 AD A1

C2 CD C5  F5 DA D1  E1 EA E0  B0 BA A2
CA CX CF  FC DX DE  ED EX EB  BE BX AC
C3 CB C4  F4 DF D6  E6 EF E7  B7 BF A3

	  B4 FD F6
	  BC FX FE
	  B3 FB F7

(This transformation, like ED's, is not a member of the usual group,
but the larger one where the center faces are allowed to move.)

It is easy to check that this permutation has order 2520, the real
question is whether you can get to this configuration from a solved
cube.  I haven't actually tried to do that, but I am fairly certain
that it is possible.  If anyone would like to try it, and report to me
their success or failure, I would be very grateful.

Date: 19 JUL 1980 0346-EDT
From: ALAN at MIT-MC (Alan Bawden)
Subject: OOPS
To: ACW at MIT-AI
CC: CUBE-LOVERS at MIT-MC

Let me suggest the following corrections to your cube language
message.  If any of these are not actual errors, but are actually
misunderstandings on my part, then I apologize for causing undue
confusion, but I thought it would be helpfull to people trying to
understand you system to have these corrected as soom as possible.

    ACW@MIT-AI 07/19/80 01:42:56
    ...

    One more piece of notation: For any manipulations M and N, I write
    M'NM as N[M], reading this as if N were a function: "N of M".

I think you must intend N[M] to mean M N M' .
    ...

    T[I] T'[I'] J' means "Rotate the floor-parallel center-slice a
	quarter turn counter-clockwise as seen from above."

T[I] and T'[I'] don't touch the floor-parallel center-slice at all and
J' rotates it clockwise (as seen from above), my guess is that the
description should simply read: "Rotate the floor-parallel
center-slice a quarter turn clockwise as seen from above."
    ...

    Set Q= (TT)[J] (TT)[ZJ] "Half turn body-slicing center-slice."

This must be a typo or some discarded notation.  My guess is that this
should read Q= (TT)[J] (TT)[J'].  I don't think that the description
here ("Half turn body-slicing center-slice") fits, but it works in
producing the pattern anyway.  If you had said Q= (TT)[J] (TT)[J'] II
then you could get the pattern and fit the description too!

Date: 19 Jul 1980 11:27 PDT
From: McKeeman.PA at PARC-MAXC
Subject: Re: Where to get them in the Boston Area, Cube Language.
In-reply-to: Your message of 07/19/80 01:42:56
To: ACW@MIT-AI
cc: CUBE-LOVERS at MIT-MC, Lynn.ES, Horning, Sturgis

Your proposal for a language leads me to suspect that you have not seen:

NOTES ON THE 'MAGIC CUBE'
David Singmaster
Mathematical Sciences and Computing
Polytechnic of the South Bank
London, SE1 0AA, England

which you can get by sending him $3.  The fourth printing runs 36 pages.  It is
definitely worth the money.

Cube quality:  I have handled about a dozen Hungarian import versions of the
cube with enormous variation in quality.  They were all the colortab-on-black
variety.  They sell out here in the discount stores like K-Mart for less than $9.  I
have also seen a colortab-on-white version which is not substantially different
in quality from the Hungarian versions.  My advice is to pick and choose as best
you can from the ones on the shelf, regardless of what kind you get.

As for your language proposal, I like the RLUDFB abbreviations given by
Singmaster and generally used in Europe to your T plus whole cube moves.
They mean rotate the (Right, Left, Up, Down, Front, Back) surface 90o
clockwise.  I also use rludfb for the inverses of them.  However since
(fortunately) these two notations have a null intersection, there is no
disadvantage in mixing them.  For consistency, however, I suggest ijk for
counterclockwise moves, and IJK for clockwise ones.  This does little violence to
the quaternion origins where lower case is generally used anyway.  The ' for
inverse notation is fine; in fact I frequently prefer R' to r when I write down
algorithms.  The mappings then are, substituting f for your T:

F = f'
R = F[J]
r = f[J] = R'
L = F[j]
l = f[j] = L'
U = F[i]
u = f[i] = U'
D = F[I]
d = f[I] = D'
B = F[ii]
b = f[ii] = B'

It also solves the problem that I was having of saying IJK in English since there
is no way to say it in "Singmaster".

I didn't follow your Pons Asinorum checkerboard solution because I didn't
understand the Z in Q, but in the mixed notation I think it is:

rrllffbbuudd

or, if Q = rrll;  QjQkQ

Your "quite pretty" is  (lrfb)^3 = (lr (lr)[j])^3

I propose the following grammer for RCML (Rubik Cube Manipulation
Language):

algorithm ::= definition* move*
definition ::= letter = algorithm ;
move ::= letter
   | move'
   | move^digit
   | move[algorithm]
   | (algorithm)
digit ::= 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9

where the letters RLUDFBrludfbIJKijk are predefined as noted above.

--Bill



Date:	  19 July 1980 1455-edt
From:	  Bernard S. Greenberg	     <Greenberg at MIT-Multics>
Subject:  General remarks
To:	  CUBE-LOVERS at MIT-MC

While watching these various "linguae cubisticae" make the rounds,
I note one categoric deficiency, whose perception may indeed be
due to the idiosyncracies of my own meta-algorithms.

In specific, my algorithms (including my 200-line procedure of the
other day) include DECISIONS of the form "find a face that has
a pattern like-so" or "look around for the target cubie, and
categorize its place like so".  Now clearly, a general
solution algorithm must include some expression of such decisions,
although I find Singmaster's procedures deficient in this regard
(he tends to say "Use enough of FOO transform to get all the..."
often). However, the generation of specific patterns from another
canonical state NEED not involve "decisions", although most of my
procedures (and hence the lisp-macro language of :cube) do.
I offer that looking for certain patterns during the course of
even one of these pretty-pattern-generations leads to more
rememberable (^= memorable) algorithms than even a well-subroutinized set of
absolute moves.

On the Physical Reality of cubes:

I have seen three species of the genus Cubi, the original Hungarians
(C. Hungaricus), which is black faced,
cubes sold by Ideal (C. Americanus), the American
Black-faced cube, and the American Whiteface (C. Albus).
The last-mentioned was first shown to me by the lady in the
Cambridge cube-store, who called me at home to make me aware of
them when they first appeared (she says they are done by
some Friend of Rubik in Virginia).

C. Americanus seems a good deal lighter in weight and build than
C. Hungaricus. While C. Hungaricus takes two to three weeks of
constant use before they are loose enough for Concert performances,
C. Americanus can be turned with one hand, yea, one finger
while held by the rest of the hand WHEN NEW.  C. Americani
do not seem to bind at all; the Hungarians not only bind, but
decompose and bind more.  I confess a certain sentimental attachment
to C. Hungaricus, on which I mastered the Art. The stiff,
clean solidity of a new Hungarian is indeed reassuring, but speed
of movment and non-binding is something else.  I have not subjected
Americanus to truly extensive use, and I do not know how
it ages, but so far, they  seem to show less CHANGE per time than
the Hungarians, and are usable from the start.  I think they
will be a win.

Of the American White-faced Cube, I have little to say; C. Albus
is a curio item, and has nothing to recommend it over either of the
others.

Date:	  19 July 1980 1524-edt
From:	  Bernard S. Greenberg	     <Greenberg at MIT-Multics>
Subject:  :cube feature
To:	  CUBE-LOVERS at MIT-MC

Ok, everybody wants a way to tell :CUBE "I have a cube like
so and so, solve it".  The reason I have avoided doing this
(other than laze) is that you may specify a non-reachable
state.  (Duplicate cubies, or purple faces, etc. are easy
to check for).  I am not convinced that there is any
better or faster check for an inconsistent/unreachable
cube than trying to solve it and blowing out: lest
we find such a check (other possibility: run the program once
silently and once loudly; if the first time fails
(no, i will not make it list all cubes it cant solve)
give an error), all internal breakpoints become
user errors.

The second issue is what is the best language
to specify a configuration in? Comments?

Date: 19 JUL 1980 1548-EDT
From: ALAN at MIT-MC (Alan Bawden)
To: Greenberg at MIT-MULTICS
CC: CUBE-HACKERS at MIT-MC

    Date:     19 July 1980 1524-edt
    From:     Bernard S. Greenberg	 <Greenberg at MIT-Multics>

    ...		    I am not convinced that there is any
    better or faster check for an inconsistent/unreachable
    cube than trying to solve it and blowing out: ...

You mean I haven't convinced you of that yet?  Show me how your cube
is represented and I will write one for you.  What would be so bad
about blowing out anyway?  What happens when you hand a person an
impossible cube? his algorithm "blows out" eventually!  There is no
shame in being tricked into trying to solve a bad cube.

Date:  19 July 1980 15:53 edt
From:  Greenberg.Multics at MIT-Multics
To:  ALAN at MIT-MC (Alan Bawden)
cc:  CUBE-HACKERS at MIT-MC
In-Reply-To:  Message of 19 July 1980 15:48 edt from Alan Bawden

I certainly believe you have shown me sufficient conditions for
inconsistency, but until now I wa not aware that you claimed
that they were necessary...

Date: 19 July 1980 16:31-EDT
From: Ed Schwalenberg <ED at MIT-AI>
Subject: Nope, I guess we don't understand
To: ALAN at MIT-MC
cc: CUBE-LOVERS at MIT-MC

First of all, I was sitting here inventing a GOOD cube notation,
when McKeeman's message about his notation came in.  I begin
to fear that ACW was right, and we will all die before agreeing
on a notation.  I dislike all proposed notations so far, for the
following reasons:
ALAN's labeling of the cube for describing a POSITION is excellent;
however it is not useful for describing transforms, which are
operations and not positions.
ACW's language I find baffling, principally because it lacks
verbosity.  I would much rather have a notation like
[doubleswap] than F[X,[Y']] since it is descriptive.  An analogy
with Life is useful here; I think that the adoption of names
like "traffic lights" is far more usable in the long run than
any of [the-oscillator-of-period-two-composed-of-three-blots-placed-
orthogonally-adjacent] or	  |    .
			    ...   |    .
				  |    .
or "if you put 3 blots in a row, the center dot remains alive forever
while the two outer dots appear first horizontally then vertically".
Creating names like "The Christman Cross" is fun, and makes for
interesting wordplay, even if you don't resort to Latin.  So my
proposed language is Augmented English, which has the great feature
of being able to put in comments (a feature notably absent in the
others).

I urge people to describe the transform and its result in any
message describing a nifty transform or pattern (provided both are
known, of course!).  RP's pretty pattern may not be what I think it
is, since "pretty" is not a good description.  I propose that this
be called "swapping-centers-in-triplets" (procedural notation) or
Twelve Squares (which is not a movie by Mel Brooks, but the positional
notation).

Bernie's comment about decision-making I think is important:  it
seems to me that cubemeisters do in fact approach the problem with
a set of "subroutines", which are defined to NOT contain
conditional branches.  Doubleswap and checkerboard-all-faces are
examples of subroutines.  Conditional branching is generally simply
the matter of selecting an orientation of the cube before applying
a transformation, "setting up" the subroutine if you will.  I think
that in the case of generating patterns from a solved cube,
branches are unnecessary but helpful:  rather than say "doubleswap
then rotate the cube 90 clockwise in Z and doubleswap again"
saying "doubleswap, then doubleswap the remaining solid face" much
more clearly indicates what is going on.  (the examples above are
spurious).

I herein announce two patterns I have independently invented; I do
not know if they are elsewhere available.  The first is called
Ten Squares by analogy with Twelve Squares, since it is highly related.
Twelve Squares causes the 3 centercubies of three mutually adjacent
faces to move to an adjacent face; three iterations of "swapping-
centers-in-triplets" suffice to return to the solved state.  Ten
Squares, on the other hand, is a configuration wherein two opposite
faces are solid, while the other 2 sets of opposing faces each possess
the centercubie belonging to its opposing face.  This is created by
first swapping-centers-in-triplets, then swapping-centers-in-triplets
again, only with the cube rotated 90 degrees away from you.  Note that
this results in the final centerslice rotation of the first transform
and the first centerslice rotation of the second effectively combining
into a single 180-degree centerslice rotation.  To resolve the cube,
simply do swapping-centers-in-triplets without regard to the orientation
of the cube, then you are back to the trivially soluble Twelve Squares.

The second I call Laughter, after the use of the string \/\/ in comlinks
to signify laughter.  It leaves the top and bottom faces solid, while
causing pairs of opposing faces to have a diagonal stripe of the
opposing color.  (I propose that the color of the face opposite a given
face be called the complementary color; my cube has complementary pairs
of red-white, orange-yellow and blue-green.)  To create Laughter, select
a top-bottom pair, which will remain solid.  Rotate the left and right
sides clockwise (in "opposite" directions as viewed from the top, thus
resulting initially in the top being 3 differently-colored stripes).  I
call this "splaying".  Then rotate the cube 90 degrees while preserving
the top/bottom orientation (i.e., rotate it about the Z axis.).  Six
iterations of splay/rotate suffice.  Laughing the cube again solves it.
Laughing the cube, then Frowning it (same as laughing, only rotate the
faces anticlockwise) results in Four Crosses:  2 complementary pairs
of crosses with the top and bottom solid.

Date: 22 Jul 1980 1211-PDT (Tuesday)
From: Mike at UCLA-SECURITY (Michael Urban)
Subject: Yet another checkerboard
To: cube-lovers at mit-mc

   Nobody has mentioned (or discovered?) the fully-interlocked
checkerboard pattern obtained by applying the Pons Asinorum
transformation to the Plummer's Cross as suggested
by Greenberg during his description thereof, and then performing
a few triple-center swaps (unfortunately, I was playing rather randomly
with the 3c swaps at the time and must leave them as an "exercise for
the reader").  The six colors are interlocked in pairs of AB, BC, CD, DE,
EF, and FA on each checkerboard.  Quite mysterious.
	Mike
-------


Date: 23 Jul 1980 4:10 am PDT (Wednesday)
From: Woods at PARC-MAXC
Subject: Xerox cube-lovers
To: Cube-Lovers at MIT-MC

This message is of (possible) interest only to Xerox members of Cube-Lovers, but
there's no easy way for me to send it only to them; apologies to the rest of you.

I've created a Laurel-format message file containing the Cube-Lovers mail (from
the archive at MC) up through approximately last weekend.  The message is in
chronological order (instead of reverse chronological the way MIT's stupid mailer
defaults) and has otherwise been cleaned up slightly (e.g., to get rid of lines just
barely too wide).  If you're interested in looking through this archive, it's in
[maxc]<Woods>Cube.mail.  It'll probably be around for several days at least.

	-- Don.


Date: 23 JUL 1980 1044-EDT
From: RP at MIT-MC (Richard Pavelle)
To: CUBE-LOVERS at MIT-MC

There is a short article on the cube written by Singmaster. The
reference is Mathematical Intelligencer, Vol.2, #1, pp.29, 1979.

Date: 23 Jul 1980 1640-PDT (Wednesday)
From: Mike at UCLA-SECURITY (Michael Urban)
Subject: Clarification
To: cube-lovers at mit-mc

   My description of the checkerboard pattern was clearly
inadequate, due to haste and the fact that my cube was "borrowed"
and scrambled before I had a chance to see precisely what was
going on.
   The pattern in question is formed from Crux Plummeri by
applying the "Pons Asinorum" transformation; from the
resulting almost-checkerboard, a simple "12-squares" transformation
will provide the 6 checkerboards; it isn't hard to inspect
the almost-checkerboard to find a trio of center cubies to
rotate.
   The resulting checkerboards are a completely interlocked
set.  In the following unfolded cube, A/B means that the center
and corners are color A, and the edges are color B.

       A/B

C/A    B/D    D/E    F/C

       E/F

   I don't THINK anyone has mentioned this pattern; the checkerboard
pattern mentioned by Greenberg consists of two cycles of three colors
each, something like

      A/B

B/C   C/A    D/E   E/F

      F/D

  although I may have the handedness wrong.

	Mike
-------


Date: 23 Jul 1980 5:23 pm PDT (Wednesday)
Sender: Woods at PARC-MAXC
From: Woods at PARC-MAXC, Boyce at PARC-MAXC
Subject: 216 vs 1260 vs . . .
To: Cube-Lovers at MIT-MC

Regarding finding a sequence of twists that requires a maximal number of
repetitions to restore the original state, it is indeed true that 1260 is the
maximum, but the sequence suggested by ED (on July 18) is not it.  (He
thought its period was 216; ALAN thought it was 1260.  It's actually 315.)
This assumes that we don't consider the orientation of the center-face cubies
nor the orientation of the cube as a whole.

First off, note that the transformation given by ED was ill-specified, since he
didn't say which way to rotate the cube relative to the rotations of the faces.
In Singmaster's notation the two operations would be, say, LBRF and LFRB.
It turns out these are each period-315, though it's not immediately obvious
that the two should be at all similar.  Having done (LBRF)^1, you'll find
there are 7 edge cubies that have cycled through each other's positions, and
the other 5 edge cubies have done likewise, 5 corner cubies have rotated 120
degrees in place, and the other 3 corner cubies have cycled through each
other's positions and one of them has rotated.  Hence, if we do (LBRF) a
multiple of 9 times, the corner cubies will be back where they started, and if
we do it a multiple of 35 times, the edge cubies will be restored, so if we do
(LBRF)^315, all of the cubies will be back to the solved state.

Perhaps the reason ALAN thought this was period 1260 was because it takes
1260 twists to restore the original cube.  But since the transformation that we
are repeating is a sequence of four twists, it actually has period 315.  If you
count total twists, you can obviously get "identity" sequences with lengths
much greater than 1260.  Or perhaps ALAN was considering the transformation
to be LJ (twist left face, rotate cube about vertical axis).  This indeed has
period 1260, but it "cheats" in that we're not supposed to consider reorienting
the entire cube as a legitimate operation.  If we do, then it is again possible
to get periods greater than 1260.

To get a period-1260 transformation, we need to observe whence the period
arises.  If we do a sequence of twists, and then look at the cycles of the form
"face X moved to where face Y used to be, face Y moved to where face Z was,
. . . , face <mumble> moved to where face X was", we take the least common
multiple of the lengths of those cycles.  If we repeat the sequence of twists
that many times, the faces will moved around those cycles an integral number
of times and end up where they started; if we stop short of that many
repetitions, at least one of the cycles will be left stuck in the middle.
(Apologies to any readers who are familiar with group theory and are bored
by this verbosity.)

To get a period of 1260, we need to have a cycle of 7 edge cubies, another
cycle of 2 edge cubies where one of them is also rotated (so that the four
faces on those cubies form a single period-4 cycle), a cycle of 5 corner cubies,
and a cycle of 3 corner cubies where again one of them is rotated (forming
a face-cycle of length 9).  If, instead of the 2 edge cubies, we could get 4
edge cubies cycling with one of them rotated, that would be a period-8 cycle
and the overall transformation would have period 2520, but due to the fact
that it's impossible to swap exactly two cubies it turns out you can't get that
combination of subcycles.

Based on this analysis, it's not hard to construct a period-1260 sequence,
although this one is almost certainly not the shortest such (it's 24 twists):

	F U F^2 u r f b l r f D^2 F b R U (R^2 F^2)^3 u r B


Date: 23 Jul 1980 1757-PDT
From: DDYER
Subject: impossible cubes
To: cubr-lovers at MIT-MC
Remailed-date: 23 Jul 1980 1758-PDT
Remailed-from: Dave Dyer <DDYER at USC-ISIB>
Remailed-to: cube-lovers at MIT-MC


 According to Singmaster, there are 12 disjoint orbits (his term) of cube
positions.  The article I read doesn't give a derivation, but the basic
idea is clear.  It should be possible to characterize the orbit of a given
position as a function of the transformations that would have had to
be done on each cubie ( as if it could be moved alone ) to bring it
to its proposed position.  Given such a charaterization, one could
declare positions fed to :CUBE illegal without trying to solve the
problem.

 This relates to my thoughts on the problem of inversion of an unknown
transformation.  Suppose you are given a Cube that is N unit moves
from home, determine the correct sequence to get it there.  I think
a representation of the Cube in terms of combined tranformations on
the cubies is the place to start.
-------

Date: 24 JUL 1980 2115-EDT
From: RP at MIT-MC (Richard Pavelle)
To: CUBE-LOVERS at MIT-MC

Many of you have seen this message before. However, since the mailing
list of CUBISTS is growing rapidly I shall repeat this on occasion.
     A file of past cube mail is on ALAN;CUBE MAIL on MC.

Date: 25 JUL 1980 0845-EDT
From: ZIM at MIT-MC (Mark Zimmermann)
Subject: "blindfold" cube?
To: CUBE-HACKERS at MIT-MC

How long does it typically take for one to be able to work with only
a mental image of the cube?  (As in"blindfold" chess, soma, pentominoes....)

Date: 25 JUL 1980 0906-EDT
From: RP at MIT-MC (Richard Pavelle)
Subject: "blindfold" cube?
To: ZIM at MIT-MC
CC: CUBE-LOVERS at MIT-MC

    Date: 25 JUL 1980 0845-EDT
    From: ZIM at MIT-MC (Mark Zimmermann)

    How long does it typically take for one to be able to work with only
    a mental image of the cube?  (As in"blindfold" chess, soma, pentominoes.
I do not know of anyone who can do it blindfolded (I cannot). However,
a few days ago mine fell into a swimming pool and I did it underwater with
five gulps of air.

Date:  25 July 1980 10:53 edt
From:  Greenberg.Multics at MIT-Multics
Subject:  Re: "blindfold" cube?
To:  RP at MIT-MC (Richard Pavelle)
cc:  ZIM at MIT-MC, CUBE-LOVERS at MIT-MC
In-Reply-To:  Message of 25 July 1980 09:06 edt from Richard Pavelle

This interchange very elegantly points out the difference between
the Mathematician's view of the Cube and the
Hacker(System Programmer-type)'s view.  While an algorithm
that looked at the initial state and categorized it
("Perform the folliwng 152 steps for configuration
106xy205a (left-handed) and it will be solved") would thrill mathematicians, practical
cube-solving ALGORITHMS require iteration, conditionals,
subroutines, and other program-like techniques.  Ineed, many
non-cognoscenti have watched me solve the cube, noting that
I look at it only a very small percent of the time
(seeing which subroutine to invoke, as you will, )
but they don't know that, and conclude that I "solve cubes
basically without looking".  The whole notion of algorithmic
cubism is based upon "I will do this hairy thing and it will
have this desired effect, and I need not thinK about the
intermediate states, unless, god forbid, the phone rings, etc."
To contain a cube map and intermediate states of transforms
in one's head woud, in my mind, involve a greater skill
than blindfold chess; it is not a normal human-memory
capability, although doubtless institutiOnalized freaks
with pathological intelligence/visulization problems
(in the positive direction) exist who could conceivable do such a thing.

Date: 26 Jul 1980 1315-PDT
From: Davis at OFFICE-3
Subject: Some (perhaps well-known) Transformations
To:   cube-lovers at MIT-MC

I have just recently joined the cube-hackers mailing list, but I have read over
all the old mail.  I suspect, however, that some of what I have to say will be
well-known already; on the other hand, I suspect that some of it is new.

When I first got my cube, I had the advantage and disadvantage of working in a
vacuum.  Hence, my original solution was more complicated than it should have
been, but I did go about it another way, and discovered a few interesting facts
along the way.  Having only seen one cube, I foolishly assumed that the color
pattern was the same on all cubes, and my notation for a move was simply a
color.  A R (or Red) move meant that you look at the red face, and turn it 90
degrees courter-clockwise.  I wrote a little hack based on that notation, and
since all my notes are in that form, I will use it here.  It should be trivial
enough to convert it to the Left, Right, ... form.  For reference, here is my
color pattern in the "solved" condition:

       GGG
       GGG
       GGG

   OOO RRR YYY WWW
   OOO RRR YYY WWW
   OOO RRR YYY WWW

       BBB
       BBB
       BBB


When I purchased my cube, the store was out of all but the demonstration model,
so I never got a chance to mess around with a virgin cube, and I approached the
problem as follows:  My program would simply take moves and print out the
condition after so many moves.  It would also, given a sequence of moves, find
the order of the move as a group element.  I then tried a number of patterns,
some from my experience with the cube, and some completely at random, and found
the orders of those moves.  If the order came out to be, say, 90, then I would
do 30 and 45 moves to see what the cube looked like after that many moves.
These half- and third- way patterns were often fairly simple,  and after trying
aout about 20 or 30 likely candidates, I generated most of the primitive
transformations I used to solve the cube originally.  I also generated (as half
way positions) a large number of nice patterns, which are pretty to look at,
but not of much use for solving a cube.

In particular, laughter, mentioned by ALAN, is gotten by:

   3(OYRW)

I find 3(RYYR) a useful solving transformation, as well as 3(RYRRRYYY) and
3(RYYYRRRY) (these last two being commutators).  I have not yet gotten a look
at any of Singmaster's stuff, but these last two may be well known.  Believee
it or not, one of the transformations I used to solvee the cube originally was
45(RGWY), which has the net effect (after 180 moves) of flipping the R-Y and
G-Y side cubies in place.  I'm glad I found a better one than that.

In fact, if you haul out any book on group theory, it is interesting to find
group equations, and plug in random cube moves as the group elements to see
what happens.  Needless to say, things which group theorists find interesting
usually tend to do interesting things to the cube.

One other thing I discovered which also does interesting things to the cube is
the following transformation:  Leave the body-slicing center slice in place,
and rotate the other two sides, one toward and one away from you.  Then rotate
the top and bottom 180 degrees, and rotate the sides back.  This transformation
also gives a cube configuration which looks like something in the center-slice
group, but is not.

   -- Tom Davis

-------

Date:  26 July 1980 22:25 edt
From:  Greenberg.Multics at MIT-Multics
Subject:  Re: Some (perhaps well-known) Transformations
To:  Davis at OFFICE-3
cc:  cube-lovers at MIT-MC
In-Reply-To:  Message of 26 July 1980 16:15 edt from Davis

Out of curiosity (I am responsible for :cube), I'd be interested
in seeing your code.  Is it someplace accessible?  -Bernie Greenberg

Date: 27 Jul 1980 1004-PDT
From: Davis at OFFICE-3
Subject: Re: Some (perhaps well-known) Transformations
To:   Greenberg.Multics at MIT-MULTICS
cc:   cube-lovers at MIT-MC, DAVIS

In response to your message sent   26 July 1980 22:25 edt

Hi,
   You're welcome to look at the code, although it is hardly fit for
human consumption.  I wrote it essentially in one sitting, and stopped
as soon as I solved the cube.  There are some poorly thought out
design features, etc.  Anyway, you can get it at SAIL.  It is written
in PASSCAL, and is called MCUBE.PAS[1,TRD].  I scribbled in a few
comments this morning, so maybe there is enough information to
figure out how to use it.  Unfortunately I am at home on a typewriter
terminal, and I refuse to try to edit without a display.  Good luck.

  By the way, I have never used :cube, and have no idea exactly what
it does.  Is there some documentation associated with it somewhere?

  Thanks,

    -- Tom DAvis
-------

Date: 31 JUL 1980 1006-EDT
From: RP at MIT-MC (Richard Pavelle)
To: CUBE-LOVERS at MIT-MC

IS IT POSSIBLE?

I just got a note from Martin Gardner who plans an article about the
cube in Sci.Am. sometime. But he claims that a person names
THISTLETHWAITE has proved a restoration procedure of 50 moves
and believes he can reduce it to 41.

Gardner also says that a 2x3x3 solid is selling in Hungary, also
by Rubic.


Date:  31 July 1980 10:52 edt
From:  Greenberg.Multics at MIT-Multics
To:  RP at MIT-MC (Richard Pavelle)
cc:  CUBE-LOVERS at MIT-MC
In-Reply-To:  Message of 31 July 1980 10:06 edt from Richard Pavelle

OK, your'e on the spot..... you'd better produce these artifacts
or we're all not gonna let you log out....

Date: 31 July 1980 13:06-EDT
From: Alan Bawden <ALAN at MIT-MC>
To: RP at MIT-MC
cc: CUBE-HACKERS at MIT-MC

    Date: 31 JUL 1980 1006-EDT
    From: RP at MIT-MC (Richard Pavelle)

    IS IT POSSIBLE?

The Singmaster notes claim that Thistlethwaite had an 85 twist
algorithm in an addenda dated November 30, 1979.  I presume that since
then Thistlethwaite has continued to cube-hack, so why not 50 (or even
41)?

It should be noted that Singmaster insists on counting a 180 twist as
ONE twist, so I presume that the 85 number is measured that way.  How
is Gardner counting?

It is certainly possible.  If you count twists Singmaster's way, you
can show that there are positions at least 18 twists away from home.
There is nothing to suggest that this might not in fact be the
maximum.  So there might be room for Thistlethwaite to lower his
number all the way to 18!

(If you count 180 twists as TWO twists, then a similar proof shows
that there are positions 21 twists away from home.  In a past message
I reported that some of us had proved the existence of positions as
far away from home as around 30.  I believe that the reasoning that
led to such a high number was incorrect.  (Although I cannot prove
that there AREN'T positions that far away, I now believe that I have
never seen a proof that there ARE.))

Date: 31 Jul 1980 10:34 am PDT (Thursday)
From: Woods at PARC-MAXC
Subject: Re: 180 degree twists
In-reply-to: ALAN's message of 31 July 1980 13:06-EDT
To: CUBE-HACKERS at MIT-MC

It appears that Singmaster, Thistlethwaite, and just about all the cube
hackers I know at and around Stanford consider anything you can do
with one wrist motion to be a single twist.  Since this gives a more
accurate measure of how complicated a sequence is, I'm happy with it.
Why do you folks at MIT insist that you're right and the world is wrong?

(I admit it complicates the notation.  My own cube notation uses two chars
per twist, one being the face and the other being the direction: left-arrow
for counterclockwise, right-arrow for clockwise, down-arrow for 180 -- isn't
extended ASCII wonderful?)

	-- Don.


Date: 31 July 1980 1413-EDT (Thursday)
From: Sandeep.Johar at CMU-10A
Subject: cube group theory.
To:  cube-lovers at mit-mc
Message-ID: <31Jul80 141352 SJ61@CMU-10A>

      In one of the previous letters on the subject of cubing
there was a letter from some one offering to put together a piece
on group theory and cubing, I for one would certainly be interested in
seeing it. When I took my cube apart and tried to put it together
I wondered as to how many ways there are of putting it together
so that the cube is unsolvable, i.e. how many equivalence classes
are there which are 'wrong'.
Any one worked it out, I would but group theory is unfortunately not
my strongest subject.
				Sandeep

ACW@MIT-AI 07/31/80 14:28:38
To: cube-lovers at MIT-MC
Yeah, we have a prejudice against regarding
180-degree twists as atomic.  I understand
your feeling that a 180-degree twist is
intuitively a single operation.

Many of the cube-hackers at MIT became
interested in the mathematical aspects of
the cube, and the preference for counting
quarter twists arose from this (admittedly
rather Spartan) mathematical viewpoint.

When the cube first appeared, the mathematicians
among us instantly exclaimed, with great
delight, "Wow, here we have a group, whose
elements are possible manipulations of the
cube, and whose binary operation consists
of following one manipulation with another."
We immediately got interested in group-
theory questions like, "What is the order
of this group?"  "Does it have any interesting
subgroups?" and, in general "What kind of
object is this group?  Does understanding
it help us solve the cube better?"

There are several common ways of representing
groups.  One is as a subgroup of a permutation
group.  This doesn't really help in the case of
the Hungarian Cube, because it is too close
to what the cube really is: few new facts or
insights are revealed.  Another way is with
generators and relations.  This means, to list
a few basic group elements from which the whole
group may be derived by multiplying them together.

We soon figured out (along with hundreds of other
mathematically-inclined cube-hackers) that the
whole group of possible manipulations could be
generated from six elements: the quarter-twists
of each of the six faces.  This observation
later turned out to be crucial in calculating
the order (number of possible states) of the group.

Hence our predilection for counting quarter-turns.
The half-turns were already accounted for, and
we thought of them as two juxtaposed quarter-turns.
I guess some of us believe that the mathematical
structure of the cube group is built on quarter-
turns.  Those whose delight in the cube is not
mathematical will not agree: after all,
a half-twist is as easy as a quarter-twist
to perform.  But you will miss things like
the fact that many useful manipulations
are 8, 12, or 24 quarter-turns long.  If you
count half-turns, you get a whole spectrum
of random move counts, thus missing some
fundamental  (and as yet little-understood)
kinship between these manipulations.

Of course, if you are not interested in such
things, any measure of complexity (why not
count equator twists?  why not penalize for
counter-clockwise twists, since they are
marginally harder for right-handed people to
do?) will suffice.

   ---Wechsler


Date: 31 JUL 1980 1431-EDT
From: RP at MIT-MC (Richard Pavelle)
To: CUBE-LOVERS at MIT-MC

Sorry to spoil your day, but.....

As you may know Ideal Toy has international rights to the manufacture
and distribution of the cube. I just spoke to a fellow from Ideal
who is writing up a solution booklet for them and he told me the
following:
1) When Ideal first bought the rights (and by the way Rubik gets
   nothing- his "institute" gets it all) they discussed various
   national promotion schemes. One of them was to offer 1000K bucks
   to the first person to solve it less than 1/2 hour- that right
   1 million.
2) One of the first stores in NY to carry it put adverts in the paper
   offering $50 to anyone who could solve 1 face in 1/2 hour. They
   lost $3,000 before they ended the offer the next day.
3) A fourth grade girl from Florida solved the cube in 3 weeks and can
   now do it in under 5 minutes.
4) The 2x3x3 solid does exist and will be marketed by Ideal later this
   year. However, instead of colors it has "dominoe" like faces, whatever
   that means.

Date: 31 Jul 1980 16:44 PDT
Sender: McKeeman.PA at PARC-MAXC
Subject: Re:  The shortest solution?
In-reply-to: ALAN's message of 31 July 1980 13:06-EDT
To: Alan Bawden <ALAN at MIT-MC>
From: (Bill) McKeeman
cc: CUBE-HACKERS at MIT-MC, Lynn.ES

A lower bound on the number of twists can be derived as follows:  There are
4.3*10^19 distinct reachable arrangments of the cube.  Suppose the moves are
restricted to the (more than sufficient) set RLFBUD.  Then there are at most six
independent choices at each step and the number of reachable places is bounded
by 6^n.  That gives
	6^25 < 4.3*10^19 < 6^26,
or 26 moves as the (probably unachievable) minimum.  If all single-hand-motion
twists,  R RR RRR L LL .... DDD are allowed, there are 18 choices, giving
	18^15 < 4.3*10^19 < 18^16,
or 16 moves as a minimum.  This isn't very interesting since Singmaster has
examples 18 twists away.  If the orientation of the center squares is also
considered, then the combinatoric is 8.8*10^22, and the minima are, respectively,
30 and 19.

Date: 31 Jul 1980 5:13 pm PDT (Thursday)
From: Woods at PARC-MAXC
Subject: Re: The shortest solution?
In-reply-to: McKeeman's message of 31 Jul 1980 16:44 PDT
To: CUBE-HACKERS at MIT-MC

You can do better than that for a lower bound!  Say you consider all
single-hand-motion twists to be okay.  Then yes, there are 18 such,
but there's no point in twisting the same face twice consecutively, so
after the first twist the tree branching factor is only 15.  In fact, there's
no point in twisting a face twice if the only intervening twist was done
on the opposite face; if we look at the operations of the form "twist face
X thusly and the opposite face thusly", there are 45 initial such operations,
and 30 at each succeeding branch, but since some branches now represent
two twists and some only one twist, it's much harder to compute the
minimum depth of the tree.

	-- Don.


Date: 31 JUL 1980 2159-EDT
From: ALAN at MIT-MC (Alan Bawden)
Subject: The shortest solution?
To: McKeeman at PARC-MAXC
CC: CUBE-HACKERS at MIT-MC

    Date: 31 Jul 1980 16:44 PDT
    Sender: McKeeman.PA at PARC-MAXC
    In-reply-to: ALAN's message of 31 July 1980 13:06-EDT
    From: (Bill) McKeeman

    A lower bound on the number of twists can be derived as follows: There
    are 4.3*10^19 distinct reachable arrangments of the cube.  Suppose the
    moves are restricted to the (more than sufficient) set RLFBUD.  Then
    there are at most six independent choices at each step and the number
    of reachable places is bounded by 6^n.  That gives
	    6^25 < 4.3*10^19 < 6^26,
    or 26 moves as the (probably unachievable) minimum.

This is not an improvement on my result.  I (and the rest of the cube
hackers I know) consider a unit move to be a 90 degree twist in EITHER
direction.  You are only considering CLOCKWISE 90 degree twists.

Let me point out that if we were to count twists your way, we would
no longer have a metric.  Both the quarter twist method and Singmaster's
method result in a measure of distance that is a true metric.

    Date: 31 Jul 1980 5:13 pm PDT (Thursday)
    From: Woods at PARC-MAXC
    Subject: Re: The shortest solution?
    In-reply-to: McKeeman's message of 31 Jul 1980 16:44 PDT
    To: CUBE-HACKERS at MIT-MC

    You can do better than that for a lower bound!  Say you consider all
    single-hand-motion twists to be okay.  Then yes, there are 18 such,
    but there's no point in twisting the same face twice consecutively, so
    after the first twist the tree branching factor is only 15.  In fact, there's
    no point in twisting a face twice if the only intervening twist was done
    on the opposite face; if we look at the operations of the form "twist face
    X thusly and the opposite face thusly", there are 45 initial such operations,
    and 30 at each succeeding branch, but since some branches now represent
    two twists and some only one twist, it's much harder to compute the
    minimum depth of the tree.

	-- Don.

Singmaster's notes are aware of these factors, that is how he improves
on the 16 count computed by McKeeman to arrive at 18.  Similarly I
used the same factors to improve on the 12^n argument for quarter
twists (which gives 19 as a lower bound) to arrive at the number 21.

I also compute 24 as the quarter twist lower bound for the extended
cube (considering orentations of the center faces).

Date: 1 Aug 1980 11:53 PDT
From: Lynn.ES at PARC-MAXC
Subject: Re: cube group theory.
In-reply-to: Sandeep.Johar's message of 31 July 1980 1413-EDT (Thursday),
 <31Jul80 141352 SJ61@CMU-10A>
To: Sandeep.Johar at CMU-10A
cc: cube-lovers at mit-mc

Singmaster ("Notes on the 'Magic Cube'", p. 12 of edition 4) gives 12 classes or
"orbits" of assemblies, each one such that the other 11 cannot be reached without
disassembling the cube.  To quote Singmaster, "This also means that if we
reassemble the cube at random, there is only a 1/12 chance of being able to get
back to START," and, "It is advisable to reassemble in the starting pattern."
Singmaster's notes have several pages of relevant group theory for those
interested.
/Don Lynn

Date: 1 Aug 1980 12:20 PDT
From: Lynn.ES at PARC-MAXC
Subject: Re: Sorry to spoil your day, but.....
In-reply-to: RP's message of 31 JUL 1980 1431-EDT
To: RP at MIT-MC (Richard Pavelle)
cc: CUBE-LOVERS at MIT-MC

2)  Another TRUE story of rubiking:

The May Company department stores in the Los Angeles area held contests at
four of their stores when they started carrying the cube, a couple of months
ago.  The object was to solve one face in three minutes with their scrambled
cube.  Reward was a 50 buck gift certificate.  They allowed about 6 people every
fifteen minutes to enter for two days.  The store where I won (and my 10 year
old daughter, my wife, and three of my next door neighbors) had 21 winners.  I
assume the other three stores had similar numbers.  They also gave away lots of
cube tee shirts for "good tries".

I got my cube the night before the contest.  But it took me another month to
solve the whole thing.  I might have worked faster at it if the megabuck prize
had been offered.  But then you all would have also.

4) Singmaster (edition 4, p. 34) reports on the Rubik domino (2x3x3 cubies), and
says each cubie has spots like a domino (1 up to 9).  They are to be lined up in
numeric order.  Top and bottom (the 3x3 faces) are two different colors.
Unfortunately, Singmaster laments, magic square type patterns (all directions add
to same number) are not possible, since all edges are even, all corners odd.  The
mechanics of the device are said to be "more complicated than for the Magic
Cube."

/Don Lynn


Date: 1 Aug 1980 15:47 PDT
From: McKeeman at PARC-MAXC
Subject: Re: The shortest solution?
In-reply-to: ALAN's message of 31 JUL 1980 2159-EDT
To: ALAN at MIT-MC (Alan Bawden)
cc: McKeeman, CUBE-HACKERS at MIT-MC

Alan,

Sorry I missed your earlier words of wisdom on the subject.  Anyway, I am
interested in how you show any of these is, or is not, a metric.  Can you
forward same?

Thanks,  Bill


Date: 2 August 1980 01:55-EDT
From: Alan Bawden <ALAN at MIT-MC>
Subject:  a metric for the cube group.
To: CUBE-HACKERS at MIT-MC, McKeeman at PARC-MAXC

First off, a metric is a function (call it D) that assigns a
non-negitive number to every pair of points in some set.  This number
is to be thought off as the distance between those two points.  The
function must satisfy the following:

For all a, b and c
1) D(a,b) >= 0
2) D(a,b) = D(b,a)
3) D(a,b) = 0 if and only if a = b
4) D(a,b) + D(b,c) >= D(a,c)

(Number 4 is usually called the "triangle inequality".  It is the
constraint that most makes D act like a distance, and not something
random.)

We wish to construct a metric on the set of all attainable cube
configurations.  So from now on, those lower case letters will
represent cube configurations.

Now we have recently been talking a lot about methods of counting the
number of "twists" that it takes to perform certain manipulations of
the cube.  We have been looking for a function (call it T) that
assigns a non-negitive integer to each manipulation.  I claim that it
is obvious that any such function should satisfy the following:

For all M and N
1) T(M) >= 0
3) T(M) = 0 if and only if M = I	(I is the identity manipulation)
4) T(M) + T(N) >= T(MN)

(We adopt the convention of using upper case letters to represent
manipulations.  Also we shall use M' to denote the inverse manipulation
from M.)

Now manipulations can be applied to configurations to yeild other
configurations.  We use aM to denote the configuration resulting from
applyint the manipulation M to the configuration a.  (Note that
(aM)N = a(MN), so we may omit the parens and simply write aMN.)

Now how may we use our twist measuring function T to obtain a metric
on the configurations?  Again I think it is obvious that we wish the
relationship D(a,aN) = T(N) to be true for all configurations a, and
all manipulations N.

It is easy to show that given that D(a,aN) = T(N), metric property
number 1 is equivalent to twist measure property number 1.  Similarly
for numbers 3 and 4.  But what about metric property number 2?

Well, if T(N) = D(a,aN), and D(a,aN) = D(aN,a) (property 2!), and
a = aNN', then we have that T(N) = D(aN,aNN') = T(N').  So the missing
property of twist measures must be that T(N) = T(N').

So this means that if we agree that T(L) = 1, and we like metrics
(how can we use words like "distance" unless we have a metric?), then
T(LLL) = T(L') = T(L) = 1.  We can argue about T(LL) some other time!

Date: 2 Aug 1980 12:26 PDT
From: McKeeman at PARC-MAXC
Subject: Re: a metric for the cube group.
In-reply-to: ALAN's message of 2 August 1980 01:55-EDT (yawn)
To: Alan Bawden <ALAN at MIT-MC>
cc: CUBE-HACKERS at MIT-MC

Alan,  I enjoyed your presentation.  I am convinced that counting the RLFBUD
manipulations will not give a metric.  I do not, however, see an easy way to
compute twists T(M).  I think one gets a metric only if one takes the minimum
over some set of manipulations.  That is, take a set, AM, of atomic moves
including their inverses, let AM* be the strings of AM, and |M| be the length
of M in AM*.  Then
	D(a, b) = min |M| such that a = bM
defines a metric.  D(a,b) would sometimes be undefined if AM did not generate
the whole group.  The recent discussion on shortest solutions is in fact about the
maximum of such a T(M) for all M in some AM*.

Bill


Date: 3 August 1980 02:20-EDT
From: Alan Bawden <ALAN at MIT-MC>
Subject: more on metrics
To: McKeeman at PARC-MAXC
cc: CUBE-HACKERS at MIT-MC

Yes, it is true that the four conditions I gave for twist measures
don't guarentee that the function will behave anything like the kind
of complexity measure we are looking for.  I was only trying to show
how some of the properties you might expect of a twist measure could
be used to generate a metric, so I didn't actually need strong enough
properties to ensure reasonable twist measures.  The additional property
I have been using to assure reasonability is the following:

5) For all M, if T(M) > 1, then there exists an N such that
0 < T(N) < T(M) and T(N) + T(N'M) = T(M).

Note that N'M has the property that 0 < T(N'M) < T(M) (easy to show)
so the two manipulations N and N'M are both "simpler" than M.  We can
thus easily show that any manipulation M can be expressed as the
product of T(M) twists (where a twist is defined as a manipulation
such that T(N) = 1).

Date:  4 Aug 1980 1156-PDT
From: Dave Dyer <DDYER at USC-ISIB>
Subject: cube permutations
To: cube-lovers at MIT-MC


 I'm not quite satisfied with the numbers that have been quoted
for various cube groups.  I believe they are probably correct,
but I haven't seen anything resembling a satisfactory sketch
of a proof.

Number of Reachable Positions:

  The standard calculation runs like : we have 8 corners that can
be in all 8! arrangements, and 12 sides that can be in all 12!
arrangements, except that the total permutation must be even.
( I'll talk about orientations later)
  I can accept this figure as an upper bound, but can anyone demonstrate
that all the positions not ruled out can actually be reached? The only
way I can think of is an actual construction, which means finding
a generator for the group (for instance of corner cube positions) and
showing that its order is 8!.  This is somewhat less than elegant,
and requires an unspecified bit of magic to 'find' a generator
for the group.
  I also don't like the recourse to geometric arguments ( the corners
and sides can't be interchanged ) but I am willing to accept it.

Number of Reachable orientations:

  A similar line of reasoning is used to argue the number of reachable
orientations : the amount of twist on all corners and (independantly)
on all sides is a multiple of 360 degrees.  I haven't seen a demonstration
that all the not-forbidden orientations are actually reachable.

Finally, I haven't seen a demonstration that the orientation and permutation
subgroups are indepentant, that is, that you can get to an arbitrarily
selected location and an arbitrarily selected orientation at the same
time.  This assumption is the basis for Singmasters claim that there
are 12 orbits of cubes : ( even-spacial-permutation X even-side-orientation
X one-of-three-corner-orientation = 2 X 2 X 3 = 12 )
-------

Date: 4 August 1980 20:36-EDT
From: Alan Bawden <ALAN at MIT-MC>
Subject: cube permutations
To: DDYER at USC-ISIB
cc: CUBE-LOVERS at MIT-MC

The part of the proof that shows that you can actually reach all the
configurations in a particular equivalence class is not particularly
elegant.  Basically, you have to appeal to the details of a particular
cube solving algorithm.

For example, I have a tool that "flips" two edge cubies in place,
without desturbing anything else.  This tool shows that I can orient
the edge cubies in ANY even permutation of the edge cubie faces.  The
fact that I cannot obtain any odd permutations is a result of the fact
that a quarter twist is itself an even permutation of the edge cubie
faces.

Most people can examine their own cube solving tools and see that in
fact, they are capable of obtaining all the configurations not
forbidden by the familiar constraints.

Date:  4 Aug 1980 2204-PDT
From: Davis at OFFICE-3
Subject: Cube Permutations
To:   ddyer at USC-ISIB
cc:   cube-lovers at MIT-MC

The following two transformations are useful in demonstrating that all of the
claimed elements of an equivalence class of positions can be reached.  Using
the RLFBUP notation, the transformation RB'R'B'U'BU reverses two of the corner
cubies and leaves all other corner cubies in place (It does, however, shuffle
around the side cubies, and does some random twists to the corners).  If the
above transformation is repeated four times in a row, everything is left
exactly fixed, except that three corner cubies are each rotated one-third of a
turn.  By then performing the inverse of this operation on two of the three
corners which were turned and a new corner, it is not hard to see that any two
corners can be the only ones moved, and that they are each rotated one-third of
a turn in opposite directions.

If we look at just the corners alone, and ignore in-place rotation, since we
can exchange any adjacent pair, we can obviously get to all permutations of the
corners.  A similar argument can be made to show that all the edge cubies can
be arranged arbitrarily (permuted arbitrarily, that is).  An easy
transformation rotates three of them (among themselves) on a face, and since we
are also allowed to rotate the face, it is easy to generate a transposition of
any pair.

Unfortunately, the two operations described above are not independent.  If we
just look at the blocks and label them ignoring color, a primitive (one-quarter
turn) transformation moves four corners into four corners, and four edge cubies
into four edge cubies.  If this is viewed as a member of a permutation group,
it is obviously even (the set being permuted is all the movable cubes).  Thus,
at least half of the positions are impossible.  If we ignore the corner cubies,
and look at the colors of the edge cubies, every primitive rotation rotates the
four front colors, and the four colors around the outside, again, an even
permutation.  Since one of the above used coloring and no corner cubes, and the
other did not, there must be at least a factor of 4 impossible positions.  A
much more complicated argument shows the necessity of a factor of three in the
set of impossibles.  (Does anyone know of a simple way to see this?  I just did
the obvious thing of defining "standard" orientations of every cube in every
corner, and showed that all the primitive transformations caused the total
rotation away from standard to be a multiple of 2 pi.)

Using the transformation which flips any two in place, and the two discussed
above, it is not hard to see that the factor is at most and at least 12.

Boy,  It sure is hard to prove things on a computer.  On myy notes with
diagrams and all, this is perfectly clear, but I get confused trying to read my
online proof.  I hope that someone can make some sense out of it.

-------

Date:  6 AUG 1980 0939-EDT
From: JURGEN at MIT-MC (Jon David Callas)
Subject: Random Notes on Cubism
To: CUBE-LOVERS at MIT-MC
CC: JURGEN at MIT-MC

I am beginning to feel competant enough to jump into the fray, so ...
1)	language: I guess that RLUDFB has become the de facto lg. for cubing.
(by the way, how IS 'RLUDFB' pronounced?) Is this to include the IJK rotations
that ACW described on 7-19? They are very handy, but I know that there is a desire tokeep the lq. brief. Also what (if anything is being done with the center
slices that Bernie used while describing the higher crosses? Given RLUDFB+IJK
notation,  they are easy algorithms (subroutines?) that can be described as:

	FPC {floor-parallel center} := udK
	BPC {body-|| center}	    := fbI
	BSC {body-slicing center}   := rlJ

without IJK, they're much more complex. Perhaps this is a case for using
RLUDFB & IJK.

	2) I have been playing a little with the group of rotations, twists
 and so forth. I thought that it mighht be very wieldy until I remembered that Don Woods has given us a transform with period 1260. (Is it just my
imagination or is 1260 important? it seems like a familiar number) This means
(by la Grange's thm ) that the order of this group is some integral multiple
of 1260 (ugh!) That eliminates hand-enumerating (at least by me) it. How
much does anyone else know about this, and are thre any interesting subgroups?
It also seems that this thing might be related in some way to the dihedral gps.
but I'm not sure.

	3)Somewhat connected with (2) above, it would be nice if we compiled
"Famous Transforms I Have Known" or something to that effect. Not only would
it help in finding my desired sbgp, but it would help keep some of the
folklore out of cubing (while there is nothing wrong with folklore,
there are problems when falsities (especially VERY inobvious ones) creep
into the body of lore. Folklore also does tend to cause duplication of
effort and that most of us have probably been wasting time re-inventing
the Pons Ansinorum (which, by the way, IS 'rrllffbbuudd', isn't it?)
Since I'm suggesting this nonesense, I guess it's only fair to
volunteer to do this compendium (unless someone else is dying to).

	4) Would it be possible to modify :cube to work on a
printing tty? I really don't care if it produces large volumes of paper.
I am avoiding implementing my own form of a Cube program on my
APPLE ][ (see my notes above on duplication of effort), if I can
get away with using :cube, but at least the Apple display will be pretty.

	5) There MUST be a better way to check for an impossible cube
than solving for it. While there is no shame in solving a bad cube,
there IS a little embarassment, and there certainly is no honor in it , either.
Besides, the anti-intuitionist part of me finds solving-as-proof distasteful.

	6) Has any one taken 'Rubik's Cube ' apart? I have tried prying
the edge cubie as mentioned, and all I did is hurt my thumb. I'm afriad
to use any sort of tool, lest I break the little beast.

	7) Can we get a Cubing Bibliography, including Singmaster's articles,
and the Games & Puzzles article and anything else anyone finds? (ONce
again, since I suggested it, I'll do it if need be.)

	8) Are Singmaster's 'orbits' REAL permutation-group type orbits,
or just equivalence classes? I'm confused by his choice of words.

				-happy cubing,
					Jurgen@mc


Date: 6 August 1980 12:22-EDT
From: Alan Bawden <ALAN at MIT-MC>
Subject: Reply to Random Notes on Cubism
To: JURGEN at MIT-MC
cc: CUBE-LOVERS at MIT-MC


1) I guess the RLUDFB notation is pretty standard, but the part about
using lower case letters to represent inverses isn't.  I find it
pretty distasteful and much prefer using "'".  This has the added feature
that you can write: (RLU)' to indicate the inverse if RLU.  (I KNOW that
it is just U'L'R', but sometimes one wants to make the distinction.)

2) Yes, lots of things are known about the group and its subgroups.
Start by reading old cube-hackers mail while you wait for your copy
of Singmaster to arrive.

5) Yes, there is an easy way to tell if a cube is solveable without
actually trying it.  Here it is:

Three conditions must be true: a) The sum of the rotations of the
corner cubies must be a multiple of 360 or 2 pi (depending on how you
count!).  b) The permutation of the faces of the edge cubies must be
even.  c) The permutation of all the cubies must be even.

8) Yes, they are real orbits.

Date: 6 Aug 1980 09:50 PDT
From: Lynn.ES at PARC-MAXC
Subject: Re: Random Notes on Cubism
In-reply-to: JURGEN's message of 6 AUG 1980 0939-EDT
To: JURGEN at MIT-MC (Jon David Callas)
cc: CUBE-LOVERS at MIT-MC

I have taken my cube apart several times (several friends wanted to see its
innards), with knife, screwdriver, letter opener, etc.  It doesn't seem to hurt it.  I
assume you are referring to the "twist the top 45 degrees, then pry up on an
edge" instructions, which I used.  The springiness in the way the 6 centers are
held together seems to allow this prying without damage.

I think the compilation of transforms and bibliography is a great idea, and I will
back your candidacy for the job.  Several references of both types are buried in
the previous messages to cube-lovers.  I would favor using Singmaster's
extensions of the RLUDFB to describe the results of a transform.  Combinations of
two of the set RLUDFB specify edges, and three give the corners.  Then
statements like "this transform moves URF to DBL" (or briefly "URF => DBL")
imply not only the position within the cube, but the rotation of the cubie also.
That is, the U face of URF corner moves to the D face of position DBL.

/Don Lynn


Date: 6 Aug 1980 1131-PDT (Wednesday)
From: Mike at UCLA-SECURITY (Michael Urban)
Subject: Ideal Toys
To: cube-lovers at mc

   Has anyone learned whether Ideal will be "supporting" Rubik's Cube
with some vehicle like Parker's "Soma Addict" newsletter of some years
past?  They will certainly be missing a bet if they don't.  Just the
number of people who would pay for a solution...
 ...hmm...maybe we shouldn't suggest it to Ideal after all..
	Mike
-------


Date:	  6 August 1980 1453-edt
From:	  Bernard S. Greenberg	     <Greenberg at MIT-Multics>
Subject:  Parsing cubes
To:	  CUBE-LOVERS at MIT-MC

I had been taking my cubes apart by hand, as described.  I have
found that they INDEED seem to suffer from it, at least
the C. Hungaricus (see earlier letter, I have taken neither
of the other species apart).  They become disconnected and
loose, and turn into a clattering collection of colligenous
junk, to paraphrase the famous wizard.

A fitting solution to answering visitors' queries about the inside of
cubes lies in a retired cube (which broke due to impact with the floor,
the central cross shattered) which lies in a little box in dissociated
cubies for perusal of passers-by.  This cube which served me well in
life, now serves even in retirement, by saving me from having to take
apart others.

Date:  6 Aug 1980 1223-PDT
From: Dave Dyer <DDYER at USC-ISIB>
Subject: 'cube lovers digest'
To: cube-lovers at MIT-MC


 A U.S. Mail Cubists publication, in the grand traditon of Lifeline
and the Soma newsletter, is a good idea.  Some Cubist with time on his
hands could be in business instantly, if he could convince Martin Gardner
to plug it in his (allegededly forthcoming) Mathematical games column.
-------

Date:  6 Aug 1980 1247-PDT
From: Steve Saunders <SAUNDERS at USC-ISIB>
To: RP at MIT-MC, Cube-Hackers at MIT-MC

We ought to adopt a distinctive name for cube-devotees. Several terms have
been used in messages to this list; unfortunately, most of them are already
taken by other groups:

	a "Cuban" is a citizen of Cuba,
	a "Cubist" is an artist of a certain school.

I propose a new word for us:

	a "Cubik" (from Cube and Rubik) is interested in Rubik's Cube.

This should be taken to mean anyone strongly interested in, or addicted to
playing with, the cubes in question.

	a "Cubemeister" is a full expert in the manipulation of R'sC.

This term, also used on this list, should remain for the restricted meaning of
"one who has truly mastered the art of Rubik's Cube".

		Steve
-------

Date: 6 Aug 1980 1241-PDT (Wednesday)
From: Lauren at UCLA-SECURITY (Lauren Weinstein)
Subject: Random Notes on Cubism
In-reply-to: Your message of  6 AUG 1980 0939-EDT
To: JURGEN at MIT-MC
CC: CUBE-LOVERS at MC

I have taken my cube apart.  Simply rotate the "top" surface so that
it is offset by 45 degrees to the rest of the cube.  Then pry up one
of the edge cubies (I used a small, thinbladed screwdriver.)  The
cubie pops right up, and the rest of the cube will come apart easily.

It is really a spectacular design... it is so elegant and yet so simple.

--Lauren--
-------


Date:  6 August 1980 16:46 edt
From:  Greenberg.Multics at MIT-Multics
Subject:  Re: Random Notes on Cubism
To:  Lauren at UCLA-Security (Lauren Weinstein)
cc:  JURGEN at MIT-MC, CUBE-LOVERS at MIT-MC
In-Reply-To:  Message of 6 August 1980 16:14 edt from Lauren Weinstein

I do not feel I am being overcautious or overobvious
in warning the potential cube dismantler to only reconstruct
the cube SOLVED after dismantling, lest it be placed
(11 to 1 odds against you) in an unsolvable
orbit by accident!!!

Date:  6 August 1980 16:48 edt
From:  Greenberg.Multics at MIT-Multics
To:  Steve Saunders <SAUNDERS at USC-ISIb>
cc:  RP at MIT-MC, Cube-Hackers at MIT-MC
In-Reply-To:  Message of 6 August 1980 15:47 edt from Steve Saunders

I have generally been willing to award the designation "cubemeister"
to anyone who can reproducibly solve the cube from randomness
, in an organized fashion (i.e., some small number of minutes).
I feel it is reasonable to grant this title to those  who
have achieved this, even if their methods are highly unorthodox, or they have not
familiarized themself with any other persons cube knowledge, etc.

Date:  6 August 1980 16:52 edt
From:  Greenberg.Multics at MIT-Multics
Subject:  Re: 'cube lovers digest'
To:  Dave Dyer <DDYER at USC-ISIb>
cc:  cube-lovers at MIT-MC
In-Reply-To:  Message of 6 August 1980 15:23 edt from Dave Dyer

I am really interested in when the _ Gardner is going to get his
act together and publish THE cubing article-- each month for the last
n I have eagerly taken the cover of SciAm and been disappointed.
This is clearly THE mathematical game (siNce the inception of
SciAm, with the possible excpetion of Conway's LIFE), and
I wonder what he's waiting for.  About a week ago, I actually dreamt
that I opened the SciAm wrapper and found the Cube on the cover,
introducing a WHOLE ISSUE about it (Social implications,
Ancient Cubing, Cubing in the Soviet Union, etc...)

Date: 6 Aug 1980 1:51 pm PDT (Wednesday)
From: Woods at PARC-MAXC
Subject: Singmaster to talk at Stanford
To: Cube-Lovers at MIT-MC

For those Stanford/Xerox/nearby people who aren't on the "AFL" mailing
list (and haven't heard about this from Frances Yao, who arranged it):

------------------------------------------------------------
Date:  5 Aug 1980 1626-PDT
From: CSD.ULLMAN at SU-SCORE
Subject: AFL
To: Algorithms for Lunch Bunch: ;

The speaker on Aug. 12 will be David Singmaster.  He will talk about the
manipulation of the Hungarian magic cube.
------------------------------------------------------------

(AFL meets 12:30 Tuesday in Margaret Jacks Hall room 301.  Bring a lunch.)


Date:  6 Aug 1980 1909-PDT
From: CSD.VANDERSCHEL at SU-SCORE
Subject: Orbit Classification
To: CUBE-LOVERS at MIT-MC
cc: CSD.VANDERSCHEL at SU-SCORE

Having read the CUBE-LOVERS mail, I note a recurring theme of dismay
that there is no clearly stated procedure for deciding whether a given
configuration is reachable (or solvable) or not.  Furthermore, I share
Dyer's view that there has been no persuasive presentation of the fact
that there are precisely 12 equivalence classes of permutations under
the transformations permitted by the cube.  In this note, I intend to
clear up this situation for the benefit of the less sophisticated
readers of this material.

I should apologize at the outset for my ignorance.  I have not read
Singmaster's pamphlet, nor have I communicated with any of the
knowledgeable people around here at Stanford on the subject of cubes.
My knowledge stems entirely from my own personal efforts at solving
cubes and what I have gleaned from CUBE-LOVERS mail.  So I hope
readers will be tolerant of any lengthy explanations of well-known
cube-lore.

COMMENTS ON NOTATION

I will start by offering my two-bits-worth on notation.  It is
relevant because I will use some of it later.

I prefer to call the cubies in the centers of the faces "face cubies",
because "center" and "corner" both start with a "C".  (Also, I have
worked with other 3X3X3 cubes and tended to use "center" for the one
in the middle that you cannot see.)

I do not agree that there is any need for notation to describe
orientation of the whole cube or reorientation of it.  For discussion
purposes, you should pick one orientation of the cube (based on the
direction in which its face cubies are oriented) and leave it that
way.  All transformations can be described relative to that
orientation.

Rotating a center-slice should not be considered to be a move, since
it changes the orientation of the cube.

Since it is not the case that all cubes have the same color pattern,
no direct reference should be made to color.  Instead the exposed
faces of any cubie should be identified in terms of the direction
(LRFBUD) they will face when the cubie is in its home (or at START)
position relative to the chosen orientation of the cube.

When I was learning to work cubes, the concept of complementarity
played a critical role.  (I do not claim my methods are good. It takes
me about 20 min.)  It is certainly handy to have a way of talking
about pairs of opposing faces.  I think most would agree that the top
and bottom can be referred to as "horizontal" faces, and left and
right can be called "lateral" faces.  For front and back, I really do
not know a good word; I suggest calling them "extremal", since their
face cubies are the closest and most distant from the observer's point
of view.


ORBIT CLASSIFICATION (for the uninitiated)

Introduction -

It is possible to define some "parity" concepts that simplify stating
the characterization of the equivalence classes of cube
configurations.  The precise definitions will follow; but to start
with, we will name them:

	Edge Permutation Parity (EPP = 0 or 1)
	Edge Orientation Parity (EOP = 0 or 1)
	Corner Permutation Parity (CPP = 0 or 1)
	Corner Orientation Parity (COP = 0, 1, or 2)

A configuration is reachable if COP and EOP are zero and CPP=EPP.

More generally, setting TPP=CPP+EPP mod 2, we can define

	Total Permutation Parity (TPP = 0 or 1).

Then the Parity Vector, defined by (TPP, EOP, COP), can be used to
represent the equivalence class of a configuration.

Permutation Parities Defined -

Background: A permutation of [1,n] is considered to be even (0) or odd
(1) depending on whether an even or odd number of pair swaps is
required to restore the set to original order.  You can compute it by
counting the number of reversals in the sequence - ie., the number of
pairs, (p,q) such that p>q and p precedes q.

For the cube, you can assign a position number (1-8) for each of the
corners and also a number (1-12) for each of the edge positions.  For
each cubie, you can the speak of its home position number and its
current position.  Writing down home position numbers in order of
current position gives a permutation of natural numbers in which you
can count reversals to see if it is odd or even.  This is done without
regard to orientation.

Properties: A quarter turn is an odd permutation of the four edges
involved and also of the four corners.  Thus a quarter turn changes
both EPP and CPP but leaves TPP unchanged.  TPP is preserved by all
twists.

Edge Orientation Parity Defined -

For each edge cubie consider its Oriented Distance from Home, defined
to be the smallest number of quarter turn twists required to put it in
its home position with correct orientation.  It is no bigger than 4.
As an example, an edge cubie at home with the wrong orientation is at
an oriented distance of 3.  EOP is the sum modulo 2 of the Oriented
Distances from Home for all edge cubies.

Properties: For each edge cubie affected, a quarter turn either
increases or decreases its Oriented Distance from Home by 1.  Since 4
edge cubies are affected, the net effect must be even.  Thus all
twists preserve EOP.

Corner Orientation Parity Defined -

Looking head-on at the apex of any corner you can consider twisting it
to any one of three positions.  For any given corner cubie we define
its individual orientation parity to be the number of 120 degree
counter-clockwise twists required to bring the horizontal face of the
cubie into horizontal position relative to the whole cube.  COP is the
sum modulo 3 of the individual parities.  (Since it is three-valued,
it could be argued that COP ought not to be called a "parity", but
that fouls up the parallelism in the discussion.)

Properties: Twisting a horizontal face does not change the orientation
of any corner.  Twisting an extremal or lateral face does alter
orientations.  Consider a counter-clockwise quarter turn of either
kind of vertical face.  For each of the two corner cubies that remain
in the same horizontal face, the orientation parity decreases by 1
modulo 3.  For the other two, it increases by 1.  Again, the net
effect is zero.  COP is preserved by all twists.

Equivalence Classes -

There must be at least 12 orbits, since all transformations preserve
the Parity Vector, (TPP,EOP,COP), and it has 12 possible values.  To
show there cannot be more, you must show, given two configurations
with the same Parity Vector, that one can be transformed into the
other.  The first paragraph of the note from Davis to DDYER, dated
Aug. 4, indicates how the corners can be permuted and reoriented.  We
need to be a little more careful about the interaction between the
processes that rearrange corners and edges.  Suppose we are
considering going from a configuration A to a configuration B, and
consider the intermediate stage C at which we have permuted and
reoriented the corners to agree with the target configuration B.
First we observe that the EPP of C must agree with that of B, whether
or not A and B have the same CPP.  Thus there is an even permutation
of the edges of C that will put them in the desired target positions
of B.  This permutation can be generated by swapping appropriate pairs
of edge cubie pairs.  Consider the intermediate stage D achieved after
all edge cubies are in their desired positions.  Now the EOPs of D and
B agree, so the number of edge cubies with wrong orientation must be
even.  This can be corrected by flipping an appropriate set of edge
pairs in place.

The Extended Problem -

If the orientations of face cubies are to be considered then we
introduce

	Face Orientation Parity (FOP = 0 or 1).

It is the number of quarter turns modulo 2 required to get all the
face cubies back to starting position.  Adding this fourth component
to the Parity Vector yields 24 equivalence classes.  Existence of
appropriate transformations for moving about within equivalence
classes was indicated by ALAN and CMB in their notes of Jul. 15.

Complementary Configurations -

A configuration may be said to be complementary if every color exposed
on a face of the cube either agrees with that of the face cubie on
that cube face or comes from the parallel opposing face.  There seems
to be a fair amount of interest in such positions.  It is easy to see
for such configurations that COP and EOP are 0.  Thus the
configruation is reachable if and only if its TPP is 0.  This can be
determined easily by counting "wrong" cubie faces - ie., edge or
corner cubie faces with the color complementary to that of the face
cubie on the same cube face.  TPP is 0 if and only if this number is a
multiple of four.  (It is always even.)

If you restrict yourself to 180 degree twists, you can generate only
complementary configurations.  Thus the more interesting complementary
configurations tend to be those that require quarter turns to achieve.


A REDUCED PROBLEM ?

Suppose you lock up four of the axles and consider turning just two
adjacent faces.  This Two-Faced puzzle deals with only 15 of the
cubies.  I got the bright idea of playing with a logically jammed cube
after I had already developed a crude methodology for working cubes.
I figured that with a simpler problem I might gain some new insights.
Actually, I got confused.  In retrospect, this is not so surprising.
I think that most of us would grant that if you can work Two-Faced
cubes, you can certainly work the whole thing.  The point is that you
tend to encounter many of the same sorts of problems, but you have
fewer degrees of freedom for dealing with them.

The argument already given shows that the Two-Faced cube must have at
least 12 equivalence classes.  It is not clear to me that there are
not more in this case.  I have not discovered all the tools to show
that two members of the same equivalence class (in the regular sense)
can be transformed into one another.  Can anyone out there resolve
this issue?
-------

Date:	  7 August 1980 0930-edt
From:	  Bernard S. Greenberg	     <Greenberg at MIT-Multics>
Subject:  Singmaster Talk
To:	  CUBE-LOVERS at MIT-MC

I was wondering, given Woods' message, if anybody on either coast
would like to corral Singmaster into speaking in Massachusetts,
and if someone on the West coast would let us know what he says......

Date: 7 Aug 1980 09:49 PDT
From: McKeeman at PARC-MAXC
Subject: Proposed Glossary
To: Cube-lovers at MIT-MC

Cube  -- Any interesting cube-shaped puzzle.
To cube -- Doing anything with a Cube.
Cubology -- The science of Cubes.
Cubologist -- One who (seriously) studies Cubology.
cuber  -- One who cubes.
cubnik  -- A fanatik cuber.
cubelet  -- A (conceptually) cubic subpiece of a Cube.
cubie  --  A cute cubelet.
Cubemaster  -- An exceptional cuber or Cubologist.
Cubophobia -- A mental disease found in mates and co-workers of cubniks.
Cubomania -- A mental disease found in cubniks.
Cubotomy, Cubectomy, Cubotherapy -- Cures for Cubophobia, Cubomania.


Date:  7 Aug 1980 1246-PDT
From: Dave Dyer <DDYER at USC-ISIB>
Subject: improve your cube
To: cube-lovers at MIT-MC


 I just squirted a tiny amount of dry graphite lubricant into the cracks
of my cube, and it now works MUCH more smoothly.
-------

Date:  7 Aug 1980 1345-PDT
From: Stuart McLure Cracraft <McLure at SRI-KL>
Subject: Where to find them?
To: cube-lovers at MIT-MC

Does anyone know where I can find one of these cubes?
The old cube mail mentions May Co. and K-Mart but we don't have
any of those around here.
-------

Date:  7 Aug 1980 1438-PDT
From: CSD.VANDERSCHEL at SU-SCORE
Subject: Re: Where to find them?
To: McLure at SRI-KL, cube-lovers at MIT-MC
cc: CSD.VANDERSCHEL at SU-SCORE
In-Reply-To: Your message of 7-Aug-80 1345-PDT

A sure way, but not so fast and not so cheap (about $14), is MARKLINE
in Mass.  Call 1-800-225-8390 and use your plastic money.
-------

Date:  7 August 1980 1751-EDT (Thursday)
From: Sandeep.Johar at CMU-10A
Subject: Re: Where to find them?
To:  Stuart McLure Cracraft <McLure at SRI-KL>
CC:  cube-lovers at mit-mc
Message-ID: <07Aug80 175123 SJ61@CMU-10A>
In-Reply-To: Stuart McLure Cracraft's message of 7 Aug 80 15:45-EST

    The cube is available via mail order from:
     LOGICAL GAMES, Inc
     4509 Martinwood Dr.
     Haymarket, VA 22069

  I have not tried it, so there is no guarantee. I found an ad
in some magazine (i do not remember, but could find it if some one
wants to know the type of mag.). They ask you to send $9 for the cube,
and/or $4 for "notes on the magic cube". VA residents to add sales tax.

   Or you can do what I did, telephone every toy store in the area,
till you find one that stocks it.
- Sandeep

Date: 7 Aug 1980 2:58 pm PDT (Thursday)
From: Woods at PARC-MAXC
Subject: Re: Where to find them?
In-reply-to: McLure's message of 7 Aug 1980 1345-PDT
To: Stuart McLure Cracraft <McLure at SRI-KL>
cc: cube-lovers at MIT-MC

Also try: Stanford Bookstore and Gemco.


Date:  8 Aug 1980 1702-PDT
From: Isaacs at SRI-KL
Subject: where to buy them
To:   cube-lovers at MC

  In the Palo Alto/Stanford area, the cheapest place to buy the cubes
seems to be Toys-R-Us (in Redwood City) at $7.85. They say they sell
the dIdeal version.  Logical Games sells the US made version; they are
the little guys, and perhaps should be supported.  L.G. price is
actually $7.50, plus $1.50 postage.  Two, therefore, are 'only'
$16.50.  They were, I believe, the first U.S. importer; the owner is
Hungarian and had bought two for his children after visiting Hungary.

The children didn't like them, but he did, and he imported 1000 more and
sent them around (including 10 to Stanford Book Store).  When Ideal took
over the rights (by offering the Hungarians more money, I assume), he
started manufacturing them here - the white plastic versions.  He is also
a source for the Singmaster pamphlet, and an 'Anginvine' solution.
-------

Date:  9 Aug 1980 1610-PDT
From: CSD.VANDERSCHEL at SU-SCORE
Subject: Two-Faced Cubing
To: CUBE-LOVERS at MIT-MC

Having posed the question about Two-Faced cubes, I could not resist
thinking about it some more and I have pretty well resolved it.  If
you want to try your hand at Two-Faced Cubing, DON'T READ THIS !


Suppose we only allow twists of the Right and Back faces.

Edge Orientations: You cannot change the orientation of any edge
cubie.  So there are 2^7 different ways to orient them.

Edge Permutations: Any permutation is possible. (Must be even if
corner permutation is.)

Corner Orientations: Just like the complete cube.

Corner Permutations: The twists generate a subgroup containing only
one-sixth of the 6! possible permutations of the 6 corners.  I do not
know a simple way to see why this has to be so.  (I have a messy
argument that is no more persuasive than enumerating them.)



Solving It -

Do whatever it takes to put the two right-front corners in their home
positions (without regard to orientation).  Rotate the back face to
bring (say) the URB corner to correct position.  Of the six possible
permutations of the remaining three corners, you must have the correct
one.  (That is, assuming you started from a reachable configuration.
Otherwise, this is a way to find a unique representative of the
equivalence class of corner permuations.)

(RB')^3(R'B)^3 may be used to fix the orientations of the corners.

(RRBB)^3 may be used to permute the edges.  (You can always get the
two edge cubie pairs you want to swap in vertical opposition.)



Counting Things -

There are 6!X3^6X12!/(6X3X2) reachable configurations.

There are 6X3X2X2^7 equivalence classes.




Anyone for Three-Faced Cubing?
-------

Date: 10 Aug 1980 1709-PDT
From: Davis at OFFICE-3
Subject: A solution subroutine
To:   cube-lovers at MIT-MC

Someone in an earlier message suggested that it might be a good idea
to keep a list of various interesting move combinations.  My
particular set of algorithms for solving the cube includes one which
flips two side cubies in place, and leaves all the others fixed.  The
method I had been using to do this up until now was 26 twists long,
and I think that it was fairly standard -- find a set of moves which
flips one edge in place and randomizes the rest of the faces, then move
another side into the same position and invert it.  Others that I
talked to since then have indicated that they used the same method.

Anyway, today I found a 22 quarter-twist version which does the same thing.
I would be interested if anyone has found shorter ones; I am also
interested in seeing the various other primitive transformations that
others use to solve the cube.  Using the FBUDRL notation, here is my
sequence:

   (F'U')^3(FU)^2L'U'LR'FFL'RU'LFF

  -- Tom Davis
-------

Date: 10 AUG 1980 2154-EDT
From: MJL at MIT-MC (Matthew Jody Lecin)
To: CUBE-HACKERS at MIT-MC

In the September issue of OMNI, in the "Games" section is a 2 page
introduction to the cube, complete with a few sketches showing a
solved cube, an example of rotating one side, and an example of a really
RANDOMIZED cube.
	For those of us not familiar with the cube (I still can't find
one...sigh) this article was very enlightening.  It also includes the
address from which one can get Singmaster's "Notes on the Magic Cube".

							{Matt}

Date:	  10 August 1980 2252-edt
From:	  Bernard S. Greenberg	     <Greenberg at MIT-Multics>
Subject:  Edge-cubie flipping algorithms
To:	  Davis at OFFICE-3, CUBE-LOVERS at MIT-MC

Here is my (and my apprentice :cube's) 2-edge-cubie-flipping subroutine,
which was named the "Spratt Wrench technique" after the fellow I was
talking to while cubing when I discovered it.  It is 24 quarter turns,
but is frumiously easy to perform rapidly blindfolded.

It is impossible to express it in FDRUBL (or FBURDL or FLUDBR or
whatever we're calling Singmaster-notation; let the euphony of the
acronym be the issue).  Let F be the usual, front clockwise.  Let Q
(please dont pounce on me for introducing a notation) be rotating the
floor-parallel centerslice clockwise as viewed from heaven above.

(It is clearly not IMPOSSIBLE to express it in FLURBD or BRULFD or
whatever, but it would be like analyzing a 4-voiced composition in terms
of frequencies instead of harmonies.)

Formula:    (FQ)^4(QF)^4

The sequence (FQ)^4 is called a right-handed Spratt Wrench;
(FQ^-1)^4 is a left-handed one.  Studying the effect of both of these
(which are used non-paired in other phases of my cube solution (the
sequence given above is used only during the final cadenza)) is
worthwhile; the equivalence of the formula above to ((rh wrench)(turn
the cube over)(lh wrench)), the form in which I first used it,
is due to ALAN.

The elegant formula above is best executed by a right-handed person, who
will grasp the Cube from above with his/her left hand, and use his/her
right hand to alternately turn the FRONT (which we will allow to face
right, for ease of manipulation), and the lower layer and two layers
alternately (while holding the top) to accomplish the Q moves.  This
procedure can be done with great grace and rapidity, esp.  on the Ideal
cubes.

Date: 11 Aug 1980 09:51 PDT
From: McKeeman at PARC-MAXC
Subject: Re: Edge-cubie flipping algorithms
In-reply-to: Greenberg's message of 10 August 1980 2252-edt
To: Bernard S. Greenberg <Greenberg at MIT-Multics>
cc: Davis at OFFICE-3, CUBE-LOVERS at MIT-MC

Bernard,

Actually, using Alan's IJK convention and the macro facility I proposed, your 24
move edge-flip is:

    Q=U'DJ';  -- horizontal center slice 90o cw macro
    (FQ)^4(QF)^4
and
    (FQ')^4

You didn't introduce notation, you just weren't very formal.  You may some
trouble getting the metric people agree that it only takes 24 moves however.

As for what we call the extended Singmaster notation, (1) I do not know that it
originated with him, (2) I do not know that he cares, (3) it could be called
RubikSong, CubeTalk, TwistSpeak, etc.  More to the point, who bothers to name
the notation of calculus, or group theory, or tensors.  Once standarditis sets in, a
notation just gets used, not named.  For time being, I nominate RubikSong,
giving a bow to both known sources of our pleasures.

Bill


Date:  11 August 1980 15:17 edt
From:  Greenberg.Multics at MIT-Multics
Subject:  Re: Edge-cubie flipping algorithms
To:  McKeeman at PARC-Maxc
cc:  Davis at OFFICE-3, CUBE-LOVERS at MIT-MC
In-Reply-To:  Message of 11 August 1980 12:51 edt from McKeeman

Yes, indeed, you can express Q as U'DJ', that's legit.  I can't see any way
you can count (FQ)^4 as more than 24 quarter turns, for thats all it is.
In the eyes of the people who count turns, not quarter turns, the
QQ in the middle (4 quarter turns) can be expressed as U^2D^2,
where U^2 and D^2 each count as one, which could be thus contributory
to counting the formula as 22 turns, but surely no more than 24 by
any kind of metric.
By the way, my description of the left-handed wrench in the
annotations was in error, it should have been (F'Q)^4, not (FQ')^4.
(The formula (FQ)^4(QF)^4 is correct as stands).

Date: 12 Aug 1980 0823-PDT
From: Davis at OFFICE-3
Subject: The Spratt Wrench
To:   greenberg at MIT-MULTICS
cc:   cube-lovers at MIT-MC

Thanks for sharing the wrench with us.  It is a truly wonderful tool.
Another question I had had is now answered.  It was basically:  Once I
get near the end of a solution, I often have more than one set of
edge flips to do.  Is there some scheme which will easily allow me to
do more than one.  After staring at how the wrench works it is clear
that in many cases, with one or two preliminary flips, or flips between
the applications of the two halves of the wrench, wonderous things can
be accomplished.

Also, having worked on my own for quite awhile, and being more of a
mathematician than a computer scientist, I decided that a standard
rotation was CCW, and most of my algorithms favor turns in that direction.
In converting the wrench to my own particular quirks (I can't turn things
CW very well anymore), I discovered that essentially any combination of
F and F', and of Q and Q' work just fine.  In other words, if "f" stands
for F or F' and "q" stands for Q or Q', then (fq)^4 (qf)^4 is a wrench.
(Better leave f and q fixed through the whole transformation, however.)

This next comment is off the subject, but I had been meaning to ask it
for some time.  I have one of the white cubes, and after reading the
message some time ago about underwater cubing, my cube "accidentally"
fell into a pool too.  Well, the white ones float, which I found very
annoying.  Does anyone know if the specific gravity of the hungarian
or Ideal cubes is greater than 1?  I'm also not so sure that I would
recommend that others try to teach their cubes to swim -- mine now
has a disturbing squeak as it turns -- I fear that something has rusted
inside.
Tom Davis
-------

Date: 12 Aug 1980 15:10 PDT
From: McKeeman at PARC-MAXC
Subject: Singmaster's talk at Stanford
To: Cube-lovers at MIT-MC

David talked at ICME (Int. Conf. Math. Ed.) in Berkeley on 8/11 about using the
cube to teach group theory.  He talked to some Rubniks at Stanford this noon.
Among the tidbits:

Notes on Rubik's 'Magic Cube', Fifth Edition, Preliminary Version, 75 pgs., $5.
Including: A Detailed Step-By-Step Solution, Thistlewaite's Best Algorithm (52
moves), Conway's Monoswop, Rubik's Duotwist and much more.   Write:
David Singmaster, Polytechnic of the South Bank, London, SE1 0AA, UK.

He brought a 3x3x2 domino version, and a 2x2x2 Stanford homebrew which is
apparently nearly identical to a Japanese patent showed up.  The 2x2x2 is
conceptually just a pasting of big overlapping corners on the standard 3x3x3
version although the one we saw was nicely machined in brass and some kind of
ivory-like material.

Singmaster counts double twists, e.g., R^2, as single moves.  He doesn't see much
use for the IJK whole-cube moves.

The Thistlewaite algorithm goes from subgroup to subgroup as follows:

Starting with a random cube, reachable by
      closure(F,B,R,L,U,D) = the full group
7 moves to a cube reachable by
      closure(F,B,R,L,U^2,D^2)
13 moves to a cube reachable by
      closure(F,B,R^2,L^2,U^2,D^2)
15 moves to a cube reachable by
      closure(F^2,B^2,R^2,L^2,U^2,D^2)
17 moves to a cube reachable by
      the identity
----
52 moves total.  Singmaster expects the 17 to be 15 by the time he returns to
London.

The Hungarians have cube races. A contestant take his/her cube out of its box
and unscrambles the judges' randomizing in about 50 seconds.   Apparently they
file and lubricate their cubes with loving care to reach such speeds.

The U.S. made white cubes violate no patents because Rubik never applied for
foreign rights.  Hmmmm.   Is that ethical?

Bill





Date: 15 Aug 1980 1558-PDT
From: CSD.VANDERSCHEL at SU-SCORE
Subject: Re: A solution subroutine
To: Davis at OFFICE-3, cube-lovers at MIT-MC
cc: CSD.VANDERSCHEL at SU-SCORE
In-Reply-To: Your message of 10-Aug-80 1709-PDT

I am surprised about all this talk about edge flipping algorithms.
Once I learned the concept of mono-ops, I immediately proceeded to
invent a very simple, and quite obvious, edge monoflip.  You can do it
by manipulating only the front and the horizontal center slice
(hereinafter referred to as the "slice").  A virtue of the procedure I
am going to describe is that you can think of it in geometrical terms
and explain it that way.  The result is that it is easy to remember
and invert, and you do not need any notation.

Hold the cube with the two edge cubies you want to flip in the U face.
Have one of them, the current "target", in the UF position.  You can
think of its "socket" as moving with the F face.  Now turn the target
cubie over to the left side.(F') Then move it to LB position by
turning the slice.  Next flip its socket over to the right.(FF) Now
put it back in its socket by giving the slice a half turn.  Finally,
return it to the U face.  Don't worry about the fact that you have
changed the orientation of the cube, as that will be fixed when you
reverse the process for the other edge you want to flip.

Conceptually, this is a 5 move sequence.  Since two moves are slice
moves, you have to count it as 7.  If you want to count quarter-turns,
it would be 10.  In any case, it is simple.  (Singmaster has published
even shorter monoflips.)  But to me the most important thing is that
it is obvious on inspection that it has to do what you want.  The way
I look at it, you are using the slice for manipulation and storage.
You remove the target from between its neighboring corners and put it
back with the opposite orientation.  (Is this as obvious to everyone
else as it is to me?)

The same way of looking at things allows you to create edge monoswaps.
Just put the two edge cubies you want to swap in diagonally opposite
corners of the slice and give it a half turn before putting them back.
You can arrange it so that they will or will not be flipped.  Such
moves are their own inverses.

Using the D face for storage and manipulation, you can also easily
invent monotwists and monoswaps for corners in the U face.  For all
such moves that I have created, it is apparent to me what needs to be
done and why it will work.  No memorization of obscure move sequences
with magical effects.
-------

ACW@MIT-AI 08/15/80 19:27:57 Re: "Monoflips"
To: CUBE-LOVERS at MIT-MC

M. van der Schel has come up with a very elegant move,
indeed.  Easy to do, easy to think about.  But he sets
no records.  A 22-qtw diflip was reported some time
ago to this mailing list.  It is true that the monoflip
takes only 10 qtw.  Unfortunately, this does not allow you
to do a 20-qtw diflip, since the monoflip is only useful
as a tranform: in other words, when accompanied by its inverse.

Monoflip U Monoflip' U' still has 22 qtw.  I suspect this to be
a minimum.

   ---Wechsler


Date: 15 Aug 1980 2347-PDT
From: Stuart McLure Cracraft <McLure at SRI-KL>
Subject: Undocumented position?
To: cube-lovers at MIT-MC

I've been attempting to use the 'method' Singmaster offers in his 5th
edition pamphlet for solving my cube, and have arrived at an
undocumented position. Perhaps I'm overlooking something and a
cubeophile out there can offer an explanation of where I'm going
wrong.  I have reached the position described after step 6 just before
proceeding to step 7. The bottom and top faces are one color, the
center slice cubies are corrrectly colored and the upper edge cubies
(the + pattern) are correctly oriented. Two of the upper adjacent
corners are correctly oriented but the opposite two are not.  At this
point Singmaster deals with putting the rest of the U corners in their
correct slots although not orienting them correctly necessarily. If we
represent my U corners with

		1	2

		3	4

then the cubies want to be rearranged as follows

		1 and 2 want to be exchanged

I don't see any way of getting this operation out of the cases
documented by Singmaster. What gives?
-------

Date: 16 Aug 1980 0010-PDT
From: Mclure at SRI-KL
Subject: Aha.
To:   cube-lovers at MC

Looks like one of his alternatives does appear when the cube is
turned upside down.
-------

Date: 16 AUG 1980 0338-EDT
From: MJL at MIT-MC (Matthew Jody Lecin)
To: CUBE-HACKERS at MIT-MC

Seems the ever popular CUBE is getting places.  In this month's issue
of one of America's popular men's entertainment magazines there is a picture
of Zsa Zsa Gabor holding a cube, and a small explanation of what the cube is,
and who's marketing it, etc.  (Isn't Zsa Zsa Hungarian?) The way she is
holding her cube you see 2 faces - orange and green - and it is right.

							{Matt}

Date:  16 August 1980 12:45 edt
From:  Greenberg.Multics at MIT-Multics
Subject:  Re: Undocumented position?
To:  Stuart McLure Cracraft <McLure at SRI-KL>
cc:  cube-lovers at MIT-MC
In-Reply-To:  Message of 16 August 1980 02:47 edt from Stuart McLure Cracraft

Would somebody out there who has Singmaster version 5 benefit us with
a brief (or as long as you can stand) description of the
algorithm for solution contained therein, as the rest of us
wait for International Snail Mail? Thanks, -Bernie.

Date: 16 Aug 1980 1053-PDT
From: Stuart McLure Cracraft <McLure at SRI-KL>
Subject: Singmaster's method
To: Greenberg.Multics at MIT-MULTICS
cc: cube-lovers at MIT-MC
In-Reply-To: Your message of 16-Aug-80 1245-PDT

Well, it does have a copyright on the method itself so I doubt
it would be reasonable to type the thing in. It is rather long too.

Briefly, he first puts the U edges in place and then the U corners.
One face is a single color now. He turns over the cube so that it
is on the bottom and puts the middle layer edges correcty in place
and then makes the U edges into an even permutation.
Then he puts the U edges in place and then the U corners.
Finally he orients the U corners. And supposedly it is solved
at that point. I haven't had much luck with the method though.

He claims that this method takes less then 200 moves.
-------

BSG@MIT-AI 08/17/80 17:22:24
To: CUBE-LOVERS at MIT-MC
OK, you guys. You asked for it, you got it.  :CUBE will
now accept typed-in cube configurations. The ^F command
(control f) to installed :cube prompts for a standard
ITS file name. If you give ? as the file name, it tells you
the thing I'm about to tell you.

The hack is as follows. You copy AI:BSG;CUBE TEMPLT into
some choice name of your own, and follow what it says
inside of it, editing it to look like your cube configuration.
Then enter :cube, and give the file name in response to ^F.

We do only the most rudimentary checking, for syntax,
center duplications, only six colors, overambitious file munging.
We have not yet put in conservation of cubies, let alone
ALAN's necessary and sufficient solubility conditions. So
it's quite possible to make :cube go blarmy in this way, so
be careful.

Occasionally, :CUBE may give an ambiguous description of a 180-degree
center-slice twist: to disambiguate it, let it do it, and see
what it did, and do it on your cube.

Go nuts in mid-August, -bsg

BSG@MIT-AI 08/18/80 06:58:28
To: cube-hackers at MIT-MC
The ^F command (read in file) of :CUBE has now been augmented to make
cubie conservation checks; it will tell you by way of what cubie
is missing if there are duplications. The Bawdenian solvability
criteria have not yet been implemented.  There is also a brief
form of the input file in AI:BSG;CUBE BFTPLT.


BSG@MIT-AI 08/18/80 19:47:49
To: cube-lovers at MIT-MC
As per request of CLIVE, :CUBE ^F now accepts R for RED, Y for
YELLOW and so forth. See AI:BSG;CUBE TEMPLT for reference.

Date: 19 AUG 1980 0927-EDT
From: ZIM at MIT-MC (Mark Zimmermann)
Subject: Singmaster 5th ed. via LOGICAL GAMES
To: CUBE-HACKERS at MIT-MC

Bela Fzalai at LOGICAL GAMES (Haymarket, VA) told me last night that he
is out of the 4th edition booklets, but that he has a bunch of the 5th on
order.  Maybe it would be quicker to get them via him than from
England directly; I don't know....		Mark

Date: 21 Aug 1980 17:14 PDT
From: Wastebasket at PARC-MAXC (not authenticated)
Subject: Re: please add me to your mailing list
In-reply-to: CSD.CVW's message of 21 Aug 1980 1536-PDT
To: CSD.CVW at SU-SCORE
cc: cube-lovers at MIT-MC

Please send requests like "please add me to your mailing list" only to the
maintainer of the list, not to everyone on the list.  Thank you.


Date: 24 AUG 1980 1623-EDT
From: RP at MIT-MC (Richard Pavelle)
Subject: Plumbers Cross?
To: CUBE-LOVERS at MIT-MC

There have been discussions of Plummer's Cross which I believed to be
unique. Lets have an almost visual representation such as the following
for one face:

X Y X
       def
Y Y Y  ===  (Y,X)

X Y X
					(O,B)
Then Plummer's Cross looks like   (W,G) (G,R) (B,Y) (Y,O)
					(R,W)

(and this gives the coloring of my cube as well). The point about this
configuration which I am stressing is that opposite sides have no colors
in common.

Now I find there is a second cross for which the point above is not valid.
This cross takes a form			(Y,G)
				  (R,B) (O,W) (W,O) (B,R)
					(G,Y)
Is this known by any Cubists out there?

Date:  24 August 1980 17:52 edt
From:  Greenberg.Multics at MIT-Multics
Subject:  Re: Plumbers Cross?
Sender:  BSG1.SIPB at MIT-Multics
To:  RP at MIT-MC (Richard Pavelle)
cc:  CUBE-LOVERS at MIT-MC
In-Reply-To:  Message of 24 August 1980 16:23 edt from Richard Pavelle

That is Christmans crross, describedin my earlier letter in which
I described Plummers cross.

Date: 25 AUG 1980 0907-EDT
From: ZIM at MIT-MC (Mark Zimmermann)
Subject: anecdotes from visit to Logical Games Inc.
To: CUBE-HACKERS at MIT-MC

I visited Bela Szalai on Saturday; his country home near Manassas battlefield
("Bull Run") is LGI, and he and his family are the employees.  I learned a few
interesting things:
--he saw the cube in 1978 Aug, during a trip to visit relatives in Hungary;
after many delays was able to get some wholesale from the Hungarian gov't., but
they wanted $1 million for exclusive rights to distribute the things in the
Western world.  Ideal may have learned of the cube from Bernie DeKovon,
GAMES magazine editor who is also a toy consultant for them; Ideal paid the
$10^6 in 1979 Sep.
--the cube is not patentable in the USA bnecause it was sold publicly for
over a year in Hungary before patents were applied for; in England, however,
it is "copyrighted" (equivalent to US patent+trademark) and Ideal has a legal
monopoly.
--Ideal will run nationwide TV advertisements for the cube beginning in a month
or so; something to do with Newton, and including an animated cube which solves
itself....
--Bela took out a second mortgage on his home to pay for the plastic molds for
his cube parts.  He uses white plastic so that it will be possible to print the
colors on ("pad printing", the same process whereby labels are put on some
shampoo bottles).  If all goes well, LGI will start making printed-color cubes
within a month.
--Bela ordered 300 copies of the 4th edition of Singmaster's booklet in 1980 Jun;
Singmaster informed him in July that the 4th edition was out of print, but that
he could have 300 of the 5th edition for the same price as soon as they come
out.  LGI has >90 orders already pending, and while the remaining <210
copies of the 5th edition last, Bela is willing to sell them for $4 each...
whenever they arrive.
--LGI wholesale prices start at orders for 12 cubes, $6 each plus shipping;
individuals may want to consolidate their orders to save money.
--Rubik developed the cube partly as an aid to teaching 3-dimensional
visualization in students.
--cube manufacturing is VERY labor-intensive:
   4 minutes to tap in and glue the caps to cover the internal 6th face of edge
	     and corner cubes for the 20 pieces necessary to make one 3x3x3
   1 minute to assemble with screws and springs 5 out of six swivels (central
	    cube faces) onto the middle cross
   1 minute to assemble the pieces and screw in the last central face
   4 minutes to perform final adjustment:  silicone grease, torque up all screws
	     evenly, rotate every which way to test each cube out, tap in and glue
	     6 central face caps over screws.
   6 minutes to apply the color squares to the faces
Bela can read while performing most of the final assembly, now that his hands have
had several thousand cubes' practice.  The color application step will be
eliminated if/when they begin to use pad printing for face coloring.
--about 4% of cubes are rejected for mechanical reasons, so far
--could work without gluing in internal faces of sub-cubes, but then about 1 cube
in 20 would fail...so, they don't
--READERS' DIGEST phoned to confirm some data, presumably for a story someday....

Date: 26 Aug 1980 1729-PDT
From: CSD.VANDERSCHEL at SU-SCORE
Subject: [CSD.VANDERSCHEL: Re: "Monoflips"]
To: cube-lovers at MIT-MC

Date: 25 Aug 1980 2041-PDT
From: CSD.VANDERSCHEL
Subject: Re: "Monoflips"
To: ACW at MIT-AI
cc: CSD.VANDERSCHEL
In-Reply-To: Your message of 15-Aug-80 1927-PDT

When I offered my monoflip, I tried to make it clear that I was
claiming no superlatives except, perhaps, that of conceptual
simplicity.  Nor can I claim originality, for I recently noticed that
Singmaster lists essentially the same move I thought of in his first
supplement.  It is presented as FUD'LLUUDDR and attributed to David
Seal.  (Besides, if I claimed originality, it would refute my claim
of obviousness.)

You indicated that you believe 22 qtw is the best one can do for a
diflip.  The only sense I know of in which your claim could be valid
would be a diflip generated by a monoflip and its inverse, where that
monoflip preserves the set of cubies in a face.
B'UUBBUB'U'B'UUFRBR'F' is a 16 qtw process that flips a pair of
adjacent edges, and Singmaster attributes it to Morwen
Thistlethwaite's computer program.

The simplest mono-ops are those that preserve the set of cubies in a
center-slice.  For example, FF could be viewed as a monoswap of the
right and left front edge cubies in the horizontal center-slice.  If
we denote by "S" a quarter turn of that slice, then FFSFFS' will
produce any three cycle you might like of edge cubies in the slice
while leaving the rest of the cube intact.  It also becomes more clear
how FFSSFFSS produces the well-known double swap through opposing
faces.

U'FR'UF' is a 5 qtw monoflip that Singmaster attributes to Frank
Barnes.  It preserves the set of cubies in the RL center-slice.  I
think it is clear how it works and that you could not possibly do it
in fewer moves.  Using this monoflip, you can generate a 14 qtw diflip
for an opposing pair of edge cubies.  I still prefer the move I
thought of because I seem to be less likely to make a mistake using
it, it is more readily adapted to any pair of edge cubies, and it is
about as easy to perform.

				David Vanderschel
-------
		---------------
-------

Date: 27 Aug 1980 0128-PDT (Wednesday)
From: Dal at UCLA-SECURITY (Doug Landauer)
Subject: Notation, transforms I like to use ...
To: cube-lovers at mit-mc,alan at mit-mc


After 5 beers and two Drambuies, finally, here we go ...





			Here is all of my knowledge and
			notation regarding the cube ...
				SPOILER warning
			these transformations are enough
			to solve any cube.   I look forward
			to anyone publishing a compendium;
			please go ahead and use anything
			in here.


Two notes about my particular cube ...
it worked fine until I decided to lubricate it (as is recommended in
one of these cube-lovers messages ... they say to use a graphite lubricant
(I believe) ... do so.)      DO NOT use "dri-slide".   It is a Molybdenum
Disulfide lubricant for use with metals and it tends to eat plastic, making
it harder to turn.   Also (this may be unrelated), three weeks or so after I
lubricated mine (an Ideal from Toys-R-Us in the San Fernando Valley (in LA))
one of the facies popped off and now it's out of commission until I try to
glue it back together.

I found that the descriptions of its inner workings do not do it justice.
It is indeed a work of 3-dimensional mechanical genius.  To elaborate on
the descriptions ... the face cubies are about 7 mm thick, and they are
what holds the other pieces in.  The corner cubies have a growth on their
"opposite" corners which is roughly cubical but the "inside" corner is
rounded to allow them to rotate.
The side cubies (sidies?) have a 7-mm thick disk-like thing sticking out
of the center of their least visible edge (the one opposite the one that
the sidie's two colors border on).     It's actually more like a thick
square with one rounded corner (the inside one).

Anyway, here's a description of my notation followed by a set of
transformations I use; this is indeed sufficient to fix a cube and
if you wish to remain self-sufficient in this matter, read no further;
but then what're you doing on this list anyway?

	happy cubing ....
		-dalgorf-

flubrd Notation:   (pronounce it `flubberd')  Object names:
     Faces: f l u b r d stand for the front, left, up, back, right
	  and down faces respectively.  (up==top and down==bottom)
     Little cubies
	  The 6 face cubies (one at the center of each face)
	  are seldom referred to; use the face name (e.g the f-face cubie)
	  Side (edge) cubies:	  fl  fu  fr  fd  ul  dl
				  bl  bu  br  bd  ur  dr
	  Corner cubies:      flu fur bul bru fdl frd bld bdr
	  For side and corner cubies, the order of its faces only matters
	  in permutation descriptions (see below), but by convention, the
	  corner cubies are all named clockwise starting in front or back.
     Center Slices
	  H (`Horizontal'==`floor-parallel') runs through faces f,r,b,l
	  P (`Parallel' for `body-Parallel') runs through faces u,r,d,l
	  S (`Slicing' for `body-Slicing') runs through faces u,f,d,b
Moves:	       The ambiguity between object names and moves is cleared up
	       by always surrounding a move in angle brackets (<>).
     A move can be any of the following:
	  - a single letter, meaning twist a single face or centerslice, or
	  rotate the entire cube one quarter-turn (see below);
	  - a move followed by a move indicates those two moves in sequence;
	  - a number n followed by '(', a move, and a ')' is the same as
	  repeating the move n times;
	  - a move followed by a prime ("'") indicates the inverse of the
	  move --- by convention also all one-letter moves are defined
	  so that changing the letter's case also inverts the move.
	  A move may be grouped by parentheses, so that e.g. <(FUDRB)'> is
	  equal to <B'R'D'U'F'> (ie, undo each twist in reverse order
	  to undo this whole move).
     <F> twists f clockwise; similarly for <L>,<U>,<B>,<R>,<D> & l,u,b,r,d
     <f> twists f counter-clockwise; similarly for <l>,<u>,<b>,<r>,<d>.
	   Note <F'>=<f>, <f'>=<F>, and so on.
Extended moves (centerslices and cube-turning)
     cube-turning:  I,J,K=front-top-right clockwise
	  <I> move cube's front to top	     <I'> or <i> top to front
	  <J> move top to right side	     <J'> or <j> right side to top
	  <K> move right side to front	     <K'> or <k> front to right side
     centerslice moves
	  <H> (`Horizontal') move H clockwise as seen from above = <uDk>
	  <P> move P clockwise as seen from the front = <fBJ>
	  <S> move S clockwise as seen from the right side = <rLi>
     Move combinations are two-letter names beginning with C, E or A;
	  C1-C9 and (CA-CZ, Ca-Cz if needed) describe corner-combinations;
	  E1-E9 and EA-EZ (Ea-Ez if needed) describe edge-combinations;
	  A1-A9, AA-AZ & Aa-Az describe combinations that mess with both.
	  I also use V,W,X,Y,Z and v,w,x,y,z as temporary macros
Permutations = { list of cubies -> list of new spots }.  The permutation
     described puts the first list into the new LOCATIONS described by the
     second list, moving each face of each cubie in the order specified.
     E.g., <F> does {flu fu fur fr frd fd fdl -> fur fr frd fd fdl fu fdl}

Useful basic corner combinations:
C1: <6(RurFU)> :: {fdl flu bul -> dlf luf ulb} twist corners clockwise
C2: <6(rFRuf)> :: {fdl flu bul -> lfd ufl lbu}	 "	"    ccw (lf2 + bul)
C3: <3(RUru)> :: {frd fur bul bru -> rfu rdf ubr ulb}
     switch those pairs of corners, twisting each
C4: <3(fuFU)> :: {frd fur flu bul -> rfu rdf bul flu} (ditto)
C5: <3(FrfR)> :: {frd fur flu bru -> rfu rdf bru flu} (ditto)

compound ones:
C6: <lC4L> = <l3(fuFU)L> :: {frd fur fdl flu -> rfu rdf luf lfd}
     switch those pairs of corners all in the front face
C7: <C1MC2m> = <6(RurFU)M6(rFRuf)m> :: {fdl bld -> dlf dbl}
     twist fdl cw and bld ccw.
C8: <C5C3> = <3(FrfR)3(RUru)> :: {flu bul bru -> bru ufl ulb}
     cycle three top corners (left two & bru) "cw"
C9: <C5C4> = <3(FrfR)3(fuFU)> :: {flu bul bru -> bul bru flu}
     cycle three top corners (left two & bru) "ccw"

Useful basic edge (side) combinations:
E1: <4(FH)4(HF)> :: {fl fr -> lf rf} twist two sides in place
E2: <2(ssdd)> :: {fu bu fd bd -> bu fu bd fd}
     "doubleswap" swaps front for back sides on top and bottom
E3: <3(FRRF)> :: {fl fr ru rd -> fr fl rd ru}
E4: <3(ffrr)> :: {fu fd ru rd -> fd fu rd ru}
E5: <3(URRU)> :: {ul ur rf rb -> ur ul rb rf}
E6: <3(ffuu)> :: {ul ur fl fr -> ur ul fr fl}
E7: <3(FLLF)> :: {lu ld fl fr -> ld lu fr fl}
E8: <3(ULLU)> :: {lf lb ul ur -> lb lf ur ul}
E9: <E3E4> = <3(FRRF)3(ffrr)> :: {fl fr fu fd -> fr fl fd fu}
EA: <f(rE3R)E5F> = <fr3(FRRF)R3(URRU)F> :: {fu ul ur -> ul ur fu}
     cycle top (left right and front) sidies clockwise
EB: <F(LE7l)E8F> = <FL3(FLLF)l3(ULLU)F> :: {fu ul ur -> ur ul fu}
     cycle top (left right and front) sidies counterclockwise
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Pons Asinorum: HHSSPP (180 degrees on each centerslice)
Laughter: <4(lrfb)> = <6(lrK)> (but in the latter the cube is KK'd)...
     Another laugh resolves the cube; a K first produces the 4 crosses.
Crux Christmani:  <SSHiE2fE2FBE2bE2iE2KjE2> or, call E2 (the doubleswap) `Z',
 <SSHiZfZFBZbZiZKjZ>.  This is 6 crosses in 3 pairs:  colors go top with
 bottom, front with right side and back with left side.
Crux Plummeri (the fancy one): <SHiE2fE2FBE2biE2fE2FBE2bE2jE2>; or (again):
 <SHiZfZFBZbiZfZFBZbZjZ> or if Y := <iZfZFBZb>, it's simply <SHYYjZ>!
 This is 6 crosses in 2 sets of three each:  Front, Right and Top each with
 each other; and Back, Left and Bottom each with each other.
Greenberg's "well-known center-cubie rotation algorithm" has 12 varieties:
 Pick two centerslices (say X and Y) and a direction (cw or ccw):
 <xYXy> does ccw and <YxyX> does cw.	Details:
     (xyz cw) means the center (face) cubies of the faces x y and z are
     rotate clockwise; ... ,ccw means counterclockwise.
     fur cw :: <SpsP> or <sHSh> or <PhpH>, ccw :: <pSPs> or <HshS> or <hPHp>
     frd cw :: <PspS> or <SHsh> or <PHph>, ccw :: <sPSp> or <HShs> or <HPhp>
     fdl cw :: <hSHs> or <spSP> or <hpHP>, ccw :: <ShsH> or <psPS> or <phPH>
     flu cw :: <HphP> or <PSps> or <shSH>, ccw :: <pHPh> or <SPsp> or <hsHS>
Outstanding questions mentioned in the note about Singmaster's talk:
(I.e., I ain't seen these yet ... )
1. Conway's Monoswop
2. Rubik's Duotwist

(ignore the two lines above beginning with Pick ... and ending with
... Details:	they're wrong.	 I'd go edit them but my Unix is out
of space on /tmp and on my filesystem and I only have 300-baud access to
it anyway).

		Have fun
		DalGorf

Date: 27 August 1980 1:58 pm PDT (Wednesday)
From: Woods at PARC-MAXC
Subject: Lexicon
To: Cube-Lovers@MIT-MC

We need a term to refer to the fear that someone will randomise your cube.  Any suggestions?

	-- Don.


ACW@MIT-AI 08/29/80 13:24:04
To: Cube-hackers at MIT-MC

    Date: 29 AUG 1980 1316-EDT
    From: ACW at MIT-AI (Allan C. Wechsler)


      On this coast we have a colorful slang term for randomizing the
    cube.  Hint: the command to randomize the cube in Bernie's
    :CUBE program is "F".

       ---Wechsler


Date: 29 Aug 1980 10:55 PDT
From: Lynn.ES at PARC-MAXC
Subject: Re: Lexicon
In-reply-to: Woods's message of 27 August 1980 1:58 pm PDT (Wednesday)
To: Woods.PA
cc: Cube-Lovers@MIT-MC

Incongruphobia seems nice, but may imply more of a fear of lack of order, rather
than of the change from order to lack thereof.  It doesn't specify the objects that
are disordered either.  Cubic discombobulaphobia might be a little better in these
respects.

/Don Lynn


MJA@MIT-AI 08/29/80 21:45:41 Re: Knowledge-based Rubik's Cube solver?
To: cube-hackers at MIT-MC
Please excuse the previous badly munged versions of this message i accidently
sent.

	I am taking an artificial intelligence course at the University of
Illinois at Urbana-Champaign and I need to do a term project for the
course that involves developing and/or using a knowledge based system.
Since I am interested in the Rubik's cube and since it seems like
there are human expert cube solvers (as is evident from msgs on this mailing
list) that use heuristic knowledge (for example the various macro's (as I like
to call them, they are more properly called composite elements of the
group, i guess) that people have come up with to do some particular operations
on the cube), it seems natural to me that a knowledge based system would
be just right for this task. I would like to hear:

	(1) Pros/cons on why this can/can't be done and how effective
such a system would be.

	(2) Considering that I have but one semester to work on this project
(while concurently taking 5 courses, incl. this AI course) is it reasonable to
attempt this for a term project? (The actual system would not have to be
100% complete at the end of the term, but I would at least like to have
something to show for a semester's worth of work.)

	(3) If this is too ambitious for a term project for a course, what
about the use of this as a topic for a master's thesis. (I would have until
the summer of '81 to finish since thats how long my company will wait for me
to get a master's degree if im going to continue working for them.)

	(4) Of course in addition to all of the above trivia, I would like
some ideas for such a system (even if it isnt feasable to do in this limited
amount for time, im still interested in at least thinking about such a system
with the idea in mind that someday it can be implemented).

	I dont know how interested the majority of the people on this mailing
list are in knowledge based systems, so we might consider a seperate mailing
list for knowledge based cube solvers to spare those on the current mailing
list from mail they dont want.

		/ MJA


ZILCH@MIT-AI 08/30/80 01:39:38
To: CUBE-LOVERS at MIT-MC
Subjects covered:random notes on the cube and cubing

1> To enable new and old cube-lovers to communicate on an equal basis
I propose that a file be established that describes FBLRUD
(or whatever),the I,J,K and centerslice move and any other concepts
that might be usefull in describing transformations.(I can not do this as I know nothing about establishing files or editting them, and as a tourist I am
unsure of my status as file-creater.)

2> Today having nothing better to do, I fiddled with the 3x3x2 version
of the cube (actually I just didn't allow certain moves on my
3x3x3). At this point I have two transforms which would enable me
to solve the 3x3x2 version if and when I ever see it. I derived the number of orbits and the reasons behind them, but will not describe them here because
the 3x3x2 is a novel idea and I don't want to ruin all the fun.
Also I came up with the idea of a 2x2x3 version which basically operates on
the same principles as the  3x3x2 , but it looks different.
   solving the 3x3x2 can give a slight amount of insight about
Thistlewaite's algorithim ,described by
McKeeman on 12 Aug. @15:10 PDT.


3> On July 15 @1413 EDT, ALAN asked if anyone had learned to solve the extended cube problem independently. I had known the faces could assume different
orientations when the cube was solved but hadn't done any work on this
and didn't know what the possible transforms might be. In a later note
cmb described what his transforms accomplished and from that idea
I recently derived similar (if not equal) transforms.With a little foresight
I find that only one of these transforms might be needed.
 This evening a friend and I came up with a modification to the cube
construction that would facilitate the extended problem. this is simply affixing  another cubie to each face cubie and coloring appropriatly.

4> As suggested by CSD.VANDERSCHEL on 9 Aug. @1610 pdt has anyone given any
thoughts to 3-faced cubing (2 types) or 4-faced cubing (again 2
types). As far as I can tell 5-faced cubing is not profitable.
Also does CSD.VANDERSCHEL know a good way (at this time) to show
why only  1/6 of the 6! possible permutations of the corners are possible?
There is also an extended problem for the 2-faced problem
involving face orientations.


5>An addendum to WOOD's message of 23 JUL. @5:23pm
regarding repetitive sequences to get to the identity. As it turns out,
by my calculations the number 1260 is sufficient even including the prblem
where the faces are permitted to move. The only subcycles of the faces
themselves have lengths 4,3,2,and of course 1.
These subccles may immediatly be generated by I (4 times), I^2
(2 times),<top-to-front>^2<left-to-front> (also 2 times)
and finally <top-to-front><right-to-top> (3 times).
(I do not use IJK notation on these last two because I have noticed a little
disagreement on exactly what these mean.)
These subcycles make no difference in the total number because subcycles of
lengths 4,3,and 2 have already been taken into consideration in the computation
of 1260.
I have no idea whether th e face cubies might change their orientations in a
sequence of length 1260,when the faces are allowed to move,
but I know it does nothe number above 1260 when the faces are not allowed to
move.

6> Singmaster's solution from his version 5, as reported by McLure
on 16 Aug. @1053 PDT sounds almost exactly like my solution
except that I keep the first face completed in the up poition at
all times. He reported that his method takes less than 200
(of his) moves. My transformations yield a solution in a maximum of 190
quarter-twists. My actual solving length (from when I bothered counting)
averages about 125 qtws , and my time (when I bother) is almost
consistantly 2.5 minutes, occasionaly under 2, with worst case about 3 min. 10sec when I messed
up. Usually my fast times do not use all of my algorithms techniques ,
because they are new to me and I don't know them by heart.

Breakdown of moves:
1. Top edges in  proper position and orientation 20 (3+6+6+5)
2. " corners " "	"	"      "	 36 (9+9+9+9)
3> 3 middle edges " "	  "	   "	   "	   45 (15+15+15)
4. 4th middle edge in proper position and orientation
	   and bottom edges in proper orientation  23
5. bottom edges in proper position		  18
6.bottom corners in proper position			     20
7.  "	     "	 "    "     "				 28

       Total						       190


Note: in step 4 proper orientation means that say if the down face is
white , after step 4 all bottom edges will be showing white on their
down sides.

This algorhitim has not been optimized much and uses lookahead only in step
4. Step 2 gives the worst case for any corner, but if only worst cases are
present then they cancel each other out somewhat.

Replys, questions, and comments are welcome.

Chris


DMM@MIT-ML 08/30/80 18:41:24 Re: DAL@ucla-security's msg of 27-Aug
To: cube-lovers at MIT-MC
Genereally, I have found themacros supplied by DAL to be quite useful,
but t|ere are a few bugs i've noticed. First of all, oC8&C9,
I believe DAL has the cw & ccw notations reversed.
Regarding C7:	What is M? I have not seen it referenced anywhere.
and last, in maco EB: the last move is f, not F.
(I`guess it was the 5 beers and 2 Drambuies.)


Date:  3 September 1980 2108-EDT
From: James.Saxe at CMU-10A (C410JS30)
Subject: Orbit classification revisited
To:  cube-lovers at MIT-MC
Message-ID: <03Sep80 210846 JS30@CMU-10A>

Hi, folks.

Having read Vanderschel's msg of Aug. 6, it appears to me that the
explanations of the orientation parities are unnecessarily complex,
though the material on permutation parities is presented in a way
that should be immediately convincing to anyone familiar with the
notion of even and odd permutations from elementary group theory (and
anyone who isn't should be!).  Here's my attempt at a more elegant
demonstration.

In what follows, I will use the term "facelet" to denote any (visible)
face of a cubie.  Thus, each edge cubie has two facelets and each
corner cubie has three facelets.

I will address Edge Orientation Parity (EOP) first.  Consider the
following diagram:

		+-------+
		|   1	|
		|0  U  0|
		|   1	|
	+-------+-------+-------+-------+
	|   1	|   0	|   1	|   0	|
	|0  L  0|1  F  1|0  R  0|1  B  1|
	|   1	|   0	|   1	|   0	|
	+-------+-------+-------+-------+
		|   1	|
		|0  D  0|
		|   1	|
		+-------+

The numbers label absolute positions, not facelets, and therefore
remain in the same configuration when the cube is manipulated.
Imagine that a mark is placed on each facelet that occupies a
position labeled "0" when the cube is in the solved configuration.
Thus, each edge cubie will have one marked and one unmarked facelet.
(Unlike the numbers, the marks are attached to the facelets and will
move as the cube is manipulated).  The parity of an edge cubie in an
arbitrary configuration is defined as the number labelling the
position occupied by its marked face, and the EOP is defined as the
sum of the parities of all edge cubies modulo 2.  A quarter turn of
any face reverses the parity of 4 edge cubies, thus leaving the EOP
fixed.  By induction, no sequence of manipulations starting from the
solved configuration can produce a configuration EOP = 1.  So much
for EOP.

[Just for grins, here's a cute way to determine the parity of an edge
cubie without consulting (or reconstructing) the diagram:  Assign
each of the cubie's two facelets a number in {0,1,2} according as it
is oriented parallel to its home position ("Self" -> 0), parallel
to the other facelet's home position ("Other" -> 2) or perpendicular
to both home positions ("Neither" -> 1); add these two numbers and
reduce modulo 3 (yes, three!) giving the parity of the cubie.  The
sum can never be 2 because that would imply that the two facelets
were parallel to each other.  Here's an addition table with
double-ended arrows showing the possible transitions:

	      (S) 0   (N) 1   (O) 2
	    +-------+-------+-------+
	    |	    |	    |no     |
      (S) 0 |	0 <-+-> 1   |	+   |
	    |	^   |	^   |	 way|
	    +---+---+---+---+-------+
	    |	v   |no |   |	    |
      (N) 1 |	1 <-+---+---+-> 0   |
	    |	    |	|way|	^   |
	    +-------+---+---+---+---+
	    |no     |	v   |	v   |
      (O) 2 |	+   |	0 <-+-> 1   |
	    |	 way|	    |	    |
	    +-------+-------+-------+

]

Now for Corner Orientation Parity (COP).  Consider the diagram below:

		+-------+
		|0     0|
		|   U	|
		|0     0|
	+-------+-------+-------+-------+
	|2     1|2     1|2     1|2     1|
	|   L	|   F	|   R	|   B	|
	|1     2|1     2|1     2|1     2|
	+-------+-------+-------+-------+
		|0     0|
		|   D	|
		|0     0|
		+-------+

Once again, imagine that we mark all (corner) facelets which occupy
positions labeled "0" when the cube is in the solved configuration.
The parity of a corner cubie in an arbitrary configuration is defined
as the number which labels the position of the cubie's marked face.
The COP is the sum of all the parities of all corner cubies modulo 3.
Inspection of the diagram will reveal that the twists U, u, D, and d
leave the parities of all corner cubies unchanged.  Any of the other
possible quarter twists will increment (modulo 3) the parities of two
corner cubies and decrement the parities of two others, thereby
leaving the COP unchanged.

[As Vanderschel pointed out, one way to compute the parity of a
corner cubie by looking at it is to note the number of clockwise (as
viewed from outside the cube) 120 degree twists of the cubie that it
would take to bring the marked facelet parallel to its home position.
Note that the parity of a corner cubie, unlike that of an edge cubie,
depends on the selection of a particular pair of opposing colors for
the marked facelets.  While this lack of symmetry may be considered
unfortunate, it is an inevitable result of the fact that four is not
divisible by three.  It is easy to show that the COP as a whole is
independent of the choice of distinguished faces.]

Vanderschel also mentions the extended problem, but does not quite
make it clear that the FOP changes every time a qtw is done.  This
constrains all three of {FOP, EPP, CPP} to be the same, so that only
two of the eight plausible states of the <FOP, EPP, CPP> vector are
actually achievable by twisting.

				Jim


Date:  3 SEP 1980 2123-EDT
From: DCP at MIT-MC (David C. Plummer)
Subject:  New problem
To: CUBE-HACKERS at MIT-MC


This was inspired by Tanya Sienko (if that means anything!!)

It is possible for each of the six faces to have a capital "T" on them.
That is, each face looks like

	X X X

	O X O

	O X O

QUESTIONS:
	1) How many classes are there?
	2) How many members are in each class?
	3) How easy is it to get to them from solved?

My answers:
	[in a few days, I don't want to spoil the fun]

Good luck, and happy hacking.

					- DCP


Date:  3 September 1980 2149-EDT
From: James.Saxe at CMU-10A (C410JS30)
Subject: Re: Lexicon
To:  Woods at Parc-Maxc
CC:  Cube-Lovers at MIT-MC, James.Saxe at CMU-10A
Message-ID: <03Sep80 214930 JS30@CMU-10A>

My suggested term for the fear that someone will randomize one's cube
is "nobility," since that is what sufferers of this phobia typically
have to solve their cubes.

				Jim

Date:  3 September 1980 2328-EDT (Wednesday)
From: Dan Hoey at CMU-10A
To: Cube-Lovers (and Hackers) at MIT-MC
Subject:  Addictiveness/Disassembly/Taxonomy/Lubrication/Spoilers
Message-Id: <03Sep80 232808 DH51@CMU-10A>

Hello.  I saw the notice announcing the formation of this list a couple
months ago, and decided that it was one of those things I could forgo
-- until I got my hands on a physical cube.  It was an immediate
necessity to own one.  I bought an Ideal brand cube, which, I
understand, is of the species C.  Americanus in spite of its "Made in
Hungary" label.

I had owned the cube less than ten minutes before a facie cover fell
off, without the aid of chemical additives.  This was not very
destructive; just about any gummy material (I used gluestick) suffices
to hold it on.  However, the screw head revealed by this unusual
transformation leads to a new method for disassembly.  Unscrewing does
not stress the cube as does prying, and probably avoids the deleterious
side-effects observed by Greenberg (16 Aug 1453).  This method is not
without its hazards, however.  It is EXTREMELY easy to strip the threads
on the plastic X that holds the cube together.  I have paper shims in
the two threads I stripped and they seem to suffice.  Still, it is
probably better just to loosen the screw until the cube comes apart
with gentle prying.

There is at least one good reason for taking a screwdriver to the cube.
Mine had been assembled with several of the screws so tight that the
springs were completely compressed.  Due to mfg inaccuracies in the
cubies, this made the cube difficult to twist.  By prying each of the
facies off with a fingernail I was able to correct the tension.

Jim Saxe's cube, putatively of the same species but purchased in a
different store at 80% the price, has facies which seem impervious to
this prying even after disassembly a la Greenberg (17 Jul 2118).  Jim
and I exhausted our fingernails to no avail, and careful prying with a
knife was unsuccessful.  Additionally, this cube has a strong tendency
to jam, due either to its uncorrectable looseness or to its edge
cubies, which have oversized tongues with extremely sharp edges.  The
differences lead us to believe that our cubes may belong to different
species in spite of their outward similarity.

I am amazed that anyone would put molybdenum disulfide on their cube.
Isn't that stuff poisonous?  Graphite works well but is messy if you
overdo it.  Silicone lubricant was mentioned by Zimmerman (25 Aug 0907)
-- has anyone any experience with this?  Merrick Furst recommends soap.

For anyone who cubes in public, the only word for LACK of fear that
someone will F your cube is Cubemeistership.  I made a mistake in taking
the cube to one session of a recent conference.  The sequence
4(Borrow)4(Borrow') appeared to have an entropic effect on the cube and
a negative effect on the transfer of information.

SPOILER WARNING:  I have one transform which I haven't seen here, and
which I find useful:  an 8-qtw move to permute three corners of a face.
Specifically, for {fdl flu fur -> ufl rfu lfd} do <fdFUfDFu>.
Mnemonically, you move a socket back and forth between flu/fdl with the
f/F transforms, alternating with moving one of the three pointies
(cornies?)  to be permuted into the socket.  Why it works is a mystery to
me, but it's useful.

Another, which should be obvious but improves Landauer's (27 Aug 0128)
C7, is the cw monotwist {flu -> luf in u-face} <fDFdfDF>=<T1>.  Then
{flu fur -> luf rfu} is <T1U(T1)'u>=<fDFdfDFUfdFDfdFu> taking 16qtw
instead of 60 (assuming <M>=<i>, <m>=<I>).

All for now. -- Dan


Date:  4 SEP 1980 0852-EDT
From: JURGEN at MIT-MC (Jon David Callas)
Subject: Lexicon...
To: woods at PARC-MAXC
CC: CUBE-LOVERS at MIT-MC

Worse yet is the fear that someone will put your cube in an
unsolvable orbit......
			Happy phobias,
				Jurgen at mit-mc


Date: 6 Sep 1980 0021-PDT (Saturday)
From: Dal at UCLA-SECURITY (Doug Landauer)
Subject: errata
To: cube-lovers at mit-mc

Attn:  DMM@mit-ml:
     No, unless my message got garbled somewhere in the midwest, it was
     correct about c8 and c9, the three corner-cubie movers.  I.e., c8
     ( <3(FrfR)3(RUru)> ) cycles them cw, and c9 ( <3(FrfR)3(fuFU)> ) ccw.
     (Much shorter ways to do similar things exist, e.g. <fdFUfDFu> to
     move around three front corners (left two and fur) cw).
	In these descreptions, the `cw' and `ccw' do not refer
     to individual cubies but rather to the triangle of three cubies involved,
     that's why I call it "cycle" == move cubies around, rather than
     "twist" == re-orient a corner cubie in place, or "flip" == do the
     same for an edge (side) cubie.
     You're right about EB.
Attn: Dan Hoey at CMU-10A
     Regarding C7:  When I first made this list up, I used m,n and o to
     represent i,j and k because it was unclear which of i,j and k was
     which.  m is the same as i, and M == I.  My cube-turning moves are:
      I,J,K=front-top-right clockwise
	  <I> move cube's front to top	     <I'> or <i> top to front
	  <J> move top to right side	     <J'> or <j> right side to top
	  <K> move right side to front	     <K'> or <k> front to right side
	thank you...
	... dalgorf
-------


Date:  8 SEP 1980 0853-EDT
From: ZIM at MIT-MC (Mark Zimmermann)
Subject: Singmaster 5th; 2x3x3 "domino"
To: CUBE-HACKERS at MIT-MC

I received the 5th edition of Singmaster's booklet a few days ago, from Logical
Games, Inc.--presumably they have some $4 copies left (they sent mine 1st
class mail)....Bela Szalai (LGI) loaned me a 2x3x3 to play with.  Conceptually,
of course, it's the same as the 3x3x3 with moves restricted to <U,D,F^2,R^2,
B^2,L^2>...physically, however, it was MUCH less aesthetic to play with...
turned poorly, and looked ugly.  Bela had a couple of them, both of which
were rather more "delicate" than the 3x3x3 (in that they tended to get jammed
or come apart).    --  Mark

Date: 10 Sep 1980 1500-PDT
From: Alan R. Katz <KATZ at USC-ISIF>
Subject: randomizing
To: cube-lovers at MIT-MC

Here's a question for all of you.  How do you maximally randomize a cube?
In other words, suppose you were going to have a cube solving compitition
and wanted to make it as hard as possible to solve a cube, what precautions
would you take?

One thing I would do would be to make sure the corners are not correct in
relation to each other.  Anyone have other ideas??

				Alan (Katz@isif)

-------

Date:  10 September 1980 18:12 edt
From:  Greenberg at MIT-Multics (Bernard S. Greenberg)
Subject:  Re: randomizing
Sender:  Greenberg.Multics at MIT-Multics
To:  Alan R. Katz <KATZ at USC-ISIf>
cc:  cube-lovers at MIT-MC
In-Reply-To:  Message of 10 September 1980 18:00 edt from Alan R. Katz

My personal theory on making a cube as hard to solve as possible
is as follows: Put the cube in
some very complex pretty, regular, pattern, such as the Plummer's
Cross with centers trebly rotated, or the Pons Asinorum on that
which looks like the simple Pons:  the challenged cubemeister
will INVARIABLY say, "Oh,
this, this is just one of those and solved like this" and go through
a long hairy procedure, hopefully the wrong one,
or make an error, or try hard at understanding it, and so forth.
In short, s/he will take MORE time trying to undo what s/he perceives
as a "simple" hairy hack than a straightforward application of
solve-from-random technology.

Date: Wed, 10 Sep 1980 1854-CDT
Message-id: <337474464.2@DTI>
From: aramini at DTI (Michael Aramini)
To: cube-hackers@mit-mc
Subject: randomizing

	There are two types of aproaches to what can be meant by total
(or maximal) randomizing. One is to consider states of the cube that
are maximally far away from being solved (that is the minimal number of
quarter twists needed to solve from a given state is maximallized.

the other, which may be more useful for making the cube more difficult
to solve, is based on maximallizing the amount of time it would take
a person (or a program, for that matter) to solve the cube. Of course,
since this amount of time is both dependent on the state of the cube, and
how the person (or program) goes about solving it. Thus the set of most
randomized positions is dependent on who is solving the cube.

the advantage of the first definition is that it seems to be of more
theorectical significance (the number of quarter turns needed to
solve from this position gives you the diameter of the state graph).
The second approach seems more kludgy since it much less well defined
since the amount of time it takes to solve the cube if a function of
many variable besides simply the initial state.

This distinction is much like differing strategies for writing chess
playing programs, you could write it assuming the oponent to be a
"perfect" player (as if a complete look ahead to the leaves of the tree
appraoach were being used by the oponent) or by considering how the
oponent is likely to play (have the program try to confuse the oponent
by taking advantage of something that the oponent wont recognise).

the second method thus will take into account the knowledge-base
available to the solver (for example trying to trick the solver into
thinking that the cube is in one of the easy to solve classes but really
isnt, thus leading the solver down a blind alley)

This gives me an idea for a game where oponents are each trying to bring
a randomized cube into two different final states (each of which
is hopefully equally far away from the initial state, just to make the
game fair) by alternately taking turns making quarter turns. of course
one must make up some rules so that that the game terminates (ei going
around cycles in the state graph an indefinite number of times is
disallowed), and these rules might not be very obvious,although
they would prob be similar to some of the stalemates rules of chess.
-----


Date:	  10 September 1980 2303-edt
From:	  Bernard S. Greenberg	     <Greenberg at MIT-Multics>
Subject:  Randomizing
To:	  CUBE-LOVERS at MIT-MC

I enter the following tidbit into the transcript of this society,
apropos to the current discussion:

  RMS at AI (Richard Stallman) has suggested the following paradigm
  for cubing tournaments:

     (Assuming physically standard cubes are used)  The two participants
     acquire (or create) solved cubes.  Each one randomizes it his/her
     favorite way, and hands it to the other to solve.  The resources
     expended by the solver (either qtw or real-time) are added to his/her
     resource expenditure in solving the other's cube.  Minimum resource
     expenditure wins.

Date: 11 SEP 1980 0016-EDT
From: ALAN at MIT-MC (Alan Bawden)
Subject: How do you maximally randomize a cube?
To: CUBE-HACKERS at MIT-MC


I am interested in maximally distant states of the cube.  I have often
wondered just what a maximally distant state would look like.  I also
wonder HOW MANY of them there are!

Interesting fact (offered without proof (it's not hard)):

Assuming we are counting quarter-twists.  If I hand you a cube in a
maximally distant state, and ask you to solve it in as few twists as
possible, you don't have to think at all in order to know what to do
first!  ANY first twist will bring it closer to home (after that it
gets harder).

Call a state with this property a "local maximum".  Any maximally
distant state is also a local maximum.  Also, any symmetric state is a
local maximum.  This doesn't mean that a maximally distant state is
symmetric, but it does get you thinking along those lines!


Date: 10 Sep 1980 9:37 pm PDT (Wednesday)
From: Woods at PARC-MAXC
Subject: Re: How do you maximally randomize a cube?
In-reply-to: ALAN's message of 11 SEP 1980 0016-EDT
To: ALAN at MIT-MC (Alan Bawden)
cc: CUBE-HACKERS at MIT-MC

It is not obvious to me that any twist in a maximally distant state
brings you closer to the home position.  How do you know there
aren't two equally distant states that are a qtw apart?  There is
probably a parity argument that proves this can't be so, but if you
count half-twists as single operations I'd be much surprised if you
could find a simple proof.

	-- Don.


Date: 11 September 1980 01:04-EDT
From: Alan Bawden <ALAN at MIT-MC>
Subject: Re:"Assuming we are counting quarter-twists."
To: Woods at PARC-MAXC
cc: CUBE-HACKERS at MIT-MC


I said:

"Assuming we are counting quarter-twists."

I realize that there are people out there who like to count
half-twists as a single twist.  I don't.  I'm not trying to force my
way of counting twists on anyone.  I always try to be carefull to make
this assumption explicit out of courtesy to those who might want to
count otherwise.  (Sort of like the Axiom of Choice.)

In fact I DON'T know if that result is true if you count half-twists
as well.  I suspect it is, but I don't have a proof.

You needn't jump up and down and point every time one of us
quarter-twisters reveals himself.  This one, at least, is well aware
of his assumptions.

Date: 10 Sep 1980 11:09 pm PDT (Wednesday)
From: Woods at PARC-MAXC
Subject: Re: "Assuming we are counting quarter-twists."
In-reply-to: ALAN's message of 11 September 1980 01:04-EDT
To: (Alan Bawden and the rest of) CUBE-HACKERS at MIT-MC

If you'll reread my message, you'll note that I claimed the parity
argument wasn't obvious to me even in the case of a qtw metric;
my reference to half-twists was intended along the lines of "this
is even less obvious".  I'd be interested in seeing your "obvious"
proof in the qtw case, if you'd care to send it to me (no need to
bother the whole mailing list with it).

	-- Don.


Date:  11 September 1980 22:06 edt
From:  Greenberg.Multics at MIT-Multics
Subject:  Singmaster v.5
To:  cube-lovers at MIT-MC

(To MIT area cubists only:)
I have received "Notes on Rubik's Magic Cube" version 5
from Singmaster, if anybody wants to see it.  It
is 75 pages long and AWESOME.

Date: 15 Sep 1980 1842-PDT
From: Alan R. Katz <KATZ at USC-ISIF>
Subject: number of reachable states
To: cube-lovers at MIT-MC

I have seen the number 4.3 * 10^19 for the number of reachable states
for the cube,  can anyone tell me how you calculate it?  This may have
been answered before in this list, but I couldn't find it.

Also, someone mentioned that one can make a checkerboard pattern from
the Pons Asinorum by trebly rotating the centers by a simple
transformation. Can anyone tell me this transformation? (again I may
have missed reading it)

Reply to either me or the list.

			Alan
-------

Date: 15 SEP 1980 2156-EDT
From: ALAN at MIT-MC (Alan Bawden)
Subject: Where to find old mail.
To: CUBE-HACKERS at MIT-MC

Just a reminder to you all that ALL of the old cube-lovers mail is
archived in the file ALAN;CUBE MAIL on MIT-MC.

Date: 16 SEP 1980 0746-EDT
From: RP at MIT-MC (Richard Pavelle)
Subject: number of reachable states
To: KATZ at USC-ISIF
CC: RP at MIT-MC, CUBE-LOVERS at MIT-MC

    Date: 15 Sep 1980 1842-PDT
    From: Alan R. Katz <KATZ at USC-ISIF>
    I have seen the number 4.3 * 10^19 for the number of reachable states
    for the cube,  can anyone tell me how you calculate it?  This may have
    been answered before in this list, but I couldn't find it.
The number is (12! * 2^12 * 8! * 3^8)/12. This comes from the following.
There are 8 corners and there are 3 positions- hence 8!*3^8. There are
12 edges with 2 positions hence 12!*2^12. Finally, the /12 comes from
parity considerations. Only 1/4 of the positions in the flippling of
two edges are possible while 1/3 of the toppling of two edges are possible.
    Also, someone mentioned that one can make a checkerboard pattern from
    the Pons Asinorum by trebly rotating the centers by a simple
    transformation. Can anyone tell me this transformation? (again I may
    have missed reading it)
The moving of centers is easy- 4 moves of the center slice while rotating
the cube 90 degrees in your hand between moves.  With the transformation
in hand you can move the centers easily to possible positions.

Date: 16 SEP 1980 0946-EDT
From: DCP at MIT-MC (David C. Plummer)
Subject: number of reachable states
To: KATZ at USC-ISIF
CC: CUBE-HACKERS at MIT-MC


    Date: 15 Sep 1980 1842-PDT
    From: Alan R. Katz <KATZ at USC-ISIF>

    I have seen the number 4.3 * 10^19 for the number of reachable states
    for the cube,  can anyone tell me how you calculate it?  This may have
    been answered before in this list, but I couldn't find it.

    Also, someone mentioned that one can make a checkerboard pattern from
    the Pons Asinorum by trebly rotating the centers by a simple
    transformation. Can anyone tell me this transformation? (again I may
    have missed reading it)

    Reply to either me or the list.

			Alan
    -------

Consider the corners. There are 8 of them, and they can go
anyplace. This leads to 8 factorial permutations. Each corner can
take on three orientations, so this is another factor of 3^8. But
the corners have three possible states (trarity [three way
parity]) so divide by 3. Now do the same with the edges. 12 edges
gives 12 factorial arrangements, times 2^12 oreintations. But the
edges have two parities involved, so divide by four (thus giving
rise to the 12 states of the cube, one of which has the solved
configuration as a member). So if you evaluate

	    8	   12
	8!*3 *12!*2
	-----------
	   3*4

you will get 4.3 * 10^19.

DPC@MIT-AI 09/17/80 17:02:00 Re: cube mode on lisp machines
To: cube-lovers at MIT-MC
cube mode has been fixed (at least temporarily) to work in
the new window system on the lisp machines.  to invoke it:

(load "bsg;cubpkg")
(cube)


it is self explanitory once you get in.  have fun.

-dpc



Date: Wed, 17 Sep 1980 2047-CDT
Message-id: <338086051.13@DTI>
From: aramini at DTI (Michael Aramini)
To: cube-hackers@mc
Cc: boken@mc
Subject: cubing hazordous to your health?

	I saw someone cubing near the digital computer lab at the Univ. of
Ill. today. Considering how much traffic there is on springfeild av
(the street in front of said bldg.) i though it might be a good idea
to be crossing the street while intently solving a cube (i didnt see the
person try crossing the street, but he was at least walking without looking
where he was going).
-----


Date:	  20 September 1980 2144-edt
From:	  Bernard S. Greenberg	     <Greenberg at MIT-Multics>
Subject:  Cubesys/LISPM fixed
To:	  CUBE-LOVERS at MIT-MC

:cube's Lisp Machine instantiation has now been fixed to work in color
on Color LISPM's again.

   To invoke, (load "bsg;cubpkg"), wait till it loads, and (cube).
Should be self-documenting.

Date: 23 SEP 1980 0851-EDT
From: ZIM at MIT-MC (Mark Zimmermann)
Subject: report on DC area cube contest
To: CUBE-HACKERS at MIT-MC

Hecht's (a local dept. store chain) offerred $100 gift certificate to
any one who could get the cube done in 5 minutes (and T-shirts promoting
the cube to anyone getting 1 face right in that time), for a 4-hour period
last Saturday...I was the 7th to get the cube done (there were a bunch of
mathematicians from the University of Maryland there before me)...the store
took everyone's name & address & social security numbers...haven't heard
anything from them yet, though...there may have been more solvers after
I left.		 --Mark


Date: 25 SEP 1980 0859-EDT
From: ZIM at MIT-MC (Mark Zimmermann)
Subject: search-from-both-ends ultimate cube algorithm
To: CUBE-HACKERS at MIT-MC

There are some standard algorithms involving time-memory tradeoffs for solving
problems like Knapsack in N^(1/2) steps instead of N...Hellman had a paper
in some IEEE journal about an application to breaking cryptosystems.  The same
could be applied to solving the cube "exhaustively", given several hundred
billion memory locations storage and a willingness to go through a similar
number of table-lookups.  Just make a table of all the positions that can
be reached within 10 meoves or less (with the route to that position recorded
too!)...then in order to solve an arbitrary cube set-up, begin working out
from that set-up, looking for a position included in the table.  If Singmaster
is right and any position can be reached in 20 moves or less, one will
always succeed within 10 moves from the arbitrary start....
The several hundred billion entries in the table are a bit too large for my
home computer, but perhaps a smaller sub-group of the cube (slice, or
anti-slice, or such) could be done this way in a reasonable amount of time....
Hellman seems to think that solving the DES (which has 2^56 keys, I think) is
not impractical, given enough money and a few years.
How about a competition to find the shortest ways to get to the Crux
Christmanni & Plummeri(?)...a simpleminded corner-shifting I tried took 84
moves, I think, to get whichever one has two cycles of 3 colors...that leaves
lots of room for improvement! --Mark

Date: 23 Oct 1980 2155-PDT
From: KATZ at USC-ISIF
Subject: whats happening?
To:   cube-lovers at MC
cc:   katz


There hasn't been much activity on this list for long time, whats happening?

Maybe we are all cubed out??  Anyway, heres a question for the LA people.
Does anyone know where I can get a Hungarian version of Rupic's cube
(NOT the ideal version) in the L A area??


				Alan
-------




Date:  3 NOV 1980 0945-EST
From: DCP at MIT-MC (David C. Plummer)
Subject:  Christman Cross
To: CUBE-HACKERS at MIT-MC


Greetings, to those who are still on this list!!

I have discovered a rather fast method of doing the Christman
Cross. It can be done in 22 quarter twists, and I think it can be
optomized down into 20. When I ask somebody what the current
representation for transforms is, I will send the HOW along.

((NOTE: This means the Plummer Cross can be done in 44 QTwists))

((((REMARK: For those who count Half Twists [[**sigh**]] as one
MOVE, this algorithm is a mere 16 MOVES))))

Date:  4 NOV 1980 0023-EST
From: DCP at MIT-MC (David C. Plummer)
Subject:  Christman Cross algorithm described
To: CUBE-HACKERS at MIT-MC


Let the following notation exist:
the faces:
	T=Top	B=Bottom
	F=Front	P=Posterior (Rear, Back, but those letters are taken)
	R=Right	L=Left
The rotations:
	+ Clockwise
	- Clockwise
	++ 180 Rotation (Two moves)

Note that the following algorithm exchanges corners diagonally on
top and bottom: (Parens for major functional components)

	(+F+P)(++T++B)(-F-P)(-R-L)(++T++B)(+R+L)(++T++B)

This is now a tool. To do the Christman Cross, we do

	-P [(+F+P)(++T++B)(-F-P)(-R-L)(++T++B)(+R+L)(++T++B)] +P

but since transforms associate, and F and P commute, and -P+P = I, we get 

	(+F)(++T++B)(-F-P)(-R-L)(++T++B)(+R+L)(++T++B) +P

Which is 20 quarter twists. 

David Christman thinks he read in Singmaster, Edition 5 that this
cross can be done in less than 20 quarter twists. Can somebody
confirm or deny this recolection.

Date: 4 Nov 1980 09:28 PST
From: McKeeman at PARC-MAXC
Subject: Re: Christman Cross algorithm described
In-reply-to: DCP's message of 4 NOV 1980 0023-EST
To: DCP at MIT-MC (David C. Plummer)
cc: CUBE-HACKERS at MIT-MC

Using Rubik Song,

	(+F+P)(++T++B)(-F-P)(-R-L)(++T++B)(+R+L)(++T++B)
becomes

            FB UUDD F'B'  R'L' UUDD RL  UUDD


Date:     12 November 1980 1849-est
From:     Paul Schauble              <Schauble.Multics at MIT-Multics>
Subject:  Cube lubrication
To:       cube-lovers @ mit-mc

Pardon me if you've seen this before, I'm not sure it got mailed.

The best material I have found for lubricating a cube is a product
called Tufoil. This is a suspension of fine Teflon particles in a
synthetic base. From the appearance, it is mostly Teflon. The only
drawbacks are that the quantity needed is fairly critical (too much will
make the outside of the cube sticky until it all wipes off) and the
stuff only comes in 8 oz bottles. This should be enough to lubricate
over 1000 cubes. Available in the automotive department at Bradlees in
Boston.


Date: 24 NOV 1980 0451-EST
From: ALAN at MIT-MC (Alan Bawden)
Subject: Old cube mail no longer on-line.
To: CUBE-HACKERS at MIT-MC

There is a klutz amoung us.

This is twice that the file of old cube lovers mail has been deleted
from my directory.  I retrieved it last time and put it in
ALAN;CUBE OMAIL and let the new messages continue to accumulate in
ALAN;CUBE MAIL.  But somebody just deleted the OMAIL again.  I will
retrieve it just once more, and if it goes away AGAIN I will just stop
bothering.

Date: 24 NOV 1980 1743-EST
From: ALAN at MIT-MC (Alan Bawden)
To: CUBE-HACKERS at MIT-MC

ALAN;CUBE OMAIL is back on disk as ALAN;CUBE XXMAIL .

Date: 24 NOV 1980 1926-EST
From: DCP at MIT-MC (David C. Plummer)
Subject:  Plumme Cross
To: CUBE-LOVERS at MIT-MC


I have an algorithm to get to the Plummer Cross from solved in
thirty quarter twists. It may get reduced if I notice I can
combine different parts of the algorithm. Description to come in
a few days.

Date: 25 NOV 1980 1308-EST
From: DCP at MIT-MC (David C. Plummer)
Subject:  30 move Plummer Cross algorithm described
To: CUBE-LOVERS at MIT-MC


Using the {Up Down Left Right Front Back} notation with clockwise
normal and ' (single quote) indicating counter-clockwise:

(B D)^7		that's right, 14 moves. this sets the back face
		up and does a couple things to the front.

	The next two are general tools to play with three corners
	and nothing else, and people should play with them for
	their utility.

(L D L') U (L D' L') U'

	This fixes front upper left, which leaves front
	upper-right, lower-right, lower-left. At this point I
	would rotate the entire cube CCW about the front-back
	axis so that these become front upper-left, upper-right,
	and lower-right, in which case the following transform
	becomes (in the new coordinates)
		U' (R' D' R) U (R' D R)
	but in the old coordinates, we get

R' (D' L' D) R (D' L D)

(+ 14. 8. 8.) ==> 30.

I don't think I made a mistake. Have fun. 

IF: Plummer == 2*Christman
 THEN Christman=14 (must be even)
	and also Plummer=28.


Date: Tuesday, 2 December 1980  10:37-EST
From: Jonathan Alan Solomon <SOLOMON at RUTGERS>
To:   CUBE-LOVERS at MIT-MC
cc:   Solomon at RUTGERS
Subject: Source to CUBE to run it locally (!)

Could someone please point to where the source to the CUBE
program lives? Some users here want to use it and don't
have access to the net. I have the newest MACLISP, so there
is no problem there.

Thanks!
JSol

Date:     2 December 1980 1040-est
From:     Bernard S. Greenberg       <Greenberg at MIT-Multics>
Subject:  Re: Source to CUBE to run it locally (!)
To:       SOLOMON at Rutgers
Cc:       CUBE-LOVERS at MIT-MC, Solomon at Rutgers

I repeat for all those interested, the :CUBE program, designed
and implemented by yours truly, lives in the BSG directory on
AI. The file .INFO.;CUBE INFO describes, at its end, the organization
of the source modules.  I will give anyone any reasonable amount
of help bringing it up elsewhere, and would very much like
to know if you succeed with it.  -bsg

 Date: 12/02/80 1059-EDT
From: PLUMMER at LL
Subject: Cube lube
To: CUBE-LOVERS at MIT-MC

I have found that talcum powder works.  Take apart your cube and
sprinkle generously.  For a day or two after some powder will leak
out, but this brushes off easily.  --Bill
-------

Date:  3 December 1980 0050-EST
From: James.Saxe at CMU-10A (C410JS30)
To: CUBE-LOVERS at MIT-MC
Subject:  28 qtw Plummer Cross
CC: James.Saxe at CMU-10A
Message-Id: <03Dec80 005058 JS30@CMU-10A>

First,

    F (LL RR) F B (LL RR) F.

Now, rotate the whole cube 1/4 turn clockwise about the up-down axis
(so that the top stays on top and the right side becomes the front)
and finish up with

    F (LL RR) F B (LL RR) F  (UU DD).

Date:  4 DEC 1980 2300-EST
From: DR at MIT-MC (David M. Raitzin)
To: CUBE-LOVERS at MIT-MC

Hi!  I'm new to this list, since I just found out about it, so I might
repeat some stuff that has already been said.  I guess I want to say
two things.  First of all, I saw Greenberg's message saying that some
guy who has written a book on the cube or something does not believe
that solutions in two minutes are possible.  Well, even though I can
not do it in two minutes (it takes me just under four minutes to do
it), my roommate does it in a minute and a half (I timed it, so it's
no lie).  Second, I've never heard of anyone using the same algorithm
that me and my roommate use.  We get the top row first, then we get
the second row, and finally get the third.  The only other solution
that I've seen (and/or heard of), and that is from anyone else who has
solved the cube, is getting all the eight corners, and then getting
the middles.  (Greenberg's Lisp machine program also solves it in this
manner.  In fact, I was quite surprised to see it done in that manner,
but as time went on, I realized that everyone I know of does it in
that manner too.)  Is that true?  Does anyone have any other algorithm
to the two I've described?  Also, I've heard that the most optimum
solution to a randomized cube is at most 41 moves.  Is that true?  And
if so, what is the algorithm it uses.  I immagine that algorithm was
achieved on a computer is that true?  As you can see, I have a lot of
questions, but this is my first letter to the mailing list.  

					Dave

Date:  4 Dec 1980 2118-PST
From: Dave Dyer <DDYER at USC-ISIB>
Subject: solution sequence
To: cube-lovers at MIT-MC


My original algorithm was:

(1) bottom corners	by random twisting about

(2) bottom sides	by random twisting about

(3) middle sides	mostly by variants of [( r u -r -u )^2 ]^3
		with extra "u" operations between the inner and outer loops

(4) top sides		mostly by variants of ( r u -r -u )^2 ( -f -u f u)^2
		with extra reps of phase 3 to swap positions around

(5) top corner position	by variations on ( r u -r -u )^2 ( r u -r -u )z
		with extra "b" operations between the major groups

(6) top corners orientation
		mostly by repitions of (gasp!)
			A = ( r u -r -u )^2
			A f A f A f A -f A -f A -f
		... which I called the "long walk"

My typical solution times with this methodology was 15-20 minutes.  I suspect
If I operated ar warp speeds (like BSG) it would still be 5 minutes.

FORTUNATELY, I have developed better since then.
-------

Date:     5 December 1980 0022-est
From:     Ronald B. Harvey           <RHarvey at MIT-Multics>
Subject:  Your mail of 4 DEC 1980 2300-EST
To:       DR at MIT-MC
Cc:       CUBE-LOVERS at MIT-MC


Welcome to the Cube-hackers !!

Note that this is my first contribution, although I have read all
previous mail.

    First of all, there are other people who solve the cube (I, myself,
average about a minute and three quarters).  It is also part of the lore
that there are people (overseas) who do it in 50 seconds or less.

Second, while I personally get eight corners first, and then randomly go
about fixing the edge cubies two at a time or so, David Singmaster, who
publishes some notes in England (availability in previous mail),
included, in his latest edition his algorithm, which proceeds as
follows:
     Solve TOP edges
     Solve TOP corners
     Flip cube so that TOP is BOTTOM from now on
     Solve MIDDLE edges
     Solve TOP edges
     Solve TOP corners

Finally, in regards to the 'most optimum solution' (also known as God's
Algorithm), a person by the name of Morwen B Thistlethwaite (that's from
memory - I left my notes at work) is mentioned all over Singmaster's
notes as contributing algorithms.  Yes, he does use a program, and his
algorithm at the time of the 5th Edition took somewhere around 50 moves.

Date:  5 DEC 1980 0132-EST
From: SHL at MIT-MC (Stephen H. Landrum)
Subject: ADDITION
To: CUBE-LOVERS at MIT-MC

COULD I BE ADDED TO THIS LIST?
THANX.

			STEPHEN LANDRUM
			(SHL@MC)

P.S. I DIDN'T KNOW WHO TO CONTACT SINCE THERE IS NO 'CUBE-LOVERS-REQUEST'

Date:  5 DEC 1980 0135-EST
From: SHL at MIT-MC (Stephen H. Landrum)
To: CUBE-LOVERS at MIT-MC

I ALSO SOLVE THE CUBE TOP ROW, MIDDLE, THEN BOTTOM.
ONE TIME I MANAGED IT IN 58 SEC, BUT I THINK THAT WAS JUST A FREAK CASE.
MY SECOND BEST TIME IS 1 MIN. 42 SEC, I AVERAGE  ABOUT THREE MINS.

			STEPHEN LANDRUM

Date:  5 DEC 1980 0250-EST
From: ALAN at MIT-MC (Alan Bawden)
To: SLH at MIT-MC
CC: CUBE-LOVERS at MIT-MC

I have added you to the cube-lovers mailing list.
Old mail is archived in MC:ALAN;CUBE MAIL.

I'll Cc this message to CUBE-LOVERS so that anyone else looking
through the (recent) old mail, will find out that I am a person to
mail "add me" requests to.

Date: 5 Dec 1980 00:29 PST
From: Woods.PA at PARC-MAXC
Subject: Re: DR's message of 4 DEC 1980 2300-EST
To: DR at MIT-MC (David M. Raitzin)
cc: CUBE-LOVERS at MIT-MC

I've hardly ever found two people who solve the same way.  The most
common method I've found around Stanford is to go for all the edges,
then the corners.  My own method is to get the top corners, then the
top edges (via operations like -l r f^2 l -r & similar tweaks) then three of
the middle edges (via things like l^2 u^2 r^x u^2 l^2, where x=1, 2, or -1),
then get the other four corners in the right place (requiring at most one
corner-swap, for which I have a fairly simple macro), then get those four
corners oriented using the "rFUfuR" commutator macro, which also randomly
alters the remaining middle edge, then get the last five edges, then flip
edges as necessary.  My typical time is about 5 minutes.  "Best" times are
meaningless, since anybody can luck out once or twice; the best measure
of your solving speed (in my opinion) is your WORST solving time over
your most recent ten or so attempts.  Of course, for real fun, pick some
pretty pattern and solve to it without going through the normal solved state!

	-- Don.


Date:  5 DEC 1980 1240-EST
From: DCP at MIT-MC (David C. Plummer)
Subject:  My 6 faces worth
To: CUBE-LOVERS at MIT-MC


The technology exists to solve the cube in any way possible.
I.E., tools exists to do the most primitive of operations to
individual cubies (rotating corners, flipping edges, moving three
corners around, and moveing trhee edges around). Using these, any
method of solving is possible (even solving a corner, then the
edges around that corner, then the corners around those edges,
etc, all the way toward the other corner). Since everybody tends
to develop their own technology, they develop their own favorite
way of solving.

Personally, at present, I solve

	the top edges
	three of the top corners
	three of the middle edges
	the last top edge

Up until here the algorithms are quite simple and now I only have
the bottom and one middle edge, which I solve in whatever way
looks most promising. Using this sequence, I am consistently on
the order of three minutes (tending toward 2.5 as I get more
familiar with the method)

Intuitively, I believe that God's algorithm is less than 35 or
so, and I would not be surprised if it is less than 30. Under 25
would really surprise me.

Date: 5 Dec 1980 1038-PST (Friday)
From: Mike at UCLA-SECURITY (Michael Urban)
Subject: Cube approach survey?
To: cube-lovers at mc

  Top edges, top corners, middle edges, bottom edges, bottom
corner placing, bottom corner orientation.  Takes me about 4
minutes, but I'm out of practice.   None of the steps requires
any real unusual tools.  I suspect I got in the habit of
solving the cube in that order because the initial challenge
was "get one face".  Then I realized that the next step must
surely be "get one face so that it matches the center
faces of the sides".  
  The beauty of the cube is that, unlike Instant Insanity
or its cousins, the Cube's solutaion can not be stumbled
across by accident.  Instead, it's a learning experience,
as you learn to manipulate things, then to manipulate
cubies without messing up what you already have
done.  This gets progressively more sophisticated as
more of the cube comes into place.  Definitely the
most satisfying puzzle I've ever dealt with.  However,
I'd be interested to find out how many as-yet-unsolved
Cubes are sitting even now on the shelves of disgusted customers
who thought they were great puzzlists because they solved Soma,
but weren't patient enough to figure out how to get those
last two corner cubies oriented right.

	Mike
-------


Date: 5 December 1980 16:28-EST
From: Alan Bawden <ALAN at MIT-MC>
Subject: cube-lovers-request
To: CUBE-LOVERS at MIT-MC

OK, I have created the cube-lovers-request mailing list.

Date:     5 December 1980 1848-est
From:     David C. Plummer           <Plummer.SIPBADMIN at MIT-Multics>
Subject:  That 28 move Plummer Cross
To:       Cube-Lovers@MIT-MC

Something I noticed about that 28 move algorithm to get to the Plummer
Cross.  Any twist can be the first one made to take the Cross back to
home.  The last instructions are (UU DD) which can obviously be (UU D'D'),
(DD UU), and (DD U'U'), so possible backward applications of the
algorithm can start with U,D,U',D'.  Add in the three way rotation
symmetry of the cross and you can get all 12 moves as viable first
moves.  So, any move takes you closer to home (by this algorithm), so it
has the chance of being the most distant from home.  This requirement
(any moves takes you closer) was mentioned by ALAN some time ago.  (I
don't hold that the Plummer Cross is the most distant.)

Date:  5 December 1980 19:10 est
From:  Greenberg.Symbolics at MIT-Multics
Subject:  Re: That 28 move Plummer Cross
To:  Plummer.SIPBADMIN at MIT-Multics (David C. Plummer)
cc:  Cube-Lovers at MIT-MC
In-Reply-To:  Message of 5 December 1980 18:48 est from David C. Plummer

It is a corollary of what Plummer just said, that
the "maximally distant" configuration, which is God's number of
moves (the maximal length of God's algorithm) would then be
no more than 28 moves from home (bounding God's number thusly).

Date:  5 Dec 1980 1624-PST
From: Dave Dyer <DDYER at USC-ISIB>
Subject: Re:  Re: That 28 move Plummer Cross
To: Greenberg.Symbolics at MIT-MULTICS
cc: cube-lovers at MIT-MC
In-Reply-To: Your message of 5-Dec-80 1910-PST


 Not true!  It may be just a local maximum.
-------

Date:  5 December 1980 21:15 est
From:  Greenberg.Multics at MIT-Multics
Subject:  Re:  Re: That 28 move Plummer Cross
To:  Dave Dyer <DDYER at USC-ISIb>
cc:  cube-lovers at MIT-MC
In-Reply-To:  Message of 5 December 1980 19:24 est from Dave Dyer

No.  It is DEFINITIONALLY a local maximum, because any move moves it
"closer". DCP was asserting its candidacy for the "ultimate" maximum, if
not its fact.

Date: 6 Dec 1980 13:40 PST
From: McKeeman.PA at PARC-MAXC
Subject: Re: That 28 move Plummer Cross
In-reply-to: Plummer.SIPBADMIN's message of 5 December 1980 1848-est
To: David C. Plummer <Plummer.SIPBADMIN at MIT-Multics>
cc: Cube-Lovers@MIT-MC, Greenberg.Symbolics at MIT-MULTICS, DDYER at
 USC-ISIB, Hofstadter at SU-AI

David,

Interesting observation!  Your argument about the Plummer Cross being a local
maximum in QTW metric holds for any completely symmetrical configuration of
the cube, independent of the algorithm used to reach it.  There are a lot of them
(including "home").

It raises the question:  Can the maximally distant point be proven to be
symmetric?  If so, the search for a bound is much simplified.

Bill


Date: 6 Dec 1980 13:57 PST
From: McKeeman.PA at PARC-MAXC
Subject: A Top-Middle-Bottom solution
To: CUBE-LOVERS at MIT-MC

Since I had this writeup in "sendable" form, I thought I would put it out.  Most
of you already have this one or better, so just delete it if you are not interested.

---------------------------

Date: 19 Jun. 1980 5:57 pm PST
From: McKeeman
To: McKeeman


File: [Ivy]<McKeeman>Rubik.txt

Kertesz' Algorithm for the Rubik Cube.  McKeeman  --  June 19, 1980

Rubik's Cube is a 3x3x3 cube such that each of its 3x3 planes can rotate freely
about its perpendicular axis.  There are 54 visible squares, colored in six sets of
9.  There are 26 visible cubelets, each colored uniquely.  (Commercially available
cubes are not colored consistently.)  The edge cublets have two colors; the corner
cublets have three.

The inventor is a Hungarian architect; my solution comes from Adam Kertesz, a
Hungarian programmer I met in Poland.   It may be less precise than you like,
but without pictures it is just too much to give all the details.  With a little
struggle this should get you to a solution.

The Problem:  From an arbitrary starting position, arrange the cube so that each
3x3 surface is of a uniform color.

Notation:  One can limit moves to 90o rotations of the six surfaces without losing
anything.  It is convenient to describe the moves in terms of the orientation of
the cube rather than the color.  The names of the faces are:
  U=up
  D=down
  R=right
  L=left
  F=front
  B=back.
A move of R, for example, is a 90o clockwise rotation of the right face, RR = R^2
is 180o, and R'=RRR= R^3 is 270o, or 90o counterclockwise.  For any move M,
M'M is a no-op.

Macros:  There a lot of macro-moves that are useful and interesting.  I will use
six here:
  X = R'F'UFUF'U'FRUU
  Xr = LFU'F'U'FUF'L'UU
  Y = RFFDFD'FFR'
  Yr = L'FFD'F'DFFL
  Z = R'DRFDF'
  Z' = FD'F'R'D'R

It is not trivial to use these things.  The general idea is to decide on the
orientation of the cube, and then follow the directions mechanically.  Note that
the color of the central square of every face is unchanged during a macro.  That
is the easiest way to keep from getting messed up.  I mumble to myself "Red is
Up, Blue is Front" or some such thing each time just before I dive into a macro. 
I find it is even helpful to close my eyes during a macro when doing it from
memory!  

Kertesz' Algorithm:  Generally speaking the idea is to get the cube right one
layer at a time.  They get increasingly hard (tedious) because you must be
careful not to destroy work already done.

----------------
Layer 1:  Pick Up color.

  Step 1.  Do Up edges.  The four Up edges must be moved between the two
           center squares that have the same colors:

    For each edge do
      (a) hold target position UF  (i.e., up front, nearest you),
      (b) get needed cubelet into bottom layer,
      (c) rotate it under target position using D^n,
      (d) do F^2, in which case you are done or it needs flipping,
      (e) in which case F'UL'U' finishes it.

You should now have the top "+" shape correct and matching the side center
squares.

  Step 2.  Do Up corners.  The four Up corners must be moved so that all three of
           their colors match their neighbors.

    For each corner do
      (a) hold target position FUR  (i.e., up right, nearest your right hand),
      (b) get needed cubelet into bottom layer,
      (c) rotate it under target position using D^n,
      (d) either DFD'F' or D'R'DR will bring it up,
      (e) it may need rotation, in which case use Z for 120o clockwise, Z'
          for counterclockwise.

You should now have the top layer correct and matching the side center squares.

----------------
Layer 2:   The four remaining edges on layer two are next.  Y and Yr leave Up
surface invariant.

  For each layer 2 edge target do
    (a) get needed cubelet into down layer (using Y or Yr if needed),
    (b) rotate it 90o from under target using D^n,
        matching color of nearest center square,
    (c) hold needed cubelet DF
    (d) if it needs F, do Y to bring it into position,
    (e) if it needs F', do Yr.

----------------
Layer 3:   Turn puzzle over.  The macros X and Xr are used over and over
again.  The sequence of events is to place the four up corners, place the four up
edges, flip the edges if necessary, and finally rotate the corners if necessary.

 Step 1.  Place Up corners.  One can always be rotated into position with U^n. 
Among other things, X will exchange the RFU and RBU corners.  At most four
applications of X will place all four corners.  It takes a little forethought.

  Step 2.  Place Up edges.  X will also exchange the LU and FU edges.  Xr will
exchange the RU and FU edges and the LFU and LBU corners.  Used in pairs X
and Xr can exchange edges without changing corners.  That is, one use
exchanges corners, the next carefully chosen one restores them.

  Step 3.  Flip Up edges as necessary.  X^2 will flip both LU and FU edges. 

  Step 4.  Rotate Up corners as necessary.   The macro ZU^nZ'U^-n will rotate
two Up corners, one 120o and one 240o.  Place the one needing 120o in FUR, do
Z, do U^n to bring one needing 240o into FUR, do Z'U^-n.  Don't look at the
mess after the first rotation; it will destroy your morale.

-----------------

You should now be done.  It took me a couple of days to be able to solve the
thing reliably.  Enjoy.

P.S. Real Rubniks will want to write for:

Notes on Rubik's 'Magic Cube', David Singmaster, Mathematical Sciences and
Computing, Polytechnic of the South Bank, London SE1 0AA, England (send $5).


------------------------------------------------------------


Date: 6 Dec 1980 14:16 PST
From: McKeeman.PA at PARC-MAXC
Subject: Re: That 28 move Plummer Cross
In-reply-to: Greenberg's message of 6 December 1980 1644-est
To: Bernard S. Greenberg <Greenberg at MIT-Multics>
cc: McKeeman, Cube-Lovers at MIT-MC, DDYER at at MIT-Multics,
 Plummer.SIPBADMIN at MIT-Multics

I do not follow the reasoning.  It seems quite possible that there is a
non-symmetric local maximum.  In any case, it is not a definition, but rather a
proof that needs doing.  It is certainly true that a move from a non-symmetric
configuration will either 
  a.  get closer to home
  b.  stay the same distance from home
  c.  get further from home.
Furthermore, it is obvious that there are usually both (a) and (c) cases.   What I
don't see is the argument that there must always be a (c) case.

One way of looking at it is that there is an enormous graph connecting all
solutions by one QTW moves.  Nearly all nodes are non-symmetric.  You argue
that from every non-symmetric node there is a move to a node that is further
from home.  I am willing to be convinced, but am not yet, and mere probability
favors the following: 

Conjecture:  There exists a non-symmetric configuration from which every move
leads to a position that is closer to home.

Bill


Date:  6 Dec 1980 1428-PST
From: Dave Dyer <DDYER>
Subject: Re: That 28 move Plummer Cross
To: McKeeman.PA at PARC-MAXC
In-Reply-To: Your message of 6-Dec-80 1340-PST
Remailed-date:  6 Dec 1980 1429-PST
Remailed-from: Dave Dyer <DDYER at USC-ISIB>
Remailed-to: cube-lovers at MIT-MC


 While I believe it is true that all symmetric positions are local maxima,
It is definitely NOT true that all local maxima are symmetric.  As
a counterexample, consider that a repeated permutation is at a local
maximum 1/2 way through its period.  So for example, the Idempotent
( r u -r -u )^6 is at a familiar but unsymmetric local maximum
halfway through.
-------

Date:  6 DEC 1980 1745-EST
From: DCP at MIT-MC (David C. Plummer)
Subject: Re: That 28 move Plummer Cross
To: McKeeman.PA at PARC-MAXC
CC: CUBE-LOVERS at MIT-MC

    Date: 6 Dec 1980 13:40 PST
    From: McKeeman.PA at PARC-MAXC
    In-reply-to: Plummer.SIPBADMIN's message of 5 December 1980 1848-est
     USC-ISIB, Hofstadter at SU-AI

    David,

    Interesting observation!  Your argument about the Plummer Cross being a local
    maximum in QTW metric holds for any completely symmetrical configuration of
    the cube, independent of the algorithm used to reach it. There are a lot of them
    (including "home").

    It raises the question:  Can the maximally distant point be proven to be
    symmetric?  If so, the search for a bound is much simplified.

    Bill

I don't know exactly where to start my comments. For one thing,
the Plummer cross is not totally symmetric. What I stated
(actually ALAN, but I seem to be the culprit now):
	It is necessary for the maximal state to have the quality
	that any quarter twist brings you closer to home.
It is also true that any symmetric state also has this quality.
What I noted was that the 28 move algorithm given shows that the
Plummer Cross also fulfills this. HOWEVER, there may exist a 26
or 24 move algorithm such that only 6 of the 12 possible moves
may be done first in order to fix it.

About your question, even if you could prove the maximal distant
point is symmetric, we still cannot prove how far away a
configuration is away from home. If you could prove that, you
would also God's Algorithm.

Date:  6 DEC 1980 1748-EST
From: DCP at MIT-MC (David C. Plummer)
Subject:  food for twisting
To: CUBE-LOVERS at MIT-MC


Here is a piece of food for thought: why can't there be two (in
general more than one) maximally distant point). For example, the
24 Plummer Crosses and the "every edge flipped" configuration.

And the obvious question is: how far apart are these maximally
distant types. Between different types is probably more
interesting than between different "reflections" of the same
type.

Date:  6 DEC 1980 1846-EST
From: DCP at MIT-MC (David C. Plummer)
Subject: Re: That 28 move Plummer Cross
To: McKeeman.PA at PARC-MAXC
CC: CUBE-LOVERS at MIT-MC


    Date: 6 Dec 1980 14:16 PST
    From: McKeeman.PA at PARC-MAXC
    In-reply-to: Greenberg's message of 6 December 1980 1644-est
     Plummer.SIPBADMIN at MIT-Multics

    I do not follow the reasoning.  It seems quite possible that there is a
    non-symmetric local maximum.  In any case, it is not a definition, but rather a
    proof that needs doing.  It is certainly true that a move from a non-symmetric
    configuration will either 
      a.  get closer to home
      b.  stay the same distance from home
      c.  get further from home.
    Furthermore, it is obvious that there are usually both (a) and (c) cases.   What I
    don't see is the argument that there must always be a (c) case.

    Bill

Except from solved, there always exists a move taking you closer
to home. Always: There is NEVER (by the QTW metric) a move that
keeps you the same distance, and from the maximally distant state
it is IMPOSSIBLE to get further from home. Notice I have said
nothing about symmetry.

Date:  6 DEC 1980 1841-EST
From: DCP at MIT-MC (David C. Plummer)
Subject: Re: That 28 move Plummer Cross
To: DDYER at MIT-MC
CC: CUBE-LOVERS at MIT-MC

    Date:  6 Dec 1980 1428-PST
    From: Dave Dyer <DDYER>
    In-Reply-To: Your message of 6-Dec-80 1340-PST
    Remailed-date:  6 Dec 1980 1429-PST
    Remailed-from: Dave Dyer <DDYER at USC-ISIB>
    Remailed-to: cube-lovers at MIT-MC


     While I believe it is true that all symmetric positions are local maxima,
    It is definitely NOT true that all local maxima are symmetric.  As
    a counterexample, consider that a repeated permutation is at a local
    maximum 1/2 way through its period.  So for example, the Idempotent
    ( r u -r -u )^6 is at a familiar but unsymmetric local maximum
    halfway through.
    -------

I think what most of mean by LOCAL MAXIMUM is that ANY twist will
bring you closer to home. The proposed counterexample can be
taken farthur away from home (eg (R U -R -U)^3 then D)

Date: 6 Dec 1980 16:30 PST
From: McKeeman.PA at PARC-MAXC
Subject: Re: That 28 move Plummer Cross
In-reply-to: Greenberg's message of 6 December 1980 1836-est
To: Bernard S. Greenberg <Greenberg at MIT-Multics>
cc: McKeeman, Cube-Lovers at MIT-MC, Plummer.SIPBADMIN at MIT-Multics

OK.  There may be non-symmetric local maxima.  I feel better; thought maybe I
was just so dense I couldn't grok the obvious.

As for a definition of symmetric: For the purposes of local maximum argument,
each of the 6 basic QTW must have the same effect.  That is, the configuration
must be isomorphic under the rotation group of the whole cube (of order 24).

Bill


Date: 6 Dec 1980 16:42 PST
From: McKeeman.PA at PARC-MAXC
Subject: Re: That 28 move Plummer Cross
In-reply-to: DCP's message of 6 DEC 1980 1745-EST
To: DCP at MIT-MC (David C. Plummer)
cc: McKeeman, CUBE-LOVERS at MIT-MC

David,

Suppose one could prove local maxima had configurations that were invariant
under the rotation group of the whole cube.  (I am not at all sure it is even
true.)

There are a small number of such symmetric configurations, and they could
probably be easily tabulated.  One of them would have to be maximally distant
from home.  Thus if we had a QTW solution for each of them, the maximum
over that set would bound God's Algorithm.

I see no reason to believe that a QTW cannot take you between two solutions
that are at the same distance.  As DPC pointed out, there are a lot of even
identity paths.  E.g., (RUR'U')^6.  The two furthest points on the path are (by
symmetry) necessarily equally distant, yet connected by a QTW. 

Bill 


Date: 7 December 1980 00:47-EST
From: Alan Bawden <ALAN at MIT-MC>
Subject: Maximally distant states
To: McKeeman.PA at PARC-MAXC
cc: CUBE-LOVERS at MIT-MC

    Date: 6 Dec 1980 16:42 PST
    From: McKeeman.PA at PARC-MAXC

    I see no reason to believe that a QTW cannot take you between two solutions
    that are at the same distance.  As DPC pointed out, there are a lot of even
    identity paths.  E.g., (RUR'U')^6.  The two furthest points on the path are (by
    symmetry) necessarily equally distant, yet connected by a QTW. 

I am not sure I understand what you are trying to say here.  But I do
know that a single quarter twist can never leave you the same distance
from anything.  This is because a single quarter twist is a odd
permutation of the "stickers".  Thus if you are N quarter twists away
from something, a single quarter twist will leave you N-1 or N+1
quarter twists away.  (And hence the proof that any quarter twist will
bring you closer from a maximally distant state.)

I'm not sure how to apply this to your statement that perhaps a "QTW"
can take you "between two solutions that are at the same distance".

Date:  7 DEC 1980 0108-EST
From: MJA at MIT-MC (Michael J. Aramini)
Subject: maximally distant state
To: CUBE-LOVERS at MIT-MC

well it is possible that to maximally distant states are half twist apart
also if you count half twists as one twist (i dont, but its
still worth thinking about) does that change the set of
maximally distant states?

also it is possible that there exists states for which all directions
lead closer to home (and twist put the cube in a state closer to home)
but the state is not necessarily maximally distant (to use
a continous analogy, think of think of a hill in a funtion of
two variables, that is not necessarily the maximum value of the
function)

Date:  7 DEC 1980 0724-EST
From: DCP at MIT-MC (David C. Plummer)
Subject: maximally distant state, setting the record straight
To: MJA at MIT-MC
CC: CUBE-LOVERS at MIT-MC

    Date:  7 DEC 1980 0108-EST
    From: MJA at MIT-MC (Michael J. Aramini)

    well it is possible that to maximally distant states are half twist apart
WRONG! (I assume you meant "two" for "to" and typo'ed). Read 
ALAN's previous message. In the half twist metric, there exist 
odd distances away, and there exist even distances away. A QTW
takes the cube from odd to even or from even to odd. The maximally
distant state is the state such that the fewest number of QTW
required to solve it is maximized. This must be odd OR even, and
thus, two states that are maximally distant must be both odd or
both even, which means the distance between them is even, or an
EVEN number of QTW. A single QTW is ODD, and thus cannot separate
maximal states.
    also if you count half twists as one twist (i dont, but its
    still worth thinking about) does that change the set of
    maximally distant states?
Maybe it does, maybe it doesn't. It is much harder to tell because
counting half twists has no analog to the QTW odd/even property of
distance, and this is one reason several of us don't count half twists.
For example, (R L R) and (L [RR]) are equivalent manipulations, but in
half twist counting, one is three and the other is two moves. (assume []
means grouping two moves into one.)

    also it is possible that there exists states for which all directions
    lead closer to home (and twist put the cube in a state closer to home)
    but the state is not necessarily maximally distant (to use
    a continous analogy, think of think of a hill in a funtion of
    two variables, that is not necessarily the maximum value of the
    function)
We have been saying this all along! Simple example: (RRLL UUDD FFBB)
is a local max (any twist takes you closer), and it is definitely not
absolute max (abs max must be at least 21 from combinatoric arguments).

Date:     6 December 1980 1836-est
From:     Bernard S. Greenberg       <Greenberg at MIT-Multics>
Subject:  Re: That 28 move Plummer Cross
To:       McKeeman.PA at PARC-Maxc
Cc:       Cube-Lovers at MIT-MC, Plummer.SIPBADMIN at MIT-Multics

I confess to not only having no proof to this proposition, but no
longer believing it.  The false equivalence of "symmetric" and
"locally maximal" seemed to me  visually obvious, given that all symmetric
positions are in fact locally maximal.  If one believes this
subconsciously, it is but a short step to the conscious "proof" that
all asymmetric positions are but "on the way" to a symmetric, maximal
one; this of course, is bogus.  Now a good definition of symmetric
is also needed here; I assume that what we have been meaning is that
all rotations of the cube are Isomorphic.  Note that the Plummer and
Christman crosses don't qualify under this definition.  What is a good
definition?

Date:     6 December 1980 1644-est
From:     Bernard S. Greenberg       <Greenberg at MIT-Multics>
Subject:  Re: That 28 move Plummer Cross
To:       McKeeman.PA at PARC-Maxc
Cc:       Cube-Lovers at MIT-MC, DDYER at at MIT-Multics,
     Plummer.SIPBADMIN at MIT-Multics

The maximally distant point is definitionally symmetric. It was
Bawden (ALAN) who first stated this; were it NOT symmetric, there
would be some twist you could make that would make it WORSE
as opposed to the ones that would make it better. Were this
the case, it would then not be a maximum.

Date:     7 December 1980 1218-est
From:     Bernard S. Greenberg       <Greenberg at MIT-Multics>
Subject:  Symmetry of configurations
To:       McKeeman.PA at PARC-MAXC
Cc:       CUBE-LOVERS at MIT-MC

The definition of symmetry proposed by you does not seem adequate.
Plummer's Cross is NOT symmetrical by that definition.  Given
Plummer's Cross in any orientation, there are plenty of elements
of the 24-move whole-cube rotation group which will shift it to
something NOT isomorphic through a substitution of colors to the
original.  Yet, we know, intuitively, that the CP is "highly
symmetric", and it is a local maximum.  It is as though I were
saying that this equivalence of all twists "out" of this position
almost defined symmetry.....

Date:     7 December 1980 1401-est
From:     Bernard S. Greenberg       <Greenberg at MIT-Multics>
Subject:  Re: That 28 move Plummer Cross
To:       DDYER at MIT-Multics
Cc:       McKeeman.PA at PARC-Maxc, CUBE-LOVERS at MIT-MC

I do not believe that (r u r' u')^3 is a local maximum.
An move other than r, u, r', u', (e.g., l) seems to pervert
it further than any of the four which bring it back.

Date:     7 December 1980 1620-est
From:     Bernard S. Greenberg       <Greenberg at MIT-Multics>
Subject:  Re: Maximally distant states
To:        McKeeman.PA at PARC-Maxc
Cc:       CUBE-LOVERS at MIT-MC, ALAN at MIT-MC

King of bogosity! Fencepost errors abound. Here's an "even"
identity transform for you: (R R)^2.  Halfway through
this transform is ONE state.  There are no "two furthest points
on the path", there is one.  There ARE, however, two ways out of it.
And similar to all even identity transforms.

Date: 7 December 1980 16:58-EST
From: Alan Bawden <ALAN at MIT-MC>
To: CUBE-LOVERS at MIT-MC, Greenberg at MIT-MULTICS
cc: ELLEN at MIT-MC

To sum up everything we know:

1) A Symetric position is a local maximum.
2) A maximally distant state is a local maximum.
3) If Plummer's cross is 28 Qs away from home, then it is a local maximum.

All the other implications that have been flying around are unproven
as far as I know.  I welcome PROOFS of anything else that I might not
be aware of.  The volume of air-headed speculation on this subject has
reached the point where the MC mailer is so overloaded with gubbish
that a "CUBE-LOVERS Digest" is being considered.  Now that is just
plain silly given that NORMALLY we only get one or two pieces of mail
a week through this list.  If you don't understand why one of the
three facts above are true, DON'T send mail to the whole list!  Look
through the old mail (they are all explained there) or send JUST ME a
question, and I will try and answer it.

Date:  8 December 1980 1309-EST (Monday)
From: Guy.Steele at CMU-10A
To: McKeeman.PA at PARC-MAXC
Subject:  Re: That 28 move Plummer Cross
CC: cube-lovers at MIT-MC
In-Reply-To:  McKeeman.PA@PARC-MAXC's message of 6 Dec 80 19:42-EST
Message-Id: <08Dec80 130910 GS70@CMU-10A>

A QTW cannot take you between two positions of equal distance, I believe
-- is there not some parity quantity obeyed by QTW's (net twist of
the six center cubies)?  If so, then there cannot be two positions of
equal distance separated by a QTW, for then there would be an odd
length identity transformation, which would violate parity.
(In the example you gave, there are *not* two positions of
equal distance, but only one -- the halfway point.)

Date:  8 December 1980 1117-EST (Monday)
From: Guy.Steele at CMU-10A
To: ddyer at USC-ISIB
Subject:  Re: That 28 move Plummer Cross
CC: cube-lovers at MIT-MC
In-Reply-To:  Dave Dyer's message of 6 Dec 80 17:28-EST
Message-Id: <08Dec80 111740 GS70@CMU-10A>

The position halfway through the cycle of a permutation need not
be a local maximum, since moves other than the permutation cycle
might take it further away.  It is more like a saddle point (most
positions probably are in that class: at least two moves take
it closer, and at least two take it further).
There is one symmetric position which is not a local maximum:
the solved state!

Date: 8 Dec 1980 17:03 PST
From: McKeeman.PA at PARC-MAXC
Subject: A Proposed Definition of Symmetry
To: cube-lovers at MIT-MC

The discussion of local maxima for the Q measure of distance led to an informal
use of symmetry.  It is not clear to me just what symmetry is needed to carry
through the maxima argument but I suggest the following is sufficient (although
perhaps too restrictive).

Let C by the rotation group of the cube (closure of IJK: order 24)
Let G be Rubik's group (closure of UDRLFB: order 10^19 or so)

Both groups can be represented as a permutation group on [0, 1, ...53] for some
arbitrary numbering of the 54 faces.  We can also use the names UDRLFB for the
six colors; where the association is made once and for all for any given physical
puzzle.  Like U=red, F=blue, etc.).

The elements of g are 1-1 with the observable configurations of the standard
cube; and in fact are the recipes to reach the configurations from "home".  g' is
the "solution" that returns g to home.

The elements of G*C are also 1-1 with the observable configurations except now
the correspondence must also take into account the observed orientation of the
cube.

Each g in G is represented by a permutation of the cubelet faces.  Each face in g
is a fixed color.

For color X, let X[g] be the set of faces of g colored X.  |X[g]| = 9.
Let Coloring[g] = {U[g], D[g], R[g], L[g], F[g], B[g]}.

Then g is totally symmetric if for all c in C, Coloring[gc] = Coloring[g].

----

It is true that "home" and UUDDRRLLFFBB are totally symmetric by this
definition.  "home" is a minimum (special case).  UUDDRRLLFFBB is a local
maximum.

Questions:

Is there a simpler equivalent definition?

How many totally symmetric configurations are there?

Is there a less restrictive definition that guarantees local maxima?



Date: Mon, 8 Dec 1980 1956-CST
Message-id: <345171394.17@DTI>
From: aramini at DTI (Michael Aramini)
To: cube-lovers@mc
Subject: "notes on rubikics cube" booklet

can someone please tell me the address of the seller of this booklet in
England (I know a can find it in the archive file, but i prefer not
to try to edit it at a mere 300 baud, without the benefit of being
able to do an i-search (local host traps ^s's)
also, is it better to order it from this english address, or from the one
in virginia?
also, what is the current price for the booklet from each of these 2 sources?
-----


Date:  9 December 1980 1638-EST (Tuesday)
From: Dan Hoey at CMU-10A
To: McKeeman.PA at PARC-MAXC
Subject:  Re: A Proposed Definition of Symmetry
CC: Cube-Lovers at MIT-MC
In-Reply-To:  McKeeman.PA@PARC-MAXC's message of 8 Dec 80 20:03-EST
Message-Id: <09Dec80 163821 DH51@CMU-10A>

Given a subroup G of permutations, we may define a cube position P to be
G-symmetric if every permutation in G preserves P up to color
relabelling.  Explicitly, if x and y are two facelets which have the
same color in position P, then g(x) and g(y) also have the same color
in P.

Consider, for instance, the group R of whole-cube moves.  The Pons
Asinorum is R-symmetric.  Any whole-cube move moves the set of blue
facelets to positions occupied (before the move) by facelets of some
single other color.  In fact, even if the cube were reflected, this
would be true.  This can be verified by looking at one blue facelet of
the cube and realizing that you know from that alone where all the
other blue facelets are.  Letting the M denote the group composed of R
augmented by the set of whole-cube mirror reflections.  The Pons
Asinorum is also M-symmetric.

McKeeman has stated that any R-symmetric position (with the exception
of the solved state) is a local maximum.  I do not know if this is true,
but I can show that any M-symmetric position (with the same exception)
is a local maximum.

Consider a robot cubenik which knows how to do whole-cube moves, but
can only QTW the "up" face.  Clearly, this robot has no problems,
because it can move any face to the top, perform U or U', and move it
back.  The robot could get along without U' by doing U^3 instead, but if
we're counting QTWs it's not going to win any races.  Solution?  Build a
transdimensional robot which can perform whole-cube reflections.  This
robot performs U' by reflecting the cube, performing U, and reflecting
it back.  The point of this parable is that for any two QTWs x and y,
there is an element m in M for which y = (m x m').  Note also that for
any m in M, the permutation (m x m') is a QTW.

Let P be an M-symmetric position, and let (x1 x2 ...  xn) = P' be the
shortest solution of P in QTWs.  Assume that P is not the solved
position, so n > 0.  For any QTW y, I will demonstrate an n-step
solution of P which begins with y.  Write y = (m x1 m').  Since P is
M-symmetric, (P m) is a relabelling of P.  This implies that (P m P')
and therefore (P m P' m') are relabellings of I.  Therefore the sequence
((m x1 m') (m x2 m') ...  (m xn m')) = (m P' m') will essentially solve
P, up to a whole-cube move.  The existance of an n-step solution
starting with an arbitrary QTW y implies that P is a local maximum.

There is a twelve-element subgroup T of M which will suffice instead of
M for this argument.  Representing elements as permutations of faces, T
is generated by the permutations (represented as cycles):

	(F L U)(R D B)	-- Rotating the cube about the FLU-RBD axis
	(F B)(U R)(L D) -- Rotation exchanging corners FLU and RBD
	(L U)(R B)	-- Reflection in the LU-RB plane

Question:  Does there exist a position other than the solved position
and the Pons Asinorum which is T-symmetric or R-symmetric?


Date: 9 December 1980 23:57-EST
From: Alan Bawden <ALAN at MIT-MC>
Subject: Re: A Proposed Definition of Symmetry
To: Hoey at CMU-10A
cc: CUBE-LOVERS at MIT-MC

    Date:  9 December 1980 1638-EST (Tuesday)
    From: Dan Hoey at CMU-10A

    There is a twelve-element subgroup T of M which will suffice
    instead of M for this argument.  Representing elements as
    permutations of faces, T is generated by the permutations
    (represented as cycles):

    	(F L U)(R D B)	-- Rotating the cube about the FLU-RBD axis
        (F B)(U R)(L D)	-- Rotation exchanging corners FLU and RBD
        (L U)(R B)	-- Reflection in the LU-RB plane

AH!  Excellent!  (I believe you mean that last permutation to be
(L U)(R D).)  It took me a while to realize that this is the subgroup
of M that leaves the FLU-RBD "diagonal" fixed.

    Question: Does there exist a position other than the solved
    position and the Pons Asinorum which is T-symmetric or
    R-symmetric?

Hmm.  I hadn't realized that we don't really know that many symmetric
positions.  I have another favorite pattern that happens to be fully
M-symmetric.  It is the pattern obtained by "flipping" all of the edge
cubies:

       U B U
       L U R
       U F U

L U L  F U F  R U R  B U B
B L F  L F R  F R B  R B L
L D L  F D F  R D R  B D B

       D F D
       L D R
       D B D

This pattern has another interesting property, it is the only other
permutation besides the identity that commutes with every other
element of the cube group!  I have often thought that this position is
a good candidate for maximality.  Dave Plummer has shown that this
position can be also be reached in 28 moves...

Date: 10 DEC 1980 0157-EST
From: DCP at MIT-MC (David C. Plummer)
Subject:  The 28 QTW all-edge-flipper mentioned by ALAN
To: CUBE-LOVERS at MIT-MC


Indeed it can be done in 28.
Tool: flip the top edges and the bottom edges:

	(R L) (U D) (F B) (R L) (U D) (F  B) 
					/\
					||
To do the other four, insert the following things at the indicated
place
	(u b') (u' D b' u' d l' u' d f' u' d r') (b u')

There may be another place to put this to find a shorter path.
Those with time are free to try and find one.

Date: 10 DEC 1980 0834-EST
From: DCP at MIT-MC (David C. Plummer)
Subject:  A configuration symmetric wrt the first QTW
To: CUBE-LOVERS at MIT-MC


I believe the following is symmetric in the sense that any QTW
will bring you closer to home:
	All corners are rotate, and letting + indicate clockwise
	and - counterclockwise, each face has the following
	corner configuration
		+-
		-+
	(and all the edges are intact)
	a total map of the cube corners might look like
		   +-
		   -+
		+- -+ +- -+
		-+ +- -+ +-
		   +-
		   -+
Each face is essentially the same: edges OK and all corners
rotated so that opposite corners are rotated in the same
direction. It is rather intuitive to me that rotating a face
clockwise is the same as rotating the face counterclockwise. This
fulfills the condition needed for maximality, but what flavor of
symmetric is it (if the symmetry is easily describable). Also,
does anybody have a 28 QTW algorithm (OR LESS!!) to go between
solved and this position.

Date: 14 December 1980 1916-EST (Sunday)
From: Dan Hoey, Jim Saxe
To: Cube-lovers at MIT-MC
Subject:  Symmetry and Local Maxima (long message)
Reply-To: Dan Hoey at CMU-10A
Message-Id: <14Dec80 191649 DH51@CMU-10A>

                      Symmetry and Local Maxima
				-- Dan Hoey
				   Jim Saxe

1. Introduction
===============

	In this note, we attempt to give a uniform treatment of the
issues raised in the recent discussions of symmetry and local
maxima.  We have attempted to restate and justify the correct
observations on these subjects that have been made in mail to
cube-lovers in recent days and also to refute a number of incorrect
ones.  We include a description of 71 local maxima, which we believe
to be all of the local maxima that can be proven using known
techniques other than exhaustive search.

	We let G denote the "Rubik group", consisting of all
transformations of the cube which can be achieved by twisting
faces.  G does not include transformations which require movements
of the whole cube.  Also, G does not take account of the
orientations of the face centers.  We will defer discussion of the
Supergroup, in which face center orientations are significant (but
whole-cube motions still excluded), to Section 5 of this message.
We will sometimes (particularly towards the end of this message)
take the liberty of identifying a transformation with the position
reached by applying that transformation to SOLVED.

	We let Q = {U, D, L, R, F, B, U', D', L', R', F', B'} be
the set of all possible quarter-twists.  Q is a subset, but not a
subgroup, of G.  The set {U, D, L, R, F, B} of all clockwise
quarter-twists is called Q+, and the set {UU, DD, LL, RR, FF, BB}
of all half-twists is called Q2.  The "length" of an element s of G
(denoted |s|) is the length of the shortest sequence of quarter-
twists whose product is s.  The "distance" between two elements, s
and t, of G is the length of the shortest r such that t = s r.  Note
that the length of s is same as the distance between s and the
identity permutation.  Note also that we are measuring distance in
the "quarter-twist metric."  We defer discussion of the "half-twist
metric" to Section 5.

	Two elements, s and t, of G are "neighbors" if there is
some q in Q such that t = s q (i.e., if the distance between s and
t is 1).  An element, s, of G is said to be a "local maximum" if no
neighbor of s is longer than s.  It is a consequence of parity
considerations that all neighbors of any local maximum, s, have the
same length, namely |s| - 1.  Conversely, any element s of G (except
for the identity permutation) whose neighbors are all equally long
must be a local maximum.  [Anyone who *still* doesn't understand why
neighbors cannot be equally long in the quarter-twist metric should
either send mail to one of the authors, or learn about even and odd
permutations from a book on group theory and think about how a
quarter-twist permutes the positions of the corner cubies.]

2. Symmetry
===========

	It has been asserted that any "symmetric" element of G must
have all its neighbors equally long and must therefore be a local
maximum (or the identity).  The first occurrence of this assertion
in cube-lovers mail was by Alan Bawden in his message of 10 Sep
1980, 11:09 pm PDT.  The recent spate of messages on this subject
has made it clear that Bawden's notion of symmetry was not clearly
defined.  In what follows, we make the notion of symmetry more
precise and categorize those kinds of symmetry for which the above
assertion is correct.

	We let M denote the group generated by all rotations and
reflections of the whole cube and C denote the subgroup of M which
contains only the rotations.  C has 24 elements (any of the six
faces can be put on top, after which any of the four adjacent faces
can be put in front, uniquely determining the positions of the
remaining faces).  M has 48 elements (six choices for U, then four
for F, then two for L).  [Our use of G and C agrees with that in
McKeeman's note of 8 Dec 1980, 17:03 PST.  Hoey's note of 9 Dec
1980, 16:38 EST used the letter R rather than C for the latter
group, a practice which we hereby retract.]

	Let G+M be the group of all transformations achievable by
any sequence of face twists and/or whole cube moves, including
reflections.  Note that G is a subgroup of G+M and that the elements
of G are precisely those elements of G+M which leave the positions
of the face centers fixed (to forestall possible confusion, we
remark, at certain risk of belaboring the obvious, that the phrases
"face center positions" and "face center orientations" have
different meanings).  We say that two elements, s and t, of G+M are
"M-conjugates" of each other if there exists some m in M such that
t = m' s m.  We assert the following results without proof because
they are obvious.

    Fact 1:  Any M-conjugate of an element of Q is an element of Q,
	and any M-conjugate of an element of G is an element of G.

    Fact 2:  If two elements of G are M-conjugates, they are equally
	long.

[Anyone who does not consider the preceding facts obvious is urged
to direct further inquiries to one of us rather than bothering
everyone on the mailing list.  The same goes for anyone who believes
that any other assertions made in this message are in error; this
procedure will help to reduce either duplication of erratum notices
or proliferation of false counterexamples.]

	Let W be a subgroup of M.  Then an element s of G is said to
be W-symmetric iff s w = w s (or, equivalently, w s w' = s) for
every w in W.  [This definition is equivalent to Hoey's definition
(9 Dec 1980, 16:38 EST) as we will now show.  By a "recoloring" of
the cube, we intuitively mean an operation which "consistently"
changes the colorings of all facelets of the cube.  Note that the
permutation performed by a recoloring depends not only on the
chosen mapping of colors to colors, but also on the configuration
the cube is in when we start recoloring.  Thus, a particular mapping
of colors to colors doesn't appear to correspond to any fixed
element of G+M.  This is not a particularly satisfying situation.  We
can rectify this situation by always doing the recolorings when the
cube is in the SOLVED state, that is, by thinking of recoloring a
configuration as pre-multiplication of the permutation that
achieves that configuration (from SOLVED) by a recoloring of SOLVED.
More succinctly, two elements, s and t, of G+M are equivalent up to
recoloring iff there is some m in M such that t = m s.  Thus, by
Hoey's definition, an element s of G is W-symmetric iff for every w
in W there is some m in M such that
		s w = m s.
But s doesn't move the face centers.  So the only way m can be chosen
so that m s will leave the face-centers in the same places as s w is
to pick m = w.]

3. Transitivity
===============

	For which choices of W can we guarantee that W-symmetric
patterns are local maxima?  To approach this question, we introduce
the notion of Q-transitivity.  Recall that Q is the set (not a
group) of all quarter-twists.  We define a subgroup W of M to be
"Q-transitive" iff for each two elements p and q of Q there is some
w in W such that q = w' p w.  (Q+-transitivity and Q2-transitivity
are defined analogously) We now come to our principal result:

    Theorem 1:  Let W be any Q-transitive subgroup of M and let s be
        any W-symmetric element of G.  Then any two neighbors of s
	are m-conjugates of each other.

    Proof:  Let p and q be any two elements of Q.  We must show that
        sq is an M-conjugate of sp.  Since W is Q-transitive, produce
        some w in W such that q = w' p w.  Thus,
			s q = s w' p w
			    = w' (s p) w.
        Since W is a subgroup of M, w must be an element of M.  So
	the definition of M-conjugacy is satisfied.
	Q.E.D.

    Corollary:  Let W be any Q-transitive subgroup of M and let s be
        any W-symmetric element of G other than the identity.  Then
        s is a local maximum.

	If an element s of G is both V-symmetric and W-symmetric,
where V and W are subgroups of M, then it follows that s is
V+W-symmetric, where V+W is the closure of the union of V and W
(that is, V+W is the group of all elements of M which can be
expressed as the product of a sequence of elements of the union of
V and W).  Thus we may unambiguously define the "symmetry group" of
any element s of G as the largest subgroup W of M such that s is
W-symmetric.  The elements of this group will be precisely those
elements of M which commute with s.  To see that this set of
elements forms a group, simply note that for any elements v and w
of M that commute with s,
	1.  (v w) s = v (w s) = v (s w) = (v s) w = (s v) w = s (v w),
	    and
	2.  v' s = v' s v v' = v' v s v' = s v'.

	By the corollary to Theorem 1, any element of G (except the
identity) whose symmetry group is Q-transitive is a local maximum.
In order to find local maxima, we will first find the Q-transitive
subgroups of M.  The search for Q-transitive subgroups is simplified
by realizing that the number of elements of any Q-transitive
subgroup W must be a multiple of twelve.  To show this, it will be
useful to introduce another way of looking at the group M, namely
as a group of permutations on the set Q.  We associate each element
m of M with a 1-1 function <m> from Q to Q defined by the rule
		<m>(q) = m' q m    for all q in Q.
These functions form a group under the operation of functional
composition, which we will write in the left-to-right manner so
that (<m>*<n>)(q) is definitionally equivalent to <n>(<m>(q)).  Call
this group <M>.  The function <> which maps each element m of M to
the corresponding element <m> of <M> is an isomorphism as may be
seen by noting that for any two elements m and n of M and for any
element q of Q,
		(<m>*<n>)(q) = <n>(<m>(q))
			     = n' (m' q m) n
			     = (m n)' q (m n)
			     = <m n>(q)
The isomorphism <> maps each subgroup of W of M into a subgroup <W>
of <M>, while <M> is in turn a subgroup of the group of all
permutations of Q.  If W is Q-transitive, then <W> is transitive on
Q in the usual group-theoretic sense that for any two elements p
and q of Q there is some element <w> of <W> such that <w>(p) = q.
This may be taken as motivation/"justification" for our use of the
term "transitive" in the former context.  It is a well-known result
of group theory that every transitive group of permutations on a
set X has a number of elements that is a multiple of the number of
elements of X.  This proves our claim that each Q-transitive
subgroup of M has a multiple of twelve elements.  [For those not
familiar with the group-theoretic result mentioned above, the proof
for this specific case goes like this.  Let <W> be a transitive
subgroup of <M>.  We must show that <W> has a multiple of twelve
elements.  Let q be any element of Q and let V be the subgroup of
<W> containing all elements <v> of <W> such that <v>(q) = q.  For
any element <w> of <W>, the right coset V<w> of V is the set
		{<v>*<w> | <v> is in V}
Since (<x>*<w>)(q) = <w>(<x>(q)) is equal to <w>(q) iff <x>(q) = q,
V<w> consists of precisely those elements of <W> which map q to
<w>(q).  All cosets must have the same size, so the number of
elements of <W> must be a multiple of the number of cosets, which
is the number of distinct values of <w>(q).  Since <W> is transitive
on Q, the set of such values is all of Q, which has twelve
elements.]

	We generated the 98 subgroups of M by computer, and found
that only 9 of them had a multiple of 12 elements.  We examine these
subgroups below.  In the preceding proof, we found it useful to
think of elements of M as permutations on the set Q.  In what
follows, we will sometimes find it useful to think of elements of M
as permutations of the set of face center positions.  An element of
M will be called "even" or "odd" according as it induces an even or
odd permutation on the six face centers.  Also, we refer to elements
of M which are not rotations of the cube as "reflections".  This
applies even to permutations which are not simple geometric
reflections but must be expressed as the composition of a
reflection and a rotation.  An example is the element of M which
permutes face centers in the cycle (U,B,R,D,F,L).

	The largest subgroup of M is, of course, M itself (48
elements).  M has three subgroups of size 24:  the group C of
rotations of the cube, the group of all even elements of M, which
we call AM (for alternating M), and the group of elements which are
in either both or neither of C and AM.  We will call this third
group H.  Thus the group H contains the even rotations and the odd
reflections.  M has five subgroups of size 12.  One of these is the
group of all even rotations, which we call AC.  The other four are
of the kind called "T" by Hoey in his message of 9 Dec 1980, 16:38
EST (corrected by Bawden, same date, 23:57 EST).  Let Z be any of
the four long diagonals of the cube (e.g., UFL-DRB).  Then T(Z) is
the contains all elements of M which map the corner cubies at the
ends of Z either to themselves or to each other.  Of these nine
groups, all except C and AC are Q-transitive.

4. A Catalog of Local Maxima
============================

In this section, we examine the seven Q-transitive subgroups of M,
and describe the 72 corresponding symmetric positions.  In general,
given a geometric interpretation of a subgroup W of M, verifying
that a position is W-symmetric is immediate, and no proof will be
given.  To prove that our catalog contains all W-symmetric positions
is more difficult, and we defer this to a later message.

	There are four M-symmetric positions:  SOLVED, the Pons
Asinorum (reached by RRLL UUDD FFBB), SOLVED with all edges flipped
(Bawden, 9 Dec 1980, 23:57 EST), and the Pons Asinorum with all
edges flipped.  The Pons Asinorum is interesting in that it is our
only example of a PROVEN local maximum which has been PROVEN NOT to
be a global maximum (it is known to have a length of at most 12,
while the global maximum must be longer because |G| > 12^12).  This
was pointed out by Plummer (7 Dec 1980, 07:24 EST).

	The only AM-symmetric elements of G are those which are
also M-symmetric.  Since the symmetry group of a position is the
largest subgroup W of M such that the position is W-symmetric,
there is no position which has AM as its symmetry group.

	Plummer (10 Dec 1980, 23:27 EST) has already presented an
example of an H-symmetric position which is not M-symmetric.  The
position is SOLVED with adjacent corners rotated in opposite
directions.  Another position, whose H-symmetry leads to our choice
of nomenclature, is shown below.

 	       U U U
 	       D U D
 	       U U U
 
  	L L L  F B F  R R R  B F B
  	R L R  F F F  L R L  B B B
 	L L L  F B F  R R R  B F B

	       D D D
	       U D U
	       D D D

There are two such "six-H" positions; composing the two yields the
Pons Asinorum.  This gives four possibilities for edge cubie
positions.  The corners of an H-symmetric position may be in any of
three orientations, all home, Plummer's configuration, or Plummer's
configuration with the twists reversed.  In any position, the edges
may all be flipped or not.  Composing the choices yields twenty-four
H-symmetric positions, twenty of which are not M-symmetric.

	There are four groups of the form T(Z).  To make our
presentation more specific, we will fix Z as the (UFL-DRB)
diagonal.  We define the "girdle" of the cube as the set containing
all the corner cubies other than UFL and DRB and all edge cubies
which are NOT adjacent to either UFL or DRB.  Thus,
		Girdle = {ULB, LB, DBL, DL, DLF, DF,
		          DFR, RF, URF, UR, UBR, UB}
A position which is T-symmetric but not M-symmetric (T has no
proper supergroups in M except for M itself) may be obtained by
flipping all edges on the girdle, as shown.

	       U B U
	       U U R
	       U U U

	L L L  F F F  R U R  B U B
	B L L  F F R  F R R  B B L
	L D L  F D F  R R R  B B B

	       D F D
	       L D D
	       D D D

Also, each edge on the girdle may be swapped with the diametrically
opposite edge, provided that the corners on the girdle are swapped
with their opposites as well.

	       R D D
	       U U D
	       U U B

	D L L  F F D  L L F  L F F
	R L L  F F B  L R R  B B F
	R R B  R B B  U R R  B B U

	       U U L
	       U D D
	       F D D

These positions may be composed with each other and with the four
M-symmetric positions to yield sixteen T-symmetric positions,
twelve of which are not M-symmetric.  Counting the positions
symmetric with respect to the four different T groups yields 48
positions whose symmetry groups are T groups.

	This completes the catalog of positions with Q-transitive
symmetry groups.  Summarizing the numbers of positions of each kind,
we have

    M-symmetric					 4
    AC-symmetric but not M-symmetric	 4-4  =  0
    H-symmetric but not M-symmetric     24-4  = 20
    T-symmetric but not M-symmetric  4*(16-4) = 48

for a total of 72 positions, one of which is the identity and 71 of
which are local maxima.

5.  Generalizations
===================

	The group of whole cube rotations, C, is Q+-transitive, but
not Q-transitive, because U = c U' c' has no solution for c in C.
This means that McKeeman's suggestion (8 Dec 1980, 17:03 PST) that
C-symmetry was a sufficient condition for being a local maximum is
not an immediate corollary of Theorem 1.  However, it happens that
all C-symmetric positions are also M-symmetric and they are
therefore local maxima with the exception of the identity.  Thus
McKeeman's claim turns out to be true "by accident".  However, the
case for the Supergroup is a different story.

	Analysis of the Supergroup, in which the orientations of
the face centers are significant, is trivial given the analysis for
G.  The only operations on the face centers which yield Q-transitive
symmetry groups are to leave them all alone or two rotate them all
by 180 degrees.  Thus there are a total of 72 * 2 = 144 elements of
the Supergroup which have Q-transitive symmetry groups.  One of
these is the identity and the other 143 are local maxima.
Considering face orientations as significant also allows us to
construct a position which is C-symmetric but not M-symmetric,
namely Big Ben, the position reached from SOLVED by turning all the
face centers 90 degrees clockwise.  Big Ben is a good candidate for
a counterexample (in the Supergroup) to McKeeman's (8 Dec 1980,
17:03 PST) suggestion that C-symmetric positions are local maxima.
Possibly Big Ben is a local maximum, but it sure isn't obvious to
us that, say, U and U' will lead to positions equally near to
SOLVED).

	Those who are interested in counting half-twists as single
moves may be pleased to hear that all 71 (143 in the Supergroup)
positions described above are also local maxima in the half-twist
metric.  To see this, first note that every Q-transitive subgroup of
M is also Q2-transitive.  This means that for any position p among
those 71 (or 143), all positions reached by quarter-twists from p
are M-conjugate (and thus equally far from SOLVED) and all positions
reached by half-twists are also M-conjugate.  The positions in one
of these two sets must all be one step closer to SOLVED (in this
metric) than p.  The positions in the other set cannot be further
from SOLVED than p since they are only one move away from positions
in the first set.  Note that this proof depends on BOTH
Q-transitivity and Q2-transitivity.  We do NOT make the claim that
any position whose symmetry group is Q2-transitive must be a local
maximum in the half-twist metric (in fact, we suspect that the
six-spot pattern mentioned below is a counterexample).

6.  On Conjectures
==================

	The point of this section is not to make conjectures, but
to examine conjectures which have recently appeared in the light of
our results.  As an example, we will first discuss a conjecture that
has not been made, but which would likely have been baldly stated
as fact had anyone thought to do so.

	"Of course, the inverse of a local maximum is also a local
maximum."  Easily said, but is it true?  All local maxima we know
about have Q-transitive symmetry groups, and the symmetry groups of
an element and its inverse are equal.  But suppose the local maximum
were not symmetric.  Consider the position reached from SOLVED by
performing the twists (U F F).  From this position, either F or F'
will move the cube closer to SOLVED.  From its inverse, (F' F' U'),
only U will move closer to SOLVED.  Is it not conceivable that from
some position, any of the twelve twists would move closer to SOLVED,
yet only eleven or fewer would move its inverse closer?  Such a
position would be a counterexample to the first statement of this
paragraph.  Of course, the example we provide is not a local
maximum, and indeed there may exist no local maxima except the
(symmetric) ones we have found.  But there is also no reason to
believe they can't exist, and there is no reason to believe that
their inverses are local maxima.  Of course, the inverse of a global
maximum is also a global maximum.

	The symmetry group of a Plummer cross has six elements and
is the intersection of H with T(Z) for an appropriate choice of
diagonal Z.  This group is not Q-transitive, but is Q2-transitive.
Consequently if the algorithm presented by Saxe in his message of 3
Dec 1980, 00:50 EST, is optimal, then the Plummer cross is a local
maximum.  The reason for this is that Saxe's algorithm ends with a
half-twist.  This means that, if the algorithm is optimal, then, by
virtue of the Q2-transitivity of the symmetry group, performing any
half-twist on a Plummer cross brings you two qtw closer to SOLVED.
This implies that performing any quarter-twist on a Plummer cross
would bring you one qtw closer to SOLVED, since the quarter-twist
could be continued into a half-twist for a total gain of two qtw.
This observation (Saxe's algorithm optimal => Plummer cross a local
maximum) was first made by David Plummer (5 Dec 1980, 20:29 EST),
who offered a slightly different, but correct, proof.  We emphasize
that this is all based on the purely speculative conjecture that
Saxe's algorithm is optimal.  The Plummer cross is NOT known to be a
local maximum merely by virtue of its symmetry, Greenberg's (bogus)
statement of 7 Dec 1980, 12:18 EST ("Yet, we know, intuitively,
that CP is 'highly symmetric', and it is a local maximum.")
notwithstanding.  To drive this point home, consider the six-spot
configuration (Pavelle, 16 Jul 1980, 20:51 EDT) produced by moving
(L R' F B' U' D L R').  This position has exactly the same symmetry
group as the Plummer cross, but is not a local maximum.  Any of the
six quarter-twists L', R, F', B, U, or D' will bring you closer to
SOLVED (obvious), any of the other six quarter-twists will take you
further away (based on exhaustive search by computer).  An even more
symmetric position is the twelve-L's [not to be confused with
Singmaster's less symmetric but visually similar 6-2L, obtained
from SOLVED by (F B U D R' L' F B)]:

	       L R R
	       L U R
	       L L R

	B F F  U U U  F B B  U U U
	B L F  U F D  F R B  U B D
	B B F  D D D  F F B  D D D

	       L R R
	       L D R
	       L L R

The symmetry group of this position is AC, the group of even
rotations.  AC is Q+-symmetric, so all clockwise twists have the
same effect, and all counterclockwise twists have the same effect.
If the two sets of neighbors should happen to have the same
lengths, then this position would be a local maximum.  Need we say
that there is no reason to believe this to be the case?

	Michael Aramini (7 Dec 1980, 01:08 EST) mentions the
possibility that two maxima might be one half-twist apart in the
half-twist metric.  This was claimed impossible by Plummer (7 Dec
1980, 07:24 EST).  We do not follow the reasoning, and we conjecture
that he misread "half" as quarter and then mistyped "quarter" as
half.  We also do not see anything to prevent two (local or global)
maxima in the half-twist metric from being a quarter-twist apart or
two maxima in the quarter-twist metric from being a half-twist
apart.  Parity considerations do not stand in the way of such
occurrences and, while none of the known (symmetric) local maxima
are so close to each other, we have no proof that either local or
global maxima must be symmetric.  We have no proof that such closely
neighboring maxima (or any non-symmetric maxima at all) *do* exist,
either.

	While we know 71 local maxima, we know only 25 distinct
ones up to M-conjugacy (3 having symmetry group H, 12 having
symmetry group T, and 10 [yes, 10] having symmetry group H).
McKeeman (6 Dec 1980, 16:42 PST) has correctly (provided we
substitute our corrected definition of symmetry) noted that, if we
could show that maxima had to be symmetric, then the maximum of the
best known solutions to these configurations would bound the length
of the global maximum.  Unfortunately, we have no proof of this
conjecture, nor any strong reason to think it true.

				Dan Hoey (Hoey @ CMU-10A)
				Jim Saxe (Saxe @ CMU-10A)


Date: 15 DEC 1980 1851-EST
From: DCP at MIT-MC (David C. Plummer)
Subject:  Administrata and an Algorithm
To: CUBE-LOVERS at MIT-MC
CC: ELLEN at MIT-MC, DUFFEY at MIT-MC


First the Administrata. CUBE-LOVERS came close to a crisis this
weekend. There were thoughts of having it digested, but DUFFEY
came through and made some very valuable suggestions. What
happened is that the Hoey-Saxe message was long enough to confuse
the mailer, and it ended up sending it twice. It took at least an
hour and a half to send, so the mailer response was terrible.
Things have been reorganized to help make things more efficient,
and we really thank DUFFEY for doing this. In the future, please
limit your messages to one to two thousand characters. If you
must send long messages, please break it up into sections and
send a piece every hour or so. Please keep this mailing list
winning. Thanks.

Now for the algorithm. Hoey and Saxe mentioned an H pattern. By a
simple method of doing it, it can be done as follows:

    FF (LLRR) BB (LLRR) RR (UUDD) LL (UUDD) DD (FFBB) UU (FFBB)
and after removing the NOPs
==> FF (LLRR) BB (LL)      (UUDD) LL (UU)      (FFBB) UU (FFBB)
which is another 28 mover. But I have found another way to do IT:

(UD LLRR FFBB UD) (FFBB LR FFBB L'R') ==> 24 QTW

Hoey and Saxe said that the Pons Asinorum is the only local
maximum that is PROVEN not to be a global maximum. It still is,
but if somebody can prove that Global Max must be larger than 24
(it is currently at 22), then this would be another example. The
other possibility is to find a faster algorithm to this pattern.

I have an intuitive sense that the global maximum is an even
distance away. I cannot prove it. Can anybody?

Date: 16 December 1980 1841-EST
From: James.Saxe at CMU-10A (C410JS30)
To: Cube-Lovers at MIT-MC
Subject:  16 qtw algs for CC and H patterns
CC: James.Saxe at CMU-10A
Message-Id: <16Dec80 184127 JS30@CMU-10A>

	In his note of 4 Nov 1980, 00:23 EST, David Plummer gives a
20 qtw algorithm for the Christman cross based on the following tool
for producing a 4-cross pattern:
	4+ = FB UUDD F'B'  R'L' UUDD RL  UUDD
This can be reduced to 16 moves as follows:
	4+ = FB UD LLRR UD FB UUDD
Consequently, Plummer's 16 qtw Christman cross algorithm, conceptually
B' 4+ B, can be reduced to
	B' [FB UD LLRR UD FB UUDD] B
	=    F UD LLRR UD FB UUDD B   (16 qtw).
[Note:  There is another 4-cross pattern besides the above, namely
	LLRR F LLRR FB LLRR B' LLRR F'B'.]

	The H pattern which Dan Hoey and I described in our earlier
message (14 Dec 1980, 19:16 EST, Sec. 4) can be achieved in 16 qtw
as follows:
	FF LL DD FF BB DD RR FF
This makes it the second proven example of a local maximum which is
not a global maximum.  Of course this applies equally to the second
H pattern which is Pons Asinorum away from the above.  I count these
two as only one example since they are M-conjugates.
				--Jim Saxe

Date: 30 DEC 1980 0109-EST
From: DCP at MIT-MC (David C. Plummer)
Subject:  Hackery (92 line message) (first of three)
To: CUBE-LOVERS at MIT-MC


Greetings...I'm back from vacation. I took a copy the old mail
home and read it. It was an interesting experience. 

Existence proofs:
	Saxe and Hoey described (14 December 1980, 19:16 EST)
several local maximums that were previously not described. I have
algorithms for a couple that are 28 moves long, showing that
THESE PARTICULAR local maximums are no further than 28 away from
SOLVED. 

==========
The edges flipped along the girdle (hereafter refered to as
GIRDLE EDGES FLIPPED [until somebody else thinks of a better
name]) can be done as follows:
	(F D'U  L D'U  B D'U  R D'U ) (U'L'F)
	(L'R F' L'R U' L'R B' L'R D') (F'L U)
[This is a Sprat Wrench, get-ready, another Sprat Wrench,
un-get-ready.] At first glance this is 30 moves, but notice that
the U and U' near the end of the first line cancel, making it 28.

==========
The edges not along the girdle flipped (hereafter refered to as
OFF GIRDLE EDGES FLIPPED) (NOTE: This is GIRDLE EDGES FLIPPED
compounded with ALL EDGES FLIPPED), can be done as follows:
	(L B'F  U B'F  R B'F  D B'F ) (F F U)
	(R'L U' R'L F' R'L D' R'L B') (U'F F)
[Again, Sprat-mung-Sprat-unmung] And again, 30 at first glance,
but this time the FFF near the end of the first line becomes F',
so 30-2 is again 28.

==========
==========
I noticed something peculiar about the second kind of girdle
configuration the mentioned. Hoey and Saxe describe it as "each
edge on the girdle may be swapped with the diametrically opposite
edge, provided that the corners on the girdle are swapped with
their opposites as well" and they give a picture. It is indeed
reachable, but the most interesting thing (in my opinion) is that
it is an odd distance away!! (Of course so are the related
configurations obtained by performing on it: PONS, ALL EDGES
FLIPPED, and PONS WITH ALL EDGES FLIPPED.) ODD!! Most of (if not
all) the previously described "nifty patterns" were even. And on
top of that, this odd permutation is a local max. Most of the
tools people seem to use are even. Several of them fall along
the lines of MUNG/DO SOMETHING/UNMUNG, and DO SOMETHING is often
even, and two MUNGS is even. Now the question is: HOW FAR AWAY
ARE THESE, AND HOW DO YOU (THE HUMAN) TRY TO FIND THE ALGORITHMS
TO DO THEM? My suggestion: Do something, Mung, Do something
(maybe different). I also suggest that Do somethingbe of length
12, and that mung be 3 or 5. I would really apprecitate it if it
were less than 28 -- I happen to like 28 as the maximal length.

==========
The local max which I described (Plummer 10 December 1980, 08:34
EST) which is all corners rotated such that adjacent corners are
rotated in opposite directions (shall we call this ALL CORNERS
ROTATED? I know it isn't explicit, but it is descriptive enough
for our purposes) can be done in 40 by a Christman Cross (16,
thanks to Saxe 10 December 1980, 18:41 EST), two half twists (4),
another Christman Cross (16), and undoing the two half twists
(4). I have been trying for the better part of a week to get this
down to 28 (my favorite number), but have not yet succeeded. What
I want is a 14 move algorithm to exchange all corners with their
diametrically opposite corner in such a way that the top forms a
cross of the top as the cross and the bottom at the corners. Two
of these with a whole cube manipulation in between will do the
trick. Alternately, if somebody could come up with a 12 mover to
move FDL to FUR (and do this so that it sould look this way if
viewed from F,L,R or B), you could do U <transform> U' and the
opposite corners would be swapped as desired. I have an almost
algorithm, but it falls short in that the four edges on the
center horizontal slice are swapped with their diametric
opposites. The algorithm is
	R'L' F'B' R'L' FB RL FB
If anybody can find the desired algorithms, please publish.

==========
Reading cube mail, I saw several things about "this Sprat Wrench
is best executed by a right handed person." The way I do the
Wrench does not descriminate, and I think it is worth it to
describe it:
(0 QTW)	Holding the right face in the right hand,
(0 QTW)	holding the left  face in the left  hand,
(2 QTW)	rotate the center verticle slice toward you,
(1 QTW)	rotate the NEW front face clockwise (either hand will do)
Do this a total of four times, giving 4*3=12 QTW. Try it, and
also try turning the NEW front face counterclockwise.

==========
That's all I can think of on this subject. Coming soon:
	Design of the higher order cubes
	Thoughts on manipulating the higher order cubes.

Date: 30 DEC 1980 0144-EST
From: DCP at MIT-MC (David C. Plummer)
Subject:  The higher order cubes (just the 4x4x4 for now) [93 lines]
To: CUBE-LOVERS at MIT-MC


And you thought the 3x3x3 was a complicated beastie...

I have plans for the 4x4x4 which I think can work. A first order
approximation is to take the 3x3x3 and split the edges in two.
This doesn't take care of the centers or axis, but those are
perhaps the most complicated anyway, and will get considerable
thought. 

WARNING: If you really want to see this, get out the graph paper
and correct the aspect ratio that the characters here will have.

The way I have labeled the cubies is as follows:
        Corners are labelled A (there are 8 of these)
        Edges are labeled both B and D (this defines the
                orientation) (there are 24 of these)
        Centers are labeled C (there are 24 of these)
So a face would look like:
        ABDA
        DCCB
        BCCD
        ADBA

Take a slice down the center of one of the planes and open it up.
It should look something like:

        .....&&&&&&&&++++++++
        .....&&&&&&&&++++++++           where & and @ are pieces of
        ..xxxxxx&&&++++++++++            the C type of cubie
        ..xxxxxx&&&++++++++++           + is a B/D type
        ....xx&&&&&++++++++++           x is the central axis
        ....xx&&&&&++++++++++
        ....xx&&&&&++++++++++
        ....xx&&&&&++++++++++
        ....xx&&&&&++++++++@@
        ....xx&&&&&++++++++@@
        ....xx&&&&/@@@@@@@@@@
        \...xx&&&/@@@@@@@@@@@
        .\..xx&&/@@@@@@@@@@@@
        ..\.xx&/@@@@@@@@@xx@@
        ...\xx/@@@@@@@@@@xx@@
        xxxxxxxxxxxxxxxxxxx@@
        xxxxxxxxxxxxxxxxxxx..
        .../xx\..........xx..
        ../.xx.\.........xx..
        ./..xx..\............

Gross, isn't it? There are a few things going on here that are
hard to show. That central cross must be able to rotate, so the
innermost parts of C and B/D are carved in somewhat. There is
another hairy constraint: the central axis MUST BE RIGIDLY
CONNECTED TO ONE AND ONLY ONE OF THE C TYPE CUBIES. The reason is
hard to describe. Hint: suppose you rotate a half cube portion.
It is possible that when the rotation is finished, the central
axis is misaligned. Connecting it to one face cubie forces it to
win. This may also make the central cross somewhat more fragile.

Forgetting about the central axis and the corner cubies (they
shouldn't be too able to work into the picture) we have the
following diagrams for the B/D and C type cubies. The numbers
indicate elevations:

                B/D                             C

        44444444                        4444
        44444444        ACTUALLY THEY   4444
        4443333333      ARE CURVED,     4443333333
        4443333333      BUT CHARS       4443333333.
        4433333333      HAVE POOR       4433333333.2
        4433333333      RESOLUTION      4433333333.2.
        4433333333                      4433333333.2.1
        4433333333                      4433333333.2.1.

I seriously think these will work. I hope to find somebody in
a material science type of lab that can make plastic molds so I
can actually try and build one of these. I hope to use a soft
plastic that is machinable, carvable, sandpaperable, etc, and
when I have a good one, make several pieces out of harder plastic
and see what happens. Perhaps I'll have something by the end of
January. 

I think that the 5x5x5 cube (though the tolerances might be
tighter) may actually be easier to construct. It may also be a
more interesting cube to work with.

Next: Notes on transforms on the 4x4x4 and 5x5x5 cubes.

PS: A mind blower: a DODECAHEDRON frob (can't call it a cube). I
thought of this one coming back on the bus. I haven't put
anything down on paper, but my minds eye tells me it has a
chance.

Comments welcome.

Date: 30 DEC 1980 0152-EST
From: DCP at MIT-MC (David C. Plummer)
Subject:  Notes on transforms on the 4x4x4 and 5x5x5 cubes. (65 lines)
To: CUBE-LOVERS at MIT-MC


First we have to define what a twist is. I propose that a twist
is a twisting of face or faces about an axis such that there is
only one plane on which sliding friction happens. This means that
slice turns are not single turns, but rather multiple turns. This
is consistent with the 3x3x3 definition, and it is generalizable
to nxnxn. Also, I think it is wisest to consider quarter twists
as the only single twist "move" (don't count half twists as one!!)

One thing to note: Even order cubes should have a STANDARD
coloring. This is nice to have since there is now visible axis to
align the corners on. I know it isn't necessary, but if I own a
cube and you own a cube, and they are colored differently, and we
swap for a day, both of us will probably have a hard time trying
to get used to a different color arrangement than what each is
used to. May I suggest
	Front	Right	Up	Back	Left	Down
	RED	WHITE	BLUE	ORANGE	YELLOW	GREEN

The general PONS can be described as follows (for cubes, not
necessarily dodecahedra):
	Pick an axis. Make 180 twists along all twistable planes.
	Pick another axis. Do the same.
	Do the same on the last axis.
This will produce a checkerboard pattern on all faces using
complementary (from the opposite side of the cube) colors. On the
4x4x4  (and in general on even order cubes), to corner cubies do
not move (but then again it is hard to tell without a fixed
reference). 

I have done more thinking about the 5x5x5 than the 4x4x4 since
the 5x5x5 offers several immediate advantages. First of all, from
solved we can do things like: do a transform as we would on a
3x3x3 and consider the 3x3 set of cubies in the center of the
faces as the center of our favorite 3x3x3 Hungarian. Another way
to view this is to do normal Hungarian rotations using only one
level deep of cubies. Then run the algorithm backwards doing
twists using two levels deep (considering theface center as a
center and 2x2 at the corners as corners, and 1x2 around the
edges as edges. (Bad explanation, but it is 1:30 in the morning).

I believe doing the traditional SIX-DOTS pattern forward then
backward using this method will produce something that looks
like:
	+++++			+++++
	+@@@+			+@@@+
	+@+@+			+@x@+
	+@@@+			+@@@+
	+++++			+++++
The one on the right is what happens if you do it again in the
same direction. I think the Plummer's Cross will look like:
	+@+@+
	@@+@@
	+++++
	@@+@@
	+@+@+
And if I am not mistaken, the approriate EXTENDED-SIX-DOTS
mentioned above performed on this will produce a 3-cycle
checkerboard. And the PONS on top of that will produce a 6-cycle
checkerboard. Of course, I could be wrong, but we wont know until
(a) somebody carfully works it out on paper (b) somebody writes a
computer program to simulate the crazy thing, or (c) somebody
actually builds one.

That's it for tonight folks, aren't you glad!!

Date: 29 Dec 1980 23:54 PST
From: McKeeman at PARC-MAXC
Subject: Re: The higher order cubes (just the 4x4x4 for now) [93 lines]
In-reply-to: DCP's message of 30 DEC 1980 0144-EST
To: DCP at MIT-MC (David C. Plummer)
cc: CUBE-LOVERS at MIT-MC

David,

Interesting thoughts.

It seems to me that the icosahedron is a more attractive physical target than the
dodecahedron.  It has 12 shear planes.  A move would be a 72o rotation of a
five-triangle subsection.  My intuition is that it is buildable more or less along
the line of the Rubik cube, except that the fixed axes point at vertices instead of
center faces.

The dual of the 2x2x2 cube seems to be an octahedron with shear planes along its
edges.  

More generally slice a sphere a buncha times.  The cuts are shear planes.  Don't
let it fall apart (there is the trick!).  Now paint the surface pieces.  Now twist to
mix.

The actual shape does not seem to matter much.  It is the interaction of the shear
planes that makes the puzzle.

Happy sphereing,

Bill	

Date: 31 DEC 1980 0330-EST
From: ACW at MIT-MC (Allan C. Wechsler)
Subject: Groups and Cayley Graphs
To: CUBE-LOVERS at MIT-MC

If there is any popular response I will write
a short introduction to Group Theory for Cubists.

= - = - = - = - = - = - = - = - = - = - = - = - =

Problem: the cube is known to have "small mountains",
states that are locally maximally distant from the
home state but not globally so.  The Pons Asinorum
is an example: it has been proved (a refreshing relief
from unmitigated conjecture) to be 12 qtw from home
and all its neighbors are closer.

The cube's Cayley graph in its six generators is vertex-
transitive (all the states look the same) like all Cayley
Graphs.  In addition, because all the generators are
conjugate in the big group (quarter-twists plus whole-
cube motions) the Cayley graph is edge-transitive
(all transitions between adjacent states look the same).

Can anyone find a small example of an edge-transitive
graph that has local maxima that are not global?

    ---Wechsler

Date: 31 DEC 1980 1115-EST
From: DCP at MIT-MC (David C. Plummer)
Subject:  the higher order SPHERE slices
To: McKeeman at PARC-MAXC
CC: CUBE-LOVERS at MIT-MC

Sigh, my goemetry needs to be zapped back into working order. I
really did mean icosahedron instead of dodecahedron (they both
have 12 faces). I was originally thinking of the fixed axes
pointing at the center faces, but pointing at the vertices is an
interesting idea. I shall have to think about it a little.

About the general case: "slice a sphere a buncha times." Holding
it together is REALLY the tricky part. Consider the 7x7x7 CUBE.
Do a 45 degree clockwise rotation on the top face (half a quarter
twist).  The problem is that the ENTIRE FRU corner cubie
COMPLETELY hangs over the front face. This means that the 3-D
jigsaw-puzzle idea that keeps the 3x3x3 cube together will not
work on order 7 and higher order CUBES.

Date: 31 Dec 1980 0729-EST
From: ZILCH at MIT-DMS (Chris C. Worrell )
To: CUBE-LOVERS at MIT-MC
Subject: relatives of the plummer cross
Message-id: <[MIT-DMS].177333>

   
Two relatives  of the plummer cross which people may find interesting
are given here:

The first is called a baseball.

      lL U L
      L U U
      L L L
F F F U U U B R B D B D
L L F U F F R R B D B B
F L F U F U B B B D D D
      R D R
      D D R 
      R R R
 

I do this one in 50 qtw though I know how to shorten it to 42.

The second is called a cube in a cube.

      L L L
      U U L
      U U L
F L L F F U B B B D D D
F L L F F U B R R B B D
F F F U U U B R R B B D
      R R R
      R D D
      R D D


I do this one in 70 qtw, though I could reduce it to 62.

These configurations both have two minor variants
which leave the essential nature of the configuration the same.
these other patterns are obtained by twisting the FUL AND BDR
corners. Both of these patterns were devised by people at Caltech.

               Chris Worrell (ZILCH @MIT-AI)




Date: 31 DEC 1980 1210-EST
From: DCP at MIT-MC (David C. Plummer)
Subject:  the 5x5x5 [133 lines]
To: CUBE-LOVERS at MIT-MC


OK, folks, I'm considering going further than 4x4x4 and entering
the realm of the 5x5x5. 

Cubies:
        C := Corner
        X := aXis (center)
        E := Edge (outside center)
        L := Left (external edge)
        R := Right (external edge)
        D := Diagonal (internal [to the face] corner)
        A := Adjacent (to the center, thanks to WER)
                (internal [to the face] edge)

A 3-D view would look like this


                z=-5   +---+---+---+---+---+
                      /   /   /   /   /   /|
                     / C / L / E / R / C / |
                    +---+---+---+---+---+  |
                   /   /   /   /   /   /|C +
                  / R / D / A / D / L / | /|
                 +---+---+---+---+---+  |/ |
        z=0     /   /   /   /   /   /|R +  |
               / E / A / X / A / E / | /|L +
              +---+---+---+---+---+  |/ | /|
             /   /   /   /   /   /|E +  |/ |
            / L / D / A / D / R / | /|D +  |
           +---+---+---+---+---+  |/ | /|E +
          /   /   /   /   /   /|L +  |/ | /|
         / C / R / E / D / C / | /|A +  |/ |
y,z=5   +---+---+---+---+---+  |/ | /|A +  |
        |   |   |   |   |   |C +  |/ | /|R +
        | C | L | E | R | C | /|D +  |/ | /|
        |   |   |   |   |   |/ | /|X +  |/ |
y=3     +---+---+---+---+---+  |/ | /|D +  |
        |   |   |   |   |   |R +  |/ | /|C +
        | R | D | A | D | L | /|A +  |/ | /
        |   |   |   |   |   |/ | /|A +  |/
y=1     +---+---+---+---+---+  |/ | /|L +
        |   |   |   |   |   |E +  |/ | /
y=0     | E | A | X | A | E | /|D +  |/
        |   |   |   |   |   |/ | /|E +
y=-1    +---+---+---+---+---+  |/ | /
        |   |   |   |   |   |L +  |/
        | L | D | A | D | R | /|R +
        |   |   |   |   |   |/ | /
y=-3    +---+---+---+---+---+  |/
        |   |   |   |   |   |C +
        | C | R | E | L | C | /
        |   |   |   |   |   |/
y=-5    +---+---+---+---+---+
     x=-5  -3  -1   1   3   5

LOVE THAT ASPECT RATIO !!!!

All in all there are
	 6 aXis faces
	 8 Corners
	12 Edges
	24 Left/Right type edges
	24 Diagonals
	24 Adjacents
	--
	98 = 5^3 - 3^3 = 125-27 visible cubies

Computation (inaccurate, but within a couple orders of magnitude)
of the number of reachable positions:

Axes:	lets not hack the extended problem yet	->	1
Corners:8 of them anywhere			->	8!
	each can take 3 orientations		->	3^8
	parity of the corner			->	1/3
Edges:	12 of them anywhere			->	12!
	each can take 2 orientations		->	2^12
	position/orientation restriction	->	1/4
L/R:	24 of them anywhere			->	24!
	orientation defined (they cannot flip)	->	1
	parity (cannot swap only two)		->	1/2 (I think)
Adjac:	24 of them anywhere:			->	24!
	one edge always touches a face center	->	1
	parity					->	1/2 (at least)
Diags:	24 of them anywhere			->	24!
	one corner always touches a face center	->	1
	parity					->	1/2 (at least)

It may not be accurate, but this computation gives
	1.291318 * 10^90

A slice through the center (z=0) probably looks something like

y=5\
   /    ..XXXXXXXXXX++++++++++EEEEEEEEEE
        ..XXXXXXXXXX++++++++++EEEEEEEEEE
        .....XXXX++++++++EEEEEEEEEEEEEEE
        .....XXXX++++++++EEEEEEEEEEEEEEE        X is an axis cubie
y=4\    .....XXXX++++++++EEEEEEEEEEEEEEE        E is an edge cubie
   /    .....XXXX++++++++EEEEEEEEEEEEEEE        + is one adjacent cubie
        .....XXXX++++++++EEEEEEEEEEEEEEE        ~ is another adjacent
        .....XXXX++++++++EEEEEEEEEEEEEEE
        .....XXXX++++++++EEEEEEEEEEEEEEE
y=3\    .....XXXX++++++++EEEEEEEEEEEEEEE
   /    .....XXXX++++++++EEEEEEEEEEEEE~~
        .....XXXX++++++++EEEEEEEEEEEEE~~
        .....XXXX++++++++EEEEEEEEEEEEE~~
        .....XXXX++++++++EEEEEEEEEEEEE~~
y=2\    .....XXXX++++++++EEEEEEEEEEEEE~~
   /    .....XXXX+++++++/~~~~~~~~~~~~~~~
        .....XXXX++++++/~~~~~~~~~~~~~~~~
        .....XXXX+++++/~~~~~~~~~~~~~~~~~
        \....XXXX++++/~~~~~~~~~~~~~~~~~~
y=1\    .\...XXXX+++/~~~~~~~~~~~~~~~~~~~
   /    ..\..XXXX++/~~~~~~~~~~~~~~~~~~XX
        ...\.XXXX+/~~~~~~~~~~~~~~~~~~~XX
        ....\XXXX/~~~~~~~~~~~~~~~~~~~~XX
        XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
y=0\    XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
   /    XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
        XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
        ..../XXXX\....................XX
        .../.XXXX.\...................XX
y=-1\   ../..XXXX..\..................XX
    /
         /\   /\   /\        /\        /\
       x=-1    0    1         3         5

This time the central axis is rigid in the sense that it does
form a cross, but each of the six spokes can rotate as in the
3x3x3 cube. The curvature and tolerances of some of the pieces
gets a little hairy, but I'm working with graph paper and looking
at the other slices through the cube. Wish me luck -- I have
thoughts of construction.

Date: 31 December 1980 1511-EST (Wednesday)
From: Guy.Steele at CMU-10A
To: DCP at MIT-MC (David C. Plummer)
Subject:  Icosahedron
CC: cube-lovers at MIT-MC
In-Reply-To:  DCP@MIT-MC's message of 31 Dec 80 11:15-EST
Message-Id: <31Dec80 151120 GS70@CMU-10A>

The icosahedron has twenty faces, not twelve.  It and the
dodecahedron are duals: each has the same number of vertices
as the other has faces.  If you join the centers of the faces
of one with lines, you get the other.  They have the same
number of edges.

Date: 31 DEC 1980 1230-PST
From: WOODS at PARC-MAXC
Subject: Re: relatives of the plummer cross
To:   ZILCH at MIT-DMS, CUBE-LOVERS at MIT-MC

In response to the message sent  31 Dec 1980 0729-EST from ZILCH@MIT-DMS 

Your "baseball" is listed in Singmaster's booklet as "the Worm" (though
I must admit I like your name for it), with a 28-qtw solution (23 if
you count half-twists=1).  I prefer the pattern named the "Snake":

    DUD
    DUD
    DUD

BBB RFR FFF LBL
BLL FFR FRR BBL
BLB RRR FRF LLL

    UUU
    DDD
    UUU

which is known to be soluble in 20 qtw (16 if htw=1).  Meanwhile, your
cube-in-a-cube has a 24-qtw (20-htw/qtw) solution in Singmaster, though
I generally find it easy enough to make in 34 qtw using the commutator
operation [spoiler warning!] nicknamed "rFUfuR":  rFRfUfuFuRUr.

	-- Don.
-------

Date: 31 Dec 1980 1545-PST
From: Dave Dyer <DDYER at USC-ISIB>
Subject: bigger&better cubes
To: cube-lovers at MIT-MC


 Clearly, we must not be deterred by mere mechanical limitations!
I'm sure a color bit-mapped raster could do a dandy job of displaying
cube-like objects in any size and geometry you can imagine.  If you
must have a physical cube, how about a box with LCD's on all faces
(in six colors?), pressure sensors to detect what ytou want to twist,
and a uP inside to do all the work.
-------

Date: 31 Dec 1980 1634-PST
From: Dave Dyer <DDYER at USC-ISIB>
Subject: Singmasters book
To: cube-lovers at MIT-MC


 Does anyone know of a domestic source for it?
-------

Date: 31 Dec 1980 1651-PST
From: MERRITT at USC-ISIB
Subject: Bigger & better cubes
To: cube-lovers at MIT-MC


In light of all the talk of cubes greater than 3 x 3 x 3, such as
4 x 4 x 4, or 5 x 5 x 5,  perhaps you would like to comtemplate a
cube that is 3 x 3 x 3 x 3, or 4 x 4 x 4 x 4.  The idea of adding
dimensions is really reasonable, since it needn't be mechanically
possible.  This Rubics hypercube could probably be represented by
a video display in some form.

This is at least a matter for further contemplation.
					-IHM
-------

Date: 31 DEC 1980 1740-PST
From: WOODS at PARC-MAXC
Subject: Re: Bigger & better cubes
To:   MERRITT at USC-ISIB, cube-lovers at MIT-MC

In response to the message sent  31 Dec 1980 1651-PST from MERRITT@USC-ISIB

And then of course there's the megacube, formed by taking a standard 3x3x3
cube and replacing each cubie with a 3x3x3 Rubik's cube.  The idea would be
that each megaface could be twisted as usual, but the megafacies (the faces
of the smaller 3x3x3 cubes) could be twisted only if they were visible.
(Otherwise all you really have is a collection of independent 3x3x3 cubes.)
The orientation of the colors on all the microcubes is the same -- that
is, the beast starts with all 81 microfacies on a given side being the same
color.

	-- Don.
-------

Date: 31 December 1980 22:35-EST
From: Ed Schwalenberg <ED at MIT-AI>
Subject: bigger&better cubes
To: DDYER at USC-ISIB
cc: cube-lovers at MIT-MC

Obviously the age of mechanical polyhedra is long past.  I propose
that the center cubies (polyhedries?) of the electronic marvels
be equipped with fixed arrows so that those who are hacking the
higher-order problem can be happy.  (I really like this idea because
it closely resembles one of my favorite ideas:  a keyboard which has
small character displays in each keytop in lieu of engraving; when you
hit SHIFT the legends change from qwertyiop to QWERTYUIOP and when you
hit CONTROL+META they change to ETAIONSHRDLU, etc.)

The all-electronic cube has many other potential features:  instant
resettability, a stack of saved states, subroutines.... Given 
extensible, customizible, self-documenting editors, I don't see why
we should settle for polyhedra that are any less featureful.

Speaking of higher-order polyhedra, I hereby nominate the tetrahedron
as being appropriate to my own level of expertise.

Date:  1 JAN 1981 0221-EST
From: ACW at MIT-MC (Allan C. Wechsler)
Subject: How much intellect does it take to solve the cube?
To: CUBE-LOVERS at MIT-MC

Well, happy new year.

I solved it while not exactly sober.

   ---Wechsler

Date:  1 JAN 1981 1315-EST
From: DCP at MIT-MC (David C. Plummer)
Subject:  several subjects
To: CUBE-LOVERS at MIT-MC

 
One last try!! What I meant was the 12 sided frob built out of
pentagons. And after refering to better and better dictionaries I
discovered this thing is called a pentagonal dodecahedron, and I
meant the faces to be the points of rotation.  Perhaps McKeeman
thought I meant the rhombic dodecahedron, and subsequent messages
got me confused and I jumped the gun without thinking very
heavily.

Woods: Could you please send the manipulations for "baseball,"
"snake," and "cube-in-cube" for the benifit of those who do not
have Singmaster. Please use prime notation (R' instead of (lower
case) r) for counterclockwise twists since that seems to be the
notation currently in use in this list.

In general, my opinion is that it would be nice if people would
send along the short algorithms that are known. ZILCH's 50 and 70
qtw algotithms are a little too long, but anything under 36
should probably be sent. I know it may be a spoiler, but (1)
there seem to be several configurations mentioned and perhaps
some people don't have time to find nice fast ways to get there,
(2) it reduces needless duplication of effort, (3) parts of the
algorithm (or the algorithm itself) might serve as a subroutine
for other algorithms under development.

On the concept of the higher order "cubes:"
	N dimensions
	has a reasonable geometric interpretation (maybe it
		doesn't have to have this condition?)
	built out of some number of "cubies" of dimension N
	Each "cubie" is in turn a "cube," perhaps of a different
		order than the larger cube (eg, a 3x3x3 cube
		whose cubies are 5x5x5)
	Each "cubie" of the "cubie" is a "cube," ad infinitum as
		desired 
	In addition to all this, each faclet is a "cube" of
		dimension N-1, ad infinitum (at least until the
		dimensions run out!!)

The thing I am doing is trying to PHYSICALLY construct a higher
order (flavor: dimension 3, cubical, order 5, cubies are "atomic"
(ie, not cubes in themselves), faclets are atomic). Personally, I
think half the fun is being able to hold one of the beasties and
mung it by twisting.

ACW: I think it would be instructive to have an short intro to
Group Theory for Cubist. This would benefit newcomers to the
mailing list, and people who hack the cube and want to know some
of the theory behind the cube. (5-10K characters if you can keep
it down to that. If not, send it when machines are generally
lightly loaded.) I vote: plase do.

Date:  3 JAN 1981 0248-EST
From: ZILCH at MIT-MC (Chris C. Worrell )
Subject: How to play with the corners of your cube (long message)
To: CUBE-LOVERS at MIT-MC

At this point I have found sufficient Algorithms such that given a
cube with everything correct except for possibly the four
corner cubies on one side, I can (with a little thought and
reference to my notes) solve the cube in 24qtws.

[SPOILER WARNING]

  If all the cubes are in the right place, but possibly oriented
wrong, the following transforms are used to TWIST the corners to
the proper orientations:

T1:  F' (R' D' R D' F D F' D)^2  F        18qtws
    (FDL,RDF,BDR,LDB) => (DLF,FRD,RBD,DBL)

T2:  L D (D L' D' L)^2 D' L' R' D' (D' R D R')^2 D R     24qtws
    (FDL,RDF,BDR,LDB) => (LFD,DFR,RBD,DBL)

T3:  (D' L' D R D' L D R') (B' L D^2 L' B L B' D^2 B L')
    (FDL,RDF,BDR) => (DLF,DFR,DRB)                20qtws

T4:  (R D' L' D R' D' L D)  (L B' D^2 B L' B' L D^2 L' B)
    (FDL,RDF,BDR) => (LFD,FRD,RBD)                 20qtws

  Note: T3 and T4 are inverses based on the same components, which
        happen to commute. (see C1 and C2 below)

T5:  (L' U L F U F') D' (F U' F' L' U' L) D          14qtws
    (FDL,RDF) => (DLF,FRD)

T6:  L' F' D' L' D R D' L D F L F' R' F            14qtws
    (FDL,RDF) => (LFD,DRB)

    Along the same line as T5 and T6, but not usefull in the
      present discussion, shown to me in a private message from
      Dan.Hoey@CMU-10A.

T7:  F' R' D' R U R' D R F D F' U' F D'             14qtws
    (FDL,BUR) => (LFD,RBU)

  If all of the corner cubies are not in the proper positions
it is more profitable to execute several corner moving transforms
rather then one corner moving one then one corner twisting one.

  As presented here all transforms cycle the cubies in the same
manner (clockwise) ,  though twisting the cubies in all possible
ways.  Their inverses (counter-clockwise) should also be kept
in mind.

C1:  D' L' D R D' L D R'                            8qtws
    (FDL,RDF,BDR) => (FRD,RBD,LFD)      (twist all clockwise)

C2:  L B' D^2 B L' B' L D^2 L' B                     12qtws
    (FDL,RDF,BDR) => (DFR,DRB,DLF)        (twist all clockwise)

  Note: make reference in c1   (twist all counter-clockwise)

C3:  F L^2 D' R' D L' D' R D L' F'                   12qtws
    (FDL,RDF,BDR) => (RDF,BDR,FDL)       (don't twist at all)

C4:  F' R' B' R F R' B R                             8qtws
    (FDL,RDF,BDR) => (FRD,BDR,DLF)

C5:  F L F' R F L' F' R'                              8qtws
    (FDL,RDF,BDR) => (RDF,RBD,DLF)

  Note: C4 and c5 have been adopted from those presented by DCP in
        his message of 25 nov. 1308-EST

C6:  L F L' D^2 L F' L' F D^2 F'                   12qtws
    (FDL,RDF,BDR) => (FRD,DRB,FDL)

C7:  R' D^2 R B' R' B D^2 B' R B                       12qtws
   (FDL,RDF,BDR) => (DFR,RBD,FDL)

C8:  (C5)' (C1)'=                                       16qtws
     (R F L F' R' F L' F') (R D' L' D R' D' L D)
    (FDL,RDF,BDR) => (RDF,DRB,LFD)

C9:  (C1)' (C4)'=                                       16qtws
     (R D' L' D R' D' L D) (R' B' R F' R' B R F)    (FDL,RDF,BDR) => (DFR,BDR,LFD)


These nine transforms are the only possible legal ones (along with
their inverses) which exchange three corners on a face (with the
possibility of twists0, though I can't guarentee minimum lengths for
any of them.

If all of the corners are not in their proper positions then there
  are three possibilities:

1) One of the corners is in the right position and has the correct
   orientation.
     to fix: do the appropriate transform, or its inverse from the
             list given above.
             max length=16qtws

2) None of the corners is in the proper position
     to fix: using C1,C4, or C5 (the shortest ones) move one of the
              corners to the proper position and orientation, then
             continue as in case 1.
              max length=8+16=24qtws

3) One of the corners is in the correct position, but is in the wrong
     orientation.
     to fix: Preferably using C1,C4, or C5 move theproperly
             positioned corner out of its spot and at the same time move
             another corner into its proper position and orientation, then
             procede as in case  1.
              If none of C1, C4, or C5 will do the proper thing then
             a combination of C2 and C3 must be used, C2 first, to orient
             the corners correctly (with respect to the bottom) then use C3 to
              position the corners correctly.
             max length=8+16=12+12=24qtws

 These algorithms may be usefull to someone making a sides first,
corners second cube solving algorithm.

 If anyone has any shorter algorithms for any of these transforms, please
send them to the list.

Unfortunatly I probably won't be able to answer any questions about this
method as I am going back to school (Caltech) tommorrow (today?)(Sat. 3rd)
and I don't have decent net access from there.

               Chris Worrell (ZILCH@MIT-MC)

p.s.: sorry about the length of this message.

Date:  3 JAN 1981 0433-EST
From: ZILCH at MIT-MC (Chris C. Worrell )
Subject: oops....
To: CUBE-LOVERS at MIT-MC

Sorry folks, I missed one possibility.
In case 2,my algorithm does not account for the configuration reached by
    L R D^2 R' L' F' B' D^2 F B D^2          14qtws

    (FDL,RDF,BDR,LDB) => (BDR,LDB,FDL,RDF)

Either add this transform to the list given, or when you get to
this particular configuration execute  C3 twice  24qtws, but still inmy
range and still along the general lines of the rest of the possibilities.

            Chris Worrell (ZILCH@MIT-MC)


ACW@MIT-AI 01/05/81 00:01:05 Re: Group theory and Cayley graphs.
To: CUBE-LOVERS at MIT-AI

I have some other short term resposibilities,
but will get around to this in a couple of 
days.

    ---Wechsler


From: Don Woods <DON at SU-AI>
Subject: excerpts from Singmaster
To:   CUBE-LOVERS at MIT-MC 

[In reply to message from Plummer sent 1 JAN 1981 1315-EST.]

	** SPOILER WARNING!! SPOILER WARNING!! **

This message gives the shortest solutions known to Singmaster (as of
the fifth edition of his booklet) for three pretty and moderately
complex patterns, recently referred to as "baseball" (or "worm"),
"snake", and "cube-in-a-cube".  People wishing to investigate these
patterns (as described in earlier messages) may not wish to read
further.

Notes:  Singmaster uses the half-twist measure.  His notation also
includes special representation for the slice (center-twist) and
antislice (such as L+R as opposed to L'+R) operations, which he counts
as two twists; I have expanded this to strict "befuddler" (BFUDLR)
notation in this message.

BASEBALL or WORM

	RUFFD'RL'FB'D'F'R'FFRUUFRRF'R'U'F'UUFR

SNAKE

	BRL'D'RRDR'LB'RR + UBBU'DBBRLUUR'L'BBD'

It's not too hard to get to the snake, if you don't mind wasting few
twists.  Start with DLLRRD' and you're most of the way there.  This
assumes you know simple macros for swapping two pairs of edge cubies
and for flipping two edge cubies.

CUBE IN A CUBE

	BL'DDLDF'DDFD'B' + F'RUUR'U'BUUB'UF

Again, the cube-in-a-cube is not hard to generate using two instances
of the commutator macro -- R'FRF' UF'U'F U'RUR' -- plus a few simple
extra twists.  This is left as an exercise to the reader.

	-- Don.



Date:  7 January 1981 12:59 est
From:  Greenberg.Symbolics at MIT-Multics
Subject:  Lisp Machine Color cubesys
To:  cube-lovers at MIT-MC

Lisp machine color cubesys has been fixed.

Date:  7 January 1981 1352-EST (Wednesday)
From: Dan Hoey at CMU-10A
To: Cube-Lovers at MIT-MC
Subject:  Pons Asinorum -- Part 1: Optimality
Message-Id: <07Jan81 135220 DH51@CMU-10A>

        The Pons Asinorum (obtained by UUDDFFBBLLRR, and known as
6-X in Singmaster) is well-known to readers of this list.  It is
perhaps surprising that this so well-known position has anything
more to teach us.

        The first surprise is that the 12-move sequence given above
is provably optimal in the quarter-twist metric.  Proofs were sent
to me by David C. Plummer, who attributed it to Alan Wechsler, and
by Chris C. Worrell.  While it is well-known (See Alan Bawden's
messages of 31 July 1980 13:06-EDT and 31 JUL 1980 2159-EDT) that
some positions require at least 21 moves, the longest sequence
which has previously been proven optimal is LR'FB'DU'LR' for the
six-spot configuration.  It is good to see a 12-move sequence
proven optimal -- and in a way not dependent on the vagaries of
programming errors and cosmic rays.

        The proof of optimality relies on the "Oriented Distance
from Home" (ODH), used by Vanderschel (6 Aug 1980 1909-PDT) in his
proof of edge orientation parity conservation.  The ODH of an edge
cubie (in some position and orientation) is defined to be the
minimum number of quarter-twists required to move that cubie to its
home position and orientation.  A table of possible ODH values of
the UF cubie is given below, indexed by the position of that
cubie's F tab.

                        + 3 +
                        2 U 2
                        + 3 +

                + 1 +   + 0 +   + 1 +   + 2 +
                2 L 2   1 F 1   2 R 2   3 B 3
                + 3 +   + 2 +   + 3 +   + 4 +

                        + 3 +
                        2 D 2
                        + 3 +

The Pons Asinorum moves every edge cubie to a position and
orientation which has an ODH of 4.  To move all twelve cubies in
this way requires a total of 48 edge moves, and only four edge
moves can be accomplished by each quarter-twist.  Thus the Pons
Asinorum requires twelve quarter-twists.

        Unfortunately, this seems to be the only really impressive
result to be derived from counting ODH.  All edges flipped
(Singmaster's 12-Flip) can be shown to require at least ten
quarter-twists, but this is a far cry from the 28-qtw process which
is the best known (Plummer, 10 Dec 1980 0157-EST).  A brute force
technique for deriving such results is of course possible, but to
show a twelve-move lower bound seems to require sorting and merging
two lists of over one hundred thousand positions each, an act which
is viewed as unsociable (or, more usually, impossible) on the
systems to which I have access.  Anyone who has a gigabit to spare
should get in touch -- there are several good problems for which
brute force seems attractive if there is enough of it.

	Surprise number two -- Pons Asinorum in the Supergroup -- in
an hour or two.
			Dan

Date:  7 January 1981 1615-EST (Wednesday)
From: Dan Hoey at CMU-10A
To: Cube-lovers at MIT-MC
Subject:  Pons Asinorum -- Part 2: Pons in the Supergroup
Message-Id: <07Jan81 161515 DH51@CMU-10A>

        The second observation I would like to make regarding the
Pons Asinorum involves the Supergroup (also known as "the extended
problem") in which the orientation of face centers is considered.
The process UUDDFFBBLLRR turns each of the face centers 180
degrees, so Pons Asinorum is symmetric in the Supergroup as well.
(Turning each face center 180 degrees is the M-symmetric position
Big Ben Squared, which I will call Noon.)  There is another optimal
way of making a (pseudo-) Pons Asinorum, (UD'FB')^3, which differs
from the true Pons only in the face center orientations.  According
to an exhaustive search I carried out by hand, this is the only
pseudo-Pons (up to M-conjugacy) that can be obtained with six slice
moves.  I would be very interested in hearing about any other
twelve-qtw positions which differ from the Pons Asinorum only in
the Supergroup.

	I have found a 16-qtw process for Pons Asinorum composed
with Noon, (UD FB FB UD)(FB UD UD FB), which looks like Pons
Asinorum, but does not rotate the face centers.  This in turn gives
a 20-qtw process for Noon itself:
        LLRR UUDD (UD FB FB UD) (FB UD UD FB) FFBB
      = LLRR    (U'D' FB FB UD) (FB UD UD F'B').
Of course, there's no assurance of optimality here.

	It occurs to me that many readers of this list may find
details of the Supergroup uninteresting.  I have more on this
subject, so if you would or wouldn't like to know more about the
Supergroup, send a vote to Hoey@CMU-10A and we'll see what to do.

				Dan

Date:  8 JAN 1981 1515-EST
From: DR at MIT-MC (David M. Raitzin)
To: CUBE-LOVERS at MIT-MC

I've been reading the old mail from this mailing list, and I've
noticed some discussion on the definition of what is a single twist.
As far as I can see, the definition of a twist should be the change in
position of all the cubies in a single plane performed by a single
motion.  In other words that should include a 90 degree twist, a 180
degree twist, and a twist of the middle slice (the last one can be
thought of as holding the U and D parts of the cube in place with one
hand, while moving the middle slice with the other).  

Date:  8 JAN 1981 1523-EST
From: DR at MIT-MC (David M. Raitzin)
To: CUBE-LOVERS at MIT-MC

The other day, having nothing to do, I started optimizing my
transformations.  Counting 90-degree twist, 180-degree twist and a
twist of the middle slice as a single move, I reached the following
numbers: for Plummer cross: 24 moves, and for what I call inverting
the cube (that means flipping every non-corner cubie) 29 moves.  Now
realizing that neither of those comes close to the optimum # of moves,
I'd like to know how far of the so far achieved optimum am I?

Date:  8 January 1981 20:06 cst
From:  VaughanW.REFLECS at HI-Multics (Bill Vaughan)
Subject:  Befuddled by BFUDLR
To:  CUBE-LOVERS at MIT-MC
cc:  VaughanW.REFLECS at HI-Multics

This is a plea for another notation.

BFUDLR is sufficient to describe anything. So what? It's about  as
readable  as  a  LISP  s-expression,  as  rich as the average grad
student, and (my particular gripe) it's impossible to  express  an
elegant sequence in it and keep any of the elegance.

I want to keep this short, so I'll only give a few  examples.  The
first  is the Sprat Wrench. BFUDLR calls it: RL'URL'BRL'DRL'F. But
the way most everyone does it, it's: (XU)*4, where X  means  "move
the LR slice clockwise as viewed from the left."

The next example (I don't know its name) flips all top and  bottom
edges.  BFUDLR  calls  it:  LRUDFBLRUDFB. Interesting, but this is
nicer: (R-B-)*3, where "foo-" means "foo antislice" and is done by
twisting  foo and its adjacent slice clockwise 1qtw, then twisting
foo another qtw.

A move yielding (RF, BL, RB) has been published (Don Woods  6  Jan
81)  in  BFUDLR  as: BRL'D'RRDR'LB'RR. Now where's the symmetry in
that? But annotate the same move BXB'RR.BX'B'RR  (X  as  in  first
example)  and  you can see how pretty it is. And it's a lot easier
to remember.

The key is  that  the  fixed  orientation  of  the  center  cubies
shouldn't  be  a  sacred  cow.  Often, keeping a corner cubie as a
fixed point will yield a far more natural notation.

The commonest  compound  moves:  slice,  antislice,  and  possibly
Singmaster's   Y   and  Z  commutators,  should  have  specialized
notations. A move that I use commonly in solving  the  cube  is  a
monotwist:   Y(f,r)*2.L.Y'(f,r)*2.L'.   That's  a  lot  harder  to
understand as: FR'F'RFR'F'RLR'FRF'R'FRF'L'.

I don't have good notations to offer for slice and antislice. What
I  do with paper and pencil involves overstrikes that my CRT can't
handle.  Something  nice  and  linear  is  needed,  with  all  the
characters in ASCII. Any suggestions, anyone?

Date:     8 January 1981 2235-est
From:     Bernard S. Greenberg       <Greenberg at MIT-Multics>
Subject:  Cubesys and FLUBRD
To:       CUBE-LOVERS at MIT-MC

A new hack for interpreting transforms expressed in FLUBRD
notation has been added to ITS and LISPM cubesys.
(As I am no longer with Honeywell, I can no longer modify
Multics cubesys, but I will give instructions below for
how to use this on Multics).

Transforms are defined in Lisp, with the "defxform" macro.
"defxform" is followed by a name to assign to the transform,
and one to any number of "cube operations".  A cube operation
is either

    1) a flubrd syllable
    2) (apply <transform-name>), meaning do that transform
    3) (inverse <transform-name>), meaning "undo" that transform
    4) any other list, which is interpreted as a list of cube-operations

A flubrd syllable is an atomic symbol whose name is a character string
consisting of one to any number of flubrdniks.
A flubrdnik is either
    1) F L U B R D f l u b r d  (case doesnt count)
    2) F* L* U* .... etc, meaning counterclockwise turn
    3) F2 L2 U2 .... etc, meaning 180 degree turn.

Here is the provided, automatically-loaded library of
transforms:

(defxform monotwist-op
          (ld l*d*) ldl*)

(defxform monotwist
          (apply monotwist-op) u (inverse monotwist-op) u*)

(defxform quark
          r2 (apply monotwist) r2)

(defxform pons f2 b2 r2 l2 u2 d2)

(defxform christman-cross       ;Saxe 16 dec 1980
        f ud llrr ud fb uudd b)

(defxform plummer-cross         ;Saxe 3 dec 1980
        f (ll rr) f b (ll rr) f
        l (bb ff) l r (bb ff) l (uu dd))

To load new transforms...
   ITS:  ^X break, load a file full of defxforms, ^G back.
   Lispm: <BREAK>, load the file,^Z
    (oh yes, Lispm cubesys now has a ^Z handler)
   Multics: ESC X loadfile the file.

To run a transform:
   ITS: Use the X command. the transform name will be prompted for.
   Lispm. Use the X command. A MENU OF KNOWN TRANSFORMS WILL POP UP!
      Mouse at it.
   Multics: ESC X run-xform <transformname>


On Multics, you will have to load the new package before doing
any of this, with esc-x loadfile. Currently, it is
>udd>sym>bsg>cxfrm .... it may move or go away. Multicians, contact 
me if you have any interest in it.

Date:  9 January 1981 0551-EST (Friday)
From: Dan Hoey at CMU-10A
To: Cube-lovers at MIT-MC
Subject:  The Supergroup -- Part 1: Physical reality
Message-Id: <09Jan81 055144 DH51@CMU-10A>

	Well, whoever doesn't like the Supergroup didn't send me
any messages, and several who do did, so here goes.  This message
is part one of three, separated for the benefit of MIT's
notoriously fragile mailer.

	In case anyone has managed to miss it, the Supergroup is
the group underlying the cube when face center orientation is taken
into account.  By the "orientation" of a face center, we refer to
the number of 90o twists of that face center from the position
designated "solved."  To visualize this, Singmaster suggests
replacing the solid colors on the sides of the cube by some
nine-piece pictures, so that the centers must be restored to their
initial (untwisted) state to solve the cube.  He reports that this
was done (on two sides only) by a company in England, which printed
its logo on cubes for a promotion.  The term "Supergroup" is also
due to Singmaster, and I adopt it in favor of the term "the
extended problem," which has appeared in Cube-lovers.

	To make the face center orientation visible on my cube, I
first used magic marker, which rubbed off, then paint, which
attacked the colortabs and looked and felt awful.  [Then I went to a
stationery store and got plastic tape and replaced my colortabs --
no orange, so my cube now has a tan face.]  I marked the face center
orientation by cutting out circles from the plastic colortabs to
let the black plastic show through.  I like it, though some people
think it looks like the cube has been used for target practice.
Each cutout circle has a diameter of about 3/8" (1/2 the side of a
cubie) and is centered at the corner of a face center, overlapping
two edge cubies and one corner cubie.  The orientation is then
visible if either the corners or the edges are in the home
position.  It doesn't particularly matter which corners of the face
centers are used; I chose the pattern which has the same symmetry
group as Plummer's cross (unique up to M-conjugacy).

	There will be a short intermission while we change reels.

Date:  9 January 1981 0629-EST (Friday)
From: Dan Hoey at CMU-10A
To: Cube-lovers at MIT-MC
Subject:  The Supergroup -- Part 2: At least 25 qtw, and why
Message-Id: <09Jan81 062915 DH51@CMU-10A>

	Alan Bawden (31 JUL 1980 2159-EDT) calculated that it must
take at least 21 quarter-twists to solve an ordinary cube, and 24
qtw to solve a cube in the Supergroup.  This message explains how
the first bound can be obtained, improves the second, and points
toward a technique for possible further improvements.

Express any (optimal) sequence of twists as a sequence of segments,
where each segment is a sequence of twists on two opposite faces,
and no two adjacent segments operate on the same pair of faces.
Because the quarter-twist has period four, and opposite faces
commute, a segment operating on faces X and Y has one of the forms
	X, X', Y, Y'				(1 qtw -- 4 ways)
	XX, YY, XY, YX, X'Y, Y'X, XY', Y'X	(2 qtw -- 6 ways)
	XXY, XXY', XYY, X'YY			(3 qtw -- 4 ways)
	XXYY					(4 qtw -- 1 way).
There are 3 ways of choosing X and Y for the first segment, and two
ways of choosing them for every succeeding segment.  Let P[n] be the
number of positions that are exactly n qtw from SOLVED.  Then
bounding P[n] by the number of n-qtw sequences,
	P[0]   = 1
	P[1]  <= 4*3*P[0]
	P[2]  <= 4*2*P[1]   + 6*3*P[0]
	P[3]  <= 4*2*P[2]   + 6*2*P[1]   + 4*3*P[0]
	P[4]  <= 4*2*P[3]   + 6*2*P[2]   + 4*2*P[1]   + 1*3*P[0]
	P[n]  <= 4*2*P[n-1] + 6*2*P[n-2] + 4*2*P[n-3] + 1*2*P[n-4]
						for n > 4.
Evaluating this recurrence, substituting strict (in)equality where
I have personally verified it, and rounding truthfully yields:

P[0]  =           1	P[9]  < 724,477,008	P[18] < 4.048*10^17
P[1]  =          12	P[10] < 6.792*10^9	P[19] < 3.795*10^18
P[2]  =         114	P[11] < 6.366*10^10	P[20] < 3.557*10^19
P[3]  =       1,068	P[12] < 5.967*10^11	P[21] < 3.334*10^20
P[4]  =      10,011	P[13] < 5.594*10^12	P[22] < 3.125*10^21
P[5]  <=     93,840	P[14] < 5.243*10^13	P[23] < 2.930*10^22
P[6]  <     879,624	P[15] < 4.915*10^14	P[24] < 2.746*10^23
P[7]  <   8,245,296	P[16] < 4.607*10^15	P[25] < 2.574*10^24
P[8]  <  77,288,598	P[17] < 4.319*10^16

	There are 4.325*10^19 positions in the standard cube; since
P[0]+P[1]+...+P[20] < 3.982*10^19, there must be a position at
least 21 qtw from SOLVED (The number 22 has appeared in Cube-lovers
recently, but it was an error).  There are 8.858*10^22 positions in
the Supergroup; since P[0]+P[1]+...+P[23] < 3.280*10^22, there must
be a position at least 24 qtw from SOLVED.  But this can be
improved:  half of the positions in the Supergroup are an odd number
of qtw from SOLVED, and since P[1]+P[3]+...+P[23] < 2.963*10^22 is
less than half the Supergroup, there must be some odd-length
elements of the Supergroup at least 25 qtw from SOLVED.  QED.  (If
you think there's something fishy here, mail to Hoey@CMU-10A for
clarification.  I am responsible for any cruft that has crept into
the original, elegant, formulation due to Jim Saxe.)

	The recurrence on which this bound relies is due to the
relations F^4 = FBF'B' = I (and their M-conjugates.)  It may be
possible to improve the recurrence by recognizing other short
relations.  Exhaustive search has shown that there are none of
length less than 10.  The most promising ones I know of come from
Singmaster:
		I = FR'F'R UF'U'F RU'R'U	(12 qtw),
		I = (FFBB RRLL)^2		(16 qtw),
and a 14-qtw relation which holds only in the standard group, since
it twists a face center 180o (see part 3).  Unfortunately, the
number of intermediate terms grows too large to be comfortably
hand-computable, and there are a few conceptual problems to hacking
it out.  If you can improve this, or know of any other relations
shorter than 24qtw, I'd like to hear about it.

	Coming up next: SPOILERS

Date:  9 January 1981 0756-EST (Friday)
From: Dan Hoey at CMU-10A
To: Cube-lovers at MIT-MC
Subject:  The Supergroup -- Part 3: A Super-H and Spoilers
Message-Id: <09Jan81 075610 DH51@CMU-10A>

			IMPROVEMENT

	Jim Saxe (16 December 1980 1841-EST) gave a 16-qtw process
for the H position:  FF LL DD FF BB DD RR FF.  Unfortunately, the
position obtained by this process is not H-symmetric in the
Supergroup:  the F, L, B, and R face centers are rotated 180
degrees, but the U and D face centers are in the home orientation.
I have a 16-qtw process which leaves all face centers in the home
orientation:  (FB UD)^2 (UD FB)^2.  This may look familiar --
conjugation by FBUD yields the 16-qtw "Bridge at Midday" I sent day
before yesterday.

			SPOILERS

	I developed a solution method on my own, but it was a long
one.  The following good methods, which affect only face centers,
are from Singmaster.  It seems simplest to solve the cube before
applying them, since some of the most popular processes (e.g., the
Spratt wrench (but not mono-ops!)) change face center orientation.

	The first method can be used to perform any multiple of
four face-center quarter-twists on the faces in a centerslice.  At
most two applications are necessary to accumulate any remaining
twist in one face center.  The method is given in Singmaster as two
examples, but he doesn't explain how they work.  Hopefully the
following discussion will make them easy to use.

	Choose a face center X that needs to be twisted, and a
centerslice containing X and other face centers to be twisted (call
these FCT's).  Place the cube with X up and with the FCT's in the FB
slice (i.e., among R, D, and L).  The basic move has much of the
flavor of a mono-op:
	1: Move the LR centerslice toward you. (Move a face
	    center from U to F.  X and the FCT's are now in
	    the UD centerslice.
	2: Select an FCT, and move it to the F face with UD
	    centerslice moves.
	3: Move the LR centerslice away from you. (Undo step 1.
	    The U face now contains the selected FCT and all the
	    U edge and corner cubies.
	4: Twist the U face the amount required by the selected
	    FCT.
Repeat the basic move until all FCTs have been selected.  Then
perform the move one last time, but select X instead of an FCT
in step 2 and twist the U face to fix the cube in step 4.

[After sending the previous part, I realized that in the case of
 one FCT, this is another 14-qtw relation in the standard cube:
	RL'FB'UD' R DU'BF'LR' U'
 But the one alluded to is in the next paragraph.]

	Since the total number of 90o face center twists must be
even (see Vanderschel's message of 6 Aug 1980 1909-PDT), the
preceding will solve the cube up to a 180o twist.  The process
(URLUUR'L')^2 twists the U face center 180o.  This is the only short
transform given in Singmaster for twisting face centers by a
nonmultiple of four; I'd be interested in any others you know.

That's it for now.  Happy Supercubing!

				Dan


Date: 9 Jan 1981 14:59 PST
From: McKeeman at PARC-MAXC
Subject: Re: Befuddled by BFUDLR
In-reply-to: VaughanW.REFLECS's message of 8 January 1981 20:06 cst
To: VaughanW.REFLECS at HI-Multics (Bill Vaughan)
cc:  CUBE-LOVERS at MIT-MC

Bill,

Good points.  I am not sure how to solve the problem either.  But here are some
comments.  RubikSong is simply assembly language to drive the human
computer.  That has both its advantages and disadvantages.  The original
suggestions also included parameterless subroutines and whole cube moves.

Your point it, I think, that there are some easy manipulations that are obscure, at
best, in FLUBRD + IJK.  One could take the position that definitions need not be
mnemonic since in steady state their names will be used with a higher level of
semantics.  Then we get

Slice = L' R J'
SpratWrench = (Slice U)*4

That seems tolerable but still not well matched to the human at the lowest level.

A nice property the notation could have would be to be concise where the
human manipulation is concise, without introducing a large number of
unmnemonic primitives.  For example, we might use Xnnn for multiple parallel
twists.

"R antislice" = L012 = L' R' J

Then we have, for example,

LRUDFBLRUDFB = (L012 U012)*3, 

and

R' = L001
R = L001'
Slice = L010.

Commutators are pretty common (i.e., X Y X').  Somebody suggested X[Y] for
them.  The idea is that X is a function to apply to Y.  The main problem is that
the human machine isn't very good at doing complicated inverses without some
special practice.  But, assuming the practice, then

BRL'D'RRDR'LB'RR
  = B[Slice] (RRB)[Slice']
  = (B[Slice])[RR]  RR
  = (B[Slice]RR)[]

Feedback appreciated...

Bill  


Date:  9 Jan 1981 1648-PST
From: Dave Dyer <DDYER at USC-ISIB>
Subject: nomenclature
To: cube-lovers at MIT-MC


 The real problem with RLUDFB is that it is nearly impossible for
the casual observer to determine if two "transformations" are the
same or not.   Transformations should be described such that there
is ONE description of each transformation.  We can't really get all
the way to the ideal, because there are discinct sequences that lead
to the same termination.  But we can eliminate a lot of the confusion.

I Propose that all transformations be described in terms of twists
done by both hands without changing the overall cube orientation.

my proposed primitives are
	1	"forward" twist of left or right face
	2	1/2 twist of left or right face
	3	"backward" of left or right face
	URF	orient the cube to work on named face,
		with your right hand. Names are lambda-bound from the
		start of the transformation.  To instantiating a transformation
		one can simply substitute the color for the orientation label

notation is <letter><right hand><(optional) left hand>


For example, a sprat wrench:
	 UDLRFB calls it:   RL' U  RL' B   RL' D   RL' F.
		I call it:  R11 U1 R11 F03 R11 U03 R11 F1

-------

Date:  9 January 1981 19:40 cst
From:  VaughanW.REFLECS at HI-Multics (Bill Vaughan)
Subject:  Re: nomenclature
To:  Dave Dyer <DDYER at USC-ISIb>
cc:  CUBE-LOVERS at MIT-MC
In-Reply-To:  Msg of 01/09/81 18:48 from Dave Dyer

It would seem that there are two needs immediately visible.

One need is for a notation to communicate a metamorphosis of the cube
from one cube-lover to another hexahedrophile.  This notation should
capture the spirit of the thing being described; it should be rich and
chunky (like soup?).

The other need is for an archival notation, for reference and
cataloguing. It must be spare and canonical.  There must be one and only
one way to describe a sequence of moves. (Implication: a way to get
reflections, inversions etc. out of the notation? how?) No nourishment
for the spirit there, but when you need to look something up...

Well, clearly they can't be the same notation. (Though even the archival
notation could invoke functions.)  Or can they? Food for thought...

Anyway, plain ole BFUDLR can't do either job decently. Not rich or
expressive enough for the first need, too expressive for the second
(gee, I wonder how many ways there are to annotate PONS?)

Well, other people can muddle this out - I'm going home to get some
dinner.




Date: 10 Jan 1981 13:19 PST
From: McKeeman at PARC-MAXC
Subject: Re: nomenclature
In-reply-to: DDYER's message of 9 Jan 1981 1648-PST
To: Dave Dyer <DDYER at USC-ISIB>, VaughanW.REFLECS at HI-Multics (Bill
 Vaughan)
cc: cube-lovers at MIT-MC

I agree there are two problems:
  1.  Neat "programs" that allow the recording and carrying out of manipulations.
  2.  Neat "configurations" that allow the recording of the results of carrying out
manipulations.
In both cases uniqueness, transparanecy, conciseness and all other notational
goodies are appropriate.

Generalizing a bit on Dave's suggestion, how about:

Manipulation = Macro*
Macro = MacroName "=" Move*
Move = Move "*" Integer                  -- power
        | Move '                              -- inverse
        | Move / Move                      -- conjugate
        | MacroName
        | Face (Near (Middle Far?)?)?      -- hand moves
        | "(" Move* ")"                       -- parenthesization
Face = F | U | R
Near = Middle = Far = 0 | 1 | 2 | 3

with considerate use of spaces and carriage control.  Face signifies "temporarily
move that face into the right hand, do the moves, then move it back where it
came from".  For compatibility, the digits count QTW clockwise (away).

The FLUBRD equivalences are:

R=R=R1=R10=R100
RR = R*2
RRR = R*3 = R'
L=R003
U=U
D=U003
F=F
B=F003
Slice=R01
AntiSlice=R103
I=F111
J=R111
X Y X' = Y/X

Note this has some chance of generalizing to other slice puzzles.  (E.g. 4x4x4)   It
also subsumes FLUBRD with a few appropriate Macro definitions like those
above.

------

For configurations the problem can be attacked by specifying which cubies go to
which cubies.  Singmaster does some of this.  The problem is to find a way to
specify patterns of change without just listing all the changes.  There are
positional change, flipping and rotating to be accounted for.

For instance, we would like to say (in some much neater way)

EdgeFlip = For all X and Y in FLUBRD, edge XY goes to edge YX.
The corners each take one step on a Hamilton path.
Each corner is rotated 120o.
Each center exchanges with its opposit.
Each edge XY goes to UV where U is the opposit of X and V is the opposit of Y.
Etcetera, etcetera, etc.


BFUDRLy yours,

Bill

Date: 10 JAN 1981 1405-PST
From: WOODS at PARC-MAXC
Subject: Re: nomenclature
To:   McKeeman, DDYER at USC-ISIB, VaughanW.REFLECS at HI-MULTICS
cc:   cube-lovers at MIT-MC

In response to the message sent  10 Jan 1981 13:19 PST from McKeeman@PARC-MAXC

I object!  Your proposed notation discriminates against lefthanded cubists!

	-- Don.
-------

Date: 10 Jan 1981 14:16 PST
From: McKeeman at PARC-MAXC
Subject: Re: nomenclature (discrimination against lefties)
In-reply-to: WOODS' message of 10 JAN 1981 1404-PST
To: WOODS
cc: cube-lovers at MIT-MC

Don,

The discrimination is actually self-imposed.  Merely rename your hands and you
will never know you were discriminated against.

Bill

Date: 12 Jan 1981 0913-PST
From: Isaacs at SRI-KL
Subject: Stanford Rubik's Cube Club
To:   cube-lovers at MIT-MC

  There is a newly formed Rubik's Cube Club, meeting at Stanford,
every Thursday, 7 p.m., Crother Memorial Hall, Conference Room.
For information, call Kersten (415)321-7725 or Paul 446-0729.
Open to all - beginners and experts.  First meeting was 1/6/81.
Second will be 1/20.
-------

Date: 12 Jan 1981 0929-PST
From: Isaacs at SRI-KL
Subject: nomenclature
To:   cube-lovers at MIT-MC

  I gave my father a cube a year ago (for which my mother may never forgive
me), and he has been working with it ever since. He has developed his own
notation, based on Angevine:  Basically, take a corner as an X-Y-Z
co-ordinate system, and call the planes (e.g.) X1, X2, and X3 (where
x2 is the slice).  L is X1, R is X3, F is Y1, etc.  + is a clockwise
rotation, looking at the -1 face (thus the X3,Y3, and Z3 faces rotate
backwards from bfudlr); - is CCW. Half twists are (on a typewriter) +/-
(he doesn't have a computer terminal).
     Anyway, his feeling on Singmaster nomenclature (with wich I disagree)
are as follows:
   "I do feel that Singmaster's limited cubist vocabulary impairs
communication of his knowledge and insights.  His general use of only
six of the nine groups available for rotation is like using only twenty
of the twenty-six letters in our alphabet.  With an unabridged dictionary 
and a thesaurus practically anything could still be said, but not as
well as using the whole alphabet. ... Perhaps mathematicians just don't
care about communication with ordinary people."
    My father is a lawyer.
-------

Date: 12 Jan 1981 10:05 PST
From: McKeeman at PARC-MAXC
Subject: Re: nomenclature
In-reply-to: Isaacs' message of 12 Jan 1981 0929-PST
To: Isaacs at SRI-KL
cc: cube-lovers at MIT-MC

Well, lawyers have themselves occasionally been subject to some criticism for
their "communication with ordinary people".

My dictionary says "Angevine"  has something to do with the line of
Plantagenet Kings.  I guess the connection is too subtle for me.

More constructively, the Isaacs senior notation seems 1-1 with FLUBRD except
that it has additional primitives for the slices.  The macro facility some have
used fills that hole.

The real trick is to find notations with (lots of) formal properties reflecting cubik
realities.  Partly that is a matter of notation design, but mostly it is a matter of
deeper understanding of the subject matter.  I do not believe it is an accident
that great science and great notations have frequently come from the same hand.

Bill


Date: 12 Jan 1981 1353-PST
From: Isaacs at SRI-KL
Subject: Re: nomenclature
To:   McKeeman at PARC-MAXC
cc:   cube-lovers at MIT-MC, ISAACS

  You'd have to ask him about his relationship to the Plantagenets (or
Anjous), but James Angevine (with an 'e') wrote out an early(?) solution
the the cube which was then sold by Logical Games, Inc, one of the first 
distributers of (what they called) the Magic Cube.  Singmasters first
published solution seemed dificult to communicate, so I sent the 'Angevine
Solution' to my father.
   Logical Games, Inc, incedentally, is currently manufacturing the cube in
the U.S.A., in white plastic with (I think) a slightly more pleasent color
scheme. 
   You're right that it's essentially the BFUDLR + slice, plus the 
different use of CW and CCW on the face furthest away.
   I hope to see you at the Rubik's Cube Club meeting.  (Actually, I may
have to miss this weeks - maybe I'll see you there the 22nd.)
                     --- Stan
-------

Date:  15 January 1981 18:30 cst
From:  VaughanW.REFLECS at HI-Multics (Bill Vaughan)
Subject:  Weird Algorithm - spoiler warning?
To:  Cube-Lovers at MIT-MC

On going through the old mail, I was a little surprised that
nobody uses the same algorithm that I do to solve the cube. But
since mine isn't terribly efficient, that's not much of a wonder.
Anyway, here it is.

1. Do bottom edges.  Honest to god.  I put all the bottom edges on
top by random dithering, then for each one, turn it so its
attached side facie abuts the color-matching face cubie, then
rotate that face 180o. That gets the bottom edges right, but
random hacking is almost as easy...

2. Do middle edges. (Getting colors right) I only use two moves
here.  FR'F'R and R'FRF'.  I pick a cubie that's on a top edge but
belongs on a middle edge - put its side facie adjacent to the
matching color face cubie (deja vu) and use FR'F'R if it has to
rotate right-and-down, or R'FRF' if it has to rotate
left-and-down.

3. Get top edges in correct places. Essentially as in Singmaster,
but I use only one of two moves. Align top edges so either: all
are OK (skip rest of this step) or one is OK and 3 are wrong.  (if
that's impossible, use one of this step's moves at random and
restart step - guaranteed to work.)  Now either FURU'R' or FRUR'U'
can be used to get everything OK.

4. Flip top edges as required. I use two different moves for this
according to whether adjacent or opposite edges need to be
flipped.  Opposite: let Q = "turn body-slicing slice 1 qtw
clockwise as seen from right". Then 4(QU) 4(UQ) flips FU and BU. 
Adjacent: FR'F'R.RU'R'U.UF'U'F flips RF and UF - you must reorient
the cube to do this on two U edges.  (I like this move because of
its symmetry and - somehow - completeness. It also rotates the
corner cubies adjacent to the edge cubies that it flips.)

5. Get the corners right.  Here I have some fun, but the basic
moves are: 3(FR'F'R) = (LFU,RBU) (RFU,FRD); 3(FRF'R') which also
does a double interchange - tho' it's asymmetrical and I don't use
it much; and B'FR'F'RB which stirs 3 of the top 4 CW or CCW - I
never remember because I just use it twice for the "wrong"
direction.

6. Tumble any corners that need it. Usually not many because of
the nice color flipping properties of 3(FR'F'R) -try it. My
tumbling move is a monotwist 2(FR'F'R).L.2(R'FRF').L' -- or
replace the L and L' with LL if necessary - sometimes it's nastier
and you have to do it twice.

I've never counted worst-case moves. The algorithm is based almost
entirely on Singmaster's Y commutator, and once you get that into
your finger bones, you hardly ever make a mistake.  On the other
hand, this algorithm is bad enough it hardly deserves a spoiler
warning.
			Bill q

Date:  15 January 1981 19:13 cst
From:  VaughanW.REFLECS at HI-Multics (Bill Vaughan)
Subject:  "Swirl Patterns"
To:  Cube-Lovers at MIT-MC

I've been recently investigating a set of patterns that I call
Swirl patterns for lack of a better name. In a Swirl, each face
looks like one of these:

     X X X                      X X X
     X Y Z (Left-hand swirl)    Z Y X (Right-hand swirl)
     Z Z Z                      Z Z Z

where X and Z are complementary colors, and Y is something else. 
I've classified them roughly into 6 classes, based on handedness
of swirl and relative alignment of faces.

If you look at the 3 faces adjacent to a corner, they may have the
same handedness, or they may have mixed handedness.  In addition,
two adjacent faces may have parallel or perpendicular swirls. 
(Parallel swirls have their XYZ columns parallel; perpendicular
swirls have thir XYZ columns perpendicular.)

Here are my 6 classes; there are another 6 which are mirror images
of these, but I don't count them. Nor do I care (at the moment)
about color pairings - though I know they are important - or about
the colors of the face cubies, which probably aren't important.

1.  Same handedness.  Two of the three faces have parallel swirls.

2.  Same handedness. All three faces have mutually perpendicular
swirls.

3. Mixed handedness. The two same-handed faces are parallel, with
thir XYZ columns in contact (i.e. forming a belt around the cube).

4. Mixed handedness. The two same-handed faces are perpendicular;
the opposite-handed face is perpendicular to both.

5. Mixed handedness. The two same-handed faces are perpendicular;
the opposite-handed face is parallel to one.

6. Mixed handedness. The two same-handed faces are parallel, with
their XYZ columns pointing towards the third face.

Four of these classes are in the antislice group and are a short
distance (8 qtw) away from SOLVED. They are classes 1, 3, 5 and 6.
They are also the antislice group's analogues of the slice group's
"6-spot" or "twelve-square" patterns.

What got me started on this is a problem that I still have. One
day while playing aimlessly in the antislice group (I thought I
had remained in it) I ran across a class 2 Swirl, which was (a)
quite pretty (when looked at from  the correct corner it looks
like a pinwheel) and (b) a bear to solve. (Clearly I thought it
was one of the "easy" Swirls.)

Having solved it, I wanted to get back to it and found I didn't
know how.  I tried solving to it and came up with an impossibility
- that's how I know the color arrangements must be important - and
I haven't found it in my searches yet - nor have I found class 4.

Questions: what's the fastest way to get to a class 2 Swirl?  What
color arrangements are permissible? Is it really in the antislice
group? (I now believe not.) Is any class 4 Swirl achievable?  How
quickly? Is there anything else interesting about Swirls?

I'm still playing with these - will give more data as I get it.
			Bill

Date:  16 January 1981 12:09 cst
From:  VaughanW.REFLECS at HI-Multics (Bill Vaughan)
Subject:  more on Swirl patterns: the Pinwheel
To:  Cube-Lovers at MIT-MC

I now have a class 2 swirl that looks like this:

    uud
    uld
    udd
 bbfrrrffbrrr
 buflfrfdblbr
 bfflllfbblll
    uud
    urd
    udd

It's pretty, and looks like a pinwheel from 4 of the corners, so i
call it a Pinwheel.

My current algorithm to get from SOLVED to Pinwheel is 84 qtw and
not worth publishing - I expect to reduce that substantially in
the near future.
			Bill

 Date: 01/16/81 1322-EDT
From: PLUMMER at LL
Subject: Ad in Popular Science
To: Cube-lovers at MIT-MC

Seems like a guy will tell you how to solve the cube for only $5.
Check the classified ads in current Popular Science!  --Bill
-------

Date: 20 Jan 1981 1632-PST
From: Isaacs at SRI-KL
Subject: Rubiks Cube Club meeting
To:   cube-lovers at MIT-MC

  The next meeting of the Stanford University Rubiks Cube Club (SURCC??)
will be thursday, Jan. 22.
        Meyer Library
        Room 145
   7:30     Puzzling with the cube, introduction to cube solving algorithms
   8:00     Discussion on how to design  and build new 3-D puzzles (magic
            tetrahedron, etc.)
   8:30     Cube theory and Pretty Patterns.  Until 9:30 or so.
    
      Further information: Kersten, (415)321-7725. 
      Also see article in Stanford Daily.  See you all there.
                         -- Stan
-------

Date: 21 January 1981 1246-EST (Wednesday)
From: Guy.Steele at CMU-10A
To: Isaacs at SRI-KL
Subject:  Re: Rubiks Cube Club meeting
CC: cube-lovers at MIT-MC
In-Reply-To:  Isaacs@SRI-KL's message of 20 Jan 81 19:32-EST
Message-Id: <21Jan81 124647 GS70@CMU-10A>

Stanford University Cube Kludge Society??
(Sorry, just kidding.)
--Guy

Date: 21 Jan 1981 10:07 PST
From: McKeeman at PARC-MAXC
Subject: Re: Rubiks Cube Club meeting
In-reply-to: Guy.Steele's message of 21 January 1981 1246-EST (Wednesday),
 <21Jan81 124647 GS70@CMU-10A>
To: Guy.Steele at CMU-10A
cc: Isaacs at SRI-KL, cube-lovers at MIT-MC

No.  Actually

Stanford University Rubik Environment For University Nuts.

Bill


Date: 22 January 1981 0010-EST (Thursday)
From: Dan Hoey at CMU-10A, James Saxe at CMU-10A
To: Cube-Lovers at MIT-MC
Subject:  Correction to "Symmetry and Local Maxima"
Reply-To: Dan Hoey at CMU-10A
Message-Id: <22Jan81 001000 DH51@CMU-10A>

	In our message "Symmetry and Local Maxima" (14 December
1980 1916-EST) we examined local maxima both in the Rubik group and
in the Supergroup.  David C. Plummer has discovered a flaw in our
argument for the Supergroup, which we now correct.

	Plummer has previously noted (30 DEC 1980 0109-EST) that
the T-symmetric position GIRDLE CUBIES EXCHANGED, depicted near the
end of section 4, is an odd distance from SOLVED.  This is also true
of the composition of GIRDLE CUBIES EXCHANGED with GIRDLE EDGES
FLIPPED, ALL EDGES FLIPPED, PONS ASINORUM, or any combination of
the three, for a total of eight positions.  In addition, there are
four different T groups, each corresponding to a choice of opposite
corners of the cube.  Thus 32 of the 72 positions with Q-transitive
symmetry groups are an odd distance from SOLVED.

	The discussion of the Supergroup in S&LM noted that the
only face-center orientations which yield Q-transitive symmetry
groups are the home orientation and all face centers twisted 180o
(called NOON in Hoey's message of 7 January 1981 1615-EST).  Any
position with either of these face center orientations must be an
even distance from SOLVED, so that any reachable position which is
T-symmetric in the Supergroup must be an even distance from SOLVED.

	In our earlier note, we erroneously calculated the number
of Supergroup positions with Q-transitive symmetry groups by simply
doubling the number of such positions in the Rubik group to allow
for the two allowable face-center orientations.  What we failed to
notice--until Plummer pointed it out--is that neither of the
allowable face-center orientations can occur in conjunction with an
odd position.

	The corrected count of known Supergroup local maxima is
determined by counting the 40 *even* symmetric positions, multiplying by
two, and subtracting 1 for the identity, yielding 79.  As Plummer notes,
this is surprisingly close to the number of known local maxima in the
Rubik group, which stands at 71.  The number of known local maxima
modulo M-conjugacy is 25 for the Rubik group and 35 ( = 2*(26-8) - 1 )
for the Supergroup.

Date: 21 Jan 1981 14:42:58-PST
From: microsoft!zibo at Berkeley
Gentlepeople: 
I have moved and would like to have my CUBE-LOVERS mail correctly sent:
My former destination was: ZBIKOWSKI@MARKET.  I now reside in:
CSVAX.MICROSOFT!ZIBO.  Could you make the necessary changes?  Gracias.



Date: 27 January 1981 0102-EST (Tuesday)
From: Jim Saxe, Dan Hoey
To: Cube-Lovers at mit-mc
Subject:  Pretty Patterns and Solutions
Sender: Dan Hoey at CMU-10A
Reply-To: Dan Hoey at CMU-10A
Message-Id: <27Jan81 010221 DH51@CMU-10A>

	We are disappointed at Chris C. Worrell's use of the term
"Baseball" for the position known in the literature as the "Worm".
Worrell's term propagates the apparently popular misconception that
baseballs are covered with three-lobed pieces of leather.  The
position which *we* call "Baseball" reflects the construction much
more accurately:

		D D D
		U U U
		F F F

		U U U	L R B	R B L	B L R
		F F F	L R B	R B L	B L R
		D D D	L R B	R B L	B L R

		F F F
		D D D
		U U U

Currently, our best process for this position is 34 qtw.  The
corners are fixed with FRLUDB, two edge four-cycles are inserted in
the middle, and a Spratt wrench is conjugated inside that:
FRL (RRLL UUDD F' U (F' LUD' BUD' RUD' FUD' F) UDD RRLL F) UDB.

	The class of patterns which Bill Vaughan calls "Swirl
Patterns" (15 January 1981 19:13 cst) are also known as "6-2L"
patterns in Singmaster, and the particular one he calls the
"Pinwheel" (on 16 January 1981 12:09 cst) is an M-conjugate of the
AC-symmetric "Twelve-L's" mentioned in our message on Symmetry and
Local Maxima (14 December 1980 1916-EST, Section 6).  [Incidentally,
the diagram he displays in the message of 16 January is in error;
the left and right face centers have been swapped.  This is made
less obvious by the unusual orientation of the cube in that
diagram.]  We have found a totally magical 12 qtw process for the
Pinwheel:  FB LR F'B' U'D' LR UD.

	Vaughan's definition of Swirl Patterns seems unduly
restrictive to us on one count:  he requires the two L's on each
face to be of "complementary" (evidently meaning opposite) colors.
This is not necessary for an L pattern.  According to our analysis,
however, at least two of the faces of any L pattern must have L's
of opposite colors, and five is easily seen to be impossible.  We
know of no patterns having three or four such pairs.  But there are
several with two pairs.  Our favorite example is a relative of the
Baseball which we name for Linda Lue Leiserson, who has the
appropriate initials.

		D D D
		F U D
		F F F

		U U U	L B B	R L L	B R R
		D F U	L R B	R B L	B L R
		D D D	L L B	R R L	B B R

		F F F
		U D F
		U U U

We have a 24 qtw process for Linda Lue's L:
	F'BB L  (B U'LR' F'LR' D'LR' B'LR' B')  R UD B'FF
This has a Spratt wrench, conjugated by B', embedded in magic.

	David C.  Plummer (3 SEP 1980 2123-EDT) reported that it is
possible for each of the six faces of the cube to show a capital
"T".  Our analysis indicates that there are two sorts of T patterns:

		D U D				D D U
		D U D				U U U
		U U U				D D U

	L L L	F F F	R R R		L R R	F F F	R L L
	R L R	B F B	L R L		L L L	B F B	R R R
	R L R	B F B	L R L		L R R	B F B	R L L

		D D D				D U U
		U D U				D D D
		U D U				D U U

		B B B				B B B
		F B F				F B F
		F B F				F B F

	      Tanya's T			    Plummer's T

Tanya's T is named for Tanya Sienko (who inspired the problem) and
for euphony.  Plummer's T is named for Plummer's Cross (which has
the same symmetry group) and for homophony.  There are 24
M-conjugates of Tanya's T, while Plummer's T has 8 M-conjugates.  By
adapting a process due to David C.  Plummer, we have developed a
16-qtw process for Tanya's T:  (FF UU)^3 (UU LR')^2.  The first part
swaps two pairs of edge cubies, and the second part is magic.  We
have found a 28-qtw process for Plummer's T, which is entirely
magical:  FF UD' F'B' RR F'B U'D RL FF RL' UD' RL FF R'L U'D'.

	A position which is not so visually striking, but which is
important in the symmetry theory we have discussed earlier, is "All
Corners Twisted":
			B U B
			U U U
			F U F

		U L U	L F R	U R U	R B L
		L L L	F F F	R R R	B B B
		D L D	L F R	D R D	R B L

			F D F
			D D D
			B D B

	This can be achieved in 30 qtw with FLU (LRRFFB')^4 U'L'F'.
The process is a conjugated from a 24 qtw process invented by
Thistlethwaite.  Unfortunately, Thistlethwaite's process twists the
wrong corners, and no cancellation can be performed in the
conjugation.  If any process can be found which twists four corners
clockwise and four counterclockwise, leaving the rest of the cube
fixed, then any such pattern can be made by adding at most 6 qtw.

Date: 29 JAN 1981 0344-EST
From: BSG at MIT-AI (Bernard S. Greenberg)
Mailed-by: BSG @ MIT-Multics
Subject:  Lisp Machine Cubesys Improvements
To: CUBE-LOVERS at MIT-MC


Largely due to the newfound coincidence of relentless Lisp Machine
hacking with my job, I have added a slew of features to Lisp Machine
Cubesys, viz., all kinds of New Window System and flavor hacking.  It is
now completely mouse-oriented, all moves are made by mousing menu items,
OR by mousing at cube-sides on the display (either display) to turn
faces, left mouse button for left (ccw) turn, etc.!  To find out more
about it, load it in the usual way, (load "bsg;cubpkg >"), invoke it
in the usual way ((cube)), and type the HELP key (or mouse the HELP
menu item). Have fun, -bsg

Date:  1 February 1981 0539-EST (Sunday)
From: Dan Hoey at CMU-10A
To: Cube-lovers at MIT-MC
Subject:  Algorithm for finding cube group orders
Message-Id: <01Feb81 053933 DH51@CMU-10A>

	David C.  Plummer (31 Dec 1980 1210-EST) gave a preliminary
analysis of the 5x5x5 cube.  I complete it here.  Let a move consist
of twisting any of the six faces, at a depth of 1 or 2.  It will be
necessary to consider the two depths as distinct; M1P will refer to
the number of depth 1 moves (mod 2), while M2P will refer to the
number of depth 2 moves (mod two).  It is important to realize that
the two parities vary independently.  The tabs on each face are
assigned types
	C L E R C
	R D A D L
	E A X A E
	L D A D R
	C R E L C
as in Plummer's analysis.

	Let COP ("C" Orientation Parity) and CPP ("C" Permutation)
parity be defined as before.  As before, COP=0 (mod 3).  We must be
explicit about the CPP this time:  Since either kind of move is an
odd permutation of the "C" faces, CPP=M1P+M2P.

	As in the 4x4x4 case, "R"'s may be ignored and "L"'s have
no orientation.  The permutation parity (LPP) is important, however.
Depth 1 moves are an even permutation of the "L"'s (two 4-cycles),
so they do not affect the LPP, but Depth 2 moves are an odd
permutation of the "L"'s (three 4-cycles).  Therefore LPP=M2P.  Note
that while LPP and CPP may vary independently, they together
determine both M1P(=LPP+CPP) and M2P(=LPP).

	The "E" faces act as in the 3x3x3 case, with orientation
and permutation parity.  Orientation changes on four "E"'s with
every move, so EOP=0 (mod 2).  Permutation parity changes with every
move, so EPP=M1P+M2P.  This has already been determined by CPP, so
only half of the "E" permutations are possible.

	Every move is an odd permutation of the "D" faces, so
DPP=M1P+M2P.  Since M1P+M2P=CPP is determined, only half of the "D"
face permutations are possible.

	Moves work differently on "A" faces depending on depth:
Depth 1 moves are odd permutations of the "A"'s, and depth 2 moves
are even.  Thus APP=M1P, which is determined by CPP+LPP, so only
half of the "A" permutations are possible.

	Finally, the "X" faces have orientation which changes on
every move, so XOP=M1P+M2P, and only half of the "X" orientations
are possible.

	Thus there are 96 orbits, corresponding to COP (mod 3) and
EOP, EPP+CPP, DPP+CPP, APP+CPP+LPP, and XOP+CPP (mod 2).  The basic
combinatoric is as Plummer described:
	8!    C Permutations
	3^8   C Orientations
	24!   L Permutations
	1     R Permutation
	12!   E Permutations
	2^12  E Orientations
	24!   D Permutations
        24!   A Permutations
        4^6   X Orientations
which when multiplied together and divided by 96 yields about
5.289*10^93.  [This differs from Plummer's result by a factor of
4096 because (4^6/2) he didn't count X Orientations, and (2) he did
not realize that LPP and CPP are independent.]  My implementation of
Furst's algorithm claims that all of these are reachable.

To count the number of reachable color patterns, divide this note
that there are by (4!)^6/2 invisible D permutations, (4!)^6/2
invisible A permutations, and 4^6/2 invisible X orientations that
satisfy the invariants.  While there are pairs of L/R edges that
look the same, they cannot be interchanged, for that would entail
putting an L tab into an R position.  So there are 2.829*10^74
different color patterns achievable.

----------------------------------------------------------------


Date:  1 February 1981 0612-EST (Sunday)
From: Dan Hoey at CMU-10A
To: Cube-lovers at MIT-MC
Subject:  Algorithm for computing cube group orders
Message-Id: <01Feb81 061255 DH51@CMU-10A>

	Oops... you have just received part 3.  This is part 1....

	This note is in somewhat delayed response to the note by
David C.  Plummer (31 DEC 1980 1115-EST) regarding the 5x5x5 Rubik
cube, and some related ideas.  In that note he tried to calculate
the size of that cube's Rubik group, but left several of the values
open to conjecture.  I will complete the answer, and answer a few
others that haven't been addressed here.

	Computing the size of a Rubik group is a special case of
computing the size of a permutation group, given generators for
that group.  The technique we have already seen in these pages is in
two parts.  The first part seems relatively easy:  certain invariants
must be observed in the generators, such as "Corner Orientation
Parity" and "Total Permutation Parity."  [In this general setting,
such invariants as "Colortabs on the same cubie move together" must
also be considered.]  It may take some thought to dig out the
invariants, but once you have seen them demonstrated for Rubik's
Cube, you have an idea of what to look for.  The second part is the
devil:  it must be demonstrated that every permutation satisfying
those invariants is actually generated.  This involves developing a
solution method for the puzzle.  Given the days or weeks (or
eternity) it takes most people to develop such a method--with cube
in hand!--it is hardly surprising that few answers have been
developed.

	Well, the second part is no longer a hard problem.  The
answer lies in a paper by Merrick Furst, John Hopcroft, and Eugene
Luks, entitled "Polynomial-Time Algorithms for Permutation Groups,"
which was presented at the 21st Annual Symposium on Foundations of
Computer Science, October 1980.  Among the results is an algorithm
which takes as input a set of permutations on n letters, and
reports the size of the group G which is generated by those
permutations.

	The algorithm operates by decomposing G into a tower of
groups I=G[0], G[1], ..., G[n]=G, where G[i] contains those
permutations p in G for which p(k) = k whenever i < k <= n.  The
index of G[i-1] in G[i] is developed explicitly by the algorithm;
in fact, a representative g[i,j] of every coset of G[i-1] in G[i]
is exhibited.  These coset representatives generate G; in fact,
every element of G is representable as a product of the form
(g[1,j1])(g[2,j2])...(g[n,jn]).  For this reason the coset
representatives are called "strong generators" for G.  There is a
good deal of structure that can be learned from the strong
generators, in addition to the size of G.

	I have coded this algorithm in Pascal, and offer the
program for the use of anyone who needs to find group orders.  The
relevant files are on CMU-10A, from which other sites may FTP
without an account.  The relevant files are
    all:group.pas[c410dh51]	The source
    all:rubik.gen[c410dh51]	A sample input -- the supergroup
    all:rubik.lst[c410dh51]	Sample output.
Of course, CMU Pascal is probably slightly different from yours,
and OS-dependent stuff like filenames is likely to be wrong.  I'll
be glad to help out in cases of transportability problems.  The
other problem you may run into is resource availability.  The
running time of the algorithm is proportional to (nm)^2, where m is
the total number of strong generators; the supergroup (n=72, m=279)
takes 639 cpu seconds on a KL-10, and bigger problems grow rapidly.
The program also requires 47000+47m words.

	It might seem that the problem has been answered, but I
find that simply knowing the size of a group is not very
satisfying.  There doesn't seem to be a better way of demonstrating
lower bounds, but the upper bounds that come from invariants are
much more elegant than a simple numerical answer.  Unfortunately, I
know of no mechanical way of finding the invariants.  Furthermore,
using group theory does not help much when we ignore the
Supergroup.  Consider the 4x4x4 cube.  If we are only concerned with
the color pattern on the cube, then a twist may or may not affect
the four face centers--it depends on whether they are the same
color or not.

	In summary, the algorithm has inverted the hard and easy
parts of cube analysis.  The size of the group is now easy to
determine, making invariant-finding the hard part.  Further, the
algorithm works on the Supergroup, making counting distinct color
patterns the part which requires further analysis.  Two messages
follow, supplying these answers for the 4x4x4 and 5x5x5 cubes.

Date:  1 February 1981 0651-EST (Sunday)
From: Dan Hoey at CMU-10A
To: Cube-lovers at MIT-MC
Subject:  Analysis of the 4x4x4 cube
Message-Id: <01Feb81 065108 DH51@CMU-10A>

	The first problem for the 4x4x4 cube is in eliminating
positions that arise from whole-cube moves.  This was done with the
3x3x3 by keeping the face-center positions fixed, but there are no
face centers on the 4x4x4--or there are, but they don't maintain a
fixed orientation relative to each other.  I standardize the spatial
orientation by keeping the DBR corner in a fixed position and
orientation.  A move then consists of twisting one, two, or three
layers parallel to the U, F, or L face.  Thus F3' is equivalent to
twisting the B face.  I will refer to the number of layers twisted
as the "depth" of the move.

	Following David C.  Plummer's notation (31 DEC 1980 1210-EST),
organize each face of the cube as
	C L R C
	R X X L
	L X X R
	C R L C.
I will assume familiarity with David Vanderschel's analysis of the
3x3x3 case, which was presented in his message of 6 August 1980.

	"C" faces act as they do in the 3x3x3 case, except that one
of them does not move.  Corner Orientation Parity (COP) is preserved
and Corner Permutation Parity (CPP) changed by every quarter-twist.

	Depending on the depth, a quarter twist can permute the "L"
faces in an odd or an even permutation.  Also, "L" faces do not
change orientation (or move to "R" positions).  Every "R" face is
determined by the "L" face (on an adjacent side of the cube) with
which it shares a cubie.  Thus the arguments for EOP and EPP do not
apply.

	Every quarter-twist is an odd permutation of the "X" faces:
either one, three, or five four-cycles, depending on the depth.
Letting XPP be the permutation parity of the "X" faces, the Total
Permutation Parity TPP=XPP+CPP (mod 2) is preserved by every
quarter-twist.

	Thus the 4x4x4 cube group has at least six orbits,
according to COP (mod 3) and TPP (mod 2).  The basic upper bound of
	7!    Corner Permutations
	3^7   Corner Orientations
	24!   L Permutations (which determine the R permutations), and
	24!   X Permutations,
divided by six, yields an upper bound (of about 7.072*10^53).  I have
run Furst's algorithm on the problem, and my program claims that all
these positions are reachable.

	To calculate the number of reachable color patterns, note
that there are 4!  permutations of each quadruple of "X" faces which
are indistinguishable.  However, the TPP constrains the XPP so as to
reduce this by a factor of two.  Dividing 7.072*10^53 by (4!)^6/2
yields 7.401*10^45.

	[At this point, you may find it instructive to view the
message before last, which analyzes the 5x5x5 cube in the context
of this message and the one immediately preceding.  I regret the
accidental disorder.  These three are all for now, although I have
results on tetrahedra, octahedra, and a dodecahedron which I am in
the process of writing up.]

Date:  6 Feb 1981 at 1330-CST
From: korner at UTEXAS-11 
Subject: cube lube
To: cube-lovers at mit-mc

	After trying almost all the cube lubricants suggested (with
the notable exceptions of the plastic eating varieties)- I would like
to suggest that a local maxima seems to be silicon gel (of the sort
used to lubricate SCUBA O rings or food processors- not the spray, the
gel).
	To use this stuff, one must disassemble the cube. As long as it's
apart, take a fine flat file to the cubies and remove any seams from the
molding process and any imperfections from the glue job (cement beads or
protruding plates). Cubus hungarius finishes well with just a file, 
cubus americanus (the white one) may need work with wet fine sandpaper
to restore a smooth surface after filing. If you're really fanatic, 
adjust the screw tension (ala Singmaster). Clean off the debris and apply
liberal coats of the gel to all tab faces. Reassemble the cube and enjoy-
one handed cubing not guaranteed but definitely possible.
		-KMK
-------

Date:  9 FEB 1981 2345-EST
From: JURGEN at MIT-MC (Jonathan David Callas)
Subject: True Stories of Cubism
To: CUBE-LOVERS at MIT-MC

I was at the Hirshhorn (Smithsonian Modern Art Museum) last Sunday to see the
exhibit of Avant-garde Russians, and lo, I saw in the museum shop what could 
only be Cubus Albus! I played with it
for awhile (I solved the top 2 tiers before my girlfriend said "That's enough!
You can do that at home!") and it worked more smoothly than my C. Americanus!
So now, I guess, not only is the Cube a source of mathematical inspiration,
but an objet d'art as well.

		Happy Cubing,
		Jurgen at MIT-MC


Date: 11 FEB 1981 1600-EST
From: RP at MIT-MC (Richard Pavelle)
To: CUBE-LOVERS at MIT-MC

The March issue of Scientific American is out. Guess what is on the
cover as well as in the interior? 

Date: 12 Feb 1981 0816-PST
Sender: OLE at DARCOM-KA
Subject: The England Scene
From: Ole at DARCOM-KA (Ole J. Jacobsen)
To: Cube-Lovers at MIT-MC
Message-ID: <[DARCOM-KA]12-Feb-81 08:16:37.OLE>

Yes, cubes are indeed very big here in England due mostly to the fact
that they have been featured on television several times recently.
About 3 weeks ago a kid solved a cube in 37 secs on the Saturday mor-
ning BBC1 show "Multicoloured Swap Shop" (very appropriate name). In
a follow up a group of people challenged him, but "only" managed it
in 57 secs. Nothing was however said about local maxima etc, so it
wasn't a very scientific exercise. The recordholder's solution se-
quence was shown in slow motion (his hands still seemed to move very
fast) and as far as I could determine he uses Kertezs's Algorithm,
i.e layer-by-layer, but with some clever shortcuts rather than just
using the macros blindly. At the moment cubes are impossible to get
but we are hoping for a new shipment to arrive soon. A cube club
will probably be formed here at Newcastle University, Newcastle upon
Tyne and I would be surprised if other universities won't be doing
the same.

By the way, has anyone ever tried turning  cube when the temperature
is 5-6 degrees C? I have, because that is the temperature my room is
at when I come home at night. English houses are VERY cold.
Ole

Date: 16 February 1981 1229-EST (Monday)
From: Guy.Steele at CMU-10A
To: bug-lispm at MIT-AI, cube-lovers at MIT-MC
Subject:  Scientific American
Message-Id: <16Feb81 122922 GS70@CMU-10A>

Congratulations to cubemeisters, LISP Machinists, and Symbolicists
alike for making *Scientific American*.

Now that the LISP Machine has been used to serve the cause of
cubing, has any thought been given to the converse?  For example,
perhaps a mouse/joystick-like device could be built based on
cube technology?

Also, anyone thought about the limiting case of odd-shaped
polyhedra: the continuous cube (or, Rubik's sphere)?  There are
three possible places to introduce continuity.  For a given
twist, one must choose an axis, choose a depth of slice,
and choose an angle of twist.  For the cube all three are
quantized.  What are the geometric/topological properties
of an object where some subset of these three choices are
given a continuous domain?  (I haven't the mathematics undert
my belt to attack this problem -- sorry.)
--Guy

Date: 16 February 1981 2327-EST (Monday)
From: Jim Saxe, Dan Hoey at CMU-10A
To: Cube-Lovers at MIT-MC
Subject:  Four colors suffice
CC: Mary Shaw at CMU-10A, Paul Hilfinger at CMU-10A, Bill Wulf at CMU-10A,
    Dorothea Haken at CMU-10A
Sender: Dan Hoey at CMU-10A
Reply-To: Dan Hoey at CMU-10A
Message-Id: <16Feb81 232721 DH51@CMU-10A>

	Douglas Hofstadter, in the Metamagical Themas column in
Scientific American this month, shows two alternate ways of
coloring a cube.  Both suffer from two drawbacks:  They fail to
distinguish all cube positions, and they use more than six colors.
This seems inefficient to us, since there is a coloring of the cube
which distinguishes all elements of the Supergroup and uses
only four colors (and which, like Hofstadter's colorings and
the standard coloring, satisfies the restriction that every
whole-cube move is a color permutation, as discussed in point 2
below).

	Our coloring, called the Tartan, is formed by assigning the
colors blue, green, red, and yellow to the four pairs of antipodal
corners of the cube.  Thus for each face of the cube, the four
corners of the face are assigned four different colors.  We use the
term ``plaid'' to denote such an assignment of colors to the
corners of a square.  To color the cube, divide each facelet of each
cubie into four squares, and color the squares so all facelets on a
side of the cube display the plaid associated with that face.  The
result is shown below, with the initial assignment of colors to
corners in lower case.

			  (r)---------------(y)
			   | R Y   R Y   R Y |
			   | B G   B G   B G |
			   |		     |
			   | R Y   R Y   R Y |
			   | B G   B G   B G |
			   |		     |
			   | R Y   R Y   R Y |
			   | B G   B G   B G |
	(r)---------------(b)---------------(g)---------------(y)
	 | R B   R B   R B | B G   B G   B G | G Y   G Y   G Y |
	 | G Y   G Y   G Y | Y R   Y R   Y R | R B   R B   R B |
	 |		   |		     |		       |
	 | R B   R B   R B | B G   B G   B G | G Y   G Y   G Y |
	 | G Y   G Y   G Y | Y R   Y R   Y R | R B   R B   R B |
	 |		   |		     |		       |
	 | R B   R B   R B | B G   B G   B G | G Y   G Y   G Y |
	 | G Y   G Y   G Y | Y R   Y R   Y R | R B   R B   R B |
	(g)---------------(y)---------------(r)---------------(b)
			   | Y R   Y R   Y R |
			   | G B   G B   G B |
			   |		     |
			   | Y R   Y R   Y R |
			   | G B   G B   G B |
			   |		     |
			   | Y R   Y R   Y R |
			   | G B   G B   G B |
			  (g)---------------(b)
			   | G B   G B   G B |
			   | R Y   R Y   R Y |
			   |		     |
			   | G B   G B   G B |
			   | R Y   R Y   R Y |
			   |		     |
			   | G B   G B   G B |
			   | R Y   R Y   R Y |
			  (r)---------------(y)

	To understand the importance of the Tartan, there are
several points to consider:

	1.  By reading off the four colors of a plaid in clockwise
order, starting at an arbitrary point, we obtain four permutations
of the four colors.  Quadruples read from different faces are
disjoint, so all 24 permutations of the four colors appear on the
Tartan, once each.

	2.  Every motion in the group C of whole-cube rotations is a
permutation of the pairs of antipodal corners, and so corresponds to
a recoloring of the Tartan.  Some restriction of this sort is
necessary to prevent us from simply drawing a different
black-and-white picture on each facelet and calling that a
two-coloring.

	3.  Point 2 implies that C is isomorphic to a subgroup of
S4, the group of permutations on the four colors.  But both C and S4
have 24 elements, so C is isomorphic to S4 itself (a fact
well-known to crystallographers).

	4.  Since every color permutation is realizable by a
whole-cube move, there is only one Tartan (up to whole-cube moves).
This is why we use colors as labels, rather than some FLUBRDoid
positional scheme.  [The actual choice of colors and the name
``Tartan'' arise from the DoD Ironman project.]

	5.  Every reflection of the Tartan is color-equivalent to a
rotation.  In particular, the identity is color-equivalent to a
reflection through the center of the cube.  If you were to lend your
Tartan to someone who ran it through a looking-glass, you could not
discover the fact except by removing the face-center caps and
examining the screw threads!

	We have constructed a Tartan from a Rubik's cube and
colored tape.  Due to the similar appearance of the plaids, it takes
us several times as long to solve the Tartan as it takes to solve
Rubik's cube.

	Our search for pretty patterns has not been particularly
rewarding.  Part of the reason seems to be that the cube's
appearance is strongly constrained by the Tartan's coloring.  On
Rubik's cube one may make a particular face pattern (e.g.  orange T
on white background) using any of several identically colored
facelets.  On the Tartan, however, the plaid on any facelet of a
cubie, together with the orientation of the plaid relative to the
cubie, determines the plaid and orientation of the other facelet(s)
of the cubie.

	The one nice pattern we have is in fact the conceptual
precursor to the Tartan.  It is Pons Asinorum (FFBBUUDDLLRR) applied
to the position shown in the diagram above.  In this position, the
plaids of adjacent facelets line up with each other to display the
same arrangement of plaids, magnified by a factor of two.  Each face
looks like the following, for some assignment of colors to the
numbers 1 through 4:

			(1)---------------(2)
			 | 1 2   2 1   1 2 |
			 | 4 3   3 4   4 3 |
			 |		   |
			 | 4 3   3 4   4 3 |
			 | 1 2   2 1   1 2 |
			 |		   |
			 | 1 2   2 1   1 2 |
			 | 4 3   3 4   4 3 |
			(4)---------------(3)


Date: 17 Feb 1981 07:54:00-PST
From: microsoft!zibo at Berkeley

Is it possible???  In the SciAm article they mentioned 4x4x4 cubes...
Has anyone seen them??


Date: 17 Feb 1981 10:49 PST
From: McKeeman at PARC-MAXC
Subject: Re: 4x4x4 cube
In-reply-to: Your message of 17 Feb 1981 07:54:00-PST
To: microsoft!zibo at Berkeley
cc: cube-lovers at MIT-MC

Zibo,

All things are possible in the computer.  My undertanding is that a 4x4x4 is
being built, but I have not heard that it is yet complete.

The real mind-bender is the continuous Rubik Sphere.  Take a sphere, slice it
arbitrarily many times.  Now each slice makes a plane of rotation and gives a
degree of freedom.   The Rubik Cube is a special case for which there is a
mechanical implementation.

Bill


Date: 17 Feb 1981 1445-PST (Tuesday)
From: Mike at UCLA-SECURITY (Michael Urban)
Subject: Rubik's Sphere
To: cube-lovers at mit-mc

   While it shouldn't be TOO hard to program an arbitrarily-fine
simulation of a Rubik sphere, one wonders how it's colored.
do the latitute/longitude coordinates on the sphere correspond
to hue/intensity?  While you can settle for any coordinate 
labelling of points, different coloring schemes will dictate
what constitutes a "pretty" pattern, yes?

	Mike
-------


Date:  17 February 1981 18:03 cst
From:  VaughanW at HI-Multics (Bill Vaughan)
Subject:  Rubik's Sphere
Sender:  VaughanW.REFLECS at HI-Multics
To:  Mike at UCLA-Security, cube-lovers at MIT-MC

Do you color a Rubik's Sphere discretely or continuously?  And if
continuously, it's probably too easy to solve - seems as though
the discontinuities would show you where to turn it; and by always
turning along a discontinuity until it vanishes, you can get to
SOLVED by what seems to be God's Algorithm.

			Bill

Date: 17 Feb 1981 16:12 PST
From: McKeeman at PARC-MAXC
Subject: Re: Rubik's Sphere
In-reply-to: Mike's message of 17 Feb 1981 1445-PST (Tuesday)
To: Mike at UCLA-SECURITY (Michael Urban)
cc: cube-lovers at mit-mc (I wish Hofstadter were on the net)

Mike,

Are you proposing a truly continuous Rubik sphere with an infinite, nay
uncountable, number of slicings with continuously varying hue to distinguish
"slices"?

Such cubes could differ in the "function" that connects the motion of
"neighboring slices".  We could have linear, quadratic, and even
hyperexponential axes of rotation.  Then giving the cube a spin about each of its
(many) axes, we would have a continuously shifting pattern of color.  Maybe
would should leak this idea to George Lucas for the visuals of StarWars III?  Or
maybe one of the LISP machine folks can whip up a simulation overnite?

Bill


Date: 17 Feb 1981 1622-PST
From: Steve Saunders <SAUNDERS at USC-ISIB>
Subject: Rubik-like sphere
To: Cube-Lovers at MIT-MC

And if you color it continuously, why not have continuous moves, too? For
instance, a smooth twist about an axis (like twisting a rubber ball that's
glued to sticks at its poles -- carries meridians into spirals), or a smooth
bending (like pushing one of those poles sideways while holding the other
fixed -- makes parallels not parallel).  I suspect that the groups resulting
from some sets of smooth motions would be very simple, but some might have
interesting interactions.  A problem with all this smoothness (a feature?) is
that it would enable approximate solutions, iterative converging infinite
"solutions", and disputes about whether SOLVED has in fact been reached --
none of which occur with the real Rubik's.

		Steve

-------

Date: 17 Feb 1981 1716-PST (Tuesday)
From: Lauren at UCLA-SECURITY (Lauren Weinstein)
Subject: Sphere carrying case
To: CUBE-LOVERS at MC

I know where I'm going to keep my Rubik's Sphere when I get one:
inside my Klein Bottle!

--Lauren--
-------


Date: 17 Feb 1981 18:00 PST
From: McKeeman at PARC-MAXC
Subject: Re: Rubik's Sphere
In-reply-to: VaughanW.REFLECS's message of 17 February 1981 18:03 cst
To: VaughanW at HI-Multics (Bill Vaughan)
cc: Mike at UCLA-Security, cube-lovers at MIT-MC, SAUNDERS at USC-ISIB

Bill,

Well,  My idea is that the continuous slicing is constrained by some function so
that you basically only have one degree of freedom per axis.  The trick is to
scramble on several axes, and then try to get back.  If the coloring is continuous
before any twisting, then it is always continuous.  (I think)

Another Bill


Date:  20 February 1981 20:17 est
From:  Greenberg.Symbolics at MIT-Multics
Subject:  A lighter note
To:  CUBE-HACKERS at MIT-AI

in this increasingly hirsute forum:

A man called us at Symbolics today, having seen our name in the
Scientific American article.  He was having trouble getting
cubes in the Chicago area, and wanted to know if we could sell
him some.... (a true story)...


Date:  20 February 1981 20:19 est
From:  Greenberg.Symbolics at MIT-Multics
Subject:  A lighter note
To:  cube-hackers at MIT-MC

in this increasingly hirsute forum:

A man called us at Symbolics today, having seen our name in the
Scientific American article.  He was having trouble getting
cubes in the Chicago area, and wanted to know if we could sell
him some.... (a true story)...

Date: 20 FEB 1981 2108-EST
From: RP at MIT-MC (Richard Pavelle)
Subject: Rubik
To: CUBE-HACKERS at MIT-MC

When speaking to the editor of Scientific American yesterday, the subject
of the cube came up. I mentioned that Rubik did not get dollar much less
a forint for his effort. Guess what! A subscription to Sci. Am. is on the
way to him (lifetime I suppose).

Date: 21 February 1981 00:14-EST
From: Ed Schwalenberg <ED at MIT-AI>
Subject: A lighter note
To: Greenberg.Symbolics at MIT-MULTICS
cc: CUBE-HACKERS at MIT-AI

Isn't this the C-Machine that you all are working on?

 Date: 03/05/81 0839-EDT
From: PLUMMER at LL
Subject: another article
To: CUBE-LOVERS at MIT-AI

Check today's Wall Street Journal: front page, center.  --Bill
-------


DAN@MIT-ML 03/05/81 20:58:59 Re: Rubiks Cube info needed
To: cube-lovers at MIT-AI
I have a few questions which you may be able to help me with...

1.  Could you please add me to this mailing list

2.  I am looking for a rubiks cube solver to play with on my microcomputer,
    and would like to know if such a program exists.  Would prefer Pascal
    or "C" (as I most readily hack these), but Lisp, et.c would be fine.

3.  Is there an archive of "Cube" info, back letters, documents,
    bibliographies, etc. lying around on one of the ITS machines?

Thanks      - Dan



Date:  6 MAR 1981 0849-EST
From: JURGEN at MIT-MC (Jonathan David Callas)
Subject: Cube Solver
To: DAN at MIT-ML
CC: CUBE-LOVERS at MIT-MC

I have a program written in pascal that won't *SOLVE* the cube but
will manipulate it. It will also find the order of a given move.
It was written by Tom Davis (of this list) and modified by me.
It should run on any USCD system with no hassles, and very minor
ones for any other sort of pascal. Since Tom was giving the program out before,
I shall assume that there is no problem with distributing this version.
If you (or anyone else) wants acopy, write me (Jurgen at MC) & I'll send
you a copy.
			-- Happy cubing,
			-- Jurgen at MC

Date:  7 Mar 1981 0224-PST
From: Alan R. Katz <KATZ at USC-ISIF>
Subject: how about...
To: cube-lovers at MIT-MC
cc: katz at USC-ISIF

How about a Braile cube (with dots instead of colors) for the blind,
or so one could solve it with both eyes closed??? (dont ask me why
you would want to solve it with both eyes closed).


				Alan
-------

Date: 8 Mar 1981 1834-EST
From: JURGEN at MIT-DMS (Jonathan David Callas)
To: KATZ at USC-ISIF, CUBE-LOVERS at MIT-MC
Subject: Braille Cube
Message-id: <[MIT-DMS].189113>

I'm sure that it's all ready been done. (Blind people are very clever that way)
There are the equivalents of dymo label-makers that print in Braille, and any
random sighted person could label a cube (even randomized). I'd bet that it would very very slow, though. The eyes have a much greater informational bandwidth than the fingers.

		-Jurgen at MC



Date:  9 MAR 1981 0855-EST
From: JURGEN at MIT-MC (Jonathan David Callas)
To: CUBE-LOVERS at MIT-MC

I have sent out copies of the cube program I mentioned earlier to (I think)
everyone who asked for it. If you didn't get it, or it was munged, or
you would like it, I saved a copy of the msg in:
DM:USERS1;JURGEN CUBE

			--Happy Cubing

		--Jon

Date: 9 Mar 1981 10:02 PST
From: McKeeman at PARC-MAXC
Subject: Re: how about a Braille cube...
In-reply-to: KATZ's message of 7 Mar 1981 0224-PST
To: Alan R. Katz <KATZ at USC-ISIF>
cc: cube-lovers at MIT-MC

Wonderful idea!   Probably even fundable by some gov't agency for the
handicapped.  As to why do it with your eyes closed, that was in some sense the
original intent of the cube: spatial visualization.  Besides, it confuses me to look
at during a macro.

Bill


Date:  9 Mar 1981 at 1721-CST
From: korner at UTEXAS-11
Subject: edge cubie rotation
To: cube-lovers at mit-mc

	does anyone have a nifty edge cubie rotation algorithm that doesn't
do a di flip in the process. I'm getting tired of
	f r 3(F R R F) R 3(U R R U) F
	There must be something better- I just haven't found it.
		-Kim Korner
-------

Date: 10 MAR 1981 0556-EST
From: ACW at MIT-AI (Allan C. Wechsler)
Subject: edge cubie rotation
To: korner at UTEXAS-11
CC: CUBE-LOVERS at MIT-AI


My basic triple-edge tool is FFRL'UUR'L.  It rotates three edges that
all lie in one equator.  Manipulation hint: move that equator instead
of doing RL' and R'L.

Something like Kim's tool can be obtained by setting up with RL'U and
finishing with U'LR'. All together: RL'U FF RL' UU R'L U'LR'.

   ---Wechsler


Date: 10 Mar 1981 1910-PST
From: CSL.JHC.DAVIS at SU-SCORE
Subject: Edge Cubie Rotation
To: korner at UTEXAS-11
cc: cube-lovers at MIT-AI


  I have been using an even shorter tool to do Kim Korner's
transformation.  In the Befuddler notation, it is:

  R' L B L' R D D R' L B L' R

It is much easier to do than this notation makes it seem.  I
think of it as pushing a center cubie down to the bottom,
turning it off to the side, and bringing back the old top.
Then I move it around to the other side of the bottom, and
go back down to pick it up.

If you begin the transformation with RR LL instead of R' L, and
end it with RR LL instead of L' R, it does the same thing
except with no flipping of center cubies.

  -- Tom Davis

PS.  As in Wechler's tool, think of center-slice moves.
-------


Date: 12 MAR 1981 2317-EST
From: ATTILA at MIT-MC (Sean N Levy)
Sent-by: ATTIL0 at MIT-MC
Subject: Re: Rubik's Sphere carrying case
To: CUBE-LOVERS at MIT-MC
CC: Lauren at UCLA-SECURITY

	Instead of putting it in a klien bottle, how about using one of
Escher's impossible boxes (decorated with an Escher on the outside,
of course...)

 -- Attila

Date: Monday, 16 March 1981  19:28-EST
From: Pat O'Donnell <PAO at MIT-EECS>
To:   cube-lovers at mc
Subject: other orbits

Has anyone investigated what kinds of patterns exist in the other 11
orbits?

Date: 18 March 1981 22:34-EST
From: Alan Bawden <ALAN at MIT-MC>
To: CUBE-HACKERS at MIT-MC

Check out this week's TIME magazine.  (The "Living" section, I believe.)

ISRAEL@MIT-AI 03/20/81 15:55:21 Re: two-person games using the cube
To: CUBE-LOVERS at MIT-AI
Folks,
  I was examining my cube the other day and I noticed that each side
looks like a tic-tac-toe board and I realized that we've never considered
the idea of two-person games using the cube.  Here are some games that I've
come up with.  Some of these may be trivial and uninteresting (i.e. obvious
wins for the first or second player) and some may be too easy to draw with,
but I'll throw them out anyway.

The first game I thought of was Rubik's tic-tac-toe.  This is just regular
tic-tac-toe with a twist (pun intended).  Each person takes turns first
writing his symbol on one of the 54 facelets on the cube.  After doing
that he twists one face and passes the cube to his opponent.  There are a
number of different variations of this game.

    1) The first person to win any side of the cube wins.  This seems to be
       a very easy game so to make it more interesting we add the rule
       that for a person to win, he must do it before executing a twist to
       the cube.

    2) To win, a person must win a majority of the faces on the cube. This
       game has the interesting property that if the cube is full (a draw
       in normal tic-tac-toe) twists can continually be made until a win is
       reached, both people agree on a draw, or some arbitrary upper limit
       on the number of moves beyond a full cube is passed.

    3) One person must fill up all nine facelets of any face with his symbol.
       This game may be too difficult to win.

    4) Each person has pattern of X's, O's, and don't cares which his opponent
       doesn't know and has to get one face to look like that pattern.

Each of these games can be modified by adding restrictions on the twist such
as; a) only quarter turns CW and CCW are allowed; b) a player cannot turn the
same face his opponent just turned; c) a player cannot turn the same face that
he turned last turn; d) if a player made a quarter turn last turn he must make
a half turn this turn and vice versa; or any  combinations of the above    
restrictions or others.  Does anyone know of good erasable writing utensil to
use on your cubes or have a metal cube that can be used for these games
with magnetic X's and O's  Another version of these games could be played
without writing on your cube by allocating one, two or three colors to each
person and starting from a randomized cube, try to play any of the above games
with each turn being taken by twisting a side and using the players set of 
colors as his symbol.

- Bruce


^_


Date: 21 MAR 1981 1454-EST
From: LSH at MIT-MC (Lars S. Hornfeldt)
To: CUBE-LOVERS at MIT-MC, RP at MIT-MC


What are the best CUBE-times nowadays?
A young guy Kimmo Eriksson in Stockholm yesterday solved 10 (ten)
cubes in a series,  with an AVERAGE of 52 seconds,
and with individual times varying between 47 sec and 61 sec.
-lsh


Date: 22 March 1981 0829-EST (Sunday)
From: Dan Hoey at CMU-10A
To: Cube-Lovers at mit-mc
Subject:  No short relations and a new local maximum
Message-Id: <22Mar81 082919 DH51@CMU-10A>

	Well, the gigabyte (well, 300Mb) came in, and brute force is
having its day.  I have a little program that generates all positions
accessible from a given position in a given number of quarter-twists.
With the increased storage available here, I was able to run it to five
quarter-twists.

	The first important fact to emerge is that there are exactly
105046 different positions at a distance of at most 5 qtw from START.
This has two consequences to the argument given in my message on the
Supergroup, part 2 (9 January 1981 0629-EST).  Note that the results
here pertain to the usual group of the cube, rather than the
Supergroup, since the program does not keep track of face-center
orientations.

	The first consequence is that there are exactly 93840 positions
exactly 5 qtw from START.  The message cited above proved the
inequality P[5] <= 93840; this is now known to be an equality.

	The second consequence is that there are no relations
(sequences that lead back to START) of length 10, with the exception of
those that follow from the relations FFFF = FBF'B' = I (and their
M-conjugates).  This is because relations of length 10 would reduce
P[5], which is not the case.  There are, however, relations of length
12; the only known ones are FR'F'R UF'U'F RU'R'U [given in Singmaster]
and its M-conjugates.

	These results can be extended to the Supergroup, by noting that
the set of observed positions places a lower bound on the number of
Supergroup positions at a distance of 5 qtw, while the upper bound
given in the cited message relies on the relations FFFF = FBF'B' = I,
which are relations in the Supergroup.

	A particular result which may be of greater interest to readers
of this list concerns the relation between symmetry and local maxima.
In our message on the subject (14 December 1980 1916-EST) Jim Saxe and
I mentioned that the six-spot pattern is not a local maximum, as
verified by computer.  [The same program was used, but only four-qtw
searches were needed.]

	With five-qtw searches, it became possible to check another
conjecture, using an approach that Jim suggested.  The four-spot
pattern

		U U U
		U U U
		U U U

	R R R	B B B	L L L	F F F
	R L R	B F B	L R L	F B F
	R R R	B B B	L L L	F F F

		D D D
		D D D
		D D D

is solvable in twelve qtw, either by (FFBB)(UD')(LLRR)(UD') or by its
inverse, (DU')(LLRR)(DU')(FFBB).  It is immediate that a twelve qtw
path from this pattern to START can begin with a twist of any face in
either direction.  The program was used to verify that there are no ten
qtw paths.  (It generated the set of positions attainable at most five
qtw from START and the set of positions obtainable from the four-spot
in at most five qtw, and verified that the intersection of the two sets
is empty.)  Thus the four-spot is exactly twelve qtw from START and all
its neighbors are exactly eleven qtw from START, proving that the
four-spot is a local maximum.  (Worried that there might be an eleven
qtw solution to the four-spot?  Send me a note.)

	This is the first example of a local maximum which cannot be
shown to be a local maximum on the basis of its symmetry.  To be more
precise, let us define a "Q-symmetric" position to be a position whose
symmetry group is Q-transitive.  This extends the terminology developed
in "Symmetry and Local Maxima".  In that message, we showed that all
Q-symmetric positions, except the identity, are local maxima.  Until
now, these were the only local maxima known.  The four-spot, however,
is not Q-symmetric; the position obtained by twisting the U or D face
of the four-spot is not M-conjugate to the position obtained by
twisting any of the other faces.  This lays to rest the old speculation
that one might find all local maxima, and thereby bound the maximum
distance from START, by examining Q-symmetric positions.

Date: 28 MAR 1981 1259-EST
From: DCP at MIT-MC (David C. Plummer)
Subject:  New toy (long message, but read it anyway!!)
To: CUBE-LOVERS at MIT-MC


Tanya Sienko is visiting me, and she says that the cube is the
craze of Japan. She also presented me with a new toy, given to
her by some Japanese. (I don't know if is in this counrty --
yet.) 

The thing is shaped like a barrel mounted on a supporting
structure. The barrel can move one UNIT up or down in the
structure. Around the circumference of the barrel there are five
equally distributed columns. Two of the columns have four rows,
and three of them have five. The ones with five have a plunger on
the associated part of both the top and bottom (or left and
right) parts of the supporting structure. Two plungers are next
to each other, and the third is opposite their midpoint. There
are 23 balls in the device: four each of green, yellow, blue,
red, orange (one for each column) and three black balls. (in a
minute you will see where these black balls go). The barrel is
divided into four parts. The left- and right-most parts are fixed
with respect to the supporting structure. Each has three cavities
either to hold a ball or one of the plungers. The barrel moves,
so either the left has balls in the cavity and the right has the
plungers, or vice versa. The middle two sections of the barrel
have two cavities in each row, and these rotate around the
circumference, taking balls with them.

I have been trying to say left and right, because I think the
corect way to thing of this devices is as follows: Hold it
horizontally, with the barrel centered in the supporting
structure. This means that each plunger is half way into its
cavity. A MOVE consists of moving the barrel one half unit right
or left, then moving one of the rotating middle sections forward
or backward one unit, and then returning the barrel to center
position. This creates four generators: move barrel [left,right],
then move middle section-[left,right] forward (or backward, which
is the inverse). Visually:

	|   |                       |   |
	A   A   A   A   A	B   B   B   B   B  
	 \   \                     /   /
	  A   A   A   A           B   B   B   B
	 /   /                     \   \
	A   A   A   A   A	B   B   B   B   B  
	 \   \                     /   /
	  A   A   A   A           B   B   B   B
	 /   /                     \   \
	A   A   A   A   A	B   B   B   B   B  
	|   |                       |   |


		|   |			    |   |
	C   C   C   C   C	D   D   D   D   D  
		 \   \			   /   /
	  C   C   C   C           D   D   D   D
		 /   /			   \   \
	C   C   C   C   C	D   D   D   D   D  
		 \   \			   /   /
	  C   C   C   C           D   D   D   D
		 /   /			   \   \
	C   C   C   C   C	D   D   D   D   D  
		|   |			    |   |

Where	A is move barrel left , move left  section
	B is move barrel right, move left  section
	C		 left ,	     right
	D		 right,	     right

The top and bottom of these drawings are connected, cavities
(filled with the balls) move along the lines. All balls move in
the same direction the same number of units (i.e., the middle
sections are rigid). I hope this is a good enough description, if
not send me mail and I will send an addendum. 

The object, so I hear, is to get each column (row in these
pictures) a single color, and if there are five slots (of which
there are three), the fifth has a black ball in it, when the
barrel is pushed all the way to one side, the plungers take up
three of the outside-barrel-sections, and the black balls take up
the opposite three. from a symmetric point of view, I think it
would be more general to SOLVE it so that the black ball is in
the middle of the five balls (this may not be solvable though)..

If we ignore the obvoius left-right symmetry of the above
pictures, the first assumption of the combinatorics of this beast
is simply P(23;4,4,4,4,4,3)=numbers of ways to permute 4 balls of
each of 5 colors and 3 balls of another color=

		23!
	------------------- = 541111756185000 = 541 trillion
	 4! 4! 4! 4! 4! 3!

Until I have played with it for a while, I can't even guess on
how many orbits there are. Perhaps only one -- I don't know.

Super-groups come in a few classes:
(1)	Each non-black ball gets a second label (1-4) 
	giving size 23!/3! = 4.3*10^21
(2)	Each black gets a second label (1-3)
	giving size 23!/(4!)^5 = 3.25*10^15
(3)	(1) and (2), all balls distinct
	giving size 23! = 25.8*10^21

If anybody sees one in this county, please let me know. Tanya
believes they are only in Japan at the moment. She has donated
the one I have seen to me/SIPB, so people at MIT and area are
free to come to 39-200. PLEASE BE CAREFUL with it. It is plastic
and it looks breakable -- especially the outer part of the
supporting structure looks like it dould break. I think a better
construction would be to have them be plates which are attached
to the axis with screws. This might lead to a temptation to
disassemble, which may be epsilon below breakage.

Date: 28 March 1981 16:13-EST
From: Carl W. Hoffman <CWH at MIT-MC>
Subject: Also from Japan ...
To: CUBE-LOVERS at MIT-MC

Cubes of different sizes and colors.  There is one 3 centimeters on an edge
sitting in the SIPB office.

Date: 31 Mar 1981 2133-PST
From: Gary R. Martins <gary at RAND-AI>
Subject: B E W A R E !!
To: cube-lovers at MIT-MC
cc: gary at RAND-AI


Bought a new cube today.  One of Ideal's "Rubik's Cube"s.  Same
price as first cube, bought about a month ago.  Packaging looks
same.  Ditto cube, except that the center-white face has
 "Rubik's Cube (tm)" printed on it in various fonts. Also, closer
inspection of the package shows that a stick-on stripe acknowledges 
manufacture in Hong Kong.

The cube itself is INFERIOR in various ways.  I'd recommend you
not buy them, unless the vendor will offer you a refund.  

The worst and most obvious feature of this cube is that is seems
to have NO lubricant in it.  The faces seem more vulnerable to
fingernail damage etc. and the colors and materials seem shoddier. 
The cube has a flimsy feel to it, and seems poorly finished in general.

Anybody else notice this, or have I just caught a lemon ?


Gary
-------

Date:  1 APR 1981 0104-PST
From: MAXION at PARC-MAXC
Subject: Re: B E W A R E !!
To:   gary at RAND-AI, cube-lovers at MIT-MC
cc:   MAXION

In response to the message sent  31 Mar 1981 2133-PST from gary@RAND-AI

I had the exact same experience.  The first one was wonderful; the
second (just as you described) was awful.  Same packaging, same story,
same observations as yours.

				Roy
-------

ZEMON@MIT-AI 04/01/81 07:45:10 Re: B E W A R E !!
To: CUBE-LOVERS at MIT-AI
	I have one of the offending cubes -- yes, it \is/ falling apart.
The colored faces have developed crinkles, holes and some are even
peeling off.  This is after only 4 weeks of use.

	Taking the cube apart and sprinkling its insides liberally
with baby powder will effectively lubricate it (although it will smell
for a while) and make it essentially noiseless, I have heard.

			-Landon-

Date:  1 Apr 1981 1459-PST
From: Gary R. Martins <gary at RAND-AI>
Subject: Cube Lube
To: cube-lovers at MIT-MC
cc: gary at RAND-AI


Is there a consensus on the best lube for one's cube ?  White
lithium grease, silicon grease, silicon spray, and baby powder
have all been mentioned.  Anybody know what's really 'best' ?

Gary
-------

Date:  2 Apr 1981 0132-CST
From: Clive Dawson <CC.Clive at UTEXAS-20>
Subject: Re: Cube Lube
To: gary at RAND-AI, cube-lovers at MIT-MC
In-Reply-To: Your message of 1-Apr-81 2203-CST

I suspect that "best" in this case is probably a matter of
personal opinion...Also note that a lot depends on trimming,
filing, sanding, etc.

Besides the lubes mentioned by Gary, I can also recommend dry graphite
powder (I used "Mr. Zip Extra Fine Graphite") which gave me very good
results on my cube.  Then I finally got a chance to examine Kim
Korner's cube (Korner@UTEXAS) and must admit his is much much better.
He used silicon gel, of the sort used to lubricate "o" rings in Scuba
equipment.  See his message to Cube-lovers of 6-Feb-81 for more
information.  About the only shortcoming I noticed was a very slight
"slimy" feeling to the cube which I'm sure will wear off with time...


By the way, on the subject of the declining quality of Ideal Toy's
version of the cube--  I too was surprised when I examined one
which was bought last month by a friend of mine.  The first thing
I noticed was that the some of the interior faces of each cubie
were missing.  My first reaction was that they'd found a way to
skimp on plastic;  then I thought that maybe it was a way to
cut down on internal friction.  Judging from some of the other
recent reports, it sounds like my first hunch was correct.  Another
annoying characteristic was the shoddy work in attaching the
colored faces.  Most were not only crooked, but also liberally
sprinkled with air bubbles throughout.

Happy cubing,

Clive
-------

Date:  2 Apr 1981 1723-PST
From: Hopper at OFFICE  
Subject: Re: B E W A R E !!
To:   gary at RAND-AI, cube-lovers at MIT-MC
cc:   hopper at OFFICE

I've bought cubes recently as follows:

1-FEB (approx) , Ideal's with "Rubik's CUBE tm" on the center white cubie.  
Pakaged in cardboard and cellophane.  Hollow edge and corner cubies.  Squeekie,
but fine after lube with graphite.  No problems with facies.  Good size 
tolerances--very smooth operation after lube.

20-FEB (approx) , bought 3  that looked identical to the pevious one, except 
they were packaged in cylindrical plastic packages.  

   Two of the three turned out the same as the one purchased 1-FEB, except that
   size tolerences were very poor and operation was very rough, even after 
   lubrication.  No problems with the faces.

   The third cube was packaged the same and has the same "Rubik's CUBE tm" on 
   the center white cubie, but is quite different.  Although the cubies are 
   hollow, it is heavier then the others.  The plastic isn't so squeeky and 
   seems more like earlier cubes from last year.  The edge cubies have casting 
   ridges visible through the middle of the faces like the early, early cube I 
   got before last summer.  The shoulders on the edge cubies were flat so the 
   action was very rough.  The corner cubies were loose.  Filing down the 
   inside surfaces of the edge cubies cured the loose corners, and rounding 
   their shoulders made the action quite acceptable.  LASTLY, ONE (just ONE!) 
   of the red facies is inferior, darker in color, and crinkling!

20-MARCH (approx) , bought a cube with no acknowledged manufacturer, with 
cylindrical plastic pakage very similar to Ideal's, labeled "Made in Taiwan".  
Cubies are hollow very much like the third one from the 20-FEB batch, but the 
corner cubies have covers glued in the openings of the inside surfaces.  The 
quality of the facies seems good, but it may be too soon to tell.  The orange 
facies are not brilliant like Ideal's--more of a peach color.  The size 
tolerances seem quite good compared to Ideal's recent cubes.  Biggest drawback 
(and possible overriding factor) is the plastic seems much softer than Ideals's
and lubrication (at least with graphite--I haven't tried other recommended 
lubes such as silica gel) doesn't seem to make it any easier to turn.  It 
remains quite stiff. Also, some of the centers were screwed in much to tightly.

I'd be interested to hear any other experiences with recently-bought cubes.  
I'm curious about their availability in the Bay area and elsewhere.  --Dave--

-------

Date:  2 Apr 1981 2325-PST
From: Gary R. Martins <gary at RAND-AI>
Subject: New Yorker
To: cube-lovers at MIT-MC
cc: gary at RAND-AI


Current issue of 'New Yorker' magazine has some cubic discussion
in the opening 'Talk of the Town' section.  Also mentions Marvin
Minsky!  P. 29, March 30, 1981 issue.

Gary
-------

Date:     3 April 1981 0500-est
From:     Allan C. Wechsler          <Wechsler at MIT-Multics>
Subject:  Magic barrel.
To:       CUBE-LOVERS at AI

Haal yawm!  I am a barrel-solver this day!
   ---Wechsler


Date: 3 Apr 1981 0750-PST
Sender: OLE at DARCOM-KA
Subject: Cube lube (yet again)
From: Ole at DARCOM-KA (Ole J. Jacobsen)
To: Cube-lovers at MIT-MC
Message-ID: <[DARCOM-KA] 3-Apr-81 07:50:47.OLE>

The following is a collection of thoughs and experiences on
cube lubrication. It only applies to the Ideal cubes which
are the only ones I have played with, but should have some
applicability to other brands.

Getting a smooth turning cube seems to me to be a combination
of the right lubrication with the spring/screw tension. If the
screws are too lose, the cube will turn easily, but frequently
jam since the lose cubies tend to get in each other's way. On
the other hand a very high tension without any lubrication would
mean a very stiff and fast wearing cube. My solution is simply to
use candle wax. I have taken my cube apart and rubbed each cubie
with a standard candle. (My friend from the Chemical Eng. Dept.
says parafine wax would be even better. This is what we used to
rub on our skis back in Norway before all the fancy ski-waxes
became available. I don't know how easy it is to get these days.)
I also "fine-tune" the cube by adjusting the screws on each face.
A couple of strips of double-sided tape stops the caps from acci-
dentally falling out during use. You may need to take the cube
apart a couple of times after the initial lubrication to allow
superfluous wax to fall out. I also found that turning a newly
waxed cube under a hot tap seems to make the wax settle nicely.
This lubrication has the advantage of not (seemingly) coming out
on your hands or otherwise disappear,- one treatment will last
you very long indeed. The only slight problem is that the cube
needs some "warming up" when it has been left idle for some time
especially in cold places. (Ref. my earlier message) But a couple
of minutes of random twisting produces a smooth and silent cube.
Good luck
                               OLE

Date: 4 Apr 1981 1727-EST
From: JURGEN at MIT-DMS (Jonathan David Callas)
To: Cube-lovers at MIT-MC
Subject: Cube preferences
Message-id: <[MIT-DMS].192744>

I have two cubes, a C. Americanus ("Rubik's Cube") which I bought last summer,
And a C. Albus that I got from Logical Games in Haymarket Va. The white
cube came lubricated with something resembling musician's cork grease, and
has not needed to be lubricated. The Rubik's cube has never been lubricated 
either, but hasn't seemed to need it. Iprefer the white cube to the black one
for some nebulous reason. It is not nearly as smooth-turning as the black
one, but in a perverse way, I like that. It seems to be better built, but I
can't substantiate that with facts, that's just gut-feeling. I *DO* like
the fact that they are uncommon, and now that people don't go "Ooh, what's
*THAT*" when they see the cube, and now people do get amazed at the sight
of the white-faced cube. Now that it seems that Ideal is going for the bucks (a
friend of mine has also gotten one of the cheap cubes, but I thought it was 
my imagination. I guess now there's real reasons for getting the white cubes.

		--Happy cubing,
		--Jurgen




Date:  6 APR 1981 1501-EST
From: DCP at MIT-MC (David C. Plummer)
Subject:  Japan frob revisted (180+ lines)
To: CUBE-LOVERS at MIT-MC


This is a long overdue re-explanation of the Japanese frob. Hoey
and Saxe at CMU gave several comments and suggestions.

PART I -- Try again
===================

Take a hollow cylinder (like a doubly unlidded coffe can), cut it
open and unravel it. We now have something like

	xxxxxxxxx
	|	|
	b	|	where the cut was made along the x
	o	t	and top and bottom are where the
	t	o	lids used to be
	t	p
	o	|
	m	|
	|	|
	xxxxxxxxx

The supporting structure corresponds roughly to the lids of the
can.


	LL    "   '   ""   '   "	RR
	LLLLLL" B ' B "" B ' B " B  RRRRRR
	LL    "   '   ""   '   "	RR
	LL    "   '   ""   '   "	RR
	LL    "   '   ""   '   "	RR
	LL    " B ' B "" B ' B "	RR
	LL    "   '   ""   '   "	RR
	LL    "   '   ""   '   "	RR
	LL    "   '   ""   '   "	RR
	LLLLLL" B ' B "" B ' B " B  RRRRRR
	LL    "   '   ""   '   "	RR
	LL    "   '   ""   '   "	RR
	LL    "   '   ""   '   "	RR
	LL    " B ' B "" B ' B " 	RR
	LL    "   '   ""   '   "	RR
	LL    "   '   ""   '   "	RR
	LL    "   '   ""   '   "	RR
	LLLLLL" B ' B "" B ' B " B  RRRRRR
	LL    "   '   ""   '   "	RR
	LL    "   '   ""   '   "	RR

(In the three dimensional case, the top and bottom of this
picture are connected together.)

L is the left part of the supporting structure, and R is the
right. They are firmly connected to each other, and are therefore
fixed in space with respect to each other. They are really
circular, but this is a view of the outside. B are the balls. The
balls can move left or right by being PLUNGED. The only allowed
plunge in the above diagram is to move the supporting structure
to the left (or equivalently the barrel to the right). With
respect to the barrel, only the balls in the first, third and
fifth rows (refered to as columns in previous message) are
affected. The diagram would now look like

	LLL B B B B B RRR
	L     B B B B   R
	LLL B B B B B RRR
	L     B B B B   R
	LLL B B B B B RRR

Balls move vertically by by moving either of the two " ' "
sections vertically, and the balls within that section stay fixed
in space with respect to each other and the section, but not
fixed with respect to the other balls (within the context of one
turn) or the supporting structure.

Thus the moves are:

	PLUNGE RIGHT or LEFT, whichever is appropriate
	  (or move barrel LEFT or RIGHT)
	and MOVE LEFT or RIGHT section UP or DOWN

My suggestion was that in the between move state, the barrel was
centered in the plungers, so the PLUNGE move is HALF-PLUNGE LEFT
or RIGHT (both of which are apporpriate), then do the vertical
move, then UN-HALF-PLUNGE.

PART II -- Hoey's comments to my original message
=================================================
[Hoey  28 March 1981 1500-EST]

Is the move you designate by

	|   |
	A   A   A   A   A
	 \   \
	  A   A   A   A
	 /   /
	A   A   A   A   A
	 \   \
	  A   A   A   A
	 /   /
	A   A   A   A   A
	|   |
really the permutation that takes
	 |   |				 |   |
	 1   2   3   4   5		 6   7   3   4   5
	  \   \				  \   \
	   6   7   8   9		  10  11   8   9
	  /   /				  /   /	
	10  11  12  13  14	to	15  16  12  13  14
	  \   \				  \   \
	  15  16  17  18		  19  20  17  18
	  /   /				  /   /
	19  20  21  22  23		 1   2  21  22  23
	 |   |				 |   |				?
Do you mean to imply that moves of the form
	|   |					|   |
	X   X   X   X   X		X   X   X   X   X
	 \   \				       /   /
	  X   X   X   X			  X   X   X   X
	   \   \			       \   \
	X   X   X   X   X	or	X   X   X   X   X
	   /   /			       /   /
	  X   X   X   X			  X   X   X   X
	 /   /				       \   \
	X   X   X   X   X		X   X   X   X   X
	|   |					|   |
(whatever they mean) are prohibited (as primitives, at least) by
the construction of the barrel?

[Both answers are YES]

PART III -- Comments later that night
=====================================

[In response to the updated description
Hoey  20 March 1981 1836-EST]

	First, it should be made clear that in (either) plunged
position, the two " ' " sections rotate freely; i. e. it is not
necessary to plunge in between.  For instance, one could solve by
counting plunges, but not rotations.  Jim suggested that it might
be "neater theoretically", but I think it smells of the half-twist
metric.

	Second, the inclusion of permutation diagrams will make the
puzzle clear to anyone who doesn't understand the mechanics.
Something like I gave in the last message, but with all
permutations given, the note that "\|/" are only comments, and the
description of the goal:  move Black to 5,14,23, and make the sets
1-4, 6-9, 10-13, 15-18, and 19-22 each a solid color.

	I ran this through the Furst/Hopcroft/Luks algorithm, and
found that in the Supergroup (all balls distinct) you get the
alternating group on 23 balls:  all even permutations.  Thus if any
two balls are indistinguishable, you can get all configurations.
Saxe remarks that there is only fourfold symmetry:  Reflection
left-to-right and up-to-down.  Their composition is in fact
achievable:  turn the whole puzzle upside down, while continuing to
face the front of it.  Strangely enough, this is an ODD permutation:
it takes you to the other orbit!

[Hoey  28 March 1981 2143-EST    Subject:  Simpler and harder toy]

	Try taping the center two rings together.  Thus A is
always performed with C, and B with D.  The same set of
permutations is achievable!

[I assume the proof is an enumeration of states by the above
algorithm.]

PART IV -- Developments by Alan Bawden (ALAN@MC), Allan Wechsler
(ACW@AI) and myself.
================================================================

Alan Bawden sat down patiently one night (Tuesday March 31 1981,
I  think) and discovered the necessary TOOL (or concept)
(singular !!) that is needed to solve the toy. I will not give a
spoiler here. Getting most of it is rather easy. The last few
balls take a little extra work. Alan told me the concept, and the
next day I successully solved it. Alan solved it later that day,
and soon Allan Wechsler solved it a few days later (signified by
his yelp to this mailing list). The three of us solve it slightly
differently (s)o like the cube, there are personal sovling
styles). We now solve it reliably, including the last few balls.

Happy what-ever-ing...

Date: 8 Apr 1981 16:07 EST
From: Marshall.WBST at PARC-MAXC
Subject: Please add me to the distribution list
To: Cube-Lovers at MIT-MC
cc: Marshall.WBST

Please add my name to the rubik's cube distribution list. I have a copy of
Kertesz' solution but am interested in better solutions and/or insights into the
underlying group.

Thank you

--Sidney (Marshall.WBST at PARC-MAXC)


Date: 18 April 1981 08:52-EST
From: Lars S. Hornfeldt <LSH at MIT-MC>
To: CUBE-LOVERS at MIT-MC, gary at RAND-AI


Kimmo Eriksson is 14 years old (a good age for cubism),
and in his series of 10 consecutive cubes, the average time was
52 sec, and average number of moves was 95, varying between
70 to 120 (half-turns and slices counted as one move).

He uses 5 macros with uncountable longer variants.
The longest of the macros are 11 moves.
-lsh

Date: 20 Apr 1981 0906-PST
From: Isaacs at SRI-KL
Subject: (Response to message)
To:   LSH at MIT-MC
cc:   ISAACS, cube-lovers at MIT-MC

Who and where is Kimmo Eriksson?  Where was this timing done? What are
his macros?  In what order does he solve it? The same way each time? etc.
                ---Stan Isaacs
-------

CMB@MIT-ML 04/23/81 13:04:49
To: cube-lovers at MIT-MC
From the Boston Globe:

  Abbie Hoffman, the former Yippie leader who managed to escape the toils of
the law for seven years by living incognito in upstate New York, has finally
gone to prison, but not for having been a fugitive.  Hoffman surrendered in
New York yesterday to begin a three-year prison term for selling cocaine and
jumping bail.  In the photo, he is shown being frisked.  In his left hand he
holds a magic cube puzzle, which he said he will solve in prison.  In his
right hand he holds a copy of the book, "Fire in the Minds of Men," that had a
bookmark looking suspiciously like a hacksaw blade.  This was whisked away from
him.  Hoffman denied the props, including the hacksaw, were a publicity stunt.


Date: 23 Apr 1981 1105-PST
From: Gary R. Martins <gary at RAND-AI>
To: CMB at MIT-ML
cc: cube-lovers at MIT-MC, gary at RAND-AI
In-Reply-To: Your message of 23-Apr-81 1304-PST


He should have offered those narcs a snort of vodka !

G
-------

Date: Sunday, 26 April 1981  10:54-EDT
From: Pat O'Donnell <PAO at MIT-EECS>
To:   Cube-Lovers at MC
cc:   PAO at MIT-EECS

May issue of Reader's Digest has a (very) short article on the cube.
It includes a claim for a French fellow solving the cube in an average
of 32 seconds.  The article contains almost no technical
information--mostly historical.

Date: 27 Apr 1981 0923-PDT
From: Isaacs at SRI-KL
Subject: cubes, barrels, and stuff
To:   cube-lovers at MIT-AI

   I was just at the fourth international puzzle party in L.A. and saw
several offshoots of the cube.  The barrel, previously mentioned in this
digest, is the best.  It is called "The Ten Billion Puzzle" (I think).
(Note to people in the Palo Alto area - come to the Rubiks Cube Club And
Other Puzzles at Stanford on Thursday night if you want to see a couple.)
   Also there were two small cubes, about 2/3 size, one from Japan, and
the other from (I think) Taiwan.  There was a 2x2 version (about half the 
size of Rubiks), with things like hearts, stars, etc on it.  There was also
a Rubik type, but with figures instead of colors.
   The Missing Link is now out from Ideal, and should be easily findable
(as of this week).  But, though they treat it as a follow-up on the cube,
it is MUCH simpler, and closer in principle to a sliding block puzzle.  
Nice, but simple.
    There were also several other types of cylinders, but mostly related
to the Missing Link, or to Instant Insanity type, rather than cube type.
   By the way, Jerry Slocum, puzzle collector extraordinaire and the
puzzle party host, thinks the magic cube will have a real impact on 
society - that it will lead to a resurgance of interest in puzzles in
general, and in thinking-type games.  Let us hope he is right.  (Send
a puzzle to your congressman - make him think!)
                         --- Stan Isaacs
-------


Date:  27 April 1981 12:15 cdt
From:  VaughanW.REFLECS at HI-Multics
Subject:  Re: cubes, barrels, and stuff
To:  Isaacs at SRI-KL
cc:  cube-lovers at MIT-AI
In-Reply-To:  Msg of 04/27/81 11:23 from Isaacs

*nothing* could make my congressman think!


Date: 27 Apr 1981 1322-EDT
From: IC.RAG at MIT-EECS
Subject: Re:  Re: cubes, barrels, and stuff
To: VaughanW.REFLECS at HI-MULTICS, Isaacs at SRI-KL
cc: cube-lovers at MIT-AI
In-Reply-To: Your message of 27-Apr-81 1315-EDT

*NOTHING* could make me think of my congressman!
-------


Date: 27 Apr 1981 1139-PDT
From: Gary R. Martins <gary at RAND-AI>
Subject: Re:  Re: cubes, barrels, and stuff
To: VaughanW.REFLECS at HI-MULTICS
cc: Isaacs at SRI-KL, cube-lovers at MIT-AI, gary at RAND-AI
In-Reply-To: Your message of 27-Apr-81 1215-PDT


Try   *M*  *O*  *N*  *E*  *Y*  !!  Worked wonders for the
FBI !

G
-------


Date: 28 Apr 1981 0233-PDT
From: Peter D. Henry <PDH at SU-AI>
Subject: mailing list add request
To:   cube-lovers at MIT-MC 

please add me to the mailing list... thanks
			Peter D. Henry
			PDH@sail



Date: 28 April 1981 08:34-EST
From: David C. Plummer <DCP at MIT-MC>
Subject: mailing list add request
To: PDH at SU-AI
cc: CUBE-LOVERS at MIT-MC

Done.

Date: 29 April 1981 1334-EDT (Wednesday)
From: Guy.Steele at CMU-10A
To: cube-lovers at MIT-MC
Subject:  New member for mailing list
Message-Id: <29Apr81 133430 GS70@CMU-10A>

Please add  Paul.Haley @ CMUA  to the cube-lovers mailing list?

Date: 6 May 1981 2030-EDT
From: ROBG at MIT-DMS (Rob F. Griffiths)
To: cube-lovers at MIT-MC
Message-id: <[MIT-DMS].196841>

  While visiting the local gaming shop today, I heard a rumour
about a pending lawsuit between (I think) The Original maker
of Rubik's cube and Ideal.. Anyone know anything about this?
			-Rob.


Date:  8 May 1981 1036-PDT
From: Isaacs at SRI-KL
Subject: Non-twisting corner moves
To:   cube-lovers at MIT-MC

   Does anyone know good move sequences for exchanging a pair or corner
cubies on a face without twisting?  (Of course, a pair of edges will have
to exchange also.)  Or of cycling 3 corners without twisting?  I'm looking
for the "shortest" sequence, and the "easiest to remember" sequence.
Most of the moves I've seen are long and complicated.
            --- Stan Isaacs
-------

Date: 8 May 1981 14:30-EDT
From: David C. Plummer <DCP at MIT-MC>
Subject: Non-twisting corner moves
To: Isaacs at SRI-KL
cc: CUBE-LOVERS at MIT-MC

How about

	L' [(R' DD R) U (R' DD R) U'] L

for moving the top three corners around perserving the top color.

Or,
	(R' D' R) U' (R' D R) U

For moving three front pieces around, preserving a couple colors.

12 and 8 moves is pretty short...

Date: 9 May 1981 08:47-EDT
From: Lars S. Hornfeldt <LSH at MIT-MC>
Sender: LSH0 at MIT-MC
To: CUBE-LOVERS at MIT-MC, isaacs at SRI-KL


Kimmo Eriksson is a 14 year old computer fan who lives in Stockholm.
  As mentioned, he solved a series of 10 cubes in 47-61 s, average 52,
using 70-120 moves, average 95 (half-turns and slices counted as one).

* He always starts with the WHOLE yellow layer (regardless of ini-
  tial state, probably because the regularity allows faster reflexes)
* Then the middle layer (betw. yellow and white)
* Then all top-corners into place, then into correct orientation.
* Finally turn and move the top-edges (requires 0-3 macro-moves).

  He keeps strictly to this scheme, but uses a large set of macros,
that are different longer varities of the following basic five:
  For middle: RUR'U'F'U'F
  Move corner: RU'L'UR'U'LU
  Turn corner: RUR'URU2R'U2
  Move edge:   MU2M'UMU2M'UMU2M'  (M moves the Mid-line of the Bottom
  Move and turn edge:MUM'U2MUM'       up Front, ie = LR' )

  For timing, he starts a stopwatch, grabs the cube, solves it
- while watching (easy) the watch during the last macro in order to
read off the time exactly as the last macro is completed.
  After re-mixing the cube, the procedure is repeated (10 times).
-lsh

Date: 10 May 1981 1258-EDT
From: Jerry Agin <AGIN at CMU-20C>
Subject: Counting moves
To: Cube-Lovers at MIT-MC

Frequently when I play with the cube, I try to solve it in as few
moves as possible.  I find this to be more intellectually challenging
than going for speed.  Does any one else do this?  I'd be interested
in comparing notes.  Presently it takes me between 70 and 100 quarter-
twists, provided I don't make gross errors.  (My guess is that if I
were counting slices and half-twists as one, the number would be
between 50 and 70.)
-------

Date: 15 May 1981 0456-PDT
Sender: OLE at DARCOM-KA
Subject: New pseudo-cube
From: Ole at DARCOM-KA (Ole J. Jacobsen)
To: Cube-lovers at MIT-MC
Cc: Oyvind
Message-ID: <[DARCOM-KA]15-May-81 04:56:31.OLE>

Pardon me if this object has been described before, but I don't remember
seeing it. While browsing for cubes in a local store here in Newcastle
the other day, I came accross a new "cube" which I shall try to describe
and I invite you all to think of a good name for it.

First of all let me point out that this new toy is really just a Rubik's
Cube with some modification with respect to coloring and construction.
Imagine taking your normal cube and making 4 vertical slices along the
corner diagonals. Your top and bottom faces would now look like:

                           ---
                          /   \
                          I   I
                          \   /
                           ---

(sorry about poor ratio, but I hope you get the idea) Now recolor the new
faces and voila, your new toy is complete. The construction has the following
consequences:      1. The object is no longer symmetrical, U and D faces
                      are different from L R F and B.
                   2. The "corners" have only got TWO colors, but act as
                      corners of the Rubik's Cube, the mechanics is identical.
                   3. Four new "edges" which I will call wedges have appeared
                      in the middle layer. These have only ONE color, but as
                      you will discover when using your edge moves: the orient-
                      ation matters. Edges and wedges may be interchanged.

I will now describe the coloring of my particular cube, note that there are 10
different colors. The U face is blue and the D face is white. Then starting at
the 6'o clock edge column (i.e 1/3 of the F face) we have: GOLD(e),ORANGE(w),
RED(e),PURPLE(w),YELLOW(e),PINK(w),GREEN(e),LIGHT BLUE(w). (Where e=edge and
w=wedge colums respectively). I chose this particular orientation because it
makes red=left and green=right which is nice. Note however that this "cube"
may be reassembled in various legal patterns since the edge column/wedge column
neighbouring properties are not forced. This further complicates solving since
there in no way of knowing which sequence the "corners" and edges go in layer1
unless you have a map. Once this is known, solving is straight forward, but
as said the wedges will confuse you. Qestion: Is there some way of deter-
mining the parity etc, such that the object may be solved without a map
of layer one?  I invite comments from Jim and Dan. If it is the case
that these pseudo-cubes (how about Rubik's Drum) are not available in
the US, I can send one or two samples. The drums are made in Taiwan and
are not as well finished or as smooth turning as Ideals cubes. Random
twisting produces very strange shapes and the Cruxi Plummeri et Cristmani
are simply out of this world.
OK, thats it. Hope this made sense, but this thing is more difficult to
describe that its predecessor,- so forgive me if I haven't succeeded.
Cheers      OLE

Date:  15 May 1981 19:04 edt
From:  Greenberg.Symbolics at MIT-Multics
Subject:  Re: New pseudo-cube
To:  Ole at DARCOM-KA (Ole J. Jacobsen)
cc:  Cube-lovers at MIT-MC, Oyvind at DARCOM-KA
In-Reply-To:  Message of 15 May 1981 07:56 edt from Ole J. Jacobsen

I have seen this thing under
the name "space shuttle" - in a cylinder like rubiks cubes used to
come in.  One way to determine the color layou is to make
an assumption (as I did) and proceed to solve-- if you lose, you
get the "impossible" single-pair-swap.  Then swap two edges
of your assumption and try again. Incidentally, the plural
of crux is cruces, not cruxi.

Date: 18 May 1981 1046-EDT (Monday)
From: Dan Hoey at CMU-10A
To: Cube-Lovers at MIT-MC
Subject:  Drum info
Message-Id: <18May81 104634 DH51@CMU-10A>


	Most of this note is pretty straightforward application of
the known cube properties, but if you want to know about the
drum....

	The drum shows everything you see on a regular cube except
the orientation of the four truncated edges, or wedges.  Because the
(invisible) edge parity is preserved, each visible position of the
drum corresponds to 2^4/2 cube positions.  Thus there are
5.406x10^18 drum positions.

	To count the number of solutions, note that as in the
normal cube, the face centers force each edge to its home position
and orientation.  In addition, each corner has a facelet that says
whether it is top or bottom and fixes the corner's orientation.
This means that solved positions are obtained from each other by
permuting top corners, bottom corners, and wedges.  But the three
cubies on a diagonal face must match, and so the three permutations
are the same.  Only even permutations are achievable in this way
(since the cube of an odd permutation is odd) and there are 4!/2=12
of these.  One easy process that goes from one solved position to
another is FF RR FF BB RR BB.

	I asked Ole Jacobsen what he meant when he said of the
wedges that "as you will discover when using your edge moves:  the
orientation matters."  It turns out this is because he solves by
layers:  top-middle-bottom, and doesn't know which way to orient the
edges in the middle so that the edges on the bottom will have the
right parity.  There are several ways out of that problem; one is to
turn the drum sideways and solve left-middle-right.

	The problem of solving without knowing the order of the
wedges is trickier.  Solving sideways is one method:  do the left
side any way; on the right side there two possibilities, one of
which will work.  (This is Bernie Greenberg's suggestion, modified
so you don't need to memorize the whole map.)

	One interesting thing to do with a drum is to turn it into
baseball.  Using colored tape and disassembly, change the colors and
positions so that the wedges appear in the UF, DF, BL, and BR
positions when the colors match.  On a baseball, there are only two
solved positions.

Date: 18 May 1981 17:59-EDT
From: Richard Pavelle <RP at MIT-MC>
Subject: the magic barrel := scraping the bottom thereof
To: CUBE-LOVERS at MIT-MC

Now from Ideal comes  "MAGIC PUZZLE" ta ta..ta ta...   It is a new barrel 
puz (not worth the extra typing) and sells for about $5.00 in Boston. It is 
not deserving of a discussion but so you may recognize and avoid it let me 
say the following: Barrel shaped, it has 6 slots perpendicular to 3 
frames which rotate about the axis. Five slots contain 3 colored 
panels each and the 6th has only two. The puz comes solved with 
all 6 slots the same color. One then tries to disarrange it (more 
difficult than actually solving it). The greatest challenge might be to
run over it with a car and humpty dumpty it.

Date: 10 Jun 1981 2327-PDT
From: Alan R. Katz <KATZ at USC-ISIF>
Subject: edge flip anyone...
To: cube-lovers at MIT-MC
cc: katz at USC-ISIF

Does anyone know of a way to flip two edges without changing anything else,
ie interchange the FR and FB edges, keeping their same orientation.  If not,
how about one that will interchange FR and FB edges without disturbing the top
layer, but can mess up the bottom??

This may have been given a while back, but I couldn't find it looking through
the old CUBE-LOVERS mail.

Also, GAMES magazine says they have 100 of the old cubes left for $10.95 plus
$2.50 for postage and handling.  These are the Ideal ones WITHOUT the IDEAL
stamped on the white face.  Order from:
			Games Shop
			515 Madison Ave.
			New York, NY  10022
(first come, first serve)



					Alan
-------

Date: 11 June 1981 0323-EDT (Thursday)
From: Dan Hoey at CMU-10A
To: Cube-Lovers at MIT-MC
Subject:  Re: edge "flip" anyone...
Message-Id: <11Jun81 032344 DH51@CMU-10A>


    Does anyone know of a way to flip two edges without changing
    anything else, ie interchange the FR and FB edges, keeping their
    same orientation.

	The term "flip" generally means "change orientation" (of an
edge).  I take it you mean "exchange" (or, acceptably,
"interchange").  This is provably not possible, because an exchange
is an odd permutation.  I will send you the two messages which deal
with what can be done with a cube.  (Anyone else who wants a copy,
request of Hoey@CMU-10A rather than the list.)  

	If you don't know what an even (or odd) permutation is, or
how it acts, any elementary algebra text should help.  If you try
and fail, drop me a line and I'll try to help.

    If not, how about one that will interchange FR and FB edges
    without disturbing the top layer, but can mess up the bottom??

	There is no FB edge.  I'd guess you mean UR and UB, for
which RU'R' UUFFBB DDFFBB RUR' suffices.


Date: 12 Jun 1981 1022-PDT
From: ISAACS at SRI-KL
Subject: Re:  Re: edge "flip" anyone...
To: Dan Hoey at CMU-10A
cc: cube-lovers at MIT-MC
In-Reply-To: Your message of 11-Jun-81 0023-PDT

  It is simpler in actually manipulating the cube to avoid moves that
use the back; in particular, the FFBB slice move is a difficult one to
actually do.  So I would recommend   BUB' and BU'B' at the ends of
a  RRLL slice.  In addition, if you define X = RL' (or R'L) followed by an
entire cube rotation (ie, move the slice and allow the orientation 
to change - turn the ham, not the bread )  then the sequence becomes:
    BUB' (UUXX)**2 BU'B'.
      ---- Stan
-------

Date: 12 Jun 1981 1522-PDT
From: Dolata at SUMEX-AIM
Subject: Set of operators
To:   cube-lovers at MIT-MC


I am new on cube-lovers mailing list, so please excuse this message
if it is a repeat of one in the past...    Does anyone know where I
can find a good dictionary of operators?   HAs one been compiled
and is it available online?

Thanks for your help
Dan Dolata (dolata@sumex-aim)
-------

Date: 14 Jun 1981 1404-PDT (Sunday)
From: Mike at UCLA-SECURITY (Michael Urban)
Subject: Howtodoit books
To: cube-lovers at mit-ai

   There are presently at least TWO books on how to solve
Rubik's Cube (TM).  One is from Bantam Books, and the other
is from a more obscure publisher.  They run about $2 each.
I haven't given them more than a cursory look as yet.  The
nonBantam one seems to use BEFUDLR notation, though.
-------



Date: 15 Jun 1981 0954-PDT
From: ISAACS at SRI-KL
Subject: re. howtodoit books
To: cube-lovers at MIT-AI

  I haven't seen the Bantom book, but I will keep an eye open.  Sigmasters
"Notes on Rubik's Magic Cube" is now out and available (at least in the
Bay Area), published by Enslow Publishers (Bloy St and Ramsey Ave, Box 777,
Hillside, N.J. 07205) , lists for $5.95.  Seems to be EXACTLY the "Fifth
Edition Preliminary Version", as in a photo-copy, although it may have
some minor corrections and changes.  There is also a solution booklet by
J.G.Nourse, published by Storc, marketed by Paul N. Weinberg, Mountain View,
Ca.  And still another solution book privately printed by Kirsten Meiers.
The solutions (happily) proliferate; each book has variations, most use
non-BFUDLR notation - in many cases, just diagrams.  Norse intends his
solution to be "interruptable" - that is, if you drop the cube in the middle,
or get a telephone call (or your computer beeps), you shouldn't have to start
over.  Meiers is meant to be easy to do.  Singmasters ( I think), is meant
to be easy to remember - mostly variations on a few basic algorithms.
    It would be nice to analyse all these solutions and arrange them for
speed, ease of action, minimal twists, minimal time, better for right-handers
etc.etc.
                 --- Stan
-------


Date: 18 Jun 1981 2101-PDT
From: Alan R. Katz <KATZ at USC-ISIF>
Subject: My previous message
To: cube-lovers at MIT-AI
cc: katz at USC-ISIF

I didn't quite get what I wanted.  I would like to interchange the
RB and RD edges, the following does it:

	RD'R'D  RD'D'R'D   R D'D'R'D  RD'R'
This will mess up the bottom, but leaves the top and middle alone.  I 
was wondering if there is a way to do it in less steps.


Also,  there is a mini-cube on a keychain out, its called the mini-
Wonderful puzzler.  The normal size Wonderful puzzler is a Rupic's
cube of inferior quality;  I don't know who makes them, but they
are not from Ideal.



				Alan
-------


DENG at MIT-AI (Dave English, University of Newcastle, UK)@MIT-AI (Sent by DENG@MIT-AI) 06/25/81 05:34:38 Re: Royal Wedding
To: cube-lovers at MIT-MC
CC: DENG at MIT-AI


Someone here in Newcastle upon Tyne, England is selling
cubes to celebrate the forthcoming marriage of His Royal
Highness, Charles Mountbatten, Prince of Wales to Lady
Diana Spencer.  Said cubes have a portrait of Prince
Charles on one face & Lady Di on the opposite.  The
four faces have identical Union Jacks.  Consequently the
orientations of only two centres are significant.

Each cube comes complete with a leaflet describing a
solution.  It gives away nearly all my fastest spells
Presumably the solution has been included in order to
avoid spoiling THE DAY for the poor unfortunates who
scramble the cubes - and therefor the portraits.

The case is marked "Made in England", & so I claim this
as the first observation of c. Britannicus.  The
material used is white plastic, but not the brittle
stuff of c. Albinus.  Movement is quite free, but not
as swish as c. Americanus.

                    Yours,
                       Dave E. English.



Date: 4 July 1981 06:17-EDT
From: Richard Pavelle <RP at MIT-MC>
Subject: 4 layered cube
To: CUBE-LOVERS at MIT-MC

I heard that a 4x4x4 is being marketed in Boston. The person who told
me claimed to have seen and tried one but did not know who was selling 
them. Can anyone add anything to this?

Date:     5 July 1981 2217-edt
From:     Ronald B. Harvey           <RHarvey at MIT-Multics>
Subject:  re: 4 layered cube
To:       CUBE-LOVERS@MIT-MC

If anybody does find out where these are being sold, please try
to get an address for a distributor so that those of us not
lucky enough to live in Boston can have a local store stock them.

Thanks

Date: 12 July 1981 1343-EDT (Sunday)
From: Dan Hoey at CMU-10A
To: Cube-Lovers at MIT-MC
Subject:  Happy birthday
In-Reply-To:  ALAN@MIT-MC's message of 12 Jul 80 13:42-EDT
Message-Id: <12Jul81 134343 DH51@CMU-10A>

                                   \
                                   /\
                                  / /
                                  \ \
                                   \I\
                                    I/
                                   ###
                                   ###
                                   ###
                                   ###
                                   ###
         --------------------------###-------
         | --          --          ###        --
         |    ---------------------###-------------
         |    |  --          --    ###   --         --
         |    |     -------------------------------------
         |    |     |  --          --          --         --
         | -- |     |     -------------------------------------
         |    --    |     |           |           |           |
         |    |  -- |     |    H a p p y   B i r t h d a y    |
         |    |     --    |           |           |           |
         |    |     |  -- |           |           |           |
         | -- |     |     |-----------+-----------+-----------|
         |    --    |     |           |  C U B E  |           |
         |    |  -- |     |           |           |           |
         |    |     --    |           |           |           |
         -    |     |  -- |           |L O V E R S|           |
           -- |     |     |-----------+-----------+-----------|
              --    |     |           |           |           |
                 -- |     |          O n e   Y e a r          |
                    --    |           |           |           |
                       -- |   342 messages -- 73000 words     |
                          -------------------------------------

Date: 13 Jul 1981 11:33 PDT
From: McKeeman at PARC-MAXC
Subject: Missing Link
To: Dan Hoey at CMU-10A
cc: Cube-Lovers at MIT-MC, Ramshaw

Have you seen the Missing Link?  ["By the people who brought you Rubik's
Cube"; oh well, if they say so]  It is a cylindrical 15-puzzle, and about as hard
to solve.  Here are some notes on it.

The actual puzzle is a square tower of height four.  When solved, the puzzle
has the appearance of linked chains.  It has 15 square pieces leaving one hole
through which you can peek to see the innards of the puzzle.  The hole can
move up or down by sliding the piece next to it down or up.  The top and
bottom slices rotate about the vertical axis which permits the hole to move to
different columns.  A rough idea of the pattern is:

			|~|n|  can rotate
edge-on view	|n|X|
			|X|X|
			|u|u|  can rotate


I use two transformations, r and R to do things with it.  If you look at an edge of
the puzzle with the hole showing, there is a little cycle and a big cycle as noted
below.  Both leave the back unchanged.

transformation	from:		to:
      name

	r		|::|1|     	|::|2|
			|7|2|  =>	|1|7| 
			|6|3|     	|6|3|
			|5|4|     	|5|4|

	R		|::|1|     	|1|2|
			|7|2| =>	|::|3| 
			|6|3|     	|7|4|
			|5|4|     	|6|5|

The move set is:
U move the hole up
D move the hole down
T twist the top clockwise 90 degrees
B twist the bottom clockwise 90 degrees

for convenience, /T = TTT = T inverse, etc.

r = D TU/T D /TUT
RR = /TDT DD /BUB U TU/T DD BD/B U

Actually, I can "feel" my way to a solution easier than I can figure my way to
one.  It feels very much like the 15 puzzle.





Date: 13 July 1981 20:36-EDT
From: Alan Bawden <ALAN at MIT-MC>
Subject: Forwarded message
To: CUBE-LOVERS at MIT-MC
cc: LSR at MIT-XX

Date: 13 Jul 1981 1441-EDT
From: Larry Rosenstein <LSR at MIT-XX>
Subject: Contest

Thought you might be interested:

This past weekend I heard a radio ad (WBZ-AM 1030) advertising a
Rubik's Cube contest being held this weekend at Jordan Marsh in
Burlington, Mass.  The contest is being sponsored by Ideal Corp. and
apparently will be world-wide; at least, the national winner is
supposed to receive a trip to Monte Carlo for the world competition.

At the time, I did not pay too much attention to the ad, so I am not
positive about the details (specifically, the time and place).  You
might try calling Jordan Marsh in Burlington for more details.

    Larry
-------

Date: 14 Jul 1981 1129-PDT
From: ISAACS at SRI-KL
Subject: HOWTODOIT books again
To: cube-lovers at MIT-MC

  An update on How To Do It books on the Magic Cube:
     There now seem to be 3 "officially" published books on the Cube, besides
an unknown number of privately printed books, pamphlets, and single-page
solutions.  The three are:
1) "Notes on Rubik's Magic Cube", by David Singmaster, Enslow Publishers,
   Bloy Street and Ramsey Avenue, Box 777, Hillside, New Jersey, 07205($5.95).
      This is, of course, THE book on the magic cube.  It has the history,
   anecdotes, the math theory, and a solution. It is almost exactly the 
   same as the fifth printing of his privately printed pamphlets.  His 
   solution is top, turn over, middle, top.  The final face is done: orient
   edges, position edges, position corners, orient corners.
2) "The Simple Solution to Rubik's Cube(TM)", by James G. Nourse, Bantam Books,
   1981($1.95). Illustrated by Dusan Krajan. This is a somewhat revised edition
   of Nourse earlier publication "Solution to Rubik's Magic cube", Storc 
   Publications.  It contains a solution, plus a little other information on
   the cube and some "other games to play" (speed, competition, pretty 
   patterns).  The main thing new to me is the idea of pretty-patterning the
   alphabet, and then spelling out 3 or 4 letter names around the cube.  Some
   are pretty stretched, but it's a nice idea.  
      His solution goes top-middle-bottom, and bottom is: place corners,
   orient corners, place edges, orient edges.  His moves tend to be longer 
   than "normal", but with the purpose of being able to recover fairly easily
   from a mistake or from a dropped cube.  He tries to move no more than one
   cube from a previously solved position at a time, and to make it always
   possible to back up only one step (rather than to the beginning).  Each 
   section has error correction (in case you go wrong), and short cuts (to
   speed things up once you get used to it).  His notation uses Top, Bottom,
   Front, Posterior, Left, Right.  He tries to use Posterior as little as
   possible (which I think is very good - but for the same reason, its 
   easier to turn the whole cube over after the first couple of steps to be
   able to work on the top.)
      The book is roughly paper-back size (though of course, much thinner),
   and on crummy paper.  But the layout and pictures are good.  I noticed a
   couple of errors - on page 45, you might have to error correct back to the
   end of step 2, since an upper corner might be misplaced.  On p. 46, Short
   Cut 2 is a replacement for step 4C, not 4D, and the sequence should end
   with B+ or B-.
3) "Mastering Rubik's Cube(TM), The solution to the 20th century's most
   amazing puzzle", by Don Taylor, an Owl Book, Holt, Rinehart and Winston,
   1980, $1.95.  This is a straightforward solution book, with a few
   additional games and pretty patterns, but not much.  He does it top, turn
   over, middle, top; and the final top is place corners, place edges, twirl
   corners, and flip edges.  He uses BFUDLR, place(to move), and position(to
   orient).  The moves are similar to Singmasters, but with some reductions.

     This note is getting long.  Does anybody know of any other published books
on the Cube?  I notice that almost all the published solutions I've seen
are top-middle-bottom (or that turned over in the middle).  Except that 
Singmasters original solution was first all the corners, then all the edges.
Does anybody do edges before corners?  I nowadays use a variant which is 
easier for me:  First I do all the top except one corner.  Then I use that
corner to make it easier to get in the middle layer (except for one). Then
I finish off the top, and put in the last middle.  I turn the cube over, and
orient first the edges, then the corners.  With the top face now showing the
right colors, I find it easier to see where I am.  I position the corners
(with a move that doesn't twist), and finally position the edges. What other	
variants are around?
      -- Stan
-------

Date: 18 July 1981 17:15-EDT
From: David C. Plummer <DCP at MIT-MC>
Subject:  Speed cubing contest
To: CUBE-LOVERS at MIT-MC


Second hand info: (first hand in the Boston Phoenix (day unknown))

                    SPEED CUBING COMPETITION

Date:  Saturday, July 25, 1981
Place: Jordan Marsh, Burlington, Massachusetts

Times:

1100-1500  3 minute qualifiers
1530-1800  Registration and Playoffs
1800-....  Finals

First prize is N hundred dollars, where N is about 5.

(Sorry for the lack of info, we got this over the phone in a
hurry.) 

Date: 18 Jul 1981 14:28 PDT
From: Pasco at PARC-MAXC
Subject: Please remove my name
To: CUBE-LOVERS at MIT-MC

Sorry to bother all of you, but it's not obvious to whom I should address this
request: Please remove my name (Pasco at PARC-MAXC) from CUBE-LOVERS.

Thank you.

	Rich Pasco


Date: 18 July 1981 17:39-EDT
From: Alan Bawden <ALAN at MIT-MC>
Subject: Conserve computrons!
To: CUBE-LOVERS at MIT-MC
cc: Pasco at PARC-MAXC

cube-lovers-request@mit-mc is the mailing list to send adminitrivia
to.  Since mailing to cube-lovers causes machines all over the ARPA
net to have to deal with the mail, you should all take a moment to
remember that fact, so that when YOU want to be removed, or want
someone new added, you will know who to mail to.

Date: 20 July 1981 23:14-EDT
From: Alan Bawden <ALAN at MIT-MC>
Subject: problem
To: CUBE-HACKERS at MIT-MC

Here's a problem I don't think I have seen asked before:

Suppose you left your cube out in the rain and one of its six axles
froze so that you were unable to turn that face.  How badly is your
cube damaged?  Can you still reach all the positions you could before?
or are you now limited to some proper subgroup?  (how big is it?)

I can see that all of the configurations of corner cubies are still
easily attainable.  But without a cube at home to fool with, I can't
figure out if I can reach all the edge positions, and I can't figure
out how the edge and corner groups interact given this restriction.

Another way of asking the problem would be:  How large is the subgroup
generated by just five of the six quarter twists.

Date: 20 July 1981 2332-EDT (Monday)
From: Dan Hoey at CMU-10A
To: Cube-hackers at MIT-MC
Subject:  Re: One stuck axle
In-Reply-To:  Alan Bawden's message of 20 Jul 81 22:14-EST
Message-Id: <20Jul81 233257 DH51@CMU-10A>

	Singmaster gives the answer.  Let A = RL'FFBBRL', then AUA = D.
Thus it is possible to dispense with D moves entirely.  Jim Saxe also
has a solution, but I don't know if it's the same one.

Date: 20 July 1981 23:35-EDT
From: David C. Plummer <DCP at MIT-MC>
Subject: problem
To: ALAN at MIT-MC
cc: CUBE-HACKERS at MIT-MC

    Date: 20 July 1981 23:14-EDT
    From: Alan Bawden <ALAN at MIT-MC>

    Here's a problem I don't think I have seen asked before:

    ....

    Another way of asking the problem would be:  How large is the subgroup
    generated by just five of the six quarter twists.

According to the Fusrt-Hopcroft-Luks algorithm, they are the same
size. I would appreciate Hoey and Saxe to confirm this (I think I
have all the bugs out of my code). Sorry for such a brute force
method, but it only took a few minutes. 

Date: 21 Jul 1981 0908-PDT
From: ISAACS at SRI-KL
Subject: More problem
To: CUBE-HACKERS at MIT-MC

  Well, then.  How about four of the six quarter twists? Three?
   -- Stan
-------

Date: 21 July 1981 2350-EDT (Tuesday)
From: Dan Hoey at CMU-10A
To: Cube-Hackers at MIT-MC
Subject:  The ten stuck-axle subgroups
In-Reply-To:  ISAACS@SRI-KL's message of 21 Jul 81 11:08-EST
Message-Id: <21Jul81 235052 DH51@CMU-10A>


1.  No faces stuck.  The familiar cube group.

2.  D face stuck.  As previously noted, all positions can be reached.
In addition, all Supergroup positions that fix the orientation of
the D face center are achievable.

3.  B and D faces stuck.  All Supergroup positions that fix the BD
edge and the B and D face centers are achievable.

4.  U and D faces stuck.  Edges cannot be flipped.  If we define edge
orientation by marking the F and B facelets of the F and B edges,
and the U and D facelets of the others [cf Jim Saxe's message of 3
September 1980], then all Supergroup positions that fix the
orientation of all edges and the U and D face centers are
achievable.

5.  L, B, and D faces stuck.  All Supergroup positions that fix the
BLD corner, the LB, BD, and DL edges, and the L, B, and D face
centers are achievable.

6.  U, B, and D faces stuck.  Again, edges cannot be flipped.  All
Supergroup positions that fix the orientation of all edges, the
position of the UB and BD edges, and the orientation of the U, B,
and D face centers are achievable.

7.  U, L, B, and D faces stuck.  Singmaster has a very nice
description of this group [indexed as Group, Two Generators].  The
group of achievable permutations of the six movable corners is
isomorphic to the group of all permutations on five letters.  All
Supergroup positions that permute the corners in an achievable
permutation, fix edge orientation, and fix the unmovable two
corners, five edges, and four face centers are achievable.

8.  U, L, D, and R faces stuck.  Sixteen positions

9.  U, L, D, B, and R faces stuck.  Four positions.

10. All faces stuck.  One position.

Date: 24 July 1981 2220-EDT (Friday)
From: Dan Hoey at CMU-10A
To: Cube-Hackers at MIT-MC
Subject:  A new problem
Message-Id: <24Jul81 222049 DH51@CMU-10A>


        Suppose you buy a new cube and the arrangement of the
colors is different from your old cube.  Naturally, you want the
new one to be like the old, so you decide to switch the colortabs
around.

A.  What is the smallest number of faces you have to recolor?

B.  What is the smallest number of colortabs you have to move?

        Note the hidden variable:  the permutation of the new cube
with respect to the old one.  This variable has thirty values,
including the identity.  There are two kinds of answers I am
interested in.

1.  A minimax value -- a recoloring algorithm and a proof of its
        optimality.

2.  A probability distribution of optimal recolorings.

        Any takers?

Date: 27 July 1981 10:29-EDT
From: Dennis L. Doughty <DUFTY at MIT-MC>
Subject:  Cube Championship
To: CUBE-LOVERS at MIT-MC
cc: DUFTY at MIT-MC


Does anyone know what the winning time was in the recent regional cube
contest? (Last saturday at Jordan Marsh)?  --Dennis

Date: 27 Jul 1981 1111-EDT
From: PDL at MIT-DMS (P. David Lebling)
To: DUFTY at MIT-MC
Cc: CUBE-LOVERS at MIT-MC
In-reply-to: Message of 27 Jul 81 at 1029 EDT by DUFTY@MIT-MC
Subject: Regional Cubing Championship
Message-id: <[MIT-DMS].205137>

According to the Boston Globe, the fastest times were;

	48.31 sec. - Jonathan Cheyer, 10
	51.16 sec. - Jeffery Varafano, 14
	51.59 sec. - Peter Pezaris, 11
(these are for the "junior" division; under 17).  The fastest
"seniors" were;

	69.64 sec. - Herbert H. Thorp, 17
	69.83 sec. - Charles Hawes
	77.26 sec. - Rick Miranda

Jordan Marsh says they sell about 2000 cubes per week.  As the
Jordan Marsh V.P. who was standing next to me said, "You can't
buy this kind of publicity!"

The competition was organized reasonably well, consisting of
three rounds:

	1) The qualifying round consisting of being able to 
solve a cube in under three minutes.  No official timing other
than "under three minutes" was done in this round.  About 20
people were tested per qualifying round, and from 20-30% qualified.
The cubes were allegedly "broken in" in advance, and all had the
same color orientation.  They were re-randomized between rounds.

	2) Those who qualified in the first round were given two
tries to solve a random cube in under two minutes.

	3) Three "patterned" cubes were solved (presumably everyone
got the same patterns).  I didn't see this round so I don't know
the details of it.

	My impression of the qualifying rounds was that those who
qualified differed from those who didn't largely in speed.  They
didn't seem to use any macros I haven't seen, they just did them
extremely fast and rarely paused more than fractions of a second
to decide what to do next.  The fact that the three top finishing
juniors all had better times than the three top-finishing seniors
indicates that competitive cubing is a young person's game.



Date: 29 JUL 1981 2103-EDT
From: RMC at MIT-MC (R. Martin Chavez)
To: CUBE-LOVERS at MIT-MC



Date: 29 JUL 1981 2107-EDT
From: RMC at MIT-MC (R. Martin Chavez)
To: CUBE-LOVERS at MIT-MC

     If it's at all possible, I would very much like
to be added to the Cube-lovers mailing list.  Though
I will not go so far as to call myself a "cube-meister", 
I have spent a great deal of time developing my own
personalized algorithm (without the book.)  I can solve
the cube in about five minutes (don't laugh), yet I
still need to discover an operator for swapping a pair of
edges and a pair of corners.
     Thanks....

Date: 30 July 1981 16:53-EDT
From: Carl W. Hoffman <CWH at MIT-MC>
Subject: Another Japanese cube
To: CUBE-LOVERS at MIT-MC

Sitting in the SIPB office is a super-group cube.  At least I think it's a
super-group cube.  Four sets of facies show the symbols on playing cards,
and the other two sets show five pointed stars of different types.  So the
center facie carrying the diamond can have one of two orientations.
Is it possible to rotate one center facie 180 degrees?
(The cube is currently in the solved state, thanks to Jim Saxe.)

Date: 30 Jul 1981 16:59 PDT
From: Woods at PARC-MAXC
Subject: Misc Cube replies
In-reply-to: recent messages
To: RMC at MIT-MC (R. Martin Chavez), CWH at MIT-MC (Carl W. Hoffman)
cc: CUBE-LOVERS at MIT-MC

To RMC:

The proper way to get yourself added to any of the large lists at MIT is to
send a message to <name-of-list>-REQUEST, e.g., CUBE-LOVERS-REQUEST. 
Likewise for getting yourself removed from the list.  Also, a macro for
swapping two corners and two edges (all on the same face, and without
flipping any of them), is (UUFLF')*5.  There are probably shorter ones,
but I like this one because it requires remembering only a very short
sequence, making it easier to do when stoned/drunk/tired/etc.

To CWH:

I don't believe you can rotate a single center facie 180 degrees
without changing anything else, which means you can make a lot of
confusion for someone by inverting the diamond and coming up with
a cube that appears to have only one center facie flipped.

	-- Don.


Date: 31 July 1981 00:04-EDT
From: Steve B. Waltman <SBW at MIT-MC>
To: CUBE-LOVERS at MIT-MC

   It is possible to rotate one center face any multiple of 90
degrees. In fact, many apparently 'null operations' rotate one
or more center faces....
				Steve

Date: 30 Jul 1981 23:01 PDT
From: Woods at PARC-MAXC
Subject: Re: Supergroup
To: Steve B. Waltman <SBW at MIT-MC>
cc: CUBE-LOVERS at MIT-MC

It was my understanding that the size (number of possible positions) of
the supergroup was only (4^6)/2 times as large as the normal group, i.e.,
that there was another level of parity involved.  This would imply that
you can rotate a center facie 180 degrees, but not 90 (at least, not without
changing something else).  I'd be most interested in seeing one of your
"many apparently 'null operations'" that rotates one center face 90 degrees. 

	-- Don.



Date: 31 July 1981 0626-EDT (Friday)
From: Dan Hoey at CMU-10A
To: Cube-Lovers at MIT-MC
Subject:  Re: Supergroup
In-Reply-To:  Woods@PARC-MAXC's message of 31 Jul 81 01:01-EST
Message-Id: <31Jul81 062607 DH51@CMU-10A>


	Don Woods's answer is correct.  When the edges and corners
are in their solved positions, the total face center twist is zero
modulo 180 degrees.  Thus the number of face centers that are
twisted 90 degrees must be even.  A proof of this appeared in
Cube-Lovers last year.

	A process to twist the U face center 180 degrees without
changing anything else is (U LR UU L'R')^2.  This process comes from
Singmaster's book.  I sent Cube-Lovers a general procedure for
solving the supergroup in January.

	If you want a copy of either of these messages, send a note
to Hoey@CMU-10A.


Date: 2 August 1981 03:43-EDT
From: Alan Bawden <ALAN at MIT-MC>
Subject: Administrivia and an assault on God's number.
To: CUBE-LOVERS at MIT-MC

First the administrivia.  Starting with this message mail to
Cube-Lovers is no longer automatically redistributed to everyone on
the list.  This was done because the "please add me to this list" type
of message is now almost as frequent as messages discussing topics of
general interest.  Now the "editors" (Dave Plummer and myself) will
have a chance to catch these boo-boos.  Hopefull this change will be
completely invisable to the rest of you except that the mail headers
will contain our names as the senders, and the turn-around time will
be a little slower.  If this really offends anyone then we can put it
back the way it was, but lets try it this way for a while.
You should continue to send your messages to Cube-Lovers at MIT-MC.

Now, on to God's number!

As you may recall, a somewhat complicated counting argument sets a
lower bound on God's number (the worst case counting of the best
possible algorithm) at 21 Quarter twists.  Dan Hoey's message of
January 9 1981 (the second message in a series about the "Supergroup")
contains an excellent summary of that argument, so I won't repeat any
of it here.  That argument takes into account certain trivial
identities such as FFFF=I and FB=BF in order to reduce the amount by
which the counting overestimates the number of configurations a
certain number of twists away from "solved".  The same argument
ignoring the identities only leads to a lower bound of 19.  It is thus
natural to expect that taking even more identities into account would
lead to an even higher lower bound.

Well, the next smallest identities are those of the form
FR'F'RUF'U'FRU'R'U=I.  It is known that there are none smaller that
aren't a consequence of those "trivial" identities mentioned above
(See Hoey's message of 22 March 1981), although there might be others
of the same length.  What happens if we take these additional no-ops
into account?  The conceptual problems in applying these new
identities to the counting have had me stumped for quite some time
now, but last week I finally figured a way that would cover at least
some (maybe even all, I haven't worked on a proof of that yet) of the
consequences.

Well, all right, I'm keeping you in suspence, what did I learn?

Nothing.  The lower bound still stands at 21 (similarly the Supergroup
lower bound still stands at 25).  Even after taking
FR'F'RUF'U'FRU'R'U=I and his many friends into consideration it seems
that the numbers that fall out of the counting scheme (and it is
amazingly complicated!) are only slightly smaller than those we
already knew.  

The relevant numbers are:

Size of the cube group: 43252003274489856000
Under old counting:
positions 20 Q's or less away from start: 39812499178877773072
positions 21 Q's or less away from start: 373188814849923987472
Under new counting:
positions 20 Q's or less away from start: 39726356237445007600
positions 21 Q's or less away from start: 372326146413193718032

As you can see the numbers are depressingly close.  This seems to shut
the door on any further improvements of this kind to this argument.
It is hard to imagine that the effects of any other identities
(remember they have to be at least 12 Q's long) could be signifigantly
greater than the effect here.  (Of course, if we knew ALL of them...
but then we would understand the group completely!)

It is of course possible that some deeper property, deeper than just
the knowledge of one identity, could improve this style of counting
argument. 

It is of course also possible that I have screwed up somewhere.  I
sould really let some of the rest of you into the details of this
thing.  As you can guess, I am not very excited at the idea of
having to explain the details of the argument to you all.  The proof is
complicated and kludgey, and I at least am convinced that it leads
nowhere.  People who are interested in the gory details can contact me
and we can work something out.

ZILCH@MIT-MC 08/02/81 06:06:43 Re: Identities (galore)!
To: CUBE-LOVERS at MIT-MC
CC: ALAN at MIT-MC
This message may be overdo, but I figuered it ought to be 
sent now since ALAN has done some work on this subject and may be able 
to use my results.
   Recently I have found many Identity transformations and this 
message is basically a catalog of them.

I12-1   FR'F'RUF'U'FRU'R'U
I12-2   L'D2F'D'FLD2BDB'
I12-3   FR'F'RUF'UL'U'LFU'

I14-1   D'L'DRD'LR'DF'D'RDR'F
I14-2   D'L'D'F'DFLDF'R'D'RDF
I14-3   LD2RL'F2L'F2LR'D2
I14-4   F'R'D'RDFUF'D'R'DRFU'
I14-5   (LRD2L'R'D')^2           NOT GOOD IN  SUPERGROUP  DOES <D2>
I14-6   LR'FRL'U'DFB'R'F'BUD'     "     "      <FR'>

I16-1   (LRD2L'R'D2)^2
I16-2   F'R'D'RDFU2F'D'R'DRFU2
I16-3   F'R'D'RDFD2B'D'L'DLBD2
I16-4   F'R'D'RDFUDR'D'B'DBRU'D'
I16-5   F'R'D'RDFU'DR'D'B'DBRUD'
I16-6   UF2U'DR2D'U'R2UD'F2D
I16-7   LR'D2RL'F2RL'F2LR'D2
I16-8   FDLD'F'LDL'F'L'D'LFD'L'D
I16-9   LD'L'D'F'DFUF'D'FDLDL'U'
I16-10  FL'F'LF'D'FUF'DFL'FLF'U'
I16-11  F'D2R'D'RD'FLD2BDB'DL'
I16-12  (F2B2R2L2)^2
I16-13  LR'FRL'U2D2L'RB'LR'U2D2     NOT IN SUPERGROUP   <FB'>
I16-14  LR'F2RL'U'DFB'R2F'BUD'        "    "      <F2R2>

I18-1   F'BD2F'D'FD2B'LBDB'L2FL
I18-2   LRD'L'R'D'LRDL'R'DLRDL'R'D'
I18-3   D'RD'R'DBDB'DBD2B'D'RD2R'
I18-4   LR'F2RL'U2D2L'RB2LR'U2D2
I18-5   LDR'L'D'LRDL'R'D'RLDR'L'D'R
I18-6   (RBL'R'B'L)^3
I18-7   LDR'L'D'LRDL'R'DRLD'R'L'DR     NOT IN SUPERGROUP     <D2>
I18-8   (F'D'FD'RD2R'D)^2                 "     "       <D2>


These are not all of the identities that I have found but are generators
of them.

How a generator generates other identities:
1. Inversion      (2)
2. Rotation       (24)
3. Reflection     (2)
4. Shifting       (N)       where N is the length of the identity

The numbers in () are the number of different ways that can be gotten
for each of these methods.  Combining gives 96N.
For example an identity of length 12 generates a possible 96*12=1152.
However this number is usually not reached becauseof inherent symmetry.
If you take the inverse of I12-1 --> U'RUR'F'UFU'R'FRF'
Then its reflection  -->UL'U'LFU'F'ULF'L'F
Then a rotation U->F,L->R,F->U     -->FR'F'RUF'U'FRU'FR'U
You get back what you started with.
When shifting is included in this process there are a total of 6 different
ways this can be done  giving  1152/6=192 different identities generated by
I12-1.

Shifting:
 Basically you chop the transform in its interior and append the first
part to the second part.
For instance.  I12-1              FR'F'R / UF'U'FRU'R'U
Becomes                            UF'U'FRU'R'U  FR'F'R
Note that this is just a rotation away from the origional.
From these some equivilences may be deduced:
DL'F'D2R'D'R = L'D2F' = BD'B'D2L'F'D
DF'R'DRD'FD' = FL'F'L =D'LD'BDB'L'D
FR'F'RUF' = U'RUR'F'U = UF'L'ULU'

Unfortunatly these equivilences only generate the 3 identities of 
length 12 , using the idea that midpoint of an identity must be the
unique maximum along the path of the transform.

ZILCH@MIT-MC 08/02/81 06:36:52 Re: 2edge-2corner swappers
To: CUBE-LOVERS at MIT-MC
   I know of 4 different transforms that swap 2 edges and 2 corners.
LD'BRDR'B2LBL2D    (FD,RD),((DLF,DFR)
R'DRD2R'L'DRD'LD2      (FD,LD),(FDL,FRD)
FD2F'B'DFD'BD2F'D       (FD,LD),(LFD,BDR)
FLFL'UBLB'U2FUF     (FD,DL),(DLF,DFR)

I found the first 3 of these before I saw Singmaster and modified one of 
Singmaster's into the fourth one which he didn't even have recorded.

Date: 2 August 1981 19:41-EDT
From: Alan Bawden <ALAN at MIT-MC>
To: CUBE-LOVERS at MIT-MC

This is a test message that should NOT get forwarded!

Date: 2 August 1981 19:42-EDT
From: Alan Bawden <ALAN at MIT-MC>
Subject: again
To: CUBE-LOVERS at MIT-MC

This an another test message to see if COMSAT really sucks rocks as big
as I think!

ALAN@MIT-AI 08/02/81 19:44:44 Re: and finally.
To: CUBE-LOVERS at MIT-AI
Here is the last test in the series.

Date: 2 August 1981 23:51-EDT
From: Steve B. Waltman <SBW at MIT-MC>
Subject:  Supergroup
To: Woods at PARC-MAXC
cc: CUBE-LOVERS at MIT-MC

  Oops.. Woods is right and I was out-to-lunch. Sorry...
				Steve Waltman

Date:  3 Aug 1981 0934-EDT
From: Jerry Agin <AGIN at CMU-20C>
Subject: One more identity
To: zilch at MIT-MC
cc: Cube-Lovers at MIT-MC

U'F'UBU'FUB'URU'L'UR'U'L
-------

Date:  3 Aug 1981 1250-PDT
From: ISAACS at SRI-KL
Subject: another new cube
To: cube-lovers at MIT-MC

   Last week I saw still another type of new cube that a friend had picked
up on a chinese boat.  Each face was a magic square - that is, each facelet
contained a number from 1 to 9, and, when solved, each line of 3 facelets
added to 15.  This particular cube was also colored, so wasn't very
interesting (you could solve it by colors and the magic-ness was automatic),
but it should be a very difficult problem with the numbers colorless.
   Apropos to this, my son says he saw two other new types (but by the
time he went back to get them for me, they were gone).  They were both
related to the 10-sided one that essentially cut off 4 parallel edges;
one cut off all EXCEPT 4 parallel edges (how?), and the other cut off
corners.  Has anybody else seen thesed?  Can you describe them more
accurately?
   By the way, there are lots of "do it yourself" variations on the cube.
For instance, a good present (though it takes a lot of work to make, and
I still don't know what the best glue to use is) is a picture cube.  Find
6 family photos (or other appropriate pictures) of the right size and glue
away.  Or (and this is based on earlier discussion of a braille cube), make
a tactile cube - put a different material on each face.  I've tried cloth
(which doesn't work too well because of the glue and the fraying), misc
stuff (washers, broken toothpicks, etc), and next want to try with different
sandpaper grades (the ideal tactile cube would be ONLY solvable by touch, and
have the faces look the same to the eye).  Another, more difficult to make
suggestion, would be to make each face a different thickness, or perhaps
a different contour.  Any ideas on how to actually construct these?  Any
ideas on a really good glue for cube faces?
       ---  Stan Isaacs
-------

Date:  3 Aug 1981 1304-PDT
From: ISAACS at SRI-KL
Subject: Another Cube Book
To: cube-lovers at MIT-MC

   Just got still another book on solving the cube: "Solve That Crazy
Mixed-Up Cube", by Don Frederick, Frederick Enterprises, P.O. Box 1016,
Oceano, Ca, 93445.  This one has a sense of humor, and likes to make up
new names.  He talks about "slabs" for a layer, "cranking the keepers"
to save some already positioned cubies while you do something else, a
"magic move", etc.  For instance, his top slab moves are: the STICK-UP,
the HANG-UP, the PUT-DOWN, the PICK-UP, the DOUBLE-DIP, DUMP TRUCK, and
the SLING-SHOT GOTCHA!  He goes top-middle-bottom, bottom done: twist edges,
position edges, position corners, twist corners.  He includes hints, memory
aids, short-cuts, patterns, a two person game called "Widow's Revenge", etc.
Even cartoons ('if all the unsolved cube puzzles were put in one pile, they'd
stay that way.').  All in all, a very nice addition to the litterature. $3.99.
      --- Stan Isaacs
-------

Date: 3 August 1981 16:47-EDT
From: Allan C. Wechsler <ACW at MIT-AI>
Subject:  God's Number
To: CUBE-LOVERS at MIT-AI


We know that some positions are "global maxima".  We don't
know how many such positions there are.  We would dearly love
to know how far away they are.

Suppose that God's Number is N. (For some obscure theological
reason I have the irrational belief that N=28, but we'll
leave such hunches out of the discussion.)  Let's say there
are K global maxima at that distance from solved.

What if we could show that there are at least 12K states at
distance N-1?  This is a little bit reasonable.  All it
means is that global maxima are all separated from each other by
more than 2q.

If that were true, mightn't we be able to increment our lower
bound on N?

Can anybody prove that it's true?

I would also like to hear from ZILCH where he gets those identities,
and whether any of those impressive lists are exhaustive.

Alan, do the other two order 12 nulls enable you to stretch your
results?

I am busy wading through Wielandt's "Finite Permutation Groups" and
will report if I learn anything applicable.

   --- Wechsler


Date: 3 August 1981 18:07-EDT
From: Alan Bawden <ALAN at MIT-MC>
To: CUBE-LOVERS at MIT-MC

The answer to everybody's questions about whether
the additional identities help my proof is: no.
My idea depended heavily on the peculiar properties
of that one identity.  Dan Hoey has compiled a list
of ALL the 12Q identities by brute forcing down to 6
twists, we suspect that ZILCH may have found them all,
but it will take a while to sort throught them to be sure.
We will let you know the results soon.



Date: 3 August 1981 22:34-EDT
From: Alan Bawden <ALAN at MIT-MC>
Subject: The Archive
To: CUBE-LOVERS at MIT-MC

Those of you who look through the archives of old Cube-Lovers mail
will notice that I have split off a new section of the archive.  The
mail now lives in:

MC:ALAN;CUBE MAIL0	;oldest mail in foward order
MC:ALAN;CUBE MAIL1	;next oldest mail in foward order
MC:ALAN;CUBE MAIL2	;more of same
MC:ALAN;CUBE MAIL	;recent mail in reverse order

Date: 4 August 1981 01:21-EDT
From: Chris C. Worrell  <ZILCH at MIT-MC>
Subject: Poll
To: CUBE-LOVERS at MIT-MC

   In line with the recent cubeing contest in Boston, I suggest a poll.
Such questions as, will be included:
1. age
2. average solving time
3 occupation  (if student, undergrad or grad and major)
4. solving method
5. how long it took you physically  working (playing) with the cube  (or some
   facimile, like a program) to solve the cube the first time, and reliably
6. how long have you been working with the cube  (say since Jan. '81)
7. max #of qtw. in method, avg. # of qtw. in method
 
I would not suggest anybody send their answers to the list, because
1. it would generate a lot of semi-useless garbage in the list,and
2. you would lose anonymity (which is one of the aspects of a poll)

If anyone can suggest how this could be implemented please pass it on.

                                             Chris Worrell

Date: 4 August 1981 01:58-EDT
From: Chris C. Worrell  <ZILCH at MIT-MC>
Subject: Identities...
To: CUBE-LOVERS at MIT-MC

 A few corrections to my list:
I16-13   has 18 qtw not 16
I18-4   has 20 qtw not 18 and has supergroup effect <F2B2>
I18-5 is not distinct from I18-6 so may be taken off the list

Additions:
I16-15 U'F'UBU'FUB'URU'L'UR'U'L        courtesy of Jerry Agin
I16-16 FLF'RFL'F'R'B'LBR'B'L'BR
I16-17 LD'R'DL'D'RUR'DLD'RDL'U'
I18-9  SAME AS I16-13
I18-10 RDLD'R'DU'L'D'LUD'R'DL'D'RD


Concerning Sources:
I found none of these by exhaustive search so I believe that this is far
from a complete list esp. in 18 region.
I used several methods to derive these identities
1. from two processes which have same effect , I concatanate the inverse of
the second to the first, and remove noops.
2. from 1 process which affects only the d face or not the U face,
I twist both the D and U faces, run the reverse process (maybe with an 
orientation change) then untwist the D and U faces.
3. examine old identities for similarities shift/rotate/invert/reflect
appropriatly concatanate and remove noops.


                                          Chris Worrell

Date:  4 Aug 1981 1123-PDT
From: Alan R. Katz <KATZ at USC-ISIF>
Subject: 10 sided "cube"
To: cube-lovers at MIT-MC
cc: katz at USC-ISIF

I have a 10 sided "cube", which is made by "Wonderful Puzzler" (they also
make crappy cheapo regular cubes).  The guts are essentially the same as
the regular cube, but the corners are cut off.

If you look down from the top, you see an octogon.  The edges are all the 
same as a regular cube, but since the corners are cut, they are one color
(thus the 10 colors).  For example, there are red, blue, and orange
faces.  On an ordinary cube you would have a red-blue-orange corner cubie,
but on this in its place is a pink face.  To make this clearer, here is
the coloring of the thing:

				red
				 *
      light-blue**gold**yellow**blue**pink**orange**violet**green
				 *
			       white

(red on top, white on bottom, looking at the blue face, back face is green,
right face is orange, left face is gold).

The interesting thing about this is that unlike the ordinary cube, every cube 
does not have a place,  you dont know that the pink corner goes between the 
blue and orange faces (in an ordinary cube it is the red-blue-orange corner 
so you know where it goes).  To solve it, you just put the corners in some
order, solve it using the usual transformations, and then if you get a 
"parity error" you must go back to the top layer but switch two of the corners
and solve it again.  Thus in general you have to solve it almost 2 times!
(almost because you dont have to redo the top layer or half of the second
layer (this assumes you solve top down, which I do.)).

What I mean by parity error is that if the corners are switched you can get 
a configuration that in an ordinary cube would tell you the cube is put
together wrong.  For example, you can be solving it and get to a point where
an odd number of edges must be fliped.  


There may be a transformation to flip an odd number of edges with this cube,
but I have not found it.  Anyway its more interesting to solve and it
changes it shape in general with each transformation.  (unlike the cube 
which stays a cube; this is a octagonal prism).


						Alan

-------

Date:  5 Aug 1981 0948-PDT
From: ISAACS at SRI-KL
Subject: 10-sided cube
To: cube-lovers at MIT-MC

  The 10-sided cube was discussed a couple of months ago.  The main result
was that the easiest way to solve it was "side across" - that is, don't
start from the octagonal face, but from one of the "sides".  Then the
last layer should be solvable except for a possible edge-flip.  Note that
there are two new difficulties with this cube: the one mentioned, and the
single edge flip.  SPOILER! This is, of course, an optical illusion, brought
about by the fact that the second edge, which of course is also flipped, is
a uni-colored edge, and you can't see it.
         --- Stan Isaacs
-------

Date:  7 Aug 1981 1005-PDT
From: ISAACS at SRI-KL
Subject: BARREL AVAILABLE
To: CUBE-LOVERS at MIT-MC

   The "Magic Barrel" or "Ten Billion" or whatever its name is, that was
talked about in this digest several months ago, seems to have finally gotten
on the American market.  In the Bay Area, at the Stanford shopping center,
they have them in stock at Macy's, and will soon have them at Games and Things.
Games and Things also has the smallest cubes I've seen - about 3/4 inch on
a side.  That makes at least 4 different small sizes, from 1 1/2 inch down.
       --- Stan 
-------

Date: 10 Aug 1981 0841-PDT
From: Tom Davis <CSL.JHC.DAVIS at SU-SCORE>
Subject: Barrel Puzzle instructions
To: cube-lovers at MIT-MC

I just purchased one of the "Wonderful Barrel" puzzles mentioned in
Stan Isaac's note, and I found the enclosed set of instructions to be
a classic.  I figured I'd pass them on, in case some of the members of
this list don't buy the puzzle itself.  The first couple of paragraphs
are not so great, but it gets better toward the end.

@begin(verbatim)
"Wonderful Barrel" is a kind of mental game plate also can be regarded
as an indoor ornament.  The red, orange, yellow, green, blue, black
color balls which is inner in barrel can create more than 10 billion
combination to reach a solution and finally be aligned on 5 straight
lines, yet each straight line are with 4 same color balls.

			* PLAY RULE *

Original alignment circumstances is as follow:  there are three black
balls are set on SHELTER but each black ball is above a shore of
PLUNGER; the 5 straight lines are up the SHELTER with different color
from one another, each line contains 4 same color balls.  The 5
straight lines are aligned in clock-wise with blue, green, yellow,
orange, red color.

As you know, the original alignment is orderly, now you can
circumvolve DRUM, then operate plunger up and down for taking the ball
out from SHELTER.  While you put the ball into SHELTER again, the
order of "each straight line with 4 same color ball" has been broken
to confused state.

Now, a challenge is ahead of you, you often successful in this line
but irregularly in that line, in a word, it always can not come to a
satisfactory arrangement of the whole lines fluently.

At the beginning, it is not easy to arrange the 5 straight lines
return to initial order in short time, so please try to solve this
puzzle from one line first, then completion of two, three, four even
to whole lines are returned to initial order finally.

At last, maybe you can analyse the mystery out, but maybe you can not
solve it today; however, how about tomorrow?  one month later? one
year later? or ten years later you may still continued to challenge to
it singlely and unwilling.
@end(verbatim)

P.S.  I @i(tried) to proofread this, but as you can see, it was difficult.
-------

Date: 14 August 1981 0111-EDT (Friday)
From: Dan Hoey at CMU-10A
To: Cube-Lovers at MIT-MC
Subject:  Results of an exhaustive search to six quarter-twists
Message-Id: <14Aug81 011137 DH51@CMU-10A>

	The first answer is that there are exactly 878,880 cube
positions at a distance of 6 quarter-twists from solved, and so
983,926 positions at 6qtw or less.  These figures reflect a decrease
of 744 from the previously known upper bounds.

	It turns out that the twelve-qtw identities reported by
Chris C. Worrell <ZILCH at MIT-MC> are complete, in a sense.  The
only reservation here is that a fifth rule must be added to his
list of the ways in which ``a generator generates other
identities.''  This rule is substitution with shorter identities,
and it's not too surprising that it was left out, since the only
shorter identities are the ``trivial'' ones like XXXX=XYX'Y'=I,
where X and Y are opposite faces.  In the case of the twelve-qtw
identities, this means that identities of the form aXXb and aX'X'b
generate each other.

	The structure of the 12-qtw identities is clearer if we
write them in a transformed way:

I12-1	FR' F'R    UF' U'F    RU' R'U
I12-2	FR' F'R   UF'   F'L FL'   U'F
I12-3	FR' F'R   UF'   UL' U'L   FU'

	The fifth rule is necessary so that I12-2 may generate the
identities

I12-2a	 FR' F'R   UF   FL FL'   U'F	and
I12-2b	F'R' F'R   UF   FL FL'   U'F'.

To see that this rule is necessary, it need only be observed that
inversion, rotation, reflection, and shifting all preserve the
number of clockwise/counterclockwise sign changes between
cyclically adjacent elements.

	In what sense are the ``trivial'' identities trivial?  I
have come to believe that they are trivial only because they are
short and simple enough that they are well-understood.  The only
identities for which I can find any theoretical reasons for calling
trivial are the identities of the form XX'=I.  In spite of the
simplicity of the ``trivial'' identities, their occurrence is one
of the major reasons that Alan Bawden and I were unable to show
earlier that I12-1-3 generated all identities of length 12.  I fear
that the combination of 4-qtw and 12-qtw identities may turn out to
be a major headache when dealing with identities of length 14 and
16.



Date: 14 August 1981 03:12-EDT
From: Chris C. Worrell  <ZILCH at MIT-MC>
Subject: CUBE-POLL@MIT-MC
To: CUBE-LOVERS at MIT-MC

The first CUBE-LOVERS poll has now begun. Send your answers to:

CUBE-POLL@MIT-MC


The questions:

1.   Occupation (if student, undergrad or grad and major)
2.   Age
3.   Average solving time
4.   How long it took you physically working with the cube in order to solve
     it reliably.
5.    How long have you been working with the cube   (say since Jan. '81)
6.   How many cubes do you own, and how many have died on you.
7.   Solving methods:
  A.    Time-efficient,  describe method in general.
  B.    Move-efficient,  describe in more detail, including number of qtws
           for each stage.
           Include a maximum number of qtws for this method.
8.   Transformations you have found or know of.
        Include a description of what the transform does.
        Include transforms you use in your method, such as top to middle slice 
        edge movers, and  transforms which affect only one face.
         The results of this question will be compiled into a dictionary of
         transfors and may also aid people in their investigations of the cube.
9.    Any specific intrests you have in the cube,  such as applications to
       group theory, or investigation of identities or whatever.
10.   Any other intrests you might have, relating or not relating to puzzles.


                                            Chris Worrell (ZILCH@MIT-MC)


Date: 17 August 1981 11:34-EDT
From: Allan C. Wechsler <ACW at MIT-AI>
Subject:  12q relations.
To: CUBE-LOVERS at MIT-AI


    I12-1	FR' F'R   UF' U'F    RU' R'U
    I12-2	FR' F'R   UF' F'L    FL' U'F
    I12-3	FR' F'R   UF' UL'    U'L FU'

I can't believe I'm the first person to notice this:

Suppose we only know I12-1 and I12-2.  Then we have

I12-1'			U'RUR'F'U (FR'F'RUF')'
(I12-1')(I12-2)		U'RUR'F'U (FR'F'RUF')' (FR'F'RUF') F'LFL'U'F
Reduce:			U'RUR'F'U F'LFL'U'F
Conjugating by (U'RUR'F'U)', we get 
			F'LFL' U'FU'R UR'F'U
But this is just the RL mirror image of
			FR'F'L UF'UL' U'LFU'

This is exactly I12-3.  So there are really only two independent 12q
identities, and the third can be deduced from them.


Date: 19 Aug 1981 09:48 PDT
From: Lynn.ES at PARC-MAXC
Subject: Re: 10 sided "cube"
In-reply-to: KATZ's message of 4 Aug 1981 1123-PDT
To: Cube-Lovers at MIT-MC
cc: Lynn.es@PARC

I last week picked up one of the same octagonal cubes (Wonderful Puzzler
brand) as Alan R. Katz in his message.  I might point out that it shares the same
workmanship and twistability (lack thereof) which Katz attributes to their brand
of standard cube.  

After some experimenting, I found that the "parity error" involved is always a
pair of edges in reverse locations on the bottom.  My solution algorithm is top,
equator, bottom [corner locations, edge locations, edge orientations, corner
orientations].  On an ordinary cube, one pair of edges reversed never happens
when you go to position edges.  I suspect that if your algorithm does edge
locations before corners, then two corners in wrong locations would be the parity
error.  Incidentally, the parity of Katz's cube is the reverse of mine, though the
center cubie colors are in the same relation to each other.  

Edge orientation (I assume that is what Katz meant by "edges must be fliped")
parity errors are irrelevant to getting the cut-edge-colors rightly paritied.  Edge
orientation parity errors happen because one of the cut-edges may have been
oriented wrongly.  Or more precisely, an odd number of them reversed from the
way they were virginally (but they look right either direction).  This is easily
cured by any of the "flip a pair of edges" macros.  

The change in cube shape is actually helpful in solving, except for being
difficult to grab sometimes.  Corners that need orientation really stand out.  The
Greenwich meridian edges (assuming you solve it from the top down with the
octagonal faces left and right), being different shape, are instantly located.  

The best part of the Wonderful Puzzler is the instructions, which I quote here:

THE CHALLENGE:
"ORIGINAL PUZZLER" presents not only a unique challenge but offers the
possibility of countless hours of relaxation.  Your mental ingenuity may be tested
for a few hours -- many days -- several weeks -- or even a period of much
longer duration.  If you can determine the key to unlock the knack for solving
the PUZZLER, the final trimph can be the psychological turning point in your
life.  Mathematicians may be tested to the limit and cry over this one -- and you
may, too!  You will gain a measure of satisfaction when you align one plane. 
You will be delighted with the completion of two.  You will be elated with the
completion of three or four!  The completion of the fifth plane will quicken your
pulse!! -- and you will have scaled the peak once the last unit of the sixth plane
falls into place!!!!

PREPARATION & CAUTION:
*Spin the PUZZLER several times, as indicated on the cover, until all color units
are randomly distributed on each of six planes.
*Do NOT remove any color unit in the process of play.
*Initially, activate the random distribution GENTLY.  With little use, this can be
accomplished easily and smoothly.
Patience and persistence will beat the Puzzler!  Good Luck!

Challenge
Can you contend with more than 18'000'000'000 combination to reach a solution?
*end of quote*

I think the errors remaining above are all theirs.  It is full of little gems: turning
point in your life, solving five planes (on the standard cube, which I believe
has the same instructions), 18 billion (better than Ideal's guess).  The GENTLY
caution is valid; a neighbor kid blew mine apart, including a center cubie, by
ignoring this.


Date: 20 August 1981 02:46-EDT
From: Alan Bawden <ALAN at MIT-MC>
Subject: Resending to the right place so that I can digest it later...
To: CUBE-LOVERS at MIT-MC

Date: 19 August 1981 23:22-EDT
From: Eric L. Flanzbaum <ELF at MIT-MC>
Subject:  A small version if the cube ...
To: CUBE-LOVERS-REQUEST at MIT-MC
cc: ELF at MIT-MC


  I don't know if this has been mentioned yet, but why
not mention it again?  I just saw Rubik's cube being sold
on a keychain.  It is about 1 1/2 inches and works/looks
like the real thing.  They sell for about $2.50 which
is considerably less than the standard version.  I live
out on the west coast, so I don't know if they sell
it back east.

			Happy solving,
				  -Eric.

Date: 22 Aug 1981 1404-EDT
From: ROBG at MIT-DMS (Rob F. Griffiths)
To: Cube-Lovers at MIT-MC
Subject: Re: ELF's Small cube
Message-id: <[MIT-DMS].207641>

  I have seen that one, and also one that is absolutely tiny.  It
is the size of one of the cubies on the full sized Rubik's, and
is fully operational.  They are really quite flexible and well
made, I don't know who manufactures them, but I will try to find
out.
			-Rob.


Date: 22 August 1981 18:46-EDT
From: Chris C. Worrell  <ZILCH at MIT-MC>
Subject: CUBE-POLL``MIT-MC
To: CUBE-LOVERS at MIT-MC

SO FAR I HAVE RECEIVE ONLY ABOUT 8 REPLIES, THIS IS HARDLY ENOUGH
TO DO STUDIES ON. PLEASE SEND MORE REPLIES. IF YOU NO LONGER
HAVE THE QUESTIONNAIRE SEND ME MAIL AND I WILL FORWARD IT TO YOU.

Date: 24 Aug 1981 1032-PDT
From: ISAACS at SRI-KL
Subject: recoloring the 10-sided cube
To: cube-lovers at MIT-MC

  A nice way to recolor the 10 sided cube is to give a side a quarter twist,
so it looks sort of like a baseball, and then make the (fairly) obvious
4 groups of 9 faces each in four colors, and two in-between stripes of 3
facies each.  Each of the 9-facie groups will have 2 triangular facies, and
3 slant rectangular facies, and 4 squares.  I think its a fairly simple
variation to solve, but I just made it last night and have not worked with
it much yet.
  By the way, I just got my first magic tetrahedron.  This one came from
Japan, but says made in Hongkong by "World-wide"(?) copyright by Meffert.
Anyone know who he is?  It seems very similar to the one invented by
Kristen Meier.
-------

Date: 25 Aug 1981 1016-PDT
From: ISAACS at SRI-KL
Subject: BAY AREA CUBE CONTEST
To: cube-lovers at MIT-MC

Date: 24 Aug 1981 1612-PDT
From: ISAACS at SRI-KL
Subject: CUBE CONTEST-BAY AREA
To: cube-lovers at MIT-MC

  There will be a Rubics Cube contest at Games and Things, at the
Stanford Shopping Center this Saturday, Aug. 29, from 10:00 am to
4:00 pm.  Speed contests, money prizes, a cube display (by me), etc.
Contestants are supposed to register at GAMES AND THINGS before 10:00
on Saturday.  Address is 128 Stanford Shopping Center, Palo Alto, Ca.,
(415) 328-4331.
    --- Stan Isaacs
-------

-------

Date: 25 Aug 1981 1019-PDT
From: ISAACS at SRI-KL
Subject: MORE TERMINOLOGY
To: CUBE-LOVERS at MIT-MC

  More possible terms (from a new cube-solver who was a biology major):
Dorsal/Ventral for front/back, Port/Starboard for right/left (left/right?).
He doesn't have a consistant term for up/down.
-------

Date: 25 August 1981 19:04-EDT
From: Alan Bawden <ALAN at MIT-MC>
Subject: Speed cubing
To: CUBE-LOVERS at MIT-MC

Does anyone have any idea what the world record for speed cube solving
really is?  The only times I can find are:

1) In the Scientific American article Hofstadter mentions an
Englishman named Nicholas Hammond who averages "down to close to 30
seconds".  Anyone have any more information on this?  Like where
Hofstadter found out about this guy?

2) In the reports about the "Regional Cubing Championship" held here
in July the best time listed is 48.31 seconds, held by a 10 year old
named Jonathan Cheyer.  (See PDL's message to Cube-Lovers dated 27 Jul
1981.)

3) I seem to remember reports that there were Hungarians who averaged
around 50 seconds.  I thought I had read this in Singmaster, but I can't seem
to find it there.

4) The best time anyone will admit to on this list (as determined by
scanning the replys to ZILCH's poll) is 2 minutes.  This time is
claimed by both Richard Pavelle (RP@MIT-MC) and Alan Katz (KATZ@ISIF).
Also Stan Isaacs (ISAACS@SRI-KL) claims that his two children take
about 1 1/4 minutes using essentially his methods.

5) Finally, Dan Pehoushek (JDP@SU-AI) tells me that a friend of his
frequently breaks the 30 second barrier.  I should have thought to ask
for his name.

Anybody know of any more good speed cubists?

Date: 25 Aug 1981 2116-EDT
From: ELF at MIT-DMS (Eric L. Flanzbaum)
To: Cube-Lovers at MIT-MC
Subject: Speed Solving
Message-id: <[MIT-DMS].208572>



	Hi Cube fans,
I have a couple of friends at school (oh about 4 or 5) who consistently
solve the cube (actually, they have races/contests) at a time of under
30-40 seconds.  I don't know if this is really unusually fast, but as
from ALAN's previous message, it looks like they all rank in there.
By the way, these people are entering the 9th and 10th grade in the 
fall.

				Happy Solving,
					-Eric L. Flanzbaum
					 ELF at MIT-AI 


Date: 27 August 1981 18:39-EDT
From: Dennis L. Doughty <DUFTY at MIT-MC>
Subject:  Speed cubing
To: CUBE-LOVERS at MIT-MC
cc: DUFTY at MIT-MC


My fastest time for the cube is 1 minute 17 seconds (everything worked out
correctly).  My average time is 1:40-1:45.  --Dennis

p.s. i'll answer the poll when I get the time.

Date: 28 August 1981 12:59-EDT
From: Robert H. Berman <RHB at MIT-MC>
To: CUBE-LOVERS at MIT-MC


You may be interested to see a cartoon about the cube on page 36 on
the August 31 issues of the New Yorker. Yes, the New Yorker.

--rhb

Date: 29 August 1981 07:54-EDT
From: Thomas L. Davenport <TLD at MIT-MC>
Subject:  Cube Song
To: CUBE-LOVERS at MIT-MC

This past week PBS broadcast the latest "Mark Russell Comedy Special"
and in it he did a funny song about the cube.  He even had one with
him on stage.
-Tom-

Date: 29 Aug 1981 1906-EDT
From: ROBG at MIT-DMS (Rob F. Griffiths)
To: cube-lovers at MIT-MC
Subject: English Whiz Kid
Message-id: <[MIT-DMS].208873>

          	From TIME:  August 31,1981:
		
		---------------------------

  Along with diet books, cat books, and  advisories on how to make
a profit from  the  coming apocalypse, there  is a  growing  shelf
concerned  solely  with  mastering  that  infuriating, six  sided,
27-part boggler with 42.3 quintillion possible  combinations known
as  Rubik's  Cube..  The  latest  entry: ''You  Can  Do The Cube''
(Penguin, $ 1.95) by  Patrick  Bossert, 13, a London schoolboy who
discovered the cube only  this spring during a family ski vacation
in  Switzerland.  Within  five days he had  mastered the  monster,
and  later began selling his schoolmates a four-page, mimeographed
tip sheet for 45 cents.  An alert editor at Penguin saw a copy and
persuaded the prodigy to turn pro.  The 112  page result  contains
3 dozen 'tricks'  for solving  the  cube (using  logic rather than
math), as well as  a chapter  on 'Cube  Maintenance'  (to loosen a
stiff cube, ''put  a blob  of Vaseline on  the mechanism'').  With
250,000 copies of  the cubist's book in print, a Penguin executive
marvels:  ''It's the biggest,  runaway, immediate  success we have
had  since we published 'Lady Chatterley's Lover' in  paperback.''

		---------------------------


			-Rob.


Date:  1 Sep 1981 1442-EDT
From: Bob Clements <CLEMENTS@BBNA>
Sender: CLEMENTS at BBNA
Subject: [Bob Clements <CLEMENTS@BBNA>: Rubik's cube sale]
To:   Cube-lovers at MC

I didn't know of the cube-lovers list, but it was suggested to me that
I re-send this msg to cube-lovers. If this sale has already been
mentioned on this list, sorry for the repeat.
/Rcc

		---------------
Mail-from: MIT-AI
Received-Date: 1-Sep-81 1343-EDT
Date:  1 Sep 1981 1221-EDT
From: Bob Clements <CLEMENTS@BBNA>
Sender: CLEMENTS at BBNA
Subject: Rubik's cube sale
To:   Info-Micro at ai

I don't want to start a whole Rubik's discussion, but for those
in the Boston area who need to replace their worn Cubes, or get
their first one, the Caldor chain has them on special for $4.66
thru Saturday.
/Rcc
-------


		---------------
-------

Date: 3 September 1981 15:40-EDT
From: Dennis L. Doughty <DUFTY at MIT-MC>
Subject:  Practical use of the Rubik's cube/Speed cubing
To: CUBE-LOVERS at MIT-MC
cc: DUFTY at MIT-MC


This rush week, my fraternity extended a bid to a freshman by the name of
Larry Singer who generally solves the cube in 1:13 or thereabouts (my
average time is 1:40).  All day Monday, our bids were playing such games as
"Pledge Pong" or "Pledge Pool."  The idea here is that the bid challenges
an active to a game of pool or ping-pong, and if he loses, he pledges.
Well, Monday night, Larry challenged me to "Pledge Cube."  One of the
brothers uniformly scrambled two cubes, and we were to compete in a
head-to-head competition.  Well, we both were under considerable pressure,
naturally, and we both made several mistakes while solving the cube.  I
won, but my winning time was 2:09.  Larry and his close friend then both
pledged.

So now, no one in the house can tell me that there's no practical use for
the Rubik's cube.

--Dennis

Date:  5 September 1981 1446-EDT (Saturday)
From: Bob.Walker at CMU-10A
To: cube-lovers at mit-mc
Subject:  Snake: **SPOILER**
Message-Id: <05Sep81 144605 BW80@CMU-10A>

I am new to the list, and in reading the archives, I found that
the transform for my favorite pattern was listed incorrectly.
Specifically, I refer to the Don Woods' message of 6 January which
listed transforms for the Snake, Worm, and Baseball (I think).
Anyway, the transform listed for the Snake was incorrect.  What is
listed IS a pretty pattern, merely the wrong one.  The blurb about
how to "hack" your way to the Snake, however, is correct.  The
proper transform to achieve the snake (from Singmaster) is:
 
            * * * S P O I L E R * * *
 
Snake:     B R L' D' R R D R' L B' R R + U B B U' D R R D'

Date:  5 September 1981 01:08 edt
From:  Greenberg.Symbolics at MIT-Multics
Subject:  TV special

Friday night at 8 NBC Magazine will cover Rubik's cube.  The coverage
will doubtless include footage of the July 25 Boston Area Cubeathon,
at which many MIT-Area cubists were present.

Date:  8 Sep 1981 0942-PDT
From: ISAACS at SRI-KL
Subject: misc
To: cube-lovers at MIT-MC

   Some short notes on various cube stuff - will try to expand on some of it
later:
1. The Stanford Shopping Center/Games and Things cube contest of a couple of
   saturdays ago:  was won by Paul cunningham, 16 yrs old. There were about
   40 entries; there were something like 8 rounds to get to a winner, double
   elimination, each pair fighting it out with the best average of 3 cubes,
   all cubes in a round scrambled exactly the same way.  Best average-of-three
   time was 56 seconds.  Best time was 41 seconds. David Tabuchi, the Games
   & Things speedster, has a best average-of-ten speed of 43 seconds!  Brian
   Robinson, with whom he works on the cube, has a best average-of-ten of 41
   seconds.  Davids fastest time was 24.98 seconds!!!
2. Cubes are multiplying like hotcakes.  Not only changes in labeling, but
   also changes in shape.  I have seen or heard reports of about 8 shape
   variations, and multiple size variations, from about 19 mm to about 60 mm.
   And someone told me of a 12mm or so version.  The corners have been cut off
   in 3 different ways (The nicest cuts them halfway through the neighboring
   edge, and uses 14 colors).  The magic Tetrahedron is readily available now.
   There is a build-it-yourself cube kit.  There are still reports of the
   elusive 4x4x4 cube, but no actual sightings as far as I know.  By the
   way, I collect puzzles, and am trying to find many of these cube variations.
   If anyone knows where I can actually buy some, or would get me some, please
   let me know.  I will be happy to pay or trade for them.
3. The snake is not a cube, but it is a toy/puzzle/art object that should a
   appeal to cube-lovers.  It also comes in a variety of colors and sizes.
4. Also lots of new books and such. Such as:
   a. Don Kolve, of Kirkland, Wash.  He solves Top-middle-bottom(position
      corners,twist corners, flip edges, position edges).
   b. L.E.Hordern, of England.  He does: bottom-middle-top ( position corners,
      orient corners, orient edges, position edges).
   c. Bridget Last, of Downham, England.  She solves: Define face colors 
      ("The easiest way of deciding which face is to be which colour is to
      define the centre faces as being correct.")  Then: position all corners,
      orient all corners, position (with orientation) all edges.
   d. Bob Easter, a friend in San Francisco, uses just one move to do
      everything. The move is  F R' F' R (an old friend). First he "walks" 
      the edge cubes to their proper corners, then rotates them into proper
      order around the corner, then flips them 2 at a time to get them aligned.
      Then he does similarly with the corners, using the same move, but 
      done 3 times to leave the edges alone.  Lots of quarter twists, but 
      little memory.  Perhaps the ultimate solution for people with strong
      wrists.
   e. Patrick Bussart, Puffin books (a division of Penguin Books).  Patrick
      has been in the news since he is only 12 or 13 years old; his solution
      is very popular in England. He does top corners, top edges, bottom
      corners, position rest of the edges, rotate rest of the edges.
5. I have been re-labeling cubes to make new puzzles, and would appreciate
   suggestions.  I have made tactile cubes of various types (that is, with
   various materials).  I'm trying to make one with 5 or 6 grades of sandpaper,
   but find my touch cannot distinguish between the two middle grades. I
   have made a magic square cube (each face is a 3x3 magic square; the
   problem is to decide what the relative orientations and forms of the squares
   should be; any suggestions?) and a magic cube cube (the center 14 cube is,
   of course, invisible).  And 2 word cubes - one has word squares (there are
   three 3 letter words in each direction), and the other, six 3 word sentences
   (FIX THE BOX, YES YOU CAN, FUN FOR ALL, etc.  I haven't had these long
   enough to know if they are "solvable" without trial and error; I think
   that the magic square cube, especially, is difficult unless you have
   the order in advance.  So far, I have made 2 of them, the first had one
   square marked, which makes it fairly easy (the other 5 squares are forced
   by the first); in the second, I made sure each corner was unique, and the
   edges as different as possible (2 have to be the same); but I haven't had
   time to try it yet.
Enough - this message is too long.
-------

Date: 10 Sep 1981 1447-PDT
From: ISAACS at SRI-KL
Subject: cube query
To: cube-lovers at MIT-MC

   The 2x2x2 cube is solved in the corner sub-group, ignoring the (non-existent)
edges.  The so-called "Dinman Style" cube (probably meant to be "diamond")
has the corners cut off and everything stretched to make a somewhat distorted
rhombicuboctahedron (the six center facies are square; the corners are now
triangles, and the old edges are rectangles).  Solving this involves only
positioning moves - all orientation (twisting) is invisible.  Thus these
two cubes involve two "pure" subgroups.  Can anyone design (by either cutting,
recoloring, or even inventing new mechanisms) cubes or pseudo-cubes which
only involve edge-type moves, or which only involve twisting, with
positioning ignored? 
   --- Stan Isaacs
-------

Date: 11 Sep 1981 0917-PDT
From: ISAACS at SRI-KL
Subject: half query answer
To: cube-lovers at MIT-MC

  First of all, I was inaccurate about the "Dinman" style "cube" - it doesn't
only eliminate twisting, but also some positioning (ie, after the top and middle
are solved, the bottom is automatically solved).  Perhaps by the judicious
adding of numbers to the facies, the pure position cube can be made.  Also,
the answer to the edge-only subgroup is easy - just remove all the labels from
the corners, so they are all identically monochromatic.  Is there a more
elegant solution?
   --- Stan
-------

GENTRY@MIT-AI 09/11/81 20:43:40
To: CUBE-LOVERS at MIT-MC
It appears that due to time restrictions, the segment about the RUBIK's
cube on NBC Magazine has been postponed until next week. Check the listings
for your area to see when it will be televised.


Date: 15 Sep 1981 1553-PDT
From: ISAACS at SRI-KL
Subject: lower bounds
To: Hoey at CMU-10A
cc: cube-lovers at MIT-MC

  [This message is being sent to Dan Hoey, and refers to his message of
9-Jan-81, subject: The Supergroup -- Part 2: at least 25 qtw and why]
   Appended to this message is a longish message I recieved, which has
some good ideas to use.  In particular, what about using your technique
on a 2x2x2 cube, or an (idealized) edge-only cube?  And then comparing
it with his clculations for the 2x2x2.
   I'm not sure without a 2x2x2 in front of me, but I think there are
only 2 distinct 1 qtw per set of opposite faces, and only one 2qtw move.
And that the period is only 2.  Is that true?  However, there should be
more low-number-of-twists identities.
   I'm distrustful of the actual calculations in the message below, because
I don't see the 9 new configurations after only 1 twist.  I think there are
only 6. Or am I missing something?
  Also, Dan or someone else on the cube-lovers network: how about compiling
all the messages about lower bounds and identities (after a while) into 
one file we can ftp and look at all together.
11-Sep-81 12:26:52-PDT,6785;000000000001
Mail-from: ARPAnet host BERKELEY rcvd at 11-Sep-81 1223-PDT
Date: 11 Sep 1981 11:43:07-PDT
From: ARPAVAX.sjk at Berkeley
To: isaacs@sri-kl
Subject: in case you haven't seen this ...

Article 16:
>From csvax:mhtsa!harpo!chico!esquire!psl Wed Sep  9 17:16:32 1981
Subject: Rubik's Cube
Newsgroups: net.games

Want to knoe how far away you can get from the solution on a Rubik's Cube?


   A Simple Lower Bound

As everybody knows, the number of discrete configurations of the 3x3x3
Rubik's Cube is:

(8! * 12! * 3^8 * 2^12) / 12 = 4x10^19 = 43,252,003,274,489,856,000

One approach to a lower bound is to calculate the maximum possible number
of configurations you can reach with a particular number of moves and then
see how many moves you would have to make to reach the number above.

With no moves at all you get 1, the starting position.  The first move gets
you 18, (any one of six faces turned one of three ways).  The next move
gets you 18*15, (no point in turning the same face twice in a row), for a
total of 1+18+270 configurations reached after two moves.  A table of these
values looks like:

	      ---------possible configurations---------
moves                 new   % max                 total
 0                      1   0.0%                      1
 1                     18   0.0%                     19
 2                    270   0.0%                    289
 3                   4050   0.0%                   4339
 4                  60750   0.0%                  65089
 5                 911250   0.0%                 976339
 6               13668750   0.0%               14645089
 7              205031250   0.0%              219676339
 8             3075468750   0.0%             3295145089
 9            46132031250   0.0%            49427176339
10           691980468750   0.0%           741407645089 Notice that
11         10379707031250   0.0%         11121114676339 not until 17
12        155695605468750   0.0%        166816720145089 moves has the
13       2335434082031250   0.0%       2502250802176339 total number
14      35031511230468750   0.1%      37533762032645089 of possible
15     525472668457031250   1.2%     563006430489676339 configurations
16    7882090026855468750  18.2%    8445096457345145089 exceeded the
17  118231350402832031250 273.4%  126676446860177176339 maximum.

So there is no possible way to reach some configurations in fewer than
17 moves.  However, this analysis has assumed that each configuration
generated was a NEW one, but there are MANY cases where this will not
be so.  A simple example is turning one face 180 degrees, the opposite
face 180 degrees, and then repeating those two moves -- four moves that
get us to an old, familiar configuration.  If we factor out the sequences
that involve these opposite face identities the minimum number of moves
becomes 18.  Needless to say there are still lots of useless move sequences,
but detecting them becomes a lot trickier.


   A Rumored Upper Bound

Rumor has it that a computer program exists, (attributed to Thistlethwaite),
that provably will solve any Cube configuration in at most 41 moves.


   Narrowing it Down

So the answer is somewhere between 18 and 41.  How do you get further?  One
way is to write a computer program that tries every sequence of moves until
it has generated every possible configuration at least once.  That sounds
easy, and it is, but such a program would take a \\\L O N G/// time to run.
However, if we limit the problem a little by considering a Cube that is two
squares on a side (2x2x2), we have a chance of learning something.


   2x2x2 Cube

By the same considerations stated above we can get a lower bound for the
2x2x2 Cube.  There are 7! * 3^6 = 3,674,160 configurations and, since we
can limit ourselves to moving only three "orthogonal" sides of the 2x2x2
cube, on the n-th move you could reach 9 * 6^(n-1) new configurations thus
we find that with 8 moves you could reach at most 3,023,307 and with 9 you
could reach at most 18,139,851.  (Note that this doesn't have the problem with
opposite side moves that the 3x3x3 cube has.)

Because the 2x2x2 cube is relatively simple we can actually run a program
to try all the possible move sequences and compare our bound with fact.
Listed below are the findings


	 ------new configurations-------    total configurations
moves    -----possible----  ---actual---    ---possible --actual
	   number      %    number   %          number    number
 0              1     0.0%       1  0.0%             1         1
 1              9     0.0%       9  0.0%             9        10
 2             54     0.0%      54  0.0%            63        64
 3            324     0.0%     321  0.0%           387       385
 4           1944     0.0%    1847  0.0%          2331      2232
 5          11664     0.3%    9992  0.3%         13995     12224
 6          69984     1.9%   50136  1.4%         83979     62360
 7         419904    11.4%  227536  6.2%        503883    289896
 8        2519424    68.6%  870072 23.7%       3023307   1159968
 9       15116544   411.4% 1887748 51.4%      18139851   3047716
10       90699264  2468.6%  623800 17.0%     108839115   3671516
11      544195584 14811.4%    2644  0.071%   653034699   3674160

Interestingly enough there are 2,644 configurations that require eleven
moves to reach a solution; this is less than one tenth of one percent of
the total configurations!

It's also interesting that it's better than a 50-50 bet that a randomly
ordered 2x2x2 cube can be solved in exactly nine moves, (it's not clear
how to turn this into a profitable bar bet, however).

Noticing that there are only 321 new configurations after three moves
instead of 324 leads us to guess that there are six non-trivial sequences
of six moves that end with the original configuration, (why?).

These results came from a C program running on a VAX 11/780 and even
though the 2x2x2 cube is simple compared to the 3x3x3 it took a lot of
time.  The figures for 11 moves took over 51 hours of cpu time.

If you'd like to make a 2x2x2 cube with which to experiment you can simply
take all the little labels off a 3x3x3 cube except the ones on the corners
and then ignore the unlabeled cubes.

Here's one sequence that gets you to one of the 2,644 configurations:
    f r f r f d2 f d- f d2 r2    f = rotate front face 90 degrees
				 r = rotate right face 90 degrees
				d2 = rotate "down" face 180 degrees
				d- = rotate "down" face 270 degrees


   So Where's That Leave Us?

I just thought of a dandy way to get the answer for the 3x3x3 cube, but
the margins on this news item are a little too small for me to include it ...

-------

Date: 15 September 1981 21:54-EDT
From: Alan Bawden <ALAN at MIT-MC>
Subject: Editor's note to the last message.
To: CUBE-LOVERS at MIT-MC

I will look into collecting all of the relevant messages on God's
number into one place.  If you want to poke around in the archives
yourself (please be carefull, and don't delete them again) I will
remind you all that old cube-lovers mail is archived in the following
places:

MC:ALAN;CUBE MAIL0	;oldest mail in foward order
MC:ALAN;CUBE MAIL1	;next oldest mail in foward order
MC:ALAN;CUBE MAIL2	;more of same
MC:ALAN;CUBE MAIL	;recent mail in reverse order

(I someone else wants to attempt the compilation, there is a better
chance it will get done.  Let me know and I will be happy to lend a
hand.)

Some of the seeming inconsistencies in the message included by Isaacs
in his message are a result of the usual half versus quarter twist
screw.  The reason the writer can see 9 configurations after a single twist
is because he has a different definition of a "single twist".

I also am not sure, but I also think that the counting argument given
here suffers from the some confusion Singmaster had when he computed a
lower bound of 17 htw.  I think, in fact, that a lower bound of 19 htw
results if the argument is executed correctly (Singmaster corrected
himself about this by the fourth edition, I think).  Someone with a
copy of Singmaster handy should look this up.

The 41 move count for Thistlethwaite's algorithm is probably a half
twist count given that it was reported by Singmaster.

Date: 16 September 1981 0003-EDT (Wednesday)
From: Dan Hoey at CMU-10A
To: ISAACS at SRI-KL, Cube-Lovers at MIT-MC
Subject:  Re: lower bounds
In-Reply-To:  Stan Isaacs's message of 15 Sep 81 17:53-EST and Alan Bawden's
             message of 15 Sep 81 20:55-EST
Message-Id: <16Sep81 000353 DH51@CMU-10A>

	Hi.  I'm really pressed for time, but I'll drop a couple of
comments.

	Alan pretty well said it--there are half-twisters and there
are quarter-twisters and the included message is one of the former.
I strongly favor the latter, since then all the moves are
equivalent, (M-conjugate, to you archive-readers).  But Singmaster's
book, though in the other camp, is too good to ignore.

	To extend the argument I gave on 9 January to the case
where quarter-twists and half-twists are counted equally (we call
such a move a `htw' whether it is quarter or half) let PH[n] be the
number of (3x3x3-cube) positions at exactly n htw from SOLVED.  Then

	PH[0]  = 1
	PH[1] <= 6*3*PH[0]
	PH[2] <= 6*2*PH[1]   + 9*3*PH[0]
	PH[n] <= 6*2*PH[n-1] + 9*2*PH[n-2] for n > 2.

Solving yields the following upper bounds:

htw        new        total      htw        new         total
 0           1            1       10    2.447*10^11   2.646*10^11
 1          18           19       11    3.267*10^12   3.531*10^12
 2         243          262       12    4.360*10^13   4.713*10^13
 3        3240         3502       13    5.820*10^14   6.292*10^14
 4       43254        46756       14    7.769*10^15   8.398*10^15
 5      577368       624124       15    1.037*10^17   1.121*10^17
 6     7706988      8331112       16    1.385*10^18   1.497*10^18
 7   102876480    111207592       17    1.848*10^19   1.998*10^19
 8  1373243544   1484451136       18    2.467*10^20   2.667*10^20
 9 18330699168  19815150304

	At least 18 htw are required to reach all the 4.325*10^19
positions of the cube.  This is the same argument that was used in
Singmaster's fifth edition, p. 34, and is the best I know.  Lest ye
be tempted to pull the trick I did in the January message, remember
that half-twists are even permutations, so there is no assurance
that half the positions are an odd distance from SOLVED.  This is
illustrated in the 2x2x2 case, where more than half of the
positions are at a particular odd distance.

	And yes, all of Thistlethwaite's analysis seems to use the
half-twist metric.  I am quite surprised, however, to hear the rumor
of 41 htw.  As of Singmaster's fifth edition, the figure was 52 htw
``... but he hopes to get it down to 50 with a bit more computing
and he believes it may be reducible to 45 with a lot of
searching.''  If anyone has harder information on the situation, I
would like to hear it.

	Well, back to real work.  I saw a Rubikized tetrahedron in a
shop window earlier this evening; I'm not sure whether I'm relieved
or infuriated that the store was closed for the day.

Date: 21 Sep 1981 08:49 PDT
From: Eldridge.ES at PARC-MAXC
Subject: An incomplete solution
To: Cube-Lovers@MC
cc: XeroxCubeLovers^.pa
Reply-To: Eldridge

For months now I have been living in blissful ignorance thinking that I too
could solve "the cube".  To my horror and dismay I have found that my solution
is not complete.  There are some cubes on the market that have pictures of fruit
on the faces rather than solid colors.  I wondered if the solution I use would get
the pictures on the faces all lined up in the proper direction.  I found that it
didn't!  The problem is that some of the center cubies do not line up in the same
direction as all the other cubies on the face.  I am currently working on finding
some macros that rotate the center cubies without affecting the rest of the cube. 
I would suggest that you might find one of these fruit cubes and try it.  Or you
can do as I did and mark the faces of an original cube so that you can tell the
orientation of the cubies.  Good Luck!

George


Date: 21 September 1981 15:50-EDT
From: Richard Pavelle <RP at MIT-MC>
Subject: 4x4x4
To: CUBE-LOVERS at MIT-MC

A person in sales at Ideal Toy Corp told me today that they will begin 
selling the elusive 4x4x4 next May or June. The die casts are being made 
now. They will be showing it at Toyfair (a trade show?). 

Date: 22 Sep 1981 0552-EDT
From: ZILCH at MIT-DMS (Chris C. Worrell )
To: CUBE-LOVERS at MIT-MC
Subject: Article on Erno Rubik
Message-id: <[MIT-DMS].210992>

Yesterday in a store I noticed that this week's
PEOPLE magazine has an article on
"Erno Rubik: The Inventor of that !#&%$*@ Cube"
(this title is approximatly right)
The article is on page 30 (40?) and runs for a total of
4 pages, however I didn't read it so I don't know what it's 
content is.



Date: 23 September 1981 0005-EDT (Wednesday)
From: Guy.Steele at CMU-10A
To: cube-lovers at MIT-MC
Subject:  The Cube in the Comics
Message-Id: <23Sep81 000559 GS70@CMU-10A>

The "Ferd'nand" comic strip for 9/22/81 features the cube.
(For those who don't see this strip: it's pure mime, with no word balloons.
Panel 1: Ferd'nand sits in an armchair, struggling with a (scrambled) cube.
Panel 2: In disgust he throws it out of a window.
Panel 3: The cube sits on the lawn.  Ferd'nand turns away, but still is
	 glaring over his shoulder at the cube.
Panel 4: Fred'nand is sitting on the grass where the cube had landed,
	 struggling away again.
Know the feeling?)
--Guy

Date: 23 Sep 1981 1554-PDT
From: HOROWITZ at USC-ISIF
Subject: question for cube-lovers
To:   cube-lovers at MIT-AI

Minh Tai, a high school senior in the Los Angeles area who can do
the cube in average 50 seconds  proposes the following problem:

Is it possible to make every face have either a Z or an S
(two colors, one forming the Z pentomino)?

He signs himself with a quick routine for six T's as follows:
(U2 R' L)2 (B2 D2)3


   ------!----!-------
   ! x   ! x  !      !
   !                 !
   ------!----!-------
   !     ! x  !      !
   !                 !
   ------!----!------
   !                 !
   !        x      x !
   ------!-----!------
-------


Date: 23 September 1981 2236-EDT (Wednesday)
From: Dan Hoey at CMU-10A
To: Cube-Lovers at MIT-MC
Subject:  Re: question for cube-lovers (S and T patterns)
CC: HOROWITZ at USC-ISIF
In-Reply-To:  Alan Bawden's message of 23 Sep 81 19:31-EST
Message-Id: <23Sep81 223642 DH51@CMU-10A>

	It is impossible to put Z patterns on all six faces of the
cube, just as it is impossible to extend the Laughter (or Zig-Zag)
pattern to six faces.  The problem is with the corners.  If every
face is to have a pair of opposite corners that agree with the face
center and a pair of opposite corners that agree with each other
but not with the center, then the only constructible pattern is
like (i.e. M-conjugate to) the following:

		F - U
		- U -
		U - F

	D - L	F - R	U - R	B - L
	- L -	- F -	- R -	- B -
	L - D	R - F	R - U	L - B

		B - D
		- D -
		D - B

But this pattern has incorrect corner orientation, and so is not
achievable.

	The T patterns were introduced to this list by David C.
Plummer [3 September 1980 2123-EDT], who assigns credit to Tanya
Sienko for the idea.  Jim Saxe and I [27 January 1981 0102-EST] gave
a process for Tanya's T, the pattern that Minh Tai uses to sign
with.  Our process, (FF UU)^3 (UU LR')^2, is four quarter-twists
shorter than Tai's ``quick routine'' because of the cancellation in
the middle.  The other T pattern, Plummer's T, can be achieved in 28
qtw:  FF UD' F'B' RR F'B U'D RL FF RL' UD' RL FF R'L U'D'.

Date: 24 Sep 1981 0835-PDT
From: Chris Van Wyk <CSD.CVW at SU-SCORE>
Subject: Soviet view of the cube
To: cube-lovers at MIT-MC

September 1981 Atlas World Press Review reports

	Americans may call the Rubik Cube a fad,
but to Izvestia's Melor Sturua it is a "new
psychosis."  He reports [July 12] the interpretations
of unidentified sociologists:  "Some assert that
the Rubik Cube reflects the philosophy of the Reagan
Administration--to build and destroy aimlessly
in a futile search for a solution to the world
situation.  Others see Americans' efforts to solve
the puzzle as the desire to escape from a disordered
life.

-------

Date: 25 September 1981 03:11-EDT
From: Chris C. Worrell  <ZILCH at MIT-MC>
Subject: Poll results
To: CUBE-LOVERS at MIT-MC

Here are the results of the CUBE-POLL.

DATE      AVG.   WORK    WHEN  #   #
REPLY AGE SOLVE  TIME    GOT   OWN DIED   OCCUPATION
          (MINS)
------------------------------------------------------------
8/28  15  --     --      4/81  1   0      STUDENT
9/7   16  --     --      9/81  1   0      STUDENT
9/8   18  3-3:15 1.5-2m  9/80  2   .75    STUDENT
9/24  19  1:40   <10h    7/80  0   0      UG-ENG
8/17  20  4      1w      12/79 1   0      UG-CS/EE
8/29  20  1:50   1w      7/81  2   0      UG
8/25  22  1:30   ?       6/80  7   0      COMP. ANALYST
8/19  22  10 unreliably  7/81  2   0      HACKER
8/15  24  4      2d      1/80  1   0      PROGRAMMER/GRAD-CS
8/17  25  2      40h     6/80  4   0      PROGRAMMER/GRAD-PHYS
8/17  27  4      2-3w late78/79 2  0      HACKER/RECENT G.-CS
8/15  28  10-15  30h since start 2 0      PROGRAMMER
8/17  30  10     ?       5/80  1   0      TEACH/PROGRAMMER
9/11  30  5-10   50-60h  12/80 1   0      TEACHER-CS
8/15  31  2:15   8d      2/80  4   0      HACKER
8/18  31  5      2w      5/80  2   0      COMP. SCIENTIST
8/24  38  2      100h    5/80  4   3      PROFFESSOR
8/31  39  3      3d    fall80  1   0      ROBOTICS RESEARCH
8/24  41  3      ?    early79  24  0      COMP. ANALYST
(the 24 cubes owned is mostly non-standard ones)

DEAD CUBES:
Many people replied that none of their cubes have died on them
though some admit that some of their cubes are in pretty bad
shape, or they have gotten additional cubes to keep their old
ones from dying.
One person reported that he has a cube which still works, even
though it got left in a hot car, and got partially melted
so is now some what surrealistically distorted.
(I wonder what the temperature was)


METHODS (INDEXED BY AGE(DATE-REPLIED)):
18.    CORNERS,"MIDDLE-RING",TOP-SIDES,BOTTOM(MAY MESS UP
       TOP),?
19.    TIME-EFFICIENT:
        U-EDGES,U-CORNERS,MIDDLE-SIDES,D-EDGES-ORIENTED,
        D-EDGES-POSITIONED,D-CORNERS-POSITIONED,
        D-CORNERS-ORIENTED
       MOVE-EFFICIENT
        U-EDGES   20
        U-CORNERS 28
        MIDDLE-SIDES    46 (primary target for improvement)
        D-SIDES   14
        D-CORNERS 20
           TOTAL=128qtw
20.(8/17)
       2 METHODS
        BOTTOM,MIDDLE,TOP SLICES
        4-CORNERS,8-CORNERS,BOTTOM,TOP,MIDDLE
20.(8/29)
       SOLVE GREEN FACE,OPPOSITE CORNERSIN CORRECT ORIENTATION
       OPPOSITE CORNERS IN CORRECT POSITIONS,OPPOSITE SIDES IN

       CORRECT POSITIONS,MIDDLE LAYER
       ABOUT 150qtw MAX.
22.(8/25)
       TOP 4 CORNERS,BOTTOM 4 CORNERS PLACED, ..ORIENTED
       EDGES PUT INTO PLACE IN PAIRS
22.(8/19)
       TOP(EDGES THEN CORNERS),MIDDLE,BOTTOM
24.    MOVE AND TIME EFFICIENT:
        TOP CORNERS  22
        BOTTOM CORNERS POSITION 26
        " " ORIENTATION  24
        TOP EDGES   40
        BOTTOM EDGES   68
        CENTERS 8 (initial posn of corners is by convenience)
        MIDDLE EDGES POSITION 20
        " "  24
         TOTAL=232qtw
25.    TOP,MIDDLE,BOTTOM CORNER ORIENTATION,BOTTOM EDGE
       ORIENTATION,BOTTOM CORNER POSITION,BOTTOM EDGE POSITION
27.    TOP CORNERS,TOP EDGES,3 MIDDLE EDGES(POSITION,POSSIBLY
       ORIENTATION),POSITION OTHER CORNERS,ORIENT OTHER
       CORNERS,POSITION OTHER EDGES,ORIENT OTHER EDGES
28.    BOTTOM LAYER,MIDDLE SIDES,ORIENT TOP SIDES,TOP CORNER
       POSITION, ..ORIENTATION,POSITION TOP SIDES
30.(8/17)
       SOLVE TWO LAYERS SIMULTANEOUSLY, THEN THIRD
30.(9/11)
       UPPER FACE EXCEPT ONE EDGE,LOWER CORNERS POSITIONED,
        ..ORIENTED,LOWER EDGES,LAST UPPER EDGE,MIDDLE
       POSITIONED, ..ORIENTED
31.(8/15)
       TOP CORNERS,TOP EDGES,BOTTOM CORNERS ORIENTED,
        ..POSITIONED,BOTTOM EDGES,MIDDLE POSITIONED,
        ..ORIENTED
31.(8/18)
       TOP CORNERS,TOP EDGES,(TURN CUBE OVER),TOP CORNERS
       POSITIONED, ..ORIENTED,REMAINING EDGES POSITION,
        ..ORIENTED
38.    2-METHODS
        3 CONSECUTIVE LAYERS
        CORNERS FIRST (FOR SPEED)
       ONE OF THESE IS 120qtw
39.    TIME
        CORNERS,BOTTOM,MIDDLE,TOP
       MOVE
        CORNERS,EDGES IN RANDOM ORDER
        AVERAGES 75qtw    <WORST CASES WERE NOT MENTIONED>
41.    2 METHODS
        UPPER EDGES,UPPER CORNERS,(TURN OVER),MIDDLE,
         ORIENT EDGES,ORIENT CORNERS,PLACE CORNERS,PLACE
         EDGES
        UPPER EDGES,3 UPPER CORNERS,3 MIDDLE EDGES
         LAST UPPER CORNER,LAST MIDDLE EDGE,(TURN OVER)
         ORIENT EDGES,ORIENT CORNERS,PLACE CORNERS,PLACE
         EDGES




As far as I can tell no two people who reported in detail
solve the cube in quite the same way.


INTRESTS IN THE CUBE:
 investigation of identities to calculate GOD's number
 application to the teaching of group theory
 training of mind and logic,  something to do
 variations (tetrahedron, 10-sided, interesting faces)
 what different solving methods are
 analogy with quark confinement
 Combinatorial algorithims
 introducing others to the cube
 how people learn to solve cubes (AI)
 pretty patterns,  cube graphics(for fun)
 determining GOD's number
 Showmanship,  flaming about cubes in public
 cubing to attract or repel people
 analytic, as opposed to heuristic or exhaustive aids to
   solve the cube



NON-CUBING INTRESTS OF PEOPLE:
 almost any types of puzzles,  science fiction  
 COMPUTERS!!!, ADVENTURE <Computerized Fantasy Simulation)
 soccer, electronics, Folk dancing, juggling, Chess
 hiking, hacking, biking, photography
 collecting/studying/sometimes even solve puzzles
 Lewis Carroll, archeolgy, linguistics, magic
 recreational math, D&D, wine-tasting
 SEX an DRUGS an ROCK an ROLL
 ANSWERING SILLY QUESTIONS ON QUESTIONAIRES
 guitar, various aspects of Computer Science
 mathematical logic, set theory, topology
 food, the investigation of earth people
 musician, GO
 computer programs for GO/chess/othello/similar 2-player games
 computer assisted or mediated simulation games


RANDOM COMMENTS:
Some people felt that the reason I might not have gotten
many responses was because of the tediousnes and difficulty
of entering a detailed description of their algorithm and
also transformations they know of.
One person suggested that lack of responses may have been
due to the fact that it is summer, and some people would not
have been at school.
This is the results(or most of them anyway) of the poll up to
this point, but I feel that this should by no means be
considered the end of the poll, please mail any new responses
addition to old responses, or updates of old responses to
CUBE-POLL@MIT-MC. If you need to know the questions, the
POLL is in the archives dated 14 AUG.,3:12EDT, or you could
probably get a copy by asking CUBE-MAIL-REQUEST.
Also feel free to enter any new poll questions into the
discussion by announcing them on CUBE-LOVERS.
I personally will probably be unable to do anything about
new responses until Christmas, but don't let this stop you
from sending them in.

I hope that i have not grossly misrepresented anybody
because of my abbreviations of the results.


             Chris Worrell -   ZILCH@MIT-MC



Date: 25 September 1981 03:25-EDT
From: Chris C. Worrell  <ZILCH at MIT-MC>
Subject: 4x4x4 TRANSFORMATION
To: CUBE-LOVERS at MIT-MC

On the normal cube the relative positions of the faces
is immutable, however this is not so on the 4x4x4.
I have found a trransformation which i beleive will facilitate
the switching of any two of the cental blocks of 4 on a face.
I am not sure as I did this totally in my head, without
the benifit of a 4x4x4. This trandsformation would be 
extremelly nasty to pull on someone who has learned how to 
solve the standard cube from someone else, and as such does
not know much about cubolgy.

Date: 25 Sep 1981 21:41:35 EDT (Friday)
From: Roger Frye <frye at BBNP>
Subject: Cube in Political Cartoon
To: Cube-Lovers at MIT-MC
Cc: frye at BBNP

The Boston Globe editorial page for 9/25/81 has a cartoon entitled
"Stockman's CUBE".  I can't make out the signature on the cartoon.
It could be M. H. Beane. It shows much struggling to solve the cube
followed by a realization.  The character, presumably Stockman,
Reagans's economist, then smashes off a corner or two of the cube
with a hammer.  In the last frame, he has his suit jacket back on
(which he had lost while struggling).  He holds up a solved cube
and says, "Simple... I just applied my economic method...

It's everywhere.  Last week we argued about it in couples therapy;
this week our therapist said she had an argument with her husband
because he just got one and wasn't listening to her.

I join Dame Ollerenshaw on the casualty list; a few nights ago
I sprained my wrist twisting too long, too late, in too cramped
a position, and sinning too fast against a jam.

Happy Cubing,
Roger Frye


Date: 25 Sep 1981 22:03:27 EDT (Friday)
From: Roger Frye <frye at BBNP>
Subject: spinning not sinning
To: Cube-Lovers at MIT-MC
Cc: frye at BBNP

Spelling corrections for my last message.
Add a quote mark after the second ellipsis.
Change "sinning" to "spinning".

Happy sinning,
Roger


Date: 26 September 1981 14:08-EDT
From: David C. Plummer <DCP at MIT-MC>
Subject:  ZILCH's 4x4x4 center block exchange
To: CUBE-LOVERS at MIT-MC

Just a clarification (and a discussion promoter, perhaps).  I can
believe it is possible to put the center 2x2 blocks of a 4x4x4
cube into any relative position.  But, there is still only one
way to solve it.  This is because the relative position in the
solved state is determined by the corners.  I guess this allows
for a new class of pretty patterns, especially DOTS:
	DOTS as we know it in  the 3x3x3 (two sets of three; one
	  set clockwise, the other couterclockwise) 
	DOTS (two sets of three; both (counter)clockwise)
	DOTS (two sets of three; similar to BASEBALL)
	DOTS (three sets of two; like Christman's Cross)
	DOTS (three sets of two; pairs opposite) (a local max)
	DOTS (three sets of two; all pairs adjacent)
	DOTS (one set of six; similar to the checkboard obtained
	  by Plummer's Cross + dots + pons)
	etc.

Anybody want to take a shot at trying to catalog all the local maxima?

Date: 28 Sep 1981 1722-PDT
From: ISAACS at SRI-KL
Subject: notation
To: cube-lovers at MIT-MC

  Anybody have a good notation for 4x4x4?  How about for the pyramid?
  Also, anybody have a good (less than 8 moves; QTW or HTW) to move a 
cubie from the bottom layer to a top corner, without disturbing top or middle,
and making bottom facie into top facie (eg DLF --> UFL or suchlike).
   --- Stan
-------

Date:  9 Oct 1981 10:55:04 EDT (Friday)
From: Roger Frye <frye at BBNP>
Subject: Cube-a-thon
To: Cube-Lovers at MIT-MC
Cc: Frye at BBNP

Does anyone have any more details on the following (as clipped
from Boston Globe 10/8/81):
  JEFF VANASANO, 15, of the Bronx, beat
  more than 2000 competitors in a Rubik's
  cube-a-thon in Jackson Township, N.J.,
  by solving a Rubik's Cube in 24.67 sec-
  onds.  Vanasano will go on to the US finals
  in Los Angeles in November and there will
  be a world championship next spring.
I wonder who sponsored the competition, what form the contest
took, who Vanasano is, what techniques he uses.  I found the
analysis of Kimmo Eriksson's technique by Lars S. Hornfeldt
(9 May 1981 08:47-EDT) to be most enlightening, and would
love to see similar analyses.
-Roger Frye


Date: 11 October 1981 13:26-EDT
From: Andrew Tannenbaum <TRB at MIT-MC>
Subject:  New Jersey Cube-a-thon
To: CUBE-LOVERS at MIT-MC

The NJ Cube-a-thon was at (Six Flags Over) Great Adventure
amusement park, sponsored, I believe, by Ideal.  I saw the kid
who won on television, zip, zip, zip, presto.  He said he was
solving 300 cubes a day as practice, and has been improving his
speed five seconds a month for the past six months or so.  I
think there were about 1500 entrants.

How about this for a bumper sticker?
"Cubists do it in under a minute"
	Andy Tannenbaum
	Bell Labs Whippany, NJ

Date:  11 October 1981 17:52 edt
From:  Greenberg.Symbolics at MIT-Multics
Subject:  Under a minute
To:  CUBE-LOVERS at MIT-MC

Several people have noted that there must be something sick about
a bunch of people whose "do it" bumperstickerisms are all
UNCOMPLIMENTARY, viz., "faster", "in under a minute"....

Date: 12 October 1981 0028-EDT (Monday)
From: Guy.Steele at CMU-10A
To: cube-lovers at MIT-MC
Subject:  Cubists do it...
Message-Id: <12Oct81 002811 GS70@CMU-10A>

Okay, then, how about: Cubists do it...

	... with both hands.
	... in groups.
	... symmetrically (!).
	... first on top, then on the bottom.
	... by swapping.
	... with their fingers.
	... while twisting their faces.
	... rectilinearly.
	... by conjugation (tautologous?).
	... despite odd orientations.

--Quux

Date: 12 Oct 1981 0919-PDT
From: HOROWITZ at USC-ISIF
Subject: Latin square question
To:   cube-lovers at MIT-MC

Herb Taylor, a research associate at USC in the EE dept. 
has proved that it is impossible to put a latin square
on every face of the 2 x 2 x 2  cube.
Can anyone answer the question on page 15 of his book
UNSCRAMBLING THE CUBE?  It asks whether or not it is
possible to put a latin square on every face of the 3 x 3 x 3 
cube.

Ellis Horowitz

p.s. Minh Thai of Los Angeles did ten scrambles with average
time 40 seconds
-------

Date: 14 Oct 1981 13:39:30 EDT (Wednesday)
From: Roger Frye <frye at BBNP>
Subject: Latin Square Answer
To: Cube-Lovers at MIT-MC
Cc: Frye at BBN-UNIX

Exhaustive search shows that there are several ways to fill all faces
of Rubik's 3^3 with Latin squares, but none lie in the primary orbit.

Here are two arrangements in the orbit where one corner is twisted
1/3 turn anticlockwise:

         UFB                           UBF        
         BUF                           FUB        
         FBU                           BFU        
     				                  
     LUD RLF RUD LRB               LDU RLF RDU LRB
     DLU LFR DRU RBL               ULD LFR URD RBL
     UDL FRL UDR BLR               DUL FRL DUR BLR
     				                  
         DBF                           DFB        
         FDB                           BDF        
         BFD                           FBD        

I did the search with pencil and scissors on quadrille lined paper.
The following observations speed the search:
 1) The 3*3 Latin square whether reduced or not must be some
    rotation or relabeling of the following pattern:  ABC
			                	      BCA
                                                      CAB
 2) The diagonal bars of the squares must be arranged as in
    the pretty pattern called "Laughter" because of the shape
    of the corner cubies.  (See the bars in the patterns above.)
 3) When you attempt to place one of the remaining four corner
    cubies, the corner color propagates to two edges which restricts
    the other color on those edge cubies to not be that color and also
    not that color's complement (e.g. U and D).  This restriction
    then propagates to another corner.


-Roger Frye


Date: 15 October 1981 10:05-EDT
From: Andrew Tannenbaum <TRB at MIT-MC>
Subject: maybe we are a little sick.
To: CUBE-LOVERS at MIT-MC

    Date: 11 October 1981 23:08-EDT
    From: Alan Bawden <ALAN at MIT-MC>

    Date:  11 October 1981 17:52 edt
    From:  Greenberg.Symbolics at MIT-Multics

    Several people have noted that there must be something sick about
    a bunch of people whose "do it" bumperstickerisms are all
    UNCOMPLIMENTARY, viz., "faster", "in under a minute"....

    -----
It so happens that cubists like to do it faster instead of
slower.  Or by twisting faces.  There are those who allege that
anyone who would try and try to solve the dumb puzzle is sick, or
that anyone who would solve and scramble a cube tens of thousands
of times is sick.  Lots of people can't imagine anyone finding
joy in hacking.  It can't give the satisfaction of the personal
communication and conquest of marketing, for instance.  This goes
for many interests.

My mother always told me that I shouldn't play with my cubies.
	Andy Tannenbaum
	Bell Labs Whippany, NJ

Date: 17 October 1981 22:54-EDT
From: Mark K. Lottor <LOTTOR at MIT-AI>
To: CUBE-HACKERS at MIT-AI

Where is old mail located?


Date: 19 October 1981 12:23-EDT
From: Andrew Tannenbaum <TRB at MIT-MC>
Subject: assembling your own
To: CUBE-LOVERS at MIT-MC

What ever happenned to Bela Szalai of LOGICAL GAMES (mentioned in
mid 1980 in this list)?  He re-mortgaged his house to produce white
cubes, did he succeed?

Does anyone make high quality cubes?

Is it possible to buy unassembled cubes which you can tweak and
lube yourself?  I don't want to try to figure out how to get at
the screws in my good cubes, I fear I'll destroy them.

What is the current edition of Singmasters?  Has anyone else
written a worthwhile document (aside from the compilation of this
digest)?
	Andy Tannenbaum
	Bell Labs Whippany, NJ

Date: 20 Oct 1981 10:26:57 EDT (Tuesday)
From: Roger Frye <frye at BBNP>
Subject: Do It in the Dark
To: Cube-Lovers at MIT-MC
Cc: frye at BBNP

I have constructed a tactile cube by epoxying buttons, snaps, paper
clips and Q-tip stems to a store-bought cube.

It takes me about 20 minutes to solve with my eyes closed as opposed
to about 5 minutes with my eyes open.  Maybe Conway cheats in his
5 look solution by feeling colors.

I minimize search times by always working faces in a fixed order
and by orienting colors (i.e. textures) before moving cubies on the
last face.  But the ability to see (i.e. feel) the back of the cube
doesn't compensate for my as yet poorly developed tactile pattern
recognizer.

- Roger Frye


Date: 22 October 1981 14:02-EDT
From: Alan Bawden <ALAN at MIT-MC>
Subject: fowarding
To: CUBE-LOVERS at MIT-MC

Date: 22 Oct 1981 0928-PDT
From: ISAACS at SRI-KL
Subject: Re: assembling your own
To: ALAN at MIT-MC
In-Reply-To: Your message of 19-Oct-81 1735-PDT

  There seem to be at least two high quality cubes, one called "Deluxe" and
I think sold by Ideal for about $14.00 around here (S.F.), the other
packaged differently (in a half-cardboard box) and sold for about $8.00.
Last night I heard that there may be still another version (perhaps identical)
for about $4.00.  Anyway, they all have the colors on the facies put on with
thin plastic squares instead of stickers.  The one I have also turns very well.
But the plastic seems
 to be what makes them "deluxe".  
   There is also a "build it yourself" cube kit sold; I've seen it, but don't
have one.  It comes with the pieces and screws, and each facie has a small 
hole in the center.  Plastic colored faces are snapped in.  Maybe it is
an early version of the "Deluxe".
   Documents on the cube:  There seem to be more books published on the cube
every day.  Almost all of them are just solutions, with a little extra
information (pretty patterns, anecdotes, other types of cubes, etc.). I
know of not other book as interesting mathematically or theoretically as
Singmasters.  An article by Dame Kathleen Ollerenshaw, called "The Hungarian
Magic Cube" in Bulletin, The Institute of Mathematics and its Applications, Vol. 16, aPril 1980, may have interesting information (I haven't seen it), and
the book? article? by Conway and others, mentioned in Singmaster, ought to
be interesting if/when it comes out.
   --Stan
-------

Date: 29 Oct 1981 0932-PST
From: ISAACS at SRI-KL
Subject: Re:  "Deluxe" cube
To: Bob.Walker at CMU-10A, cube-lovers at MIT-MC
In-Reply-To: Your message of 22-Oct-81 1953-PDT

  Last night my friend brought me the "deluxe" cubes.  They came from:
   Ckoach House Cards and Gifts
   Sunnyvale Town Center
   2502 Town Center
   Sunnyvale, Calif.        (address from phone book - it may be inaccurate)
  (408)736-7244
    They had the cubes there for $3.95 per cube. They were made in Korea, I
forget the brand name.  Coach House was also apperently selling the regular
cheap-type cube (Wonderful Puzzler?) for $1.10 apiece.  A good price.
   ---Stan
-------

Date: 1 November 1981 19:12-EST
From: Andrew Tannenbaum <TRB at MIT-MC>
Subject: Trick or Treat
To: CUBE-LOVERS at MIT-MC

Last night I dispensed treats to two little walking cubes.  Both solved.
They made my evening, God bless 'em.

Also one of those missing link frobs.
	Andy Tannenbaum
	Bell Labs Whippany, NJ

Date:  2 Nov 1981 1153-PST
From: ISAACS at SRI-KL
Subject: more trick or treat
To: cube-lovers at MIT-MC

   In my neighborhood, too, there were at least two Rubik's Cubes at the
door.  Also solved - did anybody see any walking unsolved cubes?  Next
someone ought to develop a walking, solvable, cube costume.
   --- Stan
-------

Date: 3 Nov 1981 0942-EST
From: ELF at MIT-DMS (Eric L. Flanzbaum)
To: Cube-Lovers at MIT-MC
Subject: Trick or treat (more of)
Message-id: <[MIT-DMS].214352>


Of course I, too, saw some Rubik's cubes wandering around. And
they were solved.

Just yesterday I talked to someone who is going to a costume party
next weekend (11/7) as Rubik's cube.  Wow, everyones getting
into the act.  Next thing you know they'll have already made constumes
sold in stores!


Date:  3 November 1981 10:41 est
From:  Greenberg.Symbolics at MIT-Multics
Subject:  Re: more trick or treat
To:  Cube-Lovers at MIT-MC
In-Reply-To:  Message of 2 November 1981 23:40 est from Alan Bawden

A party I went to Sat Nite had a walking Rubik's cube, but alas, the
wearer was cube-gnorant, and it had multiple center cubies
of the same color and cubies with identical faces on two sides, etc.

Date: 3 Nov 1981 1452-EST
From: ROBG at MIT-DMS (Rob F. Griffiths)
To: cube-lovers at MIT-MC
Subject: Walking Talking Cubes
Message-id: <[MIT-DMS].214388>

  Down on Boulder's mall Halloween night, I saw about 4 cubes.
All of these were solved, except one.  I asked the man why he
had designed his cube with only one face complete.  His reply
was that that was the closest to sloving his cube he had ever
come, and he figured looking at a bigger model might help him!
  Speaking of larger cubes, does anyone know of any cubes that
are larger than the standard Ideal Toy size cubes? And where
could I obtain one?
			-Rob.



Date: 3 Nov 1981 1608-PST
Sender: OLE at DARCOM-KA
Subject: It's bigger, but its round!
From: Ole at DARCOM-KA (Ole J. Jacobsen)
To: Cube-Lovers at MIT-MC
Message-ID: <[DARCOM-KA] 3-Nov-81 16:08:12.OLE>


Following Rob's question about bigger cubes I would like to report 
on "Wonderful Puzzler"'s latest: The Rubik Sphere. Yes, it is here
and is about 90mm in diameter with six circular faces. Solving is
hence just as for the cube apart from some confusion caused by the
fact that there are no edges to hold onto, the thing has no distinct
orientation. This also makes twisting a bit tricky and I would think
those 24 second champions would have trouble with this one. The ball
or sphere has a smaller brother (keyring size) and this one has a
different coloring scheme, "stripes" instead of "faces" and is in
this respect very similar to the 10 sided drum discussed here earlier.
The mechanics of the sphere is very much as you would expect, just
an extension of the original cube mechanism,- or if you like: imagine
heating your cube up to near melting point and squeezing it into
a ball (not recomended!)

Happy.. eh, well Sphering!

OLE

Date: 4 Nov 1981 1133-PST
Sender: OLE at DARCOM-KA
Subject: Global Cube
From: Ole at DARCOM-KA (Ole J. Jacobsen)
To: Cube-Lovers at MIT-MC
Message-ID: <[DARCOM-KA] 4-Nov-81 11:33:04.OLE>


Having seen the cover of next weeks 'Newsweek' (November 9)
I got the idea of continuously coloring the Sphere (see my
previous message) with a map of the world, i.e make it into
a globe. This would also incorporate the Supergroup and per-
haps help people learn geography (if they already know how 
to solve the cube/sphere...)

What will they think of next?

OLE

Date:  5 Nov 1981 0954-PST
From: ISAACS at SRI-KL
Subject: Rubic's Sphere
To: CUBE-LOVERS at MIT-MC

   One of my favorite pretty patterns on the Sphere is the "universe", from
"Unscrambling the cubE": every edge and corner flipped/twisted in place; the
each adjacent pair of corners twisted towards each other.  Forms a nice, 
symmetric pattern of each face "exploded" outward from its center, which
is more visible on the sphere than the cube.
---Stan
-------

Date: 8 Nov 1981 1151-EST
From: ROBG at MIT-DMS (Rob F. Griffiths)
To: Cube-Lovers at MIT-MC
Message-id: <[MIT-DMS].214863>

  I don't know how many of your newspapers carry this strip, but in
today's 'Goosemeyer', there is a set of pictures of a general going
through many secturity checks and M.P.'s.  He then walks into a
'Top Secret' room and says 'Sorry Im late...Whats the situation,
men?'  One of the others replies 'It looks hopeless sir...Everything
we have tried has failed!' The general coutners 'Thers precious time
left to come up with the right solution...' The last frame shows
them standing around a table looking at a scrambled cube, and the
general saying 'Ive got to get the back to my son by three o'clock'!!
			-Rob.


Date: 3 December 1981 15:19-EST
From: Richard Pavelle <RP at MIT-MC>
Subject: MASQUERADE
To: CUBE-LOVERS at MIT-MC

As you may have noticed, cubism seems to be stagnating. Therefore
I suggest that we start discussions on a different sort of puzzle 
called MASQUERADE!

MASQUERADE is a book of about 15 pages beautifully illustrated by
the author and artist Kit Williams. The story and pictures contain
clues which supposedly will lead the solver to a "treasure" described
by Williams as "... a golden hare adorned with precious stones and 
faience". It is hidden somewhere in Britain. The book is published by
Shocken Books in Manhattan and sells for about $10. There is also an
ad on Page 35 of Scientific American for December (North American 
Edition).

If any of you have interest in this idea send mail to me, RP@MC, and I
will keep accumulate comments and names. 


Date: 4 Dec 1981 1459-PST
Sender: OLE at DARCOM-KA
Subject: Diversion
From: Ole at DARCOM-KA (Ole J. Jacobsen)
To: Cube-Lovers at MIT-MC
Message-ID: <[DARCOM-KA] 4-Dec-81 14:59:07.OLE>

Folks,

Masquerade might be a good continuation for cube-lovers, I have not seen 
it so I can't really say. However, I am having sleepless nights about
another of our puzzle friends the Ten Billion Barrell described herein
by David Plummer a few months back. No-one ever seemed to come up with a 
good notation nor any hints on how to solve this toy, an as far as I can
tell no good books are out which offer any help either. So, could we
perhaps have a discussion on the barrell or would the masters of this
time stealer come forward and advice me. I am much to young to be getting
grey hairs, but found a few this morning!


OLE

Date:  4 Dec 1981 1802-PST
From: ISAACS at SRI-KL
Subject: Globe Cube
To: cube-lovers at MIT-MC

  A friend of mine brought me a Spherical "Rubiks cube" with a world map on
it from Hawaii.  Gotten in a clothing store (Casablanca?) on Maui, not a
puzzle place.  She also got me some Hawaiian cubes - each face with one of
the islands.
  There's a great proliferation of Cube (and related items) books.  "Not 
Another Cube Book", a 'humorous' book (not very good). A new book from
Nourse which has solutions to the barrel, the tetrahedron, the Missing Link,
and the Snake.  A cube smasher.  And many more.  I've heard there is or will
be a book on what to do with a smashed cube.  I've been thinking of trying
to cash in on this craziness and write my own.  Q: What's Red and Orange and
Green and Blue and White and Yellow and lives at the bottom of the sea?
Answer:  Moby Cube!  Oh well.  Maybe I'll stick to computers and puzzles.

--- Stan
-------

EH@MIT-AI 12/05/81 19:20:20 Re: Tiny cubes
To: cube-lovers at MIT-MC
Does Ideal Toy Corp. make those tiny cubes? (about a inch square)
Beware of stuff made in Tiawan as they break QUICKLY..as what
happened to me with a 3/4 inch cube..and I heard from my friend
that his tiawan-made cube (standard size 2 1/2 inch) broke
in a few days....
The only RUbic cube is the Ideal Toy corp cube and it's great!

	Edward


Date: 5 Dec 1981 1654-PST
Sender: OLE at DARCOM-KA
Subject: Rubikmania
From: Ole at DARCOM-KA (Ole J. Jacobsen)
To: Cube-Lovers at MIT-MC
Message-ID: <[DARCOM-KA] 5-Dec-81 16:54:23.OLE>

Check out TIME magazine, December 7th page 35, for an update on the
cube and its many spin-offs and solution guides. The article mentions
the Cube Smasher, "...guaranteed to pound the puzzle to bits", as well
as the "Not Another.." book that Stan was talking about. Most interesting
is perhaps the figure 10 MILLION (!) which is how many cubes have been
sold Worldwide so far. This is presumably the official Ideal figure, so
one wonders how many have been sold all together, Wonderful Puzzler
seem to have done very well, at least here in the UK.

If you are interested in computers, and presumably you are since you're
on this list,- then check out page 33 of the same magazine for a very
interesting article on Texas Instruments.

OLE

Date: 6 December 1981 03:14-EST
From: Eric L. Flanzbaum <ELF at MIT-MC>
Subject: Sticky cubes ...
To: CUBE-LOVERS at MIT-MC
cc: ELF at MIT-MC


Today, I just purchased a Rubik's cube (The real thing,
not a fake!).  I would like to know if there is any way to
make the cube move more freely and loosely w/out having to
wait for it to loosen up itself by using it.  Any suggestions?

				 --- Eric ---

EH@MIT-AI 12/06/81 16:34:03 Re: Sticky Cubes...
To: Cube-Lovers at MIT-MC
CC: ELF at MIT-AI
From: "You can do THE CUBE" by Patrick Bossert / published by Puffin.

To loosen up the cube if it's sticky to turn, turn the top slice 45 degs
and insert a screwdriver under one of the edge pieces (the one in the 
middle) ....

(1) [] [] [] <- turn this slice 45 deg
    [] [] []
    [] [] []

(2) [] [] []
       ^ insert a screwdriver blade under that part and lift up until
it pops out.

(3) Then you can remove the two corner pices. Looking inside, you
can see the + shape of the cube mechanism. Put a BLOB of vaseline
inside.

(4) put the corner pieces back and the middle piece too. MAKE SURE
you get them in exactly as they were before you opened them up.
Otherwise,you WILL NOT be able to solve the cube!!

	Hope this helps,
	Edward

ps: PLEASE feel free to write to me if you're confused as this
is the best I can do without graphics.... (use graphics to
show but you can't here)


Date:  7 December 1981 1911-EST (Monday)
From: Dan Hoey at CMU-10A
To: Cube-Lovers at mit-mc
Subject:  Brute Force Report: The fourteen-qtw identities
Message-Id: <07Dec81 191129 DH51@CMU-10A>

	Several messages in August of this year [mail to
Hoey@CMU-10A for copies] concerned short identities of the cube,
i. e. processes which return the cube to solved.  Later in that
month I assisted David Plummer in a brute force attack on the
problem.  We had plans to investigate all the positions up to eight
qtw, but unfortunately became busy on other projects.  I have
finally come up with enough time to analyze and report the data
from the seven-qtw search.

	There are 8,221,632 cube positions at a distance of seven
quarter-twists from solved, and 9,205,558 positions at seven or
fewer qtw.  By recording cases of different seven-qtw processes
yielding the same position, a complete list of fourteen-qtw
identities is obtained.  The task is then to reduce the list to
exclude multiple instances of equivalent identities.

	We call two identities equivalent when one can be obtained
from the other by some combination of the following operations:

	- uniformly relabeling the twists according to a rotation
		or reflection of the cube,
	- cyclically permuting the twists,
	- reversing the order of the twists and inverting each one,
		and
	- substituting a sequence x for a sequence y, where xy' is
		a shorter identity.

The first three criteria are easily implemented on a computer.  The
fourth can be performed for the shortest identities, those
equivalent to F^4 and FBF'B', but I know of no algorithm to detect
all cases of equivalence due to substitution of the longer
identities.  My strategy was to reduce the (several thousand)
identities by computer for the simple kinds of equivalence, and
then to look by hand for substitution equivalence between the
fourteen identities then remaining.  I found three equivalences,
listed at the end of this note, but the possibility remains that
some of the following identities are equivalent.  The list is,
however, complete (modulo bugs and cosmic rays).

	Identities equivalent to the first six on this list were
independently discovered by Chris C.  Worrell; I follow his
numbering for them.  Identities I14-5 through I14-7 do not hold in
the Supergroup, because they twist face centers as noted in the
brackets.

I14-1   BF' UB'U'F UL' BU'B'U LU'
I14-2   B UBL' B'D'BD LB'U' L'B'L
I14-3   BB U BB UD' RR U' RR U'D
I14-4   BUB'U' L'FL UBU'B' L'F'L
I14-5   (BB UD B U'D')^2                    [Supergroup BB]
I14-6   BF' U B'F  LR' UD' F' U'D  L'R      [Supergroup UF']
I14-7   BF  U B'F' LR' UD  F' U'D' L'R      [Supergroup UF']
I14-8   BF' UFRU'R'B'U'B'RBUR'
I14-9   BF' UFRU'B' UD' F'U'R'FD
I14-10  (BUBU'L'B') R (BLUB'U'B') R'
I14-11  (BUBU'L'B') D'R'B' DLD'RD

	The twelve-qtw identity I12-2 = (BUBU'L'B') (B'D'B'DLB) can
be substituted into identities I14-10 and I14-11 to yield:

I14-10a  (B'L'D'BDB) R (BLUB'U'B') R'
I14-10b  (BUBU'L'B') R (B'D'B'DLB) R'
I14-10c  (B'L'D'BDB) R (B'D'B'DLB) R'
I14-11a  (B'L'D'BDB) D'R'B' DLD'RD

Identity I14-10c can be obtained from I14-10 [by shifting seven
places and reflecting the cube through the UD plane] but I14-10,
I14-10a, and I14-10b are mutually inequivalent when twelve-qtw
identities are ignored.  The same holds for I14-11 and I14-11a.
Strangely enough, I14-11a can also be transformed to I14-11 by
substituting with the identity (BDBD'R'B') (B'U'B'URB), which is
equivalent to I12-2.

Date: 8 Dec 1981 12:45 PST
From: Hoffman at PARC-MAXC
Subject: Re: Diversion 
In-reply-to: Ole's message of 4 December 1981 
To: Cube-Lovers at MIT-MC


   About the "Wonderful Barrell".  I, too, had been searching for a good notation
and some ideas about how to solve it.  I finally found it!  James Nourse, author
of 'The Simple Solution to Rubik's Cube'  has a new book out entitled 'The
Simple Solution to Cubic Puzzles'  (or close to that title).  It goes for just $2, and
I picked mine up at a Waldenbooks.  It includes a complete solution for the
barrell, as well as the pyramid, the picture cubes, the Missing Link, and some
items about the Rubik Snake.  A good deal, but BEWARE of some most
aggravating typographical errors. (I know of two in the barrell solution.)

  --  Rodney Hoffman

Date: 8 Dec 1981 14:30 PST
From: Lynn.ES at PARC-MAXC
Subject: Re: Diversion 
In-reply-to: Ole's message of 4 Dec 1981 1459-PST
To: Cube-Lovers at MIT-MC
cc: Lynn.es

I just got my first look at a borrowed copy of Nourse's new 
"solve everything" book, specifically to look at the barrel puzzle 
solution he gives.  His notation is:

The two rotating pieces are "rings".  
They rotate through 5 "notches".

"Layers" (horizontal slices containing 5 balls each) are 
numbered 1 thru 4 from bottom to top, plus T for the 3 
balls immediately under "plungers" (when they are up).  
No notation is used for positions within the puzzle, 
except this numbering of the layers.  Positions are given 
by word descriptions or drawings.  

Moves: 

Up-arrow         move the plungers up
Down-arrow       move the plungers down
Two boxes, one above the other, which represent movement 
of the two rings.  A "T" above, or inverted "T" below the boxes 
shows the current plunger position.  Each box may contain:
	blank           don't move this layer at all
	Left-arrow      rotate the ring left (clockwise, as seen from the top) one
notch
	Left-arrow 2    rotate the ring left two notches
	Right-arrow     rotate the ring right one notch
	2 Right-arrow   rotate the ring right two notches

All moves assume the lonesome plunger is toward the puzzle solver.  
All move sequences begin assuming the plungers are in the up position.  

There is a great deal of room for improving this notation.  
Particularly, the boxes should be dispensed with in favor of 
some single-printed-line way of differentiating between top 
and bottom rings.  The current plunger position is not needed.  
Because many terminals or typewriters do not have four 
directional arrows in their character sets, these should be 
replaced.  I like thinking in terms of left and right rotations 
rather than clockwise, since this puzzle is usually solved 
without turning the puzzle as a whole.  Position notations 
are needed.  For these reasons I propose the following: 

Move notations:

U      move the plungers Up
D      move the plungers Down
T      rotate the Top ring right (counter-clockwise from the top) one notch
B      rotate Bottom ring right one notch
-T     rotate Top ring left one notch
-B     rotate Bottom ring left one notch
T2, B2, -T2, -B2 	two notch versions of the above four respectively

Positional notation:

Same naming for "rings", "layers", "notches", "plungers", 
except call the top layer "5".

The ball positions in layer one are, clockwise, as seen from 
the top, beginning with the one under the lonesome plunger: 
1F (front), 1L (left), 1LL, 1RR, 1R.  

Correspondlingly for the other layers.  
There is no 5L or 5R, because of the orientaion of the puzzle.  

During a move, at times when the plungers are down, there 
will temporarily exist positions 0F, 0RR, 0LL.  


Any suggestions or arguments about the notation are welcome.  


Nourse's solution has two typos in it.  Page 33: upper rightmost 
move on page should be in the bottom ring, not top.  Page 36: 
second move on the page should have a 2, not single notch.  


If people are interested, I can summarize Nourse's approach to solving.  

/Don Lynn
 

Date: 8 Dec 1981 1952-EST
From: ELF at MIT-DMS (Eric L. Flanzbaum)
Reply-to: ELF at MIT-DMS
To: Cube-Lovers at MIT-MC
Subject: That's incredible!
Message-id: <[MIT-DMS].217487>

Last night on the show "That's Incredible!" they held the finals of
the Rubik's Cube competition.  The contest was for the U.S.A. and the
winner wa then to go onto world competition.  If I remember correctly,
the fastest time (by the winner, of course) was ~26 seconds (it was
actually between 27 and 26, but I can't remember that part).  The show
also mentioned that there were 43 quintillion combinations possible
(is this true?).  The second place time was between 28 and 29.

-- Eric
*******


Date: 9 December 1981 19:23-EST
From: Alan Bawden <ALAN at MIT-MC>
Subject: Luks
To: CUBE-LOVERS at MIT-MC

I don't know any more about this, but it sounds like it might be
interesting to some of us cube group hackers:

    MSG: SEMINA 3     
    DISTRIB:  *DM, *MC, *ML, *AI
    EXPIRES: 12/17/81 16:21:19
    BJ@MIT-ML 12/09/81 16:21:19 Re: Luks
    Gene Luks from the Mathematics Department at Bucknell University will 
    give a talk on Tuesday, December 15 at 4:00 (refreshments at 3:45) in 
    room NE43-512A.  The talk is "ALGORITHMS FOR PERMUTATION GROUPS."

Date: 10 Dec 1981 1903-PST
From: ISAACS at SRI-KL
Subject: Ten Billion Puzzle (the Barrel)
To: CUBE-LOVERS at MIT-MC

  It just happens that the day before I got the Nourse solution to the
Ten Billion (Barrel, Magic Barrel, etc), I got another solution by Naoaki
Takashima, that he presented to the ARMJ (??) meeting in Japan.  His 
notation:  
Columns and stages are defined             5-   x   x x
as shown on the right.                 s   4- x x x x x
                                       t   3- x x x x x
                                       a   2- x x x x x
                                       g   1- x x x x x
                                       e
                                       s      | | | | |
                                              1 2 3 4 5

                                             columns
He moves the drums, rather than the plunger.
   U--move drums up.       D--move drums down.
   Ru (R sub u) Rotate upper drum right one column; superscripts for number
   Rl (R sub l)   "    lower  "     "    "    "     of columns
   R for upper and lower together to the right
   L for left similarly.

 At any rate, a cumbersome notation for computer writing.  What's more, unit
move people (who don't like slice moves or half turns on the cube) won't like
combining the two drums (rings).  BUT since I hold my barrel by the bottom
plunger, I do find it easier to think of moving the rings up and down, rather
than pushing the plunger.  I like rings better than drums, but columns better
than notches.  And I would rather take standard numbering to start from 
what is above defined as column 2, ie, the one under the neighborless plunger.

  So, all in all, I would like to propose accepting Don Lynns' notation
with the following changes:  U and D would refer to the rings moving up
and down, instead of the plunger.  "Notches" would become "columns".
The picture above, moved to reflect the new columns, could be used when it
is needed for clarity.  

* * * * Small spoiler * *
  The Takashima solution is really based on one move, which cycles 3 balls
more-or-less vertically:
  let X = U T D -T / U B D -B / (TB) U -(TB) D
  then  X2 moves 3F TO 4F TO 2L TO 3F;  ie, the 3-cycle (3F,4F,2L).
* * * * End of small spoiler * * *

 I already propose apostrophe instead of -, so the above becomes:
   UTDT'-UBDB'-(TB)U(TB)'D    which I think is a little clearer.

Notice that the above can be simplified to:
     UTDT'UBDTUB'DT'UBDTUTBD  (I think.  I find it hard to think about in
                               such small pieces; I like the less efficient
                               chunks at first.)
* * * * * * * * SPOILER WARNING * * * * * * * * * * *SPOILER WARNING * * * *

Anyway, with this move alone (plus some playing at the beginning), I can
solve the puzzle as follows:  First get the three black balls at the top,
under the plungers, any old way.  Then put 5 different colors in the first
layer, moving only the bottom ring (to leave the black balls alone).  Then
use the move above to move in the matching second layer from the top (layer
4).  When necessary, use the same move to get from 3 to 4 first, or even from
2 to 4.  Using the inverse move can speed things up, but just doing it twice
is easier on the memory.  Move the space in layer 2 you want to fill, to column
5 (new numbering; col. 1 in above diagram, or "2L" in notation), by moving
the lower ring; move the column containing the ball you want in that space
to col. 1 (F), apply the move once or twice, and move the rings back.  Rest
position has the barrel down (and the plunger up).  Once the second layer
is complete (or, for efficiency, complete except for 1), do similar things
for layers 3 and 4, using the second layer for temporary storage.

* * * * * * * END OF SPOILER * * * * * * * * * * * * * * * * * * * * * * * * *

  Don Lynn:  I would be very interested if you would summarize Nourse's
approach, and either send it to me or to cube-lovers.  I have not had the
time to break it down into chunks to see what basic moves he really uses.
I find that to be the main trouble with most of the solution books to 
the cube as well as this - that they offer long sequences to memorize, with
little thought to learning what is going on, and what the logic of the moves
are.  In Nourse's solution, what are the basic pieces of his moves. Can they
be expressed (as above) as a "simple to comprehend" set as well as a "shortest
number of unit moves" way.  By use of parenthesis (or maybe slashes), U and D
could probably be eliminated; at each slash, switch from up to down or vice
versa.  Would this be useful notationally.  What are some basic algorithms
to exchange 2 balls on a single layer or in a single column?  To cycle a
layer or column?  
  Enough questions for tonight.  When's the first Barrel contest?
   --- Stan
-------

Date: 11 Dec 1981 1643-EST
From: ROBG at MIT-DMS (Rob F. Griffiths)
To: cube-lovers at MIT-MC
Subject: Re: Ultimate Cube
Message-id: <[MIT-DMS].217732>

   I have seen the 'Ultimate Cube', except here in Boulder, it is 
done a little differently.  They are all Black cubes, and come
in the standard (Ideal standard) container. And (no offense from
me, this is what they are called) they are labeled 'Polish Cubes'
with 'No wrong move'. Also, we have a Pyraminx puzzle (which is rather
trivial compared to the cube) and it has a problem: if you twist the
top cube on a certainm side too many times, (like 3 twists) it comes
off and the whole mechanism falls apart.  Has anyone else had this problem?
How can I correct it?
			-Rob.


Date: 11 December 1981 18:20-EST
From: Alan Bawden <ALAN at MIT-MC>
Subject: fowarded
To: CUBE-LOVERS at MIT-MC

Date: 11 Dec 1981 at 1630-CST
From: korner at UTEXAS-11
Subject: Pyramid
To: ALAN at MIT-MC

	Someone mentioned the pyramid recently. Are solutions really worth
discussing? I had the solution within 15 min. and the group structure seemed
trivial. Is this as simple as it seems or have I missed something?
		-Kim Korner
-------

Date: 12 December 1981 02:12-EST
From: Alan Bawden <ALAN at MIT-MC>
Subject: Puzzle-Lovers?
To: CUBE-LOVERS at MIT-MC

Richard Pavelle has suggested that since cubism seems to be
stagnating, we might consider expanding the scope of this discussion
to include other sorts of puzzles.  He wondered (in a letter to this
list a few days ago) if anyone was interested in the puzzle/book
called "MASQUERADE".  The response he got was enthusiastic.

The discussion on this list has never been exclusively about cubes,
but it has never really strayed very far either.  Also, recently the
discussion has slowed somewhat.  So we were wondering how people would
feel about changing the purpose of this list from cube hacking to more
general puzzle hacking.  This would still be the place to talk about
cubing, but everyone would have to at least tolerate other puzzles as
well.  Note that there is another option for general puzzle hackers
which is the formation of a separate Puzzle-Lovers (or something)
mailing list.  This second option would be the thing to do if there is
enough interest, but also some objections from people currently on
Cube-Lovers.  (If a new list is formed, they will have to find someone
else to be responsible for it.  I won't deal with two!)

To decide this tricky question I would like to hear from anyone who
objects to the expansion idea.  With enough objections, we simply
won't do it.  Don't bother to send me a note if you are simply in
favor of, or don't mind the idea.  I really only need to hear from the
NO votes.  I'd be interested to hear any other thoughts you have on
the subject as well.  Mail to Alan@MIT-MC.

Date:  13 December 1981 00:27 est
From:  Schauble.Multics at MIT-Multics
Subject:  Russian views
To:  cube-lovers at MIT-AI

    "Some assert that the Rubik Cube reflects the philosophy of the
    Reagan Administration -- to build and destroy aimlessly in a
    futile search for a solution to the world situation. Others see
    Americans' efforts to solve the puzzle as the desire to escape
    from a disordered life."
    
        Izvestia reporter Melor Sturua, reporting the views of
        unidentified Soviet sociologists that the Rubik's cube fad in
        the U.S. is a "new psychosis"  July 12, 1981



Date: 13 December 1981 0150-EST (Sunday)
From: Paul.Haley at CMU-10A
To: cube-lovers at mit-mc
Subject:  Leaving the mailing list
CC: Paul.Haley at CMU-10A
Message-Id: <13Dec81 015008 PH71@CMU-10A>

Please remove me from the cube-lovers mailing list. Paul

Date: 16 Dec 1981 1055-PST
From: ISAACS at SRI-KL
Subject: general puzzles
To: cube-lovers at MIT-MC

  In order to perhaps start some discussion of puzzles in general, I will
include some first steps in attempting to come up with some kind of
classification scheme for puzzles. I have included both my own, and Jerry
Slocums', though I haven't put in the details of either at this time (I'm
lazy - if there is enough interest, I will fill out the outlines a little
more.) Anyway, any suggestions for diferent ways, or changes, will be
appreciated.  Can you put all YOUR favorite types of puzzles somewhere in
the two lists?

-------

Date: 16 Dec 1981 1057-PST
From: ISAACS at SRI-KL
Subject: inclusion for previous message
To: cube-lovers at MIT-MC

              PUZZLES

I. LIFE
  A. Technical (science, social science, etc)
  B. Psychological (ie, people problems)

II. CREATED
  A. Mental
     1. Language (word) puzzles
        Crosswords, Acrostics, word squares, etc.
     2. Logical puzzles
        Bridge crossing, truth tellers, etc,etc -- all those pencil and
        paper puzzles that appear in most non-word puzzle books.  Mostly
        mathematical.
  B. Physical  (Also sometimes called "Mechanical Puzzles")
     (THESE ARE THE KINDS I AM MOST INTERESTED IN, AND PROPOSE GOING INTO MORE
      DETAIL ABOUT.)
     1. Geometrical (including 2 and 3 dimensions; jig-saws; things like that.)
     2. Topological (rope and string puzzles, that involve deformation.  
                     Possibly including wire puzzles.)
     3. Combinatorial (box filling puzzles; Rubics Cube)
     4. Manipulation (dexterity?) (mazes, rolling balls, ballance puzzles)
     5. Physical? Miscellaneous? (Other types, like puzzle jugs; optical
                                  puzzles, centrifical force.)

I'll try to expand on this classification at a later time - I just want to
get something started now.  You can see that there are many problems in this -
a lot of puzzles don't fit easily in any of these, and others fit clearly into
more than one.  The trouble is that you can look at puzzles by their form, or 
by how they are constucted, or how they are solved.  And no one way seems
satisfactory for all.  Anyway, following is another classification, just
of Mechanical (Physical) puzzles, by Jerry Slocum, who has one of the biggest
puzzle collections in the world.

1. PUT-TOGETHER PUZZLES - Putting the object together is the puzzle
      Tangrams, jigsaw, soma, instant insanity, magic squares, puzzle rings
2. TAKE-APART PUZZLES - Taking the object apart or open is the puzzle
      Trick or secret opening
3. INTERLOCKING SOLID PUZZLES - Disassembly and Assembly is the puzzle
     Burrs, 3-D jigsaws, keychain, geometric object and figures (the Japenese
     wooden puzzles)
4. DISENTANGLEMENT PUZZLES - Disentanglement and entanglement is the puzzle
     Wire puzzles, nails, string and rope puzzles
5. SEQUENTIAL MOVEMENT PUZZLES - Moving Parts of the Object to a Goal While
   Following Rules is the Puzzle
     Solitaire, sliding block puzzles, rotating cube (Rubik) puzzles, maze and
     route.
6. DEXTERITY PUZZLES
     Rolling Ball, cup and ball, etc
7. MISCELLANEOUS 
     Puzzle jugs, folding, paper and card, matchstick, trick, vanish puzzles, 
     optical puzzles, find-the-object, rebus, reversibles, etc.
-------

Date: 16 December 1981 22:47-EST
From: Alan Bawden <ALAN at MIT-MC>
Subject: puzzle-lovers
To: CUBE-LOVERS at MIT-MC

Please note that despite the previous message we are NOT all agreed
that cube-lovers should devolve (?) into a discussion of puzzling in
general.  Those of you who object to this idea are still welcome and
encouraged to mail me an objection.  Currently I have recieved only a
couple of replies, so if you don't say something now, this is likely
to become the Puzzle-Lovers mailing list in a week or so!

Date: 16 December 1981 22:58-EST
From: "Martin B. Gentry, III" <GENTRY at MIT-AI>
To: CUBE-LOVERS at MIT-AI


Considering the recent suggestion that this list be used for discussing
puzzles of all types, but not trying to force it in that direction, the
following is submitted for your interest:

	In the Games section of the January OMNI there
	is an article about some currently unsolved puzzles such as
	\Masquerade/, \The Will: A Modern Day Treasure Hunt/ and
	the Beale ciphers. Included is a listing of Beale Cipher #1.


Date:  16 December 1981 23:32 cst
From:  VaughanW at HI-Multics (Bill Vaughan)
Subject:  puzzles of a different kind
Sender:  VaughanW.REFLECS at HI-Multics
To:  cube-lovers at MIT-MC

the puzzle classification is interesting - but i propose another
main heading:  game puzzles (chess problems, bridge puzzles etc.)
These probably fit in somehow but I don't know exactly where.

(my 2 cents' worth: vivat puzzle-lovers, &  keep up the digest
format.)

Date: 18 December 1981 00:37-EST
From: Alan Bawden <ALAN at MIT-MC>
Subject: Merry Christmas
To: CUBE-LOVERS at MIT-MC

While I am visiting my relatives in Philadelphia and while Dave
Plummer is doing something-or-other similar in honor of the Christmas
break, we have put the Cube-Lovers list back on direct distribution.

				-Alan

Date: 18 Dec 1981 1550-PST
From: ISAACS at SRI-KL
Subject: Authors' signing
To: cube-lovers at MIT-MC

 Don Frederick, author of "Solve That Crazy Mixed-Up Cube", will be
at the OLD MILL Shopping Center in Palo Alto/Mountain View Friday evening
from 5 pm to 9 pm, and Saturday, Dec. 19 from 11 am till about 5 pm. His
is one of the few books that at least tries to help you remember a solution
to the cube.  He'll be selling and signing his book.  He claims plain language,
cartoons, solving other shapes, and a diploma in the back of the book.
  -- Stan
-------

Date: 21 Dec 1981 1522-PST
From: ISAACS at SRI-KL
Subject: new cube game
To: cube-lovers at MIT-MC

  I just got a new cube puzzle, called "Color Cube Mental Game".  Not a
Rubik's cube, but still combinatorial, and a lot easier.  It consists
of 8 cubes in a 3x3 box, arranged around the perimeter (so there is a
cubical hole in the center).  The cubes are colored on all 6 sides 
(identically in mine).  They roll by tilting them over an edge to an
adjacent vacant space, and they change colors in doing so.  The puzzle,
of course, is to roll them and randomize the pattern, then restore the
original color (or a new color).  What patterns can be made?  What is the
minimum number of rolls to solve from some pattern to another?  This type
of puzzle was talked about in Journal of Recreational Mathematics, and
similar types in Mathematical Games by Martin Gardner (I can give more
accurate references if anyone wants).  
  Here's the last paragraph on the "instruction" sheet:  "It should be
finished.  please don't dishearten and try your best till you can complete
the all test.  Good Luck!!!"

--- Stan
------- Happy Holidays to all
-------

Date: 29 December 1981 23:48-EST
From: Alan Bawden <ALAN at MIT-MC>
Subject: Happy New Year!
To: CUBE-LOVERS at MIT-MC

With this message we become a digest again.

Date: 30 Dec 1981 2017-EST
From: ELF at MIT-DMS (Eric L. Flanzbaum)
To: Cube-Lovers at MIT-MC
Subject: Query
Message-id: <[MIT-DMS].218800>


Cube Fans;

  Why on the package of the original Rubik's cube, does it say "Over 3
billion combonations ... just one solution" when in reality there are over
43 quintillion? Is it because the manufacturers of it thought nobody would
believe them?

				    Happy Cubing,
				the ELF


Date: 31 December 1981 16:14-EST
From: David C. Plummer <DCP at MIT-MC>
Subject: Query
To: ELF at MIT-DMS
cc: CUBE-LOVERS at MIT-MC

      Why on the package of the original Rubik's cube, does it say "Over 3...

Something like that.  I believe there is a small discussion about
this somewhere in the archives.

Date: 3 January 1982 02:11-EST
From: Alan Bawden <ALAN at MIT-MC>
Subject: nuts
To: CUBE-LOVERS at MIT-MC

There is a puzzle that has been around for several years called
something like "Nuts To You" or "Drives You Nuts" or some variation
thereof.  It consists of 7 hexagonal pieces (each resembling a nut,
hence the name) each with the numbers 1 through 6 inscribed around its
perimeter in some order.  The idea is to play a kind of hexagonal
dominoes with these pieces; a solution is found when all 7 pieces are
arranged in a hexagon (the puzzle comes complete with a hexagonal
frame for arranging them, it also resembles a nut) such that every
pair of adjacent pieces are labeled the same along their common edge.
For example one piece might look like this:

   _____
  /  6  \
 /3     1\
/         \
\         /
 \2     5/
  \__4__/
                                6
which I shall abbreviate by:  3   1
                              2   5
                                4

which might participate in the following solution:

           5
         6   4
   6     1   2     3
 3   1     3     2   4
 2   5     3     1   6
   4     5   1     5
   4     2   6     5
 5   2     4     6   1
 3   1     4     3   2
   6     1   3     4
         5   2
           6

Each piece can be rotated (of course), but it cannot be flipped over.
(This restriction is enforced by simply not printing the numbers on
the other side!)

The seven pieces I have represented in my sample solution are
definitely NOT the ones that make up the commercially sold puzzle.  I
don't have that anymore, I just made these up.  It did have the
property that no two pieces were alike, just like my hypothetical set.

The commercial set of pieces seemed to have the property that there
was JUST ONE possible solution, although I cannot be 100% certain that
this was the case.

Now the puzzle itself is a bit dry.  But what I wonder about is the
following meta-problem:

Given that there are many (how many?) different sets of 7 pieces that
can be chosen, how "interesting" a property is it for there to be only
ONE solution?  Do you have to try hard to achieve that property?  Or
do most of them have it?  Indeed, do ANY of the have it (remember I
only SUSPECT that the commercial set has it!).

How about if we relax the restriction about duplicate pieces?

				-Alan

Date: 3 Jan 1982 0232-EST
From: ELF at MIT-DMS (Eric L. Flanzbaum)
To: Cube-Lovers at MIT-MC
Subject: The Pyramid
Message-id: <[MIT-DMS].219009>


Is there any written (i.e., a book like "The Simple solutio....")
solution for the pyramid? Do any of you people know a solution?
People tell me the pyramid is easier than the cube (that is of
course if you know how to do both). Is this true?

					Happy Cubing,
				    the ELF


Date: 3 January 1982 17:00-EST
From: Alan Bawden <ALAN at MIT-MC>
Subject: The Pyramid
To: ELF at MIT-DMS
cc: CUBE-LOVERS at MIT-MC

I know of no books about the pyramid, but I can vouch for the fact
that the pyramid is much easier than the cube.  I have essentially ONE
tool that is sufficient for all manipulations.  It isn't even a very
long one.  It is in fact such an obvious tool to try that many people
have discovered it independently.  I'll tell you what it is after I
insert the following spoiler warning:

*** SPOILER WARNING !!! ***  If you wish to solve the pyramid
yourself, stop reading now!

First off, notice that the points of the pyramid can be rotated
independently of all the other pieces.  Thus we can safely ignore
those pieces and simply twist them into position as the last step.  So
from here on in, when I say to twist about a particular point, I mean
to grasp the larger sub-pyramid that shares that point with the entire
pyramid and rotate it.  This motion always leaves the face opposite
the designated point fixed.  

This is not analogous to the way we think about the cube.  When
dealing with the cube we rotate a face, leaving the rest of the cube
fixed.  When dealing with the pyramid I am proposing to always rotate
everything BUT a face, a move that arranges to leave the four points
in the same position.

Having how described what I mean by a move, the actual tool is easy.
It's only four twists long.  Chose any two points, call them A and B.
The tool is simply AB'A'B.  It permutes three edge pieces that share
a common face, it flips two of those pieces over, and everything else
is untouched.

Since everything except the edge pieces can be placed in position
trivially (both the point pieces and the third kind of piece I haven't
mentioned yet), this tool might be sufficient (after arranging
everything else).  I don't know if it is in fact sufficient since I
also employ some conjugates (like CAB'A'BC') in my solution.

I leave the details of actually applying this tool for you to
discover by yourself.

				-Alan

From: Woods.pa @ PARC-MAXC
Date:  3-Jan-82 18:34:41 PST
Subject: Re: nuts
In-reply-to: ALAN's message of 3 January 1982 02:15-EST
To: Alan Bawden <ALAN at MIT-MC>
cc: Cube-Lovers at MIT-MC

I've seen the "Nuts" puzzle; I think it is indeed "Drive you nuts",
but I'm not sure.  I seem to recall that not every (any?) piece
had all six numbers on it; that is, I think one or more numbers
were duplicated on one or more pieces.  This of course opens up
the search space even more in trying to find a set of pieces with
exactly one solution.  Also, rather than consider which sets
of pieces (or what fraction of possible sets of pieces) have unique
solutions, I think it makes more sense to choose a solution and see
how likely it is that the pieces forming that solution have another
solved orientation (other than trivial rotations of the whole puzzle).
That way you at least eliminate all the sets that have NO solution.

	-- Don.


Date: 3 January 1982 23:05-EST
From: Alan Bawden <ALAN at MIT-MC>
To: CUBE-LOVERS at MIT-MC, Woods.pa at PARC-MAXC

Well, my memory is very clear on the point that each nut had all 6
numbers since my solution depended on that fact.  But sure, if we can
answer your extended problem all the better.

You mention rotations as a symmetry of a solution, which reminds me of
another kind of symmetry that is relevant to exactly what is meant by
"a set of nuts".  Clearly the values of the integers inscribed on the
pieces have no effect of the character of the solution.  If someone
broke into your house one night an erased all the 1's and replaced
them with 2's and replaced all the 2's with 1's, they wouldn't have
damadged your puzzle in any way (ignoring that you may have memorized
the solution by number, but that's your fault for chosing a bad way to
remember the solution).  So applying any of the 6! permutations of 6
things to the numbers leaves your "set of nuts" fixed (relative to
eachother).  The manufacturers's choice of numbering is thus at least
partially arbitrary.

Date: 5 Jan 1982 09:38 PST
From: Lynn.ES at PARC-MAXC
Subject: Re: The Pyramid
In-reply-to: ELF's message of 3 Jan 1982 0232-EST
To: Cube-Lovers at MIT-MC

Nourse's new book "The Simple Solutions to Cubic Puzzles" has a pyramid
solution.  Aside from the (almost) obvious stuff about the little and big portions
of each corner, it has specific macros for moving edge pieces from any edge
position to the top-front (he solves it with a flat surface up), and for flipping an
edge in place (actually a pair of edges, but the second one is by design an edge
not yet solved).  Then there are two macros that jumble the edges of the bottom
large corner until it is right.  The solution is simple to follow as a cookbook
operation, but not so simple to memorize, involving several macros rather than
one or two that many people are using.  

/Don Lynn


Date:  6 Jan 1982 1024-PST
From: ISAACS at SRI-KL
Subject: drive ya nuts
To: cube-lovers at MIT-MC

  The numbers on my "Drive Ya Nuts" are as follows (starting from 1 and reading
clockwise: 123456, 143652, 162453, 164253, 165432, 165324, and 146235.  As you
can see, each "nut" is unique.
  There are several related puzzles, including:
  "Japanese Mind Bender", which is logically identical to "Drive Ya Nuts", e
except it used colors instead of numbers, and each hexagon looks like a small
circus tent.
  "Super Dominos", 4x6 squares, each divided diagonally into four sections, and
colored in all possible ways with 3 colors (yellow, orange, and brown).  Object
is to arrange the squares in a 4x6 array, with adjacent edges of matching
colors, and the border to be only one color.  It claims there are about
12000 different solutions, a computer generated number.  I find it difficult to
find even one.
  "Colored Squares Puzzle", A magnetic version of Super Dominos.  They also
suggest a second puzzle as above, but with 2 colors around the edge.
  "Colored Triangles Puzzle", also magnetic (a match for "Colored Squares"),
this one has the 3 colors on 24 triangles, arranged in a hexagon.  Same problem,
though.
  "Try Nine" is octagonal pieces with numbers (similar to "Drive Ya Nuts"),
to be arranged in a 3x3 lattice such that numbers match vertically, horizontally
, and diagonally.  That is, the square spaces between the octagons have opposite
 edges with the same number.
  "Square Crazy" or "Le Carre fou! fou!", a new puzzle from Montreal (If
anyone wants one, I can give you the address).  It is 9 cardboard square
"cards", each of which has half a fish on each edge (either the head or the
tail), and each fish is one of 3 or 4 colors.  So direction and color has to
match up at each edge of the 3x3 square.  I met the inventer, and he says
he used a computer to ensure there is only one solution.
  "Its Knot Easy" has 16 square plastic pieces to be arranged in a 4x4 array,
so that the picture of a rope running through the pieces forms a continuous loop.
  And many more.  Including a dodecahedron whose faces turn, to try to get a
continuous loop running through it, and a version of 3x3 squares which uses
heights instead of colors.
  I think all the "Instant Insanity" type of puzzles are also related to this.
What we need are some non trial-and-error methods of solution, as well as
algorithms to tell if the solution is unique.  
  By the way, I am planning to make a Rubik's cube with 9 colors, with a
solution having 9 different colors on each face.  The question is, how to 
design it so it is solvable (preferably not just by trial-and-error), and
so there is a unique solution.
--- Stan Isaacs
-------

Date: 6 January 1982 19:09-EST
From: Alan Bawden <ALAN at MIT-MC>
Subject: fowarding this...
To: CUBE-LOVERS at MIT-MC

Date:  6 Jan 1982 0955-PST
From: ISAACS at SRI-KL
Subject: Re: Masquerade
To: RP at MIT-MC, TRB at MIT-MC, mike at UCLA-SECURITY, SWG at MIT-DMS,
    HAGERTY at RUTGERS, CC.Clive at UTEXAS-20
cc: ALAN at MIT-MC
In-Reply-To: Your message of 6-Jan-82 0404-PST

  How should we do this?  List all the things we see? By "chapter", or
in some other order.
  I don't have my book here right now, so I will start from memory, by 
listing some of the obvious, just to make sure we all start from the
same place.  
  1. Each picture has a hare in it.  Sometimes hidden, sometimes in the open.
     I've always wondered if some of the obvious hares also have a second, 
     hidden one somewhere.  The man in the rabbit ears and with rabbit feet
     bothered me the most.
  2. Each picture has around the outside edge, letters in red, which can
     be re-arranged to make some appropriate word.  In addition, there are
     letters marked with a spike, which form a second word.
  3. Several of the chapters have riddles; I think I know the answers to all
     but one.
   Besides these characteristics of each picture, there are many symbols.  We
should just start to make a list of what they are, and guesses as to what they
mean.  Again, I don't have my book handy, but two examples:
  1. Many pictures have a number square.  What is its' significance?  If I 
remember correctly, all but one of them were missing the same number.
  2. In one of the pictures, the moon(?) has her fingers arranged to say
"love" in sign language.

  Somebody should put in the answers to the riddles and pictures in 1-3 above;
just to have them on record, and to make sure we all agree.
  --- Stan Isaacs
-------

Date: 6 January 1982 19:10-EST
From: Alan Bawden <ALAN at MIT-MC>
Subject: Puzzle-Lovers
To: CUBE-LOVERS at MIT-MC

Well, it looks like the discussion is going to start wandering a bit.
It is not to late to register your complaints with me if you don't
like this turn of events.

Where can one obtain this "Masquerade" object anyway?  Is this
something I find in any bookstore?

Date:  6 Jan 1982 2328-PST
From: Alan R. Katz <KATZ at USC-ISIF>
Subject: Folding puzzle
To: cube-lovers at MIT-MC
cc: katz at USC-ISIF

There is a puzzle out which looks to be a book of paper.  The purpose seems
to be to fold over the paper to make various patterns or shapes (various
pages are folded over but the patterns or two dimensional).  Does anyone
know anything about this and is it any good??


				Alan
-------

Date:  8 Jan 1982 1009-PST
From: ISAACS at SRI-KL
Subject: FOLDING PUZZLE
To: CUBE-LOVERS at MIT-MC

  The folding puzzle book (I can look up the name and reference at home if
anyone is interested) is original, and a "cute" idea, but not very difficult
to solve.  It is a pleasant pastime, and good to do with older children, I
found.  It seems to be solved almost purely by trial and error.  I think with
some work, a truly interesting and difficult puzzle could be made using
the idea of paper folding to make patterns.
--- Stan
-------

Date: 25 Jan 1982 0952-PST
From: ISAACS at SRI-KL
Subject: SILENCE
To: CUBE-LOVERS at MIT-MC

  HELLO! Anybody still there?  Its been awfully quiet lately.
  --- Stan
-------

Date: 8 February 1982 20:36-EST
From: Alan Bawden <ALAN at MIT-MC>
Subject: St. Valentine's Day Massacre?
To: CUBE-LOVERS at MIT-MC

I wondered out loud to some of the Masquerade-Lovers why no one was
discussing Cubes or Masquerade or anything.  And got the following
response which I foward to you all:

Date:  8 Feb 1982 1639-CST
From: Clive Dawson <CC.Clive at UTEXAS-20>
Subject: Re: Masquerade

Masquerade is still on my stack, but actually I've come across another
amusement which has been taking up my time during the last couple of
weeks.  Does anybody know about the St. Valentine's Day Massacre?
It is a cross-country road rally which takes place on paper--the Rand McNally
road atlas to be precise.  You are given hundreds of very detailed and
tricky instructions and puzzles as you make your way from the Golden Gate
Bridge to the Statue of Liberty.  You must answer questions along the
way about what you see, time and distance between various points, etc.
I've found the whole thing to be a lot of fun and quite challenging, even
though the entry fee was $24.00 (includes a copy of the Road Atlas).
Trophies will be given out in 3 different classes of competition (Class C=
first timers, Class B= previous participants, Class A= Previous trophy
winners).

If anybody is interested, I'll be happy to provide more info--deadline for
entry is Feb. 14, answers must be sent in by March 1.

Also, I'd like to hear from anybody already involved if they'd like to
compare notes, etc.

Clive
P.S.  Alan--go ahead and forward to Cube-Lovers if you wish.
-------

Date:  9 Feb 1982 1624-CST
From: Clive Dawson <CC.Clive at UTEXAS-20>
Subject: More on St. Valentine's Day Massacre
To: cube-lovers at MIT-MC

Several people have inquired about this further, so I thought I'd go
ahead and send more info to the group...


Write to:

	St. Valentine's Day Massacre
	P.O. Box 53
	La Canada, CA 91011

Since time is short, you may want to give them a call between noon and
3PM PST at (213)790-4937.  Deadline for entering is Feb. 14.

For $24, they will send you the Rand McNally Road Atlas, and a 50-page
booklet with all of the rules, course instructions, questions, and
answer sheet.  Here's a sample of what it's like:

1.  Begin the 1982 St. Valentine's Day Massacre at the interchange
north of the Golden Gate Bridge (NC-19 on page 11), where you find
two cars from which to choose your first Massacre conveyance:
a 1960 Falcon and a 1930 L-29 cabriolet.  Hmmm: Ford or Cord.
Ponder the selection for something under a second, then hop in and
head south on state highway.

   Q1.  How many "San Gregorio"s do you see?

	a) 0	b) 1	c) 2

2.  Left on 152

    Q2.  How many of these do you see?

		Bell
		Bells				a) 0	d)3
		Bell's Station			b) 1    e)4
		"Bells Station"			c) 2

3.  South onto 33.

     Q3.  Do you see Devils Den?		a) yes   b) no

4.  Left on 58

     Q4.  How many of these do you pass?

		Calloway
		Oak				a) 0   b) 1  c) 2

5.  South on 99

6.  Southeast on Intersate 5

     Q5.  Which of these do you see first?

	   a)  Los Angeles    b) Burbank    c) Gorman
- - - - - - - -


By the time you start answering these questions, you will have read
through the 6 pages of rules which spell out VERY PRECISELY just what
it means to "see" something (be within 1/4 inch of it on the map),
the difference between going ONTO a highway and ON a highway, and
the difference between "Bell" and Bell.  The course following rules
can be very tricky, telling you when to try to stay on a given road,
when you can turn, when you can switch maps, when to start and stop
consideration of a given question, etc.

Later in the course you will be given various puzzles (visit all 6
National Monuments in 4 Arizona counties by travelling only on U.S.
and state highways with no U-turns, for example.)  At one point you
actually travel by balloon at 120,000 feet--this turns out to be about
1 inch above the paper for purposes of "see"!  The instructions are
filled with various tricks and traps which will take you miles off
course if you're not careful.  But they are also cleverly constructed
to eventually get you back on the right track again so that you don't
realize your mistakes.  One of their favorite tricks is exemplified by
Question 5 above: you're tempted to say Gorman since it is north of
L.A., until you realize that you see Los Angeles COUNTY first!

They give you plenty of help on the first leg so that you get an idea
of what it's all about.  You can stop after 4 legs and compete for the
Class C trophies, or else go onto legs 5-7 to try for a Class B
trophy as well as one for Class C.  Legs 8 and 9 are for the Class A
trophies, and are the hardest of all.

I've entered a couple of similar contests put on by the same people
(The Great Maltese Circumglobal Trophy Dash) and even though they're
a little slow with scoring and sending results, I've found them to
be very fair and reasonable.  As soon as the contest deadline passes,
they will send you an answer booklet with a complete explanation of
the entire course and reasons for each answer.  You then have the right
to challenge their reasoning if you disagree with an answer.  If they
agree with your challenge, they will eliminate that question from the
scoring.  After the protest period ends, they do the final scoring,
sending out a complete list of scores for all participants as well
as trophies.  The top 10% in each class all get something, which
gets progressively fancier for the higher percentiles and classes.

Depending on how careful you are, you can expect to spend anywhere
between 30 and 100 hours if you complete the whole course.
Personally I find the entry fee a little high, but considering the
free road atlas and the number of hours of tortu...er, entertainment,
I guess it's worth it!

If you have any other questions, let me know.

--Clive
-------
-------

Date: 16 Feb 1982 0931-PST
From: ISAACS at SRI-KL
Subject: MAGIC OCTAHEDRON
To: CUBE-LOVERS at MIT-MC

  The Magic Octahedron (Octahedron Cube?) is out.  A friend got one for me
over the weekend.  Each face is 9 triangles; it works sort of like two
tetrahedrons base to base, except each tetra is really a 4-sided pyramid.
The solution is not too hard, and not too different from the tetrahedron.
Like the tetra, the corner and next-to-corner pieces don't travel-they
only twist.  The remaining edges are isomorphic to the edges of a cube.
In fact, if you corner-center a Cubes coloring (ie each face has 4 colors,
each corner has one; see the Scientific American article), and then peel 
off the labels from the corners, you have an exact isomorphism of the
Octahedron.
   By the way, this version twists around the corners, using the same
type mechanism as the original Cube (I think).  It should be possible to 
use the mechanism of the Tetrahedron (with its 4 axes of rotation) to 
build an Octahetron whose faces twist.  Any mechanical engineers out there
to do so?
   --- Stan
-------

Date: 14 Mar 1982 0703-PST
Sender: OLE at DARCOM-KA
Subject: Poison
From: Ole at DARCOM-KA (Ole J. Jacobsen)
To: Cube-Lovers at MIT-MC
Message-ID: <[DARCOM-KA]14-Mar-82 07:03:13.OLE>

Folks,

You may love your cubes, but don't kiss them or lick your fingers
after having twisted them! According to a recent article in one
of the big papers here, the colotrtabs contain exceptionally high
levels of lead. The worst is the yellow face, closely followed by
the green, orange, red and blue. You have been warned.

<OLE>

Date: 14 March 1982 10:32-EST
From: Richard Pavelle <RP at MIT-MC>
Subject:  MASQUERADE- too little, too late
To: CUBE-LOVERS at MIT-MC

We were beaten by a dog. According to an article on Page 2 of the
Sunday Globe (3/14/82) the Golden Hare was found "by a man whose
dog accidentally stumbled across it"...  "The hare was buried just
beneath the surface in the shadow of an ancient cross at Ampthill
which commemorates Catherine of Aragon..". One of the major clues
in the book pointed to her.

Date: 23 Mar 1982 1032-PST
From: ISAACS at SRI-KL
Subject: new book
To: CUBE-LOVERS at MIT-MC

  "Handbook of Cubik Math", by Alexander Frey and David singmaster, is
now out from Enslow Publications.  Most of the contents, but not all,
are from Singmasters other opus, but this is better organized and
clearer.  I have not finished it yet, so I can't give it a full review.
   By the way, over the weekend I found that Meffert, inventer of the
Pyraminx Tetrahedron has also invented about TWENTY (20) other variations!
Diferent versions of Icosahedra and dodecahedra; a triangular prism, a cube
cut along the three planes that make hexagonal cross-sections, etc. I
have high hopes that they will be available in this country soon.
   We need a name for this type of puzzle.  Clearly, "Cube" is not
very satisfactory any more - not with spheres, tetrahedra, octahedra, etc.
"group theory puzzles" is too broad - it also includes sliding block
puzzles and others.  Perhaps something like "Axial" or Axially Rotation
or some such.
   David Singmaster is publishing a "Cubic Circular" (nice name) now.  If
anyone wants his address, I'll be glad to supply it.  He will also be
selling Mefferts puzzles, I understand.
    Has anyone seen a 4x4x4 or 5x5x5 cube yet?  Anynew information on
them?  Does anyone know if the Masquerade clues have been published now
that the hare has been found?
   The Subject of this message is out of date. Oh well.
     ---Stan
-------


Date: 23 March 1982 15:51-EST
From: David C. Plummer <DCP at MIT-MC>
Subject:  rumor confirmation
To: CUBE-LOVERS at MIT-MC

The 4x4x4 has been confirmed by two stores in the Boston area
(Games People Play [Harvard Square] and The Name of the Game
[Quincy Market]).  Unfortunately Ideal is only taking orders now;
delivery is not expected until the fall (****sigh****).  I have
heard nothing of the 5x5x5. 

ALAN thinks we should solve the beast before we actually get one.
This could done in our heads, on paper, by machine assistance, or
by machine.  I have thought about it a little, and I think I have
all the necessary tools to do it.  It is not as easy as saying,
"Well if we just consider these two planes as one, it looks like
the 3x3x3, and..."  That doesn't work for the majority (the
hardest parts) of the puzzle, for reasons I'll be happy to
explain.

Before people start asking a lot of questions about the
properties of the 4x4x4, they should look in
	MC:ALAN;CUBE 4X4X4
which contains the parts of the archives that discuss the 4x4x4.

If there is any interest I can try and create a discussion...

Date: 25 Mar 1982 1714-PST
Sender: OLE at DARCOM-KA
Subject: Singmaster and IBM
From: Ole at DARCOM-KA (Ole J. Jacobsen)
To: Cube-Lovers at MIT-MC
Message-ID: <[DARCOM-KA]25-Mar-82 17:14:12.OLE>

Singmaster has as you probably know set up his own company selling
cubes and other puzzles of all shapes and sizes. The latest price-
list includes Rubik's Calendar Cube, a true Executive Toy. The object
is, if I understand it correctly, to get one or more faces displaying
the correct date/month. He also sells Braille cubes,- Ideal cubes
with raised dots. His "Cubic Circular" as mentioned by Stan, is well
worth getting, as it contains alot of interesting information.

A Marketing Consultant from IBM gave me a cube which has IBM4331
printed on all the six center faces and the words: Value,Function,
Support,Reliability,Performance,Productivity on each facie of each
face respectively. The orientation of the IBM 4331 relative to the
other 8 repeated words on each face makes it a supergroup variety.
In my opinion, this cube is the best product to come from IBM in
years!

<OLE>

Date: 31 Mar 1982 1821-PST
From: ISAACS at SRI-KL
Subject: 4**3,5**3
To: cube-lovers at MIT-MC

    I should think you could solve a 4x4x4 cube by applying 3x3x3 moves, using
different combinations for the "center slice" - ie, rotating the center
two slices together allows all the regular corner moves; it also allows
switching adjacent edge pieces (in pairs) by using whatever you use to flip
edges in 3x3x3.  Using only one of the center slices as "center", and 
turning the other with a face should allow flipping a pair of L-R pieces,
so that L becomes R and vice versa.  The center 4 pieces (which I haven't
thought about carefully) probably can be changed around at
(even parity) will, sometimes treating them as edges (since they can be
carried along on certain of the "edge" moves), and sometimes as centers
(for instance, to rotate a group of 4 of them halfway around in place).
   I don't know what new parity limitations exist; nor do I know if
the same type of sequences are efficient for solving (ie, top-middle-bottom,
etc), but I shouldn't think the new cubes will be so very much more
difficult.
   I would assume the 5x5x5's could be handled similarly, except, of course,
they have a real center.
--- Stan
-------

Date: 31 March 1982 22:26-EST
From: David C. Plummer <DCP at MIT-MC>
Sender: DCP0 at MIT-MC
Subject: 4**3,5**3
To: CUBE-LOVERS at MIT-MC

NO, NO, NO !!!!  You CANNOT treat 1 of the center slices of a
4x4x4 as a center of a 3x3x3.  Suppose you did this for one axis,
and for the other two axes you treated both "centers" as a unit
(and therefore the center slice of a 3x3x3).  Now take one of the
axes with a double width center, and rotate an outer slice 180
degrees.  Suppose the front face looked like this:

	+####+####+####+####+
	#    #    #    |    #
	#    #    #    |    #
	+####+####+####+####+
	#    #    #    |    #
	#    #    #    |    #
	+----+----+----+----+
	#    #    #    |    #
	#    #    #    |    #
	+####+####+####+####+
	#    #    #    |    #
	#    #    #    |    #
	+####+####+####+####+

You rotate the top slice and the front face now looks like:

	+####+####+####+####+
	#    |    #    #    #
	#    |    #    #    #
	+####+####+####+####+
	#    #    #    |    #
	#    #    #    |    #
	+----+----+----+----+
	#    #    #    |    #
	#    #    #    |    #
	+####+####+####+####+
	#    #    #    |    #
	#    #    #    |    #
	+####+####+####+####+

Notice that the top layer does not go very well with the bottom 3
layers.  The 5x5x5 has similar problems.  

I think the right way to solve both the 4x4x4 and 5x5x5 at first
is to use mono-flips.  Once conceptually understood, they are
very powerful and easy to visualize.

Date: 9 Apr 82 15:00-PDT
From: mclure at SRI-UNIX
To: cube-lovers at mc
Subject: Peking newspaper criticizes cube

a058  0436  09 Apr 82
PM-China-Cube,100
Newspaper Decries Rubik's Cube Craze
    PEKING (AP) - Rubik's Cube, the colorful brain-teaser that has
started to captivate the Chinese, is being criticized by a Peking
newspaper as a dangerous pastime that can lead to divorce, abnormal
behavior, high blood pressure and aching fingers.
    ''The Magic Cube possibly is beneficial to sharpening intelligence,
but don't forget that its side effects might bring danger,'' the
Peking Evening News said, using the puzzle's Chinese name.
    Last November, the Chinese press reported people were standing in
line to buy the cube in Shanghai, and that city already had held its
first Magic Cube contest although the toy hit the market only a few
months earlier.
    
ap-ny-04-09 0721EST
**********


Date: 22 April 1982 1703-EST (Thursday)
From: Dan Hoey at CMU-10A
To: Cube-Lovers at MIT-MC
Subject:  Cubebot
Message-Id: <22Apr82 170345 DH51@CMU-10A>

Caption from a photograph said to be from the Washington Post,
April 13, 1982:

	University of Illinois engineering student Daniel Talken
	adjusts the mechanical hands of a robot built by students
	at the school in Urbana to solve the Rubik's Cube puzzle.
	The robot's computer brain can work out the solution in
	two-tenths of a second, but it takes the hands about 12
	minutes to make average 110 moves to solve the puzzle.

The photo shows a guy diddling a complicated machine in which an
unsolved cube is visible.  Eyes are painted on the front of the
machine's support above a bulge that may be functional as well as
vaguely resembling a nose.  No camera equipment is apparent;
presumably they tell the machine how the cube is scrambled, or
cheat and have the machine itself scramble the cube.


Date:  3 May 1982 1724-PDT
From: ISAACS at SRI-KL
Subject: RUBIK'S CUBE CLUB
To: CUBE-LOVERS at MIT-MC

  I just got a flyer from Ideal announcing their Rubik's Cube Club,
address Box 72, Hollis, NY 11423, $5.00 for a year.  They also have
"merchandise" (no prices included), including cube shirt, tie, patch,
button, bumper sticker, etc.  They also have a poster with many of
Ideal's cube-related products on it, including the 4**3 (called
Rubik's Revenge), a 2**3, Rubik's Race, Rubik's Challenge, etc.  I
can't tell quite what all of them are.
  Also, there are several new group-theory puzzles on the market (here
or in England), sort of related to the barrel.  Orb-it is a sphere 
with beads that rotate around 4 parallels.  In addition, the whole 
sphere rotates around a meridian, to bring different halves of 
the parallels into contact, changing from 4 separate circles of 
beads to one continuous "spiral", or two disjoint closed paths.
Equator Puzzle is a sphere with 3 intersecting (of course) great
circles, each consisting of 12 squares which move around the circles.
The coloring is into 4 segments, sort of like orange slices.
The Trillion Puzzle (will be available from Ideal) is a cross of colored
pieces, 17 of them, four each of 4 colors plus an extra red piece.
They can be arranged with each arm of the cross monochromatic, or in
4 circles around the center.  "The cross lies in a circle divided into 
3 concentric rings which are independently rotatable, though the outer
ring cannot move when the plunger is pushed in.  The middle ring is 2
pieces wide."  (From the description in Singmasters catalogue). The 
plunger moves one arm (9 pieces) over 1 piece.  About 1 billion 
distinguishable patterns.  And many more - Hungarian Rings, and
Gears, for two.  I also saw some which used polarized light, so
it was hard to tell where the pieces should go.
   The latest cube I've seen has Pac-man and other electronic game
figures on it.
       --- Stan Isaacs
-------



Date: 4 May 1982 00:15-EDT
From: Alan Bawden <ALAN at MIT-MC>
Subject: The Archives
To: CUBE-LOVERS at MIT-MC

Those of you who look through the archives of old Cube-Lovers mail will notice
that I have split off a new section of the archive.  The mail now lives in:

MC:ALAN;CUBE MAIL0	;oldest mail in foward order
MC:ALAN;CUBE MAIL1	;next oldest mail in foward order
MC:ALAN;CUBE MAIL2	;more of same
MC:ALAN;CUBE MAIL3	;still more of same
MC:ALAN;CUBE MAIL	;recent mail in reverse order

(I have also corrected a problem with the tabs in MAIL0 so that the diagrams
contained therein are readable once again.)

Date: 10 May 1982 0004-EDT
From: ROBG at MIT-DMS (Rob F. Griffiths)
To: cube-lovers at MIT-MC
Subject: Games magazine...
Message-id: <[MIT-DMS].231410>

There is a conteset in the cureent issue of Games magazine called
Air Race. Its basically modeled after the type of puzzle where you
try to run through once and rack up as many points as possible
without retracing your steps. I am wondering if nay of you out
there subscribeto Games, and are trying the contest? What types
of scores are you achieving? Any info appreciated. (I would like
to try to find out if my scores are higher/lower than average)
			Thanks;
			-Rob.


Date: Thursday, 13 May 1982, 14:48-EDT
From: Bernard S Greenberg <BSG at SCRC-TENEX>
Subject: 4 x 4 x 4
To: alan at MIT-MC, dcp at MIT-MC, cube-lovers at MIT-MC

4 x 4 x 4's are here.  Games People Play on Mass Ave sell's 'em at $15.00 apiece,
called "Rubik's Revenge".  I have two here (at Symbolics).  No progress yet, other than
the obvious corner moves.

Date: 13 May 1982 23:44-EDT
From: David C. Plummer <DCP at MIT-MC>
Subject: 4 x 4 x 4
To: BSG at SCRC-TENEX
cc: ALAN at MIT-MC, DCP at MIT-MC, CUBE-LOVERS at MIT-MC

Indeed, I have one also.  I thought about solving it for a couple
of months now.  Never wrote anything down, I think I just have a
working knowledge of the types of transforms needed.  It took me
about 1.5 hours to solve it.  I thought I had ALL the necessary
tools, but there is still one tool that alludes me.  It happens
half the time, and when I get it, I try to mess it up just enough
to allow it to be solved again.  As ACW pointed out, most
intuition about edges and centers is WRONG.  The first thing I
tried to the original solved cube was the extended Pons.  Much to
my dismay, it didn't work.  So if you think you know how to
extend your 3x3x3 transforms to the 4x4x4, put your fingers where
your mouth is!!

Date:     14 May 1982 0413-edt
From:     Ronald B. Harvey           <RHarvey at MIT-MULTICS>
Subject:  4 x 4 x 4
To:       cube-lovers @ mit-mc

I just got mine (two) today.  I had a friend who was in Boston on a
business trip get them (to Phoenix) for me.  They are also available at
The Name of the Game in Faneuil Hall-Market Square also for $15 (14.99).

I spent a few hours playing this evening, wondering what tools others
have come up with.  I seem to be able to get three faces which share a
vertex into perfect shape, and then kind of falter.  Oh well, it will
have to wait until after work tomorrow (Friday).

 - Ron

Date: 14 May 1982 14:35-EDT
From: Alan Bawden <ALAN at MIT-MC>
Subject:  cube talk
To: CUBE-LOVERS at MIT-MC

Date: 14 May 1982 13:53-EDT
From: Alias for WGD <bil>
Sender: ___100
To:   ALAN
cc:   CWH
Re:   cube talk

There is a talk at Northeastern on the group of the cube this Monday
by Ramshaw. I don't recall the details wrt to time and place but
there is a notice on the seminar bulletin board on the 2nd floor
of building 2 (last corridor).

Date: 14 May 1982 21:46-EDT
From: Alan Bawden <ALAN at MIT-MC>
Subject:  cube talk
To: BIL at MIT-MC
cc: CUBE-LOVERS at MIT-MC, CWH at MIT-MC

Well, I looked for the announcement to get the details, but I couldn't find it.
The closest thing I could find was an announcement for a conference this summer
in Canada on finite groups (hot stuff these days).  Perhaps someone else can
supply details?

Date: 16 May 1982 21:24-EDT
From: Richard Pavelle <RP at MIT-MC>
Subject: 4 x 4 x 4 = C^4
To: CUBE-LOVERS at MIT-MC
cc: RP at MIT-MC

I played with C^4 and I conjecture that the tools from C^3 are 
sufficient to solve it. I have not completely finished it but I 
think it is just a matter of time. 

One transformation I use repeatedly, in generic terms, is
S = top 180, 2nd slice up or down, top 180, 2nd slice down or up. This
is the verticle slice which is facing you. This is just the move in 
C^3 to move 3 edges in a plane whereas in C^4 the outcome is far more 
complicated.

The steps for C^4 are then
1) Align the centers with a map. Some use of S is necessary.
2) Do all corners as in C^3.
3) Use S repeatedly to position the edges and this is very laborious.
4) Use the C^3 edge flip (Rubik's transformation) to finish it off.

After several hours of C^4 I find C^3 looks like a toy. 

Date:  17 May 1982 04:07 edt
From:  RHarvey.Multics at MIT-MULTICS
Subject:  Re: 4 x 4 x 4 = C^4
To:  cube-lovers at MIT-MC
In-Reply-To:  Message of 17 May 1982 00:11 edt from Alan Bawden

I have also been using the tools described by Richard Pavelle as S to
do things to edges on the C^4.  However, there is nothing
that I have found that will flip to edges in place (that is, two
halves of a C^3).

Also, for aligning centers, I have been doing it AFTER edges.
The tools I use do not really map into C^3, and I have not
yet mastered them to the point wwhere I can finish them off. I
sit and stare at the damn thing a lot.  This behavior was
considered strange even at the science fiction convention
that I attended this weekend.

Speaking of strange, what kind of reactions have people
been getting from folks who see one for the first time?  I have
gotten the equivalent of double-takes from ones who do notice a difference.
Most, however, tend to say "Oh, I have a friend who can do that in
X minutes/seconds", in which case you know that they never learned
(or probably even played) to solve.

 - Ron

Date: 17 May 1982 11:39:22 EDT (Monday)
From: Bernie Cosell <cosell at BBN-UNIX>
Subject: solving the 4x4x4x4 cube
To: cube-lovers at mit-mc
Cc: frye at BBN-UNIX

I've managed to solve C4 in a way similar to those mentioned, but using a
different transform.  I solve C3 a little differently than most folk because I
use only one (!) transform to do ALL of the edge work.  On C3 the result of
this transform is (looking at the edges only):
               a                            a
from	    b     c             to      ~d     b
	       d                           ~c

Using it and its inverse and a few simple conjugates of it I can do all the
work that is necessary for the edges.  For me, at least, it has the twin
advantages of 1) being easy to remember (since it is fairly short and there is
only one of it), and 2) it has such bounded consequences that it is easy to
fix a cube without requiring a lot of planning (in the picture above, nothing
that is not shown changes: no other edges, no corners).

Anyhow, since that little move is a favorite of mine, I tried it on C4.  On C3
it comes in two flavors (the move and its inverse, or as it turns out, the
right-handed and left-handed versions).  On C4 it comes in four versions: the
move and its inverse, but each in a `left central slice' and `right central
slice' version.  Now for the fun part:  First off I started ignoring the
centers and I noticed that the move (lets call it `M') only moves around edges
in a single plane.  As I tried to figure out what the damn thing did I
discovered that it is a move of order 5!!!.  I find it truly hard to plan out
what happens when five cubes move around a little orbit, but I'm getting better
at predicting it.  The result is:
from:                       to:
	a   b                           a   b
    c           d                  ~d           e
    e           f                  ~f           g
	g   h                           c   h

With some pain I have been able to use ONLY M (and conjugates and powers of it,
of course) to get all the edges in place.  Then I looked at M a little more to
see what it did with centers.  This one is NOT planar, unfortunately, but is
simple enough to be useful: only two sets of centers are affected.  If you are
doing M on the top (to get the above edge transform), only the top centers and
the rear centers are affected.  And what happens is that there are two disjoint
three-cycles each involving two of the top center cubes and one of the rear
center cubes.  thus, you can easily use conjugates of M to move cubes, one or
two at a time, into place on all of the faces around any particular top.
Voila: done!

An amusing thing about M: since its edge permutation is of order 5 and the
center permutation is of order 3, the damn thing is an order 15 move.

    /Bernie


Date: 17 May 1982 17:30-EDT
From: Allan C. Wechsler <ACW at MIT-AI>
Subject: Reactions to 4^3.
To: CUBE-LOVERS at MIT-AI


Bernie and I have divided the double-takers into three major
classes:

1. "Oh, a four-by-four cube!"

2. "Oh, a four-sided cube!"

(and the best of all)

3. "Oh, a four-dimensional cube!"

One of the contributors to this list inadvertently made
error #3 by calling the 4x4x4 a "C^4".  Which brings me to
an interesting question.  What would a four-dimensional
cube really be like?  Let's just start with a 2x2x2x2.  It
would have sixteen hypercubies, all of the "corner" type.
Each hypercubie presents to the outside world four
three-dimensional hyperstickers.

The 2^4 has eight three-dimensional hyperfaces, presumably
each its own color.  I like the idea of using black and
violet along with the traditional red, yellow, orange,
blue, green, and white.  We can call the hyperfaces Back,
Front, Up, Down, Left, Right, In, and Out.  In is across
the puzzle from Out.  I doesn't touch O, but does touch
all the  other  hyperfaces.

The 2^3 can be seen as two square slabs stuck together.  Of
course, they aren't really "square" since they have
thickness.  A move consists of rotating one of these
squares with respect to the other.

Similarly, the 2^4 is two cubical hyperslabs stuck
together.  Of course, they aren't really cubical since they
have hyperthickness.  A move consists of rotating one of
these cubes with respect to the other.  

While the two slices of a 2^3 can only have four relative
positions, the hyperslices of a 2^4 can have twenty-four
different alignments.  There are twenty-four corresponding
twists.  One is the null twist, which consists of just
sitting and looking at the thing.  (In this case, this is
not such a trivial operation!)  Then there are six
different quarter-twists, as opposed to two in the 3D case.
Then there are, get this, eight "third-twists", which when
repeated three times bring the slice home.  These have no
analog in the 3D puzzle.  The remaining nine twists are
half-twists, three of one kind and six of another.

As in the three-dimensional case, the half-twists and
third-twists are all products of quarter-twists.  If we
regard one of the sixteen hypercubies as fixed (without
loss of generality, if you'll believe that) then there are
twenty-four different quarter-twists in all.  Twelve of
these are inverses of the other twelve, but selecting the
twelve "clockwise" ones is a lot harder than it is in the
three-dimensional case.  My intuition fails me.

I haven't tried to apply the Furst-Hopcroft-Luks algorithm
to this monster.  At most there are 6^15*15! =
4.7*10^11*1.3*10^12 = (very roughly) 6*10^22.  I suspect
that some kind of parity, trinity, or quaternity argument
will reduce this by a factor of two, three, or four.  

Yours with a headache,

   --- Allan

Date:     17 May 1982 2218-edt
From:     Ronald B. Harvey           <RHarvey at MIT-MULTICS>
Subject:  4x4x4 solution
To:       cube-lovers @ mit-mc

Amazing! I log in after working on my cube for an hour or so tonight in
order to announce my solution and I find another solution!

As previously stated, I use S to put edges into place.  This is done to
totality after I have put all of the corners together.

Now comes the hard part - all of the centers.  The tools that I have to work
on centers work on 4 or eight center cubies on opposite faces, plus, on
the 4-cubie swap version, two pair of edge cubies get swapped.  After
having a few edges swap (these are now treated like 3x3x3 edges...),
I usually replace these so I can undo the complex conjugations more
reliably.
Now that I can solve it, I going to search for tools that do finer
things, like double swaps for centers or edges without munging
the other centers or edges.

BTW, I have had the tools all weekend - it has just taken this long to
get the proper perspective in order to set up the proper conjugations...

Has anyone solved the construction problem?  Is Ideal's version similar
to Plummer's design of last December? Who DID do Ideal's version?

Any truth to the rumor (that I just started) that the reason that it has
taken Ideal so long to actually start selling the things was that they
were working on a solution booklet all this time? (They DO offer one)

 - Ron

Date: 18 May 1982 09:12-EDT
From: Richard Pavelle <RP at MIT-MC>
Subject: C^4
To: CUBE-LOVERS at MIT-MC

I solved it after about 10 hours. The very tricky part, as I mentioned
in my last message, is to move the edges about in a plane. I still do
not have a move which moves 3 in one slice (nicely), rather 2 in one slice 
and one in the adjacent slice. This makes the process very time consuming.
I also believe centers first is probably the way to approach it although
it is just a feeling. I think that moving only some of the center cubies 
will require very complicated transformations.

I would guess this cube requires about 10 times more moves than C^3.
Does anyone else care to speculate? I will be surprised if anyone
will ever solve C^4 regularly in under 10 minutes. 

Two questions:
1) Any comments about cube-lube. This cube is massive and needs some.
2) I heard there are about 10^50 configurations, true?

Date:     19 May 1982 0107-edt
From:     Ronald B. Harvey           <RHarvey at MIT-MULTICS>
Subject:  4^3 Cube
To:       cube-lovers @ mit-mc

In response to Richard Pavelle's message, there are about 7.4*10^45
color combinations.  See "mc:alan;cube 4x4x4" for more details on the
subject (especially the derivations) by Dan Hoey.

I bought a solution book from my favorite bookstore today. The title is
"The Winning Solution to Rubik's Revenge". It is a sequel to "A Winning
Solution to Rubik's Cube" by the same author (Minh Thai), who is billed
as the U.S. National Champion of the Rubik's Cube-A-Thon.  I assume the
Revenge book is not REALLY a winning solution... at least not yet.

His method is to put all corners together, go for two opposite centers,
then the edges of said centers.  Next he goes for the remaining edges
(numbering eight), and then the last four centers.

He also goes into many patterns, and has developed a notation which I
have not found immediately obvious.  I haven't looked at the book more
than to scan it yet, however.  He does list a few pretty-patterns, none
of which are checkerboards of any sort.

On a different subject, Paul Schauble accidentally found out how to take
apart the 4^3 today - twist an outer layer about 30 degrees so that an
edge of the twisted face is directly over the edge cube that now forms a
corner on the rest of the cube.  Pop out the cube from the outer layer,
twist the face again so that the popped cube's partner is in a similar
position for popping, and then pop it.  The corners now come out fairly
easily.

Unlike Plummer's design, the insides of this beastie is a sphere with
grooves running along it to make a kind of universal joint.  The center
cubies ride in these grooves and hold all of the other pieces in.  We
only took out about a half-dozen cubies because the cube was NOT in an
initialized  state, but closer to a pretty pattern.  I was definitely
not interested in getting it ALL apart in order to get it back together
correctly. (we did it correctly the first try!).  Unlike the 3^3 cube,
the 4^3 does not seem to come apart easily after the first few are out.

I seem to have gone on for a bit longer than I intended.  I will study
the pamphlet (published by Dell/Banbury by the way) and report in more
detail on notation later.

I just noticed that in the section on Cubology, the author lists the
number of "arrangements is something in excess of 3.7*10^45".  Since
Hoey's number is larger, I guess the statement is correct.

 - Ron

Date: 21 May 1982 1439-EDT
From: SWG at MIT-DMS (S. W. Galley)
To: cube-lovers at MIT-MC
Subject: Newsweek 4/19 (in case you missed it)
Message-id: <[MIT-DMS].232584>

"Rubikmania: Lots Of New Twists"
Newsweek, 19 April 1982, pp. 16f

  It has been hurled out of moving buses, dumped into trash mashers and
pounded to bits with blunt instruments.  With more than 43 quintillion
possible arrangements, Rubik's Cube puzzle may be the most infuriating
plaything ever marketed, as well as one of the most popular.  Seven
years after Hungarian architecture Prof. Erno Rubik constructed the
first cube to help his students understand three-dimensional objects,
world sales have passed the 30 million mark.  "It's phenomenal," says a
vice president for the Ideal Toy Corp., which makes the plastic cube
under an agreement with a Hungarian manufacturing company.  "Every month
we pinch ourselves and say it won't last, but the cube is still selling
like nothing else."
  Ideal has capitalized on the cube's success with a number of
spin-offs.  There's Rubik's Revenge, which has sixteen tiles on a side,
instead of nine; Rubik's Pocket Cube, a simpler version intended for
children; Rubik's World, a globe made of 26 sections that twist apart;
Rubik's Game, a three-dimensional pegboard, and Rubik's Race, a
two-player game in which the multicolored tiles must duplicate various
patterns.  True masochists might also want to try something called the
Calendar Cube---which requires twiddling the tiles every day to form the
correct date.
  Rubikmania has also spread to the publishing industry.  Ideal's
solution booklet and another one written by a 13-year-old London
schoolboy ("You Can Do the Cube") are both big sellers.  "The Simple
Solution to Rubik's Cube," which runs to 64 pages, has sold 7 million
copies; it is the fastest-selling title in the history of Bantam Books. 
There are even books for people who are fed up with the craze.  Among
them: "Not Another Cube Book," "You Can Kick the Cube" and "101 Uses for
a Dead Cube."
  Rubik receives about 5 percent of the puzzle profits, making him
perhaps the only self-made millionaire in Hungary.  "With the money I
earn, I can afford to buy myself a new Fiat every two days," he jokes. 
"A little Fiat."  He has taken a leave of absence from his teaching post
to help Ideal organize a world cube-twisting championship in Budapest
this spring.  To qualify for the contest, you must be able to solve the
puzzle in less than one minute---which eliminates Rubik himself.  It
takes him at least twice that long.



Date: 21 May 1982 1710-PDT
From: ISAACS at SRI-KL
Subject: frustration and Alexander's Star
To: cube-lovers at MIT-MC

  Aauugghh - no one I've been able to find in the Bay ARea has the 4^3.
If anyone is flying in from Boston, BRING SOME!  Several other of
Ideals new stuff is out here - including the game "Rubik's Race", some
of the Puzzle pens, and the 2^3.  The latter uses the sliding disks
on an internal sphere mechanism, by the way.  If one comes apart, it
seems difficult (but not impossible) to get back together. 
  Anyway, the only really worthwhile new Ideal thing is Alexander's
Star, which is a worthy addition to the sliding axis puzzle field.  It
is a Great Dodecahedron in form, and each "star", that protrudes from
a pentagonal face, rotates (5 positions).  Each face (the pentagons, not
the stars) is monochromatic in the solved state.  It is colored with 6
colors, opposite faces the same.  (Look in a decent geometry book for
a picture of a great dodecahedron. It's one of the 4 Kepler-Poinsot
regular concave polyhedra, this one having 12 pentagonal faces 
interpenatrating each other in a star-like manner).  It's not too
hard to solve, the main difficulty is figuring where each piece goes.
(Each moving piece is the triangular wedge which can be found between
the points of a stellated dodecahedron, which "turn it into" a great
dodecahedron).
   If anyone knows where to get the 4^3 out here, or is coming
visiting and can bring one, my phone number is (415)326-7788,
if you can't get at a terminal.  My work number is (415)497-2577.
   -- Stan Isaacs
-------

Date: 22 May 1982 19:13-EDT
From: Richard Pavelle <RP at MIT-MC>
Subject: C^4
To: CUBE-LOVERS at MIT-MC

I take it back. There is more to C^4 than I first thought. I encountered
a final move today and do not have the tools and perhaps this is what
DCP meant in his message of May 14th. The final configuration requires
flipping only ONE pair of edges. Can anyone explain this difference
between C^3 and C^4?

Date: 22 May 1982 22:53:41 EDT (Saturday)
From: Bernie Cosell <cosell at BBN-UNIX>
Subject: A missing transformation for C^4
To: cube-lovers at mc

I take back my statement about solving the cube, also: I do have one transform
that is the product of an order-3 centers-only cycle and an order-5 edges-only
cycle, and I can use that transform to do very nearly everything...except for
one maneuver.  If the normal layout of the edges on one face is:
        a b
      h     c
      g     d
        f e
Then, when I try to solve the cube, once in a while I end up with one of the
following two equivalent forms:
        ~a ~b                 c  b
       h      c     or      h      a
       g      d             g      d
         f  e                 f  e
I have not yet fully determined exactly which of the two identical edges of
each color is where.  In particular, I strongly suspect that the first form
is really `~b ~a' (that is, if you think of the two edges together as a single
C^3 edge, then that edge was flipped), while I suspect that the second form is
correctly rendered.

I have found that if I do sensible things to the cube, I can mostly work until
the cows come home and not fix the thing, but if I screw up a transform and
then have to recover from a somewhat scrambled mess, more often than not it
comes out OK when I get the cube back in shape.  sigh.  right now I'm working
on a group-theoretic analysis to try to get some idea for what type of move or
conjugate I have to be looking for in order to make the thing work out.

    /Bernie



Date: 23 May 1982 16:15:19 EDT (Sunday)
From: Winston Edmond <edmond at BBN-UNIX>
Subject: Re: A missing transformation for C^4
In-Reply-to: Your message of 23 May 1982 01:48 EDT
To: Alan Bawden <ALAN at MIT-MC>
Cc: Cube-Lovers at MIT-MC, Edmond at BBN-UNIX, Mann at BBN-UNIX

   The 4x4x4 cube does indeed have a "parity" problem.  It may be described
approximately thusly:
   Look at one face of the cube.  Number the cubies across one edge as 1, 2,
3, 4.   It can be shown that an edge cubie of type two can be either in
a type 2 position unflipped or in a type 3 position flipped.
   Next, assuming that the top three planes (or slices) are correct, that
the corners of the last face are correct, and that the proper color is up for
all of the edge cubies, then the edge cubies in group 2 and those in group 3
either have the same "parity" or opposite parity.  The parity groups for the
edge cubies are defined as 
            a                           a
	b       d                   b       c
	    c                           d
and any of the twelve variations of each obtainable by one or more rotations
of three elements at a time.  (Each letter represents a different color.)
   When solving the cube, the cubies of group 2 and group 3 will either end up
with the same or opposite parity.  If the parity is the same, the cube can be
solved straightforwardly.  (I assume people have discovered the transforms that
invert the parity of both groups at the same time.)
   However, when the parity is opposing, there is only one "transform" that
will correct the problem.  People accustomed to thinking of a useful transform
as one which performs some limited rearrangement of cubies while leaving
everything basically unchanged will find that none of their transforms is
sufficient to solve the problem.
   Credit goes to Bill Mann for discovering the "transform" that solves the
problem.  I leave it up to him to describe the solution if he wishes to.
 -WBE



Date: 24 May 1982 09:33-EDT
From: Richard Pavelle <RP at MIT-MC>
Subject: Ideal and the C^4
To: CUBE-LOVERS at MIT-MC

I just spoke to Ideal Toy. They are aware of the edge flip problem
and give a tool to do it in their published solution. However, it
requires 32 moves and they ask us to provide them with a shorter one if
found.  They also say they are unable to do a complete checkerboard
of any kind on C^4 and would like to know if we can do it. 

Date: 24 May 1982 10:56-EDT
From: David C. Plummer <DCP at MIT-MC>
Subject: Ideal and the C^4
To: RP at MIT-MC
cc: CUBE-LOVERS at MIT-MC

I have done a complete search on paper for a complete
checkboard.  It cannot be done for any cobe of even side.  The
restrictions are in the corners.  I would like somebody to double
check this though.  The standard problems to run into are:
	You need two of one type of corner, or
	You have to rotate exactly one corner, which is impossible.

There are a couple other amazing things I found.
As it is known, any two edges can be exchanged (or appear to be
exchaged because of center arbitrariness).  It is ALSO possible
to (appear to) exchange two corners, for about the same reason.
This is impossible on the 3^3 becuase it requires the exchange of
two edges.  But in the 4^3 there are two cubies per 3^3 edge.
Therefore, we just do a double exchange, which does not violate
any parity arguments.  Combining these two moves, you can flip a
1x1x4 edge. 

Happy revenge.

Date: 24 May 1982 08:47 PDT
From: Hoffman at PARC-MAXC
Subject: Re: frustration and Alexander's Star
In-reply-to: ISAACS at SRI-KL's message of 21 May 1982 1710-PDT
To:  Cube-Lovers at MIT-MC


   I shared your frustration.  I'm in L.A., and, upon calling Ideal's office here, I
learned that they are following a rigid schedule of TV ads first before
introducing the toy in any area.  Here in L.A. (and probably in the Bay Area),
the TV ads are not scheduled until mid-June!  Curiously, Minh Thai's solution
book is already in some bookstores here.

   Anyway, I wound up calling Games People Play in Cambridge, and
mail-ordering one ($15 + $2 handling + UPS mailing cost) using a credit card. 
A two-minute transcontinental call before 8 am PDT is about 40 cents.  I hope to
have my Rubik's Revenge in a day or so now.

--Rodney Hoffman


Date: 24 May 1982 19:43-EDT
From: Martin Minow <MINOW at MIT-AI>
To: CUBE-LOVERS at MIT-AI

Please add me to the mailing list, using the address
	decvax!minow at berkeley
(are back issues available?)

The 4x4 Rubic's revenge has just come out.  Does anybody
have a good nomenclature for positions and moves yet?

Thanks very much.
Martin Minow
decvax!minow


Date: 24 May 1982 17:19-EDT
From: Allan C. Wechsler <ACW at MIT-AI>
Subject:  |4^3| = 1.7*10^55
To: CUBE-LOVERS at MIT-AI


The corner group of 4^3 is exactly like that of the 3^3.  It has
8!*3^7 elements.

Now to calculate the order of the whole group, we have to find
the order of the corner stabilizer  group.  That's the group of 
moves that leave the corners fixed.  

SO let's start by thinking about the edge group of the corner
stabilizer.  I personally have a tool that exchanges two edge cubies
without moving corners.  Since the edge group is transitive, I can
exchange any two edges without moving corners.  

You cannot flip any edge of the 4^3 without moving it.  So there's no
edge-flippage in the 3^3 sense on the 4^3.  That means that the edge
group in the corner stabilizer has order 24!.

Now all there is left is compute the order of the edge AND corner
stabilizer.  The inside twists (slices) are center-even.  The outside
twists (faces) are center-odd, but they are also corner-odd, and so if
we want to bring the corners home we have to make an even number of
outside twists.  That means that all the center-permutations in the
edge-and-corner stabilizer are even.  But how many of these can be
achieved?

I have a tool that exchanges two pairs of centers.  I think (but I
haven't yet proved) that the center group is 4-transitive, so that
ANY two pairs of centers can be exchanged.  That means that all the
even permutations of centers can be done without perturbing edges and
corners.  Hence the size of the edge-and-corner stabilizer is 24!/2.

As for supergroupiness, Dave Plummer just pointed out to me that
once you know a center's position, you know its orientation, since
center cubies always keep one corner pointing to the middle of the
face.  So 4^3 has no supergroup.

So the order of the 4^3 group is 8!*3^7*24!*24!/2.  In numbers, if you
insist,

|4^3|=

16,972,688,908,618,238,933,770,849,245,964,147,960,401,887,232,000,000,000

or about 1.7*10^55.

Now that number is a little deceptive, because it includes whole-cube
rotations.  The 4^3 has no nice fixed reference frame like the 3^3 has.
If you don't want to count whole-cube rotations you have to divide by
24, to get

707,195,371,192,426,622,240,452,051,915,172,831,683,411,968,000,000,000

or about 7.1*10^53.

Finally, we have to face the fact that the center cubies are
in six nonintradistinguishable sets of four.  (All the edge cubies are
distinguishable by color or orientation).  So we have to divide our
last result by 4!^6/2.  That leaves

7,401,196,841,564,901,869,874,093,974,498,574,336,000,000,000

or about 7.4*10^45 distinguishable color patterns.  Remember that these
do not form a group.

I leave it as an exercise for Hoey and Saxe to find a lower bound for
the diameter of the group.

   ---Wechsler


Date: 24 May 1982 22:54:03 EDT (Monday)
From: Bernie Cosell <cosell at BBN-UNIX>
Subject: The missing C^4 transform found
To: cube-lovers at mc

I have to admit to not quite yet fully understanding all of the
parity issues involved, however I have managed to apply some VERY
SIMPLE logic to the mysterious transform and since it worked I'd
like to share the insight with you-all. 
As I mentioned in an earlier note I have been trying to do some
analysis to try to sneak up on the missing transform.  Mostly
I've been unsuccessful at fully identifying the parity issues
(my understanding is basically at the same not-so-useful level as
wbe's is: I kind of know that some moves seem to come in pairs and
there are two `classes' of edge cubes and if you interchange one from each
class they both flip... but that mostly didn't help me see what had to be done
to make the `missing transform' happen, nor would it let me look at a scrambled
cube (or even a nearly-done cube) and guess which parity class it was in).

However, I did make one observation: all of my normal transforms contained an
EVEN number of quarter-twists.  I fumbled around a bit to try to find a
limited-change transform that used an ODD number of twists and mostly I
couldn't.  So I decided to simply take the `simplest' odd-twist transform I
could: a single twist!  IT WORKS!!  As far as I can tell it must be a slice
twist (but ANY one), a face twist doesn't do it.  If I get the cube edges all
solved except for a pair that must be flipped, I simply make a SINGLE quarter
slice and then re-solve the cube from there.  Since all of my normal solving
maneuvers are EVEN, when I get the thing solved again I will have preserved the
ODD parity of the configuration, and poof! the edges are solved.

Now to start on the search for pretty patterns and elegant (and short)
operators...

  /Bernie


Date:  27 May 1982 00:08 edt
From:  Senft.Multics at MIT-MULTICS (HVN 341-7244)
Subject:  Cube-lovers
To:  Cube-Lovers at MIT-MC
cc:  Senft.Multics at MIT-MULTICS (outgoing.sv)
Acknowledge-To:  Senft.Multics at MIT-MULTICS

Please add "Senft.Multics@PCO-Multics" -at MIT-Multics to the cubelovers
list.


Date: 28 May 1982 1149-EDT
From: SWG at MIT-DMS (S. W. Galley)
To: cube-lovers at MIT-MC
Subject: Moleculon v. Ideal
Message-id: <[MIT-DMS].233206>

Cambridge firm seeks $60m in Rubik's Cube suit
By Christy George
Special to The [Boston] Globe [5/27/82]

  Moleculon Research Corp., a Cambridge chemical, research and
development firm, yesterday filed a $60 million patent infringement
lawsuit against the Delaware-based Ideal Toy Corp., which manufactures
and markets the highly-lucrative puzzle known as Rubik's Cube.
  Moleculon holds a 1972 US patent for a similar mathematical cube
puzzle invented by Dr. Larry Nichols of Moleculon.
  The prototype of the Nichols Cube is a 2x2x2 cube held together by
magnets, while the original Rubik's Cube is a 3x3x3 cube held together
mechanically.  However, the 1972 Moleculon patent also lists
specifications for a mechanically-constructed cube as well for cubes of
other varying sizes.  Ideal has no US patent for Rubik's Cube, although
its inventor, Erno Rubik, was issued a Hungarian patent in 1978.
  In its complaint, Moleculon alleges that Ideal "willfully and
maliciously" continued to manufacture and sell Rubik's Cube despite
having been notified a year ago that Moleculon holds the only valid US
patent on the invention.
  According to Moleculon's president, Dr. Arthur Obermayer, the
Cambridge firm unsuccessfully tried to interest Ideal in its cube puzzle
in 1969.  In 1970, the firm applied for a patent at the suggestion of
other national toy manufacturers, who also were uninterested in
marketing the puzzle.
  Lawyers for Ideal would not comment on the pending suit because, they
said, they have not yet received the complaint.  But Ideal's general
counsel, Samuel Cohen, dismissed the notion that Ideal might have stolen
the idea from Moleculon.  "It's ridiculous on its face," Cohen said,
"we're paying heavy sums to the Hungarians for use of their patent.  It
would be stupid to do that if we had it in our pocket all the time."
  Obermayer says the company didn't realize how similar Rubik's Cube was
to the Nichols Cube until the phenomenal success of the product began to
make headlines.  Rubik's Cube was first marketed worldwide in 1980. 
Because of the puzzle's complexity, inventor Nichols wasn't surprised
that toy firms were unimpressed.
  "I'm delighted people enjoy solving the puzzle," Nichols said, "but
I'm also frustrated that neither I nor Moleculon has seen any financial
reward all these years."
  Obermayer is confident of success in the lawsuit, citing the fact that
three prominent patent law firms have agreed to collaborate on
Moleculon's behalf on a contingency basis.
  Moleculon is asking for $20 million in estimated lost royalties and
$40 million in damages, as well as interest on other costs.  Even more
could be at stake in terms of future royalties.
  According to an article in Newsweek magazine last month, annual world
sales for the basic 3x3x3 Rubik's Cube, which retails for between $7 and
$8 in the Boston area, have exceeded the $30 million mark.  And the
company also markets a range of spin-off products, including cubes in
three other sizes and types, a Rubik's Cube solution booklet, a Rubik's
Cube world globe and a Rubik's Cube board game.
  "Sales are just incredible," said Carol Monica, owner of the Cambridge
store, The Games People Play.  Monica said she already has sold more
than a hundred of Ideal's latest spinoff, a 4x4x4 verison of Rubik's
Cube which costs $16 and went on the market only three weeks ago.
  Ironically, in an action pending before the International Trade
Commission, Ideal claims it is besieged by countless "knockoffs" of
Rubik's Cube by unscrupulous foreign competitors.  The Delaware company
is asking the commission to squelch international imports of similar
mathematical puzzles.


Date: 30 May 1982 04:52:01-PDT
From: decvax!duke!uok!mwm at Berkeley
Date-Sent: Wed May 26 19:09:15 1982
To: duke!decvax!ucbvax!cube-lovers@ai
Subj: Diameter of the group

I just saw a letter on the list that talked about finding the diameter
of the group for 4^3.  This implies (to me, anyway) that somebody out there
knows something about the diameter of the group for 3^3. Since I just started
getting mail from the list, I missed any discussion that may have happened on
that topic, and would appreciate it if someone would send me the apporpriate
information, or tell me which archive the discussion is in.

	Thanx in advance,
	mike meyer





Date: 30 May 1982 10:52:24-PDT
From: decvax!duke!uok!mwm at Berkeley
Date-Sent: Wed May 26 19:09:15 1982
To: duke!decvax!ucbvax!cube-lovers@ai
Subj: Diameter of the group

I just saw a letter on the list that talked about finding the diameter
of the group for 4^3.  This implies (to me, anyway) that somebody out there
knows something about the diameter of the group for 3^3. Since I just started
getting mail from the list, I missed any discussion that may have happened on
that topic, and would appreciate it if someone would send me the apporpriate
information, or tell me which archive the discussion is in.

	Thanx in advance,
	mike meyer





Date: 30 May 1982 16:30-EDT
From: Alan Bawden <ALAN at MIT-MC>
Subject: God's number
To: CUBE-LOVERS at MIT-MC

OK, it's been some time since I pointed out where I keep archives and
things...  Old cube-lovers mail is archived in the following places: 

MC:ALAN;CUBE MAIL0	;oldest mail in forward order
MC:ALAN;CUBE MAIL1	;next oldest mail in forward order
MC:ALAN;CUBE MAIL2	;more of same
MC:ALAN;CUBE MAIL3	;still more of same
MC:ALAN;CUBE MAIL	;recent mail in reverse order

(Files can be FTP'd from MIT-MC without an account.)

In addition, I have the following two excerpts from the archives sitting on my
directory since they contain some of the more asked-for material:

MC:ALAN;CUBE 4X4X4	;Contains some pre-release speculations on the 4x4x4
			;cube.  Some are out of date, but it contains the only
			;analysis this list has seen of the 4x4x4 group, I
			;believe.
MC:ALAN;CUBE S&LM	;While most of the speculation about the diameter of
			;the 3x3x3 group is scattered randomly through the
			;archives, this file contains the single message with
			;the highest content.  Hoey and Saxe's message on
			;Symmetry and Local Maxima.

To briefly remind you all of ALL that we know about the diameter of the 3x3x3
group (refered to as "God's number" in many of our discussions):

We know that God's number is greater or equal to 21 quarter twists.
 (See Hoey's message of January 9 1981: "The Supergroup -- Part 2 ..." in MAIL1
 for a good explanation of this, as well as some other interesting bounds.)
We know that God's number is greater or equal to 18 half twists.
 (See Singmaster.)
We know that God's number is less than or equal to 52 half twists.
 (See Singmaster again, this is Thistlethwaite's algorithm of several years 
 ago.  I'll bet it's been improved upon by now.  There is a persistent rumor
 that he was trying for 41.)

We have never bothered to figure out an upper bound on God's number in quarter
twists ("Q"s).  It must be less than 104 Qs because of the half twist result,
but we could probably do better than that if we took the trouble to understand
Thistlethwaite's algorithm.

Proofs of these numbers, and a great deal of other discussion can be found by
sifting through the archives (unfortunately they are spread all throught the
files).  I would urge people to sift through the archives before starting any
new discussions on the subject.

Date:  1 June 1982 2220-EDT (Tuesday)
From: Dan Hoey at CMU-10A
To: Cube-Lovers at MIT-MC
Subject:  Lower bounds for the 4x4x4
Message-Id: <01Jun82 222052 DH51@CMU-10A>

Well, since you insist, here are my lower bounds for the 4^3.  For
the "colored" 4^3, where only the color pattern matters, some
positions require at least 41 qtw to solve.  For the "marked" 4^3,
where center facets of the same color are distinguished so as to
force a unique home position for each, some positions require at
least 48 qtw.  The proof is in MC:ALAN;CUBE4 LB and is about 5K
characters long.

A qtw of the 4^3 is either a quarter-twist of a face relative to
the rest, or a quarter-twist of half of the puzzle relative to the
other half.  Note that this makes a slice twist into two moves.  I
like this metric because it is consistent with our conventions for
the 3^3.  One of these days I'll explain why I like those
conventions.


Date:  1 June 1982 2222-EDT (Tuesday)
From: Dan Hoey at CMU-10A
To: Cube-Lovers at MIT-MC
Subject:  Lower bounds for the 4x4x4 (Long)
Message-Id: <01Jun82 222230 DH51@CMU-10A>

		Lower bounds for solving the 4^3 puzzle

In order to prevent counting positions which arise from whole-cube
moves, I do not move the DLB corner.  So the generators I use are
	U1	U1'	R1	R1'	F1	F1'
	U2	U2'	R2	R2'	F2	F2'
	U3	U3'	R3	R3'	F3	F3'
where the digit indicates how many layers are being moved and the
prime indicates a counterclockwise quarter-twist.

Let us break up any process into "syllables", where a syllable is a
maximal nonempty string of generators with the same letter.  Note
that the order of the generators within a syllable is irrelevant.
We may make a syllable canonical by simplifying so that
	1. A generator and its inverse do not both appear,
	2. Clockwise generators appear at most twice,
	3. Counterclockwise generators appear at most once, and
	4. Generators appear in numerical order.

For each letter there are 63 canonical syllables.  Omitting the
letter, they are:

1,1',2,2',3,3'				Six of length one

11,12,12',13,13',1'2,1'2',1'3,1'3',22,
23,23',2'3,2'3',33			Fifteen of length two

112,112',113,113',122,123,123',12'3,
12'3',133,1'22,1'23,1'23',1'2'3,1'2'3',
1'33,223,223',233,2'33			Twenty of length three

1122,1123,1123',112'3,112'3',1133,1223,
1223',1233,12'33,1'223,1'223',1'233,
1'2'33,2233				Fifteen of length four

11223,11223',11233,112'33,12233,1'2233	Six of length five

112233					One of length six

(Exercise for the reader:  There is a reason these are binomial
coefficients.  Why did we skip a row?)

The number of canonical syllable strings containing N generators is

P(-N) =  0 if -N < 0,
 P(0) =  1,
 P(N) =  6 C(N-1) P(N-1) + 15 C(N-2) P(N-2) + 20 C(N-3) P(N-3)
      + 15 C(N-4) P(N-4) +  6 C(N-5) P(N-5) +    C(N-6) P(N-6)
	   if N > 0,

where C(N) is the number of ways of choosing a new letter after N
generators:  C(-N) = 0 if -N < 0, C(0) = 3, C(N) = 2 if N > 0.

Evaluating this recurrence yields
P( 0)=       1    P(13)<1.338E15    P(26)<1.404E30    P(39)<1.472E45
P( 1)=      18    P(14)<1.914E16    P(27)<2.007E31    P(40)<2.105E46
P( 2)=     261    P(15)<2.737E17    P(28)<2.871E32    P(41)<3.011E47
P( 3)=    3732    P(16)<3.915E18    P(29)<4.106E33    P(42)<4.307E48
P( 4)=   53379    P(17)<5.599E19    P(30)<5.873E34    P(43)<6.160E49
P( 5)=  763506    P(18)<8.009E20    P(31)<8.400E35    P(44)<8.811E50
P( 6)=10920771    P(19)<1.146E22    P(32)<1.202E37    P(45)<1.261E52
P( 7)<1.563E 8    P(20)<1.639E23    P(33)<1.719E38    P(46)<1.803E53
P( 8)<2.235E 9    P(21)<2.344E24    P(34)<2.459E39    P(47)<2.579E54
P( 9)<3.196E10    P(22)<3.353E25    P(35)<3.516E40    P(48)<3.688E55
P(10)<4.572E11    P(23)<4.795E26    P(36)<5.029E41    P(49)<5.275E56
P(11)<6.539E12    P(24)<6.858E27    P(37)<7.194E42    P(50)<7.545E57
P(12)<9.352E13    P(25)<9.810E28    P(38)<1.029E44    P(51)<1.080E59

The number of positions exactly N qtw from SOLVED is at most P(N),
so the number of positions within N qtw of SOLVED is at most
P(0)+P(1)+...+P(N). Since there are 7.07E53 marked and 7.40E45
colored positions, this would give us lower bounds of 40 qtw for
the marked cube and 47 qtw for the colored cube.

But we can improve these lower bounds by one each, using the parity
hack I described some time ago.  Every qtw is an odd permutation of
the corner cubies, so the number of positions with even corner
permutations within 2N+1 qtw of SOLVED is at most
P(0)+P(2)+...+P(2N) and the number of positions with odd corner
permutations within 2N+2 qtw of solved is at most
P(1)+P(3)+...+P(2N+1).

There are 3.70E45 colored positions with odd corner permutations,
and fewer than 1.583E45 odd-length canonical processes with length
at most 39.  So some odd colored positions require at least 41 qtw
to solve.

There are 3.53E53 marked positions with even corner permutations,
and fewer than 1.939E53 even-length canonical processes with length
at most 46.  So some even marked positions require at least 48 qtw
to solve.

		Directions for future hacks

Note that we could also distinguish between positions with even or
odd edge permutations.  The recurrence gets hairier, but my
analysis of that problem indicates that the numbers get very close
very quick, so no luck.

We could divide the positions into buckets based on other quotients
of the group.  For instance, count the number of positions with
each of the 729 possible corner orientations.  This should be
feasible, but probably no help for the 4^3 puzzle.  For the 3^3,
where there are 2187 corner buckets, some of which stay empty for a
fair number of moves (I seem to recall 8), and our current lower
bound is small, I think there is a chance of improvement.  Any
God's numerologist willing to hack the good hack?

Date: 2 June 1982 10:33-EDT
From: Richard Pavelle <RP at MIT-MC>
Subject: Moleculon/Ideal, cont.
To: CUBE-LOVERS at MIT-MC

I spoke to a person at Moleculon and was given the U.S.
patent number, 3655201.  I sent for a copy and will report
again when I have read it. It will likely be several weeks
before I receive it so if someone is near DC and feels
motivated to get the patent sooner, please do so. I will
also check out the Patent Gazette and see if there is anything
of interest there.

Date:  4 Jun 1982 1006-EDT
From: PDL at MIT-XX (P. David Lebling)
Subject: Re: Moleculon/Ideal, cont.
To: ALAN at MIT-MC, Cube-Lovers at MIT-MC
In-Reply-To: Your message of 3-Jun-82 1933-EDT

There was a report on the Moleculon suit on WCVB, including a closeup
of the patent's picture page. The picture is of a 2x2 cube with the
cubies held together by magnets (the patent shows how the magnets are
to be arranged so the object will hold together).  The inventor of the
"Nicholls Cube" was shown working it.  It doesn't turn; you pull it
apart, turn it, and then stick it back together.  He never quite
claimed to have made a 3x3 version, and carefully tried to fudge the
distinction between his 2x2 frob and Rubik's 3x3.

My impression was that the guy is a sleazo trying to get an out-of-court
settlement from Ideal for $100k or so.

	Dave
-------

Date:  4 Jun 1982 1508-PDT
From: ISAACS at SRI-KL
Subject: 4^3
To: cube-lovers at MIT-MC

  Well, I finally got one (UPS from Boston).  I agree with the preceding
messages; basically, I could solve most of it with 3^3 techniques,
and work out a couple of variations to get edge halves together; then
if a 2-edge monoflip or exchange is left over, do an odd slice move and
recover.  Some of the tools are quite long, however; does anyone have
a short edge flip or exchange, or a short half-center exchange?
-------

Date:  4 Jun 1982 1531-PDT
From: ISAACS at SRI-KL
Subject: Winning Ways
To: cube-lovers at MIT-MC

  The book "Winning Ways" by Berlekamp, Conway, and Guy is finally out.
(A puzzlists' dream - to get 4^3 one day, and Winning Ways the next.)
It seems, on quick scanning, to be all rumor had it to be.  It's sort
of a sequel to Conway's "On Numbers and Games", and contains a more
understandable of his Nimbers (Surreal Numbers), with lots of examples
applied to game analysis.  There are chapters on many puzzles, including
one on Rubik's Cube.  Also included are sliding block puzzles, Instant
Insanity, Life, Sprouts, Polyominoes and Polyiamonds, Wire puzzles, 
Chinese Rings, peg jumping, etc, etc.  An almost encyclopaedic 
collection of puzzles and games (of certain types) from a mathematical
standpoint.  With plenty of humor and puns.  The only problem is its
price - about $20 for each of the two volumes, in paperback. (When the
hardback comes out later this year, it's supposed to be about $60 per
volume.)  The 4 sections are "Spade Work" and "Change of Hearts" in the
first volume (on "Games in General"), and "Games in Clubs" and "Solitaire
Diamonds" (puzzles) in Vol. 2. ("Games in Particular).  Academic Press.
Don't miss it.
   --- Stan
-------

Date: 5 June 1982 01:07-EDT
From: Alan Bawden <ALAN at MIT-MC>
Subject: 4x4x4 mechanics
To: CUBE-LOVERS at MIT-MC

I had a small accident the other day.  I dropped Dave Plummer's 4x4x4 cube and
broke one of the center cubies (sorry about that Dave).  This gave me an
opportunity to closely examine the insides of the beastie.  I can't possibly
describe it through the mail (I can barely describe it in person), but there is
an interesting problem raised by the insides:

Inside of a 4x4x4 cube is a 57th piece.  It is not permanently connected to any
of the cubies.  (The center cubies are free to slide back and forth in slots
cut in the center piece.  The rest of the cubies are held in by the centers.)
It is impossible to determine by examining the outside of a cube exactly what
orientation the center piece has.  However, it IS deterministic how the center
piece will move under a certain twist.  When twisting a face, the center piece
stays fixed with respect to the other 3 layers of the cube.  When doing an
"equatorial" twist, the center piece can follow only one of the halves of the
cube (as determined by it's orientation).  (I believe I have just constrained
things enough so that you can figure out exactly how the thing moves on your
own.) 

Thus there is an even larger permutation group to the 4x4x4 (beyond the
supergroup problem where the identities of the center cubies are considered)
that includes the center piece.  Call this the "hypergroup".  And since the
center piece has a 3 element symmetry group there is another group beyond that
("superhypergroup"?) that takes that into account.

Now the first question to consider about the hypergroup is:  Is it really
larger than the original group or supergroup?  In other words:  When you have
solved the 4x4x4, does the center piece necessarily return to its original
position?  How about if you solve the cube in the hypergroup?

A problem with this problem, is that you cannot learn how to manipulate the
orientation of the center piece without taking your cube apart to look at the
thing.  (I DON'T recommend that.  My experience with Plummer's cube has taught
me that those center pieces are fragile.)  

Anybody have any insights into the problem?

				-Alan

Date: 5 June 1982 08:34-EDT
From: Martin Minow <MINOW at MIT-AI>
Subject: A notation and a simple edge/center process
To: CUBE-LOVERS at MIT-AI

I have been using the following notation for the 4x4 cube:

Name the faces L, R, F, B, U, D.  This follows the 3x3 notation.  The inner
edges are then l, r, f, b, u, d.  The front, upper, right quadrant
consists of one corner cube, FUR, two edge cubes, FUr and FRu, and one
center cube, fur.

Moves follow 3x3 notation:  F moves the front face clockwise, f moves
the front slice clockwise.  F' moves counterclockwise.

** SPOILER WARNING ** SPOILER WARNING ** SPOILER WARNING **

The remainder of this message describes two processes for the 4x4 cube.

** SPOILER WARNING ** SPOILER WARNING ** SPOILER WARNING **

The following process rotates three edge cubes.  It preserves parity.

(FUl UBl URb)		l' RUR'U' L URU'R'

The following process rotates three center cubes.  It can be used to
move one center cube between two faces.

(Ful Ulb Urb)		l' rur'u' l uru'r'

While the center process is sufficient, the edge process will not
organize all edges.  It can easily be extended to a process that
moves (FUl to UBl) but that process messes up centers, corners, and
other edges.  It also does not do 4-way moves.  So the bloody thing
still takes an hour.

Martin Minow
decvax!minow@berkeley


Date: 5 Jun 1982 10:44:47-PDT
From: decvax!minow at Berkeley
To: ucbvax!cube-lovers@mit-ai

Subject:  Rubic's Domino
I found one (for $11.00) at Games People Play in Cambridge.  Made in
Hungary, not by Ideal.

Haven't solved it yet.  The notes say you should be able to solve it
in 15 minutes.  It is a 3x3x2 construction; one face white; one black.
Each cube is numbered as on a domino.  The object is to get the numbers
and colors arrainged.

Martin Minow
decvax!minow at berkeley



Date:  5 Jun 1982 2332-EDT
From: Joseph A. Bowbeer <LS.BOWBEER at MIT-EECS>
Subject: 4x4x4 solution book
To: cube-lovers at MIT-MC

Jeff Adams (MIT math graduate student) and another person have coauthored
a 4x4x4 solution book which should be on the shelves shortly.  I'll try to
get more information.  The 3x3x3 solution posters which are on sale at the 
MIT math undergraduate office, among other places, were designed by these
two cubists, also.
-------

Date: 6 Jun 1982 17:11:14-PDT
From: decvax!minow at Berkeley
To: ucbvax!cube-lovers@MIT-AI

Note: this is a resubmission of my note of Saturday June 6 which
was bit by typo's.  Thanks to RP@MIT-MC for pointing them out.

I have been using the following notation for the 4x4 cube:

Name the faces L, R, F, B, U, D.  This follows the 3x3 notation.  The inner
edges are then l, r, f, b, u, d.  The front, upper, right quadrant
consists of one corner cube, FUR, two edge cubes, FUr and FRu, and one
center cube, fur.

Note that the Ideal book uses Lx etc for the mid-slices.

Moves follow 3x3 notation:  F moves the front face clockwise, f moves
the front slice clockwise.  F' moves counterclockwise.

** SPOILER WARNING ** SPOILER WARNING ** SPOILER WARNING **

The remainder of this message describes two processes for the 4x4 cube.

** SPOILER WARNING ** SPOILER WARNING ** SPOILER WARNING **

The following process rotates three edge cubes.  It preserves parity.

(FUl UBl URb)		l' RUR'U' l URU'R'

The following process rotates three center cubes.  It can be used to
move one center cube between two faces.

(Ful Ulb Urb)		l' rUr'U' l UrU'r'

While the center process is sufficient, the edge process will not
organize all edges.  It can easily be extended to a process that
moves (FUl to UBl) but that process messes up centers, corners, and
other edges.  It also does not do 4-way moves.  So the bloody thing
still takes an hour.

Martin Minow
decvax!minow@berkeley



Date:  7 Jun 1982 15:34:28 EDT (Monday)
From: Roger Frye <frye at BBN-UNIX>
Subject: 4^3 breakage, notation, and solution
To: Cube-Lovers at MIT-MC
Cc: decvax!minow at BERKELEY, ISAACS at SRI-KL, frye at BBN-UNIX

About broken 4^3 center cubies:

The center cubies take all the stress of holding in the edges and corners.
The post which breaks is 4mm square compared to the 16mm square cubie.
When I called Games People Play about my broken cube, another cube had
just been returned with the same problem.  So I traded my broken cubie
for an unbroken one and picked up a spare.

About Minow notation for 4^3:

     Name the faces L, R, F, B, U, D.  This follows the 3x3 notation.  The inner
     edges are then l, r, f, b, u, d.  The front, upper, right quadrant
     consists of one corner cube, FUR, two edge cubes, FUr and FRu, and one
     center cube, fur.  [decvax!minow at Berkeley]

Isn't there still a mistake here?  I think of "fur" as being a spot buried
in the cube.  If you mean the four cubies in the upper right quadrant of
the front face, then the name of the center cubie would be "Fur".
If you mean the front, upper, right octant then the two other center
cubies are "Urf" and Rfu".

About 4^3 processes:

** SPOILER WARNING ** SPOILER WARNING ** SPOILER WARNING **

My process to exchange three edge cubies is a U face commutator:

(RDf URf UFl) =    R'd'R  U'  R'dR  U

I think of my other 4^3 process as exchanging two half centers:

(Rfu Flu Rbd)          \
(Rub Fur Rdf)           =   R2 u' R2 u
(RBd FRd BRu RFu FLu)  /

You can get lost trying to read the permutations.  Just think of it
as injecting the two upper cubies from the front face into the right
face in exchange for the two upper cubies from the right face.
This is a move ISAACS requested.

My solution strategy:

** SPOILER WARNING ** SPOILER WARNING ** SPOILER WARNING **

1) Solve the top center using random methods.
2) Solve the corners using 3^3 moves.  (See \\Jeff Conquers the Cube//.)
3) Solve all centers with my center process.
4) Solve all edges with my edge process.
5) If a 2-edge monoflip or exchange is left over:
5a) In 2^3 mode, turn top and rotate three corners back; i.e.
    Uu   (Rr)2 (Bb)2 Rr Ff R'r' (Bb)2 Rr F'f' Rr.
5b) In 3^3 mode, undo 5a; i.e.
    U'   L2 B2 L' F' L B2 L F L.
5c) Repeat steps 3 and 4.

 -Roger Frye


Date:  7 Jun 1982 1606-PDT
From: Dolata at SUMEX-AIM
Subject: I quit!
To:   cube-lovers at MIT-AI


I have decided to give up cubing for my sanity!  Therfor, I have
2 ideal cubes in good tight condition for sale.  They go to the
highest bidder, assuming the bid is sufficiently above postage
to make it worthwhile.

Dan (ex-cubeist) Dolata
-------


Date: 8 June 1982 19:34-EDT
From: Alan Bawden <ALAN at MIT-MC>
Subject:  [DANIEL: FYI.  [dave: Rubic's Cube World Championship]]
To: CUBE-LOVERS at MIT-MC

Just for the record:

    Date: 8 June 1982 11:44-EDT
    From: Daniel Weise <DANIEL at MIT-AI>
    Re:   FYI.  [dave: Rubic's Cube World Championship]

       Date: 7 June 1982 2054-PDT (Monday)
       From: dave at UCLA-Security (David Butterfield)
       Re:   Rubic's Cube World Championship

       Ming Thai, the Vietnamese student who I met a couple of weeks ago,
       won the World Rubic's Cube Contest in Budapest on June 5.  His time
       was around 23 seconds.

    Very impressive time.

Date:  9 Jun 1982 1001-PDT
From: ISAACS at SRI-KL
Subject: S&M and other moves
To: cube-lovers at MIT-MC

BACKGROUND
    On the 3^3, many people use variants of a 3-cycle of edges,
more-or-less as follows:

      let Rs = Right slice = R'L + roll cube so front is up

      T  (for tri-cycle?) = Rs U2 Rs' U2  = (UF,DF,UB)  
                                  to cycle 3 edges on a slice

Some people use it in the form
      R2 D' - T - D R2 = (UF,UR,UB) to get 3 on a face, without

flips, and some use

      Rs' U - T - U' Rs = Rs' U Rs U2 Rs' U Rs = (UF,LU,RU)

for 3 on a face, with flips, since it saves a move due to 

       U2 U' = U.
END BACKGROUND

   Now, this move, when transferred to the 4^3, seems to be the basis
of both Richard Pavelle's "S" move of 16-May, and Bernie Cosell's "M"
move of 17-May (Bernie using a left slice version of the third form).
   The move Roger Frye describes on 7-June is also the same as this
(re-oriented to a different face), and so is the "quite long" tool I
use.  To move only edges requires S3, and to move only centers
requires S5!!!  I still would like a "nice" (preferably short) center
pair move.

SPOILER          SPOILER             SPOILER

We have now several 3-cycles:

 (UBl,UFr,LDf)  =  U2 f' D f - U2 f' D' f      (from Minh Thai's book)

 (UFl,BUl,RUb)  =  l' - R U R' U' - l - U R U' R' (Minow, 6-June)

 (RDf,URf,UFl)  =  R' d R - U' - R' d R - U       (Frye, 7-June)

    Note that all of these have corresponding 3-cycles of centers, by
simply substituting a slice for its' corresponding edge in each of the
moves.  For instance, in Thai, substitute d for D; in Minow, r for R,
etc. There is also a version which cycles an edge an center together,
by rotating the appropriate face and slice together.
    In fact, it looks as though any simple move on the 4^3 will have a
left and right version, a forward and back version, and a slice and
edge version or two, along with combinations of these.  Is there any
way to put these into a canonical form so we can recognize related
moves?  What about canonical form on the 3^3?
    One more spoiler: to switch opposite centers (all 4 cubies):

  (F,B)  = (r2 U2 l2 U2)3.

END SPOILER          END SPOILER        END SPOILER

    Can the notation be usefully extended so as to talk about UB as
the pair of upper back edges, F as the front 4 center cubies, and Fu
as the upper two of F, etc.  It might make notating the permutations
easier.  Is there any notation to make various repeats easier;  for
instance, some expansion of "squared" to indicate  "primed"
(commutator) type repeat, or a repeat with all quarter twists in the
opposite direction.  Maybe (  )2 means repeat; (  )i2 means repeat
with sub-parts "primed", and (  )-2 repeats with qtws opposite.  The
purpose is to try to make it more obvious what each move is really
doing, and to be able to compare moves easier.
    Has anybody given any thought to notation the 5^3?
    By the way, if Bill Mann is listening, would you describe your
transform (mentioned by Edmond on 23-May)?
---  Stan
-------

Date: 10 Jun 1982 0940-PDT
From: ISAACS at SRI-KL
Subject: correction
To: cube-lovers at MIT-MC

   At least 2 errors appeared in my previous message:
    In Minh Thai's 3-cycle, substitute "r" for "f" to do the permutation
reported. 
    In the Frye 3-cycle, the first "d" should be "d'".
 
    I thought I had a half center move:
    
       (r2 u2)2  =  (Fur,Bur)

   but when repeated, I got

      (r2 u2)4  =  (Fdr,Bur),

    so I realized that what I really had was the 6-cycle of centers:

      (r2 u2)2  = (Ful,Fdr,Bul,Bdl,Fur,Bur)   (I think).

  Oh, well.


----   Stan
-------

Date: 14 Jun 1982 11:17:40 EDT (Monday)
From: Bernie Cosell <cosell at BBN-UNIX>
Subject: Magic Domino
To: cube-lovers at mc

The magic Domino (available, at least, at Games People Play here
in Cambridge) is a delightful little puzzle.  I would guess that it
is the easiest of the `Rubik-type' puzzles to date -- I think it is
even easier than the 2x2x2.  I solve it with two basic moves:
one fixes the corner (but screws up the edges), and then the other
fixes up the edges.

An interesting sidelight on the domino is the ability to interchange
ONE pair of edges.  The trick here (for reasons that I don't
understand yet) is that the supergroup rotation seems to invert the
parity of the edges.  That is: if you have the whole domino solved
and then re-solve the thing with both faces rotated by a quarter turn
around the center (5) cube.  You'll discover that you can't do this hardly
at all on just one face, since you wont be able to get the corners fixed.
But if you rotate both the white and black faces, then the corner parity
is OK, BUT.... the edge parity has been flipped and so you have done a
single interchange of two edges.
  
    /Bernie


Date: 14 Jun 1982 16:10:19-PDT
From: decvax!minow at Berkeley
To: ucbvax!cube-lovers@mit-ai

Subject:  Some simple edge processes for the 4x4 cube.


** ** ** ** SPOILER WARNING ** ** ** **
This message describes some edge processes for the 4x4 Rubic's Cube.

My collegue, Marty Hiller, came up with two simple processes
which she uses to order edges on the 4x4 cube.  They are sufficient
to solve all edge permutations:

(FUl UBl FDr DBr)		(r U2)^4 r

(FUl UFr UBl BUr FDr)		(r U2 r' U2)^3

The ^n notation means repeat the process n times.
Note that these processes probably modify the center cubes as well
but this is invisible if you have already solved the center.

Martin Minow



Date: 15 June 1982 1045-EDT (Tuesday)
From: Dan Hoey at CMU-10A
To: Cube-Lovers at MIT-MC
Subject:  Patterns for Rubik's Revenge
Message-Id: <15Jun82 104503 DH51@CMU-10A>

			CHECKERBOARDS

I confirm Dave Plummer's result that it is impossible to make a
checkerboard on all faces of a cube of even side.  Proving this for
the 2^3 is sufficient, and implies the corollary I sent last year
that it is impossible to make S or Z or Zig-Zag patterns on all
sides of the 3^3.  This time I will outline my analysis of the
problem.

What we are asking for in each case is a pattern in which every
face has matching diagonally opposite corner facets and contrasting
adjacent corner facets.

First, consider any position in which diagonally opposite corner
facets match on each face.  If we connect every pair of corner
cubies that have diagonally opposite corner facets we get two
tetrahedrons of corner cubies.  A quick examination will show that
the four cubies in such a tetrahedron must either be

	1) Not all from the same cube,
	2) Four copies of the same cubie, or 
	3) Four different cubies from the same cube, in the correct
		position and orientation relative to each other.

Assuming case 3, we may place one tetrahedron in the home position
and consider how the other tetrahedron has been rigidly rotated
with respect to the first.  There are 12 rotations of the
tetrahedron.  One is the identity, three are 180-degree rotations
about the midpoints of two opposite edges, and six are 120-degree
rotations about a vertex.  

In the identity every face's adjacent corner facets fail to
contrast.  The 180-degree rotations have the corners like the 3^3
Zig-zag pattern, where two faces have noncontrasting corners.  The
120-degree rotations would make checkerboards, but violate the
corner twist invariant.

Approximate checkerboards can be made from the 180-degree rotation
and from the 120-degree rotation with one corner twisted.  In the
180-degree approximation, two faces have two wrong facets each.  In
the 120-degree approximation, three faces have one wrong facet
each.  Both are achievable with the 2^3 and 4^3, and I think these
are as close as you can get on any even-sided cube.

			SPOTS

Spot patterns of the 4^3 are those which have the pattern
		X X X X
		X Y Y X
		X Y Y X
		X X X X
on all nonblank faces.  There are a lot of them:  every permutation
of the centers is possible.  There are thus 6! = 720 spot patterns.
We may cut this number down by identifying positions that are
M-conjugates (recolorings) of each other.

As long as I am listing the permutations by conjugacy class, I may
as well break them down by recoloring type.  Last July I asked a
question about the possible rearrangements of colors on the cube.
I worked on the solution long enough to find that given a
``standard'' cube, there are five kinds of recoloring up to
M-conjugacy.
	Identity   -- The standard coloring.
	Reflection -- Identity in a mirror.
	Swap	   -- Identity with two adjacent colors exchanged.
	Wrench	   -- Identity with three adjacent colors cycled.
	Befuddler  -- Wrench in a mirror.
The Identity and Reflection are unique colorings.  There are twelve
Swaps and eight each of the Wrench and Befuddler recolorings.  Each
recoloring corresponds to 24 face permutations, achievable by
whole-cube moves.

Here, then, are the spot patterns.  Columns correspond to the
number of spots in the pattern.  The row groupings show the kind of
cube coloring the spot pattern comes from.  The patterns are given
as a permutation of faces:  (...XY...) and (Y...X) both mean that
face X has a spot colored Y, as shown at the beginning of this
message.  The number of permutations in each conjugacy class is
also given.

				Number of spots
	   0     2	 3	      4	            5		   6
Coloring
--------------------------------------------------------------------
Identity  :8                   (BUFD):6                 (BUL)(FDR):8
	    		     (BF)(UD):3               (BF)(UL)(DR):6
--------------------------------------------------------------------
Reflection  (BF):3	     (BU)(FD):6                   (BULFDR):8
					                (BF)(ULDR):6
					              (BF)(UD)(LR):1
--------------------------------------------------------------------
Swap	   (BU):12 (BFU):24   (BFUD):12    (BUFDL):48    (BFULDR):48
			    (BF)(UL):12  (BF)(UDL):24  (BF)(UDLR):12
				         (BU)(FDL):48  (BU)(FDLR):48
--------------------------------------------------------------------
Wrench		   (BUL):16   (BUFL):24    (BFUDL):48    (BFULRD):24
				         (BU)(FLR):48   (BUL)(FRD):8
						       (BU)(FLDR):24
--------------------------------------------------------------------
Befuddler		      (BFUL):48    (BFULD):48    (BFUDLR):16
			    (BU)(FL):24	                 (BUFLDR):24
						       (BFU)(DLR):24
						      (BU)(FL)(DR):8
--------------------------------------------------------------------

The answer: Fifteen six-spot patterns, six five-spots, eight
four-spots, two three-spots, two two-spots, and one no-spot.

			CROSSES

What cross patterns are possible on the 4^3?  We must first ask
what a cross pattern on the 4^3 should look like.  I consider the
following two kinds of cross.

		Thick Cross		Thin Cross

		  X O O X		 X O X X
		  O O O O		 O O O O
		  O O O O		 X O X X
		  X O O X		 X O X X

Every thick cross pattern is a rigid rotation or reflection of the
edge and face center pieces with respect to the corners.  This is
just like the 3^3 case except that the 3^3 has face centers that
are fixed relative to each other, and so does not allow reflec-
tions.  So in addition to the thick versions of Plummer's Cross and
Cristman's Cross, there are three new crosses.

    Thick Pons Cross                        Thick Fliptwist Cross
                       Thick Interlaced Cross
         U D D U                                   U L L U
         D D D D              B U U B              L L L L
         D D D D              U U U U              L L L L
         U D D U              U U U U              U L L U
                              B U U B
L R R L  F B B F  R L L R                 L D D L  F B B F  R U U R
R R R R  B B B B  L L L L     U L L U     D D D D  B B B B  U U U U
R R R R  B B B B  L L L L     L L L L     D D D D  B B B B  U U U U
L R R L  F B B F  R L L R     L L L L     L D D L  F B B F  R U U R
                              U L L U
         D U U D                                   D R R D
         U U U U     L F F L  F D D F  R B B R     R R R R
         U U U U     F F F F  D D D D  B B B B     R R R R
         D U U D     F F F F  D D D D  B B B B     D R R D
                     L F F L  F D D F  R B B R
         B F F B                                   B F F B
         F F F F              D R R D              F F F F
         F F F F              R R R R              F F F F
         B F F B              R R R R              B F F B
                              D R R D

For thin crosses, we first examine those in which the arms of the
crosses meet at the edges.  Again the figure facets are rigidly
rotated and reflected with respect to the ground.  This time cubie
conservation becomes an issue, because of the impossibility of
flipping an edge cubie, so there are only three such thin crosses.

     Thin Pons Cross			      Thin Plummer Cross
		       Thin Interlaced Cross
	 U U D U				   U U R U
	 U U D U	      B B U B		   U U R U
	 D D D D	      B B U B		   R R R R
	 U U D U	      U U U U		   U U R U
			      B B U B
L L R L  F F B F  R L R R		  L L B L  F F U F  R F R R
R R R R  B B B B  L L L L     U U L U	  B B B B  U U U U  F F F F
L L R L  F F B F  R L R R     U U L U	  L L B L  F F U F  R F R R
L L R L  F F B F  R L R R     L L L L	  L L B L  F F U F  R F R R
			      U U L U
	 D D U D				   D D L D
	 U U U U     L L F L  F F D F  R B R R	   L L L L
	 D D U D     F F F F  D D D D  B B B B	   D D L D
	 D D U D     L L F L  F F D F  R B R R	   D D L D
		     L L F L  F F D F  R B R R
	 B B F B				   B B D B
	 B B F B	      D D R D		   B B D B
	 F F F F	      R R R R		   D D D D
	 B B F B	      D D R D		   B B D B
			      D D R D

When we relax the constraint that thin cross arms must meet at the
edges, the figure is no longer rigidly transformed with respect to
the ground.  Indeed, we might expect that adjacent crosses whose
arms do not meet might have colors that are opposite on the cube.
I carried out a long examination of the cases, and found that this
does not happen.  In fact, only one new color permutation arises.

	   Thin Fliptwist Cross

		 U U B U
		 B B B B
		 U U B U
		 U U B U

	L R L L  F F U F  R L R R  B D B B
	L R L L  F F U F  L L L L  B D B B
	R R R R  U U U U  R L R R  D D D D
	L R L L  F F U F  R L R R  B D B B

		 D D F D
		 D D F D
		 F F F F
		 D D F D

The only other thin cross patterns in which not all crosses meet at
the edges are 43 versions of the Thin Pons Cross (modulo my missing
a case or two in the analysis).

Date: 24 June 1982 16:28-EDT
From: Richard Pavelle <RP at MIT-MC>
Subject: Moleculon vs Ideal, cont.
To: CUBE-LOVERS at MIT-MC

I have read the patent of Larry Nichols who assigned it to Moleculon
and it looks to me like they have a very strong case against Ideal.
One should keep in mind, however, that more than 70% of patent
infringement cases go against the patentee.

The basic drawings and descriptions in the patent deal with the 2^3 
held together (non-rigidly) by magnets. Nichols discusses the possibility
of fixing the cubies rigidly in his description in a manner not unlike
Rubik. But the fact that he does not mention this aspect in his claims
may be his undoing. He mainly stresses the magnetic attachment. I would
not be surprised if the PTO told him that the rigid attachment would
comprise another implementation and he wished to avoid the extra 
expense (I am speaking from experience here). He discusses the puzzle 
aspects and some of the higher order cubes and non-cubes we have seen 
on the market.

In conclusion, I doubt he will get his 60M from Ideal but I think
he will get a non-trivial percentage and a continuing royalty. 

Date: 25 Jun 1982 0916-PDT
From: ISAACS at SRI-KL
Subject: Scientific American
To: CUBE-LOVERS at MIT-MC

  The July issue is out, with an update on Cubes by Hofstadter.  Don't
miss it.
  Also, the first Rubik's Cube Newsletter from Ideal came.  It's an 8
page glossy, mostly about Cube contests and Ideal's new puzzles.  So
far, mostly an Ideal advertising sheet.
  I got a letter from Ideal which said that "Revenge" won't hit California
until August!
  Adam Alexander is doing a promotional tour for Ideal with his "Alexander's
Star".  He said it is not out in the East yet - we get the star, East coast
gets the revenge, and the meet somewhere in the middle.  He lives in N.Y.,
was (is) a mathematician, and has lots of Cube-like inventions and models,
mostly built out of cardboard so far.
  -- Stan
-------

Date: 25 June 1982 1225-EDT (Friday)
From: Dan Hoey at CMU-10A
To: Cube-Lovers at MIT-MC
Subject:  New Scientific American article
Message-Id: <25Jun82 122513 DH51@CMU-10A>

    Date:  6 August 1980 16:52 edt
    From:  Greenberg.Multics at MIT-Multics
    Subject:  Re: 'cube lovers digest'

    I am really interested in when the _ Gardner is going to get
    his act together and publish THE cubing article-- each month
    for the last n I have eagerly taken the cover of SciAm and
    been disappointed. This is clearly THE mathematical game
    (since the inception of SciAm, with the possible exception of
    Conway's LIFE), and I wonder what he's waiting for.  About a
    week ago, I actually dreamt that I opened the SciAm wrapper
    and found the Cube on the cover, introducing a WHOLE ISSUE
    about it (Social implications, Ancient Cubing, Cubing in the
    Soviet Union, etc...)

Well, there finally was a Metamagical Themas column on the cube in
March '81.  Unfortunately, that article was mostly old hat to the
readers of Cube-Lovers.  The July '82 issue, just out, is a
different story.  Hofstadter discusses about twenty different
``Cube'' puzzles, based on all the regular polyhedra (and the
tesseract), several arrangements of twist axes and several coloring
schemes.  It even mentions cubing in the Soviet Union!  Stan Isaacs
<ISAACS at SRI-KL> is the Cube-Lovers representative for the new
article.

At last I can believe that Hofstadter's column will replace
Gardner's.

Date: 29 Jun 1982 2232-PDT
From: ISAACS at SRI-KL
Subject: Twisting Cube
To: CUBE-LOVERS at MIT-MC

  Take a cube(3x3x3) and re-color it in two colors, as follows:
Make 2 opposite faces with color 1 in the center, and color 2 on the
rest.  Call them "dots".  Take another pair of opposite faces and make
the 4 corners color 2 and the cross color 1.  Call them "pluses".
Finally take the last pair and make an "H" with color 1 and the 2
edges NOT adjacent to "dots" color 2.  Call them "H's".  In this cube,
every edge is the same, so is each corner, and the centers.  The result
is a cube where you only have to solve orientation problems, and never
need to position any cubie.  However, it's not quite the entire orientation
sub-cube - if you flip all 12 edges, you can't tell the difference.  Can
anybody come up with a coloring (2 or 3 colors - the centers could all be
colored with a different color from the edges and corners.  In fact, I guess
you could use 6 colors; the only necessity is for all edges to be the same,
and all corners the same. )... come up with a coloring which uses the
complete twisting sub-group.
   By the way, an elegant solution to the "edge-only" cube is to recolor
an Octahedron Cube so it is vertex centered.  Much nicer than peeling all
the corners of a regular cube.
   --- Stan
-------

Date:  6 Jul 1982 2318-PDT
From: Zellich at OFFICE-3 (Rich Zellich)
Subject: Mailing-list for "List of lists" update notices
To:   All mailing-lists:
cc:   ZELLICH

For those of you not previously aware of it, I maintain a master list
of ARPANET mailing-lists/digests/discussion groups (currently 756
lines or ~29,000 characters) on OFFICE-3 in file:

   <ALMSA>INTEREST-GROUPS.TXT

   For ARPANET users, OFFICE-3 supports the net-standard ANONYMOUS
   login within FTP, with any password.

To keep people up to date on the large number of such lists, I have
established a mailing list for list-of-lists \update notices/.  I do
not propose to send copies of the list itself to the world at large,
but for those ARPANET users who seriously intend to FTP the updated
versions when updated, I will send a brief notice that a new version
is available.  For those counterparts at internet sites who maintain
or redistribute copies for their own networks (DECNet, Xerox, etc.)
and can't reach the master by ARPANET FTP, I will send out the
complete new file.  I do \not/ intend to send file copies to
individual users, either ARPANET or internet; our system is fairly
heavily loaded, and we can't afford it.

There is no particular pattern to the update frequency of INTEREST-
GROUPS.TXT; I will occasionally receive a burst of new mailing-lists
or perhaps a single change of address for a host or mailing-list
coordinator, and then have a long period with no changes.

To get on the list, send requests to ZELLICH@OFFICE-3, \not/ to the
mailing-list this message appears in.

Cheers,
Rich
-------

Date: Friday, 16 July 1982, 10:20-EDT
From: Clark M. Baker <CMB at SCRC-TENEX>
Subject: Supplement to the Oxford English Dictionary
To: cube-lovers at MIT-AI

contains Rubic's Cube, described as an international craze of 1981 and
invented in 1975 by Erno Rubik of Hungary.


Date:      28 Jul 82 22:20:52-PST (Wed)
From:      Scott.uci at UDel-Relay
To:        cube-lovers.uci at UDel-Relay
Subject:   the 4x4 cube (a.k.a. Rubik's Revenge)
Via:  UCI; 29 Jul 82 5:43-EDT

   Has anyone seen a 4x4 cube anywhere in Orange County, or even L.A.?  I'd
appreciate any leads on where/how to get one.
   I saw on (a friend's) in Santa Barbara, but he got it in Boston.  I've
also seen solution books (e.g. Crown Books).  But the object itself is elusive.
   If you like the 3x3, I highly recommend the 4x4.  If you have a scrambled
one, I can solve it for you.

-- Scott Huddleston


Date: 4 Aug 1982 08:41 PDT
From: Mendelson.es at PARC-MAXC
Subject: Supercube
To: Cube-Lovers at MIT-MC
cc: Mendelson.es

I have been patiently(?) waiting for the 4^3 Supercube (Rubik's Revenge) to
show up in the stores in the Los Angeles area.  To my knowledge it has not
done so.  Does anyone know where it can be obtained in Los Angeles?  If it
cannot, is there anyone out there who will volunteer to send me one, all costs to
be borne by me?  Thanks for your responses.

Jerry Mendelson


Date: Friday, 6 August 1982, 18:51-EDT
From: Jonathan L. Handel <JLH at SCRC-TENEX>
Subject: Reagan cubed
To: cube-lovers at MIT-MC, bsg at SCRC-TENEX
Cc: jlh at scrc-tenex

From The Nation, August 7-14, page 103 (a continuing column of Reagan's
slip ups, called There He Goes Again:  Reagan's Reign of Error):

           Reagan: Elizabeth Drew wrote in the June 21 New Yorker that
         in a White House meeting, the President lauded the inventor of
         Rubik's Cube as exemplifying the virtues of American free enterprise.
	 
	  Truth: Erno Rubik, the inventor of the cube, is a Hungarian
	professor living in Communist Budapest.

Date: 6 August 1982 19:51-EDT
From: Allan C. Wechsler <ACW at MIT-AI>
Subject: Invisible group of the 4^3
To: CUBE-LOVERS at MIT-AI


The 4^3 doesn't have a supergroup in the sense of the 3^3 --
the orientations of the ceter cubies are determined by their positions.

However, there is one fairly natural adjunct group that people might 
try thinking about and solving.  A 4^3 shows 24 center cubies,
24 edge cubies, and eight corner cubies.  But if it were really
a solid cube chopped up by parallel slices, it would have eight
more cubies buried inside.  Call them stomach cubies.  The eight
stomach cubies form a 2^3 buried in the 4^3.  They move when you
twist slices.  Can people come up with tools to frob the
stomach cubies without disturbing the visible cubies?  What is
the order of the adjunct group?

   --- Allan


Date:  7 Aug 1982 1446-PDT
From: ISAACS at SRI-KL
Subject: 4^3 Supergroup
To: CUBE-LOVERS at MIT-MC

  Of course there is a supergroup on the 4^3.  Just number the 4 centers
to see it.  I think you can exchange any 2 pairs of centers.
  -- Stan
-------

Date:  7 Aug 1982 1501-PDT
From: ISAACS at SRI-KL
Subject: POLYCUBE (on IBM PC)
To: cube-lovers at MIT-MC

  I got the POLYCUBE program mentioned in the latest Scientific American
cube article, which has from 1x1x1 up to 7x7x7 cubes on the IBM Personal
Computer.  I found it disappointing.  It doesn't show the back of the
cube, making solving it VERY difficult.  It should also allow user-defined
shorthand, so one could build macros (or simply define better notation).
The notation is good for the general case, but hard for the 3^3 case and
down - it is a general X-Y-Z notation, R or L direction, 1-n layer.
Thus RX1 is our "R"; "ZL1" = U', etc. The colors are pretty.  You can
save a cube on disk if you haven't finished solving it, but only one.
   Why doesn't someone design and write a general group-theory puzzle
simulation program.  Draw any pattern (2 or 3 dimensions) on a screen, 
associate it with a matrix, name some permutations in the matrix for
moves, and you should have any conceivable (drawable) rotating axis
puzzle modeled.
  -- Stan
-------

Date: 7 August 1982 20:02-EDT
From: Yekta Gursel <YEKTA at MIT-MC>
Subject: 4^3 is available in LA area in some of the CHESS and GAMES stores...
To: CUBE-LOVERS at MIT-MC

A friend of mine just got his for about $11.  Apparently they were having an
opening sale.  I got mine a month ago from a friend in Boston.

                                               YEKTA@MC

Date:  9 August 1982 0737-EDT (Monday)
From: Dan Hoey at CMU-10A
To: Cube-Lovers at MIT-MC
Subject:  Invisible Revenge
Message-Id: <09Aug82 073740 DH51@CMU-10A>

When you have seen Rubik's Revenge, have you seen everything?
Maybe not.


                    Supergroups

There is some confusion about the meaning of the term Supergroup
for the larger cubes.  There are two issues at stake:

        (1) In the 4^3 and larger cubes, there are pieces that may
            be permuted but are colored the same.  Are positions
            that differ only in the permutation of identically-
            colored facets to be considered distinct?

        (2) In all odd-sized cubes, each face has a center facet
            that may be twisted.  Are positions that differ only in
            the twist of face centers to be considered distinct?

If the answer to (1) is `no', the puzzle is not a group, but a
collection of cosets.  I find this interesting only in that it is
understood by the masses, and then the answer to (2) is `no'.

If the answer to both (1) and (2) is `yes', we have what I would
call the Supergroup.  I mark my cubes so that I can distinguish
patterns in this group.  I discussed a way of doing this for the
3^3 in my message of 9 January 81 0551-EST.  For Rubik's Revenge, I
use a similar procedure, except now two spots must be cut out on
each face:
        +-----------------------+
        |     |     |     |     |
        |     |    *|*    |     |
        |-----+-----+-----+-----|
        |     |    *|*    |     |
        |     |     |    *|*    |
        |-----+-----+-----+-----|
        |     |     |    *|*    |
        |     |     |     |     |
        |-----+-----+-----+-----|
        |     |     |     |     |
        |     |     |     |     |
        +-----------------------+
Also, I now arrange the faces T-symmetrically, with the spots
toward the girdle.

Jim Saxe first brought it to my attention that we may answer `yes'
to (1) and `no' to (2).  We get a group that ignores face center
orientation.  This is probably what people mean when they say there
is no Supergroup for Rubik's Revenge:  This is the group of the
puzzle, and the Supergroup (as I have defined it) is the same for
the 4^3.  But in the 5^3 this group is distinct both from the
Supergroup and from the Color Cosets.


                    The 57th Piece

Alan Bawden, in his message of 5 June 82, despaired of explaining
the mechanical workings of Rubik's Revenge.  I will now rush in.

If you take your Rubik's Revenge apart (as described in Ronald B.
Harvey's message of 19 May 82, not recommended by Alan Bawden), you
will find that the cubies ride around on a sphere:  the mysterious
57th piece.  Only the face centers are connected to the sphere;
they have flanges that hold the edges in place, and the centers and
edges have flanges that hold the corners in place.  The tricky part
is the linkage between the sphere and the face center cubies.

The four center cubies of a face have extensions that together form
a mushroom-shaped plug.  Each plug extends from the center of the
face inward and is cut in quarters lengthwise.  There are six
sockets on the sphere in which these plugs will fit and may rotate,
but may not be pulled out.  This is sufficient to implement a
puzzle isomorphic to the 3^3, namely the 4^3 where we allow only
face twists.

The other necessity for a 4^3 is to be able to twist the six center
slices, which we name after their adjacent faces.  To accomplish
this, there are grooves in the sphere that form three orthogonal
great circles.  Each groove has the cross section of half of a
socket, and includes the four half-sockets that correspond to a
center slice.  When we twist one of those center slices, the
half-plugs formed by pairs of face centers in that slice ride
around the sphere in the grooves.  Of course there is an adjacent
center slice; when it is moved, it takes the sphere with it.  The
reason the grooves cannot have the cross-section of a socket is
that then, when we twisted half the cube with respect to the other,
the sphere might turn forty-five degrees, preventing center slice
twists along the other great circles.

When you twist the U center-slice of your cube, does the sphere
move or stay fixed?  To find out without taking the cube apart,
hold the cube by the D center slice and repeatedly twist the U
center slice clockwise. Don't touch the U or D faces while doing
this.  Mostly, the U face will turn and the D face will stay fixed,
because the cubie-cubie friction is greater than the cubie-sphere
friction.  Eventually, however, you will either see the U face lag
behind the U center slice, or the D face move to follow the U
center slice.  If the U face lags, the sphere is not moving; if the
D face moves, the sphere is moving.

I will take this opportunity to mention another feature of the
interior sphere:  It has screws in it.  I took my screws out, but
the sphere didn't come apart.  Then I put them back in, on the
`don't screw with it' principle.  Perhaps they are there so Rubik's
Revenge won't float?  This issue was raised by Tom Davis (12 August
80) back when people were interested in solving the 3^3 underwater.
The last I heard, Richard Pavelle (25 July 1980) was able to solve
the cube with only five gulps of air.  I imagine some of the
30-second whiz kids can solve Rubik's Cube while completely
submerged.  But Rubik's Revenge?  Don't hold your breath.


                    The Mechanical Invisible Group

Alan Bawden's message posed an interesting question about the 57th
piece, which I will state somewhat differently.  Suppose we paint
the sockets of the sphere according to the colors of the face
centers that inhabit them.  Then we mix up Rubik's Revenge and
solve it.  Must the sockets still match their face centers?

The answer is no.  In fact, the sphere may be in any of the
twenty-four positions consistent with it having once matched the
face centers.  To show this, we will show how to perform a
ninety-degree whole-cube move of the outside without moving the
sphere.  This is equivalent to turning the sphere ninety degrees,
and we can repeat the procedure and its conjugates to move the
sphere to any of its twenty-four positions.

In order to twist Rubik's Revenge without moving the sphere, we
place the cube in a position such that the U, F, and R center
slices do not move the sphere and restrict ourselves to the moves:

U1                      Clockwise quarter twist of the U face
U2                      Clockwise quarter twist of the U half
                                (face and center slice together)
D1                      Clockwise quarter twist of the D face
U1',U2',D1'             Counterclockwise quarter twists
R1,R2,L1,R1',R2',L1'    Likewise for R
F1,F2,B1,F1',F2',B1'    Likewise for F

The move is actually fairly simple.  The tricky part is moving the
L center slice cubies without moving the sphere.  To do this,
remember the eight-flip

        X = (R1 L1 U1 D1 F1 B1)^2

from the 3^3.  This exchanges the L and R center slices in the 4^3,
allowing us to cycle the L center slice cubies in the R center
slice.

        R2 L1' X R2 R1' X

is a sphere-fixing whole-cube move taking 24 qtw after cancellation.


                    The Theoretical Invisible Group

So much for tawdry reality.  As Allan C. Wechsler pointed out on 6
August 82, we can imagine a 4^3 puzzle that contains a 2^3 on the
inside.  If we solve the outside, must the inside be solved?  The
process shown above for the 57th piece implies not, for that
process performs the RL antislice on the 2^3 with respect to the
outside.  Can any move of the 2^3 be accomplished?  Elementary
group theory says no, for odd permutations of the Rubik's Revenge
edge cubies are also odd permutations of the 2^3's cubies.

I ran the Furst, Hopcroft, and Luks algorithm on the problem.  It
turns out that the permutation parity is the only restriction on
what can be done with the 2^3.  Thus if we solve the outside, the
inside may be in any one of the (8! 3^7)/2 positions obtainable in
an even number of quarter twists on a 2^3 puzzle.  This in
particular includes all whole-cube moves of the 2^3.
Unfortunately, I don't know any simple processes for whole-cube
moves or for turning two adjacent faces.

Date: 11 Aug 1982 2204-PDT
From: ISAACS at SRI-KL
Subject: Tsukuda's Square
To: CUBE-LOVERS at MIT-MC

  Just got a new group-theory puzzle, called Tsukuda's Square.  It's
sort of like the 15/16 puzzle, but harder.  It consists of a 4x4 matrix
in the center with the numbers from 1 to 16.  On the left side are
4 plungers, one for each row.  At the top is one plunger, which pushes
down columns 2, 3, and 4, all at once.  Both the top and the side plungers
push the rows or columns over 1 square; releasing them causes the same row(s)
or columns to slide back.  When the top plunger is pushed down, the number
1 row plunger cannot be pushed because of interference; that is the only
interference.
   A typical move, for instance, would be to push in the #2 row plunger,
push the top, release the #2, release the top.  This would change
           1 2 3 4
           5 6 7 8  (only looking at the top 2 rows)
  to
    
           1 3 4 8
           2 5 6 7.
  You can push 2 or 3 (or even4) side plungers at once.  Pushing an
adjacent pair is even useful in solving this thing.
   At any rate, it has the normal even parity problem - that is, you 
cannot exchange a single pair, but you can exchange 2 pairs, or permute
three squares.  Because of the limited moves, it is not nearly so simple
as it looks.  I can permute 3 in one specific position, and it takes me
12 moves to do so (where a move is a push of 1 or more side plungers,
then push the top, then release them, or the reverse [push the top first]).
Can anyone develop some more efficient algorithms?
   -- Stan
-------

Date: 12 Aug 1982 08:49 PDT
From: Mendelson.es at PARC-MAXC
Subject: Re: Tsukuda's Square
In-reply-to: ALAN's message of 12 August 1982 02:38-EDT
To: ISAACS at SRI-KL
cc: Cube-Lovers at MIT-MC


Sorry, I don't have any answers for you.  I'm strictly a hacker and not a theorist,
but I always find puzzles such as the one you describe to be a lot of fun.  I
would like to suggest that whenever a new one such as this shows up the
announcement of its arrival includes a reference to where it can be obtained.

Jerry Mendelson


Date: 12 Aug 1982 1129-PDT
From: Isaacs at SRI-KL
Subject: re: Mendelson
To:   cube-lovers at MIT-MC

  In the south Bay Area, Tsakuda's Square can be bought at Tex's Toys at
the San Antonio Shopping Center.  They also, at last report (about 2 days
ago) had Rubik's Revenge
  -- Stan
-------

Date: 12 Aug 1982 1725-CDT
From: Clive Dawson <CC.Clive at UTEXAS-20>
Subject: Rubik's Revenge
To: cube-lovers at MIT-MC

For those of you tracking the progress of Ideal's Rubik's Revenge into
stores across the country, it has made it to Texas.  Walgreen's drugstores
here in Austin have put them on sale for $9.99, which is the best price
I've heard of so far.  They say the regular price after the sale ends will
be $12.99, but stickers on the boxes show $15.99.
-------

Date: 14 August 1982 17:23-EDT
From: Alan Bawden <ALAN at MIT-MC>
Subject:  [Hoffarth.wbst: Rubik's Revenge]
To: CUBE-LOVERS at MIT-MC

Date: 13 Aug 1982 10:41 EDT
From: Hoffarth.wbst at PARC-MAXC
To:   Alan Bawden <ALAN>
Re:   Rubik's Revenge

Is it best to send messages to you for forwarding?


Last week Gold Circle, Rochester, N.Y. area, started advertising Ideal's
Rubik's Revenge for $9.99.

Date: 14 August 1982 17:27-EDT
From: Alan Bawden <ALAN at MIT-MC>
Subject:  Where to send your messages.
To: Hoffarth.wbst at PARC-MAXC
cc: CUBE-LOVERS at MIT-MC

    Date: 13 Aug 1982 10:41 EDT
    From: Hoffarth.wbst at PARC-MAXC

    Is it best to send messages to you for forwarding?

No it isn't.  It is best to mail your messages directly to Cube-Lovers@MIT-MC,
otherwise I have to forward your messages there myself before I can make a digest.

Mail-from: SU-NET host SU-SHASTA rcvd at 16-Aug-82 1204-PDT
Date: Monday, 16 Aug 1982 12:04-PDT
To: cube-lovers at Mit-mc
Subject: Cube Mechanics
From: Tom Davis <davis@Shasta at Sumex-Aim>

I finally found a 4x4x4 cube a couple of days ago, and have a couple of
interesting observations.  Forgive me if I repeat anything said so far,
but I have been ignoring everything on this list having to do with the
4^3 cube for fear of any sort of spoilers.

Using my 3^3 knowledge, I found it fairly easy to get it almost solved.
Half the time, however, I got it to the state where everything was
solved except that two adjacent edge cubies were flipped.  I finally
convinced myself by means of a somewhat involved (and probably
fallacious) "proof" that I would have to exchange them before I could
solve the cube.

My first observation is simply a trivial proof of that fact that I
discovered immediately after I took the cube apart for the first time
to see what was inside -- it is mechanically impossible to put the cube
back together with the cubies flipped (but not exchanged).  Some
similar parity-type arguments can be made about possible configurations
of the center cubies.  What is interesting is that this presents a new
method of proving things about configurations -- if one can dream up a
mechanical model of a cube with different guts, it may be obvious that
some sorts of things are impossible.  The cube simply has to behave the
same way externally.  I wonder if there are nice ways to look at the
various parity-trinity features of the three-cube by looking at it
using a different model of the internal mechanics.

My second observation is that although a 5^3 and a 6^3 may someday
appear on the market, the 7^3 will be pretty tricky to build.  When one
of the faces of a 7^3 is turned 45 degrees, the corner will lie
completely outside the original cube.  Any mechanical linkage will be
complicated indeed.  Maybe a cube could be built with little
microprocessors inside each cubie controlling little arms and hooks to
grab adjacent cubie faces ...

  -- Tom Davis


Date: 17 Aug 1982 2103-PDT
From: ISAACS at SRI-KL
Subject: RUBIK's Magazine
To: CUBE-LOVERS at MIT-MC

  I just got my first issue of "RUBIK'S", International Game Magazine,
subtitled "Logic & Fantasy in Space 1/82".  This is a magazine published
in Hungary, Editor in Chief: Erno Rubik.  "Responsible Editor": Norbert
Siklosi.  It contains "The Order of Disorder", by Peter Gnadig, likening
the Cube to Entropy; a sketch of Rubik; the words to the song Mr. Rubik,
"Games and Mathematics" by Gerzson Keri, articles on competitions, clubs,
fans; on the company that manufactures the Cube, book reviews (Bosserts'
book and Singmasters "Notes"), puzzles, advertisements, and more.  I haven't
finished reading all of it yet, but it seems to be an attempt at a serious
cube magazine, a little stilted because of translation from the Hungarian.
It's a quarterly, single issues are US $2.00, I guess the yearly
subscription is $8.00.  The address:
    Lapkiado Vallalat
    Kereskedelmi Iroda

   H-1906 Budapest
   P.O.B. 223
 
   HUNGARY

The Chorus and first verse of "Mr. Rubik" (out in England, by the Barron Knights

  Mr. Rubik Rubik Rubik
  Is your Cube from outer space?
  Mr. Rubik Rubik Rubik
  He got three sides then lost his place
  Mr. Rubik Rubik Rubik
  He just twists your cube all day
  This ain't my idea of child's play.

  Being the kind of guy I am, I told him I would try
  To help him solve the secrets of the cube that made him cry
  Well that was thirty days ago and half a million moves
  My wife's black and blue 'cos I keep dreaming she's a cube.

----Stan
-------

Date: 18 Aug 1982 2226-PDT
From: ISAACS at SRI-KL
Subject: KERI's Article
To: CUBE-LOVERS at MIT-MC

  The "Games and Mathematics" article in "Rubik's" Magazine (mentioned
yesterday) asks an interestion question: how can you characterize the
random coloring on a cube in order to determine if the cube is 1) solvable
by twisting, or 2) solvable by  dismantling and reassembling.  
  The obvious criteria are 6 colors, 9 of each, 4 on edges, 4 on corners, 1
on a center, no 2 facies of a cubie the same color.  For case 2, Keri
claims you need 4 more tests.  For instance, he gives test 1: Given 1
corner with colors A, B, and C, let the other 3 colors be a, b, and c.
Then you can't have a capital and small of the same letter on one corner,
and the 8 corners are exactly the 8 combinations.  What are the other
tests?  Are 4 really necessary?  What are the tests for case 1?
   By the way, he (from some other article) classifies the 3 unscrambling
methods as follows:
   1) Chemical unscrambling: repaint the sides.
   2) Physical unscrambling: dismantle and reassemble
   3) Mechanical (or mathematical): normal way, by twisting.

  --- Stan
-------

Date: 20 August 1982 0242-EDT
From: James.Saxe at CMU-10A (C410JS30)
To: Cube-Lovers at MIT-MC
Subject:  Rubik's Revenge problem--deep & shallow hypermoves
Message-Id: <20Aug82 024233 JS30@CMU-10A>

Consider all manipulations of Rubik's revenge as consisting of two
sorts of moves, namely (1) shallow moves, which turn an outer layer
with respect to the remaining three layers, and (2) deep moves, which
turn an outer layer and the adjacent inner layer with respect to the
remaining two layers.  [For the purposes of this problem, we will
regard a manipulation that turns only an inner layer--resulting, for
example, in faces that look like
				XXXX
				OOOO
				XXXX
				XXXX
when applied to a solved cube--as consisting of two moves, one deep and
one shallow, in opposing directions.]  If only shallow moves are
permitted, the 4x4x4 simulates a 3x3x3.  If only deep moves are
permitted, the 4x4x4 simulates a 2x2x2.  Define a shallow (deep)
hypermove as an arbitrary sequence of shallow (resp. deep) moves.
My question is:

	What is the maximum number of hypermoves required
	to solve the 4x4x4?

Notice that the answer to this question may depend on whether or not
one considers identically-colored face centers to be distinct (as
Hoey points out, the puzzle is not a group if identically-colored
face centers are not distinguished) and on whether or not one worries
about the positions of the eight hypothetical stomach cubies.  Also,
if the minimal number of hypermoves is odd, then it might be
important to start with one class of move.  That is, it is plausible
that sequences of the form SDSDS may be sufficient while sequences of
the form DSDSD may not.
				Jim Saxe

Date: 23 August 1982 1623-EDT (Monday)
From: Dan Hoey at CMU-10A
To: Cube-Lovers at MIT-MC
Subject:  Hypermove lower bounds
CC: James Saxe at CMU-10A
Message-Id: <23Aug82 162336 DH51@CMU-10A>

There is an easy lower bound on the number of hypermoves needed to
solve Rubik's Revenge.  If we distinguish like-colored face
centers, let us fix the BL center of the D face, permitting only
the shallow moves B1, F1, U1, U3, L1, R1, and their inverses, and
the deep moves F2, U2, R2, and their inverses.  Let us compute the
number of hypermoves needed to solve just the face centers of
Rubik's Revenge.  A shallow hypermove can achieve SF = 4^6 = 4096
different face center positions.  A deep hypermove can achieve
DF = 7! 3^6 = 3674160 different face center positions.  So in four
hypermoves, at most
	1 + (SF + DF) + 2 SF DF + (SF + DF) SF DF + 2 SF DF SF DF
	       = 453,021,789,719,303,692,337
face center positions can be achieved.  Since this is fewer than
the
	23! = 25,852,016,738,884,976,640,000
face center positions of Rubik's Revenge, some face-center
positions will require at least five hypermoves.

If like-colored face centers are not distinguished, the best lower
bound I can find using this method is three hypermoves.  If stomach
cubies are considered, I think both bounds increase by one, since
only deep moves can touch them.

It seems strange that this method relies only on the face center
solution.  Similar arguments about edges are not as good, because
so many edge positions are achievable using shallow hypermoves.
Corners are practically irrelevant, since they can be fixed using
only shallow hypermoves.

With respect to the question of odd sequences of hypermoves, Jim
Saxe mentions that ``it is plausible that sequences of the form
SDSDS may be sufficient while sequences of the form DSDSD may
not.''  I would like to add the further plausibility that both
types may be sufficient, while neither may suffice alone.

Date: 31 Aug 1982 1022-PDT
From: ISAACS at SRI-KL
Subject: re:hypermove lower bounds
To: Hoey at CMU-10A
cc: cube-lovers at MIT-MC, isaacs at SRI-KL

  Dan:  Could you explain to me how you got the formula of you message
of 23 Aug.  I can see more-or-less what you're doing, but I haven't been
able to parse the formula.  Also, I haven't seen your notation before.
I take it that "U3" is equivalent to "D1'", except hold the D1 in place.
How do you represent half twists?  Only by two quarters, or is there a 
shorthand?  From a group theory perspective, is it easier to talk about
hypermoves than slice moves?  Will that also be true on the 5^3, 6^3, etc?
From a solving perspective, it seems clumsy.
  -- Stan
-------

Date:  2 September 1982 0755-EDT (Thursday)
From: Dan Hoey at CMU-10A
To: ISAACS at SRI-KL
Subject:  Re: Hypermove Lower Bounds
CC: Cube-Lovers at MIT-MC
In-Reply-To:  ISAACS@SRI-KL's message of 31 Aug 82 12:22-EST
Message-Id: <02Sep82 075524 DH51@CMU-10A>

    Date: 31 Aug 1982 1022-PDT
    From: ISAACS at SRI-KL

    Could you explain to me how you got the formula of your message
    of 23 Aug.

There are SF different shallow hypermoves and DF different deep
hypermoves, and two similar hypermoves in succession can be
collapsed into one.  The formula expresses the number of
alternating sequences of length at most four, which is the sum of
the ``Number'' column below.

        Length   Type    Number
        0                1
        1        S       SF
        1        D       DF
        2        SD      SF * DF
        2        DS      SF * DF
        3        SDS     SF^2 * DF
        3        DSD     SF * DF^2
	4        SDSD    SF^2 * DF^2
	4        DSDS    SF^2 * DF^2

    I can see more-or-less what you're doing, but I haven't been
    able to parse the formula.

Well, I did leave out the multiplication signs.

    Also, I haven't seen your notation before.  I take it that "U3"
    is equivalent to "D1'", except hold the D1 in place.

Right.  I used this notation in "Lower Bounds for the 4x4x4" on 2
June and "Invisible Revenge" on 9 August.

    How do you represent half twists?  Only by two quarters, or is
    there a shorthand?

There is U3^2, not much of a shorthand.  For the U slice move, I
hinted at U21'.

    From a group theory perspective, is it easier to talk about
    hypermoves than slice moves?

Hypermoves are a curiosity that Jim Saxe dreamed up.  Any sequence
of depth 1 (or 3) moves is a single hypermove, as is any sequence
of depth 2 moves.  I assume you mean to ask whether it's easier to
talk the way I usually talk, in terms of what I will call "twist
moves" to distinguish them from "slice moves".

The question boils down to what set of generators (moves) you want
to use when counting the length of a process.  This topic was
endlessly rehashed in 1980 when people were trying to decide
whether to call a half-twist a single move or two.  Jim Saxe nearly
sent a message in 1980 about using only two generators to solve
Rubik's cube.  [As I recall, computer failure trashed the message
and he never retyped it.]  Certainly we can all do slices and
half-twists.  The question is how many moves to charge for such an
operation.  The richer the set of generators, the fewer the number
of moves, but the more complex the explanation of the generators.

I use the "quarter-twist" convention for the 3^3.  The generators
are 90-degree rotations of faces.  This seems natural, because it
is the minimal set that satisfies the following criteria.

    1.  Every possible cube position can be created using these
            generators, up to whole-cube moves.  This is a basic
            criterion.
    2.  The inverse of every generator is a generator.  This is
            necessary so that we have a metric.
    3.  Any position that can be reached by performing part of a
            generator is a generator.  This criterion ties the
            mechanical operations used in the cube to the
            permutation group.  Otherwise we could have generators
            like FUF and F'U' and perform F with their composition.
            Charging two moves for F in that circumstance is
            somewhat bizarre.  
    4.  Every M-conjugate of a generator is a generator.  This is
            an aesthetic consideration.  We could leave out the D
            and D' twists and still solve the 3^3, but that breaks
            up the symmetry of the puzzle.

Why do I want the generator set to be minimal?  Well, we could make
it maximal, but then we would have ``over 3 billion'' generators.
What I am looking for is a canonical set, and minimality seems like
the best way of choosing among metrics.  Thus we exclude slice
moves as generators because they are not necessary.

For the 3^3, this set is particularly fortunate, because the
converse of criterion 4 holds:  Every two generators are
M-conjugate.  This allows us to identify some local maxima without
long computations (14 December 1980) and to tighten lower bounds
using parity principles (9 January 1981).

    Will that also be true on the 5^3, 6^3, etc?

I would just as soon stick with a compatible metric.  This is not
to say that there cannot be abbreviations for these moves, simply
that for the sake of asking ``how many moves does this take'' we
count the number of quarter-twists.

We unfortunately don't have the converse of criterion 4 for cubes
larger than 3^3.  For the 4^3, for instance, there are two flavors
of move:  deep and shallow.  Dave Plummer (26 September 1981)
described certain positions of the 4^3 as local maxima, but I have
convinced him that we cannot demonstrate the truth of that
assertion using known techniques other than exhaustive search.  My
note of 2 June was able to use only one kind of parity in the lower
bound argument.  Both problems are due to the lack of the converse.

    From a solving perspective, it seems clumsy.

I'll agree that the generators are few in this scheme, but it is
possible to generate macros.  For instance, consider describing a
slice move on the 3^3 cube.  Singmaster uses the notation Fs to
denote the F1B1' = F12'3 slice move.  We can now also talk about
the F12' slice move, which is how everyone actually does the move.
I think this is much easier to remember than Allan Wechsler's IJK
notation (introduced 18 July 1980) or Doug Landauer's HPS notation
(27 August 1980) for dealing with whole-cube moves.

I'm still not too happy about the state of RubikSong, but I think
it's a matter of human engineering, and I like to stick with the
mathematics.

Date:  9 Sep 1982 1453-PDT
From: ISAACS at SRI-KL
Subject: IMPOSSI-BALL
To: CUBE-LOVERS at MIT-MC

  If you live in the SF Bay area, BESTs now has the Impossi-ball, as
well as most of the other better group-theory puzzles at fairly
reasonable prices.  Rubik's Revenge is $9.00.  The Impossi-ball
was in the Scientific American column; its the icosahedron-like ball
with 2 each of 6 colors (as Alexanders Star).  If fact, I think it 
is equivalent to the star.  It doesn't look as good, but the mechanism
is more clever; what's more, you can remove one face and use it as a
sliding block puzzle.  Has anybody seen the other new puzzles from the
column?  
  -- Stan
-------

Date: 15 September 1982 12:36-EDT
From: Richard Pavelle <RP at MIT-MC>
To: CUBE-LOVERS at MIT-MC

Ideal is sponsoring another cubathon (3x3) on October 9 at the
South Shore Plaza in Braintree. Prizes will be awarded in 
various categories but unless you are under 40 seconds you will
not be good enough.

I understand that Ideal has sold about 22 million cubes but their
world-wide rights have not helped too much. They estimate that
3 times that number have been sold by illegal competition. 

Date: 17 Sep 1982 0918-PDT
From: Dave Dyer       <DDYER at USC-ISIB>
Subject: Cubes in L.A.
To: cube-lovers at MIT-MC


 I picked up my 4x4x4 at Toys R Us for $10.  They also have a good
selection of 2x2x2, 3x3x3 "cubes" in various shapes and other cube
inspired puzzles.
-------

Date: 22 Sep 1982 1425-PDT
From: ISAACS at SRI-KL
Subject: OMNI cubes
To: CUBE-LOVERS at MIT-MC

  The current OMNI (Oct.) has mention of various "cubes" from Meffert, in
the Games section.  Doesn't say much, but has nice pictures.  The main one
of interest is the Megaminx, a cube-like cutting of a dodecahedron, so
each face has 5 corners, 5 edges, and one fixed center (which only rotates,
but does not move relative to the others), exactly as in the cube.  Scot
Morris claims the Cube has 10^19 combinations, the Revenge has 10^48, and
that the Megaminx has 10^69.  Is this correct?  Meffert claims the Megaminx
is harder than the Revenge, but I would bet it is easier, in the sense
that cube moves should move over easier, and do closer to the same things
as before.  
  -- Stan
-------

Date: 23 September 1982 1206-EDT (Thursday)
From: Dan Hoey at CMU-10A
To: Cube-Lovers at MIT-MC
Subject:  MegaMinx and orientation theory
In-Reply-To:  ISAACS's message of 22 Sep 1982 1425-PDT
Message-Id: <23Sep82 120644 DH51@CMU-10A>

Funny you should ask about the positions of the MegaMinx.  I
analyzed this puzzle with the FHL algorithm.  It was the longest
run of that algorithm that I have heard of, something like 20 CPU
hours.  Necessity was the mother of checkpointing.

The standard MegaMinx has 20 corners and 30 edges.  Both the edge
and corner permutation parities must be even.  Corner and edge
orientation parities are zero (mod 3) and (mod 2), respectively.
These are the only invariants, so there are twenty-four orbits and
	(20! 3^20 30! 2^30)/24 ~ 1.007E68
positions.  In the SuperMegaMinx, a 1/5 twist of a single face
center is possible, so there are
	(20! 3^20 30! 2^30 5^12)/24 ~ 2.458E76
positions.  I think the MegaMinx is going to be easier than Rubik's
Revenge, not because there is more carryover from the cube, but
because there is more of the puzzle that is not affected by a
single generator.  Computing a lower bound that takes account of
generators that commute is going to be difficult, however: there
are triples of commuting generators, and we can find commuting
triples {A, B, C} and {A, B, D} such that C and D do not commute.

A curious question:  What do we mean by the corner orientation of
the MegaMinx when the corner permutation is not the identity?  On
the cube, we took the corner colortabs on two opposite faces of a
solved cube and marked both the tab and its position.  Then in a
scrambled cube we counted the position of each marked corner
colortab relative to the marked position on its cubie.  This
procedure doesn't work on the dodecahedron:  where should we mark
the tabs on those corner cubies that are not on the two opposite
faces?

It was a year before I realized that the choice of placing the
marks on opposite faces of the cube was arbitrary.  The marked tabs
and positions can be chosen any way that gives us one marked tab
and one marked position per cubie, and has zero orientation parity
in a solved cube.  It is customary to enforce the second criterion
by requiring that marked positions be the positions of marked tabs
in a solved cube.  The same argument holds for edges, and both
generalize to the MegaMinx.

Why is there orientation parity?  When we twist an n-gon face, the
edge cubies move in an n-cycle and their colortabs move in two
n-cycles.  We call this an ``untwisted cycle''.  But we could
conceive (see below) of a puzzle where the edge colortabs would
move in a single 2n-cycle.  This would be a ``twisted cycle'', and
would violate orientation parity.  Similar arguments hold for
corner cubies, whose kn tabs must move in k n-cycles rather than
k/d nd-cycles, for d a divisor of k (note that k=4 for the
icosahedron (but the icoshahedron has two corner tab orbits, so
quite a different argument applies)).  To summarize, any puzzle
that moves pieces in untwisted cycles must preserve orientation
parity.

I wonder if some argument of this type can be made for the
tesseract.  The argument depends on the orientation group being
Abelian (it has in fact been cyclic in our examples), and at least
some of the hypercubies have nonabelian orientation groups.
Perhaps we have to use Abelian quotients of the orientation groups.
Has anyone seen the paper on the tesseract that was mentioned by
Hofstatder?

Retreating to three dimensions, let's consider a variant on Rubik's
cube.  Suppose you had a gear (heh) embedded in the URF and BLD
corners, such that whenever an edge cubie passed by one of these
corners (e.g. from UR to UF by twisting U) it would flip (go from
UR to FU).  This would violate EOP on every quarter-twist!  Of
course, you know what positions would be possible in such a puzzle.
So now let's consider gears in the FU, BL, and RD edges that twist
corner cubies as they go by.  I don't know about this puzzle, but I
suspect that there are only four orbits.

Date: 26 September 1982 22:12-EDT
From: Martin Minow <MINOW at MIT-AI>
Subject: Solving Rubic's Cube
To: CUBE-LOVERS at MIT-AI

The fall catalog of the Cambridge Center for Adult Education
had a course (8 1-hour sessions) on solving Rubic's Cube.

It was apparently cancelled for lack of interest.

Martin Minow
decvax!minow @ Berkeley



Date: 28 Sep 1982 2102-PDT
From: ISAACS at SRI-KL
Subject: Meffert Catalogue
To: cube-lovers at MIT-MC

  I recieved a copy of the Meffert "Pyraminx" catalogue a couple of days
ago.  The only solid axis-rotation puzzles not already mentioned in the
Scientific American article or the OMNI article are the Pyriminx Magic
Triangle, Pentagon, and Hexagon, 3 puzzles, all prisms of the named shape,
each side face with 9 squares, and the top and bottom the appropriate n-gon
with corners, edges, and a center.  Ie, the Magic Hexagon has a hexagon
on top and bottom, with 6 corners and 6 edges. These should be similar to
a cube, except, of course, the side faces cannot make quarter turns.  Also
in the catalogue are a junior and senior "Magic Barrell", a 3-layered
cylinder with 2 (jr) or 3 (sr) arcs cut down from top to bottom, which look
as if they allow half twists; a Double Pyraminx, an octagon with 4 triangles
per side, which I would assume twists about the 4 central hexagons, and
is mechanically similar to the Skewb.  Also shown are some textured pyraminxes,
the Pyraminx Star, with small tetrahedra on several Pyraminx faces (it looks
strange, but is equivalent to the Pyraminx, I assume); some bead puzzles
(a version of the Hungarian Rings, another related to the Orb), some
sliding block puzzles, a group of tetrahedron dissections (into 2, 3, 4, 5, and
6 pieces),and a couple of jigsaw-like puzzles.  It's quite a catalogue, but
he says most of the items are not yet in production! More waiting.
  -- Stan
-------

Date: 28 Sep 1982 2118-PDT
From: ISAACS at SRI-KL
Subject: Cubic Curcular and Scientific American
To: cube-lovers at MIT-MC

   Ole Jacobson is visiting, and he brought the new issue of David
Singmasters Cubic Circular.  It is labeled Issues 3&4; neither of us
ever recieved issue 2. Did anyone else?  This issue contains more anecdotes,
reports on contests, new cube products (try combining pieces from octagonal
prism-cubes and regular cubes in various ways..someone even made "Siamese
Cubes", gluing parts of 2 cubes together), some processes for the Revenge,
some information on the 5^3 (a prototype exists by 21-year-old Gaston Saint-
Pierre of Quebec; Singmaster claims 2.83E74 patterns), an analysis and 
theorems about magic polyhedra in general, more pretty patterns, a magic
disc, and a table of the 73 possible orders of elements with how many
elements are in each order (average order is 122, median is 67.3, and
the order with the most elements is 60, with 4.6e18 elements or about
10.6%).
   Also, this months Scientific American has the cube again in Hofstadters'
"Metamagical Themas", this time as an example of creativity via variations
on a theme.
   Enough writing for tonight.
  -- Stan
-------

Date: 29 Sep 1982 18:16:54 EST (Wednesday)
From: Mike Meyer <mwm at OKC-UNIX>
Subject: Re: Meffert Catalogue & Cubic Curcular and Scientific American
In-Reply-to: Your message of 29 Sep 1982 18:48 EDT
To: Alan Bawden <ALAN at MIT-MC>
Cc: Cube-Lovers at MIT-MC

I have issue number 2 of the cubic circular, and have had it since late
apr. If you are interested, I can forward a summary.

	mike


Date: 22 Oct 1982 19:09-EDT
From: Dan Hoey at CMU-10A
Subject: The 2x2x2x2 magic tesseract
To: Cube-Lovers at MIT-MC

Allan Wechsler's message of 17 May 1982 contains some interesting
comments on the four-dimensional hyper-cube, or tesseract.  I will
expand on them, and offer a correction.

The tesseract has eight cubical sides, labeled Back, Front, Up,
Down, Left, Right, Out, and In.  Each side may be twisted in any of
the twenty-four ways that a cube may be rotated in three-space.
Since these twenty-four twists are generated by repeated
application of the six quarter-twists of the cube, I consider a
move to be a single quarter-twist of one of the cubical sides.

I have picked three of the quarter-twists of the Out side to be the
``clockwise'' twists, given as Of, Ou, and Or below.  Given the
constraint that clockwise twists must be conjugates of each other
with respect to the movement group of the tesseract in four-space,
the remaining clockwise quarter-twists are determined.  In the
following list, the upper-case letter denotes the side to be
twisted, and the quarter-twist is displayed as a permutation on the
(square) faces of that (cubical) side.

Of=(URDL)  Ou=(RFLB)  Or=(FUBD)     If=(RULD)  Iu=(FRBL)  Ir=(UFDB)
Ro=(UFDB)  Rf=(OUID)  Ru=(FOBI)     Lo=(FUBD)  Lf=(UODI)  Lu=(OFIB)
Ur=(OFIB)  Uo=(FRBL)  Uf=(ROLI)     Dr=(FOBI)  Do=(RFLB)  Df=(ORIL)
Fu=(ORIL)  Fr=(UODI)  Fo=(RULD)     Bu=(ROLI)  Br=(OUID)  Bo=(URDL)

These twists have the satisfying property that when two different
twists move an edge from position E1 to position E2, then one of
the twists is clockwise and the other counterclockwise.  For
instance, both the Dr and the Fr' twists move an edge from FID to
FOD.  Another property that mimics the three dimensional cube is
that clockwise twists on opposite sides are reversed:  The action
of Of on the O side is the inverse of the action of If on the I
side.

To see how the table above is constructed we must describe the
movement group of the tesseract (the group of whole-tesseract moves
in four-space).  I look at it as operating on quadruples VWXY of
mutually adjacent sides.  To see if VWXY->V'W'X'Y' is in the group,
replace all occurrences of B, D, L, and I with F, U, R, and O,
respectively.  The resulting permutation must have the same parity
as the number of replacements performed.  Thus FLOD->UROB is in the
group, because we perform three replacements to form FROU->UROF, an
odd permutation.

To tell whether a quarter-twist is clockwise or not, take the side
V to be twisted, two consecutive letters WX from the permutation,
and a fourth, orthogonal letter Y from {F, U, R, O}.  If VWXY->OURF
is in the movement group of the tesseract, then we have the
clockwise quarter-twist Vy, otherwise the counterclockwise
quarter-twist Vy'.  For instance, if we twist the U side as (LFRB),
then VWXY=ULFO->OURF is in the movement group (one replacement
creates the four-cycle URFO->OURF), so we have the clockwise twist
Uo.

Let us now examine the reachable configurations of the corners of
the tesseract.  Every quarter-twist moves eight corners in two
four-cycles, so only even permutations of the corners are
achievable.  The orientations of the corners are more complex.  If
we move corner VWXY to V'W'X'Y', then VWXY->V'W'X'Y' must be in the
movement group of the tesseract.  Thus only half of the twenty-four
permutations of {V', W', X', Y'} are achievable, because of the
permutation parity constraint.

To define the orientation of the corners, we label the sides of
each corner and each corner position with the letters VWXY, and
read the orientation of a cubie as the letters it has in the V, W,
X, and Y sides of its position.  It is important here to obey the
the permutation parity constraint when doing the labelling, so that
each cubie may be placed in the home (VWXY) orientation in any
position.  For instance, one possible labelling is as follows,
where each column refers to a corner:

V   F  F  F  F  F  F  F  F  B  B  B  B  B  B  B  B
W   U  U  U  U  D  D  D  D  U  U  U  U  D  D  D  D
X   R  O  L  I  O  L  I  R  O  L  I  R  R  O  L  I
Y   O  L  I  R  R  O  L  I  R  O  L  I  O  L  I  R

Thus if the FURO corner (column 1) is in the FLUO position (column
2), then its orientation is VXYW.  Any orientation that is an even
permutation of VWXY is possible.  The group of orientations, A4, is
the same as the movement group of the tetrahedron (with vertices
labeled VWXY) in three-space.

As I suggested in my message of 23 September 1982, this orientation
group is not Abelian, so the orientation of the last corner is not
completely determined by the orientations of the other fifteen.  To
see what is determined, let us look at the tetrahedron with
vertices at half of the corners of a three-dimensional cube, say
the FUR, FDL, BUL, and BDR corners.  As I reported on 15 June 1982,
the twelve movements of the tetrahedron consist of the identity,
three 180-degree rotations, and eight 120-degree rotations.  The
June message also mentions that if those corner cubies are moved as
a unit, preserving their positions and orientations relative to
each other, then the 180-degree rotations are achievable on the 3^3
(they are the corners of the Zig-Zag pattern) but the 120-degree
rotations violate the corner twist invariant.  Of course, four of
them perform a net clockwise twist, and four of them perform a net
counterclockwise twist.  Define the twist of a tetrahedron movement
to be the net clockwise twist it applies to the corners of the
cube.  Thus the twist of the movements of a tetrahedron VWXY is
given by the following table.

        Twist 0: VWXY, WVYX, XYVW, YXWV
        Twist 1: VXYW, WYXV, XVWY, YWVX
        Twist 2: VYWX, WXVY, XWYV, YVXW

By reasoning about the actions on corners of the cube, it is clear
that the twist of the product of two movements is the sum of their
twist, modulo three.  Thus the twist group is an Abelian quotient
of A4, isomorphic to the cyclic group on three elements.

Since the orientation group of the tesseract corners is also A4, we
may use the twist group to construct an orientation invariant of
the corners of the tesseract.  As described in the September
message, each qtw moves the corners in untwisted cycles, so the sum
of the twists of the orientations of the corners must be zero,
modulo 3.

I ran the Furst, Hopcroft, and Luks algorithm on the 2^4 tesseract
and found that this is the only invariant of corner orientation.
Therefore, the number of reachable positions of the tesseract is
(15! / 2) (12^15 / 3) ~ 3.358 x 10^18.  This is larger than Allan
Wechsler's upper bound because he thought there were only six
orientations of each corner.

Date: 25 Oct 1982 0846-PDT
From: ISAACS at SRI-KL
Subject: megaminx; octahedron
To: cube-lovers at MIT-MC

  The Megaminx is now on sale in the San Francisco Bay Area.  It is,
as pictured, a dodecahedron with each face twistable in fifths, and
containing 5 corners, 5 edges, and a center.  Instead of 12 different
colors it seems to have 10, with the red and yellow duplicated; that 
means you can have a parity problem at the end if you exchange the
duplicate edges.  Solving seems to be pretty straightforward, except
new edge moves must be developed - the "slice" moves aren't very
effective.
  I also found an octahedron much more analogous to the cube then the
one with the 9 triangular faces (and independent vertices).  ON this
one there is a triangular center, 3 diamond shaped corners, and longish
edges on each face.  The centers are equivalent to the corners of a cube,
but monochromatic; the corners of the octahedron are equivalent to the
centers of a cube, but have 4 colors.  You can solve it with supergroup
cube moves, but it can be solved more efficiently, I think, by doing
the corners first (matching up the colors), then the edges, then the centers.
Needed are moves that move edges without twisting corners (do we
have any good corner moves that don't rotate centers on a cube?)
  One final puzzle that has come out - it's called Inversion, and it's
a sliding block cube.  There are 19 identical cubies, each colored half
red and half blue (3 faces each) and arranged around the edges of a cube
with one extra empty space.  They are held in place by a Rubiks-like
mechanism through the centers of the big cube.  Thus each edge of the big
cube has 3 of the little cubes, or 2 of them plus a space.  The little
cubes are each in one of the 8 possible orientations; One orientation
is represented by one cubie, 3 orientations by 2 cubies each, and 4
orientations by 3 cubies each.  The idea is to slide the cubies around
the edges of the big cube so that the outside is all red or all blue,
or some other regular pattern;  Inverting from (say) red to blue means
sliding all cubies to more-or-less the diagonal opposite positions.
   -- Stan
-------

Date:  1 Nov 1982 0902-PST
From: ISAACS at SRI-KL
Subject: Re: megaminx; octahedron
To: BSG@SCRC-TENEX at MIT-MC
cc: cube-lovers at MIT-MC, Isaacs at SRI-KL
In-Reply-To: Your message of 29-Oct-82 0922-PDT

  Yes, I solved it.  It didn't take much work at all - mainly deciding
which cube moves transfered well.  The only really diferent moves involved
switching edges on the last face, or flipping them.  Both had fairly easy
solutions.  I do one face first, and then work upwards from there.  There
is a 7 move sequence on the cube, which moves an edge from the upper face
to a middle edge, which transfers nicely to the Megaminx.  Corners just
work, without any special algorithms.  The 4 move commutator F R' F' R
moves to the Megaminx and will move corners around.  Of course, none of
the slice moves work, so new edge moves are needed.  We also need an
easy way to move centers, and to move stars around, to make pretty patterns.
  By the way, the one two I got, and several a friend got in L.A., all had
10 colors, with red and yellow duplicated.  However, last Friday I saw
ones with 12 colors (I think - they were in boxes), with dark green and
brown as the new colors.  I don't know if the 10 colors are a mistake or
intentional.  The presence of a duplicate edge piece makes the solution
slightly more difficult.
  -- Stan
-------

Date: 5 Nov 1982 16:54:01-EST
From: acw at scrc-vixen
To: rp at scrc-vixen
Subject: "The Cube Lovers at MIT"
Cc: CUBE-LOVERS@mit-oz


The article could be salvaged without
damaging it too much.  Cube-Lovers
could be described as a group based
at MIT which communicates via
electronic mail.  All mention of
ARPA and even the word "network"
should be eschewed, and the nationwide
nature of Cube-Lovers should be
downplayed.  The rest of the
article can stand, as far as I
am concerned.  Although I think the
style is a little dippy, I'm sure
he doesn't want my advice about it.

   --- Allan

Date: 11 Nov 1982 1421-PST
From: Carolyn Tajnai <CSD.TAJNAI at SU-SCORE>
Subject: Pentominoes
To: cube-lovers at MIT-MC

My husband, Joe, has requested a set of pentominoes on his Christmas
wish list.  Don Woods suggested I sndmsg to cube-lovers for pointers.

Where can I buy a set?  

Help!
Carolyn

(please snd answers, solutions, suggestions to CSD.Tajnai@SU-SCORE)
-------

CEL@MIT-ML 11/23/82 16:00:18 Re: Pentominoes request
To: Cube-lovers at MIT-MC
Hexominoes are easier to find than pentominoes.  If you treat four of
the basic trianglular pieces of the Snake puzzle as a unit square, a
24-piece Snake can be configured to make any hexomino.  Just buy
enough snakes.  If you want to break the rivets on the snake, you can
make pentominoes by removing four pieces.  In fact, I wish there were
a modular Snake consisting of snap-together pieces so that different
length Snakes could be realized more simply.

Date: 29 November 1982 22:51-EST
From: Alan Bawden <ALAN at MIT-MC>
Subject:  [YEKTA: "Pentominoes"...]
To: CUBE-LOVERS at MIT-MC

Date: 24 November 1982 19:57-EST
From: Yekta Gursel <YEKTA>
To:   ALAN, CSD.Tajnai at SU-SCORE
Re:   "Pentominoes"...

   GABRIEL toy company markets a nice set of pentominoes under the name
"HEXED".  They are the makers of the puzzles "Hi-Q" and "PYTHAGORAS".  The 
latter is a nice set of "tangrams" if you  are interested in them.  You should
be able to get any of these in any well stocked toy store ( like Toys 'R Us ).
If you are interested in getting solutions for them, me ( Yekta Gursel ) and
a friend of mine ( Douglas Macdonald ) wrote a program on a microcomputer to 
solve any pentomino problem.  The description of the program appeared in
November 1979 issue of BYTE.  It is written for a COMMODORE PET, but can be
adopted toi any microcomputer relatively easily.  If you do not have access to
a microcomputer we can print the solutions out for you just for the cost of
paper and postage ( a few dollars ).  If you are interested please let me
know.  My network address is YEKTA@MIT-MC.


[ ALAN: If anybody asks about pentominoes, you can refer them to me... Thanks!]

Date: 25 November 1982 05:12-EST
From: Yekta Gursel <YEKTA>
To:   ALAN
Re:   Addition to my pentominoes message...

Sorry, I accidentally sent the pentominoes message to you instead of the 
CUBE-LOVERS.  

   What I wanted to add is that one can also get a set of hexaminoes, though
this is a little bit harder.  They are marketed in England under the name 
"Spear's Multipuzzle".  The set comes with a few extra pieces, but nevertheless
contains a full set of hexaminoes.  The quality is not as good as Gabriel's
"HEXED".

Date: 30 Nov 1982 18:16-EST
From: hoey
Subject: Pentominoes
To: CSD.TAJNAI at SU-SCORE, Cube-Lovers at MIT-MC
Message-Id: <82/11/30 1816.433@NRL-AIC>
Origin: NRL-AIC

A friend of mine has a set of pentominoes called ``Quintillions'',
available from

	Kadon Enterprises, Inc.
	1227 Lorene, Suite 16
	Pasadena MD  21122

Basic Quintillions is a set containing the twelve planar
pentominoes for $29.00.  Super Quintillions is a set containing the
nonplanar pentominoes for $39.00.  (The literature says there are
eighteen pieces, but I think there are only seventeen nonplanar
pentominoes.  Perhaps they include some extra piece so you can make
a 3x5x6 rectangular prism)  The basic set is a high-quality puzzle,
each piece laser-cut from a single piece of wood, with a velour
box.

If you want to make your own, you can glue little cubes together.
Here are patterns for the planar pentominoes:

X   X X     X X         X           X   X X   X X    X X X X X
X     X X   X X   X   X X X     X   X     X     X               X
X X   X     X     X     X       X   X X   X X   X   X   X     X X
X               X X X       X X X     X         X   X X X   X X

For the nonplanar pieces, let U, D, and B stand for cubes on the
upper level, lower level, and both levels.  The pieces I know of
are as follows, where an asterisk denotes a piece that can be
reflected to form a new piece:

 *                 *         *         *     *     *

B     D     D     D B   D   D D       B     D     U     B
D D   B D   D B   D     D   D     B   D D   B D   B D   D B
D     D     D     D   D B   B   D B     D     D     D

Date: 30 Nov 1982 18:16-EST
From: hoey at CMU-10A
Subject: Pentominoes
To: CSD.TAJNAI at SU-SCORE, Cube-Lovers at MIT-MC

A friend of mine has a set of pentominoes called ``Quintillions'',
available from

	Kadon Enterprises, Inc.
	1227 Lorene, Suite 16
	Pasadena MD  21122

Basic Quintillions is a set containing the twelve planar
pentominoes for $29.00.  Super Quintillions is a set containing the
nonplanar pentominoes for $39.00.  (The literature says there are
eighteen pieces, but I think there are only seventeen nonplanar
pentominoes.  Perhaps they include some extra piece so you can make
a 3x5x6 rectangular prism)  The basic set is a high-quality puzzle,
each piece laser-cut from a single piece of wood, with a velour
box.

If you want to make your own, you can glue little cubes together.
Here are patterns for the planar pentominoes:

X   X X     X X         X           X   X X   X X    X X X X X
X     X X   X X   X   X X X     X   X     X     X               X
X X   X     X     X     X       X   X X   X X   X   X   X     X X
X               X X X       X X X     X         X   X X X   X X

For the nonplanar pieces, let U, D, and B stand for cubes on the
upper level, lower level, and both levels.  The pieces I know of
are as follows, where an asterisk denotes a piece that can be
reflected to form a new piece:

 *                 *         *         *     *     *

B     D     D     D B   D   D D       B     D     U     B
D D   B D   D B   D     D   D     B   D D   B D   B D   D B
D     D     D     D   D B   B   D B     D     D     D

Date: 2 December 1982 06:14-EST
From: Richard Pavelle <RP at MIT-MC>
To: CUBE-LOVERS at MIT-MC

The times for REVENGE are now under 90 seconds!!


Date: 2 December 1982 06:42-EST
From: Richard Pavelle <RP at MIT-MC>
Subject:  material for Ripleys'
To: CUBE-LOVERS at MIT-MC


The following are true:

The times for REVENGE are now under 90 seconds!
The one-handed record for the 3x3x3 is 89 seconds!!
The time for the 3x3x3 with two feet is 2 minutes and 50 seconds!!!

Does any have times for no hands and no feet? 

Mail-From: CMUFTP host CMU-EE-AMPERE received by CMU-10A at 3-Dec-82 03:24:38-EST
Date: 3 Dec 1982 03:19:57-EST
From: Phil-Servita-H at CMU-EE-AMPERE at CMU-10A
To: Cube-Lovers@mc@a
Subject: puzzle search

 does anybody out there know where i can get either Megaminx (otherwise 
known as Pyraminx Ultimate) or the Pyraminx Cube (known as the Skewb to 
all you Scientific American readers) ?  Ive been trying for months. Any
help will be greatly appreciated.

                                                   -phil
                                          ( pxs@cmu-ee-ampere@cmu-10a )

Date:  3 Dec 1982 1820-PST
From: ISAACS at SRI-KL
Subject: Megaminx
To: CUBE-LOVERS at MIT-MC

  Megaminx, under that name, should be available at your local stores (you
don't say where you live).  It is out in the San Francisco Bay Area, and
L.A., I know, including in such chains as GEMCO and TOYS R US.  It is
distributed by TOMY Corp.  I have not seen the Skewb anywhere, except in
the hands of a couple of people that got it from Meffert personally; he
did not send me one by mail, though I asked.  He did send his Double
Pyramid, though, a nice version of an octahedron, cut as Rubik's Cube, e.g.
with centers. (The other octahedron, the "Star Cube", has no centers and a
lot of extra corners.)
  -- Stan
-------
Mail-From: CMUFTP host CMU-EE-AMPERE received by CMU-10A at 11-Dec-82 11:15:41-EST
Date: 11 Dec 1982 11:08:22-EST
From: Phil-Servita-H at CMU-EE-AMPERE at CMU-10A
To: cube-lovers@mc@a
Subject: 4x4x4 competition


            can you solve the 4x4x4?

            can you solve it fast?

            dont you wish that SOMEBODY would hold a 4x4x4 competition?

    
    if you answered yes to any of those questions, then read on.
    
    i have been in contact with Ideal/Gabriel about this topic. this is
    all that they want: people. if enough people seem interested in a 
    4x4x4 competition, they may hold one sometime between spring and fall '83.

                          so here's what you do:

    if you are interested in a 4x4x4 competition, mail your response to:

                    pxs@cmu-ee-ampere@cmu-10a

              Cc:   meister@ccc@mc, ps0k@topsf@cmuc@cmu-10a

     ill keep a list of responses and send it to Bob Weissman at Ideal.

                                                               -phil



Date: 11 December 1982 20:43-EST
From: Alan Bawden <ALAN at MIT-MC>
Subject:  The Archive & Administrivia
To: CUBE-LOVERS at MIT-MC

Those of you who look through the archives of old Cube-Lovers mail
will notice that I have split off a new section of the archive.  The
mail now lives in:

MC:ALAN;CUBE MAIL0	;oldest mail in foward order
MC:ALAN;CUBE MAIL1	;next oldest mail in foward order
MC:ALAN;CUBE MAIL2	;more of same
MC:ALAN;CUBE MAIL3	;yet more of same
MC:ALAN;CUBE MAIL4	;still more of same
MC:ALAN;CUBE MAIL	;recent mail in reverse order

Those of you who are wondering if Cube-Lovers might be temporarily off the
air when TCP is installed on January 1 can rest assured that there is an
ITS TPC implementation in progress, and come January 1 the ITS machines
will probably be no worse off than the rest of the net.  My biggest worry
is not that ITS won't be able to send mail (since I can always move the
list to another machine if that proves necessary), but that so many of you
all will be having difficulty recieving mail that I will be up to my neck
in error messages from various mailers around the net.

That is one of my reasons for splitting off a section of the archive at
this time.  I'm willing to bet that a fair number of you will be requesting
the back mail that you have missed during the storm at the beginning of the
year, and this way I have a small file that I can easily send off to you.

Finally let me remind you all to please address your requests to
Cube-Lovers-Request at MIT-MC.  If that address, or Cube-Lovers at MIT-MC
ever seems to be giving you trouble, you might also try Cube-Lovers at
MIT-Multics and Cube-Lovers-Request at MIT-Multics.

				-Alan

Date: 21 December 1982 23:50-EST
From: Alan Bawden <ALAN @ MIT-MC>
Subject: Correction to previous message.
To: CUBE-LOVERS @ MIT-MC

It being so difficult to set up a Multics mailing list I'm retracting the 
promise that those Multics addresses will ever work and will instead
just give out my personal Multics address as an emergency contact.

Date: Wednesday, 22 December 1982, 15:33-EST
From: Allan C. Wechsler <acw at SCRC-VIXEN>
Subject: Hinton's cubes.
To: Cube-Lovers at MC

In September I received a query from Phil Servito, who
has since joined this mailing list, about a game or puzzle
called "Hinton's cubes".  Does anybody have any information
about this toy besides the following?

Charles Howard Hinton was an American mathematician who
believed that anybody could learn to do four-dimensional
visualization if properly trained.  He invented a set of
colored blocks which were supposed do this training.

Philip -- you said Martin Gardner knows more.  Is this because
the cubes were once discussed in a Sci. Am. Mathematical
Games column?  If so, the Sci. Am. cumulative index may help.

   --- Allan

Date: 29 Dec 1982 0954-PST
Subject: article of interest.
From: Dave Dyer       <DDYER@USC-ISIB>
To: cube-lovers@MIT-MC


 The Christmas issue of NEW SCIENTIST has an article by Singmaster
about the basics and metaphysics of the cube.  No great revelations,
but a literate discussion of cubeology.  
-------

Date: 29 Dec 1982 19:01:04-EST
From: meister at mit-ccc
To: cube-lovers@mit-mc
Subject: 4x4x4 poll


  i am interested i getting information on the various solving methods
used on the 4x4x4, and the times that it is being done in. (average) 
my own method is to solve it as follows:

       1) top layer (no general order to this)
       2) the rest of the centers by layers 
       3) the edges on the center slices
       4) bottom corners
       5) bottom edges

 average time: 1 min 55 sec - 2 min 5 sec depending on my blood-alcohol level. 

   please send me your general method and solving times.
   i'll compile the results and put them on the list.

                                              thanks for replying,
                                                          -phil
 p.s. -  send replies to  meister@ccc@mc 



Date: 30 December 1982 00:41-EST
From: Jonathan David Callas <JURGEN @ MIT-MC>
Subject: Hinton's cubes.
To: ALAN @ MIT-MC
cc: CUBE-LOVERS @ MIT-MC


I remember reading about Hinton's cubes in an old Mathematical Games anthology,
so a cumulative index would be the thing. Gardner wrote about them and 
apparantly they worked, but there were stern warnings not to play with them,
because (!) they cause insanity in some people(!!). Gardner said something
to the tune of that the human brain is not meant to do 4-dimensional geometry,
and these things will make your brain bugcheck. Ever since then, I have been
fascenated by the thought of a puzzle that would blow your mind  literally! Of
course, this was in the days before  it was fashionable to use chemical means
to do this. Anyone else know anything about them?

	Jon

Date:  3 Jan 1983 0915-PST
From: ISAACS at SRI-KL
Subject: Hinton's Cubes
To: cube-lovers at MIT-MC
cc: isaacs at SRI-KL

   Hinton's Cubes are mentioned in Martin Gardners "Mathematical Carnival",
Chapter 4, Hypercubes, a reprint of the November, 1966, Mathematical Games
column from Scientific American.  He doesn't say much, mainly that "Hinton...
devised a system of using colored blocks for making three-space models of 
sections of a tesseract."  Hinton believed that they would help develop
an intuitive grasp of 4-space.  The fullest account of them mentioned is
the book k"A New Era of Thought", by C. Howard Hinton, Swan Sonnenschein, 
1888, apparantly reprinted in 1910.  
  Gardner also prints a letter he received from Hiram Barton, of England,
including "A shudder ran down my spine when I read your reference to Hinton's
cubes.  I nearly got hooked on them myself in the nineeen-twenties.  Please 
believe me when I say that they are completely mind-destroying.  The only
person I ever met who had worked with them seriously was Francis Sedlak, a 
Czech neo-Hegelian philosopher...who lived in an Oneida-like community near
Stroud, in Gloucestershire.
   "As you must know, the technique consists essentially in the sequential
visualizing of the adjoint internal faces of the poly-colored unit cubes 
making up the large cube.  It is not difficult to acquire considerable 
facility in this, but the process is one of autohypnosis and, after a while, 
the sequences begin to parade themselves through one's mind of their own 
accord."
   It goes on, but doesn't say a great deal more.  I don't have my copy of
Hintons book "The Fourth Dimension", but maybe it will give some more 
description.
  If anybody does find some marketed versions of the cubes, put it on the list.
   --- Stan
-------

Date: 4 January 1983 03:54-EST
From: Alan Bawden <ALAN @ MIT-MC>
Subject: Humor
To: CUBE-LOVERS @ MIT-MC

From "Special Deliverance" a recent novel by Clifford Simak:

   "He also mentioned a cube," said Sandra.  "I wonder what it could be.
   Never before have I heard anything described simply as a cube."

Well I certainly have!

(And no, the book actually has nothing to do with Hungarian Cubes.  It just
struck me as strange that someone could write that line at a time when an
object that frequently WAS described simply as "the cube" was enjoying some
popularity.  Perhaps Sandra's ignorance is excusable in light of the fact
that she comes from an alternate universe...)

Date:  5 Jan 1983 1802-PST
From: ISAACS at SRI-KL
Subject: Other news
To: CUBE-LOVERS at MIT-MC

   The new RUBIK'S magazine is out.  Not as interesting (to me) as the first.
Several articles on the Snake, on cube contests, and some on other games.
A computer flow-chart for the game Awari, and an analysis of Chinese Rings
are included.  The most interesting cube articles are one on how disordered
a cube can get; and one on pretty patterns in other cube universes (ie, when
an edge is flipped, what symmetric patterns are there). 
  If you haven't seen it, the book "INside Rubik's Cube and Beyond" by
Christoph Bandelow is very good.  It's translated from German, and contains
a fairly technical discussion of the Mathematical Model, solving in the
supergroup, other magic polyhedra, a detailed flowchart of a cube-solving
program (which also checks the input for legality), and a comprehensive 
Maneuver Index, which contains, among other things, a complete catalog of
3-cycles.  It includes an index, a bibliography, and a good section of
colored pictures in the center (including one of a 5x5x5!!!). Recomended.
   I've also heard a rumor that the solution to "Masquerade" has been
written (and maybe published).  Has anyone heard anything of this?
  -- Stan
-------

Date:  5 Jan 1983 1806-PST
From: ISAACS at SRI-KL
Subject: more on Hinton's Cubes
To: cube-lovers at MIT-MC

  I got my "Speculations on the Fourth Dimension, Selected Writings of
Charles H. Hinton", Dover, 1980, back.  It doesn't reprint Hinton's
description of his cubes, but Rudolf v.B.Rucker, in the introduction 
describes something more of them:
  "The second part of "A New Era of Thought" consists of a description 
of how to visualize a tesseract by looking at various 3-D cross sections
of it.  On is to construct a set of 12 cubes, coloring the faces, edges
and corners all manner of different colors. (Eighty-one different colors
are used, and some rather unfamiliar ones are resorted to. ...)Eight of
these cubes make up the boundaries of the hypercube, and the four others
are cross sections taken between pairs of opposite cubes.  The way in
which all the cubes fit together is really explained rather well, if one 
has the will to endure not only 81 colors, but the 81 Latin names which
Hinton assigns to the parts of the tesseract.
   "In addition to the set of 12 large cubes, there was also to be a set of 81 
small monochrome cubes, each representing a part of the tesseract.  By
moving theese little cubes about one could better comprehend the fact that
rotation through the fourth dimension corresponds to mirror image reflection
in 3-D space."
   In Hinton's book "The Fourth Dimension", published n 1904, he has a
streamlined version of the tesseract section models.  "There were actually 
three parts to the complete set of tesseract models.  There was a set of 27
"slabs," actually cardboard squares; a set of 81 one-inch monochrome cubes,
each a different color; and a set of 12 multicolored "catalogue cubes,"
which were depicted in a color plate bound into "The Fourth Dimension".  When
the book came out, one could buy a set consisting of the 27 slabs and the
81 little cubes for 16 shillings, or a set consisting of the 12 catalogue
cubes for 21 shillings."
  If anyone on this list can find a copy of "The Fourth Dimension", I, for
one, would like to buy it.  I wonder if any of the models are around anywhere?
     --- Stan
-------

Date:  7-Jan-83 10:19:31 PST (Friday)
From: Pettit at PARC-MAXC
Subject: 4-D Graphics Program
In-reply-to: ALAN's message of 5 January 1983 23:40-EST
To: Alan Bawden <ALAN @ MIT-MC>
cc: Cube-Lovers at MIT-MC

Scott Kim, the author of Inversions, wrote a program about 4 or 5 years ago which ran on Tenex, and allowed one to rotate 4-D skeletal objects and view the 2-D projections (on an Imlac display).  I don't know what language it was written in.  It was apparently quite good at allowing one to achieve an intuition for 4-D space.  I'll send him a message and ask if his program is available anywhere for playing with.

--Teri Pettit at Xerox OSD   

Date: Monday, 24 January 1983  13:57-EST
From: Leonard N. Foner <TK.FONER @ MIT-OZ>
To:   Cube-Lovers @ MIT-MC
Subject: Shortest sequence of moves?
cc:   Tk.Foner @ MIT-OZ
Reply-to: Foner at MIT-MC

I remember hearing about a program, probably running on the color LISP
machine, that could take an arbitrary cube and try to see what the
shortest move sequence to its slution is.  Does anyone remember where
this program is and how to use it?

Specifically, I have been given the following cube and asked to find
the shortest solution:

    GBG
    BGW
    GGG
RRR WYW ORO YYY
ORO GWY OOO WYB
RRR WWW ORO YWY
    BBB
    GBY
    BGB

If anyone can either find the program I think I remember, or can
determine what they think is the shortest solution, I'd be \very/
grateful!  Speed is of the essence.

Thanx much, everyone!

						<LNF>

Date: 24 Jan 1983 1122-PST
From: ISAACS at SRI-KL
Subject: Masquerade
To: cube-lovers at MIT-MC

   A small paperback version of Masquerade has come out in England, which
contains a history, who solved it, and what the solution is.  I don't know
if it is available in the U.S. yet, but I have seen a copy, and got the
following info:  The full title & subtitle is "The Treasure Hunt of the 
Century / Masquerade  / Kit Williams Tells the Answer to the Riddle",
ISBN 0 224 02937 1; Jonathan Cape (pub), Thirty Bedford Square, London.
I will check around to see if it is available   ---  Just did; At lease
one local book store (Printers Inc, in Palo Alto, Ca.) says it's on order,
and they expect in in a day or two.
  If anyone wants a rough idea of how to find the solution, I can put it
on the net.
 -- Stan
-------

Date: 24 January 1983 16:51 EST
From: David C. Plummer <DCP @ MIT-MC>
Subject: Shortest sequence of moves?
To: FONER @ MIT-MC
cc: CUBE-LOVERS @ MIT-MC

    Date: Monday, 24 January 1983  13:57-EST
    From: Leonard N. Foner <TK.FONER @ MIT-OZ>
    Reply-to: Foner at MIT-MC

    I remember hearing about a program, probably running on the color LISP
    machine, that could take an arbitrary cube and try to see what the
    shortest move sequence to its slution is.  Does anyone remember where
    this program is and how to use it?

Sorry, no such program.  Don't we wish though!!  If we understood
the mathematics of the cube well enough to write such a program,
we probably wouldn't have this mailing list anymore.

Date: 28 Jan 1983 3:33-EST
From: Dan Hoey <HOEY @ CMU-CS-A>
Subject: The shortest sequence of moves.
To: Foner at MIT-MC
Cc: Cube-Lovers at MIT-MC
In-Reply-To: Leonard N. Foner's message of Monday, 24 January 1983  1357-EST

Leonard,

The process (R U^2 B^2 L')^2 will restore your cube in twelve
quarter-twists when executed with the Green face Up and the White face
Front, and twelve is the minimum sufficient number of quarter-twists.

Dave Plummer's discouraging word is usually right--we know of no
algorithm to let us find optimal processes for most positions.  This is
because the only known algorithms involve exhaustive breadth-first
search, and there are far too many positions of the cube to make this
practical in either time or space.  But when the optimal process is
sufficiently short, some headway can be made.  Having some megabytes
and CPU-hours at my disposal, I was able to list
    (A) all positions reachable in five qtw from your cube, and
    (B) all positions reachable in five qtw from SOLVED.
Finding that sets (A) and (B) are disjoint, I conclude that there is no
ten qtw process for the pattern, so the twelve qtw process is optimal.

I discovered the optimal process by hand.  Of course, I could have just
run the program one more qtw and it would give me the process, along
with any other twelve-qtw processes that may exist.  The problem with
that approach is that I don't have that many megabytes and CPU-hours.

My program, by the way, is written in C and runs under Unix.  It trades
time and storage efficiency for programmer laziness, making extensive
use of the Unix sort utility.  Dave Plummer has written a much
optimized program, in assembler language for the PDP-20, that uses very
clever hacks (some of them my own).  As I recall, he and I estimated
that with about 150 megabytes and a day or two elapsed time on an
unloaded machine it could take the search three more quarter-twists.
Does anyone need to settle a bar bet on an eighteen qtw process?

				Dan

Date: 28 Jan 1983 1838-PST
From: ISAACS at SRI-KL
Subject: Circle Puzzles
To: cube-lovers at MIT-MC
cc: isaacs at SRI-KL

   I just got some very nice examples of circle (group-theory) puzzles -
similar to the Lorente Grill puzzles shown in Hofstadters' Sci.Am.
article in July, 1982 (p. 26).  They are made by Douglas Engel, under
the name of General Symmetrics, Inc, 2935 W. Chenango, Englewood, Co,
80110.  He is selling some as experimental items, and one version he
hopes to have marketed "soon".  They all consist of two intersection
circles, each of which rotate; I have 3 kinds: the "21 piece", similar
to 2/3 of Lorente's Trebol puzzle; the "35 piece", similar to the
one in the upper right corner of the article "Another Grill puzzle by
Lorente", and a "19 piece", similar to the 21 piece, except the circles
intersect in the center.  Each of these are quite different from each other;
the 35 piece has 3 different pieces which rotate in 4 different cycles
(compare to the Cubes' 3 kinds of pieces - edge, corner, and center).
In the 35 piece, a straightforward "in in out out" (R'LRL') type move
will move the centers in a 3-cycle, the "square" pieces in a pair of
2-cycles, and the "triangles" in 2 separate 5-cycles.
   (Doug is selling these at $10.00 for the 19 or 21 piece, $20 for the
35 piece.)
   Doug Engel has also written a paper called "Some Problems Suggested
By Circle Puzzles", mostly asking questions about various kinds of
symmetry in this type, whether there are an infinite number of them,
etc.  I wonder what the best way to talk about them and analyze them
from a group theory perspective is.  I'll try to put in more information
and theories as I work on them; I'd like to hear any thoughts or
suggestions.  Right now, I want to go home for dinner.
  -- Stan
-------

Date: Saturday, 29 January 1983  23:33-EST
From: Leonard N. Foner <TK.FONER @ MIT-OZ>
To:   Dan Hoey <HOEY @ CMU-CS-A>
Cc:   Cube-Lovers @ MIT-MC, Tk.Foner @ MIT-OZ
Reply-to: Foner at MIT-MC
Subject: The shortest sequence of moves.
In-reply-to: The message of 28 Jan 1983  03:33-EST from Dan Hoey <HOEY at CMU-CS-A>

Thanx for the solution...  we determined independently that 12 qtw
seemed to be minimal, but couldn't be absolutely sure.

Since you say with a couple of days of CPU time you could deal with 18
qtw processes, doesn't this mean that \any/ cube process can be
checked for minimality?  I can't recall whether the maximum number of
moves from solved is 17 or 19...  if the former, 18 qtw may be
unnecessary.  If the latter, then almost every cube process can be
checked directly, save the absolute longest ones (which are then
provably 19 anyway, since that's all they can be).  Is my reasoning
faulty here?  In any case, exactly how much more work in involved for
each qtw in solving the cube?  In other words, how fast does it grow
(what's the power on the exponent, if describable this way)?

Thanx much folks.

						<LNF>

Date: 30 January 1983 00:40 EST
From: Alan Bawden <ALAN @ MIT-MC>
Subject:  The shortest sequence of moves.
To: FONER @ MIT-MC
cc: CUBE-LOVERS @ MIT-MC, Tk.Foner @ MIT-OZ, HOEY @ CMU-CS-A
In-reply-to: The message of 29 Jan 1983  23:33-EST from Leonard N. Foner <TK.FONER at MIT-OZ>

The maximum number of moves (qtw) from solved is certainly not 17 or 19.
All we know about that number is that it is greater than or equal to 21.
It could be even larger.  There is a fair amount of sentiment that it is
probably around 28, but that's just intuition.

Date: 30 Jan 1983 3:15-EST
From: Dan Hoey <hoey@cmu-cs-a>
Subject: Finding optimal processes
To: Cube-Lovers at mit-mc

The best known bounds put the maximum number of qtw at between 21
and 104.  At most about 1 percent of the cube's positions are
within 18 qtw of solved.

At each qtw, the number of positions increases by a factor of
almost sqrt(12+5*sqrt(6))+sqrt(6)+2 ~ 9.374.  [I say ``almost''
because this rate of increase does not take into account the
identities of length 12 qtw or greater.  Eventually, long identies
reduce this factor to zero.  But between six and seven qtw, the
factor is about 9.356].  If we want to show a 2n qtw process
optimal, we need only scan the table of n-1 qtw processes, so the
increase is more like a factor of 3 per qtw.

Finally, I would like to call attention to the fact that showing an
18 qtw process to be optimal requires about 150 megabytes and two
days on a PDP-20 for each such proof.  Finding someone to lend you
a big machine and a big disk for two days is not something you can
take for granted.  And if you want to do it on your Apple, be
prepared to sit around swapping 2400 floppies in and out for the
next decade or so, as that two day figure is mostly disk latency
rather than CPU time.

Date: Monday, 31 January 1983  21:48-EST
From: Leonard N. Foner <TK.FONER @ MIT-OZ>
To:   Alan Bawden <ALAN @ MIT-OZ>
Cc:   CUBE-LOVERS @ MIT-MC, HOEY @ CMU-CS-A, Tk.Foner @ MIT-OZ
Reply-to: Foner at MIT-MC
Subject: The shortest sequence of moves.
In-reply-to: The message of 30 Jan 1983 00:40 EST from Alan Bawden <ALAN>

Ugh.  I see the problem is far more intractable than I had first
assumed.  I suppose it'll be quite some time indeed before we see any
sort of complete optimality solution if the maximum number of moves
from solved is anything larger than the low twenties.  I'll hunt
through the list archives for what's been accomplished on proving what
that number is...  if anyone feels like giving me a brief refresher
instead, I'd appreciate it, though the whole list might not like to
see it all again.

Thanx again everyone.

						<LNF>

Date: 1 February 1983 13:20 EST
From: David C. Plummer <DCP @ MIT-MC>
Subject: The shortest sequence of moves.
To: FONER @ MIT-MC
cc: CUBE-LOVERS @ MIT-MC, ALAN @ MIT-OZ, Tk.Foner @ MIT-OZ,
    HOEY @ CMU-CS-A

What you really want is some deep mathematical insight.  You
know, the kind Gauss and Fermat had...

Date: Tue 15 Feb 83 15:04:02-PST
From: ISAACS@SRI-KL.ARPA
Subject: "6-T" pretty pattern
To: cube-lovers@MIT-MC.ARPA

  A friend of a friend says he can do the 6-T pattern in 16 moves
(he may call 1/2 turns a move, I'm not sure).  Does anybody know if
this is a minimum?  I'll try to get the sequence.
 -- Stan
-------

Date: Wednesday, 16 February 1983, 09:24-EST
From: Bernard S. Greenberg <BSG at SCRC-TENEX>
Subject: 5 x 5 x 5
To: cube-lovers at MIT-MC

I have seen and touched a 5 x 5 x 5 Rubik cube device.  It was 
allegedly a prototype by Meffert, the people who created the
Megaminx.  It was about the same size as a 4 x 4 x 4, with
slightly smaller cubies.  It had never been devirginized, nor
did its owner want it to be.  The face coverings are of the
same material as the coverings of the Megaminx.  No clue as to
how it operates, although looking at the black surface exposed
by partial twists, there seemed to be new techniques involved, of 
protuberances coming down from cubies sliding behind other
cubies, at least.  Although stated to be a prototype, it seemed
completely developed, and I would expect to see it available 
soon.

Date: 24 Feb 1983 0618-PST
From: JAY <JAY@USC-ECLB>
Subject: NxNxN: Finite N
To: cube-lovers@MIT-MC

I have a program, not finished, which allows manipulation of a NxNxN
(N^3) cube.  Presently it WILL manipulate any NxNxN cube (i tried up
to 23) but it has a VERRY BAD user interface.  I plan to improve the
interface in the near future (this morning?)  I hope to be able to
parse the Cube-Talk mentoined in the first message in the archive
alan;cube mail1 (or 2?), or a super-set of it (like being able to save
cube's on disk, save move seq's (macros), defining macros with
parameters, and make logfiles of especialy hairy session).  As for
output... For small cubes (up to 8^3) i can use the normal air-plane
pattern of display, for medium cubes (8^3 <= size <= 12^3) i can put it
all on the normal terminal, but for larger cubes (24^3 >= cube >=
12^3) all i can do is display a face, or maybe two.

[Implementation]

 I am interested in a, WORKING, N^3 simulator, not speed or space.  As
a result my representation of the cube is loosing on both counts.
(yes it realy is a NxNxN array of cubes (a record of six integers!))
It is written in CLU for TOPS-20 os. 

However my rep. brought up an interesting super-groop, immagin a cube
realy made of N^3 cubies.  Each of these cubies would be a 'Miniature'
N^3 in color scheme.  Now this new device (a 'compleatly colored'
cube) is solved only when ALL cubies are oriented correctly (ie. all
have red up and blue forward), and positioned correctly.  In the 3^3
we would not only have to solve the centres, but also the imaginary
inside cubie (is it ever un-solved?)

[Questions]
  What do you (readers) think is a good display scheme for a N^3,
remember it should be useable on a 24x80 h19?  Is a N^3 simulator even
interesting?  What sort of speed improvement could be gained from a
comp-cube? with or without macros?

  Is the 'compleatly colored' cube interesting?  For what sizes is it
similar to the 'partialy colored' (normal) cube (1^3 and 2^3 for
sure...)?  Could the solution of compleat-5^3 be a solution of the
outer shell, and the inner 3^3?

  Is a simulation of a 'compleatly colored' cube interesting?  How
would one view the manny inside faces?

  What other reps for the cube are there? (other than the obvious two;
an array of color tabs, or a 3d array of cubies..)  Which reps
optimize storage, time, or simplicity to compute a twist, or even ease
of compairison?   Just occured to me that each cubie could be rep'd as
a number twixt 1 and 24 (as a cube has only 24 orientations).  This
optimization would reduce storage by a factor of 6 (not too bad!)

enough.....  enough.....

j'  <jay@eclc>
-------

Date: 24 February 1983 1815-EST (Thursday)
From: Dan Hoey at CMU-CS-A
To: Cube-Lovers at mit-mc
Subject: Whole-Cubing
Message-Id: <24Feb83 181536 DH51@CMU-CS-A>

When we describe cube positions, we typically fix the position of
the face centers.  This avoids counting different positions of the same
pattern more than once.  But suppose we wished to distinguish between
different positions of a pattern?  How should we form this group G*
of spatially oriented patterns?

A simple way of generating G* is to adjoin generators for C, the
motion group of the cube, to generators for G, the usual Rubik
group.  Generators for C were discussed in early Cube-Lovers mail
as I, J, and K, three orthogonal quarter-twists of the whole cube
in space, although there was some disagreement about which was
which.  We generate G with B, F, U, D, L, and R as usual.
Unfortunately, the two kinds of generators are not conjugates,
which was a nice thing about our generators for G.  Also the
generators do not interact very strongly.  If we have an identity
on {IJKBFUDLR}, the substring on {IJK} must be an identity in C,
and there is a simple way of transforming the substring on {BFUDLR}
to be an identity in G.

In @i<Winning Ways>, Berlekamp, Conway, and Guy described another
way of generating G*, by appending the slice moves, which they name
by mnemonic greek letters; this labeling scheme was also reported
in Hofstadter's column.  Here we have some stronger interaction
between the two kinds of generators, but they are still two
different kinds:  they are not conjugate.  This has never really
bothered the English, though; they gratuitously include half-twists
as generators as well.

I thought up a scheme involving what I will call depth-2 moves
named B2, F2, U2, D2, L2, and R2.  Readers of my reports on the
4^3 and 5^3 cubes will find these familiar.  Essentially, a depth-2
move is performed by turning a face together with its adjacent
center slice.  Thus F2 involves holding the B face immobile and
turning the rest of the cube a quarter-turn clockwise, as seen
from the front.

Clearly the depth-2 moves are M-conjugate.  Unfortunately, they do
not generate all of G*, nor even all of G.  This can be seen easily
by noticing that each depth-2 move is an even permutation of the
edge cubies, while G includes odd permutations of edges.  We can
look at this a different way by observing that a depth-2 move is
the same as a vanilla depth-1 move of the opposite face together
with a whole-cube quarter-twist.  Thus we can perform any depth-1
process using depth-2 moves, and observe the spatial orientation of
the cube at the end.  If we performed an odd number of depth-2
moves, then there were an odd number of whole-cube moves, so the
cube cannot be in its home orientation at the end.

So just what group do the depth-2 moves generate?  It turns out
that they generate precisely half of G*, the half that contains
even edge permutations.  Suppose we want to produce such a position
in G*.  First operate in G, simulating depth-1 moves as described
above, to produce a position that is an even number of whole-cube
moves from the desired one.  Then use M-conjugates of the process
	F2 L2 D2' B2' D2' F2 L2 B2 L2 U2' F2' D2' R2' B2,
which performs a whole-cube third-twist.  [This was derived from
identity I14-4].  Since all the even whole-cube moves are generated
from third-twists, this is all you need.

What can we do about generating all of G* with a set of conjugate
generators?  Sad to say, that is impossible.  For supposing we had
such generators, their cycle structures would have to be the same;
in particular, they would have to have the same permutation
parities.  Applying an odd number of such generators would yield
those parities, and applying an even number would yield even parity
on every kind of cubie.  But G* includes four different parity
classes:
		Face centers	Edges	Corners
		    even	even	even
		    even	odd	odd
		    odd		even	odd
		    odd		odd	even
so at most half of G* can be generated with a set of conjugate
generators.

Date: 24 May 1983 2218-EDT
From: James.Saxe@CMU-CS-A (C410JS30)
To: Cube-Lovers@MIT-MC
Subject: Cubists do it . . .
Message-Id: <24May83.221845.JS30@CMU-CS-A>

. . . in over three billion positions.

Date: 25 May 1983 1001-PDT
Sender: OLE at SRI-CSL
Subject: Anybody there?
From: Ole at SRI-CSL (Ole J. Jacobsen)
To: Cube-Lovers at MIT-MC
Cc: Alan at MIT-MC
Message-ID: <[SRI-CSL]25-May-83 10:01:13.OLE>

Several months have gone by without a word from this list.
Have we lost our interest (or have I been removed from the
list)!??  What about the 5x5x5 that somebody reported having
actually touched, anyone seen any on sale?
 
Regards,
              |-+-+-|
              |O+L+E|
              |O+L+E|
              |O+L+E|
              |-+-+-|

Date: 25 May 1983 20:17 EDT
From: David C. Plummer <DCP @ MIT-MC>
Subject: Anybody there?
To: Ole @ SRI-CSL
cc: CUBE-LOVERS @ MIT-MC, ALAN @ MIT-MC

    Date: 25 May 1983 1001-PDT
    From: Ole at SRI-CSL (Ole J. Jacobsen)

    Several months have gone by without a word from this list.
    Have we lost our interest (or have I been removed from the
    list)!??  
I don't know that much of interest has happened.  Maybe I'll find
a little time some year and try a couple things. 
    What about the 5x5x5 that somebody reported having
    actually touched, anyone seen any on sale?
It will never be sold in mass.  MANY stores are stuck with 4x4x4s
that nobody will buy.

Date: Thu 26 May 83 09:27:30-PDT
From: ISAACS@SRI-KL.ARPA
Subject: anybody there; 5x5x5; circular
To: cube-lovers@MIT-MC.ARPA

  There doesn't seem to be much happening with the cube, but the (at least)
3 newletters are still being published. Ideals "Rubik's Cube Newsletter" no.
3 came, marked winter, 1983.  Mostly about contests, also an article on 
computer cubing and the calendar cube.  Rubik's magazine from Hungary is 
also on issue 3, with article on contests, other games and toys, and a little
bit of mathematics.  Singmasters "Cubic Circular" is on issue 5/6 (phisically,
the forth issur), and is the best of the lot mathematically.  He describes
and talks some about rotating rings, about Tsukuda's square (and how to 
solve it), generalized Hungarian rings, the Incredi-ball (equivalent
to the corners of a Magic Dodecahedron without centers?? anyway, about
2*10**25 positions), rotating rings of cubes and tetrahedra (STarburst, 
Kaleido Puzzle, etc), some notes on the 4x4x4 (his suggested order:  2
opposite face centers, say L and R; Pair up and place the F,C,B edge
pairs on the L & R faces; D corners; Pair up FD & store them, then pair BD
edges, then put these edge pairs in place; check parity of U corners - if
odd, apply U; 3-cycle U edges to get UR & UL edge pairs;  EXAMINE, carefully,
the parity of the 4 edge pieces at UF&UB - if odd, r2D2l'D2r2, which 4 cycles
these edges; get all edges together and place; do U corners; do centers with
3-cycles like [[r,b],U].  phew.  If people are interested, he gives 
slightly more information on his method, also "Useful moves", and some other
strategies taken from the literature.  He talks about "Evisceration", 
that is, the principle of Duality, that if you replace capitals with small
letters and vice versa, in any move sequence, you turn the cube inside out -
corners become invisable centers, edges become face centers, etc.  Singmaster
uses this to transfer knowledge of 3x3x3 to 4x4x4.  He guesses the minimum
for God's Algorithm on the 4x4x4 to be about 44 "hand moves" or 66
quarter turns.  (question, from DS: how should we count moves?  "r" is
really 2 hand moves - Rr followed by R' - while Rr and even Rrl' could be
counted as a single move.)  The circular talkes about Pyraminx, and gives
God's Algorithm as between 8 and 17 moves; there was some work on that done
in the Journal of Recreational Mathematics (a minimum solution was
presented), but I can't find the issue right off hand.  Also a bunch of
Pyraminx patterns are given.  There is an article on the prehistory of
the cube,and a little on the legal problems.  Anyway, this parenthetical
paragraph is way out of hand - if anybody wants more, or more organized,
on the Circular, let me know.)
    The 5x5x5:  Yes, I saw one and played with it (a very little bit) at
Jerry Slocums' puzzle party.  I also got an add from Meffert for one, and
have sent him my money, but it hasn't come yet.  Neither has my Skewb, for
which I sent money half a year ago or so.
    What else is new?  Well, nothing much.  Sales in the stores are slow,
and the number of puzzles available has lessened.  Doug Engels rotating
circles are ready for production, and his manufacturer is mainly waiting
for the "right time" to release them.  (These are 2 overlapping rotating
circles, with 3 piece overlap and 21 pieces all together.)  He is trying
to decide between 2 colorings; if he choses the more difficult one, it 
is a fairly hard puzzle, very cube like, but harder to get the already
solved pieces "out of the way".  
   Sorry this turned out so long and rambling - I was just going to 
write a short note, and got carried away.  Happy puzzling.
   -- Stan
-------

Date: 26 May 1983 23:44 EDT
From: Alan Bawden <ALAN @ MIT-MC>
Subject:  [DCP: 5x5 cubes]
To: CUBE-LOVERS @ MIT-MC

Date: Thursday, 26 May 1983  13:20-EDT
From: DCP at SCRC-TENEX
To:   Bernard S. Greenberg <BSG at SCRC-TENEX>
cc:   ALAN, cmb at SCRC-TENEX
Re:   5x5 cubes

    Date: Thursday, 26 May 1983, 10:02-EDT
    From: Bernard S. Greenberg <BSG@SCRC-TENEX>

        From: David C. Plummer <DCP @ MIT-MC>
    	What about the 5x5x5 that somebody reported having
    	actually touched, anyone seen any on sale?
        It will never be sold in mass.  
    What, pray tell, are the details and source of this information?
    As if there were anything I could do with it other than stare at it...
A person at Games People Play believes Ideal will never produce a
5x5x5 commercially.  He believes the only people that seriously
bouhght the 4x4x4 were those that solved the 3x3x3 without a solution
book.  He finally said he is stuck with many 4x4x4 that he fears he
won't be able to get rid of.  The economics of making a 5x5x5 just
don't sound very good.

Date: Tue 31 May 83 09:52:44-PDT
From: ISAACS@SRI-KL.ARPA
Subject: economics
To: cube-lovers@MIT-MC.ARPA

     Meffert claimed in an early letter, that he could produce his strange
products (Skewb, 5x5x5, and many others) if he had some not-to-big number
of guaranteed sales, on the order, if I remember, of 1000 or so.  He has
a "club", that is, we paid about $10 to him up front.  Then, when he is
"ready" with something, he sends an order form; they cost about $20 or so.
So far, he has done this with the Skewb, about 5 months ago, and the 5x5x5
recently.  However, we have not actually received either of these yet.
I talked to about 3 other people who ordered them and have not received 
them.  So we don't actually know whether this system will work.  We know
he has produced real, working prototypes, though, since we have seen them.
    If they ever do come, I'll put up a message here.  (Assuming I still
have access to this list.)
  -- Stan
-------

Date:  1 June 1983 1939-EDT (Wednesday)
From: Dan Hoey@CMU-CS-A
To: cube-lovers@MIT-MC
Subject: Eccentric Slabism, Qubic, and S&LM
Message-Id: <01Jun83.193917.DH51@CMU-CS-A>

I don't know whether Isaacs or Singmaster know just what a
bombshell was contained in the Cubic Circular.  I am somewhat
frightened at the possibilities.

Section 1 discusses the history of metrics for N^3 puzzles and
proposes a new one.  Section 2 describes new symmetries of the
generators of the 4^3 puzzle.  Section 3 outlines a theory of
symmetry and local maxima for the 4^3 puzzle.  Section 4 indicates
directions for further work.


1. Cutism, Slabism, and Eccentric Slabism
=========================================

Let's start with the question that was posed by Stan Isaacs in his
message of 26 May:  what is a quarter-turn on the 4^3 or larger
cube?  On the N^3 puzzle, each set of N-1 parallel ``cuts'' divides
the cube into N ``slabs''.  There seem to be two straightforward
metrics.  The Slabist defines a move to be a turn of one slab with
respect to the rest of the cube.  The Cutist defines a move to be a
turn of a connected part of the cube with respect to another
connected part, the two parts being separated by a cut.  [In the
terminology I used on 2 September 1982, the Slabist counts ``slice
moves'' while the Cutist counts ``twist moves.'']

I have heretofore espoused Cutist theory.  For one thing, it agrees
with our current theories on the 3^3 in disallowing a turn of a
center slice as a single move.  This seems to be a good idea, since
the current quarter-turn theory has the advantage of conjugate
generators, which it would lack if we allowed center-slice moves.
[This is presumably not a problem for Singmaster, who allows the
squared moves, which are not conjugate.]  Another reason for Cutism
is that it makes it easier to equate positions that arise from a
whole-cube move of the N^3.  A third reason is that it makes the
parity hack (see my message of 1 June 1982) easier.  The last two
reasons are for convenience only; the arguments can still be made
in a Slabist formulation.

But as I admitted, I solve the cube as a Slabist.  Slabs are
probably convenient because they minimize the degree of each
generator.  I casually dismissed this tawdry practicality until I
was struck by Evisceration.

In the course of my examination of Evisceration I have experienced
an epiphany which converted me to Eccentric Slabism.  I now define
a move to be a turn of any slab except one whose interior contains
the center of the cube.  In other words, any slab except a center
slice.

At first glance, Eccentric Slabism looks like a hack, since there
is an excluded slab only in the case of a puzzle of odd size.  I
believe that the truth is more complicated, but the explanation is
partly beyond the scope of this note and partly beyond my
knowledge.  If you really want an answer I suggest you study
tic-tac-toe.


2. Evisceration, Inflection, and Exflection
===========================================

The (Eccentric) Slabist moves of the 4^3 puzzle form the 24-element
set Q4={B,B',b,b',...,r'}, where upper-case refers to turning a
side (an outslab move) and lower-case refers to turning the
adjacent internal slab (an inslab move).  We consider these moves
as generators of G4, the ``Theoretical Invisible Group'' [Invisible
Revenge, 9 August 1982] in which the inslabs turn the eight stomach
cubies like a 2^3 puzzle.  Thus two positions in G4 are equal if
and only if all sixty-four pieces of the cube are in their home
position and orientation.  [Actually, this is not quite the
Theoretical Invisible Group, since we do not equate positions that
differ by a whole-cube move.  I feel confident that the
identification can be performed, but I will speak of the
unidentified group here.]

Consider the following permutations on Q4:

Rotations:
  I=(FRBL)(F'R'B'L')(frbl)(f'r'b'l'),
  J=(FUBD)(F'U'B'D')(fubd)(f'u'b'd'),

Reflection:
  R=(FB')(F'B)(RL')(R'L)(UD')(U'D)(fb')(f'b)(rl')(r'l)(ud')(u'd),

Evisceration:
  V=(Ff)(F'f')(Bb)(B'b')(Rr)(R'r')(Ll)(L'l')(Uu)(U'u')(Dd)(D'd'),

Inflection:
  N=(fb')(f'b)(rl')(r'l)(ud')(u'd),

Exflection:
  X=(FB')(F'B)(RL')(R'L)(UD')(U'D).

Permutations I, J, and R are familiar generators of M, the group of
rotations and reflections of the cube.  Singmaster introduced
Evisceration, which consists of swapping each outslab with the
adjacent inslab.  I extend the list with Inflection and Exflection.
Inflection consists in swapping each inslab with the inverse of its
parallel inslab;  Exflection swaps each outslab with the inverse of
its parallel outslab.

It is well known that M is a group of automorphisms on G4.
Singmaster observed that Evisceration is also an automorphism.  I
observe that Inflection and Exflection are automorphisms, too.
Thus M4, the 192-element group generated by <I,J,R,V,N,X> is a
group of automorphisms on G4.  [Actually, since R=NX and X=VNV,
M4=<I,J,V,N>.  The group M4 is also the automorphism group of the
game of Qubic, or 4^3 tic-tac-toe.]

I began to doubt Cutism when I noticed that M4 sometimes maps cut
moves to pairs of cut moves.  I went home last night wondering why
this might be so.  I nearly got to sleep before I realized the big
news:  M4 is Q4-transitive!  Eccentric slabs are conjugate!


3. Symmetry and Local Maxima
============================

This section relies especially heavily on ``Symmetry and Local
Maxima'' [14 December 1980; available as file "MC:ALAN;CUBE S&LM"
on MIT-MC].

Following the argument in S&LM, consider the symmetry group of the
Pons Asinorum (with the face-centers each half-twisted, as is
customary).  We already know Pons is M-symmetric; by examination,
the symmetry group of Pons also contains Evisceration and
Inflection.  Thus Pons is M4-symmetric.  The result is that Pons is
a local maximum in G4.  This is the first local maximum to be found
in a close relative of Rubik's Revenge.

It is not hard to show that we can dispense with fixing the Pons in
space, and it is only slightly harder to carry out in general.
Unfortunately, I see no way of showing that Pons is a local maximum
if we ignore the stomach cubies without ignoring the corners.


4. Open problems
================

This is a pretty random collection of directions for further work.
Some of these were posed in the text.  The ones I think likely to be
impossible are labeled (*).

Conjecture:  The automorphism group of the Eccentric Slabs of the
N^3 puzzle is the same as the automorphism group of N^3
tic-tac-toe.  I don't believe this has been rigorously done for
any N>1.

Stronger conjecture:  The automorphism groups of the N^D puzzle and
N^D tic-tac-toe are the same.  (Hint: There are at least two
definitions of the N^D puzzle.  I think both work.)

Extension:  The relation between the automorphism groups is too
amazing to be accidental.  What is really going on here?

Search:  There is published literature on tic-tac-toe
automorphisms; in particular the group of automorphisms of N^D
tic-tac-toe is well known.  I seem to recall seeing some horribly
theoretical work, approaching the problem from the standpoint of
algebraic geometry or some such.  At that time I settled for
scanning the results.  Now I have questions that need a general
treatment.  If the world's leading expert on Qubic has his
bibliography on line, I think there's a reference I'd appreciate.
Actually, I'll take references from anybody and send the
compilation to any requestors.

Query:  Why must slabism be eccentric?

Query:  Can Cutism be saved?  Are cut moves conjugate in some sense?

Easy extension: Equate positions that differ by whole-cube moves.

Hard extensions (*): Equate positions that differ only internally.
Equate positions that differ only in the permutation of
like-colored face cubies.

Problem:  Prove that the Pons requires 12 quarter-turns in the 4^3
puzzle.  Ditto for 12 qtw in the N^3 puzzle(*?).  Prove or disprove
for 4D qtw in the N^D puzzle (*).

Problem:  Find the Q4-transitive subgroups of M4, then list all the
Q4-symmetric local maxima in the 4^3.

Problem:  Describe all symmetric local maxima of the N^3(*), or
place useful conditions on them.

Problem:  Demonstrate an infinite class of local maxima (Ponses?).

Final query:  Did someone ask if Cubism was dead?

	Dan Hoey
	HOEY@CMUA.ARPA


Date: Wed 22 Jun 83 09:54:45-PDT
From: ISAACS@SRI-KL.ARPA
Subject: 5x5x5 and "bricks"
To: cube-lovers@MIT-MC.ARPA

    I got my 5x5x5 today, from Meffert in Hong Kong.  Whether they will 
appear in regular (puzzle) stores, I don't know, but I would assume they will
be available from him.  His address is 
         Puzzles Club
         P.O.Box 31008
         Causeway Bay
         Hong Kong.
    I haven't had a chance to do much with it yet; it seems well and strongly
made.  I have not taken it apart because I don't know how to do it safely.
The corners seem to have the usual flanges.  The center edges seem to have
a very deep tab, perhaps to go completely through the second layer.  The
non-center middle pieces seem to be some kind of "caps" over the edge
extensions.  That's all I can tell from "peeking" in the cracks.
    One of the suggestions in Singmaster's "Cubic Circular" was to take
a regular 3x3x3 and tape over some cracks to limit the movement.  The idea
is to take each pair of cubies and tape them together into one "brick", 2x1x1.
If you do it right, it makes a very interesting variation.  For taping, I used
labels from another cube; that way the colors can be make to look right. 
The pattern I used is as follows:
   Choose a corner.  It will be the only single cubie (except for a center).
Make the other 2 cubies along each edge projecting from the chosen corner,
into one brick (by putting tape on both cracks separating the cubies).
Now, in each of the 3 faces bordering the chosen corner are 4 untaped
facies in a square.  Tape to make 2 adjacent bricks (don't forget the
other side of the edge bricks).  Note that when you have done the first
face, you will have 3 parallel bricks; the other 2 faces should have
their parallel bricks with the "same handedness" - If you look down from
the chosen corner, the 3 faces should have rotational symmetry. (I haven't
tried an irregular pattern yet; it might have different properties than
this one).  Anyway, all that is left is a 2x2x2 sub-cube, diagonally
across from the original chosen corner.  Picture that as 4 1x1x2 bricks
piled with one pair crossways from the other.  One brick, of course, includes
the very center of the cube (invisible), and will leave the only other untaped
facie - namely one center.  Which one is the center doesn't matter, but I
think there is a clockwise/counter-clockwise (relative to the outer handedness)
choice here.  
    When all that is done, you will have a very interesting new puzzle. Any
twist must include the chosen corner!  Home position is easy to recognize -
it is when the chosen corner is back in its original place, and the bricks
around it are in their positions (not necessarily with the correct colors
together).  I have not solved it yet, but it seems that the sub-cube is
both hard to mix up, and easy to fix once you have.  It looks as if all the
interesting moves can be done with just the 3 faces surrounding the chosen
corner.
   Have fun..
     -- Stan
-------

Date:     23 Jun 83 12:20:43-EDT (Thu)
From: Stuart Stirling <silver.emory@Rand-Relay>
Return-Path: <Silver%emory.Emory@Rand-Relay>
Subject:  Singmaster's "Cubic Circular"
To: cube-lovers@Mc
Cc: silver.Emory@Rand-Relay
Via:  Emory; 23 Jun 83 23:31-PDT

Where is this available?

	Stuart Stirling			{sb1,sb6,msdc,gatech}!emory!silver
	Math and Computer Science	silver.emory@rand-relay
	Emory University
	Atlanta, GA 30322 		404/329-5637

Date: 27 June 1983 19:04 EDT
From: Alan Bawden <ALAN @ MIT-MC>
Subject:  Bizarre Bricks
To: CUBE-LOVERS @ MIT-MC

A bizarre variant of the "bricks" described by Isaacs last week is
currently sitting on the front desk of the SIPB office at MIT.  Imagine
that you took three cubies along any edge of a cube and welded them
together (perhaps by taping over the cracks as Singmaster suggests).  Now
imagine that you perform the same mutilation on a second cube.  Now imagine
that you arrange to \share/ the same 1 x 1 x 3 cubie between the two cubes.
The resulting arrangement looks like this from above:

XXX
XXX
XXXXX
  XXX
  XXX

A moment's thought should convince you that sharing the 1 x 1 x 3 cubie has
done \nothing/ to increase the difficulty of the puzzle.  Amazing what
lengths puzzle manufacturers will go to to try and squeeze a few more bucks
out of the Cube Craze.

Date: Tue 28 Jun 83 09:04:29-PDT
From: ISAACS@SRI-KL.ARPA
Subject: Cubic Circular address
To: silver.emory@RAND-RELAY.ARPA
cc: cube-lovers@MIT-MC.ARPA

   The address for the "Cubic Circular" is indeed that of David Singmaster:
     David Singmaster
     87 Rodenhurst Road
     London, SW4 8AF  ENGLAND.
   Hope this helps.
     -- Stan
-------

Date: Thu 30 Jun 83 10:07:06-PDT
From: ISAACS@SRI-KL.ARPA
Subject: double cube
To: alan@MIT-MC.ARPA
cc: cube-lovers@MIT-MC.ARPA

  I've heard of the double cube - do you have any idea where to buy one?
Another strange cube I've seen, is a normal 3x3x3 with extra cubies glued on
three faces so that it looks like a 4x4x4; of course, it twists in a very
lopsided manner.  The main problem in making one (which I have not solved)
is how to grind or cut the flanges off cubies before gluing.  Without
equipment, can anyone think of a good, practical way to do this?
 -- Stan
-------

Date: 30 June 1983 20:47 EDT
From: Alan Bawden <ALAN @ MIT-MC>
Subject:  [Barry Margolin: double vision]
To: CUBE-LOVERS @ MIT-MC

Date: 30 June 1983 16:57 edt
From: Barry Margolin at MIT-MULTICS
To:   Alan Bawden <ALAN>
Re:   double vision

Well, I got it at the Harrod's department store in London (they say you
can buy ANYTHING at Harrod's).  I was in Games People Play last week and
didn't see it, so maybe it hasn't gotten to this country yet.

Sorry I can't be of more help.
                                        barmar

Date: 21 July 1983 1450-PDT (Thursday)
From: cline at AEROSPACE (Ken Cline)
Subject: put me on your mailing list!
To: cube-lovers at mit-mc

  Please include me (CLINE@AEROSPACE) in your Rubik's Cube mailing-list.
My address will change in September to SCGVAXD!4CCVAX!KCLINE@CIT-VAX,
but I will let you know when this occurs.

			Thank you,
			Ken Cline


Date: Monday, 21 November 1983  10:08-EST
To:   cube-lovers at mc
From: BSG
Subject: Meffert cube availability

Someone asked me from afar where to get the hairy Meffert
cube-like things.  It was a long time ago that I ordered a
five-cube from somewhere in Los Angeles.  Would somebody care to
repeat the name and address of that store, or some other stores?
Thank you.

Date: 21 November 1983 20:30 EST
From: Alan Bawden <ALAN @ MIT-MC>
Subject:  Rubik's cube cornered
To: CUBE-LOVERS @ MIT-MC

>From mit-eddie!genrad!decvax!dartvax!andyb Tue Oct 25 00:03:36 1983
Relay-Version: version B 2.10.1 6/24/83; site mit-vax.UUCP
Path: mit-vax!mit-eddie!genrad!decvax!dartvax!andyb
From: andyb@dartvax.UUCP
Newsgroups: net.puzzle
Subject: Rubik's cube cornered
Message-ID: <307@dartvax.UUCP>
Date: Tue, 25-Oct-83 00:03:36 EDT
Article-I.D.: dartvax.307
Posted: Tue Oct 25 00:03:36 1983
Date-Received: Tue, 25-Oct-83 08:53:25 EDT
Lines: 7

It is possible to solve the corners of the Rubik's Cube in no more
than 11 turns.  Over 4/5 of all possible patterns can be solved in
9 turns or fewer, but over half take a full 9 turns.  Ten turns are
enough to solve more than 99.9% of all scrambled states.  (Only
2,644 states require 11 turns).
 
This from L. John Kelley, solver of the Magic Pyramid.

Date: 23 November 1983 17:01 EST
From: Alan Bawden <ALAN @ MIT-MC>
Subject:  [genrad!decvax!dartvax!andyb: Rubik's cube cornered]
To: CUBE-LOVERS @ MIT-MC

Date: 23 Nov 1983 06:56:09-EST
From: genrad!decvax!dartvax!andyb at mit-eddie,
Received: by decvax.UUCP (4.12/4.2)
	id AA26306; Wed, 23 Nov 83 03:10:45 est
Date: Wed, 23 Nov 83 03:10:45 est
From: decvax!dartvax!andyb
Message-Id: <8311230810.AA26306@decvax.UUCP>
To: decvax!genrad!mit-eddie!ALAN@MIT-MC, genrad!decvax!dartvax!andyb@MIT-EDDIE
Subject: Rubik's cube cornered
Cc: ALAN@MIT-MC

Alan:

Feel free to rebroadcast any of this discussion to the cube-lovers list.
(Could you enroll me on the list, also?)  

Both solutions were generated by "brute force", using a fairly short PL1
program running on a VAX.  The solutions are tables which express the
information "if the pyramid/cube's state is X, the best move is Y" for all
possible states.  In that sense, they are optimal solutions, but if you wanted
to know the best solution for any particular state, you would have to carry the
entire table around with you.  Perhaps we could publish the list :-)   
 
The Magic Pyramid is the one sold under the name "Pyraminx" -- it's the 
tetrahedron that rotates in only 4 planes.
 
The cube solution is for corners only, with no reference to centers or edges.
You could think of it as a complete solution to the "Pocket Cube", which is
a 2x2x2 cube.
 
Glad to help.
                             Andy Behrens (& L. Kelley)
                             decvax!dartvax!andyb

Date: Sat 26 Nov 83 22:35:59-EST
From: Tai <Jin@COLUMBIA-20.ARPA>
Subject: cube solvers
To: cube-lovers@MIT-MC.ARPA
cc: Jin@COLUMBIA-20.ARPA


hi, are there any programs out there that solve the cube?  i would like
to get a copy of the source if possible.  also, are there any general
methods for solving the cube?

thanks...tai
-------

Date: 29 February 1984 23:08-EST
From: Alan Bawden <ALAN @ MIT-MC>
Subject:  [CL.BOYER: A Programming Language for Group Theory (Dept. of Math)]
To: CUBE-LOVERS @ MIT-MC

Some of the group theory hackers amoung us might be interested in this
abstract.  Of course the talk has already happened, but perhaps one of you
knows more about this?

Date: Sun 26 Feb 84 17:06:23-CST
From: Bob Boyer <CL.BOYER at UTEXAS-20.ARPA>
To:   AIList
Re:   A Programming Language for Group Theory (Dept. of Math)

        [Forwarded from the UTexas-20 bboard by Laws@SRI-AI.]

            DEPARTMENT OF MATHEMATICS COLLOQUIUM
          A Programming Language for Group Theory
                        John Cannon
        University of Sydney and Rutgers University
                 Monday, February 27, 4pm

     The past 25 years has seen the emergence of a small but vigorous branch of
group theory which is concerned with the discovery and implementation of
algorithms for computing structural information about both finite and infinite
groups.  These techniques have now reached the stage where they are finding
increasing use both in group theory research and in its applications.  In order
to make these techniques more generally available, I have undertaken the
development of what in effect is an expert system for group theory.

     Major components of the system include a high-level user language (having
a Pascal-like syntax) and an extensive library of group theory algorithms.  The
system breaks new ground in that it permits efficient computation with a range
of different types of algebraic structures, sets, sequences, and mappings.
Although the system has only recently been released, already it has been
applied to problems in topology, algebraic number theory, geometry, graphs
theory, mathematical crystalography, solid state physics, numerical analysis
and computational complexity as well as to problems in group theory itself.

Received: From ti-csl.csnet by csnet-relay;  25 Jun 84 20:44 EDT
Date:     Mon, 25 Jun 84 14:30 CST
From:     Gil Gamesh <dnichols%ti-csl.csnet@csnet-relay.arpa>
To:       cube-lovers@mit-mc.arpa
Subject:  Recent Activity?

Is this mailing list still alive? Does anyone know of any
recent activity in the cube world? Are there still any
Cube Newsletters that anyone knows of? Has anyone ever seen
any of the many puzzles I read about in places such as 'Scientific American' and 'Omni', ex.
the Skewb, the Double Pyramid, the 5x5x5 cube, etc.?

Dan Nichols
Dnichols.ti-csl@csnet-relay

Date: Tue 26 Jun 84 11:00:18-PDT
From: ISAACS@SRI-KL.ARPA
Subject: cubes dead?
To: cube-lovers@MIT-MC.ARPA

I don't see much about cubes these days.  And, as a puzzle collector, I'm still
looking.  Singmaster and Ideal have both discontinued their newsletters; 
however, the Hungarian "Rubics Magazine" still is being published.  It has
broadened somewhat to talk more about puzzles in general.  I also found last
Saturday, that there is a childrens cartoon about Rubics Cube - in fact
the cube is the hero.  It is truly awful.
   As Alan says, most of the objects mentionsed have been seen.  But they
are hard to get nowadays.  Meffert, in Hong Kong, is silent.  I ordered
a dozen or so 5x5x5's more than 6 months ago, and have yet to hear anything.
Or get my money back.  And he never sent me any skewbs.  
  Well, it was an exciting couple of years in the puzzle world; now, I guess,
things are more normal.  Stores no longer have puzzle sections, and I have to
write to specialty places to find new puzzles.  For me, wood puzzles are back
in.  I've been digging up old puzzle patents, and hope to find someone to make
some of the puzzles described for me.
  -- Stan Isaacs
    (isaacs@hplabs)
-------

Received: from SCRC-CONCORD by SCRC-STONY-BROOK via CHAOS with CHAOS-MAIL id 54973; Tue 26-Jun-84 10:22:30-EDT
Date: Tue, 26 Jun 84 10:20 EDT
From: "Bernard S. Greenberg" <BSG@SCRC-QUABBIN.ARPA>
Subject: Recent Activity?
To: ALAN@MIT-MC.ARPA
cc: cube-lovers@MIT-MC.ARPA
In-Reply-To: The message of 25 Jun 84 23:23-EDT from Alan Bawden <ALAN at MIT-MC>
Message-ID: <840626102050.3.BSG@CONCORD.SCRC.Symbolics>

    Date: 25 June 1984 23:23-EDT
    From: Alan Bawden <ALAN @ MIT-MC>

    Date:     Mon, 25 Jun 84 14:30 CST
    From:     Gil Gamesh <dnichols%ti-csl.csnet@csnet-relay.arpa>
    Subject:  Recent Activity?

    Is this mailing list still alive? Does anyone know of any recent activity
    in the cube world?
More significantly, has anyone heard of any progress in the
important theoretical problems (God's number/algorithm)?

Date: 28 June 1984 00:17-EDT
From: Alan Bawden <ALAN @ MIT-MC>
Subject:  The Cube meets Massive Parallelism
To: CUBE-LOVERS @ MIT-MC
cc: CM-I @ MIT-MC
In-reply-to: Msg of Tue 26 Jun 84 10:20 EDT from Bernard S. Greenberg <BSG at SCRC-QUABBIN.ARPA>

Since I spend most of my time these days thinking about designing and
programming massively parallel computers, it occured to me to think about
applying such machines to exploring the Cube's Group.  Here are some
preliminary thoughts.

When we say "massively parallel" we are talking at least a quarter million
simple processors.  This is enough processors to give all of the positions
5 or fewer quarter twists away from home their own processor.  A million
processors would be enough to get up to 6 Qs, but lets not push our luck.
Given a machine like the MIT Connection Machine we could set up a database
in which every processor representing a configuration knew the addresses ot
the 12 other processors representing its 12 closest neighbors, in almost no
time at all.  (Processors 5 Qs away from home would have null pointers for
their unrepresented 6 Q neighbors.)  A conservative statement would be
that operations like generating a list of all identities of length 10 or
less (which has previously taken us hours to accomplish) could be done so
fast that the machine could generate output faster than you could read it.

Since this is all so absurdly easy, there must be ways to go beyond this to
generate significant new results using this (promised) new kind of
hardware.  Perhaps Dave Christman, who is both a cube hacker, and a
designer of algorithms for massively parallel machines, could be persuaded
to devote some spare cycles to figuring out ways to brute-force the Cube
using such a machine.

Date: Fri 13 Jul 84 15:24:23-PDT
From: Micheal Hewett <HEWETT@SU-SCORE.ARPA>
Subject: Please add me to your mailing list
To: cube-lovers@MIT-MC.ARPA

Thanks.

Mike Hewett    (HEWETT@SU-SCORE.ARPA)
-------

Date: Tue 24 Jul 84 15:30:18-PDT
From: Haym Hirsh <HAYM@SU-SCORE.ARPA>
Subject: God's number
To: cube-lovers@MIT-MC.ARPA

Can anyone verify a rumor - that some Princeton student, as an
undergraduate thesis, solved the problem of how far from start one
can get (i.e., the longest sequence God's algorithm would give for
any position)?  The number I heard was 26.

Haym
-------

Date: 1 Aug 1984 11:57-EDT
From: Dan Hoey <hoey@NRL-AIC>
Subject: Pocket cube program
To: cube-lovers at mit-mc

The UNIX-SOURCES@BRL mailing list recently forwarded this note from
Usenet.

    Date:     Tue, 31 Jul 84 13:47:22 EDT
    From:     news@BRL-TGR.ARPA
    Subject:  /usr/spool/news/net/sources/397

    Path: brl-tgr!seismo!hao!hplabs!zehntel!ihnp4!ihnet!eklhad
    From: eklhad@ihnet.UUCP (K. A. Dahlke)
    Subject: solve the 2x2x2 Rubix cube in a minimum number of moves
    Date: Mon, 30-Jul-84 10:49:48 EDT
    Article-I.D.: ihnet.142
    Organization: AT&T Bell Labs, Naperville, IL

    After solving the Rubix cube 4 years ago, I turned my attention to
    more interesting (and more difficult) questions.
    How can one find the minimum path solution for an arbitrary position?
    How far away is the farthest positions?
    Is there one position diametrically opposed to start,
    or does it fan out into billions?
    Recently, I have started playing again, and have made some progress.
    Here is a computer program (C/unix) which solves the 2x2x2 cube
    in a minimum number of moves.
    The 2 cube is not as common as the 3 cube, but it is commercially
    available.
    If you only have a 3 cube (standard), just ignore the sides and centers,
    and use the corners.
    This effectively simulates a 2 cube.
    Thanks to ATT-BL for the use of their computing facilities.
    Later versions may come, if i am ambitious.
    Unfortunately, my program cannot be expanded to handle the 3 cube.
    Nobody has that much memory/CPU time.
    I will have to come up with something better.
    Feel free to contact me with any ideas about this subject.

----------------------------------------------------------------

The note is followed by about 1000 lines of c code that I can make
available if you want it.

Unfortunately, the program seems to believe that there are 870 * 729 =
634230 positions of the 2^3, while assiduous cube lovers realize the
number is actually (7! / 2) * 3^6 = 2520 * 729 = 1837080 positions.
The number 870 = 29 * 30 is strange.  I guess it is an approximation of
6! + 5! + 4! + 3! + 2! = 872, since the code for encoding a position as
an integer contains a table of those factorials.

Dan

Date: 7 Aug 1984 19:24-EDT
From: Dan Hoey <hoey@NRL-AIC>
Subject: The pocket cube: correction, calculation, and conjectures
To: cube-lovers at MIT-MC

Well, maybe this list is dead after all, if I can tell you there are
(7!/2)(3^6) positions of the pocket cube, and have it stand for a week.
The correct number is of course (7!)(3^6) = 3674160, since the
generators are odd.

But not being one to eat crow with a straight face, I have hacked the
good hack, so I can give you the exact number P(N) of pocket cube
positions exactly N quarter-turns from solved.  (This was done in
September 1981 for the half-twist metric; see the archives.)  I have
also computed the number L(N) of local maxima at each distance.  These
numbers are given below.

         N     P(N)     L(N)	
         0        1        0
         1        6        0
         2       27        0
         3      120        0
         4      534        0
         5     2256        0
         6     8969        0
         7    33058       16
         8   114149       53
         9   360508      260
        10   930588     1460
        11  1350852    34088
        12   782536   402260
        13    90280    88636
        14      276      276

An approach for dealing with these numbers (suggested to me by Dale
Peterson) is to form the Poincare polynomial

         p(x) = SUM P(i) x^i
                 i

in hopes that it can be factored nicely.  Unfortunately, this doesn't
work out--with the exception of the obvious factor (x+1), p(x) is
irreducible.  I have also tried to decompose p(x) using the power
             (1+x)^2
series for ------------, which agrees with the first five terms of p(x)
           3 - 2(1+x)^2
due to the lack of non-trivial identities.  I haven't found any good
ways of expressing p(x), but there may be something there.  The point
of all of this is that it could conceivably lead to a conjecture--or
even a derivation--of God's number for the 3^3 puzzle.

I might pass along another fuzzy recollection from a year and a half
ago, in hopes that it is more informative than incorrect.  Dale
mentioned another classical method for dealing with group diameters.
It seems there is a class of groups, called reflection groups, for
which tight diameter bounds can be derived.  A reflection group is a
group of matrices with eigenvalues of plus and minus one.  Some
properties generalize to pseudo-reflection groups, where the
eigenvalues all have complex magnitude one.  We managed to construct
isomorphisms between the 3^3 edge group and a reflection group, and
between the corner group and a pseudo-reflection group.  As I recall,
he was fairly certain that the full cube group did not qualify, but
that was beyond my depth.

So if you think cubes are dead, remember it's not because the
results are all in.

Dan

Date: 20 Aug 1984 4:34-EDT
From: Dan Hoey <hoey@NRL-AIC>
Subject: The pocket cube and corners of the full cube
To: cube-lovers at mc
Cc: umcp-cs!seismo!ihnp4!ihnet!eklhad at NRL-AIC

Karl Dahlke, the author of the pocket cube program I mentioned on 1
August, sent me a note about the appearance of the unusual constant 870
in his program.  It turns out that the program is correct, and the
constant arises in an interesting way.

Recall that the pocket cube has 729 orientations and 5040 permutations
of the pieces.  Dahlke had noticed that the ``reflections and
rotations'' of a position need not be stored, since they are the same
distance from start.  By reflections and rotations, he means the
S-conjugates, where S is the six-element symmetry group of the pocket
cube with one corner fixed.

It turns out that the pocket cube has 2 permutations with a six-element
symmetry group, 16 permutations with a three-element symmetry group,
138 permutations with a two-element symmetry group, and 4884
permutations with a one-element symmetry group.  Thus the number of
permutations that are distinct up to S-conjugacy is 2 + 16/2 + 138/3 +
4884/6 = 870.

This discussion of symmetry recalls a question I have meant to propose
to Cube-Lovers for some time:  How many positions are there in Rubik's
Cube?  We know from Ideal that the number is somewhat over three
billion.  Most cube lovers will tell you a number of about 43
quintillion.  But I really don't see why we should count twelve
distinct positions at one quarter-twist from solved--all twelve are
essentially the same position.  So the question, suitably rephrased, is
of the number of positions that are distinct up to conjugacy in M, the
48-element symmetry group of the cube.  I think this is an interesting
question, but I don't see any particularly easy way of answering it.
My best guess is that it involves a case-by-case analysis of the 98
subgroups of M, or at least the 33 conjugacy classes of those
subgroups.  In ``Symmetry and Local Maxima'', Jim Saxe and I examined
five of the classes, which we called M, C, AM, H, and T.

Even finding the numbers for the pocket cube is a little tricky.  If we
limit ourselves to symmetry in S, I believe the pocket cube has 2
positions with a six-element symmetry group, 160 positions with a
three-element symmetry group, 3882 positions with a two-element
symmetry group, and 3670116 positions with a one-element symmetry
group, for 613062 positions distinct up to S-conjugacy.  But the
numbers for M-conjugacy are still elusive; I am not even sure how to
deal with factoring out whole-cube moves in the analysis.  I hope to
find time to write a program for it.

I expanded my pocket cube program to deal with the corner group of
Rubik's cube.  This group is 24 times as large as the group of the
pocket cube, having 3^7 * 8! = 88179840 elements.  The number of
elements P(N) and local maxima L(N) at each (quarter-twist) distance N
from solved are given below.

                 N         P(N)        L(N)
                 0            1           0
                 1           12           0
                 2          114           0
                 3          924           0
                 4         6539           0
                 5        39528           0
                 6       199926         114
                 7       806136         600
                 8      2761740       17916
                 9      8656152       10200
                10     22334112       35040
                11     32420448      818112
                12     18780864     9654240
                13      2166720     2127264
                14         6624        6624

The alert reader will notice that rows 10 through 14 contain values
exactly 24 times as large as those for the pocket cube.  This is not
surprising, given that the groups are identical except for the position
of the entire assembly in space, and each generator of the corner cube
is identical to the inverse of the corresponding generator for the
opposite face except for the whole-cube position.  Thus when solving a
corner-cube position at 10 qtw or more from solved, it can be solved as
a pocket cube, making the choice between opposite faces in such a way
that the whole-cube position comes out right with no extra moves.

Dan

Date: 22 Aug 1984 18:06-EDT
From: Dan Hoey <hoey@NRL-AIC>
Subject: An outer automorphism of the cube group
To: cube-lovers at mit-mc

If you have a Rubik's cube where all the edges flip on each
quarter-turn, you can solve it by using bifocals when it's odd.

I noticed this while drawing a Hasse diagram of the subgroups of M.  It
turns out that M has a similar automorphism, where the odd elements are
reflected through the center of the cube.

If anyone wants the Hasse diagram, I can send it--it takes about 30
minutes to draw in the lines, for which directions are included.

If you know whether there are other outer automorphisms of M, please
let me know.

Dan

Date: Fri, 7 Sep 84  9:16:44 EDT
From: Michael Frishkopf <mfrishko@BBNCCY.ARPA>
Subject: list request
To: CUBE-LOVERS@mit-mc.arpa
Cc: mfrishkopf@BBNCCY.ARPA

Dear sir/ms:

Please place me on your arpa mailing list:

mfrishkopf@bbnccy
    
Thank you,
Michael Frishkopf


Received: by gyre.ARPA (4.12/4.7)
	id AA09303; Mon, 1 Oct 84 21:10:44 edt
Date: Mon, 1 Oct 84 21:10:44 edt
From: Rehmi Post <rehmi@gyre>
Message-Id: <8410020110.AA09303@gyre.ARPA>
To: cube-lovers@mc
Subject: Archives

Does anyone have copies (partial or otherwise) of the cube-lovers mailing
list? I need it as a reference generator/thought provoker, and if I could
ftp it all from someone, it'd be nice...

			- rehmi at maryland

Date: 5 October 1984 17:07-EDT
From: Alan Bawden <ALAN @ MIT-MC>
Subject:  Moleculon wins a round!
To: CUBE-LOVERS @ MIT-MC

From today's Boston Globe:

DOVER, Del.  ---  A massachusetts firm that claimed it had the first patent
on the Rubik's cube puzzle has won a patent infringement suit against CBS
Inc. and its subsidiary Ideal Toy Corp., which manufactures the puzzle.  US
District Court Judge Walter Stapleton issued the ruling Tuesday, validating
the patent held by Moleculon Research Corp. of Cambridge, Mass.  The ruling
said CBS infringed on the patent with its various Rubik's cube products.
Moleculon's attorney, Robert Perry, said if no appeal is filed, the case
would continue to a second trial to determine monetary damages.  Moleculon
is seeking $60 million in damages and an undetermined amount in profits
from Rubik's cube and related sales.

Received: from CheninBlanc.ms by ArpaGateway.ms ; 08 OCT 84 08:57:05 PDT
Date: 8 Oct 84 08:58:06 PDT (Monday)
From: Lynn.es@XEROX.ARPA
Subject: Re: Moleculon wins a round!
In-reply-to: ALAN's message of 5 Oct 84 18:01 EDT
To: Cube-Lovers@MIT-MC.ARPA
cc: Lynn.es@XEROX.ARPA

A similar article in the LA Times went on to say that the fellow at
Moleculon had patented a 2x2 cube held together with magnets, but
mention was made in the patent of larger cubes and mechanical linkages.
How specific the mentions were is not clear.  He was also quoted to the
effect that they do not dispute that Rubik invented the present cube,
but it infringes on their patent.  Moleculon apparently offered to sell
rights to their toy to Ideal before Rubik and were turned down.
/Don Lynn

Received: from SCRC-DUPAGE by SCRC-STONY-BROOK via CHAOS with CHAOS-MAIL id 185428; Tue 26-Feb-85 11:09:54-EST
Date: Tue, 26 Feb 85 11:00 EST
From: Richard Pavelle <RP@SCRC-QUABBIN.ARPA>
Subject: cube-freak
To: cube-lovers@MIT-MC.ARPA
cc: rp@SCRC-QUABBIN.ARPA
Message-ID: <850226110043.5.RP@DUPAGE.SCRC.Symbolics.COM>

I received a letter from

Georges Helm
76, Bomicht
4936 Bascharage
Luxembourg

who seems to have outdone most of us. He has collected 14000
pages of material on the 3x3 and its variations and has composed
a bibliography on them. He would like to communicate with anyone
who is still interested and is willing to swap material. 

Received: from Gamay.ms by ArpaGateway.ms ; 01 MAR 85 08:55:07 PST
Date: 1 Mar 85 08:54:47 PST (Friday)
From: Hoffman.es@XEROX.ARPA
Subject: 5-by-5-by-5
To: Cube-Lovers@MIT-MC.ARPA
cc: Hoffman.es@XEROX.ARPA

Anyone have a written solution algorithm for the 5-by-5-by-5 cube?  I
got mine a year and a half or so ago from Meffert in Hong Kong by mail
order.  After a couple of weeks of just trying out "pretty patterns" on
it, I finally got up the nerve to mess it up.  Well, of course, it's
never been altogether solved since.  (I was never a great cube wizard.)
It sits on display as the base of my cube tower, with those few nagging
cubies out of place.  I haven't touched it in many months.

--Rodney Hoffman

Received: from SCRC-NEPONSET by SCRC-VALLECITO via CHAOS with CHAOS-MAIL id 371; Fri 1-Mar-85 17:16:32-EST
Date: Fri, 1 Mar 85 17:16 EST
From: David C. Plummer in disguise <DCP@SCRC-QUABBIN.ARPA>
Subject: 5-by-5-by-5
To: Hoffman.es@XEROX.ARPA, Cube-Lovers@MIT-MC.ARPA
In-Reply-To: The message of 1 Mar 85 11:54-EST from Hoffman.es@XEROX.ARPA
Message-ID: <850301171626.2.NFEP@NEPONSET.SCRC.Symbolics.COM>

    Date: 1 Mar 85 08:54:47 PST (Friday)
    From: Hoffman.es@XEROX.ARPA

    Anyone have a written solution algorithm for the 5-by-5-by-5 cube?  I
    got mine a year and a half or so ago from Meffert in Hong Kong by mail
    order.  After a couple of weeks of just trying out "pretty patterns" on
    it, I finally got up the nerve to mess it up.  Well, of course, it's
    never been altogether solved since.  (I was never a great cube wizard.)
    It sits on display as the base of my cube tower, with those few nagging
    cubies out of place.  I haven't touched it in many months.

It took me about half an hour to solve the 4x4x4 by generalizing
techniques for the 3x3x3.  I think many of us convinced ourselves that
the 5x5x5 is just a similar extension.  The generalization is one of
those "Aha!" insights. 

Unfortunately for me, I've never seen a 5x5x5 to prove myself.

Received: from WISCVM.ARPA by MIT-MC.ARPA; 11 MAR 85 23:00:01 EST
Received: from (ENG130)BOSTONU.BITNET by WISCVM.ARPA on 03/11/85 at
  21:58:19 CST
Date: 11 Mar 85   22:51 EST
From:      <eng130%BOSTONU.bitnet@WISCVM.ARPA>
Subject: Please remove me from this list
To:  cube-lovers@mit-mc

-----
  Please take me off this mailing list.  Thanks alot.
-----

Received: from csnet-relay by MIT-MC.ARPA;  3 APR 85 17:23:55 EST
Received: from ti-csl by csnet-relay.csnet id aa14999; 3 Apr 85 17:00 EST
Date:  3 Apr 1985 1028-CST
From: Dan <NICHOLS%ti-csl.csnet@csnet-relay.arpa>
Subject: articles help
To: cube-lovers@mit-mc.ARPA
Received: from csl60 by ti-csl; Wed, 3 Apr 85 13:41 CST


I would very much like to obtain a copy of the following
articles. If anyone can send me a copy or give me more
information about where and how to obtain them, I will
really appreciate it. Thanks.

Eidswick, Jack - How to solve the nxnxn cube.
    Mathematics and Statistics Dept - Univ of Nebraska 1982

Kamack, HJ & T R Keane - The Rubik Tesseract -
    Unpublished manuscript 1982

Kim, Scott E - The Impossible Skew Quadrilateral: A Four-Dimensional
   Optical Illusion - Proceedings of 1978 AAAS Symposium on Hypergraphics:
   Visualizing Complex Relationships in Arts & Sciences

Marx, George, Eva Gajzago, & Peter Gnadig - The Universe of Rubik's Cube
   European Journal of Physics 3 (1982) pp 34-43

 Daniel Nichols
 1721 E. Frankford Rd
 Apt. 1514
 Carrollton, TX 75007

 ARPA: dnichols%ti-csl@csnet-relay
 CSNET: dnichols@ti-csl
 UUCP: {ut-sally,smu,texsun,rice}!waltz!dnichols
 VOICE: 214-492-3275
-------



Received: from csnet-relay by MIT-MC.ARPA; 24 APR 85 18:42:16 EST
Received: from ti-csl by csnet-relay.csnet id ab16901; 24 Apr 85 14:45 EST
Date: 24 Apr 1985 1051-CST
From: Dan <NICHOLS%ti-csl.csnet@csnet-relay.arpa>
Subject: information requests
To: cube-lovers@mit-mc.ARPA
Received: from csl60 by ti-csl; Wed, 24 Apr 85 12:49 CST

1.  Does anyone know of anywhere I can obtain any of the following
    puzzles:
           skewb
           pyraminx crystal
           pyraminx ball
           pyraminx ultimate
           double pyramid

2. I would very much like to obtain copies of Singmaster's Cubic
   Circular. I have the first 4 issues, but saw a reference to there
   being 8 volumes. I would be glad to pay for copying and postage. Can
   anyone help?

3. Does anyone know the status of Meffert's Puzzle Club?

4. Do Singmaster or Doug Hofstadter have net addresses?


   Thanks for any help.


 Daniel Nichols
 1721 E. Frankford Rd
 Apt. 1514
 Carrollton, TX 75007

 ARPA: dnichols%ti-csl@csnet-relay
 CSNET: dnichols@ti-csl
 UUCP: {ut-sally,smu,texsun,rice}!waltz!dnichols
 VOICE: 214-492-3275




-------


Received: from CISL-SERVICE-MULTICS.ARPA by MIT-MC.ARPA; 30 APR 85 09:51:49 EDT
Received: FROM HIS-PHOENIX-MULTICS.ARPA BY CISL-SERVICE-MULTICS.ARPA WITH dial; 30 APR 1985 09:47:08 EDT
Acknowledge-To:  "Craig L. Senft" <Senft@HIS-PHOENIX-MULTICS.ARPA>
Date:  Tue, 30 Apr 85 06:44 MST
From:  "Craig L. Senft" <Senft@HIS-PHOENIX-MULTICS.ARPA>
Subject:  Re: information requests
To:  Alan Bawden <ALAN@MIT-MC.ARPA>
cc:  Cube-Lovers@MIT-MC.ARPA
In-Reply-To:  Message of 30 Apr 85 01:09 MST from "Alan Bawden"
Message-ID:  <850430134407.956530@HIS-PHOENIX-MULTICS.ARPA>

Please remove me from Cube-Lovers mailing list.  Thank you,


 ==cls==

Date: Tue, 30 Apr 85 17:01:09 EDT
From: Alan Bawden <ALAN@MIT-MC>
Subject:  information requests
To: Senft.Multics%PCO-Multics@MIT-MULTICS
cc: CUBE-LOVERS@MIT-MC
In-reply-to: Msg of Tue 30 Apr 85 06:44 MST from Craig L. Senft <Senft at HIS-PHOENIX-MULTICS.ARPA>
Message-ID: <[MIT-MC].477537.850430.ALAN>

    Date: Tue, 30 Apr 85 06:44 MST
    From: Craig L. Senft <Senft at HIS-PHOENIX-MULTICS.ARPA>
    Please remove me from Cube-Lovers mailing list.  Thank you,

Removed.

Received: from Xerox.ARPA by MIT-MC.ARPA 31 May 85 18:33:35 EST
Received: from CheninBlanc.ms by ArpaGateway.ms ; 31 MAY 85 15:31:50 PDT
Date: 31 May 85 15:31:46 PDT (Friday)
From: Hoffman.ES@Xerox.ARPA
Subject: Re: information requests
In-reply-to: <[MIT-MC].476352.850430.ALAN>
To: Alan Bawden <ALAN@MIT-MC.ARPA>
cc: Cube-Lovers@MIT-MC.ARPA
Message-ID: <850531-153150-1581@Xerox>

Dan,

Did you learn anything about Skewb availability or the status of
Meffert's club and company?  In January, 1983, I sent Meffert a money
order for a skewb and followed up twice with letters in Fall '83 and
January '84.  I never heard a thing.  However, in August '83 (after Stan
Isaacs said it worked for him), I sent Meffert money for the
5-by-5-by-5, and it arrived very promptly!  I would LOVE a Skewb.  Have
you even seen one?

I do have a pyraminx ball and a small double pyramid.  I don't recall
what stores I found those in a couple of years ago, but they were local
toy or department stores.

Doug Hofstadter usually is connected to the net, but I haven't
corresponded with him in a year and a half, and I know he's moved since
then.  Isn't he visiting UMass just now?  If you ever get hold of Scott
Kim, he usually knows how to reach DRH.

--Rodney

Received: from Xerox.ARPA by MIT-MC.ARPA 31 May 85 18:36:52 EST
Received: from CheninBlanc.ms by ArpaGateway.ms ; 31 MAY 85 15:34:09 PDT
Date: 31 May 85 15:34:05 PDT (Friday)
From: Hoffman.ES@Xerox.ARPA
Subject: Re: information requests
In-reply-to: Your message of 24 Apr 1985 1051-CST
To: Dan <NICHOLS%ti-csl.CSNet@CSNet-Relay.ARPA>
cc: Cube-Lovers@MIT-MC.ARPA
Message-ID: <850531-153409-1583@Xerox>

Dan,

Did you learn anything about Skewb availability or the status of
Meffert's club and company?  In January, 1983, I sent Meffert a money
order for a skewb and followed up twice with letters in Fall '83 and
January '84.  I never heard a thing.  However, in August '83 (after Stan
Isaacs said it worked for him), I sent Meffert money for the
5-by-5-by-5, and it arrived very promptly!  I would LOVE a Skewb.  Have
you even seen one?

I do have a pyraminx ball and a small double pyramid.  I don't recall
what stores I found those in a couple of years ago, but they were local
toy or department stores.

Doug Hofstadter usually is connected to the net, but I haven't
corresponded with him in a year and a half, and I know he's moved since
then.  Isn't he visiting UMass just now?  If you ever get hold of Scott
Kim, he usually knows how to reach DRH.

--Rodney

Date: Mon, 15 Jul 85 16:02:41 EDT
From: Alan Bawden <ALAN@MIT-MC.ARPA>
Subject:  [alice!amo: Cube-lovers]
To: CUBE-LOVERS@MIT-MC.ARPA
Message-ID: <[MIT-MC.ARPA].576851.850715.ALAN>

Date: Mon, 15 Jul 85 05:54:21 pdt
From: alice!amo at Berkeley
To:   alan
Re:   Cube-lovers

You might like to post the following message to the Cube-lovers list:

One of the best sources for various "twisting puzzles" (and purportedly
the only remaining supplier of the 5x5x5 cubes, which he is
selling for $ 15 plus postage) is

  Dr. Christoph Bandelow 
  Haarholzer Strasse 13
  4630 Bochum-Stiepel
  West Germany
  phone: 0234-797794

I have just obtained two 5x5x5 cubes from him.

Received: from mitre-bedford by MIT-MC.ARPA 30 Jul 85 12:50:06 EDT
Date: Tuesday, 30 Jul 1985 12:47-EDT
From: pes@Mitre-Bedford
To: ALAN@MIT-MC.ARPA
Cc: Cube-Lovers@MIT-MC
Subject: Re: The elusive 5x5x5
In-reply-to: Your message of Monday, 22 Jul 1985 07:05-EDT.
             <[MIT-MC.ARPA].583932.850722.ALAN>


	How should I go about buying a 5 cubed cube?  Can I just
mail 20 or so American dollars to Dr. Bandelow in W. Germany?
Would a personal check be acceptable?  Thanks for any info.

"Just because you're a cube lover doesn't mean you're square"

-Paul Silvey
 pes@mitre-bedford.arpa
 The MITRE Corporation


Received: from Xerox.ARPA by MIT-MC.ARPA 18 Sep 85 18:42:54 EDT
Received: from CheninBlanc.ms by ArpaGateway.ms ; 18 SEP 85 15:38:16 PDT
Date: 18 Sep 85 15:35:09 PDT (Wednesday)
From: Hoffman.es@Xerox.ARPA
Subject: Puzzlers Club resurrected?
To: Cube-Lovers@MIT-MC.ARPA
cc: Hoffman.es@Xerox.ARPA
Message-ID: <850918-153816-1795@Xerox>

This came Tuesday (spelling and grammar as you see it).  See my comments
at the end.

	Dear Sirs,
	
	It has come to my attention that apparently quite a lot 
	of the puzzles we sent to Puzzlers Club members were not 
	received by them.  Upon careful investigation of this matter, 
	it now appears that they were confiscated by the Customs who 
	must have thought at the time that they were Taiwan copies.  
	As we did send the goods and paid the Airmail postage and 
	then had so many claims to send the goods a second and third 
	time, the Puzzlers Club suffered a very big loss and 
	consequently had to closed down.
	
	Due to the suppressed Puzzle market, we are not producing new 
	Puzzles at this time.  However, we will be releasing later this 
	year the "I.Q. DIE" which is the Skewb with Die markings, and a 
	Game called 'King/Ace'.  It is the Pyraminx without the 4 apexes,
	decorated in the 4 Card Suits, the Game is similar to Back-Jack.  
	The "I.Q. DIE" and 'King/Ace' including instructions are now 
	available to Ex-Puzzlers Club members and collectors at US$25 
	each including registered Airmail postage.
	
	Also a few 5 x 5 cubes and Regular Skewbs, Timber finished 
	Pyraminx and Immposi-ball are still available as per price list 
	below, including registered Airmail postage.  A copy of the 
	registered receipt will be sent by separate airmail.  Stocks, 
	however are limited, as there will be no additional production 
	for several years at least.
	
	Please let me know if you wish some of the above items.  I do 
	apologize if you were one of the unfortunate ones who were 
	affected by the customs action.
	
	However if you would like to send us your new order with payment, 
	I personally guarantee that you will get the goods this time.  
	(The customs have been notified that we are the legal copyright 
	holder.)
	
	Once again my apologizes for any inconvenience caused.
	
	With warm regards,
	
	Mr Uwe Meffert
	
		
			1985 PRICE-LIST
	Descriptions		Item No.	Unit Price
	
	5 x 5 x 5 cube		CE8426		US$25/Air
	Skewb			CE8431		US$15/Air
	Pyraminx		CE8423-24	US$20/Air
	Timber finished Pyraminx		US$20/Air
	Impossi-Ball		CE8429		US$29/Air
	3D Nought's + Crosses	CE8434		US$ 9/Air
	
	*All plus US$1 for registered mail.
	
	Please rush me with Item No...............Amount:...............
	
	PRICEWELL (FAR EAST) LIMITED
	Excellente Commercial Bldg., (Fifteenth Floor) 
	456 Jaffe Road
	Hong Kong.
	Postal Address: P.O. Box 31008 Causeway Bay, Hong Kong.
	Tel.: 5-8939944
	Cable: "REFOUNDEL"
	Telex: 75600 REF HX

------------------------------------------------------

It was accompanied by one color page (sloppily torn out of a brochure)
containing pictures of the puzzles.

Well, I'm one of those who got burned, though I never was a member of
the Puzzlers Club.  In January 1983, I sent Meffert a Cashier's Check
for a Skewb.  I followed up by letter late that Spring.  In Summer '83,
I paid for and received very promptly a 5x5x5 cube from him.  In January
1984, I wrote a second letter about the missing Skewb.  I never got the
Skewb nor any replies to my letters until this arrived!  So now it's all
the  (U.S.) government's fault.  Hmph.  (Maybe I should attend those
occasional Customs sales I've heard about!)

I'm just a Skewb-hungry sucker, for I'm going to try again.  But I'm not
very reassured by his "personal guarantee".  What good will that do me?
The man doesn't even answer his mail.  Oh, well.  Order at your own
risk.

--Rodney Hoffman  

Date: Wed, 18 Sep 85 21:38:00 EDT
From: Alan Bawden <ALAN@MIT-MC.ARPA>
Subject:  Administrivia
To: CUBE-LOVERS@MIT-MC.ARPA
Message-ID: <[MIT-MC.ARPA].650223.850918.ALAN>

Several years ago, when Cubes were a big rage, the great volume of mail
sent to this mailing list forced us into making it into a digest.  For most
of the last few years the list has been fairly low-volume, nevertheless I
have refrained from converting the list back to direct-distribution because
some of the messages sent directly to Cube-Lovers were really add-me
requests.  Now I find that I am experiencing a new phenomena:  There is a
growing tendency for people to address replies to messages I forward to
Cube-Lovers to -me-, rather than to the original sender.  This is a bit of
a pain for me, so as of right now, I am converting Cube-Lovers back to
direct-distribution.

Let me take this opportunity to remind you of two bits of administrative
trivia:

1.  Please continue to send all administrative requests (like if you want
    to get off of the list) to Cube-Lovers-Request@MIT-MC.

2.  Old cube-lovers mail is archived in the following places:

    MC:ALAN;CUBE MAIL0	;oldest mail in forward order
    MC:ALAN;CUBE MAIL1	;next oldest mail in forward order
    MC:ALAN;CUBE MAIL2	;more of same
    MC:ALAN;CUBE MAIL3	;still more of same
    MC:ALAN;CUBE MAIL	;recent mail in reverse order

    Files can be FTP'd from MIT-MC without an account.

Date: Wed, 18 Sep 85 22:18:28 EDT
From: Alan Bawden <ALAN@MIT-MC.ARPA>
Subject:  Puzzlers Club resurrected?  &  Administrivia
To: CUBE-LOVERS@MIT-MC.ARPA
Message-ID: <[MIT-MC.ARPA].650261.850918.ALAN>

Date: 18 Sep 85 15:35:09 PDT (Wednesday)
From: Hoffman.es@Xerox.ARPA
Subject: Puzzlers Club resurrected?
To: Cube-Lovers@MIT-MC.ARPA
cc: Hoffman.es@Xerox.ARPA

This came Tuesday (spelling and grammar as you see it).  See my comments
at the end.

	Dear Sirs,
	
	It has come to my attention that apparently quite a lot 
	of the puzzles we sent to Puzzlers Club members were not 
	received by them.  Upon careful investigation of this matter, 
	it now appears that they were confiscated by the Customs who 
	must have thought at the time that they were Taiwan copies.  
	As we did send the goods and paid the Airmail postage and 
	then had so many claims to send the goods a second and third 
	time, the Puzzlers Club suffered a very big loss and 
	consequently had to closed down.
	
	Due to the suppressed Puzzle market, we are not producing new 
	Puzzles at this time.  However, we will be releasing later this 
	year the "I.Q. DIE" which is the Skewb with Die markings, and a 
	Game called 'King/Ace'.  It is the Pyraminx without the 4 apexes,
	decorated in the 4 Card Suits, the Game is similar to Back-Jack.  
	The "I.Q. DIE" and 'King/Ace' including instructions are now 
	available to Ex-Puzzlers Club members and collectors at US$25 
	each including registered Airmail postage.
	
	Also a few 5 x 5 cubes and Regular Skewbs, Timber finished 
	Pyraminx and Immposi-ball are still available as per price list 
	below, including registered Airmail postage.  A copy of the 
	registered receipt will be sent by separate airmail.  Stocks, 
	however are limited, as there will be no additional production 
	for several years at least.
	
	Please let me know if you wish some of the above items.  I do 
	apologize if you were one of the unfortunate ones who were 
	affected by the customs action.
	
	However if you would like to send us your new order with payment, 
	I personally guarantee that you will get the goods this time.  
	(The customs have been notified that we are the legal copyright 
	holder.)
	
	Once again my apologizes for any inconvenience caused.
	
	With warm regards,
	
	Mr Uwe Meffert
	
		
			1985 PRICE-LIST
	Descriptions		Item No.	Unit Price
	
	5 x 5 x 5 cube		CE8426		US$25/Air
	Skewb			CE8431		US$15/Air
	Pyraminx		CE8423-24	US$20/Air
	Timber finished Pyraminx		US$20/Air
	Impossi-Ball		CE8429		US$29/Air
	3D Nought's + Crosses	CE8434		US$ 9/Air
	
	*All plus US$1 for registered mail.
	
	Please rush me with Item No...............Amount:...............
	
	PRICEWELL (FAR EAST) LIMITED
	Excellente Commercial Bldg., (Fifteenth Floor) 
	456 Jaffe Road
	Hong Kong.
	Postal Address: P.O. Box 31008 Causeway Bay, Hong Kong.
	Tel.: 5-8939944
	Cable: "REFOUNDEL"
	Telex: 75600 REF HX

------------------------------------------------------

It was accompanied by one color page (sloppily torn out of a brochure)
containing pictures of the puzzles.

Well, I'm one of those who got burned, though I never was a member of
the Puzzlers Club.  In January 1983, I sent Meffert a Cashier's Check
for a Skewb.  I followed up by letter late that Spring.  In Summer '83,
I paid for and received very promptly a 5x5x5 cube from him.  In January
1984, I wrote a second letter about the missing Skewb.  I never got the
Skewb nor any replies to my letters until this arrived!  So now it's all
the  (U.S.) government's fault.  Hmph.  (Maybe I should attend those
occasional Customs sales I've heard about!)

I'm just a Skewb-hungry sucker, for I'm going to try again.  But I'm not
very reassured by his "personal guarantee".  What good will that do me?
The man doesn't even answer his mail.  Oh, well.  Order at your own
risk.

--Rodney Hoffman  

-------

Date: Wed, 18 Sep 85 21:38:00 EDT
From: Alan Bawden <ALAN@MIT-MC.ARPA>
Subject:  Administrivia
To: CUBE-LOVERS@MIT-MC.ARPA

Several years ago, when Cubes were a big rage, the great volume of mail
sent to this mailing list forced us into making it into a digest.  For most
of the last few years the list has been fairly low-volume, nevertheless I
have refrained from converting the list back to direct-distribution because
some of the messages sent directly to Cube-Lovers were really add-me
requests.  Now I find that I am experiencing a new phenomena:  There is a
growing tendency for people to address replies to messages I forward to
Cube-Lovers to -me-, rather than to the original sender.  This is a bit of
a pain for me, so as of right now, I am converting Cube-Lovers back to
direct-distribution.

Let me take this opportunity to remind you of two bits of administrative
trivia:

1.  Please continue to send all administrative requests (like if you want
    to get off of the list) to Cube-Lovers-Request@MIT-MC.

2.  Old cube-lovers mail is archived in the following places:

    MC:ALAN;CUBE MAIL0	;oldest mail in forward order
    MC:ALAN;CUBE MAIL1	;next oldest mail in forward order
    MC:ALAN;CUBE MAIL2	;more of same
    MC:ALAN;CUBE MAIL3	;still more of same
    MC:ALAN;CUBE MAIL	;recent mail in reverse order

    Files can be FTP'd from MIT-MC without an account.

-------

Date: Wed, 25 Sep 85 10:14:28 EDT
From: Alan Bawden <ALAN@MIT-MC.ARPA>
Subject:  Well, it kept me entertained for an evening.
To: CUBE-LOVERS@MIT-MC.ARPA
Message-ID: <[MIT-MC.ARPA].658952.850925.ALAN>

Dewdney's column in this month's Scientific American presents a puzzle
which he claims to be comparable in difficulty to a Rubik's cube.  (Like he
claims there are people who can do the cube but haven't done this thing.)
Interested Cube-Hackers might find it diverting to give it a try.  It
didn't take me long to devise a sufficient set of tools for solving it
using only pencil-and-paper.  Personally I think it was significantly
easier than a cube, but perhaps it is harder than the majority of other
permutation puzzles I have gotten my hands on in the last few years.
(But perhaps not.  I did this without having a physical model of it in my
hands, so perhaps that has caused me to overlook something.)

You know, I don't recall ever hearing anybody speculate about just what
makes a permutation puzzle interesting and/or difficult.  I guess the group
has to be large and have a large diameter, and there should be a scarcity
of short identities...

Date: Thu, 10 Oct 85 20:47:33 EDT
From: Alan Bawden <ALAN@MIT-MC.ARPA>
Subject:  Has anybody heard of this?
To: CUBE-LOVERS@MIT-MC.ARPA
Message-ID: <[MIT-MC.ARPA].675661.851010.ALAN>

    Date: Thu 10 Oct 85 16:21:40-PDT
    From: Haym Hirsh <HIRSH at SUMEX-AIM.ARPA>
    ...
    3.  (forward to cube-lovers if necessary)  I hear there is a textbook
    on group theory using Rubik's Cube for all its examples (or some such
    gimmick).  Have you heard anything about it?
    ...

Not I.  Anyone else?  I'd be interested in seeing this myself.

Received: from Xerox.ARPA by MIT-MC.ARPA 15 Oct 85 01:27:00 EDT
Received: from CheninBlanc.ms by ArpaGateway.ms ; 14 OCT 85 22:24:05 PDT
Date: 14 Oct 85 22:23:48 PDT (Monday)
From: Hoffman.es@Xerox.ARPA
Subject: Skewbs
To: Cube-Lovers@MIT-MC.ARPA
cc: Hoffman.es@Xerox.ARPA
Message-ID: <851014-222405-4931@Xerox>

Hey, it worked!  On Sept. 18, I sent an order for a skewb with a
cashier's check for US$16 to Meffert in Hong Kong, in response to his
apology/order form.  I included a cover letter mentioning that I was one
of those who had never received an earlier order (that is, my skewb
order of January 1983).

On Wednesday, Oct. 9, I received the promised airmail letter containing
the registered mail receipt showing the package was mailed Oct. 2, and
on Saturday, Oct. 12, I received a package containing TWO skewbs!  The
thing seems rather easy, I think, but it's great fun because it's so
weird -- so odd looking in mid-turn, so awkward to manipulate quickly,
so difficult to keep your bearings (no face centers that stay put to
navigate by).  I haven't fully explored it, but I've found a few handy
"macros" and identities that seem to get me out of most trouble.

Anyway, if you have wanted a skewb ever since first seeing the drawings
of them in Hofstadter's "Scientific American" column, go ahead and try
Meffert.  In case you've lost my note of Sept 18, here's the relevant
portion again:

			1985 PRICE-LIST
	Descriptions		Item No.	Unit Price
	
	5 x 5 x 5 cube		CE8426		US$25/Air
	Skewb			CE8431		US$15/Air
	Pyraminx		CE8423-24	US$20/Air
	Timber finished Pyraminx		US$20/Air
	Impossi-Ball		CE8429		US$29/Air
	3D Nought's + Crosses	CE8434		US$ 9/Air
	
	*All plus US$1 for registered mail.
	
	Please rush me with Item No...............Amount:...............
	
	PRICEWELL (FAR EAST) LIMITED
	P.O. Box 31008 
	Causeway Bay
	Hong Kong

--Rodney Hoffman

Received: from CSNET-RELAY.ARPA by MIT-MC.ARPA 17 Oct 85 12:49:05 EDT
Received: from ti-csl by csnet-relay.csnet id ab07525; 17 Oct 85 12:46 EDT
Date: 17 Oct 1985 0923-CDT
From: Dan <NICHOLS%CSL60%ti-csl.csnet@CSNET-RELAY.ARPA>
Subject: group theory book
To: cube-lovers@mit-mc.arpa
Received: from csl60 by ti-csl; Thu, 17 Oct 85 09:40 CST

10-Oct-85 20:47:33-EDT,00000540;000000000001
Date: Thu, 10 Oct 85 20:47:33 EDT
From: Alan Bawden <ALAN%mit-mc.arpa@CSNET-RELAY>
Subject:  Has anybody heard of this?
To: CUBE-LOVERS%mit-mc.arpa@CSNET-RELAY

    Date: Thu 10 Oct 85 16:21:40-PDT
    From: Haym Hirsh <HIRSH at SUMEX-AIM.ARPA>
    ...
    3.  (forward to cube-lovers if necessary)  I hear there is a textbook
    on group theory using Rubik's Cube for all its examples (or some such
    gimmick).  Have you heard anything about it?
    ...

Not I.  Anyone else?  I'd be interested in seeing this myself.



There was a book published a few years ago which had one chapter
dealing with the cube.

Groups and Geometry - Peter Neumann, Gabrielle A. Stoy, & E. C. Thompson
  The Mathematical Institute - Oxford University
  The last chapter is "The Group Theory of the Hungarian Magic Cube".

Dan Nichols
-------


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Date: 17 Oct 1985 0923-CDT
From: Dan <NICHOLS%CSL60%ti-csl.csnet@CSNET-RELAY.ARPA>
Subject: group theory book
To: cube-lovers@mit-mc.arpa
Received: from csl60 by ti-csl; Thu, 17 Oct 85 09:40 CST

10-Oct-85 20:47:33-EDT,00000540;000000000001
Date: Thu, 10 Oct 85 20:47:33 EDT
From: Alan Bawden <ALAN%mit-mc.arpa@CSNET-RELAY>
Subject:  Has anybody heard of this?
To: CUBE-LOVERS%mit-mc.arpa@CSNET-RELAY

    Date: Thu 10 Oct 85 16:21:40-PDT
    From: Haym Hirsh <HIRSH at SUMEX-AIM.ARPA>
    ...
    3.  (forward to cube-lovers if necessary)  I hear there is a textbook
    on group theory using Rubik's Cube for all its examples (or some such
    gimmick).  Have you heard anything about it?
    ...

Not I.  Anyone else?  I'd be interested in seeing this myself.



There was a book published a few years ago which had one chapter
dealing with the cube.

Groups and Geometry - Peter Neumann, Gabrielle A. Stoy, & E. C. Thompson
  The Mathematical Institute - Oxford University
  The last chapter is "The Group Theory of the Hungarian Magic Cube".

Dan Nichols
-------


Received: from Xerox.ARPA by MIT-MC.ARPA 25 Oct 85 12:14:59 EDT
Received: from CheninBlanc.ms by ArpaGateway.ms ; 25 OCT 85 09:14:40 PDT
Date: 25 Oct 85 09:14:33 PDT (Friday)
From: Hoffman.es@Xerox.ARPA
Subject: Rubik's Revenge:  The Group Theoretical Solution
To: Cube-Lovers@MIT-MC.ARPA
cc: Hoffman.es@Xerox.ARPA
Message-ID: <851025-091440-1219@Xerox>

I only yesterday came across this from a couple of months ago:  "Rubik's
Revenge:  The Group Theoretical Solution" by M.E. Larsen, in the
'American Mathematics Monthly', Vol. 92, No. 6, 1985, pages 381-390.

Received: from ddn2 by MIT-MC.ARPA  5 Nov 85 11:36:09 EST
Date: 5 Nov 85 11:04 EST
From: mccliman @ DDN2.ARPA
Subject: mailing list
To: cube-lovers @ mit-mc.arpa
CC: mccliman @ DDN2.ARPA

I recently discovered your mailing list involving  the CUBE, and 
would very much like to be added to it.  Thanks.
'mcclimans at ddn2'


Received: from ngp.UTEXAS.EDU by MIT-MC.ARPA 14 Nov 85 18:06:20 EST
Date: Thu, 14 Nov 85 14:27:38 cst
From: jknox@ngp.UTEXAS.EDU (John W. Knox)
Posted-Date: Thu, 14 Nov 85 14:27:38 cst
Message-Id: <8511142027.AA02634@ngp.UTEXAS.EDU>
Received: by ngp.UTEXAS.EDU (4.22/4.22)
	id AA02634; Thu, 14 Nov 85 14:27:38 cst
To: cube-lovers@mit-mc.ARPA, mccliman@DDN2.ARPA
Subject: Re:  mailing list
Cc: mccliman@DDN2.ARPAn


q


Received: from CSNET-RELAY.ARPA by MIT-MC.ARPA 15 Nov 85 02:48:31 EST
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Date: 14 Nov 1985 1008-CST
From: Dan <NICHOLS%CSL60%ti-csl.csnet@CSNET-RELAY.ARPA>
Subject: new article
To: cube-lovers@mit-mc.arpa
Received: from csl60 by ti-csl; Thu, 14 Nov 85 10:46 CST

A new article on the cube has appeared. 
"Rubik's Groups" by Edward Turner and Karen Gold in
The American Mathematical Monthly v92n9 November 1985
-------


Received: from Xerox.ARPA by MIT-MC.ARPA 11 Dec 85 14:44:07 EST
Received: from CheninBlanc.ms by ArpaGateway.ms ; 11 DEC 85 10:35:51 PST
Date: 11 Dec 85 10:35:16 PST (Wednesday)
From: Hoffman.es@Xerox.ARPA
Subject: Gift packs
To: Cube-Lovers@MIT-MC.ARPA
cc: Hoffman.es@Xerox.ARPA
Message-ID: <851211-103551-1873@Xerox>

I never really asked to be, but I seem to be a member of Meffert's
Puzzler's Club.  The latest mailing offers several individual items plus
a Christmas special:  9 skewbs plus 9 5x5x5 cubes for US$99 registered
airmail or $US82 registered surface mail.

It says the individual prices are $25 for the 5x5x5 and $15 for the
Skewb, so this is said to be a $360 value.

If you're like me, you're the only cube-lover among those you exchange
gifts with.  But if you happen to have several cube-lovers on your list,
maybe this can solve your shopping worries.

The offer is "limited to December 15, 1985"  All orders should have
"bankdraft or money order in US Dollar Currency".  The address is:
	
	PRICEWELL (FAR EAST) LIMITED
	P.O. Box 31008 
	Causeway Bay
	Hong Kong
	
--Rodney Hoffman

Received: from LLL-MFE.ARPA by MC.LCS.MIT.EDU  6 Jan 86 21:11:01 EST
Date: Mon, 6 Jan 86 18:02 pst
From: "del tufo joseph%e.mfenet"@LLL-MFE.ARPA
Subject: "deltufo%d"@lll-mfe.arpa    
To: cube-lovers@mit-mc.arpa

requesting any information

Received: from Louie.UDEL.EDU by MC.LCS.MIT.EDU 27 Jan 86 12:33:53 EST
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	id AA09563; Mon, 27 Jan 86 11:42:48 est
Date: Mon, 27 Jan 86 11:42:48 est
From: TOMCHANY <jethro%vax1.acs.udel.edu@Louie.UDEL.EDU>
Message-Id: <8601271642.AA09563@vax1.acs.udel.edu>
To: cube-lovers@mc.lcs.mit.EDU

Please add me to your mailing list concerning THE CUBE and other puzzles.
					Thank you,
						M. Tomchany

Received: from Xerox.COM by MC.LCS.MIT.EDU 21 Mar 86 21:07:36 EST
Received: from CheninBlanc.ms by ArpaGateway.ms ; 21 MAR 86 17:36:22 PST
Date: 21 Mar 86 17:35:55 PST (Friday)
From: Hoffman.es@Xerox.COM
Subject: Eidswick article
To: Cube-Lovers@MC.LCS.MIT.EDU
cc: Hoffman.es@Xerox.COM
Message-ID: <860321-173622-1700@Xerox>

In case no one's mentioned it yet:

	Eidswick, J. A., "Cubelike Puzzles -- What Are They
	and How Do You Solve Them?", 'American Mathematical
	Monthly', Vol. 93, #3, March 1986, pp. 157-176.

From the introductory section:

	"This article is an attempt to put some algebraic order
	into the business of solving cubelike puzzles.  Ideally,
	an algebraic theory would unfold that would, in an 
	elementary way, yield highly efficient algorithms for all 
	such puzzles.  The work here is a step in that direction."

Included in the applications section are the tetrahedron, octahedron,
dodecahedron, icosahedron, and the general n X n X n cube.

--Rodney Hoffman

Received: from SCRC-STONY-BROOK.ARPA by MC.LCS.MIT.EDU 26 Mar 86 19:25:52 EST
Received: from SORA.SCRC.Symbolics.COM by SCRC-STONY-BROOK.ARPA via CHAOS with CHAOS-MAIL id 445199; Mon 24-Mar-86 10:30:16-EST
Date: Mon, 24 Mar 86 10:30 EST
From: Bernard S. Greenberg <BSG@SCRC-STONY-BROOK.ARPA>
Subject: Eidswick article
To: Cube-Lovers@MC.LCS.MIT.EDU
In-Reply-To: <860321-173622-1700@Xerox>
Message-ID: <860324103055.2.BSG@SORA.SCRC.Symbolics.COM>

Now that I saw a letter on this list....

Try to find a store that sells an ordinary
Rubik's cube.  The largest Cambridge game shop
hasn't had them for eternity....


Date: Tue, 15 Apr 86 04:40:29 EST
From: Alan Bawden <ALAN@AI.AI.MIT.EDU>
Subject:  Cube-Lovers moves!
To: CUBE-LOVERS@AI.AI.MIT.EDU
Message-ID: <[AI.AI.MIT.EDU].27214.860415.ALAN>

The Cube-Lovers mailing list has moved from MIT-MC to MIT-AI.  All this
really means is that the archives are in a new place.  Sending mail to
Cube-Lovers@MIT-MC and Cube-Lover-Request@MIT-MC will still work, although
it will be infinitesimaly faster if you mail to Cube-Lovers@MIT-AI and
Cube-Lover-Request@MIT-AI instead.

(Please do not take this move as an opportunity to start spreading false
rumors about the status of MIT-MC.  I am moving Cube-Lovers from MC to AI
because I am -personally- moving to AI, and it is thus convenient for the
mailing list to move with me.  There will be a machine named MIT-MC for
years and years to come.)

For the record:  Old cube-lovers mail is archived on MIT-AI in the
following places:

AI:ALAN;CUBE MAIL0	;oldest mail in forward order
AI:ALAN;CUBE MAIL1	;next oldest mail in forward order
AI:ALAN;CUBE MAIL2	;more of same
AI:ALAN;CUBE MAIL3	;still more of same
AI:ALAN;CUBE MAIL4	;yet more
AI:ALAN;CUBE MAIL	;recent mail in reverse order

Files can be FTP'd from MIT-AI without an account.  Unfortunately the
archives are way too large to be conveniently mailed by either computer or
physical mail.

Received: from MC.LCS.MIT.EDU by AI.AI.MIT.EDU via Chaosnet; 2 AUG 86  16:12:56 EDT
Received: from ICSE.UCI.EDU by MC.LCS.MIT.EDU  2 Aug 86 16:14:39 EDT
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To: cube-lovers@mc.lcs.mit.edu
Subject: 5x5x5
Date: Sat, 02 Aug 86 11:05:08 -0800
From: Mark Wadsworth <mw@ICSE.UCI.EDU>

On July 15 I sent a check for $20 to Christoph Bandelow in Germany, and I got 
my cube sometime last week.  He also sent me a letter in which he says,
    I shall be at the International Congress of Mathematicians at
    Berkeley this August 2 - 10 with most of my "cubes". After the
    congress and some travelling time, that is in September, I will
    send you my then updated catalogue - but I hope I can sell the
    complete rest in Berkeley.
If you happen to be around Berkeley this week, it might be worth checking out.
If he sends a catalog with anything in it, I will post info.


Received: from Xerox.COM by AI.AI.MIT.EDU  4 Aug 86 13:44:42 EDT
Received: from CheninBlanc.ms by ArpaGateway.ms ; 04 AUG 86 10:40:38 PDT
Date: 4 Aug 86 10:40:31 PDT (Monday)
From: Hoffman.es@Xerox.COM
Subject: Rubik Redux
To: Cube-Lovers@AI.AI.MIT.EDU
cc: Hoffman.es@Xerox.COM
Message-ID: <860804-104038-1367@Xerox>

From the New York Times, Sunday, August 3, Business section, p. 6:

	WILL 'SON OF CUBE' BRING ANOTHER BOUT OF RUBIKMANIA?
		     By Martin Gottlieb

... Erno Rubik... is about to mark his return with a puzzle that, if
anything, is more difficult to sove than the once-ubiquitous cube....
Professor Rubik... said his new puzzle, called Rubik's Magic, may have
even more configurations.  But, in his view, who cares? 

"You can solve the puzzle by discovering the possibilities, but it is
magic, I think," Professor Rubik said, absently fondling a prototype of
his new invention in the local offices of Matchbox International Ltd.,
the key arm of a Bermuda-based holding company, that will make and sell
it.

... Within the toy industry ... there is a fair amount of anticipation
about Rubik and his Magic.  "We hope we can create a craze," said David
C.W. Yeh, chairman of Universal Matchbox Group, Matchbox International's
parent company.

Matchbox ... has already taken on 2,000 production workers in China to
make the puzzle, and plans to market hundreds of thousands of copies
around the world this fall, for about $10....

The [Hungarian] Government two years ago approved [Rubik's] plans for a
private business, Rubik Studio, which employs 20 and develops designs
for what Mr. Rubik hopes will be a wide range of items, from buildings
to work flow charts and puzzles.  Rubik's Magic is its first commercial
venture.

... When Professor Rubik traveled to the Nuremburg Toy Fair this year
with his latest puzzle, by all accounts, he caused a commotion.

... Key to [Matchbox's successful proposal to Rubik] were a three- to
five-year game plan and concepts for developing more advanced versions
of Rubik's Magic to be introduced in later years.

Rubik's Magic is marked by the same sort of handsome design as its
predecessor.  Palm-sized, it is made of eight squares of
impact-resistant transparent plastic that in their original position
form two equal rows.  Spread across the squares are depictions of three
unconnected rainbow-colored rings printed on a black background.  The
object of the puzzle is to intertwine the three rings by rejiggering the
squares, which are linked by an ingenious hinge patented by Professor
Rubik, that flexes on the four sides of each square.

Unlike the cube, Mr. Rubik's new puzzle can be nameuvered into a
plethora of different shapes.  [Accompanying picture shows this, but
little more.  Rubik is seen leaning on a table which has four Magic toys
arrayed in different configurations.]  The multi-colored loops break
into fanciful swirls and any number of variations would probably look
pretty good on coffee tables.  "On the way to the solution," Mr. Rubik
said, "you can have wonderful discoveries because you have beautiful
shapes.  The cube was very intellectual.  This item could be more fun
and more pleasure -- it is beautiful and changeable."

... "You sort of expected a lot from him after the cube and I think he
delivered," said Rick Anguilla, editor in chief of Toy & Hobby World,
the leading industry journal, who believes the new puzzle is marked by
the same quality as The Cube.  Mr. Anguilla is one of the few industry
analysts to get a glimpse of the new puzzle and he has a hunch it could
be this year's Teddy Ruxpin.

.... Matchbox has devised a plan for producing and promoting the puzzle
aimed in no small part at heading off the knockoff artists who polluted
the Cube market with counterfeits.  It has applied for patents the world
over, and plans to kick off a simultaneous international sales campaign
in October.  Professor Rubik's signature is on the puzzles, and both
professor and puzzle will be promoted with an exceptionally high sales
budget...

... For Professor Rubik, filming commercials that might well place him
second only to Soviet chief Mikhail S. Gorbachev as the most recognized
Eastern bloc citizen in America, is another adventure.

Received: from ARDEC-LCSS.ARPA.ARPA by AI.AI.MIT.EDU 14 Aug 86 15:44:23 EDT
Date: 14 Aug 86 15:34:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: CUBE EXCHANGE
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>
I have been a cube person for 5 or 6 years now and I am upset by the almost
total disappearance of the cube and its asociated memorabilia and paraphernalia.
 Therefore, I propose to establish a "CUBE EXCHANGE" to buy, sell and trade cube
items in suport of the greater development of cubing.   To this end it will be
necessary to catalog what was available, what is avaiable and from whom it can
be obtained.  Your assistance in this endeavor will be much appreciated,
including pointers through the archives.  So far I have located a source for
3x3x3 cubes (solid colors, card suits, flowers, numbers, domino dots, fruits,
sports balls, and ball).  Your assistance in who has what for sale would also be
much appreciated.


peter beck
BECK@ARDEC-LCSS.ARPA
------

Received: from PROPHET.BBN.COM by AI.AI.MIT.EDU 14 Aug 86 16:36:18 EDT
Date:     Thu, 14 Aug 86 16:28:18 EDT
From:     Bernie Cosell <cosell@prophet.bbn.com>
To:       cube-lovers@ai.ai.mit.edu
cc:       cube-lovers <cube-lovers@ai.ai.mit.edu>
Subject:  Re:  CUBE EXCHANGE

``The Games People Play'' in Cambridge, MA  (617-492-0711) had, and still
has, lots of cubes.  Carol (who owns the place) continually complains about
the basement-full of 4x4x4 cubes that she's ended up stuck with...

   /bernie

Received: from harvard.HARVARD.EDU by AI.AI.MIT.EDU 19 Aug 86 10:45:41 EDT
Received: by harvard.HARVARD.EDU; Tue, 19 Aug 86 10:44:51 EDT
Date: Tue, 19 Aug 86 10:44:51 EDT
Received: by h-sc4.HARVARD.EDU; Tue, 19 Aug 86 10:44:36 edt
From: mazzarel%h-sc4@harvard.HARVARD.EDU
To: cube-lovers@mit-ai.arpa
Subject: Paraphernalia


I bought a book about four years ago in Boston that had procedures for
generating all sorts of patterns on the 3x3 cube.  I think it was about
five to ten 8-1/2 x 11" pieces of paper folded in half and stapled to
form a book.  The cover was blue paper.

Unfortunately, I lost it moving about half a year after that.  By that
time I guess interest in the cube was waning; I have not been able to
find another copy.  If anyone else has a copy or just knows where to
find them, could you please post it to this newsgroup?  I thought it
was quite a find, and think no cube lover should be without one.

                               --Paul (pm@harvard.HARVARD.EDU)

Received: from ARDEC-LCSS.ARPA.ARPA by AI.AI.MIT.EDU 25 Aug 86 10:34:12 EDT
Date: 25 Aug 86 10:15:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: singmaster's book
To: "cube-lovers" <cube-lovers@mit-ai.arpa>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>
RE:  Paraphernalia, 19 aug 86

The book you are talking about is Sinmaster's
"Notes on Rubik's Cube".  It was available as you
describe in the 1981 time frame.  I believe it is 
currently avaialble in hardcover with an expanded
bibliography from "Enslow Publishers" .  They are 
located in northern NJ.  If you can't get their 
address send me a message and I will look it up
for you, if you just want the patterns and not the 
book I can send them to you since I have a copy
of the book.   They also have a second book by Singmaster
that was oriented towards teaching math concepts
with the cube.


pete <beck@ardec-lcss.arpa>
------

Received: from ARDEC-LCSS.ARPA.ARPA by AI.AI.MIT.EDU 25 Aug 86 13:59:23 EDT
Date: 25 Aug 86 13:33:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: CUBE AVAILABILITY
To: "cube-lovers" <cube-lovers@mit-ai.arpa>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>
CUBE AVAILABILITY:

1.  REVENGE - on 21 aug 86 I bought a revenge for
$1.99 at the KAY BEE toy store in the Gallery mall,
phila, pa. NOTE: kay bee is a national chain, so check
your area if interested.

2.  GLOBE PUZZLE -  This is a 3x3x3  sphere with the
following difference,  the edge pieces are broken 
into two pieces. Thus the edges and centers are 
interchangeable.  (does this make it a planar puzzle,
for edges and centers)  I haven't received mine yet.  Is 
there any discussion of this variant in the archives.
AVAILABLE from:  the nature company
                             pob 2310
                             berkley, ca 94702
                             800/227-1114
                             globe puzzle, cat # 1449, $12.50
------

Received: from CSNET-RELAY.ARPA by AI.AI.MIT.EDU 26 Aug 86 00:54:03 EDT
Received: from dartmouth by csnet-relay.csnet id ah03666; 25 Aug 86 22:00 EDT
Received: by dartmouth.EDU (4.13D/2.3D)
	id AA19439; Mon, 25 Aug 86 21:13:13 edt
Date: Mon, 25 Aug 86 21:13:13 edt
To: cube-lovers@AI.AI.MIT.EDU
From: Andy Behrens <andyb%dartmouth.edu@CSNET-RELAY.ARPA>
Subject: Re: Paraphernalia
References: <8608191515.AA23076@EDDIE>
Organization: Dartmouth College, Hanover, NH

The blue-covered book that Paul (pm@harvard.HARVARD.EDU) refers to is
probably David Singmaster's pamphlet Notes_on_Rubik's_'Magic_Cube'.  It
is 60 pages long, and includes sections on the basic mathematical
problem, some of the simple subgroups, pretty patterns, algorithms to
restore the cube in the shortest number of moves.  It lists many
processes for flipping and rotating edges and corners.

If the book is still in print, it is available from

      Enslow Publishers
      Bloy St. and Ramsey Ave.
      Hillside, New Jersey 07205     (USA)
      -- ISBN 0-89490-043-9

      or

      David Singmaster & Co.
      66 Mount View Road
      London N4 4JR     (UK)
      -- ISBN 0-907395-00-7

					Andy Behrens

{astrovax,decvax,ihnp4,linus,harvard}!dartvax!andyb.UUCP
andyb@dartmouth.EDU
andyb%dartmouth@csnet-relay.ARPA


Received: from ARDEC-LCSS.ARPA.ARPA by AI.AI.MIT.EDU 26 Aug 86 08:05:35 EDT
Date: 26 Aug 86 07:55:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: HOFSTADTER
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>
HOFSTADTER

I haven't noticed any reference to his 3rd book
which is a reprint of his scientific american
column with postscripts.  I.E. there are 8pgs of
new info; trivia, distribution of patterns from
start, reverse scramblings, measuring
scrambleness.

METAMAGICAL THEMAS
by DOUG HOFSTADTER
BASIC BOOKS
1985 COPYRIGHT
ISBN 0-465-04540-5
PRICE ABOUT $20

PETE     <BECK@ARDEC-LCSS.ARPA>
------

Received: from ARDEC-LCSS.ARPA.ARPA by AI.AI.MIT.EDU 26 Aug 86 08:05:54 EDT
Date: 26 Aug 86 07:57:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: MEFERT
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>
MEFFERT

I have written mefert in march and april 86
 and received no response.  Does anybody out
there know if he is still in business.  By the 
way I ordered and received his XMAS special
of 9 5x5s and 9 skewbs by boat (about 2
months delivery time) with no problems.

PETE     <BECK@ARDEC-LCSS.ARPA>
------

Received: from STONY-BROOK.SCRC.Symbolics.COM by AI.AI.MIT.EDU 26 Aug 86 08:21:27 EDT
Received: from PEGASUS.SCRC.Symbolics.COM by STONY-BROOK.SCRC.Symbolics.COM via CHAOS with CHAOS-MAIL id 87545; Tue 26-Aug-86 08:19:07 EDT
Received: by scrc-pegasus id AA00685; Tue, 26 Aug 86 07:40:08 edt
Date: Tue, 26 Aug 86 07:40:08 edt
From: Bernard S. Greenberg <bsg@scrc-pegasus>
To: cube-lovers%ai.ai.mit.edu@stony
Subject: Paraphernalia

    Date: Mon, 25 Aug 86 21:13:13 edt
    To: cube-lovers@AI.AI.MIT.EDU
    From: Andy Behrens <andyb%dartmouth.edu@CSNET-RELAY.ARPA>
    Subject: Re: Paraphernalia
    References: <8608191515.AA23076@EDDIE>
    Organization: Dartmouth College, Hanover, NH

    The blue-covered book that Paul (pm@harvard.HARVARD.EDU) refers to is
    probably David Singmaster's pamphlet Notes_on_Rubik's_'Magic_Cube'.  It
    is 60 pages long, and includes sections on the basic mathematical
    problem, some of the simple subgroups, pretty patterns, algorithms to
    restore the cube in the shortest number of moves.  
It most certainly made no claim to minimal algorithms.  Minimal
cube algorithms are still an unsolved problem.
						       It lists many
    processes for flipping and rotating edges and corners.

    If the book is still in print, it is available from


(**PLEASE DO NOT REPLY TO THE ABOVE ADDRESS, IT WILL HANG YOUR MAILER.
 I AM SORRY I AM FORCED TO SEND THIS FROM A BROKEN COMPUTER.
 I CAN'T PUT IN A FROM: or REPLY-TO: Either.
 I UNDERSTAND IT's MY PROBLEM, NOT YOUR PROBLEM, PLEASE DON'T FLAME
 AT ME.  Reply to BSG@SCRC-STONY-BROOK not about mailers.  Sorry.)


Received: from ARDEC-LCSS.ARPA.ARPA by AI.AI.MIT.EDU 29 Aug 86 08:02:18 EDT
Date: 29 Aug 86 07:54:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: globe puzzle
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>
 GLOBE PUZZLE -  I made a mistake.  The "Globe Puzzle" is NOT a cube.  It looks
like a cube with the edge pieces broken up into its two faces but the corners
are fixed and only the three equators are free to move.  It has a metal surface
painted as a globe and each surface area has a dimple.  These dimples do not
make it any easier to work the puzzle.  It is the same 3" diameter as a cube
ballThe puzzle is made in Hungary.

Thus the globe puzzle is three intersecting rings and appears to be simpler then
the cube.  If the corners could move like the cube would that puzzle also be no
more difficult then the cube?  Any ideas on how to engineer such a puzzle? 

AVAILABLE from:  the nature company
                             pob 2310
                             berkley, ca 94702
                             800/227-1114
                             globe puzzle, cat # 1449, $12.50

peter beck  <beck@ardec-lcss.arpa>
------

Received: from ARDEC-LCSS.ARPA.ARPA by AI.AI.MIT.EDU 10 Sep 86 15:09:04 EDT
Date: 10 Sep 86 13:38:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: 
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>
1 - I have an incomplete reference to a PUZZLE MUSEUM in the Wash DC area.  
Anybody have any details.
2  -  Is there any good (complete) mailorder catalog for puzzles.
3 -  "INTERLOCKING" CUBE,  this is a puzzle with 27 cubies held together in a
series connection by a rubber band.  The  joint that connects cubies is not
universal.  The cubies all have the same six colors, but not in the same
orientation.  THE OBJECT OF THE PUZZLE IS TO FOLD UP THE CUBIES to make a 3x3x3
cube with solid colored faces.  ANYBODY DONE THIS PUZZLE, ANY REFERENCES.


           /                       \
          |\                      / |
          |   \         /      /   |
          |\   |\     /\    /|  /|
          |   \|   \/    \/  |/  |
           \  |\   |\    /|  /|  /                <BECK@ARDEC-LCSS>
             \|   \|  \/  |/  |/
               \   |\  |  /|  /
                  \|  \|/  |/
                    \  |  /
                      \|/    
------

Received: from ARDEC-LCSS.ARPA.ARPA by AI.AI.MIT.EDU 19 Sep 86 11:28:14 EDT
Date: 19 Sep 86 10:48:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: interlocking cube
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>
RE:  msg from Randel Shane requesting clarification of the  "INTERLOCKING" CUBE
PUZZLE, 
 this is a puzzle with 27 cubies held together in a series connection by a
rubber band, like a snake.  
The cubies faces each have a different color, ie, 6 per cubie.  The blue and
green colors are along the connecting axis while the red,white,yellow and orange
colors are parallel to the connecting axis.  The cubbies are not all identical. 

The  joint that connects cubies is not universal.  The joint is made by a slot
in the blue and green faces.  These slots are perpendicular to each other.  This
is limits the possible orientations.
THE OBJECT OF THE PUZZLE IS TO FOLD UP THE CUBIES to make a 3x3x3 cube with six
solid colored faces.  
HAS ANYONE DONE THIS PUZZLE, ANY REFERENCES.

PS Randell, I hope this description is somewhat better.  If you would like to
see it first hand I can send you one for $2.  

PPS I need a better ARPANET address to send you msgs directly.


           /                       \
          |\                      / |
          |   \         /      /   |
          |\   |\     /\    /|  /|
          |   \|   \/    \/  |/  |
           \  |\   |\    /|  /|  /                <BECK@ARDEC-LCSS>
             \|   \|  \/  |/  |/
               \   |\  |  /|  /
                  \|  \|/  |/
                    \  |  /
                      \|/    
------

Received: from ARDEC-LCSS.ARPA.ARPA by AI.AI.MIT.EDU  6 Oct 86 08:44:03 EDT
Date: 6 Oct 86 08:19:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: rubiks magic
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>
RUBIK'S MAGIC:

1.  In the NYC area the puzzle is available from Macy's for $10.
2.  Matchbox Toys         "is the distributor"
     141 West Commercial Ave
     Moonachie, NJ 07074
     212/696-5400
3.  HOTLINE:  sat & sun from 9am to 5pm until April 26, 1987.  1800-843-1202, NJ
RES 1800-843-1203
4.  Repairs from matchbox $3
5.  The mechanism is similar to a 2-dimensional Jacobs ladder.
6.  Trivial observations:
   a.  The puzzle has a front and back.
   b.  For three intersecting rings the 8 squares have to be arranged as a 2x3x3
and not as a 2x4.  
7.  It is a good puzzle.

           /                       \
          |\                      / |
          |   \         /      /   |
          |\   |\     /\    /|  /|
          |   \|   \/    \/  |/  |
           \  |\   |\    /|  /|  /                <BECK@ARDEC-LCSS>
             \|   \|  \/  |/  |/
               \   |\  |  /|  /
                  \|  \|/  |/
                    \  |  /
                      \|/    
------

Received: from ARDEC-LCSS.ARPA.ARPA by AI.AI.MIT.EDU 10 Oct 86 08:40:07 EDT
Date: 10 Oct 86 08:32:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: rubiks magic
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>

RUBIK'S MAGIC:  >>> UPDATE  <<<

ADMINISTRIVIA
1.  In the NYC area the puzzle is available from Macy's for $10.  In your area
it may available from "TOYS R US".
2.  Matchbox Toys         "is the manufacturer"
     141 West Commercial Ave
     Moonachie, NJ 07074
     212/696-5400
3.  HOTLINE:  sat & sun from 9am to 5pm until April 26, 1987.  1800-843-1202, NJ
RES 1800-843-1203
4.  Repairs from matchbox $3

PUZZLE NOTES:
1.  It took about 4 days for my crack puzzle team to solve Rubik's Magic.  
2.  The mechanism (similar to a 2-directional Jacobs ladder) is durable and the
puzzle is fun to manipulate.  There are many 3-dimensional shapes/configurations
that the squares can be arranged as.
3.  It is necessary to establish a notation scheme and establish moves to do the
puzzle systematically.  The puzzle is similar in difficulty to the Skewb or
Pyramid.  I would say about 10 - 15 moves.  
4.   Observations:
   a.  The puzzle has a front and back.
   b.  Each piece always has a hinge with two (2) other pieces.  This causes the
pieces when rotated to rotate as if they were gears, ie, adjacent pieces rotate
in opposite directions.
    c. The 2x4 layout can go either up and down or left and right.
    d.  It is possible to turn the open/loop layout inside out, ie, the
non-intersecting rings can be either on the inside or outside.  This is the same
as changing how the hinge works on the 2x4 layout.  This also exchanges the top
and bottom of the 2x4 layout.
  e.  For three intersecting rings (the object of the puzzle) the 8 squares have
to be arranged as a 2x3x3 and not as a 2x4.  

           /                       \
          |\                      / |
          |   \         /      /   |
          |\   |\     /\    /|  /|
          |   \|   \/    \/  |/  |
           \  |\   |\    /|  /|  /                <BECK@ARDEC-LCSS>
             \|   \|  \/  |/  |/
               \   |\  |  /|  /
                  \|  \|/  |/
                    \  |  /
                      \|/    
------

Received: from MX.LCS.MIT.EDU by AI.AI.MIT.EDU via Chaosnet; 10 OCT 86  13:42:48 EDT
Date: Fri, 10 Oct 86 13:41:11 EDT
From: Richard Pavelle <RP%MX.LCS.MIT.EDU@MC.LCS.MIT.EDU>
Subject: Rubiks Magic
To: CUBE-LOVERS%MX.LCS.MIT.EDU@MC.LCS.MIT.EDU
Message-ID: <[MX.LCS.MIT.EDU].952133.861010.RP>

It is available at Toys-R-Us in the Boston area.

Received: from DIAMOND.S4CC.Symbolics.COM (TCP 20024231403) by AI.AI.MIT.EDU  4 Nov 86 17:11:18 EST
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Date: Tue, 4 Nov 86 17:06 EST
From: David C. Plummer <DCP@QUABBIN.SCRC.Symbolics.COM>
Subject: rubiks magic
To: "CLSTR1::BECK" <beck@clstr1.decnet>,
    cube-lovers <cube-lovers@MIT-AI.ARPA>
In-Reply-To: The message of 10 Oct 86 09:32 EDT from "CLSTR1::BECK" <beck@clstr1.decnet>
Message-ID: <861104170610.5.DCP@KOYAANISQATSI.S4CC.Symbolics.COM>

Rubik's Magic is available at "Games people play" near Harvard Square
for a little under $15.  As with Beck, it took me about 4 days to solve
it.  My resulting solution is strickingly simple (3 macros), but I took
a lot of interesting paths to get there.  I think it could have been
made harder be repainting the back surface, but doing so may take it out
of the reach of the general public.  (It would require solving it to an
intermediate configuration that doesn't look anywhere near solved, and
then frob it again.)

It's a good puzzle for playing with and relieving the fidgets.
Mathematically it isn't very interesting, nor are there any "pretty
patterns" other than solved1 and solved2.


Received: from SUMEX-AIM.ARPA (TCP 1200000070) by AI.AI.MIT.EDU  4 Nov 86 17:48:42 EST
Date: Tue 4 Nov 86 14:43:05-PST
From: Haym Hirsh <HIRSH@SUMEX-AIM.ARPA>
Subject: SF Bay area availability of Rubik things
To: cube-lovers@AI.AI.MIT.EDU
cc: hirsh@SUMEX-AIM.ARPA
In-Reply-To: <861104170610.5.DCP@KOYAANISQATSI.S4CC.Symbolics.COM>
Message-ID: <12252334415.75.HIRSH@SUMEX-AIM.ARPA>

Rubik's magic was available at the local Sears (Palo Alto/Mtn View) for
around $8, but they are out of stock now.  It is currently available
at Norney's in the same mall for around $13.

A few 5x5x5 Rubik's cubes were available at Games and Things in the
Stanford shopping center about a week ago for around $20.
-------

Received: from WAIKATO.S4CC.Symbolics.COM (TCP 20024231532) by AI.AI.MIT.EDU  4 Nov 86 18:25:14 EST
Received: from ROCKY-MOUNTAINS.S4CC.Symbolics.COM by WAIKATO.S4CC.Symbolics.COM via CHAOS with CHAOS-MAIL id 71267; Tue 4-Nov-86 18:23:59 EST
Date: Tue, 4 Nov 86 18:22 EST
From: Allan C. Wechsler <acw@WAIKATO.S4CC.Symbolics.COM>
Subject: rubiks magic
To: DCP@QUABBIN.SCRC.Symbolics.COM, beck@clstr1.decnet,
    cube-lovers@MIT-AI.ARPA
In-Reply-To: <861104170610.5.DCP@KOYAANISQATSI.S4CC.Symbolics.COM>
Message-ID: <861104182207.9.ACW@ROCKY-MOUNTAINS.S4CC.Symbolics.COM>

    Date: Tue, 4 Nov 86 17:06 EST
    From: David C. Plummer <DCP@QUABBIN.SCRC.Symbolics.COM>

    Rubik's Magic is available at "Games people play" near Harvard Square
    for a little under $15.  

As usual, Carol Monica is charging unreasonable prices.  Boycott this
moderately obnoxious woman and buy your Magic from Zayre or Bradlee's at
about $7-8.

    It's a good puzzle for playing with and relieving the fidgets.
    Mathematically it isn't very interesting, nor are there any "pretty
    patterns" other than solved1 and solved2.

The "pretty patterns" are more sculptural than geometrical.  Can you
make a cube?

Mathematically, there are a couple of interesting points.  There are
sixteen achievable 2x4 configurations, linked by a nice little rosette
of generators.

Unsolved stumper:  The pattern

	XXX
	X X
	XXX

appears unreachable, but we haven't been able to prove it.  Can someone
come up with a proof?  (Or -- hope against hope -- has anyone achieved
it?)

I find the mechanical aspect more pleasing than the cube.  "Magic" is more
satisfying to manipulate than almost any of its predecessors.  The
linking principle can be generalized to any number of squares.  I am
considering breaking a couple and wiring them together to make Big
Magic.


Received: from PROPHET.BBN.COM (TCP 20026200117) by AI.AI.MIT.EDU  4 Nov 86 19:23:09 EST
Date:     Tue, 4 Nov 86 19:12:58 EST
From:     Bernie Cosell <cosell@prophet.bbn.com>
To:       cube-lovers@ai.ai.mit.edu
cc:        beeler@prophet.bbn.com, alatto@prophet.bbn.com, jr@prophet.bbn.com, 
           lcosell@prophet.bbn.com
Subject:  Re:  rubiks magic

On the other side, GPP is generally the *only* comprehensive games-hackers
store about and we would do well to go out of our way to patronize it.
I hardly care about the $5.00 one way or the other on something like
Magic, but where else can you get Monkey puzzles, and the Pentagle series
(are they still in business?  I haven't seen any new ones recently),
go/shogi sets/books, imported jigsaw puzzles and on and on.  If she doesn't
get a decent profit on the things she expects to sell a fair number of,
then she won't be able to stock the esoteric and fun stuff.

  /Bernie

Bernie Cosell                       Internet:  cosell@bbn.com
Bolt, Beranek & Newman, Inc         USENET:    bbnccv!bpc
Cambridge, MA   02238               Telco:     (617) 497-3503

Received: from MC.LCS.MIT.EDU by AI.AI.MIT.EDU via Chaosnet; 6 NOV 86  15:06:54 EST
Received: from nrl-aic (TCP 3200200010) by MC.LCS.MIT.EDU  6 Nov 86 15:02:30 EST
Date: 6 Nov 1986 13:41:48 EST (Thu)
From: Dan Hoey <hoey@nrl-aic.ARPA>
Subject: Magic
To: Cube-lovers@mit-mc.ARPA
Message-Id: <531686509/hoey@nrl-aic>

I have three macros for transforming 2x4 rectangles.  To solve the
puzzle, I use two of them followed by a seven-flip macro that changes a
2x4 shape into a 3x3-1 shape (beats me how BECK can call this a
``2x3x3'').  Took me a couple of half-days to solve it.

I have found 32 different 2x4 rectangles.  I think that is all of them,
but I haven't got any proofs, nor even a decent mathematical model for
deciding when a flip is possible.

I am trying to understand how the strings work.  First, it looks like
there is twice as much string as necessary; each string is doubled.  I
guess that this duplication has no effect on the puzzle except for
durability, but until I can dyke one out can't be sure.  I'm concerned
that the string may be one double loop, so I'm looking for a good way
to make sure the thing doesn't unstring entirely when I cut one.

Each side of each piece has four short channel segments and four long;
half are occupied with string.  If you continue each segment across
each hinge to the next piece, you get eight channels composed of
alternating long and short segments.  Again, four of the eight channels
are occupied.  In the positions I've seen, each of the channels
contains eight pairs of segments.

But Magic is more complicated than that--the strings do not always
follow a channel from piece to piece.  On half of the pieces, there is
an extra loop of string that wraps back onto the piece without
following the channel to the next piece.  I don't know what function
this serves.

If a good model of the string interactions can be developed, we may be
able to make an attack on the doughnut problem based on the length of
string channels.  In the doughnut, there are four ten-pair channels and
four six-pair channels.  We may be able to show that the string wouldn't
reach one, and would exceed the other.  More likely, the model will
prohibit the doughnut more directly.

There is another string-related question I am wondering about.  I have
noticed some of the string-pairs getting twisted.  I wonder how bad
this can get.  Does anyone have an operation that can be repeated to
make the twists tighter and tighter?  Are these puzzles built for
obsolescence?

I have been considering Magic metrics, but it's a difficult problem.
Counting flips is easy enough, but how do you count a move that skews a
parallelogram?  Are such skew moves necessary?

Dan

Received: from MC.LCS.MIT.EDU by AI.AI.MIT.EDU via Chaosnet; 6 NOV 86  18:17:21 EST
Received: from WAIKATO.S4CC.Symbolics.COM (TCP 20024231532) by MC.LCS.MIT.EDU  6 Nov 86 18:15:02 EST
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Date: Thu, 6 Nov 86 18:13 EST
From: Allan C. Wechsler <acw@WAIKATO.S4CC.Symbolics.COM>
Subject: Magic
To: hoey@NRL-AIC.ARPA, Cube-lovers@MIT-MC.ARPA
In-Reply-To: <531686509/hoey@nrl-aic>
Message-ID: <861106181326.6.ACW@ROCKY-MOUNTAINS.S4CC.Symbolics.COM>

    Date: 6 Nov 1986 13:41:48 EST (Thu)
    From: Dan Hoey <hoey@nrl-aic.ARPA>

    I have found 32 different 2x4 rectangles.  I think that is all of them,
    but I haven't got any proofs, nor even a decent mathematical model for
    deciding when a flip is possible.

I think I have a proof.  Wait a few paragraphs.

    If a good model of the string interactions can be developed, we may be
    able to make an attack on the doughnut problem based on the length of
    string channels.  In the doughnut, there are four ten-pair channels and
    four six-pair channels.  We may be able to show that the string wouldn't
    reach one, and would exceed the other.  More likely, the model will
    prohibit the doughnut more directly.

Here is my model.  It might be wrong.

The puzzle is a cycle of eight squares.  Their underlying adjacency
relationships never change.

Each pair of squares is bound together by two loops of string (nylon
fishing wire, actually).  Those two loops are dedicated to holding those
two squares together -- they never migrate to other squares, although
parts of a loop may sometimes lie on one of the pair, sometimes on the
other.

I need a diagram of the channels:

         ---- ----
        |/\/\|/\/\|
        |\/\/|\/\/|
	|/\/\|/\/\|
        |\/\/|\/\/|
         ---- ----

I offer the usual apologies about aspect ratio.  Now, ignore the two
sided nature of the puzzle.  Imagine that the two squares form a 1x2
unit billiard table, and think of the channels as trajectories of
billiard balls, and you will see that the channels form two disjoint
"orbits", each (* 4 (sqrt 2)) units long.  The two loops of string
follow these two orbits, crossing from the obverse to the reverse sides
of the puzzle and back again at every chance they get.  If you work it
out, you see that half the channels are empty at any given time.

The details are too mindbending, but the result is clear: those two
squares are bound together.  They don't depend on synergy from the rest
of the puzzle to bind them.

Now close the pair of squares, putting the right on top of the left as
if you were finishing reading a book.  You can open the pair again
vertically, but only in one direction.  The front square can flip up, or
it can flip down, but not both.  Without loss of generality, let's say
it can flip up.  If you had folded the right square behind the left
instead of in front, it would be able to flip down.

Now gedankenfollow these gendankendirections.  

Put the squares back as above.  Only the square on the right (the rotor)
will move. Keep the left square (the stator) fixed.

Fold the rotor in front of the stator.  Flip the rotor up, and over, and
behind the stator.  Now it will open to the left.  Flip the rotor around
the left edge of the stator until it is in front again.  Then it will
flip down.  Do so, until it is in back again.  Now it will open to the
right.

You have moved the rotor all the way around the stator.  At some point,
each edge of both served as the hinge.  This amazing orbit is the basis
for the bewilderingness of the puzzle.  The path is a bizarre
three-dimensional cloverleaf.

I think that I have now given all the "laws of motion" of the puzzle.

Since the laws are all local, governing the motion of one adjacent pair
of squares, there is no obvious invariant that forbids the doughnut.

    There is another string-related question I am wondering about.  I have
    noticed some of the string-pairs getting twisted.  I wonder how bad
    this can get.  Does anyone have an operation that can be repeated to
    make the twists tighter and tighter?  Are these puzzles built for
    obsolescence?

I am convinced not.

    I have been considering Magic metrics, but it's a difficult problem.
    Counting flips is easy enough, but how do you count a move that skews a
    parallelogram?  Are such skew moves necessary?

    Dan

Your 32 configurations are characterized as follows:  Turn the puzzle so
the unlinked rings are in front.  Rotate it so Rubik's signature is in
the bottom row.  The signature square could be in any of four positions.
The "matchbox" square could be clockwise from Rubik, or
counterclockwise.  And Rubik himself could be in any of four
orientations.  This determines the orientations of all the other
squares.  There are no more degrees of freedom.

It is more instructive to consider the "supergroup", which includes
180-deg rotations but excludes turning the puzzle over.  This group has
sixty-four elements in thirty-two similar pairs.  I have a table.


Received: from nrl-aic (TCP 3200200010) by AI.AI.MIT.EDU 13 Nov 86 15:00:34 EST
Date: 13 Nov 1986 13:16:44 EST (Thu)
From: Dan Hoey <hoey@nrl-aic.ARPA>
Subject: Magic
To: acw%WAIKATO.S4CC.Symbolics.COM@scrc-stony-brook.ARPA, jr@bbncc5.ARPA,
        alatto@bbncc5.ARPA, cube-lovers@mit-AI.ARPA
In-Reply-To: <861106181326.6.ACW@ROCKY-MOUNTAINS.S4CC.Symbolics.COM>
Message-Id: <532289805/hoey@nrl-aic>

    Date: Thu, 6 Nov 86 18:13 EST
    From: Allan C. Wechsler <acw@WAIKATO.S4CC.Symbolics.COM>

    ... Here is my model.  It might be wrong.

From what I have seen of the puzzle, your model seems to correctly
specify necessary conditions for puzzle movement.  I have some
additional constraints, below.

    Each pair of squares is bound together by two loops of string (nylon
    fishing wire, actually).

This is acceptable for a model, though the physical puzzles I have seen
are have twice as much wire on half of the squares as on the others.  I
don't know why, and I am nearly ready to take one apart just to verify
my several curiosities about the construction.

    ... Only the square on the right (the rotor) will move. Keep the left
    square (the stator) fixed.  Flip the rotor ....
    You have moved the rotor all the way around the stator.  At some point,
    each edge of both served as the hinge.  This amazing orbit is the basis
    for the bewilderingness of the puzzle.  The path is a bizarre
    three-dimensional cloverleaf.

This corresponds with my understanding of two-square interaction.
Thanks for nailing it down.

    I think that I have now given all the "laws of motion" of the puzzle.

There is something more, because each square has *two* neighbors.
Consider a stator with two rotors.  There are two ``bizarre three-
dimensional cloverleaves'' about the stator, but both rotors move in
the same b3dc.  Furthermore, they cannot pass each other in their
common b3dc.  Thus if one goes around more than a complete revolution,
the other must make a net advance in the same direction.

    Since the laws are all local, governing the motion of one adjacent
    pair of squares, there is no obvious invariant that forbids the
    doughnut.

I think I can demonstrate one, given the fact that the pieces are
joined in a cycle.  Let us tie each pair of adjacent squares together
as in your model.  We notice that in the starting position, each of the
occupied channels is occupied by two wires, one for each neighbor of
the square.  Suppose we dyke out a loop from the left neighbor, and
splice the left neighbor's connections into the right neighbor's loop.
Then do this all the other duplicate loops.  I believe we end up with a
puzzle held together by four pieces of wire, each (* 8 (sqrt 2)) units
long.  One of them is drawn below, assuming transparent squares.  I
believe that this puzzle will move in the same way as the original one.

         ---- ---- ---- ---- 
        |/\  |    |/\  |    |
        |\ \ |   /|  \ |    |
        | \ \|  / |   \|    |
        |  \ |\/  |    |\   |
         ---- ---- ---- ---- 
        |   \|    |  /\| \  |
        |    |\   | /  |\ \ |
        |    | \  |/   | \ \|
        |    |  \/|    |  \/|
         ---- ---- ---- ---- 

Now put away the dykes and try to fold the puzzle into a doughnut.  If
you succeed, you will have formed two (* 10 (sqrt 2)) wires and two
(* 6 (sqrt 2)) wires:

         ---- ---- ----          ---- ---- ----
        |/\  |    |/\  |        |    |/\  |    |
        |\ \ |   /|  \ |        |   /|  \ |    |
        | \ \|  / |   \|        |  / |   \|    |
        |  \ |\/  |   /|        | /  |    |\   |
         ---- ---- ----          ---- ---- ----
        |   \|    |  / |        |/   |    | \  |
        |   /|    | /  |        |\   |    |  \ |
        |  / |    |/   |        | \  |    |   \|
        | /  |    |\   |        |  \ |    |   /|
         ---- ---- ----          ---- ---- ----
        |/   |  /\| \  |        |   \|    |  / |
        |\   | /  |\ \ |        |    |\   | /  |
        | \  |/   | \ \|        |    | \  |/   |
        |  \/|    |  \/|        |    |  \/|    |
         ---- ---- ----          ---- ---- ----

Changing the string lengths is a little more magic than I expect from
Rubik.

My thanks to John Robinson for encouraging me to believe that the
string-length criterion has merit.  He tells me Andy Latto has a
similar proof.

Dan

Received: from MC.LCS.MIT.EDU by AI.AI.MIT.EDU via Chaosnet; 30 NOV 86  15:36:27 EST
Received: from cc2.bbn.com (TCP 1000200022) by MC.LCS.MIT.EDU 30 Nov 86 15:36:55 EST
Date: Sun, 30 Nov 86 15:34:25 EST
From: Buz Owen <ado@cc2.bbn.com>
Subject: 5 cube?
To: cube-lovers@mc.lcs.mit.edu
Cc: ado@cc2.bbn.com

Can someone on this list supply me with the address of a store
that can sell me a 5x5x5 cube?  I am not on the list so please
reply directly.  Thanks/Buz



Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU  3 Dec 86 10:27:14 EST
Date: 3 Dec 86 08:55:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: 5X5X5 AVAILABILITY
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>
RE:  Buz Owen's request for a source for 5x5x5 cubes

1.  Jerry Slocum
     257 South Palm Drive
     Beverly Hills, CA 90212
     213/273-2270
has both 5x5x5s and skewbs available for $20 each.

2. Jerry Slocum has also authored a book "Puzzles Old and New, How to make and
Solve Them" which he sells for $22 outside of CA, autographed if requested.  
The table of contents is:
chap 1  PUT-TOGETHER PUZZLES - Putting the object together is the puzzle.
chap 2  TAKE-APART PUZZLES - Opening or taking the object apart is the puzzle.
chap 3  INTERLOCKING SOLID PUZZLES - Disassembly and assembly is required to
solve the puzzle.
chap 4  DISENTANGLEMENT PUZZLES - The puzzle is to disentangle and re-entangle
the parts of the puzzle.
chap 5  SEQUENTIAL MOVEMENT PUZZLES -  The puzzle is to move parts of the object
to a goal (The cube, et al)
chap 6  PUZZLE VESSELS - Drinking or pouring liquid, or filling the vessels
without spilling is the puzzle.
chap 7  DEXTERITY PUZZLES -  Manual dexterity is required to solve the puzzle.
chap 8  VANISH PUZZLES - The puzzle is to explain a vanishing or changing image.
chap 9  IMPOSSIBLE OBJECT PUZZLES - The puzzle is to discover how the object is
made.
chap 10  FOLDING PUZZLES - The puzzle is to achieve a specified gola by folding.

3.  Richard Hess
     4100 Palos Verdes Dr. East
     Rancho Palos Verdes, CA 90274
has available for sale many of the puzzles from Jerry Slocum's book.

4.  Another book of interest is:
        536 Puzzles & Curious Problems by  Henry Ernest Dudeney,
        1967 by Charles Scribner's Sons
        Library of Congress Catalog Card NUmber 67-15488.

5.  There are sources for other variations of the cube, if anybody is interested
I will compile a list.

CUBING IS FOREVER ---     <BECK@ARDEC-LCSS>
------

Received: from REAGAN.AI.MIT.EDU by AI.AI.MIT.EDU via Chaosnet; 3 DEC 86  17:04:45 EST
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Date: Wed, 3 Dec 86 17:04 EST
From: Alan Bawden <Alan@AI.AI.MIT.EDU>
Subject: BITNET Redistribution of Cube-Lovers
To: Cube-Lovers@AI.AI.MIT.EDU
Message-ID: <861203170452.2.ALAN@PIGPEN.AI.MIT.EDU>

I'm looking for someone on BITNET to volunteer to maintain a BITNET
redistribution list for the Cube-Lovers mailing list.  This is necessary
because the machine that's acting as the Internet/BITNET gateway,
WISCVM.WISC.EDU, is swamped with mailing list mail.  List administrators
(like me) are being asked to set up redistribution lists to help lighten
the load.  If we can set one up for Cube-Lovers then WISCVM will only have
to relay each message to one host instead of the current 20 or so.

If you're willing and able to take this on, please let me know.

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU  5 Dec 86 09:20:22 EST
Date: 5 Dec 86 09:06:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: 
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>
From:	CLSTR1::SYSTEM        4-DEC-1986 11:27
To:	CLSTR1::BECK
Subj:	Undeliverable mail

 ----Transcript of session follows----
"mit" is an unrecognized hostname/address

 ----Unsent message follows----
Date: 4 Dec 86 11:18:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: CUBE AVAILABILITY
To: "cube-lovera" <cube-lovera@mit>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>
1.  MY MSG TO BUZ OWEN was not delivered (RE: 5x5x5 availability).  If anybody out there knows him or how to forward the msg to him (he is not on the cube lovers list) please do so.

2.  It will take a bout a week but I will compile a list where and what is available in the world of cubes.  It will be a compilation of Helm's 
, Bandelow's, Hess's and Beck's supplies.  Please be patient.  If you are looking for a specific cube let me know I can get you that answer faster.

<beck@ardec-lcss>
------
------

Received: from Xerox.COM (TCP 1200400040) by AI.AI.MIT.EDU  7 Dec 86 19:21:54 EST
Received: from CheninBlanc.ms by ArpaGateway.ms ; 07 DEC 86 16:21:13 PST
Date: 7 Dec 86 16:21:12 PST (Sunday)
From: Hoffman.es@Xerox.COM
Subject: Puzzle show
To: Cube-Lovers@AI.AI.MIT.EDU
Reply-To: Hoffman.es@Xerox.COM
Message-ID: <861207-162113-1746@Xerox>

PUZZLES OLD AND NEW:  Head Crackers, Patience Provers, and Other Tactile
Teasers

[including more variations on Rubik's cube than I've ever seen before]

EXHIBITION TOUR:
  	
	Craft and Folk Art Museum, LA	Nov. 26, '86 - Feb. 22, 1987
	MIT Museum			April 6 - June 15, 1987
	Hudson River Museum		July 22 - Sept. 27, 1987
	Science Museum of Minnesota	Oct. 19, '87 - Jan 3, 1988
	Ontario Science Center		Jan. 25 - March 6, 1988

Accompanied by the book PUZZLES OLD AND NEW: HOW TO MAKE AND SOLVE THEM
by Jerry Slocum and Jack Botermans ($20).

The exhibit discusses
	The Art in Puzzles
	The Social Experience
	Cultural Values
	Puzzles in the Industrial Age
	Puzzles in Education
	Puzzles and Science

It aims "to explore the history, meanings, and design of mechanical
puzzles."  The extensive displays categorize puzzles according to their
object:
	Put-Together Puzzles
	Take-Apart Puzzles
	Interlocking Solid Puzzles
	Disentanglement Puzzles
	Sequential Movement Puzzles
	Puzzle Vessels
	Dexterity Puzzles
	Vanish Puzzles
	Impossible Object Puzzles
	Folding Puzzles

Puzzles have been lent by collectors and museums around the world.  The
exhibit consultants are Benjamin Kilborne and Martin Gardner.

There are hands-on puzzles, but too simple and too few.  It's a
wonderful and tantalizing display of puzzles.  I was SOOO frustrated not
to be able to handle all the beautiful, enticing pieces.  Of course, if
I had been permitted to, I would never leave.... 

  --  Rodney Hoffman

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU  9 Dec 86 14:05:07 EST
Date: 9 Dec 86 13:53:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: 
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>
From:	CLSTR1::SYSTEM        9-DEC-1986 09:22
To:	CLSTR1::BECK
Subj:	Undeliverable mail

 ----Transcript of session follows----
"mit.ai" is an unrecognized hostname/address

 ----Unsent message follows----
Date: 9 Dec 86 09:18:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: 
To: "cube-lovers" <cube-lovers@mit.ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>
From:	CLSTR1::SYSTEM        8-DEC-1986 15:52
To:	CLSTR1::BECK
Subj:	Undeliverable mail

 ----Transcript of session follows----
"ai.ai.mit.edu" is an unrecognized hostname/address

 ----Unsent message follows----
Date: 8 Dec 86 15:32:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: slocum's book
To: "cube-lovers" <cube-lovers@ai.ai.mit.edu>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>
RE:  Slocum's book.
I have not read it yet but I have ordered a copy and hope to be able to answer
your questions soon.  In issue #6 of "World Game Review" (avaialble for $8 for 4
issues from Michael Keller, 3367-I North Chatam Road, Ellicott City, MD 21043)
Slocum's book is reviewed.  It says that this book is a collaboration with Jack
Botermans author of "creative puzzles of the world" which was the best puzzle
book but is now out of print.  The new book has much of the older books material
on mechanical puzzles plus new material.  The book was liked in general.

<beck@ardec-lcss>
------
------
------

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU  9 Dec 86 14:27:43 EST
Date: 9 Dec 86 13:56:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: 
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>
From:	CLSTR1::SYSTEM        8-DEC-1986 15:52
To:	CLSTR1::BECK
Subj:	Undeliverable mail

 ----Transcript of session follows----
"mit.ai" is an unrecognized hostname/address

 ----Unsent message follows----
Date: 8 Dec 86 15:35:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: buz owens
To: "cube-lovers" <cube-lovers@mit.ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>

1.  MY direct MSG TO BUZ OWEN was not delivered (RE: 5x5x5 availability).  If
anybody out there knows him or how to forward the msg to him (he is not on the
cube lovers list) please do so.
 
2.  It will take a bout a week but I will compile a list where and what is
available in the world of cubes.  It will be the highlights from Helm's
, Bandelow's, Hess's and Beck's supplies.  If you want a complete classification
guide request it separately and I will mail you a copy of Bandelow's collection
or if you are more patient wait for Cecil Smith to complete his book.  Please be
patient.  If you are looking for a specific cube let me know I can get you that
answer faster.
 
<beck@ardec-lcss>
------
------
------

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU 10 Dec 86 08:41:01 EST
Date: 10 Dec 86 08:28:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: 
To: "cube-lovers" <cube-lovers@mit-ai>
cc: beck
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>
RE:  Slocum's book.
I have not read it yet but I have ordered a copy and hope to be able to answer
your questions soon.  In issue #6 of "World Game Review" (avaialble for $8 for 4
issues from Michael Keller, 3367-I North Chatam Road, Ellicott City, MD 21043)
Slocum's book is reviewed.  It says that this book is a collaboration with Jack
Botermans author of "creative puzzles of the world" which was the best puzzle
book but is now out of print.  The new book has much of the older books material
on mechanical puzzles plus new material.  The book was liked in general.
 
<beck@ardec-lcss>
------
------
------

Received: from MC.LCS.MIT.EDU (CHAOS 3131) by AI.AI.MIT.EDU 10 Dec 86 17:15:52 EST
Received: from navajo.stanford.edu (TCP 4402000060) by MC.LCS.MIT.EDU 10 Dec 86 17:00:36 EST
Received: by navajo.stanford.edu; Wed, 10 Dec 86 12:56:25 PST
Received: from jack-jr.edsel.uucp by edsel.uucp (2.2/SMI-2.0)
	id AA27267; Wed, 10 Dec 86 12:51:04 pst
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	id AA12942; Wed, 10 Dec 86 12:52:32 PST
Date: Wed, 10 Dec 86 12:52:32 PST
From: edsel!jack-jr!jeb@navajo.stanford.edu (Jim Boyce)
Message-Id: <8612102052.AA12942@jack-jr.edsel.uucp>
To: navajo!cube-lovers%mit-mc.ARPA@navajo.stanford.edu
Subject: Magic: Construction

A friend on mine applied too much force in an odd direction and
produced a tangled mess.  I was nominated to fix it.  My approach 
was to take it apart completely and then put it back together.

Tools required: 
Well, I used a paper clip.

Components:
A Magic is made out of 16 loops of string (actually nylon fishline)
and 8 panels.  The loops of string are all the same length.  The
loops of string are not tangled in any way.  The panels each
decompose into two clear plastic covers and a piece of paper
(actually plastic).  The panels are held together by the string.

Construction:

Each loop of string is twisted around three panels.  It follows a
path like these:

 ---- ---- ----           ---- ---- ----
|/\  |  /\|/\  |         |  /\|/\  |  /\|
|\ \ | / /|\ \ |   or    | / /|\ \ | / /|
| \ \|/ / | \ \|         |/ / | \ \|/ / |
|  \/|\/  |  \/|         |\/  |  \/|\/  |
 ---- ---- ----           ---- ---- ----

For each string, there is another string that lies in the same
channels.  When stringing a loop through the channels, there is a
choice at the points where the string passes from one panel the
next: Which string is closer to the center of the panel?  That
question is answered differently for the two strings running throught
the same channels.

[I believe that this is done so that the net force trying to twist
the toy at that point is near zero.  The redundancy also probably
strengthens the toy.]

The panels can be divided into two sets of four: The panels that are
centers in these triples and the panels that are ends.  Each "end panel"
is an end for two different triples.

Strings don't lie in crossing channels on the same side of a panel.  
(That describes how the two pairs of loops go on the same triple and 
how two triples interact on their common end panel.)

-jim boyce


Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU 11 Dec 86 09:03:53 EST
Date: 11 Dec 86 08:58:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: 
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>
From:	CLSTR1::MAILER        3-DEC-1986 10:50
To:	BECK
Subj:	[TCP/IP Mail From: <@AI.AI.MIT.EDU,@ardec-lcss.arpa:beck@clstr1.decnet>] 5X5X5 AVAILABILITY

Return-Path: <@AI.AI.MIT.EDU,@ardec-lcss.arpa:beck@clstr1.decnet>
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Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU  3 Dec 86 10:27:14 EST
Date: 3 Dec 86 08:55:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: 5X5X5 AVAILABILITY
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>
RE:  Buz Owen's request for a source for 5x5x5 cubes

1.  Jerry Slocum
     257 South Palm Drive
     Beverly Hills, CA 90212
     213/273-2270
has both 5x5x5s and skewbs available for $20 each.

2. Jerry Slocum has also authored a book "Puzzles Old and New, How to make and
Solve Them" which he sells for $22 outside of CA, autographed if requested.  
The table of contents is:
chap 1  PUT-TOGETHER PUZZLES - Putting the object together is the puzzle.
chap 2  TAKE-APART PUZZLES - Opening or taking the object apart is the puzzle.
chap 3  INTERLOCKING SOLID PUZZLES - Disassembly and assembly is required to
solve the puzzle.
chap 4  DISENTANGLEMENT PUZZLES - The puzzle is to disentangle and re-entangle
the parts of the puzzle.
chap 5  SEQUENTIAL MOVEMENT PUZZLES -  The puzzle is to move parts of the object
to a goal (The cube, et al)
chap 6  PUZZLE VESSELS - Drinking or pouring liquid, or filling the vessels
without spilling is the puzzle.
chap 7  DEXTERITY PUZZLES -  Manual dexterity is required to solve the puzzle.
chap 8  VANISH PUZZLES - The puzzle is to explain a vanishing or changing image.
chap 9  IMPOSSIBLE OBJECT PUZZLES - The puzzle is to discover how the object is
made.
chap 10  FOLDING PUZZLES - The puzzle is to achieve a specified gola by folding.

3.  Richard Hess
     4100 Palos Verdes Dr. East
     Rancho Palos Verdes, CA 90274
has available for sale many of the puzzles from Jerry Slocum's book.

4.  Another book of interest is:
        536 Puzzles & Curious Problems by  Henry Ernest Dudeney,
        1967 by Charles Scribner's Sons
        Library of Congress Catalog Card NUmber 67-15488.

5.  There are sources for other variations of the cube, if anybody is interested
I will compile a list.

CUBING IS FOREVER ---     <BECK@ARDEC-LCSS>
------

------

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU 11 Dec 86 09:27:35 EST
Date: 11 Dec 86 08:53:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: SOME CUBES
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>
------
HIGHLIGHTS OF CUBE TYPES WITH AVAILABILITY ( Bandelow's complete compilation is
9 pages. Supposedly both Slocum and Cecil Smith have more extensive lists.)

I will follow Bandelow/Helm classification guide, ie, 
  A.  3x3x3 cubes of standard size (57mm edge)
  B.  size variations of 3x3x3 cube
  C.  shape variations of the 3x3x3 cube
  D.  NxN (N not 3) cubes and their relatives
  E.  other magic polyhedra
  F.  some other logical puzzles
  G.  various cube rel;ated products

SOURCES:  PB - Peter Beck, 54 Richwood place, Denville, NJ 07834, add $1
shipping per order
                CB  - Christoph Bandelow, Haarholzer Strasse 13, D-4630 Bochum
1, W Germany
                GH  - Georges Helm, 22A rue Bommert, 4716 Petange, Luxembourg
                RH  - Richard Hess, 4100 Palos Verdes Dr. East, Rancho Palos
Verdes, CA 90274

  A.  3x3x3 cubes of standard size (57mm edge)
ORIGINAL HUNGARIAN RUBIK'S CUBE (CB $8)
DELUXE CUBE, bonded plastic instead of stickers, Ideal (PB $3)
SIAMESE CUBES, 2 standard cubes joined by having one edge in common, Ideal (CB
$15)
THERE ARE MANY STICKER VARIATIONS; some are one to one subsitions of cubie
stickers,  some are pictures that cover a whole cube face, some are puzzles
(like the calendar cubes), some divide the coloring of cubies so that the 3x3x3
simulates other cubes (see Hofstadter's articles).  I have not seen any
references to Penrose Tiling cubie designs.
NUMBERS, FRUITS, CARDS, FLOWERS, SPORTS BALLS,DIE 1, FEMALE NUDES wonderful,
taiwan (PB $2)

  B.  size variations of 3x3x3 cube
50mm, 38mm, 33mm, 31mm, 30mm, 25mm, 20mm - made into keychains and necklaces.

  C.  shape variations of the 3x3x3 cube
because the cubies always maintain the same in-out orientation relative to the
center of the cube, cubies can have any surface extension desired.  Therefore
there are cubes are spherical cubes, cubes with corner pieces trimmed to
equilateral triangles, some edge pieces are trimmed so that they make a single
plane presentation (eg, makes a barrel), I HAVE NOT HEARD OF ANY THAT MAKE A
SOLID SCULPTURE OF THE CUBE, eg, a bust of RUBIK.
OCTAGONAL PRISM - four parallel edges are trimmed giving an octagonal prism, (CB
$8)
MEDIUM MAGIC GLOBE - Sold as Rubik's World, (CB or GH $10)
MEDIUM MAGIC BALL - sp-herical cube, (PB $2)

  D.  NxN (N not 3) cubes and their relatives
56mm 2x2x2 RUBIK'S POCKET CUBE, (CB $8)
OCTAHEDRON 2X2X2 each rotation axis runs through two opposite corners of the
octahedron, (CB $8)
TETRAHEDRAL 2X2X2, each rotation axis runs through the centers of two opposite
edges, (CB $8)
STAR PUZZLE 2X2X2, similar to tetrahedron with a unicolored tetrahedron attached
to the center of the big tetrahedron.
TRICKHAUS 2X2X2, four edges are trimmed leaving a prism, can be arranged to look
like a house, ie, square base and and roof, (CB $8)
4x4x4, RUBIK'S REVENGE, if you are lucky $2 at KAYBEE TOY STORES
5x5x5, ($20 from jerry slocum, see previous msg)

  E.  other magic polyhedra
SKEWB
PYRAMINX
MEGAMINX
ALEXANDER'S STAR
IMPOSSI-BALL
ORB
HUGARIAN GLOBE, FROM NATURE CO., NIT A CUBE
MISSINK LINK

  F.  some other logical puzzles
  G.  various cube related products

<BECK@ARDEC-LCSS>
------

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU 11 Dec 86 15:15:58 EST
Date: 11 Dec 86 15:06:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: slocums book
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>
RE:  Slocum's book.
I have not read it yet but I have ordered a copy and hope to be able to answer
your questions soon.  In issue #6 of "World Game Review" (avaialble for $8 for 4
issues from Michael Keller, 3367-I North Chatam Road, Ellicott City, MD 21043)
Slocum's book is reviewed.  It says that this book is a collaboration with Jack
Botermans author of "creative puzzles of the world" which was the best puzzle
book but is now out of print.  The new book has much of the older books material
on mechanical puzzles plus new material.  The book was liked in general.
 
<beck@ardec-lcss>
------
------

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU 24 Dec 86 11:41:58 EST
Date: 24 Dec 86 11:38:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: magic review article
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>
I have written a review of MAGIC for the  "World Game Review", edited and
published by Michael Keller, 3367-I north Chatam Road, Ellicott City, MD 21043,
$8 for 4 issues.   It as you can tell is based on the CUBE-LOVERS list dialog. 
If anybody would like to comment or make recomMend any additions or deletions
please forward your suggestions to me  ,<BECK@ARDEC-LCSS>.

A REVIEW OF RUBIK'S MAGIC
BY Peter Beck,  Dec 24, 1986

The Hungarian Government two years ago approved Rubik's plans for a private
business, Rubik Studio, to develop designs for what Mr. Rubik hopes will be a
wide range of items including puzzles.  Rubik's Magic is its first commercial
venture.  It is being manufactured and marketed by Matchbox and is generally
available at prices ranging from $9-$15, $7 on sale.  Matchbox's deal with 
Rubik was based on a three-to-five year plan that includes the development and
marketing of more advanced versions of Magic.

Like the cube, MAGIC is good for playing with and relieving the fidgets.  It is
palm-sized and made up of eight 2"x1/4" squares ,of impact-resistant transparent
plastic, folding up into a 1"x2"x4" block which easily fits into a shirt or
jacket pocket.  Unlike the cube it can be maneuvered into a plethora of
different geometrical shapes which makes it more fun and pleasureful to
manipulate than the cube.    Any of the various geometrical variations look 
good on the coffee table and your guests can make magical discoveries as they
play with it.

The object of the puzzle is to manipulate the squares from their original
pattern (henceforth known as pattern #1) to pattern #2.  In pattern #1 the
squares form two equal rows (i.e., 2x4 arrangement) and spread across one side
is a depiction of three unconnected rainbow-colored rings printed on a black
background.  By folding, flipping, flopping and flapping the squares, which are
linked by an ingenious hinge, we arrive at pattern #2 which is three
intersecting rings on the reverse side.

Even though Magic has many interesting 3-dimensional geometric shapes, e.g.,
cube, A-frame house, 1x2 box with lid, both named patterns occur when the puzzle
is in its planar or flat state.  To go from pattern #1 to #2 there are two
operators necessary in the 2x4 arrangement to position the squares for the
operator that transforms the puzzle into the 3x3 minus a corner arrangement that
displays the three intersecting rings pattern.  When the puzzle is in the 2x4
arrangement with pattern #1 correct, the squares can be rearranged by either
folding it on the long axis to make a loop which can be rotated or by flipping
the 2 ends towards the center and then by un-flipping the puzzle on the opposite
side in a perpendicular direction to the flipping.  In order to display the
solved puzzle it is now necessary to flip and flap and flop the puzzle (6 moves)
into the 3x3 minus a corner arrangement.  It should be noted that it is possible
to be in a 2x4 arrangement where you cannot get to pattern #1 with only the two
2x4 operators of above.  You will either need to use the operator that changes
the puzzle to the 3x3 or develop another operator.

For those of you who would like to take it apart and then put it back together
here are some hints.  Tools required:  a paper clip.  Each loop of string is
twisted around three squares in a path like this:
 
               ---- ---- ----            ---- ----  ----
              |/ \  |  / \ | / \  |           |  / \ |/  \  |  / \ |
              |\  \ | /   /|\   \ |   or     | /   /|\   \ | /   /|
              | \  \|/   / | \   \|           |/   / | \   \|/   / |
              |  \ /|\  /  |  \  /|           |\  /  |  \  /|\  /  |
               ---- ---- ----            ---- ----  ----
 
16 loops of string (actually nylon fishline) are used to make the hinges that
hold the squares together.  The loops of string are all the same length.and are
not tangled in any way.  For each string, there is another string that lies in
the same channels.  When stringing a loop through the channels, there is a
choice at the points where the string passes from one square to the next: Which
string is closer to the center of the square?  That question is answered
differently for the two strings running throught the same channels.  Strings
don't lie in crossing channels on the same side of a square. (That describes how
the two pairs of loops go on the same triple and how two triples interact on
their common end square.)

Now that you have decomposed the puzzle into its component parts why don't you
customize it before reassembly.  (The squares each decompose into two clear
plastic covers and a piece of paper with the design printed on it which are held
together by the strings and not glue.).   So what about some original designs,
maybe even Penrose tiling?  How about adding additional squares?  

For the more mathematically inclined it has been noted that the sameness of the
string length is what restricts the arrangements of the squares.  Can you prove
that the Donut, 3x3 with center missing, is impossible?  Can you invent a
nomenclature and metric for counting moves?   

The future exists, first in the imagination, then in the will and fianlly in
reality.

ACKNOWLEDGEMENT:  The above was written with the passive participation of with
CUBE-LOVERS computer bulletin board at MIT;  	MILNET ADDRESS
<CUBE-LOVERS@MIT-AI>.  Thanks to all who participated in the MAGIC dialog to
date.

------

Date: Mon, 29 Dec 86 00:31:25 EST
From: Marty Connor <MDC@AI.AI.MIT.EDU>
To: CUBE-LOVERS@AI.AI.MIT.EDU
cc: BOTTLENECK@AI.AI.MIT.EDU
Message-ID: <134690.861229.MDC@AI.AI.MIT.EDU>

There is a new puzzle named the "Bottleneck" from a company called
Groton Ltd. in Fitchburg, MA. It looks impossible to solve but they
claim it can be done. It consists of an ordinary glass bottle and a 
solid plastic rod which fits closely within the bottle. There is a 
marble in the bottle but the rod is held in the bottle by a screw 
and nut as shown below:

          |   |
          |   |
          |   |
         ||   ||
         ||   ||
         ||   ||
        / |   | \
       /  |   |  \
      /   |   |   \
     |    |   |    |
     |    |   |    |
     |    |   |    |
     | |==|===|=|| |    ----  Threaded screw and nut go thru threaded rod
     |    |   |    |
     |    |   |    |
     |             |
     |   __        |
     |  (  )       |    ----  Marble
     ---------------

The object is to remove the marble. It is fairly easy to remove the nut
but it is very difficult to turn the screw. It apparantly can be done
without inserting a tool between the bottleneck and the rod. 

It costs about $10.00 and is available from Games People Play in Cambridge 
and Miscellanea in Acton. Has anyone tried it?

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU  6 Jan 87 14:57:48 EST
Date: 6 Jan 87 14:43:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: 
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>
From:	CLSTR1::SYSTEM        5-JAN-1987 16:10
To:	CLSTR1::BECK
Subj:	Undeliverable mail

 ----Transcript of session follows----
"ai.ai.mit.edu" is an unrecognized hostname/address

 ----Unsent message follows----
Date: 5 Jan 87 16:02:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: bottleneck
To: "mdc" <mdc@ai.ai.mit.edu>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>
The puzzle is trivial.  It takes about 5 minutes to disassemble and 10 to assemble.  No sopecial tools are needed.

<beck@ardec-lcss>
------
------



Date: Sat, 10 Jan 87 02:56:40 EST
From: Alan Bawden <ALAN@AI.AI.MIT.EDU>
Subject:  The Archive & Administrivia
To: CUBE-LOVERS@AI.AI.MIT.EDU
Message-ID: <138605.870110.ALAN@AI.AI.MIT.EDU>


Those of you who look through the archives of old Cube-Lovers mail will
notice that I have split off a new section of the archive.  The mail now
lives on MIT-AI in the files:

AI:ALAN;CUBE MAIL0	;oldest mail in forward order
AI:ALAN;CUBE MAIL1	;next oldest mail in forward order
AI:ALAN;CUBE MAIL2	;more of same
AI:ALAN;CUBE MAIL3	;still more of same
AI:ALAN;CUBE MAIL4	;yet more
AI:ALAN;CUBE MAIL5	;more still
AI:ALAN;CUBE MAIL	;recent mail in reverse order

As always files can be FTP'd from MIT-AI without an account.  (And yes, the
spaces in those filenames are a significant part of our filename syntax.)

While I have everyone's attention let me remind you all that last year
Cube-Lovers moved from its original home on MIT-MC to MIT-AI.  Our new
addresses are Cube-Lovers@AI.AI.MIT.EDU for submissions and
Cube-Lovers-Request@AI.AI.MIT.EDU for administrivia.

If you have occasion to send mail to Cube-Lovers, you will generally find
that a fair number of copies of your message will be returned to you by
various mailers around the world for various reasons.  This is always a
problem with old, and fairly quiet mailing lists.  If you would like to be
helpful, you can collect these errors and forward them to
Cube-Lovers-Request (I will eventually flush anyone who is consistently
unreachable), but under no condition should you forward the error message
to Cube-Lovers itself.  Thank you.

				-Alan

Received: from MIT-MULTICS.ARPA (TCP 1200000006) by AI.AI.MIT.EDU 12 Jan 87 01:15:27 EST
Date:  Mon, 12 Jan 87 01:07 EST
From:  Paul Schauble <Schauble@MIT-MULTICS.ARPA>
Subject:  Sci. Am. reference needed
To:  Cube-Lovers@AI.AI.MIT.EDU
Message-ID:  <870112060752.852656@MIT-MULTICS.ARPA>

Perhaps someone on this list can help me locate an item that appeared in
the Mathamatical Games section of Scientific American.  If memory
serves, the primary subject of the article was Erno Rubik and the
Hungarian School of Architecture.  The particular item I am looking for
is a puzzle that waas given to entering students.  They were shown a
picture of a monument and were told to duplicate it using paper and
scissors.  The article contained the picture.  Can anyone give me the
issue that this appeared in?

And while we're at it, does any have a machine-readable copy of the
current rules for Eleusis?  If not, the issue they last appeared in.

          Thanks much,
          Paul <Schauble -at MIT-Multics>

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU 20 Jan 87 15:05:42 EST
Date: 20 Jan 87 14:47:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: recreational math
To: "cube-lovers" <cube-lovers@mit-ai>
cc: gerritsen,mailer!<fuzzy@ardec>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>
BOOKS:  Singmaster is editing a series of books for Oxford University Press on
Recrreational Mathematics and is requesting input on the following:
1) " I (Singmaster) have embarked on a project to find the sources of classical
problems in recreational mathematics.  .....
    The initial object of this project was to produce a book of sources,
translated into english with annotation, for .... However, it now appears that
the first stage must be the prpearation of an annotated bibliography of the
material.  ...  draft of paper which outlines the project and some of the
material is available.  I would be delighted to hear from anyone interested in
this project, particularly anyone able to provide info."
2)  "I am also compiling a list of mathematical monuments and have a draft
article on this."

ADDRESS;  DAVID SINGMASTER, POLYTECHNIC OF THE SOUTH BANK, LONDON, SE1 OAA, UK

>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
BOOKS IN SERIES SO FAR:
1)  MATHEMATICAL BYWAYS IN AYLING, BEELING, AND CEILING by Hugh ApSimon,  128
pages; 30 illus,  853201-6, $10
2) THE INS AND OUTS OF PEG SOLITAIRE by John Beasley, 300 pages; 571 illus,
853203-2,  $17
3) RUBIK'S CUBIC COMPENDIUM by Erno Rubik et al, 200 pages, 183 illus, 853202-4,
$15
CONTENTS:  Intro: the fascination of rubik's cube - david singmaster, 1. in play
-rubik, 2 the art of cubing - varga, 3. restoration methods and table of
processes - keri, 4. mathematics - keri & varga, 5. the universe of the cube -
marx, 6. my fingers remember - vekerdy, 7. afterword - singmaster, bibliography
& index.
4)  SLIDING PIECE PUZZLES by Edward Horden - in preparation.
AVAILABLE FROM:  Science and Medical Marketing Manager, OXFORD UNIVERSITY PRESS,
200 Madison ave, NY, NY 10016,  212/679-7300
ADD $1.50 for shipping
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

from  <beck@ardec-lcss>
.........................................
------

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU 22 Jan 87 08:39:00 EST
Date: 22 Jan 87 08:25:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: magic variants
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>
1.  My crack puzzle team has added four squares to "MAGIC".  This puzzle has
much more variability in making planar patterns (I count doubled up squares) ,
while retaining the same flavor as the original puzzle.  Some examples:

     XXXXXX            XXXX              XXXX                     XXXX
     XXXXXX            XXXX              X    X                     XXXX
                            XXXX             X    X                     XX
                                                XXXX                     XX

Has anybody else changed the number of squares?  

2.  Has anybody speculated on Rubik's next puzzle based on this hinge mechanism?
 My puzzle team thinks a equilateral triangles instead of squares has potential.
 Any comments?

<beck@ardec-lcss>
..............................
------

Received: from PROPHET.BBN.COM (TCP 20026200117) by AI.AI.MIT.EDU 23 Jan 87 14:11:26 EST
Date:     Fri, 23 Jan 87 14:06:33 EST
From:     Bernie Cosell <cosell@prophet.bbn.com>
To:       cube-lovers@ai.ai.mit.edu
cc:       jr@prophet.bbn.com, beeler@prophet.bbn.com, alatto@prophet.bbn.com
Subject:  Postscript on the Oxford RecMath series

I just talked to Oxford and the Rubik's book is expected any moment now,
but is apparently not yet available (in the states, at least).  The
fourth in the series (sliding piece puzzles) is expected to be available
in April or May.

  /Bernie\

ps, the number for phone orders is 201-796-8000 /b

Bernie Cosell                       Internet:  cosell@bbn.com
Bolt, Beranek & Newman, Inc         USENET:    bbnccv!bpc
Cambridge, MA   02238               Telco:     (617) 497-3503

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU 23 Jan 87 16:06:52 EST
Date: 23 Jan 87 15:52:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: MEFFERT CORRECTION
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>


RE:  MEFFERT  (this is a duplicate, I forgot to give his address)

I received amil from meffert yesterday, 1/22/87.  He says "he will be releasing
later this year the 'I.Q. DIE" ( which is the Skewb with Die markings on the
corner pieces), and a GAME called 'KING/ACE (which uses the Pyraminx tetrahedron
dercorated in the 4 card suits and truncated by removing the four apexes.  The
game is similar to black jack.  They are  available for US$25 each including
registered airmail postage.  He also has standard Skewbs available for US$16.

The mailing included a 2-sided glossy color flyer with pictures of various
puzzles.  The ones new to me are:  The Crystal Ball -  it looks like a
Babylonian Tower on a sphere with the capablity to move groups of balls at one
time;  Space Grenade -   a cylindrical version of the crystal ball.

PRICEWELL (FAR EAST) LIMITED

*business address* 
EXCELLENTE COMMERCIAL BLDG (15TH FLOOR)
456 JAFFE ROAD
HONG KONG

*postal address*
POB 31008 CAUSEWAY BAY,  HONG KONG

<beck@ardec-lcss>
....................................
------

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU 23 Jan 87 16:51:04 EST
Date: 23 Jan 87 15:42:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: meffert
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>



RE:  MEFFERT

I received amil from meffert yesterday, 1/22/87.  He says "he will be releasing
later this year the 'I.Q. DIE" ( which is the Skewb with Die markings on the
corner pieces), and a GAME called 'KING/ACE (which uses the Pyraminx tetrahedron
dercorated in the 4 card suits and truncated by removing the four apexes.  The
game is similar to black jack.  They are  available for US$25 each including
registered airmail postage.  He also has standard Skewbs available for US$16.

The mailing included a 2-sided glossy color flyer with pictures of various
puzzles.  The ones new to me are:  The Crystal Ball -  it looks like a
Babylonian Tower on a sphere with the capablity to move groups of balls at one
time;  Space Grenade -   a cylindrical version of the crystal ball.

<beck@ardec-lcss>
....................................
------

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU 18 Feb 87 08:14:06 EST
Date: 18 Feb 87 07:51:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: new magic version
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>



RE:  New versions of MAGIC

1.  The 2/14/87 edition of the "Toy & Hobby World's Show (ie, the NY Toy  Show)
Daily" announced the planned 1987 additions to the MAGIC line.  It said "Rubik's
MAgic puzzle, which debuted in October, 1986, is updated with a Masters Edition
puzzle, Unlink The Rings, featuring 12 panels and multi-color, multi-graphic
designs on a silvery Mylar-foil background.

Rubik's Magic Strategy Game features the colors of the original puzzle in a
tic-tac-toe game; the difference is that the pieces are colored black on one
side and silver on the other, and players can flip opponents' pieces before
making a move."

2.  As previously noted on this board it is fairly simple to make your own 12
piece MAGIC.   I am using MAcDraw to experiment with graphic patterns.  I have
not decided on which graphic(s) I like best yet.

3.  As an aside there are "CLONES"  of MAGIC around and Matchbox is prosecuting.
 I am not aware of exactly what is protected so using a different number of
squares or different graphics might be legal.

<beck@ardec-lcss>

............
------

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU 19 Feb 87 08:36:06 EST
Date: 19 Feb 87 08:18:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: magic construction
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>



RE: CONSTRUCTION

Since I have suggested that people might want to take their MAGICs apart I have
prepared the following directions.  I would appreciate comments as to their
clarity and completeness.  

...................................
For those of you who would like to take MAGIC apart and then put it back
together here are some hints.  First get out your tools, a heavy duty paper clip
or a nut pick will do (black electrical tape is helpful for keeping the strings
in place when putting it back together) and then pull the string over the corner
of a square (strings do break, the weak point is the crimp so minimize the
pulling and stretching you do by the crimp also when you reassemble put the
crimp in the middle of a long channel).  Keep doing this until the puzzle is
completely disassembled. 

If you failed to take notes you may have missed the following.  The loops of
string (they are actually nylon fishline and they are redundant, ie, each path
is taken by two strings, with 16 strings in all) are threaded through the
channels, one set of strings takes the long path on the front face and the other
set of strings takes the short path on the front face (the opposite is true on
the back face OR adjacent square) with both sets of strings going in the same
direction on the same face.    Thus the strings on the front faces are
perpendicular to those on the back face of the same square.   NOTE:  The strings
are not really redundant.  They are placed to maximize lateral stability (twist
of the squares).  This is done by having the strings (there are two) of a given
channel routing form the same sandwiching order where they cross over to the
next square.  The string that uses the long channel and the string that uses the
short channel cross at separarte points.  Each string criss crosses itself at
this point (making 4 string segments at the cross over point) with one part of
itself in the NE channel and its other part in the NW channel.  The stability is
gained by having the NE going string sandwiched between the NW going string (or
vice versa) for both crossover points, ie , sets of strings.   Both patterns
shown below are used on the same set of three squares (THIS UNIT IS CALLED A
TRIPLET.).  

               string #1 in                    string #2 in
        SHORT channel ON TOP       LONG channel ON TOP 
             for squares 1&3              for squares 1&3
               ---- ---- ----             ---- ----  ----
              |/ \  |   / \|/ \   |            |   / \|/ \   |   / \|   TRIPLET
HAS BOTH
              |\  \ | /   /|\   \ |   AND   | /   /|\  \  | /   /|   STRING
PATTERNS
              | \  \|/   / |  \  \|            |/   / |  \  \|/   / |
              |  \ /|\ /   |   \ /|            |\ /   |   \ /|\ /   |
               ---- ---- ----             ---- ----  ----

After having made two triplets there will be two squares free.  They are used to
join the triplets.  Place one of this extra squares between the two triplets,
ie, where the "AND" is in the diagram above and thread the strings through the
channels as if this square was the middle square of a triplet (REMEMBER THAT the
STRINGS GO ONLY ONE WAY ON each face OF A SQUARE).  THEREFORE, THE ENDS OF THE
PREVIOUSLY MADE TRIPLETS WILL BE THE ENDS OF THIS NEW TRIPLET ALSO.  THIS WILL
CAUSE THESE ENDS TO HAVE TWICE AS MANY STRINGS AS THE MIDDLE SQUARES OF THE
TRIPLETS AND IN FACT IF YOU LOOK AT MAGIC YOU WILL SEE THAT THE NUMBER OF
STRINGS IN THE CHANNELS ALTERNATES FROM SINGLE DENSITY TO DOUBLE DENSITY, ie,
either 2 or 4.



CUSTOMIZATION OF MAGIC

In the disassembly process an easy thing to do is to break the circularity of
the puzzle by removing one square, leaving a chain of seven squares.  This can
be done by lifting the strings off a single density square.  The square will
come out but its strings will stil be entangled with the puzzle.  You will now
have to temporarily lift strings off the adjacent squares to disentangle them. 
This can be done easily.  You now have a chain of seven squares.  Each hinge can
be manipulated without the constraint of being connected as a loop. A basic
hinge between two squares has the following motions:
NOTE:  The flipping of the pieces changes the direction of the squares as shown
by the arrows.

       POSITION  1                  folded  A            folded  B
________  _______          _________         _________
sq 1 TOP      sq 2 top           sq 1 on bot          sq 2 on bot
  >>>>>>          >>>>>>             sq 2 on top          sq 1 on top
________  _______          _________         _________


      POSITION  2       >>>>>       A unfolded          B unfolded
                                              ________          _________
                                               SQ 1 TOP             SQ 2 TOP
                                                <<<<<<<<               <<<<<<<
                                               ________          _________
                                               ________          _________
                                                SQ 2 TOP             SQ 1 TOP
                                                  >>>>>>>                >>>>>>>
                                               ________          _________

The robustness of this hinge permits the making of all possible planar patterns
that has each square butting up to the edge of another square.


<beck@ardec-lcss>
------

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU 20 Feb 87 08:35:41 EST
Date: 20 Feb 87 08:08:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: rubik's books
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>


I have been reviewing issues of the "CUBIC CIRCULAR" ,a set of which I recently
obatined form Singmaster, and would like references to  information on the
following:

1-   a book Rubik edited called  "A BUVOS KOCKA"  which was to be also done in
english by PENGUIN.

2 -  Rubik's newsletter "RUBIK'S - LOGIC AND FANTASY DIMENSIONS"

3-  "TEN BILLION BARREL" puzzle invented by Gumpei Yokoi.  Possibly a reference
to a published?? solution by Trevor Hutton.  This is supposed to be harder than
the cube and is available for about $6 at Child World in northen NJ.

4 -  "TRILLION" a flat version of 3.

<beck@ardec-lcss>
./..................
------

Received: from ARDEC-AC3.ARPA (TCP 30003004017) by AI.AI.MIT.EDU 10 Apr 87 08:28:52 EDT
Date:     Fri, 10 Apr 87 8:15:46 EST
From:     Peter Beck (LCWSL) <pbeck@ARDEC.ARPA>
To:       cube-lovers@AI.AI.MIT.EDU
Subject:  dc puzzle shops
Message-ID:  <8704100815.aa28501@ARDEC-AC3.ARDEC.ARPA>

 
I am taking a trip to Wash DC the week of April 13 and expect to have some extra
time while there to pursue puzzling.  If anybody out there knows of a good
puzzle store or possible an exhibit I would greatly appreciate the reference.
 
                    THNX
                 Pete Beck   <beck@ardec-lcss>
 
PS In FEb an article was published in TELECRAN , author Henri Leyder, titled
"610 Denkspiele im Regal"  about Marcel Gillen's puzzle collection.  The article
had several black an d white pictures of his puzzle collection.  The article is
in german, which I do not read, so I cannot paraphrase it.
 
..............

Received: from note.nsf.gov (TCP 1202200024) by AI.AI.MIT.EDU 27 Apr 87 16:08:22 EDT
To: cube-lovers@AI.AI.MIT.EDU
Subject: Rubik's Cube
Date: Mon, 27 Apr 87 16:07:17 -0400
From:  "Aaron R. Coles" <acoles@nsf.arpa>
Message-ID:  <8704271607.aa27081@note.note.nsf.gov>

Maybe this is not the approriate place to ask this question, but does anyone
out there know where I can purchase a Rubik's Cube Revenge from?  I would
appreciate any help.

Thanks


Received: from nrl-aic.ARPA (TCP 3200200010) by AI.AI.MIT.EDU 27 Apr 87 16:53:27 EDT
Return-Path: <hoey@nrl-aic.ARPA>
Received: Mon, 27 Apr 87 16:50:36 edt by nrl-aic.ARPA id AA20059
Date: 27 Apr 1987 16:47:22 EDT (Mon)
From: Dan Hoey <hoey@nrl-aic.ARPA>
Subject: Rubik's Cube
To: "Aaron R. Coles" <acoles@nsf.arpa>
Cc: Cube-Lovers@ai.ai.mit.edu
Message-Id: <546554843/hoey@nrl-aic>

If you happen by the Boston area, you can get Rubik's Revenge at Games
People Play in Cambridge.

A harder problem is to get an ordinary magic cube.  I haven't seen one
for sale in years.

Dan

Received: from BFLY-VAX.BBN.COM (TCP 20026200235) by AI.AI.MIT.EDU 27 Apr 87 17:56:43 EDT
To: Dan Hoey <hoey@nrl-aic.ARPA>
cc: "Aaron R. Coles" <acoles@note.nsf.gov>, Cube-Lovers@ai.ai.mit.edu, 
    dm@bfly-vax.bbn.com
Subject: Re: Rubik's Cube
In-reply-to: Your message of 27 Apr 1987 16:47:22 EDT (Mon).
	     <546554843/hoey@nrl-aic>
Date: 27 Apr 87 17:55:18 EDT (Mon)
From: dm@bfly-vax.bbn.com


> A harder problem is to get an ordinary magic cube.  I haven't seen one
> for sale in years.

Yard sales.  Even if you don't find one for sale at the sale, you can
probably ask the person running the sale if they have one they
wouldn't mind selling.

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU 28 Apr 87 08:10:41 EDT
Date: 28 Apr 87 08:06:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: 
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>
From:	CLSTR1::SYSTEM       27-APR-1987 15:30
To:	CLSTR1::BECK
Subj:	Undeliverable mail

 ----Transcript of session follows----
"mit.ai" is an unrecognized hostname/address

 ----Unsent message follows----
Date: 27 Apr 87 15:18:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: rubiks square
To: "cube-lovers" <cube-lovers@mit.ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>


HI CUBE-LOVERS, 

I understand that the "Puzzle Exhibition" tour has been changed and that it
won't be coming to MIT until the fall.  Does anybody have more information.  

One of my co-workers advised me that she saw a puzzle called "Rubik's SQUARE",
not magic, in the San Jose airport gift shop.  Does anybody know what this is
and where it can be obtained or even a picture seen.

I have been working on alternate designs for the 8 square magic with some
success.  If anybody out there is also doing the same I would like to talk, swap
designs, etc.   If anybody has suggestions for alternate designs I would
appreciate receiving them.

The Future is Puzzling and Cubing is Forever,

Pete beck                  <BECK@ARDEC-LCSS>
..................................

------
------

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU 28 Apr 87 08:30:40 EDT
Date: 28 Apr 87 08:28:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: availability of cubes
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>


RE:  SEARCH FOR CUBES

1.  The large chains are slowly clearing out the surplus stocks of cubes.  The
best for a revenge is "THRIFT DRUG" probably around $5, also try KAYBEE.  If
this fails send me your address and I will send you one a competitive price.

2.  As I have previously posted I am also trying to buy up surplus cubes and
make them available to cube collectors.  If anybody would like a list send me
your name and address and I will mail it to you.  If you are looking for a
particular cube ask for it I will tell you where to get it.  I correspond with
Singmaster, Gillen, Helm, Bandelow, Hess, Cecil Smith, Wally Webster about
cubes.

3.  If anybody knows of a good source of cubes a wholesale prices I would
appreciate the reference.

................................................................
>>>>>>  The Future is Puzzling and Cubing is Forever,  <<<<<<<<

Pete beck                  <BECK@ARDEC-LCSS>
..................................

------

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU 28 Apr 87 14:50:45 EDT
Date: 28 Apr 87 14:32:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: RUBIK'S "SQUARE"
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>


THIS MAYBE A DUPLICATE - I AM HAVING TROUBLE WITH MY MAILER

HI CUBE-LOVERS, 

I understand that the "Puzzle Exhibition" tour has been changed and that it
won't be coming to MIT until the fall.  Does anybody have more information.  

One of my co-workers advised me that she saw a puzzle called "Rubik's SQUARE",
not magic, in the San Jose airport gift shop.  Does anybody know what this is
and where it can be obtained or even a picture seen.

I have been working on alternate designs for the 8 square magic with some
success.  If anybody out there is also doing the same I would like to talk, swap
designs, etc.   If anybody has suggestions for alternate designs I would
appreciate receiving them.

The Future is Puzzling and Cubing is Forever,

Pete beck                  <BECK@ARDEC-LCSS>
..................................

------

Date: Tue,  5 May 87 17:19:21 EDT
From: Alan Bawden <ALAN@AI.AI.MIT.EDU>
Subject:  TOC Seminar -- Akos Seress -- Thursday May 28, 1987
To: CUBE-LOVERS@AI.AI.MIT.EDU
Message-ID: <195855.870505.ALAN@AI.AI.MIT.EDU>

Mathematically inclined Cube Hackers in the Boston area might find the
following seminar interesting.  All I know about this is what I read here
in the abstract.  (I'll bet its been a while since anyone on this list did
any serious thinking about the Cube as a permutation group...)


			DATE: Thursday, May 28, 1987
			 TIME: Refreshments: 3:45PM
			      LECTURE:  4:00PM
			      PLACE: NE43-512A

			  PERMUTATION GROUPS IN NC

				 AKOS SERESS
			Mathematical Institute of the
			Hungarian Academy of Sciences

Given a permutation group G on an n-element set A by a list of generators, we
present NC-algorithms (parallel algorithms using (log n)^c time and
n^c processors) for basic permutation group manipulation (membership
testing, order).  These problems have been suggested by Cook and McKenzie to
be LOGSPACE-complete for P and therefore not in NC unless NC=P.
    We shall outline previous work by Luks on the subject and focus on the
key problems left open by Luks' 1986 FOCS paper.  In particular, we shall
discuss in detail, how to construct in NC any permutation from a given set of
generators of the symmetric group.
    The presentation will be elementary, although the analysis of the
algorithms depends in several ways on consequences of the classification of
finite simple groups.
    Our methods have sequential consequences as well.  We obtain algorithms
for basic permutation group management with O(n^4(log n)^c) running
time, improving one order of magnitude from the best prevously known results
(Knuth, Babai, and Jerrum).
    This is joint work with Laszlo Babai and Eugene M. Luks.

Host:  Professor David Shmoys

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Date: Tue, 5 May 87 17:56:58 PDT
From: long@oliveb.atc.olivetti.com (Tom Long)
Message-Id: <8705060056.AA29561@oliveb.ATC.OLIVETTI.COM>
To: ALAN@ai.ai.mit.edu, CUBE-LOVERS@ai.ai.mit.edu
Subject: Re:  TOC Seminar -- Akos Seress -- Thursday May 28, 1987


       Akos,
Ez egy baratomnaatk a gepjen dolgozom.  Metg vagyokgg vagyok lepodve hogy mas
magyar is itten ir.  Te honan irs? En Californiaban vagyaok.
iIde irhatc viszasza.  A rubik kockarol beszeltel? aA z iskolaban omost
tanulok a permutation-okrol. Nagyon erdekesnek hallatcik.
				irIrjal Vissza,
					Jeno

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Date: Tue, 5 May 87 22:23:43 edt
From: Bernard S. Greenberg <bsg@scrc-pegasus>
To: cube-lovers%ai.ai.mit.edu@stony
Subject: A magyar kocka

    Date: Tue, 5 May 87 17:56:58 PDT
    From: long@oliveb.atc.olivetti.com (Tom Long)
    Message-Id: <8705060056.AA29561@oliveb.ATC.OLIVETTI.COM>
    To: ALAN@ai.ai.mit.edu, CUBE-LOVERS@ai.ai.mit.edu
    Subject: Re:  TOC Seminar -- Akos Seress -- Thursday May 28, 1987


       	   Akos,
    Ez egy baratomnaatk a gepjen dolgozom.  Metg vagyokgg vagyok lepodve hogy mas
    magyar is itten ir.  Te honan irs? En Californiaban vagyaok.
    iIde irhatc viszasza.  A rubik kockarol beszeltel? aA z iskolaban omost
    tanulok a permutation-okrol. Nagyon erdekesnek hallatcik.
				    irIrjal Vissza,
					    Jeno


1. I refuse to deal with something so full of overstrikes and corrections.
   How do you expect anyone to make sense out of it?

2. "That way, mate.  Two blocks down and to the left."


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	id AA15491; Wed, 6 May 87 14:54:29 PDT
Date: Wed, 6 May 87 14:54:29 PDT
From: long@oliveb.atc.olivetti.com (Tom Long)
Message-Id: <8705062154.AA15491@oliveb.ATC.OLIVETTI.COM>
To: ACW@waikato.s4cc.symbolics.com, ALAN@ai.ai.mit.edu,
        CUBE-LOVERS@ai.ai.mit.edu, long@oliveb.atc.olivetti.com
Subject: Re:  TOC Seminar -- Akos Seress -- Thursday May 28, 1987

Your progtranslation of dthe the message is comendable.  Yoyu did fair;luy well.
Did Adoskos write from your machine, or is there a link to Hungary?
The later I doubt.  I am a nat6itive speaker th,thougyh bortn here.

tThe reason for the simple language is that I did not know
the vocabularythere wasw no need to 
use anything more complex.  IWhile I wasw not looking for a pen-pal,
I was intderested that there was an other Hungarian on the system,
having never seen it befor.





Thanks for your repy.


Date: Thu, 14 May 87 18:24:57 EDT
From: Alan Bawden <ALAN@AI.AI.MIT.EDU>
Subject:  TOC Seminar--Adi Shamir--Friday, May 22, 2:00PM
To: CUBE-LOVERS@AI.AI.MIT.EDU
Message-ID: <200230.870514.ALAN@AI.AI.MIT.EDU>

This one is even better than the last seminar announcement I forwarded to
this list!  (And -this- time, REMEMBER:  replying to this message will
-not- send mail that Shamir will get; it will only send everyone on
Cube-Lovers a piece of junk mail.)


		       DATE: Friday, May 22, 1987
                       TIME: Refreshments: 1:45PM
                             Lecture:      2:00PM
		      PLACE: NE43-512A

			    HOW TO SOLVE THE CUBE

				 Adi Shamir
				Applied Math
		       The Weizmann Institute, Israel

   Given k generators for a permutation group G, it is easy to verify that a
permutation belongs to G but NP-complete to find a short representation of
the permutation as a product of the generators.  In this talk we describe a
new algorithm for computing the shortest representation which significantly
improves the time/space complexities of previous algorithms.  The new
algorithm is particularly interesting in the context of Rubik's cube since it
makes it possible to solve previously intractable problems such as finding
the shortest sequence of moves which fixes a given state or the optimal
subroutine for permuting certain subcubes, in just 2^40 time and 2^20 space,
compared to 2^80 time in previous algorithms.

Host:  Prof. Ron Rivest

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Date: Wed, 20 May 87 10:05:20 CDT
From: jknox@ngp.utexas.edu (John W. Knox)
Posted-Date: Wed, 20 May 87 10:05:20 CDT
Message-Id: <8705201505.AA00538@ngp.utexas.edu>
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	id AA00538; Wed, 20 May 87 10:05:20 CDT
To: ALAN@ai.ai.mit.edu, CUBE-LOVERS@ai.ai.mit.edu
Subject: Re:  TOC Seminar--Adi Shamir--Friday, May 22, 2:00PM



Date: Wed, 27 May 87 16:34:04 EDT
From: Alan Bawden <ALAN@AI.AI.MIT.EDU>
Subject:  Shamir's talk really was about how to solve the cube!
To: CUBE-LOVERS@AI.AI.MIT.EDU
Message-ID: <205924.870527.ALAN@AI.AI.MIT.EDU>

Here is a rough sketch of Shamir's algorithm, as he presented it at the
talk last Friday.  The fact that I have typed this in does not
-necessarily- indicate a willingness on my part to supply any further
details.  Nor do I guarantee that my description will enable you to
correctly reconstruct the algorithm, although I tried to make it
comprehensible.

I think Shamir gave credits for a couple of graduate students of his for
some of this, but I didn't make a note of their names.  It is always nice
to give people proper credit...

You are given a scrambling of the cube, and you want to know if the cube
can be restored in 4*N quarter twists.  (Shamir is a half-twister, but I in
this message I will rephrase everything in quarter-twist terms.  It makes
little difference, the algorithm applies equally well to any set of
generators.)

Represent a permutation as a vector that simply lists the values of the
permutation.  That is, if the permutation sends 0 to 7, 1 to 2, 2 to 4, 3
to 1, etc., then the vector [7, 2, 4, 1, ...] represents it.

We will be ordering permutations in "dictionary order".  That is, a
permutation sigma is less than a permutation tau just in case there exists
an i such that

          sigma(i) < tau(i) and for all j < i, sigma(j) = tau(j).


We start by generating a list of all permutations generated by N quarter
twists.  The algorithm requires space to store several datastructures
proportional to the size of this list.  (If N=5, this list has 93840
elements.  Its size is about 10^N for the cube group in quarter twists.
For an arbitrary group and generators it will be exponential in N.)

Now what we would like to do is generate -and- -sort- the list of all
permutations generated by 2*N quarter twists.  We could do this by simply
multiplying all possible pairs of elements from our list, and then sorting
again, but this generates an absurdly large list, that it takes an absurd
amount of effort to sort.

The trick is to generate this list of products both incrementally and
already sorted!  This gives us the ability to ask for the -next- element of
the list, which is exactly what we need:  Given two such permutation-list
generators we can easily scan through both lists to see if any element
occurs in both lists (using an algorithm that the reader can easily
reconstruct).  Thus we can build one generator for the list of all
positions 2*N twists away from solved, and another for positions 2*N twists
away from the given one, and if we find an element on both lists, then we
have a 4*N twist solution.

OK, so how do we construct this magic generator?

First we take the list of length N permutations and make it into a tree.
There will be one leaf for each permutation in the list.  All permutations
that share a common prefix will share a common internal node in the tree.
For example, given the permutations:

                              [0, 1, 2, 3, 4]
                              [0, 1, 3, 2, 4]
                              [1, 0, 3, 4, 2]
                              [1, 3, 0, 4, 2]
                              [2, 3, 1, 4, 0]

we get the tree:

                                      [0, 1, 2, 3, 4]
                                     /
                                    /2
                         [0]--[0, 1]
                        /   1       \3
                       /             \
                      /               [0, 1, 3, 2, 4]
                     /
                    /0        [1, 0, 3, 4, 2]
                   /         /
                  /         /0
                []-------[1]
                  \  1      \3
                   \         \
                    \2        [1, 3, 0, 4, 2]
                     \
                      \
                       \
                        \
                         [2, 3, 1, 4, 0]

Now consider one element of the original list, say 

                         sigma = [2, 3, 1, 4, 0].

We want to find the permutation tau, such that sigma * tau is smallest in
dictionary order.  So, what is the first entry of sigma * tau?  That is,
what is sigma(tau(0))?  Well, looking at sigma we can see that -if-
tau(0) = 4, then sigma(tau(0)) = 0.  Unfortunately their are no
permutations in our list that start with 4, but we can get
sigma(tau(0)) = 1 if tau(0) = 2.  Now there is only one such permutation on
our list, so that must be it:

              tau = [2, 3, 1, 4, 0] (= sigma as it happens).

What about the tau such that sigma * tau is -second- smallest?  We have
exhausted permutations with tau(0) = 2, what should we consider next?  Well,
if tau(0) = 0, then sigma(tau(0)) = 2, so we should next consider
permutations that start with 0.  After that, we should do those that start
with 1, followed by those that start with 3.

The exact same argument applies to tau(1).  That is, to minimize the
product, first consider permutations such that tau(1) = 4, followed by
tau(1) = 2, then tau(1) = 0, then tau(1) = 1, and finally tau(1) = 3.

Thus you can see that to generate the sigma * tau products in order, we can
just take tau to be successive leaf nodes in the above tree, where we order
the inferiors of any internal node in the order 4, 2, 0, 1, 3.  It is easy
to generate this ordering given sigma.

Now for -each- permutation sigma in our original list we will be taking a
different walk through the tree using a different ordering of inferiors.
So we maintain a queue of pairs <sigma, tau>, sorted according to
sigma * tau.  When called upon to generate the next element of the
list-of-products, we take the head of the queue (smallest) and return it.
Then we advance to the next tau for the given sigma, and insert the new
pair <sigma, tau'> back into the queue.

Modifying this construction to generate permutations of the scrambled
position, rather than solved is easily accomplished by first composing the
inverse of the scrambling permutation with each element of the list of
length N permutations.  Now we combine each element of this list with each
element of the same tree as above.

Analysis of space and time requirements left to the reader.

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU  2 Jun 87 09:48:40 EDT
Date: 2 Jun 87 09:45:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: PUZZLE TOUR EXHIBITION SCHEDULE UPDATE
To: "cube-lovers" <cube-lovers@mit-ai>
cc: beck
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>


>>>  UPDATE <<<<  to  " PUZZLES OLD AND NEW"  exhibition tour SCHEDULE
    source  Jerry Slocum,  5/87

 LOCATION:
    Craft and Folk Art Museum, LA   Nov. 26, '86 - Feb. 22, 1987
         Hudson River Museum             July 26 - Sept. 27, 1987
             Yonkers, NY    914/963-4550
         MIT Museum                            OCT 22 - JAN 3 1988
             Cambridge, MA   617/253-4444
         Ontario Science Center          Jan. 25 - March 13, 1988
             Toronto, Canada  416/429-4100
         JAPAN TOUR                           APR 28 - SEPT 1988
            (TOKYO, OSAKA, KYOTO, NAGOTA)

RE:   Rodney Hoffman's review posted in the spring.

<BECK@ARDEC-LCSS>
..........................................................


------

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU  3 Jun 87 12:01:50 EDT
Date: 3 Jun 87 11:45:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: request for puzzles sources
To: "cube-lovers" <cube-lovers@mit-ai>
cc: beck
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>

I am interested in obtaining the following puzzles ( references are from
"Sliding Pieces Puzzles" by Ed Horden,  Oxford Univ Press, 1986):
  1 -  Change the Seasons, plate IX
  2 -  Inversion, plate IX
  3 -  Great Gears, plate X

I would like to trade for them, if possible.   A source where they can be
purchased would also be appreciated.   Thanks for any and all help.

The Future is Puzzling but Cubing is Forever,

Pete beck                  <BECK@ARDEC-LCSS>
..................................
------

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU  4 Jun 87 08:25:49 EDT
Date: 4 Jun 87 08:17:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: MAZES
To: "cube-lovers" <cube-lovers@mit-ai>
cc: beck
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>


I would like to hear from anybody who has visited a "MAZE" amusement center or
read an article, hopefully with pictures on same.   

I understand that there are 20 such centers in Japan and that the "MAZE PRODUCTS
COMPANY" (a NOB enterprise) is going to bring it to the USA.  

Nob's centers have mazes designed by a New Zealander, Stuart Landsborough and
puzzle shops that sell Nob's puzzles.  There was also an article in the dec 86
Sci Amer, pg140 on Labryinths which referenced a book called "Celebration of
Mazes" available from Minotaur Designs, 247 Montgomery st, Jersey City, NJ 07302
for $9.  Has anybody read this?  Are these phenomina the same?  Are the MAZES
coming !!!

Pete beck                  <BECK@ARDEC-LCSS>
..................................

------

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	(contact usenet@ucbvax.Berkeley.EDU if you have questions)
Date: 4 Jun 87 16:35:54 GMT
From: fluke!ssc-vax!cxsea!blm@beaver.cs.washington.edu  (Brian Matthews)
Organization: Computer X Inc.
Subject: Look for two books by Dmitri Borgmann
Message-Id: <2100@cxsea.UUCP>
Sender: cube-lovers-request@ai.ai.mit.edu
To: cube-lovers@ai.ai.mit.edu

I'm looking for information about the books _Beyond Language_ and
_Language on Vacation_ by Dmitri Borgmann.  I can't find them in any
of the various Books in Print, so they're either no longer in print,
or from a small publisher.  In any event, I would appreciate it if
someone could send me the following information:

  1. The correct spelling of the titles and the author's name (if I
     don't have them correct above).

  2. The publisher, and the publisher's address if it's available in
     either of the books.

  3. The ISBN.

Thanx for any information anyone can provide!

-- 
                                                +---------+
Brian L. Matthews                               | P L A N |
...{mnetor,uw-beaver!ssc-vax}!cxsea!blm         | A H E A |
+1 206 251 6811                                 +--------D+
Computer X Inc. - a division of Motorola New Enterprises

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	id AA20441; Mon, 8 Jun 87 11:51:27 EDT
From: <eric@ATHENA.MIT.EDU>
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Message-Id: <8706081551.AA19656@THESEUS.MIT.EDU>
To: cube-lovers@ai.ai.mit.edu
Reply-To: eric@athena.mit.edu
Subject: Info on Dmitri Borgmann books
Date: Mon, 08 Jun 87 11:51:05 EDT

Brian Matthews asked for information about Dmitri Borgmann books.  Here it is.
Dmitri Borgmann published three books:

	Language On Vacation (An Olio of Orthographical Oddities)
		Copyright 1965.  Published by Charles Scribner's Sons

	Beyond Language (Adventures in Word and Thought)
		Copyright 1967.  Published by Charles Scribner's Sons

	Curious Crosswords (edited and annotated by Borgmann)
		Copyright 1970.  Published by Charles Scribner's Sons

_Language on Vacation_ is a paperback.  _Beyond Language_ is a hardcover.
_Curious Crosswords_ is a large format paperback.  So far as I know, all three
have been out of print for years and it's unlikely they'll be reprinted.
But, there's good news -- you can get them all as follows:

_Curious Crosswords_ and _Language on Vacation_ are available for $7.00 each
from

	National Library Publications
	Box 73
	Brooklyn, NY 11234

Don't forget to add 10 percent for postage and handling.  I got my copies
from this place and had no trouble with them.  _Beyond Language_ can only
be found by combing used book stores, which I did for several years.  Then
I found two of them!  If you promise to love the book, I'll send you one of
them for $10.00.  Please note that it is missing its cover, but is otherwise
in perfect condition.

If you are a maniac about the peculiarities of language, all three of these
are must-buys.  You should also consider subscribing to the journal "Word
Ways" for $15.00 a year.  It's a quarterly 64-page journal devoted to the
kind of stuff Dmitri Borgmann writes about (weird spellings, words with
greatest number of vowels, pangrams, dictionaries, etc., etc.).  You can
order a subscription from Faith W. Eckler, Spring Valley Road, Morristown,
New Jersey 07960.

By the way, Dmitri Borgmann died last year of a heart attack.  It was a
great blow to logologists everywhere.  "Word Ways" has been running
memorial issues filled with unpublished articles by him.

Hope this helps!

Eric Albert (eric@athena.mit.edu)
12 Abbott Street
Medford, MA 02155
(617) 396-2424




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Date: 21 May 87 01:24:05 GMT
From: philabs!micomvax!musocs!mcgill-vision!mouse@nyu.arpa  (der Mouse)
Organization: McGill University, Montreal
Subject: Re: Repeated words answer
Message-Id: <777@mcgill-vision.UUCP>
References: <1036@theory.cs.cmu.edu>
Sender: cube-lovers-request@ai.ai.mit.edu
To: cube-lovers@ai.ai.mit.edu

In article <1036@theory.cs.cmu.edu>, dld@theory.cs.cmu.edu (David Detlefs) writes:
> [about repeated word sentences]
>	Police!
>	Police police.
....
>	Police police police police police.

> Etc....I have to stop now.  "Police" is becoming a meaningless text
> string...

Indeed.  I find this will happen with any word, if you examine it
enough.  My favorite example is "sock".

					der Mouse

				(mouse@mcgill-vision.uucp)

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU 11 Jun 87 08:52:17 EDT
Date: 11 Jun 87 08:37:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: INFO RQST ON DOMINOES
To: "cube-lovers" <cube-lovers@mit-ai>
cc: beck
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>


I picked up the following on AI-LIST.  

<beck@ardec-lcss>
.......................
Date: 8 Jun 87 19:46:16 GMT
From: ai!gautier@rsch.wisc.edu  (Jorge Gautier)
Subject: WANTED: references on the game of dominoes
 
I am looking for references on computer implementations of the game of
dominoes.  I suspect there are many variations on the rules for this game,
but any pointers to papers, commercial products, Ph.D. theses :-), etc.
would be much appreciated.  Please reply by mail.
 
Jorge Gautier
gautier@ai.wisc.edu
 
 
.............
------

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Date: 15 Jun 87 15:20:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: DUTCH CLUB
To: "cube-lovers" <cube-lovers@mit-ai>
cc: beck
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>



SUBJECT :  Review of "Cubism For Fun" newsletter; the newsletter of the "Dutch
Cubists Club"


1.   The dutch cubists club is alive and well.   It is probably the only
organized group of people still collecting and distributing information about
Rubik's cube and related combinatorial and geometrical puzzles.  

2.  This spring in order to improve contacts between cubists they started
publishing their newsletter in english (issue #14 dated 3/87).  The newsleter is
distributed free to members of the club.   Membership for 1987 is  US$5.   A
photocopied set of the newsletters, issues 1-13, written in DUTCH is also
available for  US$7.    To order either of these send a POSTAL MONEY ORDER to: 
Anton Hanegraaf,  Heemskerkstraat,  6662 AL ELST,  The Netherlands.

3.  The table of contents from issue #14 follows (it is 14 double sided folded 8
1/2 x 11 sheets, making a 28 page newsletter):

INTRODUCTION
LETTER FROM THE PRESIDENT - KLAAS STEENHUIS
SECRETATIAL REPORT - GUUS RAZOUX SCHULTZ
DUTCH CUBISTS DAY - ANTON HANEGRAAF
CALL FOR PAPERS
MORE ABOUT RUBIK'S MAGIC - GUUS RAZOUX SCHULTZ
THE ALGORITHM USED BY MARC WATERMAN - ANNEKE TREEP & MARC                       
WATERMAN
COMPUTERS IN SRVICE OF THE CUBE-KNOWLEDGE - KLAAS STEENHUIS
UPPER-TABLE BY HOME COMPUTER - BEN JOS WALBEEHM
SHORT HISTORY OF THE UPPER TABLE - ANTON HANEGRAAF
IMPROVEMENTS TO THE UPPER TABLE
LETTERS TO THE EDITOR
LIST OF MEMBERS

4.  If anybody would like further details please ask!

CUBING IS FOREVER      PETER BECK

<BECK@ARDEC-LCSS>  OR  <PBECK@ARDEC>
....................
------

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Date: 16 Jun 87 13:15:52 GMT
From: umnd-cs!umn-cs!dayton!ems!srcsip!notch@rsch.wisc.edu  (Michael k Notch)
Organization: Honeywell, Inc. Systems & Research Center, Camden, MN
Subject: Give or Take
Message-Id: <148@altura.srcsip.UUCP>
Sender: cube-lovers-request@ai.ai.mit.edu
To: cube-lovers@ai.ai.mit.edu


Have I got an interesting puzzle to at least look at:


           GIVE or TAKE

You can take 45 from 45 and have
a remainder of 45. A trick, yes, but it
can be done.
           Give it a try
             It's fun


There is an obvious answer - You take 45 eggs from 45 chickens and you still
                             have 45 chickens left.
Can you find the other answer?
GOOD LUCK!

-- 
 "Go with the flow, Have a plan, Go with the grain, And ...
    never never let the VAX see you sweat."
					   -G Saunders  07/14/86

USENET:  {ihnp4,mmm,philabs,umn-cs}!srcsip!notch        Michael k Notch
MailSt:	 Honeywell S&RC/SIP/MVT  MN65-2300
         3660 Technology Drive
         Minneapolis, Mn 55418
--

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Date: 17 Jun 87 13:47:50 GMT
From: ihnp4!homxb!houxm!homxc!halle@ucbvax.Berkeley.EDU  (J.HALLE)
Organization: AT&T Bell Laboratories, Holmdel
Subject: Re: Give or Take
Message-Id: <515@homxc.UUCP>
References: <148@altura.srcsip.UUCP>
Sender: cube-lovers-request@ai.ai.mit.edu
To: cube-lovers@ai.ai.mit.edu

In article <148@altura.srcsip.UUCP>, notch@srcsip.UUCP (Michael k Notch) writes:
> 
> Have I got an interesting puzzle to at least look at:
> 
> 
>            GIVE or TAKE
> 
> You can take 45 from 45 and have
> a remainder of 45. A trick, yes, but it
> can be done.
>            Give it a try
>              It's fun
Just get together two score and five gunslingers, only one of whom has two
guns, and take away one of the guns (assuming you survive :-) ).

Received: from ucbvax.Berkeley.EDU (TCP 1200400116) by AI.AI.MIT.EDU 19 Jun 87 06:03:20 EDT
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Date: 19 Jun 87 03:18:06 GMT
From: duke!mps@mcnc.org  (Michael P. Smith)
Organization: Duke University, Durham NC
Subject: Palindromes
Message-Id: <9794@duke.cs.duke.edu>
Sender: cube-lovers-request@ai.ai.mit.edu
To: cube-lovers@ai.ai.mit.edu


I was waiting for someone in the recent discussion on palindromes to
refer to this book but no one did ...
	PALINDROMES & ANAGRAMS  Howard W. Bergerson, Dover 1973.
It has hundreds of palindromic sentences longer than those posted, as
well as palindromic poetry, etc.  No section on palindromes
constructed of palindromic letters, though.
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

Michael P. Smith		"Ungate me, vi, I've met a gnu!"
mps@duke.cs.duke.edu
 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

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Date: 18 Jun 87 22:45:19 GMT
From: tektronix!tekgen!tektools!gvgpsa!friday!kirkg@ucbvax.Berkeley.EDU  (Kirk M Gramcko)
Organization: Grass Valley Group Professional Video Div, Grass Valley
Subject: Re: Give or Take
Message-Id: <120@friday.UUCP>
References: <148@altura.srcsip.UUCP>
Sender: cube-lovers-request@ai.ai.mit.edu
To: cube-lovers@ai.ai.mit.edu

In article <148@altura.srcsip.UUCP> notch@srcsip.UUCP (Michael k Notch) writes:
>
>Have I got an interesting puzzle to at least look at:
>           GIVE or TAKE
>You can take 45 from 45 and have
>a remainder of 45. A trick, yes, but it
>can be done.
>There is an obvious answer - You take 45 eggs from 45 chickens and you still
>                             have 45 chickens left.
>Can you find the other answer?

I have to disagree with your obvious answer for 2 reasons.

	1.  Using that kind of logic there are innumerable answers such as:

	You take 45 donuts from 45 bakers and you still have 45 bakers!

	2.  Your answer is not correct if you consider eggs to be chickens in
their primary stage of development.  Then you really have 90 chickens!!


Here is another puzzle to solve (perhaps a bit too easy):

	How can the following equation be correct?

	80 - 50 = 0

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Date: 18 Jun 87 13:12:20 GMT
From: mtune!mtgzz!mtuxo!gertler@RUTGERS.EDU  (D.GERTLER)
Organization: AT&T, Middletown NJ
Subject: Re: Give or Take
Message-Id: <36@mtuxo.UUCP>
References: <148@altura.srcsip.UUCP>
Sender: cube-lovers-request@ai.ai.mit.edu
To: cube-lovers@ai.ai.mit.edu

In article <148@altura.srcsip.UUCP>, notch@srcsip.UUCP (Michael k Notch) writes:
< Have I got an interesting puzzle to at least look at:
<            GIVE or TAKE
< You can take 45 from 45 and have
< a remainder of 45. A trick, yes, but it
< can be done.
<            Give it a try
<              It's fun
< There is an obvious answer - You take 45 eggs from 45 chickens and you still
<                              have 45 chickens left.
< Can you find the other answer?
< GOOD LUCK!
Simple!  Your friend has $45.  You TAKE his $45.  You have $45 remaining.
(Of course, you're out a friend.)  :-)

-- 

-Don Gertler		UUCP:	...!mtuxo!gertler

   "If this works, we'll eat like kings."

Date: Fri, 19 Jun 87 13:10:57 EDT
From: Alan Bawden <ALAN@AI.AI.MIT.EDU>
Subject:  recent random mail
To: CUBE-LOVERS@AI.AI.MIT.EDU
Message-ID: <216830.870619.ALAN@AI.AI.MIT.EDU>

The recent spate of messages to Cube-Lovers having to do with wordplay,
numerology, etc. was caused by an automatic feed from some usenet newsgroup
for puzzle fans.  This was set up without without asking
Cube-Lovers-Request's permission or even informing us that it had been
done.  I believe that I have now arranged for the feed to be terminated, so
there shouldn't be any more such messages.

If you were actually interested in some of the things that were sent, you
can probably arrange to subscribe to the newsgroup directly.  But please
don't ask me how to do it, I haven't even been able to get anyone to tell
me the -name- of the newsgroup in question!

Received: from MCC.COM (TCP 1200600076) by AI.AI.MIT.EDU 22 Jun 87 12:57:33 EDT
Date: Mon 22 Jun 87 11:52:32-CDT
From: Clive Dawson <AI.CLIVE@MCC.COM>
Subject: Re: Give or Take
To: mtune!mtgzz!mtuxo!gertler@RUTGERS.EDU
cc: cube-lovers@AI.AI.MIT.EDU
In-Reply-To: <36@mtuxo.UUCP>
Message-ID: <12312563719.54.AI.CLIVE@MCC.COM>

I don't know how this nonsense got started, but I'll gladly contribute. :-)

	How can you start with 4, then take away 4, and have 8 remain?

Answer in a day or two.

Clive
-------

Received: from MCC.COM (TCP 1200600076) by AI.AI.MIT.EDU 22 Jun 87 13:08:27 EDT
Date: Mon 22 Jun 87 12:02:02-CDT
From: Clive Dawson <AI.CLIVE@MCC.COM>
Subject: Re: recent random mail
To: ALAN@AI.AI.MIT.EDU
cc: cube-lovers@AI.AI.MIT.EDU
In-Reply-To: <216830.870619.ALAN@AI.AI.MIT.EDU>
Message-ID: <12312565446.54.AI.CLIVE@MCC.COM>

Oh, so those random messages were not generated by cube lovers?!
Then I guess I should modify my last puzzle to make it more
relevant to cubes (and thereby provide a hint):

	How do you start with 8, then take away 8, and have 24 remain?

Clive
-------

Received: from note.nsf.gov (TCP 1202200024) by AI.AI.MIT.EDU 22 Jun 87 16:57:39 EDT
To: Clive Dawson <AI.CLIVE@MCC.COM>
cc: mtune!mtgzz!mtuxo!gertler@RUTGERS.EDU, cube-lovers@AI.AI.MIT.EDU
Subject: Re: Give or Take 
In-reply-to: Your message of Mon, 22 Jun 87 11:52:32 -0500.
             <12312563719.54.AI.CLIVE@MCC.COM> 
Date: Mon, 22 Jun 87 16:53:22 -0400
From: Aaron Coles <acoles@note.nsf.gov>
Message-ID:  <8706221653.aa29184@note.note.nsf.gov>

In response to your question:

    How can you start with 4, then take away 4, and have 8 remain?

Here is my answer:  First you start off with 4 cows three in each row,
                    then you take 4 cows aways, and then you have
                    8 cows remaining.


Received: from nrl-aic.ARPA (TCP 3200200010) by AI.AI.MIT.EDU 24 Jun 87 03:18:32 EDT
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Received: Wed, 24 Jun 87 03:12:52 edt by nrl-aic.ARPA id AA18187
Date: 24 Jun 1987 02:40:53 EDT (Wed)
From: Dan Hoey <hoey@nrl-aic.ARPA>
Subject: Groups of the larger cubes
To: Cube-Lovers@AI.AI.MIT.EDU
Message-Id: <551515254/hoey@nrl-aic>

Last year Rodney Hoffman cited an article by J. A. Eidswick (in the
March 1986 Math Monthly) that develops a general approach to analyzing
several magic polyhedra.  Did anyone else go read this one?  Of
particular interest is Eidswick's analysis of the larger three-
dimensional cubes.  The article shows that the only constraints on
these cubes are the permutation parity constraints implicit in the
generators and the corner and edge orientation constraints we already
know about.  Eidswick shows that this even holds for the ``theoretical
invisible group'', where we imagine that the interior of the magic
N-cube is a magic (N-2)-cube that must be solved simultaneously.  The
solution method he presents is to solve the parity problems by applying
zero or one qtw at each of the floor(N/2) depths, then to work with
commutators (aka mono-ops) to solve the rest of the cube, piece by
piece.

As a supplement to that article, here are the number of positions
G[t](N) of the N^3 magic cube, where t, a subset of {s,m,i},
indicates the set of traits we find interesting:

    s (for N odd) indicates that are working in the Supergroup, and
      so take account of twists of the face centers.
    m (for N > 3) indicates that the pieces are marked so that we
      take account of the permutation of the identically-colored
      pieces on a face.
    i (for N > 3) indicates that we are working in the theoretical
      invisible group, and solve the pieces on the interior of the
      cube as well as the exterior.  I will assume that the M and S
      traits apply to the interior pieces as if they were on the
      exterior of a smaller cube.

A formula for the number of positions is

               2^A (8!/2 3^7)^B (12!/2 2^11)^C (4^6/2)^D (24!/2)^E
    G[t](N) =  ---------------------------------------------------
                                 24^F (24^6/2)^G

The following table gives the values of parameters A-G, depending on
the traits, and on whether N is even or odd.

        Parameter   Traits  (N odd)              (N even)

(Parity)        A =         (N-1)/2              N/2

(Corners)       B = i       (N-1)/2              N/2
                    ~i      1                    1

(Edge centers)  C = i       (N-1)/2              0
                    ~i      1                    0

(Face centers)  D = ~s      0                    0
                    s,i     (N-1)/2              0
                    s,~i    1                    0

(Other cubies)  E = i       (N+4)(N-1)(N-3)/24   N(N^2-4)/24
                    ~i      (N+1)(N-3)/4         N(N-2)/4

(Whole-cube)    F =         0                    1

(Color cosets)  G = m       0                    0
                    ~m,i    (N^2-1)(N-3)/24      N(N-1)(N-2)/24
                    ~m,~i   (N-1)(N-3)/4         (N-2)^2/4

In any case, the size of the group is exponential in a polynomial in N;
the polynomial is cubic if trait "i" is present and quadratic otherwise.
Here is a table of numeric approximations for cubes up to 10^3.

              Traits excluding s
 N     {}        {m}       {i}      {m,i}
 2  3.674e6   3.674e6   3.674e6   3.674e6
 3  4.325e19  4.325e19  4.325e19  4.325e19 
 4  7.401e45  7.072e53  3.263e53  3.118e61 
 5  2.829e74  2.583e90  6.117e93  5.585e109
 6  1.572e116 1.310e148 3.077e170 2.451e210
 7  1.950e160 1.484e208 2.982e253 2.072e317
 8  3.517e217 2.335e289 3.247e388 1.717e500
 9  1.417e277 8.208e372 5.283e529 2.126e689
10  8.298e349 4.007e477 4.041e738 1.032e978

              Traits including s
 N     {s}      {s,m}     {s,i}    {s,m,i}
 3  8.858e22  8.858e22  8.858e22  8.858e22 
 5  5.793e77  5.289e93  2.566e100 2.343e116
 7  3.994e163 3.039e211 2.562e263 1.780e327 
 9  2.902e280 1.681e376 9.293e542 3.740e702 

Enough, then, of what are essentially Eidswick's results.  In my next
message, I plan to produce lower bounds for solving these cubes.

Dan

Received: from nrl-aic.ARPA (TCP 3200200010) by AI.AI.MIT.EDU 24 Jun 87 09:10:18 EDT
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Received: Wed, 24 Jun 87 09:04:34 edt by nrl-aic.ARPA id AA19977
Date: 24 Jun 1987 08:54:48 EDT (Wed)
From: Dan Hoey <hoey@nrl-aic.ARPA>
Subject: Lower bounds for the 3^N cubes
To: Cube-Lovers@AI.AI.MIT.EDU
Message-Id: <551537688/hoey@nrl-aic>

The ability to calculate the sizes of large cube groups prompts me to
generalize the lower-bound machinery we have for magic cubes, to see
how it behaves asymptotically.  The machinery we have uses the three
isomorphic Abelian groups A(N) generated by the three sets of parallel
generators for the N^3 cube.  Since the group of the N^3 cube is a
quotient of the free product of three copies of A(N), we can
upper-bound the number of positions close to SOLVED in the cube group
by the number of positions close to SOLVED in the free product.  This
implies a lower bound for the diameter of the cube group.

One useful tool for group diameter work is the Poincare series.  The
Poincare series of a finitely generated group is the power series p(z)
in which the coefficient of z^d is the number of group elements of
length d.  When the group is finite, the power series is a polynomial.
For our analysis, we will need the Poincare polynomial of A(N).

When N is odd, the analysis is straightforward, since A(N) is the
direct product of (N-1) cyclic groups of order 4.  Let S be the set of
generators for A(N), |S| = 2(N-1) including inverses.  Now suppose we
take a subset T of S.  We can construct a group element F(T) by
multiplying the elements of T together, *except* that when both a
generator G and its inverse G' appear in T we replace them with G^2.
It is easy to see that F is a bijection between the power set of S and
the group A(N).  The interesting thing about F is that the length of
F(T) is |T|.  So the number of length-d elements of A(N) is the number
of d-element subsets of S, and the binomial theorem gives us the
Poincare polynomial of A(N):  p(z) = (z+1)^(2(N-1)).

When N is even, we are in considerably murkier waters.  It's easy
enough in the Cutist analysis I presented on 1 June 1982:  There are
N-1 ways of cutting the cube into two pieces perpendicular to each
axis, and so 2(N-1) generators of A(N), and the analysis proceeds as
above.  But a year later I converted to Eccentric Slabism, and I
suppose I should present that analysis here.  In the Slabist
interpretation, the generators of A(N) are the 2N quarter-turns of
unit-thickness slabs.  But to avoid charging for whole-cube moves, we
must single out a particular slab S0 for which a turn is equivalent to
turning each of the other slabs {S1,S2,...,SN} in the opposite
direction.  The Poincare polynomial for A(N) is
        p(z) = ((z+1) (SUM[0<=i<N/2] C(N-1,i) z^i))^2
where C(N,r) is the binomial coefficient.  This can be shown by
constraining T to include fewer than N/2 of {S1,...,SN} and fewer than
N/2 of {S1',...,SN'} in the previous analysis.  Showing that this is a
length-preserving bijection is tricky, and I will spare you the gory
details, barring pleas to the contrary.  Does anyone know of a closed
form for that sum?

In general, the free product of three copies of A(N) has Poincare
series
    q(z) = p(z)/(3 - 2p(z))
where p(z) is the Poincare series of A(N).  The series q(z) can be
expanded term by term using the technique I sent in on 9 January 1981.
As in that message, we add up every other coefficient through that of
z^d to produce an upper bound on the number of positions of the
move-parity of d reachable in d qtw or fewer, and if this is less than
half the size of the group of the cube we have a lower-bound of d+2 qtw
to solve the cube.  I've done this for the cubes up to 10^3, and I get
the following lower bounds.  The traits are as given in my previous
message.

          LOWER BOUNDS ON GOD'S NUMBER FOR N^3 PUZZLES
      Traits
 N     {}     {m}     {i}   {m,i}     {s}   {s,m}   {s,i} {s,m,i}
 2     11
 3     21                              25
 4     37      44      43      50
 5     59      72      74      87      62      74      79      92
 6     81     102     118     145
 7    111     143     174     218     113     146     181     225
 8    138     183     246     316
 9    175     236     334     435     178     238     343     443
10    208     284     439     580

Asymptotically, the coefficients of q(z) grow by a factor approaching
R, where p(1/R) = 3/2.  This allows us to calculate the rate at which
the free product grows.  Here's a table of the smaller values.

    N    R (exact)                   R (approx)

    2    1/((3/2)^(1/2) - 1)          4.449
    3    1/((3/2)^(1/4) - 1)          9.371
    4    6/((4 + 6^(3/2))^(1/2) - 4) 18.519
    5    1/((3/2)^(1/8) - 1)         19.235
    6                                29.047
    7    1/((3/2)^(1/12) - 1)        29.098
    8                                38.960
    9    1/((3/2)^(1/16) - 1)        38.963
   11    1/((3/2)^(1/20) - 1)        48.828 (for even N>=10, use
   13    1/((3/2)^(1/24) - 1)        58.693  approximation for N+1).
   15    1/((3/2)^(1/28) - 1)        68.558
   17    1/((3/2)^(1/32) - 1)        78.423
   19    1/((3/2)^(1/36) - 1)        88.288
   21    1/((3/2)^(1/40) - 1)        98.153

Clearly R grows proportionally to N, so our asymptotic lower bound will
be somewhere around Log[N](G[t](n)), which is O(N^3/log(n)) for the
theoretical invisible groups (trait i) and O(N^2/log(n)) for the
surface groups.  This is as opposed to Eidswick's upper bounds, which
are O(N^3) and O(N^2), respectively.  So the gap increases, but not
terribly quickly.

It is interesting to compare this with the sort of behavior we see in
the 8-puzzle, 15-puzzle, ..., N^2-1-puzzles, as Jim Saxe suggested to
me many years ago.  The N^2-1-puzzle has (N^2)!/2 positions and 2 to 4
possible moves, so the lower bound based on this sort of counting
argument is O(log((N^2)!)) = O(N^2 log N).  Yet we know that we can put
O(N^2) pieces at a distance of O(N) from their home, so God's number
for the puzzle is O(N^3).  It is pleasant to see that our bounds on the
cubes are tight to within a log factor.

Dan

Received: from ARDEC-AC4.ARPA (TCP 30003004020) by AI.AI.MIT.EDU 29 Jun 87 10:22:30 EDT
Date:     Mon, 29 Jun 87 8:13:39 EDT
From:     Peter Beck (LCWSL) <pbeck@ARDEC.ARPA>
To:       cube-lovers@AI.AI.MIT.EDU
cc:       beck@ARDEC-LCSS.ARPA
Subject:  puzzle availability
Message-ID:  <8706290813.aa06155@ARDEC-AC4.ARDEC.ARPA>



HI CUBE-LOVERS, 

PUZZLE AVAILABILITY:

6/17 MACY'S RAN AN AD IN THE NY TIMES FOR 12 PIECE RUBIK'S MAGIC.  The puzzle
has 5 rings that have to be linked and was advertised for $15.

In Japan, associated with the New Zealand mazes are puzzle shops.  These shops
are creating a demand for puzzles.  A co-worker just back from Tokyo purchased
some puzzles at the "MATSUYA" department store in GINZA.  There is a line of
cast puzzles that cost about $7:  KEY, A-B-C, STAR, S&S, HORSESHOES (see slocums
book).  There is also a line of Ring disentanglement puzzles that cost about $6;
 BROKEN HEART, SWING, "U" RING, DEVIL, POT, LOOP, TRIO RING.  Both puzzle lines
are Bronze colored and come nicely gift boxed.

The Future is Puzzling and Cubing is Forever,

Pete beck                  <BECK@ARDEC-LCSS>
..................................
ps this msg has been delayed to "MAILER" problems with my host.
...........

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU 29 Jun 87 12:08:58 EDT
Date: 26 Jun 87 12:45:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: 
To: "cube-lovers" <cube-lovers@mit-ai>
cc: mailer!<pbeck@ardec>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>
From:	CLSTR1::SYSTEM       26-JUN-1987 12:13
To:	BECK
Subj:	Undeliverable mail

 ----Transcript of session follows----
Mail could not be delivered in 3 days to <beck@ardec-lcss.arpa>

 ----Unsent message follows----
Date: 23 Jun 87 11:48:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: 
To: "cube-lovers" <cube-lovers@mit-ai>
cc: beck
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>
From:	CLSTR1::SYSTEM       23-JUN-1987 11:22
To:	BECK
Subj:	Undeliverable mail

 ----Transcript of session follows----
Mail could not be delivered in 3 days to <cube-lovers@mit-ai>

 ----Unsent message follows----
Date: 19 Jun 87 12:44:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: puzzle avialability
To: "cube-lovers" <cube-lovers@mit-ai>
cc: beck
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>

HI CUBE-LOVERS, 

PUZZLE AVAILABILITY:

6/17 MACY'S RAN AN AD IN THE NY TIMES FOR 12 PIECE RUBIK'S MAGIC.  The puzzle
has 5 rings that have to be linked and was advertised for $15.

In Japan, associated with the New Zealand mazes are puzzle shops.  These shops
are creating a demand for puzzles.  A co-worker just back from Tokyo purchased
some puzzles at the "MATSUYA" department store in GINZA.  There is a line of
cast puzzles that cost about $7:  KEY, A-B-C, STAR, S&S, HORSESHOES (see slocums
book).  There is also a line of Ring disentanglement puzzles that cost about $6;
 BROKEN HEART, SWING, "U" RING, DEVIL, POT, LOOP, TRIO RING.  Both puzzle lines
are Bronze colored and come nicely gift boxed.

The Future is Puzzling and Cubing is Forever,

Pete beck                  <BECK@ARDEC-LCSS>
..................................

------
------
------

Received: from Xerox.COM (TCP 1200400040) by AI.AI.MIT.EDU 27 Jul 87 18:16:37 EDT
Received: from CheninBlanc.ms by ArpaGateway.ms ; 27 JUL 87 15:06:53 PDT
Date: 27 Jul 87 15:06:45 PDT (Monday)
From: Hoffman.es@Xerox.COM
Subject: Puzzle shows
To: CUBE-LOVERS@AI.AI.MIT.EDU
Message-ID: <870727-150653-1010@Xerox>

The traveling puzzle show, "PUZZLES OLD AND NEW" is reviewed in the
Sunday, July 26 New York Times' Arts & Leisure section, page 31.  It has
opened at the Hudson River Museum.

Also of possible interest:  In the same section, page 33, is a review of
the exhibit "SAFE AND SECURE: KEYS AND LOCKS", which is at the
Cooper-Hewitt Museum.

  -- Rodney

Received: from nrl-aic.ARPA (TCP 3200200010) by AI.AI.MIT.EDU 30 Jul 87 15:51:13 EDT
Return-Path: <hoey@nrl-aic.ARPA>
Received: Thu, 30 Jul 87 15:47:50 edt by nrl-aic.ARPA id AA22816
Date: 30 Jul 1987 15:46:10 EDT (Thu)
From: Dan Hoey <hoey@nrl-aic.ARPA>
Subject: Planar positions of Rubik's Magic
To: Cube-Lovers@AI.AI.MIT.EDU
Message-Id: <554672771/hoey@nrl-aic>

    PLANAR POSITIONS OF RUBIK'S MAGIC, THE 8 SQUARE PUZZLE

    by P Beck and D Hoey,  July 1987
       <BECK@ARDEC-LCSS.ARPA> or <PBECK@ARDEC.ARPA>,
       <HOEY@NRL-AIC.ARPA>

This is a catalog of the 96 planar positions of the 8-square Rubik's
Magic puzzle.  The list is based on two rules for positioning the eight
squares.

RULE 1--Placement:  Let the pieces be numbered from 1 to 8.  Any planar
position must consist of squares in the pattern

                ``2x4''       or      ``3x3''

                A B C D                A B C
                H G F E                H E D
                                       G F

where A,B,C,D,E,F,G,H is a cyclical rearrangement of 1,2,3,4,5,6,7,8.
These patterns can also be rotated or reflected.

Both the 2x4 and the 3x3 patterns have eight rotations and reflections,
and there are eight possible assignments of the numbers 1-8 to the
letters A-H.  However, a 180-degree rotation of the 2x4 is equivalent
to a reassignment of the numbers.  So there are only 32 different 2x4
positions, while there are a full 64 of the 3x3 positions.

RULE 2--Orientation:  The pieces fit together as if the four edges
of each unrotated piece were

         +-b-+                                +-d-+
labeled  a O c  for odd-numbered pieces, and  a E c  for even-numbered
         +-d-+                                +-b-+

pieces, and the small letters must match where neighbors abut.

From rule 1, it is apparent that when neighbors abut, one of them must
be an even-numbered piece and the other odd.  From rule 2, we observe
that if a piece is rotated by 0 or 180 degrees, then its top and bottom
neighbors must be rotated the same amount and its left and right neigh-
bors must be rotated 180 degrees differently.  In this catalog, piece 1
will be placed in its unrotated orientation.  Then the orientation of
each piece is determined from its position relative to piece 1, and the
entire position is determined by the choice of pattern under rule 1.

TRANSFORMATION:  Each 2x4 position can be directly transformed into any
of four 3x3 positions, by folding out either end to either side.  Each
of the 3x3's can be directly transformed into either a vertical or a
horizontal 2x4.

SYMBOLOGY:  Plain numbers indicate an unrotated piece, while numbers
followed by an asterisk indicate pieces rotated 180 degrees.

The use of numbers seems to be the most popular alternate graphics
pattern at this time, as it most clearly shows what is happening as the
puzzle is manipulated.

ACKNOWLEDGEMENT:  Thanks to Rodney Hoffman for reviewing a preliminary
version of the catalog and the inspiration to prepare it in the first
place.

THE CATALOG:

>>>>>   16  VERTICAL  POSITIONS

+--+--+    +--+--+    +--+--+    +--+--+
|1 |8*|    |1 |2*|    |8*|1 |    |2*|1 |
+--+--+    +--+--+    +--+--+    +--+--+
|2 |7*|    |8 |3*|    |7*|2 |    |3*|8 |
+--+--+    +--+--+    +--+--+    +--+--+
|3 |6*|    |7 |4*|    |6*|3 |    |4*|7 |
+--+--+    +--+--+    +--+--+    +--+--+
|4 |5*|    |6 |5*|    |5*|4 |    |5*|6 |
+--+--+    +--+--+    +--+--+    +--+--+

+--+--+    +--+--+    +--+--+    +--+--+
|8 |7*|    |2 |3*|    |7*|8 |    |3*|2 |
+--+--+    +--+--+    +--+--+    +--+--+
|1 |6*|    |1 |4*|    |6*|1 |    |4*|1 |
+--+--+    +--+--+    +--+--+    +--+--+
|2 |5*|    |8 |5*|    |5*|2 |    |5*|8 |
+--+--+    +--+--+    +--+--+    +--+--+
|3 |4*|    |7 |6*|    |4*|3 |    |6*|7 |
+--+--+    +--+--+    +--+--+    +--+--+

+--+--+    +--+--+    +--+--+    +--+--+
|7 |6*|    |3 |4*|    |6*|7 |    |4*|3 |
+--+--+    +--+--+    +--+--+    +--+--+
|8 |5*|    |2 |5*|    |5*|8 |    |5*|2 |
+--+--+    +--+--+    +--+--+    +--+--+
|1 |4*|    |1 |6*|    |4*|1 |    |6*|1 |
+--+--+    +--+--+    +--+--+    +--+--+
|2 |3*|    |8 |7*|    |3*|2 |    |7*|8 |
+--+--+    +--+--+    +--+--+    +--+--+

+--+--+    +--+--+    +--+--+    +--+--+
|6 |5*|    |4 |5*|    |5*|6 |    |5*|4 |
+--+--+    +--+--+    +--+--+    +--+--+
|7 |4*|    |3 |6*|    |4*|7 |    |6*|3 |
+--+--+    +--+--+    +--+--+    +--+--+
|8 |3*|    |2 |7*|    |3*|8 |    |7*|2 |
+--+--+    +--+--+    +--+--+    +--+--+
|1 |2*|    |1 |8*|    |2*|1 |    |8*|1 |
+--+--+    +--+--+    +--+--+    +--+--+

>>>> 16  HORIZONTAL POSITIONS

+--+--+--+--+    +--+--+--+--+    +--+--+--+--+    +--+--+--+--+
|1 |8*|7 |6*|    |1 |2*|3 |4*|    |2 |3*|4 |5*|    |8 |7*|6 |5*|
+--+--+--+--+    +--+--+--+--+    +--+--+--+--+    +--+--+--+--+
|2 |3*|4 |5*|    |8 |7*|6 |5*|    |1 |8*|7 |6*|    |1 |2*|3 |4*|
+--+--+--+--+    +--+--+--+--+    +--+--+--+--+    +--+--+--+--+

+--+--+--+--+    +--+--+--+--+    +--+--+--+--+    +--+--+--+--+
|2*|1 |8*|7 |    |8*|1 |2*|3 |    |3*|4 |5*|6 |    |7*|6 |5*|4 |
+--+--+--+--+    +--+--+--+--+    +--+--+--+--+    +--+--+--+--+
|3*|4 |5*|6 |    |7*|6 |5*|4 |    |2*|1 |8*|7 |    |8*|1 |2*|3 |
+--+--+--+--+    +--+--+--+--+    +--+--+--+--+    +--+--+--+--+

+--+--+--+--+    +--+--+--+--+    +--+--+--+--+    +--+--+--+--+
|3 |2*|1 |8*|    |7 |8*|1 |2*|    |4 |5*|6 |7*|    |6 |5*|4 |3*|
+--+--+--+--+    +--+--+--+--+    +--+--+--+--+    +--+--+--+--+
|4 |5*|6 |7*|    |6 |5*|4 |3*|    |3 |2*|1 |8*|    |7 |8*|1 |2*|
+--+--+--+--+    +--+--+--+--+    +--+--+--+--+    +--+--+--+--+

+--+--+--+--+    +--+--+--+--+    +--+--+--+--+    +--+--+--+--+
|4*|3 |2*|1 |    |6*|7 |8*|1 |    |5*|6 |7*|8 |    |5*|4 |3*|2 |
+--+--+--+--+    +--+--+--+--+    +--+--+--+--+    +--+--+--+--+
|5*|6 |7*|8 |    |5*|4 |3*|2 |    |4*|3 |2*|1 |    |6*|7 |8*|1 |
+--+--+--+--+    +--+--+--+--+    +--+--+--+--+    +--+--+--+--+

>>>> 16 NORTHWEST POSITIONS

   +--+--+       +--+--+       +--+--+       +--+--+
   |1 |8*|       |1 |2*|       |8*|1 |       |2*|1 |
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|3*|2 |7*|    |7*|8 |3*|    |6 |7*|2 |    |4 |3*|8 |
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|4*|5 |6*|    |6*|5 |4*|    |5 |4*|3 |    |5 |6*|7 |
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+

   +--+--+       +--+--+       +--+--+       +--+--+
   |7*|6 |       |3*|4 |       |2 |3*|       |8 |7*|
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|1 |8*|5 |    |1 |2*|5 |    |8*|1 |4*|    |2*|1 |6*|
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|2 |3*|4 |    |8 |7*|6 |    |7*|6 |5*|    |3*|4 |5*|
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+

   +--+--+       +--+--+       +--+--+       +--+--+
   |7*|8 |       |3*|2 |       |6*|5 |       |4*|5 |
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|5 |6*|1 |    |5 |4*|1 |    |8 |7*|4 |    |2 |3*|6 |
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|4 |3*|2 |    |6 |7*|8 |    |1 |2*|3 |    |1 |8*|7 |
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+

   +--+--+       +--+--+       +--+--+       +--+--+
   |5 |6*|       |5 |4*|       |6*|7 |       |4*|3 |
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|3*|4 |7*|    |7*|6 |3*|    |4 |5*|8 |    |6 |5*|2 |
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|2*|1 |8*|    |8*|1 |2*|    |3 |2*|1 |    |7 |8*|1 |
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+

>>>> 16 NORTHEAST POSITIONS

+--+--+       +--+--+       +--+--+       +--+--+
|1 |8*|       |1 |2*|       |8*|1 |       |2*|1 |
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|2 |7*|6 |    |8 |3*|4 |    |7*|2 |3*|    |3*|8 |7*|
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|3 |4*|5 |    |7 |6*|5 |    |6*|5 |4*|    |4*|5 |6*|
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+

+--+--+       +--+--+       +--+--+       +--+--+
|8 |7*|       |2 |3*|       |7*|8 |       |3*|2 |
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|1 |6*|5 |    |1 |4*|5 |    |6*|1 |2*|    |4*|1 |8*|
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|2 |3*|4 |    |8 |7*|6 |    |5*|4 |3*|    |5*|6 |7*|
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+

+--+--+       +--+--+       +--+--+       +--+--+
|6 |7*|       |4 |3*|       |3 |4*|       |7 |6*|
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|5 |8*|1 |    |5 |2*|1 |    |2 |5*|6 |    |8 |5*|4 |
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|4 |3*|2 |    |6 |7*|8 |    |1 |8*|7 |    |1 |2*|3 |
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+

+--+--+       +--+--+       +--+--+       +--+--+
|6*|5 |       |4*|5 |       |5 |6*|       |5 |4*|
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|7*|4 |3*|    |3*|6 |7*|    |4 |7*|8 |    |6 |3*|2 |
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|8*|1 |2*|    |2*|1 |8*|    |3 |2*|1 |    |7 |8*|1 |
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+

>>>> 16 SOUTHWEST POSITIONS

+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|1 |2*|3 |    |1 |8*|7 |    |8*|1 |2*|    |2*|1 |8*|
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|8 |7*|4 |    |2 |3*|6 |    |7*|6 |3*|    |3*|4 |7*|
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
   |6*|5 |       |4*|5 |       |5 |4*|       |5 |6*|
   +--+--+       +--+--+       +--+--+       +--+--+

+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|3 |2*|1 |    |7 |8*|1 |    |2 |3*|4 |    |8 |7*|6 |
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|4 |5*|8 |    |6 |5*|2 |    |1 |8*|5 |    |1 |2*|5 |
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
   |6*|7 |       |4*|3 |       |7*|6 |       |3*|4 |
   +--+--+       +--+--+       +--+--+       +--+--+

+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|3*|4 |5*|    |7*|6 |5*|    |4 |3*|2 |    |6 |7*|8 |
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|2*|1 |6*|    |8*|1 |4*|    |5 |6*|1 |    |5 |4*|1 |
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
   |8 |7*|       |2 |3*|       |7*|8 |       |3*|2 |
   +--+--+       +--+--+       +--+--+       +--+--+

+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|4*|5 |6*|    |6*|5 |4*|    |5 |4*|3 |    |5 |6*|7 |
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|3*|2 |7*|    |7*|8 |3*|    |6 |7*|2 |    |4 |3*|8 |
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
   |1 |8*|       |1 |2*|       |8*|1 |       |2*|1 |
   +--+--+       +--+--+       +--+--+       +--+--+

>>>> 16 SOUTHEAST POSITIONS

+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|1 |2*|3 |    |1 |8*|7 |    |8*|1 |2*|    |2*|1 |8*|
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|8 |5*|4 |    |2 |5*|6 |    |7*|4 |3*|    |3*|6 |7*|
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|7 |6*|       |3 |4*|       |6*|5 |       |4*|5 |
+--+--+       +--+--+       +--+--+       +--+--+

+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|3 |2*|1 |    |7 |8*|1 |    |2 |3*|4 |    |8 |7*|6 |
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|4 |7*|8 |    |6 |3*|2 |    |1 |6*|5 |    |1 |4*|5 |
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|5 |6*|       |5 |4*|       |8 |7*|       |2 |3*|
+--+--+       +--+--+       +--+--+       +--+--+

+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|5*|4 |3*|    |5*|6 |7*|    |4 |3*|2 |    |6 |7*|8 |
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|6*|1 |2*|    |4*|1 |8*|    |5 |8*|1 |    |5 |2*|1 |
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|7*|8 |       |3*|2 |       |6 |7*|       |4 |3*|
+--+--+       +--+--+       +--+--+       +--+--+

+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|3 |4*|5 |    |7 |6*|5 |    |6*|5 |4*|    |4*|5 |6*|
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|2 |7*|6 |    |8 |3*|4 |    |7*|2 |3*|    |3*|8 |7*|
+--+--+--+    +--+--+--+    +--+--+--+    +--+--+--+
|1 |8*|       |1 |2*|       |8*|1 |       |2*|1 |
+--+--+       +--+--+       +--+--+       +--+--+

>>>> END OF CATALOG

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU 25 Aug 87 15:14:27 EDT
Date: 25 Aug 87 14:58:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: TILING BOOKS
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>


BOOK OF INTEREST

Tilings and Patterns  by Branko Grunbaum and G. C. Shephard
from w. h. Freeman and Co,  1987,    ISBN   0-7167-1193-1
......................
FROM THE PREFACE:   
_______________________________
ORGANIZATION OF THE BOOK
------------------------------------
The book falls naturally into two parts.  The first, up to and including Chapter
7, can be used as the text for a geometry course at the undergraduate level -
...   The first few sections of chapter 1 are fundamental, however, chapter 2
deals mostly with tilings in which the tiles are regular polygons.  ...   The
general theory of tilings is presented in chapters 3 and 4, these chapters are
rather more technical than the rest of the book, and ...    In chapter 5 we
begin our discussion of the theory of patterns; this continues in chapter 7. 
...

the second part (chapters 8-12) presents detailed surveys of various aspects of
the subjects of patterns and tilings.  These include colored patterns and groups
of color symmetry, tilings by polygons, tilings in which the tiles are unusual
in a topological sense, as well as, a detailed and self-contained account of the
intriguing topic of aperiodic tilings.  ...
_____________________________
TABLE OF CONTENTS
1  BASIC NOTIONS
2  TILINGS BY REGULAR POLYGONS AND STAR POLYGONS
3  WELL-BEHAVED TILINGS
4  THE TOPOLOGY OF TILINGS
5  PATTERNS
6  CLASSIFICATIONS OF TILINGS WITH TRANSIVITY PROPERTIES
7  CLASSIFICATION WITH RESPECT TO SYMMETRIES
8  COLORED PATTERNS AND TILINGS
9  TILINGS BY POLYGONS
10 APERIODIC TILINGS
11 WANG TILES
12 TILINGS WITH UNUSUAL HINDS OF TILES
REFERENCES

...........................................

This is a great book on the subject with plenty of pictures for those of us who
can't visualize well.  

beckardec-lcss.arpa
33333333###################
------

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU 26 Aug 87 08:11:43 EDT
Date: 26 Aug 87 07:55:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: PASTIME JIGSAW PUZZLES
To: "cube-lovers" <cube-lovers@mit-ai>
cc: beck
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>


I am interested in information on the "Pastime Puzzles"  (jigsaw puzzles) made
by Parker Brothers in 1932/33/ and other puzzles of similar construction .  The
ones I have were sold by the Kohler Puzzle Exchange, 105 Roseville Ave, Newark
NJ  and if anybody has information on this firm I would also like the reference.

These puzzles are of the type currently being manufactured by "Stave", ie,
many pieces have shapes (eg, numbers, violins, fruits),  false corners and
edges, cuts where the colors change, etc


The Future is Puzzling but Cubing is Forever,

Pete beck                  <BECK@ARDEC-LCSS.arpa>
..................................

------

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU 27 Aug 87 08:10:48 EDT
Date: 27 Aug 87 08:06:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: FRACTALS
To: "cube-lovers" <cube-lovers@mit-ai>
cc: beck
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>

BOOK OF INTEREST

THE BEAUTY OF FRACTALS  by  H.-O. PEITGEN & P.H. RICHTER
  from SPRINGER-VERLAG  1986,   ISBN  3-540-15851-0  OR ISBN 0-387-15851

......................
FROM THE FLAP:   This book is an unusual attempt to publicize the field of
Complex Dynamics, ...  .  

In 88 full color pictures, and many more black and white illustrations, the
authors present variations of a theme whose repercussions reach far beyond the
realms of mathematics.  
...........................................

This is agreat first book on the subject with plenty of pictures for those of us
who can't visualize well.

beckardec-lcss.arpa
------

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU  3 Sep 87 08:58:50 EDT
Date: 3 Sep 87 08:38:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: dutch newsletter #15
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>


SUBJECT :  Review of "Cubism For Fun" newsletterissue #15; the newsletter of the
"Dutch Cubists Club";  in english starting with issue 14 

1..  The table of contents >> annotated << from issue #15, august 87 follows (it
is 8 double sided folded 8 1/2 x 11 sheets, making a 32 page newsletter):

INTRODUCTION
LIVING WITH A CUBIST  BY  LUKAS  SCHOONHOVEN 
RUBIK'S MAGIC'S CUBE  BY  RONALD FETTERMAN  < HOW TO FOLD RUBIK'S               
               MAGIC INTO A CUBE<
LENGTH DATA FOR UPPER TABLE PROCESSES  BY  ANTON  HANEGRAAF
MARC WATERMAN'S ALGORITHM , PART; CONTINUED FROM ISSUE 14 
                 - ANNEKE TREEP & MARC  WATERMAN
THE MAGIC NUMBER CUBE  BY  WALLY  WEBSTER > MARKING WITH NUMBERS A              
3X3X3 RUBIK'S CUBE SO THAT ALL OF THE 3X3 VERTICALS AND                      
HORIZONTALS ADD UP TO 42<<
THE MAGIC MOSAICS BY RONALD FETTERMAN  > SIMILAR TO THE CATALOGUE               
     OF RUBIK'S MAGIC POSITIONS POSTED ON CUBE LOVERS WITH A                    
  DIFFERENT NOTATION AND WITH THE ADDITION OF A                                 
      NOMENCLATURE AND A MOVE SEQUENCE TO GET TO EACH ONE                       
    FROM START<<
MAGIC AND AND IS NHO MAGIC BY TOM VERHOEFF  > A GROUP THEORY                    
            ANALYSIS OF RUBIK'S MAGIC<
MAGIC VARIATIONS BY PETER BECK >PREVIOUSLY POSTED TO CUBE LOVERS<
PRETTY CUBIC PATTERNS  BY  ANNEKE TREEP
NEWS AND LETTERS TO THE EDITOR  >  a list of collectors wanting to trade        
                        cubes/puzzles;  a statement that Guus Schultz has built 
                               MAGICs where the number of squares is not a
multiple                                   of "4" <<
LIST OF MEMBERS

2.   Membership for 1987 is  US$5.   A photocopied set of the newsletters,
issues 1-13, written in DUTCH (in the feature selected articles will be
available in english) is also available for  US$7.    To order either of these
send an 'INTERNATIONAL"  POSTAL MONEY ORDER to:  Anton Hanegraaf, 
Heemskerkstraat,  6662 AL ELST,  The Netherlands.

3.  If anybody would like further details please ask!

CUBING IS FOREVER      PETER BECK

<BECK@ARDEC-LCSS>  OR  <PBECK@ARDEC>

------

Received: from GTEWIS.ARPA (TCP 3201600102) by AI.AI.MIT.EDU 22 Sep 87 12:25:05 EDT
Date: Tue, 22 Sep 87 11:49:12 EDT
From: rdavenport@GTEWIS.ARPA
Subject: Where's ITC?
To: CUBE-LOVERS@AI.AI.MIT.EDU







      Greetings fellow cube lovers,

         I have only learned to solve the cube this summer, and now am working
 on R*bik's Revenge (and Magic) and I was wondering if anyone out there knows
 of any publications dealing with solving them and any dealing with the theory
 behind them - as I have really enjoyed Frey & Singmaster's _Cubik Math_.

         I have tried writing to Ideal Toy Corporation as mention in the Revenge
 product but they seem to have moved - anybody know where they are?


          thanks in advance,

                  Rob



^^^^^^^^^^^^^^^^^vvvvvvvvvvvvvvvvv^^^^^^^^^^^^^^^^vvvvvvvvvvvvvvvvv^^^^^^^^^^^^^

Rob Davenport                  Arpanet :  RDAVENPORT@GTEWIS.ARPA
GTE
Billerica, Massachusetts
(617) 671-5180

vvvvvvvvvvvvvvvvv^^^^^^^^^^^^^^^^^vvvvvvvvvvvvvvvv^^^^^^^^^^^^^^^^^vvvvvvvvvvvvv

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU 28 Sep 87 13:05:50 EDT
Date: 28 Sep 87 12:41:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: ROTA & GD TIMES, BAD TIMES
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>

1.  For the collectors, I recently obtained a new to me shape variant of the
2x2x2.  It is a cylinder, mine is half black and half white.  It was made by the
Swiss company NAEF and is called "ROTA".  It is very similar in size to the
standard cube.  It should cost about $15.

2.  Has anybody played a game called "GOOD TIMES, BAD TIMES".  It uses the
PYRAMINX in some fashion and I believe it was developed/marketed by MEFFERT.

Thanks in advance.

CUBING IS FOREVER ----    PETER BECK
                                          <BECK@ARDEC-LCSS.ARPA>
                                OR      <PBECK@ARDEC.ARPA>
......................................................................
------

Received: from Xerox.COM (TCP 1200400040) by AI.AI.MIT.EDU  2 Oct 87 10:44:46 EDT
Received: from Gamay.ms by ArpaGateway.ms ; 02 OCT 87 07:35:05 PDT
Date: 2 Oct 87 07:35:00 PDT (Friday)
From: Hoffman.es@Xerox.COM
Subject: Rubik's Magic article
To: Cube-Lovers@AI.AI.MIT.EDU
Message-ID: <871002-073505-5351@Xerox>

The October '87 issue of 'Scientific American' is completely devoted to
Advanced Computing, so it is probably of interest to many of us.  In
addition, the 'Amateur Scientist' column in that issue (by Jearle
Walker, pages 170-173) is all about Rubik's Magic (the original, 8-panel
version).  It includes tables and diagrams of permutations, and a
complete solution with pictures.

  -- Rodney Hoffman

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU 14 Oct 87 14:14:13 EDT
Date: 14 Oct 87 13:53:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: MAZES
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>



FOR FUN, JAPAN TURNS TO MAZES

The Sunday NY TImes of 10/11/87 Travel section, pg3 had a small piece on
amusement park mazes.

Some random quotes:

"A lot of people are willing to pay the $3 fee that most of the approximately 20
outdoor mazes charge for the opportunity to become confused."

"The object, ...,  is to get through the maze as fasy as possible.  On average,
it takes 45 minutes to escape or give up."

"each maze has a theme - such as the Paris-Dakar Rally Maze in Tokyo and the
Sherlock Maze in Osaka."

MORE INFORMATION:  JAPAN NATIONAL TOURIST ORGANIZATION,  630 FIFTH AVE,  SUITE
2101, NY, NY 10111;   212/757-5640.

...........................................................
PS If anybody has other references I would like them.

Thanks

pete beck

<BECK@ARDEC-LCSS>

..........................
------

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Message-Id: <kVTYcPy00Uo1gws0=j@andrew.cmu.edu>
Date: Thu, 22 Oct 87 14:10:35 edt
From: ap1a+@andrew.cmu.edu (Andrew Balen Philips)
To: Cube-Lovers@ai.ai.mit.edu
Subject: More Expensive Cubes


To anyone out there:

I am currently involved in a research project on the Rubik's Cube and expert
solving of it.  The conventional cubes sold in most stores fall apart in the
hands of the expert, because corners catch.  For awhile there were cubes
manufactured that have tiles for colored plates.  We have one of these, but
would like to have some more.  If anyone knows where we may look for these
cubes or has access to these cubes, please notify me.

								Thank you,
										Andy
Philips, ap1a+@andrew.cmu.edu

Send mail direct or post.
By the way, this bboard is very cool!



Received: from po2.andrew.cmu.edu (TCP 20000574551) by AI.AI.MIT.EDU 22 Oct 87 19:12:50 EDT
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          Thu, 22 Oct 87 19:01:12 -0400 (EDT)
Message-Id: <wVTcssy00UkZB080Uy@andrew.cmu.edu>
Date: Thu, 22 Oct 87 19:01:12 -0400 (EDT)
From: dt+@andrew.cmu.edu (David Tilbrook)
To: ap1a+@andrew.cmu.edu (Andrew Balen Philips), Cube-Lovers@ai.ai.mit.edu
Subject: Re: More Expensive Cubes
In-Reply-To: <kVTYcPy00Uo1gws0=j@andrew.cmu.edu>

The cubes to which you refer were manufactured in Korea.

I purchased three or four in Toronto at a Korean trade show about six years
ago,
and still have them in perfect working order.

Sorry I can't give you anymore information than that,
other than to say that they are trully amazing in that they
have lasted extremely well and require absolutely no maintenance.

The red's a little hard to distinguish from the orange
in the wrong light but that is the only problem.

While we are on the subject, does anyone know where to acquire
a 5x5x5?  There was a source in London two years ago but they ran out
and haven't been seen since.

Also I'd like to acquire the globe
on which the equator, greenwich and 90' meridians rotate.

david tilbrook



Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU  5 Nov 87 08:53:44 EST
Date: 5 Nov 87 08:40:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: CMU CUBE RESEARCHERS
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>

TO:  R DEUSER, A PHILLIPS, D TILLBROOK

MY MAILER SAYS THAT YOUR ADDRESSES ARE VALID BUT I SUSPECT THAT MY MSGS ARE
NOT GETTING THROUGH.  I AM TRYING TO RESOLVE THIS PROBLEM.  IF YOU HAVE ANY
SUGGESTIONS (MAYBE THE PROBLEM IS AT YOUR END) THEN HELP ME.  I HAVE RECEIVED
YOUR PREVIOUS MSGS.

SORRY CUBE LOVERS FOR THIS PERSONAL MSG.

PS  THE DELUXE CUBES YOU WANT ARE ON THE WAY.

PETE BECK  <PBECK@ARDEC.ARPA>
......................................
------

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU  5 Nov 87 08:54:10 EST
Date: 5 Nov 87 08:46:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: MAGIC VARIANT
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>
I SAW IN THE STORE THE OTHER DAY A GRAPHICS VARIATION OF "MAGIC" FROM MATCHBOX.
IT IS CALL ED RUBIK'S MAGIC CUBE.  THE OBJECT IS TO MAKE THREE DIMENSIONAL 
SHAPES THAT ARE IN HARMONY WITH THE ALTERNATE GRAPHICS.  ANYBODY OUT THERE
INVESTIGATE THIS VARIANT OR THE MAGIC GAME?  I WOULD APPRECIATE ANY IMPRESSIONS.

A REMINDER!!!   FOR ALL OF YOU BOSTON PEOPLE, THE PUZZLE EXHIBIT STILL HAS SOME
TIME TO RUN AT THE MIT MUSEUM BRFORE FADING IN TO NEVER NEVER LAND - SO GET OUT
AND SEE IT.  I THINK YOU HAVE UNTIL JAN.

PETE BECK  <PBECK@ARDEC.ARPA>
..............................
------

Received: from WILMA.BBN.COM (TCP 20026200730) by AI.AI.MIT.EDU  5 Nov 87 10:31:44 EST
Date:     Thu, 5 Nov 87 10:26:29 EST
From:     Bernie Cosell <cosell@WILMA.BBN.COM>
To:       cube-lovers@AI.AI.MIT.EDU
cc:       jr@WILMA.BBN.COM, beeler@WILMA.BBN.COM, alatto@WILMA.BBN.COM
Subject:  Deluxe Magic

I picked up a "deluxe Rubik's Magic" at Games People Play the other day.
It is a twelve-square magic.  Has anyone solved this guy yet?  My wife
has been hacking on it some and and has managed to run it from the
starting state (2x6) to the target state (as in the normal Magic, but
moreso), but not enough comprehension of it all yet to get all the
circle pieces in the right places, yet.

It seems to be more fun that the normal magic because if you ignore
the circles you can make a bunch of interesting shapes (the big-hollow-
square was neat to blunder into).

  /Bernie\

ps, a while back someone (pete?) posted a pointer to some magazine (foreign,
maybe?) that had an article about folding a Magic into a cube.  I don't
remember if I've asked this before or not, but... can anyone send me hints
about how the fold-magic-into-a-cube goes?  tnx /b\

Received: from 40700016315 by AI.AI.MIT.EDU via Chaosnet; 5 NOV 87  13:08:11 EST
Received: by mit-nc.MIT.EDU with sendmail-5.31/4.7 
	id <AA03499@mit-nc.MIT.EDU>; Thu, 5 Nov 87 13:08:45 EST
Date: Thu, 5 Nov 87 13:08:45 EST
From: meister@mit-nc.MIT.EDU (phil servita)
To: cube-lovers@ai.ai.mit.edu
Subject: in search of...


I have been in search of anyplace in the USA where i can obtain a Skewb. 
So far, in about a year, no luck. Games of Berkeley keeps claiming that they
have an order in, but it never seems to get there.  Does anybody out there have
one that they would be willing to sell me? 

                                            -meister

(reply either to the list, or to meister@eddie.mit.edu)


Received: from WAIKATO.S4CC.Symbolics.COM (TCP 20024231532) by AI.AI.MIT.EDU  5 Nov 87 14:00:38 EST
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Date: Thu, 5 Nov 87 13:55 EST
From: Allan C. Wechsler <ACW@WAIKATO.S4CC.Symbolics.COM>
Subject: Deluxe Magic
To: cosell@WILMA.BBN.COM, cube-lovers@AI.AI.MIT.EDU
cc: jr@WILMA.BBN.COM, beeler@WILMA.BBN.COM, alatto@WILMA.BBN.COM
In-Reply-To: The message of 5 Nov 87 10:26 EST from Bernie Cosell <cosell@WILMA.BBN.COM>
Message-ID: <871105135516.1.ACW@ROCKY-MOUNTAINS.S4CC.Symbolics.COM>

    Date:     Thu, 5 Nov 87 10:26:29 EST
    From:     Bernie Cosell <cosell@WILMA.BBN.COM>

    I picked up a "deluxe Rubik's Magic" at Games People Play the other day.
    It is a twelve-square magic.  Has anyone solved this guy yet?  My wife
    has been hacking on it some and and has managed to run it from the
    starting state (2x6) to the target state (as in the normal Magic, but
    moreso), but not enough comprehension of it all yet to get all the
    circle pieces in the right places, yet.

Well, /my/ wife solved it. 

    It seems to be more fun that the normal magic because if you ignore
    the circles you can make a bunch of interesting shapes (the big-hollow-
    square was neat to blunder into).

You bet!  As a matter of fact the order-6 puzzle is so much more fun 
than the order-4 that I am wondering whether higher orders might be 
even more fun.  In my opinion the order-4 cube was /less/ fun than the 
order-3, and it's a pleasure to see a puzzle where bigger really is
better.

Jenny and I have a conjecture that if a given flat shape is possible,
a flat shape that is derived from the possible one by moving a single
square one step diagonally -- is impossible.  There is probably a parity
argument lurking somewhere that can prove this.

Is a similar puzzle with triangular tiles possible?


Received: from WAIKATO.S4CC.Symbolics.COM (TCP 20024231532) by AI.AI.MIT.EDU  5 Nov 87 17:01:13 EST
Received: from ROCKY-MOUNTAINS.S4CC.Symbolics.COM by WAIKATO.S4CC.Symbolics.COM via CHAOS with CHAOS-MAIL id 141683; Thu 5-Nov-87 13:55:41 EST
Date: Thu, 5 Nov 87 13:55 EST
From: Allan C. Wechsler <ACW@WAIKATO.S4CC.Symbolics.COM>
Subject: Deluxe Magic
To: cosell@WILMA.BBN.COM, cube-lovers@AI.AI.MIT.EDU
cc: jr@WILMA.BBN.COM, beeler@WILMA.BBN.COM, alatto@WILMA.BBN.COM
In-Reply-To: The message of 5 Nov 87 10:26 EST from Bernie Cosell <cosell@WILMA.BBN.COM>
Message-ID: <871105135516.1.ACW@ROCKY-MOUNTAINS.S4CC.Symbolics.COM>

    Date:     Thu, 5 Nov 87 10:26:29 EST
    From:     Bernie Cosell <cosell@WILMA.BBN.COM>

    I picked up a "deluxe Rubik's Magic" at Games People Play the other day.
    It is a twelve-square magic.  Has anyone solved this guy yet?  My wife
    has been hacking on it some and and has managed to run it from the
    starting state (2x6) to the target state (as in the normal Magic, but
    moreso), but not enough comprehension of it all yet to get all the
    circle pieces in the right places, yet.

Well, /my/ wife solved it. 

    It seems to be more fun that the normal magic because if you ignore
    the circles you can make a bunch of interesting shapes (the big-hollow-
    square was neat to blunder into).

You bet!  As a matter of fact the order-6 puzzle is so much more fun 
than the order-4 that I am wondering whether higher orders might be 
even more fun.  In my opinion the order-4 cube was /less/ fun than the 
order-3, and it's a pleasure to see a puzzle where bigger really is
better.

Jenny and I have a conjecture that if a given flat shape is possible,
a flat shape that is derived from the possible one by moving a single
square one step diagonally -- is impossible.  There is probably a parity
argument lurking somewhere that can prove this.

Is a similar puzzle with triangular tiles possible?


Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU 10 Nov 87 12:55:04 EST
Date: 10 Nov 87 12:26:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: WORLD GAME REVIEW
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>

ARE THERE ANY SUBSCRIBERS TO THE WORLD GAME REVIEW (MIKE KELLER) OUT THERE.
IN PARTICULAR I'VE HEARD RUMORS THAT ISSUE #7 (ABOUT 6 MONTHS IN COMING) IS
OUT.  IS THIS TRUE?  ANYBODY SEEN IT?

PETE BECK  <PBECK@ARDEC.ARPA>

------

Received: from WILMA.BBN.COM (TCP 20026200730) by AI.AI.MIT.EDU 10 Nov 87 12:55:36 EST
Date:     Tue, 10 Nov 87 12:51:48 EST
From:     Bernie Cosell <cosell@WILMA.BBN.COM>
To:       cube-lovers@AI.AI.MIT.EDU
cc:        alatto@WILMA.BBN.COM, math@WILMA.BBN.COM, jr@WILMA.BBN.COM
Subject:  New "Recreations in Math" editions

Oxford U Press continues to produce entries in their "Recreations in
Mathematics" series.  I got #s 1 & 2 last year I just got vol 4.  I've
never seen vol 3.

To review, #1 was "Mathematical byways ..." by Hugh ApSimon.  I thought
it was BORING, but it did discuss one thing I've never seen: *how* you
set up a problem so it is both interesting and solvable.  He runs
through starting with some idea for a puzzle (something like the
"you put an X foot ladder up against a wall and it just touches a
box that is Y feet on a side, what's inside the box?") and gives the
"composer's problem" related to that topic: how to get the problem set
up.  Interesting, sort of, but overall pretty boring stuff (especially
since they are for the most part old, stuffy, dull problems).

#2: Ins and Outs of Peg Solitaire.  Really quite definitive reference
to the jump-the-pegs-and-leave-one-in-the-middle puzzle.  I can't remember
where, but I've actually seen most of that material before.  Maybe
Mathematics magazine, or JRM.  But in any event, this is a great book
if you're at all interested in this kind of problem.

#3: Rubik's Cubic Compendium, by Rubik, et al.  I've *never* seen this
anywhere.  I'd love to get/have/see a copy.  If any of you have a lead
to this guy, please let me know.

#4 Sliding Piece Puzzles (Hordern).  I just picked this up at the
Harvard Coop today.  Not much theory on either the design or solution
of this kind of puzzle.  Just page after page of example puzzles.  This
is more of a catalog than a math book.  One cute touch: there is a pocket
inside the back cover with push-out paper "shapes"   I guess that there
are enough miscellaneous shapes on the card (about 2"x4") so that you
can piece together a large number of the puzzles described in the
book.  My first impressions are that this book will be a definite
"Ho Hum".

  /Bernie\

Received: from BFLY-VAX.BBN.COM (TCP 20026200235) by AI.AI.MIT.EDU 10 Nov 87 14:59:58 EST
To: Bernie Cosell <cosell@cosell.bbn.com>
cc: cube-lovers@ai.ai.mit.edu, alatto@cosell.bbn.com, math@cosell.bbn.com, 
    jr@cosell.bbn.com, dm@bfly-vax.bbn.com
Subject: Re: New "Recreations in Math" editions
In-reply-to: Your message of Tue, 10 Nov 87 12:51:48 EST.
Date: 10 Nov 87 14:56:04 EST (Tue)
From: dm@bfly-vax.bbn.com


In Bernie's defense, I'll point out that he didn't name his machine
after himself -- it got named after him by the computer center staff,
who were installing workstations faster than we could think of
cute names for them, and adopted a simple, if boring algorithm for
coming up with machine names.



Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU 17 Nov 87 18:36:50 EST
Date: 17 Nov 87 10:51:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: CUBE DAY 87
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>

<><><><><><><><><><><><><><><><><><><><><><><><><><><><>
INVITATION   FROM  GUUS RAZOUX SCHULTZ
TO   ----->>  CUBE DAY  1987
..........................................
SAT 12 DEC 1987,  10AM - 17PM
AT:  GUUS RAZOUX SCHULTZ  -- RESIDENCE, PHONE 053-359617
       CORT VAN DER LINDENLAAN 30,
       ENSCHEDE, NETHERLANDS
<><><><><><><><><><><><><><><><><><><><><><><><><><><><>
PROGRAM:
  COMPETITIONS FOR CUBE AND MAGIC
  VIDEO SHOW ON PREVIOUS CUBE DAYS BY KLAAS STEENHUIS
  LECTURE ON GOD'S ALGORITHM BY GUUS RAZOUX SCHULTZ
  LECTURE ON THE SKEWB ??
  INTRO TO SUPER MAGIC (MASTER/GENUIS???) BY TOM VERHOEFF
  DEMO OF 6,10,16 PANEL MAGICS BY GUUS RAZOUX SCHULTZ
  
anyone who has interesting puzzles, books, articles, news, correspondence,
tables, posters, etc is asked:  please, don't hestitate to take it all with you!

Anyone who has computer programs for the cube, magic or other combinatorial
puzzles, please let us know what equipment is needed for demo.  

Anyone who has video tapes on puzzle events is also gladly invited! (We are
still looking for someone who has a recording of the MAGIC - championship.
...............................................................................
above received in mail.  If somebody out there stops in it would be great if
they take notes and post.

CUBING IS FOREVER!!!

PETE BECK
------

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU 10 Dec 87 10:54:43 EST
Date: 10 Dec 87 10:43:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: WHO'S WHO OF PUZZLES
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>


COLLECTORS AND PUZZLE DESIGNERS WHO WISH TO BE INCLUDED IN AN UNOFFICIAL
DATABASE CAN SEND THEIR NAME AND ADDRESS AND PHONE NUMBER WITH A BRIEF DESCRIPTION OF THEIR PUZZLE INTEREST TO:

ROBERT HOLBROOK
5225 CARROLTON RD
ROCKVILLE, MD 20853.

THIS WILL GET YOUR NAME ARROUND AND YOU WILL PROBABLY RECEIVE JUNK MAIL FROM
PEOPLE SELLING PUZZLES.  TO THE BEST OF MY KNOWLEDGE THIS IS A HARDCOPY LIST.


THE FUTURE IS PUZZLING

PETE BECK

<PBECK@ARDEC>

------

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU 10 Dec 87 10:54:43 EST
Date: 10 Dec 87 10:43:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: WHO'S WHO OF PUZZLES
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>


COLLECTORS AND PUZZLE DESIGNERS WHO WISH TO BE INCLUDED IN AN UNOFFICIAL
DATABASE CAN SEND THEIR NAME AND ADDRESS AND PHONE NUMBER WITH A BRIEF DESCRIPTION OF THEIR PUZZLE INTEREST TO:

ROBERT HOLBROOK
5225 CARROLTON RD
ROCKVILLE, MD 20853.

THIS WILL GET YOUR NAME ARROUND AND YOU WILL PROBABLY RECEIVE JUNK MAIL FROM
PEOPLE SELLING PUZZLES.  TO THE BEST OF MY KNOWLEDGE THIS IS A HARDCOPY LIST.


THE FUTURE IS PUZZLING

PETE BECK

<PBECK@ARDEC>

------

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU 10 Dec 87 16:11:44 EST
Date: 10 Dec 87 15:53:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: cff #16
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>


SUBJECT :  Review of "Cubism For Fun" newsletter issue #16; the newsletter of
the "Dutch Cubists Club";  in english starting with issue 14 

1..  The table of contents for issue #16, NOV 87 follows (it is 8 double sided
folded 8 1/2 x 11 sheets, making a 32 page newsletter):

INVITATION TO "CUBISTS DAY"  BY Guus Schultz 
MY PATTERNS COLLECTION BY CECIL SMITH
THE UPPER TABLE AVERAGED BY BEN JOS WALBEEHM
PRETTY CUBIC PATTERNS BY ANNEKE TREEP 
PRETTY MAGIC STRUCTURES BY RONALD FLETTERMAN  
NOTES ON RUBIK'S MAGIC BY Guus Schultz 
RUBIKS MASTER MAGIC BY ED HORDERN
MINIMAL SOLUTIONS FOR THE 12-MAGIC BY TOM VERHOEFF 
MARC WETERMAN'S ALGORITHM , PART 3; CONTINUED FROM ISSUE 14/15 
                 - ANNEKE TREEP & MARC  WATERMAN
THE INVISIBLES B RONALD FLETTERMAN  
NEWS AND LETTERS TO THE EDITOR  
LIST OF MEMBERS

2.   Membership for 1987 is  US$5.   A photocopied set of the newsletters,
issues 1-13, written in DUTCH (in the future selected articles wy|l be available
in english) is also available for  US$7.    To order either of these send an
'INTERNATIONAL"  POSTAL MONEY ORDER to:  Anton Hanegraaf,  Heemskerkstraat, 
6662 AL ELST,  The Netherlands.

3.  If anybody would like further details please ask!

CUBING IS FOREVER      PETER BECK

<BECK@ARDEC-LCSS>  OR  <PBECK@ARDEC>

------

Received: from ARDEC-LCSS.ARPA.ARPA (TCP 30003004013) by AI.AI.MIT.EDU 11 Dec 87 10:39:46 EST
Date: 11 Dec 87 10:19:00 EST
From: "CLSTR1::BECK" <beck@clstr1.decnet>
Subject: jigsaws
To: "cube-lovers" <cube-lovers@mit-ai>
Reply-To: "CLSTR1::BECK" <beck@clstr1.decnet>


For puzzler's who like to plan ahead...d there will be a "JIGSAW" puzzle
exhibition in the summer of 1988.  It will be at:

Bates College Museum of Art,
Lewiston, Maine 04240 (I think)
May 19-Aug 12 1988.

The guest curator is:

Anne D. Williams
49 Brooks Ave
Lewiston, ME 04240.

The exhibit will have over 100 jigsaws in it, dating back to 1766.


The Future is Puzzling, 
but Cubing is Forever.

Pete beck                  <PBECK@ARDEC>
..................................



------

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To: cube-lovers-request@ai.ai.mit.edu, cube-lovers@ai.ai.mit.edu
Subject: Subscription


[]

Please put me on the CUBE-LOVERS mailing list.

(Sorry if this goes to the whole list, but the -request address
does not seem to work!)

Thanks,
           Sergiu Hart
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Date: Sun, 27 Mar 88 16:33:28 EST
From: Alan Bawden <ALAN@AI.AI.MIT.EDU>
Subject:  Subscription
To: hart%TAURUS.BITNET@MITVMA.MIT.EDU
cc: CUBE-LOVERS-REQUEST@AI.AI.MIT.EDU, CUBE-LOVERS@AI.AI.MIT.EDU
In-reply-to: Msg of Sun 27 Mar 88 19:43:11 +0200 from hart%TAURUS.BITNET at MITVMA.MIT.EDU
Message-ID: <348312.880327.ALAN@AI.AI.MIT.EDU>

    Date: Sun, 27 Mar 88 19:43:11 +0200
    From: hart%TAURUS.BITNET at MITVMA.MIT.EDU
    Please put me on the CUBE-LOVERS mailing list.
    (Sorry if this goes to the whole list, but the -request address
    does not seem to work!)

Let's not start any rumors.  Cube-Lovers-Request works just fine.  I added
you to the list three days ago, and I mailed you an acknowledgment at that
time.  Perhaps some mailer between here and there ate my message for lunch,
but I certainly can't help that.  

Cube-Lovers is an extremely low-volume mailing list these days, so the fact
that your mailbox didn't immediately fill with Cube-Lovers mail means
nothing.  (The previous Cube-Lovers mail was sent last December 11th.)

Received: from ARDEC-AC4.ARPA (TCP 30003004020) by AI.AI.MIT.EDU 28 Mar 88 16:00:38 EST
Date:     Mon, 28 Mar 88 15:54:07 EST
From:     Peter Beck (LCWSL) <pbeck@ARDEC.ARPA>
To:       cube-lovers@AI.AI.MIT.EDU
Subject:  magic polyhedra
Message-ID:  <8803281554.aa20390@ARDEC-AC4.ARDEC.ARPA>

a puzzle fool's view of plate tectonics
by peter beck   april 1, 1988

what follows is an unfounded speculation of how "magig polyhedra" can be used to
understand the manifistations of plate tectonics.  my imagination was piqued
while manipulating the "megaminx" (a dodecahedron with flat pentagon shaped
faces, marketed in the usa by tomy) because the puzzle locks up when an attempt
is made to turn too many faces simultaneously.  this causes the surface to
distort and when too much force is excerted the puzzle comes apart in an
explosive fashion.  

i, impusively concluded that the geometric principles governing this explosion
are analogous to what happens when the surface plates of the earth are rotated
by the forces behind plate tectonics.  this analogy is useful because it
provides a macro model with physical parity constraints to study plate
tectonics, eg, by helping forecasters tie together observable events around the
world a better understanding of individual events could be obtained.  another
area of study could be the parity constraints on the motion of the plates, ie, 
the directions of plate rotation are constrained by their neighbors, because
each plate does not move independently (see fig. 465.10 in fuller's book
"synergetics").   [it should be noted that other dodecahedron magic polyhedra
may be more appropriate for the study of plate tectonics; ie, the "impossiball"
or 'alexander's star".]

now that my fantasizing is in high gear i will expand my speculation to consider
the engine that drives plate tectonics.  i have decided that if one knew some
physics it could probably be shown that a rotating sphere with a liquid center
would develop 12 local circulations which the surface plates would float on. 
thus, plate tectonics can on a macro scale be reduced to a simple problem of
fluid dynamics and some geometric parity constraints which can be displayed with
magic polyhedra.

the future is puzzling, but cubing is forever !!


distribution:
   cff
   wgr
   cube-lovers@mit

Received: from WAIKATO.S4CC.Symbolics.COM (TCP 20024231532) by AI.AI.MIT.EDU 28 Mar 88 16:50:43 EST
Received: from ROCKY-MOUNTAINS.S4CC.Symbolics.COM by WAIKATO.S4CC.Symbolics.COM via CHAOS with CHAOS-MAIL id 165953; Mon 28-Mar-88 16:47:32 EST
Date: Mon, 28 Mar 88 16:48 EST
From: Allan C. Wechsler <ACW@WAIKATO.S4CC.Symbolics.COM>
Subject: Magic Polyhedra and parity constraints
To: Cube-Lovers@MIT-AI.ARPA
Message-ID: <19880328214819.9.ACW@ROCKY-MOUNTAINS.S4CC.Symbolics.COM>

In response to Peter Beck's thought-provoking idea about connecting
Rubikoid puzzles to plate tectonics, I have two slight spoilers.

First, plate tectonics involves spreading zones, which are places where
new crust is created, and subduction zones, where crust is destroyed.
In any permutation group, the things being permuted are not allowed to
appear or disappear.  So it seems unlikely that group theory can be
directly applied to tectonics.

Second, just because a puzzle is Rubikoid does not mean it has parity
constraints.  Consider the "Magic Octohedron", which has eight
triangular faces.  You can grab any pyramidal cluster of four faces and
rotate it.  This is really the 2x2x2 Cube in disguise.  In this form, it
has no parity constraints, that is, all the 8-factorial different
configurations are achievable.  So even if group theory could be applied
to tectonics, we couldn't assume parity constraints in the general case.



Received: from ardec-lcss.arpa (TCP 30003004013) by AI.AI.MIT.EDU  2 May 88 13:29:41 EDT
Received: by ardec-lcss.arpa id <202004A3041@ardec-lcss.arpa> ;
       Mon,  2 May 88 13:29:02 EST
Date: Mon,  2 May 88 13:28:09 EST
From: BECK@ardec-lcss.arpa
Subject: new puzzles
To: cube-lovers@ai.ai.mit.edu
X-VMS-Mail-To: EXOS%"cube-lovers@mit-ai"
Message-ID: <880502132809.202004A3041@ardec-lcss.arpa>


I VISITED BY "KAY-BEE" toy store at the mall today.  


REPEAT:  For those of you who like 3-d assembly puzzles a set of 2 GEO-LOGIC
(TAURUS&CETUS) puzzles for $7 .   These puzzles are plastic, and were designed
by Stuart Coffin and manufactured by Skor-Mor (OUT OF BUSINESS).   The full
collection of these puzzzles is called the GEO-LOGIC series.  They are each made
of 6 identical pieces (different for each puzzle) that can be assemblied into an
interlocking self supporting solid.   These puzzles are hard to find.   So if
you may be interested don't delay - call up your KAY BEE now.

NEW FOR 1988
  GRIPPLE as seen on TV for $10.  This is a sequential movement puzzle more
similar to missing link then the cube.

  YOSHI'S (shortening of designers name) PUZZLE for $12 from Parker Bros.  This
is a re-release of a formerly unbranded item, circa 1982 called miraculous cube.
 It is composed of a loop of tetrahedrons taped together to form hinges.   Yoshi
designed/invented two puzzles like this. The other is called the Shisei Mystery
and it works like a Rubik's magic with depth.

  Mattel has a cube puzzle made up of magnetic cubies called "Magic Force" 
I think.  not at kay-bee.
 

The Future is Puzzling, 
but Cubing is Forever.

Pete beck                  <PBECK@ARDEC>
..................................

Received: from ardec-lcss.arpa (TCP 30003004013) by AI.AI.MIT.EDU  5 May 88 09:22:15 EDT
Received: by ardec-lcss.arpa id <2100026E041@ardec-lcss.arpa> ;
       Thu,  5 May 88 09:20:50 EST
Date: Thu,  5 May 88 09:20:15 EST
From: BECK@ardec-lcss.arpa
Subject: JIGSAW PUZZLE EXHIBIT
To: CUBE-LOVERS@ai.ai.mit.edu
X-VMS-Mail-To: EXOS%"CUBE-LOVERS@MIT-AI"
Message-ID: <880505092015.2100026E041@ardec-lcss.arpa>


UPDATE:    The "Jigsaw" puzzle exhibition will be May 19 - Aug 12, 1988.   There
will be an open house Sunday 22 May.  

It will be at the:
    Museum of Art, Olin  Arts Center, Bates College
    Lewiston, Maine 04240 

Exhibit/museum hours:
    Tuesday - Sat, 10-4PM
    Sunday  1-5PM
    Closed Mondays and holidays.

The guest curator is:
     Anne D. Williams
     49 Brooks Ave
     Lewiston, ME 04240.

The exhibit will have over 100 jigsaws in it, dating back to 1766.


The Future is Puzzling, 
but Cubing is Forever.

Pete beck                  <PBECK@ARDEC>
..................................

Received: from ardec-lcss.arpa (TCP 30003004013) by AI.AI.MIT.EDU 19 May 88 12:32:10 EDT
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       Thu, 19 May 88 12:27:29 EST
Date: Thu, 19 May 88 12:26:10 EST
From: BECK@ardec-lcss.arpa
Subject: cube museum
To: cube-lovers@ai.ai.mit.edu
X-VMS-Mail-To: EXOS%"cube-lovers@mit-ai"
Message-ID: <880519122610.2020087F051@ardec-lcss.arpa>


-->   CUBE  MUSEUM  <--

On April 29, 1988 a museum devoted to Rubik's cube opened in  Grand Junction,
CO.   The museum is run by  Cecil Smith and is located in his home at 329 Ouray
Ave;  245-6734.

Cecil is primarily a documentor of pretty patterns and has 4,900 cubes in his
collection.  SO when in Grand Junction don't miss this one of a kind museum.  If
you know of or have something that should be in this museum please contact
Cecil.

REFERENCE:  The front page of the May 1, 1988  issue of the Grand Junction, Co
Daily Sentinel (Vol 96, No 153).

the future is puzzling,
but  CUBING is forever !!

pbeck@ardec.arpa

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Subject: subscription


[]

Please put me on the CUBE-LOVERS mailing list.

Thank you,
           Sergiu Hart
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                Tel-Aviv University
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       Fri,  3 Jun 88 09:59:01 EST
Date: Fri,  3 Jun 88 09:56:08 EST
From: BECK@ardec-lcss.arpa
Subject: cube memorabilia
To: cube-lovers@ai.ai.mit.edu
X-VMS-Mail-To: EXOS%"cube-lovers@mit-ai"
Message-ID: <880603095608.2080015B041@ardec-lcss.arpa>

Hi CUBE-LOVERS,

I am a "collector" of Rubik's cubes and other magic polyhedra.  As a
collector I am not only interested in the puzzles themselves but also
in the literature about them, the packaging of them and the
merchandise/events that traded on the popularity of the cube.    Below
is a crude taxonomy with  items I have identified.  I would appreciate
criticism of the taxonomy and additions to the specific items . 
 Also, if anybody has or knows where to obtain any items of this genre
please let me know.   Since I am not personally a collector of most
books, articles or solution algorithms about the cube (Bandelow, Helm,
Singmaster, et al are doing that) please do not provide citations
unless they are especially noteworthy.

TAXONOMY for RUBIK'S CUBE ITEMS (5/26/88 REV)

0.  CUBE PATENTS
0.1  US PATENT#3,081,089,  william Gustafson, 1958
0.2  Frank Fox, 1970
0.3  US PATENT#  , LARRY Nichols, 1972  
0.4  HUGARIAN PATENT, ERNO RUBIK, 1975
0.5  JAPANESE PATENT,  Terutoshi Ishige, 1976

1.  RUBIKOID PUZZLES - see photo on page 138/139 of "PUZZLES OLD &
NEW"  (PON) by Jerry Slocum & Jack Botermans

2.  CUBE EPHEMERA  -   
2.1 ADVERTISING - 
2.1.1  counter display boxes; I have for 20mm keychain size cubes and
for the 3x3x3, Rubik's Revenge, Alexander's Star, and Missing Link ITC
solution books.
2.1.2 CATALOGS & FLYERS
2.1.2.1 BANDELOW'S CATALOG
2.1.2.2 MEFFERT'S FLYERS
2.2 PACKAGING -
2.2.1  cardboard diplay box used by ITC for original cube shrink
wrapped
2.2.2  clear plastic display cylindrical dome for black plastic base
used by ITC for DELUXE cube
2.3  ANNOUNCEMENTS OF CUBE RELATED EVENTS, EG, CONTESTS,  CONFERENCES,
2.3.1  A contest announcement -> BUDAPEST INTERNATIONAL - HAVE A GO  
  WITH RUBIK, CHALLENGE THE WORLD CHAMPION, it also has a photo Rubik
holding a cube and one of Singmaster wearing a cube T-shirt; OBTAINED
FROM DAVID SINGMASTER
2.3.2  1987 (11/18-11/28) puzzle exhibition (expsoition casse-tete) by
Marcel Gillen & Carlo Gitt in Luxembourg

3.  BOOKS - not my area of interest

4.  ARTICLES - not my area of interest

5.  CLUBS/NEWSLETTERS/MUSEUM/EXHIBITS
5.1   On April 29, 1988 a museum devoted to Rubik's cube opened in
 Grand Junction, CO.   The museum is run by  Cecil Smith and is
located in his home at 329 Ouray Ave;  245-6734.   REFERENCE:  The
front page of the May 1, 1988  issue of the Grand Junction, Co Daily
Sentinel (Vol 96, No 153).Press release of opening
5.2  Wally Webster's exhibit in Kirkland, WA.  Press release of

6.  SOLUTION ALGORITHMS INCLUDING COMPUTER PROGRAMS
6.1  A 33RPM LP record with a solution to the cube:  "How to Solve the
Cube Puzzle", The Marko Van Eckelen's Method - Guiness 36 second
record holder; Gateway Records (GSLP-4506)^), GENCOM INC, POB 5087,
FDR STATION, NY, NY 10150
6.2  COMPUTER PROGRAMS
6.2.1  cartridge for RADIO SHACK TRS-80 microcomputer called "COLOR
CUBES";  sold with the cartridge, book, and keyboard cover.

7.   PRETTY PATTERNS

8.  CUBE ACCESSORIES  -  
8.1 REPLACEMENT STICKERS
8.1.1 cube covers; PON
8.1.2 CUbe Mates (Cinderella Co., POB 265, Skykesville, MD 21784) is a
set of 54 lettered stickers to put on your cube in order to play word
games;  CC# 5&6, pg 5
8.1.3  (Eidolon LTD, Vancouver V6B 3X9); a)  computer font numbers,
 b) solid silver foil
8.1.4  (Steven Mfg Co, Hermann MO 65041); large selection
8.2 REPAIR/BUILD-A-CUBE KIT; PON

9.  FAN ITEMS  -  

9.1  DECALS - 
9.1.1  An oval shaped 5"x3" decal with a picture of the cube in the
center, saturn on the left, earth on the right and written at the top
is "I do the cube" and at the bottom "RUBIK'S CUBE CLUB"; OBTAINED
FROM DAVID SINGMASTER in 1986
9.1.2  A sticker of the cube approximately 1 1/2" by SANDYLION, 340
Alden road, Markham, Ontarion, Canada L3R 4C1,  416/475-0554.
9.2  A thin rubberized magnet approximately 1 1/2".
9.3  CAR STRIPS - A car strip (goes on inside of window) that says
'CUBISTS DO IT IN 52 MOVES"; OBTAINED FROM DAVID SINGMASTER in 1986
9.4  BUMPER STICKERS
9.5  BUTTONS- I "heart" Rubik's cube & buttons with sayings and
pictures of cubes from Singmaster in 1986;  PON
9.6  Hungarian POSTAGE STAMP & FIRST DAY COVER (6/4/82), PON &CC# 5&6,
pg 28
9.7  CLOTHING  - 
9.7.1  A childs Tee shirt with a picture of a cube above which is
 written 'RUBIK'S CUBE"; OBTAINED FROM DAVID SINGMASTER in 1986
9.7.2  A mans necktie; black with a cube on it;  PON
9.7.3  A childs halloween costume, size large (12-14), fits an 8-10
year old, from Collegeville Flag & MFG Co., Collegville, PA 19426 (I
have some for trade)
9.7.4  Sew on patch from ITC cube club; PON
9.8  Ink stamp - I have a rubber ink pad stamp of a cube
9.9  POSTER OF THE JIGSAW PUZZLE
9.10  CUBE IN BOTTLE; PON
9.11  CUBE SMASHER; PON
9.12  UTILITARIAN ITEMS TRADING ON CUBE -  
9.121 COASTERS - A set of coasters to put glasses on:  Six 3"x3"
lucite pieces with 9 silk screened squares each colored one of the
colors of a cube, comes in a lucite holder and is called "Cubics
Coasters in Six Winning Colors", a quality product from Caryl Craig
Studios, c 1982, Box 2221 Sepulveda, CA 91343 (I got mine from Greg
Stevens in 1987)
9.12.2  PENCILS with I "heart" Rubik's cube printed on them; PON
9.12.3  SHOELACES in both 27" & 40" lengths made in Taiwan for
Goodties of LA, CA.  Imprinted with a solution algorithm (Greg Stevens
has for trade)
9.12.4  LUNCH BOXES
9.12.5  BOOK BAGS - I have a red canvas briefcase type bag
9.12.6  COFFEE CUPS
9.12.6.1  "IT'S A Mugs GAme",  plain white mug with decals pasted on
One says " IT'S A Mugs GAme", another is of a mixed cube, and another
is of a cube at start. Mug made in UK;  CC#3&4, PG11 (I got mine from
Bandelow in 1988)
9.12.7  NOTE PADS;  CC#3&4, PG11
9.12.8  ERaser - I have 1" cube rubber pencil eraser
9.12.9  Duvet (comforter) cover; Amsterdam; CC# 5&6, pg 5
9.12.10  Greeting card,  HALLMARK "150B 905-3" bought in 1988;  HAPPY
BIRTHDAY, SON;  has a picture of a cat in a checkered sweater holding
a cube like item in one hand with a saying "a little bit of genius".
9.12.11  Hot plate -  white ceramic tile with a decal pasted on (BR is
registered mark).  The picture is of a women sitting on a mixed cube
thinking of how to solve the cube.  The floor is littered with paper
notes of failed solution attempts.  Tile is german, decal ?  (I got
mine from Bandelow in 1988)
9.13  GAMES/TOYS  -
9.13.1  Jigsaw puzzles
9.13.1.1  "Cube Twister", BOX COVER a single 9x9x9 cube; PON
9.13.1.2  "Cube Twister", box cover has many cubes;
one version has "rubik's cube" as part of box markings other does not
but it has a disclaimer denying any connection with ITC or Rubik
9.13.1.3  "RUBIK'S ZIG ZAW PUZZLE" from ITC
9.13.2  Rubik's Race a board game,  ITC,  CC#3&4, PG 8
9.13.3  Rubik's Challenge/Mill "GAME" is a tic-tac-toe variant played
on the surface of a cube, ITC;  CC#3&4, PG 8
9.13.4  "Color Match" a color card strategy board game marketed by
ITC, does not use a cube.  Object of game is to match colored squares
in a 3x3 grid, like doing one side of a cube.

10.  CULTURAL IMPACT
10.1     MUSIC
10.1.1  "Mr. Rubik" recorded by the Barron Knights and released by
Epic as a single EPC A 1596 and on the LP
10.1.2  "TRICK IN THE MIDDLE" by Bea Muszty and Andras Dobay:  Start
Records SP5 70537;  CC#3&4, PG 2
10.1.3  "The Cube" by Mike Brady and the Cubettes, australian, has a
video clip with dancers;  CC#3&4, PG 2
10.2    POETRY 
10.2.1  "A Rubric on Rubik Cubics" by Claude Shannon;  1st mention
CC#3&4, PG 2 - printed in CC# 7&8, pg 36
10.3    APPEARANCE IN ADVERTISEMENTS/MOVIES - 
10.3.1 movies -  MOONSTRUCK
10.3.2   cartoon strips
10.3.3  POLITICAL CARTOONS
10.3.3.1  NY Daily News Sunday magazine section of 2/7/88 cartoon by
Jeff MacNelly called "I.R.S.  TAX REFORM CUBE",  frustrated looking
person holding a cube that has IRS type things written on it in place
of stickers.
10.3.4  crossword puzzles  
10.3.5  ARVEY paper & supplies,  3351 West Addison, Chicago, IL 60618
used pictures of cube in their April 1988 sales flyer whose theme was
an "AMAZING SALE"                     
10.4   TV SHOWS ABOUT (Greg Stevens has a tape of 10.4.3-10.4.5 for
trade)
10.4.1  BBC TV appearance by Rubik and Nicholas Hammond (a cube
solver) on Swap-Shop 24 January 81,  CC#1, pg 15
10.4.2 NBC TV show "That's Incredible" covered the US speed
championship which took place on 13 November 1981 and was aired on 7
December 1981.  Minh Thai won with a time of 26.04 seconds.  CC#2, pg
4
10.4.3  Interview with Rubik on the TV show "the Rich & Famous"
10.4.4  Episode ?? of "Night Court" TV show with Bull Shannon playing
with cube.
10.4.5  An RC Cola TV commercial that uses a 4x4x4 cube as focus.
10.5  BOOKS
10.5.1  a novel entitled "Rubik's Cube" by Leela Dutt, published by
Gee & Son LTD, Dendigh, Clwyd, Wales, 1984;  CC# 7&8, pg 46
10.5.2  "THE Berenstain Bears and TOO MUCH TV",  ISBN 0-394-86570,
RANDOM HOUSE 1984, $1.95;  lesson is that there are things to do
besides watching TV like solving a cube puzzle,  2 picture plates
include a cube.
10.6  jokes
10.6.1  CUBIC SICK JOKE:  What goes "click-click -  have done it?",  A
blind man doing Rubik's cube. - Oh No! Not Another 1000 Jokes for
KIds, Ward Lock, London, 1983, (Michael Kilgarriff ???);  CC# 7&8, pg
45
10.7     EDUCATION CURRICULAR
10.8   NEW WORDS
10.8.1  CUBITIS MAGIKIA, N.  A severe mental disorder accompanied by
itching of the fingertips,  which can be relieved only by prolonged
contact with a multicolored cube originating in Hungary and Japan.
 Symptoms often last for months.  Highly contagious.   - METAMAGICAL
THEMAS, ISBN 0-465-04540-5, Douglas Hofstadter, pg 301
10.8.2  SKEWB, N.  The name given by D Hofstadter to a magic polyhedra
puzzle invented by Tony Durham.  The puzzle is cube shaped with for
planes of movement corresponding to the diagnols of the cube.  -
 METAMAGICAL THEMAS, ISBN 0-465-04540-5, Douglas Hofstadter, pg 341
10.8.3  TWOBIK



-->  thanks for any inputs/corrections or general comments.

<pbeck@ardec.arpa>


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Date: Fri, 10 Jun 88 15:12:51 -0100
From: kuling!starback@uunet.UU.NET (Per Starb{ck)
Message-Id: <8806101412.AA21284@kuling.UU.SE>
To: cube-lovers@ai.ai.mit.edu
In-Reply-To: BECK@ardec-lcss.arpa's message of Fri,  3 Jun 88 09:56:08 EST
Subject: Early patents

In his taxononomy of cube items (<880603095608.2080015B041@ardec-lcss.arpa>) 
Pete Beck (pbeck@ardec.arpa) wrote:


>   TAXONOMY for RUBIK'S CUBE ITEMS (5/26/88 REV)
>
>   0.  CUBE PATENTS
>   0.1  US PATENT#3,081,089,  william Gustafson, 1958
>   0.2  Frank Fox, 1970
>   0.3  US PATENT#  , LARRY Nichols, 1972  
>   0.4  HUGARIAN PATENT, ERNO RUBIK, 1975
>   0.5  JAPANESE PATENT,  Terutoshi Ishige, 1976
> ...

I only knew about 0.4 (of course) and 0.5 (mentioned for instance in
Singmaster's Notes). Could anyone inform me about 0.1, 0.2, and 0.3 please.
Are they the normal 3x3x3-cube, or not?
--
Quote: "Life is but a ramble! Let flipism chart your ramble!"
Per Starback			starback@kuling.UU.SE
Karlsrogatan 13:H33		 or
S-752 38  UPPSALA		starback%kuling.UU.SE@uunet.UU.NET
SWEDEN

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       Tue, 14 Jun 88 11:06:22 EST
Date: Tue, 14 Jun 88 11:05:46 EST
From: BECK@ardec-lcss.arpa
Subject: patents
To: cube-lovers@ai.ai.mit.edu
X-VMS-Mail-To: EXOS%"cube-lovers@mit-ai"
Message-ID: <880614110546.2080017F071@ardec-lcss.arpa>

CUBE PATENTS:

The best overall discussion of cube patents is in Metamagical
Themas,pg 356; D Hofstadter.  Hofstadter says "that the real reason
the Rubik-Ishige Cube took off was that it could be manufactured and
that it did hold together".   In Puzzles Old and New,
Slocum&Boetermans it is stated that the courts gave the US patent
rights to Nichol's.   The is a further discussion of Rotary Puzzle
patents in Doug Engel's self published book "Circle Puzzler's Manual".

If anybody has further patent details please send me citations or even
rumors.  Of especial interest is information on the granting of the US
rights to Nichol's.

I've added some more details to the patent section.
0.  CUBE PATENTS  (see Metamagical Themas,pg 356; D Hofstadter for an
overview)
0.1  US PATENT#3,081,089,  william Gustafson, 1958 (2x & 3x)
0.2  Frank Fox, 1970 British (3x)
0.3  US PATENT#  , LARRY Nichols, 1972 (2x) 
0.4  HUGARIAN PATENT, #170062, ERNO RUBIK, Jan 30,1975&Oct 28,1976&Dec
31,1977 (3x &2x cubes)
0.5  JAPANESE PATENTS
0.5.1  JAPANESE PATENT,#55-8192,  Terutoshi Ishige, Oct 10,1976&Apr
26,1978&Mar 3,1980,(3x)
0.5.2  JAPANESE PATENT,#55-8193,  Terutoshi Ishige, Mar 12,1977&Oct
10, 1978&Mar 3,1980; (2x)
0.5.3  JAPANESE PATENT,#55-3956,  Terutoshi Ishige, Oct 21,1978&Jan
28,1980,(3x)


THE FUTURE IS PUZZLING, 
but CUBING IS FOREVER !!


peter beck

<pbeck@ardec>  or  <beck@ardec-lcss>


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       Thu, 16 Jun 88 10:50:44 EST
Date: Thu, 16 Jun 88 10:49:35 EST
From: BECK@ardec-lcss.arpa
Subject: HEXAGON MAGIC VARIATION
To: CUBE-LOVERS@ai.ai.mit.edu
X-VMS-Mail-To: EXOS%"CUBE-LOVERS@MIT-AI"
Message-ID: <880616104935.20800176041@ardec-lcss.arpa>

HI CUBE-LOVERS, 

If you enjoyed Rubik's magic you might enjoy "BETCHA CAN'T!, FLEXIBLE HEXAGON"
 This puzzle is constructed like magic, ie, with panels jointed by nylon fishi
line.  It is made up of 6 hexagon panels that are jointed on 2 edges.  The
panels can be manipulated from a hexagon (6 panels around a hexagon hole) to a
2x3 grid.  I didn't buy mine but the price label looks like it was from Toys R


"BETCHA CAN'T !, FLEXIBLE HEXAGON"
TANDEM TOYS
ROLLING HILLS, CA
COPYRIGHT 1987


The Future is Puzzling,
but Cubing is Forever !!

Pete beck                  <BECK@ARDEC-LCSS> or <pbeck@ardec>
..................................



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       Fri, 17 Jun 88 09:17:37 EST
Date: Fri, 17 Jun 88 09:16:34 EST
From: BECK@ardec-lcss.arpa
Subject: cff #17
To: cube-lovers@ai.ai.mit.edu
X-VMS-Mail-To: EXOS%"cube-lovers@mit-ai"
Message-ID: <880617091634.20400156041@ardec-lcss.arpa>


SUBJECT :  Review of "Cubism For Fun" newsletter issue #17, may 1988;
the newsletter of the "Dutch Cubists Club";  in english starting with
issue #14 

1..  The table of contents for issue #17, may 88 follows (it is 6
double sided folded 8 1/2 x 11 sheets, making a 24 page newsletter):

WELCOME by Klaas Steenhuis
SECRATERIAL REPORT by Guus Schultz 
REVIEW OF "CUBISTS DAY"  BY ANNEKE TREEP 
REVIEW OF "CUBISTS DAY"  BY Hans Dockhorn
NKC Contests by Guus Schultz (MARC  WATERMAN had fastest time 17.1
secs)
Committee and Editing Team photo by Frankie van Dam
Hot News on the UPPER TABLE AVERAGED BY Anton Hanegraaf
Updated Averages by BEN JOS WALBEEHM
The Rank and File Pattern by Wally Webster
PRETTY CUBIC PATTERNS BY ANNEKE TREEP 
PRETTY MAGIC STRUCTURES BY Guus Schultz 
THE SKEWB, Part 1 BY RONALD FLETTERMAN  
NEWS AND LETTERS TO THE EDITOR  - new  Rubik's items, rubik's magic
strategy game,  rubik's clock;  Georges Helm's cube bibliography has
more than 500 citations;  Peter Beck is compiling a cubik memorabilia
list;  
LIST OF MEMBERS - 49 currently
--> 8 1/2 x 11 double sided insert of cubes/puzzles forsale/wanted <--

2.   Membership for 1988 is  US$8.   A photocopied set of the
newsletters, issues 1-13, written in DUTCH (in the future selected
back articles will be available in english) is also available for
 US$7.    To order either of these send an 'INTERNATIONAL"  POSTAL
MONEY ORDER to:  Anton Hanegraaf,  Heemskerkstraat,  6662 AL ELST,
 The Netherlands.

3.  If anybody would like further details please ask!

CUBING IS FOREVER      PETER BECK

<BECK@ARDEC-LCSS>  OR  <PBECK@ARDEC>


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From:     KILGORBL%SNYPLABA.BITNET@MITVMA.MIT.EDU
Subject:  Hope this is correct?


SUBSCRIBE CUBE-LOVERS Brian Kilgore

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       Wed,  6 Jul 88 10:12:00 EST
Date: Wed,  6 Jul 88 10:10:46 EST
From: BECK@ardec-lcss.arpa
Subject: oxford u's rubik book update
To: cube-lovers@ai.ai.mit.edu
X-VMS-Mail-To: EXOS%"cube-lovers@mit-ai"
Message-ID: <880706101046.2020019E042@ardec-lcss.arpa>
Subj:  New RECREATIONS IN MATH editions - Rubik's Cubic Compendium

Rubik's Cubic Compendium, by Erno Rubik, Tamas Varga, Gerzon Keri,
Gyorgy Marx, and tamas Vekerdy.  200pp; 142 illust., 61 color -
OUP#853202-4, list price $26.95.
-->  YES ITS AVAILABLE!!  I have ordered some copies, and when I
receive             them I will write a first hand review. <--

Hype from OUP copy,  
> This book co-written by the cube's inventor, and serves as a 
> comprehensive guide to the cube for both the puzzler and the 
> mathematician. The book reveals the wealth of fascinating
mathematics
> concealed within the cube's apparently simple operation, and even
those
>  who have solved the cube will discover a vast number of new ideas
and
>  possibilities.

The Oxford U Press Rec Math series (series editor David Singmaster) to
date is: 
#1  "Mathematical byways ..." by Hugh ApSimon. 
 #2: Ins and Outs of Peg Solitaire.  
 #3: Rubik's Cubic Compendium, by Rubik, et al.  
 #4 Sliding Piece Puzzles (Hordern). 


Received: from Xerox.COM (TCP 1500006350) by AI.AI.MIT.EDU 26 Jul 88 16:38:01 EDT
Received: from Gamay.ms by ArpaGateway.ms ; 26 JUL 88 13:24:12 PDT
Sender: Hoffman.es@Xerox.COM
Date: 26 Jul 88 13:25:00 PDT (Tuesday)
Subject: Rubik's Clock
To: Cube-Lovers@AI.AI.MIT.EDU
From: Rodney Hoffman <Hoffman.es@Xerox.COM>
cc: Hoffman.es@Xerox.COM
Message-ID: <880726-132412-1483@Xerox>


From the 'Los Angeles Times', July 26, 1988:

  So you figured out Rubik's Cube, eh?  Well, get ready for
  Rubik's Clock -- a puzzle the Hungarian professor says he 
  hasn't even solved yet, a spokeswoman for the toy company
  marketing the product said Monday.  But Erno Rubik's failure
  has not arisen from a lack of ability, merely a lack of time
  to puzzle out the secrets of Rubik's Clock, said Melanie 
  Bateman of Matchbox Toys Ltd.  "Really, it's not because it's
  impossible, he's just too busy to take the time to do it,"
  she said, adding that the new toy will be launched at a major
  London toy store on Saturday.  What can you expect to pay for
  more hours of mental anguish?  About $12, Bateman said.

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Sender: Hoffman.es@Xerox.COM
Date: 26 Jul 88 14:39:50 PDT (Tuesday)
Subject: Rubik's Clock
To: Cube-Lovers@AI.AI.MIT.EDU
From: Rodney Hoffman <Hoffman.es@Xerox.COM>
Message-ID: <880726-143858-1670@Xerox>


From the 'Los Angeles Times', July 26, 1988:

  So you figured out Rubik's Cube, eh?  Well, get ready for
  Rubik's Clock -- a puzzle the Hungarian professor says he 
  hasn't even solved yet, a spokeswoman for the toy company
  marketing the product said Monday.  But Erno Rubik's failure
  has not arisen from a lack of ability, merely a lack of time
  to puzzle out the secrets of Rubik's Clock, said Melanie 
  Bateman of Matchbox Toys Ltd.  "Really, it's not because it's
  impossible, he's just too busy to take the time to do it,"
  she said, adding that the new toy will be launched at a major
  London toy store on Saturday.  What can you expect to pay for
  more hours of mental anguish?  About $12, Bateman said.


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Sender: Hoffman.es@Xerox.COM
Date: 6 Aug 88 17:33:32 PDT (Saturday)
Subject: Re: Rubik's Clock
In-follow-up-to: My message of 26 Jul 88 13:25:00 PDT (Tuesday)
 <880726-132412-1483@Xerox>
From: Rodney Hoffman <Hoffman.es@Xerox.COM>
To: Cube-Lovers@AI.AI.MIT.EDU
Message-ID: <880806-173310-2387@Xerox>


Recently, I posted (twice -- sorry!) a short note from the 'Los Angeles Times'
about Rubik's Clock.  Another paper had just a little more information.  From
the 'Los Angeles Herald Examiner', July 26, 1988:


                      RUBIK'S PUZZLING NEW TWIST
		      
    Father of cube craze offers clock toy for those with time on their hands
    
    
    LONDON (AP) - The Hungarian professor [pictured] who frustrated 
    millions with his Rubik's Cube is introducing his latest mind-twister 
    -- a puzzle he says even he hasn't solved, a spokeswoman for the toy 
    company marketing the product said yesterday.
    
    But Erno Rubik's failure has not arisen from a lack of ability, 
    merely a lack of time to puzzle out the secrets of Rubik's Clock, 
    said Melanie Bateman of Matchbox Toys Ltd.  "Really, it's not because 
    it's impossible, he's just too busy to take the time to do it," she said, 
    adding that the new toy will be launched at a major London toy store on
    Saturday.
    
    Rubik's latest brain-teasing toy requires a player to get 18 clocks 
    on both sides of a plastic disc to strike midnight simultaneously by 
    twisting wheels that turn some of the hands but not others, said Bateman.
    
    Speaking of his latest invention, Rubik warned in a press release:  
    "Don't cheat by being taught how to do it by someone else.  It is 
    much more satisfying to decode the puzzle on your own."
    
    "It is important to remember that your brain needs to be kept in shape 
    ... My new puzzle can help because it enables you to focus entirely on 
    finding the formula which, although it may seem frustrating at the time, 
    will do you good," he added.
    
    Rubik's Clock will retain for about $12, Bateman said.
    
    Rubik, 44, a professor of architecture and design at the Academy of Arts 
    and Crafts in Budapest, invented the multicolored Rubik's Cube as a 
    teaching aid.
    
    After the success of the cube, which sold more than 120 million units 
    worldwide, he founded a private business, Rubik's Studio, in conjunction 
    with the Hungarian government.

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       Wed, 10 Aug 88 10:13:45 EST
Date: Wed, 10 Aug 88 10:13:00 EST
From: BECK@ardec-lcss.arpa
Subject: UPDATE ON RUBIKS BOOK
To: CUBE-LOVERS@ai.ai.mit.edu
X-VMS-Mail-To: EXOS%"CUBE-LOVERS@MIT-AI"
Message-ID: <880810101300.20200E05051@ardec-lcss.arpa>

8/9/88 update
Subj: New RECREATIONS IN MATH editions - Rubik's Cubic Compendium

Rubik's Cubic Compendium, by Erno Rubik, Tamas Varga, Gerzon Keri,
Gyorgy Marx, and Tamas Vekerdy.  240pp; 203 line drawings
OUP#853202-4, list price $26.95.
-->  YES ITS AVAILABLE!!  I have ordered some copies, and when I
receive             them I will write a first hand review. <--

Hype from OUP copy,  
> This book co-written by the cube's inventor, and serves as a 
> comprehensive guide to the cube for both the puzzler and the 
> mathematician. The book reveals the wealth of fascinating
mathematics
> concealed within the cube's apparently simple operation, and even
those
>  who have solved the cube will discover a vast number of new ideas
and
>  possibilities.

CONTENTS:
  Introduction:  The Fascination of Rubik's Cube
    1.  In Play
    2.  The Art of Cubing
    3.  Restoration Methods and Tables of Processes
    4.  Mathematics
    5.  The Universe of the Cube
    6.  My Fingers Remember
    7.  Afterword

The Oxford U Press Rec Math series (series editor David Singmaster) to
date is: 
#1  "Mathematical byways ..." by Hugh ApSimon. 
 #2: Ins and Outs of Peg Solitaire.  
 #3: Rubik's Cubic Compendium, by Rubik, et al.  
 #4 Sliding Piece Puzzles (Hordern). 
   
 TO ORDER:  Send check or credit card info (MASTERCARD OR VISA) to:
        SCIENCE & MEDICAL MARKETTING DIRECTOR,
        OXFORD UNIVERSITY PRESS
        200 MADISON AVE,
        NEW YORK, NY 10016  - 
-- >  ADD $1.50 for shipping



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       Fri, 21 Oct 88 13:24:00 EST
Date: Fri, 21 Oct 88 13:23:03 EST
From: BECK@ardec-lcss.arpa
Subject: GAMES FAIR
To: CUBE-LOVERS@ai.ai.mit.edu
X-VMS-Mail-To: EXOS%"CUBE-LOVERS@MIT-AI"
Message-ID: <881021132303.2140021D041@ardec-lcss.arpa>


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%        --->      CALL FOR PARTICIPATION       <---         %
%                     ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^       
              %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


THE NEW YORK GAMES FAIR
NOV 11-13, 1988 (VETERANS DAY WEEKEND)
ROOSEVELT HOTEL
MADISON AVE @ 45TH STREET
NY NY 10117


A PROFFESIONALLY ORGANIZED (JACK JAFFE OF UK) CONSUMER PARTICIPATION
FAIR (YES ADMISSION WILL BE CHARGED, I THINK ABOUT &5) FEATURING BOARD
GAMES, COMPUTER GAMES, ROLE PLAYING GAMES, PROTOTYPE GAMES,
PLAY-BY-MAIL GAMES, PUZZLES, GAME BOOKS, AND CONJURING APPARATUS.

IF YOU WANT TO EXHIBIT/SELL OR IN SOME OTHER WAY BE INVOLVED CONTACT:
    PAMELA JOHNSON
    THE NEW YORK GAMES FAIR
    SUITE 1121
    122 E 42ND STREET
    NY NY  10168
    212/986-3469


<PBECK@ARDEC>


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       Thu, 27 Oct 88 08:33:30 EST
Date: Thu, 27 Oct 88 08:32:25 EST
From: BECK@ardec-lcss.arpa
Subject: CFFF #18
To: CUBE-LOVERS@ai.ai.mit.edu
X-VMS-Mail-To: EXOS%"CUBE-LOVERS@MIT-AI"
Message-ID: <881027083225.20801361041@ardec-lcss.arpa>

SUBJECT :  Review of "Cubism For Fun" newsletter issue #18, SEPT 1988;
the newsletter of the "Dutch Cubists Club";  in english starting with
issue #14 

1..  The table of contents for issue #18, SEPT 88 follows:

INVITATION to Dutch "CUBISTS DAY"  BY  Jan de Geus;  (saturday, 10 Dec
1988 at his house)
Short History of Dutch Club BY ANNEKE TREEP 
Cube Museum by Cecil Smith
Cube Timer (for speed contests) by Paul Sijben
Latest Results for the UPPER TABLE  BY Anton Hanegraaf
U-Table: We came close  by Hans & Kurt Dockhorn
Notes on the preceeding by BEN JOS WALBEEHM
THE SKEWB, Part 2 BY RONALD FLETTERMAN  
The Dockhorn-Treep Production BY ANNEKE TREEP
PRETTY CUBIC PATTERNS BY ANNEKE TREEP 
Pretty SKEWBic Patterns  BY RONALD FLETTERMAN 
Book Reviews;  RUBIC'S CUBIC COMPENDIUM (RUBIK ETAL), SLIDING PIECE
PUZZLES (HORDERN) , SOLUTIONS TO VARIOUS ROTATIONAL AND MECHANICAL
SLIDING PIECE PUZZLES (HORDERN) 
NEWS AND LETTERS TO THE EDITOR  - Rubik's Clock, BetchaCan't,
 Three-dimensional labyrinths, Jan de Geus's collection.
LIST OF MEMBERS - 52 currently

2.   Membership for 1988 is  US$8.   A photocopied set of the
newsletters, issues 1-13, written in DUTCH (in the future selected
back articles will be available in english) is also available for
 US$7.    To order either of these send an 'INTERNATIONAL"  POSTAL
MONEY ORDER to:  Anton Hanegraaf,  Heemskerkstraat,  6662 AL ELST,
 The Netherlands.

3.  If anybody would like further details please ask!

CUBING IS FOREVER      PETER BECK

<BECK@ARDEC-LCSS>  OR  <PBECK@ARDEC>

Received: from ardec-lcss.arpa (TCP 30003004013) by AI.AI.MIT.EDU 27 Oct 88 08:36:25 EDT
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       Thu, 27 Oct 88 08:35:49 EST
Date: Thu, 27 Oct 88 08:34:55 EST
From: BECK@ardec-lcss.arpa
Subject: RUBIK'S CLOCK
To: CUBE-LOVERS@ai.ai.mit.edu
X-VMS-Mail-To: EXOS%"CUBE-LOVERS@MIT-AI"
Message-ID: <881027083455.20801361042@ardec-lcss.arpa>


Subj:   Re: Rubik's Clock

  >   Rubik's latest brain-teasing toy requires a player to get 18
clocks 
  >   on both sides of a plastic disc to strike midnight
simultaneously by 
  >   twisting wheels that turn some of the hands but not others.  

There are interlocking devices that make the puzzle a "SLIDING BLOCK"
sequential motion puzzle.
    
The puzzle is on sale now in the UK and probably will not see US or
europe distribution for Xmas 1988.

There are solution book already in print.

Christoph Bandelow tells me that the puzzle is not particular
challenging and that he does not expect it to be a great success.




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       Wed, 21 Dec 88 10:47:14 EST
Date: Wed, 21 Dec 88 10:46:02 EST
From: BECK@ardec-lcss.arpa
Subject: cff #19
To: cube-lovers@ai.ai.mit.edu
X-VMS-Mail-To: EXOS%"cube-lovers@mit-ai"
Message-ID: <881221104602.20400587041@ardec-lcss.arpa>

SUBJECT :  Review of "Cubism For Fun" newsletter issue #19, DEC 1988;
the newsletter of the "Dutch Cubists Club";  in english starting with
issue #14 

1..  The table of contents for issue #19, DEC 88 follows:

Welcome to Erno Rubik on becoming an honorary member of club
MEETING RUBIK  by  Guus Razoux Schultz
EVEN AND ODD PERMUTATIONS by Christoph Bandelow
RUBIK'S CLOCK by the editors
SOLUTION TO THE CLOCK by  Ed Hordern
MATHEMATICS FOR THE CLOCK by  Guus Razoux Schultz 
more PRETTY CUBIC PATTERNS BY ANNEKE TREEP 
THE SKEWB's R,r group BY RONALD FLETTERMAN 
NEWS AND LETTERS TO THE EDITOR  - U-TABLE NEWS, CUBE PROCEDURE
RESEARCH, VALUE OF cubes for exchnage, plastic fatigue of cube
spindles
LIST OF MEMBERS - 

>>  NOTE:  Rubik's clock is being distributed in europe with a planned
introduction to the USA in April 1989.

>>  Christoph Bandelow's 1989 puzzle catalog of Rubik's cube type
puzzles was sent along with this issue.  If you want a copy write him
and enclose US$1(send an actual dollar bill) for postage;
....   Christoph Bandelow
....   Haarholzer Str. 13
....   4630 Bochum-Stiepel
....   W. Germany


2.   Membership for 1988 is  US$8.   A photocopied set of the
newsletters, issues 1-13, written in DUTCH (in the future selected
back articles will be available in english) is also available for
 US$7.    To order either of these send an 'INTERNATIONAL"  POSTAL
MONEY ORDER to:  Anton Hanegraaf,  Heemskerkstraat,  6662 AL ELST,
 The Netherlands.

3.  If anybody would like further details please ask!

CUBING IS FOREVER      PETER BECK

<BECK@ARDEC-LCSS>  OR  <PBECK@ARDEC>

Received: from ardec-lcss.arpa (TCP 30003004013) by AI.AI.MIT.EDU  1 Feb 89 07:34:55 EST
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       Wed,  1 Feb 89 07:37:39 EST
Date: Wed,  1 Feb 89 07:36:50 EST
From: BECK@ardec-lcss.arpa
Subject: whats new
To: cube-lovers@ai.ai.mit.edu
X-VMS-Mail-To: EXOS%"cube-lovers@mit-ai"
Message-ID: <890201073650.21000F95041@ardec-lcss.arpa>

HI CUBE-LOVERS, 

PUZZLE AVAILABILITY:
1.  KAYBEE is closing out "YOSHI'S" puzzle for $2.  This is a linked
set of pieces folding puzzle.  Good for solid geometry perception.
 Buy more than one, there are interesting objects with 2.

2.  Binary Arts has a new puzzle (availability april 89) called
"TOP-SPIN".  This is a railway shunting/turntable puzzle.  It has a
circular channel with 20 numbered beads with a superimposed
circle/turntable that can change the direction of 4 beads.  object is
to havbe the beads consecutively number in either the clockwise or
counter clockwise direction.


The Future is Puzzling,
but Cubing is Forever,

Pete beck                  <BECK@ARDEC-LCSS>
..................................



Received: from MCC.COM (TCP 1200600076) by AI.AI.MIT.EDU  1 Mar 89 05:01:57 EST
Date: Wed 1 Mar 89 04:00:05-CST
From: Clive Dawson <AI.CLIVE@MCC.COM>
Subject: Force Field
To: cube-lovers@ai.ai.mit.edu
Message-ID: <12474493626.38.AI.CLIVE@MCC.COM>

I acquired a cube puzzle called Force Field a few weeks
ago which I don't recall having seen discussed on this
list.  It's called Force Field (by Mattel), and it consists
of 8 cubies, each measuring about 1" on a side, and each
solid black.  The idea is to arrange these 8 cubies into
a 2x2x2 cube.  This sounds and looks trivial, until you
learn that there are magnets attached to some (but not all)
of the inner surfaces of the cubies.  This means that 
the sides of two cubies may:

	a) repel each other
	b) attract each other but jog off-center (since
	    the magnets are not necessarily at the center of the side)
	c) neither attract nor repel each other (if magnets
	   don't exist on both of the sides involved)
	d) attract each other and stay perfectly aligned

The final 2x2x2 cube has to hold together perfectly, without one
or more of the cubies popping out.  Furthermore, it is not enough
to juxtapose the sides with no magnets, since the final cube has
to be placed in a special stand which balances it on one of
its corners.  This is the acid test-- the cube might look ok when
resting on the table, but in order to survive in the stand, all
internal sides must actively attract each other with perfect
alignment.

I finally had to resort to temporary labels on the cubies in order
to systematically search for a solution.  One of the first things
you learn is that each cubie has 3 magnets, precisely the number
required.  This cuts down the search space tremendously: since the
3 sides with magnets must be internal, this constrains a particular 
corner of each cubie be at the center of the large cube.  But it
still involves over 33 million possible positions (3^8)*(7!).

I'd be interested to hear if any of you folks have played with this,
and if anybody has developed a procedure for putting the cube together
which does not involve labeling, which is what I'm working on now.

Happy cubing,

Clive
-------

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Received:  by XN.LL.MIT.EDU; Wed, 1 Mar 89 23:14:14 EDT
Date: Wed, 1 Mar 89 23:14:14 EDT
From: rp@XN.LL.MIT.EDU (Richard Pavelle)
Posted-Date: Wed, 1 Mar 89 23:14:14 EDT
Message-Id: <8903020314.AA17583@XN.LL.MIT.EDU>
To: cube-lovers@ai.ai.mit.edu
Subject: Impossible object puzzle

A "former friend" of a "former friend" passed on to me a puzzle which
appears to be impossible to solve. I do not know its name nor who
makes it. It is also hard to describe but here goes: It is a clear box
about (1.5")^3. Inside are three (3/4) yellow planes which intersect
in the center of the box and extend to the sides.
____
| x |
|   -----    
|    0  |
|_______|

Here is a picture of the 3/4 plane. The "x" and "0" are holes which
will be explained. The box contains three identical blue objects which
look like half dumbbells. They will fit through the "0" hole but only
the shaft will fit into the "x" hole. The puzzle is to insert each of
the three shafts into each of the three "x" holes.

It is easy to get one in and I once got two in. But to get all three
in simultaneously seems impossible. Does anyone know whether this
puzzle can be solved and add more info on who makes it, etc?

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Return-Path: <gcs60575@zach>
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	id AA17430; Wed, 8 Mar 89 20:19:35 EST
Date: Wed, 8 Mar 89 20:19:35 EST
From: gcs60575@zach.fit.edu ( GONZALEZ)
Message-Id: <8903090119.AA17430@zach.fit.edu>
To: CUBE-LOVERS@ai.ai.mit.edu
Subject: info


hi please send me information about your activities and
what ever you can send me thanks

my user name gcs60575
my      name Luis E Gonzalez
my address   887 emerson dr N.E.
             palm bay FL 32907


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       Fri, 10 Mar 89 12:41:45 EDT
Date: Fri, 10 Mar 89 12:40:45 EDT
From: BECK@ardec-lcss.arpa.ARDEC.MIL
Subject: REINTRODUCTION OF CUBES
To: CUBE-LOVERS@ai.ai.mit.edu
X-VMS-Mail-To: EXOS%"CUBE-LOVERS@MIT-AI"
Message-ID: <890310124045.206002CA041@ardec-lcss.arpa.ARDEC.MIL>


CUBING is back??

I here from europe (the Nuremberg Toy fair) that Rubik's Cube is back
as the 'MATCHBOX CUBE" (Rubik's is associated with Matchbox Toys).  I
presume that Ideal toy owns the trademark to Rubik's cube but not the
patent.  

The cubes displayed are standard 3x3x3 cubes with Rubik's signature
and silhouette on the center cubie (center moves will now be
required).

Anybody out there know anything more about this?

THE FUTURE IS PUZZLING,
BUT CUBING IS FOREVER

Pete Beck

<pbeck@ardec>

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       Wed, 26 Apr 89 12:00:19 EDT
Date: Wed, 26 Apr 89 11:58:18 EDT
From: BECK@ardec-lcss.arpa.ARDEC.MIL
Subject: CFF #20  3/89
To: CUBE-LOVERS@ai.ai.mit.edu
X-VMS-Mail-To: EXOS%"CUBE-LOVERS@MIT-AI"
Message-ID: <890426115818.21C00184051@ardec-lcss.arpa.ARDEC.MIL>

SUBJECT :  Review of "Cubism For Fun" newsletter issue #20, MAR 89;
the newsletter of the "Dutch Cubists Club";  in english starting with
issue #14 

1..  The table of contents for issue # 20, MAR 89 follows:

COMMITEE AND EDITORS
MEMBERSHIP FEE BY Paul Sijben
SECRETARIAL REPORT by  Guus Razoux Schultz
LOOKING BACK by Klaas Steenhuis
LOOKING FORWARD by  Guus Razoux Schultz
CUBE DAY 1988 by Klaas Steenhuis
HOLLOW MAZES by Oskar van Deventer
THE MAGIC CROSS by Anton Hanegraaf
more PRETTY CUBIC PATTERNS BY ANNEKE TREEP 
THE DOCHORN THEOREM by Anton Hanegraaf
THE EQUATOT-NUMBER by Klaas Steenhuis and Anton Hanegraaf
MINIMA IN PRACTICE by  Guus Razoux Schultz
3X3X3 CUBE WITHIN A 4X4X4 CUBE BY RONALD FLETTERMAN 
PLATE TECTONICS by Peter Beck
NEWS AND LETTERS TO THE EDITOR  - Announcement of planned Puzzle     
        PArty, aug 27 in London;  Planning for new book on Hoffman
Puzzles.
MAGAZINE REVIEW of "World Game Review"
CHANGES IN THE LIST OF MEMBERS - 

>>  NOTE:  I HAVE SEEN Rubik's clock in ToysRus for $10.

2.   Membership for 1989 is  US$8.   A photocopied set of the
newsletters, issues 1-13, written in DUTCH (in the future selected
back articles will be available in english) is also available for
 US$7.    To order either of these send an 'INTERNATIONAL"  POSTAL
MONEY ORDER to:  Paul Sijben, Witbreuksweg 397-304, 7522 ZA Enschede,
 The Netherlands.

3.  If anybody would like further details please ask!

CUBING IS FOREVER      PETER BECK

<BECK@ARDEC-LCSS>  OR  <PBECK@ARDEC>

PS   Matchbox has reintroduced the cube in europe.


Received: from REAGAN.AI.MIT.EDU (CHAOS 13065) by AI.AI.MIT.EDU 29 Jun 89 22:56:41 EDT
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Received: from Burger.ms by ArpaGateway.ms ; 29 JUN 89 19:49:26 PDT
Sender: Hoffman.ElSegundo@Xerox.COM
Date: 29 Jun 89 13:24:11 PDT (Thursday)
Subject: PUZZLE PARTY / PUZZLE BOOK
From: Hoffman.ElSegundo@Xerox.COM
To: Cube-Lovers@AI.AI.MIT.EDU
cc: pbeck@PICA.ARMY.MIL
Message-ID: <890629-194926-4202@Xerox>


[I'm posting this for Pete Beck <BECK@ardec-lcss.arpa> or
<pbeck@PICA.ARMY.MIL> -- Rodney Hoffman]

>>>>>>>>>>>>>>>

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%        --->   FOR SERIOUS PUZZLERS "ONLY"   <---         %
%                     ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^       
              %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


THE TENTH INTERNATIONAL PUZZLE PARTY
SUNDAY  27 AUGUST 1989
RAMADA INN, LILLIE ROAD, EARLS COURT, 
LONDON SW6

ENTRY BY INVITITATION ONLY.   Apply to  L.E.HORDERN, CANE END HOUSE,
CANE END, READING RG4 9HH, ENGLAND.

****************************************************

PUZZLE BOOK:  I have copies of "CREATIVE PUZZLES OF THE WORLD (the
predecessor book to "PUZZLES OLD & NEW") for sale.  If interested
contact me directly.

<PBECK@ARDEC>

Received: from REAGAN.AI.MIT.EDU (CHAOS 13065) by AI.AI.MIT.EDU 15 Aug 89 09:40:10 EDT
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Date:     Tue, 15 Aug 89 9:31:25 EDT
From:     Peter Beck (LCWSL) <pbeck@PICA.ARMY.MIL>
To:       cube-lovers@ai.ai.mit.edu
cc:       pbeck@PICA.ARMY.MIL
Subject:  cff #21
Message-ID:  <8908150931.aa20728@AC4.PICA.ARMY.MIL>

subject :  review of "cubism for fun" newsletter issue #21, july 89;
the newsletter of the "dutch cubists club";  in english starting with
issue #14 

1..  the table of contents for issue # 21, july 89 follows:

introduction by the editors
membership fee by paul sijben
announcement:  cube day 1989;  sat 9 dec,  voorshoten netherlands
announcement:  10th international puzzle party, details previously
posted
hidden cubes  by tony fisher
rubik's cube:  a new solution approach by ed hordern
the anver's globe: by arie verveen, a construction of a spherical    
                               megaminx
magic cross news by anton hanegraaf
pretty  magic cross formulas by clemens de brouwer
unicolored cross patterns:  by lucien matthijsse
magic cross half-tile 3-cycles by ronald fletterman
frying pan and oskar keys by oskar van deventer
pretty cubic patterns by anneke treep
book review by anton hanegraaf;  puzzle in wood by nob yoshigahara,  
                private publication
news and letters to the editor  - jan de geus, valkenboslaan 262a,
 2563eb den haag, netherlands  is compiling a list of computer
programs for simulating or solving puzzles and games of all kinds,eg,
 mostly 2d sequential movement puzzles.  he would any assistance.
changes in the list of members - 

2.   membership for 1989 is  us$8.   a photocopied set of the
newsletters, issues 1-13, written in dutch (in the future selected
back articles will be available in english) is also available for
 us$7.    to order either of these send an 'international"  postal
money order to:  paul sijben, witbreuksweg 397-304, 7522 za enschede,
 the netherlands.

3.  if anybody would like further details please ask!

cubing is forever      peter beck

<beck@ardec-lcss>  or  <pbeck@ardec>


*   american games fair is scheduled for sept 8,9,10  1989  at the
roosevelt hotel, nyc.   tel # 212/867-5159

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Date: Fri, 18 Aug 89 12:21:29 EDT
From: rp@XN.LL.MIT.EDU (Richard Pavelle)
Posted-Date: Fri, 18 Aug 89 12:21:29 EDT
Message-Id: <8908181621.AA18188@XN.LL.MIT.EDU>
To: cube-lovers@ai.ai.mit.edu
Subject: the 3x3

I took some time off this week and began playing with the cube to
teach one of my kids how to solve it. I had not tried for perhaps 5
years. To my surprise I had forgotten a few transformations while
recalling a few which are "equally difficult". It took about 10 hours
to get back to the stage where I can solve it in about 3 minutes
except for the flip of two opposite edges. I recall that we discussed,
in this forum, a nice procedure for this move many years ago and I
wonder whether anyone recalls it.

Also, to what extent have others shared my experience of forgetting
moves?

Received: from REAGAN.AI.MIT.EDU (CHAOS 13065) by AI.AI.MIT.EDU 18 Aug 89 14:16:43 EDT
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Date: Fri, 18 Aug 89 14:17 EDT
From: Allan C. Wechsler <ACW@YUKON.SCRC.Symbolics.COM>
Subject: the 3x3
To: rp@xn.ll.mit.edu, cube-lovers@ai.ai.mit.edu
In-Reply-To: <8908181621.AA18188@XN.LL.MIT.EDU>
Message-ID: <19890818181707.3.ACW@WHIMBREL.SCRC.Symbolics.COM>

    Date: Fri, 18 Aug 89 12:21:29 EDT
    From: Richard Pavelle <rp@xn.ll.mit.edu>

    I took some time off this week and began playing with the cube to
    teach one of my kids how to solve it. I had not tried for perhaps 5
    years. To my surprise I had forgotten a few transformations while
    recalling a few which are "equally difficult". It took about 10 hours
    to get back to the stage where I can solve it in about 3 minutes
    except for the flip of two opposite edges. I recall that we discussed,
    in this forum, a nice procedure for this move many years ago and I
    wonder whether anyone recalls it.

Recall the "Extended Befuddler" language:

B, F, U, D, L, R are counter-clockwise quarter twists.
Lower-case are clockwise.

I, i, J, j, K, k are whole-cule rotations agreeing in sense and axis
with B, F, U, D, L, R, in that order.

We group together sequences that are order-independent.  These sequences
almost always correspond to intuitive "moves".

And now, to flip the FD and BU edges:

;;; First monoflip:
f	; Get FD edge into equator.
jUd	; Slice it to the back.
FF	; Turn the vacated slot over
JuDJuD	; and slice the cubie back into the inverted slot.
F	; Move the cubie to the top.
;;; Segue
UU	; Exchange it with the other edge to be flipped.
;;; Second monoflip:
f	; Move the new edge into the equator,
JuDJuD	; slice it to the back the long way,
FF	; turn the vacated slot over,
JuD	; and slice the cubie back into the inverted slot, the short way.
f	; Get it back to the top
;;; Coda
UU	; Un-segue 
FF	; and take first edge back to the bottom.
;;; Checksum of whole-cube moves: jJJJJJ = 1.
;;; 26 qtw, 13 "moves" including half-twists and slices.

I doubt if this is minimal, but it is so intuitive that I was able to
type this sequence without a cube in my hands.

    Also, to what extent have others shared my experience of forgetting
    moves?

Some.

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Date: Tue, 12 Sep 89 15:18:51 -0400 (EDT)
From: "Howard D. Look" <hl08+@andrew.cmu.edu>
To: Cube-Lovers@ai.ai.mit.edu
Subject: Solution Algorithm

Does anyone have a full-blown, step-by-step algorithm for solving an
arbitrarily messed cube that would be suitable in an interactive,
graphical computer simulation of the cube?

Thanks,
Howard Look
Carnegie Mellon Univeristy
hl08+@andrew.cmu.edu

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          20 Oct 89 10:47 EDT
Date:     Fri, 20 Oct 89 10:27:35 EDT
From:     Peter Beck (LCWSL) <pbeck@pica.army.mil>
To:       cube-lovers@ai.ai.mit.edu
Subject:  [To: PBECK:  new cube puzzle]
Message-ID:  <8910201027.aa21814@AC4.PICA.ARMY.MIL>




----- Forwarded message # 1:

Date:     Fri, 20 Oct 89 8:35:47 EDT
From:     Peter Beck  (LCWSL)  <pbeck@pica.army.mil>
To:       PBECK@PICA.ARMY.MIL
cc:       pbeck@PICA.ARMY.MIL
Subject:  new cube puzzle
Message-ID:  <8910200835.aa11532@AC4.PICA.ARMY.MIL>



Are there any patent engineers out there?

Patent 4,872,682 by Kuchimanchi (U of Maryland PHD student) and Thekur
(UC santa cruz PHD student) is for a new cube puzzle.

From the NY Times 10/14/89 Patents column:

"... ,the new brain teaser is a cube divided into squares, nine on
each face.  Each square can be rotated in both a horizontal and
vertical plane, creating billions of possible combinations.  at the
outset, all squares on each side have the same color.  the goal is to
mix up the colors and get them back in order.

In addition, however, the new puzzle contains one blank square, which
can be slid to any location on the cube. This makes the challenge
easier, because it gives the players another way to move squares from
one place to another."

So what do you think this puzzle is?  Do cubies rotate or do cubie
faces rotate?  Is this a sliding block puzzle on the equators like the
hungarian globe puzzle as sold by nature company etal?  Is this just
sam lloyd 15 puzzle on the surface of a cube?

pete beck,  <pbeck@ardec>


----- End of forwarded messages

Received: from lcs.mit.edu (CHAOS 15044) by AI.AI.MIT.EDU 21 Oct 89 13:25:36 EDT
Received: from YUKON.SCRC.SYMBOLICS.COM by mintaka.lcs.mit.edu id aa01632;
          21 Oct 89 13:08 EDT
Received: from WHIMBREL.SCRC.Symbolics.COM by YUKON.SCRC.Symbolics.COM via CHAOS with CHAOS-MAIL id 502163; Fri 20-Oct-89 16:27:22 EDT
Date: Fri, 20 Oct 89 16:24 EDT
From: "Allan C. Wechsler" <ACW@yukon.scrc.symbolics.com>
Subject: [To: PBECK:  new cube puzzle]
To: pbeck@pica.army.mil, cube-lovers@ai.ai.mit.edu
In-Reply-To: <8910201027.aa21814@AC4.PICA.ARMY.MIL>
Message-ID: <19891020202439.8.ACW@WHIMBREL.SCRC.Symbolics.COM>

    Date:     Fri, 20 Oct 89 10:27:35 EDT
    From:     Peter Beck (LCWSL) <pbeck@pica.army.mil>

    Are there any patent engineers out there?

The following is a public service announcement -- everybody ought to
know this.

    Patent 4,872,682 by Kuchimanchi (U of Maryland PHD student) and Thekur
    (UC santa cruz PHD student) is for a new cube puzzle.

Whenever you know a patent number, you can obtain a complete copy of the
patent by writing to:

Commissioner of Patents and Trademarks
Washington, DC 20231
Attention: Patent orders

Include the patent number and a check for $1.50.  Used to be you could
get the patent back in ten days.  Lately the delay is more like four
weeks.

    From the NY Times 10/14/89 Patents column:

    "... ,the new brain teaser is a cube divided into squares, nine on
    each face.  Each square can be rotated in both a horizontal and
    vertical plane, creating billions of possible combinations.  at the
    outset, all squares on each side have the same color.  the goal is to
    mix up the colors and get them back in order.

    In addition, however, the new puzzle contains one blank square, which
    can be slid to any location on the cube. This makes the challenge
    easier, because it gives the players another way to move squares from
    one place to another."

    So what do you think this puzzle is?  Do cubies rotate or do cubie
    faces rotate?  Is this a sliding block puzzle on the equators like the
    hungarian globe puzzle as sold by nature company etal?  Is this just
    sam lloyd 15 puzzle on the surface of a cube?

All these questions will be answered by the complete patent.

If you are more impatient, call the local government printing office,
and ask them where the patent depository for your area is.  Access to
patent depositories is free, although the depository (usually a public
library) can charge a fee for printing.  The librarian at the depository
can tell you how to look up the patent.  Usually it is on microfilm or
fiche.

Warning: if you are like me, you will find the patent depository
addicting.  Stay away if you have family resposibilities.  Above all,
avoid learning the seductively simple cross-reference scheme or you will
spend the rest of your life browsing through puzzle patents!

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	id AA11232; Fri, 1 Dec 89 10:02:37 est
Date: Fri, 1 Dec 89 10:02:37 est
From: Roland Zito-wolf <rjz%cs.brandeis.edu@relay.cs.net>
Posted-Date: Fri, 1 Dec 89 10:02:37 est
To: cube-lovers.2@cs.brandeis.edu, beeler.2@cs.brandeis.edu
Subject: an interesting talk


Stu Coffin, internationally known puzzle designer, will be speaking
to the Philomorphs (form-lovers) Society at Harvard next Monday
Dec 4 at 7:45 pm (Carpernter Center, Studio 2 west).
He will be talking about his new book, :The Puzzling World of
Polyhedral Dissections", and about his bizarre and clvere puzzle
designs in general. Better yet, there will probably be examples of his
work to puzzle around with. Stu's puzzles are just amazing; they combine
an amazing intuition for geometric structure with excellent craftsmanship;
they aree really worth seeing if you like puzzles, geometry, or just
nifty design and execution. See you there!!
cheers, roy


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          8 Jan 90 11:34 EST
Date:     Mon, 8 Jan 90 11:31:19 EST
From:     Peter Beck (LCWSL) <pbeck@pica.army.mil>
To:       CUBE-LOVERS@ai.ai.mit.edu
cc:       PBECK@pica.army.mil
Subject:  CFF
Message-ID:  <9001081131.aa14481@AC4.PICA.ARMY.MIL>




FEEDBACK PLEASE:  Is anybody out there interested in my continuing to
post the CFF table of contents?  1 yes and I will continue, none and I
will stop.

SUBJECT :  Review of "Cubism For Fun" newsletter issue #22, DEC 89;
the newsletter of the "Dutch Cubists Club";  in english starting with
issue #14 

1..  The table of contents for issue # 22, DEC 89 follows:

REVIEW OF THE 10th international puzzle party, London 8/28,29/89.  66
attendees from 14 countries. A puzzle fair was also held on monday at
the camden market in north central london.  Next puzzle party april 91
in california??
MY TRIP TO THE USA:  ANNEKE TREEP
SPINNING CUBE: TOM VERHOEFF
RUBIK'S CUBE IN 44 MOVES:  HANS KLOOSTERMAN
THE IMPOSSIBLE DOUBLE-DOMINO SQUARE: by Oskar van Deventer - a packing
problem; the pieces are made from 2 dominoes.
PACKING WITH CONGRUENT SHAPES (pentaCUBING): FRITS GOBEL
PUZZLE IN STAPLES:  by Oskar van Deventer - USING STAPLES to make
puzzle pieces and sculpture.
A CHECKERED STAPLE-BLOCK: Paul Sijben
THE WIRREL-WARREL MAXI CUBE: Jan de Geus - new puzzle, english name
I.Q.UBE
THE TOP SPIN PUZZLE: ED HORDERN - 89 INTRODUCTION from binary arts.
DARIO'S BLOCKED SLIDING PIECES: by Anton Hanegraaf - new puzzle,
double layer sliding block
RIK'S CUBE KIT book review by Anton Hanegraaf - H.J.M. van Grol, has 2
self published booklets on cube packing puzzles; 1-solid block
puzzles, 16PGS (US$3), 2-solutions for riks cube kit, 20 pgs (US$3).
 The cube kit is the complete set of all non-planar polycubes of
maximum 5 units, ie, 3 tetra cubes and 17 pentacubes.
PRETTY CUBIC PATTERNS BY ANNEKE TREEP
NEWS AND LETTERS TO THE EDITOR  - RUBIK'S ILLUSION a board game;
 ROUNDY "the clever disk from interconcept, w ger;  citation for cube
used as and advertisement.
CHANGES IN THE LIST OF MEMBERS - total list as an insert, 85 active
members.


2.   Membership for 1989 is  US$8.   A photocopied set of the
newsletters, issues 1-13, written in DUTCH (in the future selected
back articles will be available in english) is also available for
 US$7.    To order either of these send an 'INTERNATIONAL"  POSTAL
MONEY ORDER to:  Paul Sijben, Witbreuksweg 397-304, 7522 ZA Enschede,
 The Netherlands.

3.  If anybody would like further details please ask!

CUBING IS FOREVER      PETER BECK

<BECK@ARDEC-LCSS>  OR  <PBECK@ARDEC>




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Received: Mon, 8 Jan 90 17:25:01 EST by aic.nrl.navy.mil id AA25339
Date: 8 Jan 1990 16:55:06 EST (Mon)
From: Dan Hoey <hoey@aic.nrl.navy.mil>
Subject: Re: Cubism for Fun
To: Peter Beck (LCWSL) <pbeck@pica.army.mil>, Cube-Lovers@ai.ai.mit.edu
Message-Id: <631835706/hoey@aic.nrl.navy.mil>

Peter, 

    I'm still interested in seeing the CFF table of contents, though I
might be subscribing to it, because you write

    RUBIK'S CUBE IN 44 MOVES:  HANS KLOOSTERMAN

Does that article actually show how to solve the cube in 44 moves?
Even if they count half-turns as single moves, it is significantly
better than the 52-move Thistlethwaite solution in Singmaster.  Also,
Thistlethwaite was thinking of improving his method, and perhaps this
is a report of it.  Or maybe it's just more rumor and conjecture, but
it's nice to hear after all this time.

I was making a few patterns over the weekend for some kids, and thought
of some stuff I was thinking of trying out.  For instance, if you
restrict a face to two colors, there are only about fifty different
patterns, at least if you ignore handedness.  I wonder how many of them
can be put on every face of the cube.  We know the ones with corners
alternating colors are impossible.  We have some experience with some
of the patterns--the X's, Crosses, Spots, T's, and H's--but that still
leaves a large number of possibilities.

My Christmas present to myself this year was to order Rubik's Cubic
Compendium.  I hope to be able to report on that sometime soon.  It's
always possible we may have a Cubic renaissance, though I'm not holding
my breath.

Dan

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Date: Tue, 9 Jan 90 04:34:17 +0100
From: Per Starb{ck <starback@kuling.docs.uu.se>
Message-Id: <9001090334.AA04286@kuling.DoCS.UU.SE>
To: pbeck@pica.army.mil, CUBE-LOVERS@ai.ai.mit.edu
In-Reply-To: Peter Beck's message of Mon, 8 Jan 90 11:31:19 EST
Subject: CFF


pbeck>   FEEDBACK PLEASE:  Is anybody out there interested in my continuing to
pbeck>   post the CFF table of contents?  1 yes and I will continue, none
pbeck>   and I will stop.
I'm interested. Thanks a lot!

pbeck>   SUBJECT :  Review of "Cubism For Fun" newsletter issue #22, DEC 89;
pbeck>   1..  The table of contents for issue # 22, DEC 89 follows:
	 - - -
pbeck>   RUBIK'S CUBE IN 44 MOVES:  HANS KLOOSTERMAN

I guess that's on an algorithm to always solve the cube in at most 44
moves. Is that right? Is that the best known algorithm (best = has
minimum maximum number of moves)? Singmaster gives a broad outline of
Thistlethwaite's algorithm in his "Notes on Rubik's 'Magic Cube'".
That algorithm would always solve it in 52 moves, and I know that has
been improved to 50 moves. Is this new algorithm something like
Thistlethwaite's algorithm or is it working in a different way?

pbeck>	<BECK@ARDEC-LCSS> OR <PBECK@ARDEC>
--
Per Starback		 email:	starback@kuling.Docs.UU.SE
Flogstav. 71 C:313
S-752 63  UPPSALA
SWEDEN		Quote: "Life is but a gamble! Let flipism chart your ramble!"

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Date: Thu, 11 Jan 90 14:13 EST
From: Amin Shafie - Univ of Cincinnati Comp Ctr <SHAFIE@ucbeh.san.uc.edu>
Subject: SIGUCCS CALL for PARTICIPATION
To: 386USERS@twg.com, 9370-L%HEARN.BITNET@mitvma.mit.edu, 
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<--------------------------------------------------------------------
< 
<                 SIGUCCS User Services Conference XVIII
<                        Call For Participation
<
<                  New Centerings in Computing Services
< 
<                  September 30 through October 3, 1990
<
<                           Westin Hotel
<                         Cincinnati, Ohio
< 
<
<<>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
<< 
<< 
<<Attention Directors, Managers, Analysts, Consultants, Programmers,
<<Technical Writers, Trainers, and Librarians!
<< 
<<The higher education computing scene in the 1990s will present exciting
<<challenges.  To accommodate users' needs, computing service organizations
<<are now visibly transforming in function and structure.  The widespread
<<adoption of personal computing by all disciplines, the increasing demand
<<for desktop access to shared resources, the growth in demand for
<<supercomputing capabilities, and the proliferation of powerful desktop
<<workstations exert irresistible forces on central computing services.
<<In response, the central site grows exponentially in staff and machinery
<<at one academic institution; at another, the computing center is disbanded
<<to provide distributed computing!  At some sites increasing specialization
<<is urged; at others, generalization is required.  Regardless of the
<<transforming strategy adopted by an individual institution, one fact
<<seems clear:  the user is the center toward which all computing services
<<are directed.
<< 
<<SIGUCCS '90 invites you to participate in the examination and discussion
<<of the myriad challenges facing user services professionals as we enter a
<<new decade and of the new centerings computing service organizations are
<<discovering to meet them.  Please join us!
<< 
<<>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
<< 
<<You can Participate
<< 
<<	Presentations
<< 
<<	Papers
<< 
<<	Panel Discussions
<< 
<<	Quick Workshops
<< 
<<	Educational Materials Competition
<< 
<<	Newsletter Competition
<< 
<<	Technical Writing Competition
<< 
<<	Documentation Display
<< 
<<>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
<< 
<< 
<< 
<<Important Dates
<< 
<<	March 1, 1990		Presentation proposals due
<<	April 1, 1990		Notification of proposal acceptance
<<	May 1, 1990		Final Papers due
<<	June 1, 1990		Newsletter entries due
<<	June 1, 1990		Technical writing entries due
<<	June 15, 1990		Notification of paper/panel acceptance
<<	September 1, 1990	Deadline for materials for
<<				documentation display
<< 
<< 
<<>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
<< 
<<Presentation Topic Areas
<< 
<< 
<<Information Exchange Technology
<< 
<<Information exchange may well be the most important computing
<<activity of the 1990s. The infrastructure for information delivery, the
<<National Research and Academic Network (NREN), is presently being developed.
<<How do we meet the challenges of a world where the
<<facilitation of information delivery may be a principal user services
<<responsibility?  Topics of particular interest include:
<< 
<<	new approaches to information exchange
<< 
<<	campus activity in implementing information exchange
<<	facilities that comply with emerging international standards
<< 
<<	research and development of computer-mediated information
<<	exchange methods
<< 
<< 
<<Distributed Services
<< 
<<As the role of user services shifts to providing distributed support,
<<we must create new ways of providing traditional services as well as
<<designing new services.  Topics of particular interest include:
<< 
<<	providing support staff in departments and colleges
<< 
<<	funding issues
<< 
<<	if and how to charge back for services
<< 
<<	human networking of distributed support staff
<< 
<<	nonlabor-intensive support strategies
<< 
<<	cooperative efforts with other departments
<< 
<< 
<< 
<<Management Strategies
<< 
<<How do user services managers cooperate with other administrative and
<<academic units that use or provide computing resources?  How do they
<<meet the many and diverse demands?  Topics of particular interest include:
<< 
<<	reorganization
<< 
<<	interaction with faculty advisory groups
<< 
<<	delegating and distributing responsibility
<< 
<<	coordinating university computing resources
<< 
<<	staff professional development
<< 
<< 
<<Marketing your Services
<< 
<<Changing roles may require changing your services and, often, your image on
<<campus as you provide new services to new users.  Topics of particular in-
<<terest include:
<< 
<<	promotional strategies
<< 
<<	conducting market research
<< 
<<	designing services for unique or special audiences
<< 
<< 
<< 
<<Strategies for Small Schools
<< 
<<How can a small liberal arts college have distributed user services and
<<centralized user services?  How do distributed and centralized services work
<<together to provide computing services beyond word processing?  The
<<sciences have become computer literate; now, how do we reach out  from the
<<center to the humanities and fine arts?  Are we getting out of the
<<office and into the trenches?  Are we making too many "house calls"?
<<Should we make them at all?
<< 
<< 
<<Security and Ethics
<< 
<<As electronic mail and conferencing become more popular, computing
<<systems are widely accessible to more users.  How secure should academic
<<computing resources be?  What are the ethical guidelines provided for users
<<of electronic mail and conferencing systems?  Topics of particular interest
<<include:
<< 
<<	promoting responsible and ethical use of computing resources
<< 
<<	security strategies
<< 
<<	adopting an ethics policy
<< 
<< 
<<Serving New Audiences
<< 
<<People from the humanities, the arts, and other traditionally nontechnical
<<disciplines are discovering that computers can help in areas other than
<<word processing.  In an increasingly proactive stance in the central
<<computing facility, what do we do to attract and support these new audi-
<<ences?  Topics of interest include:
<< 
<<	providing information about off-the-shelf specialized
<<	programs for music, fine arts, and the humanities
<< 
<<	facilitating technical support of nontraditional areas
<< 
<<	serving the computing beginner who wants to do
<<	sophisticated tasks
<< 
<< 
<<Consulting, Training, and Documentation
<< 
<<Supporting those who use the computing resources that we provide re-
<<mains an important responsibility of user services organizations.  Topics
<<of particular interest include:
<< 
<<	new approaches to training
<< 
<<	providing distributed consulting
<< 
<<	documentation distribution services
<< 
<< 
<<and/or other topics that would be of interest to your national
<<and international colleagues
<< 
<< 
<<>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
<< 
<<Submitting Proposals
<< 
<< 
<<Submit proposals via electronic mail to:
<< 
<<	SIGPAPER@OHSTVMA.BITNET or
<< 
<<	SIGPAPER@OHSTVMA.IRCC.OHIO-STATE.EDU
<< 
<<If you do not have access to electronic mail, send a printed copy to:
<< 
<<		Susan Jenkins Saari
<<		Instruction and Research
<<		Computer Center
<<		The Ohio State University
<<		1971 Neil Avenue
<<		Columbus, OH 43210
<< 
<<		phone:      (614) 292-4843
<<		fax:      (614) 292-7081
<< 
<< 
<<>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
<< 
<<Accepted Proposals
<< 
<< 
<<Proposals must be received by March 1, 1990.  Any submisson received
<<after this date will not be considered by the Program Committee.  You will
<<be notified of the Program CommitteeUs decision by April 1, 1990.  If your
<<proposal is accepted, you will be asked to submit a full paper by May 1,
<<1990.  Any papers received after this date will not be considered.  You will
<<be notified of the Program CommitteeUs decision by June 15, 1990.
<< 
<<If your presentation is accepted, SIGUCCS is depending on you.  If you are
<<ker to make your presentation (not a substitute presentation).
<< 
<< 
<<>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
<< 
<< 
<<How to Participate
<< 
<< 
<<Proposals
<< 
<<For each proposal, include your name, title, affiliation, mailing ad-
<<type of  proposal (presentation or panel discussion) and its topic area.
<<In addition, you must enclose the proper materials from the following
<<requirements list:
<< 
<<Description
<< 
<<Papers		Papers will be presented in 20-minute ntervals, with
<<		three papers scheduled per 90-minute session. Speakers'
<<		papers will be published in the conference proceedings.
<< 
<<Panels		Panels will be in-depth treatments of a single topic by
<<		two to four speakers from at least two different schools,
<<		coordinated by a moderator.  Allow ample time for audience
<<		discussion.  Abstracts for panels should be submitted
<<		as a unit by the person who wishes to act as a moderator.
<<		Panelists' papers will be published in the conference
<<		proceedings.
<< 
<<Quick Workshops	Quick workshops provide 90-minute overviews of new technolo-
<<		gies, innovative applications, and creative strategies
<<		for addressing the needs of computer users on campus.
<< 
<< 
<<Requirements
<< 
<<Papers		A 250- to 300-word abstract of the paper.  Acceptance of
<<		a proposal does not automatically ensure acceptance
<<		of a paper for presentation; you must submit a full
<<		paper to be considered for the conference program.
<< 
<<Panels		A 250- to 300-word description of the panel, including
<<		each panelist's name, title, affiliation, and presentation
<<		topic.  Acceptance of a panel description does not
<<		automatically ensure acceptance of the panel for
<<		presentation; each panelist must submit a full paper
<<		to be considered for the conference program.
<< 
<<Quick Workshops	A one- to two-page outline of the presentation and a
<<		10-minute videotape excerpt from the proposed presentation.
<<		Acceptance of a proposal does not automatically ensure
<<		acceptance of a workshop for presentation; you must
<<		submit a full paper to be considered for the conference
<<		program.  Only three or four presentations will be a
<<		ccepted in this category because it is highly competiive.
<< 
<< 
<<>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
<< 
<< 
<<Other Ways to Participate
<< 
<<Education and Training Materials Competition
<< 
<<Interest in and the importance of user education and training have grown
<<with each SIGUCCS conference.  The 1990 SIGUCCS Conference offers,
<<for the first time, competition for user education and training materials for
<<colleges and universities.*  We invite you to submit no more than two
<<entries in any or all of the following categories: curriculum catalog, class-
<<room printed materials, or self-contained printed tutorials.  Although the
<<first year of this competition includes only printed materials, we would like
<<to know if there is an interest in expanding our future competitions to
<<include video, audio, and computer-based tutorials.  Deadline for entry is
<<June 1, 1990.  For more details and an entry form, or to address the issue
<<of future competition categories, contact:
<< 
<<		Diane Jung-Gribble
<<		Indiana University
<<		750 North State Road 46 Bypass
<<		Bloomington, IN  47405
<< 
<<		(812) 855-0962
<< 
<< 
<<		JUNG@IUBACS.BITNET
<<		JUNG@JADE.BACS.INDIANA.EDU
<< 
<<*NOTE:  this competition is not open to commercial materials
<< 
<<Newsletter Competition
<< 
<<Winning an award in the SIGUCCS Newsletter Competition is a mark of
<<distinction for your institution, and for your editors, writers,artists,and
<<designers.  You will be asked to submit two consecutive issues published
<<between June 1989 and May 1990.  Deadline for entry is June 1, 1990.
<<For more details and an entry form, contact:
<< 
<<		Jess Anderson
<<		Madison Academic Computing Center
<<		University of Wisconsin-Madison
<<		1210 West Dayton Street
<<		Madison, WI   53706
<< 
<<		(608) 263-6988
<< 
<<		ANDERSON@MACC.WISC.EDU
<<		ANDERSON@WISCMACC.BITNET
<< 
<< 
<<Technical Writing Competition
<< 
<<If you have written or published a particularly good article in a computing
<<newsletter, enter it in the Technical Writing Competition.  Each computing
<<center may enter one article.  Deadline for entry is June 1,1990.  To obtain
<<entry forms and more details, contact:
<< 
<<		Donald J. Montabana
<<		University of Pennsylvania
<<		Computing Resources Center
<<		1202 Blockley Hall
<<		Philadelphia, PA  19104-6021
<< 
<<		(215) 898-9085
<< 
<<		MONTABANA@A1.RELAY.UPENN.EDU
<< 
<< 
<< 
<<Documentation Display
<< 
<<The documentation room will feature an online system for submitted
<<documentation.  Conference attendees who have BITNET or INTERNET
<<access will be able to email documentation to their university or college.
<<Documentation may be submitted electronically to DOCUMENT@MIAMIU,
<<by hardcopy, or diskette (IBM or Mac formatted) and must be received
<<before September 1, 1990.  Direct inquries to:
<< 
<<		Al Kaled
<<		Academic Computing Services
<<		Miami University
<<		Oxford, OH  45056
<< 
<<		(513) 529-6226
<< 
<<		AK75STAF@MIAMIU
<< 
<< 
<<>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
<< 
<< 
<<More Information
<< 
<< 
<<General Information
<
<<Amin Shafie, Conference Chair
<<University of Cincinnati
<< 
<< 
<<		e-mail:		SHAFIE@UCBEH.BITNET
<< 
<<		phone:		(513) 556-9001
<< 
<<		fax:		(513) 556-0035
<< 
<< 
<<Call for Participation
<<Susan Jenkins Saari, Program Chair
<<The Ohio State University
<< 
<<		e-mail:		SIGPAPER@OHSTVMA.BITNET
<< 
<<		phone:		(614) 292-4843
<< 
<<		fax:		(614) 292-7081
<< 
<< 
<<Registration
<<Ken Maccarone, Registration Chair
<<University of Cincinnati
<< 
<<		e-mail:		MACCARON@UCBEH.BITNET
<< 
<< 
<<		phone:		(513) 556-9098
<<		fax:		(513) 556-0035
<< 
<< 
<<>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
<< 
<< 
<<ACM SIGUCCS
<< 
<<The Association of Computing Machinery's (ACM) Special Interest Group
<<for University and College Computing (SIGUCCS) is one of ACM's
<<organizational units devoted to the technical activities of its members.
<<SIGUCCS provides a link for guidance and the interchange of ideas among
<<computing professionals in the full range of small to large institutions.
<<Its newsletter, annual conferences, and workshops promote the discussion
<<of mutual problems. networks, user services, and computer center management.
<<This SIGUCCS conference emphasizes practical ways to improve services for
<<those who use university and college computing centers.


Amin Shafie
Assistant Director
Academic Computing Services                UCBEH::SHAFIE
University of Cincinnati                   SHAFIE@UCBEH.SAN.UC.EDU
ML 088                                     SHAFIE@UCBEH.BITNET
Cincinnati, Ohio  45221
(513) 556-9022

Received: from lcs.mit.edu (CHAOS 15044) by AI.AI.MIT.EDU; 18 Jan 90 11:59:38 EST
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Date:     Thu, 18 Jan 90 11:46:15 EST
From:     Peter Beck (LCWSL) <pbeck@pica.army.mil>
To:       cube-lovers@ai.ai.mit.edu
cc:       pbeck@pica.army.mil
Subject:  puzzle exhibit
Message-ID:  <9001181146.aa07007@AC4.PICA.ARMY.MIL>


Thanks to all of you who responded to my request for feedback.  I will
continue to post CFF Table of Contents.  The main area of interest
appears to be in minimum move solution algorithms.


>>>  UPDATE <<<<  to  " PUZZLES OLD AND NEW"  exhibition tour SCHEDULE
    source  Jerry Slocum,  1/90

 LOCATION:   the Childrens Museum of INDIANAPOLIS,  317/924-KIDS
             >>>>    1/20/90 - 4/1/90

RE:   Rodney Hoffman's review posted in the spring 87 follows.

..........................................................

PUZZLES OLD AND NEW:  Head Crackers, Patience Provers, and Other
Tactile
Teasers  [including more variations on Rubik's cube than I've ever
seen before]

Accompanied by the book PUZZLES OLD AND NEW: HOW TO MAKE AND SOLVE
THEM  by Jerry Slocum and Jack Botermans ($20), [wash univ press,
available from slocum or myself.  The reprint of the HOFFMAN book is
also available now from slocum, about $35.]
 The exhibit discusses
        The Art in Puzzles
        The Social Experience
        Cultural Values
        Puzzles in the Industrial Age
        Puzzles in Education
        Puzzles and Science
 It aims "to explore the history, meanings, and design of mechanical
puzzles."  The extensive displays categorize puzzles according to
their
object:
        Put-Together Puzzles
        Take-Apart Puzzles
        Interlocking Solid Puzzles
        Disentanglement Puzzles
        Sequential Movement Puzzles
        Puzzle Vessels
        Dexterity Puzzles
        Vanish Puzzles
        Impossible Object Puzzles
        Folding Puzzles
Puzzles have been lent by collectors and museums around the world.
 The
exhibit consultants are Benjamin Kilborne and Martin Gardner.
 
There are hands-on puzzles, but too simple and too few.  It's a
wonderful and tantalizing display of puzzles.  I was SOOO frustrated
not
to be able to handle all the beautiful, enticing pieces.  Of course,
if
I had been permitted to, I would never leave....
 
 REVIEW BY --  Rodney Hoffman

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Date: Thu, 22 Feb 90 19:16:07 EST
From: Dan Hoey <hoey@aic.nrl.navy.mil>
Message-Id: <9002230016.AA05186@aic.nrl.navy.mil>
To: Cube-Lovers@AI.ai.mit.edu
Subject: Language in Rubik's Cubic Compendium

Well, my order of Rubik's Cubic Compendium came through.  I ordered it
through Reiter's (The DC technical bookstore) and paid $30 for it.
It's basically six largely independent chapters translated from
Hungarian, with foreword and afterword by Singmaster and a
bibliography.  There's definitely some neat stuff there.

My favorite piece of the book is in Tamas Varga's ``The Art of
Cubing'', which develops some interesting new additions to what we
used to call ``Rubiksong'', the language we use to describe processes.
He starts by renaming the Up face to be the Top, the advantage of
which is to make all the face names consonants.  He then uses vowels
to indicate the direction of turn, "O" for 90 degrees fOrward (or
clOckwise), "A" for 90 degrees bAckward (or Anticlockwise) and "I" for
a 180 degree half-turn (twIce).  This works out neatly to allow a
process to be described with a syllable for each quarter- or
half-turn.  So Pons Asinorum can be done with FIBITIDIRILI and
Laughter is 3 FOBOROLOs.

But wait, there's more!  Remember how Befuddler never was able to
handle whole-cube moves neatly?  In this notation, you append a "C" to
a syllable to indicate that instead of turning the face, you turn the
whole cube.  So the way I usually do Laughter is really 6 ROLOTOCs.
This notation is not as parsimonious, since FOC=BAC, TOC=DAC, and
ROC=LAC, but it's better than having to stop in the middle and say
``then move the cube''.  For instance, Jim Saxe's 28-qt Plummer's
Cross can be done as "FOLIRIFO BOLIRIFO ROFIBIRO LOFIBIRO TIDI", but
the way he originally described it (3 Dec 1980 00:50) was "FOLIRIFO
BOLIRIFO TOC FOLIRIFO BOLIRIFO TIDI", but instead of TOC he had a
couple of lines of text.

He also has a way of talking about the slice moves, where you move
the middle layer of the cube instead of the faces.  For moving the
middle, you append "M" to the syllable.  So the way most people do a
Spratt wrench is 4 TOROMs, and we can do the Plummer's Cross as
FOLIMBO FOLIMFO TOC FOLIMBO FOLIMFO TIM.  Of course, ROM=LAM, etc.
(This could also work for Rubik's revenge, where ROM and LAM are
different, being moves of the inner layers adjacent to the R and L
faces).

It's unfortunate that he doesn't extend the language past the point of
appending "M" and "C".  I would like to have a way of talking about
slice moves where you move the faces rather than the middle.  Of
course, we could say ROLA, but I'd rather say something like ROS.
This might interfere with the use of "s" for plurals (as they do in
the book and I do above), but that could be fixed by pronouncing the
pluralizing s as "z".

Another idea is to append "N" to syllables for aNtislice moves.  So
Laughter would be six RONTOCs.  I'm a little concerned though, that "M"
and "N" might be difficult to distinguish.

Another suggestion is to append "P" to allow "deeP" moves, where we do
RO and ROM simultaneously by grabbing two layers of the cube and
turning them while keeping the remaining face fixed.

It might be nice to use "G" to denote the way we "Wring" the cube,
as with ROPRO.  So 6 ROGTOC's does an 8-Flip.

To summarize,

    F,B,T,D,R,L -- faces Front, Back, Top, Down, Right, Left.
    O,A,I       -- directions fOrward, bAckward, twIce.
    C,M,S,N,P,G -- extensions whole-Cube, Middle, Slice, aNtislice,
		    deeP, wrinG

All extensions but C are redundant, since

        ROM=ROCRALO
        ROS=ROLO
        RON=ROLA
        ROP=ROCLA
        ROG=ROCROLO

I'm going over their list of pretty patterns, and hopefully I can find
out which ones are improvements.  I did notice they don't have Saxe's
Plummer's cross process.

Dan

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To: Cube-Lovers@ai.ai.mit.edu
Subject: Re: Language in Rubik's Cubic Compendium 
In-Reply-To: Your message of Thu, 22 Feb 90 19:16:07 -0500.
             <9002230016.AA05186@aic.nrl.navy.mil> 
Date: Fri, 23 Feb 90 21:24:17 EST
From: ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++CAN WE MAKE A STATE <tazman@athena.mit.edu>

Please remove me from this mailing list.

    TAZMAN@ATHENA.MIT.EDU


thanks in advance.

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Date:     24 FEB 90 14:02:27 CDT
From: GREGORY LOVE               <LOVEGA%USMCP6.BITNET@mitvma.mit.edu>
To: <CUBE-LOVERS@ai.ai.mit.edu>
Subject:  REMOVE PLEASE
Message-Id: <900224.14022690.024312@USM.CP6>

Also remove me from this list.

  LOVEGA@USMCP6.BITNET

  thanks also in  advance..

Gregory Love <LOVEGA@USMCP6.BITNET>, University Of Southern Mississippi

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Date: Wed, 7 Mar 90 23:04:35 -0500
From: Stan Schwartz <stan@grasp.cis.upenn.edu>
Posted-Date: Wed, 7 Mar 90 23:04:35 -0500
Message-Id: <9003080404.AA03836@grasp.cis.upenn.edu>
To: CUBE-LOVERS@ai.ai.mit.edu
Subject: Looking for Chinese puzzle box


Does anyone out there in cube-land know of a domestic source for obtaining
Chinese puzzle boxes?

	Thanks,
	Stan Schwartz <stan@grasp.cis.upenn.edu>

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Date:     Wed, 14 Mar 90 16:21:26 EST
From:     Peter Beck (LCWSL) <pbeck@pica.army.mil>
To:       cube-lovers@ai.ai.mit.edu
Subject:  [To: cube-lovers:  nob newsletter]
Message-ID:  <9003141621.aa24293@FSAC1.PICA.ARMY.MIL>

second try

----- Forwarded message # 1:

Date:     Wed, 14 Mar 90 16:18:22 EST
From:     Peter Beck  (LCWSL)  <pbeck@pica.army.mil>
To:       cube-lovers@mit.ai.ai.edu
cc:       pbeck@PICA.ARMY.MIL
Subject:  nob newsletter
Message-ID:  <9003141618.aa24145@FSAC1.PICA.ARMY.MIL>


 'PUZZLETOPIA"
NOB YOSHIGAHARA has just mailed out a new issue (after 3 yrs) of his
newsletter 'PUZZLETOPIA".  With it came a 1990 promotional calendar
from PUZZLE CITY (a subsidary of Toyo Glass) a puzzle city catalog and
a catalog from PUZZLAND HIKIMI PUZZLE COLLECTION.  If you want the
whole package write Nob (its free), if you just want PUZZLETOPIA"
e-mail me your address.

NOB YOSHIGAHARA,  4-10-1-408 IIDABASHI,  TOKYO 102 JAPAN.

TRENTON (NJ) COMPUTER FESTIVAL (TCF)
On 4/21 & 22/90 (sat & sun) the oldest and largest amateur computer
festival will take place at TRENTON STATE COLLEGE;  north of trenton
nj near I-295.   I will be there in the fleamarket selling puzzles,
especially rubik's cubes.  If you are in the area stop by and say
hello.





----- End of forwarded messages

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Date: Thu, 22 Mar 90 08:57:35 -0500 (EST)
From: "Brian E. Gallew" <bg11+@andrew.cmu.edu>
To: Cube-Lovers@ai.ai.mit.edu
Subject: Cube opinions needed

My friend's group is looking at buying a NeXT.  Unfortunately, they do not have
the opportunity for hands-on evaluation and are looking for all the opinions
the can get to help them make their decision.  If you have an opinion, be it
good, bad, or ugly, please send it to:

       arnold@freezer.it.udel.edu

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Date:     Fri, 13 Apr 90 14:35:03 EDT
From:     Peter Beck (LCWSL) <pbeck@pica.army.mil>
To:       cube-lovers@ai.ai.mit.edu
cc:       pbeck@pica.army.mil
Subject:  cff #23
Message-ID:  <9004131435.aa18164@FSAC1.PICA.ARMY.MIL>


SUBJECT :  Review of "Cubism For Fun" newsletter issue #23, MAR 90;
the newsletter of the "Dutch Cubists Club";  in english starting with
issue #14 

1..  The table of contents for issue # 23, MAR 90 follows:

NEW CLUB PRESIDENT - Jan de Geus has taken over from Guus Razoux
Schultz
CUBE DAY 1989 by Lucia Dalessi
COMPUTERS AT CUBE DAY 1989 by Guus Razoux Schultz 
BUSINESS MEETING MINUTES
FIND THE IMPOSSIBILITY PROOF by Ekkehard Kunzell
A PRIZE CONTEST by Ekkehard Kunzell
RUBIK'S CUBE not yet IN 44 MOVES:  HANS KLOOSTERMAN
TRANSFORMATIONS OF CUBIC PUZZLES by Jean Claude Constantin and Dieter
Gebhardt
THE OPAQUE CUBE PROBLEM by Martin Gardner
ABOUT THE DESIGN OF TOP SPIN PUZZLE: Ferdinand Lammertink, the
designer
LOGICAL LABYRINTHS by Anneke Treep
a WIRREL-WARREL SPACE CROSS by Pieter Torbijn
THE SQUA-RING PUZZLE: by Oskar van Deventer
THE INTRACTABLE TEN by Rik van Grol
SIMPLE CHECKERED STAPLE-BLOCKS: by Rik van Grol
PLAYING MAGIC CROSS BY COMPUTER: Guus Razoux Schultz
PLAYING WITH PENTACUBES book review by Anton Hanegraaf - SPIELE MIT
PENTAKUBEN (games with pentacubes) by Ekkehard Kunzell
BLOCKED SLIDING PIECES by Dario Uri
NEWS AND LETTERS TO THE EDITOR  -
THINK ABOUT (DUTCH) CUBE DAY 1990: probably 15 & 16 dec at Willem van
der poel's place, Den Haag
THINK ABOUT THE "SILVER" ANNIVERSARY CFF: ideas, contests requested.
CHANGES IN THE LIST OF MEMBERS -  100 active members, notable new
addition Martin Gardner.


2.   Membership for 1990 is  US$8.   A photocopied set of the
newsletters, issues 1-13, written in DUTCH (in the future selected
back articles will be available in english) is also available for
 US$7.    To order either of these send an 'INTERNATIONAL"  POSTAL
MONEY ORDER (cost $3 at post office) to:  Paul Sijben, Witbreuksweg
397-304, NL-7522  ZA  Enschede,  The Netherlands.

3.  If anybody would like further details please ask!

CUBING IS FOREVER      PETER BECK

<BECK@ARDEC-LCSS>  OR  <PBECK@ARDEC>


SPECIAL REMINDER:  TRENTON COMPUTER FESTIVAL WILL BE THE 21 & 22 OF
APRIL AT A NEW LOCATION  --> MERCER COUNTY COMMUNITY COLLEGE, NEAR
PRINCETON NJ.  THIS IS THE LARGEST AND OLDEST AMATEUR COMPUTER
FESTIVAL.






From alan@ai.mit.edu  Sun Oct 14 20:01:36 1990
Return-Path: <alan@ai.mit.edu>
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From: alan@ai.mit.edu (Alan Bawden)
Received: by wheat-chex (4.1/AI-4.10) id AA02846; Sun, 14 Oct 90 20:01:37 EDT
Date: Sun, 14 Oct 90 20:01:37 EDT
Message-Id: <9010150001.AA02846@wheat-chex>
To: cube-lovers
Subject: Testing 1 2 3

This message shouldn't go anywhere except into the archive and into my own
mailbox.  I anyone else gets it, then it will be time to give up and go
home.

From alan@ai.mit.edu  Sun Oct 14 20:48:51 1990
Return-Path: <alan@ai.mit.edu>
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From: alan@ai.mit.edu (Alan Bawden)
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Date: Sun, 14 Oct 90 20:48:52 EDT
Message-Id: <9010150048.AA03070@wheat-chex>
To: cube-lovers
Subject: [alan@ai.mit.edu: Surprise!]

Now that the archive is fixed again, here is the message that I sent to
resurrect the list:

From: alan@ai.mit.edu (Alan Bawden)
Date: Fri, 12 Oct 90 16:03:05 EDT
To: cube-lovers
Subject: Surprise!

That's right.  Cube-Lovers has returned from the dead.  Due to various
hardware, software and personal crises, Cube-Lovers has been down since
sometime last spring.  I'm sure that many of you didn't even notice, given
what a low-volume list this has become.

Our "official" addresses remain Cube-Lovers@AI.AI.MIT.EDU for submissions
and Cube-Lovers-Request@AI.AI.MIT.EDU for administrivia.  (Actually, you
will find that using simply "...@AI.MIT.EDU" will work just as well.)

Since this is the first message to this list after moving it to a new
machine with a different mailer, I expect that many addresses on the list
have ceased to function.  If I were you, I wouldn't send any mail here for
about a week -- just to give me a chance to process all the bounces I'm
about to get.

The archives are currently unavailable, but I hope to have them available
for FTP soon.


From alan@ai.mit.edu  Mon Oct 15 03:11:33 1990
Return-Path: <alan@ai.mit.edu>
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From: alan@ai.mit.edu (Alan Bawden)
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Date: Mon, 15 Oct 90 03:11:35 EDT
Message-Id: <9010150711.AA05063@wheat-chex>
To: cube-lovers
Subject: Second Announcement

OK, I think I've cleaned up Cube-Lovers enough that it's safe for anyone to
use it.  (For some of you this message is the first indication that
Cube-Lovers is back -- many copies of my first announcement bounced back to
me.)

Those of you who keep asking for the archives will be pleased to know that
they are again available for anonymous FTP: Connect to AI.MIT.EDU, login as
"anonymous" (any password), and in the directory "/pub/alan" you will find
the seven (compressed) files "cube-mail-0.Z" through "cube-mail-6.Z".
Archive vital statistics:

	   File		  From		   To	     Size (bytes)
	   ----		  ----		   --	     ------------
	cube-mail-0	12 Jul 80	23 Oct 80	185037
	cube-mail-1	 3 Nov 80	 9 Jan 81	135719
	cube-mail-2	10 Jan 81	 3 Aug 81	138566
	cube-mail-3	 3 Aug 81	 3 May 82	137753
	cube-mail-4	 4 May 81	11 Dec 82	139660
	cube-mail-5	11 Dec 82	 6 Jan 87	173364
	cube-mail-6	10 Jan 87	13 Apr 90	216733

(Unfortunately, due to the way mail works here at the AI Lab, it is not
possible to have the current active archive accumulate anywhere where
anonymous FTP can pick it up.)

As you can see, things really slacked off after 1982, and we were really
quiet during the middle of the decade.

For those of you who missed my first message, let me repeat that our
"official" addresses remain Cube-Lovers@AI.AI.MIT.EDU for submissions and
Cube-Lovers-Request@AI.AI.MIT.EDU for administrivia.

				- Alan

From pbeck@pica.army.mil  Mon Oct 15 08:13:57 1990
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Date:     Mon, 15 Oct 90 8:08:48 EDT
From: Peter Beck (LCWSL) <pbeck@pica.army.mil>
To: cube-lovers@life.ai.mit.edu
Cc: pbeck@pica.army.mil
Subject:  [To: cube-lovers-incoming%csl.:  algorithm]
Message-Id:  <9010150808.aa22705@FSAC1.PICA.ARMY.MIL>

welcome back

----- Forwarded message # 1:

Date:     Fri, 1 Jun 90 12:00:37 EDT
From:     Peter Beck  (LCWSL)  <pbeck@pica.army.mil>
To:       cube-lovers-incoming%csl.ti.com@relay.cs.net
cc:       pbeck@PICA.ARMY.MIL
Subject:  algorithm
Message-ID:  <9006011200.aa13780@FSAC1.PICA.ARMY.MIL>




Date: Tue, 27 Mar 90 22:23:00 EST
From: adobe!uunet!canremote!nigel.allen@labrea.stanford.edu
Subject: PROGRAMMING NEWSLETTER

      A.K. Dewdney, Computer Recreations columnist with Scientific
American magazine, has launched a personal programming newsletter,
Algorithm. The new publication is aimed at amateur and professional
programmers alike. It extends the Computer Recreations tradition of
recreational and educational programming projects: the Mandelbrot set,
cellular automata, chaos and dynamics, weird machines, stellar
simulation, puzzles and many other topics.

      The new publication carries seven features and will expand to
include more columns. Currently, it includes Algoletter, advice from
professionals; Easy Pieces, fascinating projects for beginning
programmers by Michael Ecker of Creative Computing fame; Personal
Programs, exercises for more advanced programmers by Cliff Pickover,
IBM's computer graphics wizard; Algopuzzles, computer mind-benders by
Dennis Shasha, author of The Puzzling Adventures of Dr. Ecco; Algofact
and Algofiction, invited articles and stories from well-known
scientists and authors. A Bulletin Board advertises hosts of
recreational products by individuals and small companies.

      Algorithm puts the "personal" back in "personal computing" by
encouraging you to develop your programming skills while pursuing high
adventure on the frontiers of science and computing. Order a free
examination copy by writing Algorithm at P.O. Box 29237, Westmount
Postal Outlet, 785 Wonderland Road, London, Ontario, Canada N6K 1M6.

--- MaS Relayer v1.00.00
 Message gatewayed by MaS Network Software and Consulting/HST
 Internet: nigel.allen@canremote.uucp
 UUCP:     ...tmsoft!masnet!canremote!nigel.allen

-------  from  infomac  -----
  

----- End of forwarded messages

From alan@ai.mit.edu  Thu Oct 18 17:06:55 1990
Return-Path: <alan@ai.mit.edu>
Received: from wheat-chex (wheat-chex.ai.mit.edu) by life.ai.mit.edu (4.1/AI-4.10) id AA02418; Thu, 18 Oct 90 17:06:55 EDT
From: alan@ai.mit.edu (Alan Bawden)
Received: by wheat-chex (4.1/AI-4.10) id AA03710; Thu, 18 Oct 90 17:06:51 EDT
Date: Thu, 18 Oct 90 17:06:51 EDT
Message-Id: <9010182106.AA03710@wheat-chex>
To: cube-lovers
Subject: Archives again

I hate to bother you folks again so soon, but naturally the AI Lab chose
today to reorganize how anonymous FTP access worked.  Here are the updated
instructions for accessing the Cube-Lovers archives:

Connect to TRIX.AI.MIT.EDU, login as "anonymous" (any password), and in the
directory "pub/cube-lovers" you will find the seven (compressed) files
"cube-mail-0.Z" through "cube-mail-6.Z".  

Archive vital statistics (when uncompressed):

	   File		   From		   To	     Size (bytes)
	   ----		   ----		   --	     ------------
	cube-mail-0	12 Jul 80	23 Oct 80	185037
	cube-mail-1	 3 Nov 80	 9 Jan 81	135719
	cube-mail-2	10 Jan 81	 3 Aug 81	138566
	cube-mail-3	 3 Aug 81	 3 May 82	137753
	cube-mail-4	 4 May 81	11 Dec 82	139660
	cube-mail-5	11 Dec 82	 6 Jan 87	173364
	cube-mail-6	10 Jan 87	13 Apr 90	216733

(Unfortunately, due to the way mail works here at the AI Lab, it is not
possible to have the current active archive accumulate anywhere where
anonymous FTP can pick it up.)

From pbeck@pica.army.mil  Wed Oct 24 13:47:23 1990
Return-Path: <pbeck@pica.army.mil>
Received: from FSAC1.PICA.ARMY.MIL by life.ai.mit.edu (4.1/AI-4.10) id AA21425; Wed, 24 Oct 90 13:47:23 EDT
Date:     Wed, 24 Oct 90 13:34:45 EDT
From: Peter Beck (LCWSL) <pbeck@pica.army.mil>
To: cube-lovers@life.ai.mit.edu
Subject:  puzzling events
Message-Id:  <9010241334.aa15999@FSAC1.PICA.ARMY.MIL>



CUBING/PUZZLING EVENTS
rev  10/24/90


<-->  DUTCH CUBE DAY IS:
---- 8  dec  1990
---- Prof Willem van der Poel's new residence
---- in the netherlands


<-->  International puzzle collector's party (I think it is #11)
---- 3/31/91  Easter Sunday
---- culver city,  ca
***  Admission by invitation only!!!  Contact Mr. jerry slocum, 257
south palm drive, beverly hills, ca 90212 for an invitation.


From pbeck@pica.army.mil  Thu Oct 25 15:26:10 1990
Return-Path: <pbeck@pica.army.mil>
Received: from FSAC1.PICA.ARMY.MIL by life.ai.mit.edu (4.1/AI-4.10) id AA24321; Thu, 25 Oct 90 15:26:10 EDT
Date:     Thu, 25 Oct 90 15:17:59 EDT
From: Peter Beck (LCWSL) <pbeck@pica.army.mil>
To: cube-lovers@life.ai.mit.edu
Subject:  puzzling newsletters
Message-Id:  <9010251518.aa13527@FSAC1.PICA.ARMY.MIL>


PUZZLING NEWSLETTERS  --  Oct 90

..........................................................
"Cubism For Fun" 
The newsletter of the "Dutch Cubists Club";  in english starting with
issue #14.  Back issues are available.  The club has over 100 active
members, notable new addition Martin Gardner.

Membership for 1990 is  US$8.   A photocopied set of the newsletters,
issues 1-13, written in DUTCH (in the future selected back articles
will be available in english) is also available for  US$7.    To order
either of these send an 'INTERNATIONAL"  POSTAL MONEY ORDER (cost $3
at post office) to:  Paul Sijben, Witbreuksweg 397-304, NL-7522  ZA
 Enschede,  The Netherlands.

..........................................................
WORLD GAME REVIEW
Michael Keller publishes a newsletter that explores the mathematical
aspects of games & puzzles.  4 issues for US$11, published
erratically.  Back issues are available.

MICHAEL KELLER, 3367-1, NORTH CHATAM ROAD, ELLICOTT CITY, MD 21043,
USA

..........................................................
 'PUZZLETOPIA"
NOB YOSHIGAHARA has just mailed out a new issue (after 3 yrs) of his
newsletter 'PUZZLETOPIA".  With it came a 1990 promotional calendar
from PUZZLE CITY (a subsidary of Toyo Glass) a puzzle city catalog and
a catalog from PUZZLAND HIKIMI PUZZLE COLLECTION.  If you want the
whole package write Nob (its free outside of Japan).

NOB YOSHIGAHARA,  4-10-1-408 IIDABASHI,  TOKYO 102 JAPAN.

..........................................................

ARM Bulletin (ACADEMY of RECREATIONAL MATHEMATICS), JAPAN
This is a monthly 40-80 page newsletter of the Japanese puzzle
hobbiests club.  Dues Y8,000.

PUZZLE KONWAKAI C/O S. TAKAGI, 1-2-4 MATSUBARA, SE TAGAYAKU, TOKYO 156
JAPAN

..........................................................



From pbeck@pica.army.mil  Thu Oct 25 15:26:06 1990
Return-Path: <pbeck@pica.army.mil>
Received: from FSAC1.PICA.ARMY.MIL by life.ai.mit.edu (4.1/AI-4.10) id AA24312; Thu, 25 Oct 90 15:26:06 EDT
Date:     Thu, 25 Oct 90 15:16:29 EDT
From: Peter Beck (LCWSL) <pbeck@pica.army.mil>
To: cube-lovers@life.ai.mit.edu
Subject:  oxford press books
Message-Id:  <9010251516.aa12361@FSAC1.PICA.ARMY.MIL>



Oxford University Press publishes a series of books called Recreations
in  Mathematics.  The series editor is David Singmaster of Rubik's
Cube fame.
They are priced at about $28 each.

As of Oct 90 the series contains the following: 
#1  "Mathematical Byways ...",  by Hugh ApSimon. 
#2: "Ins and Outs of Peg Solitaire",  by John Beasley.  
#3: "Rubik's Cubic Compendium", by Rubik, et al.  
#4  "Sliding Piece Puzzles", by L.E. Hordern. 
#5  "The Mathematics of Games",  by John Beasley.
#6  "The Puzzling World of Polyhedral Dissections", by Stewart Coffin.
#7  "More Mathematical Byways",  by Hugh ApSimon. 

 TO ORDER:  Send check or credit card info (MASTERCARD OR VISA) to:
        SCIENCE & MEDICAL MARKETTING DIRECTOR,
        OXFORD UNIVERSITY PRESS
        200 MADISON AVE,
        NEW YORK, NY 10016  - 
-- >  ADD $1.50 for shipping


From pbeck@pica.army.mil  Fri Oct 26 20:05:37 1990
Return-Path: <pbeck@pica.army.mil>
Received: from FSAC1.PICA.ARMY.MIL by life.ai.mit.edu (4.1/AI-4.10) id AA19290; Fri, 26 Oct 90 20:05:37 EDT
Date:     Fri, 26 Oct 90 15:25:41 EDT
From: Peter Beck (LCWSL) <pbeck@pica.army.mil>
To: cube-lovers@life.ai.mit.edu
Subject:  puzzling events expanded
Message-Id:  <9010261525.aa18619@FSAC1.PICA.ARMY.MIL>


CUBING/PUZZLING EVENTS
rev  10/26/90

...............................................................
<-->  The 10th DUTCH CUBE DAY  <-->
...............................................................
WHEN ---- 8  dec  1990
WHERE ---- Prof Willem van der Poel's new residence, DUBLINSTRAAAT
143, ZOETERMEER, THE NETHERLANDS
TIME ---- 10:00 AM
INVITITATIONS:  Prof van der Poel,  tel # 079-211912 or Anneke Treep,
tel# 074-501181

AGENDA:  
.. LECTURES - A NEW CUBE SOLVING ALGORITHM BY HANS KLOOSTERMAN,
 POLYLINKS BY NANCO BORDEWIJK, WIRREL-WARREL CUBES BY JAN DE GEUS,
POLYSPHERES BY BERNARD WIEZORKE
.. EXHIBITIONS - PUZZLE COLLECTION OF Willem van der Poel, TRACO
PUZZLES BY GERARD TRAABACH, NEW PUZZLES BY OSCAR VAN DEVENTER AND WILL
STRIJBOS, POLYHEDRAL DISSECTIONS BY JOACHIM KRAUSE AND ANTON HANEGRAF
.. PRIZE CONTESTS - RUBIKS CUBE COMPETIYION, EKKEHARD KUNZELL'S GAME
RESERVAT
.. VIDEO SHOWS - EALIER CUBE DAYS, FAST CUBE SOLVING
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

...............................................................
<-->  11th International puzzle collector's party and fair  <-->
...............................................................
WHEN ---- 3/31/91  Easter Sunday
WHERE ---- PACIIFICA HOTEL, 6161 CENTINELA AVE, culver city,  ca,
90231-2200 USA, TEL # 213/649-1776.  This is near Los Angeles Airport
and a hotel courtesy bus will take travelers from airport to hotel.
INVITATIONS ***  Admission by invitation only!!!  Contact Mr. jerry
slocum, 257 south palm drive, beverly hills, ca 90212 for an
invitation.
AGENDA:
.. PUZZLE PARTY
.. SALES /EXHIBITS  table rental available
.. Saturday evening (3/30) dinner and magic show,  estimated cost $40

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^


From pbeck@pica.army.mil  Fri Nov  9 11:09:19 1990
Return-Path: <pbeck@pica.army.mil>
Received: from FSAC1.PICA.ARMY.MIL by life.ai.mit.edu (4.1/AI-4.10) id AA10686; Fri, 9 Nov 90 11:09:19 EST
Date:     Fri, 9 Nov 90 8:41:23 EST
From: Peter Beck (LCWSL) <pbeck@pica.army.mil>
To: Cube-Lovers@life.ai.mit.edu
Subject:  [To: cube-lovers:  CFF #24]
Message-Id:  <9011090841.aa19903@FSAC1.PICA.ARMY.MIL>





----- Forwarded message # 1:

Date:     Wed, 7 Nov 90 8:56:58 EST
From:     Peter Beck  (LCWSL)  <pbeck@pica.army.mil>
To:       cube-lovers@ai.ai.mit.edu
cc:       pbeck@PICA.ARMY.MIL
Subject:  CFF #24
Message-ID:  <9011070856.aa07569@FSAC1.PICA.ARMY.MIL>


SEND TO:  CUBE-LOVERS@AI.AI.MIT.EDU

SUBJECT :  Review of "Cubism For Fun" newsletter issue #24, July 90;
the newsletter of the "Dutch Cubists Club";  in english starting with
issue #14 

1..  The table of contents for issue # 24, july 90 follows:

TENTH CUBE DAY announcement by the secretary
GRAND PRIX editors announcement of the results of the "HIKIMI WOODEN
PUZZLE COMPETITION 1990"
PENTAKUBEN CONTEST ANNOUNCEMENT BY EKKEHARD KUNZELL
SQUA-RING by Nanco Bordewijk
BLOCKED SLIDING by Wim Zwaan
LOGICAL LABYBRINTHS PART 2 by Anneke Treep 
THE RHYTM OF MIX-BOX by Anton Hanegraaf
PRETTY CUBIC PATTERNS by Anneke Treep 
A STRING FOLDING PROBLEM by Oskar van Deventer
KEY THROUGH KEY by Oskar van Deventer
MEMEBRSHIP FEE
TOP SPIN PROCESSES by Bernhard Wiezorke and Anton Hanegraaf
"SEVEN" PUZZLES by Dieter Gebhardt and Anton Hanegraaf
THE CASCADE PYRAMID PROBLEM by joachim Krause
THE DUTCH DRAUGHTBOARD PUZZLE by Wil Strijbos
CRACKING THE (MAGIC) CROSS BY Ronald Fletterman
WIRREL-WARREL SUPER CUBE by Paul Sijben
A HEXOMINO PROBLEM by Pieter Torbijn
NEWS AND LETTERS TO THE EDITOR  -
INTERNATIONAL PUZZLE PARTY ANNOUNCEMENT BY JERRY SLOCUM
BACK ISSUES announcement
CHANGES IN THE LIST OF MEMBERS -  120 active members and growing

* Also an ad from STRIJBOS offering to sell his bolt puzzle (US$28
ppd) and a COCA-COLA BOTTLE puzzle (US$12 ppd)


2.   Membership for 1990 is  US$8.   A photocopied set of the
newsletters, issues 1-13, written in DUTCH (in the future selected
back articles will be available in english) is also available for
 US$7.    To order either of these send an 'INTERNATIONAL"  POSTAL
MONEY ORDER (cost $3 at post office) to:  Paul Sijben, Witbreuksweg
397-304, NL-7522  ZA  Enschede,  The Netherlands.

3.  If anybody would like further details please ask!

CUBING IS FOREVER      PETER BECK

<PBECK@PICA.ARMY.MIL>





----- End of forwarded messages

From hoey@aic.nrl.navy.mil  Fri Nov  9 15:01:52 1990
Received: from Sun0.AIC.NRL.Navy.Mil by life.ai.mit.edu (4.1/AI-4.10) id AA16350; Fri, 9 Nov 90 15:01:52 EST
Received: from sun13.aic.nrl.navy.mil by Sun0.AIC.NRL.Navy.Mil (4.1/SMI-4.0)
	id AA24929; Fri, 9 Nov 90 14:57:26 EST
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Received: by sun13.aic.nrl.navy.mil; Fri, 9 Nov 90 15:02:48 EST
Date: Fri, 9 Nov 90 15:02:48 EST
From: hoey@aic.nrl.navy.mil
Message-Id: <9011092002.AA00993@sun13.aic.nrl.navy.mil>
To: Cube-Lovers@life.ai.mit.edu
Subject: Rubik's Cube reassembly problem and solution
References: <3924@idunno.Princeton.EDU> <1990Nov8.182534.18625@agate.berkeley.edu>
Reply-To: Hoey@aic.nrl.navy.mil (Dan Hoey)

In rec.puzzles article <1990Nov8.182534.18625@agate.berkeley.edu>,
   greg@math.berkeley.edu (Greg Kuperberg) writes:

>Consider a standard Rubik's cube.  Disassemble it and put it back
>together at random.  Find, with proof, the probability that it can be
>solved.

It depends on how you take it apart.  If you just pull out the corner
and edge pieces then put them back in without respect to color, the
probability is one in 12 that you will put it back into the right
orbit.  I won't bore you with yet another proof of this; if you spent
the last decade in a box see the archives, Singmaster's NOTES ON
RUBIK'S MAGIC CUBE, J. A. Eidswick's article in the March 1986 Math
Monthly, or even Hofstadter's METAMAGICAL THEMAS.

Now if you take the face centers off and scramble them, then there is
only one chance in 60 of getting it right.  Of the 720 permutations of
the six face centers, only 24 can be generated by rigid motions of the
cube.  But half of these 24 permutations are odd, and leaving the cube
in an unsolvable orbit.  If you put the face centers on in the
``standard'' configuration with opposite faces ``differing by yellow''
(i.e., white opposite yellow, red opposite orange, and blue opposite
green), your chances go up to one in four--half the time you will get
an odd permutation, and half the time you will get a mirror-reversed
configuration.

But wait, if you took the face centers off you probably noticed that
the corners and edges don't stay on very well.  So, say you scrambled
all three kinds of pieces.  You will be able to solve the resulting
cube if you could solve the corner/edge permutation and the face-
center permutation.  But if the only thing keeping you from solving
the corner/edge permutation and the face-center permutation is that
both permutation parities were odd, then you will be able to solve the
two of them together.  Therefore your chances of success are one in
360 (= (1/12)*(1/60)*2), or one in 24 if you preserved opposite pairs
of face centers.

Now suppose you peeled off the 54 colored stickers and stuck them back
on at random (carefully keeping them out of the reach of children, as
there are rumors the paint contains lead, especially on the cheap
Taiwanese knockoffs), what is the probability of getting a solvable
cube?  This question was posed years ago (in Singmaster?) but I
believe it is still open.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From hirsh@cs.rutgers.edu  Sat Nov 10 18:50:06 1990
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Received: from pei.rutgers.edu by life.ai.mit.edu (4.1/AI-4.10) id AA11436; Sat, 10 Nov 90 18:50:06 EST
Received: by pei.rutgers.edu (5.59/SMI4.0/RU1.2/3.05) 
	id AA16007; Sat, 10 Nov 90 18:49:51 EST
Sender: Haym Hirsh <hirsh@pei.rutgers.edu>
Date: Sat, 10 Nov 90 18:49:48 EST
From: Haym Hirsh <hirsh@cs.rutgers.edu>
Reply-To: Haym Hirsh <hirsh@cs.rutgers.edu>
To: Hoey@aic.nrl.navy.mil (Dan Hoey), Cube-Lovers@life.ai.mit.edu
Subject: Re: Rubik's Cube reassembly problem and solution
In-Reply-To: Your message of Fri, 9 Nov 90 15:02:48 EST
Cc: Haym Hirsh <hirsh@cs.rutgers.edu>
Message-Id: <CMM-RU.1.0.658280988.hirsh@pei.rutgers.edu>

> Now suppose you peeled off the 54 colored stickers and stuck them back
> on at random (carefully keeping them out of the reach of children, as
> there are rumors the paint contains lead, especially on the cheap
> Taiwanese knockoffs), what is the probability of getting a solvable
> cube?  This question was posed years ago (in Singmaster?) but I
> believe it is still open.
> 
> Dan Hoey
> Hoey@AIC.NRL.Navy.Mil

This seems easy, so I've probably messed up on something.
Can anyone catch a mistake?

Assume each of the stickers is given a number from 1 to 54.  Then
there are 54! different labelings, ignoring rotation of stickers
(we'll ignore this throughout, so we'll never need to consider it).
Thus there are

  54!
  = 230843697339241380472092742683027581083278564571807941132288000000000000
  = 2.3*10^71

ways to randomly resticker the cube.  We want to know what proportion
of these are legal (i.e., the cube can be solved).

There are 8!*12!*8^3*2^12/12 = 43252003274489856000 = 4.3*10^19 legal
cube states.  Thus there are this many legal stickerings, if each
sticker must go back to where it was.  Since they need not (just the
color must match), there are really an additional (9!)^6 for each of
these, or 98760760257294265888495040331277846607560704000000000 =
9.9*10^52 legal stickerings.

Thus the proportion of randomly restickered cubes that can be solved,
and hence the probability that a randomly restickered cube can be solved,
is

	98760760257294265888495040331277846607560704000000000
------------------------------------------------------------------------
230843697339241380472092742683027581083278564571807941132288000000000000

= 9.9*10^52/2.3*10^71 = 4.3*10^-19

From dik@cwi.nl  Sat Nov 10 20:17:16 1990
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Received: from charon.cwi.nl by life.ai.mit.edu (4.1/AI-4.10) id AA12248; Sat, 10 Nov 90 20:17:16 EST
Received: by charon.cwi.nl with SMTP; Sun, 11 Nov 90 02:17:08 +0100
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Date: Sun, 11 Nov 90 02:17:02 +0100
From: dik@cwi.nl
Message-Id: <9011110117.AA27431@paring.cwi.nl>
To: Cube-Lovers@life.ai.mit.edu
Subject: Re: Rubik's Cube reassembly problem and solution

Aside from the disassembly/assembly problem there was another problem that
I have not yet seen answered satisfactory.  The question is:  what is the
maximum number of stickers that can bee peeled of such that there is still
an unique solution for the cube (i.e. the remaining stickers must match in
color on a face).  The only solution I have seen was along the lines (this
is from memory, but I do not think there are any mistakes):
1.  The total rotation of the corner cubes is 0, so there is one corner
    cube that can have all its stickers removed; the remaining corner
    cubes need at least one sticker.  Suppose this is the FUR cube.
    (3 stickers.)
2.  You can remove two stickers (F, R and/or U) from each of FRD, BRU, FLU;
    they still remain distinguishable.  (6 stickers.)
3.  Of the remaining corner cubes (DBL, DRB, ULB, DLF) you cannot *now*
    (emphasis mine) remove two stickers because the cube will become
    indistinguishable from one of the cubes handled in step 2.  You can
    remove sticker R, U and F from DRB, ULB and DLF respectively.  No
    other stickers can be removed.  (3 stickers.)
4.  Because of flip parity you can remove two stickers from (say) FU.
    (2 stickers.)
5.  You can remove the F sticker from all of FR, FL and FD.  (3 stickers.)
6.  Now, because of the product parity of corner cubes permutation and
    edge cube permutation you can make either two corner cubes identical or
    two edge cubes.  You must nevertheless still be able to observe both
    the corner twist parity and the edge flip parity.  This means you may
    a.  Remove a single sticker from any edge cube that still has two stickers.
    b.  Remove a single sticker from the DLB cube.
    (You can not remove two stickers from the DLB cube.  Say you remove the L
    and B sticker.  Let us denote removed sticker by lower case letters.
    In that case Dlb is indistinguishable from Dfr, which is not a problem.
    But the DLf cube can now be put in the Dlb position, leading to a 3-cycle.)
    (1 sticker.)
7.  You can remove the sticker from the front center cube.  (1 sticker.)

This leads to a total of 19 removable stickers.

This is not maximal.  There are, for instance, other ways to do corner cubes:
1.  Remove all F stickers.  (4 stickers.)
2.  From all Fxy cubes, remove the y sticker.  (4 stickers.)
3.  Also from all Bxy cubes, remove the y sticker.  (4 stickers.)
4.  From one Bxy cube remove also an x sticker.  (1 sticker.)
5.  From one Fxy cube remove also an x sticker.  (1 sticker.)
(Fxy and Bxy named clockwise.)
This leads to 14 stickers for the corners and a total of 21.
Are there other ways leading to more?  Are there better ways that we can
remove more center stickers?
--
dik t. winter, cwi, amsterdam, nederland
dik@cwi.nl

From hirsh@cs.rutgers.edu  Sun Nov 11 15:34:28 1990
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	id AA00310; Sun, 11 Nov 90 15:34:25 EST
Sender: Haym Hirsh <hirsh@pei.rutgers.edu>
Date: Sun, 11 Nov 90 15:34:23 EST
From: Haym Hirsh <hirsh@cs.rutgers.edu>
Reply-To: Haym Hirsh <hirsh@cs.rutgers.edu>
To: Cube-Lovers@life.ai.mit.edu
Subject: Re: Rubik's Cube reassembly problem and solution
In-Reply-To: Your message of Sat, 10 Nov 90 18:49:48 EST
Cc: Haym Hirsh <hirsh@cs.rutgers.edu>
Message-Id: <CMM-RU.1.0.658355663.hirsh@pei.rutgers.edu>

I just caught a bug in my reasoning.  The restickering need not yield
something equivalent to the original undestickered cube, but rather
just one that can be solved to obtain solid colors on each face.
Since there are 5*3*2 different distinguishable cubes (i.e., 30
different ways to label a die with the numbers 1-6) (6! labelings, but
rotational symmetry removes 24 -- six faces can be brought to the top,
and for each it can be rotated around the axis perpendicular to that
face in one of 4 ways), the numerator should be multiplied by 30, and
hence the probability is actually

       2962822807718827976654851209938335398226821120000000000
------------------------------------------------------------------------
230843697339241380472092742683027581083278564571807941132288000000000000

= 3.0*10^54/2.3*10^71 = 1.3*10^-17

Haym

From RGC915@uacsc2.albany.edu  Mon Nov 12 01:24:08 1990
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Date:         Sun, 11 Nov 90 22:01:08 EST
From: Robert Clark <RGC915@uacsc2.albany.edu>
Subject:      Rubik's Cube Variants?
To: cube-lovers@life.ai.mit.edu

  Does anyone know where I can find all those variations on the Rubik's
theme that popped up after the Cube came out? I mean puzzles like the
Pyraminx, Impossiball, etc. I haven't seen any place that sells them
in the area where I live, New york state.
  I would even be willing to send for them from overseas if the price
is reasonable.



                                      Robert Clark

From @mitvma.mit.edu:RCC2@VAXB.YORK.AC.UK  Mon Nov 12 09:45:25 1990
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Date:     Fri, 9 Nov 90  20:13 GMT
From: RCC2%VAXB.YORK.AC.UK@mitvma.mit.edu
To: CUBE-LOVERS@life.ai.mit.edu
Subject:  hello there

Hello there,

This is my first posting to the cube-lovers board, so I'm
probably gonna ask a couple of really obvious questions:

a) Does anyone know where I can get a copy of David Singmaster's
   book "Notes on Rubik's magic cube?"  This was THE definitive book
   on the cube about 8 years ago, but I lost my copy....does anyone
    know if it's still in print?? ( Oh yeah, maybe I should mention
   that I'm in England...David Singmaster was a lecturer at one of
   the colleges in London I think - was this book EVER published in
   the states? )

b) ( This is a real obvious one... )   Does anyone have any tips or
   advice on solving the 4*4*4 cube that appeared a few years after
   the original 3*3*3 one.  I got really close to getting it right
   a couple of years ago, but never quite made it.

Thanks in advance for any help,
  Rod Chapman

rcc2@vaxa.york.ac.uk


From hirsh@cs.rutgers.edu  Mon Nov 12 12:05:53 1990
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Sender: Haym Hirsh <hirsh@pei.rutgers.edu>
Date: Mon, 12 Nov 90 12:05:42 EST
From: Haym Hirsh <hirsh@cs.rutgers.edu>
Reply-To: Haym Hirsh <hirsh@cs.rutgers.edu>
To: Robert Clark <RGC915@uacsc2.albany.edu>
Cc: cube-lovers@life.ai.mit.edu
Subject: Re: Rubik's Cube Variants?
In-Reply-To: Your message of Sun, 11 Nov 90 22:01:08 EST
Message-Id: <CMM-RU.1.0.658429542.hirsh@pei.rutgers.edu>

Peter Beck, pbeck@pica.army.mil, has many cube spinoffs for sale.
That's where I got the last few I was missing.  Jerry Slocum in
Calif also has some items for sale -- I got his address from old
cube-lovers mailings (sent by Peter, I believe).  I seem to recall
a few other sources outside the US, but Peter probably can provide
them if there's something Slocum and Peter don't have.

Haym

From rp@xn.ll.mit.edu  Mon Nov 12 12:09:56 1990
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Date: Mon, 12 Nov 90 12:07:55 EST
From: Richard Pavelle <rp@xn.ll.mit.edu>
To: CUBE-LOVERS@life.ai.mit.edu
In-Reply-To: RCC2%VAXB.YORK.AC.UK@mitvma.mit.edu's message of Fri, 9 Nov 90  20:13 GMT <9011121445.AA02330@life.ai.mit.edu>
Subject: hello there

   Date:     Fri, 9 Nov 90  20:13 GMT
   From: RCC2%VAXB.YORK.AC.UK@mitvma.mit.edu

   Hello there,

   This is my first posting to the cube-lovers board, so I'm
   probably gonna ask a couple of really obvious questions:

   a) Does anyone know where I can get a copy of David Singmaster's
      book "Notes on Rubik's magic cube?"  This was THE definitive book
      on the cube about 8 years ago, but I lost my copy....does anyone
       know if it's still in print?? ( Oh yeah, maybe I should mention
      that I'm in England...David Singmaster was a lecturer at one of
      the colleges in London I think - was this book EVER published in
      the states? )

   b) ( This is a real obvious one... )   Does anyone have any tips or
      advice on solving the 4*4*4 cube that appeared a few years after
      the original 3*3*3 one.  I got really close to getting it right
      a couple of years ago, but never quite made it.

I have not looked at it for several years but if memory serves you need
only one extra transformation which is not applicable to the 3^3. It is
the single edge flip. I no longer recall it explicitly but it was kinda
trivial to find.

From hoey@aic.nrl.navy.mil  Mon Nov 12 18:37:55 1990
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Date: Mon, 12 Nov 90 18:38:49 EST
From: hoey@aic.nrl.navy.mil
Message-Id: <9011122338.AA00219@sun13.aic.nrl.navy.mil>
To: Cube-Lovers@life.ai.mit.edu
Subject: Re: Rubik's Cube reassembly problem and solution

This problem of counting the number of solvable restickerings seems to
be a lot easier than I had thought, but a lot trickier than you might
think.

Haym Hirsh sent in a buggy analysis, then corrected himself, but not
quite enough.  The fix was to account for cases where the stickers
corresponded to a cube recoloring, but he just multiplied by 30 (cube
recolorings up to rotational symmetry) rather than by 720 (total cube
recolorings).  We are dividing by 54!, which includes positions
differing only by a rotation, so when figuring how many are solvable
you have to count such positions also.

Another way of figuring this is 6! ways of coloring the face centers,
then (8! 3^8 12! 2^12)/12 ways of coloring the rest of the cube, then
9!^6 ways of arranging stickers among identically-colored faces, out
of 54! ways of arranging stickers randomly.

So the probability that a random restickering will be solvable is

        71107747385251871439716429038520049557443706880000000000
------------------------------------------------------------------------
230843697339241380472092742683027581083278564571807941132288000000000000

           40122452017152
 = ------------------------------  ~ 3.0803 X 10^-16.
   130253249618151492335575683325

It seems odd to me that this is not the reciprocal of an integer, but
I guess that's because we are dealing with color cosets rather than
some cube group.

Haym Hirsch also asked me how to figure out the minimum number of
stickers to fix to make an unsolvable stickering solvable.  Sounds
hard to me.  His question arises in the same way that I recall the
original problem arising: trying to clean up after someone who tried
to solve the cube by restickering.  Since the adhesive isn't designed
for moving the stickers around, this leads rapidly to Dik Winter's
problem: dealing cubes that have lost some of their stickers.

Dan
Hoey@AIC.NRL.Navy.Mil

From pbeck@pica.army.mil  Wed Nov 14 09:59:24 1990
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Date:     Tue, 13 Nov 90 7:48:47 EST
From: Peter Beck (LCWSL) <pbeck@pica.army.mil>
To: Robert Clark <RGC915@uacsc2.albany.edu>
Cc: cube-lovers@life.ai.mit.edu
Subject:  Re:  Rubik's Cube Variants?
Message-Id:  <9011130748.aa09243@FSAC1.PICA.ARMY.MIL>

I am the best general source for rubik's cube items.

If you want a list of whatr is available e-mail me your postal address.

THE FUTURE IS PUZZLING,
BUT CUBING IS FOREVER!!!!!!!!!!!

<PBECK@ARMY.PICA.MIL>


From pbeck@pica.army.mil  Fri Nov 30 07:50:59 1990
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Date:     Thu, 29 Nov 90 12:33:41 EST
From: Peter Beck (LCWSL) <pbeck@pica.army.mil>
To: cube-lovers@life.ai.mit.edu
Cc: pbeck@pica.army.mil
Subject:  bottleneck and
Message-Id:  <9011291233.aa21311@FSAC1.PICA.ARMY.MIL>


BOTTLENECK  SOURCE:

Will the designer of Bottleneck please contact 
  MIKE GREEN
  24832 144th PLACE  S.E
  KENT WASHINGTON 98042.

Mike has a puzzle business and wants to sell Bottleneck.  

If anybody else out there manufactures or deals in mechanical puzzles
and is looking for a retail or wholesale outlet please feel free to
contact Mike.

Mike manufactures and sells a line of wire disentanglement puzzles
called "PUZZLETTS".  He has opened a retail outlet in his home (the
address above).

Mike also collects puzzles and has a list of puzzle suppliers and
puzzle solution sheets.  If anybody out there is looking for something
or wants to contribute I am sure he would happy to correspond.  NO
E-MAIL,  postal or telephone only (sorry I don't have phone number
handy).

PS  If anybody wants to contact Mike through me please feel free.



From pbeck@pica.army.mil  Sat Dec  8 09:51:49 1990
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Date:     Fri, 7 Dec 90 11:41:03 EST
From: Peter Beck (LCWSL) <pbeck@pica.army.mil>
To: cube-lovers@life.ai.mit.edu
Subject:  rec & ed computing
Message-Id:  <9012071141.aa04773@FSAC1.PICA.ARMY.MIL>



Anybody have an opinion on the newsletter/magazine
"RECREATIONAL & EDUCATIONAL COMPUTING" edited by Dr. Michael Ecker.

From cosell@bbn.com  Sat Dec  8 15:48:28 1990
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Message-Id: <9012082048.AA24247@life.ai.mit.edu>
Date:     Sat, 8 Dec 90 15:42:21 EST
From: Bernie Cosell <cosell@bbn.com>
To: Peter Beck <pbeck@pica.army.mil>
Cc: cube-lovers@life.ai.mit.edu
Subject:  Re:  rec & ed computing

Sure....

  -  REC is VERY slanted toward high school students, and so there is
     very little advanced or profound stuff in there.
  -  While there is a nod to other worlds, primarily it is all in BASIC,
     and generally focused on the IBM PC.
  -  There is a fascination with mindless crunching just to print out
     numbers that I can't fathom.  A good portion of the articles
     center on a numbers with some odd property or another, or finding
     the actual _numeric_ solution to something and usually brute force
     [or close to it].  The graphics hacks, such as they are, are
     primarily crunching-based [moire patterns and such].

No real discussion of 'puzzles', for example, nor of the kinds of techniques
and such you need to partially-tame one of those awful [but real world]
exponential searches, nor of representing 3D objects or manipulations of them,
or search strategies, no word problems, etc.

/Bernie\

From pbeck@pica.army.mil  Mon Dec 10 11:35:28 1990
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Date:     Mon, 10 Dec 90 11:18:02 EST
From: Peter Beck (LCWSL) <pbeck@pica.army.mil>
To: Bernie Cosell <cosell@bbn.com>
Cc: Peter Beck <pbeck@pica.army.mil>, cube-lovers@life.ai.mit.edu
Subject:  Re:  rec & ed computing
Message-Id:  <9012101118.aa09879@FSAC1.PICA.ARMY.MIL>

thanks bernie.

PS:  do you have address & name of owner for games people p;lay.?

From @relay.cs.net:AGIN@cgi.com  Thu Dec 13 23:25:58 1990
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Date: Thu, 13 Dec 90 14:26 EDT
From: AGIN%cgi.com@relay.cs.net
Subject: Re: construction project
To: cube-lovers@life.ai.mit.edu
X-Vms-To: IN%"cube-lovers@life.ai.mit.edu"

I was successful in creating Peter Beck's Christmas Tree ornament.

The project requires 50 modules, not 120.  There are 30 outside modules and
20 inside connecting modules.  The outside modules correspond to th edges of
a dodecahedron.  The inside modules create an interior icosahedron.

I used 3M Post-It notes cut in half, each starting rectangle being 3" by 1-
1/2".  I folded the adhesive to the inside on the first step, so the
adhesive was not holding the project together.  It probably would have been
possible to use the adhesive to keep each module together.  This would have
required a lot of extra care in the assembly, but produced a much sturdier
product.  As it was, once I got the hang of it, I didn't have any major
problems with modules coming apart.  The finished construction required no
staples or extra glue.

A previous attempt using 1" x 2" rectangles cut out of graph paper kept
falling apart.  I've got a partially finished ornament made with dollar
bills, which seem to work fine.

The ideal shape for an outside module is not an equilateral triangle, but an
isosceles one with an apex angle of about 42 degrees.  I took care of this
by allowing the outside surfaces to bow outward.

To finish the assembly I left three outside modules and their common
connecting module until last.  The outside modules were threaded into place
but not closed, with the ends of the paper pointing outward.  The connecting
module was placed over the nearest ends of the three outside modules, then
the outside modules could be closed.

From j9@icad.com  Fri Dec 14 19:26:43 1990
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Date: Fri, 14 Dec 90 13:38 EST
From: Jeannine Mosely <j9@icad.com>
Subject: Peter Beck's construction project
To: cube-lovers@life.ai.mit.edu
Message-Id: <19901214183850.8.J9@MOE.ICAD.COM>

I have made something along the lines that Peter Beck describes in
his "construction project", but it does not quite fit his
description, so I don't know if it is the same thing. It uses only
50 modules and I can't for the life of me imagine where the other 70
should go.

My object looks like this.  Imagine a regular icosahedron (20
equilateral triangular faces, with 5 coming together at each
vertex).  Erect on each of these faces a triangular prism (20
modules).  At each edge of the icosahedron, two square faces of
adjacent prisms rise up from the surface of the icosahedron.  Band
each such pair together with a module (30 modules).  The reulting
form resembles the Archmidean solid most conveniently designated
(3,4,5,4), which means that each vertex contains a triangle, square,
pentagon, square, in that order.

I say "resembles" this solid, in part, because only the squares are
actually present, the triangular and pentagonal "faces" are voids.
But a more compelling reason for saying "resembles" is that the
geometry is only approximate.  If one uses the modules you describe
for the triangular prisms (that is, the height of the prism equals
the edge of the triangle) then the quadrilateral faces on the outer
surface connecting the triangular and pentagonal voids are not
squares, but rectangles whose side are in the ratio of (sqrt 5)-1 to
(sqrt 3).  This discrepancy can be fudged, by allowing the squares
to bulge outward slightly.  On the other hand, a figure could be
constructed where the outer quadrilaterals were in fact square, but
this would require the prisms to be shorter, and that cannot be
fudged.

Better results can be achieved if you do not fudge the geometry (or
at least not much).  It turns out that

	(/ (- (sqrt 5) 1) (sqrt 3)) = 5/7 

(pardon my lisp) to within one tenth of one percent. Hence I make my
modules as diagrammed below.  Dimensions given assume paper in the
ratio of 2 to 1.

This module is used to make the triangular prisms:

    _______________________________________________
   |         :             :             :         | 5/24
   |.........:.............:.............:.........|
   |         :             :             :         |
   |         :             :             :         | 7/12
   |         :             :             :         |
   |.........:.............:.............:.........|
   |         :             :             :         | 5/24
   |_________:_____________:_____________:_________|
       1/2        1/2           1/2          1/2


This module is used to band the triangular prisms together:

    _______________________________________________
   |         :             :             :         | 1/4
   |.........:.............:.............:.........|
   |         :             :             :         |
   |         :             :             :         | 1/2
   |         :             :             :         |
   |.........:.............:.............:.........|
   |         :             :             :         | 1/4
   |_________:_____________:_____________:_________|
      5/12        7/12          7/12        5/12


Natually, you might ask, how do I fold 5/12?  There is a trick.
First fold the the long edge in half, and then in quarters at one
end, but don't make the second crease go all the way across--just
nick one edge of the paper, as a marker (point B).  Now fold point B
to touch the upper left-hand corner (point A).  This would make a
diagonal crease across the strip, but again, don't make the crease
go all the way across--just nick the lower edge (point C).  The line
AC is the hypoteneuse of the old 5,12,13 right triangle, and point C is
at 5/12, as desired.  (Pretty neat, huh?)

       A _______________________________________________
        |                      :                        |
        |                      :                        |
        |                      :                        |
 12/12  |                      :                        |
        |                      :                        |
        |                      :                        |
        |                      :                        |
        |_______:______________:_____________:__________|
          5/12  C     7/12           6/12    B   6/12


A similar technique is used to make the other module.

I did not need any staples.

	-- jeannine mosely

From mindcrf!ronnie@boris.mindcraft.com  Tue Mar  5 19:29:53 1991
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From: mindcrf!ronnie@boris.mindcraft.com (Ronnie Kon)
Message-Id: <9103051945.AA33973@boris.mindcraft.com>
To: @ames.uucp:ai.ai.mit.edu!Cube-Lovers
Subject: Is Meffert still around?

	I am wondering if Meffert is still around, with his club to purchase
new and interesting Cube products.  If he is, would somebody please send me
his address and what the current membership fee is.  Also a price list.

	If he is not, would people be interested in restarting such a club?
I have a hard time believing that there aren't enough people for a cube-of-
the-month club (or perhaps cube-of-the-quarter) as these things are not that
complex (ie., expensive) to produce.  We might even be able to do runs in
rolled aluminum instead of plastic.

				Ronnie
				kon@groundfog.stanford.edu

From @po5.andrew.cmu.edu:dn1l+@andrew.cmu.edu  Fri Mar  8 13:14:12 1991
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Date: Fri,  8 Mar 91 13:13:44 -0500 (EST)
From: "Dale I. Newfield" <dn1l+@andrew.cmu.edu>
To: Cube-Lovers@life.ai.mit.edu
Subject: 5x5x5 Cube

I saw a 5x5x5 Cube (Rubik's type) in a friend's office (One of his
office-mates had it.)(I left a message, but never got a response from
him.)  I was wondering if anybody out there has seen one of these, and
could point me in a direction that would lead to one?


Thank You!

Dale Newfield
dn1l@andrew.cmu.edu

From pbeck@pica.army.mil  Mon Mar 11 11:13:02 1991
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Date:     Mon, 11 Mar 91 11:10:46 EST
From: Peter Beck (LCWSL) <pbeck@pica.army.mil>
To: kon@groundfog.stanford.edu
Cc: cube-lovers@life.ai.mit.edu
Subject:  meffert
Message-Id:  <9103111110.aa20787@FSAC1.PICA.ARMY.MIL>


SHORT ANSWER:  Meffert's puzzle club is dead!

MORE:  Meffert is alive and is trying to get back into the puzzle
 business - if you want specific information Jerry Slocum, beverly
hills has been in conatct with him.

PUZZLE CLUBS, ETC.:  
1 - the economics of a puzzle club is that well made  puzzles (both
from a design and engineering perspective) cost $20 and up.
2 - people interested in being current on whats happening in puzzles
should subscribe to CFF, Puzzletopia and possible ARM (all have been
discussed before)


I have been busy but CFF#25 (silver aniverssary; cost $18) is 5
volumes plus vendor catalogs (bandelow, constatin) and encompassses
the last 10 tears of puzzling around the world.

SOME RETAIL PUZZLE SOURCES:
  cubes:   peter beck, usa
              bandelow, germany
              constatin, germany
 other:  bits & pieces
            kadon
            jon foolery
            science museum shops

SOME WHOLESALE SOURCES:
  USA,  PUZZLETTES
  USA,  ISHI PRESS - JAPANESE PUZZLES
  USA, TUCKER-JONES TAVERN PUZZLES (what Bush does on way to camp
david)
  UK, pentangle
  French, Arjeu

SOME NEWER PUZZLES:
  magic cross, germany
  masterball, swiss
  rotos, german
  new rubiks, europe
  square one, milton bradley - in stores soon

**  INTERNATIONAL PUZZLE PARTY (200+ worldwide attendees) 3/30/91 in
LA.  Admittance by invitation.


FURTHER DISCUSSION REQUESTED.

From pbeck@pica.army.mil  Mon Mar 18 12:42:18 1991
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Date:     Mon, 18 Mar 91 12:36:36 EST
From: Peter Beck (LCWSL) <pbeck@pica.army.mil>
To: cube-lovers@life.ai.mit.edu
Cc: pbeck@pica.army.mil
Subject:  tcf 91
Message-Id:  <9103181236.aa07170@FSAC1.PICA.ARMY.MIL>



The  TRENTON COMPUTER FESTIVAL 1991 WILL BE THE 20 & 21 OF APRIL AT
the same "OLD"  LOCATION  --> TRENTON STATE COLLEGE on state highway
31 in Trenton NJ.  THIS IS THE LARGEST AND OLDEST AMATEUR COMPUTER
FESTIVAL in the country.    I have a table selling puzzles in the
fleamarket.


<BECK@PICA.ARMY.MIL>   my best address




From pbeck@pica.army.mil  Tue Mar 19 09:22:42 1991
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Date:     Tue, 19 Mar 91 8:30:13 EST
From: Peter Beck (LCWSL) <pbeck@pica.army.mil>
To: rp@xn.ll.mit.edu, cube-lovers@life.ai.mit.edu
Subject:  more on tcf
Message-Id:  <9103190830.aa28520@FSAC1.PICA.ARMY.MIL>


The  TRENTON COMPUTER FESTIVAL 1991 WILL BE THE 20 & 21 OF APRIL AT
the same "OLD"  LOCATION  --> TRENTON STATE COLLEGE on state highway
31 in Trenton NJ.  THIS IS THE LARGEST AND OLDEST AMATEUR COMPUTER
FESTIVAL in the country.    I have a table selling puzzles in the
fleamarket.

TCF is a combination retail sales show, technical symposium and
fleamarket sponsored by the amateur computer clubs in the
NYC-philadelphia metro area.  It has a state fai atmosphere with a PC
theme.  Attendees come mostly from east of the mississippi and average
10-15,000 per day.  It lasts for 2 days.

FLEAMARKET HOURS - it is outdoors and rain or shine, sat is ALWAYS
best
  sat 7am - 5pm
  sun 9am- 4pm

RETAIL COMMERCIAL SALES EXHIBITS - inside gymnasium
  9-4 both days

TECHNICAL LECTURES - inside
  this is multi track
  9-4 both days

USER GROUP MEETINGS - inside
  9-4 both days

KEYNOTE BANQUET AND LECTURE
 8pm sat - notable speaker, eg, bill gates


<BECK@PICA.ARMY.MIL>   my best address




From pbeck@pica.army.mil  Thu Apr 11 09:21:50 1991
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Date:     Thu, 11 Apr 91 9:17:33 EDT
From: Peter Beck (LCWSL) <pbeck@pica.army.mil>
To: cube-lovers@life.ai.mit.edu
Cc: brodin@pica.army.mil
Subject:  tcf correction
Message-Id:  <9104110917.aa15850@FSAC1.PICA.ARMY.MIL>


4/11/91  version

The  TRENTON COMPUTER FESTIVAL 1991 WILL BE THE 20 & 21 OF APRIL "NOT"
at the same "OLD"  LOCATION  --> ie, it is at MERCER COUNTY COMMUNITY
COLLEGE not TRENTON STATE COLLEGE.  Directions are below, phone
609/655-4898/4999 - SORRY for the previous misinformation.   THIS IS
THE LARGEST AND OLDEST AMATEUR COMPUTER FESTIVAL in the country.    I
have a table selling puzzles in the fleamarket.

TCF is a combination retail sales show, technical symposium and
fleamarket sponsored by the amateur computer clubs in the
NYC-philadelphia metro area.  It has a state fair atmosphere with a PC
theme.  Attendees come mostly from east of the mississippi and average
10-15,000 per day.  It lasts for 2 days.

ADMISSION -  $7 FOR SAT & SUN, $5 for sun only - students $3

FLEAMARKET HOURS - it is outdoors and rain or shine, sat is ALWAYS
best
  sat 7am - 5pm
  sun 9am- 4pm,  900 spots

RETAIL COMMERCIAL SALES EXHIBITS - inside gymnasium
  9-4 both days , 
      exhibitors include: microsoft, HP, ashton-tate, software publ,
micrografx, lotus, intel, adobe, borland, ast

TECHNICAL LECTURES - inside
  this is multi track, over 100 KEYNOTE SPEAKER; Fred Gibbons, CEO
software publishing corp 9:30 AM in the theater, "What lies ahead for
the software industry"
  9-4 both days

USER GROUP MEETINGS - inside
  9-4 both days

KEYNOTE BANQUET AND LECTURE
 8pm sat - notable speaker, eg, bill gates

$$$$$$$$$$$$$$$$$$$$$$$$$$$
map locator -->  MCCC is near the intersection of US #1 and I-295.

Here's how to get to a parking space at TCF!

Guaranteed parking is available at Mercer County Park, which surrounds
the MCCC campus on three sides.  MCP has two entrances, one on Rt 535
(Edinburg Rd. or Old Trenton Rd., depending on whether you're north or
south of MCCC) and one on Hughes Drive.  My advice is to turn off at a
park entrance.  HOWEVER, if you're early (on sat this means before 7
on sun before 9) or if you're daring, continue past the park entrance
to the College entrance.  If you're lucky, the gendarmes will let you
on campus to look for a parking space.  IF THEY REFUSE YOU ACCESS,
continue to the traffic light and turn (either LEFT onto Old Trenton
Rd. or RIGHT onto Hughes Drive.) and proceed to the  Park entrance.

FROM THE NORTH:

1) VIA U.S. 1: Go south on US 1 and take the Rt 533 South
(Quakerbridge
Rd.) overpass.  After about two miles, turn left onto Hughes Dr.  The
Park entrance will be on your left in about a mile, with the College
entrance about 0.5 miles further.

2) VIA N.J. TURNPIKE: Go south to Exit 8 (Hightstown) and get on Rt 33
WEST.  In downtown Hightstown, turn right onto Rt 571 and follow.
 Near
GE Astro, turn left onto Rt 535.  The Park entrance is about 4 miles
down
the road, with the College entrance about a mile further.

FROM THE SOUTH:

1) VIA U.S. 1: Go north on Rt. 1 and turn right onto Rt 546 (at Mrs.
G's
Appliances).  Just after the overpass, turn right onto Youngs Rd.
 Follow
to the end and turn right onto Hughes Drive.  The Park entrance is
about
a mile away.

2) VIA I-95: I-95 NORTH becomes I-295 SOUTH (don't ask!).  Take Exit
65A,
Sloan Rd and follow to the end (Sloan Rd. becomes Flock Rd. at the
light -
also don't ask!!).  Go left onto Old Trenton Rd.  The Campus entrance
is
by a jughandle turn, about a mile up Old Trenton Rd.  The Park
entrance
is on the left about a mile further up.

3) VIA I-295: Follow I-295 NORTH to its temporary end at Rt 130, and
go
north to Rt. 206 where you will follow signs to TRENTON and then to

I-295 NORTH. Take Exit 65A to Sloan Ave. and follow it to the end.  Go
left
on Old Trenton Rd.  The campus entrance is by a jughandle turn, about
a mile up Old Trenton Rd.  The Park entrance is on the left about a
mile further.

4) VIA N.J. TURNPIKE:  Take Exit 7A to I-195 WEST.  Take the first
exit,
5B to Rt 130 NORTH.  Left at the first light onto Rt. 526  Bear left
and take an immediate right, still on Rt 526.  At end, turn left onto
Rt 535, Old Trenton Rd.  The Park entrance is a bout a mile away; the

College, another mile.

$$$$$$$$$$$$$$$$$$

<BECK@PICA.ARMY.MIL>   my best address




From pbeck@pica.army.mil  Mon Apr 22 15:35:39 1991
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Date:     Mon, 22 Apr 91 15:20:08 EDT
From: Peter Beck (LCWSL) <pbeck@pica.army.mil>
To: cube-lovers@life.ai.mit.edu
Subject:  puzzle party review
Message-Id:  <9104221520.aa15409@FSAC1.PICA.ARMY.MIL>



Review of the 11th  INTERNATIONAL PUZZLE PARTY
 Held on March 30,31 1991 at the Pacifica Hotel Culver City, CA USA

EVENTS
..  Saturday
          daytime - puzzle exchange; admission requirement a puzzle
gift for each other attendee
          evening - dinner party and magic show;  MAGICIANS:  Max
Maven, Mark Setteducati, Mike Weber
..  Sunday:  A ballroom is set up for cash sales of puzzles.

CUBING HIGHLIGHTS
..  Minh Thai gave a demonstration of doing the cube (he is Guiness
world record holder).  His algorithm is:
     1  -  corners first
     2  -  3 of 4 edges on each face
     3  -  last face
..  Anneke Treep a founder of CFF was in attendance.  CFF will
probably host 13th party scheduled for Europe.
..  A spherical SKEWB is in production.  Very interesting puzzle.

OTHER HIGHLIGHTS
..  partial list of attendees (about 100 puzzlers attended):  
     NOB, a prominent Japanese puzzler
     Ed Hordern author of Sliding Block book published by OUP
     Jerry Slocum, party arranger and author of Puzzles Old & New
     Kathy Jones, owner of Kadon
     Solomon Golomb, polycube inventor
     Jose Grant, designer of jewelry quality puzzle rings
     Scott Kim, inversions
     Christoph Bandelow, German seller of magic polyhedra and author
     Doug Engel, designer of circle puzzler, flexagon based puzzles
     James Dalgety, a founder of Pentangle

..  4 artists in attendance:  one makes sculptures that assemble as
puzzles, one makes pattern assembly puzzles by vacuum deposition of
metals on glass squares, one makes Tiffany style lamps using tangram
pieces & silhouettes for the design, the last uses puzzles primarily
for inspiration.
..  


FUTURE PARTY SCHEDULE
-   12th  Tokyo Japan, Host NOB
-   13th  Netherlands, host CFF
-   14th  probably USA

From mindcrf!ronnie@peabody.mindcraft.com  Mon Apr 22 20:09:08 1991
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Date: Mon, 22 Apr 91 16:43:31 -0700
From: mindcrf!ronnie@peabody.mindcraft.com (Ronnie Kon)
Message-Id: <9104222343.AA22704@peabody.mindcraft.com>
To: @mindcrf:ames!ai.ai.mit.edu!Cube-Lovers
Subject: 5-cube in a game store!!!

	The GameKeeper (in the Valley Fair Mall in San Jose) actually has
5x5x5 Rubik's cubes on sale (for $37 I think).  They had three in stock on
Saturday.

	Are we seeing a renascence of cubing?  This is certainly a welcome
development.

				Ronnie

-------------------------------------------------------------------------------
Ronnie B. Kon                         |  "I don't know about your brain, but
kon@groundfog.stanford.edu            |	 mine is really bossy."
...!{decwrl,ames,hpda}!mindcrf!ronnie |			-- Laurie Anderson
-------------------------------------------------------------------------------

From ncramer@bbn.com  Mon Apr 22 23:37:39 1991
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Message-Id: <9104230337.AA04955@life.ai.mit.edu>
Date:     Mon, 22 Apr 91 21:01:31 EDT
From: Nichael Cramer <ncramer@bbn.com>
To: Ronnie Kon <mindcrf!ronnie@peabody.mindcraft.com>
Cc: Cube-Lovers@life.ai.mit.edu
Subject:  Re:  5-cube in a game store!!!

>Date: Mon, 22 Apr 91 16:43:31 -0700
>From: Ronnie Kon <mindcrf!ronnie@peabody.mindcraft.com>
>To: @BBN.COM,@mindcrf.uucp:ames!ai.ai.mit.edu!Cube-Lovers
>Subject: 5-cube in a game store!!!
>	The GameKeeper (in the Valley Fair Mall in San Jose) actually has
>5x5x5 Rubik's cubes on sale (for $37 I think).  They had three in stock on
>Saturday.
>				Ronnie

(First: personal to Ronnie: THANK YOU THANK YOU THANK YOU THANK YOU THANK
YOU THANK YOU THANK YOU THANK YOU THANK YOU THANK YOU THANK YOU THANK YOU
I've been looking for a 5by for _years_.)

I just called.  First, the cubes are at Games GALLERY (boy, you guys on the
west coast sure have friendly, helpful phone operators.  Also store owners:
when I called the nearest GamesKeeper that the operator could find, the
manager there gave the number for Games Gallery!).  $27.95.  They got them
from Dr Christopher <somethingortheother> in Germany.

And, yes, they have a couple left after taking my phone order.  ;)

Nichael

BTW, Graham (the guy that I talked to at GG --is _everybody_ this friendly
out there?) said they had about a half-dozen new Rubik's toys (i.e. "new" =
post Rubik's Clock).  Including something called "Rubik's 15", which Graham
(note how we're on a first-name basis now) described as "like the old 15-
puzzle, but *NASTY*!"

Does anybody know any of these?

From hirsh@cs.rutgers.edu  Tue Apr 23 12:26:23 1991
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Sender: Haym Hirsh <hirsh@pei.rutgers.edu>
Date: Tue, 23 Apr 91 12:26:16 EDT
From: Haym Hirsh <hirsh@cs.rutgers.edu>
Reply-To: Haym Hirsh <hirsh@cs.rutgers.edu>
To: cube-lovers@life.ai.mit.edu
Subject: rubik's magic alternate coloring
Cc: Haym Hirsh <hirsh@cs.rutgers.edu>
Message-Id: <CMM-RU.1.0.672423976.hirsh@pei.rutgers.edu>

After seeing my collection of Rubik's magics in my office, a student
came by yesterday with a variation I hadn't seen before.  It is a 2x4
version, but the 8 "tiles" are colored differently.  Each of the eight
tiles has a "four-square" pattern -- the square is divided into four
regions, each colored red, blue, yellow, or green.  The center of each
is black with Rubik's signature on it.  The tiles thus look something
like the following:

	+----+----+
	|Blue|Yell|
	|   / \ ow|
	+--+   +--+
	|   \ /   |
	|Red |Gree|
	+----+----+

(with Rubik's signature in the center)

Both the front and back tiles have this four-square pattern.  However,
on one side the order of colors on the tiles are all as in the picture
above, and on the other side four have that order and the remaining
four have yellow and green switched (so that blue and yellow are on
opposite corners).

I don't know if this description gets the idea across to those who
have never seen one like this, but I'm more interested in those who
have seen it.  Is anyone familiar with this version, and if so, what
is the goal pattern to reach?  It turns out that the student worked at
Bradlees (a downscale version of Kmart, if such a thing is possible)
four years ago, and he got it from the returns bin, without any
packaging.  I've looked at it briefly, and didn't come up with an
obvious goal pattern.

About the only other info that may be helpful is that the copyright
for this variation is 1987.  The copyright for the original 2x4 is
1986, and similarly for the 2x2 I have; the 2x6 is copyright 1987.

Finally, since I am on the topic of the magic, I have heard a
number of times about yet another version of the magic that can
be folded into a cube.  Does anyone know any sources for it?
(I thought for a while that the alternate-coloring version may
be it, but it seems to have the same connectivity as the standard
2x4.)

Thanks for any help!

Haym (hirsh@cs.rutgers.edu)

From latto@lucid.com  Tue Apr 23 18:37:44 1991
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Date: Tue, 23 Apr 91 18:40:21 EDT
From: Andy Latto <latto@lucid.com>
Message-Id: <9104232240.AA02639@boston-harbor>
To: hirsh@cs.rutgers.edu
Cc: cube-lovers@life.ai.mit.edu, hirsh@cs.rutgers.edu
In-Reply-To: Haym Hirsh's message of Tue, 23 Apr 91 12:26:16 EDT <CMM-RU.1.0.672423976.hirsh@pei.rutgers.edu>
Subject: rubik's magic alternate coloring

The version your student has is the one where the object is to fold it
into a cube (I have it, with the instructions. Yes, it does have the
same structure as the original 2x4 one---you can fold that one
into a cube (with two "flaps") too. The object is to fold it
into a cube where the colors of the three faces meeting at each corner
always match.

						Andy Latto
						latto@lucid.com

From @po2.andrew.cmu.edu:dn1l+@andrew.cmu.edu  Sat May  4 12:20:57 1991
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          Sat,  4 May 91 04:58:29 -0400 (EDT)
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Date: Sat,  4 May 91 04:58:29 -0400 (EDT)
From: "Dale I. Newfield" <dn1l+@andrew.cmu.edu>
To: Cube-Lovers@life.ai.mit.edu
Subject: Re: 5-cube in a game store!!!
Cc: Cube-Lovers@life.ai.mit.edu
In-Reply-To: <9104230337.AA04955@life.ai.mit.edu>
References: <9104230337.AA04955@life.ai.mit.edu>

I called and ordered things from this store in California, and just
recieved things over the past two days.

THE 5X5X5 CUBE!!!!!!!!

(I just started the 5X5X5 about 2 hours ago, and already have all but a
few on the bottom.  I think it will be MUCH easier than the 4X4X4.  It
comes with instructs on how to open it, and I took it apart to look. 
THIS DESIGN IS IMPRESSIVE!  You can make some REALLY neat patterns with
the 5x5x5.)

5x5x5 from:
Dr. Christoph Bandelow
Haarholzer Str.  13
4620 Bochum 1
Germany

Write (Dr. Bandelow) for a free mail order catalog with many twisting
puzzles and books about these puzzles.

The new Rubik's things:

Rubik's Dice:  
	Rubik's Dice, unlike any other dice, has nothing to do with luck.  It
has spots whose color can be changed from white to red and from red to
white.  Rbik's Dice in fact, is a hollow cube with which has 7 plates
inside it.  The plates are white with red dots on them.  The plates are
loose but adhere to the inner sides of the cube.  By shaking and turning
the cube, the postition and orientation of the plates can be changed and
this in turn alters the color of the spots of the dice.
	Object:	Re-arrange the plates within the cube in such a way that the
dice has white and only white spots.  If red is shown anywhere on the
dice even through the small controll holes -- the puzzle is not
complete.  The number of possible combinations is 7! x 4^7=82,575,360. 
There is however, ony one correct solution.

Rubik's Tangle:
	Rubik's Tngle has 25 square tiles each tile has the very same pattern
of ropes, but the color of the ropes varies.
	Object:	Lay down the tiles into a 5X5 square in such a way that each
colored rop forms it's own continuous line.
24! x 4^24 = 1746 x 10^38 ( I have a feeling it should be 25!, not 24!)
2 correct solutions.

4 different tangle puzzles.
each worrks by itself (differently) and together they form a 10X10 grid
that also works

Rubik's Triamid
	Dumb, and i'm tired, so I won't explain.  I'm disappointed in this one,
don't buy it.

Rubik's 15:
	2 puzzles: one side is a magic square
		 the other side is a fifteen puzzle
 but the mechanisms that manouver pieces are SICK!
both can't be solved at once.


Sorry, I started typing in the sheet that explained them all, but i'm
falling asleep, so I just finished by explaining them a little.

Dale Newfield

dn1l@andrew.cmu.edu


From ncramer@bbn.com  Sun May  5 05:02:30 1991
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Message-Id: <9105050902.AA19791@life.ai.mit.edu>
Date:     Sat, 4 May 91 17:58:35 EDT
From: Nichael Cramer <ncramer@bbn.com>
To: "Dale I. Newfield" <dn1l+@andrew.cmu.edu>
Cc: Cube-Lovers@life.ai.mit.edu, Cube-Lovers@life.ai.mit.edu
Subject:  Re:  5-cube in a game store!!!

>Date: Sat,  4 May 91 04:58:29 -0400 (EDT)
>From: "Dale I. Newfield" <dn1l+@andrew.cmu.edu>
>To: Cube-Lovers@life.ai.mit.edu
>Subject: Re: 5-cube in a game store!!!
>
>I called and ordered things from this store in California, and just
>recieved things over the past two days.
>
>THE 5X5X5 CUBE!!!!!!!!

Synchronicity!! I was just typing in a virtually identical message when
Dale's posting came!  (D: thanks for saving me all the typing.  ;)

>(I just started the 5X5X5 about 2 hours ago, and already have all but a
>few on the bottom.  I think it will be MUCH easier than the 4X4X4.  It
>comes with instructs on how to open it, and I took it apart to look. 

Yeah, I think anyone who _understands_ how to work a 3by or a 4by (as
opposed to merely memorizing cookbook solutions) should have no problem
with it.

I'm certainly no speed whiz.  The UPS man rang the doorbell at 1pm and I
had scrambled and "solved" it by 3[*].  (Solved is in quotes because I had
the cube completly done except that two non-central edge-cubies were
flipped.  It took me another 15-20 minutes to back out and fix this.)

  [* this includes feeding lunch to my two daughters and a few minutes of
     code-debugging over the phone.]

>THIS DESIGN IS IMPRESSIVE!  You can make some REALLY neat patterns with
>the 5x5x5.)  

(I've got one of mine completely covered in checkerboard patterns).

As Dale says, the 5by seems very solid.  Probably this is because it has an
odd number of cubes and so has the fixed center cubie.  Certainly it moves
more consistently smoothly than my 4bys.  On the other hand, at least once
I've felt one of the cubies start to pop out in my hand while I was turning
it.

Another thing: the cubies are starting to get pretty small.  The whole cube
is only about 1/4 longer on each side than my 4by.

My only concern is _where_ do these cubes actually come from?  The rest of
my cubes (2X, 3X, 4X) have _all_ had the Rubik's symbol and copyright
notices on them.  These 5bys have neither.  Could these be pirated cubes?
On the other hand they seem solidly made and the colors are bright and
distinct unlike most cheepy copy-cubes that I've seen.  But it's curious
that there are no copyright notices _anywhere_ either the cube or the
enclosing box.

Oh well.  Something to keep my hands busy during those long compiles for
the next couple of weeks.

Cheers
Nichael

From mindcrf!roadrunner.mindcraft.com.mindcraft.com!ronnie@decwrl.dec.com  Fri May 10 19:17:14 1991
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From: mindcrf!ronnie@roadrunner.mindcraft.com.mindcraft.com (Ronnie Kon)
Message-Id: <9105082018.AA21608@roadrunner.mindcraft.com.mindcraft.com>
To: @mindcrf.pa.dec.com:decwrl!ai.ai.mit.edu!Cube-Lovers
Subject: Patterns on the order 5 cube

	OK, for everybody out there with the 5-cube, this is the most difficult
pattern I have come up with to implement (which is still highly ordered).


Top:	|A|A|A|A|A|
	|A|B|B|B|B|
	|A|B|A|A|A|
	|A|B|A|B|B|
	|A|B|A|B|A|

Front:	|B|C|B|C|B|	|C|A|C|A|C|	:thgiR
	|B|C|B|C|C|	|A|A|C|A|C|
	|B|C|B|B|B|	|C|C|C|A|C|
	|B|C|C|C|C|	|A|A|A|A|C|
	|B|B|B|B|B|	|C|C|C|C|C|


Where this pattern is also present on the remaining 3 sides.  (This amounts
to twirling a 4-cube, a 3-cube, a 2-cube and a 1-cube around a pair of
diagonally opposite corners in alternating directions.  It's not difficult
to do after you've done it a couple of times, but the potential for getting
confused is surprising.

				Ronnie

From mindcrf!peabody.mindcraft.com!ronnie@decwrl.dec.com  Fri May 10 19:16:47 1991
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From: mindcrf!ronnie@peabody.mindcraft.com (Ronnie Kon)
Message-Id: <9105061814.AA22955@peabody.mindcraft.com>
To: @mindcrf.pa.dec.com:decwrl!ai.ai.mit.edu!Cube-Lovers
Subject: 5by cubes

	As far as I can tell, if you can solve the order 3 and order 4 cube,
you should be able to solve the order 5 with no additional fiddling, even if
you only know cookbook solutions.

	Spoiler follows:

	I solve the off-center edges first (just like in the order 4 cube--
the transformations are identical), then the corners (exactly like all other
orders, from 2 through 4), then the center edges (exactly like the order 3
cube, just treat the two edge faces as attached and you have an order 3
cube).  All that's left are the eight centers.  Four of these can be solved
exactly as in the order 4, and if you can't generalize your cookbook solution
to solve the remaining 4 you have no business cubing.

	I suspect this is why there are (and will probably never be) cubes of
orders greater than 5.  I believe (though have not proved) that the 5 cube 
contains all the complexity that is possible.  Adding more cubies would only
increase the amount of time needed to solve.

	On the other hand, I would be willing to pay a fair amount of money for
an order 21 cube. :-)

				Ronnie

From mindcrf!peabody.mindcraft.com!ronnie@decwrl.dec.com  Fri May 10 19:16:59 1991
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From: mindcrf!ronnie@peabody.mindcraft.com (Ronnie Kon)
Message-Id: <9105072205.AA14556@peabody.mindcraft.com>
To: @mindcrf.pa.dec.com:decwrl!ai.ai.mit.edu!Cube-Lovers
Subject: Rubik's tangle

>>> From: "Dale I. Newfield" <dn1l+@andrew.cmu.edu>
>>> 
>>> Rubik's Tangle:
>>> 	Rubik's Tngle has 25 square tiles each tile has the very same pattern
>>> of ropes, but the color of the ropes varies.
>>> 	Object:	Lay down the tiles into a 5X5 square in such a way that each
>>> colored rop forms it's own continuous line.
>>> 24! x 4^24 = 1746 x 10^38 ( I have a feeling it should be 25!, not 24!)

	No, I think the 24! is correct.  Since we don't count rotations as
different, the first tile can be placed any way you want without affecting
the outcome.

				Ronnie

From ncramer@bbn.com  Fri May 10 21:33:52 1991
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Date:     Fri, 10 May 91 20:45:08 EDT
From: Nichael Cramer <ncramer@bbn.com>
To: Ronnie Kon <mindcrf!ronnie@peabody.mindcraft.com>
Cc: cube-lovers@life.ai.mit.edu
Subject:  Re:  5by cubes

>Date: Mon, 6 May 91 11:14:02 -0700
>From: Ronnie Kon <mindcrf!ronnie@peabody.mindcraft.com>
>Subject: 5by cubes
>	I suspect this is why there are (and will probably never be) cubes of
>orders greater than 5.  I believe (though have not proved) that the 5 cube 
>contains all the complexity that is possible.  Adding more cubies would only
>increase the amount of time needed to solve.

On the other hand, a 5X (or any cube of odd order) will still have the
constraints imposed by a fixed center.  As a single example, the 4X here in
my office is completely "solved" except that two opposite corners are
swapped.  That's not something that can happen on a cube of odd order (at
least I don't think so, but I would love to be proved wrong ;).

N

From latto@lucid.com  Fri May 10 22:55:59 1991
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From: Andy Latto <latto@lucid.com>
Message-Id: <9105110258.AA29787@boston-harbor>
To: mindcrf!ronnie@peabody.mindcraft.com
Cc: Cube-Lovers@life.ai.mit.edu
In-Reply-To: Ronnie Kon's message of Mon, 6 May 91 11:14:02 -0700 <9105061814.AA22955@peabody.mindcraft.com>
Subject: 5by cubes


> On the other hand, I would be willing to pay a fair amount of money for
> an order 21 cube. :-)

You can't make an order 21 cube, or any cube of order 7 or higher.
When you turn the top layer of such a cube by 45 degrees, the corner
cubie will not touch the other layers at all, so there's
no way to keep it attached, and it will fall off.

						Andy Latto
						latto@lucid.com

From @po5.andrew.cmu.edu:dn1l+@andrew.cmu.edu  Sat May 11 03:37:00 1991
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From: "Dale I. Newfield" <dn1l+@andrew.cmu.edu>
To: Cube-Lovers@life.ai.mit.edu
Subject: Re: Rubik's tangle
In-Reply-To: <9105072205.AA14556@peabody.mindcraft.com>
References: <9105072205.AA14556@peabody.mindcraft.com>

> Excerpts from internet.cube-lovers: 7-May-91 Rubik's tangle Ronnie
> Kon@peabody.mindc (568)

> >>> From: "Dale I. Newfield" <dn1l+@andrew.cmu.edu>
> >>> 
> >>> Rubik's Tangle:
> >>> 	Rubik's Tngle has 25 square tiles each tile has the very same
> pattern
> >>> of ropes, but the color of the ropes varies.
> >>> 	Object:	Lay down the tiles into a 5X5 square in such a way that each
> >>> colored rop forms it's own continuous line.
> >>> 24! x 4^24 = 1746 x 10^38 ( I have a feeling it should be 25!, not
> 24!)

> 	No, I think the 24! is correct.  Since we don't count rotations as
> different, the first tile can be placed any way you want without
> affecting
> the outcome.

> 				Ronnie


No, Because off the rotation, the 4^25 goes down to 4^24, but again, I
still think that it should be 25!, because there are that many pieces to
be arranged.

Dale

From @po5.andrew.cmu.edu:dn1l+@andrew.cmu.edu  Sat May 11 03:48:46 1991
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Date: Sat, 11 May 91 03:47:40 -0400 (EDT)
From: "Dale I. Newfield" <dn1l+@andrew.cmu.edu>
To: Cube-Lovers@life.ai.mit.edu
Subject: Re: 5by cubes
Cc: Cube-Lovers@life.ai.mit.edu
In-Reply-To: <9105110258.AA29787@boston-harbor>
References: <9105110258.AA29787@boston-harbor>

I solve the cubes in a way much different than lots that people have explained:

(Don't read if you don't want!)

I pick a "top" side and solve it.
I put the centers together(on the order 3, this was REAL easy! :-)
I put the edges together that go from the top to the bottom.
I solve the bottom 4 corners
I solve the bottom 4 middles.

depending on which cube, the 2nd and 3rd steps are switched.

I am only having one problem with the 5x5x5 cube, though:


X|O|X|X|X
X|X|X|X|X
X|X|X|X|X
X|X|X|X|X
X|X|X|O|X

looking at the bottom of my cube, the 2 pieces marked O are swaped
sometimes, so that the face is still a solid color, but the sides are
swapped.

I also got it to have the swapped pieces near each other:

X|X|X|X|X
X|X|X|X|X
X|X|X|X|X
O|X|X|X|X
X|X|X|O|X

My question is this:

I can't figure out what causes the swapping.
Is it the because in this face,

X|X|X|X|X
X|O|I|O|X
X|I|X|I|X
X|O|I|O|X
X|X|X|X|X

the I's and the O's are in the "wrong" positions, even though they are
indestingiushable?

Dale Newfield

From ncramer@bbn.com  Sun May 12 17:18:48 1991
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Date:     Sun, 12 May 91 17:12:57 EDT
From: Nichael Cramer <ncramer@bbn.com>
To: "Dale I. Newfield" <dn1l+@andrew.cmu.edu>
Cc: Cube-Lovers@life.ai.mit.edu
Subject:  Re: 5by cubes

>Date: Sat, 11 May 91 03:47:40 -0400 (EDT)
>From: "Dale I. Newfield" <dn1l+@andrew.cmu.edu>
>To: Cube-Lovers@life.ai.mit.edu
>Subject: Re: 5by cubes
>Cc: Cube-Lovers@life.ai.mit.edu
>
>(Don't read if you don't want!)

Ditto!   ;)

>I am only having one problem with the 5x5x5 cube, though:
>
>X|O|X|X|X
>X|X|X|X|X
>X|X|X|X|X
>X|X|X|X|X
>X|X|X|O|X
   ^   ^
   1   2

>looking at the bottom of my cube, the 2 pieces marked O are swaped
>sometimes, so that the face is still a solid color, but the sides are
>swapped.  [ ... ] I can't figure out what causes the swapping.

Dale, 

What is wrong is that *one* of the inner planes [marked 1 & 2 above] are a
quarter turn [i.e. 90dgs] out of phase.

1] The way I solve this is to turn one of the planes a quarter turn, [to
get, for example the following]:

>X|O|X|o|X      <--(Where "o" is the other face of the "O" above.)
>X|X|X|Y|X
>X|X|X|Y|X
>X|X|X|Y|X
>X|X|X|Y|X
   ^   ^
   1   2

2] Then, to keep things straight in my head, I then "mark" the new position
by replacing the center cubies in the turned plane to their correct
positions (being careful not to mess with anything else --particularly the
edge pieces):

>X|O|X|o|X
>X|X|X|X|X
>X|X|X|X|X
>X|X|X|X|X
>X|X|X|Y|X
   ^   ^
   1   2

There's a relatively simple operator to do this, which I leave as an
exercise for the reader.  ;)

(It's probably better to do this to all affected four faces, but you can
save this until you have solved the edges.)

3] This leaves you five inner, non-central edges to solve.  But it should
be straightforward, so long as you be careful not to mess up anything else.

Nichael

From ncramer@bbn.com  Sun May 12 18:05:24 1991
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Date:     Sun, 12 May 91 18:01:38 EDT
From: Nichael Cramer <ncramer@bbn.com>
To: dn1l+@andrew.cmu.edu
Cc: "Dale I. Newfield" <dn1l+@andrew.cmu.edu>, Cube-Lovers@life.ai.mit.edu
Subject:  ARGGHHH!! [was: 5by cubes]

>Date:     Sun, 12 May 91 17:12:57 EDT
>From: Nichael Cramer <ncramer@BBN.COM>
>To: "Dale I. Newfield" <dn1l+@andrew.cmu.edu>
>Subject:  Re: 5by cubes
>>Date: Sat, 11 May 91 03:47:40 -0400 (EDT)
>>From: "Dale I. Newfield" <dn1l+@andrew.cmu.edu>
>>To: Cube-Lovers@life.ai.mit.edu
>>I am only having one problem with the 5x5x5 cube, though:
>>X|O|X|X|X
>>X|X|X|X|X
>>X|X|X|X|X
>>X|X|X|X|X
>>X|X|X|O|X
>>looking at the bottom of my cube, the 2 pieces marked O are swaped
>>sometimes, so that the face is still a solid color, but the sides are
>>swapped.  [ ... ] I can't figure out what causes the swapping.

[I write]:
>Dale, [...]

#$%@!! 

I just realize that I answered the wrong question!  My answer was to the
question:

    "My cube is completely solved *except* that the 2 pieces marked `O' are
     flipped."

(Sorry.)  The right answer should be:

The state of the cube is not:

X|O|X|X|X                  X|A|X|C|X
X|X|X|X|X                  X|X|X|X|X
X|X|X|X|X     But rather:  X|X|X|X|X
X|X|X|X|X                  X|X|X|X|X
X|X|X|O|X                  X|X|X|B|X

Where cubie "C" just "looks" like it's in the right place.

You need an operator that rotates  A->B->C->A.  (Left as an exercise; hints
available on request.)

This will very likely leave an inconvenient number of edges flipped.  For
the answer to _this_ problem, see my last post.  ;)

Nichael-walks-with-the-red-face

From @po5.andrew.cmu.edu:dn1l+@andrew.cmu.edu  Mon May 13 00:51:55 1991
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Date: Mon, 13 May 91 00:49:34 -0400 (EDT)
From: "Dale I. Newfield" <dn1l+@andrew.cmu.edu>
To: Cube-Lovers@life.ai.mit.edu
Subject: Re: 5by cubes
Cc: Cube-Lovers@life.ai.mit.edu
In-Reply-To: <9105110258.AA29787@boston-harbor>
References: <9105110258.AA29787@boston-harbor>

Excerpts from internet.cube-lovers: 10-May-91 5by cubes Andy
Latto@lucid.com (383)

>You can't make an order 21 cube, or any cube of order 7 or higher.
>When you turn the top layer of such a cube by 45 degrees, the corner
>cubie will not touch the other layers at all, so there's
>no way to keep it attached, and it will fall off.


There is no law that says that the cubes have to be the same size.

XXXX|XXX|XX|X|XX|XXX|XXXX
XXXX|XXX|XX|X|XX|XXX|XXXX
XXXX|XXX|XX|X|XX|XXX|XXXX
XXXX|XXX|XX|X|XX|XXX|XXXX
----+---+--+-+--+---+----
XXXX|XXX|XX|X|XX|XXX|XXXX
XXXX|XXX|XX|X|XX|XXX|XXXX
XXXX|XXX|XX|X|XX|XXX|XXXX
----+---+--+-+--+---+----
XXXX|XXX|XX|X|XX|XXX|XXXX
XXXX|XXX|XX|X|XX|XXX|XXXX
----+---+--+-+--+---+----
XXXX|XXX|XX|X|XX|XXX|XXXX
----+---+--+-+--+---+----
XXXX|XXX|XX|X|XX|XXX|XXXX
XXXX|XXX|XX|X|XX|XXX|XXXX
----+---+--+-+--+---+----
XXXX|XXX|XX|X|XX|XXX|XXXX
XXXX|XXX|XX|X|XX|XXX|XXXX
XXXX|XXX|XX|X|XX|XXX|XXXX
----+---+--+-+--+---+----
XXXX|XXX|XX|X|XX|XXX|XXXX
XXXX|XXX|XX|X|XX|XXX|XXXX
XXXX|XXX|XX|X|XX|XXX|XXXX
XXXX|XXX|XX|X|XX|XXX|XXXX

JUST AS AN EXAMPLE.

From gls@think.com  Mon May 13 12:04:08 1991
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To: latto@lucid.com
Cc: mindcrf!ronnie@peabody.mindcraft.com, Cube-Lovers@life.ai.mit.edu
In-Reply-To: Andy Latto's message of Fri, 10 May 91 22:58:29 EDT <9105110258.AA29787@boston-harbor>
Subject: 5by cubes

   Date: Fri, 10 May 91 22:58:29 EDT
   From: Andy Latto <latto@lucid.com>


   > On the other hand, I would be willing to pay a fair amount of money for
   > an order 21 cube. :-)

   You can't make an order 21 cube, or any cube of order 7 or higher.
   When you turn the top layer of such a cube by 45 degrees, the corner
   cubie will not touch the other layers at all, so there's
   no way to keep it attached, and it will fall off.

Assuming the current technology, anyway.  But imagine a less passive
approach.  Suppose each cubie had a cheap microprocessor, and some
little latches.  Normally cubies hang onto their neighbors, but when
they notice you are applying torque, they let go of their neighbors
in just that one direction and hang on for dear life in the other two
directions.  The latches can also be conducting in order to convey
the necessary actuating power from a centrally placed battery.

--Wild and Crazy Guy


From hoey@aic.nrl.navy.mil  Mon May 13 14:46:09 1991
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Date: Mon, 13 May 91 14:46:06 EDT
From: hoey@aic.nrl.navy.mil
Message-Id: <9105131846.AA12000@sun13.aic.nrl.navy.mil>
To: Cube-Lovers@ai.mit.edu
Subject: Very silly ways of building very large cubes (was Re: 5by cubes)
Organization: Navy Center for Applied Research in AI

Andy Latto <latto@lucid.com> wrote:

>You can't make an order 21 cube, or any cube of order 7 or higher.
>When you turn the top layer of such a cube by 45 degrees, the corner
>cubie will not touch the other layers at all, so there's
>no way to keep it attached, and it will fall off.

Then "Dale I. Newfield" <dn1l+@andrew.cmu.edu> responded:

>There is no law that says that the cubes have to be the same size.

and showed that by making the outer layers thicker, we can increase
the size of the cube.  There is another way around Andy Latto's con-
cern, and that is that we can--at least in theory--design a physical
cube that lets pieces overhang, such as corners that touch only two
surfaces, and yet still holds the pieces so they cannot be removed.

This idea (which came up in talks with Jim Saxe about a decade ago) is
to slice up the cube with a fresnel saw.  A fresnel saw is used to cut
a piece of glass into two fresnel lenses out of pieces of glass, and
you find them in the same stores that sell plaid paint and jelly-
doughnut cookie cutters.  (In case you don't know what a fresnel lens
(pronounced freh-NEL) is, for this note it's sufficient to think of it
as a surface with small concentric circular grooves in it.  Kind of
like those old vinyl recordings people used to listen to, except that
the grooves are circular instead of spiral, and the grooves don't
wiggle back and forth.)

Now if you have two surfaces with mating grooves--each one has a ridge
that fits in each of the other's grooves--when you put them together
you can twist one with respect to the other, but you can't slide one
across the other, because the grooves are locked together.  There is
one thing you can do that we don't want: you can lift one slab away
from the other.

The solution now is to get a *very* *sharp* fresnel saw, that cuts
hooked grooves that interlock with each other.  You get surfaces with
cross sections that look somewhat like

                            hook-in surface                    
       _________       _______________________       _________       
\     /         \     /           .           \     /         \     /    
 |   |           |   |            |            |   |           |   |
 |   |  __       |   |  __        .        __  |   |       __  |   |     
 |   |    \      |   |    \       |       /    |   |      /    |   |     
  \   \___/|      \   \___/|      .      |\___/   /      |\___/   /      
   \       |       \       |      |      |       /       |       /       
    \_____/         \_____/       .       \_____/         \_____/        
        _____           _____________________           _____           
       /     \         /          |          \         /     \          
      | ___   \       | ___       .       ___ |       /   ___ |
      |/   \   \      |/   \      |      /   \|      /   /   \|
       \__  |   |      \__  |   axis    |  __/      |   |  __/
            |   |           |    of     |           |   |       
            |   |           | rotation  |           |   |
___________/     \_________/             \_________/     \_____

                           hook-out surface

except that the surfaces are closer, so the hooked grooves are engaged
with each other.  (Now we see why we need a fresnel saw, so that we
can cut the two mating surfaces in one cut, and avoid the problem of
trying to assemble two separated pieces (though we could get around
that difficulty messily with glue)).

So we may cut up a 2n+1 x 2n+1 x 2n+1 cube with a fresnel saw, to make
a large Rubik's cube.  The only really touchy point is the need to
make the ``direction'' of each cut match the direction of the other
cuts at that ``depth.''  Here, direction refers to whether the hook-in
surface faces toward the nearest parallel side or away from that side,
and ``depth'' refers to the distance from the nearest parallel side.
This ensures that when we permute cubies around the directions of the
groove hooks will not change, so the grooves will always mate.

If n is large, then pieces of one slab will overhang at each turn, so
you can see the grooves on a whole surface of a corner, or on two
surfaces of an edge piece.  But you can't pull the piece off, because
it won't move straight with respect to the rest of the cube, only in
curved trajectories.  We have to keep the fresnelling small with
 respect to the size of the cubies, and the tolerances are pretty
tight, but that's the regime we theoretical engineers are working in.
(I'd like to mention that cubes made with this method also have the
nice feature that there's a 2n-1 x 2n-1 x 2n-1 Rubik's cube on the
inside, so you can play with the theoretical invisible group while
you're at it.)

Now what about cubes of even side?  The fresnel saw cuts two surfaces
that mate to each other but not to themselves.  How can we get a
surface that mates to itself?  I think the answer is that we can't.
But this doesn't mean we are out of luck, as there are several ways of
fixing up the center cuts of these cubes.  Perhaps easiest way is to
embed a 2x2x2 cube in the center of the original solid cube, and use
it to hold the octants together.  Unfortunately, this method requires
an appeal to the existence of even-sided cubes, rather than teaching
us how to build them.

The other ways of finessing the center cut involve the thin-center-
slab approach.  You know you can simulate a 2x2x2 pocket cube with the
corners of a regular 3x3x3 Rubik's cube, and similarly you can
simulate any even cube with a larger odd cube.  Also, we can make that
center slab very thin, so it becomes part of the supporting structure
rather than a significant part of the cube.  We also remove the cor-
ners from the center slab, so it does not protrude from the cube.  We
may even make covers for the cubies slabs adjacent to the center, to
cover up the crack where the center slab lives.  We are ready to cube!

Or are we?  The thin-center-slab suffers from the partial-twist
problem.  We can see this in the simulation of the 2x2x2 by a 3x3x3.
If you try to simply ignore the center slabs, you can end up with the
corners being aligned with each other but with a center slab twisted
by 45 degrees.  This makes it impossible to turn the corners except in
the plane parallel to the oblique slab.  If we shrink the center slab
enough that it becomes unnoticeable, we will still be unable to twist
the cube except in one direction except by breaking the center slab.

The first solution to the partial-twist-problem is to select one of
the eight near-central cubies, a cubie that abuts the center slabs on
three sides.  We then glue the adjacent parts of the center slabs to
that cubie.  Then when we turn along the center slice(s), the glued
part of the thin center slab will follow the selected cubie, and will
push the rest of the thin center slab along.  This is a modification
of the solution that is taken inside Rubik's Revenge, as I described
to this group in my Invisible Revenge article of 9 August 1982.

I like this solution except for one thing.  It destroys the symmetry
of the cube, by selecting one specialized octant that the center slab
must follow.  There is one more solution, though, that keeps the cube
symmetric, which is *even* *sillier* than the thin center slab itself.
Let us now visualize the center slab.  It has the corners removed, so
it is in the shape of a disc.  The disc is cut in a grid pattern by
the cuts from perpendicular planes.  Now suppose we cut each slab in a
second grid pattern, with the grid at a 45 degree offset from the
original.  With such a center slab, the cube can be twisted if each
slab grid is in the correct position, or if some are at a 45 degree
offset from the correct position.

And how shall we prevent turns of less than a 45 degrees?  Gears!
Embed tiny gears in each fragment of the center slab, that contact
tiny toothed tracks in the adjacent slabs on both sides.  This will
force the center slab to turn at exactly half the angular rate of one
half of the cube with respect to the other.  Thus when the off-center
slabs of the cube are aligned, the center slab will be at one of the
positions that allows twisting.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From latto@lucid.com  Tue May 14 16:46:39 1991
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Site: 
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Date: Tue, 14 May 91 13:54:02 EDT
From: Andy Latto <latto@lucid.com>
Message-Id: <9105141754.AA10376@boston-harbor>
To: gls@think.com
Cc: Cube-Lovers@life.ai.mit.edu
In-Reply-To: Guy Steele's message of Mon, 13 May 91 12:03:50 EDT <9105131603.AA01148@ukko.think.com>
Subject: 5by cubes

Should you really be posting the secret proposed architecture
for the CM-6 to a publicly available mailing list?

:-) :-)

					Andy
					latto@lucid.com

From kon@bach.stanford.edu  Tue May 14 21:14:14 1991
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Date: Tue, 14 May 91 18:14:11 PDT
From: kon@bach.stanford.edu (Ronnie Kon)
Message-Id: <9105150114.AA00195@bach.Stanford.EDU>
To: mindcrf!ronnie@peabody.mindcraft.com, ncramer@bbn.com
Subject: Re:  5by cubes
Cc: cube-lovers@life.ai.mit.edu

>>	I suspect this is why there are (and will probably never be) cubes of
>>orders greater than 5.  I believe (though have not proved) that the 5 cube 
>>contains all the complexity that is possible.  Adding more cubies would only
>>increase the amount of time needed to solve.
>
>On the other hand, a 5X (or any cube of odd order) will still have the
>constraints imposed by a fixed center.  As a single example, the 4X here in
>my office is completely "solved" except that two opposite corners are
>swapped.  That's not something that can happen on a cube of odd order (at
>least I don't think so, but I would love to be proved wrong ;).

Wow!  I could have sworn I have gotten to this position before, but you are
very definitely correct.  The state with two diagonal corners swapped is in
the orbit with edge cubies exchanged.

				Ronnie

From kon@bach.stanford.edu  Tue May 14 21:31:35 1991
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Date: Tue, 14 May 91 18:31:33 PDT
From: kon@bach.stanford.edu (Ronnie Kon)
Message-Id: <9105150131.AA00208@bach.Stanford.EDU>
To: dn1l+@andrew.cmu.edu, ncramer@bbn.com
Subject: Re:  ARGGHHH!! [was: 5by cubes]
Cc: Cube-Lovers@life.ai.mit.edu

>
>(Sorry.)  The right answer should be:
>
>The state of the cube is not:
>
>X|O|X|X|X                  X|A|X|C|X
>X|X|X|X|X                  X|X|X|X|X
>X|X|X|X|X     But rather:  X|X|X|X|X
>X|X|X|X|X                  X|X|X|X|X
>X|X|X|O|X                  X|X|X|B|X
>
>Where cubie "C" just "looks" like it's in the right place.
>
>You need an operator that rotates  A->B->C->A.  (Left as an exercise; hints
>available on request.)
>
>This will very likely leave an inconvenient number of edges flipped.  For
>the answer to _this_ problem, see my last post.  ;)

I think you must be wrong here (but would love to be proved wrong--I'm no
mathematician so group theory is very much beyond me).

Proof #1:
	We hold the cube with the red face on top, and the yellow face in
front (colors obviously don't matter, but I find it easier to discuss using
them).  We will assign a parity to the edge cubies, being defined by holding
the cube such that the red face of the cubie is on top and the yellow in
front.  If the cubie is on the left as we look at it in this position it is
parity 0, on the right it is parity 1.

	There are only two operations available which affect the cubie we
are interested in: rotating the front face 90deg; and rotating the slice the
cubie is in 90deg.  It is easy to see that neither of these moves alters the
parity (assume the cubie's frame of reference, and think of rotating the
rest of the cube around it--it is clear that it will not end up on the other
side).

	Therefore the move C->A in the above is impossible.

Proof #2:
	Take apart the order 4 cube (my falls apart depressingly easilly)
and try to reassemble it with the two edges exchanged.  It will not fit, as
they are mirror images of each other.



	Note that you get an apparant parity reversal by flipping the
cubies, but this does not actually move anything.  In other words, no amount
of flipping and moving will allow you to end up moving A->B->C->A.  That's
why I solve edges first.

				Ronnie

From ncramer@bbn.com  Wed May 15 22:52:53 1991
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Date:     Wed, 15 May 91 22:09:23 EDT
From: Nichael Cramer <ncramer@bbn.com>
To: Ronnie Kon <kon@bach.stanford.edu>
Cc: dn1l+@andrew.cmu.edu, ncramer@bbn.com, Cube-Lovers@life.ai.mit.edu
Subject:  Re:  ARGGHHH!! [was: 5by cubes]

Ronnie Kon <kon@bach.stanford.edu> writes:
>I write:
>>The state of the cube is not:
>>
>>X|O|X|X|X                  X|A|X|C|X
>>X|X|X|X|X                  X|X|X|X|X
>>X|X|X|X|X     But rather:  X|X|X|X|X
>>X|X|X|X|X                  X|X|X|X|X
>>X|X|X|O|X                  X|X|X|B|X
>>
>>Where cubie "C" just "looks" like it's in the right place.
>>
>>You need an operator that rotates  A->B->C->A.  [...]
>>
>>This will very likely leave an inconvenient number of edges flipped.  For
>>the answer to _this_ problem, see my last post.  ;)
>
>I think you must be wrong here (but would love to be proved wrong--I'm no
>mathematician so group theory is very much beyond me).
>
> [Proofs deleted.]

Hi.

I think we're in complete agreement, at least up to here.  (I particularly
enjoyed your "proof by hardware ;).

I didn't mean to imply that the A->B->C->A operator preserved flipped-ness
of the Non-Central-Edge[NCE] Cubies.  Moreover, I was being imprecise
where I said "a NCE cubie is simply flipped"; rather "the cubie *appears*
as if it were in the right place (i.e. judged by its colors) and flipped".
As you point out, *really* means that it is in the slot of its "twin".

To recap more succinctly, what I was proposing was a rather pedestrian,
two-step solution to the original problem.  Starting from the initial
state in FIG1 (where the cube is completely "solved" except that the
cubies marked "O" are swapped.  Also they are swapped in such a way that
the visible face is all a single color).

FIG1: X|O|X|X|X                      FIG2: X|Q|X|Q|X
      X|X|X|X|X                  	   X|X|X|X|X
      X|X|X|X|X     A->B->C->A gives:      X|X|X|X|X
      X|X|X|X|X                  	   X|X|X|X|X
      X|X|X|O|X                  	   X|X|X|X|X

STEP1] If we then apply the A->B->C->A operator, we end up with the state
in FIG2, where the cube is completely "solved" except that the cubies
marked "Q" "appear" to be "simply" flipped.

STEP2] We can then solve this problem, which (imo) is easier.  For example
see the method that I described in an earlier post; this involves turning
the non-central plane (containing the flipped cubie) through a quarter turn.

Of course, now that I say it, it seems that the correct course would be to
*start* with the quarter turn of the non-central plane.  This would leave
five NCE cubies out of place, but the cube would be in the right orbit.
From there the solution should be straightforward (e.g. two intersecting
3-cycles).

Finally, it seems clear that this entire problem --and all the subsequent
discussion-- maps directly onto a virtually identical problem on the 4by
cube (i.e. simply be removing the center planes).

>Note that you get an apparant parity reversal by flipping the cubies, but
>this does not actually move anything.  In other words, no amount of
>flipping and moving will allow you to end up moving A->B->C->A.  That's
>why I solve edges first.

Again, perhaps I'm missing the point, but if you don't care about how
the flipping comes out, the A->B->C->A  3-cycle is certainly doable:

For example:

[WARNING: EVEN MORE BORING STUFF AHEAD!!   ;]

(I have no idea how to show this notationally, so I'll try pictorially.)

                         |
1]                    2] V                  3]
 X|A|X|C|X             X|Y|X|C|X           ->Z|Z|Z|Z|Z
 X|X|X|X|X             X|Y|X|X|X             X|Y|X|X|X
 X|X|X|X|X             X|Y|X|X|X             X|Y|X|X|X
 X|X|X|X|X             X|Y|X|X|X             X|Y|X|X|X
 X|X|X|B|X             X|A|X|B|X             X|A|X|B|X


4]                   5]                     6]
 Z|A|Z|Z|Z             X|B|X|X|X             X|B|X|X|X
 X|X|X|X|X             X|X|X|X|X             X|X|X|X|X
 X|X|X|X|X             X|X|X|X|X             X|X|X|X|X
 X|X|X|X|X             X|X|X|X|X             X|X|X|X|X
 X|X|X|B|X             Z|Z|Z|A|Z           ->?|?|?|?|?
   ^
   |                  [Rotate Face           
                       one-half turn]

      >---
7]        \           8]                    9]
 X|B|X|X|X \           X|B|X|X|X             Z|C|Z|Z|Z
 X|X|X|X|X |           X|X|X|X|X             X|X|X|X|X
 X|X|X|X|X |           X|X|X|X|X             X|X|X|X|X
 X|X|X|X|X V           X|X|X|X|X             X|X|X|X|X
 ?|?|?|?|?             Z|Z|Z|C|Z<-           X|X|X|B|X 
      
 [Rotate next-                              [Rotate Face
  to-bottom                                  one-half turn]
  plane 1/4 Turn]

                            <----
10]                   11]        \          12]
  Z|C|Z|Z|Z             Z|C|Z|Z|Z \           Z|C|Z|Z|Z
  X|X|X|X|X             X|X|X|X|X |           X|X|X|X|X
  X|X|X|X|X             X|X|X|X|X |           X|X|X|X|X
  X|X|X|X|X             X|X|X|X|X ^           X|X|X|X|X
->?|?|?|?|?             ?|?|?|?|?             X|X|X|A|X<-

                        [Rotate next-
                         to-bottom
                         plane 1/4 turn]

    |
13] V                 14]                   15]
  Z|Z|Z|Z|Z             X|Y|X|B|X<-           X|C|X|B|X
  X|Y|X|X|X             X|Y|X|X|X             X|X|X|X|X
  X|Y|X|X|X             X|Y|X|X|X             X|X|X|X|X
  X|Y|X|X|X             X|Y|X|X|X             X|X|X|X|X
  X|C|X|A|X             X|C|X|A|X             X|X|X|A|X
                                                ^
                                                |


                         Cub.E.D


From pbeck@pica.army.mil  Tue May 21 11:10:45 1991
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Date:     Tue, 21 May 91 11:03:02 EDT
From: Peter Beck (LCWSL) <pbeck@pica.army.mil>
To: cube-lovers@life.ai.mit.edu
Subject:  bilinski rhombicdodecahedron
Message-Id:  <9105211103.aa16846@FSAC1.PICA.ARMY.MIL>




I would like help on finding info on the "BILINSKI" rhombic
dodecahedron.  

BACKGROUND:  The common rhombic dodecahedron is the  KEPLER and its
diagonals are in the ratio of 1:sqr rt of 2.   The Bilinski has its
diagonals in the ratio of 1:tau  (ie, the golden section ~ 1.618). 
 The only reference I have been able to find so far is on page 31 of
Coxeter's "Regular Polytopes".


AREAS OF INTEREST
..  Is there a proof of why there are only 2 rhombic dodecahedrons?
..  are there any interesting features of how the Bilinski fills
space?  Any interesting relationships with other polygons, eg,
triacontahedron?
..  Has anybody studied the dissections of the bilinski?  Is there any
significance that it takes both obtuse and acute rhomboids to
construct a bilinski while a kepler only requires an obtuse?
..  Is there a crystal or some other real world object that
corresponds to the bilinski?
..  Any ideas on fixturing/jigging to make bilinski's from wood?


Thanks for any help.  
Pete Beck   <pbeck@pica.army.mil>





From gdparker@nike.calpoly.edu  Wed May 22 04:25:46 1991
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Date: Wed, 22 May 91 01:25:48 -0700
From: gdparker@nike.calpoly.edu (Gene Dillon Parker)
Message-Id: <9105220825.AA837751@nike.calpoly.edu>
To: cube-lovers@life.ai.mit.edu
Subject: mailing list


  Hi there,
        Im an Aero/CSC major at Cal Poly and would like to be added to your dail
y mailing list.
 
                login:   gdparker
                where:  polyslo.calpoly.edu
 
                        Please include me in the list or E-mail me the info need
ed to do so. Thanks!
 
 
                                        Gene Parker
                                        gdparker
 
Cc:


From pbeck@pica.army.mil  Wed May 22 10:23:37 1991
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From: Peter Beck (LCWSL) <pbeck@pica.army.mil>
To: cube-lovers@life.ai.mit.edu
Subject:  slide
Message-Id:  <9105221016.aa14580@FSAC1.PICA.ARMY.MIL>



ANNONCEMENT OF "SLIDE" - A Sliding Block Puzzle Simulation Program

DESCRIPTION:  "SLIDE"  is a Sliding Block Puzzle Simulation Program
based on the book "Sliding Block Puzzles", by Ed Hordern, 1986 Oxford
University Press.  It comes on one 360K floppy with an instruction
booklet.  The booklet tells you how to install the program, how to use
it (program does have help files) and how to add your own puzzles.  It
is in color and REQUIRES a mouse to move the pieces.

It has all of hordern's puzzles including the background notes for
each, eg, name, producer.  The program gives you the minimum number of
moves, the object of the puzzle and a randomized version to test your
skill.

The version I have was obtained at the 11th International Puzzle Party
3/30/91 and is mostly bug free.  The author is in the process of
updating it and will make updates available at cost (for now anyhow).

I recommend  "SLIDE" for everyone with a PC who enjoyed Hordern's
book.

PRICE:    50 dutch Guilders
SOURCE:  H.J.M. van Grol (Rik) (the author),  
               van Hogendorpstraat 1a,  
               2515 NR DEN HAAG, 
               The Netherlands
COMPUTER REQUIREMENTS:  MS-DOS machine with one 360K floppy minimum;
can use 1.2 MB and harddisk if available.  REQUIRES a mouse.  Best
with EGA, okay with VGA, needs user defined configuration for Herc or
CGA.



From phygillen@cs8700.ucg.ie  Mon Jun 24 12:19:58 1991
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From: Patsy Gillen <PHYGILLEN@cs8700.ucg.ie>
To: cube-lovers@life.ai.mit.edu
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SUBCRIBE Rubic's Patrick Gillen

From pbeck@pica.army.mil  Mon Jul 29 18:16:19 1991
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Date:     Mon, 29 Jul 91 14:02:26 EDT
From: Peter Beck (BATDD) <pbeck@pica.army.mil>
To: cube-lovers@life.ai.mit.edu
Subject:  chess variants
Message-Id:  <9107291402.aa16288@FSAC1.PICA.ARMY.MIL>



ANNOUNCEMENT OF "World Game Review special issue #10" - CHESS
variations:  rules & sample games, reviews, index & bibliography

DESCRIPTION:  This a a special edition devoted only to chess.  It is
100   81/2 x 11 pages.   The breadth is indicated by the front cover
illustration of the 6 most popular chess variant boards (8x8, 10x10,
4x16,  most common 4 handed board,  xiang qi,  3 colored hexagonal)
and the rear cover illustration of 7 other boards (tesche's 3 handed,
petroff's 4 handed, de vasa's tricolor, rutland's, decimal oriental
chess, double rettah, petty).

TABLE OF CONTENTS BY HEADING & PAGE NUMBER:
1..  COLOPHON
2..  TABLE OF CONTENTS
3..  EDITORIAL & ACKNOWLEDGEMENTS
4..  GENERAL OBSERVATIONS
5..  APPEAL FOR INFORMATION , DAVID PRITCHARD IS WRITING A BOOK -
ENCYCLOPEDIA OF CHESS VARIANTS & TERMS
7..  NOTATION
8..  GENERAL RULES, BEST VARIANTS
9..  CV ORGANIZATIONS
10..  GAMES NEWS
12..  BOOK & MAG REVIEWS - SHOGI WORLD, CHINESE CHESS, CHINESISCHE,
SCHACH/KOREANISCHES SCHACH, CHINESE CHESS FOR BEGINNERS
14..  GAME REVIEWS - 4 WAY CHESS, FORAY, BATTLE CHESS II
15..  CV TIMELINE
---  A PANORAMA OF CHESS VARIANTS
16..  MODIFICATIONS TO FORCES
30..  MODIFICATIONS TO BOARD
40..  MODIFICATIONS TO MOVEMENT
52..  MODIFICATIONS TO RULES OF CAPTURE
63..  OTHER MODS
69..  SAMPLE GAMES
72..  COMPUTERES AND ...
73..  ADDITIONAL PIECES
74..  ADDITIONAL RULES
76..  INVENTORS
78..  BIBLIOGRAPHY  
82..  ADDRESSES
84..  INDEX OF VARIATIONS


PRICE:    US$10
SOURCE:  WGR
              C/O  MICHAEL KELLER
              3367-I NORTH CHATAM ROAD
              ELLICOTT CITY, MD, 21042


PLEASE DON'T ASK ME ANY QUESTIONS ABOUT CHESS, I AM NOT A PLAYER AND
HAVE NOT READ THE ISSUE..  IF YOU WANT ME TO LOOK SOMETHING UP BE VERY
SPECIFIC AND I WILL -  BEST YET IS TO ORDER YOUR OWN COPY.





From pbeck@pica.army.mil  Tue Jul 30 19:59:44 1991
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Date:     Tue, 30 Jul 91 15:50:01 EDT
From: Peter Beck (BATDD) <pbeck@pica.army.mil>
To: cube-lovers@life.ai.mit.edu
Cc: urban@rand.org
Subject:  world game review
Message-Id:  <9107301550.aa00806@FSAC1.PICA.ARMY.MIL>


PUZZLING NEWSLETTERS  --  Oct 90, revised 7/91
..........................................................
"Cubism For Fun" 
The newsletter of the "Dutch Cubists Club";  in english starting with
issue #14.  Back issues are available.  The club has over 100 active
members, notable new addition Martin Gardner.

Membership for 1991 is  20 Belgian francs  (US$10).   A photocopied
set of the newsletters, issues 1-13, written in DUTCH (in the future
selected back articles will be available in english) is also available
for  US$7.    To order either of these send an 'INTERNATIONAL"  POSTAL
MONEY ORDER (cost $3 at post office), no personal checks,  to:  Lucien
Matthijsse,  Loenpad 12,  3402 EP IJSSELSTEIN,  The Netherlands.

..........................................................
WORLD GAME REVIEW
Michael Keller publishes a newsletter that explores the mathematical
aspects of games & puzzles.  4 issues for US$11, published
erratically.  Back issues are available; ISSUE #10 CHESS ($10), ISSUE
ON POLYOMINOES, some magic polyhedra.

MICHAEL KELLER, 3367-1, NORTH CHATAM ROAD, ELLICOTT CITY, MD 21042,
USA

..........................................................
 'PUZZLETOPIA"
NOB YOSHIGAHARA mailed out a fall 90 issue (after 3 yrs) of his
newsletter 'PUZZLETOPIA".  With it came a 1991 promotional calendar
from PUZZLE CITY (a subsidary of Toyo Glass) a puzzle city catalog and
a catalog from PUZZLAND HIKIMI PUZZLE COLLECTION.  If you want the
whole package write Nob (its free outside of Japan).

NOB YOSHIGAHARA,  4-10-1-408 IIDABASHI,  TOKYO 102 JAPAN.

..........................................................

ARM Bulletin (ACADEMY of RECREATIONAL MATHEMATICS), JAPAN
This is a monthly 40-80 page newsletter of the Japanese puzzle
hobbiests club.  Dues Y8,000.

PUZZLE KONWAKAI C/O S. TAKAGI, 1-2-4 MATSUBARA, SE TAGAYAKU, TOKYO 156
JAPAN

..........................................................



From pbeck@pica.army.mil  Thu Aug  1 08:11:10 1991
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Date:     Thu, 1 Aug 91 7:55:41 EDT
From: Peter Beck (BATDD) <pbeck@pica.army.mil>
To: cube-lovers@life.ai.mit.edu, sonicdruid@sctnve.sct.peachnet.edu
Cc: pbeck@pica.army.mil
Subject:  last cube
Message-Id:  <9108010755.aa28998@FSAC1.PICA.ARMY.MIL>


The question was "where can I obatin the last cube made"

I am going to assume that last cube means "MAGIC POLYHEDRA", of
which there a three families:  the cube with 3 axis of rotation, the
tetrhedron group with 4 axis of rotation, the dodecahedron group
with 5 axis of rotation.

The last original addition I have seen was a spherical SKEWB, april 1991.
Jean Claude Constantine (germany) is making shape/surface variants by
hand based on all three mecahanisms, eg, take a 3x3x3 and join together
(make in operable) one 2x2x2 corner - you can now change those pieces shape
anyway you want - by the way this type of restriction on the moves available
to solve the cube are very interesting.  Also, in the spring of 1991
Christoph Bandelow has reintroduced the truncated octahedron and had the
Hong Kong factory complete from previously manufactured parts some 5x5x5's.

In my estimation there will NEVER be a last cube.

Solutions for all magic polyhedra, except the spherical skewb have been 
published.  First source is CFF ,cubism for fun newsletter of 
dutch cubists club,  some where also published in world game review.  
Puzzles popular when ideal was in business where also covered by many
popular publications which are now hard to get.  CFF has a library
and somebody has a bibliography of solution algorithms, including
Thistletwaites ??.

This is a general answer if there are specific questions please ask them!!

PS I sell much of this stuff and your US mail address will get you a listing
of what is available from me.

PPS I also collect this stuff and would like to trade and/or buy, any
quantity, ie, onesies to 100s.  Not only puzzles but all things a
associated with the cube - books, patents, solutions, accessories,
promotional items, replacement stickers, what have you.  a good overview
of this kind of stuff is in jerry slocum's book puzzles old and new.


From Hoffman.El_Segundo@xerox.com  Thu Aug  1 14:40:41 1991
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Date: 	Thu, 1 Aug 1991 08:55:42 PDT
From: Hoffman.El_Segundo@xerox.com
Subject: New from Rubik
To: cube-lovers@life.ai.mit.edu
Cc: Hoffman.El_Segundo@xerox.com
Message-Id: <" 1-Aug-91  8:55:42 PDT".*.Hoffman.El_Segundo@Xerox.com>


This is taken (without permission) from the 31 July 1991 `Los Angeles Times.'
It reads as though it's directly from a press release.  I haven't seen any of
these, nor have I called the listed phone number.

  -- Rodney Hoffman

------------------------------------------------------

       RUBIK RETURNS WITH MENTAL FITNESS GAMES

Remember Rubik's Cube, invented in 1977 by Hungarian professor of architecture
Erno Rubik?  Prof. Rubik is back with four "mental fitness" puzzles and a
redesigned version of the cube.

"One of the great misunderstandings surrounding Rubik's Cube was that I was
somehow trying to drive people crazy," Rubik says.  "In fact, the objective of
these puzzles is to help bring about a more alert and active mental condition."

Even if you can't solve the puzzles, according to Rubik, "the few moments
you've spent fiddling around with them has helped greatly in exercising your
mind and reducing everyday tensions."

Rubik's Tangle (suggested retail, $5.99), his first two-dimensional puzzle,
requires players to arrange 25 tiles of rope to create four continuous lines.

Rubik's XV ($6.99) is two puzzles in one.  The object of the first is to
arrange Roman numerals I through XV in order by sliding levers on the puzzle's
side.  In Part 2, players must create a square, lining up numbers so each
column, row and diagonal totals 15.

Rubik's Dice ($8.99) offers 82,575,360 possible combinations, with only one
correct answer.  The puzzle is a hollow cube with seven plates inside.  White
plates, which include red dots, are loose and can adhere to the inner sides of
the cube.  By shaking and turning the dice, players solve the puzzle when no
red appears through the holes.

Rubik's Triamid ($8.99) may be tougher than the original cube.  Players have to
construct a large pyramid, with each color on its own side, using 10 smaller
pyramids.  Rubik says there are "hundreds of blind alleys programmed into the
design."

All the puzzles are available nationally at game, toy and specialty stores.  In
Los Angeles, you can find them at Thrifty Drugs; in Orange County, at Toy City
and PlayCo stores.  Or you can order them by calling (800) 236-7123.

From mindcrf!peabody.mindcraft.com!ronnie@decwrl.dec.com  Thu Aug  1 21:38:30 1991
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From: mindcrf!ronnie@peabody.mindcraft.com (Ronnie Kon)
Message-Id: <9108011923.AA11139@peabody.mindcraft.com>
To: @mindcrf.pa.dec.com:decwrl!ai.ai.mit.edu!Cube-Lovers
Subject: Re: New from Rubik

	Well I have the Rubik's XV and Rubik's pyramid.  The XV is a pretty
good puzzle (ie., I haven't solved it yet after trying for an hour or so).
The pyramid is essentially a Pyraminx.  The only complication beyond the
Pyraminx that the Pyramid offers is that the vertex tetrahedrons can be
rotated such that a useless color shows and a necessary color is hidden.

	The solution becomes trivial once you have solved the Pyraminx.  This
seems like a good place to save your money.

				Ronnie


From @po5.andrew.cmu.edu:dn1l+@andrew.cmu.edu  Thu Aug 22 20:09:31 1991
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Date: Thu, 22 Aug 1991 20:07:45 -0400 (EDT)
From: "Dale I. Newfield" <dn1l+@andrew.cmu.edu>
To: Cube-Lovers@life.ai.mit.edu
Subject: New "CUBE"

I found a fun new cube, sorta.

It is called Square 1.

it rotates in WIERD ways.

it is a challenge to return to the state of being a cube, much less to
solve it.

My friend calls it "unfriendly."

the way it is set up, it is a cube, with a center band that has one
split.  the two faces on either side of it that are split into the
normal three on a side, but the pieces meet at the center, i.e.: the
side ones are wedges, and the corners are almost-squares with the point
not on the outside being the center.

it is fun.

I was told it will be out at christmas, but I bought it in a store
called Games Unlimited in Squirrel Hill, a neighborhood of Pittsburgh. 
Buy one from someone, and fry your brain.

Enjoy.

Dale

From dik@cwi.nl  Thu Aug 22 20:44:41 1991
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From: dik@cwi.nl
Message-Id: <9108230037.AA00481@paring.cwi.nl>
To: Cube-Lovers@life.ai.mit.edu, dn1l+@andrew.cmu.edu
Subject: Re:  New "CUBE"

 > I found a fun new cube, sorta.
 > It is called Square 1.
 > it rotates in WIERD ways.
Yes, it is also on sale in Europe.
 > 
 > it is a challenge to return to the state of being a cube, much less to
 > solve it.
True.  To solve it when it is a cube, knowledge of the magic domino is
sufficient.  But it might even be that restoring it to cube form is not
much more dificult than the magic domino.  I do not know yet.  When my
cube got in disorder, by some magical moves it was restored to cube form;
not by me, but by my 8 year old daughter, I still do not know how.
 > 
 > My friend calls it "unfriendly."
That is uncalled for.
 > 
 > the way it is set up, it is a cube, with a center band that has one
 > split.  the two faces on either side of it that are split into the
 > normal three on a side, but the pieces meet at the center, i.e.: the
 > side ones are wedges, and the corners are almost-squares with the point
 > not on the outside being the center.
Something is missing from this description.  I think it can best be
explained based on the MagiBall that came some years ago from Austria.
The MagiBall consists of two halves that can be rotated with respect to each
other.  But only rotates of 180 degrees make sense.  Further it has four bands
of moving pieces.  Movement of the bands is perpendicular to the rotations
of the halves.  Each band has 8 pieces.  When the halves are rotated with
respect to each other, the "left" half of an upper band is connected to the
"right" half of a lower band.  This creates two puzzles, each with 16 pieces.
Now consider that ball; remove the inner two bands (leaving only the uppermost
and lowermost band).  Next go through the bands clockwards when you look from
above.  Alternatingly glue two pieces in a band together and skip a piece.
This will leave you in each band with four double size pieces.  Ignoring
the middle layer of the new cube, this is the new cube (but the middle layer
is easily dealt with).  Now what was missing from the description is that the
centre angle of a cornerpiece is exactly twice the center angle of a wedge.

I have some algorithms to do cycles on three corner pieces and also for
three wedge pieces.  The latter are fairly long however.  And they all only
work if the puzzle is already in cube form.
 > 
 > it is fun.
 > 
 > I was told it will be out at christmas, but I bought it in a store
 > called Games Unlimited in Squirrel Hill, a neighborhood of Pittsburgh. 
 > Buy one from someone, and fry your brain.
 > 
 > Enjoy.
 > 
 > Dale
 > 

From meister@gaak.lcs.mit.edu  Sat Sep  7 00:18:53 1991
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Date: Sat, 7 Sep 91 00:18:10 EDT
From: meister@gaak.lcs.mit.edu (phil servita)
Message-Id: <9109070418.AA13124@gaak.LCS.MIT.EDU>
To: cube-lovers@ai.mit.edu
Subject: waiting for the bounces...



From meister@gaak.lcs.mit.edu  Mon Sep  9 00:48:52 1991
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From: meister@gaak.lcs.mit.edu (phil servita)
Message-Id: <9109090448.AA06357@gaak.LCS.MIT.EDU>
To: cube-lovers@ai.mit.edu
Subject: Square One


"Square One" is also available at Games People Play in Cambridge, Mass. 
I just purchased one a few days ago. It turned out to be more interesting 
than i had expected. The puzzle is comprised (essentially) of two halves 
of 8 pieces each, 4 corners, and 4 'edges'. All pieces radiate outward from 
the center. The 'edge' pieces expand outward at a 30 degree angle, while 
the corners expand outward at a 60 degree angle. The center layer is composed
of just 2 pieces. If you removed the top slice, the center slices would trace
out quadrilaterals ABIK and BDJL (or ACIJ and BDKL, etc) below.

The center slice serves no major function save as a "gate"
which allows or disallows rotation about one of the skewed axes formed by 
the two slices. To rotate about some axis you must line up the "gate" 
with the axis. While it is possible to swap the center pieces with respect 
to the corners frame of reference, doing so is really of no importance 
to the "meat" of the puzzle. 

        Top View:  (in lousy ascii resolution)
          
         A             B             C             D
          -------------*-------------*--------------
          |             *           *              |
          |              *         *               |
          |               *       *                |
        E *                *     *                 * F
          |     *           *   *            *     |
          |            *     * *     *             |
          |                   * O                  |
          |            *     * *     *             |
          |     *           *   *            *     |
        G *                *     *                 * H
          |               *       *                |
          |              *         *               |
          |             *           *              |
          -------------*-------------*--------------
          I            J             K             L

     Side View: 

          _____________________
          |      |     |      |
          |______|_____|______|  top
          |      |            |  
          |______|____________|  center "gate" lined up for left axis 
          |      |     |      |
          |______|_____|______|  bottom

All meaningful moves either rotate the top or bottom face, line up the 
center gate with some skew axis, or twist about a skew axis 180 degrees. 
The neat thing about the puzzle is that moves do not necessarily preserve 
the cube shape of the puzzle, or even the number of corner/edge pieces 
that are on each side. It is possible, for instance, to have only 3 corners
on the top face, and 6 edges, while 5 corners and 2 edges are on the bottom 
face. 

A notation for this beast is somewhat cumbersome. If anyone wants, I will try 
to describe what i am using.

After playing with it for a few hours, i *thought* i had mapped out a complete
solution algorithm; however i was later playing with the wierd shapes you could
put the puzzle into, and eventually started putting it back to "Square One", 
and was very surprised to end up in a position with just 2 of the edges 
swapped! Since this puzzle acts much like the Skewb, each move essentially
cutting the puzzle in half (ignoring the gate layer), i did not expect this 
to be possible. It took a bit of thought before the light hit, and i went 
on to construct a crude "Parity Swap" transform. Currently the best one i have
found cycles around  UF -> UL -> UB -> UR, and takes 28 'moves'. (a 'move' 
being defined as any motion of some slice, regardless of degrees turned) 
Anyone else found anything better? 

Square One is one of the better variations on the cube theme i have seen 
in a long time. Find one and have fun.

                                                -phil






From ccw@eql.caltech.edu  Fri Nov  1 19:10:26 1991
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Date:     Fri, 1 Nov 91 15:49:59 PST
From: ccw@eql.caltech.edu (Chris Worrell)
Message-Id: <911101154959.23403c64@EQL.Caltech.Edu>
Subject:  What is the smallest cube which has all possible types of pieces?
To: cube-lovers@ai.mit.edu

Some cube discussion has started in rec.games.misc.  I am trying to
re-direct it to rec.puzzles, or hopefully cube-lovers.

This is a copy of what I posted to rec.games.misc and rec.puzzles.

Message follows:
------------------------------------
Newsgroups: rec.games.misc, rec.puzzles
Subject: Re: Rubik's Wonderful Madness
Followups-to: rec.puzzles

johnf@apollo.hp.com (John Francis) writes...
>
>The 5x5x5 cube is the largest "interesting" cube from a solvers viewpoint.
>The corner cubelets of cubes of all sizes can be solved the same way.
>Similarly with the edge-centre cubelets of odd-sized cubes, etc., etc.
>The 5x5x5 does not actually require any new solving techniques (if you can
>solve a 3x3x3 and a 4x4x4 you can solve a 5x5x5), but it is the smallest
>cube that contains cubelets of all possible types.

Actually the 7x7x7 has a type of piece that the 5x5x5 does not have.
This is a "face" piece which is not on a main diagonal of the face, nor
on the main horizontal or vertical.  These pieces come in 2 flavors,
right-handed and left-handed (but I am not sure which should be called which).

----------------------
|A |B |C |D |  |  |  |      The 2x2x2 has types A
----------------------      The 3x3x3 has types A, D, K
|  |E |F |G |  |  |  |      The 4x4x4 has types A, B(C), E(I)
----------------------      The 5x5x5 has types A, B(C), D, E(I), G(J), K
|  |H |I |J |  |  |  |      The 6x6x6 has types A, B, C, E, F, H, I
----------------------      The 7x7x7 has types A-K
|  |  |  |K |  |  |  |
----------------------      B and C are distinct at size 6+, but the same
|  |  |  |  |  |  |  |          sorts of procedures work on both.
----------------------      Similarly for E & I.
|  |  |  |  |  |  |  |      Similarly for G & J.  For odd sizes 7+.
----------------------      Similarly for H & F. These are related
|  |  |  |  |  |  |  |          by mirror imaging.  Not by slice renumbering,
----------------------          like the above.

Types F & H only start appearing at size 6.  There are 24 of each of these
pieces, and a type F can not be mixed with a type H.


Discussions about the Rubik's cube and related puzzles should probably be
directed to rec.puzzles rather than rec.games.misc.

There is also a mailing list for theses topics.
cube-lovers@ai.mit.edu
Send mail to cube-lovers-request@ai.mit.edu to be added.
Some sites may also see this as a newsgroup mlist.cube-lovers. But do not
post to this group.

From dik@cwi.nl  Fri Nov  1 20:10:55 1991
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From: dik@cwi.nl
Message-Id: <9111020110.AA30522@paring.cwi.nl>
To: ccw@eql.caltech.edu, cube-lovers@ai.mit.edu
Subject: Re:  What is the smallest cube which has all possible types of pieces?

 > Some cube discussion has started in rec.games.misc.  I am trying to
 > re-direct it to rec.puzzles, or hopefully cube-lovers.
I follow up in cube-lovers.
 > 
 > Actually the 7x7x7 has a type of piece that the 5x5x5 does not have.
Actually the 7x7x7 is also the first cube that can not be made.
 >
 > (Showing a pattern like: ABCD
 >			     EFG
 >			     HIJ
 >			       K.
 > Noting that B&C, E&I, G&J, H&F are similar and can be solved by the same
 > set of procedures.)
But, all of E to J can be solved with the same set of procedures %.  So
although there appear to be 7 essentially different cubelets, actually
there are only 5.  And the 5x5x5 cube has them all.  Solving these involves
rotating the three slices they are in originally.  So if we rename cubelet
types we get:
1:  A
2:  B&C
3:  D
4:  E-J
5:  K
and we find for the different cubes:
2x2x2:	Type 1
3x3x3:	Type 1, 3 and 5
4x4x4:	Type 1, 2 and 4
5x5x5:	All types.
--
%  All of E to J occur 4 times on each face, so for each type there are 24
   cubelets.  B and C also occur 24 times, procedures working for these
   two also work for E to J (although I use different procedures both on
   4x4x4 and 5x5x5); but there are procedures for E to J that do not work
   for B to C (these two have more constraints on position).  This would be
   different if the cube was patterned such that all cubelets of type B, C
   and E to J had a unique position, in that case we would have only four
   essentially different types.

From ronnie@cisco.com  Fri Nov  1 21:24:02 1991
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Date: Fri, 1 Nov 91 17:36:38 -0800
From: Ronnie Kon <ronnie@cisco.com>
Message-Id: <9111020136.AA16266@lager.cisco.com>
To: ccw@eql.caltech.edu, cube-lovers@ai.mit.edu
Subject: Re:  What is the smallest cube which has all possible types of pieces?


I made essentially the same assertion some months back (about the 5x5x5
being the last "interesting" cube), and someone pointed out that there
is a position in even order cubes which has no parallel in odd order cubes--
that where all the corner cubies are correct, but all the edge cubies on
the top layer are wrong.  The presence of the fixed center cubie makes it
clear that the edge cubies are really all wrong.

Boy it's great to get cube mail again.

				Ronnie

From tjj@lemma.helsinki.fi  Mon Nov  4 12:00:58 1991
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From: tjj@lemma.helsinki.fi (Timo Jokitalo)
Message-Id: <9111041700.AA02272@lemma.helsinki.fi>
To: cube-lovers@ai.mit.edu
Subject: Square One


Hi everyone!

Well, now I finally had time to try and solve my Square One. After 
hours and hours of pondering it is now finally done. (For the first time ever -
I noticed the leaflet with instructions on how to get it to square one only
after I'd done a few rotations.) Now what needs to be done is to polish
the 'method' a bit (a lot), and find out if I was just lucky this time...

Altogether, it seems to be quite a nice puzzle, although at first I thought
that it was _very_ ugly and unpleasant. (But not half as ugly and unpleasant as
'Rubik's Dice'. Perhaps I'm clumsy, but It's awful: whenever I get
somewhere with it, I touch it a bit too hard, all the plates fall to the
bottom and I throw the thing back to the corner in disgustment. Opinions?)

But: does anyone have the number of configurations for it? I tried to count
them, and got something like 2.8 E 13, but very probably there are a few
factors of two or something missing. It's also a bit tricky to decide on when
two configurations are different. (I mean, should one count as different a
configuration reached by rotating, say, the top layer by 30 degrees? Sometimes,
yes, but sometimes it seems a bit funny to do so.)

	Timo (tjj@rolf.helsinki.fi)



From SCHMIDTG@astro.pc.ab.com  Wed Nov  6 14:09:10 1991
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Date: 6 Nov 91 14:04:00 EST
From: "24305, SCHMIDT, GREG" <SCHMIDTG@astro.pc.ab.com>
To: "cube-lovers@ai.mit.edu" <cube-lovers@ai.mit.edu@odin.icd.ab.com>


Please add me to the mailing list

Thanks.


-- Greg Schmidt		--> schmidtg@iccgcc.decnet.ab.com <--



From diamond@jit081.enet.dec.com  Thu Nov  7 06:49:22 1991
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From: 07-Nov-1991 2050 <diamond@jit081.enet.dec.com>
To: cube-lovers@ai.mit.edu
Apparently-To: cube-lovers@ai.mit.edu
Subject: subscribe

Please subscribe diamond@jit081.enet.dec.com to the cube-lovers mailing list.

From hoey@aic.nrl.navy.mil  Mon Nov 11 17:45:47 1991
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From: hoey@aic.nrl.navy.mil
Message-Id: <9111112245.AA03752@sun1.aic.nrl.navy.mil>
To: baggett@mssun7.msi.cornell.edu (Jeffrey Baggett)
Cc: Cube-Lovers@life.ai.mit.edu
Subject: Rubik's Cube question
Organization: Naval Research Laboratory, Washington, DC

Jeff Baggett (baggett@mssun7.msi.cornell.edu) asks on the sci.math
newsgroup:

> 1.  I am seeking a description of the group of symmetries associated
> 	with Rubiks cube.  I have some ideas but they aren't particularly
> 	elegant.  Can someone suggest a paper?

Jeff,

I have looked into this somewhat.  As far as I know, the symmetries of
the 3^3 cube are just the symmetries of the cube, but in larger sizes
we can do better.  The best way of looking at this is to imagine that
there is a (N-2)^3 cube sitting inside your N^3 cube, and smaller
cubes within, and you are trying to solve them all together.

Suppose we address each cubelet of the N^3 cube using cartesian
coordinates (x,y,z), where (0,0,0) is the center of the cube (for N
odd) and no cubelets have any coordinate zero if N is even.  The
maximum absolute value of the coordinates is [N/2].

Then for 1<=I<=[N/2], there is a symmetry
 F[I]:(x,y,z)->(f(x),f(y),f(z)),
where f(I)=-I, f(-I)=I, and f(x)=x otherwise.

Then for 1<=I<J<=[N/2], there is a symmetry
 E[I,J]:(x,y,z)->(e(x),e(y),e(z)),
where e(I)=J, e(J)=I, e(-I)=-J, e(-J)=-I, and e(x)=x otherwise.

These are symmetries of the cube group, and they map elementary moves
to elementary moves (provided we take an elementary move to be a
rotation of the slab of N^2 cubelets that have a particular nonzero
value of a particular coordinate).  Symmetries of the cube group that
preserve elementary moves are useful in the study of local minima in
the cube group.

It turns out that if you only want to consider the outside of the cube
(ignoring the (N-2)^3 cube inside) all of these symmetries are still
present except F[[N/2]] and E[I,[N/2]].

I mentioned these symmetries in a note to the Cube-Lovers mailing list
in 1983.  I called E[I,J] evisceration, F[1] inflection, and F[[N/2]]
exflection in that note (where I was dealing explicitly with only the
4^3).  The discussion of the relation to local minima took place in
1980.  Let me know if you'd like a copy of these messages.

I ran into these symmetries earlier, though.  They are symmetries of
the N^3 tic-tac-toe board!  I would not be surprised if they arise in
some other connection in mathematics, but I have never run into them.
They generalize into larger dimensions, as well.

I've also taken the liberty of Cc'ing the Cube-Lovers list with this
note.  If you'd like to be on that list, you may ask of
"Cube-Lovers-Request@AI.AI.MIT.Edu".

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From sjfc!ggww@cci632.cci.com  Tue Nov 12 07:16:07 1991
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From: sjfc!ggww@cci632.cci.com (Gerry Wildenberg)
Message-Id: <9111111418.AA08293@sjfc.UUCP>
To: cube-lovers@ai.mit.edu
Subject: mailing list

Please put me on the mailing list.

If there is a FAQ for this group, lease mail me a copy.

Gerry Wildenberg                         ggww@sjfc.uucp
St. John Fisher College                  sjfc!ggww@cci.com
Rochester, NY 14618                      ...!uunet!uupsi!cci632!sjfc!ggww

From pbeck@pica.army.mil  Fri Nov 15 13:00:10 1991
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Date:     Fri, 15 Nov 91 10:36:45 EST
From: Peter Beck (BATDD) <pbeck@pica.army.mil>
To: cube-lovers@ai.mit.edu
Subject:  puzzle shows
Message-Id:  <9111151037.aa18151@FSAC1.PICA.ARMY.MIL>



Mark Your Calendars and take advantage of the airfare wars

CUBING/PUZZLING EVENTS
rev  11/15/91, transcribe by p beck

...............................................................
<-->  The 11th DUTCH CUBE DAY  <-->
...............................................................
WHEN ----   Saturday,  14  Dec  1991
WHERE ---- Office building of ABT, Arnhemsestraatweg 358,  Velp, THE
NETHERLANDS
TIME ---- 10:00 AM
ENTRANCE FEE:  Dfl  5.00;  includes lunch and drinks
INVITITATIONS:  cut-off date 11/20/91
Anton Hanegraaf, Heemskerkstraat 9, NL-6662 Al EST  (08819 - 72402),
 or   
FAX   --    ABT Velp,  085 - 635326

AGENDA:  
.. LECTURES 
- Lee Sallows:  Serila Sided Isogons
- Oskar van Deventer:  A Mirror Problem
- Koos Verhoeff:  Design of Spatial Strucutres
- Willem van der POel:  Survey of Mechanical Puzzles

.. EXHIBITIONS 
- Wim Zwaan:  Puzzles in Wood
- Theo Bense:  Dodecahedral Compositions
Popke Bakker/Koos Verhoeff:  Sculptures in Wood
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^



...............................................................
<-->  12th International puzzle collector's party and fair  <-->
...............................................................
WHEN ---- 7/31/92  
WHERE ----  TBD target is $70 per night
INVITATIONS ***  Admission by invitation only!!!  Contact Mr. Nob
Yoshigahara,  4-10-1-408 Iidabashi,  Tokyo  102 Japan
AGENDA:
  7/31  welcoming party in the evening
  8/1    collectors and dealers
details will be available in early feb 92
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^


From pbeck@pica.army.mil  Fri Nov 15 14:59:13 1991
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Date:     Fri, 15 Nov 91 10:39:21 EST
From: Peter Beck (BATDD) <pbeck@pica.army.mil>
To: cube-lovers@ai.mit.edu
Subject:  book review
Message-Id:  <9111151039.aa18635@FSAC1.PICA.ARMY.MIL>



SHAPING SPACE; A POLYHEDRAL APPROACH
M Senechal, G Fleck eds.
284 pp, Birkhauser, 1988 
$49.95, ISBN 0-8176-3351-0

REVIEW BY:
  Peter Beck;  Rubik's Cube Collector

Other reviews:
  SCITECH BOOK NEWS, V12, APR 88, P3
  AMERICAN SCIENTIST, V77, JAN 89, P72

Shaping Space, the book, was inspired by the Shaping Space Conference
held at Smith College in April 1984.  The content of the book
addresses itself to an interdisciplinary readership and should be
considered as a cornerstone book in guiding any level of reader on an
exploration of the Polyhedron Kingdom.   It encourages readers to
experience polyhedra directly, by including recipes for construction
as well as numerous illustrations (188 line and 174 halftone
illustrations). 

The book is organized in five parts with parts 1,2 &5 of greater
interest to the beginning explorer.

Part 1 is an introduction to polyhedra which includes a chapter on
making polyhedra ,ie, Chapter 2 "Five Recipes for Making Polyhedra."
Part 2 is an interdisciplinary overview of polyhedra which includes a
chapter on "Milestones in the History of Polyhedra" (chapter 4), a
chapter on "Polyhedra and Crystal Structures" (chapter 5),and a
chapter on "Spatial Perception and Creativity" (chapter 7), and
concludes with a chapter on "Why Study Polyhedra?" (chapter 8).  
Part 3 is about the roles of polyhedra in science. 
Part 4 is the theory of polyhedra. 
Part 5 is a discussion on incorporating the teaching of polyhedra in
the curriculum.  Included in part 5 is a resource guide that is
organized in the following categories:
A.  Architecture
B.  Art
C.  Geometry
D.  Instructional and Recreational Materials
E.  Science
F.  Symmetry

In addition to the resource guide each chapter has its own list of
references.  The book has a comprehensive index of 10 pages and it
also has a list of the conference contributors addresses.



From pbeck@pica.army.mil  Tue Dec  3 11:52:11 1991
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Date:     Tue, 3 Dec 91 11:26:50 EST
From: Peter Beck (BATDD) <pbeck@pica.army.mil>
To: cube-lovers@life.ai.mit.edu
Cc: pbeck@pica.army.mil
Subject:  rubik's tangle, etc
Message-Id:  <9112031126.aa10408@FSAC1.PICA.ARMY.MIL>


looking for nyc source of rubik's tangle, xv, dice, triamid

any suggestions??

thanks in advance


From @bullet.ecf.toronto.edu:tee@ecf.toronto.edu  Tue Dec 10 18:03:38 1991
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From: TEE  LUNS <tee@ecf.toronto.edu>
To: Cube-Lovers@ai.mit.edu
Subject: 7x7x7
Message-Id: <91Dec10.180328est.8345@bullet.ecf.toronto.edu>
Date: 	Tue, 10 Dec 1991 18:03:14 -0500

 
   I was reading through the archives the other night (just signed onto the
mailing list) and one of the last posts in cube-mail-7 triggered something in
me head.
 
   The suggestion was to use a fresnel saw to cut all the cubelets out of
a single chunk of material, with the cut such that the pieces all interlock.
The interlocking however doesn't need to be quite as intricate as the diagram
given - why not a simple dovetail?
 
    That's actually to some extent what the 3x3x3 cube is - the center cubelets
dovetail into the edge cubelets, and the edges dovetail into the corners. It
just happens that there's enough reduncancy that the outside half of the
dovetail joints can be disposed of, and the edges made straight while still
allowing the cube to stay in one piece.
 
    If we have a complete (both sides) locked dovetail, we can actually assemble
almost all of the cube out of the cubelets. Since the cubelets will always
require an entry point for their dovetail grooves, there will be a few cubelets
that have to be attached differently.

    The simplest solution I can think of is to have the dovetail/cublet pair
seperate, with a countersink on the dovetail, and holes through the other
cubelets so that we can screw the dovetails (which are already in their grooves)
onto the last couple of cubelets.

    One drawback with this approach is that the boundaries between layers of
cubelets will be quite jagged. If the dovetails go right to the surface, one
has to be *VERY* careful when turning the cube to make sure that all layers
are lined up in the axes that aren't being turned (this problem plagues the
magic truncated octahedron I have - pieces pop out all the time). The solution
is to make the dovetail taper off at its ends so that if it's out of line with
the groove its going into, it can still correct itself. This will lead to holes
at the surface though, so the cube won't be too pretty.

    A novelty with this approach though is that no centre is required. We
could build a hollow 3x3x3 cube with face centres hollow, and see right
through the cube. This should be possible with larger odd-sized cubes too,
but there comes a point (probably 7x7x7 again) where mechanical play would
let middle layers shear enough to pop out cubelets.

    But, if we had the smaller odd-sized cubes trapped inside, not only
would they help hold the outer layers together, if we made the cubelets 
mostly transparent, we'd be able to see  what we've had to imagine in the
past. Now that'd be one heck of a puzzle.


From pbeck@pica.army.mil  Wed Dec 11 09:20:29 1991
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From: Peter Beck (BATDD) <pbeck@pica.army.mil>
To: cube-lovers@life.ai.mit.edu
Subject:  collectors directory
Message-Id:  <9112110909.aa13900@FSAC1.PICA.ARMY.MIL>


The world of MECHANICAL PUZZLE collecting is getting organized. 
 Jerry Slocum is compiling a directory of collectors, also designers,
buyers, sellers.  If you want to be included there is a form to fill
out (hardcopy).  The form has to be to Jerry by 1/15/92 to be
included.  This is a non-commercial venture and I am unaware of any
extended plans.

To get a form contact Jerry:
  JERRY SLOCUM
  257 SOUTH PALM DRIVE
  BEVERLY HILLS, CA 90212 USA
  PHONE & FAX  310/273-2270

I assume form can be FAXed.  I have copies so if need be I can mail
you one.

PS:  There is classification for computer puzzles but not one for
computer solutions/helper programs, eg, cube simulations & non
physically realizable cubes, like 10x10x10.

From ronnie@cisco.com  Wed Dec 11 18:04:35 1991
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From: ronnie@cisco.com
To: Cube-Lovers@life.ai.mit.edu
Subject: A Sam Loyd Rubik puzzle unearthed!!!
Date: Wed, 11 Dec 91 14:57:25 PST
Sender: ronnie@cisco.com

	An original Sam Loyd puzzle involving the Rubik's cube has
come into my hands; somewhat surprising, in that Sam Loyd died in
the early years of this century, but no more so than the truly
astounding circumstances by which the puzzle came to me, which I would
detail if I believe that anyone were interested.  However, as this
list consists only of people interesting in things Cubic, I will limit
this posting to the puzzle itself.




		Crooked Gambling in Puzzleland

				by Sam Loyd


	Tommy Riddles has challenged King Puzzlepate to a game of
dice, using Rubik's Cubes as dice.  However, Tommy is planning to
cheat by changing the ordinary Rubik's Cube into tops [ tops are dice
which are misspotted, by having only three different numbers on them,
each appearing opposite to itself, such that it is indetectable
without turning the die around  -ed ]  spotted 1-2-3, which he is able
to make from a standard cube in 14 moves.

	King Puzzlepate, however, has learned from the General of his
plans, and has figured out to convert Tommy's tops into 2-3-4 tops,
which are favorable to His Majesty, in only 3 moves.

	Can you duplicate both these feats?


[ Note that Loyd appears to consider a move as moving any of the nine
slices any number of degrees.  Thus the move we would designate as
L2R2 and count as four, Loyd would count as one move of the center
slice by 180 degrees. ]

From hoey@aic.nrl.navy.mil  Thu Dec 12 10:29:25 1991
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From: hoey@aic.nrl.navy.mil
Message-Id: <9112121529.AA02722@sun13.aic.nrl.navy.mil>
To: ronnie@cisco.com (Ronnie Kon)
Subject: Rubik's cube dice tops
Cc: Cube-Lovers@ai.mit.edu

ronnie@cisco.com (Ronnie Kon) writes:

> An original Sam Loyd puzzle involving the Rubik's cube has come into
> my hands; somewhat surprising, in that Sam Loyd died in the early
> years of this century, but no more so than the truly astounding
> circumstances by which the puzzle came to me, which I would detail
> if I believe that anyone were interested.

Hmm.  A likely story.

We are challenged to find ``tops'',

> ... dice which are misspotted, by having only three different
> numbers on them, each appearing opposite to itself ... spotted
> 1-2-3, ... from a standard cube in 14 moves.

Where he counts a half turn, a slice, and and a half-slices as one
move each.  I have found how this can be done in 13 such moves.  I
have some suspicion that it can be done in 12; I'll let you know.

We are then challenged to convert this

> ... into 2-3-4 tops ... in only 3 moves.

The second part can be solved by any person who achieves mastery of
the cube.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From hoey@aic.nrl.navy.mil  Fri Dec 13 14:50:15 1991
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From: hoey@aic.nrl.navy.mil
Message-Id: <9112131950.AA06512@sun13.aic.nrl.navy.mil>
To: TEE LUNS <tee@ecf.toronto.edu>
Cc: Cube-Lovers@life.ai.mit.edu
Subject: Big groovy cubes, revisited

Tee Luns writes:

> ... one of the last posts in cube-mail-7 triggered something in me
> head.  The suggestion was to use a fresnel saw to cut all the
> cubelets out of a single chunk of material....

Well, I'm glad that my silly ideas triggered something.  Sometimes I
wonder if they are as amusing to read as they were to write.

> ... why not a simple dovetail?

Certainly a dovetail would do it.  I guess when I got to sharpening
the fresnel saw I didn't know when to quit.

> ... have the dovetail/cubelet pair separate, ... screw the dovetails
> (which are already in their grooves) onto the last couple of
> cubelets.

Surprisingly enough, this is just how Rubik's Revenge is put together.
One of the center cubelets (perhaps always on the blue side) has a
screw that joins the outside of the cubelet to its dovetail.  You can
usually find locate it by the dimple in the colored sticker.

> If the dovetails go right to the surface, one has to be *VERY*
> careful....  The solution is to make the dovetail taper off at its
> ends....  This will lead to holes at the surface though, so the cube
> won't be too pretty.

In the 7^3 and larger, they have to go through the surface, and even
if they were squared-off dovetails they wouldn't match the color of
the adjacent square except in the solved position.  Unless of
course we make the outer layers thicker, as Dale Newfield mentioned
when we were discussing this back in May.

>    A novelty with this approach though is that no centre is
> required.  We could build a hollow 3x3x3 cube with face centres
> hollow, and see right through the cube....

>    But, if we had the smaller odd-sized cubes trapped inside, not
> only would they help hold the outer layers together, if we made the
> cubelets mostly transparent, we'd be able to see what we've had to
> imagine in the past. Now that'd be one heck of a puzzle.

Wow, I want one!  But I don't think the material really needs to be
transparent, as long as the face center pieces are hollow.  It would
help let light in, though.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From ACW@yukon.scrc.symbolics.com  Fri Dec 13 18:29:48 1991
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Date: Fri, 13 Dec 1991 18:16-0500
From: Allan C. Wechsler <ACW@yukon.scrc.symbolics.com>
Subject: Big groovy cubes, revisited
To: hoey@aic.nrl.navy.mil, tee@ecf.toronto.edu
Cc: Cube-Lovers@life.ai.mit.edu
In-Reply-To: <9112131950.AA06512@sun13.aic.nrl.navy.mil>
Message-Id: <19911213231647.9.ACW@PALLANDO.SCRC.Symbolics.COM>

I want to know whether it is feasible, with modern electronics and
electro-mechanicals, to make a 3x3x3 cube that solves itself at the
touch of a button.  How much would a prototype cost?  $10,000?
$100,000?

From hoey@aic.nrl.navy.mil  Tue Dec 17 16:38:10 1991
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From: hoey@aic.nrl.navy.mil
Message-Id: <9112172118.AA14791@sun13.aic.nrl.navy.mil>
To: Cube-Lovers@life.ai.mit.edu, ronnie@cisco.com (Ronnie Kon)
Subject: Re: Rubik's cube dice tops (Spoiler)

Last week ronnie@cisco.com (Ronnie Kon) challenged us to find Rubik's
cube patterns with dice pips for 1, 2, and 3 on the three pairs of
opposite sides.  He claimed it could be done in fourteen HST, where
one HST is a turn of a face or center slice by 90 or 180 degrees.  I
responded that it could be done in thirteen HST.  Here is how.  I will
use this opportunity to practice the enhanced Varga Rubiksong I
described (unfortunately with many typos) on 22 Feb 90.

The (only such) pattern is the composition of Four-Spot and Laughter.
We have long known the processes

          ris-fos tis-fos, or (RL)^2 FB' (TD)^2 FB', for Four-Spot and
  fon-ron fon-ron fon-ron, or (FBRL)^3,              for Laughter.

When we compose them, the F and B moves combine and cancel to produce

ris-fos tis-fi ron-fon ron-fon ron, or (RL)^2 FB' (TD)^2 F^2 (RLFB)^2 RL.

This 14 HST process is presumably something like what Ronnie had in
mind.  But since this pattern commutes with ris, or (RL)^2, we can get
the same pattern with the conjugate process

    fos tis-fi ron-fon ron-fon ran, or FB' (TD)^2 F^2 (RLFB)^2 R'L'.

This uses only 13 HST.  This is also the shortest process I know of in
the normal metric: 18 QT, which is not so bad for the combination of
two 12 QT processes.

I suggested that perhaps 12 HST would be sufficient, but I have not
found such an improvement.  Nor do I know whether 13 HST is the best
that can be done: it seems that proving 13 HST optimal would require
examining about 160 million positions, almost as many as the 200
million it would take to prove 18 QT optimal.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From raymond@cps.msu.edu  Sun Jan  5 12:54:44 1992
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From: raymond@cps.msu.edu (Carl J Raymond)
Message-Id: <9201051754.AA12647@galaxy.cps.msu.edu>
To: cube-lovers@life.ai.mit.edu
Subject: Subscribe

I would like to join the cube lover's mailing list.  My email address
is raymond@cpsin3.cps.msu.edu.  Also, I am trying to find an email
address for Peter Beck.
Thanks,
Carl Raymond

From cosell@bbn.com  Tue Jan  7 09:49:04 1992
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Date:     Tue, 7 Jan 92 7:23:29 EST
From: Bernie Cosell <cosell@bbn.com>
To: cube-lovers@life.ai.mit.edu
Subject:  Hungarian Rings solution?

Does anyone have an algorithm for solving the "Hungarian Rings" they'd
be willing to share.  I found one in a drawer that had been long
misplaced, and I've been fiddling with it some.  It seems easy enough
to find lots of 'commutators' for the thing, but all the ones I've run
into have the annoying property that they produce symmetric
perumtations [given a 180 degree rotation of the puzzle].  But of
course, my puzzle is no where near symmetrically messed up.

Any advice, hints, full-blown algorithm, etc, would be appreciated... Thanks!
 /Bernie\

ps, For those that don't remember it, the "Hungarian Rings" is a puzzle with
beads arranged in two intersecting circles:

	*       *
      *   *   *   *
    *       *       *
  *       *   *       *
 *       *     *       *
  *       *   *       *
    *       *       *
      *   *   *   *
	*       *

[Imagine these both round...:-)]  and you can slide the beads around either
cicle.    The beads come in four colors andthe object is to get the colors
all sorted out.  /b\


From dik@cwi.nl  Tue Jan  7 10:14:17 1992
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From: Dik.Winter@cwi.nl
Message-Id: <9201071514.AA05644@boring.cwi.nl>
To: cosell@bbn.com, cube-lovers@life.ai.mit.edu
Subject: Re:  Hungarian Rings solution?

You don't nood commutators for it, cycles are sufficient (because there
are so many similar colored beads).  If I remember right one useful move
is:  turn right ring clockwise two beads, turn left clockwise two beads,
turn right anti-clockwise two beads, turn left anti-clockwise two beads.
Using them properly will solve the rings.

dik
--
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
dik@cwi.nl

From johnf@apollo.com  Tue Jan  7 14:20:49 1992
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From: johnf@apollo.com
Date: Tue, 7 Jan 92 11:45:34 EST 
Subject: Square One
To: cube-lovers@life.ai.mit.edu


I got given one of these things for Christmas (well, actually I gave it
to myself). I was wondering if anyone has any good basic operators that
they would like to share.  I would imagine that the puzzle must be less
complex than a true cube,  but the restricted set of moves make solving
it more complicated than you might think!
[I currently have six corners correct, but I still have two (diagonally
opposite) corners interchanged.  The cube-solving technique that I used
for a real cube doesn't work here - I need something different].

From cosell@bbn.com  Tue Jan  7 15:48:19 1992
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Date:     Tue, 7 Jan 92 15:02:18 EST
From: Bernie Cosell <cosell@bbn.com>
To: cube-lovers@life.ai.mit.edu
Subject:  Re:  Hungarian Rings solution?

In rsponse to my request for info about the Hungarian Rings, 

Dik.Winter@cwi.nl writes:

} You don't nood commutators for it, cycles are sufficient (because there
Dik.Winter@cwi.nl writes:

} You don't nood commutators for it, cycles are sufficient (because there
} are so many similar colored beads)....

My apologies --- I meant to say "cycles" when I said I had found lots of
them... And I hate to seem dense, but but I'm still stuck...

} ... If I remember right one useful move
} is:  turn right ring clockwise two beads, turn left clockwise two beads,
} turn right anti-clockwise two beads, turn left anti-clockwise two beads.
} Using them properly will solve the rings.

The 'properly' is the part I'm finding hard.   There seem to be LOTS of
cycles, but even with that big choice I can't see, quite, how to solve
the thing.

As far as I can tell, basically ANY set of ring-turns that has a total
movement of zero seems to define a pretty small cycle.  For example,
the sequence LnA RnA LnC RnC, for n not a multiple of 5[*], does a
three-bead cycle: if you look at the upper intersection:
                        A                     C
   Intersection --->  C      ======>        B
                        B                     A
Where 'A' and 'B' are each n beads away from the intersection [and by
changing theorder of L/R you reverse the cycle, and by interchanging A
and C you move the cycle to the other side of the intersection.
BUT: the problem is that this isn't really a 3-cycle, but rahter _two_
3-cycles: you also make a central-symmetric move of the beads at the
bottom intersection.
   [*] since five is the distance between the intersections, if the
   rotate is a muiltiple of 5 the intersections interact, things get a
   little different: it makes a *two* cycle!  In the diagram above
   [with A five away from C], the move just _swaps_ A & C [and the A'
   and C' at the lower intersection, too, of course].

Given that my rings are totally non-symmetrically messed up, I can't
figure out a plan for making forward progress.  I can do lots of
diffent cycles, but I can't manage to get the rings set up so that the
cycle at both intersections is useful: if I try to fix something at the
top intersection I invariably mess up something at the bottom one.

Thanks again for you patience with my rantings.  I feel like I'm
overlooking something simple [since this wasn't supposed to be all that
hard a puzzle], but I don't see what it is ...

  /Bernie\

From dik@cwi.nl  Tue Jan  7 16:13:50 1992
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From: Dik.Winter@cwi.nl
Message-Id: <9201072113.AA06247@boring.cwi.nl>
To: cosell@bbn.com, cube-lovers@life.ai.mit.edu
Subject: Re:  Hungarian Rings solution?

 > Dik.Winter@cwi.nl writes:
 > } You don't nood commutators for it, cycles are sufficient (because there
 > } are so many similar colored beads)....
I have already been chastised that what I described are commutators.  Of course
they are.  Not only is my thinking bad late at night, but apparently my
spelling is atrocious :-).

 > As far as I can tell, basically ANY set of ring-turns that has a total
 > movement of zero seems to define a pretty small cycle.  For example,
 > the sequence LnA RnA LnC RnC, for n not a multiple of 5[*], does a
 > three-bead cycle: if you look at the upper intersection:
 >                         A                     C
 >    Intersection --->  C      ======>        B
 >                         B                     A
 > Where 'A' and 'B' are each n beads away from the intersection [and by
 > changing theorder of L/R you reverse the cycle, and by interchanging A
 > and C you move the cycle to the other side of the intersection.
 > BUT: the problem is that this isn't really a 3-cycle, but rahter _two_
 > 3-cycles: you also make a central-symmetric move of the beads at the
 > bottom intersection.
True.  But if you prefix the move by a (series of moves) that makes the upper
three of an identical color (and postfix by its inverse), you will not see
the difference between a true cycle.  At least that is how I always did the
final part.  (Correctly coloring the two lobes is in fact easy; you better
start with that.)

From Hoffman.El_Segundo@xerox.com  Wed Jan  8 12:07:08 1992
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Date: 	Wed, 8 Jan 1992 09:06:16 PST
From: Hoffman.El_Segundo@xerox.com
Subject: Re: Square One
In-Reply-To: "johnf%apollo:com's message of 7 Jan 92 08:45:34 PST (Tuesday)"
To: johnf@apollo.com
Cc: cube-lovers@life.ai.mit.edu
Message-Id: <" 8-Jan-92  9:06:16 PST".*.Hoffman.El_Segundo@Xerox.com>


I, too, gave myself Square One for Christmas, and I, too, would love to
exchange some useful moves.  Here are three that I use.  I need some more!

  -- Rodney Hoffman
     Hoffman.El_Segundo@Xerox.com
     or   rodney@oxy.edu

------------------------------------------------------

Conventions used in these descriptions:

  *  These moves start and end in a cube shape.  I always hold the
     logo to my left, with the two faces which can rotate in front
     and back.  That means the plane of the 180-degree twist is
     perpendicular to my face, angling from upper right to lower left.
     The "front" face is the one I'm looking directly at.

  *  "the smallest increment" as in "Turn the front face cw
      the smallest increment".   This means the smallest amount
      that permits the big 180-degree twist that must follow.
      Note that the angle of turn is not always the same!  Sometimes
      "the smallest increment" turn is truly the smallest possible
      increment, that is, the width of an edge piece.  At other
      times, it may be the width of a corner piece (much larger),
      or even two or three piece's widths combined.

  *  "to match" as in "Turn the back face to match".
     This means the front and back faces remain aligned with one
     another.

  *  "Now do the 180-degree twist."  I move the right half 180
     degrees, holding the left half fixed.  The logo stays fixed
     in position and orientation, never moving.

  *  To help in describing the effects of these moves, I will
     refer to the pieces by number, as follows.  Here I have
     numbered the pieces clockwise from the upper left corner
     piece on the front face.  I have numbered the back face
     similarly, ** as if looking straight through to it **,
     that is, piece 9 is directly beneath piece 1, piece 10
     is directly beneath piece 2, etc.:

            Front Face               Back Face

            1   2   3                9  10  11
	    8       4               16      12
	    7   6   5               15  14  13


------------------------------------------------------

(1) Swaps all 8 corner pieces diagonally directly across in pairs, staying on
the same faces (front and back).  In my numbering scheme, this move swaps 1
with 5, 3 with 7, 9 with 13, and 11 with 15.

  (a) Turn the front face cw the smallest increment.
      Turn the back face to match.  Now do the 180-degree twist.

  (b) Turn the front face ccw the smallest increment.
      Turn the back face to match.  Now do the 180-degree twist.

  (c) Repeat the (a)(b) combination three times.

  Note:  This entire move, repeated twice, is, of course, an identity.

------------------------------------------------------

(2) Swaps all 8 edge pieces directly across in pairs, staying on the same faces
(front and back).  In my numbering scheme, this move swaps 2 with 6, 4 with 8,
10 with 14, and 12 with 16.

  (a) Turn the front face cw the smallest increment.
      Turn the back face to match.  Now do the 180-degree twist.

  (b) Repeat (a) six times.

  (c) Turn the front and back faces 180 degrees.

  Note:  This entire move, repeated twice, is, of course, an identity.

------------------------------------------------------

(3) Although this move itself is simple to describe, its effect is not. It
moves four pieces (two large and two small) from the front to the back, and
vice-versa.

It's easiest to just give a map of the changes:

BEFORE:     Front Face               Back Face

            1   2   3                9  10  11
	    8       4               16      12
	    7   6   5               15  14  13

AFTER:      Front Face               Back Face

            11  6  13                9  10   3
	     8     12               16       2
	     7 14   1               15   4   5

(Because the back face is never turned in this move, its pieces 9, 10, 15, and
16 always stay fixed.  They are on the immobile half of the back face, the half
not moved during the 180-degree twists.)

  (a) Turn the front face cw the smallest increment.
      Do not turn the back face at all.
      Now do the 180-degree twist.

  (b) Turn the front face ccw the smallest increment.
      Do not turn the back face at all.
      Now do the 180-degree twist.

  (c) Repeat the (a)(b) combination three times.

  Note:  This entire move, repeated five times, is an identity.

------------------------------------------------------

From GOFFJEFFREYM@bvc.edu  Wed Jan  8 16:51:55 1992
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Date: Wed, 8 Jan 1992 15:51 CST
From: The Phantom <GOFFJEFFREYM@bvc.edu>
To: cube-lovers@life.ai.mit.edu
Message-Id: <01GF2UNCTVTC94DUAR@bvc.edu>
X-Vms-To: IN%"cube-lovers@life.ai.mit.edu"

	Another miscellaneous thought on the dovetail topic.  The way I
understand it is that each piece is dovetailed into the surrounding pieces. 
I.E. on the 3x3x3 the face-centers are dovetailed into the edges which are
dovetailed into the corners.  Now, the maximum radius of the dovetailing circle
permits this all the way out to 5x5x5, but at 6x6x6, a piece hangs outside the
dovetail radius.
	The master design could work still.  First of all, keep those cubes
nested together.  Second of all, dovetail the inner cubes outward, so the whole
damn thing holds.  Last, and certainly not least, in the case of the 6x6x6, I
think that we can add in another circle of dovetails at the radius which would
support a 7x7x7 cube.

	This is really hard to explain without graphics, but try to imagine
this.  Each dovetail ring will be a simple circle.  Draw a 3x3 grid, and
superimpose a circle with a radius of 1.5 grids.  Now, that will be
the inside of the dovetail, 1/3 of the way into the face.  For a 5x5 grid,
repeat the 3x3 procedure and do the same for the 5x5, except at 2.5 grids
radius.  That represents the next layer of the cube.

	Keeping the original Rubik concept of faces-edges-corners nesting in
mind, we keep building out this way.  It probably will be damn inconvenient to
build the 7x7x7 cube this way, but I am reasonably sure that it will hold
together and still retain all the necessary movements from the original cube. 

	In the case of the 5x5x5 cube, it will look like this:

abcba
bdedb
cefec
bdedb
abcba

a is held by 2 b's.
b is held by c and d.
c is held by e.
d is held by 2 e's.
e is held by f.

The only real weak spot is c, but it is surrounded by b's and an e.

	The 7x7x7 cube will look like this:

abcdcba
befgfeb
cfhihfc
dgijijd
cfhihfc
befgfeb
abcdcba

a is held by 2 b's.
b is held by c and e.
c is held by d and f.
d is held by g.
e is held by 2 f's.
f is held by h and g.
g is held by i.
h is held by 2 i's.
i is held by j.
j is the center.

Again, the problem shows up in d, g, and i.  I think that this is unavoidable,
but since the 3x3x3 cube holds together in much the same way, I think that it
should be about as a 5x5x5 cube is compared to the 3x3x3 cube.  Not much
difference.  The only problem that I have had with my 5x5x5 cube is the edge
cubies, and as I have noted, the middle and third edges should be the problems,
since everything else is buried within a face.

	I haven't built a prototype yet, but once I get finished with this
semester of college (my last one), I intend to start working on this in between
job-hunting and paying off loans.  If anyone is interested in this idea, I
would like to start correspondence on this topic.

	I have some diagrams available, but they don't come over too well in
ASCII.  If you would like a copy of my preliminaries, please E-Mail me at the
following addresses.  Thank you for your consideration.

*************************************************************************
*					*				*
*	Internet: goffjeffreym@bvc.edu	*	Hailing From:		*
*		  			*				*
*					*	Storm Lake, IA 50588	*
*	Snail: Box 260 BVC		*	Home of Happiness	*
*					*				*
*************************************************************************
*                                                    	*
*	'Dr. Floyd, you are not a very practical man.'	*
*	'Look out there.  Tell me what's practical.'	*
*                                                   	*
*	-2010						*
*							*
*********************************************************

From hoey@aic.nrl.navy.mil  Wed Jan  8 21:20:58 1992
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Date: Wed, 8 Jan 92 21:20:01 EST
From: hoey@aic.nrl.navy.mil
To: Post-NetNews@ucbvax.berkeley.edu, Cube-Lovers@life.ai.mit.edu
Cc: mkt@vax5.cit.cornell.edu (Gregory E. Dionne),
        tlg@uknet.ac.uk (Tim.Goodwin)
Subject: Rubik's Cube, the minimax number of moves
Summary: 2x2x2=11(9), 3x3x3=21(18)-97(50), 4x4x4=37/41(??)-??(??)
References: <1992Jan3.213615.9689@vax5.cit.cornell.edu> <564@uknet.ac.uk>
Message-Id: <920108.Hoey@AIC.NRL.Navy.Mil>
Keywords: Bounds, Metrics, Thistlethwaite, RCC
Organization: Naval Research Laboratory, Washington, DC

mkt@vax5.cit.cornell.edu (Gregory E. Dionne) asks:

> Does anybody know what the minimum number of moves are required to
> solve the 3x3 and/or 4x4 cubes in the worst-case scenario?

and receives some answers, some of them accurate.  The following is my
understanding of the best answers now known, which I am sending both
to rec.puzzles and the Cube-Lovers mailing list.  The latter will, I
hope, excuse some information repeated from the archives.

First, you have to decide what you mean by a move.  On grounds of
symmetry and parsimony I prefer to count each quarter-turn of a face
as a (QT) move.  However, most of the literature counts either a
quarter-turn or a half-turn of a face as a single (HT) move, and there
has been more work done on the problem by the HT contingent.

Second, you have to be careful to define what constitutes solved!
While most people are content to make each face a solid color, some
cubes have markings that display whether the face centers are twisted
with respect to the rest of the cube.
    [This has recently been done commercially in an spectacularly
     braindamaged way, in a product known as ``Rubik's cube--the
     fourth dimension'' or some such nonsense.  The mfrs have marked
     only four face centers, breaking symmetry while they fail to show
     the surprising invariant of the Supergroup.  What bagbiters!]
But that is a topic for other messages; I will not further discuss the
invisible features of cubes here, save to note that there are more
invisible features in larger cubes, and that if you take them into
account, the cube will be harder to solve and require more moves than
if you don't.

Third, you have to understand that in either case, nobody knows the
exact answer except for the tiny cubes.  It boils down to knowing
lower bounds L and upper bounds U, such that there must be some
positions requiring at least L moves, while any position can be solved
in at most U moves.  I will discuss these in turn, below.

For lower bounds, it is easy to calculate how many positions of
Rubik's cube are achievable, and we may reason that only a few
positions are within a few moves of start.  Counting them, we can
determine that at least 21 QT (or 18 HT) are needed to solve some
positions of the cube.  In fact, at least half of the positions of the
cube require at least 18 HT, and at least 90% of the positions require
at least 20 QT.  The calculations are elementary, and have been known
for over a decade.  Nobody knows any other very good way of finding
lower bounds.  In Rubik's_Cubic_Compendium (1987), Tamas Varga writes
    Experts believe that even in the worst cases--the patterns
    furthest away from start--it should be possible to restore the
    cube in 20-odd [HT] moves, maybe 25, not more.
However, such beliefs are clearly conjectural, based on the behavior
of much simpler puzzles.

The known upper bounds are constructive, consisting of a solution
procedure guaranteed to solve any cube within U moves.  The best known
bound is embodied in a procedure invented by Morwen B. Thistlethwaite,
and improved by students of Donald E. Knuth (The students are not
identified in R_C_C).  The improved procedure requires at most 50 HT
in the worst case.  Thistlethwaite was hoping (in 1980) to improve
this to 41 HT, but (rumors to the contrary) he apparently did not
succeed.  A 1989 article by Hans Kloosterman entitled ``Rubik's Cube
in 44 moves'' refers to an attempt to refine Thistlethwaite's method.
I have not heard of any success there, either.

Since any HT is at most 2QT, any Rubik's cube position can be solved
in at most 100 QT using Thistlethwaite's method.  According to my
understanding of the method, this can actually be reduced to 97 QT,
but this is still poor.  A method tailored to minimizing QT would
almost certainly guarantee a much shorter solution.

Unfortunately, Thistlethwaite's method requires enormous tables of
partial solution methods.  Gerszon Keri describes a simpler method in
R_C_C that requires at most 97 HT and which humans can memorize.  The
method is attributed to ``a Cambridge group,'' which I think must
refer to England.  According to Keri the Cambridge method has been
refined to use only 79 HT, but I do not know if the refined version is
still humanly comprehensible.

For the 2x2x2 cube, the worst-case number of moves is known to be
exactly 14 QT (11 HT).  Only 276 (2644) positions require all 14 QT
(11 HT).  Half of the positions can be solved in 11 QT (9 HT) or
fewer.

For the 4x4x4 and larger cubes, the problem of defining a move is more
complicated.  Besides the QT/HT dichotomy, there is the question of
whether a move consists of a slice (turning one part of the cube with
respect to the other) or a slab (turning a 1x4x4 section of the cube
with respect to the rest).  We might expect that, as we do not count
the center-slab moves of the 3x3x3 as a single move, we should not
count the inner-slab moves of the 4x4x4 as a single move.  However, I
have discovered excellent reasons of symmetry (send email if you want
to know) for us to consider all slabs alike, whether internal or face,
with the exception of the center slab of an odd-sized cube.  This is
known as the Eccentric Slabist metric.

According to my calculations of some years ago, some 4x4x4 positions
require at least 37 slab QT or 41 slice QT to solve.  The Eccentric
Slabist also calculates at least 59, 81, 111, 138, 175, and 208 QT for
the 5x5x5 through 10x10x10 cubes.  (And yes, I've heard the widespread
misinformation that Rubik's cubes larger than six cubies on an edge
are impossible).

I seem to recall calculating HT lower bounds, but I can't seem to find
them.  I do not recall whether upper bounds have been calculated for
the large cubes, other than that they are O(N^2)--see J. A. Eidswick's
article in the March 1986 Math Monthly.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From tjj@lemma.helsinki.fi  Thu Jan  9 03:07:28 1992
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From: tjj@lemma.helsinki.fi (Timo Jokitalo)
Message-Id: <9201090646.AA24202@lemma.helsinki.fi>
To: Cube-Lovers@life.ai.mit.edu
Subject: Re: Rubik's Cube, the minimax number of moves



I wonder how large the necessary tables for Thistlethwaite's method for the
cube are? I seem to recall reading that there were a few hundred entries, but
is this anywhere near? And, more important, have they been published, or does
anyone have them in an electronic format?

	Thanks,
	Timo Jokitalo (tjj@rolf.helsinki.fi)

From hoey@aic.nrl.navy.mil  Fri Jan 10 18:35:28 1992
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From: hoey@aic.nrl.navy.mil
Message-Id: <9201102332.AA13941@sun13.aic.nrl.navy.mil>
To: tjj@lemma.helsinki.fi (Timo Jokitalo), Cube-Lovers@life.ai.mit.edu
Subject: Re: Rubik's Cube, the minimax number of moves
Keywords: Upper-Bounds, Thistlethwaite, RCC, NoRMC

tjj@lemma.helsinki.fi (Timo Jokitalo) asks

> I wonder how large the necessary tables for Thistlethwaite's method
> for the cube are?  I seem to recall reading that there were a few
> hundred entries....

Well, this is the information from Singmaster's _Notes_on_Rubik's_
_Magic_Cube_ (1980).  Thistlethwaite's method is to work from group to
subgroup as follows:

    G0: <F,B,L,R,T,D>
    G1: <F^2,B^2,L,R,T,D>
    G2: <F^2,B^2,L^2,R^2,T,D>
    G3: <F^2,B^2,L^2,R^2,T^2,D^2>
    G4: <I>

The following table shows the number of cosets (the index of each
subgroup in the next larger group).  Then I include the number of HT
moves proven, anticipated, and best possible, from the 1980 N_o_R_M_C.
Finally, I include the number of HT claimed in the 1987 R_C_C.  It is
interesting to note that the improvement did not occur where
Thistlethwaite anticipated it.

Step | Number of Cosets  |     Number of HT, 1980      | #HT, 1987
     |                   |  Proven  Anticipated  Best  |   Proven
     |                   |                             |
  1  | G0:G1 =     2,048 |     7        7         7    |      7 
  2  | G1:G2 = 1,082,565 |    13       12        10    |     13
  3  | G2:G3 =   663,552 |    15       14 ?      13 ?  |     15
  4  | G3:G4 =    29,400 |    17       17        15 ?  |     15
-----+-------------------+-----------------------------+-----------
               Total HT  |    52       50 ?      45 ?  |     50
               Total QT  |   101       97 ?      87 ?  |     97

I had thought the tables contained one entry for each coset, and so
there would be over a million entries for step 2.  However, I was
surprised just now to notice in N_o_R_M_C that tables were only needed
in step 4, and then only 172 entries, so there must be some
abbreviation or algorithmic approach.  Of course, when Knuth's
students improved step 4, they may have changed it to use a huge
lookup table; I don't know.  Still, this is much better than the
situation I expected in my note two days ago.

In listing QT I assume that in we can require steps 1, 2, and 3 to
each end with a quarter-turn.  So the number of QT is at most three
less than twice the number of HT.

> And, more important, have they been published, or does anyone have
> them in an electronic format?

The bibliography in N_o_R_M_C mentions Thistlethwaite's algorithms as
being in typescript, but I don't know if they were available by
request, and I don't know anyone who has them.  I don't know anything
about the improvements by Knuth's students, and there's nothing in
the R_C_C bibliography that looks like a Stanford tech report.

Dan

From ACW@yukon.scrc.symbolics.com  Mon Jan 13 22:03:15 1992
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Date: Mon, 13 Jan 1992 14:38-0500
From: Allan C. Wechsler <ACW@yukon.scrc.symbolics.com>
Subject: Re: Rubik's Cube, the minimax number of moves
To: hoey@aic.nrl.navy.mil, tjj@lemma.helsinki.fi, Cube-Lovers@life.ai.mit.edu
In-Reply-To: <9201102332.AA13941@sun13.aic.nrl.navy.mil>
Message-Id: <19920113193832.8.ACW@PALLANDO.SCRC.Symbolics.COM>

I would like to see us develop a programming language for expressing
cube-solving algorithms in.  Then we could cooperate in trying to find
an algorithm with smaller and smaller numbers of moves in the worst
case.

I just completed an exercise I have wanted to try for a while, a rough
worst-case analysis of my own technique.  It's pretty scary.  It turns
out that my worst case is 236 qtw.  My algorithm is "bottom-heavy" -- it
starts with "intuitive" moves for fixing the first few corners.  These
were the hardest to analyze, but they take the fewest moves.  It
finishes up with great big macros for the last few fiddles and fixes.
For example, flipping two edges in place takes 22 qtw.  Obviously a lot
could be gained from tweaking the earlier part of the algorithm to
guarantee that I don't need to do this at the end.

From hoey@aic.nrl.navy.mil  Tue Jan 14 12:49:25 1992
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From: hoey@aic.nrl.navy.mil
Message-Id: <9201141749.AA27091@sun13.aic.nrl.navy.mil>
To: Cube-Lovers@life.ai.mit.edu
Cc: "Allan C. Wechsler" <ACW@yukon.scrc.symbolics.com>
Subject: A new upper bound: 91 QT
Keywords: Upper-Bounds, Thistlethwaite

I just wrote a quick program to count the number of QT to move from
the full cube group to the subgroup generated by <F^2,B^2,L,R,T,D>.
Thistlethwaite computed that this takes at least 7 HT in the worst
case.  The surprisingly good result is that it also takes only 7 *QT*
in the worst case.  This reduces the upper bound I posted Friday to 91
QT.

I had wondered if the worst cases could be reduced by choosing a
different pair of faces to restrict to half-twists.  Unfortunately,
the all-edges-flipped position is one of those that requires at least
7 HT (and so 7 QT), and by symmetry it cannot be improved.

Allan C. Wechsler <ACW@YUKON.SCRC.Symbolics.COM> analyzed his own
cube-solving method, finding that:

> For example, flipping two edges in place takes 22 qtw.

This can be done in 16 QT, though I don't know if that is the best
known.  Any pair can be flipped with a conjugate of the 14 QT slice
mono-op FOTAROFATO-RAM TAFORATOFA-ROM (FT'RF'T L'R B'TR'BT' LR').
Adjacent and antipodal pairs require the introduction of a
non-cancelling QT in the conjugator.

> Obviously a lot could be gained from tweaking the earlier part of
> the algorithm to guarantee that I don't need to do this at the end.

Probably, but it's hard to make that guarantee.  The problem is that
unless you flip edges in place with no other action (the very problem
you're trying to avoid) you may affect the later choices in the
algorithm, making the earlier tweaks wrong for that branch of the
algorithm.

For instance, the 7-QT method my program found solves the orientation
of all the edges (using a particular non-standard labeling of the
orientation that, when solved, is invariant under F^2, B^2, L, R, T,
and D).  But it may permute edges, and permute and twist corners, so
it may not form a useful part of an arbitrary cube-solving algorithm.
It works in Thistlethwaite's only because he is careful in all
branches of the rest of the algorithm never to mix up the orientation
of those edges.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From raymond@cps.msu.edu  Tue Jan 14 13:22:14 1992
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From: raymond@cps.msu.edu
Message-Id: <9201141822.AA01898@cpss16.cps.msu.edu>
To: cube-lovers@ai.mit.edu
Subject: Cube software


If anyone is interested, I wrote a program for examining the cycle
structure of various move sequences on Rubik's cube.  It's got
a lex and yacc front end, which let you enter moves using the
UDLRFB notation.  You give it a move sequence, and it will give
you the permutation in cycle notation, taking edge flips and
corner twists into account.  For example, you can say
  (R'D2R B'U2B)2
which is a corner twister, and it will tell you that the urf corner
is twisted clockwise, and the dlb corner is twisted clockwise.  YOu
can also give names to sequences, and refer to the sequence by its
name.  You can save and load named sequences from a file.

The code is pretty quick-and-dirty, but I'll email the source to
anyone who is interested.  I wrote it on a PC with Microsoft C 5.1, and
flex and bison, but it should work fine under Unix.

Carl Raymond

From ACW@yukon.scrc.symbolics.com  Tue Jan 14 19:20:46 1992
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Date: Tue, 14 Jan 1992 13:44-0500
From: Allan C. Wechsler <ACW@yukon.scrc.symbolics.com>
Subject: A new upper bound: 91 QT
To: hoey@aic.nrl.navy.mil, Cube-Lovers@life.ai.mit.edu
Cc: ACW@yukon.scrc.symbolics.com
In-Reply-To: <9201141749.AA27091@sun13.aic.nrl.navy.mil>
Message-Id: <19920114184432.1.ACW@PALLANDO.SCRC.Symbolics.COM>

Regarding the meta-approach of descending through a series of subgroups,
how much leverage does properly selecting the chain give you?  It seems
like the jump from <F2,B2,...> to <F2,B2,L2,R2,...> is pretty large.
There are probably other paths through the subgroup lattice.

Does anyone have a table of subgroups?

From @mitvma.mit.edu:rb%uk.ac.ic.cc@sunss1cc-gw.cc.ic.ac.uk  Wed Jan 15 10:38:59 1992
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From: rb@cc.ic.ac.uk
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To: cube-lovers <cube-lovers@ai.mit.edu>
Subject:    Minimove Solution
Sender: rb%uk.ac.ic.cc@sunss1cc-gw.cc.ic.ac.uk

I have recently read a book entitled
	"Learning To Solve Problems By Searching For Macro Operators"

	by Richard E. Korf (exact spelling not in my head).

In a nutshell the book discusses an algorithmic method of problem soving
by discovering useful operators.  The method was applied to the 3D Rubik
cube and as I remember managed on average to solve the problem in about
37 - 38 QTW.  Apparently it was slightly better than human experts.  The
tables discovered by the program weren't terribly large :-)!

Also either in that book or another I remember an approach to the minimove
problem based on measuring the disorderedness of the cube after n random
moves.  The measure of disorder was based on the number of correct colours
on each face or something like that.  From a graph of this measure averaged
over (I suppose) some number of trials it seems as though the cube can be
maximally disordered in around 18 moves.  It would seem that this means
that on average the cube should be restorable in about the same number of
moves.

Of course this doesn't help giving tight bounds,  but I guess it gives
some hope to the clever guys.

As an aside I have something called  a Supernova which is dodekahedral
(i.e. has 12 5 sided faces each of which can rotate) which is alledged
to be 10power 43 more complicated than the 3cube.  I can solve it using
fairly standard cube methods (only the last face is slightly difficult)
in about 20 times my 3-bube time.  Has anyone else seen this and or is
there any standard notation + information on this thing.
	Robin (not batperson) Becker

From raymond@cps.msu.edu  Wed Jan 15 11:07:35 1992
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From: raymond@cps.msu.edu
Message-Id: <9201151607.AA01817@cpss11.cps.msu.edu>
To: cube-lovers@ai.mit.edu
Subject: Cube software

I received 8 requests for my cube cycle program.  They are:
berson@cs.pitt.edu
bosch@mips.com
DEVO@GDLVM7.VNET.IBM.COM
keng@zcar.asd.sgi.com
tout@cps.msu.edu
palmerp@MATH.ORST.EDU
schmidtg@iccgcc.decnet.ab.com
tjj@rolf.helsinki.fi

If I omitted anyone, please send your request again.

Carl

From gkomatsu@uhunix.uhcc.hawaii.edu  Thu Jan 16 04:08:17 1992
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Date: Wed, 15 Jan 1992 23:08:12 HST
From: Galen Tatsuo Komatsu <gkomatsu@uhunix.uhcc.hawaii.edu>
To: cube-lovers@life.ai.mit.edu
Subject: Rubik's Magic Clock & Triamid
Message-Id: <CMM.0.88.695552892.gkomatsu@uhunix.uhcc.Hawaii.Edu>

...hmm new to this list, been reading some of the postings and things just
seem to go *FOOM*, right over my head.  But it didn't stop me from asking
these.....

Rubik's Magic Clock.
	Sister is Japan sent me this for Christmas (Have still yet to
translate the instructions...as if I need it!) and solved it in a day... Well,
more like "stumbled" upon the solution after a day of fiddling with it.  But
I continued to play with it, and came upon this.....
	Sometimes when I give one of the wheels a good quick "flick", one of
the gears inside slips.  Result is one (or maybe two) of the clock faces
affected is an hour "behind" (or ahead) of the others.  Deep in my mind I
concluded that this rendered the puzzle unsolveable.  And I ended up pulling
out a screwdriver and readjusting the face (or I just "zeroed" all of them.)
Was I correct in this conclusion?
	(...oh yea, she sent this for Christmas '88, not this past year.  I'm
not THAT behind the times!  THIS Christmas I recieved.....)

Rubik's Triamid.
	In the instructions it says,

       "It is physically possible to dismember Triamid into it's
	10 constituent elements and reassemble it into a complete
	Triamid.  A word of warning however--as there are 2 possible
	ways of doing this (a right and a left one) solving the
	puzzle after such a ressembly has an additional sting in
	the tail."

	What exactly is this "sting"?  And what did it mean by "right and
left"?  (if there's some joke here, I missed it...)  I was wondering, if I
took it apart and reassembled it to a completed form, the puzzle is still
solvable, I just scramble it and get back to the form I reassembled it to.
So this can't be the "sting" mentioned.  Unless it meant reassemble it to some
unfinished form.
	Next question... Sometimes when I play around with it, one of the
corner pieces pops off and lands on the floor.  I pick it up and put it back
on wondering, how was it originally oriented?  And considering the 11/12
chance that I'll have put it back on wrong way.  Have I just rendered the
Triamid (once again) "unsolveable"?
	Final question, for fun...  Anyone bought more than one Triamid, and
put 'em together to make a "Monster(a)mid"? =^)

	...also for a Square-1 for x-mas too, still fiddling around with it
too.  And have yet to lay my hands on Rubik's Tangle, Dice, and Fifteen.  But
NOT the Cube^4 (was that it?) Couls never solve the original, why should I
touch this one?  =^/

Galen Komatsu
gkomatsu@uhunix.uhcc.hawaii.edu !

From tjj@lemma.helsinki.fi  Thu Jan 16 07:15:27 1992
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Date: Thu, 16 Jan 92 14:14:15 +0200
From: tjj@lemma.helsinki.fi (Timo Jokitalo)
Subject: Re:  Rubik's Magic Clock & Triamid
To: cube-lovers@life.ai.mit.edu, gkomatsu@uhunix.uhcc.hawaii.edu
Message-Id: <9201161214.AA28342@lemma.helsinki.fi>

I haven't tried the Triamid or Tangle, but I bought the Fifteen and the Dice.
I'm still thinking of attacking the Fifteen, but I really hate the Dice after
I twice had gotten four of the faces in their right places, knocked the thing
a bit too much, which caused the plates to fall to the bottom. I hate that 
**** thing and won't touch it again for a long time!

	Timo

From pbeck@pica.army.mil  Wed Jan 22 00:26:53 1992
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Date:     Tue, 21 Jan 92 15:18:15 EST
From: Peter Beck (BATDD) <pbeck@pica.army.mil>
To: reid@math.berkeley.edu
Cc: cube-lovers@life.ai.mit.edu
Subject:  cff
Message-Id:  <9201211518.aa19961@FSAC1.PICA.ARMY.MIL>


CUBISM FOR FUN is the newsletter of the 
DUtch Cubist Club.  It is in english.
The club is alive and well and is planning
to host the 1993 International Puzzle party.

To subscribe send US$11 (in cash or
international money order) to
LUCIEN MATTHIJSSE
LOENAPAD 12
3402 EP IJSSELSTEIN
NETHERLANDS

I will be happy to answer any questions you may have.

THE FUTURE IS PUZZLING,
BUT CUBING IS FOREVER !!

From reid@math.berkeley.edu  Wed Jan 22 02:29:03 1992
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Date: Tue, 21 Jan 92 23:27:28 PST
From: reid@math.berkeley.edu (michael reid)
Message-Id: <9201220727.AA03361@math.berkeley.edu>
To: cube-lovers@ai.mit.edu
Subject: Re: Rubik's cube dice tops (Spoiler)
Cc: ronnie@cisco.com

a while ago (last month), Dan (hoey@aic.nrl.navy.mil) writes:

> Last week ronnie@cisco.com (Ronnie Kon) challenged us to find Rubik's
> cube patterns with dice pips for 1, 2, and 3 on the three pairs of
> opposite sides.  He claimed it could be done in fourteen HST, where
> one HST is a turn of a face or center slice by 90 or 180 degrees.  I
> responded that it could be done in thirteen HST.  Here is how.  I will
> use this opportunity to practice the enhanced Varga Rubiksong I
> described (unfortunately with many typos) on 22 Feb 90.

> The (only such) pattern is the composition of Four-Spot and Laughter.
> We have long known the processes

[ description deleted ]

> This uses only 13 HST.  This is also the shortest process I know of in
> the normal metric: 18 QT, which is not so bad for the combination of

here's a shorter way.  in the "flubrd" notation, use:

              D' F^2 R U^2 F^2 B^2 D^2 R^2 L' F^2 U' D^2

which is 11 "HST" (which i call "slice turns").  this is also 12
"face" turns, but 20 quarter turns.  this can also be done in only 14
quarter turns as follows:
                            F^2 U D F B U D F B U D F B'     (*)

note that this can easily be obtained from the well-known manuever for
"laughter":
                              ( F B  C_U )^6                (**)

where  C_  means "turn the whole cube" (as in Bandelow's book).  note
that this manuever reorients the cube.  then manuever (*) is just the
"flubrd" translation of the manuever  M_F (**) M_F'  (without the cube
reorientation), where  M_  means "turn the middle slice," again, as in
Bandelow's book.

here's a question for those out there with 5x5x5 cubes: have you
noticed that the stickers seem to be more happy on the floor than on
the facelets of the cube?  the more i use my cube, the more restless
they seem to become.  does anyone know of a good cure for this?  i'm
thinking of taking them all off, cleaning off the glue (or gum or
whatnot) and gluing them back on, using a stronger glue.  anyone have
any suggestions for what kind of glue?  i'll let you know how my
experiment works.

mike

From wft@math.canterbury.ac.nz  Wed Jan 29 04:27:21 1992
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Date: Wed, 29 Jan 92 17:00:01 NZD
From: wft@math.canterbury.ac.nz (Bill Taylor)
Subject: subgroups
To: Cube-Lovers@ai.mit.edu
Cc: wft@cantva.canterbury.ac.nz
Message-Id: <9201290400.AA29423@math.canterbury.ac.nz>
X-Envelope-To: Cube-Lovers@AI.MIT.EDU

On 14 Jan 1992,  Allan C. Wechsler posted

>Regarding the meta-approach of descending through a series of subgroups,
>how much leverage does properly selecting the chain give you?  It seems
>like the jump from <F2,B2,...> to <F2,B2,L2,R2,...> is pretty large.
>There are probably other paths through the subgroup lattice.
>
>Does anyone have a table of subgroups?

There hasn't been any response to this, seemingly, which is a pity.

In any event, I would like to know of any other well-known subgroups.
There are the slice group; double-slice group; U group; square group;
anti-slice group. How many others are there not mentioned here, that 
people know of ?



From pbeck@pica.army.mil  Wed Feb 26 07:50:17 1992
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Date:     Wed, 26 Feb 92 7:42:09 EST
From: Peter Beck (BATDD) <pbeck@pica.army.mil>
To: cube-lovers@ai.mit.edu
Subject:  news
Message-Id:  <9202260742.aa03535@FSAC1.PICA.ARMY.MIL>


What's UP

-  GAMES magazine is sponsoring 
  "WORLD PUZZLE TEAM CHAMPIONSHIPS" 
   in NYC June 24-28;  registration forms in 
   April issue of GAMES, on newsstands march 1


-  NEW PUZZLE
It is called "PYRIX" (retails for $10, 
   I do not have a retail source) and is from: 
   Enpros Novelty Products, Lorentzstraat 2, 
   2912 AH Niewerkerk aan den IJssel,
   The Netherlands -  
   tel 31 (0)1803-19133

DESCRIPTION:  The puzzle is an assembly folding
  puzzle based on a size 3 tetrahedron.  The 
  tetrahedron is dissected into 3 regular octahedrons
  and 11 tetrahedrons (1/3 the size of the original).
  These pieces are strung on a thread like a necklace;
  an octahedron, 3 tetras, an octahedron, etc except
  for one position that has only 2 tetras.  The octahedrons
  are threaded on the diagonal of their square cross section.
OBJECT:  The faces are colored and the object is to not 
  only assemble a tetra but of course to do it with solid
  colored faces,  the enclosure says that there are 
  2 solutions as they have colored and strung the pieces.
PRELIMINARY REVIEW:  It took about 1 hour.  The puzzle 
  is awkward to manipulate since it falls apart easily.
  Doing it on a flat surface and using tape to hold it
  together seems to be the trick.



From pbeck@pica.army.mil  Fri Mar 13 19:31:17 1992
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Date:     Fri, 13 Mar 92 13:57:41 EST
From: Peter Beck (BATDD) <pbeck@pica.army.mil>
To: cube-lovers@ai.mit.edu
Subject:  12 th puzzle party
Message-Id:  <9203131357.aa04103@FSAC1.PICA.ARMY.MIL>


...............................................................
<-->  12th International puzzle collector's party and fair " UPDATE"
 <-->
 transcribed by pbeck, 3/13/92
...............................................................
WHEN ---- 7/31/92  - 8/2
WHERE ----  Korakuen Kaikan
LODGING ---- about $70 per night at either
    1) Satellite Hotel Korakuen
    2) Koraku Garden Hotel
              ---- about $20 per night at Tokyo International Youth
Hostel, 10 minute walk away but next to Nob's Studio

***  INVITATIONS ***  Admission by invitation only!!!  
     Contact: Mr. Nob Yoshigahara,  4-10-1-408 Iidabashi,  Tokyo  102
Japan before   APRIL 15, 1992

AGENDA: (final details,ie, cost, food service & entertainment will be
sent  to registrants)
  7/31  18:00 - 20:00 welcoming party 
  8/1     10:00 - 20:00 collectors and dealers
  8/2 unfinished business
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^


From pbeck@pica.army.mil  Tue Apr 14 12:15:09 1992
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Date:     Tue, 14 Apr 92 7:48:14 EDT
From: Peter Beck (BATDD) <pbeck@pica.army.mil>
To: cube-lovers@ai.mit.edu
Subject:  puzzle rings
Message-Id:  <9204140748.aa18790@COR4.PICA.ARMY.MIL>


Jewelry quality puzzle rings are available from:

as of 4/91,

JOSE GRANT
3000 SUMMER STREET
STANFORD, CT 06905

203-327-4055
800-426-1947


I am posting this to correct mis-information I 
may have sent out.  If this is incorrect
please let me know and I will track down the
corrrect info.

THE FUTURE IS PUZZLING,
but CUBING is FOREVER!!

pete beck

From wft@math.canterbury.ac.nz  Wed Apr 15 01:19:50 1992
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Date: Wed, 15 Apr 92 17:19:18 NZS
From: wft@math.canterbury.ac.nz (Bill Taylor)
Subject: God's algorithm
To: Cube-Lovers@ai.mit.edu
Message-Id: <9204150519.AA13903@math.canterbury.ac.nz>
X-Envelope-To: Cube-Lovers@AI.MIT.EDU

In rec.puzzles,  sijben@cs.utwente.nl (Paul Sijben) writes:

> As far a I know is the maximum number of moves requierd to solve The
> Cube is just over 30 (35 by the last count a year ago, and decending).
> Someone in the NKC (Nederlandse Kubus Club= dutch cube club) was busy
> working on a system hoping to reach god's algorythm. I can dig in my
> archives if anyone want more precice infomation.

Has anyone else heard anything of this business ?


From reid@math.berkeley.edu  Wed Apr 15 01:54:00 1992
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Date: Tue, 14 Apr 92 22:53:57 PDT
From: reid@math.berkeley.edu (michael reid)
Message-Id: <9204150553.AA29772@math.berkeley.edu>
To: Cube-Lovers@ai.mit.edu, wft@math.canterbury.ac.nz
Subject: Re:  God's algorithm

> From: wft@math.canterbury.ac.nz (Bill Taylor)
> Subject: God's algorithm

> In rec.puzzles,  sijben@cs.utwente.nl (Paul Sijben) writes:
> 
> > As far a I know is the maximum number of moves requierd to solve The
> > Cube is just over 30 (35 by the last count a year ago, and decending).
> > Someone in the NKC (Nederlandse Kubus Club= dutch cube club) was busy
> > working on a system hoping to reach god's algorythm. I can dig in my
> > archives if anyone want more precice infomation.
> 
> Has anyone else heard anything of this business ?

well, it's been kicking around in both rec.puzzles and sci.math lately.
maybe you should ask if anyone has heard any follow-up to the above.
i've sent away to the dutch cubists' club for membership and info,
but they want payments in the form of international money orders
(which probably means a good month or two delay).
if/when i have some more info on this, i'll gladly share it with
cube-lovers.

mike

From ccw@eql.caltech.edu  Sat Apr 18 23:17:20 1992
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From: ccw@eql.caltech.edu (Chris Worrell)
Message-Id: <920418201450.2b6009a5@EQL.Caltech.Edu>
Subject:  Some solutions to Rubik's Tangle have been found.
To: Cube-Lovers@ai.mit.edu

(No spoilers are included)

I now know two distinct solutions to Rubik's Tangle.
Each of these solutions can be done with each 5x5 set, 1-4.

I have not yet found any solutions to the 10x10 puzzle. I do know
that if the 10 by 10 is composed of four 5x5 solutions, than my solutions
of the 5x5 do not lead to a solution of the 10x10. I am now seeking a
solution where all 100 pieces can be used anywhere in the 10x10, not just in
a corner of the 10x10 for its own 5x5 set.

Unfortunately, these solutions were found by brute force, not by any
real calculation.  I have not been able to discover any "Science" about
the puzzle which contributes to any discovery of a solution.

These solutions were found by hand, not by a computer search. So there may be
other solutions to the basic puzzle which are still unknown.
My search path has approximately 750 cases in it, of which I have
tested about 300.  One solution was found by 'accident' in that I have not
yet looked at the case which actually yields that solution, but one of the
test cases had been 'close' to a solution, so I looked outside my search path.
The other solution was found in my search path, so it was not found by
accident. (except by luck that it was at case 300 not 750)

GENERAL THOUGHTS ON RUBIK'S TANGLE AS A PUZZLE
In short, it is not a good puzzle.  It will never be popular.
Solutions might only be findable by Computer, by Luck, and by Stubborness
(brute-force).
As far as I can tell the only real method of solution is by using a computer.
I did it by hand because I am stubborn, and I did get lucky.
My search path contained 750 cases, but I had already considered search paths
with an estimated 4000 and 8000 cases. These are so large that I had never
even completed the basic enumeration of cases.
A hand search of this magnitude is almost impossible.  There are too many
places for errors, and no real ways of checking.  The amount of time is also
absurd.  I worked on between 1 and 8 of my test cases at a time, with
about 250 of these groups in my search path. (later in
the search I would have, on ocassion, looked at 16 at once).  Sometimes a
group could be disposed of within 20 minutes, but sometimes it took several
hours.

Has anybody discovered more mathematical content than I have?

---Chris Worrell (ccw@eql.caltech.edu)

From pl1x+@andrew.cmu.edu  Sat Apr 18 23:43:08 1992
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Date: Sat, 18 Apr 1992 23:41:45 -0400 (EDT)
From: Peter Andrew Lopez <pl1x+@andrew.cmu.edu>
To: cube-lovers@ai.mit.edu, ccw@eql.caltech.edu (Chris Worrell)
Subject: Re: Some solutions to Rubik's Tangle have been found.
Cc: 
In-Reply-To: <920418201450.2b6009a5@EQL.Caltech.Edu>
References: <920418201450.2b6009a5@EQL.Caltech.Edu>

I love cubes

But i'll never admit it!

cube-annonymous


From Don.Woods@eng.sun.com  Sun Apr 19 02:58:44 1992
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Date: Sat, 18 Apr 92 23:58:42 PDT
From: Don.Woods@eng.sun.com (Don Woods)
Message-Id: <9204190658.AA04381@colossal.Eng.Sun.COM>
To: ccw@eql.caltech.edu, cube-lovers@ai.mit.edu, pl1x+@andrew.cmu.edu
Subject: Re: Some solutions to Rubik's Tangle have been found.

> From: Peter Andrew Lopez <pl1x+@andrew.cmu.edu>
> I love cubes
> But i'll never admit it!
> cube-annonymous

In addition to being self-contradicting (and misspelled), the above seems
to have nothing to do with the subject of Rubik's Tangle.

Lest this message suffer the same flaw, I'll add that I too was unable to
come up with any mathematical or intuitive method for solving the Tangle.
I solved mine by computer.  (I've always been fairly good at finding ways
to prune a bushy search tree down to manageable size.)  I have Tangle #1
and can confirm it has exactly two solutions (ignoring overall rotations
of the 5x5 array, of course).

I haven't had a chance to examine closely the other Tangles.  How do they
differ from #1?  Do they use a different pattern of connectivity on the
tiles?  Do they have a different mix of the permutations?  (#1 has each
4-color permutation exactly once, except for one permutation which appears
twice.)  I hope they do not simply permute the colors relative to #1; that
would be dull since they would then be identical puzzles, and collecting
more than one would be silly except for the purpose of building the 10x10
combined puzzle.

	-- Don.

From ACW@riverside.scrc.symbolics.com  Fri Apr 24 14:34:12 1992
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Date: Fri, 24 Apr 1992 14:33-0400
From: Allan C. Wechsler <ACW@riverside.scrc.symbolics.com>
Subject: Some solutions to Rubik's Tangle have been found.
To: ccw@eql.caltech.edu, Cube-Lovers@ai.mit.edu
In-Reply-To: <920418201450.2b6009a5@EQL.Caltech.Edu>
Message-Id: <19920424183328.2.ACW@TRANTOR.SCRC.Symbolics.COM>

I hope I'm not wasting too many people's time, but... can you describe
the Rubik's Tangle puzzle for those of us who haven't seen it?  Your
description was interesting, but I wonder about your statement that it
can't be solved without a computer.  Perhaps you just didn't have the
right insight.

From Don.Woods@eng.sun.com  Fri Apr 24 17:56:26 1992
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Date: Fri, 24 Apr 92 14:56:22 PDT
From: Don.Woods@eng.sun.com (Don Woods)
Message-Id: <9204242156.AA22010@colossal.Eng.Sun.COM>
To: ACW@riverside.scrc.symbolics.com
Subject: Description of Rubik's Tangle
Cc: Cube-Lovers@ai.mit.edu

> From: Allan C. Wechsler <ACW@riverside.scrc.symbolics.com>
> I hope I'm not wasting too many people's time, but... can you describe
> the Rubik's Tangle puzzle for those of us who haven't seen it?  Your
> description was interesting, but I wonder about your statement that it
> can't be solved without a computer.  Perhaps you just didn't have the
> right insight.

I would love to hear an insight that makes this puzzle tractible in real
time (hours rather than days) by hand.  Here's a brief description of
Tangle #1; as I said in my earlier posting, I don't know how the other 3
differ, though I'm pretty sure they each have the same number of tiles.

Tangle #1 consists of 25 square tiles, each of which has four colored
ropes crossing it in the following pattern.  (Note: This may be mirror
imaged, since I'm working from memory.)

	---------------------
	|     @       #     |
	|     @       #     |
	|$$    @      # %%%%|
	|  $    @    %#%    |
	|   $    @ %% #     |
	|    $    %@  #     |
	|    $  %%  @@#     |
	|    %%%      #@@   |
	|%%%% $       #  @@@|
	|     $       #     |
	|     $       #     |
	---------------------

The connection marked with $ actually wanders around the tile a bit more,
but the connectivity is as shown.  The object is to place the tiles in a
5x5 array such that wherever two tiles touch the colors of the ropes match.
In Tangle #1 each permutation of colors occurs once, with one permutation
occurring twice.

The box says that if you get all four Tangles, you can put them together
to make a 10x10 array under the same color-matching constraints.

	-- Don.

From ccw@eql.caltech.edu  Tue Apr 28 01:59:05 1992
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Date:     Sat, 25 Apr 92 08:47:55 PDT
From: ccw@eql.caltech.edu (Chris Worrell)
Message-Id: <920425084746.2bc000e4@EQL.Caltech.Edu>
Subject:  Description of Tangle, Part 2
To: cube-lovers@ai.mit.edu
Cc: don.woods@eng.sun.com, acw@riverside.scrc.symbolics.com

Annotating Don.Woods diagram  (which is in the correct orientation)
              2       3
        ---------------------
        |     @       #     |
        |     @       #     |
      1 |$$    @      # %%%%| 4
        |  $    @    %#%    |
        |   $    @ %% #     |
        |    $    %@  #     |
        |    $  %%  @@#     |
        |    %%%      #@@   |
      4 |%%%% $       #  @@@| 2
        |     $       #     |
        |     $       #     |
        ---------------------
              1        3

The duplicate piece in each tangle is:
                1       2       3       4
Tangle 1        Blue    Red     Yellow  Green
Tangle 2        Yellow  Blue    Green   Red
Tangle 3        Green   Yellow  Blue    Red
Tangle 4        Red     Green   Yellow  Blue

All 4 Tangles are the same puzzle, just colored differently.
Each has all 24 color permutations, plus a duplicate.

Each Tile also has a letter (A-Y) on the back, and a reference to the
Tangle that it occurs in.  These letters appear to be for reference
only.  I have found no corresponddence between the letterings in one
puzzle, and the letterings in another.  I just use them as a convenience
for recording configurations, and sorting through the tiles.

------
ccw@eql.caltech.edu (Chris Worrell)

From reid@math.berkeley.edu  Wed Apr 29 04:37:32 1992
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Date: Wed, 29 Apr 92 01:37:26 PDT
From: reid@math.berkeley.edu (michael reid)
Message-Id: <9204290837.AA13934@math.berkeley.edu>
To: cube-lovers@ai.mit.edu
Subject: an upper bound on god's number

in an earlier message, i promised to pass along any information
i obtained about progress on the upper bound for the length of
god's algorithm.  i've received a copy of the article "rubik's
cube in 42 moves" by hans kloosterman, which i summarize below.

first here are some caveats:

* i haven't verified this algorithm.
* throughout, `move' refers to any turn of a single face.
  i don't know what bound is achieved in the quarter-turn metric.
* it may be the case that this algorithm has been improved.
  please let me (and cube-lovers) know if you have more information.


"rubik's cube in 42 moves"  by  hans kloosterman

the algorithm used here is a slight variant of thistlethwaite's
algorithm.  we work through a chain of subgroups:

G_0 = < F, B, L, R, U, D >   ( full group )

G_1 = < F2, B2, L, R, U, D >

G_2 = < F2, B2, L2, R2, U, D >

G_3 = subgroup of G_2 in which all U-cubies are in the U face
      (thus all D-cubies are in the D face and all U-D slice
      cubies are in the U-D slice)

G_4 = { 1 }.

there are four stages:  stage i reduces our pattern from G_{i-1} to G_i.

the indices of the subgroups are:

( G_0 : G_1 ) = 2048 = 2^11
( G_1 : G_2 ) = 1082565 = 3^9 * 5 * 11
( G_2 : G_3 ) = 4900 = 2^2 * 5^2 * 7^2
( G_3 : G_4 ) = 3981310 = 2^14 * 3^5

the maximum number of moves in the stages are 7, 10, 8 and 18
respectively, for a maximum total of 43 moves.  however, in
the worst-case scenario of stages 3 and 4, it was checked that
no position actually required 26 moves; i.e. we can arrange a
cancellation between the two stages.  thus stages 3 and 4
together require 25 moves at most, which gives the final
figure of 42 moves.

it seems to me that a lot of work was done on an algorithm
for restoring the "magic domino" (the 2x3x3 puzzle), and
these results were applied to stages 3 and 4.


mike

From tjj@lemma.helsinki.fi  Wed Apr 29 05:39:47 1992
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Date: Wed, 29 Apr 92 12:38:45 +0300
From: tjj@lemma.helsinki.fi (Timo Jokitalo)
Message-Id: <9204290938.AA20924@lemma.helsinki.fi>
To: cube-lovers@ai.mit.edu, reid@math.berkeley.edu
Subject: Re: an upper bound on god's number

Do you have the article by Hans Kloosterman in emailable form? If not,
where could I obtain a copy?

	Timo
	tjj@rolf.helsinki.fi

From dik@cwi.nl  Sun May  3 21:43:51 1992
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From: Dik.Winter@cwi.nl
Message-Id: <9205040143.AA00557.dik@boring.cwi.nl>
To: cube-lovers@life.ai.mit.edu
Subject: Are we approaching God's algorithm?

Because it might interest the readers and to be ahead of Peter Beck:
Saturday I received CFF (Cubism For Fun) # 28.
It has an interesting article by Herbert Kociemba from Darmstadt, who
describes his program to solve Rubik's Cube.  He states that he has not
yet encountered a configuration that required more than 21 moves.  A short
description follows:

Basicly the program consists of two stages, based on the groups:
G0:	[U,D,R,L,F,B]
G1:	[U,D,R^2,L^2,F^2,B^2]
G2:	I
The stages are from G0 to G1 and next from G1 to G2.  Note that the first
stage is the combination of the first two stages of Thistlethwaite, and
the last stages combine his last two stages.

His first stage can loosely be described as working in a three dimensional
coordinate system where the coordinates are resp. twist, flip and permutation.
He searches his way until the coordinate [0,0,0] is reached.  Most important
here is that he is able to find multiple ways.  The second stage is similar,
although he is not very clear here.

He uses lookup tables, but does not tell us how large his lookup tables
are.  But his program runs on 1 MByte Atari ST.  The heart is coded in
a few lines of 68k assembly, the remainder in GFA Basic.  As far as I
know GFA Basic it can be interpreted, but also compiled.  He does also
use tree pruning.

What he does is find successive solutions in stage 1 and fit solutions
from stage 2.  Letting the system run longer generally finds shorter
solutions.

His claims are on average less than 14 turns in stage 1, on average less
than 10 turns in stage 2.  But according to his article this is not completely
deterministic, so there is no proven upperbound.  Perhaps a proof can be
found; I do not know.  In practice he finds an upperbound of 21 moves, which
is indeed far better than other algorithms do obtain.

To take this in perspective here Thistlethwaites results from Singmaster's book:
Stage		1	2	3	4
Proven		7	13	15	17
Anticipated	7	12	14?	17
Best Possible	7	10?	13?	15?
(Are there configurations that require the maxima proven for Thistlethwaites
algorithm?)

Apparently the combination of stages largely reduces the number of turns
required.

In CFF 25 there was an article by Hans Kloosterman which did already improve
on the number of moves.  His stages 1 and 2 are identical to Thistlethwaites,
he has a third stage that combines the 3rd and 4th stages of Thistlethwaite.
He reports a proven maximum for his three stages of 7, 10 and 25 moves, so
slightly better than Thistlethwaites conjectured best figures.

Kociemba's algorithm appears however to be a big leap forward, although there
are no proofs yet.  One example:

In 1981 Christoph Bandelow wrote a book where he offered a prize for the
shortest sequence of moves that would flip every edge cuby and twists
every corner cuby.  The deadline was September 1, 1982; at that time the
prize was offered for a 23 move manoeuvre.  As Christoph writes:
	All solutions presented after the main deadline and shorter than
	all solutions submitted before were also proised a prize.  Using
	his very ingeneous new search program Herbert Kociemba, Darmstadt,
	Germany now found:
		DF^2U'(B^2R^2)^2LB'D'FD^2FB^2UF'LRU^2F'
	for 20 moves.
Kociemba himself writes about this:
	Our first solution had 12 moves in stage 1 and 14 moves in stage 2.
	In progress solutions 12+13, 12+12 and 12+11 were found.  However,
	as soon as we introduced manoeuvres of 13 moves in stage 1, we found
	successively 9, 8 and at last 7 moves for stage 2.  The program was
	stopped after treating about 3000 solutions.
He further states that the first solution in general takes 95 seconds, but
successive solutions take much shorter, and in general he finds one of less
than 22 moves in a few hours on his 8 MHz Atari.

What is clear is that one does not need the minimal solution in one stage
to get the minimal overall total.

dik
--
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
dik@cwi.nl

From reid@math.berkeley.edu  Mon May  4 23:38:05 1992
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Date: Mon, 4 May 92 20:37:54 PDT
From: reid@math.berkeley.edu (michael reid)
Message-Id: <9205050337.AA06009@math.berkeley.edu>
To: Dik.Winter@cwi.nl, cube-lovers@life.ai.mit.edu
Subject: Re:  Are we approaching God's algorithm?
Cc: dseal@armltd.co.uk

<dik@cwi.nl> writes:
> Because it might interest the readers and to be ahead of Peter Beck:
> Saturday I received CFF (Cubism For Fun) # 28.
> It has an interesting article by Herbert Kociemba from Darmstadt, who
> describes his program to solve Rubik's Cube.  He states that he has not
> yet encountered a configuration that required more than 21 moves.  A short
> description follows:

it would be nice to know how many patterns he has tested.

> Basicly the program consists of two stages, based on the groups:
> G0:	[U,D,R,L,F,B]
> G1:	[U,D,R^2,L^2,F^2,B^2]
> G2:	I
> The stages are from G0 to G1 and next from G1 to G2.  Note that the first
> stage is the combination of the first two stages of Thistlethwaite, and
> the last stages combine his last two stages.
>
> His first stage can loosely be described as working in a three dimensional
> coordinate system where the coordinates are resp. twist, flip and permutation.
> He searches his way until the coordinate [0,0,0] is reached.  Most important
> here is that he is able to find multiple ways.  The second stage is similar,
> although he is not very clear here.
>
> He uses lookup tables, but does not tell us how large his lookup tables
> are.  But his program runs on 1 MByte Atari ST.  The heart is coded in
> a few lines of 68k assembly, the remainder in GFA Basic.  As far as I
> know GFA Basic it can be interpreted, but also compiled.  He does also
> use tree pruning.

does he describe his method of "tree pruning"?  this would seem to
be the "intelligent" part of the program, i.e. recognizing when to
abandon a given approach.  if anyone has any insight on the tree
pruning, please let me know.

> What he does is find successive solutions in stage 1 and fit solutions
> from stage 2.  Letting the system run longer generally finds shorter
> solutions.
>
> His claims are on average less than 14 turns in stage 1, on average less
> than 10 turns in stage 2.  But according to his article this is not completely
> deterministic, so there is no proven upperbound.  Perhaps a proof can be
> found; I do not know.  In practice he finds an upperbound of 21 moves, which
> is indeed far better than other algorithms do obtain.

it's not likely that this will lead to a proof of an effective upper
bound.  perhaps he can shave a few moves off the 42 obtained by
kloosterman, but i wouldn't expect him to prove an upper bound
anywhere near 21.  probably the best bet for this would be to
exhaustively calculate the diameter of the group  G_1  (with the
given generators) and the diameter of the coset space  G_0 / G_1.
their respective sizes are:  19508428800  and  2217093120,  both of
which are out of my league.

i'm not belittling kociemba's program; it's a very impressive feat!

> To take this in perspective here Thistlethwaites results from Singmaster's book:
> Stage		1	2	3	4
> Proven		7	13	15	17
> Anticipated	7	12	14?	17
> Best Possible	7	10?	13?	15?
> (Are there configurations that require the maxima proven for Thistlethwaites
> algorithm?)

now look what happens when people use TABs!  :-}  (the "Proven" line
should be shifted to the left.)

i believe that the diameters of the respective coset spaces are exactly
those numbers listed in the "Best Possible" line.  can anyone confirm
this?  i've finally written a few programs, and those are the diameters
i get.  i'm surprised that thistlethwaite didn't just do an exhaustive
search on these coset spaces.  perhaps it's just a matter of not having
the technology when he did his work (~12 years ago).

> Kociemba's algorithm appears however to be a big leap forward, although there
> are no proofs yet.  One example:
>
> In 1981 Christoph Bandelow wrote a book where he offered a prize for the
> shortest sequence of moves that would flip every edge cuby and twists
> every corner cuby.  The deadline was September 1, 1982; at that time the
> prize was offered for a 23 move manoeuvre.  As Christoph writes:
> 	All solutions presented after the main deadline and shorter than
> 	all solutions submitted before were also proised a prize.  Using
> 	his very ingeneous new search program Herbert Kociemba, Darmstadt,
> 	Germany now found:
> 		DF^2U'(B^2R^2)^2LB'D'FD^2FB^2UF'LRU^2F'
> 	for 20 moves.

very nice.  now how about "superflip," and also "supertwist?"  these
are also reasonable candidates for antipodes of "START."  i know the
following manuever for "supertwist" (22 face / 30 quarter turns):
U F' U' (L R2 F2 B')^4  U F U'
(obtained by conjugating a manuever singmaster attributes to thistlethwaite)

> Kociemba himself writes about this:
> 	Our first solution had 12 moves in stage 1 and 14 moves in stage 2.
> 	In progress solutions 12+13, 12+12 and 12+11 were found.  However,
> 	as soon as we introduced manoeuvres of 13 moves in stage 1, we found
> 	successively 9, 8 and at last 7 moves for stage 2.  The program was
> 	stopped after treating about 3000 solutions.
> He further states that the first solution in general takes 95 seconds, but
> successive solutions take much shorter, and in general he finds one of less
> than 22 moves in a few hours on his 8 MHz Atari.

it would also be nice to know how long this first solution usually is.

from the figures i have (111207592 "different" sequences of 7 or fewer
twists and 167024 "different" sequences of 6 or fewer twists WITHIN G_1)
it's clear that exhaustive search is too cumbersome.  thus i reiterate
both my statement that the "tree pruning" algorithm is the essential
part of this program, and my desire to know more about it (i.e. for
implementation purposes.)

> dik
> --
> dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
> dik@cwi.nl

thanks for the info!

mike             reid@math.berkeley.edu

From dik@cwi.nl  Tue May  5 03:58:28 1992
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Date: Tue, 5 May 1992 09:57:53 +0200
From: Dik.Winter@cwi.nl
Message-Id: <9205050757.AA01813.dik@boring.cwi.nl>
To: cube-lovers@life.ai.mit.edu, reid@math.berkeley.edu
Subject: Re:  Are we approaching God's algorithm?
Cc: dseal@armltd.co.uk

 > > It has an interesting article by Herbert Kociemba from Darmstadt, who
 > > describes his program to solve Rubik's Cube.  He states that he has not
 > > yet encountered a configuration that required more than 21 moves.  A short
 > > description follows:

 > it would be nice to know how many patterns he has tested.

He does not say how many, but his article gives nine patterns that have been
published earlier in CFF for which he finds shorter answers.  Also more are
promised for future issues.

 > > His first stage can loosely be described as working in a three dimensional
 > > coordinate system where the coordinates are resp. twist, flip and permutation.
 > > He searches his way until the coordinate [0,0,0] is reached.
...
 > >                                                           He does also
 > > use tree pruning.

 > does he describe his method of "tree pruning"?  this would seem to
 > be the "intelligent" part of the program, i.e. recognizing when to
 > abandon a given approach.  if anyone has any insight on the tree
 > pruning, please let me know.

I can give some information.  What he does do is calculate in advance through
the three axis of his space the minimal number of moves needed to get at
[0,0,0].  This is used for tree pruning.  It obviously will not prune
everything (e.g. if you are at point [x,y,z] it may very well be that [x,?,?]
for other points requires less moves, and similar across the y and z
direction), but he tells that his pruning is very effective.  I do not know
how he prunes in the second case, because he does not completely describes
the coordinates of his second space, but I presume pruning is done in a
similar way.

 > it's not likely that this will lead to a proof of an effective upper
 > bound.  perhaps he can shave a few moves off the 42 obtained by
 > kloosterman, but i wouldn't expect him to prove an upper bound
 > anywhere near 21.
I think so too.  Moreover, it is difficult to take in account what he
found, namely that a minimal solution in the first stage does not guarantee
a minimal overall solution.

 > i believe that the diameters of the respective coset spaces are exactly
 > those numbers listed in the "Best Possible" line.  can anyone confirm
 > this?  i've finally written a few programs, and those are the diameters
 > i get.  i'm surprised that thistlethwaite didn't just do an exhaustive
 > search on these coset spaces.  perhaps it's just a matter of not having
 > the technology when he did his work (~12 years ago).
Well, apparently Thistlethwaite did not know that those were the diameters,
otherwise I have no explanation for the question marks as they appear in
Singmaster.

 >             now how about "superflip," and also "supertwist?"
I will try to contact him to see what he has to say about those.


From reid@math.berkeley.edu  Fri May  8 17:58:59 1992
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Date: Fri, 8 May 92 14:58:49 PDT
From: reid@math.berkeley.edu (michael reid)
Message-Id: <9205082158.AA21236@math.berkeley.edu>
To: Dik.Winter@cwi.nl, cube-lovers@life.ai.mit.edu
Subject: Re:  Are we approaching God's algorithm?
Cc: dseal@armltd.co.uk

>  >             now how about "superflip," and also "supertwist?"
> I will try to contact him to see what he has to say about those.

of course, these aren't exactly the patterns to test.  apply your
favorite quarter turn to either of these patterns, and you're
one move closer to START.

how do i know that we're one move closer to start?  the patterns
"superflip," "supertwist" and "superfliptwist" each have the
following three properties:

1. the group of symmetries of the pattern acts transitively on the
   set of "oriented" faces of the cube.
2. the pattern commutes with the square of each face turn.
3. the pattern is NOT in the subgroup generated by the squares of
   face turns.

now suppose that a pattern with the above properties requires
N  face turns to return to START.  let  A B C  be a minimal sequence
of face turns to solve this pattern, where  A, B,  and  C  are
subsequences such that:  A  consists only of squares of face turns,
B  is a quarter turn of some face and  C  is the rest of the sequence.
we can dissect the sequence into these three parts from hypothesis 3.
from hypothesis 2, the sequences  A B C  and  B C A  have the same
effect.  finally, from hypothesis 1, the quarter twist  B  may as well
be our favorite quarter twist.

see the hoey-saxe message on "symmetry and local maxima," for a good
discussion of this idea.  (in the archives, cube-mail-1, 14 dec 1980)

my apologies if this is obvious to everyone.

on the other hand, the kociemba algorithm isn't completely symmetric.
thus it may be wise for him to test 2 patterns: "super----" followed
by  U,  and "super----" followed by  R.  the tradeoff is testing 2
patterns for being 1 move closer.  i'd say this is probably a good trade.

now to correct something misleading i posted earlier:

) i believe that the diameters of the respective coset spaces are exactly
) those numbers listed in the "Best Possible" line.  can anyone confirm
) this?  i've finally written a few programs, and those are the diameters
) i get.  i'm surprised that thistlethwaite didn't just do an exhaustive
) search on these coset spaces.  perhaps it's just a matter of not having
) the technology when he did his work (~12 years ago).

oops!  i don't mean to say "diameter" here!  these are coset spaces, so
there's no reason to suppose that (the group of automorphisms of the)
graph is vertex transitive.  what my programs calculated was the maximal
distance from the identity coset in each of these spaces.  (i am told
that the graph theory term is the "eccentricity" of the given vertex.)

mike

From dik@cwi.nl  Sat May  9 20:43:35 1992
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From: Dik.Winter@cwi.nl
Message-Id: <9205100043.AA00802.dik@boring.cwi.nl>
To: cube-lovers@life.ai.mit.edu
Subject: More on the Cube (2x2x2 in this case).

Singmaster states that the diameter of the group for the 2x2x2 cube is not
known.  I do not know whether it has been calculated in the mean time, so
I just did calculate it.  The number of elements in the group is
3,674,160.  (Fix one corner, the others allow every permutation and one
third of all possible twists.)  The diameter is 11 if we do allow half-turns,
it is 14 if we do not allow half-turns.  The distribution is:
If we allow half-turns:
         1 with  0 moves
         9 with  1 moves
        54 with  2 moves
       321 with  3 moves
      1847 with  4 moves
      9992 with  5 moves
     50136 with  6 moves
    227536 with  7 moves
    870072 with  8 moves
   1887748 with  9 moves
    623800 with 10 moves
      2644 with 11 moves
If we do not allow half-turns:
         1 with  0 moves
         6 with  1 moves
        27 with  2 moves
       120 with  3 moves
       534 with  4 moves
      2256 with  5 moves
      8969 with  6 moves
     33058 with  7 moves
    114149 with  8 moves
    360508 with  9 moves
    930588 with 10 moves
   1350852 with 11 moves
    782536 with 12 moves
     90280 with 13 moves
       276 with 14 moves
In the first case heuristics give a diameter of at least 9.  We see that the
majority of the configuration is within distance 9 from start.  So it appears
that heuristics get close to the real value.
We see also that in both cases there is more than one diametrally opposite
configuration.  Next I will find out which those are (and if they have
something in common).

BTW, calculation did not take very long, only a few (<3) minutes on an FPS
(i.e.  an extremely fast SPARC).  But as the calculations are memory bound
rather than compute bound, the speed of the processor is not so very important.

From reid@math.berkeley.edu  Mon May 11 20:31:30 1992
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From: reid@math.berkeley.edu (michael reid)
Message-Id: <9205120031.AA22739@math.berkeley.edu>
To: Dik.Winter@cwi.nl, cube-lovers@life.ai.mit.edu
Subject: Re:  More on the Cube (2x2x2 in this case).

> Singmaster states that the diameter of the group for the 2x2x2 cube is not
> known.

in his "cubic circular," issue(s) 5/6 (pages 26, 27) he gives some info
about this, although it is (at least) third hand information, and therefore
not necessarily reliable.

> I just did calculate it.
  ...
> If we allow half-turns:
>          1 with  0 moves
>          9 with  1 moves
>         54 with  2 moves
>        321 with  3 moves
>       1847 with  4 moves
>       9992 with  5 moves
>      50136 with  6 moves
>     227536 with  7 moves
>     870072 with  8 moves
>    1887748 with  9 moves
>     623800 with 10 moves
>       2644 with 11 moves

he gives the same figures, so they are probably correct.

> If we do not allow half-turns:
>          1 with  0 moves
>          6 with  1 moves
>         27 with  2 moves
>        120 with  3 moves
>        534 with  4 moves
>       2256 with  5 moves
>       8969 with  6 moves
>      33058 with  7 moves
>     114149 with  8 moves
>     360508 with  9 moves
>     930588 with 10 moves
>    1350852 with 11 moves
>     782536 with 12 moves
>      90280 with 13 moves
>        276 with 14 moves

he does not give these, but he does mention that the diameter is 14.

> BTW, calculation did not take very long, only a few (<3) minutes on an FPS

singmaster says that the calculation took "over 51 hours of computer time"!
ouch!  this was 10 years ago, though.  (what's 51 hours / 3 minutes ?)

the "unix news item" from which singmaster apparently got his info was
included in a cube-lovers message.  it's in the archives, cube-mail-3,
sept 15, 1981 in a message from "ISAACS at SRI-KL".

dik's results show that the corners of the 3x3x3 can be "solved" (i.e.
positioned correctly with respect to one another) in 11 face (respectively
14 quarter) turns.  it would be nice to know if they can be solved with
respect to the centers within 11 face (respectively 14 quarter) turns.
this seems likely.

mike

From hoey@aic.nrl.navy.mil  Tue May 12 11:03:38 1992
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From: hoey@aic.nrl.navy.mil (Dan Hoey)
Message-Id: <9205121503.AA04073@Sun0.AIC.NRL.Navy.Mil>
To: Cube-Lovers@life.ai.mit.edu
Cc: Dik.Winter@cwi.nl, reid@math.berkeley.edu (michael reid)
Subject: Diameter of the 2^3 cube and the 3^3 corners

I sent the results of a quarter-turn analysis of these puzzles to
Cube-Lovers in several messages during August, 1984.  I modified a
program written by Karl Dahlke to get these results.  I counted both
positions and local maxima at every distance up to the diameter of 14
quarter-turns.  In case you don't have the archives handy, here are
the results:

  Quarter        2^3 Puzzle           Corners of 3^3 Puzzle
   Turns   Positions  Local Maxima   Positions  Local Maxima
____________________________________________________________
     0            1        0                1           0
     1            6        0               12           0
_____2___________27________0______________114___________0___
     3          120        0              924           0
     4          534        0             6539           0
_____5_________2256________0____________39528___________0___
     6         8969        0           199926         114
     7        33058       16           806136         600
_____8_______114149_______53__________2761740_______17916___
     9       360508      260          8656152       10200
    10       930588     1460         22334112       35040
____11______1350852____34088_________32420448______818112___
    12       782536   402260         18780864     9654240
    13        90280    88636          2166720     2127264
____14__________276______276_____________6624________6624___

The first column agrees with Dik Winter's findings.  As Michael Reid
surmised, the diameters of the two groups are the same.

My hazy recollection is that the 2^3 program ran for a few minutes on
a Vax 750, while the corners program took a couple of hours.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From dik@cwi.nl  Tue May 12 17:47:25 1992
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From: Dik.Winter@cwi.nl
Message-Id: <9205122146.AA06553.dik@boring.cwi.nl>
To: Cube-Lovers@life.ai.mit.edu, hoey@aic.nrl.navy.mil
Subject: Re:  Diameter of the 2^3 cube and the 3^3 corners
Cc: reid@math.berkeley.edu

 > I sent the results of a quarter-turn analysis of these puzzles to
 > Cube-Lovers in several messages during August, 1984.
I must have somewhere a printed stack of cube-lovers mailings, but I never
came around to read them all.  Also, my reference to Singmaster was his
notes.  The latest page of the latest printing states that the 2x2x2 case
was still unsolved, I never have seen his additional notes.

 >                                                       I counted both
 > positions and local maxima at every distance up to the diameter of 14
 > quarter-turns.
After Mike Reid's question I modified my program to do the counting on the
corners of the 3^3.  The biggest change was that it is now able to handle
that case in memory on this 32 MByte machine.  I did not count local maxima
(although that could be done).  The quarter turn case is identical to Dan
Hoey's results.  If we count half turns as a single move I get the following
results:
         1 with  0 moves
        18 with  1 moves
       243 with  2 moves
      2874 with  3 moves
     28000 with  4 moves
    205416 with  5 moves
   1168516 with  6 moves
   5402628 with  7 moves
  20776176 with  8 moves
  45391616 with  9 moves
  15139616 with 10 moves
     64736 with 11 moves

 > The first column agrees with Dik Winter's findings.  As Michael Reid
 > surmised, the diameters of the two groups are the same.
Also here the diameter is the same.

 > My hazy recollection is that the 2^3 program ran for a few minutes on
 > a Vax 750, while the corners program took a couple of hours.
My calculation took slightly less than half an hour.  The differences in
timings we see are (I think) mostly due to memory constraints on older
machines.  So we see a difference between Memory bound and I/O bound.
I could go to disk for storage of (intermediate) results, but even than the
edges can not be handled.  (980*10^9 configurations so my program would
require 245 GBytes of storage.  I think methods can be found to reduce this by
a factor of 30-40, but it is still much too large to handle, and in that case
you probably get the diameter only.)

dik
--
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
dik@cwi.nl

From dik@cwi.nl  Thu May 14 19:44:27 1992
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From: Dik.Winter@cwi.nl
Message-Id: <9205142344.AA13499.dik@boring.cwi.nl>
To: cube-lovers@life.ai.mit.edu
Subject: Kociemba's algorithm

I am now trying to implement Kociemba's algorithm.  The initialization parts
are done.  To recap, it is a two stage algorithm.  The first stage tries to
get to the subgroup generated by [R^2,L^2,F^2,B^2,U,D], the second stages
comes back to start.

The first stage uses a three dimensional coordinate system: twistyness,
flippancy and choosyness (where are the 4 middle slice edge cubies?).
The second stage uses (I think) also a three dimensional coordinate system,
all permutations: corner cubies, edge cubies not on the middle slice, slice
cubies.  I found the maximal distance along each coordinate as follows:

stage 1:
twistyness:	6
flippancy:	7
choosyness:	4
This seems not in contradiction with his 10 moves or less on average.

stage 2:
corners:	13
edges:		8
slice edges:	4
I think this contradicts his 14 moves or less, there are configurations
that require at least 13 moves to get the corners correct.  I would be
surprised if only one more move is needed to get everything correct.

*But* some of his best moves use a sub-optimal solution for stage 1!
Now if that could be quantified...

Next step is implementing the searching algorithms.

dik
--
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
dik@cwi.nl

From dik@cwi.nl  Sat May 16 21:14:18 1992
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From: Dik.Winter@cwi.nl
Message-Id: <9205170114.AA20529.dik@boring.cwi.nl>
To: cube-lovers@life.ai.mit.edu
Subject: Kociemba's algorithm

I have implemented it based on his description.  I am not yet completely
satisfied, but can give some results.  Both are the best I found after a
run of about 30 minutes.  (The numbers are first the number of moves to
get at [F^2,R^2,B^2,L^2,U,D], second the numbr of moves to complete.)

Superflip:
(11+10=21): F B R U^2 B^2 U' D' R^2 B' R L U F^2 L^2 D^2 B^2 D' F^2 D L^2 D

Supertwist:
(7+9=16): F R^2 L^2 U^2 D^2 F^2 B' R^2 U F^2 B^2 R^2 L^2 U^2 D' L^2

So clearly the supertwist is not even close to the opposite of start!

Currently the program needs still a bit of hand-tuning.  I am looking how
I can improve that.

dik
--
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
dik@cwi.nl

From dik@cwi.nl  Sun May 17 18:49:53 1992
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From: Dik.Winter@cwi.nl
Message-Id: <9205172249.AA22984.dik@boring.cwi.nl>
To: cube-lovers@life.ai.mit.edu
Subject: Kociemba's algorithm

I have made my program a bit faster.  While previously the best I found for
superflip was a 21 move solution (even after 10 hours computation time,
actually the solution was found after about 30 minutes), I have now a 20
move solution, found after only 15 minutes:

Superflip:
(13+7=20): F B U^2 R F^2 R^2 B^2 U' D F U^2 R' L' U B^2 D R^2 U B^2 U

Some more information.  First a short recap.  Phase1 brings the cube in
the group generated by [F^2,R^2,B^2,L^2,U,D], phase2 brings him back
to start.  Phase 1 searches in the space generated by the three (orthogonal)
coordinates:
Twist (2187 entries), flip (2048 entries) and slice-edge-cube placing (495
entries).
Phase 2 searches in the space generated by the three (non-orthogonal)
coordinates:
Permutations of corner cubes (40320 entries), permutations of edge cubes
not in the middle slice (40320 entries) and permutations of the middle
slice edge cubes (24 entries).

While Kociemba originally did tree pruning based on the minimal number of
moves needed along single coordinates, I now use pairs of coordinates
(except that in phase 2 the 40320*40320 pair is not used of course).
This is part of the speed-up.  (Another part is that I do now disallow
successive moves of a single face, three or more consecutive moves of
opposite faces, and a move of an opposite face if the current face moved
is B, L or D.)

Program details: the program starts with phase1 allowing for succesively
1, 2 etc. until a maximal number of moves.  As soon as phase1 hits a
solution phase2 is called, again with a maximum number of moves starting
at 1.  This means that if the program runs long enough it will ultimately
find the shortest solutions (phase 1 might just solve it!).  But that wil
take a long time (of course).  For the superflip, the program has now
checked all phase1 solutions of upto 12 moves and is busy with 13.  It
found 792256 solutions of 12 moves (and that in less than 10 minutes)!

Some additional data about minimal paths along coordinates:
Phase 1:
twist:		6
flip:		7
choice:		5
twist+flip:	9
twist+choice:	9
flip+choice:	9

Phase 2:
corners:	13
edges:		8
slice edges:	4
corners+slice:	14
edges+slice:	12

Based on this I expect a maximal distance in phase 1 of about 10/11, and
in phase 2 of about 16/17.

dik
--
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
dik@cwi.nl

From dik@cwi.nl  Sun May 17 21:03:44 1992
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Date: Mon, 18 May 1992 03:03:34 +0200
From: Dik.Winter@cwi.nl
Message-Id: <9205180103.AA23353.dik@boring.cwi.nl>
To: cube-lovers@life.ai.mit.edu, keng@zcar.asd.sgi.com
Subject: My program is too fast ;-).

I have posted a bit on my version of Kociemba's program, and I can only
conclude that my program is too fast.  After doing the superfliptwist,
the superflip and the supertwist I thought about trying those configurations
where Singmaster's notes did not give a solution better than 21 moves.
I find now that it takes more time to enter the configurations than what
the program needs to solve it!  Upto now I found the following (I will not
give the exact moves, as I think Kociemba wants to publish a bit more about
this):
superflip:	20 moves
supertwist:	16 moves
superfliptwist:	20 moves
Walker's 6+:	17 moves (was 22)
Walker's 6X:	19 moves (was 25)
Walker's worm:	14 moves (was 23)

Initialization time for the program is 2.5 minutes.  But it finds solutions
after only a few seconds!  If you have a configuration that you think is at
a large distance from start, mail it and I will disprove it ;-).

dik
--
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
dik@cwi.nl

From reid@math.berkeley.edu  Sun May 17 22:10:48 1992
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Date: Sun, 17 May 92 19:10:33 PDT
From: reid@math.berkeley.edu (michael reid)
Message-Id: <9205180210.AA08289@math.berkeley.edu>
To: Dik.Winter@cwi.nl, cube-lovers@life.ai.mit.edu, keng@zcar.asd.sgi.com
Subject: Re:  My program is too fast ;-).

stop it, you're killing me!  i also have the same idea for combining
the "coordinates" in pairs, but i'm not getting too far implementing
it.  :-(  i wouldn't suggest using singmaster's notes for pattern
maneuvers.  have you seen bandelow's book?  it has very short
maneuvers for most patterns, including two different ones for
"walker's worm" in 14 turns (assuming i've got the right pattern in
mind).  bandelow counts "slice turns" as one move, though, so
his maneuver for 6X (order 3) is 24 face turns.

what amazes me about this whole business is that the algorithm
finds very short moves when they exist.  i would have expected
the program to produce maneuvers of approximately the same
length for all patterns.  i would say that this is a major step
forward.

you'll probably be swamped with patterns to test, but here's a
couple:

stripes: (18) F3 U1 F2 U3 R1 B2 R3 U1 F2 L2 U3 L1 B2 L3 U1 L2 U3 F1.

python: (15) R1 U3 F3 B1 L1 F2 L3 F1 B3 U3 R3 L1 F2 U2 L3.

since i found these by hand, i'm curious to see how close they
are to optimal.  hopefully i'll have my program running soon.

mike

From reid@math.berkeley.edu  Mon May 18 09:02:04 1992
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From: reid@math.berkeley.edu (michael reid)
Message-Id: <9205181302.AA10968@math.berkeley.edu>
To: cube-lovers@ai.mit.edu
Subject: kociemba's algorithm

after a long struggle, it appears that i finally have kociemba's
algorithm working properly, with the improvement that dik mentioned.
the first pattern i tested was "anaconda" (bandelow's terminology),
which dik called "walker's worm" (singmaster's terminology?).
(i think that's the same pattern.)  within 10 minutes, most of which
was initializing tables, the program produced the following 14 quarter
turn maneuver, which is better (in the quarter turn metric) than the two
maneuvers bandelow gives in his book.  here it is:

anaconda: (14 qt) B1 R1 D3 R3 F1 B3 D1 F3 U1 D3 L1 F1 L3 U3.

more results as they come in.  my program isn't quite finished yet;
hopefully i won't screw it up attempting to improve it.

mike

From dik@cwi.nl  Mon May 18 19:17:44 1992
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Date: Tue, 19 May 1992 01:17:40 +0200
From: Dik.Winter@cwi.nl
Message-Id: <9205182317.AA25710.dik@boring.cwi.nl>
To: cube-lovers@life.ai.mit.edu, reid@math.berkely.edu
Subject: Re:  My program is too fast ;-).

 > stop it, you're killing me!
Stop yourself ;-).
 >           i wouldn't suggest using singmaster's notes for pattern
 > maneuvers.  have you seen bandelow's book?  it has very short
 > maneuvers for most patterns, including two different ones for
 > "walker's worm" in 14 turns (assuming i've got the right pattern in
 > mind).  bandelow counts "slice turns" as one move, though, so
 > his maneuver for 6X (order 3) is 24 face turns.
Leider haben wir nur die Deutschen Ausgabe.  There are apparently
differences between the German and the English edition.  The English
edition is later and has probably improved sequences (he has promised
also improved sequences for the second German edition, but I do not
think it ever came out).

 > what amazes me about this whole business is that the algorithm
 > finds very short moves when they exist.  i would have expected
 > the program to produce maneuvers of approximately the same
 > length for all patterns.  i would say that this is a major step
 > forward.
The point is that my program deliberately tries short sequences first in
both phases.  This is at the expense of more work when longer sequences
are tried (because you will find the shorter sequences again that you
abort).  The main advantage is that if you find an overall short sequence
you have much less work to do in the second phase, and can much faster
abort your searches.  Here some information about the searching for
superflip (the numbers are for phase 1):
	Length	Number found	Solutions found
	   8	      0		-
	   9	      0		-
	  10	   3072		10+13, 10+12
	  11	  61568		11+10
	  12	 792256		-
	  13	8695488		13+ 7
As you see when having the large number of solutions for length 13 in phase 1
we have only to look at most 7 moves deep in phase 2, and 6 deep as soon as
we found the solution (which occurs reasonably fast).  That is why it is
possible to do all that work in about 45 minutes.  (For der Mouse, the machine
is an FPS.  A SPARC processor twice as fast as a SPARCStation ELC.  But I
think the program is bounded by memory access time.  The program requires
about 12 MB memory.)  Now I think the longest distance in phase 1 is about 11
or 12 (it is at least 11, that is sure).  So in general you will find a
solution already when there is no need yet to do very much.

 > you'll probably be swamped with patterns to test, but here's a
 > couple:
 > stripes: (18) F3 U1 F2 U3 R1 B2 R3 U1 F2 L2 U3 L1 B2 L3 U1 L2 U3 F1.
 > python: (15) R1 U3 F3 B1 L1 F2 L3 F1 B3 U3 R3 L1 F2 U2 L3.
I could not improve the first but what do you think of:
(12+ 2=14): U2 B3 D3 L1 F1 B3 U3 F1 U3 D1 R3 B1 D1 F2

 > since i found these by hand, i'm curious to see how close they
 > are to optimal.  hopefully i'll have my program running soon.
14 for python is probably optimal.  But you can try as you have your program
running.

From Don.Woods@eng.sun.com  Mon May 18 19:27:15 1992
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From: Don.Woods@eng.sun.com (Don Woods)
Message-Id: <9205182327.AA04124@colossal.Eng.Sun.COM>
To: Dik.Winter@cwi.nl
Subject: Re: Kociemba's algorithm
Cc: cube-lovers@life.ai.mit.edu

> Program details: the program starts with phase1 allowing for succesively
> 1, 2 etc. until a maximal number of moves.  As soon as phase1 hits a
> solution phase2 is called, again with a maximum number of moves starting
> at 1.

I don't remember from the earlier description; are the searches being done
depth-first or breadth-first?  If breadth-first, then there is no reason to
put an upper limit on the number of moves for finding a phase1 solution,
since the algorithm HAS to solve phase1 in order to find an overall solution.

Once you have an overall solution, of course, the length of phase1+phase2
solution can be used as an upper bound on subsequent phase1 solutions.

Also, do you reduce the "maximum number of moves" for phase2 based on
solutions already found?  For instance, once you have found a combined
phase1+phase2 solution in 20 moves, then if you find an alternative phase1
solution in 14 moves there is no reason to look deeper than 5 moves in
phase2 (or 6 if you want to find all solutions that tie for minimum length).

	-- Don.


From dik@cwi.nl  Mon May 18 19:48:28 1992
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Date: Tue, 19 May 1992 01:48:26 +0200
From: Dik.Winter@cwi.nl
Message-Id: <9205182348.AA25754.dik@boring.cwi.nl>
To: Don.Woods@eng.sun.com
Subject: Re: Kociemba's algorithm
Cc: cube-lovers@life.ai.mit.edu

 > I don't remember from the earlier description; are the searches being done
 > depth-first or breadth-first?  If breadth-first, then there is no reason to
 > put an upper limit on the number of moves for finding a phase1 solution,
 > since the algorithm HAS to solve phase1 in order to find an overall solution.
I do depth-first.  I think the breadth of the tree is much too large to be
handled in breadth first fashion (currently I abort the program when it
reaches a length in phase 1 such that the number of solutions is a few
million).

 > Once you have an overall solution, of course, the length of phase1+phase2
 > solution can be used as an upper bound on subsequent phase1 solutions.
Right.

 > Also, do you reduce the "maximum number of moves" for phase2 based on
 > solutions already found?  For instance, once you have found a combined
 > phase1+phase2 solution in 20 moves, then if you find an alternative phase1
 > solution in 14 moves there is no reason to look deeper than 5 moves in
 > phase2 (or 6 if you want to find all solutions that tie for minimum length).
Also done, I do not look for ties.

dik

From dik@cwi.nl  Mon May 18 22:07:37 1992
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Date: Tue, 19 May 1992 04:07:26 +0200
From: Dik.Winter@cwi.nl
Message-Id: <9205190207.AA26407.dik@boring.cwi.nl>
To: cube-lovers@life.ai.mit.edu
Subject: Kociemba's algorithm

I did some more and found fast algorithms.  The most amazing one was for
the configuration with two 2x2x2 cubes embedded in the cube:
	( 9+ 6=15): U1 L2 D1 R1 B3 R1 B3 R1 B3 D3 L2 U1 R2 F2 U2
This looks remarkably like a random sequence of moves.  I find especially
the part R1 B3 R1 B3 R1 B3 intriguing!

(For der Mouse.  I lost your e-mail address.  I can provide you with the
program.  And for everybody, it is available, although I still do not like
the input method.)

dik
--
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
dik@cwi.nl

From reid@math.berkeley.edu  Tue May 19 13:25:44 1992
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Date: Tue, 19 May 92 10:25:38 PDT
From: reid@math.berkeley.edu (michael reid)
Message-Id: <9205191725.AA25455@math.berkeley.edu>
To: cube-lovers@ai.mit.edu
Subject: assorted results

>  > stripes: (18) F3 U1 F2 U3 R1 B2 R3 U1 F2 L2 U3 L1 B2 L3 U1 L2 U3 F1.
>  > python: (15) R1 U3 F3 B1 L1 F2 L3 F1 B3 U3 R3 L1 F2 U2 L3.
> I could not improve the first but what do you think of:
> (12+ 2=14): U2 B3 D3 L1 F1 B3 U3 F1 U3 D1 R3 B1 D1 F2

i mentioned these because i was particularly proud of them, especially
since they were constructed by hand.  my program was able to improve
"stripes", but i reoriented the pattern first.  perhaps this is why
i had better success.  stripes (17 face / 20 quarter):
R1 U1 D1 R2 F1 U3 D1 F3 B1 U3 R3 L3 B1 L2 U1 D3 B2  (14 + 3)
so if you have any efficient pattern maneuvers that you're especially
proud of, let's see if they can't be improved.

> I did some more and found fast algorithms.  The most amazing one was for
> the configuration with two 2x2x2 cubes embedded in the cube:
> 	( 9+ 6=15): U1 L2 D1 R1 B3 R1 B3 R1 B3 D3 L2 U1 R2 F2 U2

this can also be done in 18 quarter turns:
L1 F1 L1 D3 B1 D1 L2 F2 D3 F3 R1 U3 R3 F2 D1  (13 + 2)
also "cube within cube within cube"  17 face / 22 quarter turns:
L3 D1 R3 B3 D1 L2 D2 L3 D3 L1 U1 D2 R3 U3 B2 R2 D3  (13 + 4)
"twisted rings"  16 face / 18 quarter turns:
B1 R1 B3 R2 U3 F3 L1 U1 R1 D1 L1 U3 B3 L1 U1 L2  (14 + 2)
"twisted cube edges"  14 face / 14 quarter turns:
D1 L3 B1 R1 D3 R3 D1 B3 L1 B1 R3 B3 R1 D3  (13 + 1)

> Leider haben wir nur die Deutschen Ausgabe.  There are apparently
> differences between the German and the English edition.  The English
> edition is later and has probably improved sequences (he has promised
> also improved sequences for the second German edition, but I do not
> think it ever came out).

'tis a shame.  the english edition is very good.  i've never seen the
german.  in the english edition he gives several "open snakes".
here's what my program has to say about them.
"rattlesnake"  14 face / 18 quarter turns:
F3 D3 L1 D3 L2 F1 B2 U1 L3 U3 R3 B2 R1 L2  (13 + 1)
"black mamba"  14 face / 14 quarter turns:
L1 U1 R1 F3 R3 L1 U1 L3 U3 D1 B1 D3 L3 U3  (13 + 1)
"green mamba"  13 face / 14 quarter turns:
R3 L2 F1 R1 U1 L3 F3 B1 U1 B3 U3 L3 U3  (12 + 1)
"boa"  12 face / 16 quarter turns:
D1 R1 F3 R1 L1 F2 R2 D3 L2 F2 L1 F3  (12 + 0)

this last one was very nice, since it was completely solved in stage 1!

> Also done, I do not look for ties.

i am currently looking for ties, but will soon stop.  this is mainly to
catch (by eye) maneuvers that are short quarter turn-wise.  i realize
this is stupid for at least two reasons:  1: the program may well be
passing up sequences which are shorter in quarter turns;  2: a slightly
different version of the program will specifically look for the shortest
sequence in quarter turns.  i'm trying to think of the best way to do this.
unfortunately, the temptation is NOT to think, but to feed every
imaginable pattern into the program.  :-)

still need to do some polishing ...

mike

From dik@cwi.nl  Tue May 19 21:00:38 1992
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Date: Wed, 20 May 1992 03:00:27 +0200
From: Dik.Winter@cwi.nl
Message-Id: <9205200100.AA23701.dik@steenbok.cwi.nl>
To: cube-lovers@ai.mit.edu, reid@math.berkeley.edu
Subject: Re:  assorted results

	                                      My program was able to improve
	"stripes", but i reoriented the pattern first.  perhaps this is why
	i had better success.
True.  Orientation of non-symmetric patterns is important because the first
step is to get at a situation that is also non-symmetric.

	'tis a shame.  the english edition is very good.
The german edition is also good, but clearly older.  The newer patterns you
mention are not in the german edition.  (I may note that Christoph Bandelow
is still selling puzzles.  The 5x5x5 amongst others.)

	R3 L2 F1 R1 U1 L3 F3 B1 U1 B3 U3 L3 U3  (12 + 1)
	D1 R1 F3 R1 L1 F2 R2 D3 L2 F2 L1 F3  (12 + 0)

	this last one was very nice, since it was completely solved in stage 1!
Yup.  But the last but one will not improve and is optimal, and in fact solved
with stage 1 (but you did not find it because you limited your search 12 deep).

	                        I'm trying to think of the best way to do this.
	unfortunately, the temptation is NOT to think, but to feed every
	imaginable pattern into the program.  :-)
How true.  I stopped feeding new patterns.  Currently I am calculating the
maximal distance in stage 1.  It will take a bit of time because I have to
consider 2,217,093,120 possibilities.  But I think that the method I have
is feasible.  My conjecture was that the maximal distance was 11 or 12.
That was wrong.  It is at least 12 (the superfliptwist needs 12 moves in
phase 1).  My current conjecture is 12.  Work is in progress, the first
1,082,565 configurations give the following picture (i.e. all configurations
without flip):
Moves	Number
  0	     1
  1	     2
  2	    17
  3	   134
  4	  1065
  5	  8214
  6	 54919
  7	269388
  8	562427
  9	183730
 10	  2668
The pattern is similar to what was found with the 2x2x2 cube.  The majority
of configurations requires 2 less than the maximum.  But apparently flips are
harder to deal with than the rest of phase 1, so I am waiting for more results.
Note that 12 would improve Kloostermans 42 moves to 37.

dik
--
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
dik@cwi.nl

From reid@math.berkeley.edu  Wed May 20 16:06:25 1992
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Date: Wed, 20 May 92 13:06:14 PDT
From: reid@math.berkeley.edu (michael reid)
Message-Id: <9205202006.AA13051@math.berkeley.edu>
To: cube-lovers@ai.mit.edu
Subject: Re:  assorted results

> True.  Orientation of non-symmetric patterns is important because the first
> step is to get at a situation that is also non-symmetric.

next stage: have the program account for symmetries of the pattern,
face turns that commute with the pattern, ...

> 	'tis a shame.  the english edition is very good.
> The german edition is also good, but clearly older.  The newer patterns you
> mention are not in the german edition.

again, a shame.  i think that these snake patterns are among the most
interesting things in the book.

>                                             Currently I am calculating the
> maximal distance in stage 1.  It will take a bit of time because I have to
> consider 2,217,093,120 possibilities.  But I think that the method I have
> is feasible.

how much time do you anticipate the job will take?  it seems that we'd
get a much better improvement (of kloosterman's bound) by calculating
the maximal distance in stage 2.  of course, this requires going through
19,508,428,800 possibilities (nearly 9 times as many).  is this feasible?
also it seems that "god's algorithm" for stage 2 (i.e. face turns are
restricted to  D, U, B2, F2, L2, R2) would be very similar to
god's algorithm for the "magic domino," and this similarity should
become stronger as the patterns get further away from START.

from the pruning tables used in stage 2, dik gives the following maximal
distances:

> Phase 2:
> corners:	13
> edges:		8
> slice edges:	4
> corners+slice:	14
> edges+slice:	12

i was slightly surprised to see a difference between the figures for
"corners" and those for "corners + slice edges".  but the difference
between "edges" and "edges + slice edges" is shocking.  so perhaps the
two "god's algorithms" above are not TOO similar.

> Based on this I expect a maximal distance  [ ... ]
> in phase 2 of about 16/17.

i'm pretty sure i know a pattern that requires 15 face turns.  i wouldn't
be all that surprised if 15 was the maximum, although i haven't tested many
patterns.

mike

From dik@cwi.nl  Wed May 20 17:14:12 1992
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Date: Wed, 20 May 1992 23:13:11 +0200
From: Dik.Winter@cwi.nl
Message-Id: <9205202113.AA02174.dik@boring.cwi.nl>
To: cube-lovers@ai.mit.edu, reid@math.berkeley.edu
Subject: Re:  assorted results

 > >                                             Currently I am calculating the
 > > maximal distance in stage 1.  It will take a bit of time because I have to
 > > consider 2,217,093,120 possibilities.  But I think that the method I have
 > > is feasible.

 > how much time do you anticipate the job will take?
A few days.  I started yesterday night at 2:30 AM, it is now 11:05 PM.  I have
calculated the distances for approximately 145 million configurations.  The
majority of the work has been done sinc 6:00 PM.  (I am using 39 SGI Indigo's,
1 Large SGI, 2 processors of an SGI file server and the scalar (SPARC)
processor of an FPS, so it is lots of computer time!)
 >                                                     it seems that we'd
 > get a much better improvement (of kloosterman's bound) by calculating
 > the maximal distance in stage 2.  of course, this requires going through
 > 19,508,428,800 possibilities (nearly 9 times as many).  is this feasible?
Right.  But is not feasible, at least I do not see possibilities at this
moment.  My estimate is that it would take much more than 9 times as much.
The reason is that the algorithm as I have organized it allows a lot of
shortcuts.  What I do is in the space of order 2048 * 2187 * 495 assign to
each processor in turn a slice of the 2048 dimension.  Within that slice,
going from 0 moves until every configuration has been assigned a distance,
a path is searched for the given distance.  The major shortcut comes when
a path is found.  Using configurations with known distance, U, D, L and R
turns are applied and also F2 and L2 turns to calculate new distances.
For those moves the flip does not change.  This makes that I have to search
paths for only about 10% of the cases.  I see no way (yet) to do similar
things for phase 2.

dik

From reid@math.berkeley.edu  Sat May 23 01:56:38 1992
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Date: Fri, 22 May 92 22:56:34 PDT
From: reid@math.berkeley.edu (michael reid)
Message-Id: <9205230556.AA03559@math.berkeley.edu>
To: cube-lovers@ai.mit.edu
Subject: new upper bound

i've managed to reduce the upper bound for the length of god's
algorithm to 39 face turns / 56 quarter turns.  we work in three
stages:

1. from  <U, F, R, L, B, D>  to  <U, R, F>
2. from  <U, R, F>           to  <U, R2, F2>
3. from  <U, R2, F2>         to  START

where we're only allowed to use moves that keep us within the
given subgroup.

these three stages were chosen because of the moderate(!) sizes
of the coset spaces that must be considered.  the numbers are
253440,  15676416,  and  10886400.  the maximum number of moves
in each stage was calculated by exhaustively searching the space.

if we count by "face turns," these maximum numbers are 8, 13 and 19.
however, if we make any turns in stage 2, the last such is a quarter
turn of either  F  or  R, and the direction is irrelevant.  those
positions at distance 19 in stage 3 (only 24 in all) were checked
to see that they may be solved in 19 face turns starting with
either  F2  or  R2.  therefore we can arrange to combine the last
move of stage 2 with the first move of stage 3 in the event that
we must make the maximal number of moves in each stage.  this
gives the final figure of 39 face turns.

if we count by "quarter turns," the maximum numbers are 9, 16 and 33.
this time, those configurations at distance either 32 or 33 in stage 3
(only 10 and 4 positions, respectively) were tested as above.
each has minimal sequences starting with  F2  and with  R2.  as above,
we may cancel a quarter turn from the end of stage 3 with a quarter
turn at the beginning of stage 3, to get the final figure of 56
quarter turns.

the next step is to reduce the figure for stage 3 by allowing other
turns.  i only plan on allowing  D, U, B2, F2, L2, R2,  as in the
second stage of kociemba's algorithm.  after that, i may try to reduce
the numbers for stage 2.  however, in this case, i don't expect much
of a reduction (maybe 1 turn).

mike

From dik@cwi.nl  Sun May 24 07:31:21 1992
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Date: Sun, 24 May 1992 13:31:13 +0200
From: Dik.Winter@cwi.nl
Message-Id: <9205241131.AA07227.dik@boring.cwi.nl>
To: anneke@fwi.uva.nl, cube-lovers@life.ai.mit.edu
Subject: New upper bound on God's algorithm for Rubik's cube

Earlier than expected I have gotten my results.  Using a farm of workstations
I have calculated the distances for all cosets in phase 1 of Kociemba's
algorithm.  As I have conjectured already, the maximal distance is 12.
Together with Kloosterman's result for their third and fourth phase (which
together form Kociemba's second phase) the upperbound on God's algorithm
is now 37.  Below follows the set of distances for the first phase:
  0:          1
  1:          4
  2:         74
  3:       1230
  4:      18056
  5:     245902
  6:    3082221
  7:   34529024
  8:  301243996
  9: 1209021801
 10:  663711855
 11:    5238847
 12:        109
tot: 2217093120

It took quite a bit of computer time.  In all 116 processors in 112 different
machines have cooperated, although not all at the same time %.  The maximal
amount of concurrent processing was on 100 processors last night.  The
distribution was: 69 Sun's (SS 1, SS 1+, SLC, ELC), 40 SGI's, 1 SGI 4D/310VGX,
5 processors of an SGI 260S compute server and the scalar (SPARC) processor
of an FPS 500.  I expect it would have taken one or two hours only on a machine
with 1 GByte of memory.  Calculations on the second phase are still out of
the question.  Distributing the processing would take much more than 9 times
as much.  I expect it would take about a day on a machine with 4 GByte of
memory.

I conjecture that the maximal distance in phase 2 is at most 16.  There is a
lower bound on it of 14.  This would make the upperbound for God's algorithm 28.

dik
--
% These processors would have been idle most of the time otherwise!
--
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
dik@cwi.nl

From reid@math.berkeley.edu  Sun May 24 09:11:51 1992
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Date: Sun, 24 May 92 06:10:21 PDT
From: reid@math.berkeley.edu (michael reid)
Message-Id: <9205241310.AA04669@math.berkeley.edu>
To: Dik.Winter@cwi.nl, anneke@fwi.uva.nl, cube-lovers@life.ai.mit.edu
Subject: Re:  New upper bound on God's algorithm for Rubik's cube

> Together with Kloosterman's result for their third and fourth phase (which
> together form Kociemba's second phase) the upperbound on God's algorithm
> is now 37.

well, at least i had the record for a couple of days!  ;-)

>             Below follows the set of distances for the first phase:
>   0:          1
>   1:          4
>   2:         74

but i don't understand how we can get 74 positions at distance 2 from
only 4 positions at distance 1.  the 4 positions at distance 1 are
easy to see: they're the positions obtained from START by the turns
B, F, L  and  R.  with only 18 different face turns, each should
extend to at most 18 positions at distance 2.  am i missing something
obvious here?  (the numbers do seem to add up, though.)

> I conjecture that the maximal distance in phase 2 is at most 16.  There is a
> lower bound on it of 14.

the pattern (written in permutation notation)  (FR, FL) (UFL, DFR) is
at distance 15, so that's (also) a lower bound.  however, if the whole
cube is turned so that the  F  face becomes the  U face, then the new
pattern is still in the subgroup of stage 2, but is now at distance 14.

mike

From dik@cwi.nl  Sun May 24 10:23:12 1992
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Date: Sun, 24 May 1992 16:22:55 +0200
From: Dik.Winter@cwi.nl
Message-Id: <9205241422.AA07622.dik@boring.cwi.nl>
To: anneke@fwi.uva.nl, cube-lovers@life.ai.mit.edu, reid@math.berkeley.edu
Subject: Re:  New upper bound on God's algorithm for Rubik's cube

I retract my earlier message.  There is something definitely wrong!
(All that idle time wasted...)  I have to reconsider!

dik

From reid@math.berkeley.edu  Sun May 24 11:04:20 1992
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Date: Sun, 24 May 92 08:04:09 PDT
From: reid@math.berkeley.edu (michael reid)
Message-Id: <9205241504.AA04847@math.berkeley.edu>
To: anneke@fwi.uva.nl, cube-lovers@ai.mit.edu
Subject: details ...

perhaps i should give the figures i obtained when getting my upper
bound of 39 face / 56 quarter turns in case i also have an error.
recall the method: we work in three stages:

1. from  <U, F, R, L, B, D>  to  <U, R, F>
2. from  <U, R, F>           to  <U, R2, F2>
3. from  <U, R2, F2>         to  START

where we're only allowed to use moves that keep us within the
given subgroup.  the number of positions that must be considered
in each stage are  253440,  15676416,  and  10886400 respectively.

stage 1:
using face turns:          using quarter turns:
distance  number           distance     number
  0            1               0             1
  1            9               1             6
  2           90               2            39
  3          852               3           276
  4         7169               4          1899
  5        44182               5         11245
  6       131636               6         49412
  7        68940               7        117221
  8          561               8         70767
                               9          2574
stage 2:
using face turns:          using quarter turns:
distance  number           distance     number
  0            1               0             1
  1            2               1             2
  2           12               2             8
  3           72               3            36
  4          420               4           158
  5         2410               5           694
  6        13752               6          2980
  7        75796               7         12744
  8       390421               8         53646
  9      1735771               9        216354
 10      5351383              10        799868
 11      6696700              11       2477802
 12      1399195              12       5310848
 13        10481              13       5419046
                              14       1356020
                              15         26192
                              16            17
stage 3:
using face turns:          using quarter turns:
distance  number           distance     number
  0            1               0             1
  1            5               1             2
  2           14               2             3
  3           44               3             8
  4          128               4            14
  5          392               5            24
  6         1157               6            52
  7         3458               7            96
  8        10057               8           176
  9        29286               9           352
 10        82814              10           664
 11       228621              11          1248
 12       591704              12          2409
 13      1362497              13          4516
 14      2545752              14          8519
 15      3272940              15         16100
 16      2260555              16         30171
 17       484818              17         56140
 18        12133              18        102981
 19           24              19        186728
                              20        331234
                              21        563985
                              22        912719
                              23       1365051
                              24       1812011
                              25       2044832
                              26       1783956
                              27       1105488
                              28        450322
                              29         97881
                              30          7958
                              31           745
                              32            10
                              33             4

it would be nice if someone could confirm these figures.  i have done
some testing, but not extensively.

mike

From dik@cwi.nl  Sun May 24 19:38:50 1992
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From: Dik.Winter@cwi.nl
Message-Id: <9205242338.AA10405.dik@boring.cwi.nl>
To: anneke@fwi.uva.nl, cube-lovers@ai.mit.edu, reid@math.berkeley.edu
Subject: Program bug, and new ideas.  Will they work?

Well, I found the bug.  There was an initialization error which made the
program think there was a shorter path than there was in a number of cases.
Alas, as always, the bug did not reveal itself in the test runs I did, it
was only in the final totals that it was apparent.

The bad thing is that removing the bug increases the compute time considerably,
because you have in essence to calculate an explicit path for every
configuration.  I estimate the increase is by a factor of about 3, based on
some experiments.

The good thing is that it got me thinking about an old idea I had.
To calculate path lengths to the group generated by [U,D,L2,R2,F2,B2]
it is irrelevant whether you rotate the complete cube around the U-D
axis.  Also half turn rotations around the R-L axis are irrelevant.
And finally, mirroring the cube along the F-R-B-L plane is irrelevant!
But in most cases this changes twist/flip and position values.  So if
we look at the twist/flip/position space of 2187*2048*495 cosets we can
reduce the calculations along one dimension as long as we remember the
number of representations.  I did some calculations and found that
2187 can be reduced to 168 or 2048 can be reduced to 219 or 495 can be
reduced to 45.  Of these obviously the first one is the most fruitful.
Of course I have to check myself (and anybody who is willing, go ahead).
Reduction of 2187 to 168 would reduce the total space from 2,217,093,120
to 170,311,680 calculations.  But then again, dividing the latter figure
by 4, that could be done in core on a machine with about 50 MByte of memory
(to calculate path lengths in core only 2 bits per configuration are needed).

I will try around, dik



From dik@cwi.nl  Thu May 28 12:33:49 1992
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From: Dik.Winter@cwi.nl
Message-Id: <9205281300.AA00307.dik@boring.cwi.nl>
To: cube-lovers@life.ai.mit.edu
Subject: Corrected calculations are now done.

After an initial false start I have now calculated the path-lengths in
phase 1 of Kociemba's algorithm.  The figures are as follows:
path	configurations
  0:          1
  1:          4
  2:         50
  3:        592
  4:       7156
  5:      87236
  6:    1043817
  7:   12070278
  8:  124946368
  9:  821605960
 10: 1199128738
 11:   58202444
 12:        476
The figure 50 for path length 2 is easily verified by hand.  I have a list
with information about the configurations requiring a path-length of 12
(actually the paths leading to such a configurations).  As should be true
for each minimal path in phase 1, all paths start and terminate with a
quarter turn of F, R, B or L.

Some details.

Phase 1 of the algorithm brings the cube in the subgroup generated by
[F^2, R^2, B^2, L^2, U, D].  There are in this case 2,217,093,120
(2048 * 2187 * 495) cosets.  This can be (and has been) reduced largely
by observing symmetries.  In this case rotating the complete cube
along the UD axis by a quarter turn, rotating the cube along the RL
axis by a half turn and mirroring through the FRBL plane reveal
equivalent cosets.  Although it is possible to remove *all* cosets that
are equivalent to some canonical coset this was not done.  The removal
has only been done for the twists of corner cubes, reducing the factor
2187 to 168, and reducing the number of configurations to be handled to
170,311,680.  For each configuration a minimal path was calculated.
This was done starting with an absolute minimum found through the coordinate
axis and through the 2-dimensional coordinate spaces.  When a path of that
length was not found the path length was increased and a new attempt was made.
This was done until a path was found.  All searches were exhaustive.  On
average paths were searched for 3 different lengths (519,177,716 attempts
for 170,311,679 configurations).

The computations were done on a farm of workstations where each workstation
got a portion of the flip dimension (2048 cases of 83,160 configurations).
Computation time for one portion was from 1 to 2 hours (1.5 on average), so
the total computation was about 3000 hours.  On a system with enough
memory (50 MByte) it would have taken only 1 hour (this based on experiments
with the corner cubes-only part).  It could also have been with a single
processor and a 50 MByte file, in that case CP time would also be about 1
hour, but the I/O time would exceed the 3000 hours very much.

Using this result and the result by Hans Kloosterman the diameter of the
cube group is at most 37.  I conjecture the maximal path length in phase 2
of Kociemba's algorithm is 16, although the requirements on computer time
cq. memory do inhibit calculations at this moment (only in memory would be
feasible, but that requires 500 to 1000 MByte and computation time would be
about one day).  This figure of 16 would reduce the upperbound of the groups
diameter to 28.

dik
--
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
dik@cwi.nl

From @mitvma.mit.edu:hans@freyr.research.ptt.nl  Fri May 29 13:21:10 1992
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From: J.M.Kloosterman@research.ptt.nl (Hans Kloosterman)
Subject: Lower-bound Kociemba's algorithm
To: cube-lovers@life.ai.mit.edu
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Dik Winter writes:

> Using this result and the result by Hans Kloosterman the diameter of the
> cube group is at most 37.  I conjecture the maximal path length in phase 2
> of Kociemba's algorithm is 16, although the requirements on computer time
> cq. memory do inhibit calculations at this moment (only in memory would be
> feasible, but that requires 500 to 1000 MByte and computation time would be
> about one day).  This figure of 16 would reduce the upperbound of the groups
> diameter to 28.

Unfortunately Dik's conjecture for phase 2 is too optimistic.
Recall the maximum distances of the 4 stages of my algorithm:
 1.  7 moves within the group <R, L, F, B, U,D>
 2. 10 moves within the group <R, L, F2,B2,U,D>
 3.  8 moves within the group <R2,L2,F2,B2,U,D>
 4. 18 moves within the group <R2,L2,F2,B2,U,D>

(Stage 3 and 4 together requires at most 25 moves.)

These number of moves are minimal and cannot be improved within their
group of moves. (Stage 2 can also not be improved using all moves.)
From this we may conclude that the maximum path length in stage 2 of
the algorithm of Kociemba is at least 18 moves.

Taking the results of Dik Winter for stage 1 into account, the lower-bound
for the mximum of Kociemba's algorithm becomes 30 moves.

Hans Kloosterman


From dik@cwi.nl  Fri May 29 20:32:28 1992
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From: Dik.Winter@cwi.nl
Message-Id: <9205300032.AA01086.dik@steenbok.cwi.nl>
To: J.M.Kloosterman@research.ptt.nl, cube-lovers@life.ai.mit.edu
Subject: Re:  Lower-bound Kociemba's algorithm

(About my conjecture of 16 moves for phase 2:)
 > Unfortunately Dik's conjecture for phase 2 is too optimistic.
 > Recall the maximum distances of the 4 stages of my algorithm:
 >  1.  7 moves within the group <R, L, F, B, U,D>
 >  2. 10 moves within the group <R, L, F2,B2,U,D>
 >  3.  8 moves within the group <R2,L2,F2,B2,U,D>
 >  4. 18 moves within the group <R2,L2,F2,B2,U,D>
 > (Stage 3 and 4 together requires at most 25 moves.)

 > These number of moves are minimal and cannot be improved within their
 > group of moves.
Did you (since your article) do an exhaustive search?  In your article you
mentioned that you had 6 positions that still do require 18 moves.  And you
mention that you doubted that there would be 17 move solvers.  Have you
proven since then that it can not be done in less than 18?  If not, send me
your positions and I will try.

I have currently a program running that tries all phase 4 positions.  It
is possible to reduce the number of searches from 3,981,312 (the article
contains a typo here) to 428,544 by observing equivalent positions (as
I did mention in a previous article (*)).  Assuming my conjecture of 16 the
complete calculations would take about 1000 to 1500 hours (%).  Not
unprecedented ;-).  (There must be a reason that I am a member of the CWI
research group on large scale computing.)  There are now only two machines
munching at the problem, but there would be no problem to start up a few more
again.  I just did it to see what happens.

dik
--
* The equivalent positions are found by rotation of the complete cube along
the UD axis for a quarter turn, along the RL axis through a half turn and
mirroring along the FRBL plane.  When looking at one dimension only this
reduces the number from 40320 to 2768.  Restricting to Hans's initial positions
in phase 4, this reduces the number from 576 to 62.  So the count becomes:
	62 * 576 * 24 / 2
in stead of
	576 * 576 * 24 / 2 (= ((4!)^5) / 2).
--
% I found that an exhaustive search upto 16 moves takes about 10 seconds.
Increasing to 17 would up the time to 110 seconds.  So if you mail me
the situations for which you do not yet have less than 18 moves I will
have an attempt at them.
--
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
dik@cwi.nl

From dik@cwi.nl  Fri May 29 20:44:04 1992
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From: Dik.Winter@cwi.nl
Message-Id: <9205300044.AA01102.dik@steenbok.cwi.nl>
To: J.M.Kloosterman@research.ptt.nl, cube-lovers@life.ai.mit.edu
Subject: Re:  Lower-bound Kociemba's algorithm

As an afterthough, it would be interesting if it is possible to reduce the
number of moves in your fourth phase.  The main difference between your
algorithm and Kociemba's is that yours is deterministic.  Kociemba's
algorithm performs quite a number of searches before finding the optimal
solution.  And even than it is not known whether the solution is indeed
optimal, longer searches might reveal better solutions.  Your algorithm
gives an upper bound to the number of moves, and the solution is reached
in limited time.  Kociemba's algorithm is in theory unlimited in time.

My experience is that it is best to limit the first phase in Kociemba's
algorithm to 13 moves.  But that is only because of time constraints.

dik
--
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
dik@cwi.nl

From @mitvma.mit.edu:hans@freyr.research.ptt.nl  Sat May 30 14:26:53 1992
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From: J.M.Kloosterman@research.ptt.nl (Hans Kloosterman)
Subject: Re:  Lower-bound Kociemba's algorithm
In-Reply-To: <9205300044.AA01102.dik@steenbok.cwi.nl>; from "Dik.Winter@CWI.NL"
 at May 30, 92 2:44 am
To: Dik.Winter@cwi.nl
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> Did you (since your article) do an exhaustive search?  In your article you
> mentioned that you had 6 positions that still do require 18 moves.  And you
> mention that you doubted that there would be 17 move solvers.  Have you
> proven since then that it can not be done in less than 18?  If not, send me
> your positions and I will try.

I have done an exhaustive search and none of the 6 situations of 18 moves
could be reduced to 17 moves (within the group of <R2,L2,F2,B2,U,D>).
For the case you want to verify, one of them is:

   L2 U R2 B2 U2 B2 L2 D2 L2 F2 D' L2 B2 F2 L2 F2 U' D

Hans Kloosterman

From dik@cwi.nl  Sat May 30 18:12:35 1992
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From: Dik.Winter@cwi.nl
Message-Id: <9205302212.AA02915.dik@boring.cwi.nl>
To: J.M.Kloosterman@research.ptt.nl
Subject: Re:  Lower-bound Kociemba's algorithm
Cc: cube-lovers@life.ai.mit.edu

 > I have done an exhaustive search and none of the 6 situations of 18 moves
 > could be reduced to 17 moves (within the group of <R2,L2,F2,B2,U,D>).
 > For the case you want to verify, one of them is:
 >  
 >    L2 U R2 B2 U2 B2 L2 D2 L2 F2 D' L2 B2 F2 L2 F2 U' D
 >  
Of course I verified it ;-).  This one does indeed kill Kociemba's algorithm.
On a fast processor (65 MHz SPARC) with a larger limit database than Kociemba
is using himself (the database is about 5 MByte for the second phase), it took
3 hours 15 minutes to find a minimal solution.  Of 18 moves.

From dik@cwi.nl  Mon Jun  8 16:40:43 1992
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From: Dik.Winter@cwi.nl
Message-Id: <9206082040.AA12825.dik@boring.cwi.nl>
To: cube-lovers@life.ai.mit.edu
Subject: How big is the Magic Domino? (part2, data)

2x2x2 cube
             q+h    loc max            q only    loc max
 0:            1          -                 1          -
 1:            9          -                 6          -
 2:           54          -                27          -
 3:          321          -               120          -
 4:         1847         11               534          -
 5:         9992          8              2256          -
 6:        50136         96              8969          -
 7:       227536        904             33058         16
 8:       870072      13212            114149         53
 9:      1887748     413392            360508        260
10:       623800     604516            930588       1460
11:         2644       2644           1350852      34088
12:                                    782536     402260
13:                                     90280      88636
14:                                       276        276

3x3x3 corners only
             q+h    loc max            q only    loc max
 0:            1          -                 1          -
 1:           18          -                12          -
 2:          243          -               114          -
 3:         2874          6               924          -
 4:        28000         88              6539          -
 5:       205416        792             39528          -
 6:      1168516      15304            199926        114
 7:      5402628      46068            806136        600
 8:     20776176     325680           2761740      17916
 9:     45391616    9757376           8656152      10200
10:     15139616   14665856          22334112      35040
11:        64736      64736          32420448     818112
12:                                  18780864    9654240
13:                                   2166720    2127264
14:                                      6624       6624

magic domino, 1 solution
             q+h    loc max            q only    loc max
 0:            1          -                 1          -
 1:           10          -                 8          -
 2:           67          -                48          -
 3:          420          -               260          -
 4:         2335          -              1330          -
 5:        12260          -              6556          -
 6:        61038          3             31301          -
 7:       291004         12            144392          -
 8:      1327429        793            638407          2
 9:      5821374       6170           2709620         64
10:     24141784      87202          10873023       1261
11:     89480354     990826          39768668      15728
12:    262907144   13212972         124815946     214530
13:    485409604   91824956         296531984    2741192
14:    508704668  161596512         460831364   23949864
15:    232904952  175407548         435219080   72423024
16:     14508468   13668852         215035460   91647012
17:       129376     128592          38469576   35228568
18:          112        112            624320     618368
19:                                      1056       1056

magic domino, 4 solutions
             q+h    loc max            q only    loc max
 0:            4          -                 4          -
 1:           28          -                24          -
 2:          136          -               108          -
 3:          672          -               480          -
 4:         3228          -              2116          -
 5:        15072          -              9120          -
 6:        69000          -             39188          -
 7:       310784         92            166408          -
 8:      1369220       1052            691508         56
 9:      5888676       8656           2812496        192
10:     24209988      92284          11015008       1860
11:     89458152     976008          39837904      16104
12:    262772436   13124304         124673780     202940
13:    485358148   91776620         296336800    2667824
14:    508703948  161595792         460769708   23896632
15:    232904952  175407548         435217336   72421280
16:     14508468   13668852         215035460   91647012
17:       129376     128592          38469576   35228568
18:          112        112            624320     618368
19:                                      1056       1056

magic domino, 8 solutions
             q+h    loc max            q only    loc max
 0:            8          -                 8          -
 1:           56          -                48          -
 2:          272          -               216          -
 3:         1344          -               960          -
 4:         6456          -              4232          -
 5:        30144          -             18240          -
 6:       138000          -             78376          -
 7:       621568        184            332816          -
 8:      2732664       3096           1383016        112
 9:     11649816      28672           5612576        384
10:     46553800     331960          21772432       4584
11:    158726064    3909520          75752384      47792
12:    377277280   46692640         208971608     783864
13:    507933248  129847936         388348544   11790688
14:    414571632  181149888         466373488   54544928
15:    102181280   86967456         334811104   78445984
16:      3271456    3221680         114248208   79836432
17:         7312       7312           7974528    7869280
18:                                     19616      19616


From dik@cwi.nl  Mon Jun  8 16:38:55 1992
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From: Dik.Winter@cwi.nl
Message-Id: <9206082038.AA12813.dik@boring.cwi.nl>
To: cube-lovers@life.ai.mit.edu
Subject: How big is the Magic Domino?

Having done a number of calculations on maximal distances I thought about
getting at newer pastures.  The Magic Domino.  The results follow in the
next mailing, this post discusses a bit about the information found there.
I added to the next mail also the previous results for the 2x2x2 and the
corners of the 3x3x3 together with some additional results not presented
previously.  There are for each puzzle five columns.  The first one
enumerates the number of moves, the next two give the results if both
quarter and half turns are accepted as moves, the last two give the
information if only quarter turns are accepted (of course, on the Domino
this distinction is there only for the U and D faces, the others know
half turns only).  For each case there are two columns, the first giving the
number of positions requiring the stated number of moves, the second
column gives the number of local maxima (i.e. each move brings you closer
to a solution).

There are three tables for the Domino.  The one you want to pick depends
on how you view the puzzle.  The first view is that there is only one
solution with on top 1 to 3 running from left back to right back.  The
second view is that rotation of the puzzle makes different configurations
indistinguishable, so the total number of configuration is (8!)^2 / 4.
An alternative way to look at it is that there are 4 solutions.  One the
standard solution, the others obtained by rotating the domino along the
UD axis.  The distinction between the two views is only a factor of four
in the number of configurations for the different path-lengths.  Finally,
we can view as a solution the configuration with on top 1 to 3 running from
right back to left back in stead of the other way around.  Actually this
solution is not worse than the other, because, if we turn over a solved
Domino we go from one to the other.  This view can also be expressed by saying
there are 8 solutions.  I give results for all three cases.  The numbers upto
(and including) path-length 2 have been checked by hand.

Some remarkable observations.

When we compare the tables for 1 solution and those for 4 solutions we see
that for short path-lengths the number of configurations is multiplied by
4.  On the other hand, for long path-lengths the number of configurations
is equal!  We can say that rotation of the Domino has only a short range
effect.

On the other hand, if we compare both with the 8-solutions tables we see that
the latter allows shorter solutions in general, so mirroring has a long
range effect.

Each of the 6 calculations on the Magic Domino took 2 to 2.5 hours on one
processor of an SGI 4D-420S.  The program is completely memory bound (and
the cache does not help).  It needs at least 31 MByte of core (and must be
resident) otherwise you will get no results at all in reasonable time.  I
tried it on the 32 MByte FPS; while it will happily give results initially
at some stage it will not longer run.  Not only that it will not walk either,
and also not crawl.  It is just sitting there paging in and paging out (a
phenomenon known as page thrashing).  I found that the program would get
less than 0.005 % of the CPU on an otherwise unloaded machine.

The program would enable me to write a 27901440 byte file that would assist
in an optimal solver for the Domino.

dik

From dik@cwi.nl  Mon Jun  8 20:48:48 1992
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From: Dik.Winter@cwi.nl
Message-Id: <9206090048.AA13958.dik@boring.cwi.nl>
To: cube-lovers@life.ai.mit.edu
Subject: Magic Domino part3

Considering my previous fiasco, I have now incorporated the changes needed to
make the magic domino calculating program working into the program that
calculated the corners on a 3x3x3 cube.  The results still match, which
gives me confidence that the algorithms are correct indeed.  Moreover,
it reduced the time to do the 3x3x3 corner calculations to 8 minutes.

dik

From wft@math.canterbury.ac.nz  Wed Jun 10 00:37:06 1992
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From: wft@math.canterbury.ac.nz (Bill Taylor)
Subject: Name query.
To: Cube-Lovers@life.ai.mit.edu
Message-Id: <9206100436.AA27010@math.canterbury.ac.nz>
X-Envelope-To: Cube-Lovers@life.AI.MIT.EDU

Can anyone tell me:-

Why is the "Pons Asinorum" pattern so called ?

---------------------------------------------------------------------
      Bill Taylor              wft@math.canterbury.ac.nz 

Artificial intelligence beats real stupidity.
---------------------------------------------------------------------


From ronnie@cisco.com  Wed Jun 10 15:12:01 1992
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Cc: Cube-Lovers@life.ai.mit.edu
Subject: Re: Name query. 
From: ronnie@cisco.com
Date: Wed, 10 Jun 92 12:11:35 PDT
Sender: ronnie@cisco.com

> Can anyone tell me:-
> 
> Why is the "Pons Asinorum" pattern so called ?

Pons Asinorum is Latin for "Asses' Bridge," and is the name of the
proposition that the base angles of an isoceles triangle are equal.
It is more generally any test of ability imposed upon the inexperienced
or ignorant.

			Ronnie (who has xwebster)

From GOET@rcl.wau.nl  Thu Jun 11 02:46:21 1992
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Date: Thu, 11 Jun 1992 08:41 GMT +01:00
From: "Kees Goet - Landbouwuniversiteit, Afd. I&D" <Kees.Goet@rcl.wau.nl>
Subject: Unsubscribe
To: cube-lovers@life.ai.mit.edu
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Could somebody please remove me from this list.
Thanks in advance.

Kees Goet.

From STEFANO@agrclu.st.it  Thu Jun 11 10:14:42 1992
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From: STEFANO BONACINA <STEFANO@agrclu.st.it>
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Could anyone tell me how can I unsubscribe from this mailing list?
Thanks in advance.
Stefano

From alan@ai.mit.edu  Thu Jun 11 15:20:31 1992
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From: Alan Bawden <Cube-Lovers-Request@ai.mit.edu>
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To: Kees.Goet@rcl.wau.nl, STEFANO@agrclu.st.it
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In-Reply-To: "Kees Goet - Landbouwuniversiteit, Afd. I&D"'s message of Thu, 11 Jun 1992 08:41 GMT +01:00 <01GL2YSCT30W9ED95T@RCL.WAU.NL>
Subject: Please do not bother everybody with administrative requests

   Date: Thu, 11 Jun 92 14:36 CET
   From: STEFANO BONACINA <STEFANO@agrclu.st.it>
   Could anyone tell me how can I unsubscribe from this mailing list?
   Thanks in advance.
   Stefano

   Date: Thu, 11 Jun 1992 08:41 GMT +01:00
   From: "Kees Goet - Landbouwuniversiteit, Afd. I&D" <Kees.Goet@rcl.wau.nl>
   Could somebody please remove me from this list.
   Thanks in advance.
   Kees Goet.

As everyone is informed when they subscribe, administrative requests should
be directed to Cube-Lovers-Request@AI.MIT.EDU (me).  Even if you lost my
original greeting message, the "-Request" suffix is a sufficiently
widespread convention for mailing lists that you should have tried it
first, before bothering the entire mailing list.

STEFANO@agrclu.st.it, I have removed you.

Kees.Goet@rcl.wau.nl, I will be sending you separate mail about your
subscription.

				- Alan

From gls@think.com  Thu Jun 11 17:11:55 1992
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In-Reply-To: ronnie@cisco.com's message of Wed, 10 Jun 92 12:11:35 PDT <9206101911.AA01765@wolf.cisco.com>
Subject: Name query. 

   From: ronnie@cisco.com
   Date: Wed, 10 Jun 92 12:11:35 PDT

   > Can anyone tell me:-
   > 
   > Why is the "Pons Asinorum" pattern so called ?

   Pons Asinorum is Latin for "Asses' Bridge," and is the name of the
   proposition that the base angles of an isoceles triangle are equal.
   It is more generally any test of ability imposed upon the inexperienced
   or ignorant.

The term also carries the connotation that the test is in fact
of the simplest and most elementary kind.  If you can't prove
the Pons Asinorum of geometry, then you don't know even the most
elementary concept of geometry--i.e., as a geometer, you know as
much as a donkey.  And if you cannot form the Pons Asinorum
pattern, you sure don't know much about cubing.

--Guy Steele

From ACW@riverside.scrc.symbolics.com  Thu Jun 11 17:44:29 1992
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Date: Thu, 11 Jun 1992 17:39-0400
From: Allan C. Wechsler <ACW@riverside.scrc.symbolics.com>
Subject: Name query. 
To: gls@think.com, ronnie@cisco.com
Cc: wft@math.canterbury.ac.nz, Cube-Lovers@life.ai.mit.edu
In-Reply-To: <9206112052.AA18593@strident.think.com>
Message-Id: <19920611213942.5.ACW@PALLANDO.SCRC.Symbolics.COM>

    Date: Thu, 11 Jun 1992 16:52 EDT
    From: Guy Steele <gls@think.com>

       From: ronnie@cisco.com
       Date: Wed, 10 Jun 92 12:11:35 PDT

       > Can anyone tell me:-
       > 
       > Why is the "Pons Asinorum" pattern so called ?

       Pons Asinorum is Latin for "Asses' Bridge," and is the name of the
       proposition that the base angles of an isoceles triangle are equal.
       It is more generally any test of ability imposed upon the inexperienced
       or ignorant.

    The term also carries the connotation that the test is in fact
    of the simplest and most elementary kind.  If you can't prove
    the Pons Asinorum of geometry, then you don't know even the most
    elementary concept of geometry--i.e., as a geometer, you know as
    much as a donkey.  And if you cannot form the Pons Asinorum
    pattern, you sure don't know much about cubing.

    --Guy Steele

I think the metaphorical leap from geometry to cubing was probably made
by Bernie Greenberg, in whatever year it was that Hofstadter did his Sci
Am column.  Hofstadter came to MIT to talk to a bunch of cubers,
gathering material for his article.  I was in the group and my name is
mentioned in the article -- the only time I have ever gotten my name
into Sci Am.

"Pons Asinorum" has a lot of Bernie's style about it -- casual use of
Latin, whimsical metaphor, fondness for naming things.  He had a bunch
of cube operators with Latin names, and also some wacky English ones.  I
remember the Spratt Wrench (F R'L D R'L B R'L U R'L) which flips four
edges and was what everyone used before monoflips were discovered.
Bernie also had things with names like the Lesser Hammer of the Right
and the Greater Hammer of the Right; his "patter" was fabulous.  I
regret not having a videotape of Bernie solving the cube in, say, 1978.
(I hope I've got the year right.)

While I'm reminiscing, I should confess that my standard corner operator
is still the same as it was then: (FUR)^5, which exchanges two corners,
leaves the rest of the corners alone, and fucks the edges completely.
(Prudes, do not hassle me.  This has been a technical term in cubing
around MIT since The Beginning.)  Because of this property of "furry
five", I have to home and orient all the corners first, before I touch
the edges.  It's the kind of quirky algorithm you don't see among
younger cubers, because everybody these days learns how to solve the
thing from a book.  In the Beginning, there were no books, and I proudly
state that I solved the cube from scratch, by brainpower.  Later I
discovered that there were easier ways to do things than (FR)^105!  I
had pages and pages covered with little cube diagrams with arrows
showing how the stickers were permuted by a particular sequence.

I'm interested in hearing other reminiscences from people who actually
solved the cube -- you're disqualified if you learned how to solve it
from somebody else, or from a book.

From dik@cwi.nl  Thu Jun 11 18:46:53 1992
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Date: Fri, 12 Jun 1992 00:46:42 +0200
From: Dik.Winter@cwi.nl
Message-Id: <9206112246.AA22860.dik@boring.cwi.nl>
To: cube-lovers@life.ai.mit.edu
Subject: Re:  Name query.  Actually reminiscences.

 >                                                                  I
 > regret not having a videotape of Bernie solving the cube in, say, 1978.
 > (I hope I've got the year right.)
I think it must be later.  The cube was put on the market in Hungary in 1977
and first exported in 1980.  Although earlier examples were privately
exported I presume.

 > While I'm reminiscing, I should confess that my standard corner operator
 > is still the same as it was then: (FUR)^5, which exchanges two corners,
 > leaves the rest of the corners alone, and fucks the edges completely.
Happens to me also.  I still use operators I found myself in favour of
(shorter) processes found later in books.  I remember them better!

 > I'm interested in hearing other reminiscences from people who actually
 > solved the cube -- you're disqualified if you learned how to solve it
 > from somebody else, or from a book.
I got one for my birthday in 1981 (yes, I was late).  By the end of the party
it was completely scrambled.  One long night and a long day afterwards had me
solve the cube.  Although at that moment I had not completely lined up
procedures to do it.  Later I more or less procedurized it.

Much stranger was my first encounter with Square 1.  As all puzzles it was
scrambled within minutes after I brought it home.  I tried to solve it, but
for some reason I did not yet see how to bring it back in the shape of a cube.
The next day when I came home from work it was in the shape of a cube.  It
appears that my 8 year old daughter had done that!  Solving the remainder was
fairly simple.

dik
--
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
dik@cwi.nl

From sjfc!ggww@cci632.cci.com  Sun Jun 14 22:54:44 1992
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From: sjfc!ggww@cci632.cci.com (Gerry Wildenberg)
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To: cube-lovers@life.ai.mit.edu
Subject: Remove me from this list.


Please remove me from the mailing list.

Gerry Wildenberg                         ggww@sjfc.uucp
St. John Fisher College                  sjfc!ggww@cci.com
Rochester, NY 14618                      ggww@sjfc.edu (New, may not yet work.)

From ACW@riverside.scrc.symbolics.com  Tue Jun 16 16:12:32 1992
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Date: Tue, 16 Jun 1992 16:13-0400
From: Allan C. Wechsler <ACW@riverside.scrc.symbolics.com>
Subject: Re:  Name query.  Actually reminiscences.
To: Dik.Winter@cwi.nl, cube-lovers@life.ai.mit.edu
In-Reply-To: <9206112246.AA22860.dik@boring.cwi.nl>
Message-Id: <19920616201302.7.ACW@PALLANDO.SCRC.Symbolics.COM>

    Date: Thu, 11 Jun 1992 18:46 EDT
    From: Dik.Winter@cwi.nl

    Much stranger was my first encounter with Square 1.  As all puzzles it was
    scrambled within minutes after I brought it home.  I tried to solve it, but
    for some reason I did not yet see how to bring it back in the shape of a cube.
    The next day when I came home from work it was in the shape of a cube.  It
    appears that my 8 year old daughter had done that!  Solving the remainder was
    fairly simple.

Our four-year-old managed to assemble our Snafooz into a cube once.  No
one else has been able to do it, and he can't duplicate his success.
(He can't even read.)

From wft@math.canterbury.ac.nz  Fri Jun 19 03:07:51 1992
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Date: Fri, 19 Jun 92 19:07:30 NZS
From: wft@math.canterbury.ac.nz (Bill Taylor)
Subject: reminiscences
To: Cube-Lovers@life.ai.mit.edu
Cc: wft@math.canterbury.ac.nz
Message-Id: <9206190707.AA12708@math.canterbury.ac.nz>
X-Envelope-To: Cube-Lovers@AI.AI.MIT.EDU

Allan C. Wechsler asks for general reminiscences from people who solved
the cube. It's just as well hardly anyone's replied, or the list would be
swamped with boring anecdotes ! So maybe I'll add an anecdote or two of
my own.

Dik.Winter@cwi.nl writes 

> > While I'm reminiscing, I should confess that my standard corner operator
> > is still the same as it was then: (FUR)^5, which exchanges two corners,
> > leaves the rest of the corners alone, and fucks the edges completely.
>
>Happens to me also.  I still use operators I found myself in favour of
>(shorter) processes found later in books.  I remember them better!

Very true. This reminds me of what I read in (I think) the math games
column of Scientific American, about mid-to-late 80's. The cubing craze had
largely passed, and someone who had been an addict, but hadn't touched it
for some years, had occasion to try it again. He realized with horror, that he
couldn't remember a single thing! However, as he began to fiddle with the 
cube rather disconsolately, he found himself automatically doing the right
things. "I couldn't remember how to do it, but my fingers could !!", he said.
         ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~  ~~~~~~~~~~~~~~~~~~~~
This was my experience too, a few years ago. It's quite uncanny, like 
starting to ride a bike again after decades of not doing; only more so.

That's what comes of sticking loyally to your first halfway-decent discoveries
on cube-solving. From madly over-addictive use, they become burned into
your brain tissue.   My own pet method has always been to put all the edges
right first, using only common sense (except maybe at the very end some
slight method needed); then put the corners right using the basic
"8-fold way"..... R~ U L U~ R U L~ U~ . We found this eightfold way quite
easy to remember, the face order is very natural, and the sequence of
clockwise vs anti-clockwise turns, i.e. ACCA CCAA, seems somehow like a 
sonnet rhyming scheme (now burnt irrevocably into my finger-moving cortex).

This eightfold way is just a commutator of a face move and (a commutator of
two face moves)); so it turns out to be group-theoretically natural, as
commutators do "as little as possible". The eight-fold way can also be 
viewed in a natural geometric light, as just a standard 3-permutation of
corners, similaritied away from one another to avoid interference. (Don't
know what the standard technical terms for all this are, sorry; it's probably
old hat to most readers here.) Viewing it this way, one can quickly re-create
several (8-fold) variants, and some 10-fold ones, all of the same type, and
all variously useful. By similarities, one can usually put a corner into a
more useful spot, so as to get two corners done at a time, with one 8-fold.

ENOUGH; of teaching grandmothers to suck eggs. I was going to reminisce. Not
many people seem to do the cube this way, that is, edges first. It was shown
to me by my late colleague Brent Wilson (the other of the "we" refered to 
above). At first it seemed a little unnatural, but once you get used to it, 
it seems super efficient. I suppose everyone feels that way about their own 
methods. The particular 8-fold mentioned above was my own invention, so I've 
always had a soft spot for it.

    Brent and I both started out on the cube the same way, which is
I suppose standard. We spent some little time learning to do the base. Then we
spent some considerably longer time learning to do the middle layer. We found
later that we had both expected the same thing:- that when the middle and
base layers were all successfully done, the top layer would automatically
have to be right !! So of course, we were both temporarily devastated when
it turned out otherwise; and we both realized that we were in the presence 
of a mighty puzzle, and were in for some great fun. So we went ahead and
discovered all the usual group-theoretic things, one by one, over the months.

I have anothger reminiscence to tell about my colleague. I once read of
someone, (J.H.Conway ?), who was alleged to do the cube behind his back !
Well Brent practiced this trick also (unaware of anyone else having
done it, if indeed it was done the same way, even). He invented the method
after having discovered the only all-commuting position, i.e. with all
edges flipped, corners all correct. He perfected a smooth method of doing
this behind his back. The trick is, of course, merely to have a pre-prepared
cube in this position. It doesn't QUITE look random, but if you ADD to it
a couple of random twists, it now looks totally random; at first (and second)
glance. He would show this "random" cube to us, let us hold it (very briefly!),
then take it and do the "all-flips" behind his back. Keeping up a continuous 
patter, as he brought it back he would be saying "...so there's only a couple 
of twists to go", and then as it appeared he would do the last two twists by
sight, without hesitation. As the two "randomizing" twists commute with
the other position, he didn't have to memorize them; indeed he could even
let the audience do them ! Of course this would mean he would have to have
the pure "flipped" pattern to start with, which was easier to detect, alas.

Well, one time, he was to give a talk to some school kids. He wanted to do
the cube behind his back, as a piece-de-resistance. He decided to train
himself up into being able to undo FOUR random twists by sight. He duly did
this. Then when the talk came around, he had a cube prepared in "all-flip"
position, with two twists added, to make it look quite random. Then, when
the highlight of the talk came around, he would display it to the class,
let one or two handle it briefly, to agree it was just another muddled up
cube. Then, HE WOULD EVEN ASK two members of the audience to add an extra
random twist each (just to prove the cube wasn't in a prepared position!)
Then he would do the all-flips operation behind his back, keeping up his
patter. He expected to be able to handle undoing the four random flips left
over, by sight, as he was completing his patter.

When the great event came along, everything went perfectly, without a hitch.
BUT, amazingly, by a 144-to-1 chance, the two flips that the audience added
exactly undid the two that he had put on himself !  So when he brought it
from behind his back, it was already perfectly done. Without batting an
eyelid, he brought his patter to a halt then and there. Needless to say, the
kids were even more staggered than they would have been otherwise. He
resisted all imploring entreaties to tell them how it was done (like all
good conjurers); and I don't thimk he ever did the trick again!

By great good luck, however, I have a vieotape of him doing this trick, from
the demo itself. So if any of you are ever in New Zealand, you can
look me up, and ask to see this amazing event !

Like Allan Wechsler, I would be delighted to hear anyone else's reminiscences,
or cube anecdotes generally. There must be tons, so, don't be shy!

Cheers,   Bill Taylor.



From reid@math.berkeley.edu  Sun Jun 21 13:11:03 1992
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From: reid@math.berkeley.edu (michael reid)
Message-Id: <9206211711.AA08266@math.berkeley.edu>
To: Cube-Lovers@life.ai.mit.edu
Subject: Re:  reminiscences

another call for reminiscences ...

) From: sjfc!ggww@cci632.cci.com (Gerry Wildenberg)
) 
) Please remove me from the mailing list.

yeah, remember back in those days when people were actually asking to
be ADDED to the mailing list?

:-)

btw, administrivia should be sent to "cube-lovers-request@ai.mit.edu".
thank you for your cooperation.                   ^^^^^^^

>From: wft@math.canterbury.ac.nz (Bill Taylor)

> This eightfold way is just a commutator of a face move and (a commutator of
> two face moves)); so it turns out to be group-theoretically natural, as
> commutators do "as little as possible".

here's the way i describe this.  if  sigma  is a permutation on  n  symbols,
(say  1, 2, 3, ... , n), define the  "support" of  sigma  to be those
integers which are NOT fixed by  sigma.  if  tau  is another permutation
on the same set, such that   supp(sigma)  and  supp(tau)  are disjoint,
then  sigma  and  tau  commute (i.e. the commutator is the identity).
if  supp(sigma) intersect supp(tau)  has just one element, then the
commutator is a three-cycle.  as a rule of thumb, the smaller the
intersection of the supports, the smaller the support of the commutator.
in bill's example, ( R~ U L U~ R U L~ U~ ) the two permutations are
"R"  and  "U L U~", which only affect one corner in common.  (actually,
to consider the cube as a permutation group, each corner is really
3 objects, one for each orientation.)  but the analogy works well.
this idea is also helpful for creating three-cycles of corner-edge pairs
as well.  on the 5x5x5 cube, you can make three-cycles of large blocks.
in fact, a larger cube is probably a better visual aid for understanding/
explaining this concept.  another good commutator to try is with the
two sequences  "B1 D2 B3"  and  "R1 U2 R3", which affect two corners
in common.  (this is a fairly well-known maneuver.)

>         "I couldn't remember how to do it, but my fingers could !!", he said.
>          ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~  ~~~~~~~~~~~~~~~~~~~~
> This was my experience too, a few years ago. It's quite uncanny, like 
> starting to ride a bike again after decades of not doing; only more so.

i've also experienced the exact same thing.  it seems as though my hands
remember a short sequence by which i've just conjugated some maneuver,
which helps at the end of the conjugation.  also in these commutators,
after i've done  A  and working on  B, it feels as though my hands are
anxious to undo  A, almost as if the cube is spring loaded and will just
snap back.

[lame story about a friend pulling some hoax deleted]

actually, i think i FINALLY understand the story.  the idea is that not
only is "brent" going to solve the cube behind his back, but he's also
going to do it WITHOUT first looking at it.  actually, the story isn't
nearly as lame as i first thought, after realizing this.

when i visited mit in 1984(?), i saw joe killian do the real trick.
i certainly would have complained if i hadn't been allowed to scramble
the cube exactly as i wished.  in fact, i may have even insisted that
he use MY cube (not too sure, though), just to be certain that the
surface hasn't been textured in any way.  it was quite impressive.
about 5 minutes of studying, then behind the back without peeking.
he said that all it took was a good system of remembering where all
the pieces are.  but i don't know what his system was.

by the way, bill, this "brent" wouldn't happen to be your friend who can
do the cube in 0.87 seconds, would he?  :-)

and speaking of tall tales, let's see if anyone can top this one:
back in the days when i was into speed, er, speed CUBING, i'd solve
the cube maybe 200 times every day.  for some reason, i got into the
habit of scrambling it behind my back (probably from listening to too
many complaints like "you're just watching all the moves you're doing!"
yeah, such a good complaint deserves such a fine solution.)  well,
anyway, one time i stopped scrambling it, and as usual, i get 15
second to study it (standard racing rules).  however, much to my
surprise, the cube was quite UNscrambled!  how could this possibly be?

well, the only explanation is certainly that after scrambling the cube
thousands of times, my hands began to get into a rhythm (maybe even a rut).
they'd just do the same sequence over and over again.  depending upon
my concentraion level, i'd find that sometimes i needed to make a
conscious effort to vary the sequence.  in fact, at least once i got
a pattern that i'd previously seen: it was 4 dots with 6 corners twisted
(hence has order 6).  so it's not too implausible.

like bill, and unlike dik, i spent quite some time struggling with
the cube before i finally solved it; probably about 6 or 7 weeks.
in fact for some time, probably about 2 weeks, i was convinced that
it couldn't be done, except by very dumb luck (as in story above).
of course, in those days, i hadn't heard of cube-lovers, hadn't even
seen a computer, didn't know the furst-hopcroft-luks algorithm, hadn't
even heard of anyone who could solve it ...  but i was just a high school
freshman (age 14) at the time.  i didn't even know what a group was!
this was shortly before the big craze started here in the u.s. (late 1980).
at school, some friends and i talked about it, but the main questions were:
how was it made, and how many combinations did it REALLY have?  i was truly
convinced that trying to solve it would be futile.  there was also a
shortage of them at the time, so i didn't get one until xmas.  in fact,
i remember the tv commercial that ideal put out.  they didn't even make
it clear that it actually turned in all possible directions!  we had all
sorts of ridiculous diagrams and ideas of cables and magnets, but none of
them quite worked.  and how could it turn in all directions?  i heard of a
bookstore somewhere that had one on display (but were otherwise sold out),
so i went to see it.  i remember spending a few minutes twisting it to
find an axis that wouldn't turn!  in fact, i could keep turning the same
face in the same direction, around and around and around ... and the
cables inside never got caught!

sometimes i'm amazed at just how stupid i can be when i try ...

anyway, the story about how i finally figured out how to solve it isn't
nearly as interesting.  after i first heard about people that could do it,
i started to work on it more seriously.  the key ideas were: get all
the corners, (here was something that you could do, and then still do
more without destroying what's already done.  but this was hard and
usually took more dumb luck and/or persistence.)  then two opposite layers.
(again, the middle slice still can turn, even with half turns on the
sides F, B, R, L.)  it took several days to flip the last two edges on
the middle layer.  (i just kept picking a different pair of opposite
layers to solve and stumbled across a U layer monoflip in the process.
of course, it took months before i realized what was actually happening.)
also figuring out how to take it apart (and finally seeing how it was made)
was helpful, 'cause then i could experiment easily.

well, i've droned on long enough.  anyone else got any interesting stories?

mike

From STEVENS@macc.wisc.edu  Tue Jun 23 07:59:22 1992
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Date: Mon, 22 Jun 92 09:19 CDT
From: PAul STevens - MACC - 2-9618 <STEVENS@macc.wisc.edu>
Subject: Re: reminiscences
To: CUBE-LOVERS@life.ai.mit.edu
X-Vms-To: IN%"cube-lovers@life.ai.mit.edu",STEVENS

reid@math.berkeley.edu writes:
 
>well, i've droned on long enough.  anyone else got any interesting stories?
 
 
I think my solution may be considered cheating; but I was pretty proud of
it.  I had almost decided to give up on the thing.  But I had just
designed and built an 8080 'computer'.  It had 2k bytes of 2102's and
had to be programmed with binary switches and whenever the program
clobbered itself the entire program had to be re-entered in binary.
 
So I wrote a program to look for combinations of moves that left most
of the cube alone and only moved a few cubelets.  I studied the best of
these at great length and managed to combine some into 'better' moves,
eventually finding some that moved only three or so cubelets.  These
were then combined into a solution.  A rather god-awful solution I
think.  But my fingers learned the moves and I have never abandoned them
for fear of becoming totally confused.  The same ugly solution has been
passed on for at least one generation and perhaps will persist for
hundreds of years.
 
I still don't know what a group or commutator or ... is except what I
have deduced from reading mail from this group.  I get the front face
corners exactly right, the back corners in the proper position, and then
the back corners rotated properly.
 
Finally the edges go where they belong one at a time, first on the
front and back and finally the four on the sides/top/bottom.
 
I have noticed a lack of discussion of cubes that have pictures on
them such that the entire cube can be right except that a single
center can be upside-down.  I have also painted a 4x4x4 so that the
center 4 squares on each face have to be in the proper position.  Every
time I solve this cube I have to rediscover how it is done.  My fingers
refuse to learn it for me.
 
Behind the back?  You gotta be kidding!
 
                   PAul

From MONET01@mizzou1.missouri.edu  Tue Jun 23 13:31:40 1992
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Date: Tue, 23 Jun 92 12:21:27 CDT
From: MONET01@mizzou1.missouri.edu
To: cube-lovers@life.ai.mit.edu
Subject: Ultimate cube

The recent posting about cubes with photos has prompted me to post about my
favorite cube.  I picked this one up around the end of the BIG cube craze and
have kept it in my desk every since.  The cube looks like someone took a knife
to a normal solved cube and cut a diagonal 'x' through each face and folded the
flaps back down the sides.  This leads to a cube where opposing centers have an
'x' that has four colors in a mirror image.  (It is hard to describe, sorry.)
This cube has to be solved and then the centers oriented properly.  The slick
thing about the cube is that part way through the solution (fairly early on),
you may have to swap top for bottom and start over.  I like to fiddle with it
because at first glance it looks impossible to determine which cubelet is which
to a novice and to a semi-experienced cubist it is not as easy as it looks.
The cube was made by ULTRACO and is called ULTIMATE CUBE (copyrighted 1982).
Unfortunately, when I went back a few weeks later to buy a couple more cubes,
they were all gone and the sales people had no idea what I was talking about.
I think I got this cube at a Mall toy store.  If anyone knows where I can get
a replacement, I would be interested as the printing on a few squares has faded
just like my youth.

From hoey@aic.nrl.navy.mil  Wed Jun 24 15:45:56 1992
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Date: Wed, 24 Jun 92 15:45:51 EDT
From: hoey@aic.nrl.navy.mil (Dan Hoey)
Message-Id: <9206241945.AA09989@Sun0.AIC.NRL.Navy.Mil>
To: MONET01@mizzou1.missouri.edu, cube-lovers@life.ai.mit.edu
Subject: Re: Ultimate cube

MONET01@mizzou1.missouri.edu writes of a cube that

    ``looks like someone took a knife to a normal solved cube and cut
    a diagonal 'x' through each face and folded the flaps back down
    the sides.  This leads to a cube where opposing centers have an
    'x' that has four colors in a mirror image.  (It is hard to
    describe, sorry.)''

I would appreciate a few more details.  I think the color scheme of
each face you describe is something like

+-----+-----+-----+
|.1111|11111|1111.|
|44.11|11111|11.22|
|4444.|11111|.2222|
+-----+-----+-----+
|44444|.111.|22222|
|44444|44.22|22222|
|44444|.333.|22222|
+-----+-----+-----+
|4444.|33333|.2222|
|44.33|33333|33.22|
|.3333|33333|3333.|
+-----+-----+-----+

where 1,2,3, and 4 are distinct colors, but there are still several
ways to make the colors on different faces match up.  Look at a
corner, where the colors are

    +-------+
   /a.bbbbb/c\
  /aaa.bbb/ccc\
 /aaaaa.b/ccccc\
+-------+.......+
 \fffff.e\ddddd/
  \fff.eee\ddd/
   \f.eeeee\d/
    +-------+

That is, one corner is colored a/b, another c/d, and the third e/f,
where I expect some of a,b,c,d,e,f will be the same color.

One possibility was pictured in Hofstatder's Scientific American
article of February, 1981.  It had b=c,d=e,f=a and used twelve colors.
Jim Saxe and I were impressed by its wasteful use of color and its
failure to exhibit edge orientation.  From your remarks about turning
it over, I suspect this isn't what you mean.

You may be talking about the cube in which a=d,b=e,c=f which uses six
colors.  I would say it is as if you cut an 'x' on a cube and
exchanged each triangle with the other triangle on the same edge of
the cube.  That is a reasonably good coloring.  It isn't really
necessary to solve it twice, though.  To find out whether a given
corner goes on the top or bottom, look at the two colors that the
corner shares with the top face center.  Either the corner will have
the two colors in the same order as the top, or they will be reversed,
and that determines whether that corner goes on the top or bottom.
That tells you where the third color on that corner goes, and the
last color is determined by elimination.

There is an even more interesting coloring that uses only four colors.
In this coloring a=c=e and the other three colors are distinct.  Jim
Saxe and I came up with this coloring in our discussions of
Hofstatder's article.  It isn't quite symmetric enough, since its
reflection is a coloring in which b=d=f, a slightly different pattern.
Our discussions then led to the Tartan coloring we talked about in our
article of 16 February 1981.

The only cube in the archives called the Ultimate Cube is the one that
has ``over 43 quintillion solutions.''  It has all six sides colored
the same.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From pbeck@pica.army.mil  Fri Jun 26 13:39:29 1992
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Date:     Fri, 26 Jun 92 13:36:42 EDT
From: Peter Beck (BATDD) <pbeck@pica.army.mil>
To: cube-lovers@ai.mit.edu
Subject:  ultimate cube
Message-Id:  <9206261336.aa13556@COR4.PICA.ARMY.MIL>

name "ultimate cube" is a brand name or TM

my description of its coloring is:

still has six solid colors
if we visualize a standard cube as 
a box with 6 stickers, 1 to each flat face
the the ultimate cube has these 6 stickers
translated and rotated so that
4 of these squares come together on a center
cubie with the diagonal of each square
laying along the edge of the cube.


From mouse@lightning.mcrcim.mcgill.edu  Fri Jun 26 20:06:15 1992
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From: der Mouse  <mouse@lightning.mcrcim.mcgill.edu>
Message-Id: <9206270006.AA06976@Lightning.McRCIM.McGill.EDU>
To: cube-lovers@ai.mit.edu
Subject: Re:  ultimate cube

> my description of ["ultimate" cube's] coloring is:

> still has six solid colors
> if we visualize a standard cube as a box with 6 stickers, 1 to each
> flat face the the ultimate cube has these 6 stickers translated and
> rotated so that 4 of these squares come together on a center cubie
> with the diagonal of each square laying along the edge of the cube.

Hm, there are 12 edges on a cube.  That leaves half of them unaccounted
for.  What do they get?  (Note that you also have to shrink the 6
stickers, beacuse the face diagonal is longer than an edge.)

					der Mouse

			    mouse@larry.mcrcim.mcgill.edu

From pbeck@pica.army.mil  Tue Jun 30 10:05:29 1992
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Date:     Tue, 30 Jun 92 7:46:56 EDT
From: Peter Beck (BATDD) <pbeck@pica.army.mil>
To: cube-lovers@ai.mit.edu
Subject:  ultimate cube, correction
Message-Id:  <9206300746.aa20380@COR4.PICA.ARMY.MIL>


ULTIMATE CUBE  - sorry for my past misinformation this time I will try
and describe it by observation.

The cube is covered with 24 (4 to a face) 45, 45, 90 deg triangles.
 These triangles have there hypotenuse along the edge of the cube and
their 90 deg apex at the center of the center cubie.

Each opposite face has the same coloring except that the rotation of
the colors is opposite.  For example if the front face has a green,
orange, yellow and red triangle in clockwise order then in order for
the rear face to correspond it has a color rotation that is counter
clockwise.

The coloring scheme is that the top and bottom triangles on the side
faces (ie, front, right, left, back) are the same.  Six colors (
green, orange, yellow , red for the front and rear, and green,
white,yellow,blue for the sides and white, orange, blue, red for the
top and bottom)  are used and there are 4 triangles of each color. 

I think this is an accurate description, if there are questions please
ask I have my cube at my desk.



From tjj@rolf.helsinki.fi  Tue Jun 30 15:36:42 1992
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Date: Tue, 30 Jun 92 22:34:53 +0300
From: tjj@rolf.helsinki.fi (Timo Jokitalo)
Message-Id: <9206301934.AA10177@rolf.helsinki.fi>
To: cube-lovers@ai.mit.edu
Subject: Please, quick, I need the address of the new puzzle shop in Amsterdam


I believe it was with the Dutch Cubists' Club newletter that I got an
advertisement of a new puzzle shop in Amsterdam. I tried to find this
advertisement, not, but could not. I'm leaving Finland on Thursday
afternoon, and will be passing through Amsterdam, so I would be forever
grateful to any kind soul who would  send me the address!!!

	Thanks, 
	Timo (tjj@rolf.helsinki.fi)

From tjj@rolf.helsinki.fi  Tue Jun 30 21:15:05 1992
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From: tjj@rolf.helsinki.fi (Timo Jokitalo)
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To: cube-lovers@ai.mit.edu


In the mail I just sent, there were a couple of serious typing errors, but
I think the gist of the message should be clear... sorry!

	Timo

From news@cco.caltech.edu  Thu Jul  9 19:28:24 1992
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From: ph@vortex.ama.caltech.edu (Paul Hardy)
Subject: Re: Name query. 
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	<19920611213942.5.ACW@PALLANDO.SCRC.Symbolics.COM>
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In article <19920611213942.5.ACW@PALLANDO.SCRC.Symbolics.COM> ACW@riverside.scrc.symbolics.com (Allan C. Wechsler) writes:

>   While I'm reminiscing, I should confess that my standard corner operator
>   is still the same as it was then: (FUR)^5, which exchanges two corners,
>   leaves the rest of the corners alone, and fucks the edges completely.
>   (Prudes, do not hassle me.  This has been a technical term in cubing
>   around MIT since The Beginning.)  Because of this property of "furry
>   five", I have to home and orient all the corners first, before I touch
>   the edges.  It's the kind of quirky algorithm you don't see among
>   younger cubers, because everybody these days learns how to solve the
>   thing from a book.  In the Beginning, there were no books, and I proudly
>   state that I solved the cube from scratch, by brainpower.  Later I
>   discovered that there were easier ways to do things than (FR)^105!  I
>   had pages and pages covered with little cube diagrams with arrows
>   showing how the stickers were permuted by a particular sequence.
>
>   I'm interested in hearing other reminiscences from people who actually
>   solved the cube -- you're disqualified if you learned how to solve it
>   from somebody else, or from a book.

I also solved the cube alone at first.  I solved the top and middle first,
then spent some time pondering the final face.  I realized that manipulating
the corners was trickier than the edges because there were three faces rather
than two, so I solved the bottom corners and then got the bottom edges in
place.

I eventually got Singmaster's book, and found that my method of solving two
layers was faster than his.  I don't quite remember now, but I think it was
because I had found a quick method for flipping a piece on the middle edge
around if necessary (i.e., if it was in the correct position but flipped the
wrong way) without disturbing anything else on the top or middle of the cube.
Still, Singmaster's book had many patterns that were fun to go through and
see evolve.

I've long since lost my copy of Singmaster's book (one move too many); is it
still available?

				--Paul
--
This is my address:         ph@ama.caltech.edu
This is UUCP:               ...!{decwrl,uunet}!
This is my address on UUCP: ...!{decwrl,uunet}!caltech.edu!ama!ph
Any questions?

"Does Emacs have the Buddha nature?"  --Paul Hardy   "Yow!" --Zippy

From alan@ai.mit.edu  Sun Jul 19 10:51:55 1992
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From: Alan Bawden <Alan@lcs.mit.edu>
Sender: Alan@lcs.mit.edu
To: Cube-Lovers
Subject: [nick@lcs.mit.edu: In-sol-u-bil-i-tyyyy!]

Date: Thu, 16 Jul 92 13:11 EDT
From: nick@lcs.mit.edu
Reply-To: nick@lcs.mit.edu
Subject: In-sol-u-bil-i-tyyyy!
To: qotd@ghoti.lcs.mit.edu


	In a wonderful article about Claude Shannon in the April 92 IEEE
Spectrum, a few lines from his poem called "a Rubric on Rubik's Cubics"
(to the tune of Ta-ra-ra-boom-de-ay):

Respect your cube and keep it clean,
Lube your cube with Vaseline.
Beware the dreaded cubist's thumb,
the calloused hands and fingers numb.
No borrower nor lender be,
Rude folk might switch two tabs on thee.
The most unkindest switch of all,
Into Insolubility.

[Chorus]

In-sol-u-bil-i-ty!
The strangest place to be
However you persist
Solutions don't exist!


From @mail.uunet.ca:mark.longridge@canrem.com  Mon Aug  3 02:57:03 1992
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From: Mark Longridge <mark.longridge@canrem.com>
Date: 	Sun, 2 Aug 1992 20:00:00 -0400
To: cube-lovers@ai.mit.edu
Subject: cube theory


I've been doing some research to try and figure out some things
about the cube. I've also tried (unsuccessfully) to develop a sort
of CRC or checksum for the cube. With this cube "signature" I could
then find out which depth each pattern requires. I'm puzzled how
some computer types managed to find God's Algorithm for the 3x3x3
squares group. How do you keep track of all the patterns without
repeating yourself? If it is by holding all patterns in an array
the array must become huge.

                    Maximum Depth (using q and h moves)
                    -------------

2x2x2 sq group       4 (24 total states)
Pyraminx            11 (or 14 including the 3 tips)
2x2x2               11 (14 using q turns only)
3x3x3 corners only  11
3x3x3 sq group      15 (half turns only, don't know if using q improves
this)
3x3x2 domino        18 (for 1 solution)

A local maxima is a state where any possible move will bring you closer
to a solution. This can occur on the 2x2x2 at depth 4 and on the 3x3x3
at
depth 3. Note that all possible patterns at maximum depth are local
maxima,
however it is surprising that local maxima may occur in patterns much
closer to the surface.

To date, no work has been done to determine the depth of the
dodecahedron
(megaminx) or square 1.

Some questions:

What pattern is an example of local maxima? e.g. 3x3x3 at depth 3
-> 12-flip, 12-flip 8-twist

        q+h     Depth   Patterns
2x2x2             1        9
3x3x3             1       18
Dodecahedron      1       48

        Analysis of the full cube group
        -------------------------------

Moves Deep       arrangements (q+h)   arrangements (q only) *

  0                    1                   1
  1                   18                  12
  2                  243                 114
  3                3,240               1,068
  4            >  48,600              10,011

* Work by Zoltan Kaufmann

Notes:  At 1 move deep each of the 6 sides can turn 3 ways (+ - 2)
giving 18 distinct patterns

        At 2 moves deep it is redundant to turn the same side again
        so 5 sides can turn 3 ways so 18x15=270
        However, with opposite turns order is not significant,
        e.g. T,D = D,T   F,B = B,F   L,R = R,L
        since each of these can occur in 9 different ways there are 27
        redundancies so 270 - 27 = 243

        At 3 moves deep with the first 2 moves on opposite faces don't
        turn the face used in move one since:
        T,D,T = T2,D   F,B,F = F2,B   L,R,L = L2,R
        This can occur in 3x3x3=27 ways for each case so 81 are dropped
        (Remember the first 2 moves have already been weeded of
        redundancy!)
        Also when the 2nd and 3rd moves are of opposite faces
   e.g. T,R,L = T,L,R  B,R,L = B,L,R   F,R,L = F,L,R   D,R,L = D,L,R
        T,B,F = T,F,B  D,B,F = D,F,B   L,B,F = L,F,B   R,B,F = R,F,B
        F,T,D = F,D,T  B,T,D = B,D,T   L,T,D = L,D,T   R,T,D = R,D,T

    since each of these can occur 27 different ways in each of the
    cases this gives 27x12 = 324 redundancies
    Thus 243x15 = 3645, removing the redundancies gives 3645-81-324=3240

        At 4 moves deep.... still working on this one! Zoltan Kaufmann
        has done 4 moves deep using quarter turns, but has anyone
        calculated farther using q and h turns? I'd be interested in
        the source code of any programs people have written on finding
        path-lengths. Also what is an example of a local maxima close
        to the surface, e.g. 4 moves. I believe Jim Saxe and Dan Hoey
        have done some work in this regard.

        One more question: What is the maximum number of moves required
        if you do the 3x3x3 one face last? The best results I've seen
        are 19 q and h moves.
        -> Mark Longridge
--
Canada Remote Systems  - Toronto, Ontario/Detroit, MI
World's Largest PCBOARD System - 416-629-7000/629-7044

From hoey@aic.nrl.navy.mil  Mon Aug  3 11:10:35 1992
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From: hoey@aic.nrl.navy.mil
Message-Id: <9208031510.AA03178@sun13.aic.nrl.navy.mil>
O: Mark Longridge <mark.longridge@canrem.com>
Cc: Cube-lovers@life.ai.mit.edu
Subject: Re: cube theory

Dear Mark,

I sent you email on 8 June.  Did you receive it?  Are you interested
in acquiring a Tartan cube? Did you ever get Minh Thai's book on
Rubik's Revenge?

With respect to the 3x3x3 squares group,

> How do you keep track of all the patterns without repeating yourself?

there are several ways.  The most general is to use the Sims table
(aka the FHL table) for the subgroup, which gives a mixed-base
enumeration of the positions.  See my message of 1 February 1981.

> A local maxima is a state where any possible move will bring you
> closer to a solution.

That's ``local maximum''.  ``Maxima'' is the plural of ``maximum''.

> Note that all possible patterns at maximum depth are local
> maxima, ....

We call such positions ``global maxima'' because they are at the
overall maximum depth.  The statement is then that every global
maximum is a local maximum.

> To date, no work has been done to determine the depth of the
> dodecahedron (megaminx)....

Well, I've done some looking at it.  Since my initial remarks on 23
September 1982, I've figured out a way to generate a recurrence for
it, but it seems I haven't put it down anywhere.  Are you interested?
(Do I have to tell anyone to answer that question only to
Hoey@AIC.NRL.Navy.Mil, not the list?)

> What pattern is an example of local maxima? e.g. 3x3x3 at depth 3
> -> 12-flip, 12-flip 8-twist

Jim Saxe and I listed the 25 symmetric local maxima in our message on
Symmetry and Local Maxima, dated 14 December 1980.  We verified Jim's
conjecture that the four-spot is a local maximum, but not on the
grounds of symmetry, and reported that on 22 March 1981.  Do you have
access to the cube-lovers archives?

> Moves Deep       arrangements (q+h)   arrangements (q only) *

>   0                    1                   1
>   1                   18                  12
>   2                  243                 114
>   3                3,240               1,068
>   4            >  48,600              10,011

This is in the archives, too

    5                                   93,840 (22 March 1981)
    6                                  878,880 (14 August 1981)
    7                                8,221,632 (7 December 1981)

David C. Plummer and I had hoped to use his program (which counted the 7
QT positions) to extend this to 8 QT, but we got busy.  I still have
hopes....

> At 2 moves deep it is redundant to turn the same side again....
> However, with opposite turns order is not significant, e.g. T,D = D,T....

This approach appeared on 9 January 1981.  It showed how to follow the
argument to 25 QT, and to get what are still the best known lower bounds
for the ordinary cube and for the supergroup.

> Also what is an example of a local maxima close to the surface, e.g.
> 4 moves. I believe Jim Saxe and Dan Hoey have done some work in this
> regard.

It's known there are no local maxima at 7 QT or less.  The shortest
known local maxima are Pons Asinorum and the four-spot, both at 12 QT.
I know of no results between 8 and 11 QT.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From pbeck@pica.army.mil  Mon Aug 10 14:13:31 1992
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Date:     Mon, 10 Aug 92 14:11:51 EDT
From: Peter Beck (BATDD) <pbeck@pica.army.mil>
To: CUBE-LOVERS@life.ai.mit.edu
Subject:  smart alex - new puzzle
Message-Id:  <9208101411.aa10693@COR4.PICA.ARMY.MIL>


SMART ALEX  about $14 retail

from  2 MCH FUN
          777-108th Avenue N.E. #2340
          Bellevue, WA  98004
          206-453-5659

purchase source
           PUZZLETTS
           MIKE GREEN
           24843 144th Place S.E.
           KENT, WA  98042
           206-630-1432

                or

           from myself

DESCRIPTION:  This puzzle is similar to the Hungarian puzzle UFO but a
little more complex.   The puzzle has a cube at its center which is
cut in half and these two pieces can rotate with respect to each
other.  In the plane of this cut there are four hubs that can rotate,
2 on the x-axis and 2 on the y-axis.  These hubs have a hexagonal
cross section and are divided into six equilateral pie shaped wedges.
 By rotating the hubs and then by rotating the center cube 3 hub
pieces on each axle are moved at a time.

The hub pieces have 2 colors each one perpendicular to the axis and
the other on its edge.  There are 2 wedges with the same coloring.
 The center cube is also colored.  The object is to arrange the hub
pieces so that the edge pieces align with the center cube and that the
perpendicular sides are solid colored and aligned.


From pbeck@pica.army.mil  Mon Aug 10 16:22:55 1992
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Date:     Mon, 10 Aug 92 14:12:42 EDT
From: Peter Beck (BATDD) <pbeck@pica.army.mil>
To: CUBE-LOVERS@life.ai.mit.edu
Subject:  mazeland
Message-Id:  <9208101412.aa10952@COR4.PICA.ARMY.MIL>


MAZELAND GARDENS and DISCOVERY CENTER
POB 573
Alexandria Bay, NY 13607
315-482-LOST
800-585-FUNN

Alexandria Bay is in the Thousand Islands region near where I-81
crosses into Canada.

Mazeland Gardens is an entertainment center (the property was formerly
a miniature golf course) of mazes.  It has 2 mazes that are
constructed with arborvitae hedges, one a 1/2 acre in size and the
other a full acre.  In addition it has 2 mazes that are constructed
with stakes and colored tape to mark the walls.

I went without kids and had fun.  The families with kids looked like
they were having a good time.  If you are in the area you might want
to give it a try.

PS   The fellow working the desk said that there is are 2 other mazes
on the east coast that he is aware of:
  one is in North Carolina somewhere and the other is in Daytona beach
Florida/

Pete Beck





From hoey@aic.nrl.navy.mil  Thu Aug 20 13:51:38 1992
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Date: Thu, 20 Aug 92 13:51:25 EDT
From: hoey@aic.nrl.navy.mil
Message-Id: <9208201751.AA14111@sun30.aic.nrl.navy.mil>
To: Allan C Wechsler <ACW@riverside.scrc.symbolics.com>,
        wft@math.canterbury.ac.nz (Bill Taylor)
Cc: Cube-Lovers@ai.mit.edu
Subject: Re: subgroups

On 14 Jan 1992,  Allan C. Wechsler posted

>Regarding the meta-approach of descending through a series of subgroups,
>how much leverage does properly selecting the chain give you?  It seems
>like the jump from <F2,B2,...> to <F2,B2,L2,R2,...> is pretty large.
>There are probably other paths through the subgroup lattice.

>Does anyone have a table of subgroups?

As far as selecting the chain goes, I have been meaning to look into
that a bit.  Of course, since Bill posted, the results of Hans
Kloosterman, Michael Reid, and Dik Winter have shown that you indeed
get a lot of leverage.  I would like to get some idea of the possible
group towers, for a more general approach to selecting which towers
give you leverage.  But what I haven't been able to figure out is how
to figure out which coset of G1 wrt G2 you're in.  I've been able to
figure it out for specific groups, but if we wanted to do this for a
lot of chains, we would need to do coset identification given G1 and
G2 as a table of strong generators.  We could in fact ensure that the
strong generators of G1 form a subset of those of G2.  Is that a hard
thing to do?

More to the point, I've heard that the FHL algorithm should more
properly be called Sims's algorithm and that Furst, Hopcroft, and Luks
mostly analyzed the performance.  I haven't read anything by Sims on
it, though.  Is there a good reference that treats this sort of
algorithm in a more general setting?  I have toyed with implementing
the Jerrum improvements to FHL, but it is a mighty complicated beast.
Also, a talk announced in the archives mentioned 1987 work by Akos
Seress that was supposed to be an improvement, but I don't know
whether it got published.  Anyway, if not, do you know if there is a
good general way of finding out which coset a given position is in.

On 29 Jan 1992, wft@math.canterbury.ac.nz (Bill Taylor) posted

> There hasn't been any response to this, seemingly, which is a pity.

For some reason, I never saw Bill's message.  I just noticed it when
comparing my files against the archives.  [ Archives seekers note: the
archives have moved to FTP.LCS.MIT.EDU (18.26.0.36), still in
directory /pub/cube-lovers. ]

> In any event, I would like to know of any other well-known subgroups.
> There are the slice group; double-slice group; U group; square group;
> anti-slice group. How many others are there not mentioned here, that 
> people know of ?

There were some tables in Singmaster with more examples, and there are
the stuck-faces groups that I wrote about on 21 July 1981.  I seem to
recall there was some non-obvious equivalence between two groups,
perhaps the slice group and the antislice group.  But a general list
of popular subgroups would be interesting.  Of course a list of *all*
the subgroups would have, um, over three beelion of them.  I suspect
it has more than 4.3x10^19.  Does anyone know a good way of counting
how many subgroups there are?  Or even a way of estimating the number?
By comparison, the symmetries of the cube form a 48-element group with
98 subgroups.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From ACW@riverside.scrc.symbolics.com  Thu Aug 20 16:24:45 1992
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Date: Thu, 20 Aug 1992 16:25-0400
From: Allan C. Wechsler <ACW@riverside.scrc.symbolics.com>
Subject: Re: subgroups
To: hoey@aic.nrl.navy.mil, ACW@riverside.scrc.symbolics.com,
        wft@math.canterbury.ac.nz
Cc: Cube-Lovers@ai.mit.edu
In-Reply-To: <9208201751.AA14111@sun30.aic.nrl.navy.mil>
Message-Id: <19920820202540.7.ACW@PALLANDO.SCRC.Symbolics.COM>

    Date: Thu, 20 Aug 1992 13:51 EDT
    From: hoey@aic.nrl.navy.mil

[...]
						
						Of course a list of *all*
    the subgroups would have, um, over three beelion of them.  I suspect
    it has more than 4.3x10^19.  Does anyone know a good way of counting
    how many subgroups there are?  Or even a way of estimating the number?
    By comparison, the symmetries of the cube form a 48-element group with
    98 subgroups.

All we should really be interested in are conjugate classes of
subgroups.  I think.


From alan@ai.mit.edu  Thu Aug 20 20:03:23 1992
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From: Alan Bawden <Cube-Lovers-Request@ai.mit.edu>
Sender: Cube-Lovers-Request@ai.mit.edu
To: Cube-Lovers
Subject: Archives

   Date: Thu, 20 Aug 92 13:51:25 EDT
   From: hoey@aic.nrl.navy.mil
   ...  [ Archives seekers note: the archives have moved to FTP.LCS.MIT.EDU
   (18.26.0.36), still in directory /pub/cube-lovers. ] ...

No, that isn't right.  The correct address is FTP.AI.MIT.EDU (which is at
128.52.32.11 -- at least this week).  Here is the text I currently send to
people who are new subscribers or who express interest in the archives:

Using FTP, connect to FTP.AI.MIT.EDU, login as "anonymous" (any password),
and in the directory "pub/cube-lovers" you will find the nine (compressed)
files "cube-mail-0.Z" through "cube-mail-8.Z".

Archive vital statistics (when uncompressed):

	   File		   From		   To	     Size (bytes)
	   ----		   ----		   --	     ------------
	cube-mail-0	12 Jul 80	23 Oct 80	185037
	cube-mail-1	 3 Nov 80	 9 Jan 81	135719
	cube-mail-2	10 Jan 81	 3 Aug 81	138566
	cube-mail-3	 3 Aug 81	 3 May 82	137753
	cube-mail-4	 4 May 81	11 Dec 82	139660
	cube-mail-5	11 Dec 82	 6 Jan 87	173364
	cube-mail-6	10 Jan 87	13 Apr 90	216733
	cube-mail-7	12 Oct 90	 9 Sep 91	137508
	cube-mail-8	 1 Nov 91	25 May 92	171205

In addition, the file "recent-mail" contains a copy of the currently active
section of the archive.  (Unfortunately, due to the way mail works here at
the AI Lab, it is not possible to have new mail accumulate directly into
this file, so there may be some delay before a new message arrives here.)

				- Alan

From reid@math.berkeley.edu  Thu Aug 20 20:10:13 1992
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Date: Thu, 20 Aug 92 17:03:31 PDT
From: reid@math.berkeley.edu (michael reid)
Message-Id: <9208210003.AA19361@math.berkeley.edu>
To: ACW@riverside.scrc.symbolics.com, hoey@aic.nrl.navy.mil,
        wft@math.canterbury.ac.nz
Subject: Re: subgroups
Cc: Cube-Lovers@ai.mit.edu

dan hoey writes:

> On 14 Jan 1992,  Allan C. Wechsler posted

> >Regarding the meta-approach of descending through a series of subgroups,
> >how much leverage does properly selecting the chain give you?  It seems
> >like the jump from <F2,B2,...> to <F2,B2,L2,R2,...> is pretty large.
> >There are probably other paths through the subgroup lattice.

> >Does anyone have a table of subgroups?

well, i don't know ALL the subgroups, but i did some investigation
before devising my three stage algorithm.  one of the great advantages
of thistlethwaite's four stage method is that since each subgroup
restricts the motion of various faces, it is routine to exhaustively
search the cosets spaces at each stage, since we only make twists that
leave us in the given space.  so i looked at all possible ways of
restricting various faces, up to symmetry.  there are three possible
restrictions for a face:  no restriction, half turns only, no turns.
our problem is then coloring the faces of the cube with 3 colors, up
to symmetry (rigid and non-rigid).  the polya polynomial for the faces
of the cube under this group of symmetries is:

          ( x^6 + 3 x^5 + 9 x^4 + 13 x^3 + 14 x^2 + 8 x ) / 48

so there are 56 different ways to three-color the faces.  i spent the
better part of an evening and most of the night calculating (by hand)
the orders of these subgroups.  shortly thereafter, i saw an announcement
for the group theory package GAP, which specifically mentions calculating
the order of the rubik's cube group.  so i used the package to verify my
answers.  here's the list (i don't see a canonical way of ordering them):

 1. |<>|                                               =                    1
 2. |<U2>|                     = 2                     =                    2
 3. |<U>|                      = 2^2                   =                    4
 4. |<U2, D2>|                 = 2^2                   =                    4
 5. |<U, D2>|                  = 2^3                   =                    8
 6. |<U, D>|                   = 2^4                   =                   16
 7. |<U2, R2>|                 = 2 3                   =                   12
 8. |<U, R2>|                  = 2^6 3^2 5^2           =                14400
 9. |<U, R>|                   = 2^6 3^8 5^2 7         =             73483200
10. |<U2, R2, L2>|             = 2^5 3                 =                   96
11. |<U, R2, L2>|              = 2^12 3^4 5^2 7        =             58060800
12. |<U2, R, L2>|              = 2^12 3^4 5^2 7        =             58060800
13. |<U2, R, L>|               = 2^14 3^4 5^2 7^2      =           1625702400
14. |<U, R, L2>|               = 2^14 3^11 5^2 7^2     =        3555411148800
15. |<U, R, L>|                = 2^14 3^13 5^3 7^2     =      159993501696000
16. |<U2, R2, F2>|             = 2^5 3^4               =                 2592
17. |<U, R2, F2>|              = 2^8 3^5 5^2 7         =             10886400
18. |<U, R, F2>|               = 2^10 3^12 5^2 7^2     =         666639590400
19. |<U, R, F>|                = 2^18 3^12 5^2 7^2     =      170659735142400
20. |<U2, R2, L2, D2>|         = 2^6 3                 =                  192
21. |<U, R2, L2, D2>|          = 2^13 3^4 5^2 7        =            116121600
22. |<U, R2, L2, D>|           = 2^15 3^4 5^2 7^2      =           3251404800
23. |<U, R, L2, D2>|           = 2^15 3^11 5^2 7^2     =        7110822297600
24. |<U, R, L, D2>|            = 2^15 3^13 5^3 7^2     =      319987003392000
25. |<U, R, L, D>|             = 2^16 3^14 5^3 7^2 11  =    21119142223872000
26. |<U2, R2, L2, F2>|         = 2^11 3^4              =               165888
27. |<U, R2, L2, F2>|          = 2^13 3^5 5^2 7^2      =           2438553600
28. |<U2, R, L2, F2>|          = 2^14 3^5 5^2 7^2      =           4877107200
29. |<U2, R, L, F2>|           = 2^14 3^5 5^2 7^2      =           4877107200
30. |<U, R2, L2, F>|           = 2^14 3^13 5^3 7^2 11  =     1759928518656000
31. |<U2, R2, L, F>|           = 2^14 3^13 5^3 7^2 11  =     1759928518656000
32. |<U2, R, L, F>|            = 2^14 3^13 5^3 7^2 11  =     1759928518656000
33. |<U, R, L2, F>|            = 2^24 3^13 5^3 7^2 11  =  1802166803103744000
34. |<U, R, L, F>|             = 2^24 3^13 5^3 7^2 11  =  1802166803103744000
35. |<U2, F2, R2, L2, B2>|     = 2^13 3^4              =               663552
36. |<U, F2, R2, L2, B2>|      = 2^16 3^5 5^2 7^2      =          19508428800
37. |<U2, F, R2, L2, B2>|      = 2^16 3^5 5^2 7^2      =          19508428800
38. |<U2, F2, R, L, B2>|       = 2^16 3^5 5^2 7^2      =          19508428800
39. |<U2, F, R, L2, B2>|       = 2^16 3^14 5^3 7^2 11  =    21119142223872000
40. |<U, F, R2, L2, B2>|       = 2^16 3^14 5^3 7^2 11  =    21119142223872000
41. |<U2, F, R, L, B2>|        = 2^16 3^14 5^3 7^2 11  =    21119142223872000
42. |<U, F2, R, L, B2>|        = 2^16 3^14 5^3 7^2 11  =    21119142223872000
43. |<U, F, R, L2, B2>|        = 2^27 3^14 5^3 7^2 11  = 43252003274489856000
44. |<U2, F, R, L, B>|         = 2^16 3^14 5^3 7^2 11  =    21119142223872000
45. |<U, F2, R, L, B>|         = 2^27 3^14 5^3 7^2 11  = 43252003274489856000
46. |<U, F, R, L, B>|          = 2^27 3^14 5^3 7^2 11  = 43252003274489856000
47. |<U2, F2, R2, L2, B2, D2>| = 2^13 3^4              =               663552
48. |<U, F2, R2, L2, B2, D2>|  = 2^16 3^5 5^2 7^2      =          19508428800
49. |<U, F2, R2, L2, B2, D>|   = 2^16 3^5 5^2 7^2      =          19508428800
50. |<U, F, R2, L2, B2, D2>|   = 2^16 3^14 5^3 7^2 11  =    21119142223872000
51. |<U, F, R2, L2, B, D2>|    = 2^16 3^14 5^3 7^2 11  =    21119142223872000
52. |<U, F, R, L2, B2, D2>|    = 2^27 3^14 5^3 7^2 11  = 43252003274489856000
53. |<U, F, R2, L2, B, D>|     = 2^16 3^14 5^3 7^2 11  =    21119142223872000
54. |<U, F, R, L, B2, D2>|     = 2^27 3^14 5^3 7^2 11  = 43252003274489856000
55. |<U, F, R, L, B, D2>|      = 2^27 3^14 5^3 7^2 11  = 43252003274489856000
56. |<U, F, R, L, B, D>|       = 2^27 3^14 5^3 7^2 11  = 43252003274489856000

subgroups with the same order are equal (possibly after necessary rotation
of the cube) with the following exceptions: (3, 4), (11, 12), (30, 31) and
(30, 32).  equality of various pairs of subgroups can be obtained from the
three maneuvers:

     R L F2 R2 F B L F2 B2 R2 F2 B2 L F B3 R3 L3  ~  U2 ,
                so that  <L, F, R, B> = <L, F, R, B, U2>,

     F2 U2 L2 F2 R2 U2 F2 R F2 U2 R2 F2 L2 U2 F2  ~  L ,
                so that  <U2, F2, R, L2> = <U2, F2, R, L>  and

     R2 F2 B2 L2 U2 L2 F2 B2 R2  ~  D2 ,
                so that  <U2, F2, R2, L2, B2> = <U2, F2, R2, L2, B2, D2>.


thistlethwaite's filtration is  56 --> 53 --> 49 --> 47 --> 1.
kloosterman replaced  47  by a subgroup not on this list (one not obtained
by restricting face turns).  call this  56 --> 53 --> 49 --> kl --> 1.
(in his final stage, kloosterman allows all twists available in the
subgroup  49.)  my filtration is  56 --> 19 --> 17 --> 1 , which was
chosen precisely because it had the smallest size of the largest coset
space amongst all three stage filtrations with subgroups from the above.
winter's filtration is  56 --> 49 --> kl --> 1.  it may be the case that
this can be improved by replacing  kl  with  17 , and allowing all face
turns available in the subgroup  49.  i haven't had the time to look into
this yet.

using subgroups on the list above, we see that the only reasonable two
stage filtrations are:

56 --> 29 --> 1    with coset sizes     8868372480    and   4877107200
56 --> 22 --> 1    with coset sizes    13302558720    and   3251404800
56 --> 27 --> 1    with coset sizes    17736744960    and   2438553600
56 --> 49 --> 1    with coset sizes     2217093120    and  19508428800
56 --> 13 --> 1    with coset sizes    26605117440    and   1625702400

of these, the best seems to be  56 --> 49 --> 1 , since it has the most
symmetries (16).  the number of symmetries the others have is  4 (for 29),
8 (for 22), 2 (for 27) and 2 (for 13).  furthermore, aside from subgroup
49, the other intermediate groups seem to have too much restriction to
be efficient.  also, of course, dik winter has already calculated that
the stage  56 --> 49  can always be accomplished in 12 face turns.


> On 29 Jan 1992, wft@math.canterbury.ac.nz (Bill Taylor) posted

> > There hasn't been any response to this, seemingly, which is a pity.

> For some reason, I never saw Bill's message.  I just noticed it when
> comparing my files against the archives.  [ Archives seekers note: the
> archives have moved to FTP.LCS.MIT.EDU (18.26.0.36), still in
> directory /pub/cube-lovers. ]

i also seem to have missed both allen's post as well as bill's reply.
perhaps 'twas the twilight zone between the start of my subscription to
cube-lovers and the time it takes recent messages to reach the archives.

however, i don't find the archives on  ftp.lcs , but rather on
ftp.ai.mit.edu.  also i see we've spawned  cube-mail-8.Z.

> > In any event, I would like to know of any other well-known subgroups.
> > There are the slice group; double-slice group; U group; square group;
> > anti-slice group. How many others are there not mentioned here, that 
> > people know of ?

in addition to those listed above there are subgroups generated by
combinations of face turns and slice turns, e.g.  <U, M_R>, <U2, M_R>,
<U, R, M_R>, etc.  i haven't looked at these at all.  there's a lot of
work to be done here.

mike

From ronnie@cisco.com  Thu Aug 20 23:32:46 1992
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Date: Thu, 20 Aug 92 17:52:40 -0700
From: Ronnie B. Kon <ronnie@cisco.com>
Message-Id: <9208210052.AA00306@ale.cisco.com>
To: Cube-Lovers@life.ai.mit.edu
Subject: Back to the 16th century

In the Cube Lovers' archives we have files:

           From        To        Size     Time      Bytes/Month
           ----        --        ----    ------     -----------
        12 Jul 80   23 Oct 80   185037   3 months      61679
         3 Nov 80    9 Jan 81   135719   2 months      67860
        10 Jan 81    3 Aug 81   138566   6 months      23094
         3 Aug 81    3 May 82   137753   9 months      15306
         4 May 81   11 Dec 82   139660  19 months       7351
        11 Dec 82    6 Jan 87   173364  48 months       3612
        10 Jan 87   13 Apr 90   216733  39 months       5557
        12 Oct 90    9 Sep 91   137508  12 months      11459
         1 Nov 91   25 May 92   171205   7 months      24458


Ladies and Gentlemen, I believe we are witnessing a Renaissance of Cubing!


                                Ronnie

From @mail.uunet.ca:mark.longridge@canrem.com  Tue Sep  1 22:09:24 1992
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From: Mark Longridge <mark.longridge@canrem.com>
Date: 	Mon, 31 Aug 1992 20:00:00 -0400
To: cube-lovers@ai.mit.edu
Subject: rare variants


   Hello fellow cube lovers...

Although this is not a strict Rubik's cube type post I think it should
be of some interest to the other subscribers.

I collect cube variants, and although the variations in cube colours are
interesting my great interest is variants in mechanisms (which require
different solving techniques).

I've been corresponding with other cube buffs around the world in an
effort to record ALL the significant cube variants and I use the
following classification system:

        M = Manufactured in quantity, readily available
        S = Produced in small quantities only
        R = Rare, a few prototypes exist, difficult to get
        P = Prototype, the inventor has the only one!
        C = Exists only as a computer simulation and/or cardboard mockup
        I = Intellectual idea only (perhaps on pencil and paper)

In my opinion, Square 1 is the most interesting cube variant in recent
years, and it gave me the most trouble! Here are some of the tough ones
to get, and if anyone knows of any others email me and I'll maintain a
list of them...

        Trajber's Octahedron (R)  Evidently Greg Stevens owns one
        Octahedral puzzle with rotating faces

        Extended Missing Link (S)
        Missing Link with 6 tiers

        Master Pyraminx (P??)
        Looks like a normal pyraminx BUT it's edges can turn (just the
        strip) 180 degrees and 2 vertices can be swapped

        Space Grenade (P??)
        Other weird one from Uwe Meffert. Mike Green of Puzzletts showed
        me a picture of this, still not sure how it moves

        Pyraminx Disc Chess (S)
        Planar puzzle with 6 rotating discs, similar to Raba's Rotoscope

        Masterball (S)
        This seems to be a recent one, it's like a VIP sphere but it has
        8 vertical cuts instead of one (like a tangerine) and 4
        hortizontal sections

        Halpern's Tetrahedron (P) Also called Pyraminx Tetrahedron
        Like a pyraminx BUT it has face centres which are small
triangles
        and it's faces rotate. Very rare.

        Pyraminx Hexagon (C) Jerry Slocum says he got a cardboard mockup
        of this from Meffert. I wrote a computer simulation of it.
Imagine a Rubik's cube with an N-prism shape, thus the top and bottom
        are hexagons, and there are 6 (rather than 4) adjacent sides.
        The top and bottom can rotate 60 degrees and the adjacent sides
        can only rotate 180 degrees.

        Twist Torus (I) My own concept. Imagine a torus segmented 4 ways
        length-wise so it can slide around. Additionally there are
        12 rings around the circumference which can rotate at right
        angles to the segments. Still thinking of a good colour
        arrangement for this one.

        Super Skewb (I) Another idea of mine. It's a skewb and a 2x2x2
        cube! A compound of two mechanisms.

        Honourable mention: Pyraminx Star, Puck, Ufo, Trick Haus
        (Anyone got multiples of these)

Imagine my disappointment when I found out the Mach Ball, Skewb and
Moody Ball all have the same basic mechanism! Anyway if anyone has
a rare variant or puzzle idea please post here or email me...

Mark Longridge
259 Thornton Rd N
Oshawa Ontario Canada L1J 6T2
Email: mark.longridge@canrem.com
--
Canada Remote Systems  - Toronto, Ontario/Detroit, MI
World's Largest PCBOARD System - 416-629-7000/629-7044

From diamond@jit081.enet.dec.com  Tue Sep  1 23:58:01 1992
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Date: Tue, 1 Sep 92 20:58:00 PDT
From: 02-Sep-1992 1249 <diamond@jit081.enet.dec.com>
To: cube-lovers@ai.mit.edu
Apparently-To: cube-lovers@ai.mit.edu
Subject: Re: rare variants

Mr. Longridge!  Please post the addresses of where to buy that stuff!
Oh, I'm drooling all over my keyboard.

Meanwhile, Jean-Claude Constantin (Pirkach 14, D-8535 ah I forgot the name
of the city but it's near Nurnberg, Germany) makes lots of variations on
cubes and skewbs etc., but the underlying mechanisms all seem to be standard
ones.

>        Honourable mention: Pyraminx Star, Puck, Ufo, Trick Haus
>        (Anyone got multiples of these)

Hmmmm.  I might confess to owning two Trick Hauses in exchange for something
of sufficient persuasion, such as some of the others listed in Mr. Longridge's
post.  In hopes of being able to exchange for some neat stuff, I'd better not
mention that Mr. Constantin can sell Trick Hauses for something around DM 20.

>Imagine my disappointment when I found out the Mach Ball, Skewb and
>Moody Ball all have the same basic mechanism!

Yeah, but with Mach Ball you have to orient the square-like pieces.
Haven't seen Moody Ball.

-- Norman Diamond      diamond@jit081.enet.dec.com
[Digital did not write this.]

From ronnie@cisco.com  Wed Sep  2 00:50:42 1992
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Message-Id: <9209020450.AA01685@wolf.cisco.com>
To: 02-Sep-1992 1249 <diamond@jit081.enet.dec.com>
Cc: cube-lovers@ai.mit.edu
Subject: Re: rare variants 
In-Reply-To: Your message of "Tue, 01 Sep 92 20:58:00 PDT."
             <9209020357.AA21950@enet-gw.pa.dec.com> 
Date: Tue, 01 Sep 92 21:50:35 PDT
From: "Ronnie B. Kon" <ronnie@cisco.com>

> Mr. Longridge!  Please post the addresses of where to buy that stuff!
> Oh, I'm drooling all over my keyboard.
> 
> Meanwhile, Jean-Claude Constantin (Pirkach 14, D-8535 ah I forgot the name
> of the city but it's near Nurnberg, Germany) makes lots of variations on
> cubes and skewbs etc., but the underlying mechanisms all seem to be standard
> ones.
> 
> >        Honourable mention: Pyraminx Star, Puck, Ufo, Trick Haus
> >        (Anyone got multiples of these)
> 
> Hmmmm.  I might confess to owning two Trick Hauses in exchange for something
> of sufficient persuasion, such as some of the others listed in Mr. Longridge'
s
> post.  In hopes of being able to exchange for some neat stuff, I'd better not
> mention that Mr. Constantin can sell Trick Hauses for something around DM 20.

Unless the the Bundespost will deliver mail to "a city near Nurnberg" I think
you're pretty safe.  Unless of course you'll post the full address?

				Ronnie

From diamond@jit081.enet.dec.com  Wed Sep  2 00:59:23 1992
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From: 02-Sep-1992 1356 <diamond@jit081.enet.dec.com>
To: cube-lovers@ai.mit.edu
Cc: ronnie@cisco.com
Apparently-To: ronnie@cisco.com, cube-lovers@ai.mit.edu
Subject: Re: rare variants

>> Meanwhile, Jean-Claude Constantin (Pirkach 14, D-8535 ah I forgot the name
>> of the city but it's near Nurnberg, Germany) makes lots of variations on

ronnie@cisco.com writes:
>Unless the the Bundespost will deliver mail to "a city near Nurnberg" I think
>you're pretty safe.  Unless of course you'll post the full address?

I was telling the truth -- I forgot the name of the city while in the middle
of typing.  Meanwhile, I had posted the full address on rec.puzzles a few
months ago.

Just now, I have recalled the name of the city, and here is the full address
(if I don't forget again while typing :-)

Constantin Geduldspiele
Pirkach 14
D-8535 Emskirchen
Germany

-- Norman Diamond    diamond@jit081.enet.dec.com
[Digital did not write this.]

From Don.Woods@eng.sun.com  Sun Sep  6 14:40:52 1992
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From: Don.Woods@eng.sun.com (Don Woods)
Message-Id: <9209061842.AA28174@colossal.Eng.Sun.COM>
To: cube-lovers@ai.mit.edu, mark.longridge@canrem.com
Subject: Re:  rare variants

I have a puzzle not in your list; it's rare enough that I've never seen
it in a store or catalog.  I got mine from a friend.  He calls it "The
Barrel".

Imagine a transparent cylinder divided into 6 circular slices.  Slices 2-5
each have 5 pockets equally spaced around the circumference, just below the
surface (see left figure below).  Slices 1 and 6 have three pockets in
positions corresponding to 3 of the 5 (see right figure).

	       *******       		       *******       
	     ****   ****     		     ****   ****     
	    *****   *****    		    *****   *****    
	   ******   ******   		   ******   ******   
	  *****************  		  *****************  
	 ******************* 		 ******************* 
	 ******************* 		 ******************* 
	*   *************   *		*********************
	*   *************   *		*********************
	*   *************   *		*********************
	*********************		*********************
	*********************		*********************
	*********************		*********************
	*********************		*********************
	 ******************* 		 ******************* 
	 ***   *******   *** 		 ***   *******   *** 
	  *    *******    *  		  *    *******    *  
	   *  *********  *   		   *  *********  *   
	    *************    		    *************    
	     ***********     		     ***********     
	       *******       		       *******       

Through the center of the cylinder runs a piece with a cap on each end.
The caps each have 3 prongs poking into the cylinder, lined up on the
3 openings in slices 1 and 6.  However, the central piece is long enough
that if the prongs are pushed into slice 6, the prongs at the other end
are lifted out of slice 1, and vice versa.  So, at any given time, three
of the end pockets are filled by one of the endcaps.  The other 23 pockets
contain colored balls.

Originally, the 3 balls in the unpronged endcap are black, and the balls
in the other slices are lined up by color; i.e. 4 blue balls lined up
above one another in slices 2-5, 4 green balls, likewise lined up, etc.
The possible moves are:

1) Slide the endcaps up and down.  E.g., from the starting position, this
would push three balls of different colors into slice 6, and push the 3
black balls from slice 1 into slice 2 (and also push various balls of the
same color down, but that has no visible effect).

2) Turn slices 2 and 3 together; they do not move separately.

3) Turn slices 4 and 5 together; ditto.

Slices 1 and 6 are fixed, so they always line up with the end cap prongs
and with each other.

That's all there is to it.  It certainly has the "cubish" feel to me, in
that it's impossible to make single moves that affect only a small portion
of the puzzle.

	-- Don.

From diamond@jit081.enet.dec.com  Sun Sep  6 20:11:45 1992
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From: 07-Sep-1992 0906 <diamond@jit081.enet.dec.com>
To: cube-lovers@ai.mit.edu
Apparently-To: cube-lovers@ai.mit.edu
Subject: Re:  rare variants

Don.Woods@eng.sun.com writes about the barrel puzzle.  If I recall correctly,
this was made by Nintendo before they switched to computer games.  I also
believe the name was "Ten Billion Barrel."  (Mine is buried somewhere and
I couldn't solve it, but I didn't want to buy the book that was published
at one time, sigh...  I didn't need a book for the cube, so why should I
cheat for a piddly little barrel... sigh.)

Anyway, there are still a few available.  But I have to warn, if anyone wants
one, it will cost more for postage and for my train fare going to the store,
than to buy the thing.  If anyone wants one, we can arrange it by e-mail.
But I'd really prefer to trade for some of those wonderful things that
Mr. Longridge described.  (I'm drooling all over my keyboard again, just
remembering them.)

-- Norman Diamond     diamond@jit081.enet.dec.com
[Digital did not write this.]

From @mail.uunet.ca:mark.longridge@canrem.com  Tue Sep 15 17:14:06 1992
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From: Mark Longridge <mark.longridge@canrem.com>
Date: 	Mon, 14 Sep 1992 20:00:00 -0400
To: cube-lovers@ai.mit.edu
Subject: pyraminx revisited


                Notes on the Pyraminx
                ---------------------

According to Dr. Ronald Turner-Smith there are 5 different Pyraminx
puzzles, all of different complexity. The following are listed in
order from easiest to hardest (to the best of my knowledge):

Pyraminx Star: Easiest of all pyraminx??
               A simplification of the popular pyraminx because of the
               little uni-coloured (usually grey or silver) tetrahedrons
               on the 3 middle pieces of each face. Effectively all
               middle pieces on this pyraminx are the same colour!

Snub Pyraminx: Same as standard pyraminx with tips removed

Popular Pyraminx: The standard pyraminx of which appeared in vast
                  quanities after the cube caught on.

Senior Pyraminx: This is a mystery puzzle. No one seems to know anything
                 about it, yet Turner-Smith's book refers to it and
                 gives the maximum number of moves for it! It is between
                 the Popular Pyraminx and Master Pyraminx in difficulty.

Master Pyraminx: All the moves of the standard pyraminx plus 180 degree
                 turns of the edges (just the strip, not the whole face)
                 446,965,972,992,000 combinations. Interestingly in
--
Canada Remote Systems  - Toronto, Ontario, Canadas
World's Largest PCBOARD System - 416-629-7000/629-7044

From @mail.uunet.ca:mark.longridge@canrem.com  Tue Sep 15 18:37:59 1992
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From: Mark Longridge <mark.longridge@canrem.com>
Date: 	Mon, 14 Sep 1992 20:00:00 -0400
To: cube-lovers@ai.mit.edu
Subject: pyraminx revisited


                Notes on the Pyraminx
                ---------------------

According to Dr. Ronald Turner-Smith there are 5 different Pyraminx
puzzles, all of different complexity. The following are listed in
order from easiest to hardest (to the best of my knowledge):

Pyraminx Star: Easiest of all pyraminx??
               A simplification of the popular pyraminx because of the
               little uni-coloured (usually grey or silver) tetrahedrons
               on the 3 middle pieces of each face. Effectively all
               middle pieces on this pyraminx are the same colour!

Snub Pyraminx: Same as standard pyraminx with tips removed

Popular Pyraminx: The standard pyraminx of which appeared in vast
                  quanities after the cube caught on.

Senior Pyraminx: This is a mystery puzzle. No one seems to know anything
                 about it, yet Turner-Smith's book refers to it and
                 gives the maximum number of moves for it! It is between
                 the Popular Pyraminx and Master Pyraminx in difficulty.

Master Pyraminx: All the moves of the standard pyraminx plus 180 degree
                 turns of the edges (just the strip, not the whole face)
                 446,965,972,992,000 combinations. Interestingly in the
                 ads for Dr. Ronald Turner-Smith's book "The Amazing
                 Pyraminx" the Master Pyraminx is cited as a puzzle
                 superior to Rubik's Cube because there are no centre
                 pieces and it's harder! (Both points debatable IMHO)

Also Turner-Smith gives the following maximum number of moves for
each type of Pyraminx puzzle: (The popular pyraminx is now known
to be 11 moves at most or 14 moves if the tips are included)

Type 1          24 moves   ??
Type 2          28 moves   ??
Type 3          38 moves   ??
Type 4         215 moves   (Senior Pyraminx)
Type 5         255 moves   (Master Pyraminx)

Also it is known that transparent pyraminx puzzles were made. This
would be a good idea for the cube as well. Meffert also considered
a textured pyraminx for the blind, and ones with leather and wood
finishes.

All the post-cube puzzles compare themselves to the cube, such as
the Master Pyraminx, and more recently Smart Alex. It seems that
Rubik's Cube is the benchmark for all others to compare with.

Alas, Uwe Meffert's puzzle club was a bust. Barring unique prototypes
(perhaps Singmaster or Hofstadter have a Master Pyraminx, I'll check)
none of the following were produced: (most of these are documented in
the extremely rare "Pyraminx The Exciting new 1982 range" or the even
more obscure 1983 edition booklet. Both of these have full colour pages
with photos of cardboard mockups of all the variants.)

1982:

Pyraminx Star (exists in small quanities, in Constantin's catalog)
Pyraminx Pentagon, Pyraminx Hexagon (Computer Simulation/Mockup only)
  One can also imagine Septagons, Octagons etc...
Pyraminx Barrel Junior, Pyraminx Barrel Senior (Mockup only)
Pyraminx Disc Chess (Prototypes exist)
Pyraminx Ultimate (as shown in July 82 Scientific American, Mockup)
Pyraminx Crystal, Pyraminx Ball (Mockups, July 82 S.A)
  I'd really like to see the mechanism for a working crystal!
Pyraminx Assembly Puzzles, 4 types (They exist)
Pyraminx Octahedron (An octahedral skewb, I believe Braun & Bandelow
made
  some) Gerd Braun is the inventor of the Moody Ball (rare but exists)
Pyraminx Tetrahedron (Ben Halpern made a prototype)

1983:

Space Grenade (???)
Crystal Ball (Looks like an orb, definitely not the same though)

However....

Just a few days ago I got Constantin's catalog. Surprisingly there is
a picture of Josef Trajber's Octahedron inside. Also there is a picture
of what appears to be a Pyraminx Ball. Other ideas he includes are a
2x2x2 siamese cube, new variants on Fisher's cube, e.g. Fisher's Domino,
and a Pyraminx Ultimate for 180 DM!

...and so the search for new cube variants continues. Please send me
your
comments (Does anyone actually own a working Master Pyraminx??) I'm
also interested in exchanging full cube lists with other collectors.

Mark Longridge                 Email: mark.longridge@canrem.com
259 Thornton Rd N
Oshawa Ontario Canada L1J 6T2
--
Canada Remote Systems  - Toronto, Ontario, Canadas
World's Largest PCBOARD System - 416-629-7000/629-7044

From tom@goat.clipper.ingr.com  Tue Sep 15 21:28:48 1992
Return-Path: <tom@goat.clipper.ingr.com>
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Subject: Re: pyraminx revisited
To: cube-lovers@ai.mit.edu
Date: Tue, 15 Sep 92 15:56:22 PDT
In-Reply-To: <199215.104.106150@dosgate>; from "Mark Longridge" at Sep 14, 92 8:00 pm
X-Mailer: ELM [version 05.00.01.20]
Message-Id: <9209151556.AA00256@goat.UUCP>
From: tom@goat.clipper.ingr.com (Tom Granvold)

> Also it is known that transparent pyraminx puzzles were made. This
> would be a good idea for the cube as well. Meffert also considered
> a textured pyraminx for the blind, and ones with leather and wood
> finishes.

    I have one of the textured pyraminx.  I had to mail order it from
Hong Kong in the early '80s.

> Just a few days ago I got Constantin's catalog.

    Where can one get a copy of Constantin's catalog?

> ...and so the search for new cube variants continues. Please send me
> your
> comments (Does anyone actually own a working Master Pyraminx??) I'm
> also interested in exchanging full cube lists with other collectors.

    I wish I did :-)

Tom Granvold
------------------------------------------------------
Mail:    2400 Geng Rd., Palo Alto, Calif., 94303
Email:   tom@clipper.ingr.com
------------------------------------------------------


From @mail.uunet.ca:mark.longridge@CANREM.COM  Mon Sep 21 00:05:34 1992
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Message-Id: <199220.104.108416@dosgate>
From: Mark Longridge <mark.longridge@canrem.com>
Date: 	Sat, 19 Sep 1992 20:00:00 -0400
To: cube-lovers@ai.mit.edu
Subject: pyraminx revisted


                Notes on the Pyraminx
                ---------------------

According to Dr. Ronald Turner-Smith there are 5 different Pyraminx
puzzles, all of different complexity. The following are listed in
order from easiest to hardest (to the best of my knowledge):

Pyraminx Star: Easiest of all pyraminx??
               A simplification of the popular pyraminx because of the
               little uni-coloured (usually grey or silver) tetrahedrons
               on the 3 middle pieces of each face. Effectively all
               middle pieces on this pyraminx are the same colour!

Snub Pyraminx: Same as standard pyraminx with tips removed

Popular Pyraminx: The standard pyraminx of which appeared in vast
                  quanities after the cube caught on.

Senior Pyraminx: This is a mystery puzzle. No one seems to know anything
                 about it, yet Turner-Smith's book refers to it and
                 gives the maximum number of moves for it! It is between
                 the Popular Pyraminx and Master Pyraminx in difficulty.

Master Pyraminx: All the moves of the standard pyraminx plus 180 degree
                 turns of the edges (just the strip, not the whole face)
                 446,965,972,992,000 combinations. Interestingly in the
                 ads for Dr. Ronald Turner-Smith's book "The Amazing
                 Pyraminx" the Master Pyraminx is cited as a puzzle
                 superior to Rubik's Cube because there are no centre
                 pieces and it's harder! (Both points debatable IMHO)

Also Turner-Smith gives the following maximum number of moves for
each type of Pyraminx puzzle: (The popular pyraminx is now known
to be 11 moves at most or 14 moves if the tips are included)

Type 1          24 moves   ??
Type 2          28 moves   ??
Type 3          38 moves   ??
Type 4         215 moves   (Senior Pyraminx)
Type 5         255 moves   (Master Pyraminx)

Also it is known that transparent pyraminx puzzles were made. This
would be a good idea for the cube as well. Meffert also considered
a textured pyraminx for the blind, and ones with leather and wood
finishes.

All the post-cube puzzles compare themselves to the cube, such as
the Master Pyraminx, and more recently Smart Alex. It seems that
Rubik's Cube is the benchmark for all others to compare with.

Alas, Uwe Meffert's puzzle club was a bust. Barring unique prototypes
(perhaps Singmaster or Hofstadter have a Master Pyraminx, I'll check)
none of the following were produced: (most of these are documented in
the extremely rare "Pyraminx The Exciting new 1982 range" or the even
more obscure 1983 edition booklet. Both of these have full colour pages
with photos of cardboard mockups of all the variants.)

1982:

Pyraminx Star (exists in small quanities, in Constantin's catalog)
Pyraminx Pentagon, Pyraminx Hexagon (Computer Simulation/Mockup only)
  One can also imagine Septagons, Octagons etc...
Pyraminx Barrel Junior, Pyraminx Barrel Senior (Mockup only)

Pyraminx Disc Chess (Prototypes exist)
Pyraminx Ultimate (as shown in July 82 Scientific American, Mockup)
Pyraminx Crystal, Pyraminx Ball (Mockups, July 82 S.A)
  I'd really like to see the mechanism for a working crystal!
Pyraminx Assembly Puzzles, 4 types (They exist)
Pyraminx Octahedron (An octahedral skewb, I believe Braun & Bandelow
made
  some) Gerd Braun is the inventor of the Moody Ball (rare but exists)
Pyraminx Tetrahedron (Ben Halpern made a prototype)

1983:

Space Grenade (???)
Crystal Ball (Looks like an orb, definitely not the same though)

However....

Just a few days ago I got Constantin's catalog. Surprisingly there is
a picture of Josef Trajber's Octahedron inside. Also there is a picture
of what appears to be a Pyraminx Ball. Other ideas he includes are a
2x2x2 siamese cube, new variants on Fisher's cube, e.g. Fisher's Domino,
and a Pyraminx Ultimate for 180 DM!

...and so the search for new cube variants continues. Please send me
your
comments (Does anyone actually own a working Master Pyraminx??) I'm
also interested in exchanging full cube lists with other collectors.

Mark Longridge                 Email: mark.longridge@canrem.com
259 Thornton Rd N
Oshawa Ontario Canada L1J 6T2
--
Canada Remote Systems  - Toronto, Ontario, Canadas
World's Largest PCBOARD System - 416-629-7000/629-7044

From mb8d+@andrew.cmu.edu  Mon Sep 21 21:31:25 1992
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Date: Mon, 21 Sep 1992 19:46:44 -0400 (EDT)
From: Matthew John Bushey <mb8d+@andrew.cmu.edu>
To: cube-lovers@ai.mit.edu
Subject: cubes are great
Cc: 


Does anyone out there know what is the cubed root of 81?

Just wondering....


From yekta@huey.jpl.nasa.gov  Mon Sep 21 22:38:35 1992
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Date: Mon, 21 Sep 92 19:34:07 PDT
From: yekta@huey.jpl.nasa.gov (Yekta Gursel)
Message-Id: <9209220234.AA00679@huey.Jpl.Nasa.Gov>
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To: mb8d+@andrew.cmu.edu
Cc: cube-lovers@ai.mit.edu
In-Reply-To: Matthew John Bushey's message of Mon, 21 Sep 1992 19:46:44 -0400 (EDT) <0ejZvYC00WBK48jY0m@andrew.cmu.edu>
Subject: cubes are great

Little smaller than its square root.  Are you having fun yet?

--Yekta

From gls@think.com  Tue Sep 22 11:50:24 1992
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To: mb8d+@andrew.cmu.edu
Cc: cube-lovers@ai.mit.edu
In-Reply-To: Matthew John Bushey's message of Mon, 21 Sep 1992 19:46:44 -0400 (EDT) <0ejZvYC00WBK48jY0m@andrew.cmu.edu>
Subject: cubes are great

   Date: Mon, 21 Sep 1992 19:46:44 -0400 (EDT)
   From: Matthew John Bushey <mb8d+@andrew.cmu.edu>


   Does anyone out there know what is the cubed root of 81?

   Just wondering....

Well, the "root of 81" is 9 (recall that when you don't say
what kind of root you want, the default is "square"), and
9 cubed is 729.

... Eh?  Oh, you meant the "cube root", not the "cubed root"?
Well, that's another kettle of fish entirely.  The n'th root
of x is equal to x raised to the power 1/n.  I fed this to
my friendly Common Lisp system:

> (expt 81 1/3)
4.3267487109222245

If I were you, I wouldn't trust the last few digits of this
approximation, but fifteen decimal places ought to hold you
for now.

Here's how you could estimate it in your head.
Note that 81 = 3 to the fourth power, so

  1/3      4  1/3    4/3        1/3
81    = ( 3  )    = 3    = 3 ( 3    )

Now, the cube root of 3 is surely between 1 and 2,
because 1 cubed is 1 and 2 cubed is 8.  So the cube
root of 3 is 1 plus some smaller fractional amount x.
               3                2    3
So  3 = (1 + x)  = 1 + 3 x + 3 x  + x     (binomial expansion).

                  3
Let's ignore the x  term, which is probably small because
x is sort of small.  Then

             2                   2
1 + 3 x + 3 x  = 3     so   x + x  = 2/3 .

                            2
Hm... if x = 1/2, then x + x  = 3/4, which is a bit
                                               2
too big.  So figure x is about 0.4; then  x + x  = .4 + .16 = .56
which is too small.  So probably x is about 0,45 or so.

So the cube root of 3 is about 1.45, and the cube root of
81 is 3 times that, or about 4.35 -- not a bad approximation.

--Guy STeele

From bosch@smiteo.esd.sgi.com  Tue Sep 22 12:10:48 1992
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From: bosch@smiteo.esd.sgi.com (Derek Bosch)
Message-Id: <9209221610.AA02225@smiteo.esd.sgi.com>
Subject: Gaby Games address needed
To: cube-lovers@ai.mit.edu
Date: Tue, 22 Sep 92 9:10:44 PDT
X-Mailer: ELM [version 2.3 PL4]

Does anyone out there in cube-land know the address for Gaby Games?
They are an Israeli manufacturer of interesting 3-d interlocking wooden
puzzles.  I have posted this to rec.puzzles, with no help so far.

Derek Bosch
bosch@sgi.com



From ronnie@cisco.com  Tue Sep 22 13:39:12 1992
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To: Matthew John Bushey <mb8d+@andrew.cmu.edu>
Cc: cube-lovers@ai.mit.edu
Subject: Re: cubes are great 
In-Reply-To: Your message of "Mon, 21 Sep 92 19:46:44 EDT."
             <0ejZvYC00WBK48jY0m@andrew.cmu.edu> 
Date: Tue, 22 Sep 92 10:37:50 PDT
From: "Ronnie B. Kon" <ronnie@cisco.com>

> 
> Does anyone out there know what is the cubed root of 81?
> 
> Just wondering....
> 

Let's see: the root of 81 is 9.  9 cubed is 729.

				Ronnie

From azimmerm@rnd.stern.nyu.edu  Mon Oct  5 16:53:08 1992
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Date: Mon, 5 Oct 92 16:51:57 EDT
From: Al Zimmermann <azimmerm@rnd.stern.nyu.edu>
To: Cube-Lovers@ai.mit.edu
Subject: Reminiscences
Message-Id: <CMM.0.90.2.718318317.azimmerm@rnd.gba.nyu.edu>

Is everybody ready for more reminiscences?

I got my first cube at Harrad's in London in October of 1980 while I was on
vacation there with my girl friend.  I spent every non-touristy moment
working out and recording moves until, on day 13, I got the final face.

When we got back to the States, my girl friend and I broke up.  Do you think
there's a moral here?

      Al Zimmermann

From diamond@jit081.enet.dec.com  Mon Oct  5 20:54:41 1992
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From: 06-Oct-1992 0949 <diamond@jit081.enet.dec.com>
To: cube-lovers@ai.mit.edu
Apparently-To: cube-lovers@ai.mit.edu
Subject: Re: miniscences

Al Zimmermann <azimmerm@rnd.stern.nyu.edu> writes:
>I spent every non-touristy moment
>working out and recording moves until, on day 13, I got the final face.
>When we got back to the States, my girl friend and I broke up.  Do you think
>there's a moral here?

Yes, at least three:

(1)  Every non-touristy moment that you weren't recording moves, you should
       have spent with your girl friend instead of working out  :-)

(2)  You should have given equal attention to the final face and to your
       girl friend's face  :-)

(3)  You should have chosen a girl friend who could solve the cube, like
       I did.

But I thank you for warning about the dangers of getting back to the States.
Maybe I shouldn't go back after all  :-)

-- Norman Diamond    diamond@jit081.enet.dec.com
[Digital did not write this.]

From imp@kolvir.solbourne.com  Fri Oct  9 17:59:44 1992
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To: cube-lovers@ai.mit.edu
Subject: Quick question....
Date: Fri, 09 Oct 1992 15:59:40 MDT
From: Warner Losh <imp@kolvir.solbourne.com>

I was wondering if there were any X programs out there that allowed
one to play with a rubic's cube (3x3x3 ... nxnxn) on a workstation?
Archie didn't seem to know of any.

Warner

From pbeck@pica.army.mil  Fri Oct 23 07:49:03 1992
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Date:     Fri, 23 Oct 92 7:47:24 EDT
From: Peter Beck (BATDD) <pbeck@pica.army.mil>
To: cube-lovers@life.ai.mit.edu
Cc: pbeck@pica.army.mil
Subject:  13th IPP (1993)
Message-Id:  <9210230747.aa12659@COR4.PICA.ARMY.MIL>


...............................................................
<-->  13th International puzzle collector's party and fair "   
and the 1993 Dutch Cube Day party
                 transcribed by pbeck, 10/23/92
...............................................................
WHEN ----    8/20  - 8/22/93
WHERE ----  Amsterdam vicinity
LODGING ---- about $90 per night at 
  MOTEL BREUKELEN
  STATIONSWEG  91
  3621 LK BREUKELEN
  NETHERLANDS
  TEL:  03462 - 65888
  FAX:  03462 - 62894

***  INVITATIONS ***  Admission by invitation only!!!  
     Contact: Mr. W.G.H. STRIJBOS
                   BREDEROSTRAAT 18
                   5921 BM VENLO
                   NETHERLANDS
                   TEL:  +31  (0) 77 -826213
                   FAX:  +31  (0) 4704 - 4656

AGENDA: 
  8/20   13:00 - 16:30 PUZZLE EXCHANGE
             17:30 - 22:00 DINNER AND MAGIC SHOW
  8/21   10:00 - 17:00 PUZZLE PARTY AND FAIR (SALES)
      COST FOR ABOVE 150 DUTCH GUILDERS & IT WILL BE HELD AT MOTEL   
          BREUKELEN - 100 EXTRA FOR SAT SALES TABLE

  8/22   10:00 - 17:00 CUBE DAY  _ THIS WILL BE HELD AT CHESS & GO   
                    CENTER IN AMSTELVEEN AND PROBABLY HAS AN EXTRA
COST.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

***  4 NIGHT RESERVATION (8/19 - 8/23) AT " MOTEL BREUKELEN" IS 625
DUTCH GUILDERS AND INCLUDES BREAKFAST - ONE DOUBLE BED IN EACH ROOM.
--->  RESERVATION REQUEST FOR THIS PACKAGE MUST BE MADE BY
***  JAN 10 1993  ***  TO STRIJBOS


>>>>>>>>>>>>>>>>>>  If I was unclear or if you have other questions
ask them to the list since several members of the Dutch Cube Club
(party hosts) are subscribers.



From hirsh@cs.rutgers.edu  Wed Nov  4 15:18:42 1992
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Sender: Haym Hirsh <hirsh@cs.rutgers.edu>
Date: Wed, 4 Nov 92 15:18:37 EST
From: Haym Hirsh <hirsh@cs.rutgers.edu>
Reply-To: Haym Hirsh <hirsh@cs.rutgers.edu>
To: cube-lovers@ai.mit.edu
Subject: masterball
Cc: Haym Hirsh <hirsh@cs.rutgers.edu>
Message-Id: <CMM-RU.1.3.720908317.hirsh@pei.rutgers.edu>

A friend just sent me email about a new (to him and to me) puzzle
called "masterball".  Anyone know anything about it?  Is it worth
getting?

Haym

> I saw a Rubik's cube variant today called "Masterball."  Have you
> seen it?  It is a sphere with 32 faces.  If you consider the sphere
> to be a world globe, there are 8 longitudinal slices each going
> through the axis of the globe, dividing the sphere into 8 segments
> like a sliced orange (sorry for starting to mix my metaphors [actually,
> I guess I was mixing similes, but I know *you* wouldn't bring up
> such a trivial point]).
> 
> Oooops I guess there are only 4 longitudinal slices, each through
> the axis, to divide the globe into 8 segments.
> 
> There are also 3 slices of latitude, one through the equator one
> each in the northern and the southern hemisphere parallel to
> the equator.
> 
> Resultant 32 faces.  Mechanism has some similarities to Square One.
> 
> Two different versions of Masterball are available.  One has eight
> different colors, corresponding to 8 segments.  The other has only
> black and white.  I don't remember the home pattern of the black
> and white sphere, I presume it is a degenerate case of the 8 color
> sphere with black and white alternating slices.
> 
> Cost:  $24.95 each.  My source is the same store in San Francisco
> (Stonestown mall) that provided the Rubiks Tangle, Rubiks Dice, etc.


From azimmerm@rnd.stern.nyu.edu  Wed Nov  4 17:30:12 1992
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Date: Wed, 4 Nov 92 17:12:04 EST
From: Al Zimmermann <azimmerm@rnd.stern.nyu.edu>
To: Haym Hirsh <hirsh@cs.rutgers.edu>
Cc: cube-lovers@ai.mit.edu, Haym Hirsh <hirsh@cs.rutgers.edu>
Subject: Re: masterball
In-Reply-To: Your message of Wed, 4 Nov 92 15:18:37 EST
Message-Id: <CMM.0.90.2.720915124.azimmerm@rnd.stern.nyu.edu>

> A friend just sent me email about a new (to him and to me) puzzle
> called "masterball".  Anyone know anything about it?  Is it worth
> getting?
> 
> Haym
> 
> > I saw a Rubik's cube variant today called "Masterball."  Have you
> > seen it?  It is a sphere with 32 faces.  If you consider the sphere
> > to be a world globe, there are 8 longitudinal slices each going
> > through the axis of the globe, dividing the sphere into 8 segments
> > like a sliced orange (sorry for starting to mix my metaphors [actually,
> > I guess I was mixing similes, but I know *you* wouldn't bring up
> > such a trivial point]).
> > 
> > Oooops I guess there are only 4 longitudinal slices, each through
> > the axis, to divide the globe into 8 segments.
> > 
> > There are also 3 slices of latitude, one through the equator one
> > each in the northern and the southern hemisphere parallel to
> > the equator.
> > 
> > Resultant 32 faces.  Mechanism has some similarities to Square One.
> > 
> > Two different versions of Masterball are available.  One has eight
> > different colors, corresponding to 8 segments.  The other has only
> > black and white.  I don't remember the home pattern of the black
> > and white sphere, I presume it is a degenerate case of the 8 color
> > sphere with black and white alternating slices.
> > 
> > Cost:  $24.95 each.  My source is the same store in San Francisco
> > (Stonestown mall) that provided the Rubiks Tangle, Rubiks Dice, etc.
> 
> 

Games magazine seems to like the puzzle.  They included Mastermind Rainbow
(the polychromatic version) in this year's "Games 100" listing.  Their
write-up isn't very informative, but there's a picture.  It appears on page
59 of the Dec. '92 issue.

By the way, they indicate that the puzzle is available from Baekgaard at
1-800-323-5413.

From pbeck@pica.army.mil  Thu Nov  5 21:56:13 1992
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Date:     Thu, 5 Nov 92 11:11:51 EST
From: Peter Beck (BATDD) <pbeck@pica.army.mil>
To: cube-lovers@life.ai.mit.edu
Subject:  RE: MASTERBALL
Message-Id:  <9211051111.aa20544@COR4.PICA.ARMY.MIL>


Masterball is similar to the VIP Sphere and Orb puzzles.

It is nicely made and has been available in europe for
 a couple of years.

I believe ISHI is also selling it.

If you don't have the ISHI x-mas flyer call them
and get it.  This flyer has several unique items
for slae, eg, 5x5x5, skewb


From yekta@huey.jpl.nasa.gov  Fri Nov  6 11:46:52 1992
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Date: Fri, 6 Nov 92 08:41:53 PST
From: yekta@huey.jpl.nasa.gov (Yekta Gursel)
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To: cube-lovers@life.ai.mit.edu
In-Reply-To: Peter Beck (BATDD)'s message of Thu, 5 Nov 92 11:11:51 EST <9211051111.aa20544@COR4.PICA.ARMY.MIL>
Subject: MASTERBALL

Could you post ISHI's phone number?

--Yekta

From dik@cwi.nl  Sun Nov  8 16:16:16 1992
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Date: Sun, 8 Nov 1992 22:15:14 +0100
From: Dik.Winter@cwi.nl
Message-Id: <9211082115.AA21833.dik@boring.cwi.nl>
To: cube-lovers@life.ai.mit.edu
Subject: Where are the archives?

Can somebody tell me where the cube-lovers archives are maintained at this
moment?

Thanks, dik
--
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
home: bovenover 215, 1025 jn  amsterdam, nederland; e-mail: dik@cwi.nl

From dik@cwi.nl  Sun Nov  8 16:15:20 1992
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From: Dik.Winter@cwi.nl
Message-Id: <9211082115.AA21833.dik@boring.cwi.nl>
To: cube-lovers@life.ai.mit.edu
Subject: Where are the archives?

Can somebody tell me where the cube-lovers archives are maintained at this
moment?

Thanks, dik
--
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
home: bovenover 215, 1025 jn  amsterdam, nederland; e-mail: dik@cwi.nl

From alan@ai.mit.edu  Sun Nov  8 17:00:39 1992
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Date: Sun, 8 Nov 92 17:00:08 EST
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From: Alan Bawden <Alan@lcs.mit.edu>
Sender: Alan@lcs.mit.edu
To: Dik.Winter@cwi.nl
Cc: Cube-Lovers@ai.mit.edu
In-Reply-To: Dik.Winter@cwi.nl's message of Sun, 8 Nov 1992 22:15:14 +0100 <9211082115.AA21833.dik@boring.cwi.nl>
Subject: Where are the archives?

   Date: Sun, 8 Nov 1992 22:15:14 +0100
   From: Dik.Winter@cwi.nl
   Can somebody tell me where the cube-lovers archives are maintained at this
   moment?

Such questions should be addressed to Cube-Lovers-Request@AI.MIT.EDU.  But
since you asked publicly, I might as well answer publicly as well.

If you are interested in the archives of the Cube-Lovers mailing list:

Using FTP, connect to FTP.AI.MIT.EDU, login as "anonymous" (any password),
and in the directory "pub/cube-lovers" you will find the nine (compressed)
files "cube-mail-0.Z" through "cube-mail-8.Z".

Archive vital statistics (when uncompressed):

	   File		   From		   To	     Size (bytes)
	   ----		   ----		   --	     ------------
	cube-mail-0	12 Jul 80	23 Oct 80	185037
	cube-mail-1	 3 Nov 80	 9 Jan 81	135719
	cube-mail-2	10 Jan 81	 3 Aug 81	138566
	cube-mail-3	 3 Aug 81	 3 May 82	137753
	cube-mail-4	 4 May 81	11 Dec 82	139660
	cube-mail-5	11 Dec 82	 6 Jan 87	173364
	cube-mail-6	10 Jan 87	13 Apr 90	216733
	cube-mail-7	12 Oct 90	 9 Sep 91	137508
	cube-mail-8	 1 Nov 91	25 May 92	171205

In addition, the file "recent-mail" contains a copy of the currently active
section of the archive.  (Unfortunately, due to the way mail works here at
the AI Lab, it is not possible to have new mail accumulate directly into
this file, so there may be some delay before a new message arrives here.)
Finally, the file "README" contains the information you are currently reading.

				- Alan

From dik@cwi.nl  Sun Nov  8 17:35:25 1992
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Date: Sun, 8 Nov 1992 23:35:17 +0100
From: Dik.Winter@cwi.nl
Message-Id: <9211082235.AA21966.dik@boring.cwi.nl>
To: cube-lovers@ai.mit.edu
Subject: Reply to old message

Going through the archives I found a message from Mike Reid, dated
August 20, where he investigates the filtration through different
subgroups.  I missed the original, but he says:
 > winter's filtration is  56 --> 49 --> kl --> 1.  it may be the case that
 > this can be improved by replacing  kl  with  17 , and allowing all face
 > turns available in the subgroup  49.  i haven't had the time to look into
 > this yet.
For the record, the filtration is from Herbert Kociemba, and it is:
	56 --> 49 --> 1.
So there is no intermediate stage between 49 and 1.

dik

From pbeck@pica.army.mil  Mon Nov  9 11:22:43 1992
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Date:     Mon, 9 Nov 92 9:04:24 EST
From: Peter Beck (BATDD) <pbeck@pica.army.mil>
To: yekta@huey.jpl.nasa.gov
Cc: cube-lovers@life.ai.mit.edu
Subject:  ishi's phone number
Message-Id:  <9211090904.aa25819@COR4.PICA.ARMY.MIL>

ISHI PRESS

408-944-9110

From gk1k+@andrew.cmu.edu  Wed Nov 18 22:59:34 1992
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Date: Wed, 18 Nov 1992 22:57:43 -0500 (EST)
From: George Cornelius Kuhl <gk1k+@andrew.cmu.edu>
To: cube-lovers@ai.mit.edu
Subject: cube question
Cc: 

what is the cube root of 81?

George


From ACW@riverside.scrc.symbolics.com  Thu Nov 19 09:31:59 1992
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Date: Thu, 19 Nov 1992 09:32-0500
From: Allan C. Wechsler <ACW@riverside.scrc.symbolics.com>
Subject: cube question
To: gk1k+@andrew.cmu.edu, cube-lovers@ai.mit.edu
In-Reply-To: <Yf2l2rO00iV249Tug6@andrew.cmu.edu>
Message-Id: <19921119143257.8.ACW@PALLANDO.SCRC.Symbolics.COM>

    Date: Wed, 18 Nov 1992 22:57 EST
    From: George Cornelius Kuhl <gk1k+@andrew.cmu.edu>

    what is the cube root of 81?

    George

[4 /3 /16 /1 /1 /5 /1 /1 /2 /16 /1 /44 /1 /2 /1 /1 /1 /1 /1 /3 /12 ...]

From pbeck@pica.army.mil  Mon Nov 23 09:35:38 1992
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Date:     Mon, 23 Nov 92 9:32:46 EST
From: Peter Beck (BATDD) <pbeck@pica.army.mil>
To: cube-lovers@life.ai.mit.edu
Subject:  ATLANTA Puzzle Party & exhibition
Message-Id:  <9211230932.aa13919@COR4.PICA.ARMY.MIL>


1/14 - 4/10/93 there will be a puzzle exhibition
at
  THE ATALANTA INTERNATIONAL MUSEUM OF ART AND DESIGN

SPECIAL EVENTS FOR PUZZLERS:

  1/14 opening and reception at 6PM
  1/15 eve reception
  1/16 open to puzzlers only 9am - 13:00
  1/17 puzzlers only

there will be puzzle trades and sales (for puzzlers)
on either sat or sun

POCs

  TOM RODGERS  404-351-7744
  TYLER BARREET  404-998-7432


Please distribute to all interested parties.


From @mail.uunet.ca:mark.longridge@canrem.com  Fri Jan  8 00:57:44 1993
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From: Mark Longridge <mark.longridge@canrem.com>
Date: 	Thu, 7 Jan 1993 19:00:00 -0500
To: cube-lovers@ai.mit.edu
Subject: computer cubing


With thanks to Dan Hoey for getting me on the right track, I
have finally got most of the squares group evaluted. The big
breakthru was developing a checksum for a squares position.

I know it's been done before, but I wanted to prove to myself
I could do it on a mere 386 with 4 megs of memory. My
latest program (rubik5.exe) took 24 hours to number the squares
group up to 8 moves deep. The point of all this was to create
a squares group database to aid in developing an optimal
solver for the cube. Ultimately the database will have an entry
for every squares group position, along with it's optimal
solution. I would be interested in hearing from any others who
have created such a database, and what type of compression or
checksum was used for the arrangement.

Also I've received a call from Richard Schneider. He is publishing
a comprehensive book on square 1, plus a follow-up book on
pretty patterns and shapes. This will be available in the States
shortly.

I haven't been seeing anything from cube-lovers in a while,
I hope it's still up and running.

To: Mike Reid --- Hope you see this! Any further progress been
                  made on God's Algorithm? I'm still trying to
catch up. I'm still interested in that code of yours to find
improvements on some pretty patterns I've discovered.

Anyways here is what my program has found so far:

Squares group (u2, d2, l2, r2, f2, b2)

Moves Deep          Number of patterns
----------          ------------------
   0                      1
   1                      6
   2                     27
   3                    120
   4                    519
   5                   1932
   6                   6484
   7                  20310
   8                  53000 (and counting)  :->

Got to improve it's speed....

 -> Mark <-
--
Canada Remote Systems  - Toronto, Ontario
World's Largest PCBOARD System - 416-629-7000/629-7044



From raymond@cps.msu.edu  Sat Mar 20 13:24:09 1993
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From: raymond@cps.msu.edu
Message-Id: <9303201824.AA02491@pacific.cps.msu.edu>
To: cube-lovers@ai.mit.edu
Subject: Seeking magic dodecahedron

Hello cube lovers,
  I haven't seen any activity on this mailing list in a long time!
Cubing is not dead, is it?

  Anyway, I'm trying to find a good quality magic dodecahedron.  Does
anyone know where I can get one?

Thanks,
Carl

raymond@cps.msu.edu

From news@nntp-server.caltech.edu  Sun Mar 21 14:11:53 1993
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Path: joelong
From: joelong@cco.caltech.edu (Joseph Louis Long)
Newsgroups: mlist.cube-lovers
Subject: Re: Seeking magic dodecahedron
Date: 21 Mar 1993 19:10:17 GMT
Organization: California Institute of Technology, Pasadena
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References: <9303201824.AA02491@pacific.cps.msu.edu>
Nntp-Posting-Host: punisher.caltech.edu

raymond@cps.msu.edu writes:
>Hello cube lovers,
>  I haven't seen any activity on this mailing list in a long time!
I'll say... when I saw this post it reminded me that I've been watching
this group for about two months hoping to see some mention of Square-1,
but have been left disapointed.. (wimper wimper whine.) :)
So let me ask... Does anyone have a solution to Square-1?  Is there
a simple ``operator'' based method, like there is for the cube?
If it is simple enough to explain in text, could someone please
post it?  Has there been a ``solutions book'' published?

obviously in the dark on recent cubic developments,
joe

From pbeck@pica.army.mil  Mon Mar 22 07:59:36 1993
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Date:     Mon, 22 Mar 93 7:58:16 EST
From: Peter Beck (BATDD) <pbeck@pica.army.mil>
To: cube-lovers@life.ai.mit.edu
Cc: pbeck@pica.army.mil
Subject:  cubing
Message-Id:  <9303220758.aa08917@COR4.PICA.ARMY.MIL>


NO cubing isn't dead.
CFF will hold its annual meeting the
day after IPP, which is in amsterdam this year.

their newsletter is still published.

There is a renewed interest as indicated by
sales of magic solids.  In particular
ISHI PRESS, san jose is comerrcially
selling:
  5xs
  super novas - hungarian made regular dodecahedrons

I believe that there are 2 books on square 1 in
the editing stage.  I don't know when/if they will
go to print.

there was discussion and a solution to square 1 
posted to this list, in addition the test issue
of a puzzling magazine from ISHI press featured
square 1.

check the literature.

pete beck

THE FUTURE IS PUZZLING,
BUT CUBING IS FOREVER !!


From cosell@world.std.com  Mon Mar 22 15:05:44 1993
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From: cosell@world.std.com (Bernie Cosell)
In-Reply-To: <1oieipINNs0k@gap.caltech.edu>
             (from joelong@cco.caltech.edu (Joseph Louis Long))
             (at 21 Mar 1993 19:10:17 GMT)
X-Mailer: //\\miga Electronic Mail (AmiElm 1.18)
Reply-To: cosell@world.std.com
Path: world.std.com!cosell
Organization: Fantasy Farm Fibers
To: joelong@cco.caltech.edu (Joseph Louis Long)
Subject: Re: Seeking magic dodecahedron
Cc: cube-lovers@life.ai.mit.edu
Content-Length: 774

In <1oieipINNs0k@gap.caltech.edu> on Mar 21, Joseph Louis Long wrote:

} So let me ask... Does anyone have a solution to Square-1?  Is there
} a simple ``operator'' based method, like there is for the cube?
} If it is simple enough to explain in text, could someone please
} post it?  Has there been a ``solutions book'' published?

Dunno about the former, but the answer to the latter is 'yes'.  In the April
GAMES magazine there is an ad:
   BAFFLED BY SQUARE 1
Now you can solve the world's most challenging
cube puzzle.  Clear, easy to unerstand book shows
you how.  Send $5 to
  Turn to Square 1
  PO Box 1451
  Westford, MA 01886

/Bernie\
-- 
Bernie Cosell                               cosell@world.std.com
Fantasy Farm Fibers, Pearisburg, VA         (703) 921-2358

From pbeck@pica.army.mil  Thu Mar 25 08:40:16 1993
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Date:     Thu, 25 Mar 93 8:27:15 EST
From: Peter Beck (BATDD) <pbeck@pica.army.mil>
To: cube-lovers@life.ai.mit.edu
Subject:  square-1
Message-Id:  <9303250827.aa21198@COR4.PICA.ARMY.MIL>


PRODUCT ANNOUNCEMENT:

SQUARE 1 SOLUTION BOOK & PROPOSED NEWSLETTER

RICHARD SNYDER
POB 1451
WESTFORD, MA 01886
617-246-0700 VOICE
617-246-1167  FAX

has written a book.  He says it is do back
from printing on april 10.  If you are 
interested CONTACT him directly.

THe world is probably waiting for the 
difinitive book review.


From myrberger@e.kth.se  Fri Apr 16 13:08:19 1993
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From: myrberger@e.kth.se
To: Cube-Lovers@ai.mit.edu
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Subject: 3x3x3 puzzles and other lists?

Hi,

I have recently found this mailing list and have just finished reading
through the earlier postings.

Perhaps this don't really belong to this list, but I have a list of
different puzzles that in their final state is a 3x3x3 cube. Among them are,
of course, Rubik's cube. The Soma cube and related puzzles are also there.

If you'd like a copy of the list, please MAIL me.

I also wonder if you know about other mailing lists or such which deals with
puzzles (preferrable mechanical). (I know about USENET/NEWS group rec.puzzles.)

Thanks

Johan

MAIL: myrberger@e.kth.se


From cosell@world.std.com  Wed May 26 22:37:04 1993
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From: cosell@world.std.com (Bernie Cosell)
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To: cube-lovers@life.ai.mit.edu
Subject: Ishi Intternational Puzzles.  NEW! (fwd)
Content-Length: 2127

On May 20, Anton Dovydaitis wrote:

[-------------------- text of forwarded message follows --------------------]

Ishi Press International is now directly accessable via e-mail:
our on-line InterNet address is 'ishius@ishius.com' and for
European customers it's 'ishi@cix.compulink.co.uk'.

Ishi Press International sells a wide variety of puzzles from
simple glass, wood and metal puzzles, to collector's items.
Our line consists of several hundred puzzles, including:
  Toyo Puzzle City Glass, Hikimi Puzzland Wood and Cast Iron puzzles by Nob
  Arjeu wood puzzles from France
  Wood, string and wire disentanglement puzzles by Jean-Claude Constantin
  Magic Bottle puzzles
  Handcrafted English Puzzles, including handblown glass Klein bottles,
    Single and Double Hourglass Paradoxes, and Wooden Trench Puzzles
  Wooden puzzles by Bill Cutler and Stuart Coffin
  Puzzle books by Slocum and Nob 
  Yamanaka Kumiki Works wooden burr puzzles
  Traditional inlaid wood Trick Puzzle Boxes by Okiyama and Ninomiya
  Yamanaka Kumiki Works wooden burr puzzles
  Kamei puzzle boxes, including the Top Box, Die Box, Book, Cup and
    Saucer and the Fan
  Bolt puzzles by Strijbos
  Puzzling People Puzzles, wooden 3-D jigsaw puzzles from England,
    including the Flummox
  Wire Puzzle Sculptures by Rick Irby, including a 3' Dragon,
    The Hong Kong Horror and puzzle earrings

and more.   Our puzzle line is always increasing: this summer we will be
carrying wood puzzles from Pentangle.

To receive our puzzle catalog with color photographs, please e-mail your
real mailing address to 'ishius@ishius.com'.

To receive regular e-mail on our latest offerings, e-mail us at 
'ishius@ishius.com'.

Please be sure to put the word PUZZLE in the subject header, as many of
our customers are interested in GO, not puzzles.

Please write me if you have any questions.

================================================
Anton Dovydaitis	Ishi Press International
ishius@ishius.com	76 Bonaventura Drive
Tel: 800/859-2086	San Jose, CA  95134
FAX: 408/944-9110

[------------------------- end of forwarded message ------------------------]

From dik@cwi.nl  Sun Jun 13 19:53:20 1993
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From: Dik.Winter@cwi.nl
Message-Id: <9306132353.AA19557.dik@boring.cwi.nl>
To: cube-lovers@life.ai.mit.edu
Subject: Contents of CFF31

Last Friday I received issue #31 of Cubism For Fun.  A short summary of
the contents:

1.  Short articles by Jan de Geus and Frans de Vreugd about Cube Day 1992.
2.  An article by Herbert Kociemba about a classification of pretty
    patterns on the cube.
3.  Reflections by Tom Verhoeff about puzzles and computers.
4.  Announcement by Koos Verhoef and Tom Verhoeff of a contest *.
5.  A short article by Jaques Haubrich about Rubik's Tangle and how to
    position 24 parts in a cube like way (four on a side).
6.  An article by Jan Verbakel about the creation of castles with solid
    pentominoes.
7.  A short article by Trevor Wood on the pecking of octacubes.
8.  A short article by Jaques Haubrich about a difficult packing problem.
9.  An article by David Singmaster about a gathering in Atlanta in honor
    of Martin Gardner.  (Nearly the whole puzzling world appears to have
    been there.)
10. A new contest by Anton Hanegraaf.
11. Announcement of the 13th Dutch cube day on August 22 in Amsterdam.
    This day is next to the 13th International Puzzle Party.

*   This is an interesting puzzle indeed.  Consider the densest sphere
    packing in 3D.  This is the packing where you start with a lattice
    of spheres based on a triangular lattice, and put on top of it another,
    similar, lattice such that each sphere of the new layer fits in a hole
    in the lower layer.  Add more layers.  Pick from that all possible
    configurations of 4 connected spheres.  There are 25 such
    configurations.  The puzzle is to create from these 25 pieces a
    pyramid with a side of 8 spheres (which contains 120 spheres), with
    a hole at the center that consists of a pyramid with a side of 4
    spheres (remember those sums of triangular numbers!).  It is not
    known whether there is a solution.  The authors tell how they have
    a TRS-80 now running 5 years on this problem, using backtracking
    techniques.  Until now the first 6 pieces did not move.  The could
    fit 24 pieces already 521,010 times.  The puzzle was first announced
    at the previous Cube Day.

CFF is a newsletter published by the Nederlandse Kubus Club NKC (Dutch
Cubists Club).  It appears a bit irregular, but a few times a year.
Yearly membership fee is now NLG 25.- (Dutch Guilders) which amounts to
approximately $ 15.-.  Information:
	Anton Hanegraaf
	Heemskerkstraat 9
	6662 AL  Elst
	The Netherlands
(sorry, there is no e-mail address).
--
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
home: bovenover 215, 1025 jn  amsterdam, nederland; e-mail: dik@cwi.nl

From @cunyvm.cuny.edu:Matt_Drobel@Novell.COM  Tue Jun 29 11:02:15 1993
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Total-To: cube-lovers%ai.mit.edu@cunyvm.cuny.edu 

signoff cube-lovers


From @cunyvm.cuny.edu:rbm8p@darwin.clas.virginia.edu  Tue Jun 29 23:15:45 1993
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unsubscribe

From hoey@aic.nrl.navy.mil  Tue Jul  6 17:29:01 1993
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From: hoey@aic.nrl.navy.mil
Message-Id: <9307062128.AA04699@sun13.aic.nrl.navy.mil>
To: Cube-Lovers@life.ai.mit.edu
Subject: Rubik's Shrimp

I hear on Usenet that there's a show on BBC1 called Wildlife 100,
where people saw a Mantis Shrimp playing with a Rubik's Cube.  Reports
are inconclusive as to whether it was able to actually turn faces, or
whether it just waved it around, or even just took it apart.  Now if
they could get it to turn faces, presumably they could film it and
play it back in reverse....

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From CPELLEY@delphi.com  Tue Jul 20 21:30:11 1993
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Date: Tue, 20 Jul 1993 21:29:06 -0400 (EDT)
From: CPELLEY@delphi.com
Subject: New idea for a puzzle
To: cube-lovers@life.ai.mit.edu
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I have a great concept for a new variant on Rubik's Cube.  Where can I
contact the people who manufacture these puzzles today?  I understand
Jean-Claude Constantin and Uwe Meffert are still around.
The idea is for a dodecahedral puzzle that is sliced up differently than
a Skewb or Megaminx.

From ronnie@cisco.com  Wed Jul 28 19:37:15 1993
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To: Cube-Lovers@life.ai.mit.edu
Subject: Hint wanted for 4x4x4
Date: Wed, 28 Jul 1993 16:37:08 -0700
From: "Ronnie B. Kon" <ronnie@cisco.com>

I've been beating my head against the order 4 Rubik's cube for long
enough, and I want a hint.  (Not a solution--I have a solution book if
I wanted to use it).

My problem is I cannot flip a pair of adjacent edges (this is
equivalent to not being able to exchange a pair of knights-move
separated edges).  All my other transformations have no side effects,
so I can solve the edges first.  But I can't see how to just affect
two of them.

I tend to solve using commutators, but I don't see a way here.  The
move I use on the top moves the marked pieces clockwise (this pattern

		. . 0 .
		. . . .
		. . . 0
		. 0 . .

rotates and reflects, of course).  There is no way to combine these
into a pair exchange (after doing the move, you still have two pieces
out of place--nothing changed from the original).

I tried to find a move that would exchange three pieces, the third
being the correctly placed piece next to one of the incorrectly placed
pieces (ie., treat a right edge cubie as if it should be a left edge
cubie) but this can easily be shown as impossible:

	Define the parity of a piece as being left if it is a left
	edge cubie when the red facelet is up, right if it is a right
	edge cubie when the red facelet is up.  The parity is
	undefined if there is no red facelet.  There are only three
	moves available that affect an edge cubie--none of them alter
	the parity.  QED


So, what am I missing?  As I said before, I really just want a hint
here.

				Ronnie

From diamond@jit081.enet.dec.com  Wed Jul 28 20:13:19 1993
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From: 29-Jul-1993 0914 <diamond@jit081.enet.dec.com>
To: cube-lovers@life.ai.mit.edu
Apparently-To: cube-lovers@life.ai.mit.edu
Subject: Re: Hint wanted for 4x4x4

Ronnie B. Kon asked for a hint but not a solution.  So here is a hint.

If I understood correctly your descriptions of two transforms which you
asserted to be equivalent, then in fact they are not equivalent.
(Of course, if I didn't understand your descriptions correctly, then
this isn't a hint.)

-- Norman Diamond      diamond@jit081.enet.dec.com
[Digital did not write this.]

From dik@cwi.nl  Wed Jul 28 20:17:38 1993
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Date: Thu, 29 Jul 93 02:17:25 +0200
From: Dik.Winter@cwi.nl
Message-Id: <9307290017.AA08740.dik@boring.cwi.nl>
To: Cube-Lovers@life.ai.mit.edu, ronnie@cisco.com
Subject: Re:  Hint wanted for 4x4x4

 > I've been beating my head against the order 4 Rubik's cube for long
 > enough, and I want a hint.  (Not a solution--I have a solution book if
 > I wanted to use it).

Yes, it is not really simple.

 > ...
 > I tend to solve using commutators, but I don't see a way here.  The
 > move I use on the top moves the marked pieces clockwise (this pattern

 > 		. . 0 .
 > 		. . . .
 > 		. . . 0
 > 		. 0 . .

 > rotates and reflects, of course).  There is no way to combine these
 > into a pair exchange (after doing the move, you still have two pieces
 > out of place--nothing changed from the original).

Still you are halfway there if you are willing to forgo the pattern of
the centers (which can always be done later).  Turn the front face (at
the bottom in the frawing), the right face and the back face one quarter
turn before your turn, and back after.  Observe that that constitutes a
cycle of three edge cubies in a single middle slice.  Combine with a
quarter turn of that middle slice.

 > So, what am I missing?  As I said before, I really just want a hint
 > here.

I hope that is enough of a hint and not enough of a giveaway.

(I thought there was a shorter sequence, but I disremember at the moment.)

My solution for the 4x4x4 always was: first corners, next edges and finally
centers.  Because there are many identical pieces for the centers those are
reasonably simple.  It would be much more difficult if each center had its
own place.

dik
--
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
home: bovenover 215, 1025 jn  amsterdam, nederland; e-mail: dik@cwi.nl

From hoey@aic.nrl.navy.mil  Thu Jul 29 08:36:32 1993
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From: hoey@aic.nrl.navy.mil (Dan Hoey)
Message-Id: <9307291236.AA08504@Sun0.AIC.NRL.Navy.Mil>
To: Cube-Lovers@life.ai.mit.edu
Cc: Dik.Winter@cwi.nl, ronnie@cisco.com (Ronnie B. Kon)
Subject: Re:  Hint wanted for 4x4x4
Newsgroups: ml.cube-lovers
In-Reply-To: <9307290017.AA08740.dik@boring.cwi.nl>
Organization: Navy Center for Applied Research in AI
Cc: 

ronnie@cisco.com (Ronnie B. Kon) asks for hints for exchanging a pair
of edges:

> > I tend to solve using commutators, but I don't see a way here....

The key is that commutators are always odd permutations.  So do the
move that is an odd permutation of the edges, then use commutators.

Dik.Winter@cwi.nl (dik t. winter) shows a neat way of moving most of
the cubies affected by the odd permutation into the top slice, where
they can be cycled using Ronnie's commutator, which cycles the 
TB(R), TR(F), and TF(L) cubies:

> > 		. . 0 .
> > 		. . . .
> > 		. . . 0
> > 		. 0 . .

(I'm naming them by their edge and their near side.)  I suspect Ronnie
is using something like (F Ti F') T (F Ti' F) T (F Ti F') T^2 (F Ti' F).
(For this I'm using "i" to mark inside slabs).  

But you can cycle the FL(T), FR(T), RB(T) cubies directly, using a
different commutator.  With more effort, there is a commutator that
doesn't mess up face centers.  We are getting to the part where it's
hard to distinguish between the hintable and the obvious, but if
people send me email about not being able to figure out what
commutators I'm talking about I'll answer, and post them if such
nobility is common.

>My solution for the 4x4x4 always was: first corners, next edges and finally
>centers.  Because there are many identical pieces for the centers those are
>reasonably simple.  It would be much more difficult if each center had its
>own place.

As I mentioned years ago, I've made places for mine by cutting corners
of to clusters of face centers and their neighboring edges on each
face.

    +----+----+----+----+
    |    |    |    |    |
    |    |    |    |    |
    +----+---( )---+----+
    |    |    |    |    |
    |    |    |    |    |
    +---( )---+----+----+
    |    |    |    |    |
    |    |    |    |    |
    +----+----+----+----+
    |    |    |    |    |
    |    |    |    |    |
    +----+----+----+----+

It's not that hard to fix the face centers, just time-consuming.  It's
a good thing we do the edges first, though, or it would be hard to
figure where the cuts go.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

( So much discussion on this quiescent list will probably flush out
  someone who wants to unsubscribe.  Remember to send your note to 
                cube-lovers-request@ai.ai.mit.edu
  to avoid annoyance.)

From ncramer@bbn.com  Thu Jul 29 09:40:49 1993
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Date:     Thu, 29 Jul 93 8:59:44 EDT
From: Nichael Cramer <ncramer@bbn.com>
To: "Ronnie B. Kon" <ronnie@cisco.com>
Cc: Cube-Lovers@life.ai.mit.edu
Subject:  Re:  Hint wanted for 4x4x4

Hi Ronnie

Let me try to restate the problem slightly to make sure we are talking
about the same problem.

Basically, you can solve the entire cube, except that the pieces "1" and
"2" in the diagram are flipped/exchanged:

  XXXX
  XXXX
  XXXX
  X12X

Assuming this is the problem, the hint is as follows: 

The problem here is that some of the face pieces are not really in their
right places.  In short, one of the center slices is 1/4 turn out of phase.

The simplest way to proceed (at least for me) is to move to the following
state: 

  XX2X    (i.e. rotate one of the central slices 1/4 turn)
  XXOX
  XXOX
  X1OX
    ^
    |
Now, you can solve for pieces "1" and "2" and --using these pieces as a
landmark-- proceed from there.

Hint #2: You can help avoid this problem by solving the face pieces last.


Extra Credit: Actually, the state as shown in the first diagram above is
pretty interesting in that the analogous position on a 3X3X3 cube (i.e. a
single flipped edge cube) is of course impossible.

From this state it is relatively easy to get to another, very interesting
state: namely the 4X4X4 appears to be completely solved except that two
opposite corners are exchanged.  (Again, this is obviously impossible on
the 3X3X3.)

Left as an exercise for the reader.      ;)


Nichael
ncramer@bbn.com            Dr Pepper: It's not just for breakfast any more.




From hoey@aic.nrl.navy.mil  Thu Jul 29 10:11:15 1993
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Date: Thu, 29 Jul 93 10:11:13 EDT
From: hoey@aic.nrl.navy.mil (Dan Hoey)
Message-Id: <9307291411.AA09096@Sun0.AIC.NRL.Navy.Mil>
To: Cube-Lovers@life.ai.mit.edu
Subject: Oops... Re: Hint wanted for 4x4x4
Cc: Dik.Winter@cwi.nl, ronnie@cisco.com (Ronnie B. Kon)

I wrote:

? The key is that commutators are always odd permutations.  So do the
? move that is an odd permutation of the edges, then use commutators.

But I *Meant*:

! The key observation is that commutators are always even
! permutations.  So you to perform an odd permutation on edge, you
! should do the move that is an odd permutation of the edges, then use
! commutators.

Dan

From tomgm@physics.purdue.edu  Thu Jul 29 11:02:30 1993
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From: Tom G. Miller <tomgm@physics.purdue.edu>
Subject: Re: Hint wanted for 4x4x4
To: ronnie@cisco.com (Ronnie B. Kon)
Date: Thu, 29 Jul 93 10:05:22 EST
Cc: Cube-Lovers@life.ai.mit.edu
In-Reply-To: <199307282337.AA06770@lager.cisco.com>; from "Ronnie B. Kon" at Jul 28, 93 4:37 pm
X-Mailer: ELM [version 2.3 PL11]

Ronnie,

It's been so long since I've messed around with my 4x4x4 that
I can't answer your question directly, however when I
see descriptions for the method in which people solve the
4x4x4 it is usually different from the way I first solved
it:

What I did was to pair up the middle two edgies, and the
four central face cubes.  There are few enough restraints
that this is not too hard to do for someone who has never
touched a 4x4x4 cube.  One then has a cube like with faces
similar to the following:

          r b b g
          y o o b
          y o o b
          r y y w

One can then "pretend" it is a 3x3x3 cube and then solve it.
Unfortunately you will occasionally end up in an orbit of the
"pseudo-3x3x3" that is impossible to solve.  Oh well... scramble 
it and try it again.

Using this technique I was able to solve a scrambled 4x4x4 cube within an
hour or so of when I set my hands on one.  Needless to say, this is NOT a
good technique for solving a 4x4x4 cube if one is interested only in the
4x4x4.  In fact I suspect it is a pretty awful algorithm, especially since
you frequently end up in an unsolveable orbit using your standard 3x3x3
techniques. But it is a useful trick for maximizing the hard work one used in
learning the 3x3x3.  

As most people who have a 4x4x4 realize, if you never make any twists of a
solved 4x4x4 cube except along the center, you of course have a 2x2x2. And
as I described, if you only make moves 1-deep, it is equivalent to a 3x3x3,
and of course it is also a 4x4x4.


Tom Miller
tomgm@physics.purdue.edu

From ronnie@cisco.com  Fri Jul 30 00:46:55 1993
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To: Cube-Lovers@life.ai.mit.edu
Subject: Re: Hint wanted for 4x4x4 
Date: Thu, 29 Jul 1993 21:46:48 -0700
From: "Ronnie B. Kon" <ronnie@cisco.com>

Thanks to all who responded.  I haven't yet got what I consider a
solution for my problem (shift a slice and resolve is my current
method which is slow and ugly) but at least I understand my problem
slightly better.

A few questions:

1.  What is the definition of parity by which commutators are even,
but slice turns are odd?  I haven't been able to come up with a
cube-wide parity.  (I know no group theory).

2.  How many orbits does the order 4 cube have?  I can only think of
three (twirling a corner cubie).  Then again, I haven't painted the
facelets yet, so there could be orbits I haven't begun to see
involving them.

3.  Would an order 6 cube have any challenge beyond the order 4?  I
think the answer is no--if you are able to solve the 3-cube and the
4-cube you can solve any cube.



From dik@cwi.nl  Mon Aug  2 20:52:23 1993
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Date: Tue, 3 Aug 93 02:52:19 +0200
From: Dik.Winter@cwi.nl
Message-Id: <9308030052.AA23253.dik@boring.cwi.nl>
To: cube-lovers@life.ai.mit.edu
Subject: Fitting puzzle solved

Some time ago I posted an article (c.q. mailed a message) describing
the contents of Cubism For Fun, the newsletter published by the Dutch
Cubists Club (NKC).  In that article (message) I gave a more elaborate
description about a problem involving fitting pieces.

Briefly:

The base problem is as follows.  Build a tetrahedron consisting of balls,
8 balls on an edge.  When you look at the lattice induced by this
tetrahedron after some thinking you will find there are 25 ways to pick
4 connected balls.  Now take those 25 ways and make "pieces" from it.
Again, go back to the tetrahedron and inside it create a hollow
tetrahedron with 4 balls on an edge.  The remainder requires 100 balls
to fill.  Try to do that with the 25 "pieces" you just created.

This has been a fairly long-standing problem but it is now (partly)
solved.

I just had word that Jan de Ruiter from Purmerend (the Netherlands)
found a number of solutions.  Details will likely be presented in
a forthcoming issue of CFF.

An amusing side-note.  Between the 25 pieces there are two that can
be created interlocked.  It is not clear whether it is possible to
separate those two pieces by hand when interlocked, so it is not
clear whether a solution that has those two pieces interlocked
really is a solution.  The first solutions Jan de Ruiter found
*had* those two pieces interlocked.  But after some time he found
a solution with those two pieces far away from each other, so there
is really a true solution.

Remaining questions:  How many solutions are there?  How many do not
have those two pieces interlocked?  Is it possible to separate those
two pieces when interlocked?  (The last puzzle resembles one of those
chinese metal separation puzzles.)

(Information about CFF can be obtained from Anton Hanegraaf,
Heemskerkstraat 9, 6662 AL  Elst, The Netherlands.  E-mail is
now also possible: gm@phys.uva.nl.)

dik
--
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
home: bovenover 215, 1025 jn  amsterdam, nederland; e-mail: dik@cwi.nl

From dik@cwi.nl  Mon Aug  2 21:10:32 1993
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Date: Tue, 3 Aug 93 03:10:22 +0200
From: Dik.Winter@cwi.nl
Message-Id: <9308030110.AA23300.dik@boring.cwi.nl>
To: cube-lovers@life.ai.mit.edu
Subject: Diameter of cube group?

I have now running (for about 60 days already) a program that implements
Kociemba's algorithm to solve the cube.  It tries to solve random
configurations and stops when a solution of 20 turns or less is found.
The random configurations are created by doing 100 random turns.

Until now, with 9000 configurations tried, all proved to be solvable
in 20 turns or less.

This strongly suggests that the diameter of the cube group is at most
21, or perhaps 22; but not more.

The figure of 9000 configurations in 60 days indicates that solution
of one configuration takes slightly less than 10 minutes.  This is
contrary to what I thought was possible.  Whenever I tried configurations
they were mostly solved within 2 or 3 minutes.  This suggests that the
random configurations are more difficult to solve than what I and many
others brought up as possible difficult patterns.

But I still need to do some analysis on the ouput (now 3 Mb of data).

Continuing and waiting for a config that requires 21 turns, dik
--
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
home: bovenover 215, 1025 jn  amsterdam, nederland; e-mail: dik@cwi.nl

From W.Taylor@math.canterbury.ac.nz  Mon Aug  2 22:20:16 1993
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Date: Tue, 3 Aug 93 14:19:52 NZS
From: W.Taylor@math.canterbury.ac.nz (Bill Taylor)
Subject: re: Diameter of cube group?
To: Cube-Lovers@life.ai.mit.edu
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Fascinating news from Dik Winter about the solving of 9000 configurations.

Can someone please remind us exactly what Kociemba's algorithm is; or at
least a breif outline of how it works.   I know I've heard the name before,
but can't remember anything about it.

Thanks,  Bill Taylor.         wft@math.canterbury.ac.nz


From dik@cwi.nl  Tue Aug  3 18:44:13 1993
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Date: Wed, 4 Aug 93 00:44:09 +0200
From: Dik.Winter@cwi.nl
Message-Id: <9308032244.AA26535.dik@boring.cwi.nl>
To: cube-lovers@life.ai.mit.edu
Subject: Kociemba's algorithm

I have received a few requests for information about the algorithm and
for the program.  I have put the program available for ftp.  In the
set of files there is also a Description I edited from a number of
messages I mailed a long time ago to this mailing list.  They give
a reasonable description of the algorithm.

Ftp to ftp.cwi.nl.  File is: /pub/dik/cube.tar.Z.  Do not forget
to set binary mode.

dik
--
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
home: bovenover 215, 1025 jn  amsterdam, nederland; e-mail: dik@cwi.nl

From acw@bronze.lcs.mit.edu  Thu Aug  5 17:03:17 1993
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Date: Thu, 5 Aug 93 17:02:45 EDT
From: acw@bronze.lcs.mit.edu (Allan C. Wechsler)
Message-Id: <9308052102.AA22841@bronze.lcs.mit.edu>
To: Dik.Winter@cwi.nl
Cc: cube-lovers@life.ai.mit.edu
In-Reply-To: Dik.Winter@cwi.nl's message of Tue, 3 Aug 93 03:10:22 +0200 <9308030110.AA23300.dik@boring.cwi.nl>
Subject: Diameter of cube group?

I wonder about the validity of your Monte Carlo analysis.  It seems
to be based on an intuition about how fast the number of configurations
falls off with the distance from SOLVED.  I share the intuition, but
I'm not sure I can rigorize it, and that makes me cautious.

What prevents a group from having a "pointy tail", that is, a "corridor"
of elements at increasing distances from the identity?  In fact, does
the number of elements as a function of distance have to be unimodal?
Could this function have a "waist"?  Intuitively, this sounds
impossible, but I am wondering what constraints on such functions are known.


From ronnie@cisco.com  Thu Aug  5 19:55:48 1993
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To: acw@bronze.lcs.mit.edu (Allan C. Wechsler)
Cc: Dik.Winter@cwi.nl, cube-lovers@life.ai.mit.edu
Subject: Re: Diameter of cube group? 
In-Reply-To: Your message of "Thu, 05 Aug 1993 17:02:45 EDT."
             <9308052102.AA22841@bronze.lcs.mit.edu> 
Date: Thu, 05 Aug 1993 16:55:36 -0700
From: "Ronnie B. Kon" <ronnie@cisco.com>

Disclaimer: this sounds more authoritative than is intended--I really
don't know what I'm talking about.


It couldn't be very pointy.  From the most distant configuration,
there are 6 positions immediately before it.  There are 6^2 two steps
away, 6^3 three steps, etc.  (well, 6^2 - 1 and 6^3 - ?) actually.

This is necessarily so, as if any of the configurations reachable with
two twists (for example) are closer in than (max - 2) steps then you
could reach the maximum configuration by getting there and then doing
the two steps.

This gives me the feeling that Monte Carlo is fairly valid.  (How's
that for rigor?)

				Ronnie

> I wonder about the validity of your Monte Carlo analysis.  It seems
> to be based on an intuition about how fast the number of configurations
> falls off with the distance from SOLVED.  I share the intuition, but
> I'm not sure I can rigorize it, and that makes me cautious.
> 
> What prevents a group from having a "pointy tail", that is, a "corridor"
> of elements at increasing distances from the identity?  In fact, does
> the number of elements as a function of distance have to be unimodal?
> Could this function have a "waist"?  Intuitively, this sounds
> impossible, but I am wondering what constraints on such functions are known.
> 

From dik@cwi.nl  Thu Aug  5 19:55:54 1993
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Date: Fri, 6 Aug 93 01:55:50 +0200
From: Dik.Winter@cwi.nl
Message-Id: <9308052355.AA05297.dik@boring.cwi.nl>
To: acw@bronze.lcs.mit.edu
Subject: Re:  Diameter of cube group?
Cc: cube-lovers@life.ai.mit.edu

 > I wonder about the validity of your Monte Carlo analysis.  It seems
 > to be based on an intuition about how fast the number of configurations
 > falls off with the distance from SOLVED.  I share the intuition, but
 > I'm not sure I can rigorize it, and that makes me cautious.

I am not sure (that is obvious).  But when looking at other (similar)
puzzles I did I think it is a save guess.

 > What prevents a group from having a "pointy tail", that is, a "corridor"
 > of elements at increasing distances from the identity?

The groups I did calculate in full do *not* have a pointy tail.  This holds
for 2x2x2, 3x3x3 corners only, magic domino.  I think it would be a big
surprise if there is a pointy tail.  Obviously we can say a priory that
there is not a single configuration opposite from start, so the tail is
not very pointy, if it is at all.  For instance for the magic domino the
tail of the list of number of configuration a certain distance from start is:
14:    508704668
15:    232904952
16:     14508468
17:       129376
18:          112
With the maximum at 14 turns.  (Here I took the table where only a single
solution is allowed; i.e. no full rotations of the domino.)  1 in 2 (approx.)
configurations requires 14 turns or more.  1 in 100 requires 16 turns or
more.  Of course the number of configurations of the cube is quite a bit more.
Still after doing about 9000 configurations not a single one is found that
requires more than 20 turns.  If we assume a picture similar to the domino
(which in my opinion is a save guess), there might be configurations that
retuire 21 or perhaps 22 turns, but more is extremely unlikely.

However, there is a remaining question; is the random choice of configuration
unbiased?  I think it is.  To create a random configuration I do 100 random
turns chosen from 18 possible turns (quarter turns, half turns and reverse
turns).  The random number generator is (as far as I know) quite good
(Berkeley Unix's random).

dik

From dik@cwi.nl  Thu Aug  5 20:01:34 1993
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From: Dik.Winter@cwi.nl
Message-Id: <9308060001.AA05312.dik@boring.cwi.nl>
To: ronnie@cisco.com
Subject: Re: Diameter of cube group?
Cc: cube-lovers@life.ai.mit.edu

The last remark first:

 > This gives me the feeling that Monte Carlo is fairly valid.  (How's
 > that for rigor?)

Not very ;-).

 > It couldn't be very pointy.  From the most distant configuration,
 > there are 6 positions immediately before it.  There are 6^2 two steps
 > away, 6^3 three steps, etc.  (well, 6^2 - 1 and 6^3 - ?) actually.

This still can create a pointy tail; just as pointy as the front.  My
experience is that the tail is much more blunt than the front.  That there
are already more than a single configuration at maximum distance makes that
reasonable.

From CPELLEY@delphi.com  Fri Aug  6 02:47:53 1993
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Date: Fri, 06 Aug 1993 00:53:23 -0400 (EDT)
From: CPELLEY@delphi.com
Subject: Square-1 Puzzle Party
To: cube-lovers@life.ai.mit.edu
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Richard Snyder's book on Square-1 is now being published.  He sent me the
following press release about a forthcoming Square-1 Puzzle Party:

Square-1 is the most challenging puzzle since Rubik's Cube!  When you turn
it, it forms many unfamiliar shapes, and it seems impossible to get it back
to the cube shape!  And if you do somehow manage to turn it into a cube,
it is scrambled and needs to be solved Rubik-style.
It's really two puzzles in one!  Harder than Rubik's, it's so hard that
only 5 people in the whole world have ever been able to come up with a
complete solution to it!
Richard Snyder of Dracut is the only person in the USA who has written a  
book which shows how to solve Square-1!
His book is clear and easy to follow, leading you step by step from any
scrambled state to the completely solved cube!  Then he gives formulas
for over 100 colored patterns which you can make on Square-1's
symmetrical shapes, and teaches you how to make your own symmetrical
patterns!  There's no other book quite like it in the world!

Richard will be presenting his new book, Turn to Square-1, to Boston in a
great puzzle party, which will be held at 1PM on Sat., Aug. 7, 1993, at
The Games People Play, 1105 Massachusetts Ave., Cambridge, MA (near
Harvard Square)
Richard will demonstrate solving Square-1, making Square-1 patterns, and he  
will also demonstrate solving Rubik's Cube, the Skewb, and other cube puzzles.
He will be autographing copies of his new book, and presenting many other
fine puzzles and books that are carried by The Games People Play.
The Press and the Media will be there, and you are invited to come.
Bring your Square-1, your Rubik's Cube, and your other Rubik's puzzles that
you haven't been able to solve!  Richard will solve them, and show you his
solution to Rubik's Cube, the world's best, fastest, and most concise Rubik's
Cube solution!
But most of all, be prepared to be astounded as Richard shows you how you 
too can Turn to Square-1!

From ccw@eql12.caltech.edu  Fri Aug  6 22:37:43 1993
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Date:     Fri, 6 Aug 93 12:50:32 PDT
From: ccw@eql12.caltech.edu (Chris Worrell)
Message-Id: <930806124838.23c011ac@EQL12.Caltech.Edu>
Subject:  Re: Square-1 Puzzle Party
In-Reply-To: Your message <01H1EMUZ0T1E91XFD4@delphi.com> dated  6-Aug-1993
To: CPELLEY@delphi.com
Cc: ccw@eql12.caltech.edu, cube-lovers@life.ai.mit.edu

Sorry.  I can't let this one pass by without comment.

CPELLEY@delphi.com says
> Richard Snyder's book on Square-1 is now being published.  He sent me the
> following press release about a forthcoming Square-1 Puzzle Party:
<text deleted>
> It's really two puzzles in one!  Harder than Rubik's, it's so hard that
> only 5 people in the whole world have ever been able to come up with a
> complete solution to it!

Unless Snyder or his agent is talking about a God's Algorithm for
Square-1, this statement is ridiculous.
I doubt that this number includes myself, as I have only told a few family
members and friends that I have solved this.  (I expect many of you can say
the same thing.)

Harder than Rubik's?  This is a matter of opinion and definition.
Do they mean conceptually harder, harder to derive a solution method,
harder to prove a solution method, or harder to achieve an individual
solution attempt?  Or does harder just mean more time?  More time
to derive a solution method, more time to prove a solution method,
or more time to achieve an individual solution attempt?
I don't really doubt the last.  Except for the Pyraminx (and the 2-Cube),
all of the puzzles of this type take me longer to solve than the Cube.
I think that the Rubik's cube still holds the record as the puzzle that
took me longest to derive a solution method.  (Of course all of the others
borrowed substantially from the cube.)

>Bring your Square-1, your Rubik's Cube, and your other Rubik's puzzles that
>you haven't been able to solve!
Sorry, I don't have any.  Except the 10x10 Rubik's Tangle.

Chris Worrell
ccw@eql.caltech.edu

From dn1l+@andrew.cmu.edu  Fri Aug  6 23:27:13 1993
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Date: Fri,  6 Aug 1993 23:22:28 -0400 (EDT)
From: "Dale I. Newfield" <dn1l+@andrew.cmu.edu>
To: cube-lovers@life.ai.mit.edu
Subject: Tangle (Was: Re: Square-1 Puzzle Party)
In-Reply-To: <930806124838.23c011ac@EQL12.Caltech.Edu>

Excerpts from mail: 6-Aug-93 Re: Square-1 Puzzle Party by Chris
Worrell@eql12.calt 
>>Bring your Square-1, your Rubik's Cube, and your other Rubik's puzzles that
>>you haven't been able to solve!
>Sorry, I don't have any.  Except the 10x10 Rubik's Tangle.

I only have one quarter of that puzzle...(section 4).

I worked on it for a considerable amount of time, and concluded that the only
solution method was trial and error.

So I wrote a program to do it for me.
I know all 16 solutions (2 unique)*(2 identical exchanged pieces)*(4
orientations).

Has anyone come up with a method, besides trial and error, that solves
this thing?  (or the 10x10?)
(hmmm--I wonder how much the other 3 would cost?)

-Dale Newfield
dn1l@{cs,andrew}.cmu.edu


From @uccvma.ucop.edu:MJTOL@UCCMVSA.BITNET  Sat Aug  7 04:25:18 1993
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From: "Michael Thwaites" <MJTOL%UCCMVSA.bitnet@uccvma.ucop.edu>
To: cube-lovers@life.ai.mit.edu
Subject:  cube tail?


> What prevents a group from having a "pointy tail", that is,
> a "corridor" of elements at increasing distances from the
> identity?  In fact, does the number of elements as a
> function of distance have to be unimodal?  Could this
> function have a "waist"?  Intuitively, this sounds
> impossible, but I am wondering what constraints on such
> functions are known.
>
It seems to me it can't be too pointy. Working backwards, the
number of arrangements working from the end has to explode
(probably in symetry) with the number of arrangements form the
start.

From weber@src.dec.com  Sat Aug  7 17:33:08 1993
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To: "Dale I. Newfield" <dn1l+@andrew.cmu.edu>
Cc: cube-lovers@life.ai.mit.edu
Subject: Tangle (Was: Re: Square-1 Puzzle Party)
In-Reply-To: Message of Fri,  6 Aug 1993 23:22:28 -0400 (EDT)
    from "Dale I. Newfield" <dn1l+@andrew.cmu.edu>
    <cgMlzoa00VpJM3_Ux4@andrew.cmu.edu>
Date: Sat, 07 Aug 93 14:32:53 -0700
From: weber@src.dec.com
X-Mts: smtp


>>>Bring your Square-1, your Rubik's Cube, and your other Rubik's puzzles that
>>>you haven't been able to solve!
>>Sorry, I don't have any.  Except the 10x10 Rubik's Tangle.
>
>I only have one quarter of that puzzle...(section 4).
>
>I worked on it for a considerable amount of time, and concluded that the only
>solution method was trial and error.

I was thinking about the Rubik's Tangle, and what was puzzling me was
WHY there should be only one solution (apart from the obvious symmetries).
After all, all pieces are identical except for coloring, and a set consists
of all 24 possible coloring, and 1 duplicate, and this doesn't sound like
an artificial construction.  Is there any mathematical reason for the
uniqueness of the solution?  What possible "Tangle-like" puzzles have
unique solutions?

  -Sam

From dn1l+@andrew.cmu.edu  Sun Aug  8 00:21:54 1993
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Date: Sat,  7 Aug 1993 19:36:39 -0400 (EDT)
From: "Dale I. Newfield" <dn1l+@andrew.cmu.edu>
To: cube-lovers@life.ai.mit.edu
Subject: Re: Tangle (Was: Re: Square-1 Puzzle Party)
Cc: 
In-Reply-To: <9308072132.AA01481@chaucer>

Excerpts from mail: 7-Aug-93 Tangle (Was: Re: Square-1 P.. by
weber@src.dec.com 
> I was thinking about the Rubik's Tangle, and what was puzzling me was
> WHY there should be only one solution (apart from the obvious symmetries).
> After all, all pieces are identical except for coloring, and a set consists
> of all 24 possible coloring, and 1 duplicate, and this doesn't sound like
> an artificial construction.  Is there any mathematical reason for the
> uniqueness of the solution?  What possible "Tangle-like" puzzles have
> unique solutions?

The section I had (4) had 2 distinct solutions (apart from the exchange
of the 2 identical pieces, and the 4 orientations).

In fact, the box that the puzzle came in said it should have 2 solutions.

-Dale


From @mail.uunet.ca:mark.longridge@dosgate.canrem.COM  Mon Aug  9 12:02:01 1993
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To: CUBE-LOVERS@ai.mit.edu
Subject: SQUARE'S GROUP ANALYSIS
From: Mark Longridge <mark.longridge@canrem.com>
Message-Id: <60.250317.104.0C180A7E@canrem.com>
Date: 	Sun, 8 Aug 1993 15:40:00 -0400
Organization: CRS Online  (Toronto, Ontario)

After reading Dik's post I figured I'd add my 2 cents worth:

                Mark's Notes on the Squares Group
                ---------------------------------

   On studying the squares group I have found 16 antipodal cases
requiring
the maximum 15 moves. Two of these cases cycle all 8 corners and leave
the edges in place. A third case "2 DOT/Inverted T's" is pleasingly
symmetric. Also I have noted that cycling only the 4 edges in the
U or D layer requires 1 move less that cycling only the 4 corners in U
or D when using only moves in the square's group, 12 moves for edges
and 13 moves for corners.

   If we define "symmetry level" as the number of distinct patterns
generated by rotating the cube through it's 24 different orientations in
space then most known antipodes are symmetry level 6. Thus the lower the
number the higher the level of symmetry. The least symmetric positions
have level 24, and this is very common. The most symmetric positions
have level 1, the two positions START and 6 X order 2.

   I have also found positions with levels 3, 8 and 12.

   Given the fact that 8 antipodal cases have symmetry level 6
and 8 cases have symmetry level 12 we can now account for ALL
8 * 6 + 8 * 12 = 144 of the 144 cases!

Cases with symmetry level 6:

p66  Double 4 corner sw  L2 B2 R2 F2 L2 F2 T2 R2 (T2 D2 F2 T2) F2 L2 D2
(15)
p67  Antipode 2          R2 B2 D2 F2 D2 F2 T2 L2 (T2 D2 F2 T2) L2 T2 B2
(15)
p80  2 DOT, Invert T's   R2 B2 D2 R2 B2 L2 B2 L2 (T2 D2 F2 T2) F2 L2 T2
(15)
p99  2 DOT, 4 ARM        R2 B2 D2 L2 B2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2
(15)
p100 2 Cross, 4 ARCH 1   R2 B2 T2 R2 F2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2
(15)
p130 2 Cross, 4 ARCH 2   L2 B2 D2 B2 L2 D2 F2 L2 (T2 D2 F2 L2) F2 L2 T2
(15)
p135 2 X, 4 T            L2 B2 D2 F2 T2 F2 T2 L2 (T2 D2 F2 T2) L2 D2 F2
(15)
p137 2 X, 4 ARM          L2 F2 T2 B2 T2 F2 T2 L2 (T2 D2 F2 T2) L2 D2 F2
(15)

Cases with symmetry level 12:

p108 2 DOT, 2 T, 2 ARM   L2 F2 T2 R2 B2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2
(15)
p128 2 H, 2 T, 2 CRN     L2 B2 R2 F2 L2 F2 T2 R2 (T2 D2 F2 T2) F2 L2 T2
(15)
p129 2 H, 2 T, 2 ARCH    R2 F2 L2 F2 L2 F2 T2 R2 (T2 D2 F2 T2) F2 L2 T2
(15)
p131 2 H, 2 ARM, 2 ARCH  L2 F2 R2 D2 B2 L2 D2 L2 (T2 D2 F2 T2) F2 L2 F2
(15)
p132 2 Cross,2 ARCH,2CRN L2 F2 D2 R2 F2 L2 B2 L2 (T2 D2 F2 T2) F2 L2 D2
(15)
p133 2 Cross, 2 T, 2 ARM L2 F2 D2 F2 D2 F2 T2 L2 (T2 D2 F2 T2) L2 T2 B2
(15)
p134 2 CRN, 2 X, 2 ARCH  L2 F2 D2 B2 T2 F2 T2 L2 (T2 D2 F2 T2) L2 T2 B2
(15)
p136 2 H, 2 ARM, 2 CRN   R2 F2 L2 T2 B2 L2 T2 L2 (T2 D2 F2 T2) F2 L2 B2
(15)

   5 of the 16 known antipodes are within 4 and 2 face turns (or 2 and 1
slice turns) of each other:

       p66 + L2 R2 T2 D2 = p80  (allowing for whole cube rotations)
       p66 + F2 B2       = p100
       p80 + T2 D2       = p99
       P66 + T2 D2       = P128

   Using full group moves these antipodes can be reduced to:

P66a alternate method    F2 R2 U2 F2 R2 U3 D3 B2 L2 F2 B2 U1 D1
 (13)
p67a alternate method    F2 R2 F2 U3 D3 L2 B2 D2 L2 B2 U1 D1 B2
 (13)
p80a alternate method    U1 F2 R2 L2 U2 D2 F2 U2 D3
  (9)
p99a alternate method    U1 R2 F2 B2 U2 D2 R2 D1
  (8)
P100a alternate method   F2 U2 D2 F2 R2 L2 D1 F2 R2 L2 B2 U1
 (12)
p108a alternate method   R2 F2 B2 L2 D1 R2 U2 R2 L2 U2 R2 D1
 (12)
p130a alternate method   F2 R2 F2 B2 U1 D1 F2 R2 D2 F2 L2 U3 D3
 (13)
p133a alternate method   R2 U1 F2 R2 L2 U2 D2 F2 U2 D3 R2
 (11)

   Both p80a and p99a are surprisingly compact, p99a being a full 7
turns
less than it's square's group equivalent. Note that in p99a a square's
group sequence is sandwiched between 2 turns on opposite faces. It is
the final turn D1 which brings it back into a sq group state! In general
U1 (sq group sequence) D1 does not lead to a sq group sequence.

   Another interesting discovery was comparing the full group sequences:

                L1 R1 D2 L3 R3             (antislice, 5 moves)
                L1 R3 D2 L3 R1             (slice    , 5 moves)
                F1 B1 D2 F1 B1             (clockwise, 5 moves)

... to their square's group equivalents:

                R2 F2 T2 L2 B2 L2 T2 R2 F2 (9 moves)
                R2 B2 L2 D2 R2 F2 L2 T2 F2 (9 moves)
                R2 B2 T2 F2 L2 F2 T2 R2 F2 (9 moves)

   Also it was found possible to permute 3 edges only using:

                L2 T2 R2 B2 R2 T2 L2 F2    (8 moves)
          or    L3 R1 F2 L1 R3 D2          (6 moves)

   In general any sequence   L1 R1 (any squares group moves) L3 R3
     will always result in a squares group position, for example:

                         L1 R1 (D2 F2 B2)    R3 L3
                         F1 B1 (T2 B2 F2 L2) F3 B3
   p66a (F2 R2 U2 F2 R2) U3 D3 (B2 L2 F2 B2) U1 D1  (13 moves)

   The longest irreducible square's group sequence discovered so far,
     which is an embedded part of longest Phase 2 sequence (p94):
     (Thus it can't be reduced by using full group moves using current
     techniques)

                R2 B2 U2 B2 L2 D2 L2 F2          (8)

   Later on I discovered this irreducible sequence by chance:

                T2 B2 T2 B2 D2 F2 R2 T2 L2 F2   (10)

   Edges only (with corners in place) can be 14 moves at most, e.g.

           D2 L2 F2 D2 F2 L2 T2 L2 T2 D2 F2 R2 F2 R2  (14)

   This answers the question David Singmaster posed in
   "Notes on Rubik's Magic Cube" on Thistlethwaite's last stage.
   That is: "Are there any positions in the square's group with
   corners fixed of length 14 or can they be done in less moves?"

   A few observations...

    -  It is not possible to swap just 1 pair of edges and corners
    -  It is only possible to have 4, 6 or 8 corners out of place
    -  Known antipodal cases can be solved in <=13 moves using full
        group
    -  In reaching an antipode one may start with any of the 6 turns
        (since antipodes are global maxima, any turn will get you one
        move closer)
    -  If the corners are fixed, the position is NOT an antipode
    -  Longest order appears to be 12
    -  All known (probably all!) antipodes have symmetry level 6 or 12
    -  Although only conjectural, it is now believed that one turn of
        a face MUST lead to a new state which is either 1 move closer
         or 1 move farther from START

Question:     Are there any irreducible square's group sequences that
               are longer then 10 moves? Are these truly irreducible
               or only irreducible under Dik Winter's Kociemba
               inspired program?

Oh well, the full group beckons. I still want to try and come up
with my own algorithm though.

-> Mark <-

From @mail.uunet.ca:mark.longridge@dosgate.canrem.COM  Mon Aug  9 12:39:08 1993
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To: CUBE-LOVERS@ai.mit.edu
Subject: CUBES (OF COURSE!)
From: Mark Longridge <mark.longridge@canrem.com>
Message-Id: <60.250315.104.0C180A7C@canrem.com>
Date: 	Sun, 8 Aug 1993 15:34:00 -0400
Organization: CRS Online  (Toronto, Ontario)

Well, I finally know what all the square's group antipodes look like.
Next message I'll post a detailed summary on these patterns.

I wrote a colour printer driver for my cube program today, tested
it and it turned out pretty slick. I'm using a Star NX2420 rainbow
printer and I'm (out of necessity) using

   FACE         Star NX2420    Code    Real Cube
   ----         -----------    ----    ---------

  TOP/DOWN     BLACK/CYAN     0/2     WHITE/BLUE
  LEFT/RIGHT   VIOLET/ORANGE  3/5     RED/ORANGE
  FRONT/BACK   YELLOW/GREEN   4/6     YELLOW/GREEN

...which works pretty good except violet is more like blue. I could
have also used magenta (sort of a pink) for red but it does not
contrast well with orange.

Should make the DOTC (Domain of the Cube) Newsletter look a lot
better if I can ever finish the damn thing!

More notes to follow....
 -> Mark Longridge, still cubing after all of these years <-

From @mail.uunet.ca:mark.longridge@dosgate.canrem.COM  Tue Aug 10 11:07:52 1993
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To: cube-lovers@life.ai.mit.edu
Subject: Cubes (of course)
From: Mark Longridge <mark.longridge@canrem.com>
Message-Id: <60.251050.104.0C180C04@canrem.com>
Date: 	Tue, 10 Aug 1993 01:22:00 -0400
Organization: CRS Online  (Toronto, Ontario)

Well, I finally know what all the square's group antipodes look like.
Next message I'll post a detailed summary on these patterns.

I wrote a colour printer driver for my cube program today, tested
it and it turned out pretty slick. I'm using a Star NX2420 rainbow
printer and I'm (out of necessity) using

   FACE         Star NX2420    Code    Real Cube
   ----         -----------    ----    ---------

  TOP/DOWN     BLACK/CYAN     0/2     WHITE/BLUE
  LEFT/RIGHT   VIOLET/ORANGE  3/5     RED/ORANGE
  FRONT/BACK   YELLOW/GREEN   4/6     YELLOW/GREEN

...which works pretty good except violet is more like blue. I could
have also used magenta (sort of a pink) for red but it does not
(contrast well with orange.

Should make the DOTC (Domain of the Cube) Newsletter look a lot
better if I can ever finish the damn thing!

More notes to follow....
 -> Mark Longridge, still cubing after all of these years <-

From @mail.uunet.ca:mark.longridge@dosgate.canrem.com  Tue Aug 10 11:11:29 1993
Return-Path: <@mail.uunet.ca:mark.longridge@dosgate.canrem.com>
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To: cube-lovers@life.ai.mit.edu
Subject: Square's group
From: Mark Longridge <mark.longridge@canrem.com>
Message-Id: <60.251051.104.0C180C07@canrem.com>
Date: 	Tue, 10 Aug 1993 01:29:00 -0400
Organization: CRS Online  (Toronto, Ontario)

After reading Dik's post I figured I'd add my 2 cents worth:

                Mark's Notes on the Squares Group
                ---------------------------------

   On studying the squares group I have found 16 antipodal cases
requiring
the maximum 15 moves. Two of these cases cycle all 8 corners and leave
the edges in place. A third case "2 DOT/Inverted T's" is pleasingly
symmetric. Also I have noted that cycling only the 4 edges in the
U or D layer requires 1 move less that cycling only the 4 corners in U
or D when using only moves in the square's group, 12 moves for edges
and 13 moves for corners.

   If we define "symmetry level" as the number of distinct patterns
generated by rotating the cube through it's 24 different orientations in
space then most known antipodes are symmetry level 6. Thus the lower the
number the higher the level of symmetry. The least symmetric positions
have level 24, and this is very common. The most symmetric positions
have level 1, the two positions START and 6 X order 2.

   I have also found positions with levels 3, 8 and 12.

   Given the fact that 8 antipodal cases have symmetry level 6
and 8 cases have symmetry level 12 we can now account for ALL
8 * 6 + 8 * 12 = 144 of the 144 cases!

Cases with symmetry level 6:

p66  Double 4 corner sw  L2 B2 R2 F2 L2 F2 T2 R2 (T2 D2 F2 T2) F2 L2 D2
p67  Antipode 2          R2 B2 D2 F2 D2 F2 T2 L2 (T2 D2 F2 T2) L2 T2 B2
p80  2 DOT, Invert T's   R2 B2 D2 R2 B2 L2 B2 L2 (T2 D2 F2 T2) F2 L2 T2
p99  2 DOT, 4 ARM        R2 B2 D2 L2 B2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2
p100 2 Cross, 4 ARCH 1   R2 B2 T2 R2 F2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2
p130 2 Cross, 4 ARCH 2   L2 B2 D2 B2 L2 D2 F2 L2 (T2 D2 F2 L2) F2 L2 T2
p135 2 X, 4 T            L2 B2 D2 F2 T2 F2 T2 L2 (T2 D2 F2 T2) L2 D2 F2
p137 2 X, 4 ARM          L2 F2 T2 B2 T2 F2 T2 L2 (T2 D2 F2 T2) L2 D2 F2

Cases with symmetry level 12:

p108 2 DOT, 2 T, 2 ARM   L2 F2 T2 R2 B2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2
p128 2 H, 2 T, 2 CRN     L2 B2 R2 F2 L2 F2 T2 R2 (T2 D2 F2 T2) F2 L2 T2
p129 2 H, 2 T, 2 ARCH    R2 F2 L2 F2 L2 F2 T2 R2 (T2 D2 F2 T2) F2 L2 T2
p131 2 H, 2 ARM, 2 ARCH  L2 F2 R2 D2 B2 L2 D2 L2 (T2 D2 F2 T2) F2 L2 F2
p132 2 Cross,2 ARCH,2CRN L2 F2 D2 R2 F2 L2 B2 L2 (T2 D2 F2 T2) F2 L2 D2
p133 2 Cross, 2 T, 2 ARM L2 F2 D2 F2 D2 F2 T2 L2 (T2 D2 F2 T2) L2 T2 B2
p134 2 CRN, 2 X, 2 ARCH  L2 F2 D2 B2 T2 F2 T2 L2 (T2 D2 F2 T2) L2 T2 B2
p136 2 H, 2 ARM, 2 CRN   R2 F2 L2 T2 B2 L2 T2 L2 (T2 D2 F2 T2) F2 L2 B2

   5 of the 16 known antipodes are within 4 and 2 face turns (or 2 and 1
slice turns) of each other:

       p66 + L2 R2 T2 D2 = p80  (allowing for whole cube rotations)
       p66 + F2 B2       = p100
       p80 + T2 D2       = p99
       P66 + T2 D2       = P128

   Using full group moves these antipodes can be reduced to:

P66a alternate method    F2 R2 U2 F2 R2 U3 D3 B2 L2 F2 B2 U1 D1
p67a alternate method    F2 R2 F2 U3 D3 L2 B2 D2 L2 B2 U1 D1 B2
p80a alternate method    U1 F2 R2 L2 U2 D2 F2 U2 D3
p99a alternate method    U1 R2 F2 B2 U2 D2 R2 D1
P100a alternate method   F2 U2 D2 F2 R2 L2 D1 F2 R2 L2 B2 U1
p108a alternate method   R2 F2 B2 L2 D1 R2 U2 R2 L2 U2 R2 D1
p130a alternate method   F2 R2 F2 B2 U1 D1 F2 R2 D2 F2 L2 U3 D3
p133a alternate method   R2 U1 F2 R2 L2 U2 D2 F2 U2 D3 R2

   Both p80a and p99a are surprisingly compact, p99a being a full 7
turns
less than it's square's group equivalent. Note that in p99a a square's
group sequence is sandwiched between 2 turns on opposite faces. It is
the final turn D1 which brings it back into a sq group state! In general
U1 (sq group sequence) D1 does not lead to a sq group sequence.

   Another interesting discovery was comparing the full group sequences:

                L1 R1 D2 L3 R3             (antislice, 5 moves)
                L1 R3 D2 L3 R1             (slice    , 5 moves)
                F1 B1 D2 F1 B1             (clockwise, 5 moves)

... to their square's group equivalents:

                R2 F2 T2 L2 B2 L2 T2 R2 F2 (9 moves)
                R2 B2 L2 D2 R2 F2 L2 T2 F2 (9 moves)
                R2 B2 T2 F2 L2 F2 T2 R2 F2 (9 moves)

   Also it was found possible to permute 3 edges only using:

                L2 T2 R2 B2 R2 T2 L2 F2    (8 moves)
          or    L3 R1 F2 L1 R3 D2          (6 moves)

   In general any sequence   L1 R1 (any squares group moves) L3 R3
     will always result in a squares group position, for example:

                         L1 R1 (D2 F2 B2)    R3 L3
                         F1 B1 (T2 B2 F2 L2) F3 B3
   p66a (F2 R2 U2 F2 R2) U3 D3 (B2 L2 F2 B2) U1 D1  (13 moves)

   The longest irreducible square's group sequence discovered so far,
     which is an embedded part of longest Phase 2 sequence (p94):
     (Thus it can't be reduced by using full group moves using current
     techniques)

                R2 B2 U2 B2 L2 D2 L2 F2          (8)

   Later on I discovered this irreducible sequence by chance:

                T2 B2 T2 B2 D2 F2 R2 T2 L2 F2   (10)

   Edges only (with corners in place) can be 14 moves at most, e.g.

           D2 L2 F2 D2 F2 L2 T2 L2 T2 D2 F2 R2 F2 R2  (14)

   This answers the question David Singmaster posed in
   "Notes on Rubik's Magic Cube" on Thistlethwaite's last stage.
   That is: "Are there any positions in the square's group with
   corners fixed of length 14 or can they be done in less moves?"

   A few observations...

    -  It is not possible to swap just 1 pair of edges and corners
    -  It is only possible to have 4, 6 or 8 corners out of place
    -  Known antipodal cases can be solved in <=13 moves using full
group
    -  In reaching an antipode one may start with any of the 6 turns
        (since antipodes are global maxima, any turn will get you one
        move closer)
    -  If the corners are fixed, the position is NOT an antipode
    -  Longest order appears to be 12
    -  All known (probably all!) antipodes have symmetry level 6 or 12
    -  Although only conjectural, it is now believed that one turn of
        a face MUST lead to a new state which is either 1 move closer
         or 1 move farther from START

Question:     Are there any irreducible square's group sequences that
               are longer then 10 moves? Are these truly irreducible
               or only irreducible under Dik Winter's Kociemba
               inspired program?

The full group beckons....

-> Mark <-

From hoey@aic.nrl.navy.mil  Fri Aug 13 19:19:41 1993
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From: hoey@aic.nrl.navy.mil
Message-Id: <9308132226.AA28300@sun13.aic.nrl.navy.mil>
To: Mark Longridge <mark.longridge@canrem.com>, cube-lovers@life.ai.mit.edu
Subject: Re: Squares group
In-Reply-To: <60.251051.104.0C180C07@canrem.com>

Mark Longridge <mark.longridge@canrem.com> has some interesting things
to say about the antipodes of the group generated by half-turns:

>    If we define "symmetry level" as the number of distinct patterns
> generated by rotating the cube through it's 24 different
> orientations in space then most known antipodes are symmetry level
> 6. Thus the lower the number the higher the level of symmetry. The
> least symmetric positions have level 24, and this is very common.

This approach is somewhat unfortunate in two ways.  First, it would be
better to use the full 48-element symmetry group M of the cube,
because some patterns are not recognized as transformed images of each
other if you only use the 24-element group C of rotations.  For
instance, the positions reached by processes F2R2T2 and F2T2R2 cannot
be related with C, so you would see four classes of positions at
distance three rather than three.  But the antipodes you give are all
mirror-symmetric, so there is no new coalescence there.

Relating processes that are conjugates by a reflection is usually
somewhat tricky, since the moves of the process must be changed in
direction (replacing clockwise by counterclockwise) but in the squares
group this is a nonproblem.

The second deficiency of your approach is that you lose information by
specifying only the index of the symmetry subgroup (the ``number of
distinct patterns generated ...'').  It makes sense to find out
exactly which subgroup of M is the symmetry group of your positions.
I've done that, below.  Each of these symmetry groups comes in three
conjugates, so I've transformed some of the processes (marked x) so
they all use the same particular symmetry group(s).  The group
elements are given as cycles of the cube faces, so (TD)(FRBL) means to
reflect T<->D and rotate F->R->B->L->F.

> Cases with symmetry level 6:

These are cases where the symmetry group has order 8.

> p66x  Double 4 corner sw  L2 D2 R2 T2 L2 T2 F2 R2 (F2 B2 T2 F2) T2 L2 B2
> p80   2 DOT, Invert T's   R2 B2 D2 R2 B2 L2 B2 L2 (T2 D2 F2 T2) F2 L2 T2
> p99   2 DOT, 4 ARM        R2 B2 D2 L2 B2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2
> p100  2 Cross, 4 ARCH 1   R2 B2 T2 R2 F2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2

p66x, p80, p99, and p100 have symmetry group P=<(TD)(FRBL),(FB),(LR)>.

> p67x  Antipode 2          R2 D2 B2 T2 B2 T2 F2 L2 (F2 B2 T2 F2) L2 F2 D2
> p130  2 Cross, 4 ARCH 2   L2 B2 D2 B2 L2 D2 F2 L2 (T2 D2 F2 L2) F2 L2 T2

p67x and p130 have symmetry group Q=<(TD),(FRBL)>.

> p135x 2 X, 4 T            D2 B2 L2 F2 R2 F2 R2 D2 (R2 L2 F2 R2) D2 L2 F2
> p137  2 X, 4 ARM          L2 F2 T2 B2 T2 F2 T2 L2 (T2 D2 F2 T2) L2 D2 F2

p135x and p137 have symmetry group S=<(TD),(FB)(LR),(FR)(BL)>.

> Cases with symmetry level 12:

These have 4-element symmetry groups.

> p108  2 DOT, 2 T, 2 ARM   L2 F2 T2 R2 B2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2
> p128x 2 H, 2 T, 2 CRN     L2 D2 R2 T2 L2 T2 F2 R2 (F2 B2 T2 F2) T2 L2 F2
> p129x 2 H, 2 T, 2 ARCH    R2 T2 L2 T2 L2 T2 F2 R2 (F2 B2 T2 F2) T2 L2 F2
> p131x 2 H, 2 ARM, 2 ARCH  L2 T2 R2 B2 D2 L2 B2 L2 (F2 B2 T2 F2) T2 L2 T2
> p132  2 Cross,2 ARCH,2CRN L2 F2 D2 R2 F2 L2 B2 L2 (T2 D2 F2 T2) F2 L2 D2
> p136x 2 H, 2 ARM, 2 CRN   R2 T2 L2 F2 D2 L2 F2 L2 (F2 D2 T2 F2) T2 L2 D2

p108, p128x, p129x, p131x, p132, and p136x have symmetry group
HP=<(FB),(LR)>.

> p133x 2 Cross, 2 T, 2 ARM L2 T2 B2 T2 B2 T2 F2 L2 (F2 B2 T2 F2) L2 F2 D2
> p134x 2 CRN, 2 X, 2 ARCH  L2 T2 B2 D2 F2 T2 F2 L2 (F2 B2 T2 F2) L2 F2 D2

p133x and p134x have symmetry group HS=<(TD),(FB)(LR)>.

In case you have trouble forming the closure of these groups:

 P = {I, (FB)(LR), (TD)(FRBL),   (TD)(FLBR),
                   (FB),         (LR),
                   (TD)(FR)(BL), (TD)(FL)(BR)}

 Q = {I, (FB)(LR), (TD),         (TD)(FB)(LR),
                   (TD)(FRBL),   (TD)(FLBR),
                   (FLBR),       (FRBL)}

 S = {I, (FB)(LR), (TD),         (TD)(FB)(LR),
                   (TD)(FR)(BL), (TD)(FL)(BR),
                   (FR)(BL),     (FL)(BR)}

HP = {I, (FB)(LR), (FB),         (LR)}

HS = {I, (FB)(LR), (TD),         (TD)(FB)(LR)}.

I should note that the subgroup names M, C, P, Q, S, HP, and HS are
part of a general classification of subgroups of M that I worked out
some time ago.  I have a chart of them I can send; just ask by email.

>   A few observations...

>    -  It is not possible to swap just 1 pair of edges and corners

Certainly, all the generators are even permutations on the edges and
on the corners.

>    -  It is only possible to have 4, 6 or 8 corners out of place

That is a nice, concise way of putting it.  To elaborate, if you
permute one of the corner orbits in a 3-cycle, the other will also be
permuted in a 3-cycle; otherwise, any pair of cycle structures of the
same parity is possible.

> - In reaching an antipode one may start with any of the 6 turns
>   (since antipodes are global maxima, any turn will get you one move
>   closer)

Careful!  This also relies on the fact you call a conjecture, below.
Otherwise you could have two neighboring global maxima, and their
inverses would be antipodes that do not have this property.  For
instance, consider the corner group as generated by the 24 pairs of
neighboring squares (F2R2, etc).  This is a 48-element group with
diameter 2, trivial enough to be analyzed by hand.  Antipodes
(L2B2)(D2R2) and (D2R2)(T2F2) are neighbors, because
(L2B2)(D2R2)(F2T2)=(D2R2)(T2F2).  So there is no length-2 process
equivalent to (F2T2)(R2D2) that starts with T2F2.

> -  If the corners are fixed, the position is NOT an antipode
> -  All known (probably all!) antipodes have symmetry level 6 or 12

I presume these comments are left over from before you found them all.

> -  Longest order appears to be 12

Appears?  The orbits are all of size 4 (two orbits of corners, three
orbits of edges), so 12=LCM(2,3,4) is an easy upper bound.  Finding
one is easy given the processes Singmaster lists.

> - Although only conjectural, it is now believed that one turn of a
>   face MUST lead to a new state which is either 1 move closer or 1
>   move farther from START

Conjectural?  It's immediate from the fact that each generator is an
odd permutation of the corner orbit {FTR,FDL,BTL,BDR}.

> Question: Are there any irreducible square's group sequences that
> are longer then 10 moves? Are these truly irreducible or only
> irreducible under Dik Winter's Kociemba inspired program?

Well, that could be searched for; a matter of checking 600K positions
for each of the 15K or so pattern representatives.  I hope I can find
the time to hack it up.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From hoey@aic.nrl.navy.mil  Mon Aug 16 20:20:25 1993
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From: hoey@aic.nrl.navy.mil
Message-Id: <9308162205.AA12648@sun13.aic.nrl.navy.mil>
To: Mark Longridge <mark.longridge@canrem.com>, cube-lovers@life.ai.mit.edu
Subject: Squares group, correction

I should proofread these things better.  I got the processes for p130,
p135x, p137, and p136x wrong in my last message.  Here is the
corrected list of squares-group antipodes and their symmetry groups.

SG Pos   Name                Process

 P p66x  Double 4 corner sw  L2 D2 R2 T2 L2 T2 F2 R2 (F2 B2 T2 F2) T2 L2 B2
 P p80   2 DOT, Invert T's   R2 B2 D2 R2 B2 L2 B2 L2 (T2 D2 F2 T2) F2 L2 T2
 P p99   2 DOT, 4 ARM        R2 B2 D2 L2 B2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2
 P p100  2 Cross, 4 ARCH 1   R2 B2 T2 R2 F2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2
 Q p67x  Antipode 2          R2 D2 B2 T2 B2 T2 F2 L2 (F2 B2 T2 F2) L2 F2 D2
 Q p130x 2 Cross, 4 ARCH 2   T2 B2 R2 B2 T2 R2 F2 T2 (L2 R2 F2 T2) F2 T2 L2
 S p135x 2 X, 4 T            L2 D2 B2 T2 F2 T2 F2 L2 (F2 B2 T2 F2) L2 B2 T2
 S p137x 2 X, 4 ARM          L2 T2 F2 D2 F2 T2 F2 L2 (F2 B2 T2 F2) L2 B2 T2

HP p108  2 DOT, 2 T, 2 ARM   L2 F2 T2 R2 B2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2
HP p128x 2 H, 2 T, 2 CRN     L2 D2 R2 T2 L2 T2 F2 R2 (F2 B2 T2 F2) T2 L2 F2
HP p129x 2 H, 2 T, 2 ARCH    R2 T2 L2 T2 L2 T2 F2 R2 (F2 B2 T2 F2) T2 L2 F2
HP p131x 2 H, 2 ARM, 2 ARCH  L2 T2 R2 B2 D2 L2 B2 L2 (F2 B2 T2 F2) T2 L2 T2
HP p132  2 Cross,2 ARCH,2CRN L2 F2 D2 R2 F2 L2 B2 L2 (T2 D2 F2 T2) F2 L2 D2
HP p136x 2 H, 2 ARM, 2 CRN   R2 T2 L2 F2 D2 L2 F2 L2 (F2 B2 T2 F2) T2 L2 D2
HS p133x 2 Cross, 2 T, 2 ARM L2 T2 B2 T2 B2 T2 F2 L2 (F2 B2 T2 F2) L2 F2 D2
HS p134x 2 CRN, 2 X, 2 ARCH  L2 T2 B2 D2 F2 T2 F2 L2 (F2 B2 T2 F2) L2 F2 D2

Sorry if anyone was led astray.

Dan

From acw@bronze.lcs.mit.edu  Thu Aug 19 15:06:28 1993
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Date: Thu, 19 Aug 93 15:04:18 EDT
From: acw@bronze.lcs.mit.edu (Allan C. Wechsler)
Message-Id: <9308191904.AA27266@bronze.lcs.mit.edu>
To: ronnie@cisco.com
Cc: Dik.Winter@cwi.nl, cube-lovers@life.ai.mit.edu
In-Reply-To: "Ronnie B. Kon"'s message of Thu, 05 Aug 1993 16:55:36 -0700 <199308052355.AA23583@lager.cisco.com>
Subject: Diameter of cube group? 

   Date: Thu, 05 Aug 1993 16:55:36 -0700
   From: "Ronnie B. Kon" <ronnie@cisco.com>

   Disclaimer: this sounds more authoritative than is intended--I really
   don't know what I'm talking about.
Don't worry.  Mathematical reasoning stands on its own merits.
   It couldn't be very pointy.  From the most distant configuration,
   there are 6 positions immediately before it.  There are 6^2 two steps
   away, 6^3 three steps, etc.  (well, 6^2 - 1 and 6^3 - ?) actually.
Very good.  This is a necessary insight, regardless of the exact
numerical details.  (For example, you mean 12, not 6.)  But the
possible flaw is that there might be more than one maximally distant
state; if their sets of neighbors overlap viciously enough, this
effect could make the tail pointier.  You can make valence-12 graphs
(not of groups, just arbitrary graphs) that have fairly bumpy
distance-vs-population functions.  Any argument that rigorously
constrains N(d) must somehow appeal to the fact that the cube graph is
a Cayley graph, that is, the graph of a group.

[...]

   This gives me the feeling that Monte Carlo is fairly valid.  (How's
   that for rigor?)

It's a start.  But we have to use groupness somehow.

From @mitvma.mit.edu:DWR2560@TAMZEUS.BITNET  Fri Aug 20 09:47:26 1993
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Date:     Fri, 20 Aug 93 05:56 CST
From: <DWR2560%TAMZEUS.BITNET@mitvma.mit.edu>
Subject:  pointy tails
To: cube-lovers@life.ai.mit.edu
X-Original-To:  cube-lovers@life.ai.mit.edu, DWR2560

Allan C. Wechsler writes:
>   It couldn't be very pointy.  From the most distant configuration,
>   there are 6 positions immediately before it.  There are 6^2 two steps
>   away, 6^3 three steps, etc.  (well, 6^2 - 1 and 6^3 - ?) actually.
>
>Very good.  This is a necessary insight, regardless of the exact
>numerical details.  (For example, you mean 12, not 6.)  But the
>possible flaw is that there might be more than one maximally distant
>state; if their sets of neighbors overlap viciously enough, this
>effect could make the tail pointier.  You can make valence-12 graphs
[deletia]

All this misses the point (so to speak) which is that 12^N is _exceedingly_
pointy for our purposes. If one samples only 1000 positions out of ~10E19,
then one could very well miss a 12^N tail of length 14 moves!

The estimate of 22 as an upper limit relies on the intuition that
the distribution is MUCH blunter than this.

Dave Ring
dwr2560@zeus.tamu.edu

From reid@math.berkeley.edu  Mon Aug 23 04:10:58 1993
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Date: Mon, 23 Aug 93 01:10:56 PDT
From: reid@math.berkeley.edu (michael reid)
Message-Id: <9308230810.AA19047@math.berkeley.edu>
To: cube-lovers@life.ai.mit.edu
Subject: Re:  Diameter of cube group?

> Continuing and waiting for a config that requires 21 turns, dik

here's a pattern to try:
first do 6 checkerboards of order 2 (F2 B2 R2 L2 U2 D2) and then do
superfliptwist.  in other words, the group product of these two elements.

From dik@cwi.nl  Tue Aug 24 20:43:14 1993
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Date: Wed, 25 Aug 93 02:42:58 +0200
From: Dik.Winter@cwi.nl
Message-Id: <9308250042.AA11725.dik@boring.cwi.nl>
To: cube-lovers@life.ai.mit.edu, reid@math.berkeley.edu
Subject: Re:  Diameter of cube group?

 > > Continuing and waiting for a config that requires 21 turns, dik

 > here's a pattern to try:
 > first do 6 checkerboards of order 2 (F2 B2 R2 L2 U2 D2) and then do
 > superfliptwist.  in other words, the group product of these two elements.

As they commute I did it the other way around.  But I am highly
suspicious that you tried it yourself.  10 minutes and only down
to 22 turns.  But continuing, possibly for weeks/months.

On another machine I am trying to prove that 20 is minimal for
superfliptwist.  90 hours gone, still nothing.  Most of the time
is not spend with phase 1 set to 16 turns.  Phase 1 to 13 got it
doen to 20.  Nothing new with phase 1 to 14 or 15.  16 turns in
phase 1 allows at most 3 turns in phase 2.  The latter can be time
consuming.  I do not know in how many cases actually something is
done in phase 2.  When I get to 17 turns in phase 1, I suspect in
most cases in phase 2 it is immediately clear that it can not be
solved.  But I am patient.

dik

From dik@cwi.nl  Wed Aug 25 16:00:57 1993
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Date: Wed, 25 Aug 93 22:00:22 +0200
From: Dik.Winter@cwi.nl
Message-Id: <9308252000.AA14852.dik@boring.cwi.nl>
To: cube-lovers@life.ai.mit.edu, reid@math.berkeley.edu
Subject: Re:  Diameter of cube group?

 > here's a pattern to try:
 > first do 6 checkerboards of order 2 (F2 B2 R2 L2 U2 D2) and then do
 > superfliptwist.  in other words, the group product of these two elements.

Was certainly one of the hardest to do.  After 17 hours the best was 22
turns, but then results came in, after 18 hours 21 turns, and finally
after 19 hours 20 turns:
  F1 R1 L2 U3 R2 L3 U3 D2 R2 F1 D1 B1 D1 F2 U3 R3 D3 F2 D2 L2
This on an SGI R4K Indigo.

dik
--
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
home: bovenover 215, 1025 jn  amsterdam, nederland; e-mail: dik@cwi.nl

From dik@cwi.nl  Wed Aug 25 16:10:23 1993
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From: Dik.Winter@cwi.nl
Message-Id: <9308252010.AA14880.dik@boring.cwi.nl>
To: cube-lovers@life.ai.mit.edu
Subject: CFF 32 table of contents

Last Sunday (on Cube Day) I was handed issue #32 of Cubism For Fun.
A summary the contents:

1.  Short articles about the solution of the puzzle by Koos en Ton Verhoeff.
    (I described the puzzle before and announced the solution by Jan de
    Ruiter a few weeks ago.)
2.  Articles about "Bob's Binary Boxes" by Hans Dockhorn and Bob Kootstra. *
3.  Description by Harold Cataquet of "Alice"; a wooden packing puzzle.
4.  Description by Wim Zwaan of a packing puzzle he entered in the "Hikimi
    Wooden Puzzle Competition".
5.  Article by Jan Verbakel about "Wirrel-Warrel" puzzles (it has a
    different name in the US that escapes me).
6.  Article by Tom Hilligers about "Kaos", a puzzle with balls in pipes.
    The orientation of the pipes with respect to each other can change.
7.  Article by Ronald Fletterman about pretty "sculptures" with Square 1.
8.  A contest announcement by Bernard Wiezorke figuring the sliding puzzle
    Vorsicht!  (I do not know whether it is available in the US.)
9.  An article by Ralph Gasser about Orbik, a puzzle introduced by
    Edward Hordern.
10. Results of a number of contests.

*   An interesting design.  These are wooden boxes with in it binary
    switches.  On top a ball can be put in, on the bottom there are a
    number of exits.  When a ball reaches a switch it passes the
    switch in the given direction and puts the switch in opposite
    direction.  The design is such that successive balls come out in
    successive exits (in circular numerical order).  Bob Kootstra built
    a few of those switches, but now is asking for an optimal design of
    a box with 7 exits.

CFF is a newsletter published by the Nederlandse Kubus Club NKC (Dutch
Cubists Club).  It appears a bit irregular, but a few times a year.
Yearly membership fee is now NLG 25.- (Dutch Guilders) which amounts to
approximately $ 15.-.  Institutional membership is also possible.
Information is available from the editor:
	Gerald Maurice
	Groen van Prinstererstraat 7-2
	1051 ED  Amsterdam
	The Netherlands
	Phone: +31206822943
	E-mail: gm@phys.uva.nl

From alan@parsley.lcs.mit.edu  Wed Aug 25 21:13:44 1993
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From: Alan Bawden <Alan@lcs.mit.edu>
Sender: Alan@lcs.mit.edu
To: Cube-Lovers@ai.mit.edu
Subject: Tools lost in the mists of time...

Despite being your moderator for the past 13 years, it has been quite a
while since I actually played with a cube (of any order).  Until last
weekend, that is, when I picked up a 5x5x5 cube that someone had loaned me
quite some time ago.  Imagine my surprise to discover that I had in fact
forgotten one of the tools I needed to solve even a 3x3x3!  In particular
my tool for inverting two edge cubies in place in a 3x3x3 was completely
gone.  It didn't take me long to develop a replacement, but I'm certain
that it is nothing like what I was using years ago.  I'm amazed -- there
was a time when I thought I could never forget any of those tools.

And by the way, it was a lot of fun to re-learn the 3x3x3 and then go on to
solve the 5x5x5 -- if anyone else out there has, like me, been neglecting
their cube hacking for some years, go pick up your cube again, you may be
pleasantly surprised.

				- Alan (aka Cube-Lovers-Request)

--
Alan Bawden                                                Alan@LCS.MIT.EDU
MIT Room NE43-538                                            (617) 253-7328
545 Technology Square
Cambridge, MA 02139                        06BF9EB8FC4CFC24DC75BDAE3BB25C4B

From hoey@aic.nrl.navy.mil  Thu Aug 26 10:31:05 1993
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From: hoey@aic.nrl.navy.mil (Dan Hoey)
Message-Id: <9308261430.AA21721@Sun0.AIC.NRL.Navy.Mil>
To: Alan@lcs.mit.edu, cube-lovers@ai.mit.edu
Subject: Tartan reborn (Re: Tools lost in the mists of time...)
In-Reply-To: <25Aug1993.204955.Alan@LCS.MIT.EDU>
Organization: Navy Center for Applied Research in AI

Alan Bawden mentioned the joy of rediscovering his lost cube-solving
techniques.  This happened to me about three years ago for an unusual
reason.  I've become active in science fiction fandom, and fans
determine where the World Science Fiction Convention (Worldcon) is
held each year by running miniature political campaigns.  A friend of
mine was bidding for Glasgow, and she asked if I had any `plaid
things'.  I told her I had a plaid Rubik's cube, and a political
strategy was born.  The plaid cube is of course the Tartan, which Jim
Saxe and I discovered and described in this group on 16 February 1981
(see archives).  I blanked some old cubes, and figured out how to use
spray paint to efficiently create Tartan cubes.  I produced a half
dozen or so, and they make good conversation pieces at conventions.

Unfortunately, I seem to be the only convention-going science fiction
fan who can *solve* a Tartan (with the possible exception of Phil
Servita <meister@gaak.lcs.mit.edu> who as I recall figured out an
effective method but wearied in its execution).  So I would see a
scrambled Tartan at a convention party, and fix it, and put it down,
and five minutes later it would be scrambled again.  I quickly found
out how rusty I was, and through the enforced practice I've gotten
about as good as I was a decade ago.  But some of the Glasgow
promoters took Tartan cubes over to the UK, and those cubes just never
get solved.  I sent them instructions for solving it, but I don't know
if any of them have figured out the instructions.

Well, eventually they told me they really wanted something mere
mortals could deal with, and I painted some pieces of wood plaid that
they could use for doorstops.  I was surprised, though, to find that
to make a plaid pattern going around a corner, if you only have four
colors of paint, it seems the *only* thing you can do is use a
coloring locally identical to the Tartan.

As for the cubes in the UK, I expect to get there in 1995.  For it
seems the clever ploy worked, and the fans voted to have the 1995
Worldcon in Glasgow.  I'm sure they owe it all to the Tartan.  Sure.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From reid@math.berkeley.edu  Fri Aug 27 22:23:04 1993
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Date: Fri, 27 Aug 93 19:22:49 PDT
From: reid@math.berkeley.edu (michael reid)
Message-Id: <9308280222.AA22114@math.berkeley.edu>
To: Dik.Winter@cwi.nl, cube-lovers@life.ai.mit.edu
Subject: Re:  Diameter of cube group?

dik winter says

>  > first do 6 checkerboards of order 2 (F2 B2 R2 L2 U2 D2) and then do
>  > superfliptwist.  in other words, the group product of these two elements.
> 
> As they commute I did it the other way around.  But I am highly
> suspicious that you tried it yourself.  10 minutes and only down
> to 22 turns.  But continuing, possibly for weeks/months.

ok, you caught me; i'd already tried this one myself.  :-)
but apparently i wasn't as patient as you.  i just remember that it ran
for a long time without doing better than 22 face turns.

the point to be made here is the following: the length of time the
program takes for a given position depends significantly on how far it
must search in stage 1.  this seems to make any claim about how long the
program takes to crunch an average position meaningless.  my experience
is that it varies greatly depending upon the position.  i think it would
be more informative to stratify this information.  i.e., how long it
takes to search 12 moves in stage 1, and how short a solution is produced.
and then the same info for 13 turns, then 14, etc.

what i've been amazed by (and continue to be) is that searching only 13
or so moves in stage 1 is sufficient to produce very short solutions for
many positions.

something i'd thought about trying, but never got around to is trying
random positions created by twist sequences such as:

F1 R1 B1 L1 F1 R1 B1 L1 F1 R1 B1 L1 F1 R1 B1 L1 F1 R1 B1 L1

or some random string of about 20 quarter turns of the faces F,L,B,R.
a string of length 12 or 13 will be solved quickly (by the inverse of
the original string).  however, for length 17 or so, the program won't
find the inverse of the original string until it is searching 17 moves
deep in stage 1.  this suggests that perhaps these positions may be
harder for the program to handle.  but are they harder than random
positions?  i don't know.

mike

From dik@cwi.nl  Sat Aug 28 20:17:32 1993
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From: Dik.Winter@cwi.nl
Message-Id: <9308290017.AA01278.dik@boring.cwi.nl>
To: cube-lovers@life.ai.mit.edu, reid@math.berkeley.edu
Subject: Re:  Diameter of cube group?

 > ok, you caught me; i'd already tried this one myself.  :-)
 > but apparently i wasn't as patient as you.  i just remember that it ran
 > for a long time without doing better than 22 face turns.

So did it here.  22 in a few minutes, 20 in a lot of hours.

 > the point to be made here is the following: the length of time the
 > program takes for a given position depends significantly on how far it
 > must search in stage 1.

This is right, and it appears (though I have not yet thoroughly verified)
that configurations that take a long time in stage 1 are a large distance
from start.

 >                          this seems to make any claim about how long the
 > program takes to crunch an average position meaningless.

Depends on how you interpret that claim.  If the claim is that it turns
up with a sequence that is 20 turns or shorter you are right.  The claim
might even be incorrect!  The actual claim is that it takes a fairly short
time to give a fairly short sequence (where fairly short is deliberately
left unquantified).  And this is true.  For my set of >10000 random
positions the program came up with a sequence of 27 turns or less in
a short time indeed.  (Actually the first solution found was 26 turns or
less for all but three configurations.)  Bringing that down to 20 took in
a number of cases extremely long (in the order of one day).  But that is
still far less than when we had done a normal single phase backtracking
process I think.

 >                                                         i think it would
 > be more informative to stratify this information.  i.e., how long it
 > takes to search 12 moves in stage 1, and how short a solution is produced.
 > and then the same info for 13 turns, then 14, etc.

Some quantification is not so very difficult I think.  Without tree-pruning
the time would be proportional to 18^n + 10^m for a n-turn phase1 and a
m-turn phase2 solution.  The tree-pruning performed is (I think) proportional
to the number of turns in each phase; it will chop branches that are to
large according to predefined tables.  Also there are some short-cuts that
make 18 not really 18 and 10 not really 10, but the reasoning remains the
same.

 > what i've been amazed by (and continue to be) is that searching only 13
 > or so moves in stage 1 is sufficient to produce very short solutions for
 > many positions.

I do not think this is so very amazing.  Each configuration can be brought in
12 turns or less to a configuration for phase 2.  The proven diameter of the
group of phase 2 is 25, my estimate is 19-21.  So, based on my estimate a
worst case would be 12 turns required in phase 1 and 21 in phase 2 giving
33 turns in an estimated time of 18^12 + 10^21, this is less than 18^17,
and hence is found faster than if we had gone to 17 turns in phase 1.
Actually both 12 and 21 are rare; most cases do phase 1 in 10 turns or less
and phase 2 in 15 turns or less.

 > something i'd thought about trying, but never got around to is trying
 > random positions created by twist sequences such as:

 > F1 R1 B1 L1 F1 R1 B1 L1 F1 R1 B1 L1 F1 R1 B1 L1 F1 R1 B1 L1

 > or some random string of about 20 quarter turns of the faces F,L,B,R.
 > a string of length 12 or 13 will be solved quickly (by the inverse of
 > the original string).  however, for length 17 or so, the program won't
 > find the inverse of the original string until it is searching 17 moves
 > deep in stage 1.  this suggests that perhaps these positions may be
 > harder for the program to handle.  but are they harder than random
 > positions?  i don't know.

I do not know, but I think not.  Yes, asking the program to find the
reverse of the string takes a long time.  Asking the program to find
an inverse of the sequence takes much less time (although the inverse
found may both be shorter or longer than the original).  I just tried,
and after initialization it found a 10+14 turn solution in 20 seconds,
going down to 11+10 after less than a minute.  Getting this down to 20
might of course take considerable time (if the original sequence is
minimal etc.).

But I have not the time right now to check.  I am busy trying to prove
that 20 is minimal for superfliptwist.  I have already found that there
is no 19 turn solution with 16 turns in phase 1.  That took about 48
hours (distributed over 6 machines).  Now I am doing the same for 17
turns in phase 1; which wil obviously take much longer.  (And yes, I
took the precaution of allowing as the first turn only F, F2, R, R2,
U, U2 in phase 1.  When I go to 19 turns in phase 1, I can skip 18,
I need only F, F2, R and R2, I think.)

dik
--
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
home: bovenover 215, 1025 jn  amsterdam, nederland; e-mail: dik@cwi.nl

From reid@math.berkeley.edu  Sun Aug 29 04:26:41 1993
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Date: Sun, 29 Aug 93 01:26:31 PDT
From: reid@math.berkeley.edu (michael reid)
Message-Id: <9308290826.AA05585@math.berkeley.edu>
To: Dik.Winter@cwi.nl, cube-lovers@life.ai.mit.edu
Subject: Re:  Diameter of cube group?

> But I have not the time right now to check.  I am busy trying to prove
> that 20 is minimal for superfliptwist.  I have already found that there
> is no 19 turn solution with 16 turns in phase 1.  That took about 48
> hours (distributed over 6 machines).  Now I am doing the same for 17
> turns in phase 1; which wil obviously take much longer.  (And yes, I
> took the precaution of allowing as the first turn only F, F2, R, R2,
> U, U2 in phase 1.  When I go to 19 turns in phase 1, I can skip 18,
> I need only F, F2, R and R2, I think.)

in fact, you can eliminate the possibility of starting with F2, R2 or U2,
since these each commute with superfliptwist, and may be done in stage 2.
in other words, if  F2 sequence  =  superfliptwist,  then also
sequence F2  =  superfliptwist.

also, you need not consider 19 turns in stage 1.  by symmetry, you may
suppose the last face turned is U, which is done in stage 2.

if you use the fact that U and D commute, L and R commute and F and B
commute, then the number of sequences of length n in stage 1 grows
exponentially, with ratio approximately 13.35.  if the runtime is
proportional to the number of sequences tested in stage 1, (which
may or may not be the case) that would mean testing 18 turns deep
would take approximately 178.18 times as long.  (eliminating the
possibility of starting with F2, R2 or U2 would cut that in half.)

here's something you may have already considered.  if your prune tables
in stage 1 consider only pairs (flip, twist), (flip, location) and
(twist, location), some search paths may be pruned 8 turns early.
(each of these pairs had positions 9 twists from start.)  at the
expense of a lot more memory, you can do some pruning 11 turns early,
by storing tables for triples (flip, twist, location).  you'd probably
have to store these tables in very compressed form, and divide out by
symmetries of the cube that preserve the U-D axis.  it may turn out
that the overhead of processing this compressed information does not
adequately compensate for the improved foresight, but it's worth
considering.

it would be excellent if you could show that 20 face turns is minimal
for superfliptwist!  even finding a shorter solution would be great!

mike

From dik@cwi.nl  Mon Aug 30 21:21:35 1993
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From: Dik.Winter@cwi.nl
Message-Id: <9308310121.AA08781.dik@boring.cwi.nl>
To: cube-lovers@life.ai.mit.edu, reid@math.berkeley.edu
Subject: Re:  Diameter of cube group?

 > in fact, you can eliminate the possibility of starting with F2, R2 or U2,
 > since these each commute with superfliptwist, and may be done in stage 2.
 > in other words, if  F2 sequence  =  superfliptwist,  then also
 > sequence F2  =  superfliptwist.

Right.  I had not considered this in the program (it is still fairly
general), but it only does mean early termination.

 > also, you need not consider 19 turns in stage 1.  by symmetry, you may
 > suppose the last face turned is U, which is done in stage 2.

But I have now had different thoughts.  Currently phase 1 checks in 3
dimensional space.  When a solution is found the program calculates the
current position for phase two after which it finds a solution in a
different 3 dimensional space.  (I just though how I might speed up the
calculations to get to the starting position for the second phase,
but will not yet elaborate on that; I will first try it out.)  But this
does not help finding whether there are solutions of 19 turns or less.
What I am now considering is to have a phase 1 program only, where phase
1 is done in an additional dimension: the permutation of the corner cubes.
So to prove the non-existence of a solution of 19 turns or less, this
program would seek for a phase 1 solution in 4 dimensional space of at
most 19 turns and next check whether this also solves the edge cubes.
This would eliminate quite a few dead alleys where the current phase 1
finds a solution and has still things to do.

 > if you use the fact that U and D commute, L and R commute and F and B
 > commute, then the number of sequences of length n in stage 1 grows
 > exponentially, with ratio approximately 13.35.  if the runtime is
 > proportional to the number of sequences tested in stage 1, (which
 > may or may not be the case) that would mean testing 18 turns deep
 > would take approximately 178.18 times as long.  (eliminating the
 > possibility of starting with F2, R2 or U2 would cut that in half.)

If I use a single phase algorithm, I can eliminate much more!  What I
see for runtime is not entirely proportional.  When looking at the
number of configurations done in phase 2, this goes up by factors that
start in the neighbourhood of 30 and diminish to (probably) ultimately
the factor you mention.  This indicates that tree pruning is much more
effective with fewer turns in phase 1.

 > here's something you may have already considered.  if your prune tables
 > in stage 1 consider only pairs (flip, twist), (flip, location) and
 > (twist, location), some search paths may be pruned 8 turns early.
 > (each of these pairs had positions 9 twists from start.)  at the
 > expense of a lot more memory, you can do some pruning 11 turns early,
 > by storing tables for triples (flip, twist, location).  you'd probably
 > have to store these tables in very compressed form, and divide out by
 > symmetries of the cube that preserve the U-D axis.  it may turn out
 > that the overhead of processing this compressed information does not
 > adequately compensate for the improved foresight, but it's worth
 > considering.

I think the overhead is much to large computation-wise and memory-wise.
The size of the table would be uncompressed 2217093120 integers in the
range from 0 to 12.  Factoring out symmetries would reduce it by a factor
of about 32 (slightly less).  [4 for rotational symmetry, 4 for mirroring
both U-D and F-B, 2 for inversion.]  Using 3.5 bits per configuration
this means > 30 MByte.  The machines I am using currently are not able
to handle that amount of information.  But it is feasable.  If we skip
inversion (which is most difficult to do) we are at > 60 MByte.  The
problem remains to adequately number the remaining positions from 1 to
max.  Some configurations are inert with respect to the rotations and/or
mirroring.  On the other hand, we need the table in core (do not try
to do this through disk access!).  Some insightful thoughts are needed
here.

 > it would be excellent if you could show that 20 face turns is minimal
 > for superfliptwist!  even finding a shorter solution would be great!

I agree to that!  I have a number of machines, still going strong.

From ROSEJM58@snyoneva.cc.oneonta.edu  Tue Sep 21 14:33:15 1993
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cube-lovers-request@ai.ai.mit.edu"

From raymond@cps.msu.edu  Fri Oct  1 12:37:21 1993
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Subject: Seeking 5x5x5 cube
Content-Length: 115

Hello cube lovers,

  I am looking for a 5x5x5 cube.  Does anybody know where I can
get one?

Thanks,
Carl Raymond

From queiroz@eepost.uta.edu  Fri Oct  1 16:25:59 1993
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Date: Fri, 1 Oct 1993 15:15:49 -0500 (CDT)
From: Ricardoh Queiroz <queiroz@eepost.uta.edu>
Subject: Re: Seeking 5x5x5 cube
To: raymond@cps.msu.edu
Cc: cube-lovers@ai.mit.edu
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On Fri, 1 Oct 1993 raymond@cps.msu.edu wrote:

> Hello cube lovers,
> 
>   I am looking for a 5x5x5 cube.  Does anybody know where I can
> get one?
> 
> Thanks,
> Carl Raymond

Hi,
I also have interest in a regular 3x3x3 and I can't find it.
If anyone has any idea, please let us know.
Thanks,
Ricardo
queiroz@eepost.uta.edu



From raymond@cps.msu.edu  Fri Oct  1 18:06:21 1993
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To: queiroz@eepost.uta.edu, raymond@cps.msu.edu
Subject: Re: Seeking 5x5x5 cube
Cc: cube-lovers@ai.mit.edu
Content-Length: 456

During the Christmas toy season last year, a local supermarket/department
store chain (Meijer in Michigan) had 3x3x3 cubes.  I don't recall the
manufaturer, but they used the "Rubik's Cube" brand name.  They also
had a picture on the center cubies on each face that had to be
correctly rotated for a "proper" solution.  I can't recall the price,
but it was reasonable.  Maybe they will be easier to find as Christmas
gets closer.

Good luck,

Carl Raymond

From ncramer@bbn.com  Mon Oct  4 08:51:15 1993
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Date:     Mon, 4 Oct 93 8:40:53 EDT
From: Nichael Cramer <ncramer@bbn.com>
To: Ricardoh Queiroz <queiroz@eepost.uta.edu>
Cc: raymond@cps.msu.edu, cube-lovers@ai.mit.edu
Subject:  Re:  Seeking 5x5x5 cube

>Date: Fri, 1 Oct 1993 15:15:49 -0500 (CDT)
>From: Ricardoh Queiroz <queiroz@eepost.uta.edu>
>Subject: Re: Seeking 5x5x5 cube
>
>Hi,
>I also have interest in a regular 3x3x3 and I can't find it.
>If anyone has any idea, please let us know.
>Thanks,
>Ricardo
>queiroz@eepost.uta.edu

Games People Play in Harvard Square had a number of 3X3X3 that go by the
name "Fourth Dimension" or something like that (they various have logos and
a picture that looks like a profile of Rubik on four of the center faces).

It was something on the order of $10.

It also seemed more cheaply made than my other, older cubes.  It felt,
well, "lighter" in my hand and is rather more difficult to turn than I'm
used to.  It's not that they're unusable, and it's just not that they're
just stiff, rather that they seem to be slightly out of alignment.  Perhaps
if you are someone who knows how to fine tune these things...

N

From ronnie@cisco.com  Mon Oct  4 11:28:32 1993
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To: cube-lovers@ai.mit.edu
Subject: Re: Seeking 5x5x5 cube 
In-Reply-To: Your message of "Mon, 04 Oct 1993 08:40:53 EDT."
             <9310041251.AA19338@life.ai.mit.edu> 
Date: Mon, 04 Oct 1993 08:28:28 -0700
From: "Ronnie B. Kon" <ronnie@cisco.com>


Try mailing to Peter Beck (pbeck@pica.army.mil).

			Ronnie

From @mizzou1.missouri.edu:HOWSER@LUA6.LU.EDU  Wed Oct  6 01:40:53 1993
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From: <HOWSER%LUA6.LU.EDU@lugate.lu.edu>
To: <cube-lovers%ai.mit.edu@lugate.lu.edu>
Subject: Stiff and/or misaligned cubes
Comments: Automatic Return Receipt Requested
Message-Id: <BMEP0921FE5EB2 @ LUA6>

Back in the 'good old days' when cubing was very popular, I had a cube that was
very prone to hanging up when you turned it in certain directions.  I solved the
 problem by disassembling the cube and working on the cublets individually to
remove any excess plastic and to smooth any rough spots by scraping with a razor
blade and/or sanding with model car sandpaper.  I raced many people with that
cube and still have it after all these years.  I found that the time I spent
working on the bad cublets has lead to that cube wearing much more evenly than
the ones I have that I never got around to working on.  I also find that it gets
more consistant in its movements as time goes by.  As it was one of the first
cubes on the market (before the BIG craze, actually) it is rather heavy but not
as precisely made as the later cubes.

------------------------------------------------------------------------
    Gerry Howser
    INTERNET:  howser@lua6.lu.edu
               Postmaster@lua6.lul.edu
               Monet01@umcvmb.missouri.edu (Alternate)

    VOICE:  (314) 681-5400
    FAX:    (314) 681-5566
------------------------------------------------------------------------


From dml@hpfrcu03.france.hp.com  Thu Oct  7 16:48:50 1993
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From: Patrick DEMICHEL  <dml@hpfrcu03.france.hp.com>
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Subject: help
To: CUBE-LOVERS@life.ai.mit.edu
Date: Thu, 7 Oct 93 17:59:18 MET
Cc: dml@hpfrcu03.france.hp.com
Mailer: Elm [revision: 72.14]

help

From diamond@jit081.enet.dec.com  Thu Oct  7 21:54:26 1993
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Date: Thu, 7 Oct 93 18:54:25 PDT
From: 08-Oct-1993 1054 <diamond@jit081.enet.dec.com>
To: "dml@hpfrcu03.france.hp.com"@jrdmax.enet.dec.com
Cc: cube-lovers@life.ai.mit.edu
Apparently-To: cube-lovers@life.ai.mit.edu
Subject: RE: help

dml@hpfrcu03.france.hp.com (Patrick DEMICHEL) writes:
>help

First you turn one side, then you turn another side.  Keep it up, and
soon you'll be done.

-- Norman Diamond        diamond@jit081.enet.dec.com
[Digital did not write this.]

From pbeck@pica.army.mil  Fri Oct  8 08:05:19 1993
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Date:     Fri, 8 Oct 93 7:52:20 EDT
From: Peter Beck (BATDD) <pbeck@pica.army.mil>
To: cube-lovers@life.ai.mit.edu
Cc: pbeck@pica.army.mil
Subject:  golden solids
Message-Id:  <9310080752.aa13373@COR6.PICA.ARMY.MIL>

i am trying to find a copy of the following:

THE AESTHETICS OF THE SACRED, A HARMONIC GEOMETRY OF CONSCIOUSNESS &
PHILOSOPHY OF SACRED ARCHITECTURE
by  ROBERT C MEURANT
  THE OPOUTERE PRESS
  BOULDER & AUCKLAND
  ISBN  0-908809-02-6


if you know of a bookseller who might carry this
please let me know.

if you have the address of opoutere press in NZ
that would also be helpful.

thanks
       

From @mail.uunet.ca:mark.longridge@canrem.com  Thu Oct 28 19:41:34 1993
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To: cube-lovers@life.ai.mit.edu
Subject: Cube Patterns
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.305920.104.0C188656@canrem.com>
Date: 	Thu, 28 Oct 1993 19:20:00 -0400
Organization: CRS Online  (Toronto, Ontario)

        Comments on Rubik's Cube Patterns
        ---------------------------------

First some positions of theoretical interest:

(F R B L)^5 = F1 L3 D2 F3 B2 R1 L3 F2 B3 R2 B1 U2 D2 R3 D2 L2
                 B2 L2 F2  (19 moves)

So in the ht metric this is compressible.

I've been thinking about new approaches to finding new patterns.
To improve on the "old-fashioned" method of simply taking a cube
and twisting it I wrote a module to test for legality of position
and another module for arrangement entry. Thus I can doodle around with
a cube pattern much more efficiently.

This approach led to the discovery of the ML Checkerboard, which is
to date the most involved of the pretty patterns:

ML's Checkerboard =  B1 U2 R1 L1 D2 B3 L2 F2 R1 F3 U3 D3 F3 B3 R2 U1
                        R2 D3 L2   (19 moves)

Also by combining the 8 twist and the first discovered square's
group antipode, a new corner's only pattern:

Antwist = R1 F2 B2 D2 R1 L3 B2 R1 B2 U1 F2 U2 F2 D2 F2 R2
             L2 D3   (18 moves)

Also I have re-evaluated what is a complex cube position. Cube positions
have different degrees (or types) of difficulty.

A. A position is difficult if it is visually hard to recognize, e.g.
    no pattern is apparent, the cube is well mixed and random.
    However the pattern superfliptwist, although being 20 moves
    deep, IS easy to recognize.

B. A position is easy with respect to computer analysis if it is
    cyclically decomposable. That is to say it by looking at a
    position a program finds it is generated by (F R B L)^5,
    so this position is EASY.

C. A position is easy with respect to the human hand if the sequence
    required to solve the position can be executed rapidly. To a
    degree such positions are similar to positions in point (B)
    in that only a subset of all cube operators are required, and
    the sequence does not require turning all 6 sides and so the
    sequence is easier to memorize as well.

As a result of thinking along these lines I am going to write a
module to do cyclic decomposition.

-> Mark <-
Email:  mark.longridge@canrem.com

....more patterns to follow...

From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU  Sat Dec  4 09:49:17 1993
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Date:      Sat, 4 Dec 1993 09:15:56 EST
From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   First Post

This is my first post to Cube-Lovers, so I will introduce myself
briefly.  I have been cubing since about 1979 or 1980 or so when
the cubes first appeared on the market.  I have been cubing with
a computer since about 1985, and have been active on the Internet
since about 1985 (purely a coincidence of dates).  I have looked
for years for a cubing list, and never found one until now.
I always looked for "Rubik" (or sometimes "Rubic").  For some silly
reason, it never occurred to me to look for "cube".

I have long since read Hofstadter's two Scientific American articles,
as well as the reprints in METAMAGICAL THEMAS.  The reprints, by
the way, are excellent because of the additional information in the
appendices.  I also have a copy of Singmaster and Frey's
HANDBOOK OF CUBIC MATH.  I have tried unsuccessfully for years to
get copies of Singmaster's earlier work  --  the circulars, for
example.  However, I suspect that the HANDBOOK includes most if not
all of the earlier material.  Also, (and you won't believe this) I have
just read all thirteen years of the archives of Cube-Lovers.

My primary interest has been in calculating God's Algorithm.  I am
interested in brute force breadth first tree searches.  In other
words, my work is akin to the solutions of the 2x2x2 and the corners
of the 3x3x3 posted by Dan Hoey and others.  It is not akin to
Thistlethwaite's methods.  I am interested to see, however, that
major recent progress appears to have been made on Thistlewaite's
method.

I have calculated God's Algorithm for the 2x2x2 cube and the corners
of the 3x3x3.  My results agree with those that have been posted here,
with the exception that my search is 48 times smaller (24*2), due
to the exploitation of a rotation and reflection group of the cube.
I have also calculated God's Algorithm for the edges of the 3x3x3.
This is a much larger problem, and took about a year running
continuously on two machines.  The resulting output file is about
4.2 gigabytes of data, and is stored on 31 reels of magnetic tape.
This result includes the "48 times smaller" factor, else it would
have been 204 gigabytes of data stored on 1464 reels of magnetic
tape.

I understand that this list has been very quiet of late.  But
assuming some modicum of interest, I will post more details
of my results in subsequent messages.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU  Sat Dec  4 21:07:09 1993
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Date:      Sat, 4 Dec 1993 21:04:23 EST
From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   God's Algorithm for the 2x2x2 Pocket Cube

I want to post my God's Algorithm results for the 2x2x2 cube.
These results generally speaking replicate other results that have
been posted here as far back as ten to twelve years ago.

In order to make my numbers make sense, I need to explain how I count
the states of the 2x2x2 cube.  As has been posted here several times
previously, the number is (7!)(3^6)=3,674,160.  Actually, I prefer
the formulation (7!)(3^7)/3=3,674,160 because the latter formulation
clearly reflects that all the cubies can be rotated but that
rotational orientation of the last one is determined by the
rotational orientation of the others.  But in any case, this
calculation is based on the following.  Let any one cube be fixed
in location and rotational orientation.  Then, there are
7! ways to arrange the other seven cubes, and (3^7)/3 ways to
rotate them.

But there is another way to look at it.  Fix none of the cubes.
Rather,  choose one to be the upper,left,front one, pick a second
one to be the upper,right,front one, etc., so that there are 8!
ways to arrange the eight cubes and (3^8)/3  ways to rotate them.
We have 8!(3^8)/3=88,179,840, which is exactly twenty-four
times larger than 3,674,160.

The reason is that the 3,674,160 figure implicitly assumes
that cubes that differ only in orientation of the overall
cube are equivalent, and there are twenty-four ways to orient
the cube in space (i.e., the order of the rotation group of
the cube is 24).

Conversely, the 88,179,840 figure implicitly assumes that
cubes that differ only in orientation of the overall cube
are distinct.  They can be made equivalent by applying the
rotation group of the cube to form equivalence classes, and
there will be exactly 3,674,160 equivalence classes.  Hence,
the two ways of counting are isomorphic.  However, I do
prefer to characterize the "things" that the 3,674,160 figure
counts as equivalence classes, and I call 3,674,160 the
number of nodes using 24-fold symmetry.

Finally, I apply a second order-24 rotation group (I will explain
how you can have a two order-24 rotation groups on the same cube
in a follow-up post) and an order-2 reflection group.  Hence, the
number of nodes to represent the entire search tree for the 2x2x2
cube should be 88,179,840/(24*24*2)=76,545, where the 76,545 figure
represents the number of equivalence classes and each equivalence
class includes 24*24*2=1152 elements.  As it turns out, a few of
the equivalence classes contain fewer than 1152 elements, so that
the total number of nodes in the search tree is slightly larger
than 76,545, namely 77,802.

The tables of results below include figures both for 24-fold symmetry
and for 1152-fold symmetry.  My search tree was for 1152-fold symmetry
only.  I then sort of "backed in" to the results for 24-fold symmetry
by calculating the size of each equivalence class.  Calculating a
search tree with 77,802 nodes representing equivalence classes, then
calculating the size of each equivalence class, was much faster than
calculating a search tree with 88,179,840 nodes or one with
3,674,160 nodes.

The little exercise with calculating the size of each equivalence
class was very gratifying in at least two respects.  First, it let
me explain the disconcerting difference between 76,545 and 77,802.
Second, it let me confirm that my results were the same as everyone
else who had gone before.

                Results Using Both q-turns and h-turns

        Distance           Number of          Number of
        from               Nodes              Nodes
        Start              using              using
                           24-fold            1152-fold
                           symmetry           symmetry

          0                       1                  1
          1                       9                  2
          2                      54                  5
          3                     321                 19
          4                    1847                 68
          5                    9992                271
          6                   50136               1148
          7                  227536               4915
          8                  870072              18364
          9                 1887748              39707
         10                  623800              13225
         11                    2644                 77
      -----                 -------              -----
      Total                 3674160              77802



                Results Using Only q-turns

        Distance             Number of          Number of
        from                 Nodes              Nodes
        Start                using              using
                             24-fold            1152-fold
                             symmetry           symmetry

          0                       1                  1
          1                       6                  1
          2                      27                  3
          3                     120                  6
          4                     534                 17
          5                    2256                 59
          6                    8969                217
          7                   33058                738
          8                  114149               2465
          9                  360508               7646
         10                  930588              19641
         11                 1350852              28475
         12                  782536              16547
         13                   90280               1976
         14                     276                 10
      -----                 -------              -----
      Total                 3674160              77802



                Results Using Only h-turns

        Distance           Number of          Number of
        from               Nodes              Nodes
        Start              using              using
                           24-fold            1152-fold
                           symmetry           symmetry

          0                       1                  1
          1                       3                  1
          2                       6                  1
          3                      11                  2
          4                       3                  2
      -----                 -------              -----
      Total                      24                  7

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU  Sat Dec  4 23:18:20 1993
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Date:      Sat, 4 Dec 1993 23:15:30 EST
From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   God's Algorithm for the Corners of the 3x3x3

Here are my God's Algorithm results for the corners of the
3x3x3 cube.  I explained in the last post what I mean by
1152-fold symmetry and 24-fold symmetry.  The 1152-fold
symmetry is what I actually calculated.  In this particular
case, I did not do the 24-fold symmetry calculations myself
based on the size of the equivalence classes as I did
with the 2x2x2 cube.  Rather, I went back and found the
figures in the Cube-Lover archives (Dik Winter's post).

             Results Using Both q-turns and h-turns

        Distance           Number of            Number of
        from               Nodes using          Nodes using
        Start              1152-fold            24-fold
                           symmetry             symmetry

          0                       1                   1
          1                       2                  18
          2                       9                 243
          3                      71               2,874
          4                     637              28,000
          5                   4,449             205,416
          6                  24,629           1,168,516
          7                 113,049           5,402,628
          8                 433,611          20,776,176
          9                 947,208          45,391,616
         10                 316,823          15,139,616
         11                   1,481              64,736



             Results Using Only q-turns

        Distance           Number of            Number of
        from               Nodes using          Nodes using
        Start              1152-fold            24-fold
                           symmetry             symmetry

          0                       1                    1
          1                       1                   12
          2                       5                  114
          3                      24                  924
          4                     149                6,539
          5                     850               39,528
          6                   4,257              199,926
          7                  16,937              806,136
          8                  57,800            2,761,740
          9                 180,639            8,656,152
         10                 466,052           22,334,112
         11                 676,790           32,420,448
         12                 392,558           18,780,864
         13                  45,744            2,166,720
         14                     163                6,624


             Results         Using Only h-turns

        Distance           Number of
        from               Nodes using
        Start              1152-fold
                           symmetry


          0                       1
          1                       1
          2                       2
          3                       4
          4                       3


It turns out that the maximum distance from Start is the same
for the corners of the 3x3x3 cube as it is for the 2x2x2 cube.
I found this rather surprising, although the archives of
Cube-Lovers do provide a reasonable explanation.  I am just
going to have to go back and read it five or ten times until
I fully understand it.  In any case, I was curious about the
following question.  Suppose you are N moves from Start on the
corners of the 3x3x3.  How many moves from Start would you be
on the 2x2x2 if the 2x2x2 was in the same configuration as the
corners of the 3x3x3 where you currently were.  As it turns out,
I stored the results for the 2x2x2 in the same data base as
I stored the results for the corners of the 3x3x3, so the
question was easy to answer.  Here are the results.

              Corresponding Distances from Start
                Using Both q-turns and h-turns

         2x2x2       Corner of 3x3x3    Number
     Distance from    Distance from    of Nodes
         Start         Start


           0
                           0                1
                           2                2
                           4                2

           1
                           1                2
                           2                2
                           3                4
                           4                3

           2
                           2                5
                           3               12
                           4               18
                           5                3

           3
                           3               55
                           4              106
                           5               41

           4
                           4              508
                           5              457
                           6               38

           5
                           5            3,948
                           6            1,237
                           7                2

           6
                           6           23,354
                           7            1,992
                           8               20

           7
                           7          111,055
                           8            3,242
                           9               20

           8
                           8          430,349
                           9            5,460
                          10               62

           9
                           9          941,728
                          10            3,770
                          11               20

          10
                          10          312,991
                          11               45

          11
                          11            1,416



              Corresponding Distances from Start
                     Using Only q-turns

         2x2x2       Corner 3x3x3       Number
     Distance from    Distance from    of Nodes
         Start         Start

           0
                           0                1
                           2                1
                           4                2
                           6                1

           1
                           1                1
                           3                2
                           5                2

           2

                           2                4
                           4               10
                           6                6

           3
                           3               22
                           5               46
                           7                4

           4
                           4              137
                           6              145

           5
                           5              802
                           7              356

           6
                           6            4,105
                           8              474

           7

                           7           16,577
                           9               83

           8
                           8           57,326
                          10               24
                          12               24

           9
                           9          180,556
                          11              148

          10
                          10          466,028
                          12              192

          11
                          11          676,642
                          13              144

          12
                          12          392,342

          13
                          13           45,600

          14
                          14              163



              Corresponding Distances from Start
                     Using Only h-turns

         2x2x2       Corner of 3x3x3    Number
     Distance from    Distance from    of Nodes
         Start         Start


           0
                           0                1
                           2                1

           1
                           1                1
                           3                1

           2
                           2                1

           3
                           3                3

           4
                           4                3

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU  Sun Dec  5 00:03:57 1993
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Date:      Sun, 5 Dec 1993 00:01:08 EST
From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   God's Algorithm for the Edges of the 3x3x3

Here are my God's Algorithm results for the edges of the
3x3x3 cube.  I explained in the last post what I mean by
1152-fold symmetry.  All results below are for 1152-fold
symmetry.  I am working on the 24-fold case, but I am not
quite done.  The 24-fold case is just a matter of determining
the sizes of the equivalence classes in the 1152-fold case.

One item of terminology needs to be explained.  Several
people, including myself, have posted results for the
2x2x2 cube and for the corners of the 3x3x3 cube.  If you
take the term "corners of the 3x3x3 cube" absolutely
literally, it is completely isomorphic to the 2x2x2 cube.
When people have posted results for the "corners of the
3x3x3 cube", they all (including myself) really mean
"corners plus centers of the 3x3x3".  See below:

   --------------      ---------------------    -------------------
   |  x  |  x   |      |  x   |      |  x  |    |  x  |     |  x  |
   |     |      |      |      |      |     |    |     |     |     |
   |-----|-------      -------|------|------    ------|-----|------
   |  x  |  x   |      |      |      |     |    |     |  x  |     |
   |     |      |      |      |      |     |    |     |     |     |
   --------------      -------|------|------    ------|-----|------
       2x2x2           |  x   |      |  x  |    |  x  |     |  x  |
                       |      |      |     |    |     |     |     |
                       ---------------------    -------------------
                          Corners of 3x3x3       Corners + Centers

Thus, when I say I have solved the "edges of the 3x3x3", I need
to clarify what I mean.  I have solved the "edges without the
centers".  I suppose my next project will be "edges with the
centers".  Unfortunately, "edges with the centers" is a twenty-four
times larger problem than is "edges without the centers".  "Edges
without the centers" took about a year running 24 hours a day,
7 days a week, on two machines.  I am going to have to rethink
"edges with the centers" before I start.  I don't want it to take
24 years.

      ---------------------    -------------------
      |      |  x   |     |    |     |  x  |     |
      |      |      |     |    |     |     |     |
      -------|------|------    ------|-----|------
      |  x   |      |  x  |    |  x  |  x  |  x  |
      |      |      |     |    |     |     |     |
      -------|------|------    ------|-----|------
      |      |  x   |     |    |     |  x  |     |
      |      |      |     |    |     |     |     |
      ---------------------    -------------------
      Edges without Centers     Edges with Centers


            Results using q-turns only

      Distance                 Number of
     from Start                Nodes using
                              1152-fold Symmetry

          0                             1
          1                             1
          2                             5
          3                            25
          4                           215
          5                         1,860
          6                        16,481
          7                       144,334
          8                     1,242,992
          9                    10,324,847
         10                    76,993,295
         11                   371,975,385
         12                   382,690,120
         13                     8,235,392
         14                            54
         15                             1



            Results using q-turns and h-turns

      Distance                 Number of
     from Start                Nodes using
                              1152-fold Symmetry

          0                             1
          1                             2
          2                             9
          3                            75
          4                           919
          5                        11,344
          6                       139,325
          7                     1,664,347
          8                    18,524,022
          9                   167,864,679
         10                   582,489,607
         11                    80,930,364
         12                           314

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU  Sun Dec  5 00:42:45 1993
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Date:      Sun, 5 Dec 1993 00:26:07 EST
From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Addendum to God's Algorithm for the 2x2x2 Cube

I had intended to include the following table in my first
post concerning God's Algorithm for the 2x2x2 cube, but
I forgot.  It addresses the question of how large are the
equivalence classes in the search tree, where the equivalence
classes are generated by the two rotational symmetry groups
and the one reflectional symmetry group.  Most of the
equivalence classes have 24*24*2=1152 elements, but some have
fewer.


         Size of          Number          Total Number
         Equivalence       of              of
         Class            Nodes           Permutations
                                          Represented

            24                 1                   24
            48                 1                   48
            72                 3                  216
            96                 1                   96
           144                14                2,016
           192                15                2,880
           288               135               38,880
           384                32               12,288
           576             2,208            1,271,808
         1,152            75,392           86,851,584
          ----             -----             --------
         Total            77,802           88,179,840

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU  Sun Dec  5 17:57:22 1993
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From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Equivalence Classes for God's Algorithm for Edges of 3x3x3

        Number of      Size of        Total
       Equivalence    Equivalence     States
           Classes      Class

                 4       24                96
                 2       48                96
                12       72               864
                16       96             1,536
               110      144            15,840
                70      192            13,440
             1,544      288           444,672
             1,252      384           480,768
           128,858      576        74,222,208
       851,493,140     1152   980,920,097,280

       851,625,008            980,995,276,800

Note that 980,995,276,800=12!(2^12)/2, so the proper total was
obtained.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU  Sun Dec  5 21:27:02 1993
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From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   God's Algorithm, 24-fold Symmetry, Edges of 3x3x3

I just finished tabulating the results with 24-fold symmetry for
the edges of the 3x3x3 cube.  I have added them to the table
I posted earlier today which had 1152-fold symmetry.

A couple of reminders.  In the case of 1152-fold symmetry, most but not
all of the equivalence classes have exactly 1152 elements.  In the case
of 24-fold symmetry, every equivalence class has exactly 24 elements.
Thus, you can almost, but not quite, get from the 1152-fold column
to the 24-fold column by multiplying by 48.  Also, my program actually
generated the 1152-fold column.  However, it did not generate the
24-fold column.  That would have taken far too long.  Rather, I generated
the 24-fold column from the 1152-fold column by determining the sizes
of all the equivalence classes.  Finally, note that the total figure
for the 24-fold symmetry column can be calculated as
40,874,803,200 = [12!(2^12)/2] / 24, so the total is correct.

            Results using q-turns only

      Distance            Number of            Number of
     from Start           Nodes using          Nodes using
                         1152-fold Symmetry    24-fold Symmetry

          0                        1                   1
          1                        1                  12
          2                        5                 114
          3                       25               1,068
          4                      215               9,759
          5                    1,860              88,144
          6                   16,481             786,500
          7                  144,334           6,916,192
          8                1,242,992          59,623,239
          9               10,324,847         495,496,593
         10               76,993,295       3,695,351,994
         11              371,975,385      17,853,871,137
         12              382,690,120      18,367,613,703
         13                8,235,392         395,043,663
         14                       54               1,080
         15                        1                   1

       Total             851,625,008      40,874,803,200


            Results using q-turns and h-turns

      Distance            Number of          Number of
     from Start           Nodes using        Nodes using
                         1152-fold Symmetry   24-fold Symmetry

          0                        1                 1
          1                        2                18
          2                        9               243
          3                       75             3,240
          4                      919            42,359
          5                   11,344           538,034
          6                  139,325         6,666,501
          7                1,664,347        79,820,832
          8               18,524,022       888,915,100
          9              167,864,679     8,056,929,021
         10              582,489,607    27,958,086,888
         11               80,930,364     3,883,792,136
         12                      314             8,827

      Total              851,625,008    40,874,803,200

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From hoey@aic.nrl.navy.mil  Mon Dec  6 10:20:06 1993
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From: hoey@aic.nrl.navy.mil
Message-Id: <9312061519.AA01483@sun13.aic.nrl.navy.mil>
To: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
Cc: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject: Re: God's Algorithm, 24-fold Symmetry, Edges of 3x3x3

These results look very interesting, though I haven't had time to
examine them closely, nor even (in a few cases) quite understand them.
I especially like to see the categorization by symmetry class.

I was somewhat startled to see the unique antipode of the 3x3x3 edges
in the quarter-turn metric.  Do you know what pattern that is?

Dan

From punjanza@dunx1.ocs.drexel.edu  Mon Dec  6 11:37:16 1993
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signoff cube-lovers

From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU  Mon Dec  6 14:03:43 1993
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Date:      Mon, 6 Dec 1993 11:06:48 EST
From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: <hoey@aic.nrl.navy.mil>, "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Re: God's Algorithm, 24-fold Symmetry, Edges of 3x3x3
In-Reply-To: Message of 12/06/93 at 10:19:00 from hoey@aic.nrl.navy.mil

On 12/06/93 at 10:19:00 hoey@aic.nrl.navy.mil said:
>These results look very interesting, though I haven't had time to
>examine them closely, nor even (in a few cases) quite understand them.
>I especially like to see the categorization by symmetry class.

>I was somewhat startled to see the unique antipode of the 3x3x3 edges
>in the quarter-turn metric.  Do you know what pattern that is?


I was extremely surprised as well.  With all my previous work, there
was no unique antipode.  I don't know what it is yet, but I can find
out without a whole lot of trouble.  It is somewhere on the 31-st
tape, so I just need to spin that tape, looking for a record at level
15, and  print it out.  I will try to get to that sometime in the
next few days.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU  Mon Dec  6 18:34:42 1993
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From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Unique Antipodal of the 3x3x3 Edges

In answer to the question by Dan Hoey, I printed out the unique
antipodal of the 3x3x3 edges  --  the one configuration that is
15 moves from Start using only q-turns on the edges of the 3x3x3.

It is really quite extraordinary and wonderful.  I already knew
that there were only four equivalence classes with 24 elements.
Well, two of them are Start itself and its antipodal.  Without
further ado:


          *6*              *6*
          6*6              3*4
          *6*              *1*
          *2*              *5*
          2*2              3*4
          *2*              *2*
       *3**1**4*        *1**1**1*
       3*31*14*4        5*23*42*5
       *3**1**4*        *6**6**6*
          *5*              *2*
          5*5              3*4
          *5*              *5*

         Start          Antipodal

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From @mail.uunet.ca:mark.longridge@canrem.com  Mon Dec  6 19:20:24 1993
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Reply-To: CRSO.Cube@canrem.com
Sender: CRSO.Cube@canrem.com
Subject: Unique antipode of edges only
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.581.5834.0C18D396@canrem.com>
In-Reply-To: <9312061519.AA01483@sun13.aic.nrl.navy.mil>
Date: 	Mon, 6 Dec 1993 10:45:00 -0500
Organization: CRS Online  (Toronto, Ontario)

-> I was somewhat startled to see the unique antipode of the 3x3x3 edges
-> in the quarter-turn metric.  Do you know what pattern that is?
->
-> Dan

It's got to be all edges flipped in place.
I would like to see the process generating the position!

I don't understand it all either :-<

But at least we got some new cube mail.
 -> Mark <-

From alan@parsley.lcs.mit.edu  Mon Dec  6 20:16:32 1993
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From: Alan Bawden <Alan@lcs.mit.edu>
Sender: Alan@lcs.mit.edu
To: BRYAN@wvnvm.bitnet
Cc: Cube-Lovers@ai.mit.edu
In-Reply-To: Jerry Bryan's message of Mon, 6 Dec 1993 18:32:15 EST <9312062334.AA01891@life.ai.mit.edu>
Subject: Unique Antipodal of the 3x3x3 Edges

   Date:      Mon, 6 Dec 1993 18:32:15 EST
   From: Jerry Bryan <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
   ...  It is really quite extraordinary and wonderful.  I already knew
   that there were only four equivalence classes with 24 elements.  Well,
   two of them are Start itself and its antipodal.  Without further ado:...

This is very interesting indeed!  
So the next natural question would seem to be: What are the -other- two?



From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU  Mon Dec  6 21:32:55 1993
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Date:      Mon, 6 Dec 1993 21:30:25 EST
From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: <cube-lovers@ai.mit.edu>
Subject:   Re: Unique antipode of edges only
In-Reply-To: Message of 12/06/93 at 10:45:00 from mark.longridge@canrem.com

On 12/06/93 at 10:45:00 mark.longridge@canrem.com said:
>-> I was somewhat startled to see the unique antipode of the 3x3x3 edges
>-> in the quarter-turn metric.  Do you know what pattern that is?
>->
>-> Dan

>It's got to be all edges flipped in place.

I had to stare at my picture for a couple of minutes to be sure, but
yes it is.  How did you know?

>I would like to see the process generating the position!

This is doable, but it is a little harder said than done.  My "data base"
is just a simple flat file with the states and the level associated with
each state.  In the case of the 2x2x2, the file is about 625K, and
I have programs to search it readily.  If you use the file in
"Solver mode", my "Solver program" just generates all successors of the
current node, finds each successor in the data base (it is a simple binary
search, the file is sorted), chooses one at level N-1 (there is always
at least one), and makes that the new current node.  It stops when
N=0.

I have a "Solver program" for the "corners plus centers of the
3x3x3" as well, but again the data base is small.  It is actually
the original 625K file for the 2x2x2 case, plus three additional
625K files.  This simple file structure was chosen to keep the file
small.  There are no pointers, trees, or processes stored in the
data base.

The "edges of the 3x3x3 without centers" is a little tougher.  Early
in the project, I generated a data base for the first few levels
(six or seven, I think), and I have a "Solver program" that will
work up to that level.  However, the full "edges of the 3x3x3 without
centers" is a 4.2 gigabyte file on tape, so it is hard to process.
Also, the size of the equivalence classes is not in the data base,
only the level.  I have to calculate the size of each equivalence
class, and it is an expensive calculation.

I made a pass at the
file and calculated the number of equivalence classes (took
125 hours on a mainframe), but I only saved a summary.  I did not
save the number of equivalence classes for each state.  I found
the antipodal by looking for level 15, since I knew there was
only one occurrence, and since the level was in the data base.

I did save the summaries by tape, so I should only have to look
on two tapes to find the other two equivalence classes which
have 24 elements.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From mouse@collatz.mcrcim.mcgill.edu  Tue Dec  7 07:38:26 1993
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Date: Tue, 7 Dec 1993 07:38:09 -0500
From: der Mouse <mouse@collatz.mcrcim.mcgill.edu>
Message-Id: <199312071238.HAA16886@Collatz.McRCIM.McGill.EDU>
To: cube-lovers@ai.mit.edu
Subject: Re:  Unique Antipodal of the 3x3x3 Edges

> In answer to the question by Dan Hoey, I printed out the unique
> antipodal of the 3x3x3 edges [...].

> It is really quite extraordinary and wonderful.  [...].  Without
> further ado:

Someone else remarks that it's "got to be all edges flipped in place",
and Jerry Bryan remarks that it is.

>           *6*              *6*
>           6*6              3*4
>           *6*              *1*
>           *2*              *5*
>           2*2              3*4
>           *2*              *2*
>        *3**1**4*        *1**1**1*
>        3*31*14*4        5*23*42*5
>        *3**1**4*        *6**6**6*
>           *5*              *2*
>           5*5              3*4
>           *5*              *5*

I disagree.  Look at the 1-2 edge.  If it's "flipped in place", then
since it appears to be fixed, the cube must flip around it.  But then
the four 3 faces would be where the 4 faces actually are.  No, it's
more complicated than just all-edges-flipped.

"[Q]uite extraordinary and wonderful" it is.

					der Mouse

			    mouse@collatz.mcrcim.mcgill.edu

From ccw@eql12.caltech.edu  Tue Dec  7 08:25:59 1993
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Date:     Mon, 6 Dec 93 19:13:20 PST
From: ccw@eql12.caltech.edu (Chris Worrell)
Message-Id: <931206185340.20400b26@EQL12.Caltech.Edu>
Subject:  Re: Unique antipode of edges only
In-Reply-To: Your message <9312070232.AA10000@life.ai.mit.edu> dated  6-Dec-1993
To: BRYAN%WVNVM.WVNET.EDU%WVNVM.WVNET.EDU@mitvma.mit.edu,
        cube-lovers@ai.mit.edu

On 12/06/93 at 10:45:00 mark.longridge@canrem.com said:
>-> I was somewhat startled to see the unique antipode of the 3x3x3 edges
>-> in the quarter-turn metric.  Do you know what pattern that is?
>->
>-> Dan

>It's got to be all edges flipped in place.


Unfortunately, this is wishfull thinking.
This antipode is 15 qtw from Home, an odd distance.
All edges flipped is an even distance from Home in the qtw metric.

Looking at Jerry Bryan's pictures, I see 5 two edge swaps.

>
>          *6*              *6*
>          6*6              3*4
>          *6*              *1*
>          *2*              *5*
>          2*2              3*4
>          *2*              *2*
>       *3**1**4*        *1**1**1*
>       3*31*14*4        5*23*42*5
>       *3**1**4*        *6**6**6*
>          *5*              *2*
>          5*5              3*4
>          *5*              *5*
>
>         Start          Antipodal
>

If we assume face 1 is F, I get
          (FU) (BD) (FD,BU) (FL,LU) (FR, RU) (LD,BL) (RD,BR)


Is the 1152 number the result of factoring out the 24 spatial rotations
and 2 reflections of the centers?

Are there any estimates of how many distinct sequences actually generate
this Antipodal Class?
Ideally, it would be interesting to have a total list of these sequences.

From formail.TCPBRIDGE.FS3.FAA1.ERICM%smte@formail.formation.com  Tue Dec  7 09:06:46 1993
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From: <formail.TCPBRIDGE.FS3.FAA1.ERICM%smte@formail.formation.com>  (Moyer, Eric )
To: <CUBE-LOVERS@ai.mit.edu>
Subject: cube availability
Date: Tue Dec  7 09:10:15 1993

Greetings.

   I have been away from cubing since the early 80's, which was
before I went to school and before I did much computer work. 
After reading the recent archives I rushed out to find a square1
and fell in love all over again, only this time, I'm armed.  I
was somewhat amazed to find, however, that not a single other
cube puzzle was available at ToysRUs or at any of the stores I
tried first. I went back and reread Hofstadter's articles after
Jerry Bryan's recommendation and came across the address for Uwe
M'effert Novelties, Princewell (Far East), Ltd., P.O. Box 31008,
Causeway Bay, Hong Kong.  Does anyone know if this company still
exists?  Additionally, does anyone know of any mail order company
where cubes and cube-variants can be purchased?  Thanks.



From andyl@harlequin.com  Tue Dec  7 10:33:00 1993
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From: Andy Latto <andyl@harlequin.com>
Date: Tue, 7 Dec 1993 10:37:37 -0500
Message-Id: <6474.199312071537@phaedrus.harlequin.com>
To: Alan@lcs.mit.edu
Cc: BRYAN@wvnvm.bitnet, Cube-Lovers@ai.mit.edu
In-Reply-To: Alan Bawden's message of Mon, 6 Dec 93 20:16:26 -0500 <6Dec1993.195513.Alan@LCS.MIT.EDU>
Subject: Unique Antipodal of the 3x3x3 Edges

   Date: Mon, 6 Dec 93 20:16:26 -0500
   From: Alan Bawden <Alan@lcs.mit.edu>
   Sender: Alan@lcs.mit.edu

      Date:      Mon, 6 Dec 1993 18:32:15 EST
      From: Jerry Bryan <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
      ...  It is really quite extraordinary and wonderful.  I already knew
      that there were only four equivalence classes with 24 elements.  Well,
      two of them are Start itself and its antipodal.  Without further ado:...

   This is very interesting indeed!  
   So the next natural question would seem to be: What are the -other- two?

Switch each edge with its antipode, with or without flipping all twelve edges.

From tom@scumby.clipper.ingr.com  Tue Dec  7 11:07:36 1993
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From: tom@scumby.clipper.ingr.com (Tom Granvold)
Message-Id: <199312071611.AA18849@scumby.clipper.ingr.com>
Subject: Re: cube availability
To: cube-lovers@ai.mit.edu,
        formail.TCPBRIDGE.FS3.FAA1.ERICM%smte@formail.formation.com (Moyer Eric)
Date: Tue, 7 Dec 93 8:11:26 PST
In-Reply-To: <9312071401.AA19052@formail.formation.com>; from "Moyer, Eric" at Dec 7, 93 9:10 am
X-Mailer: ELM [version 07.00.00.00 (2.3 PL11)]

>After reading the recent archives I rushed out to find a square1
>and fell in love all over again,

    Congratulation.

>only this time, I'm armed.

    Watch out, he is dangerous. :-)

>I was somewhat amazed to find, however, that not a single other
>cube puzzle was available at ToysRUs or at any of the stores I
>tried first.

    Instead of toy stores, I'd try game stores.  I don't know if you'll
be able to find a Square-1 or not.  It has been a couple of years since
they come out.

>came across the address for Uwe
>M'effert Novelties, Princewell (Far East), Ltd., P.O. Box 31008,
>Causeway Bay, Hong Kong.  Does anyone know if this company still
>exists?

    I believe that this company has not been around for several years.
I did buy a couple of their "cubes" about ten years ago.  But at some
point they seemed to have disapeared.  Too bad they had some unique
variations.

>Additionally, does anyone know of any mail order company
>where cubes and cube-variants can be purchased?  Thanks.

    Yes.  It seems that just recently Ishi Press has made some of the
cube-variants available.  I saw a 5x5x5 cube in a game store recently
from Ishi Press.  Game stores are a good place to look since Ishi has
long been providing products for the games; Go and Shogi.  Note that
there is even an email address!

        Ishi Press International
        76 Bona Ventura
        San Jose Ca 95134
        (408)944-9900
        fax - (408)944-9110
        email - ishius@ishius.com

Have fun,
Tom Granvold

tom@clipper.ingr.com



From senya@rainbow.ldgo.columbia.edu  Tue Dec  7 11:15:30 1993
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Date: Tue, 7 Dec 93 11:14:55 -0500
From: senya@rainbow.ldgo.columbia.edu (Semyon Basin)
Message-Id: <9312071614.AA29505@rainbow.ldgo.columbia.edu>
To: cube-lovers@ai.mit.edu
Subject: Needed: Basic elements of solving Rubic Cube:


Could you gentelmen suggest me the place where I can find the basic
(elementary) combinations to solve the cube?

Like the sequence to swap two "internal" side's boxes while not
changing the positions of all other "internal" boxes but probably only
rotating them?    

Or could you tell me about the method to rotate some of edge elements
without swapping them?
___________________________________________________________________________
                                        ____   ______   __    __     ______     
 Semyon Basin,                         / ___\ /\  ___\ /\ \  /\ \   / ____ \    
 Lamont-Doherty Earth Observatory,    /\ \/_/ \/\ \/_/ \/\ \_\/\ \ /\ \___\ \   
 Route 9W,           	              \/\ \    \/\  _\  \/\  ____ \\/\__  __ \  
 Palisades, NY 10964     	       \/\ \____\/\ \/___\/\ \/_/\ \\/_/ /_/\ \ 
 				        \/\_____\\/\_____\\/\_\ \/\_\ /\_\ \/\_\
Internet:senya@rainbow.ldgo.columbia.edu \/_/_/_/ \/_/_/_/ \/_/  \/_/ \/_/  \/_/
________________________________________________________________________________


From dseal@armltd.co.uk  Tue Dec  7 12:02:56 1993
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Date: Tue, 07 Dec 93 16:13:51 GMT
From: dseal@armltd.co.uk (David Seal)
To: (Cube) cube-lovers@ai.mit.edu
Subject: Re: Unique antipode of edges only
Message-Id: <2D04ABBF@dseal>


> Someone else remarks that it's "got to be all edges flipped in place",
> and Jerry Bryan remarks that it is.
>
> >           *6*              *6*
> >           6*6              3*4
> >           *6*              *1*
> >           *2*              *5*
> >           2*2              3*4
> >           *2*              *2*
> >        *3**1**4*        *1**1**1*
> >        3*31*14*4        5*23*42*5
> >        *3**1**4*        *6**6**6*
> >           *5*              *2*
> >           5*5              3*4
> >           *5*              *5*
>
> I disagree.  Look at the 1-2 edge.  If it's "flipped in place", then
> since it appears to be fixed, the cube must flip around it.  But then
> the four 3 faces would be where the 4 faces actually are.  No, it's
> more complicated than just all-edges-flipped.
>
> "[Q]uite extraordinary and wonderful" it is.

It is in fact the position arrived at by flipping all edges in place, *then*
reflecting the entire configuration. I believe this also tells us what the
other two equivalence classes with just 24 elements are: they are the
results of doing each of these two operations separately.

David Seal
dseal@armltd.co.uk

From andyl@harlequin.com  Tue Dec  7 12:24:30 1993
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From: Andy Latto <andyl@harlequin.com>
Date: Tue, 7 Dec 1993 12:29:32 -0500
Message-Id: <12292.199312071729@phaedrus.harlequin.com>
To: ccw@eql12.caltech.edu
Cc: BRYAN%WVNVM.WVNET.EDU%WVNVM.WVNET.EDU@mitvma.mit.edu,
        cube-lovers@ai.mit.edu
In-Reply-To: Chris Worrell's message of Mon, 6 Dec 93 19:13:20 PST <931206185340.20400b26@EQL12.Caltech.Edu>
Subject: Unique antipode of edges only



   Unfortunately, this is wishfull thinking.
   This antipode is 15 qtw from Home, an odd distance.
   All edges flipped is an even distance from Home in the qtw metric.

   Looking at Jerry Bryan's pictures, I see 5 two edge swaps.

   >
   >          *6*              *6*
   >          6*6              3*4
   >          *6*              *1*
   >          *2*              *5*
   >          2*2              3*4
   >          *2*              *2*
   >       *3**1**4*        *1**1**1*
   >       3*31*14*4        5*23*42*5
   >       *3**1**4*        *6**6**6*
   >          *5*              *2*
   >          5*5              3*4
   >          *5*              *5*
   >
   >         Start          Antipodal
   >

The antipodal position is an interesting one. If you take the antipodal
position, and flip all the edges, you get:

        *5*                      
        5*5                      
        *5*                      
        *1*                      
        1*1                      
        *1*                      
     *3**2**4*                   
     3*32*24*4                   
     *3**2**4*                   
        *6*                      
        6*6                      
        *6*                      

Antipodal with edges flipped.

This looks like a rotation of the solved state at first glance, since
all the faces on a given side of the cube are the same color. But
look again! This is not the solved state of the original cube, but 
of the mirror image cube. If you added in the centers or the corners,
there would be no way to add them to make this a solved state.

					Andy Latto
					andyl@harlequin.com

From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU  Tue Dec  7 13:42:56 1993
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From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <cube-lovers@ai.mit.edu>
Subject:   Re: Unique antipode of edges only
In-Reply-To: Message of 12/07/93 at 12:29:32 from andyl@harlequin.com

On 12/07/93 at 12:29:32 Andy Latto said:

>The antipodal position is an interesting one. If you take the antipodal
>position, and flip all the edges, you get:

>        *5*
>        5*5
>        *5*
>        *1*
>        1*1
>        *1*
>     *3**2**4*
>     3*32*24*4
>     *3**2**4*
>        *6*
>        6*6
>        *6*

>Antipodal with edges flipped.

>This looks like a rotation of the solved state at first glance, since
>all the faces on a given side of the cube are the same color. But
>look again! This is not the solved state of the original cube, but
>of the mirror image cube. If you added in the centers or the corners,
>there would be no way to add them to make this a solved state.

Indeed.  I spoke too quickly when I said the antipodal was simply
Start with the edges flipped.  I stared at it, flipped the edges in
my mind, and it "looked" solved, so I assumed it was Start.

I am not yet for sure what they look like, but of the other two states
with order-24 equivalence classes, one is at level 9 and the other
is at level 12.  Since the only one at an even level is at level 12,
I am assuming that will be the one which is Start with the edges all
flipped.  The one at level 9 will probably be the mirror image of Start.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From hoey@aic.nrl.navy.mil  Tue Dec  7 20:13:23 1993
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Date: Tue, 7 Dec 93 20:13:08 EST
From: hoey@aic.nrl.navy.mil (Dan Hoey)
Message-Id: <9312080113.AA29049@Sun0.AIC.NRL.Navy.Mil>
To: "Cube Lovers List" <cube-lovers@ai.mit.edu>
Subject: Re: Unique antipode of edges only

"Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu> writes:

> I spoke too quickly when I said the antipodal was simply
> Start with the edges flipped.  I stared at it, flipped the edges in
> my mind, and it "looked" solved, so I assumed it was Start.

It's interesting to note that this is All-Edges-Flipped composed with
a mirror reflection of Start.  Begging the question: *which* mirror
reflection?  The answer is, it doesn't matter: since these are the
edges of a cube without centers, all reflections are the same
position.  As long as we get to choose which reflection, the canonical
one would be the central reflection.  When composed with All-Edges-
Flipped, it makes the following antipode.  (I think using BFTDLR
notation instead of 123456 makes these diagrams a lot easier to read).

               + T +                        + F +
               T   T                        R   L
               + T +                        + B +

        + L +  + F +  + R +          + D +  + D +  + D +
        L   L  F   F  R   R    =>    F   B  R   L  B   F
        + L +  + F +  + R +          + T +  + T +  + T +

               + D +                        + B +
               D   D                        R   L
               + D +                        + F +

               + B +                        + T +
               B   B                        R   L
               + B +                        + D +

> I am not yet for sure what they look like, but of the other two states
> with order-24 equivalence classes, one is at level 9 and the other
> is at level 12.  Since the only one at an even level is at level 12,
> I am assuming that will be the one which is Start with the edges all
> flipped.  The one at level 9 will probably be the mirror image of Start.

If an order-24 equivalence class means what I think it does, I'm
pretty sure those two states have to be Mirror-Start and All-Edges-
Flipped, being the only sufficiently symmetric positions.  But as to
their depth, the parity argument (which Chris Worrell also cited) is
not valid here.  Remember that the cube has no face centers, so the
position is not changed by rotating the assemblage of edges in space
(i.e., with respect to the absent face centers).  But a quarter-turn
of the cube in space is an odd permutation of the edges.  So permuta-
tion parity is not an intrinsic property of edge positions.  We can
show that there is no sort of parity here by explicitly constructing
an odd cycle.  Just use a process that would permute the edges of a
cube with faces as (FR,FT,FL,FD)(BR,BT,BL,BD)(RT,TL,LD,DR).  This has
to be an odd process, but it's an identity on the faceless cube.

My (very cheap) guess for where we will find the other two M-symmetric
positions is opposite to Jerry Bryan's.  On a cube with faces, the
central reflection of the edges with respect to the faces is Pons
Asinorum, which has the easy 12-qt tight lower bound we've seen before
(or if not, you can of course get it from me with email).  I'm
guessing that this bound happens to be tight on the cube without
faces, as well.  But I have no proof of this guess, and I'm very
grateful we won't have to settle for guesses for very long.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

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From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <cube-lovers@ai.mit.edu>
Subject:   Re: Unique antipode of edges only
In-Reply-To: Message of 12/07/93 at 20:13:08 from hoey@aic.nrl.navy.mil

On 12/07/93 at 20:13:08 hoey@aic.nrl.navy.mil said:

>My (very cheap) guess for where we will find the other two M-symmetric
>positions is opposite to Jerry Bryan's.  On a cube with faces, the
>central reflection of the edges with respect to the faces is Pons
>Asinorum, which has the easy 12-qt tight lower bound we've seen before
>(or if not, you can of course get it from me with email).  I'm
>guessing that this bound happens to be tight on the cube without
>faces, as well.  But I have no proof of this guess, and I'm very
>grateful we won't have to settle for guesses for very long.

>Dan Hoey
>Hoey@AIC.NRL.Navy.Mil

Dan Hoey is correct.  Mirror-Image-of-Start is at level 12.
Edges-Flipped is at level 9.  Mirror-Image-of-Start-and-Edges-Flipped
is at level 15.  And, of course, Start is at Level 0.  This exhausts
the list of configurations with order-24 symmetry.

I am still thinking about the easiest way to extract sequences of
operators from my data base.  I gather from Dan's comments that a
12-qt operator is known for Mirror-Image-of-Start.  Are operators
known for the other two cases?  This is going to be sufficiently
time-consuming that I don't want to try to find operators that
are already known.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From senya@rainbow.ldgo.columbia.edu  Wed Dec  8 12:21:51 1993
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Date: Wed, 8 Dec 93 12:20:38 -0500
From: senya@rainbow.ldgo.columbia.edu (Semyon Basin)
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Subscribe me please


From @mail.uunet.ca:mark.longridge@canrem.com  Wed Dec  8 14:21:29 1993
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Sender: CRSO.Cube@canrem.com
Subject: Antipodal Edge Position
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.600.5834.0C18D8B1@canrem.com>
Date: 	Wed, 8 Dec 1993 12:33:00 -0500
Organization: CRS Online  (Toronto, Ontario)

>>It's got to be all edges flipped in place.

Oops. Well I figured if all edges flipped was one of the hardest
know cube states that in the case of edges-only it would be the
antipode. I'm now sure (I think) that it is really:

       all edges flipped + 4 X
(with the 4 X on sides F, R, B, L which should match Dan's diagram)

Hmmmm, I don't know if this is a standard form of representation,
but this picture looks like a folded out cube:

               + T +                        + F +
               T   T                        R   L
               + T +                        + B +

        + L +  + F +  + R +          + D +  + D +  + D +
        L   L  F   F  R   R    =>    F   B  R   L  B   F
        + L +  + F +  + R +          + T +  + T +  + T +

               + D +                        + B +
               D   D                        R   L
               + D +                        + F +

               + B +                        + T +
               B   B           --------->   R   L
               + B +           |            + D +
                               |

                             + D +
In my program I would have   L   R  on the screen for the bottom face.
                             + T +

The idea is you are always looking at a cube face head-on (just to
clarify the difference in diagrams).

More quotes for Jerry Bryan:
>The "edges of the 3x3x3 without centers" is a little tougher.  Early
>in the project, I generated a data base for the first few levels
>(six or seven, I think), and I have a "Solver program" that will
>work up to that level.  However, the full "edges of the 3x3x3 without
>centers" is a 4.2 gigabyte file on tape, so it is hard to process.
>Also, the size of the equivalence classes is not in the data base,
>only the level.  I have to calculate the size of each equivalence
>class, and it is an expensive calculation.
>
>I made a pass at the
>file and calculated the number of equivalence classes (took
>125 hours on a mainframe), but I only saved a summary.  I did not
>save the number of equivalence classes for each state.  I found
>the antipodal by looking for level 15, since I knew there was
>only one occurrence, and since the level was in the data base.
>
>I am not yet for sure what they look like, but of the other two states
>with order-24 equivalence classes, one is at level 9 and the other
>is at level 12.  Since the only one at an even level is at level 12,
>I am assuming that will be the one which is Start with the edges all
>flipped.  The one at level 9 will probably be the mirror image of
Start.

I'd still like to see the process for all-edges-flipped (not
caring about the centres or corners). So "level" is the number of moves
required to solve the position? That means edges flipped in place
can be done in 12 qtw.

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From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   1152-fold Symmetry and 24-fold Symmetry

I guess it's time to try to explain what I mean by 1152-fold symmetry
and 24-fold symmetry.

Let me start with two or three very simple ideas.  First, consider
two equally colored and oriented cubes at Start.  To one of them,
apply F, and to the other one apply R.  The obvious solution to the
first one is then F' and the obvious solution to the second on is
then R'.  But take both cubes and toss them through the
air to someone else, so that the spatial orientation is lost.
Almost anyone would solve either cube by finding the one face that
was twisted clockwise and simply twisting it counter-clockwise.
No distinction between F and R would be made, and it would be
"obvious" to any reasonable person that the cubes were equivalent.

As a slightly more formal application of this idea, consider again
Start to which R has been applied.  We could rotate the whole
cube in space using Singmaster's script-U operation.  That is, grasp
the Up (top) of the cube and turn the whole cube in space clockwise.
Now, the cube looks like F has been applied rather than R, and the
solution looks like F' rather than R'.  If we applied F', the cube
would be solved, but the colors would be oriented wrong.  We could
restore the colors by script-U'.  Thus, (script-U F' script U') is
exactly the same thing as R' (we are just using conjugates in a
very simple way).

Continuing in this vein, take any two equally colored and oriented
cubes at Start.  To one of them, apply some long sequence of
permutations P.  To the second one, apply (script-U P script-U').
At this point, the two cubes are definitely not "equal" in some
sense  -- you could clearly tell them apart.  Yet, they are
clearly "equivalent" in some sense, because if P' is a solution to
the first cube, then (script-U P' script-U') is a solution to the
second one.  Furthermore, it should be obvious that it is not really
necessary to use the (script-U script-U') conjugate on the second
cube.  Rather we can think of some rotation as having been performed
on P to give Q, and then of Q as having been performed on Start, so
that the same rotation that was applied to P could be applied to P'
to give Q', and Q' is equivalent to (script-U P' script-U').

If I can wax sophomorically philosophical for a minute, I tend to
think of there being two kinds of permutations in mathematics.
The first is the "permutations and combinations" kind of thing you
run into in probability and statistics.  The second is the permutation
operator kind of thing you run into in abstract algebra or group
theory.  With this kind of thinking, the cube itself represents the
first kind of permutation, where the cube is an object being operated
on, and the twists of the cube are the second kind of permutation,
where the twists are permutation operators and are doing the operating.
Well, at some deep level, the two kinds of permutations are very much
the same thing, so it is very natural to think of operating on
(rotating, in this particular case) a permutation P, where P itself
is an operator.

I need one more simple idea.  Again, think of a cube in Start, and
think of Singmaster's script-U operator.  We can (informally) write
script-U = (Front --> Left --> Back --> Right --> Front).  But suppose
the cube is colored as Font=Red, Left=White, Back=Orange, Right=Blue).
We could just as well write script-U = (Red --> White --> Orange
--> Blue --> Red).  It looks as if for any fixed coloring, we can
freely interchange Singmaster's notation with a notation based on
colors.  But we can't really.  For example, colored as I described it
above, script-F is equivalent to script-Red.  Either is the same as
grasping the front of the cube and rotating the whole cube clockwise.
But first perform script-U.  Now, script-F is the same as
script-Blue).  The Front/Back/Up/Down/Left/Right notation is fixed in
space, but the color notation is not.

Now, we try to put all this together.  Once again, consider two
equally colored and equally oriented cubes in space, and apply F
to the first one and (R script-U) to the second one.  Both
cubes can now be described as "Start with the front twisted clockwise
by 90 degrees), but the colors are not the same.  They are clearly
equivalent, but under my internal computer model for the cube, they
are not equal.  Furthermore, no amount of additional application of
Singmaster's whole cube "script" operators will make them equal.
The only thing that will make them equal will be to rotate the colors.

The exact same thing applies to reflections.  Consider two equally
colored and oriented cubes in Start, and apply F to one and F' to
the other.  The cubes are equivalent but not equal.  Hold up the
cube to which F' has been applied to a mirror.  The mirror-image
now has F applied instead of F', but the colors are wrong (they
have been reflected).  To make the cubes equal, it is necessary to
reflect the colors of the mirror-image.

Hence, my program generates equivalence classes by applying
a cube rotation, a color rotation, a cube reflection, and a color
reflection.  There are 24 cube rotations and 24 color rotations
(one of each is the identity), and any cube rotation can occur with
any color rotation.  There are 2 cube reflections and 2 color
reflections (one each is the identity), but the cube reflection
identity must occur with the color reflection identity and vice
versa.  Thus, there are (in general) 24x24x2 elements in each
equivalence class.  I only store one element of each equivalence
class in my data base.  Some of the equivalence classes have fewer
than 24x24x2 elements, namely those for which the cube configuration
inherently has a high degree of symmetry.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From @mail.uunet.ca:mark.longridge@canrem.com  Wed Dec  8 15:31:43 1993
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Subject: More corrections
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.601.5834.0C18D8D1@canrem.com>
Date: 	Wed, 8 Dec 1993 13:52:00 -0500
Organization: CRS Online  (Toronto, Ontario)


Mark gaffs again:
>          I'm now sure (I think) that it is really:
>
>       all edges flipped + 4 X
> (with the 4 X on sides F, R, B, L which should match Dan's diagram)

* sign *  No, I see I entered the position into my program wrong.
A central reflection of the edges with respect to the faces is
simply 6 X or checkerboard order 2, solvable in 12 qtw or 6 htw.
So the edges-only antipode is:  all-edges-flipped + 6 X.

Jerry Byran quote:
>Dan Hoey is correct.  Mirror-Image-of-Start is at level 12.
>Edges-Flipped is at level 9.  Mirror-Image-of-Start-and-Edges-Flipped
>is at level 15.  And, of course, Start is at Level 0.  This exhausts
>the list of configurations with order-24 symmetry.

Ok, only 9 qtw.... it's got to play havoc with corners. I got it now.

* Hmmm, what are all the possible orders of symmetry? *

Also I note my "Symmetry Level" is the opposite of Jerry's Order-N
symmetry:

>   If we define "symmetry level" as the number of distinct patterns
>generated by rotating the cube through it's 24 different orientations
in
>space then most known antipodes are symmetry level 6. Thus the lower
the
>number the higher the level of symmetry. The least symmetric positions
>have level 24, and this is very common. The most symmetric positions
>have level 1, the two positions START and 6 X order 2.

Of course all-edges-flipped I never included, as at the time I was
looking at the square's group.

-------------------------------

As a small postfix to my cyclic decomposition article, I found the
following patterns. I'm fond of pattern 16 myself. I am looking for
CD-type processes for 6 X order 3 and 6 X order 6. I find when I
am physical cubing (as opposed to computer cubing or old fashioned
mental cubing!) it really helps having a CD-type process memory-wise.
Memorizing the computer generated processes is like memorizing
prime numbers.

p161 Mark's Pattern 16  (F1 R1 L1 B1) ^3 + F2 B2 D2 F2 B2 T2    (18)
p162 2 X, 4 H full      (F1 T2 B1) ^4                           (12)
p163 4 ARM Full         (F2 T1 B2) ^4 + T1 D3                   (14)
p164 4 Y's Rotated      (F1 T2 D2) ^6 + F1                      (19)
p165 2 Swap, 4 H full   (F1 L2 T2 R2 B1) ^2 + L2 R2 T2 D2       (14)
p166 2 H adj swap       (F1 L2 T2 R2 B1) ^2 + L2 T2 R2 D2 L2 T2 (16)

No doubt these are compressible and hence not as efficient, but if
you consider ease of execution....

-> Mark <-

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From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Pretty Pattern at Level 13

Here is a pretty pattern at level 13 using q-turns.  The size
of the equivalence class is 48.

               + B +
               R   L
               + F +

               + D +
               L   R
               + U +

        + B +  + F +  + B +
        U   D  R   L  D   U
        + F +  + B +  + F +

               + U +
               L   R
               + D +

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU  Thu Dec  9 04:19:46 1993
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From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Pretty Pattern (again)

Pretty pattern, 8 q-turns from Start, size of equivalence
class is 48.

               + B +
               F   F
               + B +

               + U +
               D   D
               + U +

        + L +  + F +  + R +
        R   R  B   B  L   L
        + L +  + F +  + R +

               + D +
               U   U
               + D +

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU  Thu Dec  9 04:56:35 1993
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From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   That's all the 48-Fold Symmetries

The two patterns I just posted are the only two for which the size
of the equivalence class is 48.  The next size up is 72, and there
are twelve patterns whose equivalence class size is 72.  Things
become less interesting as the equivalence class size increases because
they are less symmetrical overall.  Also, there are more patterns.
Finally, these things take a good deal of time to chase down.
Therefore, I am going to stop chasing down "pretty patterns"
for now unless somebody is just dying to see the patterns whose
equivalence class size is 72.

Finally, nobody really answered an earlier question, but is the
following true:  1) Mirror-Image-of-Start is 12 q-turns from
Start, and a sequence is known (Dan Hoey sent me a well-known
sequence), 2) All-Edges-Flipped is 9 q-turns from Start, and a
sequence is known, and 3) Mirror-Image-of-Start-and-All-Edges-Flipped
(the antipodal of Start) is 15 q-turns from Start, and a sequence
is *not* known?  If this is true, then I will start working on
the 15 move sequence.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From mouse@collatz.mcrcim.mcgill.edu  Fri Dec 10 09:16:34 1993
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Date: Fri, 10 Dec 1993 09:16:29 -0500
From: der Mouse <mouse@collatz.mcrcim.mcgill.edu>
Message-Id: <199312101416.JAA04504@Collatz.McRCIM.McGill.EDU>
To: senya@rainbow.ldgo.columbia.edu
Subject: Re:  Needed: Basic elements of solving Rubic Cube:
Cc: cube-lovers@ai.mit.edu

> Subject: Needed: Basic elements of solving Rubic Cube:

(Does anyone happen to know whether Ern  Rubik knows what's happened to
his name?  I don't mean just here, but how common a term it's become.
And I apologize for using   instead of what I think is the proper long
accent, but Latin-1 doesn't have the proper one....)

> Could you gentelmen suggest me the place where I can find the basic
> (elementary) combinations to solve the cube?

There are no *the* basic combinations.  I would guess there are as many
different sets of combinations as there are cube solvers.

> Like the sequence to swap two "internal" side's boxes while not
> changing the positions of all other "internal" boxes but probably
> only rotating them?

> Or could you tell me about the method to rotate some of edge elements
> without swapping them?

Well, for what it's worth, when I solve a cube, I do it as follows.

(Slice turns: if I write FB, I mean turn the F-B slice in the direction
one would turn F for an F turn, similarly for other slice turns: turn
the slice as if it were carried along by the turn named by the first
letter.  Thus LR and RL are inverses.  Is there some standard
representation for slice turns?)

- Solve one layer ad-hoc.  (This refers to a layer of cubies, not just
  one face of the cube.)  I often don't worry about edge flips at this
  stage.  Some simple operators I use:
	To get corners in place: F D' F, or F' D F, depending on corner
		orientation.
	To get edges in place: If the cubie is on the D face,
		FB/BF/RL/LR, D/D', inverse of the slice turn.  If it's
		on the middle layer, F/B/R/L, UD/DU, inverse of face
		turn.

- Turn the cube so the solved face is L.  Solve what then becomes the
  R-L slice layer with a combination of R2 U2 R2 U2 R2 U2, to move
  cubies around within the slice layer, and either of two sequences to
  move cubies between the R layer and the slice layer:
	R2 D R' B2 R2 B2 R2 B2 R' D'
	FB D R' B2 R2 B2 R2 B2 R' D' BF
  (The first one is a sequence that normally ends with R2, but since
  the R layer is scratch at this point I often don't bother.)  These
  are, of course, interspersed with R, R2, and R' turns to get edges in
  the correct places for them.

At this point you will have two layers solved, except possibly for some
flipped edges.

Next, corners of the "scratch" layer.  Check them for placement,
ignoring orientation.  They can be:
	1) All in place.  This is the easy case. :-)
	2) Two swapped.  Make one quarter-turn to reach case 3.  (They
	   can't be diagonal, they must be adjacent - or some joker has
	   taken your cube apart.)
	3) One in place, other three permuted.
	4) Two pairs, each swapped.  If the swaps are diagonal, turn the
	   layer a half-turn to reach case 1.
In case 3 or 4, the corners can be put in place by holding the cube
with the unsolved layer as U and repeating
L F L' F' L F L' F' L F L' F' twice, turning U so as to place
appropriate pairs of cubies in the UFL and UBL corners.

To orient the corners correctly, hold the cube with the unsolved layer
as F and use R B2 R' U' B2 U and its inverse U' B2 U R B2 R' with a
turn of the F face in between; this will allow you to twist the corners
into correct orientation.

All that remains at this point is positioning the edges on the last
layer, and possibly some edge flips.  To position the edges, I normally
use (with U as the unsolved layer)
	R2 D R' B2 R2 B2 R2 B2 R' D' R2
	FB D R' B2 R2 B2 R2 B2 R' D' BF
	R2 U2 R2 U2 R2 U2
with appropriate turns of U in between, swapping the FR and BR edges
repeatedly as auxiliaries while swapping pairs of edges on U to get
them in place.  (The difference between the first two sequences is that
the first one swaps UB and UR, the second UB and RU.)

Edge flips are all that's left at this point.  Judicious choice of
which of the two sequences above can often drastically reduce the work
to be done here, but there's often some left anyway.  The basic
sequence I use for this is RL U RL U RL U RL U, which flips four edge
cubies in-place: UB, UL, DB, and DF.  (A similar sequence U RL U RL U
RL U RL is similar but flips UR instead of UL; this can be thought of
as U X U', where X is the first-given sequence.)  My use of this
sequence is usually ad-hoc; sequences such as X F X F' will let you
flip arbitrary pairs of edges.

Presto!  You have a solved cube. :-)  In practice, I often take
shortcuts; for example, if X represents the R B2 R' U' B2 U sequence,
then X B2 X B2 X B2 will twist three corners on B....

					der Mouse

			    mouse@collatz.mcrcim.mcgill.edu

From hoey@aic.nrl.navy.mil  Mon Dec 13 22:31:38 1993
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Date: Mon, 13 Dec 93 22:31:31 EST
From: hoey@aic.nrl.navy.mil (Dan Hoey)
Message-Id: <9312140331.AA25974@Sun0.AIC.NRL.Navy.Mil>
To: Cube-Lovers@ai.mit.edu
Subject: Symmetry

I suppose it's time for a few observations on symmetry.  After all,
tomorrow is the thirteenth anniversary of "Symmetry and Local Maxima."

As Jerry Bryan notes, we can perform the "R" turn by rotating the cube
to put the R face in front, performing "F", and undoing the rotation.
But we can also perform "R'" by reflecting the cube in a left-to-right
mirror, performing "L", and undoing the reflection.  Thus conjugation
can be extended to use the 48-element group of rotations and
reflections, which we call M.

In the absence of face centers, there is another kind of reduction
that takes account of the 24 possible positions of the resulting
collection of edges in space.  So two positions X and Y are considered
equivalent if
    X = m' Y m c
where m is a rotation or reflection in M, and c is a rotation.

My understanding of Jerry Bryan's method is that he combines "m c"
into a single rotation or reflection, and factors out the reflection
on both sides.  It seems to me that what he calls a a "color rotation"
is premultiplication, while a "cube rotation" is postmultiplication.
[I am somewhat uncertain of this, because it doesn't explain how there
can be a 1252-element symmetry group when face centers are present, so
perhaps I'm missing something crucial.]

But I think we are at least conceptually better off dealing with M
when dealing with conjugation, because it takes account of the
essential similarity between clockwise and anticlockwise turns.  Alan
Bawden mentioned back in 1980 that certain positions with sufficient
symmetry were local maxima (in terms of distance from start), on the
grounds that any clockwise or anticlockwise turn gives us essentially
the same position.  Jim Saxe and I formalized the notion in a paper
entitled "Symmetry and Local Maxima" that we posted on 14 December
1980.  [You can find it in /pub/cube-lovers/cube-mail-1.Z on
FTP.AI.MIT.Edu].

We had some hope that some of these local maxima might turn out to be
global maxima.  My hopes for that have been somewhat low in recent
years.  That is perhaps my best excuse for not noticing immediately
that the single global maximum for the edge group turns out to be one
of these symmetric local maxima.  In fact, all four of the positions
with 24-element equivalence classes appear in the list of M-symmetric
positions.

The paper on Symmetry and Local Maxima also catalogues the positions
that have 48-element equivalence classes and 72-element equivalence
classes.  The The former are the H-symmetric positions, "Six-H" and
"Six-H with all edges flipped".  The latter are the twelve T-symmetric
positions.  For T-symmetry, the set of flipped edges may be any of
{none, girdle-edges, off-girdle-edges, or all}; the set of edges
exchanged with their antipodes may be any of the four as well.  But if
we choose "none" or "all" for all both choices we get one of the four
M-symmetric positions with 24-element equivalence classes, so only
twelve of the sixteen possibilities have 72-element equivalence
classes.

With regard to the edge cube, I should mention that no one has
mentioned a 9 QT process for the all-flip nor a 15 QT process for the
pons-asinorum-all-flip.  Of course, the latter would be somewhat more
interesting, being the longest optimal sequence.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From avm@bgerug51.bitnet  Tue Dec 14 02:42:21 1993
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From: Anne-Mie Vandermeeren - RUG <AVM@bgerug51.bitnet>
Subject: Unsunscribe rob@bgerug51.bitnet
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Hi,

Please remove rob@bgerug51.bitnet from your mailing list cube-Lovers

Thanks,
Anne-Mie Vandermeeren
Postmaster for BGERUG51

From anandrao@hk.super.net  Tue Dec 14 03:18:49 1993
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Date: Tue, 14 Dec 1993 16:11:47 +0800 (HKT)
From: Mr Anand Rao <anandrao@hk.super.net>
Subject: Tangle
To: Cube Lovers <Cube-Lovers@ai.mit.edu>
Message-Id: <Pine.3.07.9312141647.A14860-a100000@hk.super.net>
Mime-Version: 1.0
Content-Type: TEXT/PLAIN; charset=US-ASCII

I managed to pick up all four Tangle puzzles in an obscure shop in
Jakarta, Indonesia. The puzzles are similar, except that the extra(25th)
piece is different in each. The solutions for each puzzle are very
different and I could not see any pattern. I solved all 4 using
'intelligent brute force', i.e. made the search as efficient as I could.
But the 10*10 puzzle seems intractable. The 5*5 could be solved using a
486DX2-66 PC in about 20 minutes. The 10*10 will take several months using
my algorithm. 
Does anyone have a more intelligent, or a more brute method? Once this
puzzle has been put in the public domain, we MUST find a solution. So, any
ideas are welcome!
Thanks



From dn1l+@andrew.cmu.edu  Tue Dec 14 11:29:57 1993
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Date: Tue, 14 Dec 1993 11:28:41 -0500 (EST)
From: "Dale I. Newfield" <dn1l+@andrew.cmu.edu>
To: Cube Lovers <Cube-Lovers@ai.mit.edu>
Subject: Re: Tangle
Cc: 
In-Reply-To: <Pine.3.07.9312141647.A14860-a100000@hk.super.net>

Could you explain what your algorithm was?

I have one of the puzzles, number 4, I believe, and spent a large amount
of time trying to find a solution that was not trial and error.  I could
not.

The algorithm that I used to have the computer solve it for me was to
fill the 5x5 in the following manner, recursively, returning when no
possible pieces fit.

 1  2  4  7 11
   /  /  /  /
  /  /  /  /
 3  5  8 12 16
   /  /  /  /
  /  /  /  /
 6  9 13 17 20
   /  /  /  /
  /  /  /  /
10 14 18 21 23
   /  /  /  /
  /  /  /  /
15 19 22 24 25

(wrapping at the edges to keep incrementing properly)

I did that because given any pieces diagonal from one another, there are
at most two pieces that can fill the gap (line up with both correctly).
(When the four colors are different, there are two tiles
 When there is a single repeated color, there is one tile
 When there are 2 pairs of colors there is no tile
 And in all these cases, if the tile(s) was already used, or didn't
exist, that is the bottom of that branch of the search tree)

Is this better or worse than the algorithm you used?

Has anyone found a non-brute-force solution scheme?

-Dale


From hoch@chem.wisc.edu  Tue Dec 14 12:13:15 1993
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From: Douglas E. Hoch <hoch@chem.wisc.edu>
Message-Id: <9312141713.AA19022@ernie.chem.wisc.edu>
To: cube-lovers@ai.mit.edu
Subject: List removal


Please remove hoch@pigggy.chem.wisc.edu from your cube-lovers list.
Thanks.

From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU  Tue Dec 14 21:23:57 1993
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Date:      Tue, 14 Dec 1993 20:50:51 EST
From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Re: Symmetry
In-Reply-To: Message of 12/13/93 at 22:31:31 from hoey@aic.nrl.navy.mil

On 12/13/93 at 22:31:31 hoey@aic.nrl.navy.mil said:

>In the absence of face centers, there is another kind of reduction
>that takes account of the 24 possible positions of the resulting
>collection of edges in space.  So two positions X and Y are considered
>equivalent if
>    X = m' Y m c
>where m is a rotation or reflection in M, and c is a rotation.

>My understanding of Jerry Bryan's method is that he combines "m c"
>into a single rotation or reflection, and factors out the reflection
>on both sides.  It seems to me that what he calls a "color rotation"
>is premultiplication, while a "cube rotation" is postmultiplication.
>[I am somewhat uncertain of this, because it doesn't explain how there
>can be a 1252-element symmetry group when face centers are present, so
          ^^^^
          should be 1152

>perhaps I'm missing something crucial.]

I just reread "Symmetry and Local Maxima".  Let's see if I can make
some sense of this.  I believe "pre-multiplication" and
"post-multiplication" are correct.  In my computer model, the
corner facelets are simply numbered from 1 to 24, and any configuration
of the corners is an order-24 row vector.  The rotation and reflection
operators are also order-24 row vectors, again with each cell simply
containing a number from 1 to 24.

In almost anybody's programming language you would copy an order-24
row vector with something like

     For i = 1 to 24  B(i) = A(i)

Well, if P is a rotation operator, you could perform a rotation
two ways.  I guess one is pre-multiplication and one is
post-multiplication.

    1)  For i = 1 to 24  B(i) = A(P(i))
    2)  For i = 1 to 24  B(i) = P(A(i))

(As an aside, this illustrates the question I raised in my previous
post about "which is the operator and which is the thing being
operated on?"  Is P operating on A, or is A operating on P?)

In fact, if what I am doing is properly called pre- and post-
multiplication, then I am doing both as a part of a single,
composite operator.  I.e.,

       For i = 1 to 24  B(i) = P(A(P(i)))

More completely, there are 24 rotations, P1 through P24, so the
actual loop looks something like

       For j = 1 to 24 for k = 1 to 24
             for i = 1 to 24 Bj,k(i) = Pj(A(Pk(i)))

Finally, if Q is a reflection (actually, if Q1 is the identity and
Q2 is the reflection), then we have

      For j = 1 to 24 for k = 1 to 24 for m = 1 to 2
             for i = 1 to 24 Bj,k,m(i) = Qm(Pj(A(Qm(Pk(i)))))

I believe this loop calculates Dan Hoey's M.  In my data base, I
store the minimum of Bj,k,m over j = 1 to 24, k = 1 to 24, and
m = 1 to 2.  I tend to call the minimum of Bj,k,m a canonical
form.  I am not sure if that is the best terminology.  The
minimal element is not any simpler than any other.  It is just
that I need a function to choose an element from a set, and
picking the minimal element seems very natural.  Any other
element would do as well, provided I could always be sure of
picking the same element.

Also, my criterion for equivalence is slightly
different (but isomorphic, I think) than the one described by
Dan Hoey.  Suppose A and B are two cubes.
Rather than mapping A to B or B to A in M, I map both A and B
to their respective canonical forms.  A and B are equivalent if
their respective canonical forms are equal.

I hasten to add that the actual loop in the program is a bit
more complex than the one shown above.  The one above would
be far too slow. The actual loop is several hundred times
faster.

Now, as to the centers.  I still sometimes have a certain doubt
about the centers.  They are fixed, so how can you reduce the
problem (i.e., increase the size of the equivalence classes)
by both rotating the cube and rotating the colors (by both pre-
and post-multiplication)?

In my computer model for the centers, I simply number center facelets
from 1 to 6, and the centers are stored as an order-6 row vector.
The centers are disjoint from the corners (as well as from the edges),
so there is no problem in numbering one set of objects from 1 to 24 and
another from 1 to 6.

I define a set of 24 rotation operators P* on the centers, corresponding
to the 24 rotation operators P on the corners, and a set of 2 reflection
operators Q* on the centers, corresponding to the 2 reflection operators
Q on the corners.  Then, if C is an order-6 row vector representing
the centers, I calculate Dj,k,m = Q*m(P*j(C(Q*m(P*k)))) anytime
I calculate Bj,k,m = Qm(Pj(A(Qm(Pk)))).

   (Read the asterisks above as superscripts.  I am not intending
    the multiplication operation which the asterisk denotes in
    many programming languages.)

Hence, I rotate and reflect the centers right along with the corners.
But there are only 24 distinct states for the centers, and each can
occur with any canonical form for the corners.  Hence, the "corners
plus centers" data base is exactly 24 times larger than
the "2x2x2" data base.

My model for the cube seems to start out 24 times larger than
everybody else's.  However, by storing only the canonical form
for each equivalence class, and since most of the equivalence
classes have 1152 elements, my data base seems to end up about
48 times smaller than everybody else's.  This fact seems to
remain true, even when the "fixed centers" are added in.

I am not sure if this answers Dan's question about my model
with centers added.  Effectively, I am using a "fixed corners"
representation of the cube, and rotating the centers.  Each
equivalence class for the corners under M has (up to) 1152
elements, and each equivalence class for the centers under
M has only 24 elements.  But it doesn't seem to matter.
(Up to) 48 different configurations of the corners within
M share each configuration of the centers.

Since I am in this deep, let me finish explaining certain details
of my model.  I don't really store all 24 elements of each
row vector.  I really just store 8.  That is, I store the
facelets for the front and back face.  The other 16 facelets
can be reconstructed from the first 8.  In effect, storing
a number from 1 to 24 stores both the location of each cubie
and its twist.  Finally, I really, really only store 7 elements.
In the canonical form, the first element is always 1, so there
is no reason to store it.  Thus, a data base record for the
2x2x2 looks like

    CCCCCCC,L

where the CCCCCCC are the seven elements representing the canonical
form, and L is the corresponding level.

When you add the centers, I started out with notion that the
order-6 row vector for center only has 24 possible states.  Thus,
it can be encoded as a number from 1 to 24.  This lead to the
following

    CCCCCCC,L,R

where CCCCCCC and L are as before, and R is an index encoding
the orientation of the centers.  But this can be improved upon
even further.  With my model for the corners plus centers, each
distinct value of CCCCCCC will occur exactly 24 times, and each
distinct value of CCCCCCC is already represented in my data
base for the 2x2x2.  Hence, I can have the exact same number of
(longer) records, and encode the corners of the 3x3x3 as

    CCCCCCC,L,L1 L2 L3 .... L23 L24

where CCCCCCC is as before, L is the level of CCCCCCC in the
2x2x2, and L1 through L24 are the levels of CCCCCCC in the corners
of the 3x3x3 when the index of the position of the centers with
respect to the corners is 1 through 24, respectively.  Hence, my
data base for the corners of the 3x3x3 has the same number of
records as the data base for the 2x2x2, and is physically only
four times larger.

>With regard to the edge cube, I should mention that no one has
>mentioned a 9 QT process for the all-flip nor a 15 QT process for the
>pons-asinorum-all-flip.  Of course, the latter would be somewhat more
>interesting, being the longest optimal sequence.

I will work on these two cases, but it will take some time.  My model
is very good at storing a great many states of the cube very
compactly, but it does not encode processes at all.  I will have
to extract the processes by hand.  This is quite easy in my
data bases for the 2x2x2 and corners of 3x3x3.  But it is quite
hard for the edges because the data base is 4.2 gigabytes.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From Don.Woods@eng.sun.com  Wed Dec 15 06:04:15 1993
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From: Don.Woods@eng.sun.com (Don Woods)
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To: Cube-Lovers@ai.mit.edu
Subject: Re: Tangle
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Anand Rao <anandrao@hk.super.net> writes:
> The puzzles are similar, except that the extra(25th)
> piece is different in each. The solutions for each puzzle are very
> different and I could not see any pattern. 

Look again.  The puzzles are identical except for a remapping of the colors.
For example, if you take Tangle #1 and paint all the Blue ropes Yellow, all
the Red ropes Blue, all the Green ropes Red, and all the (originally) Yellow
ropes Green, you'll have Tangle #2.  So you can solve Tangle #1 by imagining
the ropes recolored as above, constructing your solution for #2, and then
restoring the original colors.

Note: The particular recoloring given above is based on colors given in a
message sent by CCW@eql.caltech.edu (Chris Worrell) to cube-lovers on April
27, 1992.  I own only #1 myself and so cannot confirm or deny the accuracy
of the colors.  But the basic idea applies, given that each puzzle (a) has
the same pattern of ropes on all pieces and (b) has each permutation of
colors exactly once except for one permutation which appears twice.

Solving the 10x10 is another kettle of fish, and I haven't tried it.  I do
have a program that solves the 5x5 in about 45 seconds on a SparcStation II,
but I haven't looked into how much longer it would take on the 10x10.

"Dale I. Newfield" <dn1l+@andrew.cmu.edu> writes:
> Could you explain what your algorithm was?
> Has anyone found a non-brute-force solution scheme?

My solution was brute-force.  I posted to cube-lovers (again, in April '92)
asking if anyone had found a more logical approach to the puzzle, but got no
affirmative responses.

Dale's method is a little inefficient in the order in which it tries tiles.
Mine used the sequence:			Dale's used:

	 1   3   5   7   9		 1   2   4   7  11

	 2   4   6   8  10		 3   5   8  12  16
	
	11  12  13  14  15		 6   9  13  17  20
	
	16  17  18  19  20		10  14  18  21  23
	
	21  22  23  24  25		15  19  22  24  25

The first three tiles in our two methods are equally constrained, but the
next seven in Dale's methods are constrained along 1-2-1-1-2-2-1 edges,
while mine are constrained along 2-1-2-1-2-1-2 edges.  So I suspect my
search tree gets trimmed a bit more quickly.  Another way in which the
search can be made more efficient is in finding the pieces to try in each
position.  For each pair of colors that can appear along an edge, my program
precomputes a table listing all tiles that can match that pair of colors,
including how to rotate the tiles.

	-- Don.



From @mail.uunet.ca:mark.longridge@canrem.com  Wed Dec 15 11:07:13 1993
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Sender: CRSO.Cube@canrem.com
Subject: Lib of Congress
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.618.5834.0C18E5EA@canrem.com>
Date: 	Wed, 15 Dec 1993 09:38:00 -0500
Organization: CRS Online  (Toronto, Ontario)

I Pulled this list from the Library of Congress. The hungarian
book I've never seen before (#2), and #3 and #4 are new to me.
"Zen of Cubing" may be interesting, has anyone read this one?


ITEMS 1-4 OF 11               SET 15: BRIEF DISPLAY           FILE: LOCI
                                (DESCENDING ORDER)


1. 86-8699: Buvos kocka. English.  Rubik's cubic compendium /  Oxford
     ; New York : Oxford University Press, 1987.  xi, 225 p. : ill.
     (some col.) ; 23 cm.
     LC CALL NUMBER: QA491 .B8813 1987
2. 85-109601: Mezei, Andras.  Magyar kocka, avagy, Meg mindig ilyen
     gazdagok vagyunk? /  Budapest : Magveto, c1984.  473 p. : ill.
     ; 21 cm.
    LC CALL NUMBER: QA491 .M49 1984
3. 82-72610: Feder, Happy Jack.  Zen of cubing : in search of the
     seventh side /  1st ed.  South Bend, Ind. : And Books, c1982.
     100 p. : ill. ; 21 cm.
     LC CALL NUMBER: PN6231.R78 F42 1982
4. 82-3755: O'Grady, Miles.  You can kick the cube] : the cube hater's
     handbook /  New York, N.Y. : Penguin Books, 1982.  p. cm.
     NOT IN LC COLLECTION
5. 82-1264: Bandelow, Christoph.  Inside Rubik's cube and beyond /
     Boston : Birkhauser, c1982.   120, [5] p., [6] leaves of plates :
     ill. (some col.) ; 23 cm.
     LC CALL NUMBER: QA491 .B2613 1982
6. 81-85850: Schlafly, Roger.  The complete cube book /  Chicago :
     Regnery Gateway, c1982.  vi, 51 p. : ill. ; 21 cm.
     LC CALL NUMBER: QA491 .S34 1982
7. 81-81556: Taylor, Don.  Mastering Rubik's cube : the solution to the
     20th century's most amazing puzzle /  1st American ed.  New York :
     Holt, Rinehart and Winston, 1981, c1980.  31 p. : ill. ; 22 cm.
     LC CALL NUMBER: QA491 .T38 1981
8. 81-21650: Varasano, Jeffrey, 1966-  Conquer the cube in 45 seconds /
     New York : Bell Pub. Co. : Distributed by Crown Publishers, c1981.
     48 p. : ill. ; 21 cm.
     NOT IN LC COLLECTION
9. 81-16682: Varasano, Jeffrey, 1966-  Jeff conquers the cube in 45
     seconds : and you can too] /  New York : Stein and Day, 1981. p.cm.
     NOT IN LC COLLECTION
10. 81-12525: Frey, Alexander H.  Handbook of cubik math /  Hillside,
     N.J. : Enslow Publishers, c1982.  viii, 193 p. : ill. ; 23 cm.
     LC CALL NUMBER: QA491 .F73 1982
11. 80-27751: Singmaster, David.  Notes on Rubik's magic cube /
     Hillside, N.J. : Enslow Publishers, 1981.  vi, 73 p. : ill.; 24 cm.
     LC CALL NUMBER: QA491 .S58 1981

More to follow
 -> Mark <-

From @mail.uunet.ca:mark.longridge@canrem.com  Wed Dec 15 12:23:33 1993
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Subject: 6 X order 3
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.619.5834.0C18E5EB@canrem.com>
Date: 	Wed, 15 Dec 1993 09:42:00 -0500
Organization: CRS Online  (Toronto, Ontario)

A few messages back I mentioned a cyclicly decomposable process for
the pattern 6 X order 3. Success! Those familar with Christoph
Bandelow's "Inside Rubik's Cube and Beyond" will recognize the
notation, but for those who don't:

Mr is the middle slice adjacent to face R
Mu is the middle slice adjacent to face U (or T)
Mf is the middle slice adjacent to face F

Thus Mr1 rotates the middle slice in the same direction as r1,
etc. ...fairly intuitive.

The 28 slice moves are rather lengthy, but one can follow the
progression to 6 X order 3 easily. Before the discovery of
process p1b, memorization and execution of this pattern was
difficult. By memorization I don't mean retention for days or
weeks or even months as I wanted a CD-type process with which I
could always reconstruct it in my head.

Perhaps this could be improved upon, nevertheless now
the checkerboard order 3 is easy to execute and easy to remember!

(rotates edges 120 degrees around the FTR corner and BDL corner)
p1b  alternate method 2 (Mr2 D3 Mr2 U1) ^3  TOP becomes LEFT     (28s)
                        (Mr2 D3 Mr2 U1) ^3  LEFT becomes TOP
                         Mr3 Mt1 Mr1 Mt3

From @mail.uunet.ca:mark.longridge@canrem.com  Wed Dec 15 13:04:34 1993
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Sender: CRSO.Cube@canrem.com
Subject: Part 2
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.620.5834.0C18E5F0@canrem.com>
Date: 	Wed, 15 Dec 1993 09:58:00 -0500
Organization: CRS Online  (Toronto, Ontario)

Hmmmm, a little inconsistent (sp?) with the notation there.
I usually only use U when quoting the processes of others.

I'm going to try tackling the 1152-fold symmetry idea
later tonight.

When looking for CD-type processes I find it helps to think
in terms of distinct states you pass through in approaching
the goal state. Sort of like factoring a composite pattern
into simpler ones you add together.

Rotating the cube in space definitely helps too.

-> Mark

From dn1l+@andrew.cmu.edu  Wed Dec 15 13:17:11 1993
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Date: Wed, 15 Dec 1993 13:15:45 -0500 (EST)
From: "Dale I. Newfield" <dn1l+@andrew.cmu.edu>
To: cube-lovers@ai.mit.edu
Subject: Re: Description of Tangle, Part 2
Cc: don.woods@eng.sun.com, acw@riverside.scrc.symbolics.com
In-Reply-To: <920425084746.2bc000e4@EQL.Caltech.Edu>

Just to make sure everyone knows what we are talking about, here is a
message from the archives:
Excerpts from mail: 25-Apr-92 Description of Tangle, Part 2 by Chris
Worrell@eql.caltec 
> Annotating Don.Woods diagram  (which is in the correct orientation)
>               2       3
>         ---------------------
>         |     @       #     |
>         |     @       #     |
>       1 |$$    @      # %%%%| 4
>         |  $    @    %#%    |
>         |   $    @ %% #     |
>         |    $    %@  #     |
>         |    $  %%  @@#     |
>         |    %%%      #@@   |
>       4 |%%%% $       #  @@@| 2
>         |     $       #     |
>         |     $       #     |
>         ---------------------
>               1        3
>  
> The duplicate piece in each tangle is:
>                 1       2       3       4
> Tangle 1        Blue    Red     Yellow  Green
> Tangle 2        Yellow  Blue    Green   Red
> Tangle 3        Green   Yellow  Blue    Red
> Tangle 4        Red     Green   Yellow  Blue
>  
> All 4 Tangles are the same puzzle, just colored differently.
> Each has all 24 color permutations, plus a duplicate.

I had kind of hoped that the connectivity on the different puzzles was
different, instead of just the colors.

(Actually, the sequence I sent before was slightly wrong--here is the
one I actually used. Using Don's format)
>Don used the sequence:                 Dale used:
> 
>         1   3   5   7   9               1   2   6  10  15
>         2   4   6   8  10               3   4   7  11  16
>        11  12  13  14  15               5   8  12  17  20
>        16  17  18  19  20               9  13  18  21  23
>        21  22  23  24  25              14  19  22  24  25

But yes, Don's fillpattern still gets more constraints in earlier--here
is the number of constraints at each step
Don's: 0 1 1 2 1 2 1 2 1 2 1 2 2 2 2 1 2 2 2 2 1 2 2 2 2
Mine:  0 1 1 2 1 1 2 2 1 1 2 2 2 1 1 2 2 2 2 2 2 2 2 2 2
As you can see, I had my 1's clustered more toward the beginning, which
is non-optimal.

Assuming that there is only a change in color(and not in connectivity),
as was posted by Chris in april of 92, I would think modifying code to
attempt the 10x10 would be fairly simple...(seeing as my code went poof
sometime last year, when a disk crashed(not that it was
complicated))...wanna try?

(Thanks for the pointers to the Apr 92 discussion)
I agree with the concensus expressed in the archives that this puzzle is
inherently "not that great" because no non-brute-force method has been
found/seems to exist.

-Dale


From anandrao@hk.super.net  Wed Dec 15 20:14:26 1993
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Date: Thu, 16 Dec 1993 09:09:21 +0800 (HKT)
From: Mr Anand Rao <anandrao@hk.super.net>
Subject: Re: Tangle
To: Don Woods <Don.Woods@eng.sun.com>
Cc: Cube-Lovers@ai.mit.edu
In-Reply-To: <9312142248.AA22891@colossal.Eng.Sun.COM>
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My method is essentially the same as yours - I have several intermediate
tables which cut down the processing required within the innermost loop. I
tried the same for 10*10 but it was taking eons. I tried making tables of
2*2 arrangements and solving for 5*5 of these(thereby solving the original
10*10, but the number of 2*2 arrangements makes the problem intractable on
my measly little computer. I even trie putting together all possible 5*5
solutions and assembling them in the 10*10 pattern, but the number of 5*5
solutions with the 100 tiles is in millions!
Do you have any further insight?


On Tue, 14 Dec 1993, Don Woods wrote:

> Anand Rao <anandrao@hk.super.net> writes:
> > The puzzles are similar, except that the extra(25th)
> > piece is different in each. The solutions for each puzzle are very
> > different and I could not see any pattern. 
> 
> Look again.  The puzzles are identical except for a remapping of the colors.
> For example, if you take Tangle #1 and paint all the Blue ropes Yellow, all
> the Red ropes Blue, all the Green ropes Red, and all the (originally) Yellow
> ropes Green, you'll have Tangle #2.  So you can solve Tangle #1 by imagining
> the ropes recolored as above, constructing your solution for #2, and then
> restoring the original colors.
> 
> Note: The particular recoloring given above is based on colors given in a
> message sent by CCW@eql.caltech.edu (Chris Worrell) to cube-lovers on April
> 27, 1992.  I own only #1 myself and so cannot confirm or deny the accuracy
> of the colors.  But the basic idea applies, given that each puzzle (a) has
> the same pattern of ropes on all pieces and (b) has each permutation of
> colors exactly once except for one permutation which appears twice.
> 
> Solving the 10x10 is another kettle of fish, and I haven't tried it.  I do
> have a program that solves the 5x5 in about 45 seconds on a SparcStation II,
> but I haven't looked into how much longer it would take on the 10x10.
> 
> "Dale I. Newfield" <dn1l+@andrew.cmu.edu> writes:
> > Could you explain what your algorithm was?
> > Has anyone found a non-brute-force solution scheme?
> 
> My solution was brute-force.  I posted to cube-lovers (again, in April '92)
> asking if anyone had found a more logical approach to the puzzle, but got no
> affirmative responses.
> 
> Dale's method is a little inefficient in the order in which it tries tiles.
> Mine used the sequence:			Dale's used:
> 
> 	 1   3   5   7   9		 1   2   4   7  11
> 
> 	 2   4   6   8  10		 3   5   8  12  16
> 	
> 	11  12  13  14  15		 6   9  13  17  20
> 	
> 	16  17  18  19  20		10  14  18  21  23
> 	
> 	21  22  23  24  25		15  19  22  24  25
> 
> The first three tiles in our two methods are equally constrained, but the
> next seven in Dale's methods are constrained along 1-2-1-1-2-2-1 edges,
> while mine are constrained along 2-1-2-1-2-1-2 edges.  So I suspect my
> search tree gets trimmed a bit more quickly.  Another way in which the
> search can be made more efficient is in finding the pieces to try in each
> position.  For each pair of colors that can appear along an edge, my program
> precomputes a table listing all tiles that can match that pair of colors,
> including how to rotate the tiles.
> 
> 	-- Don.
> 
> 




From anandrao@hk.super.net  Wed Dec 15 20:27:19 1993
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Date: Thu, 16 Dec 1993 09:17:27 +0800 (HKT)
From: Mr Anand Rao <anandrao@hk.super.net>
Subject: Re: Description of Tangle, Part 2
To: "Dale I. Newfield" <dn1l+@andrew.cmu.edu>
Cc: cube-lovers@ai.mit.edu, don.woods@eng.sun.com,
        acw@riverside.scrc.symbolics.com
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Just because no non-brute-force method has been found, does not make this
puzzle any less intersting. As we have been told that there is a
solution, it is exciting to search for one, even by brute force methods.
The real challenge is to find a brute-force method with sufficient insight
to solve the problem within a reasonable time-frame. All the algorithms so
far are exponential. We may never find a linear algorithm for this
problem. The idea is to find one algorithm that can be used in actual
practice. We can then bury this puzzle into the archives, for the next
generation to pick up!
  >  > (Thanks for the pointers to the Apr 92
discussion)
> I agree with the concensus expressed in the archives that this puzzle is
> inherently "not that great" because no non-brute-force method has been
> found/seems to exist.
> 
> -Dale
> 
Is this the reason why Rubik has gone into hiding? I haven't seen any
puzzle from him after this set of 4 released in 1990/1991. I tried to
contact the Hong Kong office of Matchbox which gets Rubik's puzzles in
China, but they have closed shop. Matchbox UK said that they have
discontinued this line. If anyone has found another source for Rubik's
puzzles, or discovered anyone else who has taken the responsibility of
giving us sleepless nights, please let me know!



From Don.Woods@eng.sun.com  Wed Dec 15 20:39:18 1993
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From: Don.Woods@eng.sun.com (Don Woods)
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> Is this the reason why Rubik has gone into hiding? I haven't seen any
> puzzle from him after this set of 4 released in 1990/1991.

Hm, didn't "Square-1" come out later than the Tangles?

Regarding solving the Tangle, I forgot one other minor optimisation: When
my program is picking a corner piece other than the first, it requires that
the piece "number" be less than or equal to that of the first corner.  I.e.,
it refuses to search for solutions that are rotations of other solutions.

I've modified my program to try the 10x10, but indeed, it's taking a long
time.  (Current estimate is it will take over a year to finish.)  I suspect
that fact that pieces aren't "used up" as fast -- i.e., since there's at
least four of any given piece, there will usually be at least one of whatever
you're looking for for quite a ways down the search tree -- makes this
approach intractible.

	-- Don.


From dik@cwi.nl  Wed Dec 15 21:03:56 1993
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Date: Thu, 16 Dec 93 03:03:19 +0100
From: Dik.Winter@cwi.nl
Message-Id: <9312160203.AA10975.dik@boring.cwi.nl>
To: anandrao@hk.super.net
Subject: Re: Description of Tangle, Part 2
Cc: cube-lovers@ai.mit.edu

My memory may be extremely faulty of course, but was there not more than
one single solution for the 5x5?  (Not unprecedented, I have one puzzle
that promises a single solution but there are hundreds.)  And, is there
a solution for the 10x10?  I seem to remember that there was (or I had)
a convincing argument that such a thing did not exist.  I should go through
the lesser used parts of my memory one of these days.
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
home: bovenover 215, 1025 jn  amsterdam, nederland; e-mail: dik@cwi.nl

From dik@cwi.nl  Wed Dec 15 21:40:59 1993
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From: Dik.Winter@cwi.nl
Message-Id: <9312160240.AA11474.dik@boring.cwi.nl>
To: Don.Woods@eng.sun.com
Subject: Re: Description of Tangle, Part 2
Cc: cube-lovers@ai.mit.edu

 > > Is this the reason why Rubik has gone into hiding? I haven't seen any
 > > puzzle from him after this set of 4 released in 1990/1991.
 > 
 > Hm, didn't "Square-1" come out later than the Tangles?

Square-1 is not by Rubik.  But he came this year with two new puzzles (at
least, they are in his name).  Rubik's Maze and Rubik's Hat.

In the first there are 6 connected cubes with a black/yellow pattern on
them.  The cubes can turn around each other fairly freely.  The purpose
is to get a 1x2x3 where there is a single black continuous line along
the cubes.  Not very difficult, interesting.

Rubik's Hat is in the form of a hat with six rings on it.  You can look
trough it (and through the rings by implication).  By turning rings you
see more or less rabbits.  The purpose is to see a rabbit in every position.
I think the puzzle is based on light polarization, with different
polarizations coming through the segments of the rings.
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
home: bovenover 215, 1025 jn  amsterdam, nederland; e-mail: dik@cwi.nl

From anandrao@hk.super.net  Thu Dec 16 01:12:12 1993
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Date: Thu, 16 Dec 1993 14:08:39 +0800 (HKT)
From: Mr Anand Rao <anandrao@hk.super.net>
Subject: Re: Description of Tangle, Part 2
To: Dik.Winter@cwi.nl
Cc: cube-lovers@ai.mit.edu
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On Thu, 16 Dec 1993 Dik.Winter@cwi.nl wrote:

> My memory may be extremely faulty of course, but was there not more than
> one single solution for the 5x5?  (Not unprecedented, I have one puzzle
> that promises a single solution but there are hundreds.)  And, is there
> a solution for the 10x10?  I seem to remember that there was (or I had)
> a convincing argument that such a thing did not exist.  I should go through
> the lesser used parts of my memory one of these days.
 For each Tangle, there are 2 solutions and no more. I have searched the
tree thoroughly and verified this. Counting 4 rotations and that 2 pieces
are identical, the total search gives 16 'solutions'.
The colourful little pamphlet that comes with the puzzle says that there
IS a solution to the 10*10 puzzle.



From anandrao@hk.super.net  Thu Dec 16 01:19:38 1993
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Date: Thu, 16 Dec 1993 14:12:56 +0800 (HKT)
From: Mr Anand Rao <anandrao@hk.super.net>
Subject: Re: Description of Tangle, Part 2
To: Don Woods <Don.Woods@eng.sun.com>
Cc: cube-lovers@ai.mit.edu
In-Reply-To: <9312160139.AA26306@colossal.Eng.Sun.COM>
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> 
> I've modified my program to try the 10x10, but indeed, it's taking a long
> time.  (Current estimate is it will take over a year to finish.)  I suspect
> that fact that pieces aren't "used up" as fast -- i.e., since there's at
> least four of any given piece, there will usually be at least one of whatever
> you're looking for for quite a ways down the search tree -- makes this
> approach intractible.
> 
 True. The 5*5 puzzle search truncates much faster because you run out of
pieces that could fit into a specific slot. The same does not apply to the
10*10 one.
Has anyone tried to solve the 10*10 for just 1 colour. That leaves you
with only 4 tile types with 24,25 or 26 of each type. The solution may
give some indication of the resultant pattern of the selected colour. If
there aren't too many solutions, maybe we can build the 4 colour solution
from this my permuting and rotating thr tiles. Any idea on how this will work?



From andyl@harlequin.com  Thu Dec 16 10:45:09 1993
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From: Andy Latto <andyl@harlequin.com>
Date: Thu, 16 Dec 1993 10:38:45 -0500
Message-Id: <21332.199312161538@phaedrus.harlequin.com>
To: Don.Woods@eng.sun.com
Cc: cube-lovers@ai.mit.edu
In-Reply-To: Don Woods's message of Wed, 15 Dec 93 17:39:20 PST <9312160139.AA26306@colossal.Eng.Sun.COM>
Subject: Description of Tangle, Part 2

   Date: Wed, 15 Dec 93 17:39:20 PST
   From: Don.Woods@eng.sun.com (Don Woods)
   X-Sun-Charset: US-ASCII
   Content-Length: 897

   > Is this the reason why Rubik has gone into hiding? I haven't seen any
   > puzzle from him after this set of 4 released in 1990/1991.

   Hm, didn't "Square-1" come out later than the Tangles?

Did Rubik have anything to do with Square-1?

In any case, it's a great puzzle, and I recommend it to anyone on the
list who hasn't tried it. While there's a group structure lurking here
as usual, this is the only puzzle I've seen where the set of
attainable positions is not a subgroup. This means lots of the usual
ways of thinking about puzzles like this (e.g. conjugation) don't
always work, which makes it quite challenging.

					Andy Latto
					andyl@harlequin.com

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Date:      Thu, 16 Dec 1993 15:36:58 EST
From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Duality of Operators and Operatees

I have mentioned several times my discomfort about "an operator"
as opposed to "the thing being operated on" when it comes to
groups.  I am never quite sure just which of the two it is
that people are talking about, even (or especially) when I am listening
to myself talk.  There is clearly an essential duality between the
two, but I am not sure I have quite a strong enough group theory
background to fully understand it.  I am very comfortable when the
operators form a group, but I am not very comfortable when the things
being operated on form a group.

I am presently rereading (hopefully VERY SLOWLY
AND VERY CAREFULLY) Dan Hoey and Jim Saxe's seminal paper from
December 1980 entitled Symmetry and Local Maxima.  Here is a quote
from their paper.

   We will sometimes (particularly towards the end of this message)
   take the liberty of identifying a transformation with the position
   reached by applying that transformation to SOLVED.

Well, I am beginning to think that the source of my discomfiture
is simply that everybody does the same thing all the time, and that
nobody ever makes the identification explicit.  However, I think that
maybe the duality is there, whether the identification is explicit,
implicit, or not made at all.  Let me see if I can make clear what I
mean with some non-cubing programming examples.

When I first started computer cubing, I was struck by the fact that
(at least with my model of the cube), the computer code to implement
a permutation operation looked exactly like the computer code to
translate between various character codes.  For example, I have had
frequent occasion to translate between ASCII and EBCDIC (in both
directions).  The code to translate between the ASCII string X and
the EBCDIC string Y is something like

    for i = 1 to n Y(i) = T(X(i))

where T is the translate table.  To make this clear by an example,
the ASCII code for the letter A is  decimal 33 and the EBCDIC code
for the letter A is decimal 193.  Hence, the 33-rd position of T
contains decimal 193, and the 193-rd position of T' contains 33.

Beyond this simple little loop above, many (if not most) programming
languages have a function (often called TRANSLATE or TRANSFORM) which
does exactly the same thing.  There are also hardware architectures
which implement the TRANSLATE in hardware.  For example, you might
have something like

   Y = TRANSLATE(X,T)

where X is the string to be translated and T is the translate table.

X and T are clearly not interchangeable as input to the TRANSLATE
function.  However, (and repeating myself) I think there is an
essential duality between X and T.  For example, consider what
would happen if you reversed the role of X and T as follows.  Let X
be the hexadecimal string 010201020301020403.  Then,
Y = TRANSLATE(X,' ABC') would yield the string ' A AB ACB'.  Such
a role reversal for the "permutation operator" and "permutation operatee"
can be a very powerful programming technique.  For example, I have
used it to redistribute data, creating well-formatted print lines or
well-formatted display screens (text mode) with one fell swoop (with
only a single invocation of the TRANSLATE function).

I am going to continue reading, but perhaps I could pose a question to
Dan Hoey anyway:  is reversing the role of X and T in the TRANSLATE
function above essentially the same thing as switching between
pre-multiplication and post-multiplication?

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From anandrao@hk.super.net  Thu Dec 16 20:14:55 1993
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Date: Fri, 17 Dec 1993 09:13:20 +0800 (HKT)
From: Mr Anand Rao <anandrao@hk.super.net>
Subject: Re: Description of Tangle, Part 2
To: Dik.Winter@cwi.nl
Cc: Don.Woods@eng.sun.com, cube-lovers@ai.mit.edu
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On Thu, 16 Dec 1993 Dik.Winter@cwi.nl wrote:

> Square-1 is not by Rubik.  But he came this year with two new puzzles (at
> least, they are in his name).  Rubik's Maze and Rubik's Hat.
> 
> In the first there are 6 connected cubes with a black/yellow pattern on
> them.  The cubes can turn around each other fairly freely.  The purpose
> is to get a 1x2x3 where there is a single black continuous line along
> the cubes.  Not very difficult, interesting.
> 
> Rubik's Hat is in the form of a hat with six rings on it.  You can look
> trough it (and through the rings by implication).  By turning rings you
> see more or less rabbits.  The purpose is to see a rabbit in every position.
> I think the puzzle is based on light polarization, with different
> polarizations coming through the segments of the rings.
> dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
> home: bovenover 215, 1025 jn  amsterdam, nederland; e-mail: dik@cwi.nl

Where can we get these puzzles from? Do you know of anyone who can take
credit card orders and mail?



From dik@cwi.nl  Thu Dec 16 20:22:01 1993
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Date: Fri, 17 Dec 93 02:21:54 +0100
From: Dik.Winter@cwi.nl
Message-Id: <9312170121.AA15294.dik@boring.cwi.nl>
To: anandrao@hk.super.net
Subject: Re: Description of Tangle, Part 2
Cc: Don.Woods@eng.sun.com, cube-lovers@ai.mit.edu

I would not know sources for Rubik's Maze and Rubik's Hat.  They are on
sale in the local shops here.  I have looked, the distributer is no
longer Matchbox but Parker, so that would imply availability in the
US I think.

dik
--
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
home: bovenover 215, 1025 jn  amsterdam, nederland; e-mail: dik@cwi.nl

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From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Some Additional Distances in the Edge Group

It is now known that using the qturn metric, Start has a
unique antipode in the edge group, namely Mirror-Image-
of-Edges-Flipped.  The antipode is 15 qturns from Start.
Also, I have a complete data base of equivalence classes
in the edge group documenting the distance from Start for
each configuration of the edges.

It seems to me that given these two facts, some additional
distances can be determined.  For example, it is possible
to determine the distance from any configuration to
Mirror-Image-of-Edges-Flipped.  Let Z be a sequence of operators
that converts Start to Mirror-Image-of-Edges-Flipped, and let
A be any configuration of the edges.  Then apply Z' to A, look
up the result in the data base of distances from Start, and
that will be the distance from A to Mirror-Image-of-Edges-Flipped.

The reason is quite simple.  Let P be a sequence which takes
Z'(A) to Start.  Then, Z'PZ takes A to Mirror-Image-of-Edges-Flipped.
This is a very nice use of conjugates.

Another consequence of this result is the following:  suppose you
began with Mirror-Image-of-Edges-Flipped and performed a
breadth-first exhaustive search.  Start would be antipodal, and
the number of nodes at each level of the tree would be identical
to the existing tree which begins at Start.

In addition, all of the above applies to Mirror-Image-of-Start
and Edges-Flipped with respect to each other.  They are
mutually antipodal, and are 15 qturns apart.  A tree built with
either at the root would have exactly the same number of nodes
at each level as the existing tree with Start at the root.

Finally, the distance of any configuration from Mirror-Image-of-Start
or Edges-Flipped can be determined.  Let Y be a sequence of operators
which converts Start to Mirror-Image-of-Start, and let X be a sequence
of operators that converts Start to Edges-Flipped.  Let A be any
cube.  Then, the distance of A from Mirror-Image-of-Start
is the same as the distance of Y'(A) from Start, and
the distance of A from Edges-Flipped is the same
as the distance of X'(A) from Start.

I have the sensation in describing this that the Edge group is
square, with Start and Mirror-Image-of-Edges-Flipped 180 degrees
apart, and Mirror-Image-of-Start and Edges-Flipped at the other
two corners of the square.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU  Fri Dec 17 11:24:53 1993
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Date:      Fri, 17 Dec 1993 11:21:51 EST
From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Size of the Cube Group

In 1984, Dan Hoey posed a question as follows:

>This discussion of symmetry recalls a question I have meant to propose
>to Cube-Lovers for some time:  How many positions are there in Rubik's
>Cube?  We know from Ideal that the number is somewhat over three
>billion.  Most cube lovers will tell you a number of about 43
>quintillion.  But I really don't see why we should count twelve
>distinct positions at one quarter-twist from solved--all twelve are
>essentially the same position.  So the question, suitably rephrased, is
>of the number of positions that are distinct up to conjugacy in M, the
>48-element symmetry group of the cube.  I think this is an interesting
>question, but I don't see any particularly easy way of answering it.
>My best guess is that it involves a case-by-case analysis of the 98
>subgroups of M, or at least the 33 conjugacy classes of those
>subgroups.  In ``Symmetry and Local Maxima'', Jim Saxe and I examined
>five of the classes, which we called M, C, AM, H, and T.
>
>Even finding the numbers for the pocket cube is a little tricky.  If we
>limit ourselves to symmetry in S, I believe the pocket cube has 2
>positions with a six-element symmetry group, 160 positions with a
>three-element symmetry group, 3882 positions with a two-element
>symmetry group, and 3670116 positions with a one-element symmetry
>group, for 613062 positions distinct up to S-conjugacy.  But the
>numbers for M-conjugacy are still elusive; I am not even sure how to
>deal with factoring out whole-cube moves in the analysis.  I hope to
>find time to write a program for it.
>
>I expanded my pocket cube program to deal with the corner group of
>Rubik's cube.  This group is 24 times as large as the group of the
>pocket cube, having 3^7 * 8! = 88179840 elements.  The number of
>elements P(N) and local maxima L(N) at each (quarter-twist) distance N
>from solved are given below.
>
>                 N         P(N)        L(N)
>                 0            1           0
>                 1           12           0
>                 2          114           0
>                 3          924           0
>                 4         6539           0
>                 5        39528           0
>                 6       199926         114
>                 7       806136         600
>                 8      2761740       17916
>                 9      8656152       10200
>                10     22334112       35040
>                11     32420448      818112
>                12     18780864     9654240
>                13      2166720     2127264
>                14         6624        6624
>
>The alert reader will notice that rows 10 through 14 contain values
>exactly 24 times as large as those for the pocket cube.  This is not
>surprising, given that the groups are identical except for the position
>of the entire assembly in space, and each generator of the corner cube
>is identical to the inverse of the corresponding generator for the
>opposite face except for the whole-cube position.  Thus when solving a
>corner-cube position at 10 qtw or more from solved, it can be solved as
>a pocket cube, making the choice between opposite faces in such a way
>that the whole-cube position comes out right with no extra moves.
>

I wish to propose an answer to Dan's question.  I will propose an
approximation then (hopefully) the exact answer.

The approximation is simply 4.3*(10^19) / 1152, or about
3.7*(10^16).  1152=24*24*2, and is based on my version of Dan's
M symmetry group.  I remain convinced that my version of M is
isomorphic to Dan's, but the subject deserves some more thought
and discussion.

But we can do better.  We already know (under my version of M) how
many equivalence classes there are for the corner group (namely,
77,802).  But each of the equivalence classes for the corners can
be rotated 24 ways with respect to the centers, so we have
77,802*24.  We also already know (under my version of M) how many
equivalence classes there are for the edge group (namely
851,625,008).  But each of the equivalence classes for the edges
can be rotated 24 ways with respect to the centers, so we have
851,625,008*24.  Hence, we have

   (77,802*24) * (851,625,008*24) = 38,164,682,230,511,620

This figure is gratifyingly close to 3.7*(10^16), and I believe it
is the correct answer to Dan's question.  It is slightly larger
than the approximation because some of the equivalence classes
have fewer than 1152 elements, and  consequently there are a few
more equivalence classes than the approximation suggests.

However, the alert reader should have noticed a problem.  Why did I
not divide by 2 to take into account the fact that odd edge
permutations can only occur with odd corner permutations and vice
versa?  Actually, I did, but the division by 2 cancelled.  The reason
it canceled is slightly tricky.  Also, remember that we are talking
about equivalence classes, not specific cube configurations.  Any
equivalence class has both even and odd members, depending on how
the members are rotated.  Hence, any corner equivalence class can be
matched up with any edge equivalence class, assuming the rotations
are compatible.  But you still have to worry about "dividing by 2",
as follows.

Let G be the number of states of the whole cube without M, namely
the 4.3*(10^19) figure, and similarly let C be the number of states of
the corners without M and let E be the number of states of the edges
without M.  Then, we have the trivial relation G = C * E / 2.
Here, the division by 2 does properly reflect the odd/even parity
of the corners vs. the edges.

Let Gm = G / (24*24*2), Cm = C / (24*24*2), and Em = E / (24*24*2).
Hence, G = Gm * (24*24*2), C = Cm * (24*24*2), and E = Em * (24*24*2).
What I have available (approximately) is Cm and Em, and what I want
is Gm.  Hence,

     Gm = G / (24*24*2)
     Gm = (C * E / 2) / (24*24*2)
     Gm = ((Cm * (24*24*2)) * (Em * (24*24*2)) / 2) / (24*24*2)
     Gm = (Cm*24) * (Em*24)

Therefore, I replace Cm by the real figure for the number of corner
equivalence classes, replace Em by the real figure for the number
of equivalence classes, and Gm becomes the real figure for the total
states of the cube.  The "division by 2" is in the formula, but it is
invisible because of all the cancellations.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU  Fri Dec 17 14:25:29 1993
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Date:      Fri, 17 Dec 1993 14:22:34 EST
From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Re: Size of the Cube Group
In-Reply-To: Message of 12/17/93 at 11:21:51 from ,
           BRYAN%WVNVM.BITNET@mitvma.mit.edu

On 12/17/93 at 11:21:51 Jerry Bryan said:

>However, the alert reader should have noticed a problem.  Why did I
>not divide by 2 to take into account the fact that odd edge
>permutations can only occur with odd corner permutations and vice
>versa?  Actually, I did, but the division by 2 cancelled.  The reason
>it canceled is slightly tricky.  Also, remember that we are talking
>about equivalence classes, not specific cube configurations.  Any
>equivalence class has both even and odd members, depending on how
                       ^^^^^^^^^^^^^^^^^^^^^^^^^

>the members are rotated.  Hence, any corner equivalence class can be
>matched up with any edge equivalence class, assuming the rotations
>are compatible.  But you still have to worry about "dividing by 2",
>as follows.

It is pretty bad when you have to followup with
corrections to your own posts.  I hurried to complete the previous
post before lunch, and just didn't think clearly enough  --  till I
had time to think *during* lunch.  Let's try this again.

A qturn of the whole cube (a 90 degree rotation of the whole cube)
is odd.  However, if you think of a qturn rotation of the whole cube
as disjoint between edges and corners, a qturn rotation of the
corners is even, and a qturn rotation of the edges is odd.  Hence,
for any equivalence class of the corners under M, either the whole
equivalence class is even, or the whole equivalence class is odd.
For any equivalence class of the edges under M, half of the equivalence
class is even and half is odd.  Thus, any equivalence class of the
corners can occur with any equivalence class of the edges, but with only
half the members of the edge equivalence class  --  namely those with
the same parity.

I believe my calculations were correct, but a piece of the justification
was not.  I hope I am not still missing something.  You do have to
"divide by 2", and my calculations do indeed "divide by 2" as previously
described, but the parity of edges vs. the parity of corners was
incorrect in the previous post.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU  Sat Dec 18 17:08:38 1993
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From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Second Addendum - Size of Cube Group under M

I feel like I am pestering the list to death with corrections.

I still believe that the figure that I proposed for the size of
the cube group under M is correct.  The first post included
a "correct" but I think unsatisfactory explanation.  The second post
improved upon one point that was unsatisfactory in the first post.
Now, let's see if I can get it completely correct.

The size of the corner group under (my version of) M is known.
The size of the edge group under (my version of) M is known as
well.  Let C be the size of the corner group, and E be the size
of the edge group.  Remember, the elements of the groups are
equivalence classes induced by (my version of) M.  Here is an incorrect
formula for G, the size of the entire cube group under (my
version of) M.

   G = (C*24) * (E*24) / 2

The division by 2 is introduced to account for parity between the
corner group and the edge group.  But the value for G produced by
this formula is only half as big as it should be.  The problem is
that M induces equivalence classes based on both rotations and
reflections, not just base on rotations.  Hence, we are led to the
following (still incorrect) formula:

   G = (C*24*2) * (E*24*2) / 2

As before, the division by 2 takes care of parity between the corner
group and the edge group.  In addition, the multiplication by 2 takes
care of reflecting each group.  But the value for G produced by this
formula is twice as big as it should be.  The problem is that while
any corner rotation can occur with any edge rotation (subject to
parity), you must either reflect both groups, or else reflect neither
group.  Thus, we have the following (correct) formula:

   G = ((C*24) * (E*24) / 2) * 2

The division by 2 takes care of parity between the groups, and the
multiplication by 2 takes care of reflection of the two groups
as a unit.  If we wish, we can cancel the multiplication and the
division to yield

   G = (C*24) * (E*24)

This is the same formula I originally posted, and I did say in the
original post that the division by 2 cancelled out.  However, I
think that this post provides a better explanation of the
cancellation than did the original post.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From xirion!jandr@relay.nl.net  Mon Dec 20 06:35:54 1993
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From: Jan de Ruiter <jandr@xirion.nl>
Date: Mon, 20 Dec 93 10:38:37 +0100
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X-Organization:  Xirion Unix Software & Consultancy bv
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To: cube-lovers@ai.mit.edu

To: cube-lovers@ai.mit.edu
Subject: Re: Search order of Tangle

I saw the discussion of Dale and Don about the search order
(fillpattern) for rubiks tangle come by, and wondered why they both
missed an even better search order (the best?):

Don:		 Dale:            Jan:              Equivalent to:
 1  3  5  7  9    1  2  6 10 15    1  2  5 10 17    17 16 15 14 13
 2  4  6  8 10    3  4  7 11 16    3  4  6 11 18    18  5  4  3 12
11 12 13 14 15    5  8 12 17 20    7  8  9 12 19    19  6  1  2 11
16 17 18 19 20    9 13 18 21 23   13 14 15 16 20    20  7  8  9 10
21 22 23 24 25   14 19 22 24 25   21 22 23 24 25    21 22 23 24 25

The number of constraints is illustrative:
don:  0 1 1 2 1 2 1 2 1 2 1 2 2 2 2 1 2 2 2 2 1 2 2 2 2
dale: 0 1 1 2 1 1 2 2 1 1 2 2 2 1 1 2 2 2 2 2 2 2 2 2 2
jan:  0 1 1 2 1 2 1 2 2 1 2 2 1 2 2 2 1 2 2 2 1 2 2 2 2

I disliked the irregularity in both don and dales search orders, and
in search for a more regular order, I found this one, which is better.
It is readily extendible to the 10 by 10 tangle.

- Jan D. de Ruiter

From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU  Mon Dec 20 06:43:01 1993
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From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Process for Antipodal of the Edge Group

This is the first process I have found for the antipodal of Start
in the edge group (edges without corners and without centers).
There are certainly many more, but I have not yet cataloged them
all.

  FR'LFL'B'R'FL'FRBL'BL'

Note that this process (as with any process for the antipodal) is
its own inverse.  Hence, you can use it once to get from Start to
the antipodal, and again to get from the antipodal to Start.  Also,
the "natural" inverse (namely, LB'LB'R'F'LF'RBLF'L'RF') is also
a process which will go in either direction, Start to the antipodal,
or antipodal to Start.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From xirion!jandr@relay.nl.net  Mon Dec 20 07:31:08 1993
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To: cube-lovers@life.ai.mit.edu

To: cube-lovers@life.ai.mit.edu
Subject: Re: Search order of Tangle

I saw the discussion of Dale and Don about the search order
(fillpattern) for rubiks tangle come by, and wondered why they both
missed an even better search order (the best?):

Don:		 Dale:            Jan:              Equivalent to:
 1  3  5  7  9    1  2  6 10 15    1  2  5 10 17    17 16 15 14 13
 2  4  6  8 10    3  4  7 11 16    3  4  6 11 18    18  5  4  3 12
11 12 13 14 15    5  8 12 17 20    7  8  9 12 19    19  6  1  2 11
16 17 18 19 20    9 13 18 21 23   13 14 15 16 20    20  7  8  9 10
21 22 23 24 25   14 19 22 24 25   21 22 23 24 25    21 22 23 24 25

The number of constraints is illustrative:
don:  0 1 1 2 1 2 1 2 1 2 1 2 2 2 2 1 2 2 2 2 1 2 2 2 2
dale: 0 1 1 2 1 1 2 2 1 1 2 2 2 1 1 2 2 2 2 2 2 2 2 2 2
jan:  0 1 1 2 1 2 1 2 2 1 2 2 1 2 2 2 1 2 2 2 1 2 2 2 2

I disliked the irregularity in both don and dales search orders, and
in search for a more regular order, I found this one, which is better.
It is readily extendible to the 10 by 10 tangle.

- Jan D. de Ruiter

From @cannon.ecf.toronto.edu:malone@ecf.toronto.edu  Mon Dec 20 14:17:33 1993
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From: MALONE  MATTHEW JAMES <malone@ecf.toronto.edu>
To: cube-lovers@ai.mit.edu
Subject: Please remove ...
Message-Id: <93Dec20.141719edt.7382@cannon.ecf.toronto.edu>
Date: 	Mon, 20 Dec 1993 14:17:10 -0500

Please remove malone@ecf.toronto.edu from the cube-lovers list.

Thanks

Matt

From Don.Woods@eng.sun.com  Mon Dec 20 19:21:46 1993
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From: Don.Woods@eng.sun.com (Don Woods)
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To: cube-lovers@ai.mit.edu
Subject: Re: Search order of Tangle
Cc: jandr@xirion.nl
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> I saw the discussion of Dale and Don about the search order
> (fillpattern) for rubiks tangle come by, and wondered why they both
> missed an even better search order (the best?):
> 
> Don:             Dale:            Jan:              Equivalent to:
>  1  3  5  7  9    1  2  6 10 15    1  2  5 10 17    17 16 15 14 13
>  2  4  6  8 10    3  4  7 11 16    3  4  6 11 18    18  5  4  3 12
> 11 12 13 14 15    5  8 12 17 20    7  8  9 12 19    19  6  1  2 11
> 16 17 18 19 20    9 13 18 21 23   13 14 15 16 20    20  7  8  9 10
> 21 22 23 24 25   14 19 22 24 25   21 22 23 24 25    21 22 23 24 25

I missed it on the 5x5 because my program was fast enough that I didn't
look further.  When I modified my program to try the 10x10 last week, I
did come up with the ordering Jan suggests.  It shaved about 1/3 the
running time off my 5x5 search, but it actually doesn't seem to make
that big a difference in the 10x10.

It turns out the 10x10 search isn't quite as bad as I thought, because
the tree does get trimmed rather early.  When a piece is constrained on
two edges, there are on average only 2/3 choices for that piece.  I've
got my program chugging along, and so far it has eliminated 4 of the 96
choices for piece (w/ orientation) for the upper left corner.  There are
4896 choices for the first 4 points in the search order, and it's going
through one choice per 25 minutes on average, so it'll finish in a mere
3 months, if I have the patience for it.  (I may try to dig up some
otherwise idle workstations to leave running over the holiday break.)

	-- Don.


From anandrao@hk.super.net  Mon Dec 20 20:16:04 1993
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Date: Tue, 21 Dec 1993 09:13:38 +0800 (HKT)
From: Mr Anand Rao <anandrao@hk.super.net>
Subject: Re: your mail
To: Jan de Ruiter <jandr@xirion.nl>
Cc: cube-lovers@ai.mit.edu
In-Reply-To: <3876.9312200938@xirion.xirion.nl>
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Your concept is theoretically extendable to the 10*10 tangle, but even
with this optimisation the puzzle would take a long time to solve. How
long do you take for the 5*5 Tangle on your computer?

On Mon, 20 Dec 1993, Jan de Ruiter wrote:

> To: cube-lovers@ai.mit.edu
> Subject: Re: Search order of Tangle
> 
> I saw the discussion of Dale and Don about the search order
> (fillpattern) for rubiks tangle come by, and wondered why they both
> missed an even better search order (the best?):
> 
> Don:		 Dale:            Jan:              Equivalent to:
>  1  3  5  7  9    1  2  6 10 15    1  2  5 10 17    17 16 15 14 13
>  2  4  6  8 10    3  4  7 11 16    3  4  6 11 18    18  5  4  3 12
> 11 12 13 14 15    5  8 12 17 20    7  8  9 12 19    19  6  1  2 11
> 16 17 18 19 20    9 13 18 21 23   13 14 15 16 20    20  7  8  9 10
> 21 22 23 24 25   14 19 22 24 25   21 22 23 24 25    21 22 23 24 25
> 
> The number of constraints is illustrative:
> don:  0 1 1 2 1 2 1 2 1 2 1 2 2 2 2 1 2 2 2 2 1 2 2 2 2
> dale: 0 1 1 2 1 1 2 2 1 1 2 2 2 1 1 2 2 2 2 2 2 2 2 2 2
> jan:  0 1 1 2 1 2 1 2 2 1 2 2 1 2 2 2 1 2 2 2 1 2 2 2 2
> 
> I disliked the irregularity in both don and dales search orders, and
> in search for a more regular order, I found this one, which is better.
> It is readily extendible to the 10 by 10 tangle.
> 
> - Jan D. de Ruiter




From pbeck@pica.army.mil  Tue Dec 21 00:24:43 1993
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Date:     Mon, 20 Dec 93 8:33:55 EST
From: Peter Beck (BATDD) <pbeck@pica.army.mil>
To: Cube-Lovers@ai.mit.edu
Cc: pbeck@pica.army.mil
Subject:  test
Message-Id:  <9312200833.aa09624@COR6.PICA.ARMY.MIL>


i am having trouble posting.

please excuse this message

From xirion!jandr@relay.nl.net  Tue Dec 21 02:45:28 1993
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From: Jan de Ruiter <jandr@xirion.nl>
Date: Tue, 21 Dec 93 08:43:59 +0100
Message-Id: <997.9312210743@xirion.xirion.nl>
X-Organization:  Xirion Unix Software & Consultancy bv
		 Burgemeester Verderlaan 15 X
		 3454 PE  De Meern
		 The Netherlands
X-Phone: 	 +31 3406 61990
X-Fax: 		 +31 3406 61981
To: anandrao@hk.super.net, cube-lovers@ai.mit.edu

Subject: Re: Rubiks tangle
To: anandrao@hk.super.net
Cc: cube-lovers@ai.mit.edu

>Your concept is theoretically extendable to the 10*10 tangle, but even
>with this optimisation the puzzle would take a long time to solve. How
>long do you take for the 5*5 Tangle on your computer?

I am sorry to say I haven't implemented the search yet.
The 5x5 is solved, so that isn't that interesting anymore; the 10x10
has such a huge search space, that it will need a very efficient
algorithm and/or clever representation. I just haven't decided on the
representation yet. I did decide on the search order though.

- Jan D. de Ruiter

From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU  Tue Dec 21 09:27:36 1993
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Date:      Tue, 21 Dec 1993 08:54:12 EST
From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   9 Qturn Sequence for All-Edges-Flipped in the Edge Group

RUB DRB LDB  (the spaces are for readability only).

Remember that this is for the "edges without corners and without centers"
case.  Hence, the edges are all flipped and are all properly configured
with respect to each other, but they are not flipped "in place" with
respect to a fixed coordinate system of centers.  They are rotated with
respect a fixed coordinate system of centers.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From pbeck@pica.army.mil  Tue Dec 21 18:32:39 1993
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Date:     Tue, 21 Dec 93 7:49:23 EST
From: Peter Beck (BATDD) <pbeck@pica.army.mil>
To: Cube-Lovers@ai.mit.edu
Cc: pbeck@pica.army.mil
Subject:  puzzle party
Message-Id:  <9312210749.aa08453@COR6.PICA.ARMY.MIL>



ROBERT HOLBROOK
11837 LINDEN CHAPEL ROAD
CLARKSVILLE,  MD  21029
  410-531-6135

IS Planning a puzzle party
for FEB 19,20 1994
at his home.

IF YOU ARE INTERESTED PLEASE CONTACT BOB
directly.

Clarksville is 1/2 way between DC and Baltimore.

VENUE  is low keyed with trading, buying, selling and 

TALKING.   

THE FUTURE IS PUZZLING,
BUT CUBING IS FOREVER !!!


From pbeck@pica.army.mil  Tue Dec 21 19:05:28 1993
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Date:     Tue, 21 Dec 93 7:51:39 EST
From: Peter Beck (BATDD) <pbeck@pica.army.mil>
To: cube-lovers@ai.mit.edu
Cc: pbeck@pica.army.mil
Subject:  puzzle party
Message-Id:  <9312210751.aa09222@COR6.PICA.ARMY.MIL>

ROBERT HOLBROOK
11837 LINDEN CHAPEL ROAD
CLARKSVILLE,  MD  21029
  410-531-6135

IS Planning a puzzle party
for FEB 19,20 1994
at his home.

IF YOU ARE INTERESTED PLEASE CONTACT BOB
directly.

Clarksville is 1/2 way between DC and Baltimore.

VENUE  is low keyed with trading, buying, selling and

TALKING.

THE FUTURE IS PUZZLING,
BUT CUBING IS FOREVER !!!


From hoey@aic.nrl.navy.mil  Wed Dec 22 13:58:45 1993
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Date: Wed, 22 Dec 93 13:58:42 EST
From: hoey@aic.nrl.navy.mil
Message-Id: <9312221858.AA08479@sun1.aic.nrl.navy.mil>
To: Cube-Lovers@ai.mit.edu
Subject: The 4^3 and 3^4 Rubik puzzles
Organization: Naval Research Laboratory, Washington, DC

[ Cube-Lovers,

  There has recently been a discussion on Usenet group rec.puzzles
  about some cube topics.  There were a few pieces of new information,
  such as that you can now get Ishi's 5^3 cubes in a lot of places (I
  got mine in Learningsmith's) for about $35.  Here's a message I sent
  that's relevant to some Cube-Lovers topics.

  By the way, I'm still working through Jerry Bryan's articles on his
  brute-force program and his approach to symmetry.  I hope to get a
  reply out soon.
]

eric@gsb002.cs.ualberta.ca (Holleman Eric) wrote:

> By the way, I found the Revenge somewhat easier than the Cube, and I
> don't think that it was because of my familiarity with the earlier
> puzzle.

x87bennett@gw.wmich.edu (Joe) wrote:

> From my experience, if you can solve a Rubik's Revenge, you can
> solve the Cube very easily.  Once you get each of the middle 2 cubes
> on each edge to match, and all 4 center cubes on each face to match,
> it works exactly like a Rubik's cube.

and alan@saturn.cs.swin.oz.au (Alan Christiansen) wrote:

> I have both. I solved both. The 4x4x4 is a superset of the 3x3x3.
> ie by fixing all the face centres and then pairing all edges you are
> left with a 3x3x3 cube, except that when you have solved this 3x3x3
> there may be a single pair of edges flipped. This is impossible
> on a real 3x3x3. Fixing this requires a middle layer to be rotated
> 1/4 revolution and then all the bits put back.

> I cant see how it can be [any] easier than a 3x3x3.

I, too, found the 3^3 easier than the 4^3.  But I can imagine ways in
which a solver could find the 4^3 easier.  Let us first consider a 4^3
with the faces fixed, the edges together, and the correct simulated
edge flip parity.  I would solve this as if it were a 3^3, and a lot
of people do.  But another solver might find it easier to take
advantage of the extra moves that are not possible on a 3^3.  To take
a concrete example, it could be that the solver has a hard time with
flipping edges by pairs, as is needed to solve the 3^3.  On the 4^3
you can flip one edge at a time.  So the solver would find the 4^3
position easier than the corresponding position on a 3^3.  If the
solver finds this so much easier that it overcomes the difficulty of
putting the faces and edges together--or in fact puts the faces and
edges together in the course of solving the corners and the edge
positions--then the 4^3 could be easier.  It depends on the solution
procedure.

alan@saturn.cs.swin.oz.au (Alan Christiansen) continues:

> ANyway the real reason I am writing this is that I have written
> a cube simulator. 
> It can simulate 3x3x3 4x4x4 5x5x5 .... cubes.
> I am working on 4x4x4x4 cube simulation.

This is interesting, as there is more than one way to model the
four-dimensional cube problem.  Consider the 3^4 cube.  It has eight
hyper-faces, each in the shape of a cube.  One model of this puzzle is
that you could turn any face of any hyper-face as if it were a face of
a 3^3 Rubik's cube.  In a second model, you cannot move part of a
hyper-face, but can turn each hyper-face as if it were a solid cube in
space.  A third model allows either kind of move.

These models are different from each other.  The second model permits
the face centers of the hyper-faces to move around, whereas in the
first model only edges and corners move.  In the first model, odd
permutations of corners are possible, which is not true in the second
model.  Of course, the third model is the closure of the first two.

According to Hofstatder's column reprinted in _Metamagical_Themas_,
there is an unpublished 1982 manuscript by H J Kamack and T R Keane
entitled ``The Rubik Tesseract''.  They calculated the size of the
group of the 3^4 puzzle, but I don't know which model was used.

Alan Christiansen indicates he has gone directly to the 4^4 puzzle.  I
don't know which model he plans, or if the models become more similar
with the extra possibilities inherent in the larger cube.  I don't
even know whether he plans to figure out how big the groups are or
whether they are identical.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From xirion!jandr@relay.nl.net  Thu Dec 23 02:48:30 1993
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From: Jan de Ruiter <jandr@xirion.nl>
Date: Thu, 23 Dec 93 08:47:44 +0100
Message-Id: <4326.9312230747@xirion.xirion.nl>
X-Organization:  Xirion Unix Software & Consultancy bv
		 Burgemeester Verderlaan 15 X
		 3454 PE  De Meern
		 The Netherlands
X-Phone: 	 +31 3406 61990
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To: cube-lovers@ai.mit.edu

Subject: Re: Rubiks tangle
To: anandrao@hk.super.net
Cc: cube-lovers@ai.mit.edu

>Your concept is theoretically extendable to the 10*10 tangle, but even
>with this optimisation the puzzle would take a long time to solve. How
>long do you take for the 5*5 Tangle on your computer?

Your question prompted me to actually write the program, and to
squeeze as much efficiency from the program as I could.

You wrote on december the 14th, your program took about 20 minutes
on a 486DX2-66,
Don Woods writes on the same date, that his program takes 45 seconds
on a SparcStation II,
And now I am proud to present my timing: trrrrr (drum roll)
7 seconds on a Compacq Deskpro 386/33. (and still only brute force!)
Now I am ready to try the 10x10.

Some thoughts in the mean time:

If the algorithm treats the duplicate pieces just as ordinary pieces,
i.e. as different, this will cause the program to find 4 solutions
for the 5x5 where only 2 exist (by exchanging the duplicate pieces).
This factor of 2 may not be dramatical, but if the same algorithm
tries the 10x10, then for every 1 solution that exists, the program
will find (5!)^4 x (4!)^20 identical versions (combinations of
duplicate exchanges).

My program views duplicate pieces as one, which may be placed several
times. So for some position X a piece with duplicates will only be
tried once.

Don Woods writes:
>Regarding solving the Tangle, I forgot one other minor optimisation:
>When my program is picking a corner piece other than the first, it
>requires that the piece "number" be less than or equal to that of the
>first corner. I.e., it refuses to search for solutions that are
>rotations of other solutions.

My program prevents finding rotations of solutions, by excluding the
rotations of just one piece. The list of possibilities to try on any
position includes this one piece just once, and every other piece four
times. You can choose any piece for this, except the duplicated one.
Regrettably this approach works only for the 5x5: the 10x10 will
probably have to use Don Woods method.

- Jan D. de Ruiter

From anandrao@hk.super.net  Thu Dec 23 04:29:34 1993
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Date: Thu, 23 Dec 1993 17:26:41 +0800 (HKT)
From: Mr Anand Rao <anandrao@hk.super.net>
Subject: Re: your mail
To: Jan de Ruiter <jandr@xirion.nl>
Cc: cube-lovers@ai.mit.edu
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On Thu, 23 Dec 1993, Jan de Ruiter wrote:

> And now I am proud to present my timing: trrrrr (drum roll)
> 7 seconds on a Compacq Deskpro 386/33. (and still only brute force!)

Were you travelling at the speed of .999c?

> Now I am ready to try the 10x10.
> 
With this algorithm, you should have the solution before the year is out!

Best luck!



From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU  Sat Dec 25 22:48:40 1993
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Date:      Sat, 25 Dec 1993 22:46:06 EST
From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Withdrawal of Proposal

I wish to withdraw, for the time being, my proposed answer to Dan
Hoey's question about how large is the cube group when symmetries
are taken into account.  Notwithstanding two rounds of "correction",
I believe my proposal is fundamentally incorrect, and it will take
some time to come up with something better.

I believe that my proposed approximation is incorrect by a factor
of 24.  That is, my proposed approximation would be correct for
corners plus edges (without centers), but would need to be
multiplied by 24 in order to be correct for corners plus edges
(with centers).  My proposed approximation was
4.3 * 10^19 / (24*24*2).  I now believe it should be
4.3 * 10^19 / (24 * 2), with the former figure correct only if
centers are omitted.

Secondly, I believe that my proposed procedure to calculate an
exact value from the known sizes of corner and edge groups is
incorrect.  My procedure would be correct if all equivalence classes
had exactly 1152 elements.  But they don't.  It is not presently
clear to me whether the size of the equivalence classes when
corners and edges are combined can be calculated from the known
sizes of equivalence classes for corners and edges separately,
or whether a computer search will have to be performed for the
case where corners and edges are combined.

I will get back to this in a week or two.  In the meantime, my
apologies if I have wasted your time, and I look forward to
any words of wisdom that any of you all might have.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From @mail.uunet.ca:mark.longridge@canrem.com  Mon Dec 27 02:33:51 1993
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Sender: CRSO.Cube@canrem.com
Subject: Local Maxima Revisited
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.656.5834.0C18F5BD@canrem.com>
Date: 	Mon, 27 Dec 1993 00:58:00 -0500
Organization: CRS Online  (Toronto, Ontario)

                Some thoughts on Local Maxima
                -----------------------------

   I have verified that the position I call "6 H order 2 type 2"
is a local maximum.

p175 6 H order 2 type 2  T2 B2 L2 T2 D2 L2 F2 T2              (8)

A rare example of a pattern with symmetry level 2, perhaps even
the only one of this type, and also the most symmetric of the
6 H patterns.

Nothing new here, as this was noted formerly by David Singmaster
in one of the Cubic Circulars and by Jim Saxe & Dan Hoey in the
archives.

Somewhat more interesting is the conclusion that the pattern
4 H order 2, or H's on the F,R,B,L faces (oriented like the
letter H) is also a local maximum, at least in the square's
group.

p160 4 H order 2 Type 2  B2 D2 (L2 R2 F2) ^2 T2 F2           (10)

From the archives:

>      We include a description of 71 local maxima, which we believe
> to be all of the local maxima that can be proven using known
> techniques other than exhaustive search.

Oh well, I used an exhaustive search. p160 is 10 moves long in the htw
metric, and each of the moves ( T2, D2, F2, B2, L2, R2 ) all bring
one to a position requiring nine 180 degree twists, thusly....

4 H + T2 =      L2, R2, F2   B2, T2, L2   R2, B2, F2   (9)
4 H + D2 =      L2, R2, F2   B2, D2, L2   R2, B2, F2   (9)
4 H + F2 =      B2, D2, L2   R2, F2, L2   R2, F2, T2   (9)
4 H + B2 =      F2, D2, L2   R2, F2, L2   R2, F2, T2   (9)
4 H + L2 =      R2, D2, L2   F2, B2, L2   F2, B2, T2   (9)
4 H + R2 =      L2, D2, L2   F2, B2, L2,  F2, B2, T2   (9)

----------------------------------------------------------------------

I did discover an interesting property of the "Cube in a cube" pattern
I didn't notice before.

p7a  Cube in a cube   U2 F2 R2 U3 L2 D1 (B1 R3) ^3 B1 D3 L2 U3  (15)

Let's say you are entertaining some cube guests at a cube party and
the topic is (cube) patterns. Your guests are impressed with the
efficiency of the well-memorized process. You would like to go on
to the next pattern but you don't quite remember how the inverse
goes. No problem! Rotate the whole cube so TOP becomes BACK then
BACK becomes DOWN, and finally FRONT becomes RIGHT. Simply repeat
the process p7a and your reputation as a cube expert is saved. ;->

This same idea works for the 6 X order 3 pattern as well.

And now for an unsymmetric local maximum!!

(Just kidding)

  -> Mark <-

From @mail.uunet.ca:mark.longridge@canrem.com  Mon Dec 27 12:31:47 1993
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Sender: CRSO.Cube@canrem.com
Subject: Cube Rotations
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.659.5834.0C18F603@canrem.com>
In-Reply-To: <19166.199312271606@phaedrus.harlequin.com>
Date: 	Mon, 27 Dec 1993 11:12:00 -0500
Organization: CRS Online  (Toronto, Ontario)

-> goes. No problem! Rotate the whole cube so TOP becomes BACK then
-> BACK becomes DOWN, and finally FRONT becomes RIGHT. Simply repeat
-> the process p7a and your reputation as a cube expert is saved. ;->
->
-> The faces FRONT and BACK are opposite each other. After your
-> rotation, they become RIGHT and DOWN, which are not opposite each
-> other. This would certainly establish a reputation for you, but if
-> you did it with my cube, it might not be the sort of reputation you
-> wanted to have :-)
->    Andy Latto
->    andyl@harlequin.com

Perhaps my description of the rotations was unclear...

Rotate the entire cube so that TOP -> DOWN
                               FRONT -> LEFT

Ok, before I meant rotate the cube in space in 3 steps so
that the TOP face becomes BACK, then the face that is the
BACK at this point becomes DOWN, and the face that is the
FRONT at this point becomes the RIGHT.

The reason I used this type of description is because there
are  multiple ways for the TOP to become the DOWN face....

TOP becomes BACK becomes DOWN and
TOP becomes RIGHT becomes DOWN and
TOP becomes LEFT becomes DOWN etc...

Perhaps it is better to use the form
old FACE A -> new FACE A
old FACE B -> new FACE B

Where the faces A & B are adjacent.

Mark
Email: mark.longridge@canrem.com

....wait a second, I don't think faces A & B have to be
adjacent for the rotation to be unambiguous. Any 2 faces
should do!

From hoey@aic.nrl.navy.mil  Mon Dec 27 17:52:28 1993
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From: hoey@aic.nrl.navy.mil
Message-Id: <9312272252.AA22049@sun30.aic.nrl.navy.mil>
To: Cube-Lovers@life.ai.mit.edu
Cc: Jerry Bryan <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
Subject: Group theory basics (Re: Symmetry)

Jerry Bryan <BRYAN%WVNVM.BITNET@mitvma.mit.edu> asked a bunch of
questions a couple of weeks ago, and I'll try to get to them all.
The first bunch has to do with some fairly basic stuff, that I thought
had been pretty well understood since the beginning of the mailing
list, but maybe we need a refresher, or an explicit statement.

In the message of Tue, 14 Dec 1993 20:50:51 EST, Jerry describes his
representation of cube positions and transformations.

> In my computer model, the corner facelets are simply numbered from
> 1 to 24, and any configuration of the corners is an order-24 row
> vector.  The rotation and reflection operators are also order-24 row
> vectors, again with each cell simply containing a number from 1 to
> 24.

That is the most usual way of doing it, but it's important to specify
what you represent by those vectors.  When I do it, I number the
corner facelet locations from 1 to 24, and these locations retain
their numbers through manipulations of the cube.  I use a vector A to
specify a position in which the facelet whose home location is i has
been moved to location A(i), for each i.  I use a vector P to specify
the transformation that moves the facelet in location i to location
A(i), for each i.  I'll assume you're doing the same, though you
could, for instance, be representing the inverse of the operators, or
the locations from which the facelets originate.  Note that a position
is represented by the same vector that represents the transformation
that takes SOLVED to that position.

> Well, if P is a rotation operator, you could perform a rotation
> two ways.  I guess one is pre-multiplication and one is
> post-multiplication.

>     1)  For i = 1 to 24  B(i) = A(P(i))

I would write this as B = P A, and say that A is premultiplied by P,
or equivalently that P is postmultiplied by A.  In a general group, we
could have B = P A where the multiplication is not considered to be
the composition of permutations.  But it turns out we can restrict our
attention to permutation groups without loss of generality.  For
instance, when we are dealing with the supergroup, we can consider the
orientation of a face center to be a permutation of the corners of the
face center.

>     2)  For i = 1 to 24  B(i) = P(A(i))

Here B = A P, A is postmultiplied by P, and P is premultiplied by A.
(Note that the operator or position name appears in the reverse order
from the prefix format.  Algebraists sometimes avoid this by writing
(i)B = ((i)A)P.  I kid you not.)

> (As an aside, this illustrates the question I raised in my previous
> post about "which is the operator and which is the thing being
> operated on?"  Is P operating on A, or is A operating on P?)

Well, the answer is ``both''.  I agree it's easy to get confused,
which is why proofs are a good idea.

> Finally, if Q is a reflection (actually, if Q1 is the identity and
> Q2 is the reflection), then we have

>       For j = 1 to 24 for k = 1 to 24 for m = 1 to 2
>              for i = 1 to 24 Bj,k,m(i) = Qm(Pj(A(Qm(Pk(i)))))

> I believe this loop calculates Dan Hoey's M.

On the the theory that proofs are a good idea, let's see what this
loop calculates.  I'm going to put brackets around the subscripts.
Then I'll substitute "R" for "Q", because I use Q for the set of
quarter-turns of faces.  Furthermore, I'll use "C" instead of "P",
because the P[j] are just the elements of C, the group of cube
rotations.  So you are computing

        B[j,k,m] = C[k] R[m] A C[j] R[m]                    (1)

for j in {1,...,24}, k in {1,...,24}, and m in {1,2}.  Now every
member of M (the group of cube rotations and reflections) has a unique
representation as M[n] = C[k] R[m].  Let us define Cind() and Rind()
as the functions for which M[n]=C[Cind(M[n])] R[Rind(M[n])].  So we
can write (1) as

        B[j,k,m] = M[n] A M'[n] (M[n] C[j] R[Rind(M[n])])

Note that (M[n] C[j] R[Rind(M(n))] must be an element of C.  So B is a
set of elements of the form M[n] A M'[n] C[o].  To see that we have
all such elements, first observe that (M[n]' C[o] R[Rind(M[n])]') is
an element of C, say C[j].  So equation (1) includes:

   C[Cind(M[n])] R[Rind(M[n])] A C[j] R[Rind(M[n])]
   = M[n] A (M[n]' C[o] R[Rind(M[n])]') R[Rind(M[n])]
   = M[n] A M'[n] C[o].

Thus the set of all B[j,k,m] is the set of all M[n] A M'[n] C[o].  Or
in English, that's the set of all M-conjugates of A, operated on by
all whole-cube rotations.

> In my data base, I store the minimum of Bj,k,m over j = 1 to 24,
> k = 1 to 24, and m = 1 to 2.  I tend to call the minimum of Bj,k,m a
> canonical form.  I am not sure if that is the best terminology.  The
> minimal element is not any simpler than any other.  It is just that
> I need a function to choose an element from a set, and picking the
> minimal element seems very natural.  Any other element would do as
> well, provided I could always be sure of picking the same element.

It's pretty common terminology.  You might be slightly better off
calling it a ``representative element,'' as that connotes that the
element is ordinary except in that it represents the equivalence class
(like representatives in the U.S. Congress).

> Also, my criterion for equivalence is slightly
> different (but isomorphic, I think) than the one described by
> Dan Hoey.  Suppose A and B are two cubes.
> Rather than mapping A to B or B to A in M, I map both A and B
> to their respective canonical forms.  A and B are equivalent if
> their respective canonical forms are equal.

This is straightforward once we show that M-conjugacy is an
equivalence relation, and B[j,k,m] is an equivalence class.
If A ~ Representative[A] = Representative[B] ~ B, then by transitivity
A ~ B.  Conversely, if A ~ B, then Class[A] = Class[B], and therefore
Representative[A] = Representative[B].  This shows that the criteria
are equivalent.

> Now, as to the centers.  I still sometimes have a certain doubt
> about the centers.  They are fixed, so how can you reduce the
> problem (i.e., increase the size of the equivalence classes)
> by both rotating the cube and rotating the colors (by both pre-
> and post-multiplication)?

What you have done is to increase the size of the whole cube problem
by a factor of 24, by dealing with all rotations of the cube, and the
equivalence classes expand by the same factor, from 48 to 1152.  This
has allowed you to calculate something like M-conjugacy classes for
cube problems that lack face centers.  But the size of the equivalence
classes doesn't shrink the problem for cubes that have face centers.
You could have just calculated M-conjugates and got the same answer.

> I am not sure if this answers Dan's question about my model
> with centers added.

It's clear now.  I hadn't realized you were rotating the cube in space
when the face centers were present.  I expected that to be a wasted
effort.  But I am impressed by the way it allows you to shrink the
database by storing positions together that differ only by whole-cube
moves of the face centers.  I think it should be possible to shrink
the database without the effort, though.

In your message of Thu, 16 Dec 1993 15:36:58 EST, on the ``Duality of
Operators and Operatees'':

> I have mentioned several times my discomfort about "an operator" as
> opposed to "the thing being operated on" when it comes to groups.  I
> am never quite sure just which of the two it is that people are
> talking about, even (or especially) when I am listening to myself
> talk.

It is hard to keep it straight.  Sometimes we all get it wrong.  The
best way to avoid errors, as far as possible, is to avoid such
language and talk about group multiplication.  But then we have to
explain what is going on with the cube, so we get caught into talking
about operators again.  It's a discomfort that must be endured.

> The code to translate between the ASCII string X and
> the EBCDIC string Y is something like
>     for i = 1 to n Y(i) = T(X(i))
> where T is the translate table.

Yes, or Y = X T as above.

> I am going to continue reading, but perhaps I could pose a question to
> Dan Hoey anyway:  is reversing the role of X and T in the TRANSLATE
> function above essentially the same thing as switching between
> pre-multiplication and post-multiplication?

Yes.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From dik@cwi.nl  Mon Dec 27 18:43:04 1993
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From: Dik.Winter@cwi.nl
Message-Id: <9312272343.AA25571.dik@boring.cwi.nl>
To: Cube-Lovers@life.ai.mit.edu
Subject: Re:  Group theory basics (Re: Symmetry)

One additional remark:

 > > Well, if P is a rotation operator, you could perform a rotation
 > > two ways.  I guess one is pre-multiplication and one is
 > > post-multiplication.

 > >     1)  For i = 1 to 24  B(i) = A(P(i))

 > I would write this as B = P A, and say that A is premultiplied by P,
 > or equivalently that P is postmultiplied by A.

There is quite a bit of confusion about this.  When permutation
groups are considered; even text-books do not agree.  When A and P
are permutations you can find both that P A means: apply P first, A next,
but also: apply A first, P next.  (The first meaning comes from the pure
group theorists, the second meaning more from the algebra inclined.)

Sorry to confuse the issue, but when I read such texts I have always to
think hard to get at the intended meaning.  I think the functional
notation is much clearer and leads to less confusion.

Of course, doing notations for cube rotations the group theorists
notation is applied, but when doing abstract operations...

From hoey@aic.nrl.navy.mil  Tue Dec 28 14:17:02 1993
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From: hoey@aic.nrl.navy.mil
Message-Id: <9312281916.AA25640@sun30.aic.nrl.navy.mil>
To: Cube-Lovers@life.ai.mit.edu
Cc: Jerry Bryan <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
Subject: Re: Some Additional Distances in the Edge Group

In his message of Fri, 17 Dec 1993 00:54:00 EST, Jerry Bryan
<BRYAN%WVNVM.BITNET@mitvma.mit.edu> makes some observations on the
distances between the following positions in the edge group:

    I = Solved,
    P = Pons Asinorum (or Mirror),
    E = All edges flipped, and
    PE = P E = Pons Asinorum with all edges flipped.

[I _will_ continue to use permutation multiplication as we have done
so in this group since its inception.  I realize that this agrees with
some textbooks and is backwards from others, but it would be far more
confusing to write these functionally all the time.]  Jerry's
brute-force search has shown that d(I,PE)=15, and he notes that
conjugation by E shows us that d(P,E)=15 as well.  He concludes:

> I have the sensation in describing this that the Edge group is
> square, with Start and Mirror-Image-of-Edges-Flipped 180 degrees
> apart, and Mirror-Image-of-Start and Edges-Flipped at the other
> two corners of the square.

Well, it's not quite a square, since d(I,P)=12 and d(I,E)=9, according
to Jerry's message of Wed, 8 Dec 1993 10:02:15 EST.  Conjugation will
similarly show that d(E,PE)=12 and d(P,PE)=9.  So we are dealing with
a rectangle.  The sides of the rectangle are 9 and 12, and the
diagonal is 15: a most fortuitous set of numbers, in that we can
actually embed such a rectangle in the Euclidean plane!

We can map the positions of the edge group to 4-tuples of distances.
For any position X, let

    f(X)=(d(I,X), d(E,X), d(P,X), d(PE,X)).

If f(X)=(a,b,c,d), then conjugation shows us that f(X E)=(b,a,d,c),
f(X P)=(c,d,a,b), and F(X PE)=(d,c,b,a).  So the set of quadruples has
the symmetries of the rectangle.

We know f(I)=(0,9,12,15).  What is more, the earlier results on
symmetry show us that I is at a local maximum distance from E, P, and
PE.  So, letting I1 be the unique (up to M-conjugacy) position
adjacent to I, we have F(I1)=(1,8,11,14).  (This destroys Euclidean
embeddability.)  An analogous result holds for the unique neighbor of
each corner of the rectangle.

We also have Jerry's results of Wed, 8 Dec 1993 22:41:28 EST and
23:16:50 EST that H (the 6-H pattern) and HE=H E are at distances 8
and 13 from start, respectively.  Since H is an M-conjugate of P H,
this gives us f(H)=(8,13,8,13).  [Note: there are two distinct
M-conjugates of H, call them H and Hbar.  This distinction is
important when we compose permutations: H H = I, but H Hbar = P.  So
we have to be careful when conflating M-conjugates.] We can by
symmetry find f(H1)=(7,12,7,12) for H's unique neighbor H1.

What quadruples are possible?  If f(X)=(a,b,c,d), and X is not one of
the eight corners and neighbors, we have

         max(2,9-b,12-c,15-d) <= a <= min(14,9+b,12+c)

with constraints on b, c, and d from symmetry.  A quick hack tells me
there are 7836 such quadruples.  I wonder how many of them are
realized?  If it's fairly few, I would like to see a diagram of
quadruples, with lines between those quadruples that represent
adjacent positions (adjacent quadruples differ by at most one in each
coordinate).  Maybe with the number of positions for each quadruple,
too.  I have an idea that such a diagram might tell us something about
the problem, or at least look pretty.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From hoey@aic.nrl.navy.mil  Tue Dec 28 18:42:22 1993
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From: hoey@aic.nrl.navy.mil
Message-Id: <9312282340.AA25691@sun30.aic.nrl.navy.mil>
To: Cube-Lovers@ai.mit.edu
Subject: Re: Cube Rotations
Cc: CRSO.Cube@canrem.com

mark.longridge@canrem.com (Mark Longridge) writes:

> Perhaps my description of the rotations was unclear...

Yes.

> ...Perhaps it is better to use the form
> old FACE A -> new FACE A
> old FACE B -> new FACE B

> Where the faces A & B are adjacent.

That will serve to uniquely identify a rotation, but it's somewhat
verbose.  Worse, it does not suffice to uniquely identify a symmetry
from the group of rotations and reflections, M.  I find it's far more
informative to identify a rotation or reflection as a permutation of
the faces, in cycle format.  There are only ten kinds:

Even rotations:          I=Identity (1),
                (FRT)(BLD)=120-degree rotation (8),
                  (FB)(RL)=180-degree orthogonal rotation (3).

Odd rotations:      (FRBL)=90-degree rotation (6), 
              (FB)(TR)(DL)=180-degree diagonal rotation (6).

Even reflections: (FR)(BL)=diagonal reflection (6),
                (FRBL)(TD)=90-degree glide reflection (6),
                  
Odd reflections:      (FB)=orthogonal reflection (3),
                  (FRTBLD)=60-degree glide reflection (8),
              (FB)(RL)(TD)=central reflection (1).

In case it isn't clear, the cycle notation for (e.g.) a 120-degree
rotation (FTL)(BDR) means that the F, T, L, B, D, and R faces move to
the T, L, F, D, R, and B, locations, respectively.  The only thing I'm
afraid of with this notation is that someone will think I'm describing
a magic-cube process rather than a whole-cube move.

So when you say Top->Down, Front->Left, I would say (TD)(FL)(BR) for
the 180-degree diagonal rotation, to distinguish it from (TD)(FLBR)
the 90-degree glide reflection.

> ....wait a second, I don't think faces A & B have to be
> adjacent for the rotation to be unambiguous. Any 2 faces
> should do!

No, you're back to your original bogosity.  Knowing the destinations
of two opposite faces doesn't give you any more information than
knowing the destination of one (unless you go breaking the axles).

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From hoey@aic.nrl.navy.mil  Wed Dec 29 17:43:47 1993
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From: hoey@aic.nrl.navy.mil (Dan Hoey)
Message-Id: <9312292243.AA05575@Sun0.AIC.NRL.Navy.Mil>
To: Cube-Lovers@life.ai.mit.edu
Cc: Jerry Bryan <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
Subject: Correction Re: Some Additional Distances in the Edge Group

A couple of days ago, I said that proofs are a good idea.  I'll say it
again today with a redder face.

Yesterday I discussed the edge group positions

    I = Solved,
    P = Pons Asinorum (or Mirror),
    E = All edges flipped, and
    PE = P E = Pons Asinorum with all edges flipped

and the function from the edge group to 4-tuples of distances

    f(X)=(d(I,X), d(E,X), d(P,X), d(PE,X)).

I wrote:

?? If f(X)=(a,b,c,d), then conjugation shows us that              ??
?? f(X E)=(b,a,d,c), f(X P)=(c,d,a,b), and F(X PE)=(d,c,b,a).     ??
?? So the set of quadruples has the symmetries of the rectangle.  ??

The first sentence is incorrect, though the argument as a whole is
reparable.

First, I'll do what I should have done yesterday, and define the
distance function d(X,Y).  We want the minimum length process Z such
that X Z = Y.  But premultiplying both sides by X', we have Z = X' Y.
So I define d(X,Y)=Length(X' Y).  From the properties of the length
function (Length(I)=0, Length(X)=Length(X'), and
Length(X Y)<=Length(X) + Length(Y)) we can conclude that d(X,Y) is a
metric.

Suppose f(X)=(a,b,c,d).  I claim f(E X)=(b,a,d,c), f(P X)=(c,d,a,b),
and F(PE X)=(d,c,b,a).

Proof: To show f(E X)=(b,a,d,c), first observe that I=I', E=E', and
P E = E P.

    d(I,E X) = Length(I' E X) = Length(E' X) = d(E,X),
    so d(E,E X) = d(I, E E X) = d(I,X);
    d(P,E X) = Length(P' E X) = Length((PE)' X) = d(PE,X)
    so d(PE,E X)=d(P,E E X)=d(P,X).

To show that f(P X)=(c,d,a,b), exchange P and E in the above argument.
To show that f(PE X)=(d,c,b,a), use both occurrences of the argument.
QED.

So the idea of yesterday's message is correct, but I had X E, X P, and
X PE instead of E X, P X, and PE X, respectively.  I would show you a
counterexample to yesterday's formulation, but it turns out there is
none.  I claim that f(X,E)=f(E,X), f(X,P)=f(P,X), and f(X,PE)=f(PE,X).

Proof: Recall that E commutes with every element of the Rubik cube
group, so f(X E)=f(E X).  It turns out that ``up to M-conjugacy'', P
commutes with every element of the edge group as well.  For P performs
a mirror-reflection of the edges, and so can be regarded as an element
of M acting on the edge group.  So P' X P = Xbar is an M-conjugate of
X, and X P = P Xbar.  Since Length(X) agrees on M-conjugates, so does
d(X,Y), and so f(X), so f(X P)=f(P Xbar) = f(P X).  Finally,
f(X PE) = f(X P E) = f(E X P) = f(P E X) = f(PE X). QED.

So it turns out it that the statement about f was true.  But I am no
less embarrassed for asserting it, for I had no reason to think it
would be true.  It's only rescued by the surprising commutativity of
the Pons Asinorum.

Finally, I would like to note something that I nearly included in
yesterday's message, but yanked when I decided it was false:
f(X')=f(X).  Now I'll prove it:

Proof:  For W among {I,E,P,PE}, we have X W = W Xbar, for Xbar an
M-conjugate of X.  So
     d(X,W)=Length(X'W)=Length(W'Xbar')=Length(W'X')=d(W,X').
QED.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From Don.Woods@eng.sun.com  Sun Jan  2 20:10:16 1994
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From: Don.Woods@eng.sun.com (Don Woods)
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To: cube-lovers@ai.mit.edu
Subject: 10x10 Tangle
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Hm.  Well, I split up the 10x10 Tangle exhaustive search and ran it on
several machines over Christmas break, getting the 90 days of compute
time done in about 10.

And turned up no solutions.

There could of course be a bug in my program, but the same code with
minor changes finds the same solutions as others have found for the 5x5.
I also tried adding some extra tiles for the 10x10, and it began finding
solutions okay.  I did doublecheck that the 100 tiles matched the info
posted to Cube-Lovers re which tiles are duplicated in the four 5x5s;
I have no way of checking whether that info was correct.

Has anyone out there ever heard definitely that someone has found a
solution to the 10x10?  Is it possible that the makers of Tangle (Matchbox,
using Rubik's name under license) merely claimed that such a solution
exists, without actually verifying it?  (Seems pretty sleazy if so,
but then, having Tangles 2-4 be merely color permutations of #1 is
pretty weak in the first place.)

	-- Don.

From dik@cwi.nl  Sun Jan  2 21:52:23 1994
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From: Dik.Winter@cwi.nl
Message-Id: <9401030252.AA07139.dik@boring.cwi.nl>
To: Don.Woods@eng.sun.com, cube-lovers@ai.mit.edu
Subject: Re:  10x10 Tangle

 > Has anyone out there ever heard definitely that someone has found a
 > solution to the 10x10?

As I wrote before, I have embedded in my memory that there is an easy
argument that the 10x10 is *not* solvable.  I do not know whether I
found it myself (and ever did mail it to other people) or whether I
found it somewhere on the net; it is a long time ago.  When I find the
time I will do a check.  (I know very sure that I have had a program
running at that time but that I abandoned the search because it would
be fruitless.)

 >                         Is it possible that the makers of Tangle (Matchbox,
 > using Rubik's name under license) merely claimed that such a solution
 > exists, without actually verifying it?

Yes, very probable.  You should never trust the number of solutions the
manufacturers give.  Sometimes it is much more, in this case it is less.
An actual example is a puzzle that consists of of nine rings (eh, this
is from memory, I do not have access to the puzzle at this time).  Five
rings contain digits; three rings contain operators; one ring contains
equal signs.  All in four positions around the rings.  The idea is to
create correct sums (like 5 + 1 - 4 + 1 = 3) on all four positions of the
rings.  The claim was that there was only a single solution.  Actually
there are many.  If there is interest I can hunt down the rings and
describe them in more detail.  (An interesting detail is that my father
was the first to find the puzzle; he had correct solutions like:
1 + 3 : 2 + 1 = 3.  He was a physicist.  The accomanying leaflet did not
give details about operator priorities.  Hence it actually makes two
puzzles; one with regards to priorities, the other just going left to
right.)

 >                                         (Seems pretty sleazy if so,
 > but then, having Tangles 2-4 be merely color permutations of #1 is
 > pretty weak in the first place.)

Indeed, the mass manufacturers are sleazy.

Cheers.

I will mail when I find back the argument disallowing 10x10.
--
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland, +31205924098
home: bovenover 215, 1025 jn  amsterdam, nederland; e-mail: dik@cwi.nl

From xirion!jandr@relay.nl.net  Mon Jan  3 02:29:28 1994
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From: Jan de Ruiter <jandr@xirion.nl>
Date: Mon, 3 Jan 94 08:29:37 +0100
Message-Id: <22110.9401030729@xirion.xirion.nl>
X-Organization:  Xirion Unix Software & Consultancy bv
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To: cube-lovers@ai.mit.edu

To: Don.Woods@eng.sun.com, cube-lovers@ai.mit.edu
Subject: RE: 10x10 Tangle

>Hm.  Well, I split up the 10x10 Tangle exhaustive search and ran it on
>several machines over Christmas break, getting the 90 days of compute
>time done in about 10.
>
>And turned up no solutions.

My program is a bit faster, but as I have less machines at my disposal
and I started a bit later, my programs are still running.
Up until now they did not produce a solution either. I am starting
to get worried.

>There could of course be a bug in my program, but the same code with
>minor changes finds the same solutions as others have found for the 5x5.

The same goes for me.

>                 I did doublecheck that the 100 tiles matched the info
>posted to Cube-Lovers re which tiles are duplicated in the four 5x5s;
>I have no way of checking whether that info was correct.

I have the puzzles myself, and checked the info in the message from
Dale I Newfield (15 Dec 1993), which quotes the archives.
I can assure you: those are indeed the duplicate pieces.

>
>Has anyone out there ever heard definitely that someone has found a
>solution to the 10x10?  Is it possible that the makers of Tangle (Matchbox,
>using Rubik's name under license) merely claimed that such a solution
>exists, without actually verifying it?  (Seems pretty sleazy if so,
>but then, having Tangles 2-4 be merely color permutations of #1 is
>pretty weak in the first place.)
>

I thought about that too, but considered that the choice for precisely
those four duplicate pieces could be dictated by the desire to have
a solution for the 10x10.

>I also tried adding some extra tiles for the 10x10, and it began finding
>solutions okay.

Question: did you add pieces at random, or did you add more duplicate
pieces?
In the latter case you may have found the duplications that should have
been made to get a solvable 10x10.
That in turn would show there could not exist an argument disallowing
10x10 (as claimed by Dik Winter), unless that argument is based on the
particular colouring of the four duplicated pieces...

	-- Jan


From Don.Woods@eng.sun.com  Mon Jan  3 05:45:40 1994
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From: Don.Woods@eng.sun.com (Don Woods)
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To: cube-lovers@ai.mit.edu, jandr@xirion.nl
Subject: Re: 10x10 Tangle
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> My program is a bit faster, but as I have less machines at my disposal
> and I started a bit later, my programs are still running.

Incidentally, I would be interested in seeing your program.  (And am
willing to send you mine.)  I'm always willing to learn something about
how to make combinatorial searches more efficient.

> >I also tried adding some extra tiles for the 10x10, and it began finding
> >solutions okay.
> 
> Question: did you add pieces at random, or did you add more duplicate
> pieces?

I just gave it 5 of each piece, instead of 4 of most pieces and 5 of some.
It churned out positions pretty quick that way!  But since this involved
giving it more than 100 tiles to draw from, it says nothing about Dik Winter's
claimed impossibility proof.

It's a shame, really.  I'll bet that it would be possible to come up with
four Tangles that (a) really are different instead of being simple color
permutations of each other, (b) each have a unique solution (not counting
rotations) instead of two, and (c) can be combined to form a 10x10 that has
a unique solution.  Well, strike the "unique" from (c) and I'd make the bet;
but with the "unique" I certainly wouldn't bet against it!

	-- Don.


From xirion!jandr@relay.nl.net  Mon Jan  3 08:17:10 1994
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From: Jan de Ruiter <jandr@xirion.nl>
Date: Mon, 3 Jan 94 14:16:52 +0100
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To: cube-lovers@ai.mit.edu

To: Don.Woods@Eng.Sun.COM
Subject: Re: 10x10 Tangle

>Incidentally, I would be interested in seeing your program.  (And am
>willing to send you mine.)  I'm always willing to learn something about
>how to make combinatorial searches more efficient.

Will be sent separately  (and yes, I would like to see yours too!)

>It's a shame, really.  I'll bet that it would be possible to come up with
>four Tangles that (a) really are different instead of being simple color
>permutations of each other, (b) each have a unique solution (not counting
>rotations) instead of two, and (c) can be combined to form a 10x10 that has
>a unique solution.  Well, strike the "unique" from (c) and I'd make the bet;
>but with the "unique" I certainly wouldn't bet against it!

When you limit yourself to 4 ropes with 4 colours, you always get 24 pieces,
and when you want to build a puzzle of 25 pieces, you will have to duplicate
one, which causes (a).
Using 5 colours instead, creates a set of 120 pieces, from which you could
probably pick 25 pieces (without duplication) which would satisfy both (a)
and (b), and probably 4 such sets could be found to satisfy (c) as well,
but such a puzzle would be less attractive, because the choice of 25 from
120 is somewhat arbitrary, and a puzzler would probably be more inclined
to use all 120 pieces... Of course it is all a matter of taste.

  -- Jan

From Don.Woods@eng.sun.com  Mon Jan  3 17:11:06 1994
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From: Don.Woods@eng.sun.com (Don Woods)
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To: cube-lovers@ai.mit.edu, jandr@xirion.nl
Subject: Re: 10x10 Tangle
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> >It's a shame, really.  I'll bet that it would be possible to come up with
> >four Tangles that (a) really are different instead of being simple color
> >permutations of each other, ...
> 
> When you limit yourself to 4 ropes with 4 colours, you always get 24 pieces,
> and when you want to build a puzzle of 25 pieces, you will have to duplicate
> one, which causes (a).

Not so.  There's nothing that says all permutations must be present.
Back in '92 when I first wrote the program to solve Tangle #1, I fiddled
with it a bit and found that removing a particular tile and adding a
duplicate of a second particular tile caused the solution to become
unique.  It didn't take long to find such a combination, so I'm confident
there are many many more that have unique solutions.

Hm, using just the set of 24 distinct tiles, I wonder if it's possible to
tile the faces of a 2x2x2 cube such that colors match at the edges of the
cube as well as within the faces?...

	-- Don.


From Don.Woods@eng.sun.com  Mon Jan  3 19:50:59 1994
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From: Don.Woods@eng.sun.com (Don Woods)
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To: cube-lovers@ai.mit.edu
Subject: tangled cube
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Well, a pleasant surprise!  It _is_ possible to take a set of 24 distinct
tiles from any Rubik's Tangle, and use them to tile the surface of a 2x2x2
cube such that all touching ropes match.  And the solution is unique!

I'll include the solution below, after some blank lines to avoid spoiling it
for anyone who wants to try solving the puzzle without seeing the answer...

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First, a hint.  If you look at any face of the cube, and look at the two
pairs of colors at any edge of that face, the two pairs will be the same.
That is, if one tile touches a cube edge with colors red-blue, the adjacent
tile on that face touching the same edge will also touch the edge with red-
blue.  I see no obvious reason why the solution should have this property,
but it does.

Solution below.

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Solution: Note that there is also a correlation between the color pairs that
occur within a face, and the color pairs that occur at the edges of that face.
Also, the orientation of every tile is the same relative to the adjacent of
the cube that the tile touches.  This makes it relatively easy to reconstruct
the solution manually.

           BR   BR                     
          G..G G..Y                    
          R..Y Y..B                    
           YB   GR                     
                                       
           YB   GR                     
          G..R R..Y                    
          R..B B..B                    
           YG   YG                     
                                       
 GR   GR   YG   YG   BY   BY   RB   RB 
Y..Y Y..B B..B B..R R..R R..G G..G G..Y
R..B B..G G..R R..Y Y..G G..B B..Y Y..R
 BG   YR   RY   BG   GB   RY   YR   GB 
                                       
 BG   YR   RY   BG   GB   RY   YR   GB 
Y..R R..B B..G G..R R..Y Y..G G..B B..Y
R..G G..G G..Y Y..Y Y..B B..B B..R R..R
 BY   BY   RB   RB   GR   GR   YG   YG 
                                       
           RB   RB                     
          Y..Y Y..G                    
          B..G G..R                    
           GR   YB                     
                                       
           GR   YB                     
          Y..B B..G                    
          B..R R..R                    
           GY   GY                     

From jbharris@tenet.edu  Mon Jan  3 20:04:38 1994
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Date: Mon, 3 Jan 1994 19:02:13 -0600 (CST)
From: Judi Harris <jbharris@tenet.edu>
Subject: Volunteers Requested
To: CUBE-LOVERS@life.ai.mit.edu
Message-Id: <Pine.3.03.9401031913.C16095-c100000@abernathy.tenet.edu>
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WOULD YOU BE WILLING TO SHARE WHAT YOU KNOW WITH 
PRE-COLLEGE STUDENTS AND TEACHERS BY ELECTRONIC MAIL?

Recent estimates indicate that there are now more than 
300,000 classroom teachers from primary, middle, and 
secondary schools who hold accounts on the Internet.  
This makes a very special kind of learning available to 
them: one which directly involves subject matter experts 
communicating with students and teachers about their 
specialties, via electronic mail.

With support from the Texas Center for Educational 
Technology, we (at the University of Texas at Austin) 
have piloted and are now expanding an Internet-based 
service (the "Electronic Emissary") that brings together 
pre-college students, their teachers, and subject matter 
experts (SMEs) electronically, helping them to create 
telecomputing exchanges centered around the students' 
learning in the SMEs' disciplines.  For example, 
	
*  A class studying South America could learn about 
recent global environmental research results from a 
scientist who studies rainforest deforestation in 
Brazil.  
	
*  A class studying geometry might "talk" electronically 
with Euclid, who is actually a mathematics professor.  
	
*  A class studying the future of education might 
converse with an emerging technologies specialist from 
California's Silicon Valley.  
	
*  A class studying American History might 
electronically interview Harry Truman, who is really a 
curator with the National Archives.  
	
*  A class exploring the rapidly-changing governmental 
structures that are emerging in what was once the Soviet 
Union might correspond with a group of graduate 
political science students at a university in the CIS.  
	
*  Or, a class reading _Huckleberry Finn_ might 
correspond with an African-American studies scholar 
about the repercussions resulting from the enacting of 
the Emancipation Proclamation.

In successive phases of the project, increasing numbers 
of SMEs or SME groups are needed to correspond regularly 
(approximately 4 times per week) with primary, middle 
school, or secondary students and their teachers (1 SME 
or expert group per class, study group, or "special 
student").  Each electronic exchange will begin with 
approximately 2 weeks of project planning via electronic 
mail between the SMEs and the teachers.  Communications 
with students will begin on mutually convenient dates, 
and will continue for previously-arranged periods of 
time, usually between 2 and 10 weeks.

Subject matter expert volunteers are sought in all 
disciplines, but there is immediate need for SMEs with 
expertise in:

~  gravity and satellite motion
~  heat transfer
~  Hitler's rise to power during World War II
~  the Indian Wars (1870's & 1880's)
~  20th century fragmentation due to weapons of war, 
   especially the atom bomb
~  Maya Angelou (and other women in literature)
~  Native American literature, specifically Cherokee
~  George Orwell's _Animal Farm_ & Russian revolutions
~  personal finance
~  geometry

==> If you would like to find out more about 
==> participating in this project, please send 
==> electronic mail to Judi Harris, jbharris@tenet.edu.  
==> Please include your name, institution, and areas of 
==> expertise.

==> PLEASE RESPOND ASAP; teacher-SME pairs in the 
==> specific areas requested above will be formed on 
==> 1/12/93.


From xirion!jandr@relay.nl.net  Tue Jan  4 02:30:20 1994
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From: Jan de Ruiter <jandr@xirion.nl>
Date: Tue, 4 Jan 94 08:30:56 +0100
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To: cube-lovers@ai.mit.edu

To: Don.Woods@eng.sun.com, cube-lovers@ai.mit.edu
Subject: RE tangle cube

>Well, a pleasant surprise!  It _is_ possible to take a set of 24 distinct
>tiles from any Rubik's Tangle, and use them to tile the surface of a 2x2x2
>cube such that all touching ropes match.  And the solution is unique!
>
>I'll include the solution below, after some blank lines to avoid spoiling it
>for anyone who wants to try solving the puzzle without seeing the answer...

You may have noticed it yourself, but the solution you promised was missing
from your message. But I take your word for it that you found it, because
(sorry to spoil your scoop) a solution for the tangle-cube as you described
was published before in CFF (Cubism For Fun) the periodical of the NKC
(Nederlandse Kubus Club = Dutch Cubist Club). Contrary to what the name
suggests members are not solely interested in cubes.
Membership to that club is open to anyone interested in puzzles like these
and highly recommended! The periodical CFF is published in English, and
appears three or four times a year.
Further information can be obtained via gm@phys.uva.nl

	-- Jan.

From mouse@collatz.mcrcim.mcgill.edu  Tue Jan  4 07:25:33 1994
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From: der Mouse <mouse@collatz.mcrcim.mcgill.edu>
Message-Id: <199401041220.HAA11684@Collatz.McRCIM.McGill.EDU>
To: jandr@xirion.nl
Cc: cube-lovers@ai.mit.edu

>> Well, a pleasant surprise!  It _is_ possible to [tile a 2x2x2 cube
>> with distinct Rubik's Tangle pieces, uniquely]

>> I'll include the solution below, after some blank lines to avoid
>> spoiling it for anyone who wants to try solving the puzzle without
>> seeing the answer...

> You may have noticed it yourself, but the solution you promised was
> missing from your message.

Not from the copy I got - though the lines in question weren't blank;
they each contained a dot.  Perhaps someone's mailer isn't doing the
hidden-dot algorithm correctly?

					der Mouse

			    mouse@collatz.mcrcim.mcgill.edu

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Date:      Tue, 4 Jan 1994 11:10:02 EST
From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Which is the Real Start?

The net is so wonderful about answering questions, here are a few
more:

1. Take a standard 3x3x3 Rubik's cube, and remove the corner and
   center labels to make an Edges Cube.  (I am assuming that the
   underlying plastic is black.  If the underlying plastic is
   white and one of the colors on the labels is also white, the
   Edges Cube is not so pretty).  Scramble the cube.  Give it to
   a cubemeister to solve.  How will the cubemeister know if the
   cube is solved?  In other words, how will the cubemeister
   distinguish Start from Pons Asinorum?

   One answer is that the cubemeister cannot.  Unless the cubemeister
   saw the cube before it was scrambled, or unless the cubemeister
   was told which reflection of the colors was Start, there would
   be no way to tell.  Another answer is that either one is Start --
   that there are two Starts.   However, if you like this answer,
   and if you identify the identity with Start, you are in the
   disquieting situation of having a group with two distinct
   identities (grin!).

   It is obvious that this problem does not arise if the labels are
   left on the centers.  Almost as obvious is the fact that the
   problem does not arise if the labels are left on the corners, even
   if the labels are removed from the centers.  The corner group
   cannot be turned inside out by a reflection as can be the edge
   group.

2. As silly as my second answer is, it leads to a second question.
   Just what is the 2x2x2 cube?  Or more correctly, how do you
   know when it is solved?  With any size of cube, if you restrict
   yourself to quarter-turns, by definition you cannot rotate the cube
   in space as a single operation.  Yet, a simple quarter-turn sequence
   such as RL' does rotate the 2x2x2 cube because it is faceless.  Is
   Start of the 2x2x2 operated on by RL' solved?  If so, you can argue
   that the 2x2x2 has 24 Starts.  Most people would not.  They would
   argue that there is only one Start, and that 2x2x2 cubes that differed
   only by a rotation are equivalent.

3. Combining #1 and #2, I *think* that most people would argue that
   Start and Pons Asinorum on the Edge Cube are not equivalent,
   but that simple rotations of the 2x2x2 are equivalent.  If I am
   correct about "most people", why?  Is a rotation symmetry
   intrinsically a stronger or weaker symmetry than a reflection
   symmetry?

4. When I was first posting my results about the Edge Group, and
   particularly when it first began to sink in what the four equivalence
   classes with only 24 elements really were, I had a moment of
   panic.  Since Start and Pons Asinorum differ only by a simple
   reflection, why had not my version of M-conjugation declared them
   to be equivalent?  (I speak of "my version of M-conjugation", but
   the question is no different if you look at Dan Hoey's original
   M-conjugation).  I think I know the answer, but I will leave the
   problem as an exercise for the student.  Furthermore, I think the
   answer to #4 is really the same as the answer to #3.

5. What is a reflection, really?  Here is an exercise to illustrate
   the question.  Take two identically colored and oriented 3x3x3
   cubes.  On one, perform F and on the other perform F'.  Examine
   the two cubes, plus their images in a mirror.  Why are there
   four distinct cubes rather than only two?  At one level of
   abstraction, the answer is simple.  Of the four, one is not
   reflected, one is pre-reflected, one is post-reflected, and one
   is both pre- and post-reflected.  Is this a sufficient answer,
   or is there something deeper?

   At this point, I can't help but note Martin Gardner's famous
   mirror question in Scientific American many years ago:  why
   does a mirror reverse left and right but not up and down?

6. I found Dan Hoey's postings about the four special states of the
   Edge Group to be delightful.  Some of the results were based on
   a computer search of the group, for example the fact that
   f(I)=(0,9,12,15) could only reasonably be determined from a
   computer search.  However, the thought occurred to me that most
   of Dan's results were independent of the computer search, and
   I was curious precisely which results would stand without the
   search?  For example, if we identified the group as being
   rectangular, would we be led to saying which of the four special
   states were diagonally opposed without the computer search?
   Without the search, I might be tempted to say that Start and
   Pons Asinorum were diagonally opposed.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From mouse@collatz.mcrcim.mcgill.edu  Tue Jan  4 13:48:23 1994
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Date: Tue, 4 Jan 1994 13:48:03 -0500
From: der Mouse <mouse@collatz.mcrcim.mcgill.edu>
Message-Id: <199401041848.NAA12863@Collatz.McRCIM.McGill.EDU>
To: cube-lovers@ai.mit.edu
Subject: Re:  Which is the Real Start?

> The net is so wonderful about answering questions, here are a few
> more:

I am hardly more than a dilettante of the Cube, but I can perhaps offer
a few suggestions, since this seems to me to be largely psychology and
3-D geometry rather than Cube group theory.

> 1. Take a standard 3x3x3 Rubik's cube, and remove the corner and
>    center labels to make an Edges Cube.  [...]  Scramble the cube.
>    Give it to a cubemeister to solve.  How will the cubemeister know
>    if the cube is solved?  In other words, how will the cubemeister
>    distinguish Start from Pons Asinorum?

>    One answer is that the cubemeister cannot.  [...]  Another answer
>    is that either one is Start -- that there are two Starts.

Obviously, it is correct to state that without some a priori knowledge
of the cube's coloring, the cubemeister cannot tell.  As for whether
you call them both Start:

>    However, if you like this answer, and if you identify the identity
>    with Start, you are in the disquieting situation of having a group
>    with two distinct identities (grin!).

Not at all.  All you have to do is consider the group elements to be
equivalence classes under not only whole-cube rotation but also
reflection.

If you take a(n ordinary) cube and rotate the whole thing a
quarter-turn, the result is not essentially different from the
original - most programs and virtually all humans would consider them
"the same".  Taking the stand that Start and P.A. are the same on the
Edge Cube means only that on the Edge Cube you consider a single group
element to consist of not only a position and all those reachable by
whole-cube rotations, but also those reachable by reflections as well.
The group-theoretic identity is then neither Start nor Pons Asinorum,
but rather the equivalence class containing both those (and 46 other
elements).

> 2. [...]  Just what is the 2x2x2 cube?  Or more correctly, how do you
>    know when it is solved?

When you have achieved any of the 24 elements of the class that we lump
together as Start.

>    With any size of cube, if you restrict yourself to quarter-turns,
>    by definition you cannot rotate the cube in space as a single
>    operation.

I'd argue the 1x1x1 breaks this statement :-)

What's more, it's not clear what "quarter-turn" includes: it usually
doesn't include slice turns on the 3-Cube, but on the 4-Cube and
higher, they must of necessity be included.

>    Yet, a simple quarter-turn sequence such as RL' does rotate the
>    2x2x2 cube because it is faceless.  Is Start of the 2x2x2 operated
>    on by RL' solved?

Yes, I would say so.  I would hope most people would.

>    If so, you can argue that the 2x2x2 has 24 Starts.  Most people
>    would not.  They would argue that there is only one Start, and
>    that 2x2x2 cubes that differed only by a rotation are equivalent.

Right.  I would say that RL' produces a cube that is precisely as
solved as that produced by RR' is - that on the 2x2x2, R and L are in
some sense the same thing.  My position would be that there is only one
Start on the 2-Cube, and it is an equivalence class with 24 members.

> 3. Combining #1 and #2, I *think* that most people would argue that
>    Start and Pons Asinorum on the Edge Cube are not equivalent, but
>    that simple rotations of the 2x2x2 are equivalent.  If I am
>    correct about "most people", why?

I would say that Start and Pons Asinorum on the Edge Cube can be looked
at as mathematically equivalent (though they need not be, if you
choose) but are not intuitively equivalent.  Physical objects generally
cannot be turned into reflected versions of themselves; they normally
*can* be turned into rotated versions of themselves.  Thus, rotations
"feel" equivalent, but reflections don't.

> 4. [...]  Since Start and Pons Asinorum differ only by a simple
>    reflection, why had not my version of M-conjugation declared them
>    to be equivalent?

I'm too lazy to answer this; I no longer have the messages describing
exactly what your M-conjugation operation is online.  Presumably, you
implemented some intuitively-reasonable operation, and it produced
identical results for rotations but not reflections.

> 5. What is a reflection, really?

Ouch. Mathematically, this is easy enough: given a center of reflection
P in Cartesian 3-space, one computes the reflection of a point p as
P+(P-p).  All the things one thinks of as reflections can be
represented as this operation compounded with rotation and/or
translation.

>    Here is an exercise to illustrate the question.  Take two
>    identically colored and oriented 3x3x3 cubes.  On one, perform F
>    and on the other perform F'.  Examine the two cubes, plus their
>    images in a mirror.  Why are there four distinct cubes rather than
>    only two?

There are certainly four cubes - or at least four cube images.  For
there to be only two distinct cubes, one would have to identify some of
them with one another.  However, the only operations (on the cube as a
whole) that will allow identifying two of them are (1) reflection and
(2) recoloring.  If your mathematical treatment considers reflections
or recolorings to be equivalent, then mathematically, there are only
two distinct cubes.  Neither of these operations "feels" trivial,
though, so the four cubes all "feel" distinct.

>    At this point, I can't help but note Martin Gardner's famous
>    mirror question in Scientific American many years ago:  why does a
>    mirror reverse left and right but not up and down?

(rot13 for those who would rather think about this for a while.)

Nf abgrq va jungrire vg jnf V ernq gung dhrfgvba va, vg qbrfa'g - vg
erirefrf sebag-gb-onpx (jurer "sebag" naq "onpx" ner qrsvarq va grezf
bs gur fhesnpr qbvat gur ersyrpgvat).  Jul guvf *nccrnef* gb nzbhag gb
erirefvat yrsg naq evtug vf n zber vagrerfgvat dhrfgvba, naq vg nzbhagf
gb nfxvat jung xvaqf bs fcngvny bcrengvbaf jr cresbez jvgubhg abgvpvat
(pbafvqrevat gurz abbcf).  Va gur pnfr bs n ersyrpgvba bs n crefba, gur
bcrengvba jr'er cresbezvat jvgubhg abgvpvat vf gung bs znccvat
crefba-vzntr bagb frys-vzntr ol ebgngvba, fb nf gb (1) znc urnq bagb
urnq naq srrg bagb srrg naq (2) znc obql-sebag gb obql-sebag naq
obql-onpx gb obql-onpx.  Ersyrpgvba, pbzcbhaqrq jvgu guvf ebgngvba,
*qbrf* erirefr yrsg-gb-evtug.

> 6. [...]

I'm not qualified to comment.

					der Mouse

			    mouse@collatz.mcrcim.mcgill.edu

From hoey@aic.nrl.navy.mil  Tue Jan  4 19:05:27 1994
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Date: Tue, 4 Jan 94 19:05:25 EST
From: hoey@aic.nrl.navy.mil (Dan Hoey)
Message-Id: <9401050005.AA07265@Sun0.AIC.NRL.Navy.Mil>
To: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>,
        <Cube-Lovers@ai.mit.edu>
Subject: Combining conjugacy classes

Last month Jerry Bryan <BRYAN%WVNVM.BITNET@mitvma.mit.edu> posted a
sequence of articles about counting the number of M-conjugacy classes
of Rubik's cube positions.  Having calculated the number of conjugacy
classes of the corner and edge groups separately, his idea was to
combine these to calculate the number of conjugacy classes of the
entire group.  Eventually, he withdrew the calculation, after
realizing that he had not found enough information to determine the
answer.  This article is about how we might calculate the answer from
separate searches of the edge and corner groups.

My first idea is to formalize the concept of symmetry in the conjugacy
classes that Jerry used in his searches.  Recall that the conjugacy
class of a position X is defined to be the set of all positions m'Xmc,
where m is an element of M, the 48-element symmetry group of the cube,
and c is an element of C, the 24-element subgroup of M consisting of
rigid rotations of the cube in space.  The reason some positions have
more symmetry than others is that for some positions, there are
nontrivial elements m and c such that m'Xmc=X.  The way in which this
arises can be formalized as a kind of symmetry group of X.

For an edge-group position X, let CSymm(X) be the set of all f in M
such that X'f'Xf is an element of C.  First, I'll claim that CSymm(X)
is a subgroup of M [see proof 1, below].  Second, I note that CSymm(X)
is the set of all m in M such that there exists c in C with m'Xmc=X
[proof: m'Xmc=X iff X'm'Xm=c'].  Third, I'll claim that if m'Xmc=Y,
then CSymm(X) and CSymm(Y) are conjugate subgroups of M [proof 2].  So
when Jerry says that a position X has order-N symmetry, he is saying
that CSymm(X) has 1152/N elements.  But the identity of CSymm(X) has
more information than just its size, and I believe this information is
crucial if we are to combine symmetry groups.  It looks to me as if it
would be sufficient to record the conjugacy class of CSymm(X), and
there are only 33 possibilities.

Now the usual symmetry group of X, Symm(X), is defined to be the group
consisting of all f in M such that X'f'Xf=I [or, equivalently, Xf=fX.
Symm(X) is the largest group such that X is Symm(X)-symmetric, in the
sense of the Symmetry and Local Maxima article].  The first step in
combining the corner and edge sets is to calculate the symmetry groups
of the rotations of a position X, AllSymms(X)={Symm(Xc) : c in C}.
This corresponds to computing the symmetry groups of the edges-and-
centers group from the symmetry groups of the edges group.  I suspect
there is a way of computing this from CSymm(X), but I do not know it.
I am not even sure that AllSymms(X) is determined by CSymm(X).  One
useful experiment would be to calculate CSymm(X) and AllSymms(Xc) for
all elements of the corner group and see what the correspondence is.

Barring an ability to calculate AllSymms(X) from CSymm(X), we could
calculate AllSymms(X) directly.  This involves a great number of
calculations, though: 24 symmetry group calculations for each element
of the edge group.  My first thought was to try to split the problem
up further, to deal with the group of permutations separately from the
group of orientations.  But I abandoned this when I realized there is
a problem that shows up when we try this with the corner group.  The
permutation of the corners that takes each cubie to its antipode is
clearly M-symmetric, and no matter how we decide to measure
orientation, there is a way to perform this permutation leaving the
cubies in their `home' orientation.  But there is no way to compose
the two together in an M-symmetric way.  I suspect the same problem
arises in the edge group.

But there may be some help from the edge search available in calcu-
lating AllSymms(X).  For take a position Y in the edges-and-centers
group; Y is also a rotated position in the edges group, so Y=m'Xmc for
some X in Jerry Bryan's list.  So for f in Symm(Y), Y'f'Yf=I is an
element of C, so f is in CSymm(Y).  This says that Symm(Y) is a
subgroup of CSymm(Y), which is a conjugate of CSymm(X).  So if Symm(Y)
is nontrivial, then CSymm(X) will also be nontrivial.  So to find the
symmetry groups of the edges-and-centers group we need only look at
those positions that have nontrivial groups in Jerry's list (i.e. less
than order-1152 symmetry), as all the others will have Symm(Y)=I.  So,
Jerry, do you have the data on how many positions of the edge group
have less than order-1152 symmetry, and which positions those are?

So, on to finding the symmetry groups of the Rubik's group positions.
We need to calculate Symm(X) for every element X of the edges-and-cen-
ters group and Symm(Y) for every element Y of the corners-and-centers
group, while keeping track of the permutation parity of X and Y.  (The
permutation parity will be constant over each Symm(X), Symm(Y)).  The
symmetry groups in the Rubik's group will be the intersections of
symmetry groups of edge and corner positions of the same parity.  We
need not keep track of the particular positions here, only the numbers
for each parity and each (conjugacy class of) symmetry group.  I have
a program that could produce a table easily enough.  Recently, I took
a look in Paul B. Yale's _Geometry_and_Symmetry_ and it looks like
this is the sort of problem we could use the Polya-Burnside theorem
on.  Unfortunately, I don't understand it yet, and it looks like the
number of cases here might be too large to conveniently carry out by
hand.  So it would help to go after this problem computationally.

The rest of this article has the proofs for the claims I mentioned in
the second paragraph.
================================================================
Proof 1: Suppose f, g are elements of CSymm(X); it suffices to show
that f'g is an element of CSymm(X).
    X'(f'g)'X(f'g)=X(g'f)X(f'g)
                  =X'g'(Xgg'ff'X')fXf'g
                  =(X'g'Xg) g'f (f'X'fX) f'g,
                  =(X'g'Xg) (f'g)' (X'f'Xf)' (f'g).
Since we assumed f, g in CSymm(X), X'g'Xg and X'f'Xf must be in C.
(f'g)' and (f'g) are elements of M that are either both in C or
neither.  So the product is in C, so f'g is in CSymm(X).  Therefore
CSymm(X) is a group, QED.

Proof 2: Suppose Y=m'Xmc. First let f be an element of CSymm(X), so
that X'f'Xf is in C.  I will first show that m'fm is an element of
CSymm(Y).

Y'(m'fm)'Y(m'fm)=(m'Xmc)'(m'fm)'(m'Xmc)(m'fm)
                =(c'm'X'm)(m'f'm)(m'Xmc)(m'fm)
                =c'm'(X'f'X)(mcm'fm)
                =c'm'(X'f'Xf)(f'mcm'fm)
All of which are elements of M, with an even number in C. Therefore
the expression is in C, so m'fm is in CSymm(Y).

Now let g be an element of CSymm(Y), so that Y'g'Yg is in C.  Let
f=mgm', so f is an element of M such that m'fm is in CSymm(Y).  I will
show that f is an element of CSymm(X):

X'f'Xf=(mc)(mc)'X'(mm')f'(mm')Xf(f'mcm'fm)(f'mcm'fm)'
      =(mc)(m'Xmc)'(m'fm)'(m'Xmc)(m'fm)(f'mcm'fm)'
      =(mc)Y'(m'fm)'Y(m'fm)(f'mcm'fm)'
      =(mc)Y'g'Yg(f'mcm'fm)',

which is in C, so f is in CSymm(X).

I've shown that for every element f of CSymm(X), m'fm is an element of
CSymm(Y), and that every element of CSymm(Y) is m'fm for some f in
CSymm(X).  Therefore CSymm(Y)=m' CSymm(X) m, QED.
================================================================
Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From hoey@aic.nrl.navy.mil  Tue Jan  4 21:36:19 1994
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	id AA07661; Tue, 4 Jan 94 21:36:18 EST
Date: Tue, 4 Jan 94 21:36:18 EST
From: hoey@aic.nrl.navy.mil (Dan Hoey)
Message-Id: <9401050236.AA07661@Sun0.AIC.NRL.Navy.Mil>
To: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>, Cube-Lovers@ai.mit.edu
Subject: Re: Which is the Real Start?

Jerry Bryan <BRYAN%WVNVM.BITNET@mitvma.mit.edu> has some more questions.

> Take a standard 3x3x3 Rubik's cube, and remove the corner and
> center labels to make an Edges Cube.... Scramble..., how will
> the cubemeister distinguish Start from Pons Asinorum?
> ... if you identify the identity with Start, you are in the
> disquieting situation of having a group with two distinct identities
> (grin!).

The problem is that we would not be dealing with a _group_ then, but a
collection of cosets of M.  Just as in the edge `group', we deal with
either    1) a less-symmetric group in which one of the edges never
moves, or 2) a larger group in which we distinguish positions that
differ by rigid motions of the cube, or
          3) a non-group in which we consider cosets--equivalence
classes of group #2, where group elements that differ by rigid motions
are equivalent.  You have got a lot of mileage out of working with
group #2 to save duplication among symmetries, then reducing to
non-group #3.  But what you lose is the group structure of the object
you are studying.  Instead, you have to work in the large group and
then deduce information about the cosets.  All in all, though, I'm
very glad of it, for the lost symmetries of group #1 were sorely
missed.

For most of the other questions, mouse@collatz.mcrcim.mcgill.edu
provides satisfactory answers.  However, strictly speaking we should
not call an equivalence class to be a group element (unless it is a
coset of a normal subgroup, and neither C nor M is normal in the large
group).  I'll admit I've also abused the term when considering
distances in the ``edge group'', as if all 24 rotations of a position
were the same element of some group.  But when we start dealing with
the distinction between fixed and movable cubes I think we need to
start being more careful.

[ mouse also mentions that quarter-turn ``usually doesn't include
  slice turns on the 3-Cube, but on the 4-Cube and higher, they must
  of necessity be included.''  I'll take that as an argument for
  eccentric slabism: a QT rotates any 1xNxN slab except a central slab
  of an odd-edged cube.  As opposed to cutism, where a QT consists of
  a rotation of part of the cube with respect to the other. ]

Other questions:
>  ...since Start and Pons Asinorum differ only by a simple
>  reflection, why had not my version of M-conjugation declared them
>  to be equivalent?

Your versino treats positions X,Y for which m'Xmc=Y (m in M, c in C)
as equivalent.  If you instead determine when m'Xmn=Y (m,n in M) you
would find them equivalent.  This is equivalent to changing the loop
in your version of M-conjugacy.

>      For j = 1 to 24 for k = 1 to 24 for m = 1 to 2
>             for i = 1 to 24 Bj,k,m(i) = Qm(Pj(A(Qm(Pk(i)))))

so that the two occurrences of Qm need not be the same.

> (I speak of "my version of M-conjugation", but the question is no
> different if you look at Dan Hoey's original M-conjugation).

No, I didn't use M-conjugation except for a cube with a fixed
orientation in space [or equivalently, with face centers].  So in the
original concept of M-conjugation that Jim Saxe and I put together,
Start and Pons Asinorum don't just differ by a reflection.

> I found Dan Hoey's postings about the four special states of the
> Edge Group to be delightful....  However, [without the results on
> distances] if we identified the group as being rectangular, would we
> be led to saying which of the four special states were diagonally
> opposed without the computer search?  Without the search, I might be
> tempted to say that Start and Pons Asinorum were diagonally opposed.

Well, really the `group' is in the shape of a sphenoid, a word I
learned yesterday for a tetrahedron whose three pairs of opposite
edges are equal.  [Or equivalently, a tetrahedron whose edges are face
diagonals of a rectangular prism.]  But it might be more accurate to
consider it as a large ball of string with a bunch of symmetries.
Calling it a rectangle or sphenoid may lead us to ignore the structure
that is not representable in Euclidean space.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU  Thu Jan  6 04:11:20 1994
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From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Square Brackets

This posting has very little to do with cubes of any sort, but I
think you may find it of interest anyway.  If not, you can just
delete it.

In his analysis of my operator which combines M-conjugation with
a whole cube rotation, Dan Hoey inserted square brackets to make
his exposition more readable.  And therein lies a story.

As it turns out, I had composed most of my original note at work,
and I had used square brackets.  Actually, I had used square
brackets to delimit the indexes for the rotation and reflection
operations, and I had used parenthesis to delimit the indexes for
the individual cells in the row vectors.  I wanted to make a
distinction between the two kinds of indexing.  Dan avoids the
necessity for a distinction by simply not detailing the
indexing of the individual cells.

Anyway, I completed the note at home.  Much to my dismay, all of my
square brackets had disappeared!  I pretty much understand the
problem.  My E-mail system is an IBM mainframe which uses EBCDIC as
its basic code. EBCDIC does not deal very well with square brackets.
There are at least two "standards" for encoding square brackets in
EBCDIC.

There are any number of ways to access an IBM mainframe, but the
native terminal support is 3270 terminals using EBCDIC.  Both at home
and at work, I use a PC running TN3270 to access our mainframe.
TN3270 is a 3270 version of TELNET.  However, the TN3270 I use at work
is considerably different from the TN3270 I use at home.  One TN3270
implements one of the square bracket standards, and the other
TN3270 implements the other.

For similar reasons, mail gateways often have difficulties with
square brackets.  They may have to translate EBCDIC to ASCII or
ASCII to EBCDIC, and it is difficult to know how best to set up
the translate tables.  My experience is that some gateways get it
"right" and others get it "wrong".  I therefore had a great fear
that if I posted my note with square brackets, that the square brackets
might appear as gibberish to at least some of you.  Thus, I sort
of temporized and faked the subscripts with upper and lower
case letters (e.g., Bk to mean B-sub-k), omitting square brackets
entirely.

It is probably no accident that old programming languages such
as FORTRAN and COBOL use parentheses for indexing.  These languages
originated in the 50's.  At that time, the dominant character
code was BCD, which did not include square brackets.  EBCDIC is just
extended BCD, and the original EBCDIC did not include square brackets,
either.  Square brackets are a latter day addition to EBCDIC, and
the implementation of square brackets in EBCDIC is inconsistent.

ASCII has always included square brackets.  "Modern" languages
(say, starting in the 70's) such as Pascal and C (and their
descendents) grew up in the ASCII world, and tend to use
square brackets for indexing and parentheses for function
arguments.  FORTRAN compilers to this day have difficulty figuring
out with things like Y=X(I) or Y=F(X)  -- which are functions and
which are subscripted arrays.  Also, Pascal and C tend not to
co-exist very well in the EBCDIC world because of these kinds of
character set difficulties.

There are several other characters with similar difficulties.  For
example, if G is the cube group, you might want to refer to the
size of the cube group as |G|.  But the delimiting vertical bars
can be very different between EBCDIC and ASCII.  Finally, FORTRAN
used ** for exponentiation.  More modern languages tend to use
some sort of up-arrow or carat.  But such characters don't
translate well between EBCDIC and ASCII.  For example, if I
write |G| = 4.3 * 10^19, it is highly problematic whether the
character between the 10 and the 19 which I am using to express
exponentiation will make any sense on your particular system.

For whatever it is worth, here are my home and work versions of
square brackets:

home left  square bracket  [[[[[[[ x'AD'
home right square bracket  ]]]]]]] x'BD'
work left  square bracket          x'BA'
work right square bracket          x'BB'


 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU  Thu Jan  6 13:52:03 1994
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From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   An Alternate Analysis of B

Here is a definition.  If X is any cube, then the function B
is defined as B(X) = min (C[i] R[j] X R[j] C[k]) for i in {1 to 24},
j in {1 to 2}, and k in {1 to 24}, and where C is the set of rotations
of the cube and R is the set of reflections of the cube.  Then, two
cubes X and Y are said to be B-equivalent if B(X)=B(Y).  Note that
the B function calculates the "canonical form" or the "representative
element" of an equivalence class under M-conjugation and rotation of
the cube in space.

Dan Hoey already gave a thorough analysis of this situation.  I would
like to provide an alternative analysis which I hope to be equivalent
to Dan's.  My analysis will be fairly informal as compared to Dan's.

Here are a couple of preliminaries.  First, multiplication of
permutations is generally not commutative.  For example, if X is any
cube, then it is generally not the case that m'Xm=mXm', where m
is in M, the set of all cube rotations composed with reflections.
However, we can calculate all M-conjugates of X as B[n]=m[n]' X m[n]
for n in {1 to 48}, or we can calculate all M-conjugates of X as
B[n]=m[n] X m[n]'.  Either way, B will be the same set of cubes.  It
will be in a different order, but it will be the same set.  The reason
is that the set of all m[n] is the same as the set of all m[n]',
namely just M, but m[n] is in a different order than m[n]'.  This means
that as long as we are calculating all M-conjugates as opposed to
a specific M-conjugate, we can sort of "violate" the normal rules
about multiplication commutivity.

Second, if X is any cube, consider the set of all rotations of
X, namely B[i] = X c[i] for i in {1 to 24}, and where c[i] is in C,
the set of all cube rotations.  Having generated the set of all
rotations of X, we can rotate as many times as we wish, for example
B[j] = X c[i] c[j] for i in {1 to 24} and j in {1 to 24}, or even
B[k] = X c[i] c[j] c[k] for i in {1 to 24}, j in {1 to 24}, and
k in {1 to 24}.  No matter how many times we multiply, B will be the
same set, it will just be in a different order.  Conversely, if we
have any number of adjacent rotations in the multiplication, we can
eliminate all but one rotation, and B will be the same set, and again
will just be in a different order.

With the preliminaries out of the way, we note that the set of all
M-conjugates of X is generated as B[i]=m[i]' X m[i] for i in
{1 to 48}.  But we can also generate the same set in a different order
as B[i]=m[i] X m[i]' for i in {1 to 48}.  We can decompose M and
calculate all M-conjugates as B[i,j]=c[i] r[j] X r[j]' c[i]' for i in
{1 to 24} and j in {1 to 2}.

But r[1]'=r[1] (r[1] is the identity) and r[2]'=r[2] (the reflection
is its own complement).  So we have B[i,j]=c[i] r[j] X r[j] c[i]'
for i in {1 to 24} and j in {1 to 2}.  The set of all c[i] is the
same as the set of all c[i]' (just in a different order), so we
define i' as the index for which c[i']=c[i]'.  Hence the calculation
of an M-conjugate can be written as B[i,j]=c[i] r[j] X r[j] c[i']
for i in {1 to 24} and j in {1 to 2}.

Finally, we wish to multiply the M-conjugate by the set of all rotations,
so we have B[i,j,k]=c[i] r[j] X r[j] c[i'] c[k] for i in {1 to 24},
j in {1 to 2}, and k in {1 to 24}.  But as we noted in our second
preliminary note, we can collapse multiple rotations into one, and we
have B[i,j,k]=c[i] r[j] X r[j] c[k], and B will be the set of all
M-conjugates of X multiplied by all rotations.  I guess I am overusing
the letter "B" a bit, because the "B" function is simply the minimum
of the "B" matrix.  But in any case, we have shown that the "B" loop
in my program is simply calculating the set of all M-conjugates multiplied
by all rotations.  This is the exact result already proven by Dan Hoey,
but I found the above derivation a little easier to follow.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From pbeck@pica.army.mil  Thu Jan  6 14:22:12 1994
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Date:     Thu, 6 Jan 94 14:06:54 EST
From: Peter Beck (BATDD) <pbeck@pica.army.mil>
To: Cube-Lovers@ai.mit.edu
Subject:  Mickey's Challenge
Message-Id:  <9401061406.aa23113@COR6.PICA.ARMY.MIL>


NEW PUZZLE "MICKEY'S CHALLENGE" is at your Disney
store now, price $10.

This is a legal MACHBALL, ie, a spherical
SKEWB.  It comes with a solution book.  
Christoph Bandelow (a longer time cuber)
wrote the solution.
I haven't bought one or it played with it
yet.

GOOD PUZZLING

pete beck

pbeck@pica.army.mil


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From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Character Test, Please Ignore

Forward slashes   ///////////////
Back slashes      \\\\\\\\\\\\\\\
Left Braces       {{{{{{{{{{{{{{{
Right Braces      }}}}}}}}}}}}}}}
Carat             ^^^^^^^^^^^^^^^
Tildes            ~~~~~~~~~~~~~~~

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

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Date:      Fri, 7 Jan 1994 10:33:04 EST
From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Some Proposed Terminology

I wish to propose some terminology and definitions to make certain
concepts a bit more precise.  For example, when we are talking about
"corners only", it is not always clear whether we are talking about
"corners with centers without edges" or "corners without centers
without edges".  In this note, I have tried to be consistent with
previous usage on the list, but I welcome any historical corrections
that might be deemed necessary.

Let G be the standard cube group for the 3x3x3 cube, and let |G| be
the size G.  Hence, we have |G| = (8!(3^8)/3 * 12!(2^12)/2) / 2,
which is the famous 4.3 * 10^19.

Let GC be the corners with centers without edges group for the 3x3x3
cube, and let |GC| be the size of GC.  Hence, we have
|GC| = 8!(3^8)/3.  (I welcome a suggestion other than "GC" as the
name for this group.  I did not find one in the archives.)  This group
could be modeled by removing the edge labels from a standard
3x3x3 cube.

Let GE be the edges with centers without corners group for the 3x3x3
cube, and let |GE| be the size of GE.  Hence, we have
|GE| = 12!(2^12)/2. (As before, I welcome a suggestion other than
"GE" for the name for this group.)  This group could be modeled by
removing the corner labels from a standard 3x3x3 cube.

Note that |G| = |GC| * |GE| / 2.

Let G\C be the corners with edges without centers group.  I intend
for the notation to indicate G reduced by C, where C is the rotation
group for the cube.  It should be the case that |G\C| = |G| / 24,
but I want to return to this question a little later.  This group
could be modeled by removing the center labels from a standard
3x3x3 cube.

Let GC\C be the corners without edges without centers group.  This is
the 2x2x2 cube.  We should have |GC\C| = |GC| / 24, but again I want
to return to this question a little later.  In addition to being the
2x2x2 cube, this group could be modeled by removing the center and
edge labels from a standard 3x3x3 cube.

Let GE\C be the edges without corners without centers group.  We
should have |GE\C| = |GE| / 24, but again I want to return to this
question a little later.  This group could be modeled by removing
the center and corner labels from a standard 3x3x3 cube.

Let G\M be the set of M-conjugate classes for G.  In this case,
|G\M| is approximately 48 times smaller than |G|.  I believe that
when Dan Hoey asked in 1984 the question "how big is G, really?",
that he was really asking how big is G\M, and that he was asking
for the approximation to be resolved to an exact number.

Let GC\M be the set of M-conjugate classes for GC.  In this case,
|GC\M| is approximately 48 times smaller than |GC|.

Let GE\M be the set of M-conjugate classes for GE.  In this case,
|GE\M| is approximately 48 times smaller than |GE|.

Recall that B is the function which calculates the canonical form
for a cube under the composed operations of M-conjugation plus
rotation.  My  programs calculate equivalence classes under B.

Let G\B be the set of B-classes for G.  Let GC\B be the set of B-classes
for GC.  Let GE\B be the set of B-classes for GE.  So far, my programs
have built complete search trees for GC\B and GE\B.

Let Gx denote any of G, GC, and GE.  Then, we have Gx\B=(Gx\C)\M=(Gx\M)\C.
In English, we can decompose B into a multiplication by C and M (in
either order).

Also, note that Gx\C=(Gx\C)\C=((Gx\C)\C)\C=....  Similarly,
(G\M)\C=((G\M)\C)\C=....  In English, having reduced once by C, we can
reduce again by C as many times as we wish, but we simply get the same
set back again each time.

This notation can help us address the question of whether B actually
accomplishes a "times 48" or a "times 1152" reduction in the size of
the cube.  If we are dealing with Gx, then Gx\B is a "times
1152" reduction.  However, information is lost.  For example, consider
GC and GC\B.  GC is "corners plus centers", and B-reduction of GC
removes the centers and calculates M-conjugates of the corners.
But you really don't have the same problem any more because the centers
are gone.  If on the other hand we are dealing with Gx\C, then (Gx\C)\B
is a "times 48" reduction.  All we have really done is calculate
M-conjugates. The reduction by the C that is composed into B is duplicate
effort which accomplishes nothing.

I have come to realize that my program for the 2x2x2 actually models
GC (corners with centers without edges) rather than GC\C (corners
without centers without edges).  My program does not explicitly encode
the centers.  However, it encodes all eight corner cubies, and when
it makes qturns, any of the eight cubies can move.  Hence, rotational
information is encoded, even if the centers themselves are not explicitly
encoded.  If I wanted to model GC\C, I would have had to either model
only seven of the cubies, or else modeled all eight but moved only seven
of them.  Since what I really wanted was (GC\C)\M, and since what I had
was GC, I had to invent this funny B thing, where GC\B=(GC\C)\M.  If I
had been clever enough to model GC\C in the first place, I never would
have had to invent B.  Similar comments apply to my model for the edges.

To convince yourself that eight corner cubies model GC and seven corner
cubies model GC\C, just think about calculating |GC| and |GC\C|.
For |GC|, there are eight ways to pick the first cubie, seven ways to
pick the second cubie, and so forth yielding the familiar
|GC|=8!*(3^8)/3.  For |GC\C|, we let one of the cubies be fixed,
then there are seven ways to pick the second cubie, and so forth yielding
|GC\C|=7!*(3^7)/3, and |GC| = |GC\C| * 24.  Hence, the "corners of the
3x3x3" problem is 24 times larger than the "2x2x2" problem.

I will discuss the "times 24" reduction that is accomplished by
reducing by C in a followup note.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

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From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Some Terminology Concerning B

I have started to use "B" to indicate various aspects of the
conjugacy class generated by m'Xmc.  The choice of B is sort
of an accident.  I used "B" in the program fragment which
I posted to the list, and Dan Hoey analyzed the program fragment.
I have called it "B" in my mind ever since.  However, I have used
B in several inconsistent ways.  This is a proposal to
rectify that inconsistency.

Let X be any cube.  Then the set of B-conjugacy classes of X is
the set of all m'Xmc for all m in M and all c in C.  We denote
this set as BClass(X).  B is the function B(X)=min(BClass(X)).

Note that we could have defined BClass(X) equivalently as the
set of all mXm'c, or as the set of all cm'Xm, or as the set of
all cmXm'.  It is in general not the case that
m'Xmc = mXm'c = cm'Xm = cmXm' for any fixed value of m and c.
(Quite the contrary!).  However, when we say "the set of all...",
the four ways of generating BClass(X) become equivalent.  This is
the justification for the assertion in a previous note that
Gx\B = (Gx\M)\C = (Gx\C)\M.

Two cubes X and Y are B-equivalent if BClass(X) = BClass(Y).
Equivalently, two cubes X and Y are B-equivalent if B(X) = B(Y).

|X| is the length of X (the distance of X from Start).  We have
|B(X)| = |X| for centerless cubes, but it is generally not the
case that |B(X)| = |X| for cubes with centers.  In fact, let
X and Y be cubes with centers such that B(X)=B(Y).  It is not
necessarily the case that |X| = |Y|.  For example, consider
the set GC of cubes with corners with centers without edges.
We have B(RL')=B(I), but |RL'|=2 and |I|=0.

|BClass(X)| is the number of elements in BClass(X).  If
|BClass(X)| = N, then X is said to have order-N symmetry.  (I
sincerely regret ever using this terminology.  As has been noted
on the list, it seems "backwards" somehow.  But given that this
usage exists, the value 1152/N is generally more useful than the
value N.)

We note the following:

  1. B(X) is a cube.

  2. BClass(X) is a set of cubes.

  3. B(B(X)) = B(X)

  4. BClass(B(X)) = BClass(X).

  5. Both X and B(X) are in BClass(X).

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU  Sat Jan  8 11:25:43 1994
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From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Calculating |G\M|

Armed with Dan Hoey's note of 4 January "Combining conjugacy classes",
I wish to propose once again a procedure for calculating |G\M|, the
number of M-conjugate classes of G, which I think in some sense is the
"real" size of G.  My proposal draws *very* heavily on Dan's note.

My first (incorrect) proposal was based on the following idea.  By
computer search, we already have a database of GC\B and GE\B, the
corners and edges of G, respectively, reduced by B-conjugacy.  Hence,
we know |GC\B| and |GE\B|.  For each X in GC\B and each Y in GE\B,
we calculate BClass(X) and BClass(Y).  Then, we can combine
BClass(X) and BClass(Y) in all legal ways (minding parity considerations).
Call the combinations BClass(X) * BClass(Y).  For any fixed X and Y,
BClass(X) * BCLass(Y) is a set of cubes in G.  Hence we can calculate
(BClass(X) * BClass(Y))\M and |(BClass(X) * BClass(Y))\M|.  My idea
then was just to sum |(BClass(X) * BClass(Y))\M| over all values
of X and Y to calculate |G\M|.  And we know how many X's and Y's there are!
But, alas, |(BClass(X) * BClass(Y))\M| is not the same across all X's and
Y's because, well because of symmetry.  All X's and Y's are not equally
symmetrical.  I was assuming that |(BClass(X) * BClass(Y))\M| was
constant, and of course it is not.

My next (incorrect) proposal was never posted to the list.  It was
a slight improvement on the first idea.  We have a data base of all X's
in GC\B and of all Y's in GE\B.  For each X in GC\B and for each
Y in GE\B, we know |BClass(X)| and |BClass(Y)|.  (Actually, we don't.
I have to calculate it.  I have done so, and I have posted
summaries of those calculations.  However, I did not store
the order of the equivalence classes in the data base.  I kick myself
for not doing so, but this is a minor problem, so let us continue).

There are only 10 distinct values for |BClass(X)| and for |BClass(Y)|,
namely 24, 48, 72, 96, 144, 192, 288, 384, 576, and 1152.  (By the
way, I have never figured out why it is *exactly* the same set of
values for both the corners and for the edges.  It is easy to see
why it is approximately the same set of values, but the structure
of the corners is enough different from the structure of the edges
that I see no obvious reason the set of values should be exactly
the same in both cases.)

Let GC[m] be the set of all X for which |BClass(X)| = m and let
GE[n] be the set of all Y for which |BClass(Y)| = n. Hence,
GC\B is partitioned into GC[m]\B for m=24,48..., and GE\B is
partitioned into GE[n]\B for n=24,48,...  Now, we form the sets
BClass(X)[m] * BClass(Y)[n] for all X in GC[m]\B and for all
Y in GE[n]\B, and for all legal values of m and n.  There will
be 100 such sets.

For any fixed X, Y, m and n, BClass(X)[m] * BCLass(Y)[n] is a
set of cubes in G.  Hence we can calculate
(BClass(X)[m] * BClass(Y)[n])\M and |(BClass(X)[m] * BClass(Y)[n])\M|.
My idea then was just to sum |(BClass(X)[m] * BClass(Y)[n])\M|
over all X in GC[m]\B and over all Y in GE[n]\B to calculate
|(BClass(X)[m] * BClass(Y)[n])\M|.  We know how many X's there are
in GC[m]\B and we know how many Y's there are in GE[n]\B, so the
calculation seemed possible.  I then intended to sum again over all
m and over all n, and I would be done.

But, alas, in order to perform the sum over all X and all Y, I
needed a theorem which I couldn't prove and which I now believe
is not true anyway.  I needed to be able to prove that for a fixed
m and n, that |(BClass(X)[m] * BClass(Y)[n]| had the same value
for all X in GC[m]\B and all GE[n]\B.  For a while I thought it was
true, but right now I can't think of any reason why it should be.

But perhaps Dan Hoey comes to the rescue with his CSymm function.
I still need a theorem which I cannot (yet) prove, but I believe it
is true.  If it can be proven, my basic overall scheme can be
rescued.

In my second proposal, I used the values of all possible BClass
sizes as indexes  --  24, 48, 72...  It would perhaps be more
convenient to make these sizes a set {24, 48, 72, ...},
and to think of the indexes m and n taking on the values from
1 to 10, where the values from 1 to 10 index the set
{24, 48, 72, ....}.  With this understanding, all the above results
are valid, and the indexing is more convenient.

We can now say that GC\B is partitioned into GC[1]\B, GC[2]\B, ...
through GC\B[10] and similarly for GE\B.  Unfortunately, using
the B-equivalence class sizes to partition GC\B and GE\B did not
permit us perform the calculations we wanted to perform.  However,
suppose we partition GC\B and GE\B a different way, namely using
CSymm.  Suppose, for each X in GC\B and for each Y in GE\B,
we calculate CSymm(X) and CSymm(Y). (We would have to do this
by computer).  CSymm(X) and CSymm(Y) are sets, but there are only
a (relatively) small number of such sets.  Let each distinct value
CSymm(X) and CSymm(Y) be mapped to an index.  We can call such a
mapping function CSind, and we can calculate CSind(CSymm(X)) and
CSind(CSymm(Y)).  Actually, there is no reason not to define
CSind in such a way that the domain is the set of X's and Y's, so
that we can calculate CSind(X) and CSind(Y).

Now, we use m=CSind(X) and n=CSind(Y) to form a partition of
GC\B and GE\B.  All our results from before are valid.  The only
issue is, can we now perform the sum?  In order to perform the sum,
we need the following to be true:

   For a fixed m and n, |BClass(X)[m] * BClass(Y)[n]| is constant
   for all X in GC[m]\B and all Y in GE\[n]\B, where GC[m] is
   the set of all X in GC such that CSind(X)=m and GE[n] is the set
   of all Y in GE such that CSind(Y)=n.

It really seems true to me, and I shall strive to prove it.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU  Sat Jan  8 18:47:56 1994
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Date:      Sat, 8 Jan 1994 15:10:52 EST
From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Calculating |G\M| (Three Typos)

I have found three typos in my article from this morning.  Two are
trivial, and I would not have bothered to report them.  One of them
is fundamental, and I feel obliged to report a correction.


>for all X in GC[m]\B and all GE[n]\B.  For a while I thought it was
                      and for all Y in GE[n]\B



>We can now say that GC\B is partitioned into GC[1]\B, GC[2]\B, ...
>through GC\B[10] and similarly for GE\B.  Unfortunately, using
         GC[10]\B




>   For a fixed m and n, |BClass(X)[m] * BClass(Y)[n]| is constant
                         |(BClass(X)[m] * BClass(Y)[n])\M|
>   for all X in GC[m]\B and all Y in GE\[n]\B, where GC[m] is
>   the set of all X in GC such that CSind(X)=m and GE[n] is the set
>   of all Y in GE such that CSind(Y)=n.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From CPELLEY@delphi.com  Mon Jan 10 17:31:46 1994
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Date: Mon, 10 Jan 1994 12:49:21 -0400 (EDT)
From: CPELLEY@delphi.com
Subject: Mickey's Challenge
To: Cube-Lovers@ai.mit.edu
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I visited the local Disney Store and picked up a Mickey's Challenge puzzle for
$10.  It's really cute, and the book that it comes with is excellent.  
Included are color photos of Christophe Bandelow, Uwe Meffert, and the puzzle
disassembled into all its parts.  Plus it gives a solution for the puzzle and
has a short bio on Uwe Meffert.  It also shows color photos of the Megaminx,
Pyraminx (not the Tomy version, but a black one), and the 5x5x5 which they
refer to as "Professor's Cube."
Some general notes on Mickey's Challenge.  It is a spherical Skewb, and it
actually turns much more smoothly than my cubical Skewb.  It has the same
delightful "clicking" mechanism that the Skewb and original Pyraminx had.
It is a bit easier than the Skewb, since there are a few blank pieces that
can be exchanged without noticing the difference.  In fact, the book's solution
actually leaves Mickey intact while solving Donald.  After you're bored with
solving it, the concept of making patterns takes on strange dimensions, as you
can make Mickey and Donald exchange body parts and look like Disney on acid!
All in all, it is an excellent little puzzle and I am very glad to see the
Skewb widely available to puzzle enthusiasts everywhere.
One final note:  the booklet gives no credit whatsoever to Tony Durham, who
was credited with the Skewb's invention in Hofstadter's Sci Am articles years
ago.  They instead credit Meffert, since the Skewb's mechanism is based on
the Pyraminx.

From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU  Mon Jan 10 23:08:35 1994
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Date:      Mon, 10 Jan 1994 23:06:00 EST
From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   |G\M| - Some Trivial Partial Results

It occurs to me that a small part of my incorrect attempt during
December to calculate |G\M| can be salvaged.  In particular,
for those cases where B-conjugate classes are of order 1152,
the calculations are trivial.  About 99.9923% of the edge conjugate
classes and about 96.924% of the corner conjugate classes are of
order 1152, so we can calculate the correct number of M-conjugates
of G for a very large percentage of the cases.

Consider some fixed X in GC\B and some fixed Y in GE\B where
|BClass(X)|=1152 and |BClass(Y)|=1152.  Form
BClass(X) * BClass(Y).  Now, |BClass(X) * BClass(Y)| =
|BClass(X)| * |BClass(Y)| / 2 = 1152 * 1152 / 2.
(The division by 2 takes care of parity).  Finally, form
(BClass(X) * BClass(Y))\M, and we have |(BClass(X) * BClass(Y))\M| =
|BClass(X) * BClass(Y)| / 48 = (1152 * 1152) / 2 / 48 = 13,824.
We know the number of BClasses of GC of order 1152 from computer
search (namely 75,392), and we know the number of BCLasses of GE of
order 1152 from computer search (namely 851,493,140).  Hence, for
the special case of both BClasses being of order 1152, we have the
total number of elements of G\M being 851,493,140 * 75,392 * 13,824 =
887,442,335,689,605,120.

We can derive similar results if only one of BCLass(X) and BClass(Y)
are of order 1152.  For example, there are 4 elements of GE\B
for which |BClass(Y)|=24.  Choose such a Y, and choose X in GC\B
such that |BClass(X)|=1152.  Form BClass(X) * BClass(Y).  It will
be the case that |BClass(X) * BClass(Y)| = 1152 * 24 / 2 = 13,824.
Hence, |(BClass(X) * BClass(Y))\M| = 13,824/48 = 288.  There are 75,392
values of X for which |BClass(X)|=1152, 4 values of Y for which
|BCLass(Y)|=24, and hence there are 75,392 * 4 * 288 = 86,851,584
elements of G\M of the form BClass(X) * BClass(Y) for which
|BCLass(X)| = 1152 and |BClass(Y)| = 24.

There are nineteen cases in all for which at least one of BClass(X)
and BCLass(Y) are of order 1152, and this note calculates only
two of the nineteen.  Completing the other
seventeen would be trivial but tedious.  However, a total solution
to the problem will require coming up with some way to deal with
the cases where neither |BClass(X)|=1152 nor |BClass(Y)|=1152.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From tonyd@earwax.pd.uwa.edu.au  Tue Jan 11 02:06:32 1994
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Date: Tue, 11 Jan 1994 15:08:16 +0800
To: Cube-Lovers@ai.mit.edu
From: tonyd@earwax.pd.uwa.edu.au
Subject: Rubik chaos?

On sci.nonlinear...

In article <1994Jan5.120409@oxygen.aps1.anl.gov> Thomas D. Orth,
orth@oxygen.aps1.anl.gov writes:
>A friend of mine has written a few papers on the subject of
>the Rubik's Cube Group, and the elements of Chaos within
>it, or Pseudo-chaos as she calls it.

The papers are being submitted to the journal CHAOS. 

cheers, Tony



From @mail.uunet.ca:mark.longridge@canrem.com  Tue Jan 11 16:07:24 1994
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Subject: Andras Mezei's Book
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.692.5834.0C190BA3@canrem.com>
Date: 	Tue, 11 Jan 1994 12:18:00 -0500
Organization: CRS Online  (Toronto, Ontario)

A while back I reported on the list of cube books available at the
Library of Congress. At the time, I did not realize the
significance of #2:

2. 85-109601: Mezei, Andras.  Magyar kocka, avagy, Meg mindig ilyen
     gazdagok vagyunk? /  Budapest : Magveto, c1984.  473 p. : ill.
     ; 21 cm.

Digging through some old magazines I reread the cube article in
the March 1986 issue of "Discover" magazine. In this issue
John Tierney talks to Rubik himself. The article itself is
excellent, showing pictures of Rubik's first wooden prototype,
and having discussions on the Golden Age of the Cube when
only Rubik had access to his invention.

I learned that Andras Mezei (a Budapest writer) wrote a book and a
play called "The Hungarian Cube". This chronicles the debacle
that occured when the Hungarians attempted to expand their
operations to meet the huge demand, rather than farm the work
to other factories. Andras writes: "Everyone made money on the
cube except the Hungarians".

Does anyone on Cube-Lovers have this book? Judging by the size
of the book, and the fact that it's illustrated, I think it
would be a worthy addition to any cubist's library, even more
so if there exists an english translation. If not, I feel
encouraged enough to get an English-Hungarian dictionary and
read it anyway!

  -> Mark

From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU  Wed Jan 12 23:46:28 1994
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Date:      Wed, 12 Jan 1994 23:43:54 EST
From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   |G\M| - the Nineteen Cases We Know

A - |BClass(X)| for X in GC\B (order of B-conjugate class for corners)
B - |BClass(Y)| for Y in GE\B (order of B-conjugate class for edges)
C - |GC[A]\B| (number of B-conjugate classes of size A for corners)
D - |GE[B]\B| (number of B-conjugate classes of size B for edges)
E - |BClass(X) * BClass(Y)| = |BClass(X)| * |BClass(Y)| / 2
F - |BClass(X) * BClass(Y)| * |GC[A]\B| * |GE[B]\B| / 48
    (number of M-conjugates of |GC[A]\B * GE[B]\B)

      A     B       C            D        E                       F

    1152    24  75,392            4   13,824               86,851,584
    1152    48  75,392            2   27,648               86,851,584
    1152    72  75,392           12   41,472              781,664,256
    1152    96  75,392           16   55,296            1,389,625,344
    1152   144  75,392          110   82,944           14,330,511,360
    1152   192  75,392           70  110,592           12,159,221,760
    1152   288  75,392        1,544  165,888          402,296,537,088
    1152   384  75,392        1,252  221,184          434,952,732,672
    1152   576  75,392      128,858  331,776       67,149,128,466,432
    1152  1152  75,392  851,493,140  663,552  887,442,335,689,605,120
      24  1152       1  851,493,140   13,824          245,230,024,320
      48  1152       1  851,493,140   27,648          490,460,048,640
      72  1152       3  851,493,140   41,472        2,207,070,218,880
      96  1152       1  851,493,140   55,296          980,920,097,280
     144  1152      14  851,493,140   82,944       20,599,322,042,880
     192  1152      15  851,493,140  110,592       29,427,602,918,400
     288  1152     135  851,493,140  165,888      397,272,639,398,400
     384  1152      32  851,493,140  221,184      125,557,772,451,840
     576  1152   2,208  851,493,140  331,776   12,995,229,448,765,440

   Total                                      901,082,361,368,033,280

Note that we have covered over 99.99 percent of edge positions (which
are combined with all corner positions), and over 96.9 percent of
the remaining corner positions (which are combined with all edge
positions).  Hence, we have covered about 99.99969 percent of all
positions.  However, that last fraction of a percent is going to
be devilishly difficult.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From @mail.uunet.ca:mark.longridge@canrem.com  Thu Jan 13 05:03:10 1994
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Sender: CRSO.Cube@canrem.com
Subject: 4x4x4 Cube moves
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.694.5834.0C190EF0@canrem.com>
Date: 	Wed, 12 Jan 1994 17:07:00 -0500
Organization: CRS Online  (Toronto, Ontario)

Some comments on flipping a single pair of edges on the 4x4x4 cube:

Singmaster notation on the 4x4x4 (same notation as Jeffery Adams)
--------------------------------

L left face             l inner left slice
r inner right slice     R right face
F front face            f inner front slice
b inner back slice      B back face
U up face               u inner up slice
d inner down slice      D down face

So L1 would be turn the left face 90 degrees clockwise and l1 would
 be turn the inner left slice 90 degrees clockwise.

I'll use the suffix "2" to be for 180 degree turns and the suffix
 "3" to be for 270 degree turns clockwise or 90 degree turns
 counterclockwise.

This is the shortest sequence I found for flipping 2 adjacent
edges on the 4x4x4 cube (LD pair):

 (r3 D3) ^3 + (r1 D1) ^4 + Rr3 D3 R1 D1 r3 D3 R3 D1 R1 D3

Note the use of Rr to represent both the turns R face & r inner
 slice. Counting slice turns the sequence is 25 turns, or
 24 "hyper moves". This sequence moves some centre pieces around.

However, on checking David Singmaster's Cubic Circular, in issues
 5 & 6, Autumn & Winter 1982 there is a shorter process on
 page 15, (UB pair):

 r2 D2 l3 D1 R3 U1 R3 U3 l3 U1 R1 U3 l1 R1 D1 r2

This process, although more difficult to memorize, is only 16 slice
 moves. It also disturbs centre pieces, although in a simpler way.
 I always solve the centre pieces last on the 4x4x4.

Hope this helps!

 -> Mark
Email: mark.longridge@canrem.com

From Mikko.Haapanen@otol.fi  Thu Jan 13 09:19:35 1994
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To: cube-lovers@life.ai.mit.edu
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> ...
> This is the shortest sequence I found for flipping 2 adjacent
> edges on the 4x4x4 cube (LD pair):
> 
>  (r3 D3) ^3 + (r1 D1) ^4 + Rr3 D3 R1 D1 r3 D3 R3 D1 R1 D3
> 
> Note the use of Rr to represent both the turns R face & r inner
>  slice. Counting slice turns the sequence is 25 turns, or
>  24 "hyper moves". This sequence moves some centre pieces around.
> 
> However, on checking David Singmaster's Cubic Circular, in issues
>  5 & 6, Autumn & Winter 1982 there is a shorter process on
>  page 15, (UB pair):
> 
>  r2 D2 l3 D1 R3 U1 R3 U3 l3 U1 R1 U3 l1 R1 D1 r2
> 
> This process, although more difficult to memorize, is only 16 slice
>  moves. It also disturbs centre pieces, although in a simpler way.
>  I always solve the centre pieces last on the 4x4x4.

Thank you. But what is the shortest way to flip 2 adj. edges without 
messing the center pieces? I can't find shorter than 49 turns. 

 -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
 -= Mikko Haapanen -=-=- hazard57@sulo.otol.fi -=-=- (981) 530 7768 =-
 -=-=-=-=-=-=-=-=-=-= Haapanatie 2C411 90150 OULU =-=-=-=-=-=-=-=-=-=-





From ishius@ishius.com  Thu Jan 13 12:38:38 1994
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Date: Thu, 13 Jan 1994 09:37:53 -0800
Message-Id: <199401131737.JAA25594@holonet.net>
To: cube-lovers@life.ai.mit.edu
From: ishius@ishius.com (Ishi Press International)
Subject: Skewb, 5x5x5 cubes available

I've been watching all the discussion here, and I thought some people on this
list might appreciate knowing that 5x5x5 Rubik's Cubes and the Skewb are
available from Ishi Press International.  We also have hundreds of other
mechanical puzzles.

If you would like to be on our puzzle e-mail list, write us.  If you would
like a color catalog of our puzzles, send us your snail-mail address.

I apologize for intruding with a commercial message, but it did seem to me
that at least a few people on this list would like to get their hands on
some of these puzzles, and I don't know of any other distributor for these
two items.

Anton Dovydaitis
Customer Support

========================================================================
Ishi Press International        800/859-2086 voice, 408/944-9110 FAX
76 Bonaventura Drive            ishius@ishius.com           The Americas
San Jose, CA  95134             ishi@cix.compulink.co.uk    Europe


From CPELLEY@delphi.com  Thu Jan 13 23:25:54 1994
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Date: Thu, 13 Jan 1994 23:18:47 -0400 (EDT)
From: CPELLEY@delphi.com
Subject: Mickey's Challenge
To: cube-lovers@life.ai.mit.edu
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I visited the local Disney Store and picked up a Mickey's Challenge puzzle for

$10.  It's really cute, and the book that it comes with is excellent.

Included are color photos of Christophe Bandelow, Uwe Meffert, and the puzzle

disassembled into all its parts.  Plus it gives a solution for the puzzle and

has a short bio on Uwe Meffert.  It also shows color photos of the Megaminx,

Pyraminx (not the Tomy version, but a black one), and the 5x5x5 which they

refer to as "Professor's Cube."

Some general notes on Mickey's Challenge.  It is a spherical Skewb, and it

actually turns much more smoothly than my cubical Skewb.  It has the same

delightful "clicking" mechanism that the Skewb and original Pyraminx had.

It is a bit easier than the Skewb, since there are a few blank pieces that

can be exchanged without noticing the difference.  In fact, the book's solution

actually leaves Mickey intact while solving Donald.  After you're bored with

solving it, the concept of making patterns takes on strange dimensions, as you

can make Mickey and Donald exchange body parts and look like Disney on acid!

All in all, it is an excellent little puzzle and I am very glad to see the

Skewb widely available to puzzle enthusiasts everywhere.

One final note:  the booklet gives no credit whatsoever to Tony Durham, who

was credited with the Skewb's invention in Hofstadter's Sci Am articles years

ago.  They instead credit Meffert, since the Skewb's mechanism is based on

the Pyraminx.



From ishius@ishius.com  Fri Jan 14 14:17:29 1994
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To: Cube-Lovers@ai.mit.edu
From: ishius@ishius.com (Ishi Press International)
Sender: ishius@ishius.com (Unverified)
Subject: 4x4x4 and 5x5x5 cubes.

I've been getting a lot of requests for 4x4x4 cubes, and we're looking into
getting them.  However, I have a couple questions.

1) Why are 4x4x4 cubes so interesting?  Do the additional symmetries make for
   interesting questions, are they more fun, or easier to solve?

2) It appears to me that if you know how to solve the 3x3x3 Rubik's cube,
   then you can easily solve the 5x5x5 rubiks (i.e., the solution is
   derivative).  For example, you can treat the inner 3x3 faces of the 5x5x5
   as a single 3x3x3 cube.  Alternately, you can treat the edges/faces along
   with the the middle three slices combined into a single slice as its own
   3x3x3 cube, and this would not really disturb the "inner face" 3x3x3 cube.
   Is this really so, or am I missing something? Is the 5x5x5 cube simply
   the group product of two 3x3x3 cubes and one or two sub-groups of a 3x3x3,
   or is it more complex than that?  How does this relate to the 4x4x4?

I do have a Bachelor's degree in mathematics and am familiar with abstract
algebra.  I appreciate any light you can shed on these questions.  I would
like to be able to converse intelligently about the cubes; that is why I
subscribed to this list.

Anton Dovydaitis

========================================================================
Ishi Press International        800/859-2086 voice, 408/944-9110 FAX
76 Bonaventura Drive            ishius@ishius.com           The Americas
San Jose, CA  95134             ishi@cix.compulink.co.uk    Europe


From mouse@collatz.mcrcim.mcgill.edu  Fri Jan 14 15:45:34 1994
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Date: Fri, 14 Jan 1994 15:44:48 -0500
From: der Mouse <mouse@collatz.mcrcim.mcgill.edu>
Message-Id: <199401142044.PAA00960@Collatz.McRCIM.McGill.EDU>
To: cube-lovers@ai.mit.edu
Subject: Re:  4x4x4 and 5x5x5 cubes.

> I've been getting a lot of requests for 4x4x4 cubes, and we're
> looking into getting them.

I'd (probably) buy one - I don't know what's become of the one I had.
Depends on price, of course, but I found the 5-Cube price acceptable.

> 1) Why are 4x4x4 cubes so interesting?  Do the additional symmetries
>    make for interesting questions, are they more fun, or easier to
>    solve?

I doubt it.  If one ignores the center slice in each dimension on a
5x5x5, one has a 4x4x4.  I think it's the completist instinct any
collector has. :-)

Now what I'd *really* like is something topologically equivalent to a
2x2x2x2 Cube.  Obviously a 2x2x2x2 Cube can't really be built, but it
should be possible to build something topologically equivalent.  (A
3x3x3x3 would be nice too, but perhaps too difficult.)  The hard part
is designing an emulation that has some aesthetic appeal; it's easy
enough to write a program that lets you permute appropriate overlapping
4-cycles of objects without any intuitively-obvious structure.

> 2) It appears to me that if you know how to solve the 3x3x3 Rubik's
>    cube, then you can easily solve the 5x5x5 rubiks (i.e., the
>    solution is derivative).

No, not really.  If you can do the 3-Cube *and the 4-Cube*, then the
5-Cube has no new challenges to offer (nor, I believe, does any size).

But the 4-Cube and 5-Cube do have challenges the 3-Cube doesn't, namely
edge cubies and face-center cubies.  All the 3-Cube experience in the
world won't help you if you get your 5-Cube solved except for two edge
cubies which are swapped.  (Or rather, general Cube-type-puzzle
experience will help - for example, how to use conjugates - but
3-Cube-specific experience won't.)

>    For example, you can treat the inner 3x3 faces of the 5x5x5 as a
>    single 3x3x3 cube.  Alternately, you can treat the edges/faces
>    along with the the middle three slices combined into a single
>    slice as its own 3x3x3 cube, and this would not really disturb the
>    "inner face" 3x3x3 cube.  Is this really so, or am I missing
>    something?

You're missing something, but not much. :-)  As you say, there are two
ways of emulating a 3-Cube on the 5-Cube, namely to paste slices 2-1-2
along each dimension and to paste them 1-3-1.  (I hope that's not too
abbreviated to be comprehensible - I mean, along each dimension, paste
the 5-Cube slices together into three groups, taking two, then one,
then two slices, or one, three, and one.)  However, it is entirely
possible to scramble the 5-Cube in ways that cannot be solved by
treating the 5-Cube as virtual 3-Cubes, except in the trivial sense
that any 5-Cube turn can be viewed as one or more turns of
appropriately-chosen virtual 3-Cubes.

For example, I can swap two edge cubies (and also permute center cubies
in invisible ways); alternatively, I can permute the face cubies so
that the 2-1-2 virtual 3-Cube has two identical corner v-cubies.

					der Mouse

			    mouse@collatz.mcrcim.mcgill.edu

From ncramer@bbn.com  Fri Jan 14 17:05:26 1994
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Message-Id: <9401142205.AA05939@life.ai.mit.edu>
Date:     Fri, 14 Jan 94 7:47:37 EST
From: Nichael Cramer <ncramer@bbn.com>
To: Ishi Press International <ishius@ishius.com>
Cc: cube-lovers@life.ai.mit.edu
Subject:  Re:  Skewb, 5x5x5 cubes available

>Date: Thu, 13 Jan 1994 09:37:53 -0800
>Message-Id: <199401131737.JAA25594@holonet.net>
>From: Ishi Press International <ishius@ishius.com>
>Subject: Skewb, 5x5x5 cubes available
>
>I've been watching all the discussion here, and I thought some people on this
>list might appreciate knowing that 5x5x5 Rubik's Cubes and the Skewb are
>available from Ishi Press International.  We also have hundreds of other
>mechanical puzzles.
>
>If you would like to be on our puzzle e-mail list, write us.  If you would
>like a color catalog of our puzzles, send us your snail-mail address.

Anton

Yes to both the above, please.

e-mail : ncramer@bbn.com

land-mail: Nichael Cramer
           123 B Spring St
           Cambridge MA 02141

Thanks much
Nichael

From @mail.uunet.ca:mark.longridge@canrem.com  Fri Jan 14 23:47:23 1994
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Sender: CRSO.Cube@canrem.com
Subject: Higher Order Cubes, correction
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.704.5834.0C191357@canrem.com>
Date: 	Fri, 14 Jan 1994 22:10:00 -0500
Organization: CRS Online  (Toronto, Ontario)

-> (fm2 + u2) ^2 + (fm2 + lm2) ^2    (corrects the centres)

This should be:
   (fm2 + u2) ^2 + (fm2 + l2) ^2

-> Mark

From @mail.uunet.ca:mark.longridge@CANREM.COM  Sat Jan 15 00:52:48 1994
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Reply-To: CRSO.Cube@canrem.com
Sender: CRSO.Cube@canrem.com
Subject: Higher Order Cubes
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.703.5834.0C191356@canrem.com>
Date: 	Fri, 14 Jan 1994 21:59:00 -0500
Organization: CRS Online  (Toronto, Ontario)

Anton Dovydaitis writes:
>I've been getting a lot of requests for 4x4x4 cubes, and we're
>looking into getting them.  However, I have a couple questions.
>
>1) Why are 4x4x4 cubes so interesting?  Do the additional symmetries
>make for interesting questions, are they more fun, or easier to
>solve?

A well lubed 4x4x4 cube is still relatively easy to physically
manipulate. As der Mouse suggests, it is arguably the largest
interesting cube from a solver's point of view. Once one starts
actually twisting with a 5x5x5 cube, the physical problems
become more severe, e.g. the stickers come off easier,
turning the slice you want to is more of a challenge, etc.

In the virtual realm of computer cubing the patterns you can
create are more elaborate, although I find in practice that
finding pretty patterns on the 5x5x5 can become wearisome
due to fact there are 9 centre pieces per side!

>2) It appears to me that if you know how to solve the 3x3x3 Rubik's
>  cube, then you can easily solve the 5x5x5 rubiks (i.e., the
>  solution is derivative).  For example, you can treat the inner 3x3
>  faces of the 5x5x5 as a single 3x3x3 cube.

Using the 4x4x4 cube we can produce a single exchange of centres
and an exchange of edge pairs, and we can invert a single edge pair.
Thus we can construct all the impossible 3x3x3 patterns except those
involving a twist of a single corner! That is why I think the 4x4x4
cube is a good cube to have. The individual centre cubies can
naturally wander all over the cube, and on the 3x3x3 cube they are
fixed.

In the case of the 5x5x5 cube, lots of the 3x3x3 knowledge does
help. When dealing with the 5x5x5's middlemost slice (let's call
one such slice "fm" for middlemost front slice) some of the
3x3x3 move sequences will move the appropriate edges, but now
these sequences will also move centre pieces, specifically the
ones which have no counterpart on the 3^3 and 4^3.

To solve cubes 4x4x4 and greater requires new sequences to
efficiently move centre cubies at will, and in the case of
the 5^3 there really is no standard language to interchange
move sequences and label individual cubies.

I find having a letter as a mnemonic helps, so I'll suggest the
following as an extension of Singmaster's 4x4x4 notation
for the 5x5x5 cube:


L left face          l inner left slice    lm left middlemost slice
R right face         r inner right slice   rm right middlemost slice
F front face         f inner front slice   fm front middlemost slice
B back face          b inner back slice    bm back middlemost slice
U up face            u inner up slice      um up middlemost slice
D down face          d inner down slice    dm down middlemost slice

Again, we follow the alphabetic component by a number to
signify the rotation (1 = 90, 2 = 180, 3 = 270 or -90)

This is overkill, and we can dispense with rm, bm and dm.

Thus we could flip 2 middlemost edges at FD and BD with:

(fm1 D1) ^3 + fm1 D2 + (fm1 D1) ^3 + fm1   (disturbs some centres)

followed by:

(fm2 + u2) ^2 + (fm2 + lm2) ^2    (corrects the centres)

I believe this is correct, and I will double-check on my physical
5x5x5 at home. Definitely 5^3 cubing is a sport for the
specialist ;->

-> Mark
Email: mark.longridge@canrem.com

From ncramer@bbn.com  Sat Jan 15 09:22:26 1994
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Message-Id: <9401151422.AA03599@life.ai.mit.edu>
Date:     Sat, 15 Jan 94 9:12:37 EST
From: Nichael Cramer <ncramer@bbn.com>
To: CRSO.Cube@canrem.com
Cc: cube-lovers@life.ai.mit.edu
Subject:  Re:  Higher Order Cubes

>Subject: Higher Order Cubes
>From: Mark Longridge <mark.longridge@canrem.com>
>Date: 	Fri, 14 Jan 1994 21:59:00 -0500
>
>A well lubed 4x4x4 cube is still relatively easy to physically
>manipulate. As der Mouse suggests, it is arguably the largest
>interesting cube from a solver's point of view. Once one starts
>actually twisting with a 5x5x5 cube, the physical problems
>become more severe, e.g. the stickers come off easier,
>turning the slice you want to is more of a challenge, etc.

This is interesting, because it's almost exactly the opposite of my
experience.

The problem seems to be the difference between the internal mechanisms of
the odd- and even- ordered cubes.  The 3X and 5X have that "fixed" center
piece attached to the core whereas the center face cubelets of the 4X are
held together "under tension".  My experience has been that this adjustment
is critical, but often out of whack.  As a consequence, of the four 4X's
I've owned, only one was really useable; two were so stiff they were very
difficult to turn (even with lubrication) and one was so loose that it
never lasted more than about 20 minutes before dissolving into a pile of
cubelets (it currently lives in a sack in my office drawer).  These were
all real "brand-named" cubes, not cheap twiz-o knock-offs.

On the other hand all of the 5X's I've owned have been _very_ easy to turn
without any special customization.  Except for the tendency (as Mark
mentions) for the stickers to come off of one of them, they're consistently
more comfortable to the hand than any of the 3X's I've owned.


Nichael
ncramer@bbn.com      -- Captain and left quarter guard, BBN Calvinball Team



From mouse@collatz.mcrcim.mcgill.edu  Sat Jan 15 12:33:29 1994
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Date: Sat, 15 Jan 1994 12:33:21 -0500
From: der Mouse <mouse@collatz.mcrcim.mcgill.edu>
Message-Id: <199401151733.MAA02409@Collatz.McRCIM.McGill.EDU>
To: cube-lovers@ai.mit.edu
Subject: Re:  Higher Order Cubes

>> A well lubed 4x4x4 cube is still relatively easy to physically
>> manipulate. As der Mouse suggests, it is arguably the largest
>> interesting cube from a solver's point of view.

Probably true; it's the largest cube that actually offers new
challenges.  However, bigger cubes are better in that they offer more
variety for making pretty patterns. :-)

>> Once one starts actually twisting with a 5x5x5 cube, the physical
>> problems become more severe, e.g. the stickers come off easier,
>> turning the slice you want to is more of a challenge, etc.

> This is interesting, because it's almost exactly the opposite of my
> experience.

> The problem seems to be the difference between the internal
> mechanisms of the odd- and even- ordered cubes.

This brings up an interesting point.  Perhaps it would be possible to
build a 4-Cube that was internally a 5-Cube but for which the middle
slice was not actually visible on the surface?  Or a 2-Cube that's
internally a 3-Cube?  I wonder if it might make for smoother-turning
cubes.

> [O]f the four 4X's I've owned, only one was really useable; two were
> so stiff they were very difficult to turn (even with lubrication) and
> one was so loose that it never lasted more than about 20 minutes
> before dissolving into a pile of cubelets [...].

I have owned only one 4-Cube, and it's been long enough since I knew
where it was that I don't recall how easy it was to turn.  I now have
two 3-Cubes and a 5-Cube.  One of the 3-Cubes is a joy to turn; it's
lubed enough that it turns readily and easily, even when the turn has a
good deal of skew to correct, but it's not so loose that it turns when
I don't want it to.  (The other 3-Cube is (a) missing one center cubie
face and (b) much more difficult to turn.)  The 5-Cube (one of the
recent Ishi Press cubes, btw) is mechanically quite good, though the
orange stickers did tend to come off (no other color did, and contact
cement worked just fine for putting them back on).  Not as good as my
good 3-Cube, though.

I've wondered whether it would be possible to build higher-order cubes.
The corners of the 5-Cube still catch by a respectable amount as the
face turns, but by little enough that it makes me wonder if the 6-Cube
or 7-Cube is actually feasible.  (Oh, for a really good
force-reflecting dataglove...then such a thing could be done virtually
with no problem at all!)

					der Mouse

			    mouse@collatz.mcrcim.mcgill.edu

From @mail.uunet.ca:mark.longridge@canrem.com  Sat Jan 15 17:03:53 1994
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Sender: CRSO.Cube@canrem.com
Subject: 4x4x4 & 5x5x5 processes
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.710.5834.0C1914B9@canrem.com>
Date: 	Sat, 15 Jan 1994 15:44:00 -0500
Organization: CRS Online  (Toronto, Ontario)

Here are a couple of processes for larger cubes, plus the
requested edge pair flip without disturbing centres (p2),
as well as a minor correction for the 5x5x5 process:

4x4x4 processes (measured in slice moves)
---------------

p1  Flip LD edge pair   (r3 D3) ^3 + (r1 D1) ^4 + Rr3 D3 R1 D1 r3 D3 R3
(disturbs centres)          D1 R1 D3    (25)
p2  Flip UB edge pair    r2 D2 l3 D1 R3 U1 R3 U3 l3 U1 R1 U3 l1 R1 D1
(retain centre positions ) + U2 r1 (u2 r2 l2) ^2 + r3 U2 r2     (26)


5x5x5 processes (measured in slice moves)
---------------

Flip 2 middlemost edges at FD and BD with:
p1 (fm1 D1) ^3 + fm1 D2 + (fm1 D1) ^3 + fm1   (disturbs some centres)
followed by:
   D1 + (fm2 u2) ^2 + (fm2 l2) ^2 + D3    (corrects the centres)  (25)

-> Mark
Email: mark.longridge@canrem.com

From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU  Mon Jan 17 09:09:39 1994
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Date:      Mon, 17 Jan 1994 09:06:59 EST
From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Number of M-Conjugate Classes for GC\M

On 4 December 1993, I posted results from a breadth-first exhaustive
search of GC\M, the corners of the 3x3x3, reduced by M-conjugation.
The posting included a summary of how many conjugate classes there
were at each level of the search tree (i.e., distance from Start).

It occurred to me that I had not also posted a summary of M-conjugates
for the corners of the 3x3x3 by the size of the conjugate classes.  I
searched my records, only to discover that I had never calculated
such sizes.  If I had, I probably would have been forced to analyze
properly the distinction between M-conjugation and B-conjugation,
because B-conjugation makes no sense for the corners of the 3x3x3.
B-conjugation *can* be performed for the corners of the 3x3x3, but you
end up with the 2x2x2 instead because B-conjugation effectively
removes the centers.

Anyway, I have now calculated M-conjugate class sizes for GC\M via
computer search, and here are the results.

                   M-Class     Number
                     Size        of
                              Classes

                         1        1
                         2        1
                         3        3
                         4        1
                         6       34
                         8       33
                        12      301
                        16      104
                        24     9064
                        48  1832428

                     Total  1841970

Notice that with M-conjugation, the maximum class is size is 48,
rather than 1152 as it is with B-conjugation.  Hence, my posting
of 4 December 1993 incorrectly identified the results as being
for "1152 fold symmetry".  The results are correct, but they
should be labeled as being for "48 fold symmetry", i.e., for
M-conjugation rather than for B-conjugation.

In calculating M-conjugate class sizes for GC\M, I did not "start
from scratch".  Rather, I used the existing results for B-conjugate
classes as a base.  In the case of B-conjugate classes of order
1152, no calculations are required.  Each such B-class can simply
be partitioned into 24 M-classes of order 48.  Hence, I had to
perform calculations for less than 4% of the B-classes.  Here is
a summary matrix, showing for each B-class size the number of
each M-class size which are derived.

                            M-Class Size

              1  2  3  4   6   8   12  16    24       48    Total

          24  1  0  1  0   2   1    0   0     0        0        5
B-Class   48  0  1  0  0   1   2    2   0     0        0        6
Size      72  0  0  2  0  11   0    2   0     5        0       20
          96  0  0  0  1   0   1    3   0     2        0        7
         144  0  0  0  0  20   0   42   0    30       14      106
         192  0  0  0  0   0  29    0   8    73       16      126
         288  0  0  0  0   0   0  252   0   682      406     1340
         384  0  0  0  0   0   0    0  96     0      224      320
         576  0  0  0  0   0   0    0   0  8272    22360    30632
        1152  0  0  0  0   0   0    0   0     0  1809408  1809408

    Total     1  1  3  1  34  33  301 104  9064  1832428  1841970

The first row of the matrix exemplifies the process of calculating
M-Class sizes from B-Class sizes.  In the case of corners, there
is only one B-class of order 24, namely Start.  The 24 elements of
the B-class are the 24 elements of the form Ic, where c is in C,
the 24 rotations of the cube.  Under B-conjugation, these 24
elements are equivalent  (i.e., in a centerless cube such as the
2x2x2, the 24 rotations of I are indistinguishable).  But in a cube
with centers, such as the corners of the 3x3x3, the 24 elements
are not equivalent.

For example, the M-class of order 1 is {I}.  One of the M-classes
of order 6 is {FB', UD', RL', LR', BF', DU'}.  The M-class of order 3
is {FFB'B', RRL'L', UUD'D'}.  That's as many as I can do in my head,
but I think the pattern is clear.

M-classes are a partition of the B-classes.
In the case of B-classes of order 1152, the partition is regular  --
i.e., you get exactly 24 M-classes of order 48.  However, all
partitions are not regular.  In the partition of the B-class of I
which we just discussed, there is 1 M-class of order 1, 1 M-class
of order 3, 2 M-classes of order 6, and 1 M-class of order 8, for
a total of 24 M-classes.  Many other partitions are not regular,
as well.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From dseal@armltd.co.uk  Mon Jan 17 14:14:08 1994
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From: dseal@armltd.co.uk (David Seal)
To: (Cube) cube-lovers@ai.mit.edu
Subject: Re: Higher Order Cubes
Message-Id: <2D3AD5C5@dseal>
In-Reply-To: <199401151733.MAA02409@Collatz.McRCIM.McGill.EDU>


> This brings up an interesting point.  Perhaps it would be possible to
> build a 4-Cube that was internally a 5-Cube but for which the middle
> slice was not actually visible on the surface?  Or a 2-Cube that's
> internally a 3-Cube?  I wonder if it might make for smoother-turning
> cubes.

Yes, I think you could build such a 4-Cube. Likewise, you could build a
2-Cube as a 3-Cube with invisible middle slices. But I don't believe you'd
want one: it could get completely jammed much too easily.

The reason: If you take a 3-Cube and rotate its left and right slices 45
degrees each, you cannot rotate any of its other faces. This isn't
surprising, since you don't expect to be able to perform one rotation
halfway through another. If its middle slices were hidden, however, it would
*appear* to be a 2-Cube which is not halfway through a rotation, and the
fact that you couldn't move any faces but the left and right ones would be
surprising - and undesirable.

Unfortunately, I believe such a situation could probably arise quite easily.
If you were to take the 2-Cube concerned and rotate its right face 90
degrees relative to its left face, you're going to be OK if the hidden
middle layer rotates 0 or 90 degrees relative to the left face, but not OK
if it rotates any other amount. I suspect most mechanisms would be more
liable to rotate it an intermediate amount!

There may be a way out, though. If you can anchor the place where the three
axes meet to one of the corner cubelets in some way, the problem is solved:
if the "anchor cubelet" is in the left face, then the hidden layer will
rotate 0 decrees; if it is in the right face, then 90 degrees.

David Seal
dseal@armltd.co.uk

From mouse@collatz.mcrcim.mcgill.edu  Mon Jan 17 16:23:05 1994
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Date: Mon, 17 Jan 1994 16:22:50 -0500
From: der Mouse <mouse@collatz.mcrcim.mcgill.edu>
Message-Id: <199401172122.QAA05806@Collatz.McRCIM.McGill.EDU>
To: cube-lovers@ai.mit.edu
Subject: Re: Higher Order Cubes

>> Perhaps it would be possible to build a 4-Cube that was internally a
>> 5-Cube but for which the middle slice was not actually visible on
>> the surface?  Or a 2-Cube that's internally a 3-Cube?

> Yes, I think you could build such a 4-Cube. Likewise, you could build
> a 2-Cube as a 3-Cube with invisible middle slices.  But I don't
> believe you'd want one: it could get completely jammed much too
> easily.

> The reason: If you take a 3-Cube and rotate its left and right slices
> 45 degrees each, you cannot rotate any of its other faces.

Duh, yeah; that never occurred to me.

> There may be a way out, though. If you can anchor the place where the
> three axes meet to one of the corner cubelets in some way, the
> problem is solved: [...].

Yes.  I think this may be possible, too...consider a normal 3-Cube, and
restrict yourself to R, U, and F turns.  Then ignore the center and
edge cubies - the ones that get invisibilized.  You're left with a
2-Cube.  Three edge cubies never move with respect to the center cubies
or the corner cubie they surround; glue those together.  Presto!

The same treatment is not possible for making a 4-Cube out of a 5-Cube,
but an alternative occurs to me, that I *think* will work for higher
cubes: key three of the (invisible) center cubies to the center
six-pronged piece, so that they can't turn.  Then half the face turns
will cause the invisible center slice to turn with them; non-face
slices (which don't exist on the 2/3-Cube) work normally.

I notice with this construction for (say) a 4-Cube, the puzzle core
turns whenever certain face slices do.  With the 4-Cube I owned (and
presumably still own, if I could find it), the puzzle core turns
whenever certain next-to-center slices do.  I suspect the latter would
make for a smoother-turning puzzle.  Perhaps someone will someday build
a 5-Cube-turned-4-Cube and this can be determined.

In the (IMO unlikely) event I originated any of the above ideas, I
hereby place it/them in the public domain.  Go wild, Ishi Press. :-)

					der Mouse

			    mouse@collatz.mcrcim.mcgill.edu

From Mikko.Haapanen@otol.fi  Wed Jan 19 15:09:31 1994
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Date: Wed, 19 Jan 1994 21:53:24 +0200 (EET)
From: "M. Haapanen" <Mikko.Haapanen@otol.fi>
Subject: Re: 4x4x4 & 5x5x5 processes
To: cube-lovers@ai.mit.edu
In-Reply-To: <60.710.5834.0C1914B9@canrem.com>
Message-Id: <Pine.3.87.9401192124.B25726-0100000@rhea>
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Hello cube lovers!

> requested edge pair flip without disturbing centres (p2),
> as well as a minor correction for the 5x5x5 process:

> ...
> (retain centre positions ) + U2 r1 (u2 r2 l2) ^2 + r3 U2 r2     (26)
                                                                  ^^^^
> 
> 5x5x5 processes (measured in slice moves)
> 
> Flip 2 middlemost edges at FD and BD with:
> p1 (fm1 D1) ^3 + fm1 D2 + (fm1 D1) ^3 + fm1   (disturbs some centres)
> followed by:
>    D1 + (fm2 u2) ^2 + (fm2 l2) ^2 + D3    (corrects the centres)  (25)
> -> Mark                                                           ^^^^
> Email: mark.longridge@canrem.com

The following might be trivial, but i write it here anyway. This was 
invented about 10 years ago:

5x5x5
-----

(R1 um2 R2 um1 R1) + U2 + (R3 um3 R2 um2 R3) + U2  ----> 12 (18) turns :)

 -=-=-=-=-=-=-=-=-=-=-=-=-=-= Finland =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
 -= Mikko Haapanen -=-=- hazard57@rhea.otol.fi -=-=- (981) 530 7768 =-
 -=-=-=-=-=-=-=-=-=-= Haapanatie 2C411 90150 OULU =-=-=-=-=-=-=-=-=-=-



From hoey@aic.nrl.navy.mil  Fri Jan 21 18:32:30 1994
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From: hoey@aic.nrl.navy.mil (Dan Hoey)
Message-Id: <9401212332.AA13137@Sun0.AIC.NRL.Navy.Mil>
To: Cube-Lovers@ai.mit.edu
Cc: Jerry Bryan <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
Subject: Re: Some Proposed Terminology

I welcome Jerry Bryan's <BRYAN%WVNVM.BITNET@mitvma.mit.edu> efforts to
improve the terminology of the groups associated with Rubik's cube.
But there is some additional clarification I think is necessary.

> Let G be the standard cube group for the 3x3x3 cube....
> Let GC be the corners with centers without edges group....
> Let GE be the edges with centers without corners group....

That much will do, mod quibbles about what name is best.

> Let G\C be the corners with edges without centers group.  I intend
> for the notation to indicate G reduced by C, where C is the rotation
> group for the cube....

> Let GC\C be the corners without edges without centers group....
> Let GE\C be the edges without corners without centers group....

First, these are not, strictly speaking, groups.  Well, you can make
them groups, by defining what the group operation is.  But I don't
know any way of doing that without losing the symmetrical nature of
the problem.

Second, I would suggest that G/C, GC/C, and GE/C are more standard
names for these objects.  The elements are nominally 24-element sets,
each of which is an equivalence class when two positions are
considered equivalent when they differ by their position with respect
to the corners.  The classes are called the cosets of C in G, GC, and
GE, respectively.

> Let G\M be the set of M-conjugate classes for G.....
> Let GC\M be the set of M-conjugate classes for GC....
> Let GE\M be the set of M-conjugate classes for GE....

The partition of a group into conjugacy classes is not at all the same
as the partition into cosets.  So I would prefer to use different
symbology, like "\" for conjugacy and "/" for cosets, but....

> Recall that B is the function which calculates the canonical form
> for a cube under the composed operations of M-conjugation plus
> rotation.  My  programs calculate equivalence classes under B.

> Let G\B be the set of B-classes for G [ and likewise for GE, GC ].

Well, if you are using "\" for a generic partition into equivalence
classes, then we should really do something like G\Conj(M) for
partitions into conjugacy classes.  At least then you can say
G/C=G\Cosets(C).

> Then, we have Gx\B=(Gx\C)\M=(Gx\M)\C.  In English, we can decompose
> B into a multiplication by C and M (in either order).

No, that's _multiplication_ by C and _conjugation_ by M.  A good
example of why it's important not to use confusing symbols.  M and C
are not at all treated the same, except inasmuch as they are used to
induce partitions into equivalence classes.  Say instead that
            Gx\B = (Gx/C)\Conj(M) = (Gx\Conj(M))/C.

> If I wanted to model GC\C, I would have had to either model only
> seven of the cubies, or else modeled all eight but moved only seven
> of them.  Since what I really wanted was (GC\C)\M, and since what I
> had was GC, I had to invent this funny B thing, where GC\B=(GC\C)\M.
> If I had been clever enough to model GC\C in the first place, I
> never would have had to invent B.  Similar comments apply to my
> model for the edges.

Well, the part about moving only seven (corner) cubies is the approach
that's been taken before on this list to deal with cubes that don't
have face centers.  It has the advantage that the object being treated
is a group.  But the problem is that the group is no longer cubically
symmetrical (in some vague sense).  This led me, at least, to lose
track of the structure that would allow analysis of M-conjugacy.  So I
have to admire your tackling GE as a whole, instead of trying to stick
to GE/C.  At first blush, it looks like GE/C is 24 times smaller.  But
since GE/C\Conj(M) is almost 48 times smaller still, it's important to
work in GE at least enough to be able to use the conjugation.  Which
is beside the point that I'm actually very interested in the structure
of Gx/Conj(M) itself.  And that is what I was really getting at in
1984 when I asked about how many positions there really are.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

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Date: Sat, 15 Jan 1994 12:44:15 -0300
From: walace@ntiaa.embrapa.ansp.br (Walace Sartori Bonfim)
Subject: ICSI94
To: cube-lovers@life.ai.mit.edu
Message-Id: <4CD9C4A5C000C949@fpsp.fapesp.br>
X-Envelope-To: cube-lovers@ai.ai.mit.edu

Dear reader,

Due to the wide spectrum of people that might be interested in the
subjects to be discussed during the III International Conference on Systems
Integration, we decided to post this call for papers in your mailing list.

We encourage you to participate in this event as a paper author.

The paper arrival deadline is March 3, 1994.

Please forward this message to whoever you think it might be of interest and we
appreciate your effort to post it.

Thanks,

Prof. Fuad Gattaz Sobrinho
Conference Chairman


    -----------------------------------------------------------------
			   Call for Papers

	The Third International Conference for Systems Integration
			Sao Paulo City - Brazil

		     July 30th - August 6th, 1994
    -----------------------------------------------------------------

The Integration of Society for the Social, Economical, Scientific and
Technological Development. This conference focuses on the integration
of technologies, processes and systems, and the development of mechanisms
and tools enabling solutions to complex multi-disciplinary problems dealing
with agriculture, housing, telecommunications, financing and business,
public services, education and software. The conference will provide an
international and interdisciplinary forum in which researchers, educators,
managers, practitioners and politicians, involved within the production
process, can share novel research and development, education, production,
trading, management and political experiences. Papers should deal with
recent effort in theory, design, implementation, methodology, technics,
tools and experiences of integration. Topics to be addressed include, but
are not limited to:

     Technical and Scientific Aspects:
	- Integration, Modeling, Characterization and Automation of Process
	  and Systems
	- Reengineering and Simplification of Processes
	- Computational Environments and Software Factories for Engineerind,
	  Design, Manufacturing and System Development
	- Rol of Human Engineering in Integration
	- Experiences within National or Continental Software Projects
	- The Implication of Systems Integration for Manpower Skills
	- Quality Control and Certification in Organizational and Process
	  Integration.

     Social, Political and Economical Aspects:
	- Experiences in Modeling, Development, Evolution and Integration
	  of Enterprises
	- Experiences in Management and Identification of Value-Add Chains
	  within Agriculture, Housing, Telecommunications, Financing and
	  Business, Public Services, Education and Software
	- Public Policies and City Management
	- Management of Multi-dimensional Integration.

     Infrastructure Aspects:
	- Qualified Information Resources
	- Education and Training
	- Science and Technology
	- Enterprise Development.

Information and Instructions for Authors: All papers must be in English
or  Portuguese, typed in double spaced format, and may not exceed 6,000
words. Each submission should provide a cover page containing author(s),
affiliation(s), complete address(es), identification of principal author,
and telephone number. Also include SIX copies of complete text with a
title and abstract. Notice of acceptance will be mailed to the principal
author(s) by March 15, 1994. If accepted, the author(s) will prepare the
final manuscript, in English, in time for inclusion in the conference
proceedings and will present the paper at the conference; otherwise, the
author(s) will incur a page charge. Authors of accepted papers must sign
a copyright release form. The proceedings will be published by the IEEE
Computer Society Press.

Send SIX copies of your paper(s) to:

Prof. Peter A. Ng
IIISis - USA Office - New Jersey Institute of Technology
University Heights
Newark, NJ  07102
USA

For Further Information, Contact:

Prof. Peter A. Ng		 Prof. Fuad Gattaz Sobrinho
Fone:(1) (201) 596-3387     OR   Phone:(55)(192) 41-4504
Fax: (1)(201) 596-5777		 Fax:  (55)(192) 41-3098
Email: ng_p@vienna.njit.edu	 Email: iiisis@ccvax.unicamp.br

-------------------------------------------------------------------
>>>>>>>>>> Paper Arrival Deadline: March 3rd, 1994 <<<<<<<<<<<<<<<<
-------------------------------------------------------------------

CONFERENCE COMMITTEE

Conference Chair		Fuad Gattaz Sobrinho
				IIISis
Program Chair			Peter A. Ng
				NJIT
Finance & Business Co-Chair	Alcir A. Calliari
				Banco do Brasil
Agriculture Co-Chair		Ney B. Araujo
				ABAG
European Co-Chair		Herbert Weber
				University of Dortmund
Pac!fic Co-Chair		Fumihiko Kamijo
				IPA
Middle East Co-Chair		Asuman Dogac
				METU
South America Co-Chair		Julio C. S. P. Leite
				PUC/RJ
North America Co-Chair		Bruce Berra
				Syracuse University
Tutorials Co-Chairs		Oscar Ivan Palma Pacheco
				EMBRAPA
				Murat M. Tanik
				SMU
Organization Co-Chairs		Rita de Cassia A. Marchiore
				IIISis
				Carole Poth
				NJIT
Steering Committee Chair	Peter A. Ng
				NJIT
Honorary Advisors		Raymond T. Yeh
				C. V. Ramamoorthy
				Laurence C. Seifert
Honorary Conference Chair	Irma Rossetto Passoni
				Sc&Tech, Info. and Comm. Comission of
				Brazilian Congress.

Sponsored by IIISis - International Institute for Systems Integration,
BB - Banco do Brasil, TELEBRAS, FINEP, CNPq, FBB, with colaboration of
NJIT, SUCESU, EMBRAPA, ABAG, ACM e IEEE Computer Society.

Instituto Internacional de Integracao de Sistemas - IIISis - Brazil.

From hoey@aic.nrl.navy.mil  Mon Jan 24 19:15:23 1994
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From: hoey@aic.nrl.navy.mil
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To: Cube-Lovers@ai.mit.edu
Cc: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
Subject: Concerning B, CSymm, and Symm

In his message of Sat, 8 Jan 1994 08:46:20 EST, Jerry Bryan
<BRYAN%WVNVM.BITNET@mitvma.mit.edu> considers his use of the term "B"
``to indicate various aspects of the conjugacy class generated by
m'Xmc.''

I don't think that's properly called a conjugacy class, but a
different sort of equivalence class.  A conjugacy class is a special
kind of equivalence class (just as a coset is a special kind of
equivalence class) but this B is a little bit of both, so I don't
think it is correct to call it either.

> Let X be any cube.  Then the set of B-conjugacy classes of X is
> the set of all m'Xmc for all m in M and all c in C.  We denote
> this set as BClass(X).  B is the function B(X)=min(BClass(X)).

That's a little unfortunate--I'd prefer to use B(X) for the
equivalence class, and min(B(X))--or repr(B(X))--for the canonical
representative.  That's because the representative is not the
important thing here, it's just a convenient way to represent (!) the
class in a computer.

> Note that we could have defined BClass(X) equivalently as the set of
> all mXm'c, or as the set of all cm'Xm, or as the set of all
> cmXm'.... This is the justification for the assertion in a previous
> note that Gx\B = (Gx\M)\C = (Gx\C)\M.

Not quite.  The justification for (Gx\Conj(M))/C = (Gx/C)\Conj(M)
is that instead of m'Xmc we could choose m'Xcm, a possibility you
didn't list.

In his message of "Sat, 8 Jan 1994 10:52:22 EST", Jerry continues with
discussion on combining conjugacy classes.  We've exchanged some
private email on the subject material, but in case anyone on the list
is following this stuff....

> There are only 10 distinct values for |BClass(X)| and for
> |BClass(Y)|, namely 24, 48, 72, 96, 144, 192, 288, 384, 576, and
> 1152.  (By the way, I have never figured out why it is *exactly* the
> same set of values for both the corners and for the edges.  It is
> easy to see why it is approximately the same set of values....

I'm not sure what kind of approximation you mean, but certainly those
ten values are all that are possible:

    Proof: For if m1,m2 are in the same coset of M/CSymm(X), then
    (m1 m2') is in CSymm(X), so X' (m1 m2')' X (m1 m2') = c0 in C so
    m1' X m1 = m2' X c0 m2.  It's then clear that
         { m1' X m1 c : c in C } =  { m2' X m2 c : c in C }      (*)
    are equal 24-element sets.  The same manipulation in reverse shows
    that if (*) holds for some m1,m2 in M, then m1 and m2 are in the
    same coset of M/CSymm(X).  So |BClass(X)|=24 |M/CSymm(X)|.
    |M/CSymm(X)| must be a divisor of |M|=48, QED.

It wouldn't have been all that surprising to see one of the possible
sizes of |CSymm(X)| fail to appear as a symmetry group of the corners
or edges, but it's not surprising that they all do, either.

> [For the original approach] I needed to be able to prove that for a
> fixed m and n, that |(BClass(X)[m] * BClass(Y)[n]| had the same
> value for all X in GC[m]\B and all GE[n]\B.

That is to say, that the sizes |CSymm(X)| and |CSymm(Y)| might
determine {|Symm(X*Y)|} for X in GC\B, Y in GE\B, and so (X*Y) in
G\Conj(M).  It doesn't, but the situation is even worse.  Jerry goes
on to suppose that perhaps CSymm(X) and CSymm(Y) themselves might
determine {|Symm(X*Y)|}, and even that isn't true.  I've discovered
this by a computer search of GC\B. (A search of GE\B is in progress,
but for the current result we can take Y=I in GE\B).  I have found
that AllSymms(X) is not determined, even up to subgroup sizes, by
CSymm(X).

According to the search, the following are the only positions of GC\B
for which |CSymm(X)|=16.

          X1                       X2                       X3
        +---+                    +---+                    +---+
        |F F|                    |B F|                    |F B|
        |B B|                    |F B|                    |F B|
    +---+---+---+            +---+---+---+            +---+---+---+
    |R R|D D|L L|            |L L|D T|R R|            |L L|D T|R R|
    |R R|T T|L L|            |L L|T D|R R|            |L L|D T|R R|
    +---+---+---+            +---+---+---+            +---+---+---+
        |B B|                    |F B|                    |B F|
        |F F|                    |B F|                    |B F|
        +---+                    +---+                    +---+
        |T T|                    |T D|                    |T D|
        |D D|                    |D T|                    |T D|
        +---+                    +---+                    +---+

Coincidentally, I have been (privately) calling the CSymm(Xi)
subgroups the "X subgroups" of M, an X subgroup being the subgroup
that maps an orthogonal axis of the cube (in the above examples, the
L-R axis) to itself.  X1 is a notable position, in that each corner
has been swapped with its opposite corner.  Symm(X1) is an X subgroup
as well, and there is another X subgroup in AllSymms(X1).  There is,
however, no 16-element subgroup in AllSymms(X2) or AllSymms(X3).  (We
have seen X2 before: it is the corners of the Laughter (or 4/)
position).  In fact, my program says that

     AllSymms(X1) contains two occurrences of 16-element X subgroups,
                           two occurrences of the 8-element HX subgroup,
                           two occurrences of 8-element R subgroups,
                           two occurrences of 8-element S subgroups,
                           eight occurrences of 4-element HS subgroups, and
                           eight occurrences of the 2-element HV subgroup.
                                     
     AllSymms(X2) and AllSymms(X3) each contain
                           two occurrences of 8-element CX subgroups,
                           two occurrences of 8-element AX subgroups,
                           two occurrences of 8-element P subgroups,
                           two occurrences of 8-element Q subgroups,
                           eight occurrences of 2-element ES subgroups, and
                           eight occurrences of 2-element HW subgroups.

The names of these groups are part of a taxonomy of the subgroups of M
I've developed, which I won't go into just now.  But the point that I
find surprising here is that AllSymms(X1) and AllSymms(X2) are
completely disjoint.  While that can't happen all the time (smaller
CSymm() groups have many occurrences of the one-element "I" subgroup)
I think the tendency to disjointness is too pronounced to be simple
anti-coincidence.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From cuf@aol.com  Thu Feb 10 23:09:33 1994
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To: cube-lovers@life.ai.mit.edu
Date: Thu, 10 Feb 94 22:17:05 EST
Subject: Computer & Health

The Computer User Family (CUF) is concerned about the health problem
associated with computers.  Video Display Terminals, emit UV and ELF
radiation and may cause cancer, immune system irregularities, miscarriages
and eye fatigue.
Computer noise from fans, disk and CD drives is also becoming a source of
anxiety, stress and general discomfort .  We usually don't realize how loud
our computers are: 50dB and more. 
These problems should be dealt with and add-ons should be provided for
present computers to avoid putting us at risks.  Some safe screens and quiet
power supplies are coming out but they are marginal and prices are
prohibitive.
Meanwhile the general guidelines for the users are:
1. Position yourself approximately 22 inches to 28 inches (arm's length) from
the screen and four feet from the sides and rear of other terminals. 
2. Eliminate sources of glare and lower light levels in the room.  Don't sit
facing a bright window. If necessary, use screen hoods, glare shields over
the screen or wear anti-UV/anti-glare glasses. 
3. Put a noise absorbing mat under your computer.  Pull your computer away
from the wall or any hard surface that reflects noise and vibration back to
you.
4. Rest occasionally during periods of intense concentration.  Closing your
eyes helps.
5. Turn off the VDT when not in use.

From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU  Sun Feb 13 16:59:40 1994
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Date:      Sun, 13 Feb 1994 16:59:22 EST
From: "Jerry Bryan" <BRYAN%WVNVM.BITNET@mitvma.mit.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Re: Some Proposed Terminology
In-Reply-To: Message of 01/21/94 at 18:32:15 from hoey@AIC.NRL.Navy.Mil

On 01/21/94 at 18:32:15 hoey@AIC.NRL.Navy.Mil said:
>I welcome Jerry Bryan's <BRYAN%WVNVM.BITNET@mitvma.mit.edu> efforts to
>improve the terminology of the groups associated with Rubik's cube.
>But there is some additional clarification I think is necessary.

>> Let G\C be the corners with edges without centers group.  I intend
>> for the notation to indicate G reduced by C, where C is the rotation
>> group for the cube....

>> Let GC\C be the corners without edges without centers group....
>> Let GE\C be the edges without corners without centers group....

>First, these are not, strictly speaking, groups.  Well, you can make
>them groups, by defining what the group operation is.  But I don't
>know any way of doing that without losing the symmetrical nature of
>the problem.

>Second, I would suggest that G/C, GC/C, and GE/C are more standard
>names for these objects.  The elements are nominally 24-element sets,
>each of which is an equivalence class when two positions are
>considered equivalent when they differ by their position with respect
>to the corners.  The classes are called the cosets of C in G, GC, and
>GE, respectively.

Dan Hoey's criticism's are quite valid.  I will attempt to repair
the damage as follows:  1) accept the Gx/C notation in lieu
of Gx\C,  2) define an operation within Gx/C such that
Gx/C is a group, and 3) use Gx/C as a model for cubes without
centers in such a way that the symmetrical nature of the problem
is retained.

Let C be the set of twenty-four whole cube rotations of the cube, and
let G be the standard 3x3x3 cube group.  We observe that if X is a cube
in G, then c'Xc is also a cube in G for every c in C.  We could call
this operation C-conjugancy.  However, there is seldom (if ever) any
reason to speak of C-conjugancy.  That is, C is just a subset of M,
the set of forty-eight whole cube rotations and reflections.  Indeed,
C is half of M, and the other half of M is the reflection of C.  Hence,
M-conjugancy of the form m'Xm is more powerful than C-conjugancy, and
there is normally no reason to speak of C-conjugancy.  I only bring it
up to emphasize that if X is in G, then c'Xc is in G.

On the other hand, if I understand correctly the model most people use
for G, elements of the form Xc or cX are not in G except for the
trivial case where c=I.  The problem is that C is considered to move
the centers, but G is generated by Q, the set of quarter-turns of the
faces, and Q does not move the centers.  For example, there is a
c in C such that F=c'Rc, but there is not a c in C such that
F=Rc or F=cR.  And indeed, neither Rc nor cR are in G at all unless
c=I.

As we said, G is generated as G=<Q>, where Q is the set of
quarter-turns Q={F,B,U,D,L,R,F',B',U',D',L',R'}.  Elements of Q
move the corners and edges, but Q is the identity on the centers.
C, on the other hand, is generally considered to move the centers.
Hence, the group generated as <Q,C> is a supergroup of G, and there
are elements of the supergroup which are not in G.  (This
supergroup, by the way, is not The Supergroup.  The Supergroup is
generated by Q alone, but with orientations of the (otherwise fixed)
centers considered.)  Therefore, our first order of business is to
make C into a sub-group of G.

We observe that since the elements of Q are the identity on the
centers, the primary function of the centers is to provide a
frame of reference.  But we can provide a frame of reference
without the centers actually being there.

For example, consider the group GC consisting of cube centers
and corners.  You can model this group by removing the edge
labels from a physical cube.  Establish the cube at Start and
perform RL'.  The corners will be rotated forward, and will be
positioned properly with respect to each other, but the cube is
clearly not solved.  You can tell that the cube is not at Start
because the corners are not aligned properly with the centers.

Now, do the same thing except remove both the edge and center
labels.  If you perform RL' at Start, the cube "looks"
solved but rotated forward.  However, we can adopt the convention
that the cube is solved only if the Up color is Up, the Front
color is Front, etc.  With this convention in place, RL' is
clearly seen not to be solved; it is two moves from Start.
The convention provides the fixed frame of reference.
Furthermore, RL' (which is in GC) is equal to an element of C, and
indeed all elements of C are in GC, as are all elements of the
form Xc or cX for c in C and X in GC.  Hence, we have <Q>=<Q,C>.

Similar comments apply to GE, the group of edges and centers, except
that processes composed from elements of Q to accomplish rotations
in C are not quite so short in GE as they are in GC.

G, the full 3x3x3 cube group consisting of corners, edges, and centers
is a bit more difficult.  The problem is that if X is in G, then
objects of the form Xc or cX are in G only if c is even.  Twelve
elements of C are even and twelve are odd.  Indeed, C[even] is
a sub-group of C, but C[odd] is not.

We will deal with this situation (as circumstances require) in two
different ways.  One is simply to restrict ourselves to C[even]
when dealing with G.  The other is to define a new group we will
call GS.  In our model for G in which the centers are implied by
a frame of reference convention rather than by actual physical
centers, we can easily add slice moves to the standard face moves.
If the centers were physically present, then the slice moves would
move the centers, but without the physical centers there is
no problem.

If S is the set of slice moves, then GS is generated as <Q,S>.
GS is essentially G with parity restrictions removed.  Hence we
observe that |G|=|GC|*|GE|/2, |GS|=|GC|*|GE|, and |GS|=|G|*2.
Also, if X is in G or in GS, then elements of the form cX or
Xc are in GS for all c in C.  In those occasions where we are
willing to think of GS rather than G, we can use C rather than
C[even].

At this point, we can say that GS/C, G/C[even], GC/C, and GE/C are
cosets of C in GS, C[even] in G, C in GC, and C in GE, respectively.
To be a little more conformant with standard coset notation, we will
write cube elements as lower case letters for the remainder of this
note, and hence for a particular cube x a coset of C is denoted as
Cx={y: y=cx} or xC={y: y=xc}.

Now, we propose a group operator for the cosets:  Cx Cy = C(xy) and
xC yC = (xy)C.  Showing that we have a group is easy.  I originally
included a proof in this note, but there is a proof in Chapter 8 of
Frey and Singmaster's _Handbook of Cubik Math_.  Hence, I will defer
to their proof instead.

According to Frey and Singmaster, G/C is called the factor group of
C in G, or the quotient group of G by C.  Of most significance to
us right now is the fact that the identity of the factor group is
Ci or iC, where i is the identity of G.  But Ci or iC is just C.
Hence, the identity of the factor group is C.  This justifies our
identification of G/C with a centerless cube.  In English, it means
that we can rotate a centerless cube in space without changing
anything.  I think this would comply with most people's intuitive
sense of what it means for a cube to be centerless.

Finally, as to whether this model retains the "symmetrical nature of
the problem", I will have to leave that as an open question,
depending on precisely what we mean by "symmetrical".  It seems to
me that this model does a better job of being "symmetrical" than
a model which includes only seven corner cubies or only eleven
edge cubies, but maybe not.  What does "symmetrical" mean when
it comes to centerless cubes?

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From anandrao@hk.super.net  Thu Feb 17 02:18:45 1994
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Date: Wed, 2 Feb 1994 10:10:28 +0800 (HKT)
From: "Mr. Anand Rao" <anandrao@hk.super.net>
Subject: Re: Mickey's Challenge
To: Peter Beck <pbeck@pica.army.mil>
Cc: Cube-Lovers@ai.mit.edu
In-Reply-To: <9401061406.aa23113@COR6.PICA.ARMY.MIL>
Message-Id: <Pine.3.07.9402021025.A16071-a100000@hk.super.net>
Mime-Version: 1.0
Content-Type: TEXT/PLAIN; charset=US-ASCII

I recently picked up this puzzle on a trip to Houston, TX.
It is really sad that even though this puzzle is made in China by a Hong
Kong company, I had to go to the US to get it! It is not available for
sale here in Hong Kong.
Many thanks for your tip. If any one knows about any other interesting
puzzles, they are welcome to contribute!

Cheers!

On Thu, 6 Jan 1994, Peter Beck wrote:

> 
> NEW PUZZLE "MICKEY'S CHALLENGE" is at your Disney
> store now, price $10.
> 
> This is a legal MACHBALL, ie, a spherical
> SKEWB.  It comes with a solution book.  
> Christoph Bandelow (a longer time cuber)
> wrote the solution.
> I haven't bought one or it played with it
> yet.
> 
> GOOD PUZZLING
> 
> pete beck
> 
> pbeck@pica.army.mil
> 




From xirion!jandr@relay.nl.net  Fri Feb 18 08:45:09 1994
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From: Jan de Ruiter <jandr@xirion.nl>
Date: Fri, 18 Feb 94 14:43:52 +0100
Message-Id: <2038.9402181343@xirion.xirion.nl>
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To: cube-lovers@life.ai.mit.edu

To: cube-lovers@life.ai.mit.edu
Subject: Re:  10x10 Tangle

Sorry about not reporting this earlier, but my search for solutions for
Rubiks Tangle 10x10 confirms the finding of Don Woods: no solutions!

Dik Winter writes:
>As I wrote before, I have embedded in my memory that there is an easy
>argument that the 10x10 is *not* solvable.  I do not know whether I
>found it myself (and ever did mail it to other people) or whether I
>found it somewhere on the net; it is a long time ago.  When I find the
>time I will do a check.  (I know very sure that I have had a program
>running at that time but that I abandoned the search because it would
>be fruitless.)

I am beginning to get real curious about that 'easy argument'.
Does this argument depend on the particular choice for the four
duplicated pieces or not?
If it does, there could exist a choice that does allow a solution, and
we could re-define the puzzle as follows:
  find which four pieces to duplicate in order to find solutions for
  the 10x10.
If the number of solutions varies depending on the choice, you could
even add a restriction:
  find which four pieces to duplicate in order to find a set which has
  the minimum number of solutions for the 10x10.

But if the easy argument does NOT depend on the choice, i.e.: any
choice would lead to no solutions, then the above puzzles would be
senseless as well.

So if anyone at all knows this argument, please tell us and solve the
mystery.

	Jan

From hochberg@gnumath.rutgers.edu  Fri Feb 18 14:36:47 1994
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From: hochberg@gnumath.rutgers.edu (Rob.)
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To: Cube-Lovers@ai.mit.edu
Subject: Add a name, please...


Could you add    edgemstr@orange.cc.utexas.edu 

to the cube lovers list?  Thank you kindly.  
Rob.




From matwood@peruvian.cs.utah.edu  Thu Feb 24 14:25:20 1994
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To: cube-lovers@life.ai.mit.edu
Subject: Book - "Simple Solution To Rubik's Cube"
Date: Thu, 24 Feb 94 12:25:17 MST
From: Mark Atwood <matwood@peruvian.cs.utah.edu>

(I sent my request to be added to this mailing list in to
 cube-lovers-request@ai.ai.mit.edu a few days ago and havnt
 heard back. Hope this works..)

I was going thru my stuff a while back and found my two original
Rubik's Cubes, one of which was given to me several years before
they became wildly popular. The solution I learned was in the
book "The Simple Solution To Rubik's Cube", which was a paperback
of about 20-30 pages.

I remember most of the solution steps outlined in the book, (my
hands remember better than my head does), however, I can't find
a copy of the book anywhere, to refresh my memory.

Anyone got a copy or know where I can get one?

..Mark Atwood

From anandrao@hk.super.net  Fri Feb 25 03:32:04 1994
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Date: Fri, 25 Feb 1994 16:18:56 +0800 (HKT)
From: "Mr. Anand Rao" <anandrao@hk.super.net>
Subject: Re: your mail
To: Jan de Ruiter <jandr@xirion.nl>
Cc: cube-lovers@life.ai.mit.edu
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On Fri, 18 Feb 1994, Jan de Ruiter wrote:
> 
> Sorry about not reporting this earlier, but my search for solutions for
> Rubiks Tangle 10x10 confirms the finding of Don Woods: no solutions!
> 
[snip]
> we could re-define the puzzle as follows:
>   find which four pieces to duplicate in order to find solutions for
>   the 10x10.
> If the number of solutions varies depending on the choice, you could
> even add a restriction:
>   find which four pieces to duplicate in order to find a set which has
>   the minimum number of solutions for the 10x10.
        ^^^^^^^
The kind Mr. Rubik has already done that - the minimum is - ZERO!

The revised problem can be solved fairly easily using your program ( I
don't know, though, how long it takes to run to completion for the 10*10
case) - try to place only 99 tiles  out of the 100 given tiles. You may
have several sub-solutions. It is then easy to determine for each of these
sub-solutions which tile you need to complete the 10*10 mosaic. If this
pattern has already been duplicated, i.e. you need THREE numbers of this
tile to find the complete solution, this sub-solution will not work and so
examine the next sub-solution .... Hopefully you find the solution this way.

After running the program for the 99 tiles, the additional time required
to solve the problem defined by you should not be significant because that
would be  a linear process.

Anand.




From sage@world.std.com  Sun Feb 27 20:08:15 1994
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Date: Sun, 27 Feb 1994 20:08:08 -0500 (EST)
From: Meisha n Thompson <sage@world.std.com>
Subject: Puzzle
To: cube-lovers@life.ai.mit.edu
Cc: Meisha n Thompson <sage@world.std.com>
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Please add me to your mailing list. Thank You Meisha Thompson

From xirion!jandr@relay.nl.net  Mon Mar  7 05:38:41 1994
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From: Jan de Ruiter <jandr@xirion.nl>
Date: Mon, 7 Mar 94 11:37:25 +0100
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To: cube-lovers@life.ai.mit.edu

To anandrao@HK.Super.NET
Subject: Re: your mail
Cc: cube-lovers@life.ai.mit.edu
In-Reply-To: <Pine.3.07.9402251654.A20452-b100000@hk.super.net>

>
>On Fri, 18 Feb 1994, Jan de Ruiter wrote:
>> 
>> Sorry about not reporting this earlier, but my search for solutions for
>> Rubiks Tangle 10x10 confirms the finding of Don Woods: no solutions!
>> 
>[snip]
>> we could re-define the puzzle as follows:
>>   find which four pieces to duplicate in order to find solutions for
>>   the 10x10.
>> If the number of solutions varies depending on the choice, you could
>> even add a restriction:
>>   find which four pieces to duplicate in order to find a set which has
>>   the minimum number of solutions for the 10x10.
>        ^^^^^^^
>The kind Mr. Rubik has already done that - the minimum is - ZERO!

Correct, but I think you know what I mean: minimum >= 1

>The revised problem can be solved fairly easily using your program ( I
>don't know, though, how long it takes to run to completion for the 10*10
>case)

More than a week

>      - try to place only 99 tiles  out of the 100 given tiles. You may
>have several sub-solutions. It is then easy to determine for each of these
>sub-solutions which tile you need to complete the 10*10 mosaic.

I am sorry, but I have to disagree on this. It is not that simple.

If you managed to place 99 pieces, you have already placed three or even
all four of the duplicated pieces (depending on which one is left over)
If you placed three, there are tree possibilities for the piece we need:
- it is nonexistent (illegal colour combinations): no solution
- it is one of the duplicated pieces: this means two of the four puzzles
  will be identical which is OK, but not so nice, or
- it is any other piece: we found a good solution

If you placed all four duplicated pieces already, any solution you find
will not satisfy the conditions of the puzzle (i.e. precisely four duplicated
pieces).

And in both cases you have not solved:
   which four pieces to duplicate in order to find solutions for the 10x10.
but:
   which four pieces to duplicate in order to find solutions for the 10x10,
   with the restriction that three of them must be identical to any three
   taken from the set of four duplicates given by Rubik.

Solving the puzzle without this restriction requires a different approach.
I was thinking of starting the program with 5 of each (120 pieces), and after
placing the 5th duplicated piece remove the rest of the duplicates from
the remaining pieces, and see if this leads to a solution. As soon as back-
tracking removes the 5th duplicate, all other duplicates must be made
accessible again.

	Jan

From phiscock@ee.ryerson.ca  Mon Mar 14 14:14:44 1994
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From: phiscock@ee.ryerson.ca (Peter Hiscocks)
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Subject: Anyone solved Rubik's Tangle?
To: Cube-Lovers@ai.mit.edu
Date: Mon, 14 Mar 94 14:09:18 EST
X-Mailer: ELM [version 2.3 PL11]


For those who haven't seen it, Rubik's Tangle is a new puzzle to 
drive us all nuts, break up our families, and divert us from
the things we should be working on.

It consists of 25 tiles, which form a 5x5 pattern. On each
tile is a pattern of coloured ropes, the ends of which must
match the ends of the ropes on the adjacent tiles.

Certain clues are evident: the shape of each rope pattern
is the same, there are equal numbers of each colour,
and each tile given a letter label on the back.

Before I waste my life on this, has anyone solved the
problem?

Peter
-- 
Peter Hiscocks                            Phone: (416) 979-5000 Ext 6109
Department of Electrical Engineering      Fax:   (416) 979-5280
Ryerson Polytechnical University, Toronto, Canada


From Don.Woods@eng.sun.com  Mon Mar 14 16:14:51 1994
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From: Don.Woods@eng.sun.com (Don Woods)
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> Before I waste my life on this, has anyone solved the problem?

Yes, it's been solved, and discussed at some length on this group.
However, I haven't seen anyone who claims to have come up with an
"insightful" solution, i.e. one in which you figure out a general
approach that leads to a solution.  All solutions I've heard of
have been found by exhaustive search, often by computer.

	-- Don.


From anandrao@hk.super.net  Mon Mar 14 21:27:38 1994
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From: "Mr. Anand Rao" <anandrao@hk.super.net>
Subject: Re: Anyone solved Rubik's Tangle?
To: Don Woods <Don.Woods@eng.sun.com>
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On Mon, 14 Mar 1994, Don Woods wrote:

> > Before I waste my life on this, has anyone solved the problem?
> 
> Yes, it's been solved, and discussed at some length on this group.

True. However, the 10*10 solution where you use all the four tangle
puzzles  to form a 10*10 pattern with matching edges, has been found to be
impossible( Although the puzzle leaflet says that it is solvable). Once
again there is no 'insightful' solution. Someone has posted that he has
seen an intuitive solution which evades his memory for the time being
but will try to recollect what it was ... reincarnation of Fermat's Last
Problem :). There have been some interesting postings in this group on
this topic in the last few weeks and you should read them .

 > However, I haven't seen anyone who claims to have come up
with an
> "insightful" solution, i.e. one in which you figure out a general
> approach that leads to a solution.  All solutions I've heard of
> have been found by exhaustive search, often by computer.
> 
> 	-- Don.
> 

Anand Rao.




From @mitvma.mit.edu:SHERE@SLACVM.BITNET  Tue Mar 15 14:08:27 1994
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Date: Tue, 15 Mar 1994   11:04 -0800 (PST)
From: SHERE%SLACVM.BITNET@mitvma.mit.edu
To: cube-lovers@life.ai.mit.edu
Subject: Mailing List

Hello, would you please add me.. shere@slac.stanford.edu .. to your
mailing list?  I've forgotten some of my key moves and am trying
to brush up.  I've downloaded your archived mail.  Where might I
find a GZ decompression utility?  Anyway, thanks.  Lee

From @mail.uunet.ca:mark.longridge@canrem.com  Sat Apr  2 21:21:55 1994
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To: cube-lovers@life.ai.mit.edu
Reply-To: CRSO.Cube@canrem.com
Sender: CRSO.Cube@canrem.com
Subject: Invariant Shifting
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.733.5834.0C1993C2@canrem.com>
Date: Sat, 2 Apr 1994 20:12:00 -0500
Organization: CRS Online  (Toronto, Ontario)

Something new to stop the drought of cube posts...

Example of Invariant Shifting
-----------------------------

The resultant position generated by process p8 is invariant under
shifting, specifically 2 X on the Left and Right sides.

P8   2 x ORDER 2:

shift
0                              D2 F2 T2 F2 B2 T2 F2 T2
1                           T2 D2 F2 T2 F2 B2 T2 F2
2                        F2 T2 D2 F2 T2 F2 B2 T2
3                     T2 F2 T2 D2 F2 T2 F2 B2
4                  B2 T2 F2 T2 D2 F2 T2 F2
5               F2 B2 T2 F2 T2 D2 F2 T2
6            T2 F2 B2 T2 F2 T2 D2 F2
7         F2 T2 F2 B2 T2 F2 T2 D2

This is the longest process I've found so far. Certainly this property
is not true of all squares group processes. I suspect there are no
processes in the full group with this property (of any significant
length). Perhaps the fact that the L and R faces never rotate will
give some clue on how to generate processes with this property.

Q: Is this the longest such process?

Further Notes on Antipodes in the Square's Group
------------------------------------------------

I just realized some things about sq group antipodes which
I should have seen before...

The closest 2 antipodes can be is 2 square's group moves.
Take the position produced by p66:

p66  Double 4 corner sw L2 B2 R2 F2 L2 F2 T2 R2 (T2 D2 F2 T2) F2 L2 D2

Any turn will reduce this to a position requiring 14 moves. Undoing
this move will regenerate the antipode. No single move can change
position p66 into another antipode, therefore the closest any 2
antipodes can be is 2 moves.

Futhermore any antipode can not be made into a local maximum
which is 14 moves deep with 1 half turn. I will conclude that
there are no local maxima in the square's group that
neighbour each other closer than 2 moves.

-> Mark <-
Email: mark.longridge@canrem.com

From pbeck@pica.army.mil  Mon Apr  4 08:24:04 1994
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Date:     Mon, 4 Apr 94 8:22:52 EDT
From: Peter Beck (BATDD) <pbeck@pica.army.mil>
To: cube-lovers@life.ai.mit.edu
Cc: pbeck@pica.army.mil
Subject:  puzzle boxes
Message-Id:  <9404040822.aa26168@COR6.PICA.ARMY.MIL>

I found a shop in NYC's chinatown that
stocks Japanese puzzle/trick boxes.

  TING'S GIFT SHOP
  18 DOYERS STREET
  NY, NY 10013
  212-962-1081

-  4-way $25
-  4-way w/music $36
-  10-way $42
-  12-way $46
-  20-way $52

NYC sales tax 8%

good puzzling and good eating

From ishius@ishius.com  Mon Apr  4 14:11:59 1994
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To: cube-lovers@ai.mit.edu
From: ishius@ishius.com (Ishi Press International)
Subject: puzzle boxes

Ishi Press International offers a variety of Japanese Trick Puzzle Boxes,
from 4 moves to 66 moves.  These are handcrafted, wood inlaid Okiyama
trickboxes.  We also have unique puzzle boxes by Kamei.

For a free catalog of our PUZZLES please send us your postal mailing address
and we will mail you one.  Please specify that you are interested in PUZZLES.

Always feel free to write me if you have any questions or comments.

Anton Dovydaitis
Customer Support


========================================================================
Ishi Press International        800/859-2086 voice, 408/944-9110 FAX
76 Bonaventura Drive            ishius@ishius.com           The Americas
San Jose, CA  95134             ishi@cix.compulink.co.uk    Europe



From mmoss@panix.com  Mon Apr  4 17:30:58 1994
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From: Matthew Moss <mmoss@panix.com>
Message-Id: <199404042130.AA12590@panix.com>
Subject: About Rubix's tetrahedron...
To: cube-lovers@life.ai.mit.edu (Cube Mailing List)
Date: Mon, 4 Apr 1994 17:30:46 -0400 (EDT)
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Here's a question for y'all.....
It's about the tetrahedron puzzle from Rubix (I forget the real name).
[the one with legal moves consisting of removing a 4-piece tetrahedron from
 a 10-piece and putting it back on in different orientation]

Anyway, mine is pretty loosed up, and occassionally when I am working on it,
one piece will come loose and go skitting across the floor.

There's no way I can remember the orientation it had on there. Has any study
been done or does someone know if I put that piece back on, will it still be
solveable if it's orientation is wrong (ie, different than it was before it
fell off)?

Do you understand my question? (I hope I'm not too confusing...)

I've been thinking about implementing this via computer, just to test this
out, but I thought I'd ask y'all first.

Thanx.

From @mail.uunet.ca:mark.longridge@canrem.com  Sun Apr 10 23:26:50 1994
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Subject: More Sq Notes
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.738.5834.0C19A249@canrem.com>
Date: Sun, 10 Apr 1994 17:37:00 -0400
Organization: CRS Online  (Toronto, Ontario)

Additional Notes on Squares Group Patterns
------------------------------------------

Note that p80a, p99a and p108a are 2 DOT patterns,
all of the form U1 (swap edges & corners in U and D faces)  D3
           or   U1 (swap edges & corners in U and D faces)  D1
           or   D1 (swap edges within U and edges within D) D3

P66a alternate method    F2 R2 U2 F2 R2 U3 D3 B2 L2 F2 B2 U1 D1
 (13)
p67a alternate method    F2 R2 F2 U3 D3 L2 B2 D2 L2 B2 U1 D1 B2
 (13)

p80a alternate method    U1 F2 R2 L2 U2 D2 F2 U2 D3
  (9)
p99a alternate method    U1 R2 F2 B2 U2 D2 R2 D1
  (8)

P100a alternate method   F2 U2 D2 F2 R2 L2 D1 F2 R2 L2 B2 U1
 (12)
p108a alternate method   R2 F2 B2 L2 D1 R2 U2 R2 L2 U2 R2 D1
 (12)
p130a alternate method   F2 R2 F2 B2 U1 D1 F2 R2 D2 F2 L2 U3 D3
 (13)
p133a alternate method   R2 U1 F2 R2 L2 U2 D2 F2 U2 D3 R2
 (11)

A) In general, any sequence of half turns which swaps edges and
corners in the U and D faces can be sanwiched between a single quarter
turn of U and a single quarter turn of D. Such a process would lead to
a square's group position.

B) Furthermore, any sequence of half turns which swaps edges within U
and
edges within D can be sanwiched between a single quarter turn of U or D
and a single quarter turn of U or D. Once again, such a process would
lead to a square's group position.

Here is an example of a position which takes over twice as many half
turns as full group moves:

        L2 T2 L2 T2 L2 T2 F2 L2 T2 F2 T2 R2 B2     (13)
        U1 F2 R2 L2 B2 D1                           (6)

As discussed in point A above, sequences which move all elements of
the U face to the D face and also move all elements of the D face
to the U face (excepting the centres naturally) appear as 2 DOT
patterns on the cube. This makes sense, as the initial quarter
turn in process p80a must be balanced by another quarter turn.
Since all of the elements subjected to the quarter turn are now
in the D face, we must turn that face a quarter turn to remain
in the squares group.

From @mail.uunet.ca:mark.longridge@canrem.com  Sun Apr 10 23:27:13 1994
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Sender: CRSO.Cube@canrem.com
Subject: More Shifting
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.739.5834.0C19A24A@canrem.com>
Date: Sun, 10 Apr 1994 17:39:00 -0400
Organization: CRS Online  (Toronto, Ontario)

More Notes on Invariant Shifting
--------------------------------

Let us define a process as "Shift Invariant" if it results in the same
displacement even after a series of left or right shifts. That is,
from a process of length N we can generate N-1 processes which
result in the same displacement by shifting the process. Sometimes
the processes generated are not all unique!

e.g. P8   2 x ORDER 2: (symmetry level 3)
     D2 F2 T2 F2 B2 T2 F2 T2
  (8)

Q: Is this the longest such process?

A: No.

The following processes are also shift invariant:

2 Swap                   D2 R2 D2 R2 D2 R2 (6)
(symmetry level 12, SI level 2)

p21  2 H                 L2 R2 B2 L2 R2 F2 (6)
(symmetry level 6, SI level 6)

Amazingly, the process p3 (found using Dik Winter's program) is actually
a series of 20 processes which all result in the same displacement!

p3   12 flip             R1 L1 D2 B3 L2 F2 R2 U3 D1 R3 D2 F3 B3 D3 F2 D3
                            R2 U3 F2 D3 (20)
(symmetry level 1, SI level 20)

Since p3 is shift invariant, we can easily shift the 3 consecutive half
turns to the beginning without fear of altering the end result:
  L2 F2 R2 U3 D1 R3 D2 F3 B3 D3 F2 D3 + R2 U3 F2 D3 R1 L1 D2 B3

From bagleyd@source.asset.com  Thu Apr 14 11:21:48 1994
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From: bagleyd@source.asset.com (David A. Bagley)
Message-Id: <9404141500.AA10500@source.asset.com>
To: cube-lovers@life.ai.mit.edu
Subject: Mailing List


Hi I was wondering if you can add  me to your mailing list.

I put some motif puzzles at ftp.x.org in /contrib/motif_puzzles.  They are:
rubik: a (nxnxn) rubik's cube
pyramid: a (nxnxn) pyraminx with period 2 and period 3 cuts
oct: an (nxnxn) octahedron with period 3 and period 4 cuts
skewb: a diagonal cut rubik's cube
cubes, triangles, & hexagons: sliding block puzzles
 
There are no self-solvers provided with these.
Also you may want to check out the tetris games which only use X.
altetris: polyomino version of tetris
alweltris: polyomino version of welltris
altertris: polyiamond version of tetris (they bounce of the walls)
alhextris: polyhexes version of tetris (again, they bounce off the walls)
These are found in ftp.x.org in /contrib

I recently heard that 10x10 tangle has no solution.  I was trying to solve
that one too.  What a waste! (90 days of compute time, wow thats one
efficient program, fast machine , or both!)

Oh, unfortuately the motif puzzles need motif to compile.  Is there a good
public domain substitute for Motif?
Have fun
David

From bagleyd@source.asset.com  Thu Apr 14 13:21:45 1994
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From: bagleyd@source.asset.com (David A. Bagley)
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To: cube-lovers@ai.mit.edu
Subject: Mailing List


Hi, I was wondering if you can add me to your mailing list. 
  
  I put some motif puzzles at ftp.x.org in /contrib/motif_puzzles.
  They are: 
rubik: a (nxnxn) rubik's cube 
pyramid: a (nxnxn) pyraminx with period 2 and period 3 cuts 
oct: an (nxnxn) octahedron with period 3 and period 4 cuts  
skewb: a diagonal cut rubik's cube 
cubes, triangles, & hexagons: sliding block puzzles 
  There are no self-solvers provided with these. A record keeps track of
how many move it takes you to solve them. A record of 32767 means I never
did it. (I do not follow any standard notation for a move (for example,
on the rubik's cube a move is any 1/4 turn)).
  Unfortuately the motif puzzles need motif to compile.  Is there a good
public domain substitute for Motif?
 
Also you may want to check out the tetris games which only use X. 
altetris: polyomino version of tetris 
alweltris: polyomino version of welltris 
altertris: polyiamond version of tetris (they bounce off the walls) 
alhextris: polyhexes version of tetris (again, they bounce off the walls) 
These are found in ftp.x.org in /contrib 
  
I recently heard that 10x10 tangle has no solution.  I was trying to solve 
that one too.  What a waste! (90 days of compute time, wow thats one 
efficient program, fast machine , or both!)                 
 
Have fun
David

From iavlang@cs.vu.nl  Tue Apr 19 03:52:25 1994
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Date: Tue, 19 Apr 1994 08:01:19 +0200
To: cube-lovers@life.ai.mit.edu
From: Izak van Langevelde <iavlang@cs.vu.nl>
Subject: Cubism for Fun
Message-Id:  <9404190801.aa24220@top.cs.vu.nl>

In the archive of this mailing list I found the following:
>CFF is a newsletter published by the Nederlandse Kubus Club NKC (Dutch
>Cubists Club).  It appears a bit irregular, but a few times a year.
>Yearly membership fee is now NLG 25.- (Dutch Guilders) which amounts to
>approximately $ 15.-.  Institutional membership is also possible.
>Information is available from the editor Gerald Maurice

Unfortunaty, the abovementioned editor didn't respond to my email.
Does anyone know whether CFF still exists? Who is the current editor?

                                                        Thanks,
                                                        Izak van Langevelde




From iavlang@cs.vu.nl  Tue Apr 19 09:32:59 1994
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Date: Tue, 19 Apr 1994 10:10:25 +0200
To: cube-lovers@ai.mit.edu
From: Izak van Langevelde <iavlang@cs.vu.nl>
Subject: Cubism for Fun
Message-Id:  <9404191010.aa25902@top.cs.vu.nl>

In the archive of this mailing list I found the following:
>CFF is a newsletter published by the Nederlandse Kubus Club NKC (Dutch
>Cubists Club).  It appears a bit irregular, but a few times a year.
>Yearly membership fee is now NLG 25.- (Dutch Guilders) which amounts to
>approximately $ 15.-.  Institutional membership is also possible.
>Information is available from the editor Gerald Maurice

Unfortunaty, the abovementioned editor didn't respond to my email.
Does anyone know whether CFF still exists? Who is the current editor?

                                                        Thanks,
                                                        Izak van Langevelde




From dik@cwi.nl  Tue Apr 19 18:17:58 1994
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From: Dik.Winter@cwi.nl
Message-Id: <9404192215.AA05363=dik@boring.cwi.nl>
To: cube-lovers@ai.mit.edu, iavlang@cs.vu.nl
Subject: Re:  Cubism for Fun

 > In the archive of this mailing list I found the following:
 > >CFF is a newsletter published by the Nederlandse Kubus Club NKC (Dutch
 > >Cubists Club).  It appears a bit irregular, but a few times a year.
 > >Yearly membership fee is now NLG 25.- (Dutch Guilders) which amounts to
 > >approximately $ 15.-.  Institutional membership is also possible.
 > >Information is available from the editor Gerald Maurice

 > Unfortunaty, the abovementioned editor didn't respond to my email.
 > Does anyone know whether CFF still exists? Who is the current editor?

It ought to work.  Perhaps mail got lost?  Just today I received CFF 33,
a summary of the contents will be forthcoming.

dik

From dik@cwi.nl  Tue Apr 19 18:47:54 1994
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Date: Wed, 20 Apr 1994 00:46:34 +0200
From: Dik.Winter@cwi.nl
Message-Id: <9404192246.AA05401=dik@boring.cwi.nl>
To: cube-lovers@life.ai.mit.edu
Subject: CFF33

Cubism For Fun number 33.

I just received it.  It is dated February 1994, so it is a bit late ;-).
Here a summary of the contents.

1.  Dr. Dragon's Polycons by Bernhard Wiezorke and Jacques Haubrich
    A new, apparently interesting, puzzle from Japan, a lot like
    polyonimos.  Given a rectangular grid you can make pieces from
    the horizontal and vertical lines connecting the points.  The
    writers coin the term 'monocon' for the piece consisting of a
    single line segment, 'dicon' for the two different pieces that
    consist of two connected line segment (one is angled, the other
    not).  Similarly there are 5 'tricons' and 16 'tetracons'.  The
    puzzle consists of 10 of the 16 'tetracons' that must be put on
    a 5x5 rectangular grid.  The authors also look at extensions of
    the puzzle.
2.  The Hollow Pyramid by Jan de Ruiter.
    In a previous issue there was a puzzle about a hollow pyramid made
    up of balls that must be constructed by the 25 different pieces
    that consist of 4 connected balls.  Jan de Ruiter was the first to
    solve the puzzle (with a computer).  Here he explains the program.
3.  Junior Polycubes by Jacques Haubrich.
    Pieces consist of 1 to 4 connected cubes that must be put on a
    6x6 square.  Not so much a puzzle, more like Tangram: create forms.
4.  Folding Puzzles by Leo Links.
    About puzzles where some intricate folding is needed to solve.
5.  Cross Pattern Piling by Dieter Gebhardt.
    A puzzle where you put counters on a square and its four neighbours.
    The goal is to pile up to a common height for all the squares.  The
    article also discusses a modified version where counting is done mod
    2.  Associated with it comes the 24th CFF contest.
6.  Gouge Packing Puzzle by Gaetan Gouge.
    Description of and some elaborations about a packing puzzle.
7.  Spots Puzzle by Harold Cataquet.
    Elaborations about a puzzle from A.L.Hoffman, Puzzles old and new,
    New York, 1920.
8.  Arrow-Minded by Ivan Moscovich.
    Start with a fully-connected hexagon.  Put random arrows on all edges.
    Next add nodes on all intersections.  This gives 19 nodes in all.  The
    problem is to find a Hamiltonian path along the nodes, minding the
    arrows (not always possible).  An original puzzle by the writer.
9.  Prime Pentacube Pakcings by Frits Gobel.
    Start with the pentacube consisting of a square base of four cubes and
    one cube on top of it in a corner.  Is it possible to pack a 5x5x5
    cube with 25 such pentacubes?  What other figures can be packed?
10. Contest 25 by Ekkehard Kuenzell.
    Pack a figure with all 29 different pentacubes.
11. Rubik's Rabbits by Luc de Smet.
    Discussion of this latest by Rubik.
12. Party Impressions by Gerald Maurice.
    Impressions of the Puzzle Party and Cube Day, Augustus 1993 in the
    Netherlands.
Further results of contest 23, a book review by Mark Peters and the first
of a series of columns by Edward Hordern.

Cubism For Fun is a newsletter published by Nederlandse Kubus Club NKC
(Dutch Cubists Club).  Applications for membership to the treasurer:
   Lucien Matthijsse
   Loenapad 12
   3402 EP IJsselstein
   The Netherlands
Membership fee NLG 25.- (about US$ 15.-; add transaction costs).

dik

From SCHMIDTG@beast.cs.hh.ab.com  Tue Apr 19 19:38:21 1994
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Date: Tue, 19 Apr 1994 19:38:17 -0400 (EDT)
From: SCHMIDTG@beast.cs.hh.ab.com
To: cube-lovers@ai.mit.edu
Message-Id: <940419193817.20407598@iccgcc.cs.hh.ab.com>
Subject: Re:  Cubism for Fun and Games

> > In the archive of this mailing list I found the following:
> > >CFF is a newsletter published by the Nederlandse Kubus Club NKC (Dutch
> > >Cubists Club).  It appears a bit irregular, but a few times a year.
> > >Yearly membership fee is now NLG 25.- (Dutch Guilders) which amounts to
> > >approximately $ 15.-.  Institutional membership is also possible.
> > >Information is available from the editor Gerald Maurice
>
> > Unfortunaty, the abovementioned editor didn't respond to my email.
> > Does anyone know whether CFF still exists? Who is the current editor?
>
>It ought to work.  Perhaps mail got lost?  Just today I received CFF 33,
>a summary of the contents will be forthcoming.
>
>dik

Glad to see someone has had some luck with CFF.  I sent some postage stamps
to the address posted in rec.puzzles a few years ago and never heard back.
Also, obtained no response from my email to Longridge.  Geez, I thought at
least I deserved some sort of reply!

-- Greg Schmidt		schmidtg@iccgcc.decnet.ab.com

From SCHMIDTG@beast.cs.hh.ab.com  Tue Apr 19 20:45:05 1994
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Date: Tue, 19 Apr 1994 20:45:03 -0400 (EDT)
From: SCHMIDTG@beast.cs.hh.ab.com
To: cube-lovers@ai.mit.edu
Message-Id: <940419204503.204073c2@iccgcc.cs.hh.ab.com>
Subject: Re: Cubism for Fun and Games (Correction and Apologies!)

>> > In the archive of this mailing list I found the following:
>> > >CFF is a newsletter published by the Nederlandse Kubus Club NKC (Dutch
>> > >Cubists Club).  It appears a bit irregular, but a few times a year.
>> > >Yearly membership fee is now NLG 25.- (Dutch Guilders) which amounts to
>> > >approximately $ 15.-.  Institutional membership is also possible.
>> > >Information is available from the editor Gerald Maurice
>>
>> > Unfortunaty, the abovementioned editor didn't respond to my email.
>> > Does anyone know whether CFF still exists? Who is the current editor?
>>
>>It ought to work.  Perhaps mail got lost?  Just today I received CFF 33,
>>a summary of the contents will be forthcoming.
>>
>>dik
>
>Glad to see someone has had some luck with CFF.  I sent some postage stamps
>to the address posted in rec.puzzles a few years ago and never heard back.
>Also, obtained no response from my email to Longridge.  Geez, I thought at
>least I deserved some sort of reply!
>

Whoops, I think I got my signals crossed here.  I was actually referring to
DOTC Newsletter (Domain of the Cube) not CFF.  I apologize for any misunder-
standing caused by this.


-- Greg Schmidt		schmidtg@iccgcc.decnet.ab.com

From coxj@rpi.edu  Mon Apr 25 11:13:35 1994
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Subject: Unsubscribe
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        I would like to be taken off the mailing list.
|   _____       _   _  |               Jeff "The Master" Cox                 |
|     /  __    / ) / ) |                   coxj@rpi.edu                      |
|  / /  /__)  (_  (_   |                                                     |
|  \/   \__  /   /     |   "There's a fine line between clever and stupid"   |


From ronnie@cisco.com  Mon Apr 25 12:14:29 1994
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To: coxj@rpi.edu (Jeffrey M. Cox)
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Subject: Re: Unsubscribe 
In-Reply-To: Your message of "Mon, 25 Apr 1994 11:12:43 EDT."
             <199404251513.LAA15800@mail1.its.rpi.edu> 
Date: Mon, 25 Apr 1994 09:14:05 -0700
From: "Ronnie B. Kon" <ronnie@cisco.com>

>        I would like to be taken off the mailing list.
>|   _____       _   _  |               Jeff "The Master" Cox                 
>|     /  __    / ) / ) |                   coxj@rpi.edu                      
>|  / /  /__)  (_  (_   |                                                     
>|  \/   \__  /   /     |   "There's a fine line between clever and stupid"   

Would that be the line between cube-lovers and cube-lovers-request?

				Ronnie

From alan@parsley.lcs.mit.edu  Mon Apr 25 21:41:35 1994
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In-Reply-To: Jeffrey M. Cox's message of Mon, 25 Apr 1994 11:12:43 -0400 <199404251513.LAA15800@mail1.its.rpi.edu>
Subject: Unsubscribe

   Date: Mon, 25 Apr 1994 11:12:43 -0400
   From: coxj@rpi.edu (Jeffrey M. Cox)
	   I would like to be taken off the mailing list.
   |   _____       _   _  |               Jeff "The Master" Cox
   |     /  __    / ) / ) |                   coxj@rpi.edu
   |  / /  /__)  (_  (_   |
   |  \/   \__  /   /     |   "There's a fine line between clever and stupid"

Despite the fact that you stupidly mailed your administrative request to
the entire mailing list, I have removed you from Cube-Lovers.

I thought I would also take this opportunity to remind the rest of you of
the contents of the greeting message I send all new subscribers.
Old-timers who've seen previous versions of this file can amuse themselves
by noticing that this winter we had some of our heaviest traffic ever.

------- Begin Standard Greeting -------

Don't expect to receive any mail anytime soon.  Cube-Lovers is mostly quiet
these days.  Our addresses are Cube-Lovers@AI.MIT.EDU for submissions and
Cube-Lovers-Request@AI.MIT.EDU for administrivia.

Please note that Cube-Lovers-Request is processed by a human being, not a
computer program (such as LISTSERV or Majordomo).  If your request is not
instantly processed, it is because I don't spend my entire life reading my
electronic mail.  I do know how to interpret many of the commands typically
sent to such programs, but I would prefer it if instead you can remember to
address me in complete sentences.

If you are interested in the archives of the Cube-Lovers mailing list:

Using FTP, connect to FTP.AI.MIT.EDU, login as "anonymous" (any password),
and in the directory "pub/cube-lovers" you will find the twelve (compressed)
files "cube-mail-0.gz" through "cube-mail-11.gz".

Archive vital statistics (when uncompressed):

	   File		   From		   To	     Size (bytes)
	   ----		   ----		   --	     ------------
	cube-mail-0	12 Jul 80	23 Oct 80	185037
	cube-mail-1	 3 Nov 80	 9 Jan 81	135719
	cube-mail-2	10 Jan 81	 3 Aug 81	138566
	cube-mail-3	 3 Aug 81	 3 May 82	137753
	cube-mail-4	 4 May 81	11 Dec 82	139660
	cube-mail-5	11 Dec 82	 6 Jan 87	173364
	cube-mail-6	10 Jan 87	13 Apr 90	216733
	cube-mail-7	12 Oct 90	 9 Sep 91	137508
	cube-mail-8	 1 Nov 91	25 May 92	171205
	cube-mail-9	28 May 92	 7 Jan 93	155755
	cube-mail-10	20 Mar 93	 6 Dec 93	171881
	cube-mail-11	 6 Dec 93	18 Feb 94	349772

In addition, the file "recent-mail" contains a copy of the currently active
section of the archive.  (Unfortunately, due to the way mail works here at
the AI Lab, it is not possible to have new mail accumulate directly into
this file, so there may be some delay before a new message arrives here.)
Finally, the file "README" contains the information you are currently reading.

				- Alan

------- End Standard Greeting -------

From 70410.1050@compuserve.com  Fri Apr 29 13:28:11 1994
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	id NAA28174; Fri, 29 Apr 1994 13:28:12 -0400
Date: 29 Apr 94 13:24:23 EDT
From: Jerry Slocum <70410.1050@compuserve.com>
To: "Cube Lovers @ MIT" <Cube-Lovers@ai.mit.edu>
Subject: Mechanical Puzzles
Message-Id: <940429172422_70410.1050_CHV84-1@CompuServe.COM>

Announcements

An exhibition of "MAZES and PUZZLES" opens May 27 and closes Sept.5 at the 
Museum of Science and Industry in Chicago. It includes a "people maze", 80 
hands-on mechanical puzzles of many types to challenge visitors and 640 
mechanical puzzles of all types and ages that are displayed in 21 cases.
I will send a flyer with details to anyone upon request.

A Directory of Puzzle Collectors (232), Mail Order puzzle sellers (96), Puzzle
periodicals (6), and Retail puzzle stores (147), worldwide, has just been 
published by the non-profit Slocum Puzzle Foundation. Is is available for $10.
postpaid.



                The Slocum Puzzle Foundation

The Slocum Puzzle Foundation was established on August 10, 1993 as a nonprofit
public benefit Corporation. It has been approved by the State of California 
and the U.S. Government as a charitable and educational Foundation.

The purpose of the Foundation is to educate the public on puzzles, their 
history, development, and use in various cultures of the world. The 
Foundation will actively support the use of puzzles for education.

The Foundation will educate the public on puzzles through:

Puzzle exhibitions at museums, libraries, universities, and primary and 
secondary schools, with emphasis on interactive, hands-on puzzles.

Publication of books, compendiums, and research papers on puzzles, and a 
Directory of puzzle collectors.

Supporting and encouraging study and research of the history, development, and
use of puzzles in various cultures of the world.

Supporting and encouraging communication among puzzle experts, educators, 
historians, and the public.

Building and maintaining a collection of puzzles and a library to support 
these activities and to be available for puzzle exhibitions, education, 
research and study.

The first project of the Foundation is to support a Maze and Puzzle exhibition
at the Museum of Science and Industry in Chicago, Illinois.

The Directory is the first publication of the Slocum Puzzle Foundation.

We are interested in suggestions of projects to support exhibitions, 
publications, research and educational activities. 

We would welcome volunteers to help with the activities of the Foundation.

We would welcome donations of puzzles and puzzle literature and financial 
support.

All gifts are deductible from California and Federal income taxes.


Jerry Slocum
Internet:70410.1050@compuserve.com
Address: 257 South Palm Drive, Beverly Hills, CA 90212
Phone:310-273-2270
Fax:310-274-3644

From e0f2m2wm@credit.erin.utoronto.ca  Sat Apr 30 23:59:34 1994
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Received: by credit.erin.utoronto.ca id <34069>; Sun, 1 May 1994 00:02:31 -0400
From: Do Anh Vu <e0f2m2wm@credit.erin.utoronto.ca>
To: cube-lovers@life.ai.mit.edu
Subject: Request to unsubscribe from Cube-Lovers list
Message-Id: <94May1.000231edt.34069@credit.erin.utoronto.ca>
Date: 	Sun, 1 May 1994 00:02:25 -0400

Hi, I would like to be removed from the Cube-Lovers list.
...I'm staying away from my mail for now.

Thanks in advance.

Do Anh

From bagleyd@source.asset.com  Sun May  1 00:00:57 1994
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          id AA38537; Sat, 30 Apr 1994 23:38:49 -0400
Date: Sat, 30 Apr 1994 23:38:49 -0400
From: bagleyd@source.asset.com (David A. Bagley)
Message-Id: <9405010338.AA38537@source.asset.com>
To: 70410.1050@compuserve.com, Cube-Lovers@ai.mit.edu
Subject: Re:  Mechanical Puzzles


Mr Jerry Slocum
  I just got your Directory, great job!

  The last order seems mixed up.  I ordered:
     Jug w/ <>'s
     1989 Puzzle Calendar
  I only received Jug w/ Diamonds.  (The directory was from    a previous
order).  I included $29 , I believe.

  Also , thanks for the info on        cube-lovers.
  If this is garbled its because my modem is not working so well.
David Bagley
58 Winsor Place
Glen Ridge NJ O7028

From pbeck@pica.army.mil  Thu May 19 09:09:47 1994
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Date:     Thu, 19 May 94 9:09:30 EDT
From: Peter Beck (BATDD) <pbeck@pica.army.mil>
To: cube-lovers@life.ai.mit.edu
Subject:  rubik's 94 items
Message-Id:  <9405190909.aa08091@COR6.PICA.ARMY.MIL>

NEWer RUBIK'S PUZZLES

There is a line of Rubik's puzzles, currently available that are
distributed by Western Publishing in their line of GOLDEN GAMES.  This
line was to have two new items for 1994.  They are available in europe
but Western has decided not to make them available in the USA.

1 - Rubik's Maze:  six connected cubes that lay in a plane and have
lines drawn on them.  The object is turn the blocks until a continuous
path is constructed.
2 - Rubik's Rabbits:  This looks a magician's top hat.  Looking down
at the hat it is divided into 8 wedges.  The layers(5) are turned
until a rabbit appears in each wedge or in no wedges at all.

PRE-1994 ITEMS
1 - Rubik's Cube 4: standard cube with rubik's likeness on a center
sticker.
2 - RUBIK's Fifteen:  a plunger type sequential motion puzzle
3 - Rubik's Dice:  a hollow cube with holes where the die spots go.
 Internally there are colored sheets of plastic that by flipping the
cube can be made to cover up the holes
4 - Rubik's Tangle:  comes in 4 versions, discussed at length on cube
lovers

MY PRICES:
 - Rubik's Cube 4:     $10
 - RUBIK's Fifteen:   $10
 - Rubik's Dice:        $12
 - Rubik's Tangle:     $8 each or $30 for a set of all 4

CONTINENTAL USA POSTAGE:  $2 for 1 item, $3 for 2 or more
outside of CONTINENTAL USA 20% surface , 40% 1st class.

PETER BECK
54 RICHWOOD PLACE
DENVILLE , NJ 07834

DAYS  201-724-4812


From pbeck@pica.army.mil  Thu May 19 09:12:09 1994
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Date:     Thu, 19 May 94 9:11:55 EDT
From: Peter Beck (BATDD) <pbeck@pica.army.mil>
To: cube-lovers@life.ai.mit.edu
Subject:  PIONEER PUZZLES
Message-Id:  <9405190911.aa08959@COR6.PICA.ARMY.MIL>


I was wondering if anybody has bought puzzles from:

PIONEER PUZZLES
POB 183
CHEROKEE, TEXAS 76832
1800-441-1796.

They make wire disentanglement puzzles.

thanks 

pete

From @wvnvm.wvnet.edu:BRYAN@WVNVM.WVNET.EDU  Mon May 23 16:51:56 1994
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   with BSMTP id 4733; Mon, 23 May 94 11:00:32 EDT
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X-Acknowledge-To: <BRYAN@WVNVM.WVNET.EDU>
Date:      Mon, 23 May 1994 11:00:24 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Modelling Centerless Cubes

On 13 Feb 1994, I proposed a way to model centerless cubes which would
(in Dan Hoey's words) retain the symmetrical nature of the problem.
I need to post a partial correction/retraction.

The conventional model for centerless cubes loses the symmetrical
nature of the problem.  For example, for a corners-only cube, seven
cubies are modeled rather than eight, and for an edges-only cube,
eleven cubies are modeled rather than twelve.

My proposal in February was to use cosets of the form xC to model
centerless cubes, where x is a cube and where C is the set of
twenty-four whole cube rotations.  This proposal in turn requires
an interpretation of C such that C is a subset of G, the entire cube
group.

C is a group, but normally it is not considered to be a subset
of G, hence it is not normally considered to be a subgroup of G.
That is, C moves the centers of the faces, but G does not.  The
required interpretation is obtained by removing the centers of each
face, and defining rotational orientation by convention so that the
cube is solved only when the Up color is Up, the Front Color is
Front, and so forth.

Under this interpretation, C is indeed a subset (and hence a subgroup)
of G.  More correctly, C[even] is a subset of G, C is a subset of
GC (the corners only cube), C is a subset of GE (the edges only cube),
and C is a subset of GS, where GS=<Q,S> (Q is the set of quarter turns
and S is the set of slice moves).  That is, when you start
talking about C as a subset of G, you have to worry about odd and
even permutations.  Hence, you have to say C is a subset of GS or
C[even] is a subset of G in order not to violate parity rules.

All of the above was posted in February, and I am still comfortable
with it.  However, I went on to say that GS/C, G/C[even], GC/C, and
GE/C were all groups under the operation xC * yC = (xy)C.  I find
that I must retract this claim.

In my note in February, I did not give a proof, but rather appealed
to a proof in Frey and Singmaster's _Handbook of Cubik Math_.  I now
find that I mis-applied their proof.  In order to show the nature of
the problem, I find it useful to go through an attempted proof and
determine the point at which it fails.

Note that the proposed group elements are not cubes, they are cosets.
We proceed as follows:

  1. Associativity:  (xC * yC) * zC =
                     (xy)C * zC =
                     ((xy)z)C =
                     (x(yz))C =
                     xC * (yz)C
                     xC * (yC * zC)

     Note that the associativity of the proposed group G/C derives
     directly from the associativity of G.

  2. Identity: we propose that the identity is iC

                     iC * xC = (ix)C = xC
                     xC * iC = (xi)C = xC

     Note that the identity of the proposed group G/C derives
     directly from the identity i of G.  Further note that the
     identity iC of the proposed group G/C is C, which is
     precisely what you would want for the identity of a centerless
     cube.

  3. Inverse: we propose that (xC)'=x'C

                     xC  * x'C = (xx')C = iC
                     x'C * xC  = (x'x)C = iC

     Note that the inverse of xC in the proposed group G/C derives
     from the inverse of x in G.

  4. Closure: This is where we have our problem.  We require that
     if xC * yC = (xy)C, then (xy)C must be a coset of C.  But the
     representation of xC and yC is not unique.  That is, xC=(xd)C,
     where d is in C, and yC=(ye)C where e is in C.  It is the
     case that (x(ye))C = (xy)C, but in general it is not the case
     that ((xd)y)C = (xy)C.  Hence, we can have xC=(xd)C, but have
     it be the case that xC * yC is not equal (xd)C * yC.  Hence,
     we do not have closure.

     Strictly speaking, this same problem afflicts our "proof" for
     the inverse, but I deferred discussing the problem until I got
     to closure.  If the problem is repaired for closure, it is also
     repaired for inverses (see the next paragraph for a discussion
     of normal subgroups).

Cosets of a subgroup H are said to be normal if xH = Hx for all x.
I was implicitly and incorrectly assuming that C is a normal
subgroup of G, but it is not.  For normal subgroups, closure of
coset multiplication is readily proven.  Frey and Singmaster's proof
is for normal subgroups only, and I was applying it to C, which is
not normal.

It is instructive to consider briefly what xC vs. Cx means for cubes.
We can interpret the left coset xC as simply holding a cube in your
hands and rotating it any way you wish in space without performing
any twists.

The right coset Cx is a little trickier.  The cube x
must be pre-multiplied by each c in C to form Cx.  If you have cube
x in your hands, there is no obvious thing you can do to form Cx.
The thing that is most intuitive to me personally is to think in
terms of "rotation by color", which is the way I described
pre-multiplication when I first posted some of the results of my
work back in December.  That is, think of holding the cube still, but
recoloring it by permuting the colors.  The elements of the coset Cx
look the "same" but with the colors permuted.  It is not possible to
perform this operation on a real cube (short of pulling off the stickers
and putting them back on), but the operation can be readily performed
on a computer model.

Having said all this, I keep thinking that there must be a way to define
an operation on the cosets xC so that they form a group.  However, I
have been unsuccessful in doing so.  I would welcome any advice from
the net.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From ishius@ishius.com  Thu May 26 14:32:42 1994
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Received: from DialupEudora (ishius@localhost) by holonet.net (Anton Dovydaitis) with SMTP
	id LAA17753; Thu, 26 May 1994 11:27:13 -0700
Message-Id: <199405261827.LAA17753@holonet.net>
Mime-Version: 1.0
Content-Type: text/plain; charset="us-ascii"
Date: Thu, 26 May 1994 11:28:21 -0800
To: cube-lovers@ai.mit.edu
From: ishius@ishius.com (ishius@holonet.net)
Subject: ISHI PRESS MAILING LIST!

ISHI PRESS MAILING LIST!

Earlier, I have intruded on this fine discussion to solicit e-mail addresses
for Ishi Press's retail puzzle e-mail list.  Due to some unforseen hardware
problems, however, we were cut off from the net for a couple weeks, and
have lost all of our e-mail lists.

If you would like to receive e-mailings of Ishi Press's impressive line
of mechanical puzzles, send us your e-mail address.  If you would like
a full color catalog, including puzzle reference guide, send us your
postal address as well.

PLEASE INDICATE THAT YOU ARE INTERESTED IN PUZZLES AND/OR GO (an ancient
oriental strategy game).

Retail E-mailings are sent about once per month, so we won't be stuffing
your e-mail box with more junk to slog through.  While some e-mailings
duplicate our paper sales literature, there are often descriptions, reviews,
and offers that we would never include in a general mailing (such as damaged
or second merchandise, unique items, out of print books).

Always feel free to write me if you have any questions or comments.

Anton Dovydaitis
Customer Support

===========================================================================
Ishi Press International                408/944-9900 vc, 408/944--9110 FAX
76 Bonaventura Drive                    800/859-2086 Toll Free Order Line
San Jose, CA  95134                     ishius@ishius.com (or @holonet.net)



From bagleyd@source.asset.com  Fri May 27 13:46:56 1994
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Date: Fri, 27 May 1994 13:42:15 -0400
From: bagleyd@source.asset.com (David A. Bagley)
Message-Id: <9405271742.AA64537@source.asset.com>
To: cube-lovers@ai.mit.edu
Subject: xrubik


Hi
  I just finished up "xrubik", a UNIX-X Rubik's cube.  It has been tested
on Linux, SunOS, and HP-UX.  It currently resides on ftp.x.org at
/contrib/games/puzzles.  Here's the blurb from the README in that 
directory:
-------------------------------------------------------------
xrubik has been converted from xmrubik.
  New features:
        hold down control key to move whole cube
        letters that represent colors can now be changed in mono-mode
  Bug fix:
        when xmrubik did not recognize when the cube was solved nontrivially
        (i.e. the number of cubes on an edge > 1  OOPS.)
 
The /R5contrib/xmpuzzles:
xmpyramid       xmoct           xmskewb
xmcubes         xmtriangles     xmhexagons
are currently being changed to exclude Motif dependencies.
The Motif versions will no longer be maintained.
 
The proposed collection includes:
SLIDING BLOCK PUZZLES
xcubes:         expanded 15 puzzle
xtriangles:     same complexity as 15 puzzle
xhexagons:      2 modes: one ridiculously easy, one harder than 15 puzzle
 
ROTATIONAL 3D PUZZLES
xrubik:         a nxnxn rubik's cube
xpyramid:       a nxnxn tetrahedron (a nxnxn pyraminx)
                with Period 2, Period 3, and Combined cut modes
xoct:           a nxnxn octahedron with Period 3, Period 4, and Combined cut
                modes
xskewb:         a cube with diagonal cuts
The rest of the platonic solids (the dodecahedron and the icosahedron) seem
too hard for me.
 
These programs do not have self-solvers like "magiccube" (Motif) or
"puzzle" (X).
-----------------------------
Have fun
David (the newbie)

From @wvnvm.wvnet.edu:BRYAN@WVNVM.WVNET.EDU  Mon May 30 22:48:07 1994
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Date:      Mon, 30 May 1994 21:36:07 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Branching Factors and God's Algorithm Search Trees

At various times, there have been discussions about what the maximum
distance from Start might be in God's algorithm.  One argument is made
with respect to worst/best case branching factors.  For example,
a simple calculation is that the first move has at most twelve
possibilities and that each subsequent move has eleven possibilities,
when dealing with Q-turns only.  For Q-turns plus H-turns, the same
argument would be eighteen possibilities for the first move and seventeen
possibilities for each subsequent move.

My experience is that search trees tend to develop relatively
constant branching factors after some sort of variable startup.
I expect Rubik's cube to be no different.  I just wonder if anyone
has calculated some number of levels for the full Rubik's cube,
enough levels for the hypothesized steady state branching factor
to be achieved.  I have not done so, but if anyone has, it might
shed considerable light on the question of the maximum distance
from Start.

Subsets of the cube such as corners only and edges only have been
calculated.  It is suggestive to examine branching factors for the
cases which have already been calculated.  The question of "average
branching factor" is subject to interpretation because it is not
necessarily clear when the distribution has achieved its steady
state.  I am including a number of tables giving branching factors
for the cases which have been calculated already.  I will preface
the tables with the following comments:

   1. The distributions for edges-only cubes have a variable branching
      factor during a startup phase, then have a relatively constant
      branching factor for several levels. and finally the distribution
      has sort of a tail.

   2. The distributions for corners-only cubes have a variable branching
      factor during a startup phase, and almost immediately the
      distribution has a tail.  The number of cases simply is not
      large enough to support an extended constant branching factor
      in the middle of the distribution.  It's sort of like a very
      short airplane flight where it is time to descend about the
      time the ascent is completed.

   3. I would expect the distributions for a full cube to have an
      even longer period with a constant branching factor than
      the distributions for edges-only cubes because the number
      of cases is so much larger.  There should be plenty of time
      for a plateau between the startup phase and any tail of the
      distribution.

   4. There are an equal number of odd and even permutations.  For
      the cases where you restrict yourself to Q-turns, there are
      therefore equal numbers of states an even distance from Start
      and an odd distance from Start.  Hence, the distribution tends
      either to have two adjacent levels with approximately equal
      numbers of states, or else tends to have one dominant level with
      a level on each side of the dominant level with about half
      the number of states in the dominant level.

   5. For the cases where you allow both Q-turns and H-turns, there
      tends to be one dominant level which contains most of the
      of the states.

   6. Those of you who followed all the traffic on this list in
      December and January will recall that my work with God's
      algorithm exploits symmetric conjugates in order to reduce
      the size of the problem.  It turns out that using conjugates
      does not change the average branching factor once you get
      past the startup portion of the distribution.  This effect
      can be a bit hard to see for corners-only cubes because the
      steady state portion of the distribution is so short, but
      the effect is very striking for edges-only cubes.  I would
      expect the effect to be very striking, as well, for the
      case of the full cube.

      ------------------------------------------------------------------

              2x2x2 Cube using Q-turns and H-turns

    Distance    Number of  Branching    Number of  Branching   Ratio of
        from        Cubes     Factor       M          Factor   Cubes to
       Start                           Conjugates            Conjugates

           0            1                       1                  1.00
           1            9       9.00            2       2.00       4.50
           2           54       6.00            5       2.50      10.80
           3          321       5.94           19       3.80      16.89
           4         1847       5.75           68       3.58      27.16
           5         9992       5.41          271       3.99      36.87
           6        50136       5.02         1148       4.24      43.67
           7       227536       4.54         4915       4.28      46.29
           8       870072       3.82        18364       3.74      47.38
           9      1887748       2.17        39707       2.16      47.54
          10       623800       0.33        13225       0.33      47.17
          11         2644       0.00           77       0.01      34.34

    Total/Avg     3674160     ? 4.83        77802     ? 3.54      47.22

      ------------------------------------------------------------------

              2x2x2 Cube using Q-turns

    Distance    Number of  Branching    Number of  Branching   Ratio of
        from        Cubes     Factor       M          Factor   Cubes to
       Start                           Conjugates            Conjugates

           0            1                       1                  1.00
           1            6       6.00            1       1.00       6.00
           2           27       4.50            3       3.00       9.00
           3          120       4.44            6       2.00      20.00
           4          534       4.45           17       2.83      31.41
           5         2256       4.22           59       3.47      38.24
           6         8969       3.98          217       3.68      41.33
           7        33058       3.69          738       3.40      44.79
           8       114149       3.45         2465       3.34      46.31
           9       360508       3.16         7646       3.10      47.15
          10       930588       2.58        19641       2.57      47.38
          11      1350852       1.45        28475       1.45      47.44
          12       782536       0.58        16547       0.58      47.29
          13        90280       0.12         1976       0.12      45.69
          14          276       0.00           10       0.01      27.60

    Total/Avg     3674160     ? 3.05        77802     ? 2.92      47.22

      ------------------------------------------------------------------

              Corners of 3x3x3 Cube using Q-turns and H-turns

    Distance    Number of  Branching    Number of  Branching   Ratio of
        from        Cubes     Factor       M          Factor   Cubes to
       Start                           Conjugates            Conjugates

           0            1                       1                  1.00
           1           18      18.00            2       2.00       9.00
           2          243      13.50            9       4.50      27.00
           3         2874      11.83           71       7.89      40.48
           4        28000       9.74          637       8.97      43.96
           5       205416       7.34         4449       6.98      46.17
           6      1168516       5.69        24629       5.54      47.44
           7      5402628       4.62       113049       4.59      47.79
           8     20776176       3.85       433611       3.84      47.91
           9     45391616       2.18       947208       2.18      47.92
          10     15139616       0.33       316823       0.33      47.79
          11        64736       0.00         1481       0.00      43.71

    Total/Avg    88179840     ? 4.74      1841970     ? 4.63      47.87

      ------------------------------------------------------------------

              Corners of 3x3x3 Cube using Q-turns

    Distance    Number of  Branching    Number of  Branching   Ratio of
        from        Cubes     Factor       M          Factor   Cubes to
       Start                           Conjugates            Conjugates

           0            1                       1                  1.00
           1           12      12.00            1       1.00      12.00
           2          114       9.50            5       5.00      22.80
           3          924       8.11           24       4.80      38.50
           4         6539       7.08          149       6.21      43.89
           5        39528       6.04          850       5.70      46.50
           6       199926       5.06         4257       5.01      46.96
           7       806136       4.03        16937       3.98      47.60
           8      2761740       3.43        57800       3.41      47.78
           9      8656152       3.13       180639       3.13      47.92
          10     22334112       2.58       466052       2.58      47.92
          11     32420448       1.45       676790       1.45      47.90
          12     18780864       0.58       392558       0.58      47.84
          13      2166720       0.12        45744       0.12      47.37
          14         6624       0.00          163       0.00      40.64

    Total/Avg    88179840     ? 4.48      1841970     ? 4.29      47.87

      ------------------------------------------------------------------

         Edges of 3x3x3 Cube Without Centers using Q-turns and H-Turns

    Distance    Number of  Branching    Number of  Branching   Ratio of
        from        Cubes     Factor       M          Factor   Cubes to
       Start                           Conjugates            Conjugates

           0            1                       1                  1.00
           1           18      18.00            2       2.00       9.00
           2          243      13.50            9       4.50      27.00
           3         3240      13.33           75       8.33      43.20
           4        42359      13.07          919      12.25      46.09
           5       538034      12.70        11344      12.34      47.43
           6      6666501      12.39       139325      12.28      47.85
           7     79820832      11.97      1664347      11.95      47.96
           8    888915100      11.14     18524022      11.13      47.99
           9   8056929021       9.06    167864679       9.06      48.00
          10  27958086888       3.47    582489607       3.47      48.00
          11   3883792136       0.14     80930364       0.14      47.99
          12         8827       0.00          314       0.00      28.11

    Total/Avg 40874803200    ? 12.26    851625008    ? 11.99      48.00


      ------------------------------------------------------------------

              Edges of 3x3x3 Cube Without Centers Using Q-turns

    Distance    Number of  Branching    Number of  Branching   Ratio of
        from        Cubes     Factor       M          Factor   Cubes to
       Start                           Conjugates            Conjugates

           0            1                       1                  1.00
           1           12      12.00            1       1.00      12.00
           2          114       9.50            5       5.00      22.80
           3         1068       9.37           25       5.00      42.72
           4         9759       9.14          215       8.60      45.39
           5        88144       9.03         1860       8.65      47.39
           6       786500       8.92        16481       8.86      47.72
           7      6916192       8.79       144334       8.76      47.92
           8     59623239       8.62      1242992       8.61      47.97
           9    495496593       8.31     10324847       8.31      47.99
          10   3695351994       7.46     76993295       7.46      48.00
          11  17853871137       4.83    371975385       4.83      48.00
          12  18367613703       1.03    382690120       1.03      48.00
          13    395043663       0.02      8235392       0.02      47.97
          14         1080       0.00           54       0.00      20.00
          15            1       0.00            1       0.02       1.00

    Total/Avg 40874803200     ? 8.80    851625008     ? 8.63      48.00

      ------------------------------------------------------------------

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From Joel.Franklin@altosax.reed.edu  Thu Jun  9 15:09:41 1994
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Date: 09 Jun 94 12:07:54 PDT
From: Joel.Franklin@altosax.reed.edu (Joel Franklin)
Subject: 
To: CUBE-LOVERS@life.ai.mit.edu

How do I subscribe to this list?

From rprakash@cdotp.ernet.in  Wed Jun 22 08:37:49 1994
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Date: Wed, 22 Jun 94 17:14:48+050
From: rprakash@cdotp.ernet.in (PRAKASH R.)
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To: cube-lovers@life.ai.mit.edu

Dear cube-lovers-request,
	please send me some information & problems about the cube. i haven't
been able to solve the cube fully yet. i can get about 60-70% . i need some
suggestions about how to solve the cube also.
	thanking you,
						-love
						prakash r.


From @wvnvm.wvnet.edu:BRYAN@WVNVM.WVNET.EDU  Wed Jun 29 14:11:23 1994
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Date:      Wed, 29 Jun 1994 13:45:42 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Comments on Cube Lengths (Long, 1 of 2)

As you all know, the length of a cube X is defined as the shortest
process P such that XP = I, and is denoted as |X|.  One definition
of God's Algorithm is simply that God's algorithm is the knowledge
of |X| for all cubes.  I wish to make some observations about
|X| as related to various models of the cubes and their symmetries.

The set C is the set of 24 rotations of the cube. The set M is the
set of 48 rotations and reflections of the cube, where half of M is
C and the other half of M is C reflected.

The first (obvious) observation is that |X| = |m'Xm| for all m in M.
That is, the set of all M-conjugates of a cube have the same
length.  Another way to say the same thing is that |m'Xm| = |n'Xn|
for all m and n in M.

Actually, there is an even more obvious observation that probably
should be made.  The set G is the set of all cubes, where
G is generated as G=<Q>, where Q is the set of 12 quarter-turns
of the cube.  The *really* obvious observation is that if X
is in G, then m'Xm is in G for all m in M.  Furthermore, if GC
is the set of corners-only cubes, then X in GC implies m'Xm in GC,
and if GE is the set of edges-only cubes, then X in GE implies
m'Xm in GE.  Finally, the observation that |X| = |m'Xm| remains
true whether X is in G, in GC, or in GE.

The process of forming M-conjugates in G (or in GC or in GE)
induces a partition which is an equivalence relation.  Hence, the
set {m'Xm} for all m in M is an equivalence class.  Since,
|m'Xm| = |n'Xn| for all m and n in M, it is meaningful to speak
of the length of {m'Xm}, namely |{m'Xm}| = |Y|, where Y is
any element of {m'Xm}.

Now, consider cubes of the form Xc where X is in G and c is in C.
We first observe that Xc is in G if and only if c is even.   Half the
elements in C are even, and half are odd.  An odd permutation in C is
even on the corners but is odd on the edges; hence, Xc is not in G when
c is odd.  On the other hand, Xc is in GC for all X in GC, and Xc is in
GE for all X in GE.

The fact that Xc is in G only for even elements of C is why I thought
it was important to make the "really obvious" observation that
m'Xm is in G for all X in G and all m in M.  The two cases m'Xm and
Xc are similar on the surface, but different when you dig a
little bit deeper.

With respect to lengths, we can observe that |Xc| >= |X| whenever
Xc is well-defined (that is, whenever c is even for G, or for all c
for GC and GE).

The process of performing rotations in G (even rotations in G, or any
rotation in GC or in GE) induces a partition which is an equivalence
relation.  Hence, the set {Xc} for all (or even, as appropriate) c
in C is an equivalence class.  The collection of all sets of the form
{Xc} can serve as a model for cubes without centers.

However, it is not the case that |Xc| = |Xd| for all c and d in C.
Nonetheless, it is meaningful to speak of |{Xc}|.  Namely,
|{Xc}| = min{|Xd|} for all (or even) d in C.  Hence, we have
|{Xc}| <= |Xd| for all (or even) d in C.

The definition |{Xc}| = min{|Xd|} probably requires a bit of
justification.  For a cube without centers, the solved or Start
state is {Ic} for all (or even) c in C.  Hence, Start is C (or
C[even]), and we need the shortest process P such that XP is in C
in order to calculate |{Xc}|.

Consider the set {P[1], P[2], ... P[24]} where P[n] is the shortest
process for which (Xc[n])P[n] = I.  Observe, that XP[n] is in C for
all n in 1..24.  This immediately gives us |P| <= |P[n]| for all
n in 1..24.

Conversely, if XP is in C, then there exists some c[n] in C such that
Xc[n]P = I.  This gives us |P[n]| <= |P| for some n in 1..24.
Therefore, |P| = min{|P[n]|} for n in 1..24.

Note that we have |{Xc}| <= |X| <= |Xd|.  On its face, this may seem
somewhat paradoxical, but I believe that it is entirely correct.
There is a huge difference is speaking of |{Xc}| as opposed to
speaking of |Xd|.  Xd is an (atomic) element of G;  {Xc} is a set.
Elements of {Xc} are also in G, but the *set* {Xc} is not in G.

My model for cubes without centers is really {m'Xmc} rather than
{Xc}.  However, the results from above are readily combined.  That is,
we can speak of |{m'Xmc}|, namely |{m'Xmc}| = min{|(m'Xm)d|} for all
(or even) d in C.  As before, we have |{m'Xmc}| <= |m'Xm| <= |m'Xmd|.
Note that in the middle of this last string of inequalities we could
insert any of |X| = |m'Xm| = |{m'Xm}|.

In my God's algorithm data base for cubes without centers, I store
ordered pairs of the form (Y,|{m'Xmc}|), where Y is a representative
element of the set {m'Xmc}.  Note that Y is in G (or GC or GE, as
appropriate).  It is a picky point, but the data base does *not*
consist of ordered pairs of the form (Y,|Y|).  Remember that
|Y| >= |{m'Xmc}|.

My God's algorithm data base for cubes with centers nominally consists
of ordered pairs of the form (Z,|{m'Xm}|), where Z is a representative
element of the set {m'Xm}.  Unlike the case without centers, we do
have |Z| = |{m'Xm}|, so we could also say the data base elements are
of the form (Z,|Z|).

However, the data representation is really a bit different to take
advantage of the relationship between sets of the form {m'Xmc}
and sets of the form {m'Xm}.  A set of the form {m'Xmc} can be
partitioned into (up to) twenty-four sets of the form {m'Xm},
where the (up to) twenty-four sets are indexed by C.

Let Y=Repr({m'Xmc}).  Then, the data base is ordered pairs of the form
(Yc[i],|Yc[i]|) for i in 1..24.  Note that Yc[i] is in G, but can be
said to be a representative element for sets of the form {m'(Yc[i])m},
which in turn is a set of the form {m'Xm} for some X in G.

Finally, there is no real need to store the Yc[i];  it is only
necessary to store the lengths.  Hence, a data base
element for cubes with centers is really, really of the form:

  (Y,|{m'Xmc}|,|Yc[1]|,|Yc[2]|, .... |Yc[24]|),

where Y is a representative element of {m'Xmc}.

Note that this is a very compressed format.  The representative element
Y is stored only once for the 24 different values for c.  Note also that
the solution for cubes without centers is stored in the same data base
as the solution for cubes with centers.   Finally, since
|m'Yc[i]m| = |Yc[i]|, we have stored the length of all cubes by storing
the length of only one cube for each M-conjugancy class.

It is not really necessary explicitly to store the solution for cubes
without centers to have the solution for cubes without centers in the same
data base.  That is, |{m'Xmc}|=min(|Yc[i]|) for i in 1..24.  But it takes
very little space to do so and is convenient for certain calculations.

   (to be continued)

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From @wvnvm.wvnet.edu:BRYAN@WVNVM.WVNET.EDU  Wed Jun 29 15:00:05 1994
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Date:      Wed, 29 Jun 1994 13:56:02 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Comments on Cube Lengths (Long, 2 of 2)

   (continuation)

I have described this data base structure once before (a little
less formally before), but I wanted to describe it again because
there is some interesting (to me, at least) analysis that can be
derived from the data base, over and above God's algorithm.

First, it is interesting to compare |{m'Xmc}| to the various |Yc[i]|.
Recall that |Yc[i]| >= |{m'Xmc}| for all i in 1..24.  Also, by the
definition of |{m'Xmc}|, there is at least one i in 1..24 such that
|Yc[i]| = |{m'Xmc}|.

I posted a comparison of |{m'Xmc}| to |Yc[i]| for corners-only cubes
on 4 December 1993.  (I have the "without centers" part of the edges-only
data base done, but it will take many more months to
complete the "with centers" part.  So corners-only is the only
complete data base we have to work with.)

At the time, I received a couple of comments to the effect
that people didn't understand what I was comparing to what.  I hope
this note clarifies the situation.  For example (and referring to my
previous note with respect to corners-only cubes), there is only one
element of the form {m'Xmc} for which |{m'Xmc}| = 0.  The only element
for which the length is 0 is Start, but in a corners only cube without
centers, any rotation of Start is still at Start and still has length 0.

Here is a brief excerpt from my note of 4 December 1993.

              Corresponding Distances from Start
                     Using Only q-turns

        Without           With
        Centers         Centers         Number
     Distance from    Distance from    of Nodes
         Start         Start

           0
                           0                1
                           2                1
                           4                2
                           6                1

What this means is as follows.  First, we have |{m'Imc}| = 0.  Let
Y=Repr({m'Imc}).  Then, there is 1 element of the form Yc for which
|Yc|=0, 1 element of the form Yc for which |Yc|=2, 2 elements of the
form Yc for which |Yc|=4, and 1 element of the form Yc for which
|Yc|=6.

In words, suppose you have a corners-only cube (peel off the edge
tabs, but keep the corners and centers).  Then, suppose the
corners look "solved" if you ignore the centers.  The corners will
be rotated relative to the centers.  In all, there are 24 different
ways they can be rotated, including the identity, where they
are not rotated.

But under M-conjugancy, some of the 24 rotations are equivalent.
Under M-conjugancy, there is one way they can be 0 moves from Start,
one way they can be two moves from Start (RL' is equivalent to DU',
for example ), two ways they can be four moves from Start, and one way
they can be six moves from Start.  Among other things, this says that any
rotation of the corners (ignoring the edges) can be accomplished
in no more than six quarter turns.

This example illustrates why a set of the form {m'Xmc} may be
partitioned into "up to" twenty-four elements of the form
{m'Xm}, rather than "exactly" twenty-four elements.  Normally,
a set of the form  {m'Xmc} contains 1152 elements,
where 1152=24*48.  It can in turn be partitioned into twenty-four
elements of the form {m'Xm} which contain forty-eight elements each.
But cubes which are "symmetric" reduce the number because
various M-conjugates are equivalent.

I normally think of the God's algorithm data base as a matrix, with
the rows indexed by the representative elements Y, and the columns
indexed by C (or more simply, by 1..24).  Because of M-conjugate
symmetry, there are always a few empty cells in the matrix.

M-conjugate symmetry did not cause me any computational difficulty
when I was working with cubes without centers.  That is,
suppose {m'(X1)mc} and {m'(X2)mc} are the same set for X1 not equal
X2.  My "representative element calculator" would calculate the
same representative element Y in both cases.  But in the case of
cubes with centers, the "representative element calculator" had to
calculate both a representative element Y and an associated rotation
index Cind in 1..24.

When a set {m'Xmc} had exactly 1152 elements (most of the time), the
calculation of Cind was correct.  But when a set {m'Xmc} had fewer
than 1152 elements, I would get a different Cind depending on which
element of the set I started with.  That is, the loops in the program
actually calculate 1152 elements in any case, but if the set really has
less than 1152 elements, then some of the elements are generated
multiple times.  (The loops have no way of knowing ahead of time how
many elements are going to be in the set.)  The generation of the same
set elements multiple times severely messed up the calculation of Cind
until I figured out what was going on.

I want to finish by getting back to what I started with, the lengths
of cubes.  As I said, the God's algorithm results for edges without
centers are complete (posted to the list back in December), but the
God's algorithm calculations for edges with centers are still work
in progress.  However, I noticed something striking about the
partial edges with centers results when I compared them with the
completed edges without centers results.  For example, here is a
table which compares the results when using q-turns only.


    Distance    Number of  Branching    Number of  Branching
        from   M-Conjugate   Factor    M-Conjugate   Factor
       Start     Classes                 Classes
                Without                  With
                Centers                 Centers

           0           1                     1
           1           1       1.00          1         1.00
           2           5       5.00          5         5.00
           3          25       5.00         25         5.00
           4         215       8.60        215         8.60
           5        1860       8.65       1886         8.77
           6       16481       8.86      16902         8.96
           7      144334       8.76     150442         8.90
           8     1242992       8.61    1326326         8.81
           9    10324847       8.31   11505339         8.67
          10    76993295       7.46   96755918         8.40
          11   371975385       4.83  750089528         7.75
          12   382690120       1.03      ....
          13     8235392       0.02      work
          14          54       0.00       in
          15           1       0.02    progress

    Total      851625008

As you can see, with or without centers, there are the same number
of cubes (actually, equivalence classes) at each distance from
Start from level 0 through level 4.  From level 5 on, there are more
cubes with centers than without.  Why is the number the same
through level 4, and what happens at level 5 to make the numbers
different?  Actually, overall there are about twenty-four times more
cubes with centers than without, so it is not surprising to find
more cubes with centers than without at fairly low levels in the
search tree.  So fundamentally, the question is, why does the
divergence occur at level 5?

Well, I can't explain why it is level 5 exactly, but I can explain
what is going on.  Consider level 0.  There is one row in the data
base where |{m'Xmc}|=0.  There are twenty-four cells in the same row
for |Yc[i]|, corresponding to the twenty-four rotations of the
representative element Y.  For exactly one of these cells, we have
|Yc[i]|=0.  The remainder of the cells are either undefined (meaning
the cell represents a rotation which is M-conjugate equivalent with
another rotation), or else we have |Yc[i]|>=5.  Hence, any
rotation of the edges of the cube requires at least 5 q-turns to
accomplish.  After the data base is complete, we can determine
exactly how many q-turns are required to accomplish each rotation of
the edges, just as we can already do with the corners.

Similar comments apply to level 1 through 4.  There is exactly one
rotation of the representative element that has the same length as
representative element.  All the other rotations of the representative
element are either M-conjugate equivalent to the representative
element, or else have a length greater than or equal to 5.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From jkato@tmastb.eec.toshiba.co.jp  Tue Jul  5 20:56:33 1994
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Date: Wed, 6 Jul 94 09:56:14 JST
From: jkato@tmastb.eec.toshiba.co.jp (Toshi Kato)
Return-Path: <jkato@tmastb.eec.toshiba.co.jp>
Message-Id: <9407060056.AA03691@tmastb.eec.toshiba.co.jp>
To: cube-lovers@life.ai.mit.edu
Subject: Cube-Lovers ML

I am a member of NKC(CFF) and Puzzle KONWAKAI(Academy of Recreational 
Mathematics, Japan).
I knew your Mailing List from Jerry's Puzzle Collectors Directory.
I would like to join your ML.

In this summer I am going to 14th International Puzzle collectors Party 
in Seattle.
------
                         Thank you,
                         Toshi(Junk) Kato
                         2-14-60 Hishinuma, Chigasaki 253 Japan
                         Tel/Fax: +81-467-52-1447
                         E-mail: jkato@tmastb.eec.toshiba.co.jp
   






From jkato@tmastb.eec.toshiba.co.jp  Wed Jul  6 07:00:52 1994
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Date: Wed, 6 Jul 94 20:00:14 JST
From: jkato@tmastb.eec.toshiba.co.jp (Toshi Kato)
Return-Path: <jkato@tmastb.eec.toshiba.co.jp>
Message-Id: <9407061100.AA05730@tmastb.eec.toshiba.co.jp>
To: Cube-Lovers@ai.mit.edu
Subject: SBP "Magic sQ"

                                   Sliding Block Puzzle "Magic sQ"
Fig.1 is incomplete.                      +---+---+---+
Can you complete a magic square           | 2 | 9 | 4 | 
with minimum sliding steps?               +---+---+---+
                                          | 7 | 5 | 3 |
You, very easy or not?                    +---+---+---+---+
                                          | 1 | 6 | 8 |   |  Fig.1
                                          +---+---+---+---+
------
                        Toshi(Junk) Kato from Japan
                        E-mail: jkato@tmastb.eec.toshiba.co.jp
                        Tel/Fax: +81-467-52-1447






From mouse@collatz.mcrcim.mcgill.edu  Fri Jul  8 15:24:40 1994
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Date: Fri, 8 Jul 1994 15:24:30 -0400
From: der Mouse <mouse@collatz.mcrcim.mcgill.edu>
Message-Id: <199407081924.PAA08873@Collatz.McRCIM.McGill.EDU>
To: Cube-Lovers@ai.mit.edu
Subject: Re:  SBP "Magic sQ"
Cc: jkato@tmastb.eec.toshiba.co.jp

>                                    Sliding Block Puzzle "Magic sQ"
> Fig.1 is incomplete.                      +---+---+---+
> Can you complete a magic square           | 2 | 9 | 4 |
> with minimum sliding steps?               +---+---+---+
>                                           | 7 | 5 | 3 |
> You, very easy or not?                    +---+---+---+---+
>                                           | 1 | 6 | 8 |   |  Fig.1
>                                           +---+---+---+---+

Not hard, but cutely deceptive.  The figure as supplied is a magic
square with the 1 and 6 switched.  It is not possible to switch two
adjacent tiles in a quadrilateral sliding-block puzzle of this sort
(there's an easy induction proof that only even permutations are
possible).  Thus, either it's not possible or the solution involves
some other magic square.

Since the 8 must clearly be in the lower right corner of the resulting
magic square, there are only two squares possible.  One is the magic
square that is almost present already; the other is its reflection:

        2 9 4        2 7 6
        7 5 3        9 5 1
        6 1 8        4 3 8

Fortunately, the "other" magic square is an even permutation from the
provided start position.  It's then just a straightforward tree search
to find a solution.  A simple brute-force "meet in the middle"
breadth-first search finds a solution easily.  Move the 8 aside, then
move as follows (* represents the blank space):

2 9 4    2 9 4    2 9 4    2 9 4    2 9 4    2 9 4    2 9 4    2 9 4
7 5 3 -> 7 5 3 -> 7 5 3 -> * 5 3 -> 5 * 3 -> 5 3 * -> 5 3 6 -> 5 3 6 ->
1 6 *    1 * 6    * 1 6    7 1 6    7 1 6    7 1 6    7 1 *    7 * 1
                                                    +-------+
2 9 4    2 * 4    2 4 *    2 4 6    2 4 6    2 4 6  | 2 4 6 |  2 4 6
5 * 6 -> 5 9 6 -> 5 9 6 -> 5 9 * -> 5 9 1 -> 5 9 1 -> 5 9 1 -> * 9 1 ->
7 3 1    7 3 1    7 3 1    7 3 1    7 3 *    7 * 3  | * 7 3 |  5 7 3
                                                    +-------+
2 4 6    2 * 6    * 2 6    9 2 6    9 2 6    9 2 6    9 2 6    9 2 6
9 * 1 -> 9 4 1 -> 9 4 1 -> * 4 1 -> 4 * 1 -> 4 7 1 -> 4 7 1 -> * 7 1 ->
5 7 3    5 7 3    5 7 3    5 7 3    5 7 3    5 * 3    * 5 3    4 5 3

* 2 6    2 * 6    2 7 6    2 7 6    2 7 6
9 7 1 -> 9 7 1 -> 9 * 1 -> 9 5 1 -> 9 5 1
4 5 3    4 5 3    4 5 3    4 * 3    4 3 *

Then put the 8 back, and you're done, in a total of 30 moves (28 shown,
plus the two moves of the 8).  For those who care about such things,
the boxed position is the midpoint at which the two searches met.
(This is fairly obvious - since the number of moves is even, the
configuration on which the searches met must be the middle one.)

This solution exhibits curious near-symmetries in portions of the path
taken by the blank space.  Anyone have any thoughts on this?  Perhaps I
should modify the program so it reports _all_ solutions of this length;
there may be something interesting lurking here.

					der Mouse

			    mouse@collatz.mcrcim.mcgill.edu

From jkato@tmastb.eec.toshiba.co.jp  Mon Jul 11 00:08:01 1994
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Date: Mon, 11 Jul 94 13:06:53 JST
From: jkato@tmastb.eec.toshiba.co.jp (Toshi Kato)
Return-Path: <jkato@tmastb.eec.toshiba.co.jp>
Message-Id: <9407110406.AA17550@tmastb.eec.toshiba.co.jp>
To: mouse@collatz.mcrcim.mcgill.edu
Cc: Cube-Lovers@ai.mit.edu
In-Reply-To: der Mouse's message of Fri, 8 Jul 1994 15:24:30 -0400 <199407081924.PAA08873@Collatz.McRCIM.McGill.EDU>
Subject: Re: SBP "Magic sQ"

Many thanks. 
Especially to Dan Hoey and der Mouse <mouse@Collatz.McRCIM.McGill.EDU>.

I have recieved E-mail individually from Mr.Dan Hoey too,
 
  Date: Thu, 7 Jul 94 16:43:02 EDT
  From: hoey@AIC.NRL.Navy.Mil
  To: jkato@tmastb.eec.toshiba.co.jp (Toshi Kato)
  Subject: Re: SBP "Magic sQ"

  A little hack tells me you can solve that by moving
  1,7,3,6,1,9,4,1,7,5,9,4,2,9,4,7,5,9,2,5,8.  That's 21 moves, or 30 if
  you count by the tile.  It's optimal in both metrics.

  Do you know anything about Rubik's Cube?

  Dan Hoey
  Hoey@AIC.NRL.Navy.Mil

I was surprised to recieve solutions both so quickly.
Cubes-Lover is very smart team,$@!!(JI think.

Thanks again,
Junk Kato
jkato@tmastb.eec.toshiba.co.jp



From @mail.uunet.ca:mark.longridge@canrem.com  Fri Jul 15 03:19:03 1994
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Sender: CRSO.Cube@canrem.com
Subject: DOTC 1.4 is done
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.770.5834.0C1A658F@canrem.com>
Date: Fri, 15 Jul 1994 01:32:00 -0400
Organization: CRS Online  (Toronto, Ontario)

        Domain of the Cube 1.4 is finally done!
        ---------------------------------------

   I've finally finished the new issue of the DOTC newsletter, and I'm
basically happy with it. I believe I owe my fellow cubists an apology
for taking so long, especially Greg Schmidt and Dan Hoey.

 I've enjoyed using the cube since 1981 and I wish I had more time and
energy to put into it. I was also rather ill earlier this year, and
things at work seemed to always interrupt. Nevertheless, the first
20 copies are finally ready to mail. Despite the fact these initial
copies are slightly flawed I am no longer willing to wait. This time
the issues have beige covers and are stapled like a booklet, much
the same as David Singmaster's Cubic Circular.

   I'm pleased with the printer's results, and I am mailing out the
first issues tomorrow. I have considerable work done on issue 1.5
and I expect the next issue to be ready relatively soon.

 - Mark

New Technique for Pattern Finding:

Cycle a process until you find the identity, e.g. (F1 B1 R1 D1)^24 = I
 then bisect the process if the order is even,
   ( F1 B1 R1 D1 ) ^ 12 = Pattern, naturally this process is order 2.

---------------------------------------------------------------------

Hmmmm, actually I have some questions that have been bugging me for
some time. I while back a guy was watching me use my cube program
and I explained that the reason I like studying group theory is
because it provided greater insights into the cube. He then asked
me:  "What are other uses of group theory?" and "What are the
practical uses of group theory" to which I haltingly replied (somewhat
vaguely) that it helped show relationships between geometry and
algebra. I felt this explanation unsatisfactory. I also mumbled
about symmetry and architecture. I'm sure there is a better
answer than that!

Also why is it in math that  |-11| means absolute value and can also
be the order of G, e.g. Let G be a Group, and |G| means the order
of G.

Here is another tidbit for the cube archives:

Rare 11-cycle of edges: ( L2 B1 R1 D3 L3 ) ^ 7   (35)
alternately:  F2 R3 U1 D3 B3 D1 L3 U3 D1 B1 L1 D1 B2 U2 D2 R2 B2 D1 (18)


-> Mark <-

From jkato@tmastb.eec.toshiba.co.jp  Mon Jul 18 06:11:51 1994
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Date: Mon, 18 Jul 94 19:11:24 JST
From: jkato@tmastb.eec.toshiba.co.jp (Toshi Kato)
Return-Path: <jkato@tmastb.eec.toshiba.co.jp>
Message-Id: <9407181011.AA16075@tmastb.eec.toshiba.co.jp>
To: Cube-Lovers@ai.mit.edu
Subject: A real robot solve the Rubik's Cube but...

A real robot which had artificial eyes and arms and computer brain was 
manufactured at Kawasaki Heavy Industry Co.,Ltd in Japan last year. 
He can solve the real commercial Rubik's Cube.

As he has not so intelligent, it takes about 12 minutes and 120 steps
average between starting and finishing the Cube to solve it.

Are there any other live robot like him over the world?  Do you know?

And you can help him more clever with your solving algolithm, can't you?
------
                        Toshi(Junk) Kato from Japan
                        E-mail: jkato@tmastb.eec.toshiba.co.jp
                        Tel/Fax: +81-467-52-1447









From @wvnvm.wvnet.edu:BRYAN@WVNVM.WVNET.EDU  Mon Jul 18 10:44:30 1994
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X-Acknowledge-To: <BRYAN@WVNVM.WVNET.EDU>
Date:      Mon, 18 Jul 1994 10:41:49 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   God's Algorithm, Additional Level

The following is for Q-turns only, whole cubes (both corners and
edges), and not considering M-conjugates.  I think it would be
possible to squeeze out another couple of levels by considering
M-conjugates.  The best previous result I have found in the
archives was through level 7 (reported on 7 December 1981, and again
on 3 August 1992).

            Distance              Number     Branching
              from                  of         Factor
             Start                Cubes


                 0                      1
                 1                     12     12.00
                 2                    114      9.50
                 3                  1,068      9.37
                 4                 10,011      9.37
                 5                 93,840      9.37
                 6                878,880      9.37
                 7              8,221,632      9.35
                 8             76,843,595      9.35   (new)


 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From newfield@vsl.ist.ucf.edu  Mon Jul 18 11:56:37 1994
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From: newfield@vsl.ist.ucf.edu (Dale Newfield)
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Date: Mon, 18 Jul 94 11:53:16 -0400
Message-Id: <9407181553.AA03650@flix.vsl.ist.ucf.edu>
To: Cube-Lovers@ai.mit.edu
Subject: Re: A reaal robot solve the Rubik's Cube but...


Sorry to pick nits, but if it is autonomous(sp?), which I think you implied,
wouldn't it be an android, instead of a robot?
-Dale

Dale Newfield							   I'd rather
newfield@vsl.ist.ucf.edu					   be playing
dn1l@{cs,andrew}.cmu.edu					       xlife.


From jkato@tmastb.eec.toshiba.co.jp  Mon Jul 18 22:40:44 1994
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Date: Tue, 19 Jul 94 11:40:15 JST
From: jkato@tmastb.eec.toshiba.co.jp (Toshi Kato)
Return-Path: <jkato@tmastb.eec.toshiba.co.jp>
Message-Id: <9407190240.AA17895@tmastb.eec.toshiba.co.jp>
To: Cube-Lovers@ai.mit.edu
Subject: Re: A real robot solve the Rubik's Cube but...

Dale Newfield said:
 <Sorry to pick nits, but if it is autonomous(sp?), which I think you implied,
 <wouldn't it be an android, instead of a robot?

Pardon me. I wonder if I shoudn't use such words "real" and "live". 
-----
                                  Toshi(Junk) Kato
                                  jkato@tmastb.eec.toshiba.co.jp








From diamond@jrdv04.enet.dec.com  Mon Jul 18 23:11:47 1994
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From: Norman Diamond  19-Jul-1994 1206 <diamond@jrdv04.enet.dec.com>
To: cube-lovers@ai.mit.edu
Apparently-To: cube-lovers@ai.mit.edu
Subject: Re: A real robot solve the Rubik's Cube but...
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Toshi Kato says:

>Dale Newfield said:
> <Sorry to pick nits, but if it is autonomous(sp?), which I think you implied,
> <wouldn't it be an android, instead of a robot?
>
>Pardon me. I wonder if I shoudn't use such words "real" and "live".

Half right.  It was a real dead robot :-)

-- Norman Diamond       diamond@jrdv04.enet.dec.com
[Digital did not write this.]

From jkato@tmastb.eec.toshiba.co.jp  Tue Jul 19 00:24:58 1994
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Date: Tue, 19 Jul 94 13:24:38 JST
From: jkato@tmastb.eec.toshiba.co.jp (Toshi Kato)
Return-Path: <jkato@tmastb.eec.toshiba.co.jp>
Message-Id: <9407190424.AA18202@tmastb.eec.toshiba.co.jp>
To: cube-lovers@ai.mit.edu
Subject: Re: A real robot solve the Rubik's Cube but...

Norman Diamond$@!!(Jsaid:
> >Pardon me. I wonder if I shouldn't use such words "real" and "live".
>
>Half right.  It was a real dead robot :-)

Thanx. I think the robotic machine isn't alive now too. 
-----
                                 Toshi(Junk) Kato
                                 E-mail:jkato@tmastb.eec.toshiba.co.jp




From @wvnvm.wvnet.edu:BRYAN@WVNVM.WVNET.EDU  Tue Jul 19 10:55:43 1994
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Date:      Tue, 19 Jul 1994 08:56:28 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   More on Centerless Cubes

On 13 Feb 1994, I proposed a model for centerless cubes which
I claimed met two criteria:  1) it was a group, and 2) it
maintained the symmetrical nature of the problem.  On
23 May 1994, I retracted the claim that the proposed model
was a group.

I am now of the opinion that it is impossible to satisfy
both criteria simultaneously.  I can make a very small modification
to the proposed model to make it a group, but the small
modification costs the model its cubic symmetry.

G is the full cube group, GC is the corners only cube group,
and GE is the edges only cube group.  The proposed model
for centerless cubes consisted of partitioning any of G,
GC, or GE into sets of the form {Xc} for all c in C, where
C is the set of twenty-four rotations of the cube and X is
a cube.  The sets are the elements of the proposed group.
The sets are called cosets and can also be denoted as
xC.  The partitions are denoted as G/C, GC/C, and GE/C,
respectively.

Originally, the proposed group operator was {Xc} * {Yc} = {XYc}.
This operator fails to maintain closure, and hence fails to
define a group.

In order to illustrate the slight modification which will define
a group, we will start by restricting ourselves to GC.  An
operator which works to define GC/C as a group is
{Xc} * {Yc} = {VWc}, where V is the unique element of {Xc} such
that the urf cubie is properly positioned in the urf cubicle,
and W is the unique element of {Yc} such that the urf cubie is
properly positioned in the urf cubicle.

Any other corner could have been used instead of urf, but once
you choose a corner the problem loses its symmetric nature.
Well, I guess it still has symmetry, but it is not the uniform
symmetry of the cube any more, because there is a preferred
orientation.

I have found only limited discussion in the archives, but
previous investigators have modeled a corners only, centerless
cube by leaving one corner fixed.  Such a model is clearly
a group.  For example, if we leave the urf corner fixed,
we can generate the group JC as JC=<L,L',D,D',B,B'>, where
we omit all twists of the U, R, and F faces from the set
of generators.

It is easy to find an isomorphism between GC/C and JC.  I would
express it as something like {Xc} = {Wc} <--> W, where W is
defined as before.  W is an element of JC, and as well is an
element of {Xc} = {Wc}.  {Xc} = {Wc} is an element of GC/C.
But W is a particular element of {Xc} = {Wc}, whereas X is
an arbitrary element.  Also, X is in GC, but X is not in JC
unless X = W.  The mapping {Wc} <--> W is clearly one-to-one and
onto in both directions.

For the edges GE, we need to keep one edge cubie fixed, so the
centerless cube could be generated by something like
JE=<D,D',L,L',R,R',B,B'>, where we keep the uf cubie fixed by
omitting all twists of the U and F faces from the set of
generators.  The isomorphism between GE/C and JE is expressed
as {Xc} = {Wc} <--> W, where X is an arbitrary element of
GE, and W is the unique element of {Xc} such that the uf
cube is properly placed in the uf cubicle.  As before, any
edge cube would do as well, but once chosen, it is no longer
arbitrary.

For the whole cube G, at first blush it appears we could model
centerless cubes either by keeping a corner cubie fixed, or by
keeping an edge cubie fixed.  But if we keep a corner cubie
fixed, the three immediately adjacent edge cubies are never moved
by any Q-turns.  We could solve the difficulty by admitting
slice turns.  But slice quarter-turns are odd on edges and
even on corners, so we have to restrict ourselves to slice
half-turns.  I find this ugly, plus I would prefer to generate
G with Q-turns only.  Hence, I would prefer to model a
centerless full cube as J=<D,D',L,L',R,R',B,B'>, where it
is an edge cubie which is held fixed rather than a corner cubie.

I said at the beginning that I thought it was impossible for a
model of centerless cubes both to be a group and also to
maintain cubic symmetry.  The reason is as follows:  it seems
to me that for any model which is a group, it should be
possible to find an isomorphism between the model and
J (or JC or JE, as appropriate).  But J and JC and JE
do not have cubic symmetry because there is a preferred
orientation.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From @wvnvm.wvnet.edu:BRYAN@WVNVM.WVNET.EDU  Tue Jul 19 19:09:07 1994
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Date:      Tue, 19 Jul 1994 11:43:11 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   God's Algorithm, Minor Progress, Q+H

Surprisingly, there seems not to be anything in the archives for
God's Algorithm for Q+H moves for the whole cube past level 3.
Here are some updated results:

                Distance           Number     Branching
                 from                 of       Factor
                Start               Cubes

                   0                    1
                   1                   18     18.000
                   2                  243     13.500
                   3                3,240     13.333
                   4               43,239     13.345  (new)
                   5              574,908     13.296  (new)
                   6            7,618,438     13.252  (new)
                   7          100,803,036     13.231  (new)

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From jkato@tmastb.eec.toshiba.co.jp  Sun Aug  7 22:18:28 1994
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From: jkato@tmastb.eec.toshiba.co.jp (Toshi Kato)
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To: Cube-Lovers@ai.mit.edu
Subject: HIKIMI Wooden Puzzle Competition

	To promote puzzles throughout the world
	and to convey the warmth of wood to as many people as possible
 THE 6th HIKIMI WOODEN PUZZLE COMPETITION
	will be held in October 1994.
	We invite puzzlers from around the world to enter.

 APPLICATION GUIDE
 1.Conditions
  1)Puzzle must be made of wood
  2)Puzzle must be original
  3)Puzzle has never been sold commercially
  4)Puzzle can be easily mass produced
  5)Do not submit puzzles that emphasize artistic design
  6)Puzzle may have two or more inventors
  7)Entry with more than one puzzle permitted
 
 2.Points for consideration
  1)Entry into the competition is free of charge, but each contestant must bear
    the expense for the sending the puzzle to Hikimi.
  2)We may request to purchase the puzzle within the constraints of our budget.
  3)Copyright of the puzzles entered belongs to the inventor, but  
    Hikimi Chamber of Commerce reserves the right to first negotiation.
  4)Deadline: Puzzle and application must be received by October 14,1994.

 3.For an application form, write to:

   Hikimi-cho Shokokai,Hikimi-cho,Mino-gun,Shimane Prefecture 698-12 Japan
 
    Applications in Japanese are prefered. However, since this may be a 
   difficult requirement for non-Japanese entrants, you may send your 
   application in English.

 4.Judging
   Judging will be held sometime during October,1994. All applicants will be
  notified directly of the results. The commendation ceremony will be held on
  November 12,1994 in Hikimi Town.

 5.Judges
   Chief Judge: Saburo Oguro
   Judge      : Nob Yoshigahara
   Judge      : Shigeo Takagi
   Judge      : A. Yamashita
   Judge      : T. Ohhata

 6.Prizes
   Grand Prize (one person): \500,000(about 5,000 US$)
   2nd Prize  (two persons): \300,000(about 3,000 US$)each
   3rd Prize(three persons): \100,000(about 1,000 US$)each
   Runner-ups  (several)   : \ 50,000(about   500 US$)each

 Sponsored by: Hikimicho Shokokai(Hikimi Chamber of Commerce)
                Tel:+81-856-56-0310  Fax:+81-856-56-0753

APPLICATION FORM FOR THE 6TH HIKIMI WOODEN PUZZLE COMPETITION
                                                            Date:
+--------------------------------------+--------------------------------------+
|Applicant's name:              Age(  )|Co-inventors of puzzle                |
|                                      |       Name         Age   Occupation  |
+--------------------------------------+-------------------+---+--------------+
|Address:                              |                   |   |              |
|                                      |                   |   |              |
|Tel:                Fax:              |                   |   |              |
|Occupation:                           |                   |   |              |
+--------------------------------------+-------------------+---+--------------+
|Name of company where you're employed:|Rating by Judges                     *|
|                                      +--------------------------------------+
|Address:                              |Application number                   *|
|                                      +--------------------------------------+
|Tel:                Fax:              |Remarks                              *|
+--------------------------------------+                                      |
|Name of puzzle:                       |                                      |
|                                      |                                      |
|Number of pieces in the puzzle:       |                                      |
+--------------------------------------+--------------------------------------+
|Object of puzzle:                     |Solution:                             |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
+--------------------------------------+                                      |
|Applicant's comments on the puzzle:   |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
+--------------------------------------+--------------------------------------+
[Note] 1)Please type or write in English block letters. For multiple entries, 
make copies of this form and submit a separate one for each puzzle entered.
2)If there are any co-inventors of the puzzle entered, be sure to write each
of their names. 3)Do not write in spaces marked(*).
 
   















From ybanezs%geds@mhsgate.salem.ge.com  Tue Aug  9 11:18:04 1994
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Date: Tue, 9 Aug 94 09:42:55 EST
From: Ybanez Sheldon <ybanezs%geds@mhsgate.salem.ge.com>
To: cube-lovers@ai.mit.edu
Subject: Cube Availability?
Return-Receipt-To: <ybanezs%geds@mhsgate.salem.ge.com>
X-Mailer: XGATE 2.12 MHS/SMTP Gateway


Fellow Cubers:

Does anyone know if the original 3x3 Rubik's Cube is still available?  My 
cube finally fell apart without any hope of repairing....  I still have 
my 'Revenge (which is also not too far from its last days) but I would 
still like to be able to play with the original... and keep my fingers 
nimble....  If the original cube is still available could you please post 
an address or something so that I can try to acquire one....

thanks,
                                                         ,,,
______________________________________________________  (o o)  _________
+----------------------------------------------------ooO-(_)-Ooo-------+
| Sheldon Ybanez  [ybanez-s@salem.ge.com]  GE Drive Systems  Salem, VA |
|     Always "Remember. No matter where you go, there you are." 88     |
+======================================================================+



From @mail.uunet.ca:mark.longridge@canrem.com  Tue Aug  9 15:17:23 1994
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Reply-To: CRSO.Cube@canrem.com
Sender: CRSO.Cube@canrem.com
Subject: < U, R> Group
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.783.5834.0C1A9A39@canrem.com>
Date: Tue, 9 Aug 1994 01:48:00 -0400
Organization: CRS Online  (Toronto, Ontario)

Well I decided to pull a "Jerry Byran" and take another look at
 some cube results, plus take a look at some new groups.


        Analysis of the 3x3x3 squares group
        -----------------------------------

                  (h only)      branching
Moves Deep       arrangements    factor      loc max (h only)

  0                    1           --             0
  1                    6           6              0
  2                   27           4.5            0
  3                  120           4.444          0
  4                  519           4.325          0
  5                1,932           3.722          0
  6                6,484           3.356          1  (6 X pattern)
  7               20,310           3.132          0
  8               55,034           2.709         65
  9              113,892           2.069      1,482
 10              178,495           1.567      7,379
 11              179,196           1.004     25,980
 12               89,728           0.501     50,320
 13               16,176           0.180     11,328
 14                1,488           0.092        912
 15                  144           0.096        144
                 -------                     ------
                 663,552                     97,611

        Analysis of the 3x3x3 <U, R> group
        ----------------------------------

                                          branching
Moves Deep       arrangements (q only)     factor

  0                    1                      --
  1                    4                       4
  2                   10                       2.5
  3                   24                       2.4
  4                   58                       2.416
  5                  140                       2.413
  6                  338                       2.414
  7                  816                       2.414
  8                1,970                       2.414
 program starts to really bog down after this...

  I leave it to Jerry or Dan to check my results. I checked up to 2
moves deep by hand and verified 10 different positions. What I don't
understand is how Jerry manages to look at so many cube positions:

On full 3x3x3 cube,   7     100,803,036     13.231  (new)

  Using 10 bytes to store a single cube position would still
need over 1 billion bytes, or am I missing something?

  I also used GAP (quite a good program) to calculate the size of
  < U1, R1 > on the magic dodecahedron: 7,999,675,084,800.

  Once again, I welcome any verification.

  -> Mark <-

From BRYAN@wvnvm.wvnet.edu  Wed Aug 10 08:38:50 1994
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Date:      Tue, 9 Aug 1994 23:18:06 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <Cube-Lovers@ai.mit.edu>
Subject:   Re: < U, R> Group
In-Reply-To: Message of 08/09/94 at 01:48:00 from mark.longridge@canrem.com

On 08/09/94 at 01:48:00 mark.longridge@canrem.com said:

>  I leave it to Jerry or Dan to check my results. I checked up to 2
>moves deep by hand and verified 10 different positions. What I don't
>understand is how Jerry manages to look at so many cube positions:

>On full 3x3x3 cube,   7     100,803,036     13.231  (new)

>  Using 10 bytes to store a single cube position would still
>need over 1 billion bytes, or am I missing something?

Well, when I deal with the big problems I want to solve to the
bitter end, I use the M-conjugate and centerless cube tricks
I have described at much too great length in the past.  This one
is a quick and dirty program using no conjugate tricks.  The
only real "trick" is that I externalize the data.  I decided
long ago that the problems I wanted to solve were too big to
keep in memory.  Hence, I keep everything in simple (but large)
flat files and sort and merge the files like crazy.  In this quick
and dirty program, the cube itself is 13 bytes and the level is
1 byte, for a total of 14 bytes per cube.  I guess that makes
the file size about 1.4 gigabytes (10^9).  I am
leery of using the word "billion" on E-mail forums because
E-mail is international and "billion" means 10^12 in some
countries.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From BRYAN@wvnvm.wvnet.edu  Sat Aug 13 17:17:43 1994
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From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   GC Local Maxima, Additional Information

In all my God's algorithm searches, I have never calculated local
maxima.  My search technique and data representation do not lend
themselves very well to calculating local maxima.  However, I thought
I would give it a try anyway.  I decided to start by calculating local
maxima for GC, the corners only group, since local maxima for this group
have been calculated before and I could check my answers.  I have
come up with some surprising results.

For a given cube X, there are twelve (not necessarily distinct)
neighbors of the form Xq, where q is in Q, the set of twelve
quarter turns.  Each Xq is either one move closer to Start or one
move further from Start than X.  I decided to determine for each cube
X, how many of the Xq were closer to Start and how many were further
from Start.  This is a superset of the local maxima problem.  Those
X for which I determine that all twelve of the Xq are closer to Start
are the local maxima, but I also determine for how many of the X
there are eleven neighbors Xq closer to Start, for how many of the X
there are ten neighbors Xq closer to Start, etc.  This is where the
surprising results come in.

The following table summarizes the results.  The table is a little hard
to read.  The rows (from 0 to 14) are the distances from Start.  The
columns (from 0 to 12) are the number of qturns which take you closer
to Start.  For example, row 4 column 3 contains 480.  This means that
there 480 cubes that are 4 moves from Start and for which 3 of the 12
qturns will take you closer to Start.  The table is too wide for a
computer screen, so I split it.  Columns 0 through 6 appear first, and
then columns 7 through 12 are below.

Column 12 represents the local maxima.  The "Total" column is simply
the total number of cubes at each level. The "Total" column and the
local maxima column appear several times in the archives, and my
numbers match the archives.  The first occurrence I have found is
Dan Hoey's note of 20 August 1984.

                             Number of Twists which Go Towards Start

     Level    Total  0      1        2     3        4     5        6
              Cubes

        0         1  1      0        0     0        0     0        0
        1        12  0     12        0     0        0     0        0
        2       114  0     96       18     0        0     0        0
        3       924  0    672      192    60        0     0        0
        4      6539  0   4032     1920   480       51     0       56
        5     39528  0  19104    14904  3792      984   216      384
        6    199926  0  71184    90984 16656    13212  1872     3936
        7    806136  0 123360   478008 42768   117576  7824    16656
        8   2761740  0  23328  1911312  9024   643536  2736   121872
        9   8656152  0      0  5573376     0  2327616     0   558336
       10  22334112  0      0 11167488     0  7057440     0  2818176
       11  32420448  0      0  4661568     0  8314272     0  8893248
       12  18780864  0      0    19008     0   123840     0   591744
       13   2166720  0      0        0     0        0     0        0
       14      6624  0      0        0     0        0     0        0

     Total 88179840  1 241788 23918778 72780 18598527 12648 13004408


                            7        8     9       10    11       12

        0                   0        0     0        0     0        0
        1                   0        0     0        0     0        0
        2                   0        0     0        0     0        0
        3                   0        0     0        0     0        0
        4                   0        0     0        0     0        0
        5                 144        0     0        0     0        0
        6                 528     1344     0       96     0      114
        7                1680     9096  1536     6552   480      600
        8                 384    23232    96     7584   720    17916
        9                   0   167616     0    19008     0    10200
       10                   0  1020384     0   235584     0    35040
       11                   0  6746688     0  2986560     0   818112
       12                   0  2202912     0  6189120     0  9654240
       13                   0      288     0    39168     0  2127264
       14                   0        0     0        0     0     6624

                         2736 10171560  1632  9483672  1200 12670110

The first surprising thing I noticed is the diagonals, especially
close to Start.  I am not quite sure why there should be diagonals.
I would describe the diagonals as a weak feature of the chart, but
they are surely there.  I think the diagonals mean something as follows.
Pick a cube X which is N moves from Start, and for which M qturns
will take you closer to Start.  Move to a cube Xq which is N+1 moves
from Start.  Then there is a *tendency* (not a certainty!) for there
to be M+1 moves which will take Xq closer to Start.  I can't think
of any reason for this to be so, but the chart has diagonals.

The second surprising thing is that the chart contains a preponderance
of even numbers.  There are only
two odd numbers in the whole chart.  Row 0 column 0
contains a 1, and row 4 column 4 contains a 51.  All the other cells
in the chart contain an even number.  Furthermore, by comparison to
each other, the even columns are densely populated and the odd columns
are sparsely populated.  Finally, below level 8, the
odd columns are completely empty.

It is often the case that when there are an even number of objects, there
is some natural pairing that can be performed on the objects.  In this
case, I think the pairing that can be performed to explain the even numbers
is twists of opposite faces.  R can be paired with L, R' can be paired
with L', U can be paired with D, etc.

The corners-only group GC is "almost" the same as the corners-only-
without-centers group GC/C.  (GC/C is also called a 2x2x2 cube or a
pocket cube).  In GC/C, the pairing between moves of opposite faces
is absolute.  You can always choose either of two opposite faces with
equivalent results.  In GC, the pairing is relative.  You can "almost"
solve GC the same way as GC/C, but sometimes you have to be sensitive
to which of two opposite faces you twist in order to rotate the
corners properly.  Dan's 20 August 1984 note explains this phenomenon
much better than I can:

>The alert reader will notice that rows 10 through 14 contain values
>exactly 24 times as large as those for the pocket cube.  This is not
>surprising, given that the groups are identical except for the position
>of the entire assembly in space, and each generator of the corner cube
>is identical to the inverse of the corresponding generator for the
>opposite face except for the whole-cube position.  Thus when solving a
>corner-cube position at 10 qtw or more from solved, it can be solved as
>a pocket cube, making the choice between opposite faces in such a way
>that the whole-cube position comes out right with no extra moves.

I can't fully explain why Dan's results are for rows 10 through 14,
whereas in my chart the odd columns are empty for rows 9 through 14.
Also, any rotation of the corners can be accomplished in no more than
6 qturns (see my note of 4 December 1993 concerning the corners of
the 3x3x3).  I think that the explanation is something to the effect
that for rows 10 through 14, if (for example) R will take you closer
to Start, then so too will L, and vice versa.  I don't think you have
to start choosing between (for example) R and L to accomplish the proper
rotation until you get below level 10.

Perhaps Dan can fully explain the mystery:  why is it
that a rotation of the corners only requires 6 qturns, full pairing
of opposite face turns kicks in at level 9, and GC becomes
exactly 24 times GC/C at level 10?  What is happening between level 6
and level 10?  Why don't all three phenomena kick in at the same level?

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From BRYAN@wvnvm.wvnet.edu  Fri Aug 19 16:26:59 1994
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From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Updated Upper Limits, Q-turns

On 9 January 1981, Dan Hoey provided a recursive procedure which
gives the best known upper bound on the number of
cubes at each distance from Start.  With Dan's recursive procedure,
the upper bound for any level is a function of the known value or
upper bound for the immediately preceding four levels.  Dan's procedure
takes into account identities of the form XX'=I and RL=LR.

At the time Dan performed his calculations, only level 0 through level 4
were known for sure.  We now have 8 levels, so Dan's calculations can be
updated.  I am going to give the new calculations, and I am also
going to include Dan's original calculations for comparison purposes.
In both tables, P[n] is the number of cubes which are n moves from
Start.

Dan's recursion formula is:

>   P[n]  <= 4*2*P[n-1] + 6*2*P[n-2] + 4*2*P[n-3] + 1*2*P[n-4]


Dan's calculations:

> P[0]  =           1   P[9]  <  724,477,008     P[18] <  4.048*10^17
> P[1]  =          12   P[10] <  6.792*10^9      P[19] <  3.795*10^18
> P[2]  =         114   P[11] <  6.366*10^10     P[20] <  3.557*10^19
> P[3]  =       1,068   P[12] <  5.967*10^11     P[21] <  3.334*10^20
> P[4]  =      10,011   P[13] <  5.594*10^12     P[22] <  3.125*10^21
> P[5]  <=     93,840   P[14] <  5.243*10^13     P[23] <  2.930*10^22
> P[6]  <     879,624   P[15] <  4.915*10^14     P[24] <  2.746*10^23
> P[7]  <   8,245,296   P[16] <  4.607*10^15     P[25] <  2.574*10^24
> P[8]  <  77,288,598   P[17] <  4.319*10^16


New Calculations:

  P[0]  =           1   P[9]  <= 720,627,064     P[18] <= 4.026*10^17
  P[1]  =          12   P[10] <= 6.755*10^09     P[19] <= 3.774*10^18
  P[2]  =         114   P[11] <= 6.332*10^10     P[20] <= 3.538*10^19
  P[3]  =       1,068   P[12] <= 5.935*10^11     P[21] <= 3.316*10^20
  P[4]  =      10,011   P[13] <= 5.563*10^12     P[22] <= 3.108*10^21
  P[5]  =      93,840   P[14] <= 5.215*10^13     P[23] <= 2.914*10^22
  P[6]  =     878,880   P[15] <= 4.888*10^14     P[24] <= 2.731*10^23
  P[7]  =   8,221,632   P[16] <= 4.582*10^15     P[25] <= 2.560*10^24
  P[8]  =  76,843,595   P[17] <= 4.295*10^16

I think that the two most interesting things about the new calculations
are: 1) they are nearly the same as the old calculations, and 2) they
are not exactly the same as the old calculations.  In both cases, the
question is "why?".

My interpretation is that Dan's analysis not only puts an upper
bound on the number cubes at each level, it also puts an upper bound
on the branching factor.  We trivially have an absolute upper limit
on the branching factor of 12.  After level 1, we trivially have an
upper limit on the branching factor of 11 (i.e., "don't undo the move
you just made", or "don't have a sequence of the form XX'").  As before,
moves of opposite faces commute.  Taking commutations of opposite faces
into account, the branching factor is reduced (empirically ) to an upper
limit of about 9.37.

This empirical analysis is starting with a high branching factor and
subtracting out the cubes we should not count, so that we are
dealing with identities of the form XX' and commutations of the form
RL=LR separately.  Dan's analysis deals with cubes we *should* count,
and he thereby deals with identities of the form XX' and commutations
of the form RL=LR in one fell swoop.

But Dan's analysis does not yield exact figures, only limits.  It seems
therefore that there must be other cases our empirical approach must
choose not to count.  What might those other cases be?  It seems that
there must be cases where a sequence X1 X2 ... Xn
is equal to a sequence Y1 Y2 ... Ym, but where there is no obvious
way to characterize the relationship between two sequences (e.g., they
are not simple commutations of each other), and where we cannot even
find the sequences without some sort of exhaustive search.

I would interpret that fact that the new upper limits do not equal the
old upper limits as meaning that such "duplicate" sequences do exist
close to Start.  I would interpret the fact that the new upper limits
are close to the old upper limits as meaning that there are not very
many such "duplicate" sequences close to Start.

But consider another quote from Dan in the same article:

>The recurrence on which this bound relies is due to the
>relations F^4 = FBF'B' = I (and their M-conjugates.)  It may be
>possible to improve the recurrence by recognizing other short
>relations.  Exhaustive search has shown that there are none of
>length less than 10.

I am afraid I need Dan to explain this further.  Dan's logic seems
impeccable.  But on the other hand
there must be cases where X1 X2 ... Xn = Y1 Y2 ... Ym, where
the sum of the length of the sequences is less than 10, and where
the equality is not explained by the relations F^4 = FBF'B' = I.
Otherwise, Dan's calculations would yield exact values rather than
upper limits close to Start, and the "new calculations" for upper
limits would equal the "old calculations".

Let me think out loud just for a second.  Consider relations such as
LRLRLRLR = I or RR'RR'RR'RR' = I.  These are *sequences* of length 8
but *cubes* of length 0.  Is it possible that such sequences are being
counted too many or not enough times when the recursion is four
levels deep?

Finally, I have argued on purely empirical grounds that the branching
factor will remain relatively constant from about level 3 to some
unknown level (maybe about level 18 or 19 or 20?), where the branching
factor will decay rapidly because you run out of cubes.  Well, I think
I want to argue further that during this "relatively constant" portion
of the distribution the branching factor *will* decay.  It might not
decay very much, and I don't see any easy way to calculate how much
it will decay.  The argument is very simple.  Any time a "duplicate
sequence" occurs, it reduces the branching factor at that level, but
also at subsequent levels.  That is, longer sequences can contain the
"duplicate sequence" as a sub-sequence.  Hence, any decay in the branching
factor at one level is propagated to all subsequent levels.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From mouse@collatz.mcrcim.mcgill.edu  Sun Aug 21 06:33:13 1994
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Date: Sun, 21 Aug 1994 06:33:02 -0400
From: der Mouse <mouse@collatz.mcrcim.mcgill.edu>
Message-Id: <199408211033.GAA06700@Collatz.McRCIM.McGill.EDU>
To: Cube-Lovers@ai.mit.edu
Subject: Re: Updated Upper Limits, Q-turns

> Dan's recursion formula is:

>>   P[n]  <= 4*2*P[n-1] + 6*2*P[n-2] + 4*2*P[n-3] + 1*2*P[n-4]

> Dan's calculations:

>> P[0]  =           1
>> P[1]  =          12
>> P[2]  =         114
>> P[3]  =       1,068
>> P[4]  =      10,011

Ummm.  4*2*P[4-1] + 6*2*P[4-2] + 4*2*P[4-3] + 1*2*P[4-4] =
       4*2*1068 + 6*2*114 + 4*2*12 + 1*2*1 = 10010 < P[4].

What have I missed?  Is Dan's formula not valid until n=5 or something?

					der Mouse

			    mouse@collatz.mcrcim.mcgill.edu

From BRYAN@wvnvm.wvnet.edu  Sun Aug 21 09:08:01 1994
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Date:      Sun, 21 Aug 1994 08:18:29 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "der Mouse" <mouse@collatz.mcrcim.mcgill.edu>,
        "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Re: Updated Upper Limits, Q-turns
In-Reply-To: Message of 08/21/94 at 06:33:02 from ,
           mouse@collatz.mcrcim.mcgill.edu

On 08/21/94 at 06:33:02 der Mouse said:
>> Dan's recursion formula is:

>>>   P[n]  <= 4*2*P[n-1] + 6*2*P[n-2] + 4*2*P[n-3] + 1*2*P[n-4]

>> Dan's calculations:

>>> P[0]  =           1
>>> P[1]  =          12
>>> P[2]  =         114
>>> P[3]  =       1,068
>>> P[4]  =      10,011

>Ummm.  4*2*P[4-1] + 6*2*P[4-2] + 4*2*P[4-3] + 1*2*P[4-4] =
>       4*2*1068 + 6*2*114 + 4*2*12 + 1*2*1 = 10010 < P[4].

>What have I missed?  Is Dan's formula not valid until n=5 or something?

>                                        der Mouse

I had just noticed the same thing, and intended to investigate.  I don't
know what happens to Dan's formula for n=4.

At the time Dan's chart was first published, P[n] was only known
for sure for n = 0..4.  Dan showed strict equality for these
levels, and I assume P[4] came from the known values rather than
from the formula.  It still does not explain why the formula
yields a value which is too low for P[4]  --  I could easier
understand why it yielded a value which is too high, but it seems
to me that it ought to yield the exact value that close to Start.

For P[5], Dan's original chart showed "=<".  Subsequent computer
search changed this to strict equality, which is a great victory
for Dans' formula.  The first term for which
Dan's chart is too high is P[6].  I had therefore intended to
start my investigations at that point until I discovered the
discrepancy at P[4].

Just as a reality check, let me mention some trivial points.  Suppose
it is discovered that (X1 X2 ... Xn) = (Y1 Y2 ... Ym).  Define
X = (X1 X2 ... Xn) and Y = (Y1 Y2 ... Ym).  Since X = Y, it is
immediate that XY' = Y'X = X'Y = YX' = I.  Conversely, a sequence
(X1 X2 ... Xn) = I can be decomposed into X = (X1 X2 ... Xk) and
Y = (X[k+1] ... Xn).  Then, XY = I and hence X and Y' (and also
X' and Y) are what I have called "duplicate sequences", that is
different sequences which yield the same cube.  This is why
identities are so important for bounding P[n].

I seem to do everything backwards, so I would just look for the duplicate
sequences.  However, it is probably more elegant to look for the identities.
Dan's original note said that computer
search has shown that there are not any identities other than the ones we
already know about up through length 10.  It looks to me like Dan's
formula takes care of the identities we already know about.  So
as usual, I am probably missing something obvious.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From BRYAN@wvnvm.wvnet.edu  Sun Aug 21 09:57:44 1994
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Date:      Sun, 21 Aug 1994 08:58:30 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Analysis of Turns Towards Start for Whole Cube

The following chart for level 0 through level 8 of the whole
cube using Q-turns gives the number of Q-turns for each cube which will
move towards Start.  (I recently gave the same analysis for corners only.)

Columns 9 through 12 are omitted from the chart since they contain
all zeros.  Any local maxima would appear in column 12.

This adds one new level known not to contain any local maxima.  It would
be extremely interesting to be able to extend the chart at least to
level 12 because level 12 is known to include a local maximum.

As with the corners-only case, the chart contains almost all even numbers.
However, unlike the corners-only case, the numbers do not cluster in
the even columns.  Rather, they cluster towards column 1.  This means that
(close to Start, at least) most cubes have only one "good" move that
takes you closer to Start.  It also serves to illustrate why "random"
moves so quickly scramble the cube.

                           Number of Q-turns which Move
                                Closer to Start

     Level  Total  0         1         2      3      4   5    6   7   8
            Cubes

       0        1  1         0         0      0      0   0    0   0   0
       1       12  0        12         0      0      0   0    0   0   0
       2      114  0        96        18      0      0   0    0   0   0
       3     1068  0       912       144     12      0   0    0   0   0
       4    10011  0      8544      1368     96      3   0    0   0   0
       5    93840  0     80088     12816    912     24   0    0   0   0
       6   878880  0    749376    120612   8640    252   0    0   0   0
       7  8221632  0   7001712   1135104  82152   2664   0    0   0   0
       8 76843595  0  65391504  10645824 777936  28200  48   56   0  27

  Total  86049153  1  73232244  11915886 869748  31143  48   56   0  27

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From BRYAN@wvnvm.wvnet.edu  Wed Aug 24 23:07:21 1994
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Date:      Wed, 24 Aug 1994 23:06:52 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "der Mouse" <mouse@collatz.mcrcim.mcgill.edu>,
        "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Re: Updated Upper Limits, Q-turns
In-Reply-To: Message of 08/21/94 at 06:33:02 from ,
           mouse@collatz.mcrcim.mcgill.edu

On 08/21/94 at 06:33:02 der Mouse said:
>> Dan's recursion formula is:

>>>   P[n]  <= 4*2*P[n-1] + 6*2*P[n-2] + 4*2*P[n-3] + 1*2*P[n-4]

>> Dan's calculations:

>>> P[0]  =           1
>>> P[1]  =          12
>>> P[2]  =         114
>>> P[3]  =       1,068
>>> P[4]  =      10,011

>Ummm.  4*2*P[4-1] + 6*2*P[4-2] + 4*2*P[4-3] + 1*2*P[4-4] =
>       4*2*1068 + 6*2*114 + 4*2*12 + 1*2*1 = 10010 < P[4].

>What have I missed?  Is Dan's formula not valid until n=5 or something?

I just rechecked Dan's original note of 9 January 1981.  He specifically
says the formula is good for n > 4.  Mea culpa.  However, I still do
not fully understand *why* this should be the case.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From hoey@aic.nrl.navy.mil  Thu Aug 25 14:43:40 1994
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Date: Thu, 25 Aug 94 14:43:13 EDT
From: hoey@aic.nrl.navy.mil (Dan Hoey)
Message-Id: <9408251843.AA03267@Sun0.AIC.NRL.Navy.Mil>
To: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>, <mouse@collatz.mcrcim.mcgill.edu>
Cc: <Cube-Lovers@ai.mit.edu>
Subject: Re: Updated Upper Limits, Q-turns

Jerry Bryan was looking at some formulas I had in the archives:

    >> Dan's recursion formula is:
    >>>   P[n]  <= 4*2*P[n-1] + 6*2*P[n-2] + 4*2*P[n-3] + 1*2*P[n-4]

    ...

    I just rechecked Dan's original note of 9 January 1981.  He
    specifically says the formula is good for n > 4.  Mea culpa.
    However, I still do not fully understand *why* this should be the
    case.

I thought I put the bound in....  I've been meaning to look that up
and explain it, but time's been short.

The analysis is done by breaking up the minimal processes into
"syllables", where a syllable is a maximal sequence of commuting turns.
So for each pair (x,y) in {(F,B),(T,D),(L,R)} there are
    four syllables of length 1: x, x', y, and y';
    six syllables of length 2: xx, xy, xy', x'y, x'y', and yy;
    four syllables of length 3: xxy, xxy', xyy, and x'yy;
    one syllable of length 4: xxyy.

(It's not really a coincidence that this is most of the fifth line of
Pascal's triangle.)

Now for the first syllable, we can pick any of the three pairs for
(x,y).  But for succeeding syllables, we must pick a pair that is not
equal to the preceding pair.  So each term in the recurrence refers to
the length of the last syllable:

Length of last syllable=1            2            3            4

  n=1    P[n] = 4*3 P[n-1];
  n=2    P[n] = 4*2 P[n-1] + 6*3 P[n-2]
  n=3    P[n] = 4*2 P[n-1] + 6*2 P[n-2] + 4*3 P[n-3]
  n=4    P[n] = 4*2 P[n-1] + 6*2 P[n-2] + 4*2 P[n-3] + 1*3 P[n-4]
  n>4    P[n] = 4*2 P[n-1] + 6*2 P[n-2] + 4*2 P[n-3] + 1*2 P[n-4]

The second part of each coefficient is 2, except that when the length
of the last syllable is equal to n (so that we are counting the first
syllable), the second part of the coefficient is 3.

In response to my description:
> >The recurrence on which this bound relies is due to the
> >relations F^4 = FBF'B' = I (and their M-conjugates.)  It may be
> >possible to improve the recurrence by recognizing other short
> >relations.  Exhaustive search has shown that there are none of
> >length less than 10.

Jerry continues:
> ... there must be cases where X1 X2 ... Xn = Y1 Y2 ... Ym, where
> the sum of the length of the sequences is less than 10, and where
> the equality is not explained by the relations F^4 = FBF'B' = I.
> Otherwise, Dan's calculations would yield exact values rather than
> upper limits close to Start, and the "new calculations" for upper
> limits would equal the "old calculations".

No.  The bounds fail to be exact when we have a relation r=s with
|r|=|s|=n.  This corresponds to a relation r s'=I of length 2n.  The
shortest relations of length >4 are the ones of length 12 (as I
reported on 22 March 1981) so my bounds become inexact at length 6.

Chris Worrell listed the length-12 relations on 08/02/81, and I
reported that his list was complete on 14 August 1981 0111-EDT.  The
12-qtw identities (up to M-conjugacy) are:

I12-1	FR' F'R    UF' U'F    RU' R'U
I12-2	FR' F'R   UF'   F'L FL'   U'F
I12-3	FR' F'R   UF'   UL' U'L   FU'

As Allan C. Wechsler noted on 17 August 1981, any two of them can be
combined to form the third.

> Consider relations such as LRLRLRLR = I or RR'RR'RR'RR' = I.

The first is a consequence of the relations L^4=R^4=LRL'R'=I.  The
second is a consequence of group theory; no relations are needed.  The
recurrence deals with these: it models the freeest group specified by
the given relations.

I have tried unsuccessfully to create a recurrence that will deal with
the 12-qtw identities, but it's complicated.  For instance, repeatedly
putting I12-1 in the center of another I12-1 yields identities of the
form:

     F (R'F' RU)^n F'U' (FR U'R')^n U

There are a bunch of other cases, too.

Dan
Hoey@AIC.NRL.Navy.Mil

From BRYAN@wvnvm.wvnet.edu  Wed Aug 31 17:17:08 1994
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From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@ai.mit.edu>
Subject:   Re: < U, R> Group
In-Reply-To: Message of 08/09/94 at 01:48:00 from mark.longridge@canrem.com

On 08/09/94 at 01:48:00 mark.longridge@canrem.com said:

>        Analysis of the 3x3x3 <U, R> group
>        ----------------------------------

>                                          branching
>Moves Deep       arrangements (q only)     factor

>  0                    1                      --
>  1                    4                       4
>  2                   10                       2.5
>  3                   24                       2.4
>  4                   58                       2.416
>  5                  140                       2.413
>  6                  338                       2.414
>  7                  816                       2.414
>  8                1,970                       2.414
> program starts to really bog down after this...

>  I leave it to Jerry or Dan to check my results. I checked up to 2
>moves deep by hand and verified 10 different positions.

Ok, here it is.  This search is narrower and deeper than any
I have ever done before.  Frey and Singmaster give <U,R>
a good bit of attention in their book, pointing out that it
is trickier than it might first appear.

It is called the Two-Generator Group.  The size of the group
can be calculated as (7!5!/2)(3^6/3) = 73,483,200.  The
3^6 factor accounts for twisting the corners, but there is no
2^n factor as the edges cannot be flipped.

These results are in terms of Q turns without any conjugate
class checking.  I would regard the following as open problems:
local maxima, results with Q+H turns, and results in terms of
conjugate classes.  In this particular case, it would not be
M-conjugates.  I would have to look at Dan Hoey's 98 subgroups
of M to see which subgroup applies to <U,R>.

  0                    1
  1                    4                       4
  2                   10                       2.5
  3                   24                       2.4
  4                   58                       2.416
  5                  140                       2.413
  6                  338                       2.414
  7                  816                       2.414
  8                1,970                       2.414
  9                4,756                       2.414
 10               11,448                       2.407
 11               27,448                       2.398
 12               65,260                       2.378
 13              154,192                       2.363
 14              360,692                       2.339
 15              827,540                       2.294
 16            1,851,345                       2.237
 17            3,968,840                       2.144
 18            7,891,990                       1.988
 19           13,659,821                       1.755
 20           18,471,682                       1.352
 21           16,586,822                       0.898
 22            8,039,455                       0.485
 23            1,511,110                       0.188
 24               47,351                       0.031
 25                   87                       0.002

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From @mail.uunet.ca:mark.longridge@canrem.com  Fri Sep  2 00:08:42 1994
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Subject:  < U, R > revisited
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.796.5834.0C1AD333@canrem.com>
Date: Thu, 1 Sep 1994 22:56:00 -0400
Organization: CRS Online  (Toronto, Ontario)

        Analysis of the 3x3x3 <U, R> group (continued)
        ----------------------------------

                                          branching
Moves Deep       arrangements (q only)     factor

  0                    1        1             --
  1                    4        5              4
  2                   10       15              2.5
  3                   24       39              2.4
  4                   58       97              2.416
  5                  140      237              2.413
  6                  338      575              2.414
  7                  816    1,391              2.414
  8                1,970    3,361              2.414
  9                4,756    8,117              2.414
 10               11,448   19,565              2.407
 11               27,448   47,013              2.401
ML's Conjecture: The < U, R > group is >=20 turns deep in qt metric

UR Reflective processes: (in the q metric)
A different sort of symmetry which I started to investigate, having
 been inspired by my friend who solves his cube 2 adjacent faces
 last!

These are the only UR reflective processes at 10 q turns:
     U3 R1 U1 R1 (U2) R3 U3 R3 U1 = R3 U1 R1 U1 (R2) U3 R3 U3 R1   (10)
     U1 R3 U3 R3 (U2) R1 U1 R1 U3 = R1 U3 R3 U3 (R2) U1 R1 U1 R3   (10)

Here is the obvious one we all know:
                    ( U2 R2 ) ^ 3 = ( R2 U2 ) ^ 3                  (12)

I liked this pattern in particular...
       U1 R1 U2 R3 U2 R3 U2 R1 U1 = R1 U1 R2 U3 R2 U3 R2 U1 R1     (12)

I hope to have an algorithm to plumb the depths of the < U, R > group
 soon. Amusingly my friend complained about not been able solve
 the cube completely as he was stuck in a position with 2 flipped
 edges. After watching him squirm for a few weeks I did tell him
 you can't flip edges in the < U, R > group!     ;->

Congrats to Dan Hoey, Dik Winter, Jerry Bryan and Ludwig Plutonium
 for making it into the 1994 Internet White Pages! I'm in good
 company.

        -> Mark <-
        Email: mark.longridge@canrem.com

P.S. I just read the last J.B. post and see I've been somewhat
     overshadowed. Ok let's see some antipodes! At least our
     results are the same though. So, ummmm I guess ML's
     conjecture is correct!

From BRYAN@wvnvm.wvnet.edu  Mon Sep  5 09:25:29 1994
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Date:      Mon, 5 Sep 1994 09:08:14 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@ai.mit.edu>
Subject:   Re: < U, R > revisited (with Local Maxima)
In-Reply-To: Message of 09/01/94 at 22:56:00 from mark.longridge@canrem.com

I performed a Local Maxima analysis.  (This is for Q-turns only.)  The Local
Maxima are in column 4.  The shortest local maxima (six of them) are of
length 12.  Interestingly, this is the same length which is suspected of
being the shortest Q-turn length for a local maximum in the full cube group.
Is there any connection?  Also, the global maxima are of length 25.  Does
this tell us anything about the Q-turn length of the global maxima for the
full cube group?  Finally, pick any cube X in <U,R>.  We know
|X| in G <= |X| in <U,R>.  Can anybody find a cube X such that
|X| in G < |X| in <U,R>?  Alternatively, can anybody prove
|X| in G = |X| in <U,R> for all X in <U,R>?  (This assumes Q-turns only in
all cases.  The questions would all have to be asked again for Q+H-turns.)

                         Number of Moves Which Go Closer to Start

     Level       Total   0           1           2           3          4
                 Cubes

         0           1   1           0           0           0          0
         1           4   0           4           0           0          0
         2          10   0           8           2           0          0
         3          24   0          20           4           0          0
         4          58   0          48          10           0          0
         5         140   0         116          24           0          0
         6         338   0         280          58           0          0
         7         816   0         676         140           0          0
         8       1,970   0       1,632         338           0          0
         9       4,756   0       3,940         816           0          0
        10      11,448   0       9,448       1,996           4          0
        11      27,448   0      22,584       4,836          28          0
        12      65,260   0      53,236      11,862         156          6
        13     154,192   0     125,196      28,616         360         20
        14     360,692   0     289,908      69,196       1,472        116
        15     827,540   0     652,792     168,008       6,180        560
        16   1,851,345   0   1,428,560     398,634      21,860      2,291
        17   3,968,840   0   2,938,808     934,908      84,312     10,812
        18   7,891,990   0   5,422,844   2,109,480     309,916     49,750
        19  13,659,821   0   8,065,268   4,288,796   1,068,480    237,277
        20  18,471,682   0   7,948,748   6,625,644   2,947,320    949,970
        21  16,586,822   0   3,485,748   5,507,066   4,831,060  2,762,948
        22   8,039,455   0     286,176   1,165,888   2,665,080  3,922,311
        23   1,511,110   0         740      15,202     156,432  1,338,736
        24      47,351   0           0           8         332     47,011
        25          87   0           0           0           0         87

            73,483,200   1  30,736,780  21,331,532  12,092,992  9,321,895

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?



From @mail.uunet.ca:mark.longridge@canrem.com  Mon Sep 12 01:05:56 1994
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Sender: CRSO.Cube@canrem.com
Subject: More UR Stuff!
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.799.5834.0C1AF26D@canrem.com>
Date: Mon, 12 Sep 1994 00:28:00 -0400
Organization: CRS Online  (Toronto, Ontario)

                      Notes on the UR Group
                      ---------------------

 Well, some small news about the < U, R > group. Previously I believed
that my 3-cycle of edges:

    UR1 = U3 R3 U2 R1 U1 R3 U1 R1 U1 R1 U2 R3 U3 R1 U3 R3 = 18 q, 16 h

...discovered by hand was minimal. My newly created < U, R > solver
(now being at the point of churning out correct results) as
happily proven me wrong!

    UR2 = U3 R1 U2 R1 U1 R1 U1 R2 U3 R3 U3 R2 U1 = 16 q, 13 h


Also I found 6 "UR-Reflective" processes altogether. This is all
 there are up to and including 12 q turns:

  U3 R1 U1 R1 (U2) R3 U3 R3 U1 = R3 U1 R1 U1 (R2) U3 R3 U3 R1   (10)
  U1 R3 U3 R3 (U2) R1 U1 R1 U3 = R1 U3 R3 U3 (R2) U1 R1 U1 R3   (10)
                 ( U2 R2 ) ^ 3 = ( R2 U2 ) ^ 3                  (12)
    U1 R1 U2 R3 U2 R3 U2 R1 U1 = R1 U1 R2 U3 R2 U3 R2 U1 R1     (12)
            ( U1 R1 U3 R3 ) ^3 = ( R1 U1 R3 U3 ) ^3             (12)
            ( U3 R3 U1 R1 ) ^3 = ( R3 U3 R1 U1 ) ^3             (12)


 The program is still a sluggish beast, but I think with further
refinements it should eventually find other interesting results
like antipodes and pure twists, etc.

             -> Mark <-

 Email: mark.longridge@canrem.com

From BRYAN@wvnvm.wvnet.edu  Mon Sep 12 17:56:28 1994
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Date:      Mon, 12 Sep 1994 15:35:32 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   God's Algorithm, Q-Moves Through Level 10

Distance   Number   Branching       Number   Branching   Ratio of
from         of      Factor           of      Factor     Cubes to
Start      Cubes                 M-Conjugate             M-Conjugate
                                    Classes               Classes

 0              1                         1
 1             12     12.000              1   1.000        12.000
 2            114      9.500              5   5.000        22.800
 3          1,068      9.368             25   5.000        42.720
 4         10,011      9.374            219   8.760        45.712
 5         93,840      9.374          1,978   9.032        47.442
 6        878,880      9.366         18,395   9.300        47.778
 7      8,221,632      9.355        171,529   9.325        47.931
 8     76,843,595      9.347      1,601,725   9.338        47.976
 9    717,789,576      9.341     14,956,266   9.338        47.993
10  6,701,836,858      9.337    139,629,194   9.336        47.997

Some of you may remember previous results where I calculated equivalence
classes of the form {m'Xmc} for all 48 elements m in M, the set of all
cube rotations and reflections, and for all 24 elements c in C, the set
of all cube rotations.  This is effectively calculating M-conjugate classes
for centerless cubes.  My previous data bases have contained
representative elements Y for each equivalent class {m'Xmc}.  To get
cubes with centers (where rotational orientation makes a difference),
I then calculated Yc for each c in C, forming a matrix indexed by
Y and c.

The previous approach permits a very compact representation of God's
algorithm, and I used it for corners-only cubes and am presently
using it for edges-only cubes.  However, I find that the {m'Xmc}
approach does not work well for whole cubes.  The problem is that
the matrix is extremely sparse close to Start.  With corners-only
or edges-only cube, I can calculate the entire problem.  With the
whole cube, I cannot even come close to calculating the whole problem,
and the matrix representation wastes space rather than saving space.

Hence, for whole cubes, I am calculating equivalence classes (which
are M-conjugate classes) of the form {m'Xm} for all 48 elements m in M.
My data base includes a representative element Z for each M-conjugate
class {m'Xm}.  This reduces the size of the problem by about 48 times,
and lets me calculate about two more levels of the search tree with the
same level of effort as before.

Just to reiterate some obvious points that have appeared before:

   1) X is an arbitrary element of {m'Xm}, but Z is a particular element
      of {m'Xm} chosen with a selection function.

   2) Z is in {m'Xm} and we have {m'Zm} = {m'Xm}.

   3) |Z| = |X| = |m'Xm| = |m'Zm| for all m in M and for all X in
      {m'Xm}.  This trivial equivalence is what makes M-conjugate
      classes a viable approach for brute force calculation of
      God's algorithm.

   4) Most M-conjugate classes of the form {m'Xm} contain 48 elements.
      The size of {m'Xm} can be used as a measure of the symmetry of
      X, with |{m'Xm}|=1 for the most symmetric cubes and |{m'Xm}|=48
      for the least symmetric cubes.  Conversely, Symm(X) is the
      set of all m in M such that m'Xm=X.  |Symm(X)|=48 for the
      most symmetric cubes, |Symm(X)|=1 for the least symmetric cubes,
      and |{m'Xm}| * |Symm(X)| = 48 in all cases.

   5) The ratio of cubes to M-conjugate classes is close to, but not
      exactly equal to, 48.  The reason the equality is inexact is
      symmetry (see item #4 above). The ratio is closer to 48 when
      you get further from Start because the proportion of asymmetric
      cubes is higher when you are further from Start.

I actually calculated (and previously reported) God's Algorithm directly
through level 8.  For levels 9 and 10, I only calculated the number of
M-equivalence classes directly.  I then calculated the size of each
M-equivalence class to obtain the number of cubes.  This particular data
base has 14 bytes for each cube (actually for each representative element
Z).  Hence, 14*139,629,194= 1,954,808,716 bytes are required to store
level 10 (each level is in a separate file).  This is about
2 gigabytes of storage, which is quite large, but which is by no means
outrageous.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From BRYAN@wvnvm.wvnet.edu  Wed Sep 21 11:32:38 1994
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Date:      Wed, 21 Sep 1994 11:31:42 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   M-Conjugate Classes as a Group

C is the set of twenty-four rotations of the cube.  After much
bungling (see my notes of 13 Feb 1994, 23 May 1994, and 19 July 1994),
I showed that the left cosets of C, denoted by xC or {Xc}, form
a group, and that the group is isomorphic to a subgroup of G.  I consider
this to be important because I use left cosets of C to model
centerless cubes.

M is the set of forty-eight rotations and reflections of the cube.
I often model the cube with M-conjugate classes of the form {m'Xm}.
Therefore, it seems that I should try to define an operation such that
the M-conjugate classes form a group, and such that the group is
isomorphic to a subgroup of G.

I would like to start by reviewing briefly the results for left cosets.
Two operations were defined:

   1.a.  {Xc} * {Yc} = {(VW)c}
   1.b.     V ** W   = (VW)

where V and W are representative elements of {Xc} and {Yc}, respectively.
Further, the mapping V <--> {Vc} defines an isomorphism between the
set of left cosets and the operation * on the one hand, and the set of
representative elements and the operation ** on the other hand.  Since
the ** operation is simply normal cube multiplication and since the
set of representative elements are a group under **, the set of
representative elements form a subgroup of G.

I tried to define groups without using representative elements and failed.
Not only that, the representative elements had to be selected in a
special way rather than arbitrarily.  For example, we could choose as
the representative element of {Xc} the unique element V such that the
ur cubie is positioned properly.

Positioning the ur cubie properly is not the only selection function for
a representative element which will work, but any selection function must
satisfy two criteria in order to work:

    A. It must select a representative element based on a property which
       is possessed by exactly one element of each coset.

    B. There must be closure in the sense that if V is the representative
       element of {Xc} and W is the representative element of {Yc},
       then (VW) must be the representative element of {(VW)c}.

Criterion #B merits some additional discussion.  First, it is the
criterion that really proves you have a group.  Associativity
for a subset of a group generally follows from the the associativity
of the group.  For a finite group, closure for a subset implies
the identity and the complement for the subset, so closure is the key
factor in demonstrating that a set of cubes is a group.  Second,
criterion #B will bear directly on our attempt to define a group
operation for the M-conjugate classes.

Suppose we choose not to require criterion #B.  We still need to
have closure in order to have a group.  We could obtain closure by
brute force as follows:

   2.a.  {Xc} * {Yc} = {(Repr{(VW)c})c}
   2.b.     V ** W   = Repr{(VW)c}

It is probably a little easier to see what is going on in equation
2.b. than in 2.a., but it is the identical mechanism in both cases.
Suppose we don't have closure.  That is, suppose the selection
function operates in such a way that if V is the representative
element of {Xc} and W is the representative element of {Yc} that
(VW) is not necessarily the representative element of {(VW)c}.
We can still find the representative element of {(VW)c} by simply
applying the selection function, which we have done.

Equations 2.a and 2.b define groups, where the left cosets are
a group under * and the representative elements are a group under
**.  Furthermore, the mapping V <--> {Vc} defines an isomorphism
between the two groups.  But even though the set of representative
elements is a subset of G, and even though they form a group under **,
they are not a subgroup of G.  The problem is that the operation
** as defined by equation 2.b. is not the same operation as
standard cube multiplication as it was in equation 1.b.

Now, let's look at M-conjugate classes.  By analogy with the left
coset case, there are two possibilities to define a group:

   3.a.  {m'Xm} * {m'Ym} = {m'(VW)m}
   3.b.       V ** W     = (VW)

   4.a.  {m'Xm} * {m'Ym} = {m'(Repr{m'(VW)m})m}
   4.b.       V ** W     = Repr{m'(VW)m}

As before, X and Y are any cubes in G, and V and W are the
representative elements of {m'Xm} and {m'Ym}, respectively.

In order to make 3.a. and 3.b. work, we need some characteristic
which can be used by the selection function which possesses the
properties of uniqueness and closure as defined by #A and #B
above.  But I can't think of any such property, and I don't think
such a property exists (see below).

4.a and 4.b certainly work.  That is, they define operations
* and ** under which the set of M-conjugate classes and the set
of representative elements, respectively, form groups, and the
groups are isomorphic under the mapping V <--> {m'Vm}.
However, the groups fail to be subgroups of G for the same reason
elements of left cosets fail to be subgroups of G under equation 2.b.
Namely, the ** operation is not really the same operation as
normal cube multiplication.

As to the question of whether 3.a. and 3.b. can be made to work,
I think we can prove that they cannot.  Suppose the contrary.
That is, suppose that there is some property such that it is
possessed by exactly one element of each M conjugancy class and
such that the normal cube product of two such elements also
possesses the property.  Then, it would be the case that the
set of representative elements would be a subgroup of G.  But the
number of representative elements is the same as the number of
M conjugate classes, and the number of M conjugate classes is
known not to divide the number of cubes in G evenly.  Hence,
the set of representative elements of M-conjugate classes is not
a subgroup of G.  Working backwards contrapositively, the desired
property cannot exist.

So, the final result is that the set of M conjugate classes can be
made into a group, and the set of representative elements of the
M conjugate classes can be made into a group.  But neither group
is a subgroup of G, nor is either group isomorphic to any
subgroup of G.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From BRYAN@wvnvm.wvnet.edu  Wed Sep 21 18:35:37 1994
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Date:      Wed, 21 Sep 1994 16:38:16 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@ai.mit.edu>
Subject:   Re: < U, R> Group
In-Reply-To: Message of 08/09/94 at 01:48:00 from mark.longridge@canrem.com

I wanted to get back to some of the symmetry considerations associated
with the Two-Generator Group.  In particular, I wanted to calculate
God's Algorithm in terms of something analogous to M-conjugates.
I haven't started the calculations, but I wanted to go ahead and
discuss what those calculations will consist of.

First, we can make the obvious observation that the <U,R> group is not
the only Two-Generator group.  Any two adjacent faces can serve
as generators for a Two-Generator Group, and there are twelve such
pairs of adjacent faces.  All twelve groups have identical structures,
and are isomorphic under M-conjugation.

With respect to any particular Two-Generator Group such as <U,R>, the
associated symmetry group is not M, it is a subgroup of M.  Dan Hoey
has determined that M has 98 subgroups, and that the 98 subgroups
may be grouped into 33 classes.  Dan has developed a taxonomy for the
33 classes and 98 subgroups.  I have seen bits and pieces of Dan's
taxonomy posted to the list, and he has sent me a good bit of it
via private E-mail, but I don't think I have ever seen the whole thing
all in one piece.  In any case, let's talk a little bit about the
98 subgroups and 33 classes.

Frey and Singmaster use script characters to describe whole cube
rotations.  For the purposes of this note, I will use lower
case letters.  For example, r will be used to describe grabbing
the right face and turning the whole cube 90 degrees clockwise
(to be distinguished from R, which means to turn only the right
face 90 degrees clockwise).  In this notation, <r> = {i,r,rr,rrr}
is one of the 98 subgroups of M.  (Note that <l> is the same
group as <r>.)  Similarly, <u> = {i,u,uu,uuu} and
<f> = {i,f,ff,fff} are subgroups of M.  The groups <r>, <u>, and
<f> have identical structures (isomorphic under rotation), and the
collection (<r>, <u>, <f>) is one of Dan's 33 classes.

Similarly, (<r2>, <u2>, <f2>) is another of Dan's 33 classes.  <r2>
is a subgroup of <r>, <u2> is a subgroup of <u>, and <f2> is a
subgroup of <f>.  Dan's taxonomy includes a complete description
of group-subgroup relationships within the subgroups of M, or maybe
I should say that it is a complete description of class-subclass
relationships.

The Frey-Singmaster script notation is adequate for rotations, but it
is not adequate for reflections.  Instead, I will use a notation which
I believe originated with Dan Hoey (e.g., 28 Dec 1993).  For example,
you could write r=(FUBD), where the upper case letters describe movements
of whole faces (*not* quarter-turns in this context!).  (FUBD) means Front
goes to Up goes to Back goes to Down goes to Front, which is what happens
when you perform r.  In the same notation, a Front-Back reflection
would be (FB), etc.

With that all said, the symmetry group for <U,R> is

  <(UR)(DL),(FB)> = {I, (FB), (UR)(DL), (FB)(UR)(DL)}

In Dan's taxonomy, this group is a member of the W class, and there
are six such groups  -- W1 through W6.  I am not sure which one this one
is (I only have a list of the classes, not of the groups), but let's
say for the sake of the argument it is W3.  Then for <U,R>, I will
be calculating W3-conjugate classes of the form {w'Xw} for all w in
W3.  The size of the problem will be reduced by about four times,
compared to a reduction of about forty-eight times for whole cube
problems where M-conjugation can be used.

I was initially surprised that there are twelve groups M-conjugate with
<U,R>, but only six corresponding symmetry groups in M.  This arises,
for example, because the <U,R> and <D,L> (diagonally opposed) groups share
the same symmetry group.  I really shouldn't have been surprised.
After all, we know how many subgroups of G there are, namely
"over three beelion" (Dan Hoey, 20 Aug 1992), and we know how
many subgroups of M there are, namely 98.  Since "over three beelion"
is a lot more than 98, there must be many, many subgroups of G
which share the same symmetry properties in the sense of sharing
a subgroup of M.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From BRYAN@wvnvm.wvnet.edu  Tue Sep 27 01:22:13 1994
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Date:      Mon, 26 Sep 1994 14:29:31 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@ai.mit.edu>
Subject:   Re: < U, R> Group (W-conjugate results)
In-Reply-To: Message of 08/09/94 at 01:48:00 from mark.longridge@canrem.com

I have completed the God's Algorithm calculations for the <U,R> group
in terms of W-conjugate classes (or really, in terms of representative
elements of W-conjugate classes), with the results below.  In general,
the use of W-conjugates reduces the size of the problem by about
four times.  However, I was surprised to see that for levels 1, 3, 5,
7, 9, and 11 the number of cubes was exactly four times larger than the
number of W-conjugate classes.  My interpretation is that all cubes
in <U,R> at these levels are completely "asymmetric" with respect to
W.  (They are somewhat symmetric with respect to M, of course.)  However,
when level 13 turned out not to have a ratio of exactly 4 between cubes
and W-conjugate classes, I was rescued from the task of explaining why
all cubes at an odd distance from Start were asymmetric.


Level   W-Conjugate  Branching      Total   Branching   Ratio
           Classes     Factor       Cubes     Factor      of
                                                        Cubes to
                                                        Classes

  0             1                        1              1
  1             1      1                 4     4        4
  2             3      3                10     2.5      3.333
  3             6      2                24     2.4      4
  4            15      2.5              58     2.416    3.866
  5            35      2.333           140     2.413    4
  6            85      2.429           338     2.414    3.976
  7           204      2.4             816     2.414    4
  8           493      2.417         1,970     2.414    3.996
  9         1,189      2.412         4,756     2.414    4
 10         2,863      2.408        11,448     2.407    3.999
 11         6,862      2.397        27,448     2.398    4
 12        16,324      2.379        65,260     2.378    3.998
 13        38,550      2.362       154,192     2.363    3.9997
 14        90,192      2.340       360,692     2.339    3.9992
 15       206,898      2.294       827,540     2.294    3.9997
 16       462,893      2.237     1,851,345     2.237    3.9996
 17       992,268      2.144     3,968,840     2.144    3.9998
 18     1,973,209      1.989     7,891,990     1.988    3.9996
 19     3,415,314      1.731    13,659,821     1.755    3.9996
 20     4,618,491      1.352    18,471,682     1.352    3.9995
 21     4,147,448      0.898    16,586,822     0.898    3.9993
 22     2,010,449      0.485     8,039,455     0.485    3.9988
 23       378,110      0.118     1,511,110     0.188    3.9965
 24        11,894      0.031        47,351     0.031    3.9811
 25            27      0.002            87     0.002    3.222

Total  18,373,824               73,483,200              3.9993

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From bagleyd@source.asset.com  Thu Sep 29 14:03:19 1994
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From: bagleyd@source.asset.com (David A. Bagley)
Message-Id: <9409291734.AA25973@source.asset.com>
To: Cube-Lovers@ai.mit.edu
Subject: X-Puzzles

HI

  I've been busy updating X puzzles on ftp.x.org in /contrib/games/puzzles.     
Here's the blurb from the xpuzzle.README file there. 
-------------------------------
What's new?:                   
xrubik now has a undo, save, and recall
  as well as self-solver (computer solves cube) up to 3x3x3.
  Currently it is the only one in this collection with a self-solver,  
  undo, save, and recall.
xmball recently added, atan2 problem on Suns fixed
xmlink recently added, initialize error fixed                      
xhexagons minor update  
                        
The collection includes:          
SLIDING BLOCK PUZZLES                       
xcubes:         expanded 15 puzzle                                       
xtriangles:     same complexity as 15 puzzle                             
xhexagons:      2 modes: one ridiculously easy, one harder than 15 puzzle
                                          
ROTATIONAL 3D PUZZLES                                      
  hold down control key to move whole cube                 
  letters that represent colors can be changed in mono-mode  

xrubik:         a nxnxn Erno Rubik's Cube(tm) (or Magic Cube)
                self-solves 2x2x2 and 3x3x3 (non-orient mode).
xpyraminx:      a nxnxn Uwe Meffert's Pyraminx(tm) (and Senior Pyraminx),
                a tetrahedron with Period 2, Period 3, and Combined cut modes
                and it also a sticky mode to simulate a Halpern's Tetrahedron
                or a Pyraminx Tetrahedron
xoct:           a nxnxn Uwe Meffert's Magic Octahedron (or Star Puzzler) and
                Trajber's Octahedron with Period 3, Period 4, and Combined
                cut modes and it also includes a sticky mode
xskewb:         a Meffert's Skewb (or Pyraminx Cube), a cube with diagonal cuts
xmball:         a variable cut Masterball(tm), variable number of latitudinal
                and longitudinal cuts on a sphere, where the longitudinal cuts
                permit only 180 degree turns.

COMBINATION ROTATIONAL AND SLIDING 3D PUZZLES
xmlink:         a nxm Erno Rubik's Missing Link(tm)

Future directions:
  Sorry about the lack of self-solvers, but I would rather write the puzzle
  than the tedious solution.
  The ability to take back moves, record moves, and start with a
  entered position to other puzzles besides xrubik should be done.
  Currently the saved file for xrubik is cryptic (not intentionally).
  Also xmlink and xmball need better algorithms for drawing sectors than just
  a series of arcs.
  A Billion Barrel would be nice but only with a self-solver (the puzzle is
  too hard (I confess, I never solved it)).

Newbies (especially DOS users 8-) ):
  DOS/Windows & Mac users, sorry no port currently available.
  What you need:
    80386 or better, or Risc, etc.
    UNIX: Linux and FreeBSD are freely available (it may work on VMS).
    X: XFree86 is freely available on Linux and FreeBSD distributions.
    gunzip: freely available from GNU and the above distributions.
    tar: freely available from GNU also.
  What you do:
    After transfering the PUZZLE file to your machine
        (DOS users may want to rename the file PUZZLE.tar.gz to PUZZLE.tgz)
      gunzip PUZZLE.tar.gz  (or gunzip PUZZLE.tgz)
      tar xvf PUZZLE.tar
        (tar xvzf PUZZLE.tar.gz or tar xvzf PUZZLE.tgz may work as a short cut)
    Then read the README generated by the above command.

----------
I hope you enjoy
David

From ishius@ishius.com  Fri Sep 30 18:16:33 1994
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To: ishius@ishius.com
From: ishius@ishius.com (ishius@holonet.net)
Subject: COMMERCIAL: Penrose, Cutler, Masterball

COMMERCIAL POSTING

September 30, 1994

Dear Puzzle Enthusiast:

        Sometimes the pieces of a puzzle just seem to fall into place.
For the Fall season, we have several new puzzles whose pieces may easily
fall into place for you. Or perhaps they'll prove more challenging than
watching autumn leaves.

         Chief among our Fall puzzles is our new plastic puzzle, Sneaky
Squares.  It's hard to talk about this puzzle without saying that it is
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Cutler. He calls it his finest achievement. It consists of just 4 blocks
that must be fitted into a box. What could be simpler you say? Yet it
stumps 99% of the people we show it to. But, once you have figured out
the solution, you can demonstrate it to your friends in seconds. Because
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be intimidated by a complex or esoteric puzzle.

        At just $15, Sneaky Squares is a must for your collection, and a
great gift for your friends!

        Birds of an entirely different feather comprise the Perplexing Poultry
series of puzzles. These intriguing puzzles, from England, are based on
the tiling theories of mathematician and cosmologist Roger Penrose.
Penrose became interested in the shapes of tiles that will cover a plane.
Some regular shapes (such as squares) do this, but Penrose came up with
a number of irregular tile sets that could cover a plane. These tiles
produce patterns that are non-periodic (that is, the patterns do not
repeat). They are called 'quasi-periodic' since the pattern appears
to repeat regularly until you examine it closely. (Scientists have
since found some real-world crystals that form in a quasi-periodic way.)

        The quasi-periodic tile sets make interesting puzzles. To decide the
position of the next tile you place, you must take into account more
than just the neighbor tiles -- you have to think about the 'whole-board
position.'

        Choose either a color or black and white version of the Perplexing
Poultry.
The Black & White is reminiscent of the work of M.C. Escher. (Many of
Escher's drawings, incidentally, are examples of periodic tiling.)
Additionally, four of the tile sets have been made into jigsaw puzzles.
The jigsaws are unusual and colorful. 500 die-cut pieces build to a 19"
diameter circle in each of them. Although these are conventional jigsaws,
they are quite difficult because of the quasi-periodicity of the pattern.
See the enclosed flyer for illustrations of each jigsaw and for more
information about Perplexing Poultry.

         As a special offer, to get your fingers moving as fast as your brain,
we'll throw in free shipping on orders of $50 or more, if you place them
by October 15! So hurry and dig into these new brainteasers.

        Happy puzzling,

        James W. Connelley
        President
        Ishi Press International

P.S. We picked up a limited number of Circusmaster and Dragonmaster
Masterballs at a special price. Masterball is a rotational puzzle
with the sphere divided along lines of latitude and longitude.
Masterballs usually sell for $19.95. While they last, we are offering
these at the special price of just $11.70 each. Because they came
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the puzzles are just fine. Take advantage of this special offer before
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Dear Ishi - Please send me the following puzzles to brighten my Fall days!


        Puzzle                           Price                 s/h

o Sneaky Squares                         $15.00                 1
o Puzzling Poultry, B&W (PX01)           $69.00                 6
o Puzzling Poultry, Color (PX02)         $99.00                 6
o Perplexing Poultry Jigsaw (PX05)       $15.00                 2
o Cat Amongst the Pigeons (PX06)         $15.00                 2
o Perplexing Pisces (PX07)               $15.00                 2
o Pentaplex (PX08)                       $15.00                 2
o Circusmaster Masterball                $11.70                 2
o Dragonmaster Masterball                $11.70                 2

                                Total   $_______              ____

FREE shipping on orders of $50
or more received by Oct. 15, 1994!

Please send these puzzles to:
________________________________________
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________________________________________
________________________________________

MC/VISA___________________________ exp:__________

California residents please include 8.25% sales tax.

Toll Free order line: (800) 859-2086


Always feel free to write me if you have any questions or comments.

Anton Dovydaitis
Customer Support
===========================================================================
Ishi Press International                408/944-9900 vc, 408/944--9110 FAX
76 Bonaventura Drive                    800/859-2086 Toll Free Order Line
San Jose, CA  95134                     ishius@ishius.com (or @holonet.net)



From @mail.uunet.ca:mark.longridge@canrem.com  Mon Oct  3 05:48:56 1994
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Sender: CRSO.Cube@canrem.com
Subject: < U, R > Processes
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.808.5834.0C1B2945@canrem.com>
Date: Mon, 3 Oct 1994 01:13:00 -0400
Organization: CRS Online  (Toronto, Ontario)

Alas, no antipodes yet, but some interesting results nonetheless.
Process UR8 improves on the best known process for a certain
quad-twist in the U layer at 20 q turns. Table 3 in Winning Ways
gives a 22 q turn process.

The following results should be particularly interesting to
the physical cube solver as it is easier to execute a sequence
of 2 adjacent sides compared to a sequence using 3 or more
sides, which may require some re-orienting of the cube. I will
measure the "face index" of a process by the number of different
sides used in a certain cube sequence. Such a measure could be
used to evaluate the relative elegance of two equally long
processes with respect to their face indices.


Jerry Bryan mentions:
>                         Also, the global maxima are of length 25.
> Does this tell us anything about the Q-turn length of the global
> maxima for the full cube group?

Well, that reminds me of one of the hardest patterns that Dik Winter
tried to find an optimal sequence for:

p141a alternate method   F1 R1 L2 U3 R2 L3 U3 D2 R2 F1 D1 B1 D1 F2 U3
of Superfliptwist + 6 X        R3 D3 F2 D2 L2
**This process was one of the hardest ever to reduce to 20 moves,
requiring  over 19 hours on an SGI R4K Indigo, 28 q turns**

My own $.02 worth is that an antipode for the full group of the 3x3x3
cube is probably deeper than an antipode for the < U, R > group.


        Optimal Sequences for < U, R > group elements (positions)
        ---------------------------------------------------------

Edge 3-cycle
UR1 = U3 R1 U2 (R1 U1)^2 R2 U3 R3 U3 R2 U1                 (16 q, 13 h)

Double adjacent edge swap
UR2 = U3 R3 U3 R2 U1 R1 U1 R3 U3 R1 U1 (R1 U3)^2 R3 U3     (18 q, 17 h)

Diagonal Corner twist
UR3 = U1 R1 U3 R1 U3 R2 U1 R1 U1 R3 (U3 R1)^2 U2 R3 U3 R3  (20 q, 18 h)

Double opposite edge swap, also in sq group 24 q, 12 h
UR4 = R2 U2 R3 (U2 R2)^2 U2 R3 U2 R2                       (20 q, 11 h)

Edge 7-cycle, equivalent to (U1 R1)^15
UR5 = U3 R1 U3 R3 U3 R1 U2 R3 U1 R3 U2 R1 U3 R3 (U3 R1)^2  (20 q, 18 h)

Corner Tri-Twist
UR6 = (U3 R3)^2 U1 R1 U3 R3 U3 R2 U1 R2 U3 R3 U3 R1 U1 R3  (20 q, 18 h)

Corner Quad-Twist, Flat style
UR7 = R1 U3 (R1 U1)^2 (R3 U3)^2 R2 U3 R1 U1 R3 U3 R1 U3 R3 (20 q, 19 h)

Corner Quad-Twist, Arms & Legs style                       (20 q, 20 h)
UR8 = R1 U1 R3 U1 R3 U3 R1 U1 R1 (U3 R3)^2 (U1 R1)^2 U3 R3 U3

ML Doodle Position
UR9 = (U2 R2)^2 U2 R3 U1 R2 (U3 R2)^2 U1 R1                (22 q, 14 h)
Same position found by hand: (a non-optimal 24 q, 15 h)
      (U2 R2)^3 U1 R1 (U2 R3)^2 U2 R1 U1

4 Opp Corner Swap, also in sq group at 26 q, 13 h
UR10 = U3 R3 (U1 R1)^2 U2 R3 U1 R1 (U2 R2)^2 U1 R3 U1      (22 q, 17 h)

        Other Subgroups within reach
        ----------------------------

11. |<U, R2, L2>|              = 2^12 3^4 5^2 7   =             58060800
12. |<U2, R, L2>|              = 2^12 3^4 5^2 7   =             58060800
17. |<U, R2, F2>|              = 2^8 3^5 5^2 7    =             10886400
21. |<U, R2, L2, D2>|          = 2^13 3^4 5^2 7   =            116121600
22. |<U, R2, L2, D>|           = 2^15 3^4 5^2 7^2 =           3251404800

I welcome any proposed < U, R > group antipodes. I haven't really
looked for anything exotic like < U, R > positions which are
shift invariant, or even if such a beast is possible!

Of course I already mentioned that...

        U2 R2 U2 R2 U2 R2 = R2 U2 R2 U2 R2 U2

...but aside from that nothing comes to mind.

Generally when there are elements which occur in both the square's
group AND the < U, R > group the latter is the shorter in q turns.

 -> Mark <-
 Email: mark.longridge@canrem.com

From BRYAN@wvnvm.wvnet.edu  Fri Oct  7 14:48:43 1994
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Date:      Fri, 7 Oct 1994 10:52:58 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@ai.mit.edu>
Subject:   Re: < U, R> Group  -- Q+H
In-Reply-To: Message of 08/09/94 at 01:48:00 from mark.longridge@canrem.com

Distance    Number     Branching       Number    Branching     Ratio of
 from         of        Factor           of       Factor       Cubes to
 Start   W-Conjugate                    Cubes                  W-Conjugate
           Classes                                              Classes

      0          1                            1                  1
      1          2          2                 6     6            3
      2          5          2.5              18     3            3.6
      3         14          2.8              54     3            3.857
      4         41          2.929           162     3            3.951
      5        122          2.976           486     3            3.984
      6        365          2.992         1,457     2.998        3.992
      7      1,091          2.989         4,360     2.992        3.996
      8      3,256          2.984        13,016     2.985        3.998
      9      9,627          2.957        38,482     2.957        3.997
     10     28,282          2.938       113,094     2.939        3.9987
     11     82,243          2.908       328,920     2.908        3.9994
     12    235,611          2.865       942,351     2.865        3.9996
     13    654,297          2.777     2,616,973     2.777        3.9997
     14  1,693,858          2.589     6,774,848     2.589        3.9997
     15  3,776,718          2.230    15,105,592     2.230        3.9997
     16  6,058,483          1.604    24,231,019     1.604        3.9995
     17  4,856,334          0.802    19,421,274     0.802        3.9992
     18    961,504          0.198     3,843,568     0.198        3.997
     19     11,954          0.012        47,465     0.012        3.971
     20         16          0.001            54     0.002        3.375

  Total 18,373,824                   73,483,200                  3.9993

Notice that using Q+H turns instead of Q turns reduces the maximum
distance from Start from 25 down to 20.

When I first calculated God's Algorithm for <U,R> for Q turns, I calculated
it for cubes first, then for W-conjugate classes.  In this case, I
really did it only for W-conjugate classes (problem is four times
smaller).  The "Number of Cubes" column is then derived by calculating
the size of each W-conjugate class; no real search is needed to
obtain the number of cubes if the W-conjugate classes are already
in hand.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From dik@cwi.nl  Wed Oct 12 21:55:07 1994
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From: Dik.Winter@cwi.nl
Message-Id: <9410130155.AA21510=dik@boring.cwi.nl>
To: cube-lovers@life.ai.mit.edu
Subject: CFF 34, summary of contents

CFF #34 came out, it ought to have been in June but is a bit late.
Still, the editors expect the next issue in December.

Summary of contents.

Leo Links: On Folding Puzzles.
  A discussion about folding puzzles made from paper or cardboard
  that display a particular figure or text when folded.

Frits Goebel and Bernhard Wiezorke: Problems for Einstein.
  They discuss a puzzle consisting of 8 octonimo's which is marketed
  as IQ CREATOR, MAGIC BLOCK or EINSTEIN PUZZLE.  They show a few
  pretty patterns that can be formed with the 8 pieces.

Vic Stok: Skyline Tetracubes.
  This discusses figures that can be formed from the 8 different
  pieces consisting of 4 cubes glued together.

Jacques Haubrich and Nanco Bordewijk: Cube Chains.
  This discusses a number of puzzles.  Each consists of 27 cubes
  connected to each other by an elastic string.  The objective is
  to form a 3x3x3 cube.

Bernhard Wiezorke: On Nob's L-Puzzle.
  This duscusses Nob's puzzle.  It consists of 10 L shaped pieces,
  3 squares high, 2 squares wide; all in the same orientation, one
  such piece 4 squares high (also the same orientation) and one
  3 square high piece in different orientation (i.e. turned over).
  The objective is to fill a 7x7 square, turnover of the pieces is
  not permitted.

Jacques Haubrich: Pantactic Patterns and Puzzles.
  This discusses an extension of the memory wheel.  On the wheel
  the digits 0 and 1 are written such that when you look at 3
  consecutive digits, all 8 different can be created.  This can
  be generalized to n consecutive digits.  It is well known (since
  N.G. de Bruijn) that 2^n digits are needed.  An 2-dimensional
  extension was made by B. Astle who had a 5x5 square with a
  black-white pattern such that when you look at the 16 different
  2x2 subsquares you will find all 16 different configurations.
  C.J. Bouwkamp made this into a puzzle (in the early 70's) as
  follows: You have 16 2x2 squares with all possible patterns.
  The puzzle is to put them together in a larger square such that
  the borders match.  Rotation is *not* permitted.

Torsten Sillke: Three 3x3 Matching-Puzzles.
  A discussion about three puzzles consisting of 9 squares that
  have to be put in a 3x3 square where some form of marking has
  to match.

Jacques Haubrich: Cube 216.
  The puzzle Gemini consists of 10 pieces where each piece is
  made by joining two 1x2x2 blocks together.  This is done in
  all possible ways.  It is known that there are 50 possible
  ways to pack them in a 4x4x5 block.  Yoshikatsu Hara extended
  this with 22 pieces that form all possible ways to join two
  1x3x3 blocks together.  One result is that there are (at
  least) 11 selections of 12 of these pieces so that they can
  be packed in a 6x6x6 cube in an unique way.  There are more
  results and the author asks also for input.

Chris Roothart: Polylambdas.
  Polylambdas are formed from the 30/60/90 degree triangle.
  Lambdas can be joined at corresponding edges.  Joining
  along the hypothenusa is not allowed.  There are 4 dilambdas,
  4 trilambdas, 11 tetralambdas en 12 pentalambdas.  These 31
  pieces can fill a parallelogram of 4 by 31 units (the short
  leg is the unit).  Many other problems are stated.

Columns:

Mark Peters: Books and Magazines (book reviews)
Edward Hordern: What's Up? (details some new puzzles)
------
CFF (Cubism For Fun) is the newsletter published by the
Nederlands Kubus Club NKC (Dutch Cubists Club).

Membership fee is NLG 25 individual, NLG 80 institutional.
(USD 1 ~ NLG 1.70).  Applications for membership to the
treasurer:
   Lucien Matthijsse
   Loenapad 12
   3402 PE  IJsselstein
   The Netherlands

If you write, please add an international reply coupon
(can be obtained at your post office).


From f94dk@efd.lth.se  Fri Oct 14 15:24:33 1994
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Date: Fri, 14 Oct 1994 16:43:57 MET
From: David Kaspar <f94dk@efd.lth.se>
To: CUBE-LOVERS@life.ai.mit.edu
Subject: You are my only hope...




Hi !!

My name is David. Long time ago I was able to solve Rubik's Cube but I ha=
ve
unfortunatly (Ooops the spelling) forget it now. Can you please help me??=


I would be very grateful, because you are my only hope.


					Many thankx, David

				email: f94dk@efd.lth.se =


From ma2gapen@lucano.uco.es  Tue Oct 18 08:13:23 1994
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Date: Tue, 18 Oct 94 13:12:54 +0100
From: ma2gapen@lucano.uco.es (Nicolas G. Pedrajas)
Message-Id: <9410181212.AA29888@lucano.uco.es>
To: cube-lovers@life.ai.mit.edu
Subject: help!


hello,
	I used to know how to resolve rubik's cube, but
i've forgotten it. Can anybody here help me?
	Thanks in advance for any help.

 Adios.


From @mail.uunet.ca:mark.longridge@canrem.com  Sun Oct 23 03:41:06 1994
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Reply-To: CRSO.Cube@canrem.com
Sender: CRSO.Cube@canrem.com
Subject: Cross and X's
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.814.5834.0C1B6751@canrem.com>
Date: Sun, 23 Oct 1994 02:36:00 -0400
Organization: CRS Online  (Toronto, Ontario)

-----------------------------
Possible Legal Cross Patterns
-----------------------------

Plummer Cross   (6 Cross order 3) =  8 patterns
Christman Cross (6 Cross order 2) =  6 patterns
4 Cross order 2 (sq group)        =  3 patterns
4 Cross order 4                   =  6 patterns
                                    -----------
      14 Six Cross + 9 Four Cross = 23 total legal Cross patterns

There are 0 total cross patterns in the swap orbit

-------------------------
Possible Legal X Patterns
-------------------------

6 X order 3                       =  8 patterns
6 X order 6                       =  8 patterns
6 X order 2 (sq group)            =  1 pattern
4 X order 2 (sq group)            =  3 patterns
2 X order 2 (sq group)            =  3 patterns
                                    -----------
    17 Six X + 3 Four X + 3 Two X = 23 total legal X patterns

For a while I thought that [6 x order 3] combined with the [2 x pattern]
would make a new sort of [6 x order 6], but combining [6 x order 3] with
the [2 x pattern] is essentially the same as combining 6 x order 3 with
the pons asinorum or 6 x order 2.

------------------------------
Possible Swap-Orbit X Patterns
------------------------------

6 X order 2                       =  6 patterns
6 X order 4                       =  6 patterns
4 X order 2                       =  6 patterns
4 X order 4                       =  6 patterns
                                    -----------

             12 Six X + 12 Four X = 24 total swap-orbit X patterns

Some description of the swap-orbit patterns is in order.
The 6 X order 2 pattern has a 2-cycle of opposite edges and
2 sets of 2-cycles of adjacent edges.
The 6 X order 4 pattern has a 2-cycle of opposite edges and
a 4-cycle of edges of adjacent faces.
The 4 X order 2 has 2 sets of 2-cycles of adjacent edges.
The 4 X order 4 has a 4-cycle of edges of adjacent faces.

To make any of these swap-orbit patterns one would have to first
exchange any 2 edge cubies.

Interestingly, a thin line 6 X order 3 is possible on the 5x5x5
cube. No process as yet....

-> Mark <-
Email: mark.longridge@canrem.com

From mschoene@math.rwth-aachen.de  Mon Oct 24 16:59:39 1994
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Date: Mon, 24 Oct 94 21:58 PST
From: Martin.Schoenert@math.rwth-aachen.de
To: cube-lovers@life.ai.mit.edu
Subject: Shift invariant processes

Mark Longridge wrote in is e-mail message of 1994/04/02

    The resultant position generated by process p8 is invariant under
    shifting, specifically 2 X on the Left and Right sides.

    P8   2 x ORDER 2:

    shift
    0                              D2 F2 T2 F2 B2 T2 F2 T2
    1                           T2 D2 F2 T2 F2 B2 T2 F2
    ...
    7         F2 T2 F2 B2 T2 F2 T2 D2

    This is the longest process I've found so far. Certainly this property
    is not true of all squares group processes. I suspect there are no
    processes in the full group with this property (of any significant
    length). Perhaps the fact that the L and R faces never rotate will
    give some clue on how to generate processes with this property.

I have classified all such shift invariant processes, using a little bit
of group theory and the computer algebra system GAP.

Let me first repeat the definition.
A *process* g_1 g_2 ... g_n, where the letters g_i come from the set
{U,U2,U3,D,D2,D3,...,B,B2,B3}, is called *shift invariant* if each of
the processes g_1 g_2 ... g_n, g_2 ... g_n g_1, ..., g_n g_1 ... g_{n-1}
effects the same element g in the cube group G.

In the following I will be a bit sloppy and neither distinguish between
letters and the corresponding generators of the cube group nor between
processes and the elements of the cube group they effect.
With this terminology a shift invariant processes would be one where
g_1 g_2 ... g_n = g_2 ... g_n g_1 = g_n g_1 ... g_{n-1}.

So lets assume that g = g_1 g_2 ... g_n is a shift invariant process.
Then for every letter g_i in g we have
g_i' g = g_i' (g_i g_{i+1} ... g_{i-1}) = (g_{i+1} ... g_{i-1})
       = (g_{i+1} ... g_{i-1} g_i) g_i' = g g_i'.
That means that g commutes with (the inverses) of each of its letters.

Because g commutes with each of its letters, it also commutes with all
elements of the subgroup H = < g_1, g_2, ..., g_n > generated by its
letters.  The set of those elements of H which commute with all elements
of H is called the centre of H.  Thus g lies in the centre of H.

Obviously the other direction is also true.  If g lies in the centre of
H = < g_1, g_2, ..., g_n >, i.e., if it commutes with every element of H,
then it especially commutes with its letters, and so the corresponding
process is shift invariant.

This says that if we have an element g in the centre of a subgroup
H = < g_1, g_2, ..., g_n >, then *every* process that effects g and
uses the letters g_1, g_2, ..., g_n will be a shift invariant process.

So there are finitely many such elements (after all there are only
finitely many elements in the entire cube group), but each gives
rise to infinitely many different shift invariant processes.
In particular there is *no* longest shift invariant process.

So the task is to search for subgroups H generated by subsets of
{U,U2,U3,D,D2,D3,...,B,B2,B3} that have non-trivial centres.
There are 729 = 3^6 such subgroups.  Of course we are only interested in
representatives under the operation of M (the subgroup of symmetries of
the entire cube), which leaves us with 56 subgroups.

Of those 21 have a non-trivial centre (for this computation I used GAP).
The centres are all very small and contain mostly the same elements,
i.e., the same element lies in the centre of different such subgroups.
I do not want to bore you with the details.  Allow me to jump to the
discussion of the results.

There are 5 different types of elements that give rise to shift invariant
processes.

1)  The ``trivial'' element.
    The identity element lies in the centre of every subgroup H.
    Thus every process that effects the identity is shift invariant.
    There is exactely one such element in the entire group.

2)  The ``central'' element.
    The superflip, which flips all edges, is in the centre of G.
    Thus every process that effects the superflip is shift invariant.
    There is exactely one such element in the entire group.

3)  The ``abelian'' elements.
    The subgroups < U > and < U, D > (and their conjugates under M)
    are abelian, and are therefore equal to their centre.
    Therefore every element in < U > and < U, D > is shift invariant.
    There are 45 such elements in the entire group.

4)  The ``odd'' element.
    The element (UR)^140 lies in the centre of the subgroup <U,R>.
    It is the only shift invariant element of odd order (hence the name).
    Thus this process and its inverse are shift invariant.
    There are 24 such elements in the entire group (two for each edge).

5)  The ``square'' elements.
    The following elements live in the ``square ring'' group <U2,D2,R2,L2>
    (though some of them are central in proper supergroups of it).

5a) The ``single square'' elements.
    The element (U2 R2)^3 lies in the centre of <U2,D,R2,L>.
    Thus this process is shift invariant.
    There are 12 such elements in the entire group (one for each edge).

5b) The ``edge square'' elements.
    The element (U2 R2)^3 (U2 L2)^3 = (D2 R2)^3 (D2 L2)^3
    lies in the centre of <U,D,R2,L2>.
    Thus this process is shift invariant.
    There are 6 such elements in the entire group (two for each axis).

5c) The ``diagonal square'' elements.
    The element (U2 R2)^3 (D2 L2)^3 = (U2 L2)^3 (D2 R2)^3
    lies in the centre of <U2,D2,R2,L2>.
    Thus this process is shift invariant.
    There are 3 such elements in the entire group (one for each axis).

For me the most amazing elements were the ``odd'' element and the
``diagonal square'' element.

They are special in the sense that the smallest subgroup in which they
lie and the largest subgroup in which they are central are equal.
That means that you have no choice which letters to choose to write
them (you have lots of choices how arrange those letters and how often
to repeat them of course).  You cannot use less, because they do not
lie in a smaller group, and you cannot use more, because they are not
central in a larger group.

Let me return to Mark's e-mail and discuss it in the light of the above.

Mark writes

    The resultant position generated by process p8 is invariant under
    shifting, specifically 2 X on the Left and Right sides.

    P8   2 x ORDER 2:

    shift
    0                              D2 F2 T2 F2 B2 T2 F2 T2
    ...

Believe it or not, this is a process for the ``diagonal square'' element.

Mark writes

    This is the longest process I've found so far.

How about (UR)^140 or (UR)^1400?  As mentioned above, you can make the
processes as long as you wish.

Mark writes

    Certainly this property is not true of all squares group processes.

No, only for processes that effect one of the 21 elements mentioned above
(31 if you want to count the ``trivial'' and ``abelian'' elements in the
square group).

Mark writes

    I suspect there are no processes in the full group with this property
    (of any significant length).

Not true.  The interesting ones are the processes effecting the
``central'' element and the ``odd'' elements.

Mark writes

    Perhaps the fact that the L and R faces never rotate will give some
    clue on how to generate processes with this property.

Now this remark makes me very suspicious.  Did Mark know the full story?
The squares subroup has trivial centre (containing only the identity),
you have to leave out at least two generators belonging to opposite
faces, to get a subgroup with non-trivial centre.

Mark writes in another e-mail message of 1994/04/10

    The following processes are also shift invariant:

    2 Swap                   D2 R2 D2 R2 D2 R2 (6)
    (symmetry level 12, SI level 2)

    p21  2 H                 L2 R2 B2 L2 R2 F2 (6)
    (symmetry level 6, SI level 6)

    Amazingly, the process p3 (found using Dik Winter's program) is actually
    a series of 20 processes which all result in the same displacement!

    p3   12 flip             R1 L1 D2 B3 L2 F2 R2 U3 D1 R3 D2 F3 B3 D3 F2 D3
                                R2 U3 F2 D3 (20)
    (symmetry level 1, SI level 20)

The first is obviously a ``single square'' element, the second is a
``edge square'' element, and 'p3' is the ``central'' element.

Thus at this time all non-trivial such elements had been found, except
for the ``odd'' element.

Have a nice day.

Martin.

PS: GAP is really a nice program to analyze the cube group from the
    group-theoretical side, though I would not use it to enumerate
    positions in search for god's algorithm.

PPS: Of course I am biased, because I am one of the main authors of GAP ;-)

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From dbisel@adrian.adrian.edu  Wed Oct 26 10:26:30 1994
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Date: Wed, 26 Oct 1994 10:25:43 -0400
Message-Id: <94102610254293@adrian.adrian.edu>
From: dbisel@adrian.adrian.edu
To: cube-lovers@life.ai.mit.edu
X-Vms-To: smtp%"cube-lovers@ai.ai.mit.edu"

Does the "mit" in the address stand for Michigan Tech University? 
I am a student at Adrian College.

What is the record in minutes (or seconds) for the time to solve a rubik's
cube?

Do you happen to have any brain teasers or hypothetical questions?

I would love to hear from you.

Diana Bisel

From MALONEY9146@a12t.cc.fredonia.edu  Wed Oct 26 11:03:59 1994
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Date:         Wed, 26 Oct 1994 11:02 am EDT (15:02:37 UT)
From: "Daniel P. Maloney" <MALONEY9146@fredonia.edu>
Organization: State University of New York - College at Fredonia
To: dbisel@adrian.adrian.edu
Cc: cube-lovers@life.ai.mit.edu
In-Reply-To:  Your message of 26 Oct 1994 10:25:43 -
Message-Id: <35483102694110235@FREDONIA>

I'm not sure what the record is, but I used to have a book
called "Jeff Conquers The Cube In 45 Seconds (And So Can You!)".
Needless to say, I never got cloise to that.

Dan

BTW MIT is probably Massachusetts Institute of Technology
  (a big, expensive school)

Dan

From nivek@frc2.frc.ri.cmu.edu  Wed Oct 26 11:19:52 1994
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Date: Wed, 26 Oct 94 11:18:26 EDT
From: Kevin Dowling <nivek@frc2.frc.ri.cmu.edu>
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To: dbisel@adrian.adrian.edu, cube-lovers@life.ai.mit.edu
In-Reply-To: <94102610254293@adrian.adrian.edu> (dbisel@adrian.adrian.edu)
Subject: MIT
Reply-To: nivek@cmu.edu (Kevin Dowling)

No sorry, it stands for Massachusetts Institute of Technology, a
little technical school located in Cambridge, MA, on the Charles River
near Boston.

CMU doesn't stand for Central Michigan University either.


				nivek

aka:    Kevin Dowling         
tel:    412.268.8830          Carnegie Mellon University
fax:    412.682.1793          The Robotics Institute
net:    <nivek@cmu.edu>       5000 Forbes Avenue
                              Pittsburgh, PA 15213




From bosch@smiteo.esd.sgi.com  Wed Oct 26 11:28:44 1994
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From: "Derek Bosch" <bosch@smiteo.esd.sgi.com>
Message-Id: <9410260827.ZM1705@smiteo.esd.sgi.com>
Date: Wed, 26 Oct 1994 08:27:59 -0700
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        "" (Oct 26, 11:02am)
References: <35483102694110235@FREDONIA>
X-Mailer: Z-Mail-SGI (3.0S.1026 26oct93 MediaMail)
To: "Daniel P. Maloney" <MALONEY9146@fredonia.edu>, dbisel@adrian.adrian.edu
Cc: cube-lovers@life.ai.mit.edu
Subject: Fast cubing
Content-Type: text/plain; charset=us-ascii
Mime-Version: 1.0

I too, have read the book, Jeff Conquers The Cube in 45 seconds, as well as Minh
Thai's book on the cube (he's the world record holder, with 22 seconds as an
official world record.  I used to compete back in the cubing days, and could
regularly get under 25 seconds, using a strategy of solving the corners,
solving the edges on two opposite sides, followed by the middle slice.

Several people on this mailing list have done serious analysis trying to reach
"God's Algorithm", which isn't terribly useful to me.  The operators that these
analyses generate are really slow to crank out on the cube.  I prefer slightly
longer ones, that are more optimized for speed (hand positions, etc).

Derek

-- 
Derek Bosch                   "Time flies like an arrow,
(415) 390-2115                 but fruit flies like bananas"
bosch@sgi.com                  J. Blaylock


From HOWSER@lua6.lu.edu  Wed Oct 26 11:59:14 1994
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From: HOWSER@lua6.lu.edu
Date: 26 OCT 94 11:03   
To: <cube-lovers@life.ai.mit.edu>
Subject: Record times for the cube
Comments: Automatic Return Receipt Requested
Message-Id: <CM3N02353B9ECC@LUA6>

Ouch!  I fat-fingered it, sorry.
-------  Forwarded message -------

Date: Wed, 26 Oct 1994 10:23 am CDT (15:23:26 UT)
From: Gerry Howser <HOWSER @ LUA6>
To: <cube-lobers @ LIFE.AI.MIT.EDU>
bcc: Gerry Howser <HOWSER @ LUA6>
Subject: Record times for the cube
Comments: Automatic Return Receipt Requested
Message-ID: <CM3M23241CBE63@LUA6>

I recall a demo on Johnny Carson where a guy solved seven cubes in seven minutes
and I think he had the current world record of around 21 seconds.  I have solved
a cube in 39 seconds but it was luck more than anything else.  When I was play-
ing around with making modifications to my solution to the cube I could solve
any cube in about a minute and a half, which was fast enough to earn me a few
drinks in bars.  I think that a legitimate record would be around 30-45 seconds
and would have to be an average for multiple cubes.

------------------------------------------------------------------------
    Gerry Howser
    INTERNET:  howser@lua6.lu.edu
               Postmaster@lua6.lul.edu
               howser@penny.lu.edu (Alternate)

    VOICE:  (314) 681-5400
    FAX:    (314) 681-5566
------------------------------------------------------------------------

------- End of forwarded message(s) -------

------------------------------------------------------------------------
    Gerry Howser
    INTERNET:  howser@lua6.lu.edu
               Postmaster@lua6.lul.edu
               howser@penny.lu.edu (Alternate)

    VOICE:  (314) 681-5400
    FAX:    (314) 681-5566
------------------------------------------------------------------------


From diamond@jrdv04.enet.dec-j.co.jp  Wed Oct 26 20:35:05 1994
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From: Norman Diamond  27-Oct-1994 0932 <diamond@jrdv04.enet.dec-j.co.jp>
To: cube-lovers@life.ai.mit.edu
Apparently-To: cube-lovers@life.ai.mit.edu
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Mime-Version: 1.0
Content-Type: TEXT/PLAIN; charset=ISO-2022-JP

>Ouch!  I fat-fingered it, sorry.
>From: Gerry Howser <HOWSER @ LUA6>
>To: <cube-lobers @ LIFE.AI.MIT.EDU>
>I could solve any cube in about a minute and a half,

Nonsense.  Anyone who can't hit the "v" key on their keyboard is surely
incapable of manipulating the right cubie :-)

-- Norman Diamond       diamond@jrdv04.enet.dec.com
[Digital did not write this.]

From dbisel@adrian.adrian.edu  Thu Oct 27 15:47:08 1994
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Date: Thu, 27 Oct 1994 15:46:27 -0400
Message-Id: <94102715462720@adrian.adrian.edu>
From: dbisel@adrian.adrian.edu
To: cube-lovers@life.ai.mit.edu
Subject: other games
X-Vms-To: smtp%"cube-lovers@ai.ai.mit.edu"

What other games are you interested in besides the Rubik's Cube? Do you know
of any other addresses where I can get fun, cool information at?
Diana Bisel



From ybanezs%geds@mhsgate.salem.ge.com  Thu Oct 27 16:22:05 1994
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Date: Thu, 27 Oct 94 16:20:08 EST
From: Ybanez Sheldon <ybanezs%geds@mhsgate.salem.ge.com>
To: cube-lovers@ai.mit.edu
Subject: Solution..
X-Mailer: XGATE 2.12 MHS/SMTP Gateway

I have been able to solve the Cube in under a minute... but that was 
years ago when my reflexes and memory was better in Junior High... now I 
pull the old cube out for limbering the fingers... and seeing how much I 
remember the solutions... now they are so ingrained in me... I no longer 
remember them as separate moves.... but a conglomeration of twists and 
turns... 

the book --the title I can't remember-- I originally learned from, showed 
the solution as a TOP to BOTTOM approach... doing the first top layer... 
then the center edges... and then completing the last layer...  I noticed 
then Mihn used the top, bottom, middle approach... when he won the 
World's Cube solving championship on the show 'THAT"S INCREDIBLE', which 
was also advocated by the solution book that was available from the 
address that was included with the original cubes...

so then I was able to solve it either way... finding the latter approach 
a little faster...  now the Revenge I can solve in about 5 minutes... 
maybe quicker, but I never really bothered to accurately time myself... I 
only learned the one way to solve the 4x4x4 cube... from Mihn's book.

Now since I joined this mailing list I have been inundated with all these 
algorithms.... how do I translate them?  Being a neophyte to cube 
'theory' its pretty frustrating trying to figure out what all the letters 
and numbers mean... and what they are trying to achieve....

can anyone help?

thanks in advance...


                                                         ,,,
______________________________________________________  (o o)  _________
+----------------------------------------------------ooO-(_)-Ooo-------+
| Sheldon Ybanez  [ybanez-s@salem.ge.com]  GE Drive Systems  Salem, VA |
|     Always "Remember. No matter where you go, there you are." 88     |
+======================================================================+



From BRYAN@wvnvm.wvnet.edu  Thu Oct 27 17:47:30 1994
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Date:      Thu, 27 Oct 1994 17:00:30 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Ybanez Sheldon" <ybanezs%geds@mhsgate.salem.ge.com>,
        "Cube Lovers List" <cube-lovers@ai.mit.edu>
Subject:   Re: Solution..
In-Reply-To: Message of 10/27/94 at 16:20:08 from ,
           ybanezs%geds@mhsgate.salem.ge.com

On 10/27/94 at 16:20:08 Ybanez Sheldon said:

>Now since I joined this mailing list I have been inundated with all these
>algorithms.... how do I translate them?  Being a neophyte to cube
>'theory' its pretty frustrating trying to figure out what all the letters
>and numbers mean... and what they are trying to achieve....

>can anyone help?

I was thinking of suggesting a few references, but then it occurs that
perhaps there are not very many references currently in print.  Here
is a little Cube Theory 101.

In the "Standard Model" (or maybe the "Singmaster Model") of the 3x3x3
cube, the cube is not rotated in space.  The only thing you can do is
twist one of the six faces.  Singmaster designates the faces as
Up, Down, Right, Left, Front, and Back.  The names are chosen so that
no two of the faces start with the same letter.  There have been some
latter day efforts to rename Up as Top so that all the faces have names
beginning with consonants.

Twists are designated by the first letter of their name  --  U, D, R, L,
F, and B for clockwise quarter-turns; U', D', R', L', F', and B' for
counter-clockwise quarter-turns; U2, D2, R2, L2, F2, and B2 for half-
turns (180 degrees).  In proper typography, the "2" in "U2" is
written as a superscript.  Sometimes U3, D3, etc. are used to denote
counter-clockwise quarter-turns because three clockwise quarter-turns
produce the same result as one counter-clockwise quarter-turn.

A sequence of twists is written left-to-right  -- e.g., FRU'LLR.

The complement notation which is used to convert clockwise quarter-turns
into counter-clockwise quarter-turns may also be applied to a group
of twists  --  e.g., (FRU')' is equal to UR'F' (twisting in the opposite
order and in the opposite direction).

The same sort of notation is used to describe cubies  -- the up-right
cubie is ur.  Singmaster distinguishes between cubies and cubicles
via italic and Roman text, but that is a bit hard to do via E-mail.

Things get a bit more complicated when you consider slice moves,
cubes larger than 3x3x3, and rotations of the whole cube.  Note that
most people solving "real cubes" (as opposed to mathematical
models of cubes) do indeed rotate the whole cube, for example they
move the Bottom face to the Up (or Top), to make it easier to twist.
However, the "Standard Model" does not rotate the whole cube because
mathematically it is just as easy to twist one face as any other.

Hope this helps.


 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From BRYAN@wvnvm.wvnet.edu  Fri Oct 28 08:54:11 1994
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Date:      Fri, 28 Oct 1994 08:53:52 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Ybanez Sheldon" <ybanezs%geds@mhsgate.salem.ge.com>,
        "Cube Lovers List" <cube-lovers@ai.mit.edu>
Subject:   Re: Solution..
In-Reply-To: Message of 10/27/94 at 17:00:30 from BRYAN@wvnvm.wvnet.edu

On 10/27/94 at 17:00:30 Jerry Bryan said:
>On 10/27/94 at 16:20:08 Ybanez Sheldon said:

>The same sort of notation is used to describe cubies  -- the up-right
>cubie is ur.  Singmaster distinguishes between cubies and cubicles
>via italic and Roman text, but that is a bit hard to do via E-mail.

Ooops.  I just pulled out my Frey and Singmaster.  Cubies and cubicles
are both italics, and the distinction is one of upper case italics
vs. lower case italics (still hard to do on E-mail).  Face twists are
Roman (block) letters.  Whole cube rotations are script letters.
Also, in proper typography, a complement (as in R') would normally
be a superscript "-1" rather than an apostrophe.

(By the way, even with a word processor or text processor, I have trouble
with script letters.  I can't get Word Perfect to do script letters, nor
Waterloo Script.  I used to use TeX, and I don't think it could do script
letters.  I haven't tried desk top publishing of the Pagemaker ilk.
Does anybody have any suggestions?  If so, I suspect this is the sort of
thing where private E-mail might be more appropriate than broadcasting
to the entire list.  Thank in advance.)

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From hoey@aic.nrl.navy.mil  Fri Oct 28 11:38:16 1994
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Date: Fri, 28 Oct 94 11:38:15 EDT
From: hoey@aic.nrl.navy.mil (Dan Hoey)
Message-Id: <9410281538.AA17190@Sun0.AIC.NRL.Navy.Mil>
To: "Cube Lovers List" <cube-lovers@ai.mit.edu>
Cc: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
Subject: Cube colors and face names
Keywords: Rubiksong, Varga, Colors, Humor

> Singmaster designates the faces as Up, Down, Right, Left, Front, and
> Back.  The names are chosen so that no two of the faces start with
> the same letter.  There have been some latter day efforts to rename
> Up as Top so that all the faces have names beginning with
> consonants.

Yes, this is the main reason for using Top, because of the Rubiksong
introduced by Varga that I described (unfortunately with many typos)
on 22 Feb 90 (<ftp:://ftp.ai.mit.edu/pub/cube-lovers/cube-mail-6.gz>
is a URL that I hope works--anyone who is actually able to point and
click on this, please let me know).

But there's another reason.  Remember the annoying feature that the
color assignments to faces were never standardized?  The first cube I
bought had red opposite yellow, blue opposite white, and orange
opposite green (I think).  Even though in later days most cubes are
manufactured with opposite faces ``differing by yellow''--red opposite
orange, blue opposite green, and yellow opposite white--there does not
seem to be a standard for the handedness of the coloring.  This has
long been a problem on cube-lovers, where everyone starts out asking
``I've got my cube solved except a blue sticker on the white face, a
white sticker on the green face, and a green sticker on the blue
face,'' and the puzzle becomes trying to figure out where those faces
are.  (This was fixed in Square 1, where they printed a full-color
instruction book coordinated with the puzzle).

My modest proposal is to define the Standard Earth-Tone Cube, which
has the faces in standard and easily remembered places.  The colors
are taupe, dun, fawn, beige, loam, and roan.  This supports the use of
Top over Up, because ``taupe'' is so much more evocative than
``umber''.  ``Dun'' is also a major win, and I wish I had better names
for the other faces.

I have yet not tried painting such a cube, because I can't figure out
which color is which.

Dan
Hoey@AIC.NRL.Navy.Mil

From @mail.uunet.ca:mark.longridge@canrem.com  Fri Oct 28 11:50:15 1994
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Sender: CRSO.Cube@canrem.com
Subject: Shift Invariant Part 2
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.825.5834.0C1B76FC@canrem.com>
Date: Thu, 27 Oct 1994 21:56:00 -0400
Organization: CRS Online  (Toronto, Ontario)

Continuing the previous discussion on shift invariance...

Mark writes:
>>    This is the longest process I've found so far.

Martin writes:
>How about (UR)^140 or (UR)^1400?  As mentioned above, you can make the
>processes as long as you wish.

...or (U1 R1)^35 ? And indeed, (U1 R1)^(35 * 40) is shift invariant.
I meant to say (and should have said):

 "This is the longest optimal process I've found so far."

 Although I was inspecting (U1 R1)^N patterns in the quest for shift
invariance, (U1 R1)^35 = (R1 U1)^35 escaped me. In fact it was my
mistaken belief that the < U , R > group had no shift invariant
processes.

 I did not realize the connection between the centre of a group and
shift invariance until Martin's message of Mon Oct 24 17:10:27 1994.
I actually did use GAP on the < U, R > group but I couldn't resolve
the resulting position (can GAP use letters? I should have used
letters).

The missing insight was realizing that, although the full group had
a unique centre, other subgroups have different centres.

So without further adieu:

6 Counterclockwise Twist,
Equivalent to (U1 R1)^35= (R1 U1)^35 & Shift Invariant
UR11 = U2 R1 U1 R1 U1 R3 U1 R3 U1 R3 U2 R1 U1 R1 U1 R3 U1 R3 U1 R3
       (22 q  or  20 h  moves)

(U3 R3)^35 would execute a 6 clockwise twist.

Martin writes:
> 4)  The ``odd'' element.
>  The element (UR)^140 lies in the centre of the subgroup <U,R>.
>  It is the only shift invariant element of odd order (hence the name).
>  Thus this process and its inverse are shift invariant.
>  There are 24 such elements in the entire group (two for each edge).

Is this odd due to ( U1 R1 )^35?
Actually everything about the above description appears even.
It is an even number of quarter turns...

Martin writes:
> For me the most amazing elements were the ``odd'' element and the
> ``diagonal square'' element.

 I concur completely, although the all-commuting 12-flip is definitely
interesting too. I was surprised to see the process was shift invariant.

Martin writes:
> Thus at this time all non-trivial such elements had been found, except
> for the ``odd'' element.

For which I refer to the process UR11, 22 q turns.

 Martin, you will be pleased to hear that I like GAP, but I need a
bigger hard drive for that beast!

 -> Mark <-
 Email: mark.longridge@canrem.com

From @mail.uunet.ca:mark.longridge@canrem.com  Fri Oct 28 12:08:29 1994
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Subject: Speed Cubing
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.826.5834.0C1B76FD@canrem.com>
Date: Thu, 27 Oct 1994 21:57:00 -0400
Organization: CRS Online  (Toronto, Ontario)

Derek Bosch writes:
> I too, have read the book, Jeff Conquers The Cube in 45 seconds, as
> well as Minh Thai's book on the cube (he's the world record holder,
> with 22 seconds as an official world record.  I used to compete back
> in the cubing days, and could regularly get under 25 seconds, using
> a strategy of solving the corners, solving the edges on two opposite
> sides, followed by the middle slice.


The "official" world record was set by Minh Thai at the 1982 World
Championships in Budapest Hungary, with a time of 22.95 seconds.

Keep in mind mathematicians provided standardized dislocation patterns
for the cubes to be randomized as much as possible.

I think the Guiness Book of Records dropped the entry in the 1985
edition due to the fact that the contests all dried up.

Interestingly David Allen, the #2 cubist in the United States, also
uses the Jeff Varasano method. I met him in Buffalo NY in the
a regional American Cube-a-thon on Sept 18, 1982. (Yes, that long
ago!)

Did you enter any of the tournaments Derek?

Derek continues:
> Several people on this mailing list have done serious analysis
> trying to reach "God's Algorithm", which isn't terribly useful to me.
> The operators that these analyses generate are really slow to crank
> out on the cube.  I prefer slightly longer ones, that are more
> optimized for speed (hand positions, etc).

I can't agree entirely. I use computer generated sequences for a lot
of patterns and I find them quite useable in some cases. Also the
< U, R > group processes only use 2 sides, and those I can do
without moving the cube in space. Usually I rotate them in
space first.

 -> Mark <-
 Email: mark.longridge@canrem.com

From BRYAN@wvnvm.wvnet.edu  Fri Oct 28 12:55:14 1994
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Date:      Fri, 28 Oct 1994 12:54:56 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@life.ai.mit.edu>
Subject:   Re: Speed Cubing
In-Reply-To: Message of 10/27/94 at 21:57:00 from mark.longridge@canrem.com

Has any analysis of speed cubing been performed in the sense of how
many twists were performed?  How many twists does somebody accomplish
in under 45 seconds or in 22.95 seconds?  For example, you might
video tape somebody and replay it in slow motion.  It would still be
hard to get an accurate count, I think.  You would have to ignore
whole cube rotations, and it might be hard to distinguish between
half and quarter turns, plus somebody might be using slice moves.
But if such an analysis *could* be done, it would be interesting to
to compare the results to what is known mathematically about
God's Algorithm.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From ybanezs%geds@mhsgate.salem.ge.com  Fri Oct 28 15:57:15 1994
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Date: Fri, 28 Oct 94 15:56:21 EST
From: Ybanez Sheldon <ybanezs%geds@mhsgate.salem.ge.com>
To: cube-lovers@ai.mit.edu, bryan@wvnvm.wvnet.edu
Subject: Re: Speed Cubing
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>Has any analysis of speed cubing been performed in the sense of how
>many twists were performed?  How many twists does somebody accomplish
>in under 45 seconds or in 22.95 seconds?



I too had wondered this...  and in what way was the solved cube 
'scrambled' for most of you cubists know that all scrambling is not 
equal...  I have found some of my quickest times involved arriving at the 
solution much sooner than expected by not needing to perform some 
auxiliary, but essential and long routines....   Could these world record 
times have been accomplished with random scrambling..... I have always 
pondered how fast I would have completed the cube if I was handed the 
'same' cube that Minh 'flew' on.... A move count would be very 
interesting indeed.... 

With the right equipment and a good copy of the world record video 
tape... it may be conceivable to actually count the moves.... 

In the Hey day of cubing, during a younger version myself, I also found 
by making a good guess at which way to solve the cube (ie.. Top to bottom 
or
top, bottom, then middle) I could easily cut a few seconds off my S.T. 
(solution time)...  but I can't for the life of me remember the 
criteria..


                                                         ,,,
______________________________________________________  (o o)  _________
+----------------------------------------------------ooO-(_)-Ooo-------+
| Sheldon Ybanez  [ybanez-s@salem.ge.com]  GE Drive Systems  Salem, VA |
|     Always "Remember. No matter where you go, there you are." 88     |
+======================================================================+



From mschoene@math.rwth-aachen.de  Fri Oct 28 17:54:41 1994
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Date: Fri, 28 Oct 94 22:13 PST
From: Martin.Schoenert@math.rwth-aachen.de
To: cube-lovers@life.ai.mit.edu
In-Reply-To: Mark Longridge's message of Thu, 27 Oct 1994 21:56:00 -0400 <60.825.5834.0C1B76FC@canrem.com>
Subject: Re: Shift Invariant Part 2

Mark Longridge writes in his e-mail message of 1994/01/27

    ...or (U1 R1)^35 ? And indeed, (U1 R1)^(35 * 40) is shift invariant.

Mark  kindly points out, that my process (UR)^140 for the ``odd'' element
is a strange choice, given that (UR)^140 = (UR)^35.

I can't recall how I arrived at this process.  Somehow I simply missed
that (UR)^140 = (UR)^35, which is especially strange since I know that
(UR) has order 105 since 1982.

Mark continues

    Equivalent to (U1 R1)^35= (R1 U1)^35 & Shift Invariant
    UR11 = U2 R1 U1 R1 U1 R3 U1 R3 U1 R3 U2 R1 U1 R1 U1 R3 U1 R3 U1 R3
           (22 q  or  20 h  moves)

Is UR11 the shortest process effecting the ``odd'' element in <U,R>?

Mark continues

    Is this odd due to ( U1 R1 )^35?
    Actually everything about the above description appears even.
    It is an even number of quarter turns...

The ``odd'' element o has odd order as element of the cube group,
i.e., o^3 = id.  All other shift invariant elements e have even order,
i.e., either e^2 = id or e^4 = id (for some ``abelian'' elements).

Mark continues

    I actually did use GAP on the < U, R > group but I couldn't resolve
    the resulting position (can GAP use letters? I should have used
    letters).

I assume you wonder whether GAP can find a process for a given element.
In fact GAP can do this (you define a homomorphism from the free group
on U,D,L,R,F,B to the cube group and then ask for a preimage of the
element).  But the process is usually extremly long, e.g., for
the ``central'' element GAP finds a process that has length > 2*10^6
(don't try this at home ;-).

There is an improved algorithm by Philip Osterlund, which is a lot
better, but still not good enough to help in the quest for god's
algorithm.  For example it finds a process for the ``central''
element of length 228.

Mark continues

    Martin, you will be pleased to hear that I like GAP,
    but I need a bigger hard drive for that beast!

Look at it this way:
The system costs you $200, and you even get a hard drive for free!

Seriously, you don't need the full distribution (32 MByte),
but only the executable and the library (5 MByte).
However, you should have enough real memory;
8 Mbyte is the minimum, 16 MByte is better,
and the 64 MByte that I have in my workstation don't hurt.

Have a nice day.

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From brett@math.toronto.edu  Mon Oct 31 14:36:30 1994
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Message-Id: <9410311936.AA07023@life.ai.mit.edu>
Subject: 
To: cube-lovers@ai.mit.edu (cube)
Date: Mon, 31 Oct 1994 14:35:34 -0500 (EST)
From: "Brett Stevens" <brett@math.toronto.edu>
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I joined this list in the middle of an ongoing discussion
on shift invariance.  can someone fill me in with the important
definitions and what has already been discussed  (if it is not 
too much work)

brett stevens

plus I am interested, scince seeing a part of Jerry Slokam's
collection at the exhibit in chicago in august if anyone
had a list or reference to a list of all made rubiks cube
type puzzles  ie   external shape and intenal, rotaional structure
I wouls very much like to know where I cvan get the following

1  a 2X2X2  cube
2  a the various puzzles with the pyramids internal structure but
   and also the cubes internal structue  (3X3X3) but different
   external structurs.
3  conglomeration cubes--Ideal made one of these that was two
   3X3X3  cubes sharing three cubies in a row  I have made
   one of these myself by surgery on two cubes but I know
   that there are other "conglomerates out there"
4  kitsch-cubes   (my name)   but things like royal4  kitsch-cubes   (my name)   but things like royal4  kitsch-cubes   (my name)   but things like royal4  kitsch-cubes   (my name)   but things like royal
wedding cubes,  mount rushmore etc.

thanks 

brett stevens

brett@math.toronto.edu
   

From brett@math.toronto.edu  Mon Oct 31 14:43:38 1994
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Message-Id: <9410311943.AA07328@life.ai.mit.edu>
Subject: diameter
To: cube-lovers@ai.mit.edu (cube)
Date: Mon, 31 Oct 1994 14:42:43 -0500 (EST)
From: "Brett Stevens" <brett@math.toronto.edu>
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  what is the diameter of the group of 3X3X3  rubiks cube rotations?
  ie. the longest cyclic subgroup.  what is the shortest path
  to solved?

  also I thought that red-orange   blue-white   yellow-green
  was standard.  all the ideal manufacturede c ubes were this way.

  and the two orientations available with the above colouring
  are not only an inconvienience  --Dr. Hana Bizek at
  ARgonne NAtional LAbs has used these two parities to do 
  veery intersesting cube sculptures or designs as she calls
   them

   brett stevens

   brett@math.toronto.edu


From brett@math.toronto.edu  Mon Oct 31 14:45:33 1994
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Message-Id: <9410311945.AA07473@life.ai.mit.edu>
Subject: diameter
To: cube-lovers@ai.mit.edu (cube)
Date: Mon, 31 Oct 1994 14:42:43 -0500 (EST)
From: "Brett Stevens" <brett@math.toronto.edu>
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  what is the diameter of the group of 3X3X3  rubiks cube rotations?
  ie. the longest cyclic subgroup.  what is the shortest path
  to solved?

  also I thought that red-orange   blue-white   yellow-green
  was standard.  all the ideal manufacturede c ubes were this way.

  and the two orientations available with the above colouring
  are not only an inconvienience  --Dr. Hana Bizek at
  ARgonne NAtional LAbs has used these two parities to do 
  veery intersesting cube sculptures or designs as she calls
   them

   brett stevens

   brett@math.toronto.edu


From alan@curry.epilogue.com  Mon Oct 31 16:31:07 1994
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From: Alan Bawden <Cube-Lovers-Request@ai.mit.edu>
Sender: Cube-Lovers-Request@ai.mit.edu
To: brett@math.toronto.edu, cube-lovers@ai.mit.edu
In-Reply-To: Brett Stevens's message of Mon, 31 Oct 1994 14:35:34 -0500 (EST) <9410311936.AA07023@life.ai.mit.edu>
Subject: Administrivia

   Date: Mon, 31 Oct 1994 14:35:34 -0500 (EST)
   From: Brett Stevens <brett@math.toronto.edu>
   I joined this list in the middle of an ongoing discussion
   on shift invariance.  can someone fill me in with the important
   definitions and what has already been discussed  (if it is not 
   too much work)

This is what the archives are for!

Some of you old-timers may have forgotten where the archives are, and it's
been a while (several years) since I reminded everybody about the existence
of Cube-Lovers-Request, so I have included the standard greeting message I
send to all new subscribers below.

Some quick administrative observations while I have your attention:

Today are 151 entries on the main mailing list.  Some of those entries are
local redistribution lists.  I estimate there are about 160 of us.

We celebrated our 14 birthday last July.  I'd bet we are among the top ten
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I periodically get requests for FTP archives of cube-related material other
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related resources on the network, I'd like to hear from them.

My policy on advertising: Since Cube-Lovers is an unmoderated mailing list,
I really have no control over what is sent here, but I do complain to
people who send advertising that isn't obviously Cube related.  And I
reserve the right to start complaining about -all- advertising should it
ever get out of hand.

------- Begin Greeting Message -------

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files "cube-mail-0.gz" through "cube-mail-12.gz".

Archive vital statistics (when uncompressed):

	   File		   From		   To	     Size (bytes)
	   ----		   ----		   --	     ------------
	cube-mail-0	12 Jul 80	23 Oct 80	185037
	cube-mail-1	 3 Nov 80	 9 Jan 81	135719
	cube-mail-2	10 Jan 81	 3 Aug 81	138566
	cube-mail-3	 3 Aug 81	 3 May 82	137753
	cube-mail-4	 4 May 81	11 Dec 82	139660
	cube-mail-5	11 Dec 82	 6 Jan 87	173364
	cube-mail-6	10 Jan 87	13 Apr 90	216733
	cube-mail-7	12 Oct 90	 9 Sep 91	137508
	cube-mail-8	 1 Nov 91	25 May 92	171205
	cube-mail-9	28 May 92	 7 Jan 93	155755
	cube-mail-10	20 Mar 93	 6 Dec 93	171881
	cube-mail-11	 6 Dec 93	18 Feb 94	349772
	cube-mail-12	24 Feb 94	 5 Sep 94	181193

In addition, the file "recent-mail" contains a copy of the currently active
section of the archive.  (Unfortunately, due to the way mail works here at
the AI Lab, it is not possible to have new mail accumulate directly into
this file, so there may be some delay before a new message arrives here.)
Finally, the file "README" contains the information you are currently reading.

				- Alan

------- End Greeting Message -------

From BRYAN@wvnvm.wvnet.edu  Mon Oct 31 16:48:57 1994
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Date:      Mon, 31 Oct 1994 15:39:04 -0500 (EST)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Speed Cubing Path Lengths

I have received several private E-mail messages indicating that
the algorithms used by speed cubists solve the cube in 50 or
60 moves.  On the one hand, that seems astonishingly good to me,
being fairly close to the solutions from early Thistlethwaite
programs.  On the other hand, it is roughly double (depending, I
suppose on whether H-turns are counted or not) what is probably
the true God's Algorithm.  Hence, it doesn't tell us much about
God's Algorithm except that the speed cubists are very, very
good.

On another subject, my Cube Theory 101 article said that the
apostrophe was used in E-mail to denote complements, when of
course it is used to denote inverses  --  not the same thing
as complements at all.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From dik@cwi.nl  Mon Oct 31 18:14:47 1994
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Date: Tue, 1 Nov 1994 00:14:45 +0100
From: Dik.Winter@cwi.nl
Message-Id: <9410312314.AA06573=dik@boring.cwi.nl>
To: Cube-Lovers@ai.mit.edu
Subject: Re:  Speed Cubing Path Lengths

 > I have received several private E-mail messages indicating that
 > the algorithms used by speed cubists solve the cube in 50 or
 > 60 moves.  On the one hand, that seems astonishingly good to me,
 > being fairly close to the solutions from early Thistlethwaite
 > programs.  On the other hand, it is roughly double (depending, I
 > suppose on whether H-turns are counted or not) what is probably
 > the true God's Algorithm.  Hence, it doesn't tell us much about
 > God's Algorithm except that the speed cubists are very, very
 > good.

The best current algorithm has a proven upperbound of 37 turns (q
and h).  God's Algorithm is probably much shorter.  In fact the
program that implements Kociemba's algorithms has not yet found
a configuration (out of many thousands random configurations
tested) that could not be solved in 20 turns or less.  If we look
at distributions for similar puzzles it is expected that more than
one in three configurations requires the maximum number of turns
minus 1 or 2.  So I expect God's Algorithm to be at most 22 turns.
Still a long way to go.

dik

From devo@vnet.ibm.com  Tue Nov  1 13:20:25 1994
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Date: Tue, 1 Nov 94 12:33:54 EST
From: "Dave Eaton" <devo@vnet.ibm.com>
To: cube-lovers@life.ai.mit.edu
Subject: Is there a symbolic cube program?

Is there a program that allows you to type in Singmaster-style
moves and then prints out the resultant state, something like
this (not actual results):

INPUT:   (R U2 R3 U2)2
OUTPUT:  (fur,drb,rdf) (fr,dr)

This is what I tried to write long ago, but I never had all
the tricks needed to get the program to work.

If no program like this exists, is there something similar?
I guess I would be looking for nicely-portable C or a DOS
binary.  Thanks to you all for sharing cube information.

......Dave Eaton, N2NOQ, Owego NY, devo@vnet.ibm.com


From MALONEY9146@a12t.cc.fredonia.edu  Wed Nov  2 12:56:09 1994
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Date:         Wed, 2 Nov 1994 12:35 pm EST (17:35:07 UT)
From: "Daniel P. Maloney" <MALONEY9146@fredonia.edu>
Organization: State University of New York - College at Fredonia
To: cube-lovers@ai.mit.edu
Subject:      Help ma, please!
Message-Id: <28373110294123506@FREDONIA>

I hate posting this message to the entire list, but how do you
unsubscribe from this list?

Dan

From alan@curry.epilogue.com  Wed Nov  2 14:57:30 1994
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From: Alan Bawden <Cube-Lovers-Request@ai.mit.edu>
Sender: Cube-Lovers-Request@ai.mit.edu
To: MALONEY9146@fredonia.edu
Cc: cube-lovers@ai.mit.edu
In-Reply-To: "Daniel P. Maloney"'s message of Wed, 2 Nov 1994 12:35 pm EST (17:35:07 UT) <28373110294123506@FREDONIA>
Subject: Help ma, please!

   Date:         Wed, 2 Nov 1994 12:35 pm EST (17:35:07 UT)
   From: "Daniel P. Maloney" <MALONEY9146@fredonia.edu>

   I hate posting this message to the entire list, but how do you
   unsubscribe from this list?

   Dan

As I reminded everybody just last weekend, 
so you can't say you didn't see it:

  Please note that Cube-Lovers-Request is processed by a human being, not a
  computer program (such as LISTSERV or Majordomo).  If your request is not
  instantly processed, it is because I don't spend my entire life reading my
  electronic mail.  I do know how to interpret many of the commands typically
  sent to such programs, but I would prefer it if instead you can remember to
  address me in complete sentences.

Your request to unsubscribe was sent to me less than 24 hours ago.
I'm sorry I didn't drop everything just to deal with it.  How about if I do
it later on this evening?  Can you wait that long?

From hoey@aic.nrl.navy.mil  Fri Nov  4 11:46:53 1994
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From: hoey@aic.nrl.navy.mil
Message-Id: <9411041646.AA21659@sun13.aic.nrl.navy.mil>
To: Cube-Lovers@life.ai.mit.edu
Subject: The real size of cube space

In January of this year, Jerry Bryan and I wrote of counting the
number of M-conjugacy classes of Rubik's cube.  In the sense that (for
instance) there is really only one position 1 QT from start, even
though that QT may be applied in twelve different ways, this task
amounts to counting the true number of positions of the cube.  The
earlier discussion centered on calculations involving computer
analysis of large numbers of positions.  However, a look in Paul B.
Yale's book _Geometry and Symmetry_ gave me a clue: the Polya-Burnside
theorem is a tool that allows us to perform this calculation by hand.

The Polya-Burnside theorem describes a relation between a finite group
J and a _representation_ of the group.  For our purposes, a represen-
tation is a homomorphism of J into a permutation group, R: J -> S[X].
Here, S[X] refers to the group of all permutations of a set X; the
image of J, called R(J), need not be the whole of S[X], but R(J) will
be a subgroup of S[X].  The _orbits_ of R(J) are the equivalence
classes of X under the relation x~y, said to be true if there is some
permutation p in R(J) for which p(x)=y.  The _fixed points_ of a
permutation p in R(J) are the elements x of X for which p(x)=x.  The
Polya-Burnside theorem states that the average number of fixed points
of permutations in R(J) is equal to the number of orbits of R(J).
That is,
         |R(J)| |Orbits(R(J))| = Sum[p in R(J)] |FixedPoints(p)|.
The average may also be taken over J:
            |J| |Orbits(R(J))| = Sum[j in J] |FixedPoints(R(j))|,
a nontrivial distinction, since R may not be one-to-one (though it is
for our application).  The Polya-Burnside theorem is not very
inaccessible nor hard to prove, but I will not prove it here.

For our purpose, we take the group J to be M, the 48-element group of
symmetries of the cube.  X will be the set of all cube positions,
which we usually call Gx (for GE, GC, or G, depending on whether we
consider edges, corners, or both; we are considering the positions
relative to fixed face centers in all three cases).  And the repre-
sentation R is the operation of M-conjugation: (R(m))(g) = m' g m.
Verifying that R is a homomorphism is an exercise in associativity
that Jim Saxe and I carried out in the Symmetry and Local Maxima
paper, in the archives [cube-mail-1, 14 December 1980].

R has been so chosen because we wish to calculate the number of
M-conjugacy classes of Gx, |Gx\Conj(M)|, which is be the number of
orbits of R(M).  To apply the Polya-Burnside theorem for this, we need
to calculate, for each element of m of M, the number of fixed points
of R(m).  That is the number of elements g of Gx for which m' g m = g.
Multiplying by m, this becomes g m = m g: the fixed points we wish to
count are just those elements g of Gx that commute with m.

There are several tools to make the counting easier.  First, I'll note
that just as there are M-conjugacy classes of Gx, there are
M-conjugacy classes of M itself.  The number of fixed points of R(m)
is the same for any m in a given conjugacy class.  So to calculate the
total number of fixed points over R(M), we need only calculate the
number of g in Gx commuting with each of the ten classes of cube
symmetry and multiply by the size of the class.

The fundamental principle we use in finding whether g commutes with m
can be found by examining the cycles of m.  Suppose m permutes a cycle
(c1,c2,...,ck), so that c2=m(c1), c3=m(c2),...,ck=m(c[k-1]),c1=m(ck).
For g to commute with m, we have g(c2)=m(g(c1)), g(c3)=m(g(c2)), ...,
g(ck)=m(g(c[k-1]), and g(c1)=m(g(ck)).  So (g(c1),g(c2),...,g(ck)) is
also a cycle of m.  Thus g must map each k-cycle of m to another
k-cycle of m, and in the same order.  Conversely, if g acts thus on
cycles, then g will commute with m, and so g is a fixed point of R(m).

Suppose that m has j different k-cycles of cubies.  There are j! k^j
possibilities for g's action on the cubies in those k-cycles: j!
permutations of cycles, and for each g:(c1,c2,...,ck)->(d1,d2,...,dk),
k choices for g(c1) among {d1,...,dk}.  It turns out to be a fairly
easy exercise to show that half of those possibilities are even
permutations and half odd, though the partition by parity is
surprisingly different depending on whether k is even or odd.  This
will allow us to combine the results for GE and GC simply by
multiplying together and dividing by two.

Now consider orientation of cubies.  This is similar to the case of
permutation, in that the orientation that g imposes on a cubie is a
constant for all cubies in a cycle.  I will first discuss the edge
orientation, which is fairly straightforward, and continue to corner
orientation, which has some surprising features.

For edge orientation, if all the cycles have even length, then g's
orientation parity is zero over each cycle, and so zero over the
entire cube.  So we can choose the orientation of imposed by c1->g(c1)
for each cycle (c1,...,ck) in 2^j ways.  If there are odd-length
cycles, then half of the orientations will have nonzero orientation
parity, and only 2^(j-1) possible orientations can be achieved.

For corners, we might expect there to be 3^(j-1) orientations, except
3^j for cycles of length a multiple of three, and this is often so.
But there are two important exceptions.  First, if m is a reflection
(i.e., not a proper rotation in C) then alternate cubies in each cycle
must be given the opposite orientation by g.  If the cycle has even
length, this conserves orientation, so there will be 3^j possibili-
ties.  If the cycle has odd length, this implies that the orientation
of each cubie must be its own opposite (i.e., zero twist).  Thus,
there there is only one possible orientation of the 1-cycles in the
diagonal reflections.  The second exception, an even bigger surprise,
occurs when m is either the 120-degree rotation or the 60-degree in-
verted rotation.  It turns out that the orientation constraint forbids
any permutation that exchanges the two 1-cycles in these positions.
(This constraint on permutations would throw off the equality between
even and odd permutations, except that these classes of m have other
corner cycles that restore the balance.)  The impossibility of m
commuting with an exchange of the two corners can be verified by
examining the possible orientations, but I haven't got any good way of
characterizing when it would be be a problem in general.  In fact, I
did not notice it until I investigated discrepancies with the
exhaustive computer analysis.

Using the above analysis, we may carry out the calculation as in the
three tables below.  The first two tables count the number of fixed
points of R(m) for an element m of each class, multiply by the class
size, and divide by |J|=48 to get the number of orbits as in the
Polya-Burnside theorem.  The third table calculates the number of
fixed points by combining the results of the first two tables, divided
by the class size (which was multiplied in both for edges and for
corners), and divided by 2 (because only half the combined positions
have matching permutation parity).

     Counting M-conjugacy classes of the edges of Rubik's cube.

M class            Cycles                     Total F.P.      Numeric
  (class size)      of m     Perms   Oris     in class       Total/48
==============  ===========  ======  ======   ==========  ===========
Identity   (1)  12 1-cycles  12!     2^12/2   12! 2^11    20437401600

Axis Rot/2 (3)   6 2-cycles  6! 2^6  2^6      6! 3 2^12        184320

Rot/3      (8)   4 3-cycles  4! 3^4  2^4/2    4! 3^4 2^6         2592

Diag Rot/2 (6)   5 2-cycles  5! 2^5  2^5
                 2 1-cycles  2       2^2/2    5! 3 2^13         61440

Rot/4      (6)   3 4-cycles  3! 4^3  2^3      3! 3 2^10           384

Inv Rot/4  (6)   3 4-cycles  3! 4^3  2^3      3! 3 2^10           384

Diag Ref   (6)   5 2-cycles  5! 2^5  2^5
                 2 1-cycles  2       2^2/2    5! 3 2^13         61440

Inv Rot/6  (8)   2 6-cycles  2! 6^2  2^2      2! 3^2 2^7           48

Axis Ref   (3)   4 2-cycles  4! 2^4  2^4  
                 4 1-cycles  4!      2^4/2    4! 3^2 2^14       73728

Inversion  (1)   6 2-cycles  6! 2^6  2^6      6! 2^12           61440
                                                          -----------
                                        | GE\Conj(M) | =  20437847376


     Counting M-conjugacy classes of the corners of Rubik's cube.

M class            Cycles                   Total F.P.    Numeric
  (class size)      of m     Perms   Oris   in class     Total/48
===============  ==========  ======  =====  ===========   =======
Identity   (1)   8 1-cycles  8!      3^8/3  8! 3^7        1837080

Axis Rot/2 (3)   4 2-cycles  4! 2^4  3^4/3  4! 3^4 2^4        648

Rot/3      (8)   2 3-cycles  2! 3^2  3^2  
                 2 1-cycles  1       3^2/3  3^5 2^4            81

Diag Rot/2 (6)   4 2-cycles  4! 2^4  3^4/3  4! 3^4 2^5       1296

Rot/4      (6)   2 4-cycles  2! 4^2  3^2/3  3^2 2^6            12

Inv Rot/4  (6)   2 4-cycles  2! 4^2  3^2    3^3 2^6            36

Diag Ref   (6)   2 2-cycles  2! 2^2  3^2
                 4 1-cycles  4!      1      4! 3^3 2^4        216

Inv Rot/6  (8)   1 6-cycle   6       3
                 1 2-cycle   1       3      3^3 2^4             9

Axis Ref   (3)   4 2-cycles  4! 2^4  3^4    4! 3^5 2^4       1944

Inversion  (1)   4 2-cycles  4! 2^4  3^4    4! 3^4 2^4        648
                                                          -------
                                        | GC\Conj(M) | =  1841970


    Counting M-conjugacy classes of the entire Rubik's cube

M class             Edge         Corner       Corner times edge
  (class size)      F.P.          F.P.             / (96*class size)
===============  ==========     =========    =======================
Identity   (1)   12! 2^11       8! 3^7       901,083,401,551,872,000

Axis Rot/2 (3)    6! 3 2^12     4! 3^4 2^4               955,514,880

Rot/3      (8)    4! 3^4 2^6    3^5 2^4                      629,856

Diag Rot/2 (6)    5! 3 2^13     4! 3^4 2^5               318,504,960

Rot/4      (6)    3! 3 2^10     3^2 2^6                       18,432

Inv Rot/4  (6)    3! 3 2^10     3^3 2^6                       55,296

Diag Ref   (6)    5! 3 2^13     4! 3^3 2^4                53,084,160

Inv Rot/6  (8)    2! 3^2 2^7    3^3 2^4                        1,296

Axis Ref   (3)    4! 3^2 2^14   4! 3^5 2^4             1,146,617,856

Inversion  (1)    6! 2^12       4! 3^4 2^4               955,514,880
                                             -----------------------
                             | G\Conj(M) | = 901,083,404,981,813,616

These results have been corroborated and expanded by use of
combinatorial computer programs, to be described in a later message.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From @mail.uunet.ca:mark.longridge@canrem.com  Sat Nov  5 23:49:54 1994
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Reply-To: CRSO.Cube@canrem.com
Sender: CRSO.Cube@canrem.com
Subject: Shifty Invariance
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.846.5834.0C1BB9FA@canrem.com>
Date: Sat, 5 Nov 1994 22:16:00 -0500
Organization: CRS Online  (Toronto, Ontario)

----------------------------------------
Even more thoughts on "Shift Invariance"
----------------------------------------

>>Mark continues
>>
>>    Equivalent to (U1 R1)^35= (R1 U1)^35 & Shift Invariant
>>    UR11 = U2 R1 U1 R1 U1 R3 U1 R3 U1 R3 U2 R1 U1 R1 U1 R3 U1 R3 U1 R3
>>           (22 q  or  20 h  moves)
>>
Martin asks:
>Is UR11 the shortest process effecting the ``odd'' element in <U,R>?

After a bit of computer cubing I found:

p183 6 Twist      R1 U3 R2 U3 R1 D3 U3 R1 U3 R3 D2 R3 U3 R1 D3 U3
                  (18 q  or  16 h  moves)

 This requires using the larger group of <U1, R1, D1>, although I
expected a 16 turn process. Note the fact this larger group has face
index 3 (rather than 2). But now the process is NOT shift invariant
and we see the route itself can determine whether it will be
shift invariant!

I welcome any mathematical explanation!

With even more contemplation I noticed that the process for
 the edge 3-cycle

UR1 = U3 R1 U2 (R1 U1)^2 R2 U3 R3 U3 R2 U1     (16 q, 13 h)

...was reducible to

UR1a= F1 U2 (F1 U1)^2 F2 U3 F3 U3 F2           (14 q, 11 h)]

Of course, now we are using <U1, F1> rather than <U1, R1>.

 -> Mark <-
 Email: mark.longridge@canrem.com

From BRYAN@wvnvm.wvnet.edu  Sun Nov  6 09:15:57 1994
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Date:      Sun, 6 Nov 1994 09:15:37 -0500 (EST)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@life.ai.mit.edu>
Subject:   Re: Shifty Invariance
In-Reply-To: Message of 11/05/94 at 22:16:00 from mark.longridge@canrem.com

On 11/05/94 at 22:16:00 mark.longridge@canrem.com said:

>p183 6 Twist      R1 U3 R2 U3 R1 D3 U3 R1 U3 R3 D2 R3 U3 R1 D3 U3
>                  (18 q  or  16 h  moves)
                    ^^^^^^^^^^^^^^^^^^^^^

This is not a shift invariance question, but rather two
questions about your searches.  One question is, do you perform separate
searches for q-turns and h-turns, or only for h-turns?  The reason I
ask is the obvious fact that optimal processes in q-turns need not
contain h-turns.  The second question is, how on earth do you keep track
of all those processes in your searches?  I have been asked how I search
so many positions.  I have answered the question before, but I guess
another part of the answer that I haven't mentioned is that I don't
keep up with processes at all, only positions.  If I am asked to provide
processes, I can do so, but it is a very painful task.  I have thought
about keeping up with processes, but I am quite sure that if I did
so it would reduce the number of positions I could search.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From mschoene@math.rwth-aachen.de  Sun Nov  6 17:31:30 1994
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Date: Sun, 6 Nov 94 23:29 PST
From: Martin.Schoenert@math.rwth-aachen.de
To: cube-lovers@life.ai.mit.edu
Cc: CRSO.Cube@canrem.com
In-Reply-To: Mark Longridge's message of Sat, 5 Nov 1994 22:16:00 -0500 <60.846.5834.0C1BB9FA@canrem.com>
Subject: Re: Shifty Invariance

Mark writes in his e-mail message of 1994/11/05

    After a bit of computer cubing I found:

    p183 6 Twist      R1 U3 R2 U3 R1 D3 U3 R1 U3 R3 D2 R3 U3 R1 D3 U3
                      (18 q  or  16 h  moves)

     This requires using the larger group of <U1, R1, D1>, although I
    expected a 16 turn process. Note the fact this larger group has face
    index 3 (rather than 2). But now the process is NOT shift invariant
    and we see the route itself can determine whether it will be
    shift invariant!

    I welcome any mathematical explanation!

As I tried to explain in my first e-mail message, a shift invariant
process is a process in a subgroup X of G corresponding to an element
x in the centre *of this subgroup*.

The ``odd'' element is an element in the centre of the subgroup < U, R >.
Thus any process effecting this element written in U and R is a shift
invariant process.  UR11 is one such process.

However, the ``odd'' element does not lie in the centre of the subgroup
< U, R, D > (in fact this subgroup has trivial centre).  Thus a process
effecting this element *involving D*, will *not* be shift invariant.

Some shift invariant processes are in fact in the centre of multiple
subgroups.  For example the square elements, except for the ``diagonal
square'' element, have this property.

For such elements one has some choice which generators to use.  For
example the ``single square'' elements (U2 R2)^3 lies in the centre of
< U2, R2 > and < U2, D, R2, L > (and all subgroups inbetween), so every
process effecting this element involving any subset of U2, D, D2, R2, L,
and L2, will be a shift invariant process.

For the ``odd'' element, one has now choice.  It lies in the centre of
< U, R >, but not in the centre of any larger group.  Thus a shift
invariant process effecting the ``odd'' element must involve U and R,
and cannot involve more generators.

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From mschoene@math.rwth-aachen.de  Mon Nov  7 19:20:36 1994
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Date: Tue, 8 Nov 94 01:18 PST
From: Martin.Schoenert@math.rwth-aachen.de
To: Cube-Lovers@life.ai.mit.edu
Cc: hoey@aic.nrl.navy.mil
In-Reply-To: hoey@aic.nrl.navy.mil's message of Fri, 4 Nov 94 11:46:50 EST <9411041646.AA21659@sun13.aic.nrl.navy.mil>
Subject: Re: The real size of cube space

Dan Hoey writes in his e-mail message of 1994/11/04

    In January of this year, Jerry Bryan and I wrote of counting the
    number of M-conjugacy classes of Rubik's cube.  In the sense that (for
    instance) there is really only one position 1 QT from start, even
    though that QT may be applied in twelve different ways, this task
    amounts to counting the true number of positions of the cube.  The
    earlier discussion centered on calculations involving computer
    analysis of large numbers of positions.  However, a look in Paul B.
    Yale's book _Geometry and Symmetry_ gave me a clue: the Polya-Burnside
    theorem is a tool that allows us to perform this calculation by hand.

    ...a very nice application of the Polya-Burnside theorem,
       to compute the number of M-conjugacy classes in G...

Yes, a little bit of group theory can answer many questions arising from
the cube.  In fact I have noticed that quite a few of well known results
in group theory have been rediscovered in this forum.  Note that I don't
think this is a bad thing.  At least for me results that I ``knew'' are
now, that they have been demonstrated for the cube, much easier to grasp
than they were before (grasp is certainly an appropriate term in
connection with the cube).

Dan continues

    For our purpose, we take the group J to be M, the 48-element group of
    symmetries of the cube.  X will be the set of all cube positions,
    which we usually call Gx (for GE, GC, or G, depending on whether we
    consider edges, corners, or both; we are considering the positions
    relative to fixed face centers in all three cases).  And the repre-
    sentation R is the operation of M-conjugation: (R(m))(g) = m' g m.
    Verifying that R is a homomorphism is an exercise in associativity
    that Jim Saxe and I carried out in the Symmetry and Local Maxima
    paper, in the archives [cube-mail-1, 14 December 1980].

The way I view this is as follows.  The entire cube group C is a
permutation group group on 6*9 points, generated by the six face turns U,
D, L, R, F, B; the three middle slice turns M_U, M_L, M_F; and the
reflection S.  This group has a subgroup M of symmetries of the cube (of
order 48), generated by U M_U D', L M_L R', F M_F B', and S.  Another
subgroup is G, generated by the six face turns, which has index 48 in G.
G is a normal divisor of C, G is the semidirect product of M and G.  The
same is true for GE and GC.

Obviously M operates by conjugation on G, and this implies that the
mapping R is a homomorphisms.

Another way to say this is that M is a subgroup of the outer autmorphism
group of G (which in this case can be easily represented as a supplement
of G).  Note that the elements of M are also a autmorphisms of the Cayley
graph.  That means that elements of M respects the length of operations.
That is if g_1 and g_2 are elements of G that are in one conjugacy class
under M, then the lenght of the shortest process effecting them is equal.
This follows from the fact that M fixes the set of the generators of G
and their inverses.  M is fact the largest subgroup of the outer
autmorphism group with this property, which makes it rather important.

Dan continues

    R has been so chosen because we wish to calculate the number of
    M-conjugacy classes of Gx, |Gx\Conj(M)|, which is be the number of
    orbits of R(M).  To apply the Polya-Burnside theorem for this, we need
    to calculate, for each element of m of M, the number of fixed points
    of R(m).  That is the number of elements g of Gx for which m' g m = g.
    Multiplying by m, this becomes g m = m g: the fixed points we wish to
    count are just those elements g of Gx that commute with m.

This set is called the *centralizer* of m in Gx.  Usually the centralizer
in a group X is only defined for elements in X, but it is obvious how to
extend this definition.

Dan continues

    The fundamental principle we use in finding whether g commutes with m
    can be found by examining the cycles of m.  Suppose m permutes a cycle
    (c1,c2,...,ck), so that c2=m(c1), c3=m(c2),...,ck=m(c[k-1]),c1=m(ck).
    ...nice discussion of what must happen to cycles if two
       permutations commute...

This can be used directly to compute the centralizer of an element in the
full symmetric group.  Since G's structure is very similar to a symmetric
group (or more accurately the direct product of two symmetric groups), it
allows to describe the centralizer of an element in G.  The more a group
differs from a symmetric group the less this analysis helps (for those
that know what I'm talking about: the more a group differs from the
symmetric group, the worse a backtrack computation using cycle structure
analysis is).

Dan continues

        Counting M-conjugacy classes of the entire Rubik's cube
    
    M class             Edge         Corner       Corner times edge
      (class size)      F.P.          F.P.             / (96*class size)
    ===============  ==========     =========    =======================

Minor typo.  You don't mean ``Corner times edge / (96 * class size)'' but
``Corner times edge / 96 * class size'', which is in fact what you
computed for the following table.

Dan continues

                             | G\Conj(M) | = 901,083,404,981,813,616

Here is how you compute this value in GAP (excuse me the plug).

    gap-3.4 -b -g 4m 
    gap> Sum( ConjugacyClasses( M ),
    >         c -> Size( Centralizer(G,Representative(c)) ) / 48 * Size(c) );
    901083404981813616

Have a nice day.

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From hoey@aic.nrl.navy.mil  Tue Nov  8 17:59:34 1994
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Date: Tue, 8 Nov 94 17:59:31 EST
From: hoey@aic.nrl.navy.mil
Message-Id: <9411082259.AA05868@sun13.aic.nrl.navy.mil>
To: Martin.Schoenert@math.rwth-aachen.de
Subject: Re: The real size of cube space
Cc: Cube-Lovers@life.ai.mit.edu

Wow, I didn't realize this sort of calculation had been automated.

Martin.Schoenert@math.rwth-aachen.de writes:

    The way I view this is as follows.  The entire cube group C is a
    permutation group group on 6*9 points, generated by the six face
    turns U, D, L, R, F, B; the three middle slice turns M_U, M_L,
    M_F; and the reflection S.  This group has a subgroup M of
    symmetries of the cube (of order 48), generated by U M_U D',
    L M_L R', F M_F B', and S.  Another subgroup is G, generated by
    the six face turns, which has index 48 in G.  G is a normal
                                              ^
    divisor of C, G is the semidirect product of M and G.  The same is
                  ^
    true for GE and GC.

I think two of those G's are supposed to be C's, right?  What is the
difference between a direct product and a semidirect product?

    ... [conjugation by] M fixes the set of the generators of G and
    their inverses.  M is fact the largest subgroup of the outer
    autmorphism group with this property, which makes it rather
    important.

In a 1983 Cubic Circular article (of which I know only Stan Isaacs's
summary) David Singmaster observed that the group is larger for larger
cubes, provided we work what I call the ``theoretical invisible
group''.  That is, we solve not only the surface of the cube, but the
hypothetical interior (n-2)^3 cube, and all the smaller (n-2k)^3 cubes
as well.  I blithered at length about this in my article of 1 June
1983 archived (I think I've got it right this time) at
<ftp://ftp.ai.mit.edu/pub/cube-lovers/cube-mail-5.gz>.

The idea is that a mapping called evisceration allows us to permute
the layers of the cube.  On the 4x4x4 cube, this for instance allows
us to exchange each inner slab with its adjacent outer slab.  It also
allows us to conjugate each inner slab move by central inversion,
while leaving the outer slab moves alone.  In general, evisceration of
a d-dimensional cube by f maps each feature (cubie, colortab, or
face-center arrow) at coordinates (x[1],x[2],...,x[d]) to
(f(x[1]),f(x[2]),...,f(x[d])), where f is a permutation of the
intervals between the cleavage coordinates of the cube.  I believe
that if f commutes with the central inversion, then conjugation by
evisceration is an outer automorphism of the Rubik's cube group.  (I
think I have proved this for d=3, and I think the proof in higher
dimensions should not be difficult given the right notation.)

The group of all eviscerations includes the central inversion; we can
of course augment it by the rotation group in d-space.  Is this the
maximum outer automorphism group that respects generators of the
Rubik's cube?  For this we take the generators to be turns of slabs
between adjacent cleavage planes.  (Turns are direct d-1-dimensional
isometries.)

I was already familiar with this augmented symmetry group because it
also induces automorphisms on d-dimensional tic-tac-toe.  (In fact, it
may be the maximal automorphism group on all tic-tac-toe boards of
side greater than two.  I know it's been proven for 4^3, but I don't
know of any larger results).  Do you know anything more about this
group, like whether it has been named or studied?

    Since G's structure is very similar to a symmetric group (or more
    accurately the direct product of two symmetric groups), it allows
    to describe the centralizer of an element in G.  The more a group
    differs from a symmetric group the less this analysis helps (for
    those that know what I'm talking about: the more a group differs
    from the symmetric group, the worse a backtrack computation using
    cycle structure analysis is).

But no, G's structure is actually similar to the direct product of two
_wreathed_ symmetric groups.  Does this interfere with the
backtracking as much as it interferes with my manual analysis?  Do you
know of any good treatments of finding centralizers of outer
automorphisms of wreath products?  In particular, I would very much
like to know under what conditions the centralizer of the wreath
product fails to cover the centralizer of the permutation factor, as
we saw with the corners.

As for when I wrote

    M class             Edge         Corner       Corner times edge
      (class size)      F.P.          F.P.             / (96*class size)
                                                  ^^^^^^^^^^^^^^^^^^^^^^

That's not a typo.  I was just saying that column 4 is equal to column
2 times column 3, divided by column 1, divided by 96.  Perhaps I
should have factored column 1 out of columns 2 and 3 first to avoid
this confusion.

    gap-3.4 -b -g 4m 
    gap> Sum( ConjugacyClasses( M ),
    >         c -> Size( Centralizer(G,Representative(c)) ) / 48 * Size(c) );
    901083404981813616

Well, call me John Henry.  Say, do you have gap libraries for other
magic polyhedra?  For higher-dimensional magic?

Dan Hoey
Hoey@AIC.NRL.Navy.MIl

From BRYAN@wvnvm.wvnet.edu  Tue Nov  8 21:23:01 1994
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Date:      Tue, 8 Nov 1994 21:22:37 -0500 (EST)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <Cube-Lovers@life.ai.mit.edu>
Subject:   Re: The real size of cube space
In-Reply-To: Message of 11/08/94 at 01:18:00 from ,
           Martin.Schoenert@math.rwth-aachen.de

On 11/08/94 at 01:18:00 Martin.Schoenert@math.rwth-aachen.de said:

>The way I view this is as follows.  The entire cube group C is a
>permutation group group on 6*9 points, generated by the six face turns U,
>D, L, R, F, B; the three middle slice turns M_U, M_L, M_F; and the
>reflection S.  This group has a subgroup M of symmetries of the cube (of
>order 48), generated by U M_U D', L M_L R', F M_F B', and S.  Another
>subgroup is G, generated by the six face turns, which has index 48 in G.
>G is a normal divisor of C, G is the semidirect product of M and G.  The
>same is true for GE and GC.

I have discussed a similar view of things recently, except that I was
not brave enough to include a reflection in the generators.  C is
normally used to denote the set of twenty-four rotations of the
cube (a sub-group of M), so let's call your "entire cube group"
big_G instead.  My version of big_G was generated by Q plus the
slice moves (like yours without the reflection), or alternatively
by Q plus C.  Your version of big_G is hence the same as the one
I discussed except that you added a reflection.  C (the rotations C,
that is) is a sub-group of both versions of big_G.  M is a sub-group
of your version of big_G, but not of mine.

Your big_G has the obvious advantage of including M as a sub-group.
Mine has the advantage (?) of being physically realizable on a
real cube.  That is, for X in your big_G, rX or Xr (r is a reflection)
is also in your big_G.  For X in my big_G, rX or Xr is not in
big_G, and correspondingly a single reflection is not physically
realizable on a real cube.  Of course, r'Xr is in big_G in either
case, r being in M.  Also, cX and Xc are in either version of big_G
for all c in C.

I tend to think that Singmaster's standard G=<Q> is not what people
think of when they hold a real cube in their hand.  Rather, they
tend to think of big_G/C.  That is, the cosets of C in big_G are
common sensically considered to be equivalent because rotating
a real cube in space is "doing nothing".  Also, for my version
of big_G we have |big_G/C| = |G|.

For either version of big_G, we have to re-interpret
parity arguments slightly.  In Singmaster's G=<Q>, we say that
even corners occur only with even edges and vice versa.  In
big_G, a face quarter-turn is odd on the corners and edges, and
a slice quarter-turn is odd on the edges and on the centers.
Hence, you can have odd corners with even edges and vice versa,
but only if the centers are simultaneously odd.  Therefore, the
rules concerning which configurations of edges and corners can
occur together are really preserved, even in big_G.

Finally, neither version of big_G is as big as you can go.  That is,
neither of them includes Singmaster's Supergroup, where different
orientations of the otherwise fixed face centers are considered.
Also, neither one of them considers Dan Hoey's Eccentric Slabism,
wherein invisible inner cubes are considered.

>        Note that the elements of M are also a autmorphisms of the Cayley
>graph.  That means that elements of M respects the length of operations.
>That is if g_1 and g_2 are elements of G that are in one conjugacy class
>under M, then the lenght of the shortest process effecting them is equal.
>This follows from the fact that M fixes the set of the generators of G
>and their inverses.  M is fact the largest subgroup of the outer
>autmorphism group with this property, which makes it rather important.

This of course is the basis for the large searches I have been able
to perform using M-conjugate classes.  The only trouble is, I don't
even know what a Cayley graph is (but I am working on it), the last
course I took in group theory being 25 years ago.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From devo@vnet.ibm.com  Wed Nov  9 13:42:09 1994
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Date: Wed, 9 Nov 94 13:18:13 EST
From: "Dave Eaton" <devo@vnet.ibm.com>
To: cube-lovers@life.ai.mit.edu
Subject: Re: Is there a symbolic cube program?

In response to my request for a algebraic cube simulator,
I have found out about the following:

  Rubik Algebra, a $10 shareware DOS program that displays
   a color picture of the cube on the left and a list
   of choices (rotate a face, library of moves, scramble)
   on the right.  It accepts a text string of moves
   similar to Singmaster notation and displays the
   resulting cube in 3D.  There is an option that will
   tell you the cycle decomposition of the current
   state.  So, this program provides the function I
   requested and I will have to play with it to see
   if the graphical cube and menus make this too hard
   to use.  Nonetheless, my brief trial of the program
   suggests that this is a good, straightforward tool
   to fiddle with and analyze the cube.  This was
   mentioned by Warut Roonguthai <kamala@chulkn.chula.ac.th>.

  Maple and X-Maple were suggested as symbolic algebra
   programs that could handle this type of task, but
   I have no further understanding of these to know
   how slick they would be.  This was mentioned by
   Brett Stevens <brett@math.toronto.edu>.  There are
   surely other symbolic algebra programs, but I don't
   know of them.

  Roll-your-own, the approach I (we all?) should use.
   I think we could build a suite of text-based tools
   written in "standard" portable C, that allows for:
     - input of move sequences
     - display of cyclical decomposition
     - definition of compound moves that can be
        used just like a standard move
     - one-shot execution from the commandline or
        running move mode
     - find solution(s) from current state
     - randomize/scramble
     - other analysis of current state, like some
        of the mathematics and numbers that have
        been discussed in this newsgroup
     - other size and shape cubes?
   If folks want to do this, then I suggest that the
   eager and capable coders who dive in first ought
   to try real hard to make a system that can be
   driven from other programs (such as a windowing
   GUI display program with graphical cube).  If for
   example the current state of the cube was stored
   in a file current.cub, then a program called cube
   could be called like "cube r2u3r2u" or "cube r2xr2"
   where 'x' is a defined move--defined by
   "cube define x rfr3f3" and that got stored in
   a file library.cub.  The other functions could be
   separate programs (which read the same files):
      cubehome
      cuberand
      cubesolv
   If I wrote this, I would have a hard time using
   C instead of REXX.

In the absence of finding exactly what I want, I
will be experimenting with Rubik Algebra and deciding
whether I ought to start writing something like the
roll-your-own described above.

Reply to this group if you know anything more.  Thanks.
Thanks for the pointers to this information and the
encouragement to give it a shot myself.  Maybe this would
be a good task for my holidays and my little remaining
vacation.  If I do something, I'll let you know.

......Dave Eaton, N2NOQ, Owego NY, devo@vnet.ibm.com


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From: "Derek Bosch" <bosch@smiteo.esd.sgi.com>
Message-Id: <9411091116.ZM5813@smiteo.esd.sgi.com>
Date: Wed, 9 Nov 1994 11:16:32 -0800
In-Reply-To: "Dave Eaton" <devo@vnet.ibm.com>
        "Re: Is there a symbolic cube program?" (Nov  9,  1:18pm)
References: <9411091842.AA08633@life.ai.mit.edu>
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To: "Dave Eaton" <devo@vnet.ibm.com>, cube-lovers@life.ai.mit.edu
Subject: Re: Is there a symbolic cube program?
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I have a symbolic cube program, which I didn't write (written by
raymond@cps.msu.edu).  It does pretty much what you want, although I'm
not too sure how portable it is.  It was written in C, using yacc and flex (lex
doesn't work on it tho, go figure.)

I could post it to the group, or send it to individuals that wanted it.

Derek


-- 
Derek Bosch                   "Time flies like an arrow,
(415) 390-2115                 but fruit flies like bananas"
bosch@sgi.com                  J. Blaylock


From raymond@cps.msu.edu  Wed Nov  9 15:28:24 1994
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To: cube-lovers@ai.mit.edu
From: raymond@cps.msu.edu (Carl Raymond)
Subject: Re: Is there a symbolic cube program?
X-Mailer: <PC Eudora Version 1.4>

>I have a symbolic cube program, which I didn't write (written by
>raymond@cps.msu.edu).  It does pretty much what you want, although I'm
>not too sure how portable it is.  It was written in C, using yacc and flex (lex
>doesn't work on it tho, go figure.)
>
>I could post it to the group, or send it to individuals that wanted it.
>
>Derek
>
>
>-- 
>Derek Bosch                   "Time flies like an arrow,
>(415) 390-2115                 but fruit flies like bananas"
>bosch@sgi.com                  J. Blaylock
>

  I'd like a copy, please.  I wrote it a while back, and now I can't find it 
anywhere!  Maybe the Internet is turning into the ultimate backup system :-)

Thanks,

Carl


From bosch@smiteo.esd.sgi.com  Wed Nov  9 15:53:26 1994
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From: "Derek Bosch" <bosch@smiteo.esd.sgi.com>
Message-Id: <9411091253.ZM6007@smiteo.esd.sgi.com>
Date: Wed, 9 Nov 1994 12:53:14 -0800
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To: cube-lovers@ai.mit.edu
Subject: Symbolic cube program
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Since I received many requests for it, here it is.  It is a
uuencoded, compressed, tar file.

Derek
-- cut here
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MN,5#.?SHX@"\@^D##0;&2%,81U>@+(.=<@_.L\=X"PMA.W".N[)7_LI@.2R+
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M7_[+@#DP"^;!3)@+LV$^S(@Y,2OFQ<R8&[-C?LR0.3)+YLE,F2NS9;[,F#DS
<:^;-S)D[LV?^S*`Y-(OFT4R:2[-I/LVH.34[!,R0
`
end


-- 
Derek Bosch                   "Time flies like an arrow,
(415) 390-2115                 but fruit flies like bananas"
bosch@sgi.com                  J. Blaylock


From hoey@aic.nrl.navy.mil  Thu Nov 10 08:43:46 1994
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Date: Thu, 10 Nov 94 08:43:35 EST
From: hoey@aic.nrl.navy.mil (Dan Hoey)
Message-Id: <9411101343.AA22346@Sun0.AIC.NRL.Navy.Mil>
To: "Derek Bosch" <bosch@smiteo.esd.sgi.com>
Cc: cube-lovers@life.ai.mit.edu
Subject: Re: Symbolic cube program

> Since I received many requests for it, here it is.  It is a
> uuencoded, compressed, tar file.

The cube lovers archives contain over two megabytes of readable
discussions on the cube.  In the interests of keeping them readable, I
ask that binary files and programs be distributed using some other
channel: private email, ftp, web, or whatever.  Please let Cube-Lovers
remain a _discussion_ list for cube topics.

I am aware that alternative distribution methods are not supplied
by Alan Bawden in his maintenance of the Cube-Lovers list and
archives.  Please consider that encouragement to provide such
methods, rather than an excuse to use this list for the purpose.

I am also aware that the recent program was not much larger than some
of the discussion articles on the list.  Nonetheless, it does not
contribute to the discussion, and imposes a burden on those of us who
subscribe for the discussion.  The revised versions and larger
programs to follow would impose a growing burden on the list.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From @mail.uunet.ca:mark.longridge@canrem.com  Fri Nov 11 16:55:11 1994
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To: cube-lovers@life.ai.mit.edu
Reply-To: CRSO.Cube@canrem.com
Sender: CRSO.Cube@canrem.com
Subject: <U, R> Searches
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.858.5834.0C1BE0A3@canrem.com>
Date: Fri, 11 Nov 1994 15:36:00 -0500
Organization: CRS Online  (Toronto, Ontario)


Jerry asks, in his message of Sun, 6 Nov 1994 09:15:37 (EST):

>This is not a shift invariance question, but rather two
>questions about your searches.  One question is, do you perform
>separate searches for q-turns and h-turns, or only for h-turns?

The <U, R> group searches are q-turns only, but if I see a way to
compress it a bit by eye I do so. For example:

UR2 = U3 R3 U3 R2 U1 R1 U1 R3 U3 R1 U1 (R1 U3)^2 R3 U3

...was reduced to

UR2 = U2 R3 U3 R2 U1 R1 U1 R3 U3 R1 U1 (R1 U3)^2 R3

> The second question is, how on earth do you keep track
> of all those processes in your searches?

 With the computer hardware I'm using (486-DX40 with 4 megs) and
my current algorithm I created a file "ur.dat" which is a flat
ascii file. It contains all the processes which generate distinct
positions up to 12 q turns. I also have the file "ursum.is" which
contains all the `cubesums' for each element of <U, R>.

 My program "x3bin.exe" loads the "ursum.is" database into memory
and it does a binary search on the cubesums to try and find a
match for the current position. If not found, it turns the
current cube and tries again, with longer and longer sequences
until a match is found.

 Using this method I have found a process for all <U, R> positions
with the longest being 22 q turns (so far). The hardest positions
can take as long as a couple of hours. I have no idea what the
antipodes look like at this time, but I'll probably try the
random approach soon.


Jerry continues:

>p183 6 Twist      R1 U3 R2 U3 R1 D3 U3 R1 U3 R3 D2 R3 U3 R1 D3 U3
>                  (18 q  or  16 h  moves)
                    ^^^^^^^^^^^^^^^^^^^^^

 This process was found using the Kociemba algorithm in a program
written by Dik Winter, which I ran on a Sun4. This program uses
h turns in it's searches and uses all 6 generators.

 After inspecting the original process found by the program,
I was able to manually reduce it somewhat, resulting in p183 above.

The original process....

Solution (13+ 4=17): R1 U1 D1 R3 U1 R1 D2 R1 U1 R3 U1 D1 R3 U1 R2 U1 R2
That is (or as Dan would say i.e.) 13 h turns in phase 1 and 4 turns
in phase 2. Hmmm, looks like I used the inverse of the original.


Jerry Continues:

>I have been asked how I search
>so many positions.  I have answered the question before, but I guess
>another part of the answer that I haven't mentioned is that I don't
>keep up with processes at all, only positions. If I am asked to provide
>processes, I can do so, but it is a very painful task.  I have thought
>about keeping up with processes, but I am quite sure that if I did
>so it would reduce the number of positions I could search.

 Well, that explains the fact you counted the number of antipodes but
had no processes for them, but do you know what they look like?
If you can tell me what one of the 87 positions at 25 q turns
looks like, I should be able to generate a sequence for it.

-> Mark <-
Email: mark.longridge@canrem.com

From BRYAN@wvnvm.wvnet.edu  Sat Nov 12 00:59:49 1994
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X-Acknowledge-To: <BRYAN@WVNVM.WVNET.EDU>
Date:      Fri, 11 Nov 1994 21:04:45 -0500 (EST)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Re: <U, R> Searches
In-Reply-To: Message of 11/11/94 at 15:36:00 from mark.longridge@canrem.com

On 11/11/94 at 15:36:00 mark.longridge@canrem.com said:

> Well, that explains the fact you counted the number of antipodes but
>had no processes for them, but do you know what they look like?
>If you can tell me what one of the 87 positions at 25 q turns
>looks like, I should be able to generate a sequence for it.

The 87 positions correspond to 27 antipodes which are
unique up to W-conjugacy.  Here they are:

     -----------------------------------------------------

     BRL                      BRL                      BRU
     BBU                      BBU                      BBF
     LBF                      RBF                      LBR

     UDR                      FDU                      UBD
     BUU                      BUU                      DUU
     FUB                      FUR                      UUF

 FRD RFR DRU              URD RFB URL              FRB LFD RLB
 LLL FFF RRB              LLL FFF RRB              LLL FFU RRR
 LLL FFB RLU              LLL FFD RLU              LLL FFF UBR

     DDU                      DDB                      DDR
     DDU                      DDU                      DDU
     DDB                      DDB                      DDB

  -----------------------------------------------------------

     BRU                      BRR                      BBU
     BBF                      BBU                      BBF
     DBR                      BBB                      FBD

     FDU                      RDL                      LUB
     UUB                      FUB                      UUU
     BUR                      RUL                      LUF

 RBR DFB URF              URU FFF URU              ULU BFD RRR
 LLL FFU LRR              LLL FFU LRR              LLL FFB RRR
 LLL FFL BRL              LLL FFR BBF              LLL FFU RRR

     DDU                      DDD                      DDF
     DDU                      DDU                      DDD
     DDF                      DDD                      DDB

  --------------------------------------------------------

     BRR                      BLU                      BRB
     BBU                      BBU                      BBU
     LBD                      RBF                      FBF

     UDB                      DUR                      UFD
     UUF                      FUD                      DUU
     UUR                      BUU                      UUR

 FRB LFF DRR              FRR DFR BRU              RRL FLB URR
 LLL FFB RRB              LLL FFB RRR              LLL FFU FRB
 LLL FFU RLB              LLL FFU LBL              LLL FFB URR

     DDF                      DDB                      DDL
     DDU                      DDU                      DDB
     DDU                      DDF                      DDD

  --------------------------------------------------------

     BFR                      BRR                      BRL
     BBF                      BBU                      BBU
     FBR                      UBU                      DBF

     LUB                      FFF                      FFR
     UUU                      DUU                      DUU
     RBR                      BBB                      DBR

 ULD BRU FRU              LRL URU RRR              RRB RRB URU
 LLL FFU BRR              LLL FFU BRF              LLL FFU BRF
 LLL FFB URF              LLL FFR BLF              LLL FFU LLF

     DDL                      DDD                      DDB
     DDD                      DDU                      DDU
     DDD                      DDD                      DDU

  --------------------------------------------------------

     BRR                      BRF                      BRF
     BBU                      BBU                      BBU
     UBU                      FBF                      FBF

     FFF                      RDR                      RDR
     UUU                      FUU                      FUU
     BBB                      DBR                      BBR

 LRL URU RLR              DRB RRB URU              DRR DRB URU
 LLL FFU BRF              LLL FFU BRL              LLL FFU BRL
 LLL FFR BRF              LLL FFU LFU              LLL FFL BFU

     DDD                      DDB                      DDU
     DDD                      DDU                      DDU
     DDD                      DDL                      DDL

  --------------------------------------------------------

     BRF                      BRR                      BRR
     BBU                      BBU                      BBU
     DBL                      UBR                      FBD

     RFU                      FFB                      RDB
     UUD                      DUU                      FUU
     RBR                      RBF                      UBL

 BRF DRB URB              LRD BRR UBU              DRB LRF URR
 LLL FFU BRL              LLL FFU RRL              LLL FFU FRB
 LLL FFU RFU              LLL FFB UFF              LLL FFR ULU

     DDF                      DDL                      DDB
     DDU                      DDU                      DDU
     DDL                      DDD                      DDF

  --------------------------------------------------------

     BRB                      BRU                      BRU
     BBU                      BBU                      BBU
     FBL                      RBF                      BBR

     RDF                      BFR                      UUU
     FUU                      DUU                      DUF
     UBB                      FBU                      UBD

 DRF RRL URU              DRD RRL FLU              LRL FRR FRF
 LLL FFU FRB              LLL FFU FRB              LLL FFU FRB
 LLL FFD RLU              LLL FFB URR              LLL FFD RLR

     DDB                      DDL                      DDB
     DDU                      DDU                      DDU
     DDR                      DDB                      DDB

  --------------------------------------------------------

     BRB                      BFB                      BRD
     BBU                      BBF                      BBU
     UBU                      UBD                      LBF

     BUF                      RUR                      UBR
     DUF                      DUU                      DUU
     UBU                      UBR                      UFR

 RRB LRL FRR              FRL FRB URF              FRB LRB URU
 LLL FFU FRB              LLL FFU LRR              LLL FFU FRB
 LLL FFF RLR              LLL FFB UBR              LLL FFD RLR

     DDD                      DDL                      DDB
     DDU                      DDU                      DDU
     DDD                      DDD                      DDF

  --------------------------------------------------------

     BFD                      BFF                      BFR
     BBF                      BBF                      BBF
     BBB                      UBF                      LBR

     RUL                      FUR                      UUD
     UUU                      UUU                      UUU
     RUL                      BUR                      UUD

 URU FBF ULU              LRL UBB ULU              FRB LBR FLB
 LLL FFB RRR              LLL FFB RRR              LLL FFB RRR
 LLL FFD RRR              LLL FFB DRD              LLL FFR FRB

     DDB                      DDR                      DDU
     DDD                      DDD                      DDD
     DDF                      DDR                      DDU



 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From BRYAN@wvnvm.wvnet.edu  Sat Nov 12 01:48:27 1994
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Date:      Fri, 11 Nov 1994 22:06:49 -0500 (EST)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@life.ai.mit.edu>
Subject:   Re: <U, R> Searches
In-Reply-To: Message of 11/11/94 at 15:36:00 from mark.longridge@canrem.com

On 11/11/94 at 15:36:00 mark.longridge@canrem.com said:

> With the computer hardware I'm using (486-DX40 with 4 megs) and
>my current algorithm I created a file "ur.dat" which is a flat
>ascii file. It contains all the processes which generate distinct
>positions up to 12 q turns. I also have the file "ursum.is" which
>contains all the `cubesums' for each element of <U, R>.

I have considered keeping a file of processes, either in parallel
with my file of positions, or together in the same file.  However,
I have always rejected the idea because it would take too much
storage for the very large searches I want to accomplish.
Also, since I only store representative elements of equivalent
classes in my data base, it is hard to know what a "process"
means (see below).

> Using this method I have found a process for all <U, R> positions
>with the longest being 22 q turns (so far). The hardest positions
>can take as long as a couple of hours. I have no idea what the
>antipodes look like at this time, but I'll probably try the
>random approach soon.

Interesting approach, and very different than what I do.  I simply
do a "game theory" type tree search of positions (*not* processes)
with Start at the root of the tree.  I have to search the whole
depth of the tree rather than half the depth of the tree, as you do.
(I could obviously verify the depth of one particular position with
two half-depth searches, but I confess I have never been very
interested in "particular positions".  I want to search the whole
tree.)

I tend to think of the processing required for whole tree searches
as "global" because you cannot determine anything about
one particular position except in the context of the other
positions down to the same level of the tree.  By contrast, there
is some interesting processing that you can do that is "local";
e.g.,  conjugate class sizes, symmetry group determination, etc.
can be determined on a position by position basis without
regard to any other position.  Such "local" processing is generally
much easier than the "global" processing.  "Local" processing
is O( N ) and requires essentially no memory. "Global" processing
is O( N log(N) ) if you are efficient, maybe O(N^2) if you are not,
and is very consumptive of memory.  As I have said before, I solve
the memory problem by externalizing the data to files and sorting
the files.

Saying that I have a hard time converting positions into processes
doesn't mean that I cannot do so.  The process for a given position
can be extracted from a data base of positions by backtracking
from the position back to Start.  However, I have two difficulties.

One is that I store only representative elements of equivalence
classes (of the form {m'Xmc} for centerless cubes, or M-conjugate
classes of the form {m'Xm} for cubes with centers)}.  That means
that as you backtrack, the process you are determining (backwards)
will rotate and reflect out from under you.  For example, suppose we
have Repr{m'Xm}=f'Xf for some fixed f in M.  If Xq for some q in Q
is a neighbor of X which is one move closer to Start than X,
then we might have Repr{m'(Xq)m}=g'(Xq)g for some fixed g in M.
But f and g need not be, and usually aren't, the same.

It might be noted in passing that there is duality between the
symmetry of a position X and a process P=p1 p2 ... pn which produces
X.  That is, if f is a fixed element of M, then we have
f'Xf = f'Pf = f'(p1)f f'(p2)f ... f'(pn)f.

The second problem is that my files are so large that I have to
keep them on magnetic tape.  If I need to find a process, I first
find a sequence of positions (really, a sequence of representative
elements) on the tapes (hard to search tapes!), and copy the
positions to a small disk file.  From that point, it is not
especially hard to run a program to display the positions in
sequence and determine the process.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From BECK@vax88a.pica.army.mil  Mon Nov 14 08:36:10 1994
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Date: Mon, 14 Nov 1994 8:03:53 -0500 (EST)
From: BECK@vax88a.pica.army.mil
To: Cube-Lovers@ai.mit.edu
Message-Id: <941114080353.20200253@LCSS.PICA.ARMY.MIL>
Subject: SYMBOLIC PROGRAMS


I have an article that discusses a Rubik's cube
learning program written in GPS.  I do not have
the program.

ref:  A PROGRAM THAT LEARNS TO SOLVE RUBIK'S CUBE
     RICHARD KORF
    DEPT OF COMPUTER SCIENCE
    CARNEGIE-MELLON
    PITT, PA 15213

DTIS :   ARPA ORDER  3597
        AF AVIONICS LAB CONTRACT  F33615-81-K-1539


From devo@vnet.ibm.com  Mon Nov 14 09:03:16 1994
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Date: Mon, 14 Nov 94 08:48:00 EST
From: "Dave Eaton" <devo@vnet.ibm.com>
To: cube-lovers@life.ai.mit.edu
Subject: SYMBOLIC PROGRAMS

I couldn't quickly get a LEX or FLEX working to process the source
of the symbolic cube program written by Carl Raymond
(raymond@cps.msu.edu).  Carl's note said "do what you want...",
so, I did.

I managed to convert his C-and-LEX cube program into REXX.  You
type in a move sequence and it displays the resulting cycle
structure.  Exactly what I have always wanted.  I am still
testing it and working the bugs out, but now I can return to
my "cube research" (that is, playing around).

If anyone else needs such a straightforward text-mode cube program
in REXX (instead of C), let me know and I will share it with you...
after a little more testing.  (Contact me directly, not in this
group.)  Interpreted REXX is built in on OS/2, VM, MVS, and is
available (somewhere) for Unix systems, Amiga, and probably others.

A million thanks to Carl.

......Dave Eaton, N2NOQ, Owego NY, devo@vnet.ibm.com


From @mail.uunet.ca:mark.longridge@canrem.com  Mon Nov 14 13:19:32 1994
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Reply-To: CRSO.Cube@canrem.com
Sender: CRSO.Cube@canrem.com
Subject: <U, R> Antipode!
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.864.5834.0C1BE7A7@canrem.com>
Date: Mon, 14 Nov 1994 11:59:00 -0500
Organization: CRS Online  (Toronto, Ontario)

 With Jerry's help I have found a process for one of the <U, R>
antipodes! Note the number of turns required in the h turn metric.
I applied the turn R1 to the position so the program only needed to
search 24 q turns deep.


First Antipodal Process    (25 q, 20 q)
UR13 = U2 R1 U3 R2 U3 R2 U3 R1 U2 R3 U1 R3 U3 R1 U1 R2 U1 R3 U1 R3

Also....      UR13 ^ 2 = I

Using current techniques, this required a run of about 11 hours.

 It remains to be seen how the Kociemba algorithm resolves this
position, and I will try this next.

 Although somewhat unrelated, I found a square's group process
for 6 X order 2 (the pons asinorum) by hand, which does
not move the centres:

                U2 B2 L2 U2 D2 L2 F2 T2
                F2 U2 L2 F2 B2 L2 D2 F2   (16 h, 32 q)

 -> Mark <-
 email: mark.longridge@canrem.com

From BRYAN@wvnvm.wvnet.edu  Mon Nov 14 15:28:55 1994
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Date:      Mon, 14 Nov 1994 15:05:12 -0500 (EST)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@life.ai.mit.edu>
Subject:   Re: <U, R> Antipode!
In-Reply-To: Message of 11/14/94 at 11:59:00 from mark.longridge@canrem.com

Of the 54 antipodes under Q+H processing, only
16 are unique up to W-conjugacy, with all 16 antipodes being
20 q+h moves from Start.  Also, there is only one cube position which
is both one of the 27 antipodes under Q (25 moves from Start) and also one
of the 16 antipodes under Q+H (20 moves from Start).

Just for completeness, here are the 16 Q+H antipodes.




     BBR                     BBB                      BBL
     BBB                     BBB                      BBB
     FBU                     LBD                      RBU

     RUR                     BUB                      BUR
     UUU                     UUU                      UUU
     RUU                     RUU                      DUF

 DLU FFL FRB             ULU FFL FRR              DLR FFR URB
 LLL FFF RRR             LLL FFF RRR              LLL FFF RRR
 LLL FFU LRD             LLL FFD FRU              LLL FFL URU

     DDB                     DDR                      DDF
     DDD                     DDD                      DDD
     DDB                     DDR                      DDB

     ----------------------------------------------------


     BBD                     BBU                      BBR
     BBB                     BBB                      BBB
     LBR                     LBR                      RBB

     UUF                     UUB                      UUL
     UUU                     UUU                      UUU
     BUR                     BUR                      LUR

 FLL UFU FRD             FLL UFD BRU              BLF UFF DRU
 LLL FFF RRR             LLL FFF RRR              LLL FFF RRR
 LLL FFR URB             LLL FFR DRF              LLL FFU RRD

     DDB                     DDF                      DDF
     DDD                     DDD                      DDD
     DDR                     DDR                      DDB


     ----------------------------------------------------

     BBL                     BBR                      BBU
     BBB                     BBB                      BBB
     RBR                     LBU                      FBB

     BUU                     BUR                      RUR
     UUU                     UUU                      UUU
     LUB                     LUF                      FUR

 DLF UFU RRF             ULF UFR URB              DLU LFU FRD
 LLL FFF RRR             LLL FFF RRR              LLL FFF RRR
 LLL FFD FRU             LLL FFD FRD              LLL FFL BRR

     DDR                     DDR                      DDU
     DDD                     DDD                      DDD
     DDB                     DDB                      DDB


     ----------------------------------------------------

     BBF                     BFU                      BRB
     BBF                     BBD                      BBU
     FBR                     LBR                      BBR

     LUU                     UUD                      UFB
     UUU                     UUB                      DUU
     UUR                     LUF                      UBF

 ULB LFB URF             FBU BLD RRB              LRL FRR ULU
 LLL FFB RRR             LLL FFU RRR              LLL FFU BRF
 LLL FFR BRD             LLL FFR FRR              LLL FFF RRR

     DDD                     DDU                      DDD
     DDD                     DDF                      DDU
     DDR                     DDB                      DDD

     ----------------------------------------------------


     BRL                     BRB                      BFB
     BBU                     BBU                      BBU
     DBF                     FBF                      FBD

     FFR                     RDR                      RUB
     DUU                     FUU                      UUF
     DBR                     UBB                      FDU

 RRB RRB URU             DRL FRL URU              DLU LRF RRR
 LLL FFU BRF             LLL FFU FRB              LLL FFB RRR
 LLL FFU LLF             LLL FFR ULR              LLL FFU LBU

     DDB                     DDB                      DDB
     DDU                     DDU                      DDU
     DDU                     DDD                      DDR

     ---------------------------------------------------------


     BFF
     BBU
     UBB

     LUU
     UUF
     BDR

 BLR DRF DRR
 LLL FFB RRR
 LLL FFU RBU

     DDF
     DDU
     DDL



 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From dik@cwi.nl  Mon Nov 14 17:41:11 1994
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Date: Mon, 14 Nov 1994 23:41:00 +0100
From: Dik.Winter@cwi.nl
Message-Id: <9411142241.AA02504=dik@boring.cwi.nl>
To: CRSO.Cube@canrem.com, cube-lovers@life.ai.mit.edu
Subject: Re:  <U, R> Antipode!

 > First Antipodal Process    (25 q, 20 q)
 > UR13 = U2 R1 U3 R2 U3 R2 U3 R1 U2 R3 U1 R3 U3 R1 U1 R2 U1 R3 U1 R3

 > Also....      UR13 ^ 2 = I

 > Using current techniques, this required a run of about 11 hours.

 >  It remains to be seen how the Kociemba algorithm resolves this
 > position, and I will try this next.

You mean: F2 U3 D2 L3 D3 R1 U2 B2 R1 B2 R3 D1 L1 D3 R2 U1 D3 (or rather
its inverse)?  Took Kociemba's algorithm 10 minutes.  I do not yet
know whether this is minimal.

From BRYAN@wvnvm.wvnet.edu  Tue Nov 15 08:47:23 1994
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Date:      Tue, 15 Nov 1994 08:46:59 -0500 (EST)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@life.ai.mit.edu>
Subject:   Re: <U, R> Antipode!
In-Reply-To: Message of 11/14/94 at 23:41:00 from Dik.Winter@cwi.nl

On 11/14/94 at 23:41:00 Dik.Winter@cwi.nl said:

>You mean: F2 U3 D2 L3 D3 R1 U2 B2 R1 B2 R3 D1 L1 D3 R2 U1 D3 (or rather
>its inverse)?  Took Kociemba's algorithm 10 minutes.  I do not yet
>know whether this is minimal.

Are you applying Kociemba's algorithm to the <U,R> antipodal positions
in the context of <U,R> or in the context of G?  The lengths of these
antipodal positions are already known to be minimal in <U,R>.
However (and obviously), the length in <U,R> can only be claimed to
be an upper bound for the length in G without further testing in G.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From dik@cwi.nl  Tue Nov 15 09:33:44 1994
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From: Dik.Winter@cwi.nl
Message-Id: <9411151433.AA04373=dik@boring.cwi.nl>
To: BRYAN@wvnvm.wvnet.edu, cube-lovers@life.ai.mit.edu
Subject: Re: <U, R> Antipode!

 > On 11/14/94 at 23:41:00 Dik.Winter@cwi.nl said:

 > >You mean: F2 U3 D2 L3 D3 R1 U2 B2 R1 B2 R3 D1 L1 D3 R2 U1 D3 (or rather
 > >its inverse)?  Took Kociemba's algorithm 10 minutes.  I do not yet
 > >know whether this is minimal.

 > Are you applying Kociemba's algorithm to the <U,R> antipodal positions
 > in the context of <U,R> or in the context of G?

In the context of G.  I have no idea what Kociemba's algorithm in the
context of <U,R> should be; hence my question.

dik

From BRYAN@wvnvm.wvnet.edu  Sun Nov 20 12:56:36 1994
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Date:      Sun, 20 Nov 1994 12:56:26 -0500 (EST)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <dlitwin@geoworks.com>
Cc: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Re: Antipode
In-Reply-To: Message of 11/14/94 at 13:54:31 from dlitwin@geoworks.com

On 11/14/94 at 13:54:31 dlitwin@geoworks.com said:

>        Despite having read all of the archives, I still don't know what an
>antipode is.  I suspect I'd have to know more about group theory, but can
>you briefly describe what one is (you may want to CC the cube-lovers list
>as well, in case more don't understand the term).

I guess the most limited definition is two points on the opposite sides
of a sphere, at the ends of a diameter  -- e.g., the north pole and
the south pole.  However, the definition need not be limited to
three dimensions (points on the opposite ends of a diameter of a
circle are sometimes referred to as antipodes, I think) nor to
circles and spheres (I have seen opposite corners of a square
referred to as antipodes).

Generalizing further, antipodes are "opposite" or "maximally distant"
points of any sort of structure, depending on what "opposite" or
"maximally distant" mean in the context at hand.  With respect to
Rubik's cube, antipodes of Start are states which are maximally
distant from Start, and it is a matter of great interest what that
maximal distance might be.

I have to admit to a certain discomforture with one aspect of the way
we tend to refer to antipodes in the Rubik's cube.  Most Rubik
structures that have been investigated do not have a single point which
is maximally distant from Start; rather, they have several or many
maximally distant points, and all the maximally distant points are
called antipodes.  I would be more comfortable using "antipode" only
when the maximally distant point is unique.  One example where the
maximally distant point is unique is the subgroup consisting of
edges only (no corners or centers) where only Q-turns are allowed.
In this case, the maximally distant point has been called the
"unique antipode".  The description "unique antipode" seems
redundant somehow  --  "antipode" ought to imply "unique", but that
has not been the custom on Cube-Lovers.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From BRYAN@wvnvm.wvnet.edu  Tue Nov 22 13:30:15 1994
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Date:      Tue, 22 Nov 1994 13:07:05 -0500 (EST)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   <U,R> Q vs. Q+H Distances from Start

The fact that there is only one position in <U,R> which is antipodal
for both Q moves and Q+H moves got me to thinking more generally about
the Q distance from Start as compared to the Q+H distance from Start
for any position.  Here follows a cross-tabulation table giving the
number of positions in <U,R> unique up to W-conjugacy which are m moves
from Start under Q and n moves from Start under Q+H.

The table has some interesting features.  The main diagonal contains
powers of 2 for distances from zero to nine.  Other entries in the
upper left portion of the table (close to Start) are often small
primes times powers of 2.  There are positions where allowing Q+H
moves saves nine moves (from nineteen down to ten), which is a
substantial savings in moves.  There is one "discontinuity" in the
table; a position which is 25 moves from Start under Q may be 20,
19, 18, 17, or 15 moves from Start under Q+H, but not 16 moves from
Start.  The longest "common distance" is nineteen;  there are
positions which are nineteen moves from Start under both Q and Q+H,
but for twenty moves and above there are no positions which are
a common distance from Start.



                             Q+H Distance from Start

        0 1 2 3  4  5   6   7   8    9   10    11    12     13     14
      0 1
 Q    1   1
      2   1 2
 D    3     2 4
 i    4     1 6  8
 s    5       3 16 16
 t    6       1 12 40  32
 a    7          4 40  96  64
 n    8          1 20 120 224 128
 c    9             5  80 336 512  256
 e   10             1  30 280 896 1148  508
     11                 6 140 888 2292 2532  1004
 f   12                 1  42 558 2632 5688  5455  1948
 r   13                     5 220 1976 7433 13656 11585   3675
 o   14                        50  976 6475 20158 32064  24082   6387
 m   15                         4  297 3810 19993 52672  73401  46779
     16                             47 1475 13603 58642 134127 160373
 S   17                              3  320  6291 45381 165993 327673
 t   18                                  39  1799 24125 141267 447893
 a   19                                   2   268  7822  80316 403413
 r   20                                        16  1297  26988 225778
 t   21                                              74   4300  67328
     22                                               1    148   8034
     23                                                           198
     24                                                             2
     25



                             Q+H Distance from Start

             15      16      17     18   19 20
      0
 Q    1
      2
 D    3
 i    4
 s    5
 t    6
 a    7
 n    8
 c    9
 e   10
     11
 f   12
 r   13
 o   14
 m   15    9942
     16   82146   12480
 S   17  321647  116231    8729
 t   18  740143  517578   98299   2066
 a   19 1083499 1319438  493690  26824   42
 r   20  961774 1957300 1304102 140670  566
 t   21  470457 1531160 1733845 337599 2685
     22  101326  545579 1018042 332435 4875 9
     23    5760   58037  194963 115863 3286 3
     24      23     680    4661   6032  493 3
     25       1               3     15    7 1

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From BRYAN@wvnvm.wvnet.edu  Wed Nov 23 17:02:19 1994
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Date:      Wed, 23 Nov 1994 17:02:26 -0500 (EST)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   <U,R> Q vs. Q+H Distance to Start -- All Positions

This is the same chart as before except that it is in terms of all
cube positions in <U,R> rather than in terms of W-conjugacy classes.

At the same time, I will take the opportunity to correct an error
in the previous posting.  Permitting Q+H moves rather than just
Q moves can save up to 10 moves (e.g., 25 moves with Q and 15 moves
with Q+H, 24 moves with Q and 14 moves with Q+H, and 22 moves with
Q and 12 moves with Q+H).  My thanks to Dan Hoey for spotting this error.


                         Q+H Distance from Start


      0 1 2  3  4   5   6    7    8     9    10    11     12     13      14
    0 1
    1   4
    2   2 8
Q   3     8 16
    4     2 24 32
D   5       12 64  64
i   6        2 48 160 128
s   7          16 160 384  256
t   8           2  80 480  896  512
a   9              20 320 1344 2048  1024
n  10               2 120 1120 3584  4590  2032
c  11                  24  560 3552  9168 10128  4016
e  12                   1  164 2232 10519 22736 21820   7788
   13                       20  880  7904 29732 54624  46332  14700
f  14                           192  3904 25890 80606 128248  96324   25528
r  15                            16  1184 15238 79970 210684 293572  187112
o  16                                 177  5900 54398 234536 536474  641412
m  17                                  12  1280 25164 181514 663960 1310584
   18                                       150  7192  96478 564998 1791466
S  19                                         8  1072  31288 321242 1613572
t  20                                              58   5186 107919  902990
a  21                                                    296  17192  269272
r  22                                                      1    592   32122
t  23                                                                   782
   24                                                                     8
   25

                         Q+H Distance from Start

           15      16      17      18    19 20
    0
    1
    2
Q   3
    4
D   5
i   6
s   7
t   8
a   9
n  10
c  11
e  12
   13
f  14
r  15   39764
o  16  328532   49916
m  17 1286534  464878   34914
   18 2960340 2069994  393142    8230
S  19 4333614 5277159 1974444  107256   166
t  20 3846819 7828346 5215618  562494  2252
a  21 1881664 6123940 6933978 1349768 10712
r  22  405221 2181984 4071140 1328952 19411 32
t  23   23010  232088  779418  462831 12969 12
   24      92    2714   18612   23981  1936  8
   25       2               8      56    19  2


 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From BRYAN@wvnvm.wvnet.edu  Tue Nov 29 14:24:02 1994
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Date:      Tue, 29 Nov 1994 14:00:58 -0500 (EST)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Antipodes Revisited

I received the following from David Singmaster, forwarded with
permission.

>        I thought I remembered that antipodes is actually singular!  In fact
>it is singular in Greek but English recognizes both antipodes and antipode.
>But the Antipodes means the region on the earth opposite to where one is and
>is construed as a plural though its sense is singular.  So antipodes is
>actually not too bad a word for all the points which are maximally far away
>and antipode should be reserved for the case where there is a unique
>maximally distant point.

Prior to my first post on this subject, I consulted a mathematics
dictionary.  This time, I consulted an English dictionary as well.
It indicates that "antipode" is a back formation from the Greek
"antipodes", and that the pronunciation is anglicized as "AN-ti-POHD".
There is a separate entry for "antipodes". (It seems unusual for a
dictionary to have a separate entry for a plural form.)  The plural
pronunciation retains its Greek form as "an-TIP-uh-DEEZ".

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET          illegitimati      (304) 293-5540 fax
837 Chestnut Ridge Road                nil           BRYAN@WVNVM
Morgantown, WV 26505               carborundum       BRYAN@WVNVM.WVNET.EDU

From ivan@antares.aero.org  Wed Dec  7 00:30:43 1994
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To: cube-lovers@life.ai.mit.edu
Cc: ivan@aero.org
Subject: Rubik's Cube, natch
Date: 	Tue, 6 Dec 1994 21:30:30 -0800
Message-Id: <22790.786778230@armadillo.aero.org>
From: Ivan Filippenko <ivan@antares.aero.org>

Is this a mailing list that I can join ?

I'm a "cube lover", too.

Thanks,

  -- Ivan

------------------------------------
Ivan V. Filippenko, Ph.D.
Senior Member of the Technical Staff
Computer Systems Division
The Aerospace Corporation
M/S M1-055
P.O. Box 92957
Los Angeles, CA 90009-2957

Internet:  ivan@aero.org
Phone:     310-336-1808
FAX:       310-336-5833
------------------------------------

From mschoene@math.rwth-aachen.de  Wed Dec  7 16:25:08 1994
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Date: Wed, 7 Dec 94 20:40 PST
From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: cube-lovers@life.ai.mit.edu
Subject: Group Slang in Cube Lovers

Dan Hoey and Jerry Bryan remind me, that not everybody reading my
messages does group theory everyday.  So the technical slang in my
messages will not be intelligable to everybody (or in the worst case
it will not be intelligable to anybody, including myself ;-).

If this happens, please *do ask*.  Almost everthing I write is simple.
If you don't understand it, it is not your fault, it is mine, since I
wasn't clear enough.  If you do ask, I will try to explain everything.

However, I must ask you to understand that my work and my family leave
very little free time for such leisures.  So it may take me some time to
answer your questions.

Have a nice day.

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From mschoene@math.rwth-aachen.de  Wed Dec  7 16:29:15 1994
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Date: Wed, 7 Dec 94 20:41 PST
From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: cube-lovers@life.ai.mit.edu
Subject: Cube Lovers on the World Wide Web

I have converted the Cube-Lovers archives into HTML.  You can now read
all old e-mail messages using your favorite WWW Browser.  Check out

    http://www.math.rwth-aachen.de:8000/~mschoene/Cube-Lovers/

I shall try to keep this reasonably up to date.

The conversion is done automatically with an AWK script.  It is certainly
not perfect.  If you have comments or suggestions, please write me.

Looking at this long list of e-mail messages got me wondering.
Should we try to condense the information into a few documents?
I am thinking of documents describing what is known about God's
algorithm, where those bounds come from, for which other groups
God's algorithm is known, processes for pretty patterns, etc.

The reason is I don't have the time to read all 1382 e-mail message
carefully, and I believe that this is true for others too.
I feel that there are many interesting things in Cube Lovers that
I would like to know, but won't read.  Any takers?

Have a nice day.

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From mschoene@math.rwth-aachen.de  Wed Dec  7 16:30:24 1994
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Date: Wed, 7 Dec 94 20:43 PST
From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: cube-lovers@life.ai.mit.edu
Subject: Corrections

Dan Hoey writes in his e-mail message of 1994/11/08

    Martin.Schoenert@math.rwth-aachen.de writes:

        The way I view this is as follows.  The entire cube group C is a
        permutation group group on 6*9 points, generated by the six face
        turns U, D, L, R, F, B; the three middle slice turns M_U, M_L,
        M_F; and the reflection S.  This group has a subgroup M of
        symmetries of the cube (of order 48), generated by U M_U D',
        L M_L R', F M_F B', and S.  Another subgroup is G, generated by
        the six face turns, which has index 48 in G.  G is a normal
                                                  ^
        divisor of C, G is the semidirect product of M and G.  The same is
                      ^
        true for GE and GC.

    I think two of those G's are supposed to be C's, right?

Correct (wouldn't make any sense for a group G to be a subgroup in itself
of index 48 ;-).

Dan Hoey continues

    As for when I wrote

        M class             Edge         Corner       Corner times edge
          (class size)      F.P.          F.P.             / (96*class size)
                                                      ^^^^^^^^^^^^^^^^^^^^^^

    That's not a typo.  I was just saying that column 4 is equal to column
    2 times column 3, divided by column 1, divided by 96.  Perhaps I
    should have factored column 1 out of columns 2 and 3 first to avoid
    this confusion.

Again you are correct.  But it was confusing, at least to me.

Have a nice day.

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From mschoene@math.rwth-aachen.de  Wed Dec  7 16:31:35 1994
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Date: Wed, 7 Dec 94 20:43 PST
From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: cube-lovers@life.ai.mit.edu
Subject: A lemma that is *not* Burnside's

Dan Hoey writes his e-mail message of 1994/11/04 

    In January of this year, Jerry Bryan and I wrote of counting the
    number of M-conjugacy classes of Rubik's cube.  In the sense that (for
    instance) there is really only one position 1 QT from start, even
    though that QT may be applied in twelve different ways, this task
    amounts to counting the true number of positions of the cube.  The
    earlier discussion centered on calculations involving computer
    analysis of large numbers of positions.  However, a look in Paul B.
    Yale's book _Geometry and Symmetry_ gave me a clue: the Polya-Burnside
    theorem is a tool that allows us to perform this calculation by hand.

The lemma used is indeed often referred to as ``Burnside's lemma'',
or if used to count combinatorical objects as ``Polya-Burnside lemma''.

However Peter M. Neumann in his paper ``A lemma that is not Burnside's''
(1979) pointed out that this lemma was known long befor Burnside
rediscovered it.

The first appearance seems to be Cauchy's work ``Memoire sur diverses
proprietes remarquables des substitutions regulieres ou irregulieres, et
des systemes de substitutiones conjugees (suite)'' of 1845.
But it appears in a rather obscure form.

It appears in more explicit form in Frobenius' paper ``"Uber die
Congruenz nach einem aus zwei endlichen Gruppen gebildeten Doppelmodul''
in 1887.

This is well befor Burnside's paper ``On some properties of groups of
odd order'' in 1900.

Polya refined it in his paper ``Kombinatorische Anzahlbestimmungen f"ur
Gruppen, Graphen und chemische Verbindungen'' in 1937, for use in the
part of combinatorics which deals with counting problems.

Peter M. Neumann thus proposed to call this lemma ``Cauchy-Frobenius
lemma''.  But this was somehow not accepted, and it is now usually called
``A lemma that is not Burnside's''.

Have a nice day.

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From mschoene@math.rwth-aachen.de  Wed Dec  7 18:55:27 1994
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Date: Wed, 7 Dec 94 20:45 PST
From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: cube-lovers@life.ai.mit.edu
Subject: Models for the Cube

I wrote in my e-mail message of 1994/11/08

    The way I view this is as follows.  The entire cube group C is a
    permutation group group on 6*9 points, generated by the six face turns U,
    D, L, R, F, B; the three middle slice turns M_U, M_L, M_F; and the
    reflection S.  This group has a subgroup M of symmetries of the cube (of
    order 48), generated by U M_U D', L M_L R', F M_F B', and S.  Another
    subgroup is G, generated by the six face turns, which has index 48 in G.
    G is a normal divisor of C, G is the semidirect product of M and G.  The
    same is true for GE and GC.

Jerry Bryan writes in his e-mail message of 1994/11/08

    I have discussed a similar view of things recently, except that I was
    not brave enough to include a reflection in the generators.  C is
    normally used to denote the set of twenty-four rotations of the
    cube (a sub-group of M), so let's call your "entire cube group"
    big_G instead.  My version of big_G was generated by Q plus the
    slice moves (like yours without the reflection), or alternatively
    by Q plus C.  Your version of big_G is hence the same as the one
    I discussed except that you added a reflection.  C (the rotations C,
    that is) is a sub-group of both versions of big_G.  M is a sub-group
    of your version of big_G, but not of mine.

    Your big_G has the obvious advantage of including M as a sub-group.
    Mine has the advantage (?) of being physically realizable on a
    real cube.  That is, for X in your big_G, rX or Xr (r is a reflection)
    is also in your big_G.  For X in my big_G, rX or Xr is not in
    big_G, and correspondingly a single reflection is not physically
    realizable on a real cube.  Of course, r'Xr is in big_G in either
    case, r being in M.  Also, cX and Xc are in either version of big_G
    for all c in C.

OK, I guess I have to be a little bit more precise and also to adapt
my terminology to common usage.  First a picture.

                      MG (48*|G|)
                     /|
                    / CG (24*|G|)
                   / /|
                  / / |
                 / /  |
                / /   G  (|G| = 8!*3^7 * 12!*2^11 / 2)
               / /    |
              / /     |
             / /      |
            / /       |
           / /        |
          / /        /
         / /        /
        / /        /
  (48) M /        /
       |/        /
  (24) C        /
        \      /
         \    /
          \  /
           <1>

The maximal cube group *MG* is a permutation group on 6*9 points.
It is generated by the six face turns < U, D, L, R, F, B >, the three
rotations of the entire cube < u, l, f >, and the reflection < x >.

The complete cube group *CG*, generated by < U, D, L, R, F, B > and
< u, l, f >, is a subgroup of MG of index 2.

The cube group *G*, generated by < U, D, L, R, F, B >, is a subgroup of
index 24 in CG.  G can be viewed as a permutation group on 48 points,
since it fixes the 6 center cubies.

The group *M* of symmetries of the entire cube, generated by < u, l, f >
and < x >, is a subgroup of MG of size 48.

The group *C* of rotations of the entire cube, generated by < u, l, f >,
is a subgroup of CG of size 24.

(I would have preferred S instead of M and R instead of C, but M and C
are too widely used to change that notation.  Of course MG is not called
MG because it is the maximal cube group, but as a reminder that it is the
product of M and G.  Likewise for CG.)

Jerry continues

    I tend to think that Singmaster's standard G=<Q> is not what people
    think of when they hold a real cube in their hand.  Rather, they
    tend to think of big_G/C.  That is, the cosets of C in big_G are
    common sensically considered to be equivalent because rotating
    a real cube in space is "doing nothing".  Also, for my version
    of big_G we have |big_G/C| = |G|.

True, what people really see is the complete cube group CG (what you call
big_C).  That is, patterns corresponding to two different elements of CG
are distinct.  Now if two patterns can be made equal by rotations of the
entire cube, they ``look alike'' and most people feel that they are
equivalent since rotations ``cost nothing''.  Especially they feel that
any pattern corresponding to an element in C is solved.  Mathematically
we describe this equivalence by saying that all 24 elements in each coset
of C are equivalent.

Unfortunately C is *not* a normal subgroup of CG, and therefore CG/C is
*not* a group.  If we want to apply group theory, we need a better model.
I argue that G is indeed a good model for the 3x3x3 cube.

First note that for each coset of C in CG, there is exactly one element
of G in this coset.  This follows since C and G together generate CG and
have trivial intersection.  We call this element the representative in G
of the coset.  Thus G is a set of representatives for the cosets of C in
CG.  In group theory terminology G is a *supplement* for C (if C was a
normal subgroup, then G would be called a complement of C).  An immediate
consequence is that |G| = |CG| / |C|.

Next note that, if we assume that rotations ``cost nothing'' and middle
slice turns cost (at least) twice as much as face turns, then any two
elements in a coset of C have the same *cost*, i.e., distance from the
solved cube, and this is equal to the cost of the representative in G.

This is a simple consequence of the fact that we can transform each
process p_1 p_2 ... p_l, where each p_i is either a face turn or a
rotation of the entire cube, into one which has a single rotation of the
entire cube first and then only face turns afterwards.  This can be done
using the rule <face-turn> <rotation> => <rotation> <another-face-turn>,
which obviously doesn't change the cost of the process (remember
rotations cost nothing).

Finally note that G's structure as a group is in a certain sense CG
without C.  Namely G is a normal subgroup of CG, and the factor group
CG/G is isomorphic to C.  Ideally we would like to have G be isomorphic
to CG modulo C, but this is not well defined, as C is not a normal
subgroup.  Put another way, CG is the semidirekt product of G with C.

Unfortunately the existence of this model is particular to the 3x3x3
cube.  It does not work as well for other cubes.

First take the 2x2x2 cube group CG_2 (I use a _2 to distinguish the 2x2x2
subgroups from their 3x3x3 counterparts).  Again we have a subgroup C_2,
generated by the rotations, of size 24.  But the subgroup G_2, generated
by the six face turns, is in fact equal to CG_2.  In particular it is not
a supplement to C_2.

But we can make a similar construction.  Namely in the case of the 3x3x3
we can view CG as generated by the six face turns and the three middle
slice turns < M_U, M_D, M_F > (instead of the six face turns and the
three rotations < u, d, f >).  And our supplement G was the subgroup of
CG generated by 6 of those 9 generators, were the 3 removed ones are
pairwise perpendicular.  In the case of the 2x2x2 cube we can take the
subgroup H_2 that is generated by three turns < U, L, F >.  Using the
transformations <face-turn> <rotation> => <rotation> <another-face-turn>
and D => u' U, R => l' L, B => f' F, we can again transform any process
into one which has a single rotation first and then only < U, L, F >
turns afterwards, without changing the cost of the process (again
rotations cost nothing).

Thus H_2, of size 7!*3^6, is a good model to use when one is looking for
God's algorithm for the 2x2x2 cube.  Nothing of this is really new, I
have just casted it into a different language.  For example see
'http://www.math.rwth-aachen.de:8000/~mschoene/Cube-Lovers/
Jerry_Bryan__God's_Algorithm_for_the_2x2x2_Pocket_Cube.html'.

But H_2 is *not* normal, and is not CG_2 without C_2 (in the sense in
which G was CG without C).  For example CG_2 has a factor group
isomorphic to S8, but there is no such factor in H_2.

Things get worse when we look at the nxnxn cube groups CG_n.  We can find
again find a supplement H_n for C_n, if we leave out three pairwise
perpendicular slice turns.  If n is odd and if we leave out the three
middle slice turns, then H_n is again a normal subgroup (and in the same
sense as above, it is again CG_n without C_n).  On the other hand if n is
even then H_n is never a normal subgroup.  Moreover if 3 < n, then the
transformation rules tell us to replace one of the removed slice turns by
a rotation and the product of the n-1 parallel slice turns.  Thus the
transformation would only respect the cost, if we assume that the removed
three slice turns have cost n-1.  While this is arguably true for n = 3,
where most people feel that the middle slice turns have cost 2, I don't
think anybody feels that for the 4x4x4 cube nine of the slice turns have
cost 1 and 3 have cost 3.

And now for something completely different (no, not really ;-).

I have argued that G is a good model for the 3x3x3 cube assuming that
rotations of the entire cube cost nothing, and that middle slice turns
have cost 2 (or higher).  In a certain sense, we got rid of C.  That
doesn't mean we can't use C anymore.  In fact, C is still very useful.

Namely let c be any element of C, and g be any element of G.  Then c' g c
is another element of G, because G is a normal subgroup.  Moreover the
cost of c' g c is the same as the cost of g.  This is trivial because
each process effecting g can be turned into a process effecting c' g c,
by replacing each generator x in this process by c' x c.

But C is not the largest such group.  The largest such group is M, i.e.,
the full group of symmetries of the entire cube.  This is the reason why
I prefer to view G as a subgroup of MG, which is the semidirekt product
of M and G, even though I realize that MG is not physically realizable.

Have a nice day.

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From mschoene@math.rwth-aachen.de  Wed Dec  7 19:36:43 1994
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Date: Wed, 7 Dec 94 20:46 PST
From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: cube-lovers@life.ai.mit.edu
Subject: Cayley Graphs

I wrote in my e-mail message of 1994/11/08

            Note that the elements of M are also a autmorphisms of the Cayley
    graph.  That means that elements of M respects the length of operations.
    That is if g_1 and g_2 are elements of G that are in one conjugacy class
    under M, then the lenght of the shortest process effecting them is equal.
    This follows from the fact that M fixes the set of the generators of G
    and their inverses.  M is fact the largest subgroup of the outer
    autmorphism group with this property, which makes it rather important.

Jerry Bryan answered in his e-mail message of 1994/11/08

    This of course is the basis for the large searches I have been able
    to perform using M-conjugate classes.  The only trouble is, I don't
    even know what a Cayley graph is (but I am working on it), the last
    course I took in group theory being 25 years ago.

The Cayley graph Gamma for a group G generated by a certain system of
generators < g_1, g_2, ... > is defined as follows.

The vertices of Gamma correspond to the elements of G.  From vertex v_1
draw an edge to v_2 labelled with g_i, if and only if v_1 g_i = v_2.
Also draw an edge from v_2 to v_2 labelled g_i^-1 (or g_i').

So the Cayley graph depends on the group *and* on the generating system.
Simple, isn't it.

In this terminology God's number for the cube is simply the diameter of
the Cayley graph of the cube group generated by the quarter face turns
(or quarter face turns and half face turns).

In general an autmorphism alpha of G maps the Cayley graph of G w.r.t.
the generating system < g_1, g_2, ... > to a new Cayley graph of G w.r.t.
the generating system < g_1^alpha, g_2^alpha, ... >.

But in this case, i.e., for the autmorphism of G induced by elements of
M, the sets { g_1, g_2, ... } and { g_1^alpha, g_2^alpha, ... } are
equal.  So the elements of M induce autmorphism of the unlabelled Cayley
graph.

And as I said, M is the largest subgroup of the outer autmorphism group
of G with this property.

Have a nice day.

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From mschoene@math.rwth-aachen.de  Wed Dec  7 19:49:39 1994
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Date: Wed, 7 Dec 94 20:48 PST
From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: cube-lovers@life.ai.mit.edu
Subject: Permutation Representations for Magic Polyhedra

Dan Hoey writes in his e-mail message of 1994/11/08

    Wow, I didn't realize this sort of calculation had been automated.

Hey, we do this stuff every day.  Really.
Well at least with a loose interpretation of ``this sort of''.

Dan Hoey continues

        gap-3.4 -b -g 4m 
        gap> Sum( ConjugacyClasses( M ),
        >     c -> Size( Centralizer(G,Representative(c)) ) / 48 * Size(c) );
        901083404981813616

    Well, call me John Henry.  Say, do you have gap libraries for other
    magic polyhedra?  For higher-dimensional magic?

I also have a permutation representation for the 2x2x2 and the 4x4x4
cube.  I must confess that I was never interested in other magic
polyhedra.

I once started writing a GAP function that creates a premutation
representation for any (hyper-)cube, i.e., 'Cube( 3, 3, 3, 2 )' would
create a 4-dimensional magic domino.  The largest problem was to define
what the ``faces'' and ``slices'' are, i.e., are they 2 or n-1
dimensional?

If there is interest, I would finish this project and also collect
permutation representations for other magic polyhedra and distribute
them together with future versions of GAP.

Have a nice day.

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From mschoene@math.rwth-aachen.de  Wed Dec  7 19:55:00 1994
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Date: Wed, 7 Dec 94 20:48 PST
From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: cube-lovers@life.ai.mit.edu
Subject: Distance Respecting Automorphisms

Dan Hoey writes in his e-mail message of 1994/11/08

        ... [conjugation by] M fixes the set of the generators of G and
        their inverses.  M is fact the largest subgroup of the outer
        autmorphism group with this property, which makes it rather
        important.

    In a 1983 Cubic Circular article (of which I know only Stan Isaacs's
    summary) David Singmaster observed that the group is larger for larger
    cubes, provided we work what I call the ``theoretical invisible
    group''.  That is, we solve not only the surface of the cube, but the
    hypothetical interior (n-2)^3 cube, and all the smaller (n-2k)^3 cubes
    as well.  I blithered at length about this in my article of 1 June
    1983 archived (I think I've got it right this time) at
    <ftp://ftp.ai.mit.edu/pub/cube-lovers/cube-mail-5.gz>.

Try

    http://www.math.rwth-aachen.de:8000/~mschoene/Cube-Lovers/
    Dan_Hoey__Eccentric_Slabism,_Qubic,_and_S&LM.html

instead (must be on a single line).

Dan continues

    The idea is that a mapping called evisceration allows us to permute
    the layers of the cube.  On the 4x4x4 cube, this for instance allows
    us to exchange each inner slab with its adjacent outer slab.  It also
    allows us to conjugate each inner slab move by central inversion,
    while leaving the outer slab moves alone.  In general, evisceration of
    a d-dimensional cube by f maps each feature (cubie, colortab, or
    face-center arrow) at coordinates (x[1],x[2],...,x[d]) to
    (f(x[1]),f(x[2]),...,f(x[d])), where f is a permutation of the
    intervals between the cleavage coordinates of the cube.  I believe
    that if f commutes with the central inversion, then conjugation by
    evisceration is an outer automorphism of the Rubik's cube group.  (I
    think I have proved this for d=3, and I think the proof in higher
    dimensions should not be difficult given the right notation.)

    The group of all eviscerations includes the central inversion; we can
    of course augment it by the rotation group in d-space.  Is this the
    maximum outer automorphism group that respects generators of the
    Rubik's cube?  For this we take the generators to be turns of slabs
    between adjacent cleavage planes.  (Turns are direct d-1-dimensional
    isometries.)

Allow me to reformulate your description again slightly.
Let P be a d-dimensional n*n*...*n cube.
Let m be (n-1)/2 if n is odd and n/2 if n is even.
The position of each cubie is described by its position vector
(p_1,p_2,...,p_d).  If n is odd, then each p_i comes from [-m..m].  If n
is even, then each p_i comes from [-m..-1,1..m] (no middle slice in this
case).  The orientation of a cubie is described by its orientation vector
(o_1,o_2,...,o_d), where each o_i comes from the set [-1,1].

I consider the puzzle solved, if each cubie is in its original position
and in its original orientation (this is stronger than we usually require
for the 3x3x3 Rubik's cube, where we ignore the orientation of the centre
cubies, but remember, we *see* the usually invisible faces).

If F is a bijection on [-m..m] (n odd) or [-m..-1,1..m] (n even),
then F induces a permutation of the cubies (ignoring orientation) via
F( (p_1,p_2,...,p_d) ) = (F(p_1),F(p_2),...,F(p_d)).

Let I_k be defined by I_k(k) = -k, I_k(-k) = k, and I_k(l) = l for l <> k.
The permutation of the cubies induced by I_k is the inversion of the
k-th slab of the cube.

If A is a permutation on [1..m], we write l^A for the image of l under A,
and we define (-l)^A := -(l^A) (you enforce the condition (-l)^A = -(l^A)
by requiring that the permutation commutes with the central inversion).
The permutation induced by A permutes the slabs of the cube.

The group of all eviscerations is generated by all the I_k and all
permutations A of [1..m].  Each evisceration first inverts certain slabs,
and then permutes the slabs.  Put differently, the group of all
eviscerations is the wreath product of {-1,1} and S_m.
Repeating the definition of the wreath product, this means that the
group of all eviscerations is the semidirect product of the normal
subgroup generated by the I_k and the symmetric group generated by the A.

This group, together with the group C of symmetries of the d-dimensional
cube P, is a subgroup of the automorphism group of P, which fixes the set
of generators (with any reasonable definition of what the generators of P
are), and thus respects the distance of elements in the Cayley graph.

Is this the largest such group?  In the case that d = 3, this is true.
In the case of larger d, I am quite certain it is true if the generators
are rotations of 2 dimensional subsets of P.  If we choose the generators
to be symmetries of d-1 dimensional subsets of P, then I still believe it
is true.  But I have been fooled often enough in such situations, to not
trust my intuition without a proof.

If we can agree on a precise definition of what the generators of the d
dimensional cube are, I would be happy to compute the largest subgroup of
the automorphism group that respects the distance of elements in the
Cayley graph.

Have a nice day.

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From mschoene@math.rwth-aachen.de  Wed Dec  7 21:30:12 1994
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Date: Wed, 7 Dec 94 20:47 PST
From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: cube-lovers@life.ai.mit.edu
Subject: Group Products

Dan Hoey writes in his e-mail message of 1994/11/08

                                                             What is the
    difference between a direct product and a semidirect product?

Allow to answer this in greater detail, and describe all the important
products of groups.  It is one of the marvels of Rubik's cube that all
these products arise in a very natural fashion when one investigates it.

Direct Product
--------------

We say that a group D is the *direct product* of its two subgroups
M and N, if M and N together generate the whole group D, M and N
have trivial intersection, and M and N are both normal subgroups.
This implies that D is isomorphic to the group of pairs (m,n),
where the multiplication is defined componentwise,
i.e., (m_1,n_1) * (m_2,n_2) = (m_1 * n_2, m_1 * n_2).

Here is a little picture to describe the situation.

       D
      / \
     /   \
    M     \
     \     N
      \   /
       \ /
        1

M and N are the factors, D is their direct product.
Note that D/N ~ M and D/M ~ N (X ~ Y means that X and Y are isomorphic).
So M and N appear both as subgroups and quotients of the direct product.

Direct products are very simple.  They are fully described by their
to factors M and N.  Also many properties follow easily from the
corresponding properties of the factors, e.g.., |M*N| = |M| * |M|.

For an example lets take a group G+ that is a little bit larger than the
group G of Rubik's cube, namely we also allow to exchange two edges
*without* exchanging two corners simultaneously (don't ask me how this
could be realized physically).  This group is the direct product of two
subgroups GC and GE.  GC is the stabilizer of the edges, i.e., the
subgroup of those elements that do not permute the edges at all, and thus
operates only on the corners. GE is the stabilizer of the corners, i.e.,
the subgroup of those elements that do not permute the corners, and thus
operates only on the edges.  G+ is isomorphic to the group of pairs (c,e)
with c in GC and e in GE.  That means, each element of G+ can be
described by describing how it operates on the corners and on the edges
(this is trivial).  And for any c in GC and any e in GE, there is an
element in G+ that operates like c on the corners and like e on the edges.
The latter is not true in G, which is a subgroup of index 2 in G+.

Semidirect Product
------------------

We say that a group S is the *semidirect product* of its two subgroups
H and N, if H and N together generate the whole group S, H and N have
trivial intersection, and N is a normal subgroup (but H need not be
normal).  This implies that S is isomorphic to the group of pairs (h,n),
where the multiplication is defined as follows
(h_1,n_1) * (h_2,n_2) = (h_1 * h_2, h_2^-1 * n_1 * h_2 * n_2).
Note that h^-1 * n * h is usually written as n ^ h.

A picture for this situation would be identical to the picture for the
direct product, since the only real difference to the direct product is
that H need not be normal.

Again both H and N appear as subgroups of S, but only H appears as
factor of S, namely S/N ~ H.  Note that S/H is *not* well defined,
since H is not a normal subgroup of S.

Semidirect products are almost as simple as direct products.  They
are described by their factors H and N, and the operation of H on N,
by specifying what n^h is for every h in H and n in N
(note that n^h is again in N, because N is a normal subgroup).
And again many properties of S can be easily calculated from the
corresponding properties of H and N.

For an example let S be the group generated by the six face turns and
rotations of the entire cube (or equivalently the six face turns and the
three middle slice turns).  Let G be the subgroup of S generated by the
six face turns.  Let C be the subgroup of S generated by the rotations of
the entire cube, which has size 24.  Obviously C and G together generate
S.  C and G have trivial intersection, since every element of G fixes the
orientation of entire cube, but only the trivial element in C fixes the
entire cube.  Finally G is normal, since for each generator (face turn) g
of G and each h in H the element g^h is again in G (in fact it is again a
generator or an inverse of a generator).  Thus S is a semidirect product
of C and G.

Subdirect Product
-----------------

Let M and N be two groups.  Let D be the direct product of M and N.
Let f: M -> H and g: N -> H be two homomorphisms onto a group H.
Then the subgroup S = {(m,n) in D | f(m) = g(n)} of D is called a
*subdirect product* of M and N.

Again a little picture to describe the situation.

            D
          / | \
         /  |  \
        /   |   \
       /    S    \
      /     |     \
     /      |      \
    M       |       \
     \      |        N
      \     S-      /
       \   / \     /
        \ /   \   /
         M-    \ /
          \     N-
           \   /
            \ /
             1

M and N are the two factors, D is their direct product.
S is the subdirect product for the equation f(u) = g(v).
M- is the kern of f, and N- is the kern of g.
Thus M/M- = M/kern(f) ~ image(f) = H = image(g) ~ N/kern(g) = N/N-.
S- is the direct product of M- and N- and is a subgroup of S
(namely the subgroup such that f(u) = g(v) = 1).
It is easy to see that S/N- ~ M and S/M- ~ N.
So M and N appear as quotients of S.
But note that M and N *do not* appear as subgroups of S!
Also note that S/N- ~ M and S/M- ~ N implies that S/S- ~ H.

Thus M and N have a common quotient H, and in the subdirect product
we have ``glued'' these two quotients together.

For an example lets again look at the direct product G+ of GC and GE.
I have already mentioned that G is a subgroup of index 2 in G+.
It is in fact a subdirect of GC and GE.  Namely H = {-1,1},
f(c) is the parity (sign) of the permutation of the corners by c, and
g(e) is the parity (sign) of the permutation of the edges by e.
In other words, G is the subdirect product of GC and GE with
the condition that whenever we exchange two corners we also
exchange two edges.

This is in fact no coincidence.  Whenever one has a permutation group
that has more than one orbit, it is a subdirect product of the operations
on the individual orbits.

Wreath Product
--------------

Let M be an abitrary group.  Let H be a permutation group operating on
[1..n].  For h in H and i in [1..n], we write the image of i under h as
i^h.  The *wreath product* W = M wr H, is the semidirect product of the
normal subgroup N = M^n (i.e., the direct product of n copies of M),
with the subgroup H, where H operates by permuting the components of N,
(i.e., (m_1,m_2,...,m_n)^h := (m_{1^h},m_{2^h},...,m_{n^h})).

For an example take the following permutation group W:
< ( 1, 2, 3), ( 4, 5, 6), ( 7, 8, 9), (10,11,12),
  (13,14,15), (16,17,18), (19,20,21), (22,23,24),
  ( 1, 4)( 2, 5)( 3, 6),
  ( 4, 7,10,13,16,19,22)( 5, 8,11,14,17,20,23)( 6, 9,12,15,18,21,24) >.
The first 8 generators generate a direct product N of 8 cyclic groups of
size 3, which is a normal subgroup in G.  The last 2 generators generate
a symmetric group of degree 8, operating on the 8 factors of the direct
factors of N.  Thus W = C_3 wr S_8.

W operates on the set { {1,2,3}, {4,5,6}, ..., {22,23,24} }, called a
blocksystem for W.  To describe what an element w in W does, we first say
how it operates on each block, and then how it permutes the blocks.
On the other hand, if we have a permutation group that has a blocksystem,
then this permutation group is a subgroup of the wreath product.

The group GC, operating only on the corners of the cube, is a subgroup of
index 3 in the above group W.  For each element of GC, you first say how
it changes the orientation of the 8 corners (the C_3^8) and then how it
permutes the 8 corners (the S_8).  The index 3 comes from the fact that
we cannot change the orientation of a single corner in GC; if we turn
one corner clockwise, we must turn another corner counterclockwise.

Have a nice day.

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

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Date:      Thu, 8 Dec 1994 13:59:48 -0500 (EST)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@life.ai.mit.edu>
Subject:   Re: Models for the Cube
In-Reply-To: Message of 12/07/94 at 20:45:00 from ,
           Martin.Schoenert@math.rwth-aachen.de

On 12/07/94 at 20:45:00 Martin Schoenert said:

>Unfortunately C is *not* a normal subgroup of CG, and therefore CG/C is
>*not* a group.  If we want to apply group theory, we need a better model.
>I argue that G is indeed a good model for the 3x3x3 cube.

Well, with great fear and trepidation, let's see if we can't interpret
CG/C in such a way that it is a group.  I agree that your statement
above is correct, but I believe we are interpreting C, G, and CG
somewhat differently.  I have discussed this subject before, but
armed with some better notation suggested via Dan Hoey, I think I
can do it again both more accurately and more succinctly.

Dan's suggestion is to carefully distinguish which of the various
types of cubies we are talking about.  I have done a lot of work
with (for example) corners-only-cubes-without-centers, corners-only-
cubes-with-centers, etc.  When we talk about the set C of rotations,
Dan suggests specifying such things as C[C] (Corners-only),
C[E] (Edges-only], C[C,F] (Corners-plus-Face-centers), etc.  The
C[C] thing looks funny, using C in two such different ways, but
there are only so many letters.  I want to reserve lower case c for
elements of C, so I will live with C[C].

I would suggest extending the notation to G and Q, so that (for example)
the corners-only with Face-centers group we have called GC could instead
be called G[C,F] = <Q[C,F]>, and the 2x2x2 cube could be called
G C =<Q C > because there are no Face-centers.

The "standard Singmaster model" (my terminology) would be written
as G[C,E,F] = <Q[C,E,F]>.  (Well, I think Singmaster would write it as
G[C,E,F] = <Q[C,E,F], H[C,E,F]>, since I think he prefers to
accept H turns as single moves.)

However, I tend to work with G[C,E] = <Q[C,E]> instead.  I consider
G[C,E] to be equivalent to G[C,E,F] for most purposes because G fixes
the Face-centers, as does M-conjugation.  I have described this
equivalence before as the Face-centers simply providing a frame of
reference that can be provided in other ways.  However, when you step
outside the friendly confines of G=<Q>, it does start to matter whether
the Face-centers are there or not.  As an example important to this
discussion, if you consider CG=<Q,C>, then it makes a considerable
difference whether you are talking about CG C,E  or CG C,E,F .

For example, G[C,E] = <Q[C,E]> can be simulated on a real cube
by removing the color tabs from the Face-centers, by
restricting yourself to Q moves only (no whole cube rotations or
slices), and by declaring the cube solved only when the Up color
is up and the Front color is Front.  Notice that with the Face
centers absent, you can make the cube look solved even when it
isn't.  It will be rotated instead, but it won't be solved.

This model may seem a little simple-minded.  Why are no rotations
allowed, and why don't you count it as solved when it looks solved?
But computers are simple-minded.  My programs only consider things
equal when they are literally equal, and equivalence is something
I have to program in.  As an example I have used before,
consider G[C]=<Q[C]>, modeled in the real world by a 2x2x2 pocket cube
or by removing both the edge and Face-center color tabs from a 3x3x3
cube.  Take a solved cube in G C  and perform RL'.  The cube will still
look solved, but it will be rotated.  The memory cells in my program
will not be the same for I as for RL', but I want to treat them as
equivalent, as would nearly everybody with a real world 2x2x2 cube
in their hands.

This is where I have claimed before that a model that treats RL' the same
as I is G[C]/C[C].  The idea is that G[C]/C[C] is a group with the
identity being C[C] itself (i.e., rotating the cube is "doing
nothing".)  The proof is fairly simple.  From each element (coset) of
G[C]/C[C], pick the unique permutation that fixes a particular
corner, say UFR, and form a new set G[C]* containing the one element
chosen from each coset.  The elements of G[C]/C[C]
are sets (namely cosets), but the elements of G[C]* are permutations
which are also in G[C].  In particular, G[C]* = <D[C],B[C],L[C]>.
Hence, G[C]* is a group.

Note that the generators of G[C]* are
the twists of those faces which are diagonally opposed to the
corner fixed by the selection function from G[C]/C[C] to G[C]*.
Hence, the generators fix the same corner as the selection function,
showing that <D[C],B[C],L[C]> is really the same set as G[C]*,
namely the set of all cubes in G[C] for which the UFR corner is
fixed.  Finally, there is an obvious isomorphism between G[C]/C[C]
and <D[C],B[C],L[C]>.  Namely, to multiply two cosets, map each
to <D[C],B[C],L[C]> via the selection function, perform the multiplication
there using standard cube multiplication, and map the
product back to a coset.  Hence, G[C]/C[C] is a group.

A similar argument applies to G[E]/C[E] except that we have to fix
an edge cubie instead of a corner cubie.  A similar argument applies
to G[C,E]/C[C,E] except we have to fix an edge cubie and restrict C to
even permutations.  Dan calls the set of even rotations E, so let's
call it G[C,E]/E[C,E].  (Still wish we had letters whose use
did not conflict so blatantly.)

But when we started, we were talking about CG/C, not about G/C.
However, notice that when our model does not include Face-centers,
we have <Q[C]> = <Q[C],C[C]>, <Q[E]> = <Q[E],C[E]>,
and <Q[C,E]> = <Q[C,E],E[C,E]>.  (I mean that the groups are equal, not
that the Cayley graphs are the same.)  Hence, speaking generically of
the first two cases, we have C is in G, CG=G, and both CG/C and G/C are
groups.  In the last case, we have to say E is in G, EG=G, and EG/E is
a group.  But we can go one step further.  Since there are no Face-centers,
we can admit Slice moves or C as generators (it doesn't matter which),
and we no longer have to restrict ourselves to even rotations.
We can say G+ C,E =<Q C,E ,C C,E > and we will have C is in G+,
CG+=G+, and CG+/C is a group which is the same size as EG/E. (G+ is twice
as big as G, of course.)

I guess this must mean that C C , C E , and C C,E  are all normal
subgroups of their respective CG's, but that C C,F , C E,F , and
C C,E,F  are not.  That should not be surprising.  Having the
Face-centers there only as a frame of reference and never moving
is not the same as having them there and really moving (as when you
rotate the entire cube).

After joining Cube-Lovers, I discovered that others
had solved God's algorithm for the 2x2x2 long before me.  I was expecting
my solution to be 24*48 times smaller than theirs because I was using
cosets of C and M-conjugates.  But my solution was only
48 times smaller than theirs.  By taking both cosets and M-conjugates
I really had reduced <Q C > by 24*48 times.  However, everybody else
who worked on the problem had modeled it as something like
<D C ,B C ,L C >, fixing a corner.  (Any other corner would do as well.
There are eight conjugate groups, any of which would do as well as any
other.) <D C ,B C ,L C > is 24 times smaller than <Q C > in the first
place, and as I said earlier, <Q C > is equivalent to <Q C,F > for
most purposes anyway because of the fixed Face-centers.  Hence,
everybody else had a 24 times head start on me.  (At the time,
Dan suggested that I was increasing the size of the problem by 24 and
then reducing it by 24*48 for a net reduction of 48.  But that would
only be true if the model were <Q C,F >.  Since the model was <Q C >,
there really was a reduction of 24*48.  But <Q C > does not really
model the 2x2x2 cube, and is 24 times larger than it needs to be in
the first place if you are trying to model the 2x2x2.)

Modeling cubes without centers such as the 2x2x2 is trickier than it
looks because of the requirement that rotations be treated as
equivalent.  I did it by using cosets of rotations; everybody else
did it by fixing a corner.  But before I realized all this, I went on
a Quixotic chase for a model which would simultaneously be a true
model for a 2x2x2 cube and would retain the cubic symmetry of the
problem (whatever that means).  There are articles in the archives
concerning this subject, with the conclusion that no such model is
possible because any true model would be isomorphic to <D C ,B C ,L C >,
which does not have "cubic symmetry".

I guess the "cubic symmetry of the problem" means that you should use
M-conjugate classes.  Recall that when I solved <U,R> I used what
Dan calls W-conjugate classes because W is the symmetry group
for <U,R>, and W-conjugate classes reduced the size of the problem
by four times.  This leads me to a question.  The way I modeled
the 2x2x2, I used M-conjugate classes of cosets and reduced the size of
the problem by 48 times.  If I were going to model <D C ,B C ,L C >,
I would be very inclined not to use M-conjugate classes, but rather to
use a subgroup of M which was the symmetry group of <D C ,B C ,L C >.
The subgroup would have less than 48 elements, and I would get less
than a 48 times reduction in the size of the problem.  But a fixed
corner model such as <D C ,B C ,L C > is isomorphic to a coset model
such as <Q C >/C C , and M-conjugates are appropriate to the coset
model.  Therefore, my analysis of the situation is obviously very
flawed.  Can anybody see what is wrong?

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From BRYAN@wvnvm.wvnet.edu  Thu Dec  8 15:02:35 1994
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From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@life.ai.mit.edu>
Subject:   Re: Models for the Cube
In-Reply-To: Message of 12/07/94 at 20:45:00 from ,
           Martin.Schoenert@math.rwth-aachen.de

On 12/07/94 at 20:45:00 Martin Schoenert said:

>But C is not the largest such group.  The largest such group is M, i.e.,
>the full group of symmetries of the entire cube.  This is the reason why
>I prefer to view G as a subgroup of MG, which is the semidirekt product
>of M and G, even though I realize that MG is not physically realizable.

But can't you speak of conjugates such as m'gm without regard to G
being a subgroup of MG?  I agree that MG seems like a very useful group,
and it is a very nice model of what is going on.  But doesn't g in G
imply m'gm in G whether I ever heard of MG or not?

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Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From BRYAN@wvnvm.wvnet.edu  Thu Dec  8 15:21:09 1994
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From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@life.ai.mit.edu>
Subject:   Re: Cayley Graphs
In-Reply-To: Message of 12/07/94 at 20:46:00 from ,
           Martin.Schoenert@math.rwth-aachen.de

On 12/07/94 at 20:46:00 Martin Schoenert said:

>The Cayley graph Gamma for a group G generated by a certain system of
>generators < g_1, g_2, ... > is defined as follows.

>The vertices of Gamma correspond to the elements of G.  From vertex v_1
>draw an edge to v_2 labelled with g_i, if and only if v_1 g_i = v_2.
>Also draw an edge from v_2 to v_2 labelled g_i^-1 (or g_i').
                               v_1

>So the Cayley graph depends on the group *and* on the generating system.
>Simple, isn't it.

These are fine points, but they bother me anyway.

  1. Suppose I write <Q>=<Q,H>.  If I mean that the group <Q> is equal
     to the group <Q,H>, then the equation is correct.  If I mean that
     the Cayley graph of <Q> is the same as the Cayley graph of <Q,H>,
     then the equation is incorrect.  Which is the conventional meaning?
     Is the meaning universal, or does it depend on the author and the
     context?

  2. I gather from your note and from things that Dan sent me that
     one should not list inverses of the generators.  For example,
     <U,R> is sufficient and one should not write <U,U',R,R'>.  But
     people conventionally write <Q> which includes six processes and
     their six inverses.  Is this acceptable usage, or should we write
     <U,D,L,R,F,B> instead?

As an additional comment, I have frequently written about the Q length
of a process in <U,R> or the Q+H length of a process in <U,R>.  I think
we would be better served to talk about the length of a process in
<U,R> or the length of a process in <U,U2,R,R2> if the generator
notation implies a particular Cayley graph.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From acoles@mnsinc.com  Thu Dec  8 16:31:30 1994
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Subject: Cubes
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I would like to know if there is any place where I can purchase 
"mind-boggling" cubes from.  

Also, if anyone knows Peter Beck's internet address, please let me know.

Thanks.

From mreid@ptc.com  Thu Dec  8 16:44:33 1994
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From: mreid@ptc.com (michael reid)
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jerry says:

> On 12/07/94 at 20:46:00 Martin Schoenert said:
>
> >So the Cayley graph depends on the group *and* on the generating system.
> >Simple, isn't it.
>
> These are fine points, but they bother me anyway.
>
>   1. Suppose I write <Q>=<Q,H>.  If I mean that the group <Q> is equal
>      to the group <Q,H>, then the equation is correct.

here you're using  "<Q>"  to denote the group.

>                                                         If I mean that
>      the Cayley graph of <Q> is the same as the Cayley graph of <Q,H>,
>      then the equation is incorrect.  Which is the conventional meaning?

but now you're trying to use the same symbol to denote the cayley graph.

>      Is the meaning universal, or does it depend on the author and the
>      context?

the context should make it clear which object (group or graph) the symbols
refer to.  as martin notes above, these are quite different objects.

>   2. I gather from your note and from things that Dan sent me that
>      one should not list inverses of the generators.  For example,
>      <U,R> is sufficient and one should not write <U,U',R,R'>.  But
>      people conventionally write <Q> which includes six processes and
>      their six inverses.  Is this acceptable usage, or should we write
>      <U,D,L,R,F,B> instead?

either is acceptable, whether you're referring to groups or cayley graphs.

however, for cayley digraphs (directed graphs), the two meanings are quite
different.  i don't imagine we'll have anything to do with digraphs until
someone complains that they can only turn the faces of their cube clockwise,
and wants to know some short processes.

in our case,  "<Q>"  is preferred, since it's shorter.

> As an additional comment, I have frequently written about the Q length
> of a process in <U,R> or the Q+H length of a process in <U,R>.  I think
> we would be better served to talk about the length of a process in
> <U,R> or the length of a process in <U,U2,R,R2> if the generator
> notation implies a particular Cayley graph.

yes, this terminology is more precise, but the meaning was already clear.

mike

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From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@life.ai.mit.edu>
Subject:   Re: Models for the Cube
In-Reply-To: Message of 12/07/94 at 20:45:00 from ,
           Martin.Schoenert@math.rwth-aachen.de

(With apologies, retransmitted to correct square brackets trashed by
our mailer.)

On 12/07/94 at 20:45:00 Martin Schoenert said:

>Unfortunately C is *not* a normal subgroup of CG, and therefore CG/C is
>*not* a group.  If we want to apply group theory, we need a better model.
>I argue that G is indeed a good model for the 3x3x3 cube.

Well, with great fear and trepidation, let's see if we can't interpret
CG/C in such a way that it is a group.  I agree that your statement
above is correct, but I believe we are interpreting C, G, and CG
somewhat differently.  I have discussed this subject before, but
armed with some better notation suggested via Dan Hoey, I think I
can do it again both more accurately and more succinctly.

Dan's suggestion is to carefully distinguish which of the various
types of cubies we are talking about.  I have done a lot of work
with (for example) corners-only-cubes-without-centers, corners-only-
cubes-with-centers, etc.  When we talk about the set C of rotations,
Dan suggests specifying such things as C[C] (Corners-only),
C[E] (Edges-only), C[C,F] (Corners-plus-Face-centers), etc.  The
C[C] thing looks funny, using C in two such different ways, but
there are only so many letters.  I want to reserve lower case c for
elements of C, so I will live with C[C].

I would suggest extending the notation to G and Q, so that (for example)
the corners-only with Face-centers group we have called GC could instead
be called G[C,F] = <Q[C,F]>, and the 2x2x2 cube could be called
G[C]=<Q[C]> because there are no Face-centers.

The "standard Singmaster model" (my terminology) would be written
as G[C,E,F] = <Q[C,E,F]>.  (Well, I think Singmaster would write it as
G[C,E,F] = <Q[C,E,F], H[C,E,F]>, since I think he prefers to
accept H turns as single moves.)

However, I tend to work with G[C,E] = <Q[C,E]> instead.  I consider
G[C,E] to be equivalent to G[C,E,F] for most purposes because G fixes
the Face-centers, as does M-conjugation.  I have described this
equivalence before as the Face-centers simply providing a frame of
reference that can be provided in other ways.  However, when you step
outside the friendly confines of G=<Q>, it does start to matter whether
the Face-centers are there or not.  As an example important to this
discussion, if you consider CG=<Q,C>, then it makes a considerable
difference whether you are talking about CG[C,E] or CG[C,E,F].

For example, G[C,E] = <Q[C,E]> can be simulated on a real cube
by removing the color tabs from the Face-centers, by
restricting yourself to Q moves only (no whole cube rotations or
slices), and by declaring the cube solved only when the Up color
is up and the Front color is Front.  Notice that with the Face
centers absent, you can make the cube look solved even when it
isn't.  It will be rotated instead, but it won't be solved.

This model may seem a little simple-minded.  Why are no rotations
allowed, and why don't you count it as solved when it looks solved?
But computers are simple-minded.  My programs only consider things
equal when they are literally equal, and equivalence is something
I have to program in.  As an example I have used before,
consider G[C]=<Q[C]>, modeled in the real world by a 2x2x2 pocket cube
or by removing both the edge and Face-center color tabs from a 3x3x3
cube.  Take a solved cube in G[C] and perform RL'.  The cube will still
look solved, but it will be rotated.  The memory cells in my program
will not be the same for I as for RL', but I want to treat them as
equivalent, as would nearly everybody with a real world 2x2x2 cube
in their hands.

This is where I have claimed before that a model that treats RL' the same
as I is G[C]/C[C].  The idea is that G[C]/C[C] is a group with the
identity being C[C] itself (i.e., rotating the cube is "doing
nothing".)  The proof is fairly simple.  From each element (coset) of
G[C]/C[C], pick the unique permutation that fixes a particular
corner, say UFR, and form a new set G[C]* containing the one element
chosen from each coset.  The elements of G[C]/C[C]
are sets (namely cosets), but the elements of G[C]* are permutations
which are also in G[C].  In particular, G[C]* = <D[C],B[C],L[C]>.
Hence, G[C]* is a group.

Note that the generators of G[C]* are
the twists of those faces which are diagonally opposed to the
corner fixed by the selection function from G[C]/C[C] to G[C]*.
Hence, the generators fix the same corner as the selection function,
showing that <D[C],B[C],L[C]> is really the same set as G[C]*,
namely the set of all cubes in G[C] for which the UFR corner is
fixed.  Finally, there is an obvious isomorphism between G[C]/C[C]
and <D[C],B[C],L[C]>.  Namely, to multiply two cosets, map each
to <D[C],B[C],L[C]> via the selection function, perform the multiplication
there using standard cube multiplication, and map the
product back to a coset.  Hence, G[C]/C[C] is a group.

A similar argument applies to G[E]/C[E] except that we have to fix
an edge cubie instead of a corner cubie.  A similar argument applies
to G[C,E]/C[C,E] except we have to fix an edge cubie and restrict C to
even permutations.  Dan calls the set of even rotations E, so let's
call it G[C,E]/E[C,E].  (Still wish we had letters whose use
did not conflict so blatantly.)

But when we started, we were talking about CG/C, not about G/C.
However, notice that when our model does not include Face-centers,
we have <Q[C]> = <Q[C],C[C]>, <Q[E]> = <Q[E],C[E]>,
and <Q[C,E]> = <Q[C,E],E[C,E]>.  (I mean that the groups are equal, not
that the Cayley graphs are the same.)  Hence, speaking generically of
the first two cases, we have C is in G, CG=G, and both CG/C and G/C are
groups.  In the last case, we have to say E is in G, EG=G, and EG/E is
a group.  But we can go one step further.  Since there are no Face-centers,
we can admit Slice moves or C as generators (it doesn't matter which),
and we no longer have to restrict ourselves to even rotations.
We can say G+[C,E]=<Q[C,E],C[C,E]> and we will have C is in G+,
CG+=G+, and CG+/C is a group which is the same size as EG/E. (G+ is twice
as big as G, of course.)

I guess this must mean that C[C], C[E], and C[C,E] are all normal
subgroups of their respective CG's, but that C[C,F], C[E,F], and
C[C,E,F] are not.  That should not be surprising.  Having the
Face-centers there only as a frame of reference and never moving
is not the same as having them there and really moving (as when you
rotate the entire cube).

After joining Cube-Lovers, I discovered that others
had solved God's algorithm for the 2x2x2 long before me.  I was expecting
my solution to be 24*48 times smaller than theirs because I was using
cosets of C and M-conjugates.  But my solution was only
48 times smaller than theirs.  By taking both cosets and M-conjugates
I really had reduced <Q[C]> by 24*48 times.  However, everybody else
who worked on the problem had modeled it as something like
<D[C],B[C],L[C]>, fixing a corner.  (Any other corner would do as well.
There are eight conjugate groups, any of which would do as well as any
other.) <D[C],B[C],L[C]> is 24 times smaller than <Q[C]> in the first
place, and as I said earlier, <Q[C]> is equivalent to <Q[C,F]> for
most purposes anyway because of the fixed Face-centers.  Hence,
everybody else had a 24 times head start on me.  (At the time,
Dan suggested that I was increasing the size of the problem by 24 and
then reducing it by 24*48 for a net reduction of 48.  But that would
only be true if the model were <Q[C,F]>.  Since the model was <Q[C]>,
there really was a reduction of 24*48.  But <Q[C]> does not really
model the 2x2x2 cube, and is 24 times larger than it needs to be in
the first place if you are trying to model the 2x2x2.)

Modeling cubes without centers such as the 2x2x2 is trickier than it
looks because of the requirement that rotations be treated as
equivalent.  I did it by using cosets of rotations; everybody else
did it by fixing a corner.  But before I realized all this, I went on
a Quixotic chase for a model which would simultaneously be a true
model for a 2x2x2 cube and would retain the cubic symmetry of the
problem (whatever that means).  There are articles in the archives
concerning this subject, with the conclusion that no such model is
possible because any true model would be isomorphic to <D[C],B[C],L[C]>,
which does not have "cubic symmetry".

I guess the "cubic symmetry of the problem" means that you should use
M-conjugate classes.  Recall that when I solved <U,R> I used what
Dan calls W-conjugate classes because W is the symmetry group
for <U,R>, and W-conjugate classes reduced the size of the problem
by four times.  This leads me to a question.  The way I modeled
the 2x2x2, I used M-conjugate classes of cosets and reduced the size of
the problem by 48 times.  If I were going to model <D[C],B[C],L[C]>,
I would be very inclined not to use M-conjugate classes, but rather to
use a subgroup of M which was the symmetry group of <D[C],B[C],L[C]>.
The subgroup would have less than 48 elements, and I would get less
than a 48 times reduction in the size of the problem.  But a fixed
corner model such as <D[C],B[C],L[C]> is isomorphic to a coset model
such as <Q[C]>/C[C], and M-conjugates are appropriate to the coset
model.  Therefore, my analysis of the situation is obviously very
flawed.  Can anybody see what is wrong?

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU



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From BRYAN@wvnvm.wvnet.edu  Fri Dec  9 22:20:45 1994
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Date:      Fri, 9 Dec 1994 22:20:42 -0500 (EST)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Normal Subgroup Question

On 12/07/94 at 20:45:00 Martin Schoenert said:

>Unfortunately C is *not* a normal subgroup of CG, and therefore CG/C is
>*not* a group.  If we want to apply group theory, we need a better model.
>I argue that G is indeed a good model for the 3x3x3 cube.

I responded at great length, showed a group for CG/C, and concluded
as follows.

>I guess this must mean that C[C], C[E], and C[C,E] are all normal
>subgroups of their respective CG's, but that C[C,F], C[E,F], and
>C[C,E,F] are not.  That should not be surprising.  Having the
>Face-centers there only as a frame of reference and never moving
>is not the same as having them there and really moving (as when you
>rotate the entire cube).

This just *has* to be wrong.  I just don't see any way that any
of the flavors of C are a normal subgroup of their respective
flavors of CG.  The presence or absence of the Face-centers can't
have anything to do with it.  I was jumping to the conclusion that
since I found a group for some of the flavors of CG/C, that therefore
the respective C's must be normal.

I have reread my note, and it still looks to me like I found groups
for all the CG/C's I discussed.  I would invite instruction and
correction from any of you group theory experts out there, but here
is the way it looks to me.

Using G and H generically for a group and subgroup (not necessarily
cubes at all), G/H is a group if H is a normal subgroup of G,
under the "natural" operation {Xh} * {Yh} = {(XY)h} (where {Xh}
etc. denotes all h in H.)  Coset notation would be (xH)(yH)=(xy)H.
Under these circumstances, G/H is the factor group of H in G.

My group operation on the cosets not the "natural" operation.
It gets around the fact that C is not normal by picking specific
rather than arbitrary elements of the cosets in order to perform
the group operation, namely a picking specific element which fixes
the same cubie for all cosets.  I guess this means that CG/C is not
the factor group of C in CG (such a thing being impossible), but
by golly it still looks like a group to me under the "unnatural"
operation ??

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From mschoene@math.rwth-aachen.de  Sat Dec 10 09:15:04 1994
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Date: Sat, 10 Dec 94 15:12 PST
From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: cube-lovers@life.ai.mit.edu
In-Reply-To: "Jerry Bryan"'s message of Thu, 8 Dec 1994 15:42:36 -0500 (EST) <9412082301.AA24053@life.ai.mit.edu>
Subject: Re: Re: Models for the Cube

Jerry Bryan wrote in his e-mail message of 1994/12/08

    On 12/07/94 at 20:45:00 Martin Schoenert said:

    >Unfortunately C is *not* a normal subgroup of CG, and therefore CG/C is
    >*not* a group.  If we want to apply group theory, we need a better model.
    >I argue that G is indeed a good model for the 3x3x3 cube.

    Well, with great fear and trepidation, let's see if we can't interpret
    CG/C in such a way that it is a group.  I agree that your statement
    above is correct, but I believe we are interpreting C, G, and CG
    somewhat differently.  I have discussed this subject before, but
    armed with some better notation suggested via Dan Hoey, I think I
    can do it again both more accurately and more succinctly.

I think that we agree much more than we actually realize, and that it is
mostly a matter of language.  So your clarification was most welcome, and
I hope mine is too.

Jerry continued

    Dan's suggestion is to carefully distinguish which of the various
    types of cubies we are talking about.  I have done a lot of work
    with (for example) corners-only-cubes-without-centers, corners-only-
    cubes-with-centers, etc.  When we talk about the set C of rotations,
    Dan suggests specifying such things as C[C] (Corners-only),
    C[E] (Edges-only), C[C,F] (Corners-plus-Face-centers), etc.  The
    C[C] thing looks funny, using C in two such different ways, but
    there are only so many letters.  I want to reserve lower case c for
    elements of C, so I will live with C[C].

    I would suggest extending the notation to G and Q, so that (for example)
    the corners-only with Face-centers group we have called GC could instead
    be called G[C,F] = <Q[C,F]>, and the 2x2x2 cube could be called
    G[C]=<Q[C]> because there are no Face-centers.

Let's see whether I understand this correct.

Let CG again be the complete cube group generated by the face turns and
the rotations, let G be its subgroup of index 24 generated by the face
turns, and C the subgroup of size 24 generated by the rotations.

CG can be represented as a permutation group on 54 points.  On this set
it makes three orbits, called C(orners), E(dges), and F(ace-centers), of
sizes 24, 24, and 6.

[A sidenote.  Old e-mails in Cube-Lovers often talk about the 12
``orbits'' of the cube group G in the group that you get when you
are allowed to take the cube apart.  This group has structure
(S(8) <wreath-product> 3) <direct-product> (S(12) <wreath-product> 2).
Strictly speaking, these are not ``orbits'' but ``cosets'' instead.]

We can now look at the operation of CG (and C and G) on the 8 sets [],
[C], [E], [C,E], [F], [C,F], [E,F], and [C,E,F] (where [X,Y] here means
the (disjoint) union of the orbits X and Y).  This gives us 8 groups to
look at, together with their respective subgroups induced by C and G.
Clearly CG[] is trivial, and CG[C,E,F] = CG.  Such a group, e.g., CG[C],
is a model for what we get when we remove the color tabs from the other
orbits, e.g., for CG[C] we would remove the color tabs from the edges and
the faces.

A little bit of group theory.  Each of those 8 groups CG[<orbits>] is
a homorphic image of CG. That means there is a homomorphism from
CG to CG[<orbits>].  This homomorphism is actually very easy to describe:
you get the image of an element by simply forgetting what that
element does on the other orbits.  The kernel of this homomorphism
is the subgroup of elements of CG that do nothing on <orbits> and
only permute the points in the other orbits.  What this means is that
each CG[<orbits>] is a factor group of CG.

Jerry continued

    The "standard Singmaster model" (my terminology) would be written
    as G[C,E,F] = <Q[C,E,F]>.  (Well, I think Singmaster would write it as
    G[C,E,F] = <Q[C,E,F], H[C,E,F]>, since I think he prefers to
    accept H turns as single moves.)

    However, I tend to work with G[C,E] = <Q[C,E]> instead.  I consider
    G[C,E] to be equivalent to G[C,E,F] for most purposes because G fixes
    the Face-centers, as does M-conjugation.  I have described this
    equivalence before as the Face-centers simply providing a frame of
    reference that can be provided in other ways.  However, when you step
    outside the friendly confines of G=<Q>, it does start to matter whether
    the Face-centers are there or not.  As an example important to this
    discussion, if you consider CG=<Q,C>, then it makes a considerable
    difference whether you are talking about CG[C,E] or CG[C,E,F].

Correct.  Since G fixes the faces, G[C,E,F] and G[C,E] are isomorphic.
But CG[C,E,F] and CG[C,E] are not isomorphic, and neither are
C[C,E,F] and C[C,E].

Jerry continued

    For example, G[C,E] = <Q[C,E]> can be simulated on a real cube
    by removing the color tabs from the Face-centers, by
    restricting yourself to Q moves only (no whole cube rotations or
    slices), and by declaring the cube solved only when the Up color
    is up and the Front color is Front.  Notice that with the Face
    centers absent, you can make the cube look solved even when it
    isn't.  It will be rotated instead, but it won't be solved.

    This model may seem a little simple-minded.  Why are no rotations
    allowed, and why don't you count it as solved when it looks solved?
    But computers are simple-minded.  My programs only consider things
    equal when they are literally equal, and equivalence is something
    I have to program in.  As an example I have used before,
    consider G[C]=<Q[C]>, modeled in the real world by a 2x2x2 pocket cube
    or by removing both the edge and Face-center color tabs from a 3x3x3
    cube.  Take a solved cube in G[C] and perform RL'.  The cube will still
    look solved, but it will be rotated.  The memory cells in my program
    will not be the same for I as for RL', but I want to treat them as
    equivalent, as would nearly everybody with a real world 2x2x2 cube
    in their hands.

Maybe a little convention would help.  We could say that a cube is
*completely solved* if all the up-color tabs are on the up-face,
all the right-color tabs are on the right-face, etc.  And a cube is
*solved up to rotations* if all the tabs on each face have the same
color, i.e., if it can be completely solved with a rotation of the entire
cube.  Talking about the groups, only the identity is completely solved,
but all elements in C[<orbits>] are solved up to rotations.

In this language CG[C] is a model for the complete solution of the
2x2x2 cube, and a supplement for C[C] in CG[C] is a model for the
solution up to rotations of the 2x2x2 cube.  And of course, most
of the time we are interested in solutions up to rotations.

Jerry continued

    This is where I have claimed before that a model that treats RL' the same
    as I is G[C]/C[C].  The idea is that G[C]/C[C] is a group with the
    identity being C[C] itself (i.e., rotating the cube is "doing
    nothing".)  The proof is fairly simple.  From each element (coset) of
    G[C]/C[C], pick the unique permutation that fixes a particular
    corner, say UFR, and form a new set G[C]* containing the one element
    chosen from each coset.  The elements of G[C]/C[C]
    are sets (namely cosets), but the elements of G[C]* are permutations
    which are also in G[C].  In particular, G[C]* = <D[C],B[C],L[C]>.
    Hence, G[C]* is a group.

    Note that the generators of G[C]* are
    the twists of those faces which are diagonally opposed to the
    corner fixed by the selection function from G[C]/C[C] to G[C]*.
    Hence, the generators fix the same corner as the selection function,
    showing that <D[C],B[C],L[C]> is really the same set as G[C]*,
    namely the set of all cubes in G[C] for which the UFR corner is
    fixed.  Finally, there is an obvious isomorphism between G[C]/C[C]
    and <D[C],B[C],L[C]>.  Namely, to multiply two cosets, map each
    to <D[C],B[C],L[C]> via the selection function, perform the multiplication
    there using standard cube multiplication, and map the
    product back to a coset.  Hence, G[C]/C[C] is a group.

I agree mostly but not completely.

First I claim that we are interested not in the cosets of C[C] in G[C],
but rather in the cosets of C[C] in CG[C].  Now since G[C] = CG[C] this
doesn't seem to make any difference.  But for a different set of orbits,
G[<orbits>] may be different from CG[<orbits>], and C[<orbit>] will then
not be a subgroup of G[<orbits>].  So in those cases it doesn't make
sense to speak of the cosets of C[<orbits>] in G[<orbits>].

Second your usage of G/H.  Many group theory textbooks restrict
this notation to the case when H is a normal subgroup of G.
Others use G/H in general for the set of cosets of H in G.
But whenever they write ``the group G/H'' or ``G/H is a group'',
they always mean that H is normal in G, and that G/H is the factor group.

I would be happy if you wrote about ``the set G[C]/C[C] with the
multiplication defined by ... is a group'' instead of ``G[C]/C[C] is a
group''.  The reason why I think it is important to be carefull is that
many properties carry over from G to a proper factor group G/H, but do
not carry over from G to ``the set G/C with the multiplication
defined by ...''.  I shall return to this point below.

I agree that G[C]* is indeed a group.  You do exactely the same thing
that I did in my message.  You pick a set of represtatives that forms a
subgroup, which I called a supplement for C[C].  Then you define the
multiplication using those representatives.  I think that it is easier to
work with the supplement instead of the structure G[C]/C[C] with
the induced multiplication, but that is clearly a matter of taste.

Jerry continued

    A similar argument applies to G[E]/C[E] except that we have to fix
    an edge cubie instead of a corner cubie.

Almost.  But there is a tricky problem here.  Again G[E] = CG[E],
so it doesn't matter whether we talk about G[E]/C[E] (as you prefer)
or about CG[E]/C[E] (as I prefer).  Again we can find a supplement
for C[E] in CG[E], namely the subgroup of all elements of CG[E]
that leave a particular edge cubie fixed.  Assume that we fix the
upper-right edge cubie, then this supplement is <L[E],D[E],F[E],B[E]>.

But this does *not* respect costs.  That is take an element e of CG[E].
Let r be its representative in <L[E],D[E],F[E],B[E]>, i.e., c e = r,
where c is a rotation of the entire cube.  The the costs of the two
elements, viewed as elements of CG[E] is clearly the same (remember,
rotations cost nothing).  But the cost of r, viewed as an element of
<L[E],D[E],F[E],B[E]> *with this generating system*, may be higher.

For example take R[E] * r[E]' (where r is the rotation of the entire
cube).  In CG[E] this element has cost 1.  And this element lies in
<L[E],D[E],F[E],B[E]>, since it fixes the upper-right edge cubie.
But the cost of this element *with respect to the generating system
L[E],D[E],F[E],B[E]* is not 1.

We can repair this by choosing a different generating system for
<L[E],D[E],F[E],B[E]>, for example the system
L[E],D[E],F[E],B[E],R[E]*r[E]',U[E]*u[E]'.

So in general a model for the solution up to rotations for a
certain set <orbits>, is a supplement of C[<orbits>] in CG[<orbits>],
with a generating system that respects costs.

Jerry continued

                                              A similar argument applies
    to G[C,E]/C[C,E] except we have to fix an edge cubie and restrict C to
    even permutations.  Dan calls the set of even rotations E, so let's
    call it G[C,E]/E[C,E].  (Still wish we had letters whose use
    did not conflict so blatantly.)

    But when we started, we were talking about CG/C, not about G/C.
    However, notice that when our model does not include Face-centers,
    we have <Q[C]> = <Q[C],C[C]>, <Q[E]> = <Q[E],C[E]>,
    and <Q[C,E]> = <Q[C,E],E[C,E]>.  (I mean that the groups are equal, not
    that the Cayley graphs are the same.)  Hence, speaking generically of
    the first two cases, we have C is in G, CG=G, and both CG/C and G/C are
    groups.  In the last case, we have to say E is in G, EG=G, and EG/E is
    a group.  But we can go one step further.  Since there are no Face-centers,
    we can admit Slice moves or C as generators (it doesn't matter which),
    and we no longer have to restrict ourselves to even rotations.
    We can say G+[C,E]=<Q[C,E],C[C,E]> and we will have C is in G+,
    CG+=G+, and CG+/C is a group which is the same size as EG/E. (G+ is twice
    as big as G, of course.)

This is the reason why I think that it is better to talk about
CG[C,E]/C[C,E].  As you say G[C,E] <> CG[C,E] (it has index 2),
and C[C,E] is not a subgroup of G[C,E].  That your model works
depends on the fact that their is a bijection between the set
CG[C,E]/C[C,E] and G[C,E]/E[C,E].  This follows by a standard
argument from the fact that E[C,E] = Intersection( G[C,E], C[C,E] ).

Jerry continued

    I guess this must mean that C[C], C[E], and C[C,E] are all normal
    subgroups of their respective CG's, but that C[C,F], C[E,F], and
    C[C,E,F] are not.  That should not be surprising.  Having the
    Face-centers there only as a frame of reference and never moving
    is not the same as having them there and really moving (as when you
    rotate the entire cube).

It would be most surprising.  In fact C[C], C[E], and C[C,E] are
*not* normal in their respective CG's.  I don't see what face
centers should have to do with it.

Jerry continued

    After joining Cube-Lovers, I discovered that others
    had solved God's algorithm for the 2x2x2 long before me.  I was expecting
    my solution to be 24*48 times smaller than theirs because I was using
    cosets of C and M-conjugates.  But my solution was only
    48 times smaller than theirs.  By taking both cosets and M-conjugates
    I really had reduced <Q[C]> by 24*48 times.  However, everybody else
    who worked on the problem had modeled it as something like
    <D[C],B[C],L[C]>, fixing a corner.  (Any other corner would do as well.
    There are eight conjugate groups, any of which would do as well as any
    other.) <D[C],B[C],L[C]> is 24 times smaller than <Q[C]> in the first
    place, and as I said earlier, <Q[C]> is equivalent to <Q[C,F]> for
    most purposes anyway because of the fixed Face-centers.  Hence,
    everybody else had a 24 times head start on me.  (At the time,
    Dan suggested that I was increasing the size of the problem by 24 and
    then reducing it by 24*48 for a net reduction of 48.  But that would
    only be true if the model were <Q[C,F]>.  Since the model was <Q[C]>,
    there really was a reduction of 24*48.  But <Q[C]> does not really
    model the 2x2x2 cube, and is 24 times larger than it needs to be in
    the first place if you are trying to model the 2x2x2.)

Allow me to translate this to a more group theoretic language.
You are interested in finding God's algorithm for CG[C].

If e is any element of this group, then clearly it has the same costs as
(c * e), where c is any element of C[C].  Thus you need only compute the
cost for one representative of each right coset (C[C] * e).  All those
cosets have size 24, so using cosets reduces the problem by a factor 24.

However, (c' * e * c) also has the same cost, so you only need one
representative of each conjugacy class (e ^ C[C]).  Taking those two
approaches together means, that you need only look at one representative
of the set { c1 * (c2' * e * c2) | c1, c2 in C[C] }.

But we can also write this set as { c1 * e * c2 | c1, c2 in C[C] }.  Such
a set if usually called a *double coset* and written as (C[C]*e*C[C]).
Most of those double cosets have size 576 = 24*24, but some are smaller
(but of course all sizes are always multiples of 24, since each double
coset is the union of single cosets).  Thus using double cosets reduces
the problem by a factor almost 576.

Finally e has the same cost as (x' e x), where x is the reflection.  Thus
you only need one representative from each set { y' * c1 * e * c2 * y |
c1, c2 in C[C], y in M[C] }.  This reduces the problem by an total factor
almost 1152.

Jerry continued

    Modeling cubes without centers such as the 2x2x2 is trickier than it
    looks because of the requirement that rotations be treated as
    equivalent.  I did it by using cosets of rotations; everybody else
    did it by fixing a corner.  But before I realized all this, I went on
    a Quixotic chase for a model which would simultaneously be a true
    model for a 2x2x2 cube and would retain the cubic symmetry of the
    problem (whatever that means).  There are articles in the archives
    concerning this subject, with the conclusion that no such model is
    possible because any true model would be isomorphic to <D[C],B[C],L[C]>,
    which does not have "cubic symmetry".

    I guess the "cubic symmetry of the problem" means that you should use
    M-conjugate classes.

Lets first take a look at the 3x3x3 cube, i.e., CG[C,E,F] = CG.
In this case you can use G as a supplement for C, i.e., as a system of
representatives for the cosets (C * e).  Now G is normal in CG,
thus you can conjugate the elemenents of G with elements of C and stay
in G.  So there is a bijection between the representatives of the
conjugacy classes of G under the conjugation by C and the representatives
for the double cosets (C * e * C).  In fact G is normal in MG,
and there is a bijection between the representatives of the
conjugacy classes of G under the conjugation by M and the representatives
for the sets ((C * e * C)^M).  This is the basis for applying
the ``Lemma that is not Burnside's'' to count the number of such sets,
as the real size of the cube.

But in the case of the 2x2x2 cube without centers, i.e., CG[C],
this is not possible.  Finding such a model would mean finding
a supplement of C[C] that is normal in CG[G], i.e., is fixed
under conjugation by C[C].  And no such supplement exists.

Jerry continued

                          Recall that when I solved <U,R> I used what
    Dan calls W-conjugate classes because W is the symmetry group
    for <U,R>, and W-conjugate classes reduced the size of the problem
    by four times.  This leads me to a question.  The way I modeled
    the 2x2x2, I used M-conjugate classes of cosets and reduced the size of
    the problem by 48 times.  If I were going to model <D[C],B[C],L[C]>,
    I would be very inclined not to use M-conjugate classes, but rather to
    use a subgroup of M which was the symmetry group of <D[C],B[C],L[C]>.
    The subgroup would have less than 48 elements, and I would get less
    than a 48 times reduction in the size of the problem.  But a fixed
    corner model such as <D[C],B[C],L[C]> is isomorphic to a coset model
    such as <Q[C]>/C[C], and M-conjugates are appropriate to the coset
    model.  Therefore, my analysis of the situation is obviously very
    flawed.  Can anybody see what is wrong?

Yes I can ;-).  The problem is in the statement ``and M-conjugates are
appropriate to the coset model''.  I think this problem comes from
your unusual usage of the notation G[C]/C[C].  If C[C] was a normal
subgroup of G[C] and G[C]/C[C] was the proper factor group, then
the operation of M-conjugation would carry over from G[C] to the
factor group (any such operation carries over to a factor group,
provided that the normal subgroup is fixed, which is certainly the
case here).  But G[C]/C[C] is not the proper factor group, so there
is no reason why the M-conjugation should carry over, and in fact
it does not.

Finally allow to correct a non-standard language again.  In group theory
one usually does not speak of the symmetry group of another group, but of
the *automorphism group* of another group.  Moreover you don't want to
know the ``subgroup of M which was *the* automorphism group of
<D[C],B[C],L[C]>'', but the ``subgroup of M which was *also a subgroup
of* the automorphism group of <D[C],B[C],L[C]>'', because the
automorphism group of <D[C],B[C],L[C]> is in fact much larger.  In other
word, you want to know the subgroup of M that fixes <D[C],B[C],L[C]>.
This is a cyclic group of size 3, namely the rotation along the diagonal
axis of the cube that goes through your fixed center and cyclically
permutes D[C], B[C], and L[C].

Have a nice day.

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From mschoene@math.rwth-aachen.de  Sat Dec 10 09:21:41 1994
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Date: Sat, 10 Dec 94 15:19 PST
From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: cube-lovers@life.ai.mit.edu
In-Reply-To: "Jerry Bryan"'s message of Thu, 8 Dec 1994 15:02:33 -0500 (EST) <9412082002.AA14755@life.ai.mit.edu>
Subject: Re: Re: Models for the Cube

I wrote in my e-mail message of 1994/12/07

    But C is not the largest such group.  The largest such group is M, i.e.,
    the full group of symmetries of the entire cube.  This is the reason why
    I prefer to view G as a subgroup of MG, which is the semidirekt product
    of M and G, even though I realize that MG is not physically realizable.

Jerry Bryan answered in his e-mail message of 1994/12/07

    But can't you speak of conjugates such as m'gm without regard to G
    being a subgroup of MG?  I agree that MG seems like a very useful group,
    and it is a very nice model of what is going on.  But doesn't g in G
    imply m'gm in G whether I ever heard of MG or not?

Yes I certainly could.  I think it is only a matter of taste.

You seem to favor the physical model.  There the reflection has no
real realization, and it makes sense to distinguish between the
rotations and the reflection.

I look at the problem more from the computational aspect.  I view
the whole thing as a permutation group, and then there is no real
reason to distinguish between the rotations and the reflection
(both being ordinary permutations on 54 points).

And when working with those groups in GAP, it is certainly a lot more
convenient to work in MG and treat all of M uniformly, then to work in
CG and to handle the reflection specially.

Have a nice day.

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From BRYAN@wvnvm.wvnet.edu  Sat Dec 10 10:13:39 1994
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Date:      Sat, 10 Dec 1994 10:13:33 -0500 (EST)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Cubic Symmetry of the 2x2x2 (Again)

The argument has been made that the 2x2x2 cube (or really any 2Nx2Nx2N)
cube cannot have the "symmetry of the cube".  In order for a real
2x2x2 cube to have the "symmetry of the cube", you would have to
adopt unreasonable rules, such as no rotations (or if you use
rotations they have a cost of at least 2) and the cube is only solved
when the colors are oriented properly.  But a 2x2x2 cube certainly
feels like a real cube when you hold it in your hands.  I offer the
following interpretation that "sort of" gives the 2x2x2 cube the
symmetry of the cube.  Since I will only be talking about the
2x2x2, I will simplify the notation by talking about C, G, Q, etc.
rather than C[C], G[C], Q[C], etc.

As I have discussed several times before, my favorite model for the
2x2x2 is G/C (or CG/C, if you prefer; G=CG for the 2x2x2).
The group operation is (xC)(yC)=(uv)C, where u and v are the elements of
xC and yC, respectively, which fix a particular corner.  (xC)(yC)=(xy)C
doesn't work because C is not normal.  There is an obvious isomorphism
between G/C and <q_i, q_j, q_k>, where the three q-turns are those which
fix the same corner as the selection function for u and v.

There are eight corners, and hence there are eight conjugate selection
functions and eight conjugate subgroups G_m of the form <q_i, q_j, q_k>
which fix a particular corner.  If you think of mapping G/C
simultaneously and in parallel to the all the elements in the set
{G_1, G_2, G_3, G_4, G_5, G_6, G_7, G_8}, then in a loose sense you
have preserved the cubic nature of the problem.  That is, none of
the individual G_m's have the same symmetry as the cube, but in a loose
sense the entire collection {G_m} does.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From BRYAN@wvnvm.wvnet.edu  Sat Dec 10 10:21:06 1994
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Date:      Sat, 10 Dec 1994 10:21:02 -0500 (EST)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@life.ai.mit.edu>
Subject:   Re: Re: Models for the Cube
In-Reply-To: Message of 12/10/94 at 15:19:00 from ,
           Martin.Schoenert@math.rwth-aachen.de

On 12/10/94 at 15:19:00 Martin Schoenert said:

>You seem to favor the physical model.  There the reflection has no
>real realization, and it makes sense to distinguish between the
>rotations and the reflection.

That is probably an accurate assessment.  However, it should be
pointed out that the edges can be reflected on a physical model,
even though the corners and Face-centers cannot.  Mathematically,
this means that <Q[E]> generates reflections, but <Q[C]>
and <Q[F]> do not.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From @mail.uunet.ca:mark.longridge@canrem.com  Fri Dec 16 04:19:31 1994
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Sender: CRSO.Cube@canrem.com
Subject: Cyclic Decomposition
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.897.5834.0C1C4371@canrem.com>
Date: Fri, 16 Dec 1994 01:32:00 -0500
Organization: CRS Online  (Toronto, Ontario)

 I've been working on a new algorithm to find move sequences to reach
certain positions on the 3x3x3 cube. The basic idea is to find a
sequence such that:

         (S1 S2 S3... SX) ^N = Goal State
where X is the number of moves in the fragment and
      N is the number of times the fragment is repeated.

 I call such a process to be "Cyclicly Decomposable".

 Certain states, such as the 12-flip, require quite a few moves, in
fact more moves than possible to search using brute force even when
using high-order computers. The best results using the Kociemba
algorithm need 28 q turns or 20 q+h turns for the 12-flip.

 While the cyclic decomposition algorithm (henceforth the CD algorithm
for short) usually requires more moves than the Kociemba algorithm
it does have an mnemonic advantage. The following is the best
result using the CD algorithm to date:

p192 2 Flip in face     (F1 B1 L1 T1 D1)^6           (30 q)
p195 12 Flip            (T1 B3 T1 L1 F3 R3)^6        (36 q)

 Note that both of these processes use 5 of the 6 generators.

 Some cube positions are extremely resistant to CD but flip and
twist patterns are no problem. In particular, the 6 X order 3
pattern does not yield to CD with values of X up to 7.
Naturally with N = 1 we can always find one of the optimal paths
but I am more interested in cases where N > 1. One may note that
with N > 1 using CD processes we are still free to use any number
of q turns except a prime number, for which N must be 1, but this
should not be too constraining.

 By relaxing the conditions somewhat we can conceive of sequences
which are near-CD, that is CD with a small suffix or prefix:

p169 4 Y's Rot + 2 X    (F2 B2 D1 L2 R2) ^2 + T1      (11 q+h)

 By looking at the best sequence the Kociemba algorithm can
find for a position, we can count the number of q turns and
use this as a starting point for an attempt with CD...

p1 6 X order 3   R2 L3 D1 F2 R3 D3 F1 B3 U1 D3 F1 L1 D2 F3 R1 L2 (20 q)

 Looking at p1 we can infer that possibly X=5 and N=4 may lead to
finding a CD process or    X=4, N=5
                           X=10, N=2
                           X=20, N=2 etc.

 At the very least we can discount  X * N = odd number.

-> Mark <-
Email: mark.longridge@canrem.com

From Wechsler@world.std.com  Fri Dec 16 10:23:44 1994
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In-Reply-To: Mark Longridge's message of Fri, 16 Dec 1994 01:32:00 -0500 <60.897.5834.0C1C4371@canrem.com>
Subject: Cyclic Decomposition

In the corner group, (RFU)^5 exchanges corners fur and bur.  I only
mention this because of all the tools I use, it is the only one that
involves a lot of repetition.  It's a fossil from my earliest cube
solution, c. 1980, which used _only_ repetitive processes.

(RFU)^5 is only useful to me because I solve corners first.
One-face-first solvers will find this incomprehensible.





From ccw@eql12.caltech.edu  Fri Dec 16 13:55:58 1994
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From: ccw@eql12.caltech.edu
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Subject: A comment on Cyclic Decomposition
To: cube-lovers@life.ai.mit.edu, mark.longridge@canrem.com
X-St-Vmsmail-To: ST%"cube-lovers@life.ai.mit.edu"

Mark Longridge proposes looking for processes that can be expressed
in the form
         (S1 S2 S3... SX) ^N = Goal State
 He calls such a processes "Cyclicly Decomposable".

I think that the results would be far richer if there was also allowed
to be one cube rotation in the subprocess.

I know of 2 examples.
I will use a * after a move to represent a full cube move.

I am a little rusty on this one, and I don't have a cube here to verify it,
but
        (L' R F*) ^ 6           (12q)
is (or should be, if I remember it correctly) the Pons Asinorum.
We also know that this pattern takes at best 12q, so it is actually optimum.
The existance of this process has always made me wonder how many
different ways there are to do the Pons, especially with different face
effects in the Supergroup.

The other one is my favorite process.
        (L D L' R' F'*)^4       (16q)
This twists 3 corners on one face.
I suspect this one is also optimum as I have never heard of a process that
twists 3 corners in less than 16q.
It has one very interesting feature, L' R' F'* can be done as 1 combined
two-hand motion.
A casual observer may think you are only turning the cube and not see the
face turns involved.  This makes the process look magic, achieving a state
in far fewer apparant moves then people think is possible.
This process is so fast and easy to remember that it is what I use while
solving.

From news@nntp-server.caltech.edu  Fri Dec 16 15:36:21 1994
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From: txr@alumni.caltech.edu (Tim Rentsch)
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Subject: Re: Cyclic Decomposition
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mark.longridge@canrem.com (Mark Longridge) writes:

> Certain states, such as the 12-flip, require quite a few moves, in
>fact more moves than possible to search using brute force even when
>using high-order computers. The best results using the Kociemba
>algorithm need 28 q turns or 20 q+h turns for the 12-flip.

I found Mark's post generally interesting and thought provoking.
Without detracting from his ideas I would like to comment on the
paragraph above.

If a certain state (such as the 12 flip) is known to be reachable
in no more than 20 moves, then isn't that state within reach of
a brute force search?  Start one brute force at the initial state,
one at the final state, expand the position trees one move at a time
until the trees touch.  A state 20 moves from start will require a
tree (well, two trees) 10 moves deep, which is about 10 billion states.
That seems achievable in a reasonable time on fast computers of today.
Doesn't it?

regards,

Tim Rentsch

From mreid@ptc.com  Fri Dec 16 17:24:16 1994
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From: mreid@ptc.com (michael reid)
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To: cube-lovers%life.ai.mit.edu@ptc.com
Subject: Re: Cyclic Decomposition
Content-Length: 2380

tim writes

> If a certain state (such as the 12 flip) is known to be reachable
> in no more than 20 moves, then isn't that state within reach of
> a brute force search?  Start one brute force at the initial state,
> one at the final state, expand the position trees one move at a time
> until the trees touch.  A state 20 moves from start will require a
> tree (well, two trees) 10 moves deep, which is about 10 billion states.

unfortunately, this estimate is too optimistic.  the number of positions
within 10 face turns of start is more like  2.6 x 10^11.

[ keep in mind that while "billion" means  10^9  in the u.s., it may
  mean  10^12  elsewhere. ]

> That seems achievable in a reasonable time on fast computers of today.
> Doesn't it?

i don't know, but it would be nice if it were possible.

i recall that dik winter was doing some work on this front, although i
think he was working on "superfliptwist".  also, he was using kociemba's
algorithm (first stage only).  my impression was that this would take
too long.  (any results here, dik?)

however, there's a method similar to the one tim mentions that hasn't
received much attention here.  i don't have all the details handy, but
here's a sketch:

the idea is to start with a list of permutations  L  and to
generate (on the fly!) all products  p1 p2  (with  p1, p2  in the list  L)
in (lexicographically) increasing order.  this means that while the list
itself is stored in memory, the list of products (denoted  L L) need
not be.  also, the technique for doing this (which i don't remember
offhand) is easily adapted to generate all products  q p1 p2  where
q  is a fixed permutation and  p1 p2  are in the list L  (q L L), again
in (lexicographically) increasing order, and again, on the fly.

now let  L  be the list of all configurations  within 5 face turns of
start, and let  q  be "superflip" or "superfliptwist".  now simultaneously
generate the products  L L  and  q L L  in increasing order, and look for
common configurations.  a common configuration gives

           p1 p2  =  q p3 p4   ==>  q  =  p1 p2 p4^-1 p3^-1

which gives a manuever of (at most) 20 face turns for  q.

of course, this technique can be used for quarter turns as well.

i don't know much about the practicality of implementing this algorithm,
but i'd be happy to hear from anyone who's done it, or even thought
about it.

mike

From ccw@eql12.caltech.edu  Fri Dec 16 20:47:12 1994
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From: ccw@eql12.caltech.edu (Chris Worrell)
Message-Id: <941216174701.25001b45@EQL12.Caltech.Edu>
Subject:  correction on my previous message.
To: cube-lovers@life.ai.mit.edu

My memory is indeed faulty.  I was incorrct about the repeated process
which yields the Pons Asinorum.  The process I was actually thinking of
does not need a cube rotation to make it have a repeated structure.
 (L' R U' D)^3
There is however an interpretation of the standard Process for the Poms, which
can be decomposed into a repeated process by a cube rotation.
Denote by "A",  turning by 120 degrees around any diagonal, it doesn't matter 
which one, nor in which direction.
 (L^2 R^2 A)^3  yields the Pons.

From news@nntp-server.caltech.edu  Fri Dec 16 23:24:15 1994
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From: txr@alumni.caltech.edu (Tim Rentsch)
Newsgroups: mlist.cube-lovers
Subject: Re: Cyclic Decomposition
Date: 17 Dec 1994 01:06:24 GMT
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mreid@ptc.com (michael reid) writes:

>unfortunately, this estimate is too optimistic.  the number of positions
>within 10 face turns of start is more like  2.6 x 10^11.

An upper bound for number of positions reachable after 10
turns is

    18 * 12**9

which is 92,876,046,336.  Admittedly this number is closer to 2.6e11
than 1e10, but the number is an upper bound.  It seems to me I
remember reading that the limiting branching factor (for q+h turns) is
about 9.5 and is reached rather quickly.  The value of

    18 * 12 * 12 * 12 * 9.5**6

is 22,864,298,166.0 (according to 'bc'), which should be within reach
of brute force algorithms.  Unfortunately this approach requires
several hundred gigabytes of disk space but that could be spread out
over lots of physical machines (parallelizing could also result in
speeding up the computation).  Anyone know where we could find 1000
machines with a few hundred megabytes free each?

Well, maybe not just yet.  But soon.

regards,

Tim Rentsch

From BRYAN@wvnvm.wvnet.edu  Sat Dec 17 11:10:56 1994
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From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   How Big is Big?

Some of the notes in the last day or two about whether or not ten levels
deep is too large to search reminded me of a note I have been meaning
to send for a long time.  Just how big is 4.3*10^19, and can we ever
hope to search it all?

First of all, 4.3*10^19 is really about 10^18.  That is, we could safely
confine ourselves to searching M-conjugate classes, and there are about
0.9*10^18 classes, which we might as well call about 10^18.  But how big
is that?

Suppose were trying to buy enough disk space.  I claim that you
could store each position in a byte with clever indexing.  Actually,
you could store each position in 5 bits, or 5/8 of a byte, but leave
it as a byte per position.  Let's say that you can purchase
a gigabyte for about 1,000 U.S. Dollars (10^12 bytes for about
10^3 USD).  (We are buying good quality used disks for mainframes
for about 1,000 USD per gigabyte; new prices are closer to 4,000 or
5,000 USD per gigabyte.  Both SCSI and IDE disks for the desktop,
PC or UNIX, are just now down to around 500 USD per gigabyte, and I
have seen firesale type prices closer to 300 USD per gigabyte).

At 10^3 USD per 10^12 bytes, the cost would be 10^9 USD per 10^18
bytes.  Well, 10^9 USD is a lot of money, but it is a lot less
than the cost of going to the moon, or the cost of an aircraft carrier.
In fact, Bill Gates could afford it if he so chose.

There are other ways to think about the problem.  The size of
chess is about 10^75 states, and Go is about 10^120 states.  The
standard 3x3x3 Rubik's cube is vastly smaller than either of these.
In fact, Go (and maybe chess, I can't remember for sure) is usually
described as being bigger than the universe.

A handy number in these types of comparisons and in determining "how big
is the universe" is Avogadro's number, which is about 6*10^23.
Avogadro's number is the number of molecules (or atoms, for substances
which occur atomically) in the gram molecular weight of a substance.
For example, molecular hydrogen has a molecular weight of 2, so
2 grams of hydrogen contain 6*10^23 molecules.  Iron is atomic with
an atomic weight of 56, so 56 grams of iron contain about 6*10^23
atoms.  If you had 56 grams of iron, and if you could store magnetically
each cube position in no more than 6*10^5 iron atoms, then you could
store the whole Rubik's cube.

By comparison to the size of the universe, the mass of the sun is
about 10^30 grams, consisting mostly of atomic hydrogen, so there
are about (10^30)*(10^23)=10*53 hydrogen atoms in the sun.  I can't
remember for sure, but I think there are about 10^11 stars in the
Milky Way.  If the sun is typical star, that would leave about
10^64 hydrogen atoms in the Milky Way.  I don't know how many galaxies
there are, but we are clearly getting close to the size of Chess
at 10^75 being about the same as the size of the universe, and of Go
at 10^120 being much larger than the size of the universe.  Rubik's
cube is small potatoes.

A couple of more items: the human genome is being mapped.  I cannot
remember the exact size of the problem, but I do remember when I
read about it that it was a larger problem than Rubik's cube.  Finally,
the Chronicle of Higher Education had an article in the last few weeks
about particle physicists and the Internet.  Traditionally, these people
send hundreds or thousands of magnetic tapes to each other via standard
mail (snail mail to E-mail folks  --  but mailing magnetic tapes can
yield tremendous data transfer rates if you actually calculate bytes
per second).  According to the article,
the physicists are already sending gigabytes over the Internet.  They are
planning soon to start sending petabytes (10^15) over the Internet.
10^15 is getting interesting close to the size of Rubik's cube
(never mind that I thought that the proper term for 10^15 bytes was
terabytes.)

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From mreid@ptc.com  Sat Dec 17 14:53:48 1994
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From: mreid@ptc.com (michael reid)
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To: cube-lovers%life.ai.mit.edu@ptc.com
Subject: planning in permutation groups
Content-Length: 866

here is some more info on the algorithm i briefly described yesterday.

the paper i have is "planning and learning in permutation groups", by
fiat, moses, shamir, shimshoni and tardos.  it appeared in the 30th
annual symposium on foundations of computer science, pp 274-9.

the paper mentions that they have calculated all identities of length 16
(they count face turns).  this means that the algorithm was successfully
implemented for the list  L  of all configurations within 4 face turns
of start!

also, alan gives a much more detailed description of this algorithm
in the archives.  see his message of may 27, 1987 "shamir's talk really
was about how to solve the cube!"  (cube-mail-6)

i encourage people to look up the paper and/or alan's message.  this is
an exciting development, and the lack of attention this idea has
received is a shame.

regards,

mike

From BRYAN@wvnvm.wvnet.edu  Sat Dec 17 23:44:44 1994
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From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Re: How Big is Big?
In-Reply-To: Message of 12/17/94 at 12:55:35 from dlitwin@geoworks.com

On 12/17/94 at 12:55:35 dlitwin@geoworks.com said:

>"Jerry Bryan" writes:
>> I claim that you could store each position in a byte with clever
>> indexing.  Actually, you could store each position in 5 bits, or 5/8 of a
>> byte...

>        Could you explain what you mean by this?  You can't mean each
>possible cube position because you only get 256 from a byte.  Are you
>talking about each type of operation you can perform on a cube?  I'd buy
>that, but I'd think you could store that in 4 bits (12 possible moves).
>I'm clearly missing something here.

I have talked about this before, but let's have a go at it again.
Previously, I have talked about it in terms of corners only or
edges only.  This time, let's talk about it in terms of whole cubes.
In terminology we have used recently, we will talk about
representing G[C,E]=<Q[C,E]>.  That is, we will only represent
corners and edges.  There is no need for the purposes of
this paper to include Face centers because |G[C,E]| = |G[C,E,F]|.

For each cube position, we only need to store the depth, assuming
we have some way to index to the proper cell in a data structure
containing the depth for each cube position.  As long as the depth
does not exceed 31, then 5 bits will suffice for each cell.

Start with G[C] and G[E] separately (corners only, and edges only).
Partition G[C] into equivalence classes of the form {m'(Xc)m}
for each m in M (the set of 48 rotations and reflections), for each
c in C (the set of 24 rotations), and for each X in G[C].
Partition G[E] into equivalence classes of the form {m'(Yc)m}
for each m in M, for each c in C, and for each Y in G[E].  These
tasks have already been accomplished via computer search.

For each {m'(Xc)m} choose a representative element V, and for each
{m'(Yc)m} choose a representative element W.  It is not strictly
necessary, but it will prove convenient if each representative
element is even, and such a choice is always possible.  Denote the
sets of representative elements as G*[C] and G*[E].  These sets
have already been created via computer search.  We have
|G*[C]|=77802 and |G*[E]|=851625008.  The sets G*[Cl and G*[E]
will be used as indices, and will have to be stored.  But storing
them is between 10^12 and 10^13 bytes, which is a drop in the
bucket compared to storing 10^18 depths.

We can think of a cube in G[C,E] as XY with X in G[C] and Y in G[E].
That is, X is the corners and Y is the edges.  Both X and Y are even,
or both X and Y are odd, and the choice of odd or even can be thought
of as an index which doesn't have to be stored.  V=Repr{m'(Xc)m} can
be thought of as an index for XY.  V has to be stored, but it only
has to be stored once for the whole data structure, not once very
every position XY for which V=Repr{m'(Xc)m}.  Similarly,
W=Repr{m'(Yc)m} can be thought of as an index for XY, and W only
has to be stored once for the entire data structure.

Given V, we can write X=n'(Vd)n for some fixed n in M and for
some fixed d in C.  Notice that since V is even, the parity of
d is the same as the parity of X, and hence there are 12 rather than
24 choices for d.  Notice also that while both n and d will always exist,
neither is necessarily unique, depending on how "symmetrical" is V.
Hence, a selection procedure is necessary to assure that both n and
d are unique.  d can be thought of as an index for XY, and d does
not need to be stored.  As for n being an index, see two paragraphs
below.

Given W, we can write Y=o'(We)o for some fixed o in M and for some
fixed e in C.  Notice that since W is even, the parity of
e is the same as the parity of Y, and hence there are 12 rather than
24 choices for e.  Notice also that while both o and e will always exist,
neither is necessarily unique, depending on how "symmetrical" is W.
Hence, a selection procedure is necessary to assure that both o and
e are unique.  e can be thought of as an index for XY, and e does not
need to be stored.  As for o being an index, see the next paragraph.

We could think of n and o as both being indices for XY, with both
of them having 48 different values.  The indices would not have to be
stored.  However, we can write XY as (n'(Vd)n)(o'(We)o).  Any
M-conjugate of XY has the same length as XY.  If we conjugate by nn'
we have

   n(n'(Vd)n)(o'(We)o)n'=n(n'(Vd)n)n')(n(o'(We)o)n')=(Vd)(p'(We)p),

where p=on', p'=no', and p is in M.  Hence, there is only one index
into M with 48 different values, not two.

Putting this all together, we need a table with
2*77802*851625008*12*12*48 cells, and each cell could be 5 bits.
The total number of cells is about .916*10^18.  The actual number
of M-conjugate classes is about .901*10^18.  (I am using a slightly
unusual decimal point placement in deference to the total size of
the table being "about 10^18".)   The reason that the table size
is a bit larger than the number of M-conjugate classes is that the
table will contain some empty cells due to the non-uniqueness of
some of the indexing by C and M.  The number of cells that will be
non-empty *will* in fact be exactly the same as the number of
M-conjugate classes.

I have talked about indices that would have to be stored, and indices
that would not have to be stored.  As an example of indices that
would have to be stored, consider a table of names and ages.  E.g.,

       Name           Age

       Doe, John      25
       Evans, Bill    42
       Jones, Jane    33
       Smith, Sarah   21

You can think of the names as indices into the ages, and the names do
have to be stored.

On the other hand, think of storing N floating point numbers in an
array X, with I as an index for I in 1..N.  You would write this
in a program as something like X[I].  The index I would have to be
stored once, I suppose, but it would not have to be stored with each
X.

Similarly, in the proposed structure for storing all of God's
Algorithm, the indices V and W would have to be stored, but the
parity index 1..2 would not have to be stored, the rotation index
1..12 for V would not have to be stored, the rotation index 1..12
for W would not have to be stored, and the M conjugation
index 1..48 for V or W (but not both) would not have to be stored.
But even though the indices V and W would have to be stored, they
would only have to be stored once for the whole program, not for
each cube position.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From BRYAN@wvnvm.wvnet.edu  Sun Dec 18 10:23:35 1994
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From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Re: How Big is Big?
In-Reply-To: Message of 12/17/94 at 22:46:08 from txr@alumni.caltech.edu

On 12/17/94 at 22:46:08 txr@alumni.caltech.edu said:
>In mlist.cube-lovers you write:

>>For each cube position, we only need to store the depth, assuming
>>we have some way to index to the proper cell in a data structure
>>containing the depth for each cube position.  As long as the depth
>>does not exceed 31, then 5 bits will suffice for each cell.

>I think depth modulo 3 is enough, since depth of adjacent positions
>will differ by at most one -- just move in the direction of depth
>getting less.  So we could get by with 2 bits per cell.

>regards,

>Tim Rentsch

You are certainly correct.  And as Dan Hoey pointed out to me via
private E-mail once upon a time, for Q turns you can get it down
to only one bit by storing (depth modulo 4)/2 because you can infer
the state of the low order bit from the parity of cube position.
(Parity of the cube position equals the parity of the depth for Q turns,
but not for Q+H turns.)

But I tend to think that certain kinds of interesting analyses
of a data base for the entire God's Algorithm
would be greatly assisted by storing the entire depth.

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Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From mouse@collatz.mcrcim.mcgill.edu  Sun Dec 18 16:02:19 1994
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Date: Sun, 18 Dec 1994 15:56:10 -0500
From: der Mouse <mouse@collatz.mcrcim.mcgill.edu>
Message-Id: <199412182056.PAA13839@Collatz.McRCIM.McGill.EDU>
To: cube-lovers@ai.mit.edu
Subject: Re:  How Big is Big?

> [Physicists] are planning soon to start sending petabytes (10^15)
> over the Internet.  10^15 is getting interesting close to the size of
> Rubik's cube (never mind that I thought that the proper term for
> 10^15 bytes was terabytes.)

I thought it was

	kilo	10^3
	mega	10^6
	giga	10^9
	tera	10^12
	peta	10^15
	exa	10^18

except, of course, that as applied to quantities that tend to come in
powers of two, like bytes, they normally refer to 2^10, 2^20, 2^30,
2^40, 2^50, and 2^60 instead.  (This is a common problem when buying
disks: manufacturers like to quote capacities in terms of powers of
ten, because it makes their disks seem larger than they really are.  A
"2.1G" disk, for example, typically has a capacity of about 2.1e9
bytes...which is really only about 1.956Gb.  The error can be roughly
estimated as 2.5% per power of 10^3: 2.5% for K, 5% for M, 7.5% for G,
etc.  Semiconductor memory manufacturers generally get this right,
probably because doing other than powers of two would be hard for them.)

It also means that a certain well-known manufacturer of data drives for
8mm videotape is being extremely arrogant with their choice of name. :-)

As for the 10^18 bytes of storage estimated (probably only about half
that, if we consider that we really need only 5*.9e18 bits, less if we
resort to some of the clever coding tricks recently mentioned)...that's
about a gig of storage each across a million machines.  The net's not
quite to the point where it can be done distributed.

Yet. :-)

Incidentally, someone mentioned that you need only store two bits, or
even only one if you don't use H turns, per position, because you don't
need to know more than how to get closer to Start...and then said that
it would be nice to have the full depth available nevertheless.  If you
have this enormous database of .9e18 positions available in the compact
form, all that's needed to get the full depth for a position is to take
the short walk through the tree back to Start.

Also note that the Cube database storage size requires the highest
prefix we have.  Time to get SI to think up some more, I guess :-)

					der Mouse

			    mouse@collatz.mcrcim.mcgill.edu

From dik@cwi.nl  Sun Dec 18 16:43:21 1994
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Date: Sun, 18 Dec 1994 22:43:19 +0100
From: Dik.Winter@cwi.nl
Message-Id: <9412182143.AA04820=dik@boring.cwi.nl>
To: cube-lovers@ai.mit.edu
Subject: Re:  How Big is Big?

 > Also note that the Cube database storage size requires the highest
 > prefix we have.  Time to get SI to think up some more, I guess :-)

They did:
  10^-24   y   yocto
  10^-21   z   zepto
  10^+21   Z   zetta
  10^+24   Y   yotta
so now you can talk about yotta bytes.

dik

From BRYAN@wvnvm.wvnet.edu  Sun Dec 18 17:32:44 1994
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Date:      Sun, 18 Dec 1994 17:32:40 EST
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <cube-lovers@ai.mit.edu>
Subject:   Re: How Big is Big?
In-Reply-To: Message of 12/18/94 at 15:56:10 from ,
           mouse@collatz.mcrcim.mcgill.edu

On 12/18/94 at 15:56:10 der Mouse said:
>> [Physicists] are planning soon to start sending petabytes (10^15)
>> over the Internet.  10^15 is getting interesting close to the size of
>> Rubik's cube (never mind that I thought that the proper term for
>> 10^15 bytes was terabytes.)

>I thought it was

>        kilo    10^3
>        mega    10^6
>        giga    10^9
>        tera    10^12
>        peta    10^15
>        exa     10^18

You are correct, of course.  In retrospect, the aspect of the
article in the Chronicle that waylaid me (and which I still find
puzzling) is the absence of any mention of "tera".  It is a giant
jump from "giga" to "peta", skipping "tera" on the way.  But "giga"
and "peta" were juxtaposed in the article.  I have known how big
"tera" is for years  -- can't believe I screwed it up.  It makes me
wonder if the article had it right.  It is reasonable to jump from
gigabytes to petabytes in one fell swoop?

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From mouse@collatz.mcrcim.mcgill.edu  Mon Dec 19 03:55:27 1994
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Date: Mon, 19 Dec 1994 03:49:19 -0500
From: der Mouse <mouse@collatz.mcrcim.mcgill.edu>
Message-Id: <199412190849.DAA15068@Collatz.McRCIM.McGill.EDU>
To: cube-lovers@ai.mit.edu
Subject: Re: How Big is Big?

> In retrospect, the aspect of the article in the Chronicle that
> waylaid me (and which I still find puzzling) is the absence of any
> mention of "tera".  It is a giant jump from "giga" to "peta",
> skipping "tera" on the way.  But "giga" and "peta" were juxtaposed in
> the article.  [...]  It makes me wonder if the article had it right.
> It is reasonable to jump from gigabytes to petabytes in one fell
> swoop?

IMO it is not.  Without seeing it, I can't be sure, but it seems likely
that it's Just Another Dumb Reporter.  Perhaps someone took notes and
wrote down 10^15 instead of 10^12, and then looked up the name for
10^15 and didn't notice the basic inconsistency of jumping from Gb to
Pb without stopping at Tb.

Hmmm, this is drifting off-topic for cube-lovers....

					der Mouse

			    mouse@collatz.mcrcim.mcgill.edu

From BRYAN@wvnvm.wvnet.edu  Mon Dec 19 08:48:32 1994
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From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Re: How Big is Big?
In-Reply-To: Message of 12/17/94 at 11:10:52 from BRYAN@wvnvm.wvnet.edu

One more correction to this giga, tera, peta, nonsense.  Since
10^9 bytes of disk space cost about 10^3 USD, then 10^18 bytes would
cost about 10^12 USD.  This is more than Bill Gates could afford, more
than going to the moon, more than an aircraft carrier, and indeed is
of the same order of magnitude as the entire United States federal
budget.  Disk prices need to come down several orders of magnitude
before we can think about storing God's Algorithm.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From Wechsler@world.std.com  Mon Dec 19 09:59:54 1994
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From: Wechsler@world.std.com (Allan C Wechsler)
Message-Id: <199412191500.AA05068@world.std.com>
To: mouse@collatz.mcrcim.mcgill.edu
Cc: cube-lovers@ai.mit.edu
In-Reply-To: der Mouse's message of Sun, 18 Dec 1994 15:56:10 -0500 <199412182056.PAA13839@Collatz.McRCIM.McGill.EDU>
Subject:  How Big is Big?

   Date: Sun, 18 Dec 1994 15:56:10 -0500
   From: der Mouse <mouse@collatz.mcrcim.mcgill.edu>

   > [Physicists] are planning soon to start sending petabytes (10^15)
   > over the Internet.  10^15 is getting interesting close to the size of
   > Rubik's cube (never mind that I thought that the proper term for
   > 10^15 bytes was terabytes.)

   I thought it was

	   kilo	10^3
	   mega	10^6
	   giga	10^9
	   tera	10^12
	   peta	10^15
	   exa	10^18

[...]

   Also note that the Cube database storage size requires the highest
   prefix we have.  Time to get SI to think up some more, I guess :-)

(Warning to Cube-Lovers: this is off the topic, but it's a digression
I can never resist.  Alan is going to come over to my house and soap
my windows for this, I just know it.)

They _have_ thought up some more -- this was in Science News about 18
months ago.  But the ones they thought up are absolutely awful, and I
want to take this opportunity to advertise my own suggestions.

First note the following relationships, which I believe are entirely
the result of coincidence:

te(t)ra     1000^4	
pe(n)ta     1000^5
(h)exa      1000^6

In each case, the prefix for 1000^n looks like the neo-greek prefix
for n, with the second-to-last consonant deleted.  I merely propose
that we continue this scheme:

he(p)ta     1000^7
o(c)to      1000^8
(en)nea     1000^9

I admit to a fudge with n=9, but I like neabytes better than
eneabytes, and the prefix E was already taken by n=6.  I wanted to
keep up the unique sequence of prefixes: K, M, G, T, P, E, H, O, N.

For those who care, megameters are good for measuring small planets,
gigameters for big planets and stars, and terameters for solar
systems.  A petameter is about a tenth of a light year, and so it's
good for measuring near interstellar distances; exameters are good for
the 100-ly range, galaxies should be measured with hetameters, and
intergalactic distances with otometers.  Current theory says the
universe is considerably smaller than one neameter.

From @mitvma.mit.edu:will4086@UDCVAX.BITNET  Wed Dec 21 18:48:20 1994
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Date: Wed, 21 Dec 1994 18:50:19 EST
From: will4086%UDCVAX.BITNET@mitvma.mit.edu
To: CUBE-LOVERS@life.ai.mit.edu
Message-Id: <0098948C.2AA66560.3192@UDCVAX.BITNET>
Subject: MAILING LIST

I WOULD LIKE TO BE PLACED ON THE MAILING LIST FOR THE SUBJECTS YOUR GROUPS
DISCUSS.THIS IS THE FIRST E-MAIL MESSAGE THAT I HAVE SENT,SO PLEASE DON'T
LAUGH AT THIS PARAGRAPH.

From dik@cwi.nl  Wed Dec 21 20:16:25 1994
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Date: Thu, 22 Dec 1994 02:16:03 +0100
From: Dik.Winter@cwi.nl
Message-Id: <9412220116.AA03233=dik@boring.cwi.nl>
To: cube-lovers@life.ai.mit.edu
Subject: CFF 35

CFF #35 came out, the editors expected it in December and it came
out in December!  Good for them.

Summary of contents.

Vic Stok: Paving stones.
  A new twist to polyominos.  Squares are linked in a brick-wall
  fashion.

Lee Sallows: Alphamagic squares.
  About the construction of magic squares where, if you replace each
  entry with the value of the word length, the result is magic again.
  The most surprising for me was one square that was alphamagic in
  both Welsh and Norwegian.

Torsten Sillke and Bernhard Wiezorke: Stacking Identical Polyspheres.
  Part 1: Tetrahedra.
  Discusses many possible and impossible tetrahedra that can be
  packed by polyspheres.

Jan de Ruiter: Braiding.
  An article about a contest problem issued on the Dutch 1992 Cube Day.
  It involves (amongst others) the number of ways a braid can be made
  from a varying number of bundles of hair.

Joop van der Vaart: IPP 1994 Impressions.
  Impressions from the 1994 International Puzzle Party in Seattle.

Leo Links: Cube Day Impressions.
  Impressions of the 1994 Dutch Cube Day in Stavoren.

Result of contest 24 (CFF 33, Cross Pattern Piling).

Anneke Treep: Anti-slide... a winner!
  A short note about the Hikimi Wooden Puzzle Competition.  Wil Strijbos
  from the Netherlands came second with his puzzle.  Start with 15 1x2x2
  square pieces and a 4x4x4 box.  Pack the pieces in the box so that no
  piece can slide.  Do the same with 14, 13, 12, 11 (actually the article
  has a typo here).

Columns:

Mark Peters: Books and Magazines (reviews)
Edward Hordern: What's Up? (details some new puzzles and other news)
------
CFF (Cubism For Fun) is the newsletter published by the
Nederlands Kubus Club NKC (Dutch Cubists Club).
Information can be obtained from one of the editors:
Rik van Grol <rik@dutncp8.tn.tudelft.nl>.

Membership fee is NLG 25 individual, NLG 80 institutional.
(USD 1 ~ NLG 1.70).  Applications for membership to the
treasurer:
   Lucien Matthijsse
   Loenapad 12
   3402 PE  IJsselstein
   The Netherlands

If you write, please add an international reply coupon
(can be obtained at your post office).
--
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland, +31205924098
home: bovenover 215, 1025 jn  amsterdam, nederland; e-mail: dik@cwi.nl

From jkato@tmastb.eec.toshiba.co.jp  Wed Dec 21 20:43:59 1994
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Date: Thu, 22 Dec 94 10:42:11 JST
From: jkato@tmastb.eec.toshiba.co.jp (Toshi Kato)
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Message-Id: <9412220142.AA02582@tmastb.eec.toshiba.co.jp>
To: cube-lovers@life.ai.mit.edu
Subject: IPP #15

To: International Puzzle Collectors

Dear Sirs,

15th International Puzzle collectors' Party(15th IPP) will be taken place
on April 15-16,1995 in Tokyo,Japan and optional HAKONE tour on April 17.

Have you received an invitation letter of 15th IPP from Nob Yoshigahara?
And then, you have done to reply to Nob, haven't you? So, thanks.
If you did not yet, please answer by express snail mail or fax or e-mail,
as soon as possible.

We are looking forward you come to Japan. 

Thank you,
Toshi(Junk) Kato
--------------------------------------JUNK: jkato@tmastb.eec.toshiba.co.jp
(Notice)If you interest in Puzzle KONWAKAI(Academy of recreatinal mathe-
matics,Japan), you may attend the meeting on April 22,1995 for a guest.


From mmoss@panix.com  Thu Dec 22 10:45:49 1994
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Please unsubscribe me.

mmoss@panix.com

From magnum@cyberstore.ca  Thu Dec 22 13:23:33 1994
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unsubscribe

Darryl EJ Ruff, Dir.
Magnum Results Corp.
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Fax:  604/769-6158
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"..If We Fail To Achieve Superior
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From epaytl@epa.ericsson.se  Thu Dec 22 16:34:29 1994
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From: epaytl@epa.ericsson.se (Y T - T/ZA)
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Please unsubscribe me.

epaytl@epa.ericsson.se

From alan@curry.epilogue.com  Thu Dec 22 23:38:49 1994
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In-Reply-To: Matthew Moss's message of Thu, 22 Dec 1994 10:45:48 -0500 (EST) <199412221545.AA13543@panix2.panix.com>
Subject: UNSUBSCRIBE

   From: Matthew Moss <mmoss@panix.com>
   Date: Thu, 22 Dec 1994 10:45:48 -0500 (EST)

   Please unsubscribe me.

   mmoss@panix.com

   Date: Thu, 22 Dec 1994 10:23:58 -0800
   From: Darryl EJ Ruff <magnum@cyberstore.ca>

   unsubscribe

   Darryl EJ Ruff, Dir.
   Magnum Results Corp.
   PO Box 692, Stn A
   Kelowna, BC
   Canada  V1Y 7P4
   Ring: 604/769-6169
   Fax:  604/769-6158
   Internet: magnum@cyberstore.ca

   "..If We Fail To Achieve Superior
   Results, We Won't Accept Your Money.."

   Date: Fri, 23 Dec 94 08:34:21 DST
   From: Y T - T/ZA <epaytl@epa.ericsson.se>

   Please unsubscribe me.

   epaytl@epa.ericsson.se

I have removed all three of you from the Cube-Lovers mailing list.

Note, please, that all three of you sent your requests to be removed to
Cube-Lovers as a whole.  You should have sent them to me,
Cube-Lovers-Request@AI.MIT.EDU, instead of bothering the entire list.  This
was all clearly explained in the introductory note I sent you when you
first subscribed (earlier this month for two of you).  This is, in fact, a
wide-spread Internet convention.  If you can remember it, you can often
avoid looking like an idiot in front of hundreds of people.

I'm sorry to bother everybody else on Cube-Lovers by CC'ing this note to
you all, but this way I can perhaps prevent more copy-cat repetitions of
the same annoying mistake.

REMEMBER:

  SEND SUBMISSIONS TO: Cube-Lovers@AI.MIT.EDU

  SEND ADMINISTRATIVE CORRESPONDENCE TO: Cube-Lovers-Request@AI.MIT.EDU

GOT IT?

From ba05133@bingsuns.cc.binghamton.edu  Fri Dec 23 10:24:30 1994
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Please unsubscribe me.


From BRYAN@wvnvm.wvnet.edu  Fri Dec 23 18:26:02 1994
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Date:      Fri, 23 Dec 1994 18:25:57 EST
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   <U,R> Antipodal Processes

With the help of a suggestion from Dan Hoey, I can now provide processes
for the 27 antipodal positions of <U,R> that are unique up to
W-conjugancy.  (It is the positions that are unique up to W-conjugancy,
not necessarily the processes.)  Without further ado, here are processes
which generate each of the 27 positions.

 1 R  U  U  R  R  U' R  U' R  U' R  U  R' U  R  U  R' U' R  R  U' R' U  R  U
 2 R' U  R' U  R' U  R  R  U' R' U' R' U' R  U' R' R' U' R  R  U' R' U  R  U
 3 R  R  U  R' U  R  U' R' U  R' R' U  R  R  U' U' R  R  U  R  R  U  R  R  U'
 4 U' R  U  R' U  R  U' R' U' R' R' U  R' U' R  U  U  R' U  R' U' R' U  R  U'
 5 R' R' U  U  R  U  R  U' R  U' U' R' U  R  R  U' R  R  U' R' U  R' U  R' U'

 6 U' R  R  U  R' U  R' R' U' R' U  U  R  U' R' U  R  R  U  R  R  U  R' U' R
 7 U  R' U  R' U  R  U  U  R' U' R  R  U  R' U' R' U  R  R  U  R' U  R  U' R
 8 R' U  R' U' U' R  U' R  U' U' R' U  R' U  R' R' U' R  U  R  U  R  U' R  U
 9 U  R  U' R  U  R  R  U' R  U' R' U  R  U' R' R' U  U  R' R' U' R  U' R  R
10 U' R  R  U  R  U  U  R' U' R' U  R  U' R' R' U' U' R' U' R' U' R' U  R  U'

11 R' U  R  U  R' U' R  U  R' U' U' R' U' R  R  U  U  R' U  R  U  R  U' R' R'
12 R  U  U  R' U  U  R' U  R  U' R' U  R' U' R  U' R' U' R' U  R  U' R  R  U
13 R  R  U  R  R  U' R  U' U' R  U' R  R  U' R' R' U' R  R  U' R  U' R  R  U
14 U' R' U' R' U' R' U  R  R  U' R  U' R  U' R  U' R' U  R  U' U' R  U  U  R
15 R  R  U' R' R' U' R  U  U  R' U  U  R' R' U' R' U' U' R' R' U' R  U' U' R

16 U  R' U  R' R' U' R  U' R  U  U  R  R  U' R  U  R' U' R  U' R' U' R  R  U'
17 U' R  U' U' R' R' U' R  U' R  U' R  U' R  R  U  R' R' U' R' R' U' R' R' U'
18 R  U  U  R  U  R  U' R' U  R' U  R  U' R  R  U' R' U' R  U  U  R  U' U' R'
19 R' U  R' U' U' R  U' R  U' R' R' U  U  R  U' R' U  R  U' R  U  R  U  U  R'
20 U  R  R  U' R  R  U' R' U' R' U  U  R' U  R' U  R  U  U  R' U  R  U' R  R

21 R' U  R' U' R' U  R  U  U  R' U  R' U  R' U' R  U' R' U  R  U  R' U  R  R
22 U  R  U  U  R  U' R  U  R' U' R  U  U  R  U' R  U' R' R' U  R  R  U  R  U
23 R' U  U  R  U  R' U  R' U  R  U  R' U' R  R  U' R  U' U' R' U  R  R  U  R
24 R' U  R' U' R  R  U  U  R' U  R  U' R' R' U  U  R' U  R' U' U' R  U  U  R
25 U  R' R' U  U  R  U  R  U' U' R  U' R  U  U  R' U  R' U' U' R' U  R  R  U

26 U' R' U  R  U' R' U  U  R  R  U  R' U  R' U  R' U' U' R  R  U' R' U' R' R'
27 R  U' R' U' U' R  R  U' R' U' R  U' R  U' U' R' U' R' R' U' R  U' R  R  U'

The 27 processes are in the same order as the 27 positions I posted on
11 November this year.  However, I want to repost the 27 positions anyway.
I found a formatting inconsistency in that posting.  Generally speaking,
when you unfold the cube for printing with the Back above, you can choose
to print the Back face right-side-up or up-side-down, and up-side-down
makes more sense to me.  All my screen displays work that way, and it
makes the cubies move smoothly under repeated applications of the R or L
operators.  However, I discovered that the print program I used for the
11 November posting printed the edge facelets of the Back face right-side-
up.  That wouldn't have been so bad, I suppose, but at the same time I
printed the corner facelets correctly as up-side-down.  So herewith
I reprint the 27 antipodal positions with all the Back faces correctly
up-side-down.

     BBL                      BBL                      BBU
     BBU                      BBU                      BBF
     LRF                      RRF                      LRR

     UDR                      FDU                      UBD
     BUU                      BUU                      DUU
     FUB                      FUR                      UUF

 FRD RFR DRU              URD RFB URL              FRB LFD RLB
 LLL FFF RRB              LLL FFF RRB              LLL FFU RRR
 LLL FFB RLU              LLL FFD RLU              LLL FFF UBR

     DDU                      DDB                      DDR
     DDU                      DDU                      DDU
     DDB                      DDB                      DDB

  ---------------------------------------------------------------

     BBU                      BBR                      BBU
     BBF                      BBU                      BBF
     DRR                      BRB                      FBD

     FDU                      RDL                      LUB
     UUB                      FUB                      UUU
     BUR                      RUL                      LUF

 RBR DFB URF              URU FFF URU              ULU BFD RRR
 LLL FFU LRR              LLL FFU LRR              LLL FFB RRR
 LLL FFL BRL              LLL FFR BBF              LLL FFU RRR

     DDU                      DDD                      DDF
     DDU                      DDU                      DDD
     DDF                      DDD                      DDB

  ----------------------------------------------------------------

     BBR                      BBU                      BBB
     BBU                      BBU                      BBU
     LRD                      RLF                      FRF

     UDB                      DUR                      UFD
     UUF                      FUD                      DUU
     UUR                      BUU                      UUR

 FRB LFF DRR              FRR DFR BRU              RRL FLB URR
 LLL FFB RRB              LLL FFB RRR              LLL FFU FRB
 LLL FFU RLB              LLL FFU LBL              LLL FFB URR

     DDF                      DDB                      DDL
     DDU                      DDU                      DDB
     DDU                      DDF                      DDD

  ----------------------------------------------------------------

     BBR                      BBR                      BBL
     BBF                      BBU                      BBU
     FFR                      URU                      DRF

     LUB                      FFF                      FFR
     UUU                      DUU                      DUU
     RBR                      BBB                      DBR

 ULD BRU FRU              LRL URU RRR              RRB RRB URU
 LLL FFU BRR              LLL FFU BRF              LLL FFU BRF
 LLL FFB URF              LLL FFR BLF              LLL FFU LLF

     DDL                      DDD                      DDB
     DDD                      DDU                      DDU
     DDD                      DDD                      DDU

  ------------------------------------------------------------------

     BBR                      BBF                      BBF
     BBU                      BBU                      BBU
     URU                      FRF                      FRF

     FFF                      RDR                      RDR
     UUU                      FUU                      FUU
     BBB                      DBR                      BBR

 LRL URU RLR              DRB RRB URU              DRR DRB URU
 LLL FFU BRF              LLL FFU BRL              LLL FFU BRL
 LLL FFR BRF              LLL FFU LFU              LLL FFL BFU

     DDD                      DDB                      DDU
     DDD                      DDU                      DDU
     DDD                      DDL                      DDL

  --------------------------------------------------------------

     BBF                      BBR                      BBR
     BBU                      BBU                      BBU
     DRL                      URR                      FRD

     RFU                      FFB                      RDB
     UUD                      DUU                      FUU
     RBR                      RBF                      UBL

 BRF DRB URB              LRD BRR UBU              DRB LRF URR
 LLL FFU BRL              LLL FFU RRL              LLL FFU FRB
 LLL FFU RFU              LLL FFB UFF              LLL FFR ULU

     DDF                      DDL                      DDB
     DDU                      DDU                      DDU
     DDL                      DDD                      DDF

  --------------------------------------------------------------

     BBB                      BBU                      BBU
     BBU                      BBU                      BBU
     FRL                      RRF                      BRR

     RDF                      BFR                      UUU
     FUU                      DUU                      DUF
     UBB                      FBU                      UBD

 DRF RRL URU              DRD RRL FLU              LRL FRR FRF
 LLL FFU FRB              LLL FFU FRB              LLL FFU FRB
 LLL FFD RLU              LLL FFB URR              LLL FFD RLR

     DDB                      DDL                      DDB
     DDU                      DDU                      DDU
     DDR                      DDB                      DDB

  ---------------------------------------------------------------

     BBB                      BBB                      BBD
     BBU                      BBF                      BBU
     URU                      UFD                      LRF

     BUF                      RUR                      UBR
     DUF                      DUU                      DUU
     UBU                      UBR                      UFR

 RRB LRL FRR              FRL FRB URF              FRB LRB URU
 LLL FFU FRB              LLL FFU LRR              LLL FFU FRB
 LLL FFF RLR              LLL FFB UBR              LLL FFD RLR

     DDD                      DDL                      DDB
     DDU                      DDU                      DDU
     DDD                      DDD                      DDF

  ----------------------------------------------------------------

     BBD                      BBF                      BBR
     BBF                      BBF                      BBF
     BFB                      UFF                      LFR

     RUL                      FUR                      UUD
     UUU                      UUU                      UUU
     RUL                      BUR                      UUD

 URU FBF ULU              LRL UBB ULU              FRB LBR FLB
 LLL FFB RRR              LLL FFB RRR              LLL FFB RRR
 LLL FFD RRR              LLL FFB DRD              LLL FFR FRB

     DDB                      DDR                      DDU
     DDD                      DDD                      DDD
     DDF                      DDR                      DDU

  ------------------------------------------------------------

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From BRYAN@wvnvm.wvnet.edu  Fri Dec 23 20:55:24 1994
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Date:      Fri, 23 Dec 1994 20:55:19 EST
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   <U,R,U2,R2> Antipodal Processes

Here are the 16 antipodal processes for <U,R,U2,R2>.  Also, as in the
case of <U,R>, I am including replacement position diagrams that
include a correction for the Back face so that both the corners and
edges are up-side-down.  Q+H turners will like these processes because
of the way turns of the U face alternate with turns of the R face.
I have performed a few of the processes on a real cube (as opposed to
on a computer screen), and I find the alternating faces to be
quite pleasant somehow or other.


 1 R' U  R2 U  R  U2 R  U  R  U2 R' U2 R  U2 R' U2 R2 U  R2 U
 2 U' R  U' R2 U' R2 U' R  U' R2 U2 R  U  R  U' R  U  R  U2 R2
 3 U  R  U2 R  U' R  U2 R2 U2 R2 U  R  U  R2 U2 R  U' R  U  R2
 4 R' U2 R' U  R  U' R2 U' R2 U' R' U  R' U' R' U' R  U' R' U2
 5 R' U2 R2 U' R  U2 R' U' R  U  R' U2 R' U2 R' U' R' U2 R  U
 6 R  U  R  U' R  U  R2 U  R2 U  R' U' R  U2 R  U2 R  U  R' U
 7 R2 U' R  U2 R  U2 R  U' R' U' R  U2 R  U  R2 U  R' U  R  U2
 8 U' R' U  R' U  R' U  R2 U' R' U' R' U  R' U  R' U  R2 U' R'
 9 R  U2 R' U  R' U  R  U' R  U  R' U2 R' U2 R  U' R  U' R' U2
10 U' R' U2 R2 U2 R' U  R' U2 R' U' R  U  R' U  R  U' R' U  R2
11 U' R' U' R' U2 R2 U2 R2 U  R2 U  R2 U' R2 U2 R' U' R  U2 R'
12 R2 U' R2 U2 R' U' R2 U' R  U  R2 U2 R2 U  R  U2 R  U' R' U
13 R  U2 R' U' R  U' R' U' R2 U' R  U2 R  U  R2 U  R' U2 R  U
14 U2 R' U  R2 U' R' U' R2 U' R  U  R' U2 R' U2 R2 U  R' U2 R'
15 U' R  U2 R' U2 R  U' R  U' R  U2 R' U' R  U2 R2 U  R  U  R2
16 U' R2 U2 R' U  R' U2 R  U' R2 U2 R' U2 R  U  R2 U' R2 U2 R2



     BBR                      BBB                      BBL
     BBB                      BBB                      BBB
     FBU                      LBD                      RBU

     RUR                      BUB                      BUR
     UUU                      UUU                      UUU
     RUU                      RUU                      DUF

 DLU FFL FRB              ULU FFL FRR              DLR FFR URB
 LLL FFF RRR              LLL FFF RRR              LLL FFF RRR
 LLL FFU LRD              LLL FFD FRU              LLL FFL URU

     DDB                      DDR                      DDF
     DDD                      DDD                      DDD
     DDB                      DDR                      DDB

  -------------------------------------------------------------

     BBD                      BBU                      BBR
     BBB                      BBB                      BBB
     LBR                      LBR                      RBB

     UUF                      UUB                      UUL
     UUU                      UUU                      UUU
     BUR                      BUR                      LUR

 FLL UFU FRD              FLL UFD BRU              BLF UFF DRU
 LLL FFF RRR              LLL FFF RRR              LLL FFF RRR
 LLL FFR URB              LLL FFR DRF              LLL FFU RRD

     DDB                      DDF                      DDF
     DDD                      DDD                      DDD
     DDR                      DDR                      DDB

  -----------------------------------------------------------

     BBL                      BBR                      BBU
     BBB                      BBB                      BBB
     RBR                      LBU                      FBB

     BUU                      BUR                      RUR
     UUU                      UUU                      UUU
     LUB                      LUF                      FUR

 DLF UFU RRF              ULF UFR URB              DLU LFU FRD
 LLL FFF RRR              LLL FFF RRR              LLL FFF RRR
 LLL FFD FRU              LLL FFD FRD              LLL FFL BRR

     DDR                      DDR                      DDU
     DDD                      DDD                      DDD
     DDB                      DDB                      DDB

  -----------------------------------------------------------

     BBF                      BBU                      BBB
     BBF                      BBD                      BBU
     FBR                      LFR                      BRR

     LUU                      UUD                      UFB
     UUU                      UUB                      DUU
     UUR                      LUF                      UBF

 ULB LFB URF              FBU BLD RRB              LRL FRR ULU
 LLL FFB RRR              LLL FFU RRR              LLL FFU BRF
 LLL FFR BRD              LLL FFR FRR              LLL FFF RRR

     DDD                      DDU                      DDD
     DDD                      DDF                      DDU
     DDR                      DDB                      DDD

  --------------------------------------------------------------

     BBL                      BBB                      BBB
     BBU                      BBU                      BBU
     DRF                      FRF                      FFD

     FFR                      RDR                      RUB
     DUU                      FUU                      UUF
     DBR                      UBB                      FDU

 RRB RRB URU              DRL FRL URU              DLU LRF RRR
 LLL FFU BRF              LLL FFU FRB              LLL FFB RRR
 LLL FFU LLF              LLL FFR ULR              LLL FFU LBU

     DDB                      DDB                      DDB
     DDU                      DDU                      DDU
     DDU                      DDD                      DDR

  -----------------------------------------------------------

     BBF
     BBU
     UFB

     LUU
     UUF
     BDR

 BLR DRF DRR
 LLL FFB RRR
 LLL FFU RBU

     DDF
     DDU
     DDL



 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From kotani@cc.tuat.ac.jp  Mon Dec 26 05:33:46 1994
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From: kotani@cc.tuat.ac.jp
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Cc: kotani@cc.tuat.ac.jp
In-Reply-To: ba05133@bingsuns.cc.binghamton.edu's message of Fri, 23 Dec 1994 10:21:44 -0500 (EST) <9412231521.AA01392@podsun7>
Subject: "unsubscribe"

Please unsubscribe me.

From jkato@tmastb.eec.toshiba.co.jp  Mon Dec 26 22:45:54 1994
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Date: Tue, 27 Dec 94 12:44:00 JST
From: jkato@tmastb.eec.toshiba.co.jp (Toshi Kato)
Return-Path: <jkato@tmastb.eec.toshiba.co.jp>
Message-Id: <9412270344.AA00285@tmastb.eec.toshiba.co.jp>
To: cube-lovers@life.ai.mit.edu
Subject: GreyhoundBus Puzzle

This is a sliding block puzzle that I thought.

  +---+---+---+---+---+  
  |[N]|[U]|[S]|[H]|[U]|  $@!N(JProblem$@!O(J
  +---+---+---+---+---+     Left figure is starting condition.$@!!(J
  |[O]|###|   |###|[G]|   Make a sequence,"GREYHOUNDBUS",with minimum step.
  +---+---+---+---+---+$@!!!!(J
  |[B]|[Y]|[R]|[D]|[E]|   Note: Vacant room is only one.$@!!(J
  +---+---+---+---+---+         ### are not moved.

If you can get answer, please write it to me or on Cube-Lovers ML.

Toshi(Junk) Kato, Japan 
--------------------------------------JUNK: jkato@tmastb.eec.toshiba.co.jp



From BRYAN@wvnvm.wvnet.edu  Tue Dec 27 16:55:19 1994
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Date:      Tue, 27 Dec 1994 11:23:06 EST
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Normal Subgroups of G

Recently, there was some discussion of whether the set C of twenty-four
rotations is a normal subgroup of the cube group G=<Q>.  It isn't, but
I decided to write up some information about normal subgroups as it
relates to the cube.  Most of the following is from Frey and Singmaster.
Any good stuff is theirs.  Any crud that sneaks in is mine.

If H is any subgroup of G, a right coset of H in G is a set {hX}
for some fixed X in G and for all h in H.  Similarly, a left coset
of H in G is a set {Xh}.  Right cosets may be denoted as Hx, and
left cosets by xH.

In general, a right coset Hx is not equal to a left coset xH.  But if
we have Hx=xH for all x in G, then H is a normal subgroup of G.  An
alternative definition is H is normal if x'Hx=H for every x in G.
The definitions are equivalent, and Frey and Singmaster give as a
theorem Hx=xH for every x in G if and only if x'Hx=H for every x in G.

It should be noted that H normal does not imply that the elements
h of H commute with the elements x of G.  That is, just because
Hx=xH we do not necessarily have hx=xh for every h in H (or even for
any h in H other than the identity).  However, I think it is fair
to characterize a normal subgroup as commuting "globally" with G,
even if it does not commute "locally".  On the other hand, if a
subgroup H does commute "locally" (i.e., if hx=xh for all h in H
and all x in G), then H is certainly normal.

Normal groups serve a function with respect to finite groups analogous
to the function served by prime numbers with respect to natural numbers.
First of all, any finite group always has at least two trivial
normal subgroups, namely the group itself and the group containing
only the identity.  Second, a finite group containing normal subgroups
may be "factored" in a fashion analogous to prime numbers factoring
composite numbers.  A finite group containing no normal subgroups
is called simple, analogous to numbers with no factors being called
prime.

The cube group G does not have very many normal subgroups, but it does
have a few.  The first place to look for normal subgroups is to look
for subgroups with index 2.  That is, look for subgroups that are
half as  big a G.  Such a subgroup is the subgroup A of even
permutations.  ("A" stands for "Alternating", I think.)

It is easy to see that A is normal.  If x is even, then Ax=xA=A.
if x is odd, then Ax=xA=Abar, where Abar is the set (not group!)
of odd permutations.

Similarly, any subgroup H with index 2 is normal.  If the index of H
in G is 2, then H partitions G into two equal size sets H and Hbar.
If x is in H, then Hx=xH=H.  If x is in Hbar, then Hx=xH=Hbar.

If we may digress briefly to the set M of 48 rotations and reflections,
then there are three subgroups of M with index 2.  In Dan Hoey's
taxonomy, they are called C, A, and H.  We may categorize the elements
of M as even or odd, and as rotations or reflections.  There are 12
even rotations, 12 odd rotations, 12 even reflections, and 12 odd
reflections.  If we take 12 even rotations and 12 odd rotations,
we have C.  So C is a normal subgroup of M, even if it is not a normal
subgroup of G.  If we take 12 even rotations and 12 even reflections,
we have A.  This A (a subgroup of M) is not to be confused with the
A we have already talked about which is a subgroup of G.  But I think
the name derives from the same source ("Alternating") in either case.
If we take 12 even rotations and 12 odd reflections, we have H.

Returning to G, the next two normal subgroups are Ac which leaves
the set of edges fixed, and Ae which leaves the set of corners
fixed.  Ac is even on the corners, and Ae is even on the edges, in
order to conserve parity.  Note that both Ac and Ae are normal
subgroups of A as well as of G.

I suppose that what is going on with Ac and Ae is obvious enough,
but I want to talk about it for a minute anyway.  I most typically
think of an equation such as X=RLUD'R as meaning something to the
effect that "X" is a shorthand *name* for the collection of five
processes (in order) R, L, U, D', and R.  But I still tend to think
of the processes as distinct.  However, from the point of view
of group theory, X is a single operation which exists in its own
right just as do the quarter turns.

With a physical cube, you cannot perform an operation in Ac or Ae
without making a fairly long sequence of quarter turns.  For
example, something so simple as performing FF on the corners while
leaving the edges fixed is non-trivial.  But from the point of view
of group theory, we can easily find a single permutation
X[C,E] such that X[C]=FF[C] while X[E]=I[E].  Indeed, from the
point of view of group theory, you are never more than one move
from Start.  That is, if you are at X, the one move which will always
solve the cube is X'.  It is only if you are asked to decompose X'
into generators such as quarter-turns that the question of how far
from Start you are makes any sense.

If a subgroup H of G is normal, the left cosets form a group under the
operation (xH)(yH)=(xy)H.  This group is called the factor group
of H in G or the quotient group of G by H, and is denoted as G/H.

Martin Schoenert recently clarified that while there may be more than
one way to define an operation on cosets such that they form a group,
the notation G/H is usually reserved for the case where the operation
is (xH)(yH)=(xy)H.

The factor group G/A contains two elements, and is isomorphic to any
group containing only two elements.  We may write it as
<Abar>={A,Abar}, where A is the identity of the group.

The factor group G/Ac is isomorphic to the set of all permutations
on the edges (which we have written as G[E] in the recent past).
The factor group G/Ae is isomorphic to the set of all permutations
on the corners (which we have written as G[C] in the recent past).

Since Ac and Ae are normal subgroups of A, we may write A/Ac and A/Ae
which are isomorphic to Ae and Ac, respectively.

We can find normal subgroups of Ac and Ae.  The set At of all
permutations in Ac which leave all corner locations fixed except for
twisting some of them is a normal subgroup of Ac.  The set Af of
all permutations in Ae which leave all edge locations fixed except
for flipping some of them is a normal subgroup of Ae.  (This twists
and flips have to follow the normal rules of conservation of twist
and flip, of course.)

This completes the list of normal subgroups.  I will now give Frey
and Singmaster's proof that we are done, while interposing some
questions of my own for the cube theory experts out there.  My
first question is that Frey and Singmaster do not state that At and
Af are normal subgroups of G.  It seems obvious that they are.
However, is the formal argument that (for example) At is a normal
subgroup of Ac and Ac is a normal subgroup of G; hence, At is a
normal subgroup of G?  How analogous is the factoring of groups
by normal subgroups to the factoring of composite numbers by
prime numbers?

Continuing with Frey and Singmaster, we may write Ac/At and
Ae/Af, where Ac/At is isomorphic to the group Asc which leaves
the corners sane and Ae/Af is isomorphic to the group Ase which
leaves the edges sane.  "Sane" is a term used by Frey and Singmaster
in their proof of conservation of twist and flip.  In general, it
is easy to see if a cubie is twisted or flipped when it is home,
but it is not so easy to see if it is twisted or flipped when it
is not home.  Their proof (and the others I have seen) define a
frame of reference so that you can tell if a cube is twisted or
flipped when it is not home.  A cubie which is not twisted or
flipped in this frame of reference is sane.

Asc and Ase are not normal subgroups of Ac and Ae, respectively.
(I tend to think that the reason they are not normal is related
to the fact that the frame of reference required to define sane
positions is not unique.)  However, Asc and Ase are isomorphic
to well known groups.

The group Sn of all permutations of n objects is the n-element
symmetric group.  The subgroup An of all even permutations of
n objects is the n-element alternating group (there is that word
"alternating" again!).  Asc is isomorphic to A8 (there being
eight corner cubies) and Ase is isomorphic to A12 (there being
twelve edge cubies).

A famous result from Abel and Galois is that An does not have any
non-trivial normal subgroups for n >= 5.  Hence, we have reduced
G to normal subgroups which have no more normal subgroups, and we
are done.

I guess my questions are as follows:  1) why must we restrict ourselves
to alternating groups?  2) For example, just as we found three
subgroups of M with index 2, might we not find other subgroups of
G with index 2 than the one we found?   3)  Might we not find a
normal subgroup of G with some index other than 2, e.g., with index 3?

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From mreid@ptc.com  Tue Dec 27 19:49:45 1994
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From: mreid@ptc.com (michael reid)
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To: Cube-Lovers%ai.mit.edu@ptc.com
Subject: Normal Subgroups of G
Content-Length: 3023

jerry writes

[ ... ]

>                                                             My
> first question is that Frey and Singmaster do not state that At and
> Af are normal subgroups of G.  It seems obvious that they are.

indeed.

> However, is the formal argument that (for example) At is a normal
> subgroup of Ac and Ac is a normal subgroup of G; hence, At is a
> normal subgroup of G?

but this argument is not valid.  your question might be rephrased:

if  H  is a normal subgroup of  G  and  K  is a normal subgroup of  H,
does it follow that  K  is a normal subgroup of  G ??

the answer is no.  here's an easy counterexample:

let  G  be the alternating group  A_4,  H  the subgroup of order 4
{e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)},  and  K  the subgroup
{e, (1 2)(3 4)}.  it is easy to see that  H  is normal  in  G  and
K  is normal in  H.  however, if  x  is any three cycle (for example),
xK != Kx.

[ ... ]

>                         "Sane" is a term used by Frey and Singmaster
> in their proof of conservation of twist and flip.  In general, it
> is easy to see if a cubie is twisted or flipped when it is home,
> but it is not so easy to see if it is twisted or flipped when it
> is not home.  Their proof (and the others I have seen) define a
> frame of reference so that you can tell if a cube is twisted or
> flipped when it is not home.  A cubie which is not twisted or
> flipped in this frame of reference is sane.

here's a completely different proof of "conservation" which doesn't
use any frame of reference.

instead of thinking of permutations of edge cubies, think of
permutations of the facelets of the edges.  any quarter turn
induces two four cycles of these edge facelets, which is an even
permutation.  thus, any legal position has an even permutation
of the edge facelets.  however, a single flipped edge is just
a two cycle of edge facelets, an odd permutation, and therefore
is not a legal position.

my proof for conservation of twist is slightly more sophisticated,
but i think it's worthwhile.

the group of legal corner states may be viewed as a subgroup of the
wreath product  S_8 wr C_3.  we have a natural homomorphism

               S_8 wr C_3  --->  C_3              (*)
defined  by
               (s, c_1, ... , c_8) |-->  c_1 + ... + c_8

(the cyclic group  C_3  is written additively).  it is easy to see
that this is a homomorphism, but it uses the fact that  C_3  is abelian.
(in general, we have a natural homomorphism

               G wr H  --->  H^ab  ( =  H / [H, H] )

defined in the same way.)

conservation of corner twist is equivalent to saying that all legal
corner states are in the kernel of the map given in  (*).  however,
any quarter turn has order 4, so its image in  C_3  must be the
identity.  thus all quarter turns lie in the kernel, and therefore
the same is true of all legal positions.

(actually, i've cheated slightly here.  we actually need a frame of
reference in order to view the group of corner states as a subgroup of
S_8 wr C_3.)

mike

From BRYAN@wvnvm.wvnet.edu  Tue Dec 27 22:46:45 1994
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Date:      Tue, 27 Dec 1994 14:26:45 EST
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Squares Group

On 9 Aug 1994, Mark Longridge posted God's Algorithm results for
the squares group <H> consisting entirely of 180 degree turns.
Mark invited corroboration.  The following will serve to verify
Mark's results, and will also provide the same information for
the M-conjugacy classes of <H>.  Notice that the ratio of cube
positions to M-conjugacy classes never gets very close to 48.
Hence, there are a significant number of positions in <H>
at each level of the search tree that are at least somewhat
"symmetrical".


                    M   Branching       Cube Branching  Ratio of
      Level Conjugate      Factor  Positions   Factor   Cubes to
              Classes                                  M Classes

          0         1                      1                   1
          1         1           1          6        6          6
          2         2           2         27      4.5       13.5
          3         5         2.5        120   4.4444         24
          4        18         3.6        519   4.3250    28.8333
          5        56      3.1111       1932   3.7225    34.5000
          6       162      2.8929       6484   3.3561    40.0247
          7       482      2.9753      20310   3.1323    42.1369
          8      1258      2.6100      55034   2.7097    43.7472
          9      2627      2.0882     113892   2.0695    43.3544
         10      4094      1.5584     178495   1.5672    43.5992
         11      4137      1.0105     179196   1.0039    43.3154
         12      2231      0.5393      89728   0.5007    40.2187
         13       548      0.2456      16176   0.1803    29.5182
         14       114      0.2080       1488   0.0920    13.0526
         15        16      0.1404        144   0.0968          9

      Total     15752                 663552             42.1249

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From mschoene@math.rwth-aachen.de  Fri Dec 30 09:20:21 1994
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Date: Fri, 30 Dec 94 15:17 WET
From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: Cube-Lovers@ai.mit.edu
In-Reply-To: "Jerry Bryan"'s message of Tue, 27 Dec 1994 11:23:06 EST <9412272155.AA00248@life.ai.mit.edu>
Subject: Re: Normal Subgroups of G

Jerry Bryan wrote in his e-mail message of 1994/12/27

    Recently, there was some discussion of whether the set C of twenty-four
    rotations is a normal subgroup of the cube group G=<Q>.  It isn't, but
    I decided to write up some information about normal subgroups as it
    relates to the cube.  Most of the following is from Frey and Singmaster.
    Any good stuff is theirs.  Any crud that sneaks in is mine.

Very good.  The lattice of normal subgroups of G is not too complicated
and relates nicely to the structure of this group.  Allow me to throw in
my $0.02.

Nitpicking alert!  C is not even a subgroup of G=<Q>.  It is a subgroup
of CG.  But not a normal one.

Jerry continued

    It should be noted that H normal does not imply that the elements
    h of H commute with the elements x of G.  That is, just because
    Hx=xH we do not necessarily have hx=xh for every h in H (or even for
    any h in H other than the identity).  However, I think it is fair
    to characterize a normal subgroup as commuting "globally" with G,
    even if it does not commute "locally".  On the other hand, if a
    subgroup H does commute "locally" (i.e., if hx=xh for all h in H
    and all x in G), then H is certainly normal.

In group theory there this distinction is made by using the terms
*normal* and *central* (even though globally and locally are perhaps
more descriptive names).

If xH = Hx, then x is said to *normalize* H.  The set of all elements
that normalize H is called the *normalizer* of H, usually written N_G(H).
It is easy to see that N_G(H) is a subgroup of G containing H.
If every x of G normalizes H, then H is said to be *normal* in G.
Of course H is normal in G, if and only if N_G(H) = G.

If xh = hx for every h in H, then x is said to *centralize* H.  The set
of all elements that centralize H is called the *centralizer* of H,
usually written C_G(H).  It is easy to see that C_G(H) is again a
subgroup of G, but it need not contain H (it contains H if and only if H
is abelian).  If every x of G centralizes H, then H is said to be
*central* in G.  Of course H is central in G, if and only if C_G(H) = G.
Furthermore it is easy to see that H is central, if and only if H is a
subgroup of C_G(G), which is the set of those elements in G that commute
with all elements in G.  C_G(G) is called the *center* of G.

And as you say, a central subgroup is also normal, but a normal subgroup
need not be central.

If N1 and N2 are two normal subgroups of G, then it is easy to see that
the intersection of N1 and N2 and the closure of N1 and N2 are both
normal subgroups too.  Thus the set of normal subgroups of G is closed
w.r.t. intersection and closure.  In other words, the set of normal
subgroups forms a lattice.  I shall draw the lattice of normal subgroups
of G below.

Jerry continued

    Normal groups serve a function with respect to finite groups analogous
    to the function served by prime numbers with respect to natural numbers.
    First of all, any finite group always has at least two trivial
    normal subgroups, namely the group itself and the group containing
    only the identity.  Second, a finite group containing normal subgroups
    may be "factored" in a fashion analogous to prime numbers factoring
    composite numbers.  A finite group containing no normal subgroups
    is called simple, analogous to numbers with no factors being called
    prime.

This is correct.  Allow me a few more remarks.

Groups are composed from simple groups, which correspond to primes.  The
simple groups have been classified.  There are several families (the
alternating groups A_n are one such family), and 26 sporadic simple
groups.  This classification is one of the outstanding mathematical
achievements.  It is estimated that the complete proof is about 10000
pages long (distributed over several papers, books, Ph.D. thesis, etc.).

And then there are the ways in which those simple groups can be composed.
In the case of natural numbers, this is very simple.  The fundamental
theorem tells us, that there is, up to the order, just one way in which
any natural number is composed from primes.  In the case of groups it is
much more difficult.  There is still a theorem which tells us that the
composition factors of a group are determined up to order.  But not any
order will do.  For example the symmetric group S_3 of size 6, has a
factor group C_2 over a normal subgroup C_3, but it cannot be decomposed
with a factor group C_3 over a normal subgroup C_2.  Furthermore given a
certain set of composition factors and a certain order, there may be
several groups that decompose in this way.  For example the cyclic group
C_6 can also be decomposed with a factor group C_2 over a normal subgroup
C_3.  Even in the simplest case, groups of prime power size, which
decompose as a sequence of cyclic groups C_p, is so difficult that they
have not been classified (and maybe never will).

Jerry continued

    The cube group G does not have very many normal subgroups, but it does
    have a few.  The first place to look for normal subgroups is to look
    for subgroups with index 2.  That is, look for subgroups that are
    half as  big a G.  Such a subgroup is the subgroup A of even
    permutations.  ("A" stands for "Alternating", I think.)

    It is easy to see that A is normal.  If x is even, then Ax=xA=A.
    if x is odd, then Ax=xA=Abar, where Abar is the set (not group!)
    of odd permutations.

    Similarly, any subgroup H with index 2 is normal.  If the index of H
    in G is 2, then H partitions G into two equal size sets H and Hbar.
    If x is in H, then Hx=xH=H.  If x is in Hbar, then Hx=xH=Hbar.

        ... and later on ...

    The factor group G/A contains two elements, and is isomorphic to any
    group containing only two elements.  We may write it as
    <Abar>={A,Abar}, where A is the identity of the group.

This is correct.  There is another way to obtain A, which is also very
instructive.

Let G be any group.  Let g1 and g2 be two elements of G.  Then the
element g1^-1 * g2^-1 * g1 * g2 is called the commutator of g1 and g2,
and is usually written as [g1,g2].  Now let g be any element of G.  Then

g^-1 [g1,g2] g = g^-1 g1^-1 g2^-2 g1 g2 g
               = (g^-1 g1^-1 g) (g^-1 g2^-1 g) (g^-1 g1 g) (g^-1 g2 g)
               = (g^-1 g1 g)^-1 (g^-1 g2 g)^-1 (g^-1 g1 g) (g^-1 g2 g)
               = [ (g^-1 g1 g), (g^-1 g2 g) ].

Thus the conjugate of a commutator is again a commutator.  It follows
that the subgroup generated by all commutators of all pairs of elements
of G is a normal subgroup.  This subgroup is called the *commutator
subgroup* or *derived subgroup* of G, and is usually written G'.

It is the minimal normal subgroup of G, such that the factor group G/G'
is an abelian group.  Minimal means that for each normal subgroup N of G
such that G/N is an abelian group, G' is a subgroup of N.

In the case of the cube group G' is A.  I shall use G' instead of A.

Jerry continued

    If we may digress briefly to the set M of 48 rotations and reflections,
    then there are three subgroups of M with index 2.  In Dan Hoey's
    taxonomy, they are called C, A, and H.  We may categorize the elements
    of M as even or odd, and as rotations or reflections.  There are 12
    even rotations, 12 odd rotations, 12 even reflections, and 12 odd
    reflections.  If we take 12 even rotations and 12 odd rotations,
    we have C.  So C is a normal subgroup of M, even if it is not a normal
    subgroup of G.  If we take 12 even rotations and 12 even reflections,
    we have A.  This A (a subgroup of M) is not to be confused with the
    A we have already talked about which is a subgroup of G.  But I think
    the name derives from the same source ("Alternating") in either case.
    If we take 12 even rotations and 12 odd reflections, we have H.

In the case of M, M' is a subgroup of index 4.  And the factor group M/M'
is isomorphic to the group { (), (1,2)(3,4), (1,3)(2,4), (1,4)(2,3) }.
This group is usually called C_2^2, because it is the direct product of
two cyclic groups of size 2.  This group has three (normal) subgroups of
index 2, which correspond to the three normal subgroups of M Jerry
described.

Jerry continued

    Returning to G, the next two normal subgroups are Ac which leaves
    the set of edges fixed, and Ae which leaves the set of corners
    fixed.  Ac is even on the corners, and Ae is even on the edges, in
    order to conserve parity.  Note that both Ac and Ae are normal
    subgroups of A as well as of G.

        ... and later on ...

    The factor group G/Ac is isomorphic to the set of all permutations
    on the edges (which we have written as G[E] in the recent past).
    The factor group G/Ae is isomorphic to the set of all permutations
    on the corners (which we have written as G[C] in the recent past).

This is again typical.  If we have a permutation group G that has more
than one orbit, then the stabilizer of each orbit is a normal subgroup
of G (which may or may not be trivial).  And the group G is a subdirect
product of the factor groups over those normal subgroups.

Time for the first picture (apologies to all that have no large screen,
but I want to add more to it later without cluttering it too much).


                                       GCE
                                       /|\ 2
                                      X G X
                                     / \|/ \ 2
                                    /   G'  \
                                   /   / \   \
                                  /   /   \   \
                                 /   /     \   \
                                /   /       \   \
                               /   /         \   \
                              /   /           \   \
                             /   /             \   \
                            /   /               \   \
                           /   /                 \   \
                          /   /                   \   \
                         /   /                     \   \
                        /   /                       \   \
                       /   /                         \   GE ~ G[E]
                      /   /                           \ / 2
                     /   /                            GE'
                    /   /                             /
           G[C] ~ GC   /                             /
                  2 \ /                             /
                    GC'                            /
                      \                           /
                       \                         /
                        \                       /
                         \                     /
                          \                   / 12!/2 2^11
                           \                 /
                   8!/2 3^7 \               /
                             \             /
                              \           /
                               \         /
                                \       /
                                 \     /
                                  \   /
                                   \ /
                                   <1>

Let GC be the operation of G on the corners, which is isomorphic to
the factor group G/GE' = G[C], and GE be the operation of G on the
edges, isomorphic to the factor group G/GC' = G[E] (this distinction
between GC and G[C] and GE and G[E] is subtle and not very important).
Let GCE be the direct product of GC and GE.  G is a subdirect product
of GC and GE, so it is a subgroup of GCE.  It is a subgroup of index 2.
G has a subgroup of index 2, namely G', which Jerry called A above.
The two stabilizers are GC' and GE', which Jerry called Ac and Ae above.
They are in fact the derived subgroups of GC and GE.

Note that GCE has two more (normal) subgroups of index 2, namely
<G',GC> and <G',GE>.  And G' is the derived subgroup of CGE.
So GCE/GCE' is again isomorphic to C_2^2.

Jerry continued

    Since Ac and Ae are normal subgroups of A, we may write A/Ac and A/Ae
    which are isomorphic to Ae and Ac, respectively.

G' (aka A) is in fact the direct product of GC' (aka Ac) and GE' (aka
Ae).  And the standard isomorphism theorem tells us that G'/GC' ~ GE'
and G'/GE' ~ GC'.

Jerry continued

    We can find normal subgroups of Ac and Ae.  The set At of all
    permutations in Ac which leave all corner locations fixed except for
    twisting some of them is a normal subgroup of Ac.  The set Af of
    all permutations in Ae which leave all edge locations fixed except
    for flipping some of them is a normal subgroup of Ae.  (This twists
    and flips have to follow the normal rules of conservation of twist
    and flip, of course.)

    This completes the list of normal subgroups.  I will now give Frey
    and Singmaster's proof that we are done, while interposing some
    questions of my own for the cube theory experts out there.

I haven't seen what Frey and Singmaster prove.  But this is not true.
Together with the trivial normal subgroup G and <1> you have listed
7 normal subgroups, but G has indeed 13 normal subgroups.

Jerry continued

    My first question is that Frey and Singmaster do not state that At
    and Af are normal subgroups of G.  It seems obvious that they are.
    However, is the formal argument that (for example) At is a normal
    subgroup of Ac and Ac is a normal subgroup of G; hence, At is a
    normal subgroup of G?  How analogous is the factoring of groups
    by normal subgroups to the factoring of composite numbers by
    prime numbers?

This is not true, and Michael Reid gave the smallest counterexample.
Group theory would certainly be a lot easier if this was true, but
probably also a lot less challenging.  But there is a similar argument.

If M is a normal subgroup of a group N that is invariant under all
automorphisms of N, then M is called *characteristic*.  For example N' is
always a characteristic subgroup of N.  If N is a subgroup of a group G,
and M is a characteristic subgroup of N, then M is normal in G.  Also if
N is a characteristic subgroup of G, and M is a characteristic subgroup
of N, then M is also a characteristic subgroup of G.

This actually happens here, At and Af are characteristic in GC' and GE',
so they are normal in G.  But this is not so easy to prove, it is simpler
to verify directly that At and Af are normal in G (or use the fact that
they are again stabilizers of appropriate operations of G).

Jerry continued:

    I guess my questions are as follows:  1) why must we restrict ourselves
    to alternating groups?  2) For example, just as we found three
    subgroups of M with index 2, might we not find other subgroups of
    G with index 2 than the one we found?   3)  Might we not find a
    normal subgroup of G with some index other than 2, e.g., with index 3?

I got A as the derived subgroup of G.  Since this is the minimal normal
subgroup such that the factor group is abelian, there cannot be another
normal subgroup of index 2 or 3.  But there can be other normal subgroups
with non-abelian factor groups, and indeed there are.

The rest of this message describes how one can find all normal subgroups
of G.  My approach may or may not be equal to Frey and Singmaster's.

Let us first consider GC.  This group is a subgroup of index 3
in the wreath product C_3 <wreath> S_8.

This wreath product is isomorphic to the group of the following elements

    ( c_1, c_2, c_3, c_4, c_5, c_6, c_7, c_8; p )

where the c_i are in {0,1,2} and p is a permutation in S_8.
Multiplication of those elements is defined as follows

    (c_1,c_2,...,c_8;p) (d_1,d_2,...,d_8;q) :=
    ( c_1 + d_{1^p}, c_2 + d_{2^p}, ..., c_8 d_{8^p}; p * q )

where d_{i^p} denotes d_j, where j is the image of i under the
permutation p, and c_i + d_j is the sum of c_i and d_j modulo 3.
p * q is simply the product of p and q in S_8.
So you first permute the components of the second element with
the permutation of the first element, then sum componentwise,
and finally multiply the two permutations.

Clearly C_3 <wreath> S_8 has 3^8 8! elements.  GC is the subgroup of
those elements of C_3 <wreath> S_8 for which c_1+c_2+...+c_8 = 0 (mod 3).

It is easy to see that the set of all 3^7 elements
(c_1,c_2,...,c_8,<identity>) of GC forms a normal subgroup of GC.
I shall call this subgroup VC (V because VC is in fact a vector space),
Jerry called this subgroup At.

Next I shall show that no proper subgroup of VC can be a normal subgroup
of GC.  Suppose that H is a normal subgroup of VC, and let
x = (x_1,x_2,...,x_8;<identity>) be any non-trivial element of H.
Because 1+1+...+1 <> 0 (mod 3) and 2+2+...+2 <> 0 (mod 3), there are two
components x_i and x_j which are different.  An easy calculation shows

    y := (0,0,...,0;(i,j))^-1 * x^-1 * (0,0,...,0;(i,j)) * x

has y_i = 1 and y_j = 2 (or the other way around).  Since H is supposed
to be normal, y must also be in H.  Furthermore the 6 elements

    zk := (0,0,...,0;(j,k))^-1 * y * (0,0,...,0;(j,k))

where k <> j and k <> i all have zk_i = 1 and zk_k = -1.
Again since H is supposed to be normal, they must all lie in H.
But y and those 6 elements are obviously linearly independent,
i.e., form a basis for VC.  In particular it follows that H = VC.
So no proper subgroup of VC is a normal subgroup of GC.

Now let N be any normal subgroup of GC.  Then the intersection of N and
VC must be a normal subgroup of GC contained in VC.  Thus there are only
two possibilities for this intersection; it can be VC (i.e. N contains
VC) or it can be trivial.

Assume first that N contains VC.  Then N/VC must be a normal subgroup of
GC/VC.  But GC/VC is S_8, which has only three normal subgroups, namely
S_8, A_8 (= S_8'), and <1>.  Thus we have three normal subgroups
containing VC, namely GC, GC', and VC.

On the other hand assume that N intersects trivially with VC.  Then the
closure M of N and VC must be one of the three normal subgroups given
above.  The isomorphism theorem tells us that M/N ~ VC.  GC does not have
a factor group isomorphic to VC, since its largest abelian factor group
is GC/GC' of size 2.  GC' also does not have a factor group isomorphic to
VC, since it has no non-trivial abelian factor group at all.  So M must
be VC, and N must be trivial.

All in all GC has 4 normal subgroups, GC, GC', VC, and <1>.

Basically the same argument works for GE.  But there is one exception.
Namely VE has one normal subgroup of size 2 , generated by the element
(1,1,...,1;<identity>).  You may not recognize this element, but it is
in fact the superflip, which flips all twelve edges.  I shall call this
subgroup Z.

Thus GE has 5 normal subgroups GE, GE', VE, Z, and <1>.

Now we are ready to return to G resp. GCE.  The normal subgroups of GCE
are direct or subdirect products of normal subgroups of GC and GE.

For a direct product we take a normal subgroup NC of GC and a normal
subgroup NE of GE and take their direct product, i.e., their closure
in GCE.

For a subdirect product we must take a normal subgroup NC of GC and a
normal subgroup NE of GE and ``glue'' together a common factor group F of
NC and NE.  But there are only two cases where a normal subgroup of GC and
a normal subgroup of GE have a common factor group.  Once case is where
NC = GC and NE = GE and F = C_2, and we get G as a subdirect product.
The other case is NC = GC and NE = Z and F = C_2.

All in all, we get the following lattice of normal subgroups of GCE.

                                       GCE
                                       /|\
                                      X G X
                                     / \|/ \
                                    /   G'  \
                                   /   / \   \
                                  /   /   \   \
                                 /   /     \   \
                                /   /       \   \
                               /   /         \   \
                              /   /           \   \
                             /   /             \   X
                            /   /               \ / \
                           /   /                 X   \
                          /   /                 / \   \
                         X   /                 /   \   \
                        / \ /                 /     \   \
                       /   X                 /       \   GE ~ G[E]
                      /   / \               /         \ / 2
                     X   /   \             /          GE'
                    /|\ /     \           /           /
           G[C] ~ GC + X       \         /           /
                  2 \|/ \       \       /           /
                    GC'  \       \     /           /
                      \   \       \   /           /
                       \   \       \ /           / 12!/2
                        \   \       X           /
                    8!/2 \   \     / \         /
                          \   \   /   \       /
                           \   \ /     \     /
                            \   X       \   /
                             \ / \       \ /
                             VC   \       VE
                               \   \     /
                                \   \   / 2^10
                             3^7 \   \ /
                                  \   Z
                                   \ / 2
                                   <1>

(Looks nice, doesn't it?)

Bear with me, we are almost done.  We only need one more step, namely to
show that the normal subgroups of GCE that lie in G are exactely the
normal subgroups of G.  One direction is obvious, the normal subgroups of
GCE that lie in G are also normal subgroups of G.  But we need to show
that any normal subgroup of G is also a normal subgroup of GCE.

Assume then that N1 is a normal subgroup of G that is not *not* normal in
GCE.  G is then the normalizer of N1, and because the index of the
normalizer in GCE is the number of conjugates of N1, it follows that N1
has one more conjugate subgroup in G.  Call this subgroup N2, and assume
N2 = x^-1 N1 x for an appropriate element x of GCE.  Because
N_GCE(N2) = N_GCE(x^-1 N1 x) = x^-1 NGCE(N1) x = x^-1 G x = G,
it follows that N2 is also a normal subgroup of G.

The closure and the intersection of a whole family of conjugated
subgroups are always normal.  Thus the closure and the intersection of N1
and N2 are normal subgroups of GCE (and are therefore subgroups that
appear in the above lattice).  Call them N12 and N.  Clearly N12 and N are
subgroups of G.

Now the isomorphism theorem tells us that N12/N1 ~ N2/N.
Then |N12/N| = |N12/N1| |N1/N| = |N12/N1| |N2/N| = |N12/N1| |N12/N1|.
Thus |N12/N| is a square.  But the only factor groups with square sizes
are VE/Z, (VE*VC)/(Z*VC), (VE*GC')/(Z*GC') (all of size 2^10).

We can intersect the whole situation into VE, so we can without loss of
generality assume that N1 and N2 are subgroups of VE.  But if N1 and N2
are normal subgroups of G, then they are certainly normal subgroups of
GE'.  But GE' has only the 4 normal subgroups: GE', VE, Z, and <1>.

Thus there cannot be a normal subgroup of G that is not normal in GCE.

So the following picture gives the lattice of normal subgroups of G.

                                        G
                                        | 2
                                        G'
                                       / \
                                      /   \
                                     /     \
                                    /       \
                                   /         \
                                  /           \
                                 /             \
                                /               \
                               /                 X
                              /                 / \
                             /                 /   \
                            /                 /     \
                           X                 /       \
                          / \               /         \
                         /   \             /          GE'
                        /     \           /           /
                       X       \         /           /
                      / \       \       /           /
                    GC'  \       \     /           /
                      \   \       \   /           /
                       \   \       \ /           / 12!/2
                        \   \       X           /
                    8!/2 \   \     / \         /
                          \   \   /   \       /
                           \   \ /     \     /
                            \   X       \   /
                             \ / \       \ /
                             VC   \       VE
                               \   \     /
                                \   \   / 2^10
                             3^7 \   \ /
                                  \   Z
                                   \ / 2
                                   <1>

So Jerry had it almost right.  But he missed the center Z, and all
the closures of pairs of normal subgroups.

Have a nice day.

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From BRYAN@wvnvm.wvnet.edu  Mon Jan  2 23:07:26 1995
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Date:      Mon, 2 Jan 1995 23:00:01 EST
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Re: Normal Subgroups of G
In-Reply-To: Message of 12/30/94 at 15:17:00 from ,
           Martin.Schoenert@math.rwth-aachen.de

On 12/30/94 at 15:17:00 Martin Schoenert said:
>Basically the same argument works for GE.  But there is one exception.
>Namely VE has one normal subgroup of size 2 , generated by the element
>(1,1,...,1;<identity>).  You may not recognize this element, but it is
>in fact the superflip, which flips all twelve edges.  I shall call this
>subgroup Z.

>Thus GE has 5 normal subgroups GE, GE', VE, Z, and <1>.

I am still absorbing this article, which exceeds my current
knowledge of group theory.  But at the risk of asking a dumb
question, doesn't the center of GE (and of G) in fact consist
of more than just the Superflip and the identity?  Does it
not also include the Pons Asinorum and the composition of
the Pons Asinorum and the Superflip?  Call the Pons Asinorum P
and the Superflip E.  I think you are saying Z={I,E}.  But
isn't the center {I,P,E,PE}, with subgroups {I,P}, {I,E},
{I,PE}, and {I}?   These should all be central, and hence
also normal, I would think.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From @mail.uunet.ca:mark.longridge@canrem.com  Tue Jan  3 02:39:05 1995
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Subject: Centres
From: mark.longridge@canrem.com (Mark Longridge)
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Date: Tue, 3 Jan 1995 00:05:00 -0500
Organization: CRS Online  (Toronto, Ontario)

Jerry Bryan asks:

>      ... doesn't the center of GE (and of G) in fact consist
> of more than just the Superflip and the identity?  Does it
> not also include the Pons Asinorum and the composition of
> the Pons Asinorum and the Superflip?

Hmmm, I don't think so...

 The centre commutes with every process, and the Pons Asinorum
just doesn't.

E.g.   R1 + F2 B2 U2 D2 L2 R2  <>  F2 B2 U2 D2 L2 R2 + R1


I think Martin has scoped out all the possible centres in other
subgroups.

-> Mark <-

From mschoene@math.rwth-aachen.de  Tue Jan  3 07:02:09 1995
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Date: Tue, 3 Jan 95 12:59 WET
From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: Cube-Lovers@ai.mit.edu
In-Reply-To: "Jerry Bryan"'s message of Mon, 2 Jan 1995 23:00:01 EST <9501030407.AA25455@life.ai.mit.edu>
Subject: Re: Re: Normal Subgroups of G

Jerry Bryan wrote in his e-mail message of 1995/01/02

    I am still absorbing this article, which exceeds my current
    knowledge of group theory.  But at the risk of asking a dumb
    question, doesn't the center of GE (and of G) in fact consist
    of more than just the Superflip and the identity?  Does it
    not also include the Pons Asinorum and the composition of
    the Pons Asinorum and the Superflip?  Call the Pons Asinorum P
    and the Superflip E.  I think you are saying Z={I,E}.  But
    isn't the center {I,P,E,PE}, with subgroups {I,P}, {I,E},
    {I,PE}, and {I}?   These should all be central, and hence
    also normal, I would think.

This is not a dump question.  Clearly ``Pons Asinorum'' P looks very
regular, and it is not farfetched to think that it is central.  But it is
not.  Only one out of 332640 elements of GE (and of G) centralizes P.
That is to say that the index of the centralizer of P in GE has index
332640 in GE.  Since all elements of GC commute with all elements of GE,
the index of the centralizer of P in G also has index 332640 in G.
Z is indeed the center of GE', GE, G, G', and GCE.

It is in fact not too difficult to find the centers.

Recall that GE consists of the elements ( c_1, c_2, ..., c_12; p ),
where c_1 + c_2 + ... c_12 = 0 (mod 2) and p in S_12.
Since we can permute the components c_i in any way
by conjugation with an appropriate element (0,0,...,0;p),
it follows that any central element must have c_1 = c_2 = ... = c_12.
Furthermore any central elemement must have a permutation p that is
central in S_12.  So we see that we have exactely two elements in the
center of GE, namely (0,0,...,0;<id>) and (1,1,...,1;<id>).
An easy argument shows that this is also the center of GE'.

The same argumentation works for GC, but the element
(1,1,...,1;<id>) is not in GC, since 1 + 1 + ... + 1 <> 0 (mod 3)
(since we have 8 summands).  So GC has trivial center.
Again an easy argument shows that this is also the center of GC'.

The center of the direct product GCE is of course the direct product of
the centers of GC and GE.  So we see that the center of GCE is again Z.
And again an easy argument shows that this is also the center of G.

If you have more questions, please do ask.  I have tried to make my
article selfcontained.  I think the only result that I used without
proof is that S_8 and S_12 have only one proper normal subgroup.
The problem is that in order to keep the article reasonably short,
I had to be rather terse at several places.

Have a nice day.

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From mreid@ptc.com  Thu Jan  5 17:01:22 1995
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From: mreid@ptc.com (michael reid)
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To: cube-lovers%life.ai.mit.edu@ptc.com
Subject: kociemba's algorithm for quarter turns
Content-Length: 605

for much too long now, i've meant to implement kociemba's algorithm
for quarter turns.  finally i've gotten around to it, and it's found

superflip:
   B3 L3 U3 L3 F1 U1 D1 L3 B1 U1 F1 R3 L1 F3 B2 U1 D1 F2 B2 R2 U1 D1   26q

it's interesting to note that david plummer gave a 28 quarter turn
maneuver for superflip on december 10, 1980.  as far as i know, this
is the first improvement since then.

also found:

supertwist:
   B3 L2 U1 D1 R2 B3 D2 F2 D3 R2 F1 B1 L2 D3 B2 U2   24q

superfliptwist:
   U1 B3 U3 L3 F3 U3 B3 R3 D1 F3 D3 B3 U3 F3 L3 U1 F1 U1 D3 B2 U3   22q

more patterns to follow ...

mike

From gej@spamalot.mfg.sgi.com  Thu Jan  5 20:58:47 1995
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From: gej@spamalot.mfg.sgi.com (Gene Johannsen)
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Subject: Re: kociemba's algorithm for quarter turns
To: mreid@ptc.com (michael reid)
Date: Thu, 5 Jan 1995 17:58:05 -0800 (PST)
Cc: cube-lovers%life.ai.mit.edu@ptc.com
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> 
> for much too long now, i've meant to implement kociemba's algorithm

	I'm new to the list, and I've seen Kociemba's Algorithm
	referred to several times.  Where can I find some more
	information on it?

	Thanks.

gene

From @mail.uunet.ca:mark.longridge@canrem.com  Sat Jan  7 00:14:01 1995
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Date: Fri, 6 Jan 1995 23:53:00 -0500
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        Notes on Notation and Terminology for Rubik's Cube
        --------------------------------------------------

  In the "Handbook of Cubik Math":

    cubicles are in lower case, cubies are in UPPER CASE.

 If we use the 6 letters to describe the 6 faces and the various
pieces and positions, e.g. UR, UF, UL, UB are the 4 edge pieces
of the U face and URF, UFL, ULB, UBR are the 4 corner pieces.
We agree to list the facelets at a corner in clockwise order.

 This gives the following edge & corner cubicles:

uf, ul, ub, ur, rf, fl, lb, br, df, dl, db, dr
urf, ufl, ulb, ubr, dfr, dlf, dbl, drb

 and the following edge & corner cubies:

UF, UL, UB, UR, RF, FL, LB, BR, DF, DL, DB, DR
URF, UFL, ULB, UBR, DFR, DLF, DBL, DRB

 By adhering to these conventions we can establish a standard
notation for cube positions.

The sequence  R2 U3 F1 B3 R2 F3 B1 U3 R2 (9 q+h, 12 q) generates
a 3-cycle of edges. The cycle representation of this sequence
would be ( UF, UR, UB ) in ( ur, ub, uf).

Thus cubie UF resides in cubicle ur
     cubie UR resides in cubicle ub
     cubie UB resides in cubicle uf

 If we assume that the unreferenced cubies are in proper position
and orientation we have enough information to completely
describe a cube in a way which provides more information on it's
cycle structure.

 If an edge pair is flipped we refer to
            ( FU, LU ) in ( uf, ul)
 If a corner triple is twisted clockwise we refer to
            ( RFU, FLU, LBU ) in ( urf, ufl, ulb )

 Here are a couple more examples:

 The super-flip has a cycle representation of

 ( FU, LU, BU, RU, FR, LF, BL, RB, FD, LD, BD, RD )
 ( uf, ul, ub, ur, rf, fl, lb, br, df, dl, db, dr )

 The 6 X order 3 has a cycle representation of

 (( FR, FU, UR ) ( BR, FD, LU ) (BU, RD, FL ) ( BD, DL, BL))
 (( uf, ur, rf ) ( df, ul, br ) (dr, fl, ub ) ( dl, lb, db))

-> Mark <-
Email: mark.longridge@canrem.com

P.S. I'm not certain if the previously mentioned Rubik Algebra
     uses something like this, but I am going to add it to my
     cube program.

From @mail.uunet.ca:mark.longridge@canrem.com  Sat Jan  7 00:27:55 1995
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Subject: Cube terms
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.938.5834.0C1C7919@canrem.com>
Date: Fri, 6 Jan 1995 23:50:00 -0500
Organization: CRS Online  (Toronto, Ontario)

Martin Schoenert states:

>     Only one out of 332640 elements of GE (and of G) centralizes P.
> That is to say that the index of the centralizer of P in GE has index
> 332640 in GE.  Since all elements of GC commute with all elements of
> GE, the index of the centralizer of P in G also has index 332640 in G.
> Z is indeed the center of GE', GE, G, G', and GCE.

 I get the fact that only the super-flip (or 12-flip) is the centre of
G and the centre of GE. Another way to look at it would be the centre
of the cube group must effect all the corners & edges in the same
way, and only the super-flip fits these conditions when we allow
all 6 generators < U, D, F, B, L, R > to be used.

 In the case of the smaller group < U, R > we can get 6 corners twisted
either clockwise or counter-clockwise, thus effecting all the corners
and edges the same, due to the fact we can have 6 twists the same and
< U, R > only contains 6 corners, and so this is the centre of
< U, R >.

 But I don't understand how only one out of 332,640 elements of GE
and G centralizes P. I thought that GE had:

     (12 ^ 2 / 2 ) * 12! =  980,995,276,800 elements

 That is to say that the group on the cube of edges only has
980,995,276,800 elements. To be honest I'm not sure what P
represents! Jerry refers to P as the Pons Asinorum, but I think
the term may have two meanings in the two messages.

Z is the centre of G right? I need an ANSI standard math dictionary,
but I doubt such a book exists.

I'm going to tackle some more cube terminology in my next message.

-> Mark <-
Email: mark.longridge@canrem.com

From @mail.uunet.ca:mark.longridge@canrem.com  Sat Jan  7 00:34:57 1995
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Subject: Cube with GAP
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.939.5834.0C1C791A@canrem.com>
Date: Fri, 6 Jan 1995 23:51:00 -0500
Organization: CRS Online  (Toronto, Ontario)

Dan Hoey states:

> Well, call me John Henry.  Say, do you have gap libraries for other
> magic polyhedra?  For higher-dimensional magic?

 Well, I've played with GAP for a while now and at the risk of being
incorrect, I'm going to make a few comments :-)

 As I understand it, the format Martin uses in GAP is to represent
the 3x3x3 cube by assigning each individual facelet an unique
number like so (by the way, the following part is all from the
GAP documentation).

----------------------------------------------------------------------
                      +--------------+
                      |  1    2    3 |
                      |  4  top    5 |
                      |  6    7    8 |
       +--------------+--------------+--------------+--------------+
       |  9   10   11 | 17   18   19 | 25   26   27 | 33   34   35 |
       | 12  left  13 | 20 front  21 | 28 right  29 | 36  rear  37 |
       | 14   15   16 | 22   23   24 | 30   31   32 | 38   39   40 |
       +--------------+--------------+--------------+--------------+
                      | 41   42   43 |
                      | 44 bottom 45 |
                      | 46   47   48 |
                      +--------------+

then the group is generated by the following generators, corresponding
to the six faces of the cube (the two semicolons tell GAP not to print
the result, which is identical to the input here).

gap> cube := Group(
>   ( 1, 3, 8, 6)( 2, 5, 7, 4)( 9,33,25,17)(10,34,26,18)(11,35,27,19),
>   ( 9,11,16,14)(10,13,15,12)( 1,17,41,40)( 4,20,44,37)( 6,22,46,35),
>   (17,19,24,22)(18,21,23,20)( 6,25,43,16)( 7,28,42,13)( 8,30,41,11),
>   (25,27,32,30)(26,29,31,28)( 3,38,43,19)( 5,36,45,21)( 8,33,48,24),
>   (33,35,40,38)(34,37,39,36)( 3, 9,46,32)( 2,12,47,29)( 1,14,48,27),
>   (41,43,48,46)(42,45,47,44)(14,22,30,38)(15,23,31,39)(16,24,32,40)
> );;
----------------------------------------------------------------------

 You can't use T for facelet 1, and in general you can only use numbers
as facelet identifiers, no alphabetics. Given the following
conventions a magic dodecahedron should be no problem, or say a
picture Rubik's Revenge ...  I don't know how a normal 4x4x4 could
be represented though.

-> Mark <-
Email: mark.longridge@canrem.com

From @mail.uunet.ca:mark.longridge@canrem.com  Sat Jan  7 03:15:53 1995
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Subject: Cube Twist Correction
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.941.5834.0C1C7961@canrem.com>
Date: Fri, 6 Jan 1995 23:59:00 -0500
Organization: CRS Online  (Toronto, Ontario)


> If a corner triple is twisted clockwise we refer to
>             ( RFU, FLU, LBU ) in ( urf, ufl, ulb )

Okay..... this would actually be an anti-clockwise tri-twist.
-> Mark <-

From mschoene@math.rwth-aachen.de  Sat Jan  7 10:55:20 1995
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Date: Sat, 7 Jan 95 16:52 WET
From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: cube-lovers@life.ai.mit.edu
In-Reply-To: Mark Longridge's message of Fri, 6 Jan 1995 23:50:00 -0500 <60.938.5834.0C1C7919@canrem.com>
Subject: Re: Cube terms

I wrote in my e-mail of 1995/01/03

        Only one out of 332640 elements of GE (and of G) centralizes P.
    That is to say that the index of the centralizer of P in GE has index
    332640 in GE.  Since all elements of GC commute with all elements of
    GE, the index of the centralizer of P in G also has index 332640 in G.
    Z is indeed the center of GE', GE, G, G', and GCE.

Mark Longridge answered in his e-mail of 1995/01/06

     I get the fact that only the super-flip (or 12-flip) is the centre of
    G and the centre of GE. Another way to look at it would be the centre
    of the cube group must effect all the corners & edges in the same
    way, and only the super-flip fits these conditions when we allow
    all 6 generators < U, D, F, B, L, R > to be used.

This sounds very plausible.  But I must admit that I find it notoriously
difficult to turn such plausible arguments into proper proofs.
If you try, you may in fact end up with something similar to my proof.
Because the crucial part in my proof is that a central element
must have all components in the wreath product equal, because one
has the full symmetric group S_12 acting on the 12 components.

Mark continued

     In the case of the smaller group < U, R > we can get 6 corners twisted
    either clockwise or counter-clockwise, thus effecting all the corners
    and edges the same, due to the fact we can have 6 twists the same and
    < U, R > only contains 6 corners, and so this is the centre of
    < U, R >.

This is the ``odd'' element I referred to in my message on shift
invariant processes.

Mark continued

     But I don't understand how only one out of 332,640 elements of GE
    and G centralizes P. I thought that GE had:

         (12 ^ 2 / 2 ) * 12! =  980,995,276,800 elements

     That is to say that the group on the cube of edges only has
    980,995,276,800 elements. To be honest I'm not sure what P
    represents! Jerry refers to P as the Pons Asinorum, but I think
    the term may have two meanings in the two messages.

Sorry, that is just me wrestling with English.  What I meant to say was
``... only one out of *every* 332640 elements of GE ...''.
That is, of the total 980995276800 elements in GE
only 980995276800/332640 = 2949120 elements centralize P.
And I used the definition of P from your e-mail of 1995/01/03,
i.e., P = (F2 B2) (U2 D2) (L2 R2) = (F2 B2) (L2 R2) (U2 D2) = ...
(one gets the same element independent of the order of the three pairs).

Mark continued

    Z is the centre of G right? I need an ANSI standard math dictionary,
    but I doubt such a book exists.

    I'm going to tackle some more cube terminology in my next message.

Z in this case refers to the subgroup generated by the superflip.
I wrote in my e-mail of 1994/12/30

    Namely VE has one normal subgroup of size 2 , generated by the element
    (1,1,...,1;<identity>).  You may not recognize this element, but it is
    in fact the superflip, which flips all twelve edges.  I shall call this
    subgroup Z.

I would have preferred to call it C, but C was already taken for the
group of rotations of the entire cube.  Thus I took Z instead, because
``Zentrum'' is the german word for center.  It is not uncommon to use
Z to denote the center of a group G, e.g., Huppert uses Z(G) for the
center in his ``Theory of Groups''.

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany


From mschoene@math.rwth-aachen.de  Sat Jan  7 11:08:45 1995
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Date: Sat, 7 Jan 95 17:05 WET
From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: cube-lovers@life.ai.mit.edu
Cc: gej@spamalot.mfg.sgi.com
In-Reply-To: Gene Johannsen's message of Thu, 5 Jan 1995 17:58:05 -0800 (PST) <199501060158.RAA02633@spamalot.mfg.sgi.com>
Subject: Re: Re: kociemba's algorithm for quarter turns

Gene Johannsen wrote in his e-mail message of 1995/01/05

	I'm new to the list, and I've seen Kociemba's Algorithm
	referred to several times.  Where can I find some more
	information on it?

Great timing.  This allows me to tell you all about the newest feature of
the Cube-Lovers WWW pages.  You can now search the old articles for
keywords (if you have a Browser that supports forms, e.g. 'Mosaic' or
'Netscape').  Check out

    http://www.math.rwth-aachen.de:8000/~mschoene/Cube-Lovers/Index_for_Keyword.html

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From BRYAN@wvnvm.wvnet.edu  Sat Jan  7 11:09:27 1995
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Date:      Sat, 7 Jan 1995 10:15:52 EST
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@life.ai.mit.edu>
Subject:   Re: Cube with GAP
In-Reply-To: Message of 01/06/95 at 23:51:00 from mark.longridge@canrem.com

On 01/06/95 at 23:51:00 mark.longridge@canrem.com said:

> As I understand it, the format Martin uses in GAP is to represent
>the 3x3x3 cube by assigning each individual facelet an unique
>number like so (by the way, the following part is all from the
>GAP documentation).

>----------------------------------------------------------------------
>                      +--------------+
>                      |  1    2    3 |
>                      |  4  top    5 |
>                      |  6    7    8 |
>       +--------------+--------------+--------------+--------------+
>       |  9   10   11 | 17   18   19 | 25   26   27 | 33   34   35 |
>       | 12  left  13 | 20 front  21 | 28 right  29 | 36  rear  37 |
>       | 14   15   16 | 22   23   24 | 30   31   32 | 38   39   40 |
>       +--------------+--------------+--------------+--------------+
>                      | 41   42   43 |
>                      | 44 bottom 45 |
>                      | 46   47   48 |
>                      +--------------+

Note that this model does not include the face centers.  That is, it
is G[C,E] rather than G[C,E,F].  56 numbers would be required to
include the face centers.  The distinction between 48 facelets and
56 facelets bears on the nitpicky question of whether the set C of
rotations is a subgroup of G or not.  What I don't see is how to
model the Supergroup in GAP.  It looks like you would have to
label each Face center with four numbers so you could see the
rotations of the Face centers, but that seems like overkill.

I call this kind of model a facelet model
rather than a cubie model, and the twists and flips are implicit in
a facelet model.  I would think that the twists and flips would have
to be made explicit in a cubie model.  Dan Hoey reported to me once
that he had an error wherein his corners turned themselves inside
out.   I can't totally picture how that happened, but it was related
to the fact that he was using a cubie model with a little
multiplication table for the twists.

I have always used a facelet
model, except that I number the corners from 1 to 24 and the edges
from 1 to 24 for historical reasons.  That is, I started with corners
only or edges only, and have only lately put the two together.  It
really does not create any problems for me to use the same numbers
for both edges and corners because the edges and corners
are stored disjointly, as are the edge and corner permutations
for quarter and half turns, and as are the edge and corner
permutations for rotations and reflections.

When I write the model out to disk, I only write out 8 corner facelets
and 12 edge facelets.  For example, I only write out the front and
back corner facelets.  This saves space and converts the model from
a facelet model to a cubie model, with the twists implicitly encoded
rather than being explicitly encoded via multiplication tables.  It
also automatically establishes a frame of reference by which a
proof of conservation of twist and flip can be accomplished.

>                        ...  I don't know how a normal 4x4x4 could
>be represented though.

I fail to see the problem.  Just number the facelets.  The only
problem would then lie in deciding what the generators are  -- i.e.,
which kind of slice moves do you accept.  You would also have to
decide whether to model the invisible 2x2x2 inside, but again if you
did, just number the invisible facelets and include their movements
with your generators.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From mreid@ptc.com  Sat Jan  7 16:43:14 1995
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From: mreid@ptc.com (michael reid)
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To: cube-lovers@life.ai.mit.edu
Subject: Re: kociemba's algorithm for quarter turns
Content-Length: 652

gene writes:

> 	I'm new to the list, and I've seen Kociemba's Algorithm
> 	referred to several times.  Where can I find some more
> 	information on it?

there are several places you might try.

dik winter first wrote about this new searching algorithm on may 3, 1992,
and there was a bit of discussion after that.  so you can look through
the archives (cube-mail-8).

if you are a member of nkc (nederlands kubus club) you might look
through the newsletter cff (cubism for fun) issue #28 where herbert
kociemba first writes about his algorithm.

also, martin gave some info about the cube-lovers archives being
available on the world-wide-web.
 
mike

From mreid@ptc.com  Sat Jan  7 19:32:56 1995
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From: mreid@ptc.com (michael reid)
Message-Id: <9501080043.AA07043@ducie.ptc.com>
To: cube-lovers@ai.mit.edu
Subject: two stage filtration
Content-Length: 4359

kociemba's algorithm uses the filtration

G = <U, D, F, R, B, L>        of order   43252003274489856000
H = <U, D, F2, R2, B2, L2>    of order            19508428800
1 = <>                        of order                      1

i've run an exhaustive search on the coset space  G / H.  the number
of cosets at each distance is:

 distance    quarter turns       face turns

        0                1                1
        1                4                4
        2               34               50
        3              312              592
        4             2772             7156
        5            24996            87236
        6           225949          1043817
        7          2017078         12070278
        8         17554890        124946368
        9        139132730        821605960
       10        758147361       1199128738
       11       1182378518         58202444
       12        117594403              476
       13            14072

the computation for face turns was already done by dik winter (see his
message of may 28, 1992), so in particular, this confirms his calculation.

the cosets  G / H  are described by triples  (c, e, l),  where

   c = corner orientation
   e = edge orientation
   l = location of middle layer edges (FR, FL, BR, BL)

there are  3^7   = 2187 corner configurations,
           2^11  = 2048 edge configurations, and
          / 12 \
          \  4 / =  495 location configurations,

to give a total of  2187 * 2048 * 495 = 2217093120  configurations.
to reduce this number somewhat, we can utilize symmetry.  there are 16
symmetries of the cube that preserve the  U-D axis,  and therefore
preserve the subgroup  H.  up to these symmetries, the number of distinct
corner configurations is 168, so we need only consider a mere
168 * 2048 * 495 = 170311680  configurations.

(so far, this is the same approach that dik used for his calculation.)

each configuration is stored with 2 bits of memory and thus the whole
space consumes about 42 megabytes.  each configuration is assigned
one of 4 values:

        distance is currently unknown
        distance = current search depth
        distance = current search depth - 1
        distance < current search depth - 1

from here, i just used a simple breadth first search.

unfortunately, something unpleasant happened along the way ...

at some point, i realized that the symmetries do not act on the edge
configurations.  to define edge flip, one must choose one facelet from
each of the 4 middle layer edges to correspond to the  U  or  D  facelet
of the other 8 edges.  (i chose the  F  and  B  facelets, but this is
completely arbitrary.)  but now we've lost some symmetry; these 12 facelets
are not preserved under the 16 symmetries, in particular, the rotation
C_U  does not preserve them.

therefore, we need lookup tables for the action of the symmetries on
edge x location space.  this gives 16 symmetries * 2048 edge configurations
* 495 location configurations * 4 bytes per integer = 64 megabytes of
lookup tables.  ouch!

i was too far along at this point to start all over, and i had the memory
available, so i just continued with this approach.  however, in hindsight,
i'd probably use one of the following ideas if i had to start over:

      i) only use the 8 symmetries that preserve my choice of
         12 edge facelets.

     ii) combine the two coordinates edge and location into a single
         coordinate and divide this coordinate by the 16 symmetries.

run times were improved significantly by using a simple trick that i hadn't
used in earlier programs.  during the first few depth levels, i use
"forward searching", i.e. i examine the neighbors of each configuration
found at the previous depth.  however, after at least half the search space
has been found, i switch to "backward searching", i.e. examine the
configurations (and their neighbors) that haven't yet been found.

(have others been using this same idea when running similar search programs?)

closer analysis of this technique suggests that the switch from forward to
backward searching should occur even before half the space has been found.
i didn't do this here since the run times were quite satisfactory:
40 minutes for quarter turns, 47 minutes for face turns.  this was done
on a DEC 3000 alpha 700, apparently a very fast machine.

mike

From mreid@ptc.com  Sat Jan  7 19:53:08 1995
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From: mreid@ptc.com (michael reid)
Message-Id: <9501080104.AA07071@ducie.ptc.com>
To: cube-lovers@ai.mit.edu
Subject: two stage filtration
Content-Length: 3087

i've also run an exhaustive search on the subgroup
H = <U, D, F2, R2, B2, L2>.   here are the number of
positions at each distance.

 distance    quarter turns       face turns

        0                1                1
        1                4               10
        2               10               67
        3               36              456
        4              123             3079
        5              368            19948
        6             1192           123074
        7             3792           736850
        8            11263          4185118
        9            34352         22630733
       10           102638        116767872
       11           287320        552538680
       12           810144       2176344160
       13          2261028       5627785188
       14          5941838       7172925794
       15         16291708       3608731814
       16         41973415        224058996
       17        107458884          1575608
       18        269542476             1352
       19        628442876
       20       1367654200
       21       2613422312
       22       3997726648
       23       4444701268
       24       3661653732
       25       1906936668
       26        407132392
       27         34358944
       28          1664168
       29            14840
       30              160

a position at distance 18 face turns was exhibited by hans kloosterman
on may 30 1992.  (he also found three others that differ only in the
middle layer edges.)  it was then observed by dik winter (also on may 30
1992) that kociemba's algorithm took exceptionally long for this position.
however, this does not appear to be the case for most of the antipodes.
(i will give the antipodes for each metric in separate messages.)

the 4 positions found by kloosterman are also antipodes in the quarter
turn metric, and, up to symmetry, are the only positions which are
antipodal in both metrics.  hmmm...

elements of  H  are described by triples  (c, e, m),  where

        c = corner permutation,
        e = U D edge permutation,
        m = middle layer edge permutation,

and the total parity is even.  there are

        8! = 40320 corner configurations,
        8! = 40320 U D edge configurations and
        4! =    24 middle layer edge configurations,

for a total of  40320 * 40320 * 24 / 2 = 19508428800  positions.

if we divide by symmetry along the corner coordinate, we get 2768 corner
configurations (of course we get the same number if we divide by symmetry
along the U D edge coordinate), so we can reduce to 1339269120 positions.
at 2 bits per configuration, this requires 327 megabytes, which is too
large.

however, if we also divide out by inversion, we can reduce the number of
corner configurations to 1672, the total number of positions to 808980480,
and the memory required to 200 megabytes.  this is still a lot, but is
within reach.

the calculations were done on the same machine: DEC 3000 alpha 700,
configured with 256 Mb RAM.  run times were much more modest:
10 hours for quarter turns, 7.5 hours for face turns.

mike

From mreid@ptc.com  Sat Jan  7 20:14:05 1995
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From: mreid@ptc.com (michael reid)
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To: cube-lovers@ai.mit.edu
Subject: new upper bounds
Content-Length: 2083

from these calculations, we get new upper bounds on the length of
"god's algorithm":  42 quarter turns, 29 face turns.  (no, i didn't
add incorrectly.)  the previous upper bounds were 56 quarter turns,
37 face turns.  the best known lower bounds are 21 quarter turns,
18 face turns.

here's how to get these upper bounds.  note that the last twist in
stage 1 is always a quarter turn of either  F, R, B  or  L,  and the
direction doesn't matter.  thus by choosing the direction of this
quarter turn properly, we hope to be able to avoid the positions at
maximal distance in stage 2.

the program verified that no two positions at distance 30 quarter
turns differ by  F2, R2, B2  or  L2,  so we may avoid these bad cases.
i expected to be able to avoid the positions at distance 29 quarter
turns as well, but alas, things do not always go as planned.  the
following two positions at distance 29 quarter turns differ by  B2:

position 1:
D1 R2 D3 L2 D3 R2 U3 D3 R2 U1 B2 D3 L2 D3 R2 D3 F2 D1 B2 D1   29q

position 2:
R2 U3 L2 U3 D3 L2 D1 L2 D1 R2 F2 D1 F2 D3 L2 B2 D1 B2 D1   29q

there are probably many other examples.

similarly, the positions at distance 18 face turns were checked and
no two of these differ by  F2, R2, B2  or  L2,  so these positions
may be avoided.

this gives upper bounds of  13 + 29 = 42  quarter turns and
12 + 17 = 29  face turns.

i expect to be able to reduce the 42 quarter turns slightly.  for
example, to improve it to 41 quarter turns, i just need to check
that any position in stage 2 can be solved in at most 28 quarter
turns, where we now allow all turns.  of course, this only requires
testing the positions at distance 29 and 30.  i expect this to be
straightforward, but i don't know how much improvement i can get
with this approach.

the same approach doesn't seem plausible for face turns.  in order
to get just 1 face turn improvement, all positions at distance 17
face turns would need to be solvable in at most 16 face turns.
this doesn't seem promising.  probably most of these require 17
face turns even with all turns available.

mike

From mreid@ptc.com  Sat Jan  7 20:33:56 1995
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From: mreid@ptc.com (michael reid)
Message-Id: <9501080144.AA07165@ducie.ptc.com>
To: cube-lovers@ai.mit.edu
Subject: antipodes in quarter turn metric
Content-Length: 2017

these are the antipodes of  H = <U, D, F2, R2, B2, L2>  in the quarter
turn metric, divided by the 16 symmetries.  btw, these symmetries
are generated by  C_U,  C_F2  and  reflection through the  U-D  plane.

  1)   D3 B2 D1 F2 R2 B2 R2 U1 R2 U3 F2 D3 R2 D1 R2 U3 L2 B2 U3     30q
  2)   D3 B2 D3 L2 F2 B2 D1 L2 U3 L2 F2 D1 F2 U3 B2 U1 L2 F2 U3     30q
  3)   D3 B2 L2 D3 B2 R2 L2 B2 R2 L2 D3 B2 R2 U3 D3 F2 L2 D1        30q
  4)   D3 B2 L2 D3 B2 R2 L2 B2 R2 L2 D3 B2 R2 U2 R2 F2 U1           30q
  5)   D3 B2 D3 B2 L2 B2 U1 F2 U1 R2 L2 U1 F2 L2 D3 F2 L2 F2        30q
  6)   D3 B2 D3 B2 L2 B2 D1 R2 D1 R2 L2 D1 L2 B2 U3 F2 L2 F2        30q
  7)   D3 B2 L2 D1 F2 B2 L2 F2 B2 L2 D1 F2 L2 U1 D1 B2 R2 D1        30q
  8)   D3 B2 D3 B2 U1 F2 U3 L2 D3 F2 D1 R2 B2 R2 L2 D1 B2 R2 U1     30q
  9)   D3 B2 D3 F2 L2 U3 R2 D1 B2 R2 D3 R2 U1 F2 L2 D1 B2 R2 U1     30q
 10)   D3 B2 L2 D3 B2 R2 L2 B2 R2 L2 D3 B2 R2 U3 D3 F2 L2 U3        30q
 11)   D3 B2 L2 D1 F2 B2 L2 F2 B2 L2 D1 B2 R2 U3 D3 F2 L2 U3        30q
 12)   D3 B2 L2 D1 F2 B2 L2 F2 B2 L2 D1 F2 L2 U1 D1 B2 R2 U3        30q
 13)   D3 B2 D3 F2 L2 D3 F2 U1 F2 L2 D3 L2 U1 B2 R2 U1 L2 B2 U1     30q
 14)   D3 B2 L2 D1 F2 B2 L2 F2 B2 L2 D1 B2 R2 U1 D1 B2 R2 U1        30q
 15)   D3 B2 L2 D3 B2 R2 L2 B2 R2 L2 D3 B2 R2 U2 R2 F2 D1           30q
 16) inverse of 15
 17)   D3 B2 L2 D3 B2 R2 L2 B2 R2 L2 D3 B2 R2 U1 D1 B2 R2 U1        30q
 18)   D3 B2 D3 B2 L2 B2 U1 F2 U1 R2 L2 U1 F2 L2 U3 R2 F2 R2        30q
 19)   D3 B2 L2 D3 B2 R2 L2 B2 R2 L2 D3 F2 L2 U3 D3 F2 L2 U1        30q
 20)   D3 B2 D3 B2 L2 U1 L2 U3 F2 L2 D1 R2 U3 B2 U1 B2 R2 D3 F2     30q
 21) inverse of 20
 22)   D3 B2 D3 B2 U1 R2 L2 F2 U3 B2 R2 D1 F2 L2 U1 B2 U3 F2 D3     30q
 23)   D3 B2 D3 B2 L2 U3 L2 U1 F2 D1 F2 D3 L2 F2 B2 D1 F2 R2 D3     30q
 24)   D3 B2 D3 L2 D1 L2 U3 B2 U1 B2 L2 U1 B2 R2 L2 D3 F2 U3 R2     30q
 25)   D3 B2 D3 B2 L2 D3 B2 U1 L2 D1 L2 D3 B2 R2 L2 D1 L2 F2 D3     30q

the position identified by hans kloosterman is number 14.  there are
also a few closely related positions.

mike

From mreid@ptc.com  Sat Jan  7 21:03:01 1995
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From: mreid@ptc.com (michael reid)
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To: cube-lovers@ai.mit.edu
Subject: antipodes in face turn metric
Content-Length: 8220

these are the antipodes of  H = <U, D, F2, R2, B2, L2>  in the face
turn metric, divided by the 16 symmetries.

  1)   D3 B2 D3 B2 D2 L2 D1 L2 B2 D2 F2 D1 R2 F2 U1 R2 F2 R2    18f
  2)   D3 B2 D3 B2 D2 L2 D1 L2 B2 D2 F2 U1 F2 L2 D1 R2 F2 R2    18f
  3)   D3 B2 D3 F2 D2 L2 U3 L2 F2 B2 U2 F2 U1 B2 U1 L2 B2 R2    18f
  4)   D3 B2 D3 L2 D2 R2 L2 F2 D3 B2 D2 R2 D3 F2 U1 B2 L2 F2    18f
  5)   D3 B2 D3 B2 U2 R2 D1 R2 F2 U2 F2 U1 F2 L2 D1 R2 F2 R2    18f
  6) inverse of 5
  7)   D3 B2 D3 L2 U2 R2 L2 B2 U3 R2 D2 F2 D3 L2 D1 B2 L2 F2    18f
  8)   D3 B2 D3 L2 U1 B2 U3 B2 R2 D3 B2 R2 U3 F2 D1 R2 B2 R2    18f
  9) inverse of 8
 10)   D3 B2 D3 L2 D1 L2 D3 B2 R2 U3 L2 B2 D3 F2 D1 R2 B2 R2    18f
 11)   D3 B2 D3 L2 D1 L2 U3 L2 B2 D3 L2 B2 D3 F2 D1 R2 B2 R2    18f
 12) inverse of 11
 13)   D3 B2 D3 L2 D1 L2 U3 L2 B2 D3 L2 B2 U3 R2 U1 R2 B2 R2    18f
 14)   D3 B2 L2 D2 L2 D3 B2 L2 U1 F2 U2 R2 U3 F2 B2 R2 D1 B2    18f
 15) inverse of 14
 16)   D3 B2 U2 R2 D1 L2 F2 R2 F2 D1 R2 D2 R2 B2 U1 B2 U1 B2    18f
 17) inverse of 16
 18)   D3 B2 L2 D2 L2 D3 B2 L2 D3 F2 D2 R2 U1 R2 L2 B2 U1 B2    18f
 19)   D3 B2 U2 R2 D1 L2 F2 R2 F2 U1 F2 D2 F2 R2 D1 B2 U1 B2    18f
 20)   D3 B2 D3 R2 F2 B2 U1 R2 L2 B2 D3 R2 D1 R2 F2 U1 L2 D3    18f
 21) inverse of 20
 22)   D3 B2 D3 F2 B2 R2 U1 R2 U3 L2 F2 R2 F2 D3 B2 U1 R2 D3    18f
 23)   D3 B2 D3 F2 B2 R2 U1 R2 D3 B2 L2 F2 L2 U3 B2 U1 R2 D3    18f
 24) inverse of 23
 25)   D3 B2 D3 F2 L2 U3 L2 D1 F2 U3 L2 D1 L2 B2 D1 F2 L2 B2    18f
 26)   D3 B2 D3 F2 B2 R2 D1 B2 U3 F2 R2 B2 R2 U3 F2 D1 F2 U3    18f
 27) inverse of 26
 28)   D3 B2 D2 L2 D1 L2 D1 L2 B2 D2 F2 D1 R2 U3 F2 L2 F2 B2    18f
 29) inverse of 28
 30)   D3 B2 D2 L2 D1 L2 U1 B2 R2 U2 R2 U1 R2 U3 F2 L2 F2 B2    18f
 31)   D3 B2 D3 F2 B2 R2 U1 R2 D3 B2 L2 F2 L2 D3 R2 D1 R2 D3    18f
 32)   D3 B2 L2 U3 F2 R2 B2 R2 D1 B2 U3 F2 U1 F2 D1 F2 D3 B2    18f
 33)   D3 F2 B2 L2 D1 B2 U1 R2 D3 L2 D1 L2 D1 R2 F2 U1 F2 R2    18f
 34)   D3 B2 R2 D3 F2 L2 B2 L2 D1 L2 D3 R2 U1 F2 D1 R2 U3 L2    18f
 35)   D3 B2 L2 D3 L2 F2 R2 F2 D1 R2 D3 B2 D1 L2 D1 L2 U3 B2    18f
 36)   D3 B2 L2 D3 L2 F2 R2 F2 D1 R2 U3 L2 D1 F2 D1 F2 D3 B2    18f
 37) inverse of 36
 38)   D3 B2 R2 U3 R2 F2 L2 F2 U1 L2 D3 R2 D1 R2 U1 R2 U3 L2    18f
 39)   D3 B2 L2 D3 R2 D2 B2 D3 R2 B2 U2 F2 D1 B2 D3 R2 B2 R2    18f
 40)   D3 B2 L2 D3 R2 U2 F2 U3 F2 R2 D2 R2 U1 B2 D3 R2 B2 R2    18f
 41)   D3 B2 L2 D1 B2 D2 R2 D1 R2 B2 U2 F2 U1 R2 U3 R2 B2 R2    18f
 42)   D3 B2 L2 D1 B2 D2 R2 U1 F2 R2 U2 L2 D1 R2 U3 R2 B2 R2    18f
 43)   D3 B2 L2 D3 R2 D2 B2 U3 B2 L2 D2 L2 D1 R2 U3 R2 B2 R2    18f
 44) inverse of 43
 45)   D3 B2 L2 D3 R2 D2 B2 D3 R2 B2 U2 F2 U1 R2 U3 R2 B2 R2    18f
 46)   D3 B2 D3 B2 L2 D2 F2 L2 F2 R2 D2 F2 R2 D1 F2 B2 R2 U3    18f
 47)   D3 B2 D3 B2 L2 U2 L2 F2 D2 F2 R2 U1 B2 R2 L2 B2 R2 U3    18f
 48)   D3 B2 D3 B2 L2 D2 L2 F2 D2 B2 L2 U1 B2 R2 L2 B2 R2 U3    18f
 49)   D3 B2 D3 B2 L2 U2 B2 L2 F2 L2 D2 B2 L2 D1 F2 B2 R2 U3    18f
 50)   D3 B2 U1 B2 U3 L2 U1 L2 U3 B2 R2 D3 F2 D1 L2 B2 L2 F2    18f
 51)   D3 B2 D1 B2 D3 R2 U1 F2 D3 R2 B2 D3 B2 D1 F2 R2 F2 L2    18f
 52) inverse of 51
 53)   D3 B2 D1 B2 D3 R2 D1 R2 D3 B2 L2 D3 L2 U1 F2 R2 F2 L2    18f
 54) inverse of 53
 55)   D3 B2 D1 B2 D3 R2 U1 F2 U3 B2 L2 D3 L2 U1 F2 R2 F2 L2    18f
 56)   D3 B2 D1 F2 L2 D3 R2 F2 D3 L2 D1 B2 R2 U1 F2 U3 B2 R2    18f
 57)   D3 B2 D1 F2 L2 D3 R2 F2 U3 F2 U1 B2 R2 D1 R2 D3 B2 R2    18f
 58)   D3 B2 D1 B2 U3 B2 U1 R2 D3 B2 L2 D3 L2 U1 F2 R2 F2 L2    18f
 59)   D3 B2 U1 F2 R2 U3 L2 F2 D3 F2 U1 R2 B2 U1 B2 U3 R2 B2    18f
 60) inverse of 59
 61)   D3 B2 L2 D3 R2 U1 F2 R2 L2 D3 L2 F2 U3 L2 U3 L2 D1 R2    18f
 62)   D3 B2 D2 L2 D1 R2 U3 F2 L2 U2 L2 D3 R2 D1 F2 R2 F2 L2    18f
 63) inverse of 62
 64)   D3 B2 D2 L2 U1 B2 R2 B2 L2 D3 R2 U2 L2 B2 D3 F2 U3 R2    18f
 65)   D3 B2 D2 L2 D1 L2 B2 L2 F2 D3 B2 U2 F2 L2 D3 R2 D3 R2    18f
 66)   D3 B2 D2 L2 D1 L2 B2 L2 F2 U3 L2 D2 L2 B2 D3 F2 U3 R2    18f
 67)   D3 B2 D2 L2 D1 R2 D3 L2 B2 D2 F2 D3 B2 U1 F2 R2 F2 L2    18f
 68)   D3 B2 D2 L2 D1 L2 B2 L2 F2 D3 B2 D2 B2 R2 U3 F2 U3 R2    18f
 69)   D3 B2 U2 R2 U1 B2 U3 R2 F2 U2 F2 D3 B2 U1 F2 R2 F2 L2    18f
 70)   D3 B2 D1 B2 D3 L2 F2 U1 B2 D3 F2 B2 L2 D1 B2 L2 D3 L2    18f
 71) inverse of 70
 72)   D3 B2 U1 R2 D3 B2 L2 U1 R2 D3 R2 L2 B2 D1 R2 B2 U3 L2    18f
 73)   D3 B2 D2 R2 U1 L2 F2 R2 F2 D3 F2 D2 B2 L2 D3 R2 D3 R2    18f
 74)   D3 B2 D2 R2 U1 L2 F2 R2 F2 U3 R2 U2 R2 B2 D3 F2 U3 R2    18f
 75)   D3 B2 D2 R2 D1 F2 R2 B2 R2 D3 R2 U2 R2 B2 D3 F2 U3 R2    18f
 76)   D3 B2 D2 R2 U1 L2 F2 R2 F2 D3 F2 U2 F2 R2 U3 F2 U3 R2    18f
 77)   D3 B2 D2 R2 D1 F2 R2 B2 R2 D3 R2 U2 R2 B2 U3 R2 D3 R2    18f
 78)   D3 B2 D3 F2 D2 L2 D3 F2 U2 L2 F2 B2 D3 R2 D3 L2 F2 R2    18f
 79) inverse of 78
 80)   D3 B2 D3 F2 D2 L2 D3 F2 D2 R2 F2 B2 U3 F2 U3 L2 F2 R2    18f
 81)   D3 B2 D3 F2 D2 L2 D1 R2 F2 B2 U2 F2 U1 F2 U3 L2 F2 R2    18f
 82) inverse of 81
 83)   D3 B2 D1 B2 D1 B2 U2 R2 D3 L2 F2 L2 B2 D1 B2 U2 F2 L2    18f
 84)   D3 B2 D3 F2 U2 R2 D3 B2 U2 R2 F2 B2 U3 F2 U3 L2 F2 R2    18f
 85) inverse of 84
 86)   D3 B2 D3 F2 U2 R2 U1 B2 R2 L2 U2 R2 U1 R2 D3 L2 F2 R2    18f
 87)   D3 B2 D1 B2 D1 B2 D2 L2 D3 R2 B2 R2 F2 D1 F2 D2 F2 L2    18f
 88)   D3 B2 D3 F2 D2 L2 U1 F2 R2 L2 U2 L2 D1 F2 U3 L2 F2 R2    18f
 89)   D3 B2 D3 F2 D2 L2 U1 F2 R2 L2 D2 R2 U1 R2 D3 L2 F2 R2    18f
 90)   D3 B2 D3 R2 D2 F2 D3 L2 D2 B2 U1 F2 R2 L2 U3 L2 B2 R2    18f
 91) inverse of 90
 92)   D3 B2 D3 R2 D2 F2 U3 F2 D2 L2 U1 R2 F2 B2 D3 L2 B2 R2    18f
 93)   D3 B2 D3 R2 D2 B2 R2 L2 U1 L2 U2 B2 U3 L2 D3 R2 B2 L2    18f
 94) inverse of 93
 95)   D3 B2 D3 R2 D2 B2 R2 L2 D3 R2 U2 B2 D1 L2 D3 R2 B2 L2    18f
 96)   D3 B2 D3 R2 D2 B2 R2 L2 U3 B2 U2 L2 U1 L2 D3 R2 B2 L2    18f
 97) inverse of 96
 98)   D3 B2 D3 R2 D2 B2 R2 L2 U1 L2 U2 B2 D3 B2 U3 R2 B2 L2    18f
 99)   D3 B2 D3 R2 D2 F2 U3 F2 U2 R2 D1 F2 R2 L2 U3 L2 B2 R2    18f
100)   D3 B2 D3 R2 D2 F2 D1 B2 D2 L2 D3 R2 F2 B2 D3 L2 B2 R2    18f
101)   D3 B2 D3 R2 D2 B2 R2 L2 D3 R2 U2 B2 U1 B2 U3 R2 B2 L2    18f
102)   D3 B2 D2 L2 D1 B2 R2 F2 R2 U1 F2 D2 B2 R2 D1 R2 U1 F2    18f
103)   D3 B2 D3 F2 D1 R2 U3 B2 L2 U3 B2 L2 U3 R2 D1 R2 F2 R2    18f
104)   D3 B2 U2 R2 D1 F2 L2 B2 L2 D1 L2 U2 L2 B2 U1 R2 U1 F2    18f
105)   D3 B2 U3 R2 D1 B2 D3 B2 L2 U3 B2 L2 U3 R2 D1 R2 F2 R2    18f
106)   D3 B2 D3 F2 U1 F2 D3 F2 R2 D3 L2 F2 D3 R2 D1 R2 F2 R2    18f
107) inverse of 106
108)   D3 B2 U2 R2 U1 L2 B2 R2 B2 U1 R2 D2 L2 B2 U1 R2 U1 F2    18f
109)   D3 B2 U3 R2 D1 B2 U3 L2 F2 U3 L2 F2 D3 R2 D1 R2 F2 R2    18f
110)   D3 B2 D3 B2 R2 L2 D1 F2 B2 R2 U3 B2 U1 B2 L2 D1 F2 D3    18f
111)   D3 B2 D2 L2 D1 F2 U1 F2 R2 D2 R2 D1 R2 D3 B2 L2 F2 B2    18f
112)   D3 B2 U3 L2 F2 B2 U1 F2 B2 R2 D3 R2 D1 B2 L2 U1 L2 U3    18f
113)   D3 B2 D3 B2 U1 L2 F2 L2 B2 D1 R2 U3 R2 F2 B2 D1 B2 D3    18f
114)   D3 B2 D3 B2 D1 F2 R2 F2 L2 D1 B2 D3 R2 F2 B2 U1 R2 U3    18f
115) inverse of 114
116)   D3 B2 D3 B2 D1 F2 R2 F2 L2 D1 B2 D3 R2 F2 B2 D1 B2 D3    18f
117)   D3 B2 D2 L2 U1 L2 U1 L2 F2 U2 B2 U1 R2 D3 B2 L2 F2 B2    18f
118)   D3 B2 D3 F2 U2 F2 B2 L2 D3 F2 D2 L2 D3 L2 D1 R2 F2 L2    18f
119) inverse of 118
120)   D3 B2 D3 F2 D2 F2 B2 R2 D1 R2 D2 B2 U1 F2 U1 R2 F2 L2    18f
121)   D3 B2 D3 L2 D2 B2 D1 L2 U2 F2 R2 L2 D3 R2 D1 F2 R2 B2    18f
122)   D3 B2 D3 F2 U2 F2 B2 L2 U1 B2 U2 R2 D1 F2 U1 R2 F2 L2    18f
123)   D3 B2 D3 F2 D2 F2 B2 R2 D1 R2 U2 F2 D1 L2 D1 R2 F2 L2    18f
124)   D3 B2 D3 F2 D2 F2 B2 R2 U3 L2 U2 F2 D3 F2 U1 R2 F2 L2    18f
125)   D3 B2 L2 D3 B2 D1 R2 B2 D2 B2 D1 F2 R2 U1 R2 D2 R2 B2    18f
126) inverse of 125
127)   D3 B2 L2 D3 B2 D1 R2 B2 U2 F2 U1 R2 B2 D1 L2 U2 R2 B2    18f
128)   D3 B2 L2 U3 L2 D1 B2 L2 U2 R2 U1 B2 L2 U1 L2 U2 R2 B2    18f
129)   D3 B2 L2 U3 L2 D1 B2 L2 U2 R2 D1 L2 F2 D1 R2 D2 R2 B2    18f
130)   D3 B2 L2 D3 B2 U1 F2 R2 D2 R2 U1 B2 L2 U1 L2 U2 R2 B2    18f
131)   D3 B2 L2 D3 B2 D1 R2 B2 D2 B2 U1 L2 F2 D1 R2 D2 R2 B2    18f
132)   D3 B2 R2 U3 F2 D3 R2 U1 L2 F2 L2 B2 U1 R2 D1 L2 U3 B2    18f
133) inverse of 132
134)   D3 B2 L2 D3 F2 D3 R2 B2 U2 L2 B2 U1 R2 D1 B2 L2 U2 L2    18f
135)   D3 B2 R2 D3 L2 U3 B2 R2 D2 F2 R2 U1 R2 D1 R2 F2 U2 B2    18f
136)   D3 B2 L2 U3 R2 D3 B2 L2 U2 F2 L2 D1 L2 D1 F2 R2 D2 L2    18f
137)   D3 B2 L2 D3 F2 U3 B2 L2 U2 F2 L2 U1 B2 U1 B2 L2 U2 L2    18f
138)   D3 B2 L2 D3 F2 D3 R2 B2 D2 R2 F2 U1 L2 D1 F2 R2 D2 L2    18f

kloosterman's position is number 46.

mike

From dik@cwi.nl  Sat Jan  7 21:36:02 1995
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From: Dik.Winter@cwi.nl
Message-Id: <9501080235.AA01808=dik@boring.cwi.nl>
To: cube-lovers@ai.mit.edu, mreid@ptc.com
Subject: Re:  two stage filtration

 > i've run an exhaustive search on the coset space  G / H.  the number
 > of cosets at each distance is:

Ah, well.  To think that I had the program publicly available since
February 1993 through anonymous ftp, I ought to have thought about
it when we got a machine large enough to run the program; about a
year ago.  Damn ;-).

 >  distance    quarter turns       face turns

I am running and will confirm tomorrow; no doubt about that.

...
 > to give a total of  2187 * 2048 * 495 = 2217093120  configurations.
 > to reduce this number somewhat, we can utilize symmetry.  there are 16
 > symmetries of the cube that preserve the  U-D axis,  and therefore
 > preserve the subgroup  H.  up to these symmetries, the number of distinct
 > corner configurations is 168, so we need only consider a mere
 > 168 * 2048 * 495 = 170311680  configurations.

 > (so far, this is the same approach that dik used for his calculation.)

The approach is the same, but I did avoid the embarrasment you had later.
I came up with 324 * 2048 * 495 = 328458240 configurations.

 > each configuration is stored with 2 bits of memory and thus the whole
 > space consumes about 42 megabytes.  each configuration is assigned
 > one of 4 values:
 >         distance is currently unknown
 >         distance = current search depth
 >         distance = current search depth - 1
 >         distance < current search depth - 1
 > from here, i just used a simple breadth first search.

This is similar to what I did outline about that time.  It comes from
a remark somebody not on this list (Arjeh Cohen) made to me about a
file helping solving the cube.  You store only the distance mod 3;
that will give you a simple database to solve it.  That again came
from a talk at some congress I do not remember at this time of the
night ;-).

...
 >       i) only use the 8 symmetries that preserve my choice of
 >          12 edge facelets.

I did this indeed.

 > run times were improved significantly by using a simple trick that i hadn't
 > used in earlier programs.  during the first few depth levels, i use
 > "forward searching", i.e. i examine the neighbors of each configuration
 > found at the previous depth.  however, after at least half the search space
 > has been found, i switch to "backward searching", i.e. examine the
 > configurations (and their neighbors) that haven't yet been found.

 > (have others been using this same idea when running similar search programs?)

 > closer analysis of this technique suggests that the switch from forward to
 > backward searching should occur even before half the space has been found.

Here I am a bit surprised.  I would think the time needed for a phase is
entirely dependend on the number of neighbo(u)rs you have to examine.  This
appears to be 6 times the number of configurations you visit.  So I would
think that going the other way pays when the number of configurations not
yet decided is less than the number of configurations found in the previous
step.  And no, I did not implement this; although it looks simple indeed.

Phase 2 for me needs a bit of consideration as obviously you reduced the
number of cases a bit more than I did when I wrote my program.  (Mine
still does not fit in 256 Mb.)

More tomorrow.

dik
--
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland, +31205924098
home: bovenover 215, 1025 jn  amsterdam, nederland; e-mail: dik@cwi.nl

From dik@cwi.nl  Sun Jan  8 05:56:01 1995
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From: Dik.Winter@cwi.nl
Message-Id: <9501081055.AA02279=dik@boring.cwi.nl>
To: cube-lovers@ai.mit.edu, mreid@ptc.com
Subject: Re:  two stage filtration

Mike Reid:
 > i've run an exhaustive search on the coset space  G / H.  the number
 > of cosets at each distance is:

I can confirm Mike's results on phase 1.  Here follows my table which
also contains the number of local maxima (which you will not find in
"backward" steps):

turns		 q	   loc.max	       q+h	   loc.max
    0	         1			         1
    1	         4			         4
    2	        34			        50
    3	       312			       592
    4	      2772			      7156
    5	     24996			     87236	         4
    6	    225949	         5	   1043817	        97
    7	   2017078	        32	  12070278	      2800
    8	  17554890	       730	 124946368	    110582
    9	 139132730	     39000	 821605960	  16713104
   10	 758147361	  10861351	1199128738	 750219596
   11	1182378518	 608836624	  58202444	  58196874
   12	 117594403	 117439129	       476	       476
   13	     14072	     14072

 > 40 minutes for quarter turns, 47 minutes for face turns.  this was done
 > on a DEC 3000 alpha 700, apparently a very fast machine.

I got 131 minutes for quarter turns and 186 minutes for face turns on a
measly SGI Challenge, apparently not so very fast.  (I presume it would
have been faster if it had been possible to run with 64 bit long's.)

I will try to verify phase 2 later next week.

dik

From mschoene@math.rwth-aachen.de  Sun Jan  8 07:50:49 1995
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From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: BRYAN@wvnvm.wvnet.edu
Cc: cube-lovers@life.ai.mit.edu
In-Reply-To: "Jerry Bryan"'s message of Sat, 7 Jan 1995 10:15:52 EST <9501071609.AA16758@life.ai.mit.edu>
Subject: Re: Re: Cube with GAP

Jerry Bryan wrote in his e-mail message of 1995/01/07 

    Note that this model does not include the face centers.  That is, it
    is G[C,E] rather than G[C,E,F].  56 numbers would be required to
    include the face centers.  The distinction between 48 facelets and
    56 facelets bears on the nitpicky question of whether the set C of
    rotations is a subgroup of G or not.

Absolutely right.  This part of the GAP documentation was written years
ago.  These days I represent MG, CG, etc. as permutation groups on 54
points.  I also changed the numbering, so that the [1..24] represent the
edges, [25..48] points represent the edges, and [49..54] represent the
centers.

Jerry continued

                                      What I don't see is how to
    model the Supergroup in GAP.  It looks like you would have to
    label each Face center with four numbers so you could see the
    rotations of the Face centers, but that seems like overkill.

This is also correct.  But GAP doesn't mind those 24 more points.

Jerry continued

    When I write the model out to disk, I only write out 8 corner facelets
    and 12 edge facelets.  For example, I only write out the front and
    back corner facelets.  This saves space and converts the model from
    a facelet model to a cubie model, with the twists implicitly encoded
    rather than being explicitly encoded via multiplication tables.  It
    also automatically establishes a frame of reference by which a
    proof of conservation of twist and flip can be accomplished.

In terms of computational group theory this sequence of 8 corner and
12 edgde facelets is called a *base* for the permutation group G.
That is, each element of the group is uniquely determined by the
images of those 20 facelets.  Of course if you have already proved
that no single corner can be twisted and no single edge can be flipped,
you can reduce this to 7 corner and 11 edge facelets.

Mark Longridge wrote in his e-mail message of 1995/01/03

                            ...  I don't know how a normal 4x4x4 could
    be represented though.

Jerry answered

    I fail to see the problem.  Just number the facelets.  The only
    problem would then lie in deciding what the generators are  -- i.e.,
    which kind of slice moves do you accept.  You would also have to
    decide whether to model the invisible 2x2x2 inside, but again if you
    did, just number the invisible facelets and include their movements
    with your generators.

The problem is that many different positions all look solved.  For
example, you can permute the 4 center facelets of one face or exchange
two adjacent edges, and the cube still looks solved (of course you cannot
do all this independently).  So if we take the obvious permutation group
on the 6*16 points, then a whole subgroup would correspond to what a
puzzler would consider solved states.  If by a model we mean a group
whose elements correspond to the different states a puzzler would see,
and whose identity corresponds to what a puzzler would consider solved,
then I have no good idea how to model the 4x4x4 cube as a permutation
group.

Have a nice day.

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From mreid@ptc.com  Sun Jan  8 16:14:33 1995
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From: mreid@ptc.com (michael reid)
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To: dik@cwi.nl, cube-lovers@ai.mit.edu
Subject: Re:  two stage filtration
Content-Length: 1392

dik writes:

>  > run times were improved significantly by using a simple trick that i hadn't
>  > used in earlier programs.  during the first few depth levels, i use
>  > "forward searching", i.e. i examine the neighbors of each configuration
>  > found at the previous depth.  however, after at least half the search space
>  > has been found, i switch to "backward searching", i.e. examine the
>  > configurations (and their neighbors) that haven't yet been found.
> 
>  > (have others been using this same idea when running similar search programs?)
> 
>  > closer analysis of this technique suggests that the switch from forward to
>  > backward searching should occur even before half the space has been found.
> 
> Here I am a bit surprised.  I would think the time needed for a phase is
> entirely dependend on the number of neighbo(u)rs you have to examine.  This
> appears to be 6 times the number of configurations you visit.  So I would
> think that going the other way pays when the number of configurations not
> yet decided is less than the number of configurations found in the previous
> step.

except that when searching backward, you need not visit all the neighbors
of a configuration.  you only need to find one neighbor at the previous
distance; after that, the other neighbors don't need to be examined.

i did not realize this until implementing this technique.

mike

From mreid@ptc.com  Sun Jan  8 16:28:22 1995
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From: mreid@ptc.com (michael reid)
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Subject: Re:  two stage filtration
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dik writes

> I can confirm Mike's results on phase 1.

great!

>                                           Here follows my table which
> also contains the number of local maxima (which you will not find in
> "backward" steps):

this is true.  i decided i was more interested in performance than
in knowing about local maxima.

> I will try to verify phase 2 later next week.

let me offer a suggestion here.  since i divided the corner configurations
by symmetry, it might be nicer if you divide the U-D edge configurations
by symmetry.  (the numbers involved are the same.)

it's always nice to confirm a calculation like this using a different method.
although what i'm suggesting isn't much of a change.

mike

From dik@cwi.nl  Sun Jan  8 19:08:05 1995
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From: Dik.Winter@cwi.nl
Message-Id: <9501090007.AA02883=dik@boring.cwi.nl>
To: cube-lovers@ai.mit.edu, mreid@ptc.com
Subject: Re:  two stage filtration

[ Mike: You need not visit all neighbors when going backwards. ]

Right you are; I missed that completely.

dik

From dik@cwi.nl  Sun Jan  8 19:12:24 1995
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From: Dik.Winter@cwi.nl
Message-Id: <9501090012.AA02894=dik@boring.cwi.nl>
To: cube-lovers@ai.mit.edu, mreid@ptc.com
Subject: Re:  two stage filtration

 > this is true.  i decided i was more interested in performance than
 > in knowing about local maxima.

 > > I will try to verify phase 2 later next week.

 > let me offer a suggestion here.  since i divided the corner configurations
 > by symmetry, it might be nicer if you divide the U-D edge configurations
 > by symmetry.  (the numbers involved are the same.)

I will see how I divide it now, but I can change easily.  My biggest
change is to weed out parity.  I just realized that, but that is the
reason it will not fit now.  Moreover, I will go completely forward
to get also local maxima.

I do not know yet what to do about the quarter turn version.  I never
put that in my program for this phase.  I had some form of quarter
turn version, but only for U and D.

I will see, dik

From BRYAN@wvnvm.wvnet.edu  Sun Jan  8 23:24:08 1995
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Date:      Sun, 8 Jan 1995 23:16:52 EST
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Re: kociemba's algorithm for quarter turns
In-Reply-To: Message of 01/05/95 at 17:12:18 from mreid@ptc.com

On 01/05/95 at 17:12:18 mreid@ptc.com said:

>for much too long now, i've meant to implement kociemba's algorithm
>for quarter turns.  finally i've gotten around to it, and it's found

>superflip:
>   B3 L3 U3 L3 F1 U1 D1 L3 B1 U1 F1 R3 L1 F3 B2 U1 D1 F2 B2 R2 U1 D1   26q

I read the articles in the archives about Kociemba's algorithm about
a year ago, without (I confess) fully understanding them.  In particular,
I do not fully understand what differentiates Kociemba's algorithm from
Thistlethwaite's algorithm, other than it uses a different arrangement
of nested subgroups.  I shall strive to read the articles again with
a deeper level of understanding.

But in the meantime, I wonder if you could verify that Kociemba's
algorithm does not guarantee to find a minimal process?  In particular,
is it the case that 26q is a minimal superflip, or is it only an
upper bound?

The reason I ask is that I have decided to go ahead and calculate God's
Algorithm under quarter turns for depth 11.  (Through depth 10 is
already in hand.)  Once that is accomplished, it should be a
*fairly* easy task to establish a lower bound on the superflip
at 22 quarter turns via two half depth searches.  In fact, the
second half depth search should be fairly easy to accomplish
because all I have to do is superflip each element of the data base
from the first search to establish the data base for the second
search.

I can already establish a lower bound of 14 quarter turns on the
superflip.  It may be recalled that I was able to accomplish a
complete search for edges-only (no corners, no Face centers, and
rotations considered equivalent).  There was some consternation
when I reported that the superflip was 9 quarter turns from Start
because the superflip is even.  But without Face centers and with
rotations considered equivalent, normal parity rules do not
apply.

I am now working on edges-only, either with centers, or else with
rotations *not* considered equivalent (either G[E,F] or G[E]),
depending on which way you want to think about it.
In this case, the superflip really is even.  I am working on level
13, and the superflip has not yet appeared.  Hence, it is at least
at level 14 (without corners), and will therefore be at least at level
14 when the corners are added in.  Strictly speaking, the superflip
has appeared already, and at level 9 just where it had to appear.
But in its appearance at level 9, it is composed with a non-trivial
rotation, so it isn't the superflip as the superflip is normally
understood.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From dik@cwi.nl  Mon Jan  9 17:25:44 1995
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From: Dik.Winter@cwi.nl
Message-Id: <9501092224.AA05290=dik@boring.cwi.nl>
To: BRYAN@wvnvm.wvnet.edu, Cube-Lovers@ai.mit.edu
Subject: Re: kociemba's algorithm for quarter turns

 > I read the articles in the archives about Kociemba's algorithm about
 > a year ago, without (I confess) fully understanding them.  In particular,
 > I do not fully understand what differentiates Kociemba's algorithm from
 > Thistlethwaite's algorithm, other than it uses a different arrangement
 > of nested subgroups.

The basis is similar (although Kociemba's algorithm uses searching to
get solutions while Thistlethwaite's uses tables and directly arrives
at solutions).  The main difference is that once a solution is found
Thistlethwaite's algorithm stops.  Kociemba's algorithm continues finding
newer solutions (even longer than the original solution) to phase 1 and
trying to fit them with a solution for phase 2 such that the total solution
is shorter.  This proves to be very effective.  Of course this is easier to
do with a 2 phase algorithm than with a 4 phase algorithm.

 > But in the meantime, I wonder if you could verify that Kociemba's
 > algorithm does not guarantee to find a minimal process?  In particular,
 > is it the case that 26q is a minimal superflip, or is it only an
 > upper bound?

Given time Kociemba's algorithm will find a minimal solution.  I confess
that my implementations does not if the configuration can be solved
through phase 2 only, but the cube can be rotated to avoid that.  The
reason is that ultimately Kociemba's algorithm will find longer part
solutions of phase 1 and ultimately stumble on a complete solution in
phase 1 which will be minimal because of the breadth first search.

But it can take long.  Getting a minimal solution if the length is 16
or less appears to be doable.  If the length is 19 or more it takes an
awfully long time.  What I have found until now is:
1.  There are no configurations known that require 21 turns or more,
    and I checked an awfully large number.
2.  There are known configurations that require 18 turns.
The middle part is a grey area.

From mreid@ptc.com  Mon Jan  9 18:04:52 1995
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Date: Mon, 9 Jan 1995 18:15:55 -0500
From: mreid@ptc.com (michael reid)
Message-Id: <9501092315.AA09092@ducie.ptc.com>
To: cube-lovers@ai.mit.edu, bryan@wvnvm.wvnet.edu
Subject:   Re: kociemba's algorithm for quarter turns
Content-Length: 3238

jerry writes

> I read the articles in the archives about Kociemba's algorithm about
> a year ago, without (I confess) fully understanding them.  In particular,
> I do not fully understand what differentiates Kociemba's algorithm from
> Thistlethwaite's algorithm, other than it uses a different arrangement
> of nested subgroups.

thistlethwaite's algorithm is a method which guarantees solving any cube
in at most 52 (well, now it's 44) face turns.  i don't think it was ever
really used to find short solutions (although at the time it was invented,
52 face turns may have been considered short).

kociemba's algorithm is a method for finding short solutions.  it didn't
come with any guarantees, (although i've just shown that the first solution
it finds is at most 43 quarter [30 face] turns, and in these extreme cases,
it will quickly find a shorter solution.)  kociemba gave an effective way
to navigate through the sequence of subgroups

     G = <U, D, F, R, B, L>,   H = <U, D, F2, R2, B2, L2>,   1 = <>,

without using enormous tables.  (this is how it differs from thistlethwaite's
method, and also why there are (well, were) no guarantees.)

kociemba also allows non-optimal sequences in stage 1 in exchange for
shorter sequences in stage 2.  we start by finding all length  n  sequences
in stage 1, and for each, the shortest sequence in stage 2.  then we move
on to length  n + 1  in stage 1.  kociemba's method is so effective, that
searching through length 14 or 15 in stage 1 is usually quite feasible.
also, this technique has been so successful that it's improved many of the
shortest known maneuvers.

> But in the meantime, I wonder if you could verify that Kociemba's
> algorithm does not guarantee to find a minimal process?  In particular,
> is it the case that 26q is a minimal superflip, or is it only an
> upper bound?

26q  is only an upper bound.  my program will eventually find the shortest
process, well, if my os doesn't crash first, the universe doesn't end, ...
but i've only given the shortest maneuver it's found so far.  at some
point you've gotta give up.  (there are plenty of other patterns waiting
to be stuffed into this program.  :-) )

> The reason I ask is that I have decided to go ahead and calculate God's
> Algorithm under quarter turns for depth 11.  (Through depth 10 is
> already in hand.)  Once that is accomplished, it should be a
> *fairly* easy task to establish a lower bound on the superflip
> at 22 quarter turns via two half depth searches.

you should search with the list you have right now.  (i presume you're
talking about a list of all positions within  10q  of  START.)

you will either find a maneuver of length  <= 20q  (unlikely, i'd say),
or you will show that its distance from START is  >= 22q.  this latter
possibility would be very exciting, since it would raise the lower
bound on the diameter of  G.

> I can already establish a lower bound of 14 quarter turns on the
> superflip.

my program can do this in only a few seconds!  in fact, it takes  12q
to get from superflip into the subgroup  H.  since we may also suppose
that the last twist in a shortest maneuver is  U , it follows that
superflip requires at least  13q  (and thus  14q , by parity).

mike

From TanisElf@aol.com  Mon Jan  9 18:35:45 1995
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Date: Mon, 9 Jan 1995 18:36:41 -0500
From: TanisElf@aol.com
Message-Id: <950109183639_3748705@aol.com>
To: cube-lovers@life.ai.mit.edu
Subject: remove from list

Please remove me from this mailing list
     TanisElf@aol.com




Tanis

From alan@curry.epilogue.com  Mon Jan  9 18:49:15 1995
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Date: Mon, 9 Jan 1995 18:50:54 -0500
Message-Id: <9Jan1995.184223.Alan@LCS.MIT.EDU>
From: Alan Bawden <Cube-Lovers-Request@ai.mit.edu>
Sender: Cube-Lovers-Request@ai.mit.edu
To: TanisElf@aol.com
Cc: cube-lovers@ai.mit.edu
In-Reply-To: TanisElf@aol.com's message of Mon, 9 Jan 1995 18:36:41 -0500 <950109183639_3748705@aol.com>
Subject: remove from list

   Date: Mon, 9 Jan 1995 18:36:41 -0500
   From: TanisElf@aol.com
   To: cube-lovers@life.ai.mit.edu
       ^^^^^^^^^^^
   Subject: remove from list

   Please remove me from this mailing list
	TanisElf@aol.com

   Tanis

YOU ARE AN IDIOT!  WHY DID YOU ASK THE -ENTIRE- MAILING LIST?

I told you quite clearly when I added you that you get off the list by
sending mail to Cube-Lovers-REQUEST@AI.MIT.EDU.

And I am purposely CC'ing this message to the entire Cube-Lovers mailing
list in order to remind everyone else about Cube-Lovers-REQUEST, because
invariably when one idiot makes this mistake, a whole pack of other
lemming-idiots follows the first idiot and makes the same mistake.

From ncramer@bbn.com  Tue Jan 10 09:54:58 1995
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Message-Id: <9501101454.AA28269@life.ai.mit.edu>
Date:     Tue, 10 Jan 95 9:47:23 EST
From: Nichael Cramer <ncramer@bbn.com>
To: cube-lovers@life.ai.mit.edu
Subject:  Source for Cheap 5X Cubes

...and now that Alan has _that_ out of his system...     ;-)

For the interested:

A topic that has come up a few times (and of which I was in hot pursuit for
a couple of years):  I just got a new Ishi Press flier in the mail
yesterday.  Among other toys they were selling 5X5X5 cubes for $20 (+S&H).

There order number is (800) 859-2086.

(Include standard disclaimers re non-connectedness, etc.)

Enjoy
Nichael

ncramer@bbn.com
Paradise Farm
Brattleboro VT


From news@nntp-server.caltech.edu  Tue Jan 10 12:45:58 1995
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Path: nntp-server.caltech.edu!txr
From: txr@alumni.caltech.edu (Tim Rentsch)
Newsgroups: mlist.cube-lovers
Subject: Re: kociemba's algorithm for quarter turns
Date: 10 Jan 1995 17:45:35 GMT
Organization: California Institute of Technology CCO Unix cluster
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References: <9501092224.AA05290=dik@boring.cwi.nl>
Nntp-Posting-Host: alumni.caltech.edu
In-Reply-To: Dik.Winter@cwi.nl's message of Mon, 09 Jan 95 23:03:57 GMT

Dik.Winter@cwi.nl writes:

>But it can take long.  Getting a minimal solution if the length is 16
>or less appears to be doable.  If the length is 19 or more it takes an
>awfully long time.  What I have found until now is:
>1.  There are no configurations known that require 21 turns or more,
>    and I checked an awfully large number.
>2.  There are known configurations that require 18 turns.
>The middle part is a grey area.

How hard would it be to write a program to try the following?

1) Initialize set S with a configuration that requires a large number
   of turns (the max, perhaps).

2) Test the length of all configurations one turn from any
   configuration in S.

3) If one or more of these has a minimum length longer than the
   initial position, replace the set S with the set of those
   configuration with longer length (of course print out useful 
   intermediate result).

4) If none of the test positions has a minimum length longer
   than the length of positions in S, replace the set of test positions
   with positions one more turn away, and test again until 3 works.
   (Obviously it would be useful to store something about previous
   results so that work is not redone needlessly.  I think it's
   easy to figure out what information should be stored, although
   I haven't done so.)

5) Give up when patience is exhausted.

It would be nice to get a higher lower bound, and this seems
like a plausible way of doing so.

regards,

Tim Rentsch


From ishius@ishius.com  Tue Jan 10 13:14:59 1995
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To: cube-lovers@life.ai.mit.edu
From: ishius@ishius.com (ishius@holonet.net)
Subject: Re: Source for Cheap 5X Cubes

>For the interested:
>
>A topic that has come up a few times (and of which I was in hot pursuit for
>a couple of years):  I just got a new Ishi Press flier in the mail
>yesterday.  Among other toys they were selling 5X5X5 cubes for $20 (+S&H).
>
>There order number is (800) 859-2086.

Yes, these are on sale for $20 plus S/H.  However, I must say that as they
are difficult to manufacture, the mechanism is not as smooth as I would
prefer, the center cube faces sometimes need to be reglued, and the stickers
sometimes slide.  On the other hand, these are the only 5x5x5 cubes you can
get (to my knowledge), and they are only $20.00.  Also, if you buy one, you
can buy a Toyo Glass puzzle for $8.70 (they usually go for $25 to $40).

I KNOW you guys would like a 4x4x4 cube, but I can't find them.  We do have
another rotational puzzle called the SKEWB, also for $20.00, which is a cube
with 4 diagonal cuts that pass through all six faces of the cube (so that any
move changes all six faces).  The mechanism is very smooth and very well made,
and I've NEVER had a defective one yet.  If you're familiar with the SKEWB,
I would like to know whether it's harder or easier than the classic 3x3x3
Rubik's Cube (I suspect it's simpler, but it has fewer symmetries).

If you would like a full color catalog of our puzzles, including the latest
offerings, please send me your POSTAL mailing address.


Always feel free to write me if you have any questions or comments.

Anton Dovydaitis
Customer Support
===========================================================================
Ishi Press International                408/944-9900 vc, 408/944--9110 FAX
76 Bonaventura Drive                    800/859-2086 Toll Free Order Line
San Jose, CA  95134                     ishius@ishius.com (or @holonet.net)



From dlitwin@geoworks.com  Tue Jan 10 14:26:42 1995
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Date: Tue, 10 Jan 95 11:23:44 PST
From: dlitwin@geoworks.com (David Litwin)
Message-Id: <9501101923.AA14522@geoworks.com>
To: ishius@ishius.com (ishius@holonet.net)
Cc: cube-lovers@life.ai.mit.edu
In-Reply-To: <199501101808.KAA07315@holonet.net>
Subject: Re: Source for Cheap 5X Cubes


ishius@holonet.net writes:
> Yes, these are on sale for $20 plus S/H.  However, I must say that as they
> are difficult to manufacture, the mechanism is not as smooth as I would
> prefer, the center cube faces sometimes need to be reglued, and the stickers
> sometimes slide.

	With regard to the stickers (as far as I can tell just the orange
side has problems) I had good luck following the suggetions that came with
mine (written up by Christoph Bandelow).  They recommend placing a piece of
paper over the cube and ironing (not too hot).  This melts the glue and the
stickers get a better grip.  I've not had any problems since.

	Dave Litwin

From whuang@cco.caltech.edu  Tue Jan 10 16:44:58 1995
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From: whuang@cco.caltech.edu (Wei-Hwa Huang)
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Subject: Re: SKEWB
To: cube-lovers@life.ai.mit.edu
Date: Tue, 10 Jan 1995 13:44:47 -0800 (PST)
In-Reply-To: <199501101808.KAA07315@holonet.net> from "ishius@holonet.net" at Jan 10, 95 10:08:07 am
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> 
> If you're familiar with the SKEWB,
> I would like to know whether it's harder or easier than the classic 3x3x3
> Rubik's Cube (I suspect it's simpler, but it has fewer symmetries).

It took me a while to give out a working algorithm from it.  Surprizingly
my moves have very little use on Rubik's, and if the SKEWB wasn't as
simple as it was it would be really tough to work out.  Fortunately,
due to its simplicity I need to remember only one code sequence.
(Solution posted on request.)


From brett@math.toronto.edu  Tue Jan 10 16:48:40 1995
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Message-Id: <9501102148.AA23054@life.ai.mit.edu>
Subject: seminar I have to give
To: cube-lovers@ai.mit.edu (cube)
Date: Tue, 10 Jan 1995 16:48:47 -0500 (EST)
From: "Brett Stevens" <brett@math.toronto.edu>
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I have to give a small seminar in two weeks and thought I 
might do it on something cube related.  Can anyone give
me suggestions on topics, please

thanks brett


From whuang@cco.caltech.edu  Tue Jan 10 16:48:49 1995
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Subject: SKEWB
To: cube-lovers@life.ai.mit.edu
Date: Tue, 10 Jan 1995 13:48:41 -0800 (PST)
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> 
> > ishius@holonet.net writes:
> > > Yes, these are on sale for $20 plus S/H.  However, I must say that as they
> > > are difficult to manufacture, the mechanism is not as smooth as I would
> > > prefer, the center cube faces sometimes need to be reglued, and the stickers
> > > sometimes slide.
> > 
> > 	With regard to the stickers (as far as I can tell just the orange
> > side has problems) I had good luck following the suggetions that came with
> > mine (written up by Christoph Bandelow).  They recommend placing a piece of
> > paper over the cube and ironing (not too hot).  This melts the glue and the
> > stickers get a better grip.  I've not had any problems since.
> > 
> > 	Dave Litwin
> > 
> I believe the poster is referring to the fact that the plastic faces fall
> off, revealing a screw inside, not the stickers (which are well done).
> Super-glue solved my problem easily.  One more thing:  the mechanism gets
> smoother after repeated use.
> 
> 
> 


From brett@math.toronto.edu  Tue Jan 10 17:01:33 1995
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Message-Id: <9501102201.AA23595@life.ai.mit.edu>
Subject: solid brass cube
To: cube-lovers@ai.mit.edu (cube)
Date: Tue, 10 Jan 1995 17:01:42 -0500 (EST)
From: "Brett Stevens" <brett@math.toronto.edu>
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My friend gave me a solid brass cube for the holidays.  he found it in 
a small antiques store in L.A. and thought I would like it.  Does
anyone know the origin of this novelty?

brett
brett@math.toronto.edu


From dlitwin@geoworks.com  Tue Jan 10 17:18:09 1995
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Date: Tue, 10 Jan 95 14:13:01 PST
From: dlitwin@geoworks.com (David Litwin)
Message-Id: <9501102213.AA18001@geoworks.com>
To: whuang@cco.caltech.edu (Wei-Hwa Huang)
Cc: cube-lovers@life.ai.mit.edu
In-Reply-To: <199501102148.NAA24322@accord.cco.caltech.edu>
Subject: SKEWB


Wei-Hwa Huang writes:
> Dave Litwin writes:
> > ishius@holonet.net writes:
> > > ...
> > > I would prefer, the center cube faces sometimes need to be reglued,
> > > and the stickers sometimes slide.
> > 
> > 	With regard to the stickers (as far as I can tell just the orange
> > ...
> > 
> > 	Dave Litwin
>
> I believe the poster is referring to the fact that the plastic faces fall
> off, revealing a screw inside, not the stickers (which are well done).
> Super-glue solved my problem easily.  One more thing:  the mechanism gets
> smoother after repeated use.

	If you notice the last line of the ishius@holonet.net post, they
also mention the stickers sliding, which is what I was addressing.  I've
only noticed 1 or 2 out of the at least 30 I've seen that *don't* have a
problem with the orange stickers: sliding off, dog earing etc.

	The problem of the center caps coming off is separate, and much
easier to fix.  Actually, before you glue them on you may want to adjust
the tensions of the screws on all the axes (only requiring one cap per
axis, not all six).  The few I've done this to move much smoother when
loosened (just a bit, as being too loose can cause problems of its own).

	Dave Litwin

From mouse@collatz.mcrcim.mcgill.edu  Tue Jan 10 17:20:31 1995
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Date: Tue, 10 Jan 1995 17:20:19 -0500
From: der Mouse <mouse@collatz.mcrcim.mcgill.edu>
Message-Id: <199501102220.RAA10414@Collatz.McRCIM.McGill.EDU>
To: dlitwin@geoworks.com, ishius@ishius.com
Subject: Re: Source for Cheap 5X Cubes
Cc: cube-lovers@life.ai.mit.edu

>> Yes, these [5x5x5 Cubes] are on sale for $20 plus S/H.  However, I
>> must say that as they are difficult to manufacture, the mechanism is
>> not as smooth as I would prefer, the center cube faces sometimes
>> need to be reglued, and the stickers sometimes slide.

> With regard to the stickers (as far as I can tell just the orange
> side has problems) [...] recommend placing a piece of paper over the
> cube and ironing (not too hot).  This melts the glue and the stickers
> get a better grip.  I've not had any problems since.

My experience matches yours, in that only the orange side has trouble.
I found that a tiny little dab of plain ordinary contact cement would
nail down a sticker very nicely.  Four of the 25 orange-side stickers
on my 5x5x5 (which I understand the local shop got from Ishi Press, btw
Anton :-) show evidence of my having used a little too much contact
cement; I don't recall how many others got glued with a more
appropriate amount.

As for 4x4x4s, I own one but have been unable to find it lately.  This
doesn't bother me especially, since given a 5x5x5, if you just ignore
the center slice along each axis (remove the stickers, paint white-out
over them, etc) and you have something mathematically identical to a
4x4x4.

					der Mouse

			    mouse@collatz.mcrcim.mcgill.edu

From mouse@collatz.mcrcim.mcgill.edu  Tue Jan 10 18:14:19 1995
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Date: Tue, 10 Jan 1995 18:14:15 -0500
From: der Mouse <mouse@collatz.mcrcim.mcgill.edu>
Message-Id: <199501102314.SAA10516@Collatz.McRCIM.McGill.EDU>
To: cube-lovers@life.ai.mit.edu
Subject: Difficulty of Skewb
Cc: ishius@ishius.com

> If you're familiar with the SKEWB, I would like to know whether it's
> harder or easier than the classic 3x3x3 Rubik's Cube (I suspect it's
> simpler, but it has fewer symmetries).

I think I have a skewb.  Each face is cut into a square turned 45
degrees, and four little 45-45-90 triangles, right?  And there are four
cuts you can turn it about, all passing through the center of the cube?

In that case, here's my opinion.  Mathematically, it verges on trivial
compared to the 3x3x3, but because it's deep-cut, there are no stable
portions to take the psychological place of the face centers on the
3x3x3, and each turn affects all six faces so, unlike the 2x2x2, there
are no unaffected faces to pick up the slack.  These combine to make it
difficult for psychological reasons.

I called it verging on trivial, but that is relative to the 3x3x3.
There is some structure.  If the 3x3x3 has 8!*3^8*12!*2^12 "imaginable"
configurations (of which only 1/12 are reachable from any given state),
the skewb has only 8!*3^8*6! "imaginable".  Of these, far less than
1/12 are reachable; in my estimate, not having thought about it hard,
the reachable number is (4!*4!)*(3^8/3)*(6!/2), or only one in 420 of
the "imaginable" configurations, a reduction achieved largely due to
the constraints on corner mixing: they form two tetrahedral sets, and
it is not possible to mix the sets.  Thus 8! turns into 4!*4!; this is
a factor of 70.  The additional factor of 6 comes from the usual sort
of parity constraints, and again is only my guess.

The size of the thing is thus somewhere on the order of 500 million
configurations.  This is why I called it trivial next to the 3x3x3. :-)
The group structure in terms of facicles, for what's-his-name to sic
GAP on should he care to, derived from the following facicle labeling

           +----------+
           |  6     7 |
           |     8    |
           |  9    10 |
+----------+----------+----------+----------+
|  1     2 | 11    12 | 21    22 | 31    32 |
|     3    |    13    |    23    |    33    |
|  4     5 | 14    15 | 24    25 | 34    35 |
+----------+----------+----------+----------+
           | 16    17 |
           |    18    |
           | 19    20 |
           +----------+

is:

	Cut 1: (15  7  4) (24 31 19) (17 22 35) (20 25 34) (18 23 33)
	Cut 2: (12  1 20) (10 32 25) (21  6 34) (22  7 31) (23  8 33)
	Cut 3: (11 19 22) ( 2 35  7) ( 9  4 31) ( 6  1 32) ( 8  3 33)
	Cut 4: (14 25  6) (16 34  1) ( 5 20 32) ( 4 19 35) ( 3 18 33)

(This formulation holds facicle 13 fixed at all times.)

					der Mouse

			    mouse@collatz.mcrcim.mcgill.edu

From dik@cwi.nl  Tue Jan 10 18:20:52 1995
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Date: Wed, 11 Jan 1995 00:09:40 +0100
From: Dik.Winter@cwi.nl
Message-Id: <9501102309.AA08275=dik@boring.cwi.nl>
To: cube-lovers@life.ai.mit.edu, txr@alumni.caltech.edu
Subject: Re: kociemba's algorithm for quarter turns

 > Dik.Winter@cwi.nl writes:
 > >But it can take long.

Note the word *long*.  There is a 20-turn sequence for superflip.
I have tried for shorter sequences using Kociemba's algorithm.
(The 20-turn sequence does not take so awfully long to find,
something like a day on this machine * perhaps).  I started
the program 12 November 1993 at 12:03.  At 11 December 11:58
it had searched phase1 up to length 16.  Alas, the machine
crashed on 16 December 22:14.  The problem is the awfully
large number of solutions that come out of phase 1 and that
have to be checked &.  To wit:
Length phase 1	    number	complete solutions
	 8	         0
	 9	         0
	10	      3072	L=23, L=22
	11	     61568	L=21
	12	    792256
	13	   8695488	L=20
	14	  87912832
	15	 841171136
	16	7765525280 %
The following solutions were successively found:
L=23: F1 B1 R1 F2 R2 U3 D1 F1 R1 L1 B2 D1 F2 B2 D3 B2 D1 B2 U2 R2 B2 D1 B2
L=22: F1 B1 R1 F2 R2 U3 D1 F1 R3 L1 U1 R2 F2 D3 B2 D3 R2 L2 U3 F2 D2 L2
L=21: F1 B1 R1 U2 B2 U3 D3 R2 B3 R1 L1 U1 F2 L2 D2 B2 D3 F2 D1 L2 D1
L=20: F1 B1 U2 R1 F2 R2 B2 U3 D1 F1 U2 R3 L3 U1 B2 D1 R2 U1 B2 U1

So the next step is 17 in phase 1 with at most 2 turns in phase 2.  I will
start the program again sometime.

So we find more than a month for a single configuration.  And for this
we need not check neighbors as the configuration is at a local maximum.
I have another file with longuish configuration.  That is from a period
when I tried random configurations, the results were:
turns	#conf
  16	    1
  17	   24
  18	  248
  19	 1429
  20	 8481
total	10183
This has also taken an awfully long time to do (I think it was about 2
months).  I let the program stop as soon as it had found a solution of
20 turns or less.  All random configurations were solved, but many of
the 20 turn configurations have shorter solutions.

So yes, it can be done, but:
 > 5) Give up when patience is exhausted.
this will come up before anything useful can be concluded I think.

Unless something can be done to reduce the numbers.  It is possible
because there are configurations that will come up many times during
the process, but I do not yet know what to do about that within a
reasonable amount of memory.
--
*  The machine is one processor of a Cray SMP: a 66 MHz Sparc.
&  The number is much larger than what I found with other configurations,
   for 15 turns about 800 times as large as the second largest.
%  Estimated, there was overflow.  It can be off an integer
   multiple of 4294967296.

From mreid@ptc.com  Tue Jan 10 18:46:05 1995
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From: mreid@ptc.com (michael reid)
Message-Id: <9501102357.AA18972@ducie.ptc.com>
To: cube-lovers@ai.mit.edu
Subject: superflip
Content-Length: 131

this just appeared today, after a lot of searching:

R3 U2 B1 L3 F1 U3 B1 D1 F1 U1 D3 L1 D2 F3 R1 B3 D1 F3 U3 B3 U1 D3   24q

mike

From diamond@jrdv04.enet.dec-j.co.jp  Tue Jan 10 20:02:53 1995
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From: Norman Diamond  11-Jan-1995 1001 <diamond@jrdv04.enet.dec-j.co.jp>
To: cube-lovers@life.ai.mit.edu
Apparently-To: cube-lovers@life.ai.mit.edu
Subject: Re: Source for Cheap 5X Cubes
Mime-Version: 1.0
Content-Type: TEXT/PLAIN; charset=ISO-2022-JP

>I KNOW you guys would like a 4x4x4 cube,

Last catalog that I received from the maker of the 5x5x5 and Skewb also had
a 4x4x4 in it, though that was a few years ago.

Meanwhile, I think the 4x4x4 is still on store shelves in the country where
Ishi gets its other stuff, and its name, from.  But it costs around 3,000 yen.

-- Norman Diamond     diamond@jrdv04.enet.dec-j.co.jp
[Digital did not write this.]

From ishius@ishius.com  Tue Jan 10 21:41:17 1995
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Date: Tue, 10 Jan 1995 18:38:02 -0800
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To: cube-lovers@life.ai.mit.edu
From: ishius@ishius.com (ishius@holonet.net)
Subject: Re: 5x5x5 cube

>> I believe the poster is referring to the fact that the plastic faces fall
>> off, revealing a screw inside, not the stickers (which are well done).

Actually, I see both problems.  But then, since I'm customer support, I see
all the complaints.

>> Super-glue solved my problem easily.

Yes, but if WE do it, and spend labor on it, we have to increase the price
to compensate.  Originally we were selling these things for $30.00.  What
I'm saying is that these ARE a good value, but you should be careful.
Remember: no one is manufacturing these things anymore, nor is anyone
likely to in the future.  If you know you'll want one, I'd buy it sooner
rather than later, because there won't be no later.  We bought up the last
from the manufactuer, and there's fewer than 100 left.

>> One more thing:  the mechanism gets smoother after repeated use.

That's good to know.  I'll have to abuse mine some more.


Always feel free to write me if you have any questions or comments.

Anton Dovydaitis
Customer Support
===========================================================================
Ishi Press International                408/944-9900 vc, 408/944--9110 FAX
76 Bonaventura Drive                    800/859-2086 Toll Free Order Line
San Jose, CA  95134                     ishius@ishius.com (or @holonet.net)



From BRYAN@wvnvm.wvnet.edu  Tue Jan 10 23:43:25 1995
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Date:      Tue, 10 Jan 1995 23:13:02 EST
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@life.ai.mit.edu>
Subject:   Re: kociemba's algorithm for quarter turns
In-Reply-To: Message of 01/11/95 at 00:09:40 from Dik.Winter@cwi.nl

On 01/11/95 at 00:09:40 Dik.Winter@cwi.nl said:

>Note the word *long*.  There is a 20-turn sequence for superflip.

I think we have a winner.  I compared the data base for level 0
through 9 against the same data base superflipped, with no
matches.  That should mean that there is no 18-turn or less
sequence for superflip.

Adding in level 10 will take a bit longer.  Also, there
is some minor new programming involved in this particular effort.
When I do get level 10 added in, and do get a match, that should
be a complete validation of the new programming, and complete
confirmation that the 20-turn sequence is minimal.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From mreid@ptc.com  Wed Jan 11 09:56:20 1995
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Date: Wed, 11 Jan 1995 10:07:37 -0500
From: mreid@ptc.com (michael reid)
Message-Id: <9501111507.AA19537@ducie.ptc.com>
To: bryan@wvnvm.wvnet.edu, cube-lovers@ai.mit.edu
Subject:   Re: kociemba's algorithm for quarter turns
Content-Length: 570

jerry bryan writes

> On 01/11/95 at 00:09:40 Dik.Winter@cwi.nl said:
> 
> >Note the word *long*.  There is a 20-turn sequence for superflip.
> 
> I think we have a winner.  I compared the data base for level 0
> through 9 against the same data base superflipped, with no
> matches.  That should mean that there is no 18-turn or less
> sequence for superflip.

wait a moment.  dik is talking about face turns.  what are you
talking about?  face turns or quarter turns?

and if you're talking about face turns, there's no reason an odd
length maneuver can't exist.

mike

From BRYAN@wvnvm.wvnet.edu  Wed Jan 11 10:11:10 1995
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From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <cube-lovers@ai.mit.edu>
Subject:   Re: kociemba's algorithm for quarter turns
In-Reply-To: Message of 01/11/95 at 10:07:37 from mreid@ptc.com

On 01/11/95 at 10:07:37 mreid@ptc.com said:
>jerry bryan writes

>> On 01/11/95 at 00:09:40 Dik.Winter@cwi.nl said:
>>
>> >Note the word *long*.  There is a 20-turn sequence for superflip.
>>
>> I think we have a winner.  I compared the data base for level 0
>> through 9 against the same data base superflipped, with no
>> matches.  That should mean that there is no 18-turn or less
>> sequence for superflip.

>wait a moment.  dik is talking about face turns.  what are you
>talking about?  face turns or quarter turns?

>and if you're talking about face turns, there's no reason an odd
>length maneuver can't exist.

I am talking about quarter turns.  "Quarter turns" is in the
subject of Dik's note????  If Dik's figures are for face turns,
then Mike's 26q solution is still the best so far, and we have
a long way to go.

(Oops.  I just peeked at Dik's 20 move solution.  It is definitely
for face turns.  In fact, as a quarter turn solution, it is 28
quarter turns.)

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From dik@cwi.nl  Wed Jan 11 10:25:57 1995
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Date: Wed, 11 Jan 1995 16:25:45 +0100
From: Dik.Winter@cwi.nl
Message-Id: <9501111525.AA11031=dik@boring.cwi.nl>
To: BRYAN@wvnvm.wvnet.edu, cube-lovers@ai.mit.edu
Subject: Re: kociemba's algorithm for quarter turns

Yes, indeed my remark was about face turns (although the subject
mentioned quarter turns).  I ought to have modified the subject.

dik

From BRYAN@wvnvm.wvnet.edu  Wed Jan 11 13:47:48 1995
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From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <cube-lovers@ai.mit.edu>
Subject:   Re: two stage filtration
In-Reply-To: Message of 01/07/95 at 19:43:27 from mreid@ptc.com

On 01/07/95 at 19:43:27 mreid@ptc.com said:

>each configuration is stored with 2 bits of memory and thus the whole
>space consumes about 42 megabytes.  each configuration is assigned
>one of 4 values:

>        distance is currently unknown
>        distance = current search depth
>        distance = current search depth - 1
>        distance < current search depth - 1

This little table reminded me of something I had meant to say about
storing the whole cube space in about 10^18 cells.  In principle,
under the Q turn metric it would be possible for each cell to contain
only one bit by storing (depth mod 4)/2.  However, in *building*
the solution I think you would need a value in the cell meaning
essentially "distance is currently unknown".  Hence, you would need
at least three separate values and therefore at least two bits.
After the table were complete, it might be possible to reduce the
two bits down to one.  Even then,  you might want to be able to
identify those small percentage of cells which were empty, in which
case you would need two bits anyway.  Under the Q+H metric,  you
would need two bits in any case to store (depth mod 3), and the
fourth bit pattern would be available to represent "distance is
currently unknown" or "empty cell" as appropriate.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From BRYAN@wvnvm.wvnet.edu  Wed Jan 11 14:01:19 1995
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From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@life.ai.mit.edu>
Subject:   Re: Re: Cube with GAP
In-Reply-To: Message of 01/08/95 at 13:47:00 from ,
           Martin.Schoenert@Math.RWTH-Aachen.DE

On 01/08/95 at 13:47:00 Martin Schoenert said:
>Jerry Bryan wrote in his e-mail message of 1995/01/07

>    Note that this model does not include the face centers.  That is, it
>    is G[C,E] rather than G[C,E,F].  56 numbers would be required to
>    include the face centers.  The distinction between 48 facelets and
>    56 facelets bears on the nitpicky question of whether the set C of
>    rotations is a subgroup of G or not.

>Absolutely right.  This part of the GAP documentation was written years
>ago.  These days I represent MG, CG, etc. as permutation groups on 54
>points.  I also changed the numbering, so that the [1..24] represent the
>edges, [25..48] points represent the edges, and [49..54] represent the
>centers.

Martin was kind enough not to point out that 6*9 is 54 rather than 56.

>    When I write the model out to disk, I only write out 8 corner facelets
>    and 12 edge facelets.  For example, I only write out the front and
>    back corner facelets.  This saves space and converts the model from
>    a facelet model to a cubie model, with the twists implicitly encoded
>    rather than being explicitly encoded via multiplication tables.  It
>    also automatically establishes a frame of reference by which a
>    proof of conservation of twist and flip can be accomplished.

>In terms of computational group theory this sequence of 8 corner and
>12 edgde facelets is called a *base* for the permutation group G.
>That is, each element of the group is uniquely determined by the
>images of those 20 facelets.  Of course if you have already proved
>that no single corner can be twisted and no single edge can be flipped,
>you can reduce this to 7 corner and 11 edge facelets.

With some of my models it can be reduced slightly more.  Since
representative elements of equivalence classes are what are being stored,
the representative element can be chosen to fix a cubie of your choice.
The cubie which is being fixed need not be stored.  This trick does
not work, of course, if all cube positions are being stored rather
than just representative elements.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From BRYAN@wvnvm.wvnet.edu  Wed Jan 11 16:21:44 1995
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From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@life.ai.mit.edu>
Subject:   When is 4x4x4 Solved in GAP?
In-Reply-To: Message of 01/08/95 at 13:47:00 from ,
           Martin.Schoenert@Math.RWTH-Aachen.DE

On 01/08/95 at 13:47:00 Martin Schoenert said:

>Mark Longridge wrote in his e-mail message of 1995/01/03

>                            ...  I don't know how a normal 4x4x4 could
>    be represented though.

>Jerry answered

>    I fail to see the problem.  Just number the facelets.  The only
>    problem would then lie in deciding what the generators are  -- i.e.,
>    which kind of slice moves do you accept.  You would also have to
>    decide whether to model the invisible 2x2x2 inside, but again if you
>    did, just number the invisible facelets and include their movements
>    with your generators.

>The problem is that many different positions all look solved.  For
>example, you can permute the 4 center facelets of one face or exchange
>two adjacent edges, and the cube still looks solved (of course you cannot
>do all this independently).  So if we take the obvious permutation group
>on the 6*16 points, then a whole subgroup would correspond to what a
>puzzler would consider solved states.  If by a model we mean a group
>whose elements correspond to the different states a puzzler would see,
>and whose identity corresponds to what a puzzler would consider solved,
>then I have no good idea how to model the 4x4x4 cube as a permutation
>group.

Start by numbering all the facelets and defining your generators (about
which there might be some controversy), and call the resulting group G.
Decide which permutations you consider equivalent and call this set K.
K would probably include such things as the whole cube rotations C,
as well invisible permutations of the four center facelets, etc.
In most reasonable choices for K, K would certainly be a group.  Your
model then becomes the set of cosets G/K (which is *not* a factor
group!  I am learning.).  The questions then become: 1) can you define
an operation on the cosets G/K such that G/K is a group, and 2) can you
find a mapping from G/K onto a subgroup of G such that the mapping
respects costs?  If the answer to both questions is "yes", then it is
this subgroup of G which you would want to put into GAP.

By the way, I am sensitive to the distinction between G and CG, but in
the case of any Face centerless cube such as 2x2x2 or 4x4x4, it would
seem to me that the distinction is less important than in cubes with
Face centers such as 3x3x3 and 5x5x5.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From mreid@ptc.com  Thu Jan 12 19:36:23 1995
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Date: Thu, 12 Jan 1995 19:47:42 -0500
From: mreid@ptc.com (michael reid)
Message-Id: <9501130047.AA25588@ducie.ptc.com>
To: cube-lovers@ai.mit.edu
Subject: superflip
Content-Length: 1768

i've also done some searching for short maneuvers for superflip,
although not to the extent that dik has.  i was never really
satisfied with my efforts to exploit its symmetry and centrality.
however, i've recently had some new thoughts about this which
look promising.

consider two cases:

case 1:
suppose that there is a minimal sequence for superflip which
contains a half-turn.  then, by applying  R' U2 to superflip,
we've moved  3q  (or  2f )  closer to start.

case 2:
otherwise, every minimal sequence contains only 90 degree turns.
then either  R' U'  gets us  2q  (or  2f )  closer to start,
or  R' U  gets us  2q  (or  2f )  closer to start.  (and probably
both do.)

it would be nice to reduce this latter case to only one of  R' U'
or  R' U .  can anyone do this?


this was how i found the  24q  sequence for superflip.  i figured
that case 1 was fairly likely, so i tested the position

        superflip  R' U2

kociemba's algorithm found in succession  11 + 22 = 33q,
11 + 20 = 31q,  12 + 17 = 29q,  13 + 14 = 27q,  13 + 12 = 25q,
which gives superflip in  28q.  there was no improvement for quite
some time until depth 17 in stage 1 when it found  17 + 6 = 23q.
it searched for several days more and finished depth 17 and depth 18
in stage 1.  i was about to give up when i saw that it found
19 + 2 = 21q,  to give superflip in  24q.

here's a nice (and quite surprising application) of the above
use of symmetry and centrality:

when searching for superflip in the face turn metric, it's
sufficient to search through depth 17 in stage 1!

suppose we have a  19f  sequence for superflip.  then, by considering
parity, some turn must be a half-turn.  now we may suppose (as above)
that the last two face turns are  U R2 ,  which is in stage 2!

mike

From dik@cwi.nl  Thu Jan 12 20:35:59 1995
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From: Dik.Winter@cwi.nl
Message-Id: <9501130135.AA14054=dik@boring.cwi.nl>
To: cube-lovers@ai.mit.edu, mreid@ptc.com
Subject: Re:  superflip

 > i've also done some searching for short maneuvers for superflip,
 > although not to the extent that dik has.  i was never really
 > satisfied with my efforts to exploit its symmetry and centrality.
 > however, i've recently had some new thoughts about this which
 > look promising.

I have indeed considered this, but have not yet come to a conclusion.

 > case 1:
 > suppose that there is a minimal sequence for superflip which
 > contains a half-turn.  then, by applying  R' U2 to superflip,
 > we've moved  3q  (or  2f )  closer to start.

I do not know whether this is clear for all readers.  My reasoning
was similar but the conclusion different, but someway equivalent:
If the minimal sequence contains a half-turn, we may just as well
assume that that half turn is the last, and F2.  I do not know
whether the proof has been shown on this list, but it is simple.
Suppose M is a minimal sequence, and Z is some random sequence,
in that case Z M Z' is also superflip.  Take Z the maximal
sequence at the end consisting of quarter-turns only, we end with
a sequence of equal length terminating with a half-turn.
Because of symmetry we may just as well consider it to be F2.

 > case 2:
 > otherwise, every minimal sequence contains only 90 degree turns.
 > then either  R' U'  gets us  2q  (or  2f )  closer to start,
 > or  R' U  gets us  2q  (or  2f )  closer to start.  (and probably
 > both do.)
 > it would be nice to reduce this latter case to only one of  R' U'
 > or  R' U .  can anyone do this?

This needs more than simple symmetry.  There are 12*8 different
endings, and we have 48 symmetries (24 by rotation * 2 by inversion).
Leaving 2 cases.  I considered this, but have not yet come to
conclusions.  On the other hand I do not yet know what to conclude
from M M' = I for every superflip sequence.

 > when searching for superflip in the face turn metric, it's
 > sufficient to search through depth 17 in stage 1!

 > suppose we have a  19f  sequence for superflip.  then, by considering
 > parity, some turn must be a half-turn.  now we may suppose (as above)
 > that the last two face turns are  U R2 ,  which is in stage 2!

Yes, I had seen that.  One of the major reasons I was not amused when
the system crashed doing depth 17 in stage 1!  I will restart the
program doing depth 17, but I will first redo the counting so that
counts larger than 2^32 are correct.

dik

From BRYAN@wvnvm.wvnet.edu  Thu Jan 12 21:14:10 1995
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From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <cube-lovers@ai.mit.edu>
Subject:   Re: superflip
In-Reply-To: Message of 01/12/95 at 19:47:42 from mreid@ptc.com

Another thought.  It would seem to me that under either metric,
the position halfway through ought to be "symmetric".  I am not
sure just what "symmetric" means here, and my use of the word
is largely arm waving similar to the use of "symmetric" to
describe local maxima prior to "Symmetry and Local Maxima".
But if such a "symmetric" position could be identified, it should
be quite easy to search from it back to Start or back to Superflip
in a minimal number of moves.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU



From mreid@ptc.com  Sat Jan 14 17:07:00 1995
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From: mreid@ptc.com (michael reid)
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To: cube-lovers@ai.mit.edu
Subject: more on superflip
Content-Length: 3294

recently i said:

> when searching for superflip in the face turn metric, it's
> sufficient to search through depth 17 in stage 1!

since posting this, i've realized that we can do much better.
here's my current approach.  everything below refers to the
face turn metric.  (i have similar reductions for quarter turns,
but they're not quite as good.)

proposition 1.  there is a minimal sequence for superflip of the form

                      sequence_1  sequence_2

                where  sequence_1  is in stage 1,  sequence_2  is in
                stage 2, and  sequence_1  is at most  17f  long.

proof.  consider the different possibilities for the length of a
        minimal sequence for superflip:  20f, 19f, 18f, 17f  or less.
        in the first case, we already know a maneuver of the form.
        in the second case, my discussion on thursday shows that
        we'll have such a maneuver.  in the case of  18f , we may
        suppose that the last face turned is  U , so we'll have such
        a maneuver.  and in the last case, we may take  sequence_2
        to be the empty sequence.  this proves prop. 1.

proposition 2.  there is a minimal sequence for superflip of the form

                    R1  sequence_1  sequence_2

                where  sequence_1  is in stage 1,  sequence_2  is in
                stage 2, and  sequence_1  is at most  16f  long.

proof.  consider the maneuver given by prop. 1.  by applying one of
        the 16 symmetries that fix the  U - D axis, we may suppose
        that the first turn of  sequence_1  is either  U1, U2, R1,
        or  R2.  in the case of

                    U1  sequence_1  sequence_2,

        replace this by

                    sequence_1  sequence_2  U1,

        and try again.  handle the cases starting with  U2  and  R2
        similarly.  we will either exhaust the stage 1 part of the
        sequence (which is impossible, since superflip isn't in
        the subgroup of stage 2) or we'll wind up with a manuever
        starting with  R1 , as desired.  this proves prop. 2.

there's still some more symmetry left to exploit.

proposition 3.  there is a minimal sequence for superflip of one of the
                forms

                          R1 F1  sequence_1  sequence_2,
                          R1 F2  sequence_1  sequence_2,
                          R1 F3  sequence_1  sequence_2,
                          R1 U1  sequence_1  sequence_2,
                          R1 U2  sequence_1  sequence_2,
                          R1 U3  sequence_1  sequence_2,
                          R1 L1  sequence_1  sequence_2,   or
                          R1 L3  sequence_1  sequence_2,

                where  sequence_1  is in stage 1,  sequence_2  is in
                stage 2, and  sequence_1  is at most  15f  long.

proof.  by applying the symmetry  C_R2  if necessary, we may suppose
        that the second turn of the maneuver given by prop. 2 is one
        of  F1, F2, F3, U1, U2, U3, L1, L2  or  L3.  this gives nine
        cases.  in the case

                R1 L2  sequence_1  sequence_2,

        replace this by

                R1  sequence_1  sequence_2  L2

        and try again.  this proves prop. 3.


i have these cases running right now, and i hope to have results soon!

mike

From mschoene@math.rwth-aachen.de  Sat Jan 14 19:20:51 1995
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From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: cube-lovers@life.ai.mit.edu, ishius@ishius.com
In-Reply-To: der Mouse's message of Tue, 10 Jan 1995 18:14:15 -0500 <199501102314.SAA10516@Collatz.McRCIM.McGill.EDU>
Subject: Re: Difficulty of Skewb

Der Mouse wrote in his e-mail message of 1995/01/10

    The size of the thing is thus somewhere on the order of 500 million
    configurations.  This is why I called it trivial next to the 3x3x3. :-)
    The group structure in terms of facicles, for what's-his-name to sic
    GAP on should he care to, derived from the following facicle labeling

Of course ``what's-his-name'' couldn't resist ;-).
Here are my findings about the SKEWB.


The SKEWB Puzzle
================

I took the liberty to renumber the points.  This makes the orbits easier
recognizable.

                    up
               +----------+
               |  9    10 |
               |    27    |
        left   | 12    11 |   right      back
    +----------+----------+----------+----------+
    |  5     6 | 18    17 |  1     2 | 22    21 |
    |    26    |    29    |    25    |    30    |
    |  8     7 | 19    20 |  4     3 | 23    24 |
    +----------+----------+----------+----------+
               | 13    14 |
               |    28    |
               | 16    15 |
               +----------+
                   down

I call the 8 possible turns LUB, LUF, RUB, RUF, LDB, LDF, RDB, RDB.
Here LUB denotes the turn around the corner common to the L(eft),
U(p), and B(ack) faces that turns L to U, U to B, and B back to L.
Those turns all have order 3, and I denote the inverses with <gen>^-1
(at least there is no question which metric to use for the SKEWB ;-).

With respect to the above numbering, the generators are the following
permutations.

    ##     corner     other corners                  centers
    RUF := ( 1,11,17) ( 2,12,20)( 4,10,18)(22, 6,14) (25,27,29);
    RUB := ( 2,10,22) ( 1, 9,23)( 3,11,21)(17, 5,15) (25,27,30);
    LUF := ( 6,12,18) ( 5,11,19)( 7, 9,17)(21, 1,13) (26,27,29);
    LUB := ( 5, 9,21) ( 6,10,24)( 8,12,22)(18, 2,16) (26,27,30);
    RDF := ( 4,14,20) ( 1,15,19)( 3,13,17)( 7,11,23) (25,28,29);
    RDB := ( 3,15,23) ( 2,14,24)( 4,16,22)( 8,10,20) (25,28,30);
    LDF := ( 7,13,19) ( 6,16,20)( 8,14,18)( 4,12,24) (26,28,29);
    LDB := ( 8,16,24) ( 5,13,23)( 7,15,21)( 3, 9,19) (26,28,30);

With r, u, and f I denote the 3 rotations of the rigid cube,
which generate the full automorphism group of the cube.
Here r means the rotation around the axis going through the
r and l faces, turning the r face in clockwise direction.

    ##     corners       opposite      centers
    r   := ( 1, 2, 3, 4) ( 6, 5, 8, 7) (27,30,28,29)
           (11,22,15,20)(12,21,16,19)(10,23,14,17)( 9,24,13,18);
    u   := ( 9,10,11,12) (16,15,14,13) (30,25,29,26)
           (22, 1,18, 5)(23, 4,19, 8)(21, 2,17, 6)(24, 3,20, 7);
    f   := (17,20,19,18) (21,22,23,24) (25,28,26,27)
           ( 1,14, 7,12)( 2,15, 8, 9)( 3,16, 5,10)( 4,13, 6,11);

Let G be the group generated by the 8 turns;
G = < LUB, LUF, RUB, RUF, LDB, LDF, RDB, RDF >.

Let C be the group generated by the 3 rotations; C = < r, u, f >.
Obviously |C| = 24 and C ~ S_4.  Let C' be the derived subgroup of C;
C' = <Comm(r,u),Comm(f,r),Comm(u,f)>.  Obviously |C'| = 12 and C' ~ A_4.

Then the group CG = < C, G > is the set of all positions a puzzler could
observate.  There are 24 solved position in CG (solved up to rotations).

The group CG has size 2 * 6!/2 * ((3^4*4!/2) * (3^4*4!/2) / 3^2).
The group G has size 6!/2 * ((3^4*4!/2) * (3^4*4!/2) / 3^2)
(and is a normal subgroup of CG, since |CG/G| = 2).
Note that this implies that |C <intersect> G| = 12.
This is not suprising, since G obviously contains all the commutators:
Comm(u,f) = LUB * RDF,  Comm(r,u)    = LUF * RDB,
Comm(f,r) = RUB * LDF,  Comm(u,f^-1) = RUF * LDB,
(those are the rotations of the entire cube around the 4 diagonals),
and so G must contain the derived subgroup C' of C.

The numbers imply that of the 6! * 3^8 * 8! position one can obtain by
taking the puzzle apart and randomly reassembling it, only ~ 0.04% are
solvable.  Much less than the ~ 8.3% one gets for Rubik's cube.
If you choose to puzzle that way, it is certainly a lot more difficult
than Rubik's cube ;-).


The Structure of the SKEWB Group
================================

G has three orbits on [1..30].  Namely the six face centers F = [25..30],
the odd corners C1 = [1,3..23], and the even corners C2 = [2,4..24].
C shall denote the set of all corners, i.e., the union of C1 and C2.
The two orbits on the corners are the two tetrahedral sets of corners
mentioned by der Mouse.

Let G[F] be the operation of G on the faces centers, i.e., G[F] = G/G_F,
where G_F is the stabilizer in G of the face centers (the subgroup of
those elements of G that fix all face centers).  Let G[C] be the
operation of G on the corners, i.e., G[C] = G/G_C, where G_C is the
stabilizer in G of the corners.

As usual with the operation of a group on several orbits, the stabilizers
are normal subgroups, and they intersect trivially.  On the other hand
the size of G_C is 6!/2 and the size of G_F is (3^4*4!/2 * 3^4*4!/2)/3^2.
So |G|=|G_C||G_F| and we see that G is the direct product of G_C and G_F.

What that means puzzle-wise is that there are no dependencies between the
operations on the faces and corners.  So take any achivable position x[F]
of the faces and any achivable position y[C] of the corners.  Then there
is a position z that has simultaineously z[F] = x[F] and z[C] = y[C]
(in the case of Rubik's cube this is not possible, you can flip a single
edge if you ignore what happens at the corners, but you cannot combine
this flip of a single edge with a trivial operation on the corners).

So we can fully understand the structure of G simply by understanding the
structures of G_C and G_F.  Of course we can also analyse G[F] and G[C],
since the isomorphism theorem tells us that G_C ~ G[F] and G_F ~ G[C].

Obviously G[F] is simply the alternating group A_6.  That means we can
achieve any even permutation on the center faces.  This is not suprising.
All 8 generators effect 3-cylces on the center faces.  And with 8
3-cycles, one expects the full alternating group to pop up.

G[C] has the 2 orbits C1 and C2.  Clearly G[C1] and G[C2] are isomorphic.
G[C1] is the wreath product of the cyclic group C_3 and the alternating
group A_4.  That means you can twist each corner independently and
achieve any even permutation on the four corners.  That you can only
achieve even permutations is obvious, since all generators of G[C]
permute 3 corners in a 3-cycle.  So |G[C1]| = |G[C2]| = 3^4*4!/2.

Now G[C] is the subdirect product of G[C1] and G[C1].  That means
it is isomorphic to a subgroup of the direct product of G[C1] and G[C2].
It is a subgroup of index 9.  That means
|G[C]| = |G[C1]| |G[C2]| / 9 = (3^4*4!/2 * 3^4*4!/2) / 3^2.

G[C] is the subgroup of G[C1] * G[C2] of those permutations where each
(anti)clockwise twist of a corner in C1 is combined with a
(anti)clockwise 3-cycle of corners in C2 and each (anti)clockwise twist
of a corner in C2 is combined with a (anti)clockwise 3-cycle of corners
in C1.

This was the most surprising observation for me, so allow me to talk a
little bit more about this aspect.

The subdirect product G[C] is a product where we ``glue'' together a
common factor group of G[C1] and G[C2].  Now this common factor group is
the direct product C_3*C_3 of two cyclic groups of size 3.  One of them,
K say, comes from the normal subgroup C_3^4 of those elements that only
twist the corners.  The other, L say, comes from the alternating group
A_4.  Now those factor groups are glued together crosswise, that is,
K of G[C1] is glued to L of G[C2], and L of G[C1] is glued to K of G[C2].

In case anybody cares, G has trivial center.  This is because the central
elements of G[C1] must be combined in G[C] with non-central elements of
G[C2] and vice versa.  And of course G[F] has trivial center.


The Normal Subgroups of the SKEWB Group
=======================================

I have also computed the lattice of normal subgroups of the SKEWB group.
Since the group is so small, I don't need any complicated argument.
G is a little bit too large to simply try 'NormalSubgroups(G);' in GAP.
But G[F] and G[C] are both small enough.  G[F] is almost trivial
(GAP finds in a few seconds that G[F] has no proper normal subgroups).
G[C] is a little bit more difficult (but it took me much longer to draw
a reasonable picture, then it took GAP to compute the normal subgroups).

Anyhow, allow me to first present the normal subgroups lattice of G[C1],
which is of course identical to the normal subgroups lattice of G[C2].

              G[C1] (972)
              //|\
            / / | \
          L1 L2 L3 K[C1] (324)
            \ \ | / \
              \\|/   \
       (108) G[C1]'   \
                 \   V[C1] (81)
                  \   / \
                   \ /   \
             (27) W[C1]   \
                     \     \
                      \     \
                       \     \
                        \   Z[C1] (3)
                         \   /
                          \ /
                          <1>

Here V[C1] is elementary abelian subgroup (the vector space) of those 3^4
elements of G[C1] that only twist the corners, but do not permute them.
G[C1]/V[C1] is the A_4 permuting the corners.  K[C1]/V[C1] is the
subgroup of G[C1]/V[C1] of the double transpositions, i.e, the elements
that permute the corners in two pairs of two corners each.  G[C1]' is the
derived subgroup of G[C1], i.e., G[C1]/G[C1]' is the largest abelian
factor group of G[C1] (b.t.w. note that the fact that G is a direct
product implies that G[C1]' = G'[C1]).  L1, L2, and L3 are three more
normal subgroups of index 3, which I don't know how to describe easily.
Z[C1] is the center of G[C1].

And now the lattice of normal subgroups of G[C].

                G[C] (104976)
                //|\
              / / | \
            L1 L2 L3 K[C] (34992)
              \ \ | / \_
                \\|/  | \
        (11664) G[C]'  \ \
                   \_  V1 V2 (8748)
                   | \ / X \
                    \ X / \ \
             (2916) W1 W2  \ \ 
                    / X \   \ \
                   |_/ \ \   \ \
                   /    \ \   \ \
          (729) W[C]     \ \   \ \
                   \      \ \   \ \
                    \      \ \  Z1 Z2 (324)
                     \      \ \ / /
                      \_     \ X /
                      | \    S1 S2 (108)
                       \ \   / /
                        \ \ / /
                         \ X /
                         T1 T2 (27)
                         / /
                        / /
                       / /
                      |_/
                      /
                     /
                    /
                   /
                 <1>

S1 and S2 are the stabilizers G[C]_C1 and G[C]_C2, so the normal subgroup
lattices above S1 and S2 are identical to the normal subgroup lattices of
G[C1] (= G[C2]).  Those two lattices over S1 and S2 share the normal
subgroups over G[C]' (this is where G[C1] and G[C2] are glued together).
W[C] (the intersection of W1 and W2) is the elementary abelian subgroups
(the vectorspace) of those elements of G[C] that only twist the 8
corners, but do not permute them.  G[C]/W[C] is the A_4*A_4 permuting the
two sets of corners.  G[C]' is the derived subgroup of G[C], i.e.,
G[C]/G[C]' is the largest abelian subgroup of G[C].

Now we get the normal subgroup lattice of G by taking the above picture
twice, once below G_F (because G_F ~ G[C]), and once above G_C (because
G/G_C ~ G[C]).

          G_______
        //|\      \
        LLL K      \
         D   \      \
          \  VV      \
           WW \\      G_F
         W  \\ \\    //|\
          \  \\ ZZ   LLL K
           \\ SS      D   \
            TT         \  VV
           //           WW \\
          /           W  \\ \\
        G_C            \  \\ ZZ
           \            \\ SS
            \            TT
             \          // 
              \_______ /
                     <1>


God's algorithm for the SKEWB
=============================

The next was to compute God's algorithm for the SKEWB.
G is not very large, but is is easier to use a smaller model.
Let H be the subgroup generated by the 4 turns LUB, LUF, RUB, and RUF.

Repeated application of the rules

    <turn> <rotation> -> <rotation> <other-turn>,
    LDB -> Comm(u,f^-1) * RUF^-1,  LDF -> Comm(f,r) * RUB^-1,
    RDB -> Comm(r,u)    * LUF^-1,  RDF -> Comm(u,f) * LUB^-1,

allows us to translate any process in CG to one that starts with a single
rotation and continues with a process in H, i.e., it starts with a
rotation and continues with a process that involves only LUB, LUF, RUB,
and RUF and their inverses.

Note however, that H is *not* a supplement for C in CG.  This is because
the intersection of H and C is not trivial.  Namely H contains d^2, i.e.,
the rotation by 180 degree around the axis through the down face (which
is fixed by H) and the up face.

That means that H contains 2 solved positions, and each position of H
contains to 12 positions of CG.  In other word H has index 12 in CG.

Here is the table for H.  The first column contains the lenght.  The
second column contains the number of positions of that length in H.
The third column contains the number of positions of that length that are
local maxima, i.e., the number of positions <pos> such that for no
generator <gen> the position <pos>*<gen> is longer.  The fourth column
contains the number of positions such that for one generator <gen> the
position <pos>*<gen> is longer.  And so on.  So the eleventh column
contains the number of positions <pos> such that for all eight generators
<pos>*<gen> is longer (this happens of course only for the 2 solutions).

    length #pos  #loc max
     0       2       0      0      0      0      0      0      0      0      2
     1      16       0      0      0      0      0      0     16      0      0
     2      96       0      0      0      0      0      0     96      0      0
     3     576       0      0      0      0      0      0    576      0      0
     4    3456       0      0      0      0      0    240   3216      0      0
     5   20496       0      0      0     48    729   2766  16953      0      0
     6  118608      48    161   1231   4228  14779  32993  65168      0      0
     7  630396    8266  33358  76349 121363 153892 137755  99413      0      0
     8 2450966 1025322 621763 381098 234661 128570  47822  11730      0      0
     9 2911712 2768641 126056  15344   1422    199     50      0      0      0
    10  162056  161876    180      0      0      0      0      0      0      0
    11     180     180      0      0      0      0      0      0      0      0

To get the table for CG, simply multiply each number by 12.

This was computed with a small C program in 200 seconds on a Pentium 90
system using approx 16 MByte of memory.  Since this is my first
computation of this kind, I would be glad if somebody could independently
verify this table.


The SuperSKEWB
==============

If we do not ignore the orientation of the face centers we obtain the
SuperSKEWB.  This could for example be done by drawing a line over
one corner and the center for each face of the cube.

Mathematically I modelled the SuperSKEWB by representing each face center
with a set of four numbers (instead of a single number).  That pushed
the total number of moved points from 30 to 48 (still pretty small).

The size of the SuperSKEWB group SG is a factor 32 greater than G's size,
i.e., it is (2^5*6!/2) * ((3^4*4!/2) * (3^4*4!/2) / 3^2).
The size of CSG (closure of C and SG) is also a factor 32 greater than
CG's size, i.e., it is 2*(2^5*6!/2) * ((3^4*4!/2) * (3^4*4!/2) / 3^2).

It is easy to see that you cannot turn a face center by 90 degrees in SG.
Namely the corner pieces fall into two tetrahedral sets, which are not
interchanged by SG.  Now a given edge of a face center will always be
adjacent to corners in one of those two sets of corner pieces.

Simply by looking at the generators of SG, we see that each
element of SG must turn an even number of face centers.
So we can turn at most five of the six face centers independently;
after that the orientation of the sixth face center is determined.
So there are at most 2^5 different orientations possible.  Since
|SG| = 2^5 |G|, we see that all 2^5 orientations are indeed achievable.

The structure of SG is of course very similar to the structure of G.
SG[C] is identical to G[C].  Remember that G[F] was A_6.
SG[F] is a subgroup of index 2 in the wreath product 2 <wreath> A_6.
The index 2 is because one cannot turn a single face.

Note that SG has a nontrivial center, namely the element that turns
all face centers at once.

Thus the lattice of normal subgroups of SG is very similar to the lattice
of G.  The only difference is that SG_C (and of course also SG/SG_F) has
two nontrivial normal subgroups with sizes 32 and 2 (resp. with sizes
32*104976 and 2*104976).

I cannot compute God's algorithm for the SuperSKEWB.  Would I use the
same approach that I used for the SKEWB, I would need 32 times as much
memory, i.e., ~ 1/2 GByte.  Does anybody have an idea how to tackle the
SuperSKEWB (provided anybody cares, I certainly don't)?

Have a nice day.

Martin.

PS:  No, I don't own a SKEWB.  Yes, I intent to order one.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From mreid@ptc.com  Wed Jan 18 10:02:03 1995
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From: mreid@ptc.com (michael reid)
Message-Id: <9501181513.AA00811@ducie.ptc.com>
To: cube-lovers@ai.mit.edu
Subject: superflip requires 20 face turns
Content-Length: 3124

superflip is now known to require 20 face turns.  in particular, the
diameter of the cube group is at least 20 face turns (and i conjecture
that it's larger).  as far as i can tell, this is the first improvement
to the lower bound (18f) given by a simple counting argument.

in my previous two messages, i gave a proof of the fact that there is a
minimal sequence for superflip in one of the following forms:

    R1 F1  sequence_1  sequence_2,
    R1 F2  sequence_1  sequence_2,
    R1 F3  sequence_1  sequence_2,
    R1 U1  sequence_1  sequence_2,
    R1 U2  sequence_1  sequence_2,
    R1 U3  sequence_1  sequence_2,
    R1 L1  sequence_1  sequence_2,   or
    R1 L3  sequence_1  sequence_2,

where  sequence_2  is in the subgroup of stage 2 of kociemba's algorithm,
and  sequence_1  is at most  15f  long.  as of monday morning, the first
six cases were completely searched, but the final two seemed to be much
slower.  fortunately, there is more symmetry available here (which is at
least part of the reason that these cases are so slow).

in the case starting with  R1 L1,  we have four symmetries (generated by
C_R2  and  C_U2)  which fix the subgroup of stage 2.  using these
symmetries, we may suppose that the third face turn is one of  U1, U2, U3,
F1, F2  or  F3.

in the case starting with  R1 L3,  we again have four symmetries which fix
the subgroup of stage 2.  in this case, the symmetries are generated by
C_R2  and reflection through the  R - L plane.  using these symmetries,
we may suppose that the third face turn is one of  U1, U2, F1  or  F2.

even with these reductions, the last two cases are still somewhat
stubborn.  finally they were completed this morning.

here's a summary of what i tested:

position tested:  depth tested

superflip  R1 F1:  15f deep in stage 1
best solution found:  15 + 3 = 18f

superflip  R1 F2:  15f deep in stage 1
best solution found:  15 + 3 = 18f

superflip  R1 F3:  15f deep in stage 1
best solution found:  15 + 3 = 18f

superflip  R1 U1:  15f deep in stage 1
best solution found:  11 + 8 = 19f

superflip  R1 U2:  15f deep in stage 1
best solution found:  13 + 5 = 18f

superflip  R1 U3:  15f deep in stage 1
best solution found:  12 + 7 = 19f


superflip  R1 L1 U1:  14f deep in stage 1
best solution found:  11 + 7 = 18f

superflip  R1 L1 U2:  14f deep in stage 1
best solution found:  10 + 7 = 17f

superflip  R1 L1 U3:  14f deep in stage 1
best solution found:  11 + 7 = 18f

superflip  R1 L1 F1:  14f deep in stage 1
best solution found:  12 + 5 = 17f

superflip  R1 L1 F2:  14f deep in stage 1
best solution found:  10 + 8 = 18f

superflip  R1 L1 F3:  14f deep in stage 1
best solution found:  12 + 5 = 17f


superflip  R1 L3 U1:  14f deep in stage 1
best solution found:  14 + 3 = 17f

superflip  R1 L3 U2:  14f deep in stage 1
best solution found:  10 + 8 = 18f

superflip  R1 L3 F1:  14f deep in stage 1
best solution found:  12 + 5 = 17f

superflip  R1 L3 F2:  14f deep in stage 1
best solution found:  13 + 5 = 18f


total run time was about 210 cpu hours (somewhat more than i'd hoped for)
distributed across several machines of varying ability.

mike

From mreid@ptc.com  Wed Jan 18 10:39:55 1995
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From: mreid@ptc.com (michael reid)
Message-Id: <9501181551.AA00837@ducie.ptc.com>
To: cube-lovers@ai.mit.edu
Subject: searching for superflip in quarter turn metric
Content-Length: 3083

here's my approach to searching for superflip in the quarter turn metric.
i gave a maneuver of length  24q  for superflip on january 10.  suppose
there is a maneuver of length 22q (or shorter).  consider three cases:

case 1.  there is a minimal maneuver which contains a half-turn.

case 2.  no minimal maneuver contains a half-turn, but there is a
         minimal maneuver which contains consecutive turns of
         opposite faces.

case 3.  neither case 1 nor case 2 hold.


in case 1, we may find a minimal sequence of the form

         sequence_1  sequence_2,

where  sequence_2  is at least  3q  long.  as in the face turn metric,
we may also suppose that  sequence_1  starts with one of

        R1 F1,   R1 F2,   R1 F3,   R1 U1,   R1 U2,   R1 U3,
        R1 L1 U1,   R1 L1 U2,   R1 L1 U3,   R1 L1 F1,   R1 L1 F2,
        R1 L1 F3,   R1 L3 U1,   R1 L3 U2,   R1 L3 F1,   R1 L3 F2.

furthermore, the case starting with  R1 F2  may be included in the
case starting with  R1 F1,  and similarly for other cases.  thus we
may suppose that  sequence_1  starts with one of

        R1 F1,   R1 F3,   R1 U1,   R1 U3,
        R1 L1 U1,   R1 L1 U3,   R1 L1 F1,   R1 L1 F3,
        R1 L3 U1,   R1 L3 F1.


in case 2, we may find a minimal sequence of the form

         sequence_1  sequence_2,

where  sequence_2  is at least  2q  long.  as in case 1, we may suppose
that  sequence_1  starts with one of the ten sequences above.


in case 3, the best we can do is  1q  in stage 2.  however, i claim
that we can find three consecutive turns of mutual adjacent faces.
otherwise, we'd have a maneuver for superflip using only the four faces
F, R, B, L,  (for example)  which is ridiculous, because edges can't
change orientation using only these turns.

therefore, we may suppose that a minimal sequence starts with three
consecutive turns of mutual adjacent faces.  up to symmetry, there
are eight cases for these turns:

      U1 R1 F1,   U1 R1 F3,   U3 R1 F1,   U3 R1 F3,
      D1 R1 F1,   D1 R1 F3,   D3 R1 F1,   D3 R1 F3.

replace   U1 R1 F1  sequence   by   R1 F1  sequence  U1  ,  and
similarly for the other seven cases.  thus we have a minimal
maneuver in the form   sequence_1  sequence_2 ,  where  sequence_2
is  1q  long  and  sequence_1  starts with either  R1 F1  or  R1 F3.


combining all the above cases, a maneuver for superflip in  22q  or less
(assuming one exists) may be found in one of the forms:

        R1 L1 U1  sequence_1  sequence_2,
        R1 L1 U3  sequence_1  sequence_2,
        R1 L1 F1  sequence_1  sequence_2,
        R1 L1 F3  sequence_1  sequence_2,
        R1 L3 U1  sequence_1  sequence_2,
        R1 L3 F1  sequence_1  sequence_2,

where  sequence_1  is at most  17q  long,

        R1 U1  sequence_1  sequence_2,
        R1 U3  sequence_1  sequence_2,

where  sequence_1  is at most  18q  long,

        R1 F1  sequence_1  sequence_2,
        R1 F3  sequence_1  sequence_2,

where  sequence_1  is at most  19q  long.


i don't know how feasible this is (but it sure looks formidable).
to get some idea, first i'll test for  20q  or less.

mike

From BRYAN@wvnvm.wvnet.edu  Wed Jan 18 11:54:12 1995
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Date:      Wed, 18 Jan 1995 11:53:21 EST
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@life.ai.mit.edu>
Subject:   Re: Re: Models for the Cube
In-Reply-To: Message of 12/10/94 at 15:12:00 from ,
           Martin.Schoenert@math.rwth-aachen.de

Both the original note and the reply were rather lengthy, so I will
not quote the whole thing.  We were at the point where we were
discussing models for cubes with edges only.  Such a cube can be
modeled as a set of cosets of C, or as a subset of G or of CG.  We
arrived at the following dialog which discussed the correspondence
between the two kinds of models.

On 12/10/94 at 15:12:00 Martin Schoenert said:

>Jerry continued

>    A similar argument applies to G[E]/C[E] except that we have to fix
>    an edge cubie instead of a corner cubie.

>Almost.  But there is a tricky problem here.  Again G[E] = CG[E],
>so it doesn't matter whether we talk about G[E]/C[E] (as you prefer)
>or about CG[E]/C[E] (as I prefer).  Again we can find a supplement
>for C[E] in CG[E], namely the subgroup of all elements of CG[E]
>that leave a particular edge cubie fixed.  Assume that we fix the
>upper-right edge cubie, then this supplement is <L[E],D[E],F[E],B[E]>.

>But this does *not* respect costs.  That is take an element e of CG[E].
>Let r be its representative in <L[E],D[E],F[E],B[E]>, i.e., c e = r,
>where c is a rotation of the entire cube.  The the costs of the two
>elements, viewed as elements of CG[E] is clearly the same (remember,
>rotations cost nothing).  But the cost of r, viewed as an element of
><L[E],D[E],F[E],B[E]> *with this generating system*, may be higher.

Indeed, *is* higher a large percentage of the time, I think.

>For example take R[E] * r[E]' (where r is the rotation of the entire
>cube).  In CG[E] this element has cost 1.  And this element lies in
><L[E],D[E],F[E],B[E]>, since it fixes the upper-right edge cubie.
>But the cost of this element *with respect to the generating system
>L[E],D[E],F[E],B[E]* is not 1.

>We can repair this by choosing a different generating system for
><L[E],D[E],F[E],B[E]>, for example the system
>L[E],D[E],F[E],B[E],R[E]*r[E]',U[E]*u[E]'.

>So in general a model for the solution up to rotations for a
>certain set <orbits>, is a supplement of C[<orbits>] in CG[<orbits>],
>with a generating system that respects costs.

I guess I need to understand a little more precisely what we mean
by "respecting costs", but I have a question here.  I was thinking
about this issue of representing edges only cubes by cosets vs.
representing edges only cubes as subsets of G before your note arrived,
and I thought I saw a problem.

It is well known that G[E] must have an equal number of even and
odd permutations.  If we generate G[E] as <Q[E]>, it is also the case
that there are just as many positions an even distance from Start as
an odd distance from Start because the parity of the distance from
Start is the same as the parity of the permutation if we restrict
ourselves to quarter turns.

But in the computer search for God's Algorithm for edges only cubes,
there were not equal numbers of positions an even distance from Start
as an odd distance from Start.  The computer search used the coset model
G[E]/C[E], where this notation means the set of cosets of C, *not* the
factor group.  In and of itself, the mismatch between the number of
positions at an even distance from Start and at an odd distance from
Start should not pose a problem.  It is not clear to me what it means
to speak of the "parity" of a coset of C, and half of each coset will
be even and the other half will be odd.  Hence, it is not clear that
a particular coset must *a priori* be an odd or even distance from
Start.

But if we map each coset to an element of G[E], it is meaningful to
speak of the parity of the element of G[E].  And if the elements of
G[E] to which we map the cosets form a subgroup, then there must be an
equal number of odd and even elements in the subgroup
(unless they are all even?!).  If the mapping respects
costs in the sense of maintaining the same distance from Start, then
I don't understand how we can avoid a conflict between the equal
number of even and odd permutations in the subgroup of G[E] and
the unequal number of even and odd distances from Start in the coset
model G[E]/C[E].

Perhaps you could clarify your generating system and its respecting
of costs a bit further?

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From @mail.uunet.ca:mark.longridge@canrem.com  Thu Jan 19 01:02:38 1995
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Sender: CRSO.Cube@canrem.com
Subject: Shift Invariance Recap
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1003.5834.0C1CA56D@canrem.com>
Date: Thu, 19 Jan 1995 00:08:00 -0500
Organization: CRS Online  (Toronto, Ontario)

Shift Invariance the Final Chapter??
------------------------------------

    2 x Order 2           (the diagonal square element)
    Subgroup <U2, D2, R2, L2>, order = 2
    D2 F2 T2 F2 B2 T2 F2 T2

    2 Swap                (the single square element)
    Subgroup <U2, D, R2, L>, order = 2
    D2 R2 D2 R2 D2 R2

    2 H                   (the edge square element)
    Subgroup <U, D, R2, L2>, order = 2
    L2 R2 B2 L2 R2 F2

    12 flip               (the central element)
    Subgroup <U, D, F, B, L, R>, order = 2
    R1 L1 D2 B3 L2 F2 R2 U3 D1 R3 D2 F3 B3 D3 F2 D3 R2 U3 F2 D3
    Special Property: Effects all edges the same

    6 Counterclockwise twist   (the odd element)
    Subgroup <U, R>, order = 3
    U2 R1 U1 R1 U1 R3 U1 R3 U1 R3 U2 R1 U1 R1 U1 R3 U1 R3 U1 R3
    Special Property: Effects all corners the same


Martin's message about the SuperSkewb having a non-trivial centre
reminded me that the SuperCube should have 3 more positions which
are also shift invariant:

3x3x3 cube with 6 centre pieces rotated 90, 180 and 270 degrees,
 with orders 4, 2 and 4 respectively. This time all the centres
 are effected the same!

Naturally there are 3 more positions in SG's <U, R> as well.

A pity there is no "Centre All-Twist" process in any of the cube
 literature.

-> Mark <-

I'll leave a superflip process for the Magic Dodecahedron as a
'exercise for the reader'     ;-)

From @mail.uunet.ca:mark.longridge@canrem.com  Thu Jan 19 01:33:07 1995
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Subject: Superflip 24q
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1002.5834.0C1CA56C@canrem.com>
Date: Thu, 19 Jan 1995 00:06:00 -0500
Organization: CRS Online  (Toronto, Ontario)


> this just appeared today, after a lot of searching:
>
> R3 U2 B1 L3 F1 U3 B1 D1 F1 U1 D3, L1 D2 F3 R1 B3 D1 F3 U3 B3 U1 D3
>    24q
>
> mike

Sm = Central Reflection

i.e. for operators

     F1 = B3, F2 = B2, F3 = B1
     L1 = R3, L2 = R2, L3 = R1
     U1 = D3, U2 = D2, U3 = D1

p = R3 U2 B1 L3 F1 U3 B1 D1 F1 U1 D3    (12 q)

Then  p + (p * Sm) = Superflip

This is Mike's process slightly patched, with the last two (commuting)
cube turns swapped in position.

-> Mark <-

From acoles@mnsinc.com  Thu Jan 19 12:55:11 1995
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Date: Thu, 19 Jan 1995 12:54:09 -0500 (EST)
From: Aaron Coles <acoles@news1.mnsinc.com>
To: cube-lovers@life.ai.mit.edu
Subject: Masterball
Message-Id: <Pine.ISC.3.90.950119125158.10455A-100000@mnsnet.mnsinc.com>
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Have anyone had any experience or even heard of "Masterball".  I picked this
"Mind Boggling 3-D puzzle with over 350 Quadrillion possible 
combinations" from a store out here is Washington, DC. 


From dlitwin@fusion.geoworks.com  Thu Jan 19 13:41:25 1995
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From: dlitwin@fusion.geoworks.com
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In-Reply-To: <Pine.ISC.3.90.950119125158.10455A-100000@mnsnet.mnsinc.com>
Subject: Masterball


Aaron Coles writes:
> Have anyone had any experience or even heard of "Masterball".  I picked this
> "Mind Boggling 3-D puzzle with over 350 Quadrillion possible 
> combinations" from a store out here is Washington, DC. 

	I've seen a bunch of them around, in all sorts of different color
patterns.  The two standard ones (that I bought) are black and white
stripes and a rainbow colored striped one.  I like the way they color the
plastic instead of using stickers.

	Dave Litwin

From mreid@ptc.com  Thu Jan 19 18:30:05 1995
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From: mreid@ptc.com (michael reid)
Message-Id: <9501192341.AA09110@ducie.ptc.com>
To: cube-lovers@ai.mit.edu
Subject: symmetric maneuvers
Content-Length: 1504

mark writes

> p = R3 U2 B1 L3 F1 U3 B1 D1 F1 U1 D3    (12 q)
> 
> Then  p + (p * Sm) = Superflip
> 
> This is Mike's process slightly patched, with the last two (commuting)
> cube turns swapped in position.

i'm surprised this hasn't been pointed out previously.  however, i would
write the above as

        (R3 U2 B1 L3 F1 U3 B1 D1 F1 U1 D3  C_X)^2

where i use  C_X  for central reflection.  this fits in with mark's
idea of "cyclic decomposition".

i've noticed that a number of minimal (or presumed to be minimal)
maneuvers for pretty patterns have some symmetry.  here i'll use
commutator notation:

              [ A , B ]    refers to   A  B  A'  B'

also i'll use bandelow's notation for rotations of the whole cube:

                C_U , C_RF , C_URF ,

denote rotation about a face-face axis, edge-edge axis, corner-corner
axis, respectively.

now some patterns:

anaconda:            B1 R1 D3 R3 F1 B3 D1 F3 U1 D3 L1 F1 L3 U3
                 = [ B1 R1 D3 R3 F1 B3 D1 , C_UB ]

python:              D2 F3 U3 L1 F3 B1 D3 B1 U1 D3 R3 F1 U1 B2
                 = [ D2 F3 U3 L1 F3 B1 D3 , C_UF ]

6 x's (order 3):     R2 L3 D1 F2 R3 D3 F1 B3 U1 D3 F1 L1 D2 F3 R1 L2
                 = [ R2 L3 D1 F2 R3 D3 F1 B3 , C_UB ]

my favorite example is
four twisted peaks:  U3 D1 B1 R3 F1 R1 B3 L3 F3 B1 L1 F1 R3 B3 R1 F3 U3 D1
                 = [ U3 D1 B1 R3 F1 R1 B3 L3 F3 , C_R2 ]

i'd hoped to find maneuvers for "cube within a cube" and "cube within a
cube within a cube", but no such luck.

mike

From mreid@ptc.com  Fri Jan 20 15:46:17 1995
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From: mreid@ptc.com (michael reid)
Message-Id: <9501202058.AA10072@ducie.ptc.com>
To: cube-lovers@ai.mit.edu
Subject: superflip in quarter turn metric
Content-Length: 3456

i've finished searching for superflip in  20q , and no solutions were
found.  thus superflip requires at least  22q , which gives a new lower
bound for the diameter of the cube group in the quarter turn metric.
total cpu spent on the search was 29 cpu hours.  based on this, i would
make a rough estimate of 2.5 to 3 months cpu time for an exhaustive
search through depth  22q.

this time i collected some statistics the way dik did.  this should be
helpful for troubleshooting.  it's not foolproof, but it's a reasonable
start.  i will rerun the face turn search and collect the same data
along the way.

mike


statistics follow:

depth in        number of times          solutions
stage 1        stage 2 is reached          found

superflip  R1 L1 U1:

 9q                     64               33q, 31q, 29q
10q                    272
11q                   3728               27q
12q                  26440
13q                 164664               25q
14q                 911112
15q                5516208
 
superflip  R1 L1 U3:

 9q                     64               31q, 29q, 27q
10q                    272
11q                   3728
12q                  26440
13q                 164664
14q                 911112               25q
15q                5516208

superflip  R1 L1 F1:

 9q                    288               31q, 29q, 27q
10q                   2192
11q                  13496
12q                  65280
13q                 352056               25q, 23q
14q                1810744
15q                9753608

superflip  R1 L1 F3:

 9q                    288               31q, 29q, 27q
10q                   2192
11q                  13496
12q                  65280               25q
13q                 352056
14q                1810744
15q                9753608

superflip  R1 L3 U1:

 9q                     64               33q, 31q, 29q, 27q
10q                    272
11q                   3728
12q                  26440               25q
13q                 164664
14q                 911112               23q
15q                5516208

superflip  R1 L3 F1:

 9q                    288               33q, 31q, 29q
10q                   2192               27q
11q                  13496               25q
12q                  65280
13q                 352056
14q                1810744
15q                9753608

superflip R1 U1:

10q                      6               32q
11q                    106               28q
12q                   4216               26q
13q                  30318
14q                 212208
15q                1414882
16q                9807890

superflip R1 U3:

10q                      6               32q
11q                    106               30q, 28q
12q                   4216               26q
13q                  30318
14q                 212208               24q
15q                1414882
16q                9807890

superflip R1 F1:

10q                     78               32q, 30q, 28q
11q                    352
12q                   5338               26q, 24q
13q                  35996
14q                 241230
15q                1549382
16q               10531798
17q               71358512

superflip R1 F3:

10q                     78               28q
11q                    352
12q                   5338               26q, 24q
13q                  35996
14q                 241230
15q                1549382
16q               10531798
17q               71358512


From BRYAN@wvnvm.wvnet.edu  Fri Jan 20 20:41:10 1995
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Date:      Fri, 20 Jan 1995 17:07:05 EST
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <mreid@ptc.com>, "Cube Lovers List" <cube-lovers@ai.mit.edu>
Subject:   Re: superflip in quarter turn metric
In-Reply-To: Message of 01/20/95 at 15:58:08 from mreid@ptc.com

On 01/20/95 at 15:58:08 mreid@ptc.com said:
>i've finished searching for superflip in  20q , and no solutions were
>found.  thus superflip requires at least  22q , which gives a new lower
>bound for the diameter of the cube group in the quarter turn metric.
>total cpu spent on the search was 29 cpu hours.  based on this, i would
>make a rough estimate of 2.5 to 3 months cpu time for an exhaustive
>search through depth  22q.

Rats.  You beat me by about a half hour.  I just finished comparing
Level 10 of my data base with the same Level 10 superflipped.  There
were no matches.

I just about have Level 11 completed.  This will provide interesting
new information in and of itself, because previously there has only
been an exhaustive search through level 10.  Once I complete Level 11,
I will superflip it and see what happens.

The superflip is especially amenable to a "two half depth search".
Normally, you would have to build one tree with Start at the root,
and a second tree with X at the root, where X is the position you
were testing.  But a search tree with superflip at the root is
identical to a search tree with Start at the root except that the
superflip tree has each element superflipped as compared to the
respective element of the tree with Start at the root.  Hence,
building the tree with Superflip at the root is quite easy once
the tree with Start at the root is in hand.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From BRYAN@wvnvm.wvnet.edu  Sat Jan 21 17:03:15 1995
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Date:      Sat, 21 Jan 1995 12:44:42 EST
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "der Mouse" <mouse@collatz.mcrcim.mcgill.edu>
Cc: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Re: superflip in quarter turn metric
In-Reply-To: Message of 01/21/95 at 09:05:55 from ,
           mouse@Collatz.McRCIM.McGill.EDU

On 01/21/95 at 09:05:55 der Mouse said:
>> The superflip is especially amenable to a "two half depth search".
>> Normally, you would have to build one tree with Start at the root,
>> and a second tree with X at the root, where X is the position you
>> were testing.  But a search tree with superflip at the root is
>> identical to a search tree with Start at the root except that the
>> superflip tree has each element superflipped as compared to the
>> respective element of the tree with Start at the root.

>Isn't this equally true of any other position, except that the
>conversion from a Start-rooted tree's position to an X-rooted tree's
>position is more complicated than just superflipping?  Just think of
>your tree, instead of describing positions as "cubie <x> in cubicle
><y>", as describing positions as "cubie that started in cubicle <x> in
>cubicle <y>".

I am taking the liberty of CC'ing Cube-Lovers as well.  A search tree
giving distances from Start calculates d(I,IY) for all positions IY
in the domain of inquiry.  With an X-rooted tree, distances are of
the form d(X,XZ) for all positions XZ in the domain of inquiry.
In general, it is not the case that d(I,IY)=d(X,XY).  Hence,
we cannot simply take Z=Y, and elements of the form XY do not produce
the X-rooted tree.  Therefore, to perform
two half-depth searches to connect I and X, you really do have to
construct a standard Start-rooted tree as well as an X-rooted tree.
You are looking for a position IY=XZ such that |IY|=|XZ|=|X|/2.

However, in the case of the Superflip E, it is the case that
d(I,IY)=d(E,EY).  Hence, in order to construct an
E-rooted tree, it is sufficient to calculate all elements of
the form EY from your existing I-rooted tree of the form IY.

>Or are you storing only M-conjugate-class representatives, and I'm
>exposing my ignorance of basic group theory? :-)

Yes, I am storing only M-conjugate-class representatives, but that
isn't the problem (see above).  However, it does make the processing
a bit less trivial than I have indicated.  For every Y in the
Start-rooted tree, I first form EY, then {m'(EY)m}, and finally
Repr{m'(EY)m}.  These representatives are sorted and then compared
against all the Y's (which are themselves representative elements,
of course).  We are looking for some V and W such that
Repr{m'(IV)m}=Repr{m'(EW)m}, and this will be a "half-way" position.
What I *really* want is some V such that Repr{m'(IV)m}=Repr{m'(EV)m}.
It seems to me that all half way positions should be "symmetric"
in this fashion, but I cannot prove it.  The recent discussions
by Mike Reid and Mark Longridge about the 24q superflips are
suggestive in this regard.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From @uconnvm.uconn.edu:dmoews@xraysgi.ims.uconn.edu  Sun Jan 22 01:25:28 1995
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Date: Mon, 23 Jan 95 01:24:21 -0500
From: dmoews@xraysgi.ims.uconn.edu (David Moews)
Message-Id: <9501230624.AA25808@xraysgi.ims.uconn.edu>
To: cube-lovers@ai.mit.edu, dmoews@xraysgi.ims.uconn.edu
Subject: Shamir's method on the superflip

I can also report that the superflip requires at least 19 face turns.   I
got this result using Shamir's algorithm, which Mike Reid describes briefly
in his message <9412162233.AA27627@ducie.ptc.com>.  To repeat him: Shamir's 
method allows you to generate in lexicographic order all permutations st,
where s and t are elements of lists S and T of permutations, respectively,
while using only space proportional to the sum of the sizes of the lists.
What I did was to first check that the superflip f couldn't be done in 11 or
fewer face turns (easy) and to then try to solve f=stuv, where s and v have
4 face turns and t and u have 2 to 5 face turns.  This is done by scanning
through the (ordered) lists of all st's and all f v^(-1) u^(-1)'s and checking
to see if there is a common element. Shamir's method then has to be applied to
S and T and to V and T, where T is a list of permutations with 2 to 5 face
turns, S is a list of permutations s with 4 face turns, and V is a list of
permutations f v^(-1), where v has 4 face turns.  The number of candidates
for s and v can be reduced by making use of the fact that f is central, has 
order 2, and is invariant under conjugation by the symmetry group of the cube. 
The computation took 52 hours of CPU time on an SGI Crimson (R4000 50/100 MHz
CPU.)  More than half the CPU time is spent composing permutations and updating
priority queues (see below.)

Some have expressed concern that Shamir's method is a memory hog.  Applying 
it to S and T requires a rather complicated tree of permutations in T and a 
priority queue of permutations in S.  I used the wreath product representation 
of the cube group (so `permutation' is something of a misnomer,) and my memory 
usage was then as follows:

Per element of S:
48 bytes      permutation s in S (can be shared with other S's and T's)
40 bytes      composition st   (not absolutely necessary, but speeds things up)
16 bytes      pointers internal to the queue and to an element t of T
---------
104 bytes

Per element of T:
48 bytes      permutation t in T (can be shared, as before)
8 bytes       pointer immediately above t
<=16 bytes    Amortized cost of higher-up regions of the tree
----------
<=72 bytes

The T tree is not altered during traversal, so if you are applying the method
to S and T and V and T simultaneously you just need one T tree.  
All told, my memory usage was around 46M.

Looking for a 20 face turn representation by this method would probably take
around 59M of memory and 710 hours of CPU time (on this machine.)
-- 
David Moews                                dmoews@xraysgi.ims.uconn.edu


From @uconnvm.uconn.edu:dmoews@xraysgi.ims.uconn.edu  Sun Jan 22 17:06:42 1995
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From: dmoews@xraysgi.ims.uconn.edu (David Moews)
Message-Id: <9501232205.AA27851@xraysgi.ims.uconn.edu>
To: cube-lovers@ai.mit.edu, dmoews@xraysgi.ims.uconn.edu
Subject: Symmetry reductions of the superflip


As I mentioned in my last message, I used symmetries to reduce the
number of candidate sequences for the superflip.  Here's how:

Suppose we have a sequence for the superflip that has at least 4 syllables.
(Here, a syllable is a maximal sequence of commuting face turns, i.e., a
maximal sequence of face turns on the same axis.)  The sequence of axes
in these syllables must look like

(1) Z X ... X Y, (2) Z Y ... X Y, (3) X Z ... X Y, or (4) X Y ... X Y,

for some distinct axes X, Y, and Z.  Remember that the superflip is
central, so we can cyclically permute the sequence of syllables.  If
doing this always results in pattern (4), we only use two axes, but
this can't flip any edges; hence, we can get (1), (2) or (3).  By inverting
the (order 2) superflip we can change (2) to (3).  Then we have (1)
or (3).  By applying a symmetry of the cube, we can let X, Y and Z be
the FB, UD, and LR axes, respectively.

We still have some of the symmetry group to work with, namely the set of
the eight symmetries of the cube that fix all cube axes.  If we need to,
we can apply a 180-degree rotation of the cube about the UD or LR axes,
which lets us restrict the first FB syllable to 9 of the 15 possibilities;
then, rotating about the FB axis, we can do the same for the last UD syllable.
Finally, we can reflect the cube through the plane between R and L; this lets
us restrict the first LR syllable to 9 possibilities, although it expands the
number of possibilities for the last UD and first FB syllables to 10 each.

Some more estimated runtimes for my Shamir implementation: 20 CPU hr for 
a 20 qtw superflip; 190 CPU hr for a 22 qtw superflip.
-- 
David Moews                                   dmoews@xraysgi.ims.uconn.edu


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From: dmoews@xraysgi.ims.uconn.edu (David Moews)
Message-Id: <9501232221.AA27990@xraysgi.ims.uconn.edu>
To: cube-lovers@ai.mit.edu, dmoews@xraysgi.ims.uconn.edu
Subject: Symmetry reductions of the superflip (oops)

I forgot to say: You should cyclically permute the sequence of face turns in
the superflip to begin with to ensure that the sequence does not begin and end
with turns on the same axis.  Only then can you be sure that you have one of 
the forms (1)...(4).
-- 
David Moews                                    dmoews@xraysgi.ims.uconn.edu


From mschoene@math.rwth-aachen.de  Tue Jan 24 17:15:49 1995
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From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: cube-lovers@life.ai.mit.edu
In-Reply-To: "Jerry Bryan"'s message of Wed, 18 Jan 1995 11:53:21 EST <9501181654.AA27651@life.ai.mit.edu>
Subject: Re: Re: Re: Models for the Cube

Jerry Bryan wrote in his message of 1995/01/18

    Perhaps you could clarify your generating system and its respecting
    of costs a bit further?

I shall try.  This is partly to clarify things for myself.
So excuse me if it is a bit long and formal.


The Puzzle
----------

First let me formalize what kind of puzzles we are talking about.

Assume that we have a set G of *basic states* with a unique basic solved
state <id> (we may need to add further marks to distuingish all states).

Assume that we have a set S of *simple operations*, such that each <s>
transforms each basic state <g> into another, which we write as <g> <s>.
Assume that for each simple operation <s> there is an *inverse operation*
<s>' in S, such that if <g_1> <s> = <g_2> then <g_2> <s>' = <g_1>.

A *process* is simply a sequence <s_1> ... <s_l> of simple operations.
It induces an operation on the set G of basic states, i.e., it again
transforms each basic state <g> into another, through the definition
<g> (<s_1> ... <s_l>) := ((...(<g> <s_1>) ... ) <s_l>).

I say that a process <p> *solves* a basic stage <g>, if <g> <p> = <id>.
Assume that G and S are such that for each basic state <g> there is a
process <p> that solves <g> (technically this means that the group of
operations on G operates transitively on the set G of basic states).
The goal of the puzzle is to find such a process for each basic state.

If a process <p> solves a basic state <g>, then the inverse process <p>',
that we get by reversing the sequence and replacing each simple operation
by its inverse, transforms <id> into <g>, and I say <p>' *effects* <g>.

Finally assume that G and S are such that if processes <p_1> and <p_2>
effect the same basic state <g>, then they induce the same operations,
i.e., then <h> <p_1> = <h> <p_2> for any other basic state <h> in G
(technically this means that the group of operations on G operates
regularly on the set G of basic states).

This then allows us to identify the set G of basic states and the group
of operations on G.  This in turn allows us to view G as a group, with
the generating system S.

Sometimes it is necessary to distuingish between a process, the operation
it induces, and the basic state it effects.  When such a distinction is
unnecessary, I shall simply talk about the *element* of G.

I call the group G a model for the *basic puzzle*.

For an example take Rubik's cube.  There are 24 * 43252003274489856000
basic states.  The simple operations are the face turns (or quarter face
turns if you prefer) and the rotations of the rigid cube.  Obviously each
state can be solved and it is easy to verify that two processes that
effect the same basic state induce the same operation.  The group CG is
the model for the basic Rubik's cube.


The Essential Puzzle
--------------------

Next we need to add the notion that, while all basic elements (states)
are different, some are more different than others.

Assume that there is a subgroup of *essentially free elements* F.
Assume that we shall consider two elements <g_1> and <g_2> to be
*essentially equal*, if there is an essentially free element <f> in F
such that <g_1> <f> = <g_2>.  Then the sets of essentially equal
elements are of course exactely the (left) cosets (<g> F).  We shall
call such a coset an *essential element*.

Assume that we have selected for each coset (<g> F) a representative
r(<g> F).  Define the operation '*' on the set of left cosets by
(<g_1> F) * (<g_2> F) := (r(<g_1> F) r(<g_2> F) F), i.e., the product of
two cosets is the coset containing the product of the representatives.

If the set of essential elements with this operations is a group,
then I call this group a model for the *essential puzzle*.

Note that there is no guarantee that we can select the representatives
such that '*' defines a group.  That is, for some puzzles there may not
be a group model for the essential puzzle.  This for example happens if
G = < (1,2), (1,2,3,4) > is the symmetric group on four points and
F = < (1,2)(3,4) >.

Furthermore there is no guarantee that each selection that gives us a
group, gives us the same group.  That is, for some puzzles there may be
different nonisomorphic group models for the essential puzzle.  This for
example happens if G = < (1,2), (1,2,3,4) > and F = < (1,2), (1,2,3) >
is the stabilizer of one point, in which case C4 = < (1,2,3,4) > and
V4 = < (1,2)(3,4), (1,3)(2,4) > are models.

But if F is a normal subgroup, then it doesn't matter how we select the
representatives; the operation '*' always gives us the factor group.

Also if there is a supplement E of F (i.e., a subgroup E, such that
G = E F and E <intersection> F = { <id> }), then selecting the elements
of E as the representatives of the cosets gives us a group model for the
essential puzzle.  This group model is of course isomorphic to E
(but note that there can be nonisomorphic supplements).

But there can be group models for the essential puzzle that come neither
from the factor group nor from a supplement.  This for example happens if
G = < (1,2,3,4,5,6,7,8), (2,8)(3,7)(4,6) > is the dihedral group of size
16 and F = < (1,2)(3,8)(4,7)(5,6), (1,5)(2,6)(3,7)(4,8) >, in which case
there are again two nonisomorphic models.

For example in the case of Rubik's cube we usually consider two basic
states to be equal if they are related by a rotation of the rigid cube.
That is the subgroup F of essentially free elements is the subgroup C of
rotations of the rigid cube in this case.  The usual group model for the
essential Rubik's cube is the supplement G of C.


The Costs
---------

Finally we need the notion of costs, both for the basic puzzle and
the essential puzzle.

In general let G be an arbitrary group, X a generating system for G that
is closed under taking inverses (that is, for each <x> in X, <x>^-1 is
also in X), and Z a subgroup of G.  Then roughly the cost of an element
<g> in G wrt. X and Z is the length of a shortest process effecting <g>,
where we only count the generators <x> in X, not the terms <z> in Z.

More formally define G_(0) := Z and G_(l+1) := G_(l) <union> (G_(l) X Z).
Since X is a generating system, there is a finite d such that G = G_(d)
(and the smallest such d is called the diameter of G wrt. X and Z).
We say that elements which lie in G_(l) but not in G_(l-1) have *cost* l.

For the basic puzzle group G, we obviously use the cost function
cost_G for G wrt. the generating system S and the subgroup F.
So the cost of a basic state <g> is the length of a shortest process
effecting <g>, where we count only the simple operations <s> in S,
not the elements for the essentially solved operations <f> in F.  Note
that of course the costs of two essentially equal elements are equal.

For the essential puzzle group E, we want to find a cost function
cost_E that preserves this cost.  That is, we want 
cost_E( <g> F ) = cost_G( <g> ).  So the question is, can we
choose a generating system X_E of E and a subgroup Z_E of E such that
the cost function for E wrt. to X_E and Z_E has the above property.

If you think about it for a moment, you will see, that we actually
don't have any choice if we require cost_E( <g> F ) = cost_G( <g> ).
Namely Z_E is simply the subgroup of E of the elements with cost 0.
But if cost_E( <g> F ) is 1, then cost_G( <g> ) must be 1,
so <g> must be in F, and then (<g> F) is F, i.e., the identity of E.
Likewise X_E is simply the subset of E of the elements with cost 1.
But if cost_E( <g> F ) is 1, then cost_G( <g> ) must be 1,
so <g> must have the form <f_1> <s> <f_2>, so we see that X_E must be
the set { (<f> <s> F) | <f> in F, <s> in S }.

Now it turns out that those two conditions are not only necessary, they
are in fact sufficient.  That is, if cost_E is the cost function for E
wrt. the generating system X_E = { (<f> <s> F) | <f> in F, <s> in S }
and the subgroup Z_E = { (<id> F) }, then cost_E preserves the cost,
i.e. cost_E( <g> F ) = cost_G( <g> ).

So for example in the case of Rubik's cube the cost of an element <g> of
CG is the length of shortest process effecting <g>, where we only count
the face turns (or only the quarter face turns), not the rotations of the
rigid cube.  A model for the essential Rubik's cube is the supplement G
generated by the face turns (without the rotations).  Because for each
process we can collect all the rotations of the rigid cube at the end of
a process, we see that the set X_E is simply the set of the face turns.

For another example take the edges only cube CG[E].  The cost of an
element <g> is again the length of the shortest process effecting <g>,
where we only count the face turns, not the rotations.  A model for the
essential edges only cube is E = <L[E],D[E],F[E],B[E]> (i.e., a subgroup
of CG[E] that fixes one edge) (Confusion warning: running out of letters
again, the first E stands for *e*ssential, the others for *e*dges).  If
we want the cost function for E to preserve the costs in CG[E], we must
take the generating system X_E = { (<f> <s> F) | <f> in F, <s> in S } =
{ L[E],D[E],F[E],B[E],r[E]'*R[E],u[E]'*U[E], L[E]', ... }.  Otherwise
some elements of E, will appear more expensive than they really are.


Summary
-------

Assume we have a puzzle modelled by a group G of basic elements with
a generating system S of simple elements and their inverses.

Assume that we have a subgroup F of essentially free elements, and that
we call two elements essentially equal if the lie in the same left coset
of F in G.

Given a system of representatives for the cosets of F in G, we define
the product of two cosets as the coset containing the product of the
representatives of the two cosets.  If this multiplication turns G/F
into a group, we call this group a model for the essential puzzle.

Note that such a model need not exist, i.e., it may happen that no system
of representatives gives a group.  Also such a model need not be unique,
i.e., different systems of representatives may give nonisomorphic models.

The cost of an element in G is the length of a shortest process effecting
this element, where we count only the simple elements from S, not the
essentially free elements from F.

Then the cost of an element in G is equal to the cost of its left coset
in G/F wrt. the generating system { (<f> <s> F) | <f> in F, <s> in S }.


Have a nice day.

Martin.

PS. I admit this more than a *bit* formal and long.
Count it as my submission for the understatement-of-the-year award ;-)

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From mschoene@math.rwth-aachen.de  Tue Jan 24 18:01:35 1995
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Date: Tue, 24 Jan 95 23:58 WET
From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: cube-lovers@life.ai.mit.edu
In-Reply-To: "Jerry Bryan"'s message of Wed, 18 Jan 1995 11:53:21 EST <9501181654.AA27651@life.ai.mit.edu>
Subject: Re: Re: Re: Models for the Cube

Jerry Bryan wrote in his message of 1995/01/18

    It is well known that G[E] must have an equal number of even and
    odd permutations.  If we generate G[E] as <Q[E]>, it is also the case
    that there are just as many positions an even distance from Start as
    an odd distance from Start because the parity of the distance from
    Start is the same as the parity of the permutation if we restrict
    ourselves to quarter turns.

If you view the elements of G[E] as permutations on 24 facelets, then all
elements are even.  But if you forget for the moment the orientations of
the edges, and view each element as only permuting the 12 edges, then
there is an equal number of even and odd elements in G[E].  And then,
since each quarter face turn cyclically permutes four edges, there must
indeed be just as many states an even distance from Start as there are
states an odd distance from Start.

Jerry continued

    But in the computer search for God's Algorithm for edges only cubes,
    there were not equal numbers of positions an even distance from Start
    as an odd distance from Start.  The computer search used the coset model
    G[E]/C[E], where this notation means the set of cosets of C, *not* the
    factor group.  In and of itself, the mismatch between the number of
    positions at an even distance from Start and at an odd distance from
    Start should not pose a problem.  It is not clear to me what it means
    to speak of the "parity" of a coset of C, and half of each coset will
    be even and the other half will be odd.  Hence, it is not clear that
    a particular coset must *a priori* be an odd or even distance from
    Start.

Allow me to translate this into the language I introduced in my other
message.  G[E] is the model for the basic puzzle.  C[E] is the subgroup
of essentially free elements.  We consider two elements of G[E] to be
essentially equal if they lie in the same left coset of C[E] in G[E].
The cost of an element <g> of G[E] (or the distance from the Start), is
the length of a shortest process effecting <g>, where we only count the
quarter face turns, not the rotations of the rigid cube.  It is clear
that two essentially equal elements have equal costs.

Jerry continued

    But if we map each coset to an element of G[E], it is meaningful to
    speak of the parity of the element of G[E].  And if the elements of
    G[E] to which we map the cosets form a subgroup, then there must be an
    equal number of odd and even elements in the subgroup
    (unless they are all even?!).  If the mapping respects
    costs in the sense of maintaining the same distance from Start, then
    I don't understand how we can avoid a conflict between the equal
    number of even and odd permutations in the subgroup of G[E] and
    the unequal number of even and odd distances from Start in the coset
    model G[E]/C[E].

Pick one edge, say the right upper edge.  Then each coset of C[E]
contains one representative that fixes this edge.  The set of those
representative is a subgroup U, generated by L[E],D[E],F[E],B[E].
Formally U is a supplement for C[E] in G[E].  It is a model for the
essential edges only cube.  And indeed it contains the same number
of even and odd permutations.  So far so good.

But we must now be carefull how we measure the cost of elements in U.

If we measure by taking the length of a shortest process effecting such
an element <u> in U using only the generators L[E],D[E],F[E],B[E] (and
their inverses), then some elements will appear more expensive than they
really are.  For example r[E]'*R[E] should have cost 1, but there is no
process of length 1 effecting this element using only the generators
above.  So we must take a larger generating system.  As I described in my
other message, the generating system to take is the set of all elements
in U that should have cost 1.  This gives us the generating system
{ L[E], D[E], F[E], B[E], r[E]'*R[E], u[E]'*U[E], L[E]', ... }.
Still, so far so good.

So where is the problem?  Well the new generators r[E]'*R[E] and
u[E]'*U[E] are *even* permutations.  And since not all generators
are odd permutations any more, the argument that the number of
elements with even cost and the number of elements with odd cost
must be equal, simply doesn't work anymore.

Have a nice day.

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From VIKING1969@delphi.com  Tue Jan 24 19:56:54 1995
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unsubsribecribe list viking1969

From rodrigo@lsi.usp.br  Fri Jan 27 21:05:43 1995
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From: Rodrigo de Almeida Siqueira <rodrigo@lsi.usp.br>
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Subject: Robot using the Cube - WWW
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Yes. We have a robot that knows how to do it!

The World Wide Web page with images of the 2 robots of DAIA
(Departament of Artificial Intelligence and Automation) of
LSI (Laboratory of Integrated Systems) at USP (University of
Sao Paulo, Brazil) is here:

http://www.lsi.usp.br/usp/rod/images/cube/rubik_cube.html

Of course, it's Netscape enhanced.
This page has also a link to a page I made with the X-Rubik's Cube
software (xrubik.html in the same address), with inlined images
of the software.

Have fun.

Rodrigo de Almeida Siqueira
Webmaster at Universidade de Sao Paulo, Brazil.
personal URL (full of things):
http://www.lsi.usp.br/usp/rod/rod.html


From @mail.uunet.ca:mark.longridge@canrem.com  Sun Jan 29 23:48:39 1995
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Subject: Skewb thoughts
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1021.5834.0C1CC7AE@canrem.com>
Date: Sun, 29 Jan 1995 23:40:00 -0500
Organization: CRS Online  (Toronto, Ontario)

Extract from Martin's very detailed skewb analysis:

>Then the group CG = < C, G > is the set of all positions a puzzler
>could observe.  There are 24 solved position in CG (solved up to
>rotations).
>
>The group CG has size 2 * 6!/2 * ((3^4*4!/2) * (3^4*4!/2) / 3^2)
>               |CG|     = 75,582,720

Note that:      |CG| /24 =  3,149,280

>The group G has size 6!/2 * ((3^4*4!/2) * (3^4*4!/2) / 3^2)
>               |G|      = 37,791,360

Note that:      |G|  /12 =  3,149,280


The number of positions both David Singmaster and Tony Durham
(the inventor) find for the skewb is 3,149,280.

If we use only one tetrad of the skewb then GAP also finds this
number:

      corners                                  centers
      (each turn permutes 4)           (each turn permutes 3)
skewb := Group(
    ( 1,11,17) ( 2,12,20)( 4,10,18)(22, 6,14) (25,27,29),
    ( 2,10,22) ( 1, 9,23)( 3,11,21)(17, 5,15) (25,27,30),
    ( 4,14,20) ( 1,15,19)( 3,13,17)( 7,11,23) (25,28,29),
    ( 6,12,18) ( 5,11,19)( 7, 9,17)(21, 1,13) (26,27,29)
);;

Size (skewb);
>   3149280

Mr. Singmaster had indicated in his last Cubic Circular that we may
determine the skewb's orientation if only one of the tetrads are
moved.

By moving first one tetrad and then the other we can easily
change the skewb's orientation in space.

Martin finds that the diameter of the skewb is 11 moves, with
perhaps 90 antipodes. The idea that the skewb has 2 positions
at 0 moves is rather odd, but I think if we divide Martin's
table by 2 we should get the answer for visually distinguishable
states for a skewb fixed in orientation.


------------------------------------------------------------

I'm still trying to tame the dodecahedron.
-> Mark <-

From mreid@ptc.com  Mon Jan 30 11:04:29 1995
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Subject: Re: superflip requires 20 face turns
Content-Length: 5225

as promised, i reran the face turn search and collected data along
the way.  total run time was 143.7 cpu hours (just a shade under
six days) on an HP 9000 series 715, 100MHz clock.  so this machine
is a bit faster than some of the others that helped out on the
original run.  (there was also some overlap between different
machines.)

these figures also show why the cases starting with  R1 L1  and
R1 L3  are so slow:  many more sequences in stage 1 to check.

mike


statistics below:

depth in        number of times          solutions
stage 1        stage 2 is reached          found

superflip  R1 F1:

 9f                      2               23f, 22f
10f                    942               21f, 20f
11f                  19180
12f                 255716               19f
13f                2967572
14f               32053344
15f              330809868               18f

superflip  R1 F2:

10f                    948               22f, 21f
11f                  19032               20f, 19f
12f                 251312
13f                2913516
14f               31351632
15f              321390912               18f

superflip  R1 F3:

 9f                      2               21f
10f                    942
11f                  19180               20f, 19f
12f                 255716
13f                2967572
14f               32053344
15f              330809868               18f

superflip  R1 U1:

 9f                      2               21f
10f                    826
11f                  17140               20f, 19f
12f                 231130
13f                2702062
14f               29334386
15f              303689360

superflip  R1 U2:

10f                    812               22f
11f                  17080               21f
12f                 232452               20f, 19f
13f                2735896               18f
14f               29776092
15f              307802732

superflip  R1 U3:

 9f                      2               23f, 22f
10f                    826
11f                  17140               21f, 20f
12f                 231130               19f
13f                2702062
14f               29334386
15f              303689360

superflip  R1 L1 F1:

 7f                     96               20f, 19f
 8f                   1824               18f
 9f                  21768
10f                 229616
11f                2267728
12f               21151120               17f
13f              189906448
14f             1660964664

superflip  R1 L1 F2:

 8f                    384               22f, 21f, 20f
 9f                   8448               19f
10f                 113440               18f
11f                1268896
12f               12941696
13f              124124064
14f             1141576128

superflip  R1 L1 F3:

 7f                     96               20f, 19f
 8f                   1824               18f
 9f                  21768
10f                 229616
11f                2267728
12f               21151120               17f
13f              189906448
14f             1660964664

superflip  R1 L1 F3:

 7f                     96               20f, 19f
 8f                   1824               18f
 9f                  21768
10f                 229616
11f                2267728
12f               21151120               17f
13f              189906448
14f             1660964664

superflip  R1 L1 U1:

 9f                    832               22f, 21f, 20f
10f                  16912               19f
11f                 224248               18f
12f                2597672
13f               27754280
14f              279317240

superflip  R1 L1 U2:

 8f                    384               22f, 21f, 20f
 9f                   8448               19f, 18f
10f                 113440               17f
11f                1268896
12f               12941696
13f              124124064
14f             1141576128

superflip  R1 L1 U3:

 9f                    832               21f, 20f
10f                  16912               19f
11f                 224248               18f
12f                2597672
13f               27754280
14f              279317240

superflip  R1 L3 F1:

 7f                     96               21f, 20f, 19f
 8f                   1824               18f
 9f                  21768
10f                 229616
11f                2267728
12f               21151120               17f
13f              189906448
14f             1660964664

superflip  R1 L3 F2:

 8f                    384               22f, 21f, 20f
 9f                   8448               19f
10f                 113440
11f                1268896
12f               12941696
13f              124124064               18f
14f             1141576128

superflip  R1 L3 U1:

 9f                    832               23f, 22f, 21f, 19f
10f                  16912
11f                 224248
12f                2597672               18f
13f               27754280
14f              279317240               17f

superflip  R1 L3 U2:

 8f                    384               21f, 20f
 9f                   8448               19f
10f                 113440               18f
11f                1268896
12f               12941696
13f              124124064
14f             1141576128


From mschoene@math.rwth-aachen.de  Tue Jan 31 08:58:26 1995
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Date: Tue, 31 Jan 95 11:43 WET
From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: cube-lovers@life.ai.mit.edu
In-Reply-To: Mark Longridge's message of Sun, 29 Jan 1995 23:40:00 -0500 <60.1021.5834.0C1CC7AE@canrem.com>
Subject: Re: Skewb thoughts

Mark Longridge wrote in his e-mail message of 1995/01/29

    Extract from Martin's very detailed skewb analysis:

    >Then the group CG = < C, G > is the set of all positions a puzzler
    >could observe.  There are 24 solved position in CG (solved up to
    >rotations).

    >The group CG has size 2 * 6!/2 * ((3^4*4!/2) * (3^4*4!/2) / 3^2)
    >               |CG|     = 75,582,720

    Note that:      |CG| /24 =  3,149,280

    >The group G has size 6!/2 * ((3^4*4!/2) * (3^4*4!/2) / 3^2)
    >               |G|      = 37,791,360

    Note that:      |G|  /12 =  3,149,280

    The number of positions both David Singmaster and Tony Durham
    (the inventor) find for the skewb is 3,149,280.

Right.  The SKEWB has 75582720 basic states.  Just as with the cube, we
consider two basic states to be essential equal if the differ only by a
rotation of the rigid cube.  Since there are 24 rotations of the rigid
cube, the SKEWB has 3149280 = 75582720/24 essential states.

Mark continued

    If we use only one tetrad of the skewb then GAP also finds this number:

    ##    corners                                  centers
    ##    (each turn permutes 4)           (each turn permutes 3)
    skewb := Group(
        ( 1,11,17) ( 2,12,20)( 4,10,18)(22, 6,14) (25,27,29),
        ( 2,10,22) ( 1, 9,23)( 3,11,21)(17, 5,15) (25,27,30),
        ( 4,14,20) ( 1,15,19)( 3,13,17)( 7,11,23) (25,28,29),
        ( 6,12,18) ( 5,11,19)( 7, 9,17)(21, 1,13) (26,27,29)
    );;

    Size (skewb);
    >   3149280

In my message on the SKEWB I used the subgroup H generated by LUB, LUF,
RUB, and RUF.  As I noted, this subgroup has a nontrivial intersection
with the subgroup C of rotations of the rigid cube.  Thus it is *not*
a model for the essential SKEWB.

The subgroup that Mark uses, which is generated by RUF, RUB, LUF, and
RDF is much better.  It has trivial intersection with C and is a model
for the essential SKEWB.

Note however, that the corners corresponding to the four generators for
this subgroups do *not* form a tetrad.  They are the corner RUF and the
three adjacent corners.  In particular, those four generators do not fix
the positions of the four corners; the generator RUF permutes the three
corner cubies at RUB, LUF, and RDF.  This subgroup has 7 other conjugated
subgroups, corresponding to the 7 other possible choices of the first
generator (the one that is adjacent to the other 3 generators).

So allow me to use the subgroup H generated by RUF, LUB, RDB, and LDF.
The corresponding four corners do form a tetrad.  This H also has trivial
intersection with C and also has size 3149280.  Thus it also is a model
for the essential SKEWB.  Note that those four generators never change
the positions of the four corner cubies.  This subgroup is ``almost
normal''; it has only 1 other conjugated subgroup, corresponding to the
stabilizer of the other tetrad.

Mark continued

    Mr. Singmaster had indicated in his last Cubic Circular that we may
    determine the skewb's orientation if only one of the tetrads are moved.

I am not certain that I understand what this means.  One possible
interpretation is, that for each state g of the SKEWB we can easily find
the rotation x of the rigid cube, such that (g x) is in the subgroup H.
That means that for each state g we can easily find how to rotate the
rigid cube, so that the rotated state can be solved using only the four
generators above.  But this is obvious.  Since the four generators do not
change the positions of the four corner cubies of the tetrad, the
rotation x must be the one that puts those four corner cubies to their
home positions.

Mark continued

    By moving first one tetrad and then the other we can easily change the
    skewb's orientation in space.

This comment I don't understand at all.  Could you clarify it a bit?

Mark continued

    Martin finds that the diameter of the skewb is 11 moves, with perhaps 90
    antipodes. The idea that the skewb has 2 positions at 0 moves is rather
    odd, but I think if we divide Martin's table by 2 we should get the
    answer for visually distinguishable states for a skewb fixed in
    orientation.

Right.  The diameter of the SKEWB is 11 moves and there are 90 essential
different antipodes.  The essential SKEWB does *not* have 2 states at 0
moves, only the subgroup H which I used has 2 essentially solved states.
This is not odder than the notion that the basic SKEWB has 24 essentially
solved states.  And yes, if you divide the numbers in my table by 2, you
get the table for the essential SKEWB.

I rerun the computation using the new subgroup H as a model for the
essential SKEWB.  Here is the output.

     0       1       0      0      0      0      0      0      0  0  1
     1       8       0      0      0      0      0      0      8  0  0
     2      48       0      0      0      0      0      0     48  0  0
     3     288       0      0      0      0      0      0    288  0  0
     4    1728       0      0      0      0      0    120   1608  0  0
     5   10248       0      0      0     36    377   1322   8513  0  0
     6   59304      12     87    662   2217   7561  15698  33067  0  0
     7  315198    4331  16897  37723  61161  76931  66997  51158  0  0
     8 1225483  515249 311594 186221 115830  65096  25012   6481  0  0
     9 1455856 1384909  61839   8280    708     95     25      0  0  0
    10   81028   80938     90      0      0      0      0      0  0  0
    11      90      90      0      0      0      0      0      0  0  0

Since the group is smaller it run faster and also used less memory.
Using some additional tricks, I could cut down the time to 40 seconds and
the memory needed to 2.5 megabytes.

As you can see, the numbers in the first column are exactely half of the
corresponding numbers in my previous message (as was expected).  The
numbers in the other columns are close to half of the corresponding
numbers in my previous message but not exactely identical.  I have to
rethink what those numbers mean and how they relate to the corresponding
numbers for the basic SKEWB.

Have a nice day.

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From mschoene@math.rwth-aachen.de  Wed Feb  1 07:05:22 1995
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Date: Wed, 1 Feb 95 13:02 WET
From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: cube-lovers@life.ai.mit.edu
In-Reply-To: "Martin Schoenert"'s message of Tue, 24 Jan 95 23:12 WET <m0rWtTd-00025hC@hobbes.math.rwth-aachen.de>
Subject: Re: Re: Re: Re: Models for the Cube

In my previous message I introduced the basic puzzle and the essential
puzzle and their models.  There is one more thing I would like to say
about models for the cube.

The situation is the same as in my previous message.

Assume we have a puzzle modelled by a group G of basic elements with
a generating system S of simple elements and their inverses.

Assume that we have a subgroup F of essentially free elements, and that
we call two elements essentially equal if the lie in the same left coset
of F in G.

Given a system of representatives for the cosets of F in G, we define
the product of two cosets as the coset containing the product of the
representatives of the two cosets.  If this multiplication turns G/F
into a group, we call this group a model for the essential puzzle.

Note that such a model need not exist, i.e., it may happen that no system
of representatives gives a group.  Also such a model need not be unique,
i.e., different systems of representatives may give nonisomorphic models.

The cost of an element in G is the length of a shortest process effecting
this element, where we count only the simple elements from S, not the
essentially free elements from F.

Then the cost of an element in G is equal to the cost of its left coset
in G/F wrt. the generating system { (<f> <s> F) | <f> in F, <s> in S }.


The Real Puzzle
---------------

Assume that we call two elements <g_1> and <g_2> in G to be *really
equal* if there are essentially free elements <f_1> and <f_2> in F, such
that <f_1> <g_1> <f_2> = <g_2>.  Then the sets of really equal elements
are the double cosets (F <g> F).  Obviously two really equal elements
have the same costs.  The set of all double cosets is usually written
F\G/F.  Let us call the size of F\G/F the *real size* of the puzzle.

Note that each double coset (F <g> F) is a disjoint union of single
cosets of F.  On the other hand let H be the largest subgroup of F such
that (H <g> F) = (<g> F).  Then the number of single cosets in the double
cosets is the index of H in F.  So we see that the size of each double
coset is a multiple of |F| and a divisor of |F|^2.

Furthermore note that the size of the double coset (F <g> F) is |F|,
i.e., (F <g> F) is a single coset, if and only if <g> normalizes F,
i.e., (<g> F) = (F <g>).

Now in general F\G/F is notoriously badly behaved.  For example the size
of F\G/F is in general not a divisor of the size of G.  So there is no
hope that we can turn F\G/F into a group that has anything to do with G.
That means that we cannot model the real puzzle with a group.

But that shouldn't stop us from investigating this real puzzle.  One
question we can ask is, what is the real size of the puzzle?  Another
question might be, what are the elements that lie in small double cosets.

For an example, let us again take Rubik's cube.  Here we have a very nice
description of when two states are really equal.  This is because the
premultiplication with <f_1> corresponds to a recoloring of the cube and
the postmultiplication with <f_2> corresponds to a rotation of the cube.
So two states are really equal if we can recolor and rotate one state to
get the other state.  This idea has come up several times in this list,
for example in Jerry Bryan's message about 1152 fold symmetry
(see Jerry_Bryan__1152-fold_Symmetry_and_24-fold_Symmetry of 1993/12/08).

Note that we must be a little bit more careful with the real cube than
with the essential cube.  With the essential cube it doesn't matter
whether the subgroup of essentially free elements is the subgroup C of
rotations of the rigid cube or the subgroup M of rotations and
reflections of the rigid cube.  That is the group G generated by the face
turns is a model for the essential cube in both cases, i.e., G is a
supplement of C in CG and is also a supplement of M in MG.  But for the
essential cube it does matter which subgroup we take.

Dan Hoey computed the real size of M\MG/M as 901083404981813616
(see Dan_Hoey__The_real_size_of_cube_space of 1994/11/04).
He used the fact that, since the supplement G is a normal subgroup of MG,
the number of double cosets in M\MG/M is equal to the number of conjugacy
classes in G under the operation of M.  With the same idea we can compute
the real size of C\CG/C as 1802166805653080256, which is slightly less
than 2*901083404981813616.

Dan Hoey and Jim Saxe searched for elements <g> such that the double
coset (M <g> M) has size 48, 96, or 192
(see Dan_Hoey__Symmetry_and_Local_Maxima_(long_message) of 1980/12/14).
More precisely, they classified the elements for which the subgroup H
that fixes the single coset (<g> M) operates transitively on the set
of quarter face turns, because those elements must be local maxima
(except for the identity).  They found 4 double cosets of size 48,
10 double cosets of size 96, and 12 double cosets of size 196,
or 72 local maxima.

Have a nice day.

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From BECK@vax88a.pica.army.mil  Wed Feb  1 08:19:36 1995
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Date: Wed, 1 Feb 1995 8:19:57 -0500 (EST)
From: BECK@vax40a.pica.army.mil
To: cube-lovers@ai.mit.edu
Message-Id: <950201081957.40400150@VAX40A.PICA.ARMY.MIL>
Subject: magic jack


a friend of mine sent me the following:

>  At a recent visit to Games People Play we saw a Magic Jack from
Fun Tech. The design was similar to a Rubiks Cube. Have you seen?


ANYBODY OUT THERE KNOW MORE of this puxxle ????


From GPATYK@aol.com  Wed Feb  1 11:08:10 1995
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Date: Wed, 1 Feb 1995 11:04:51 -0500
From: GPATYK@aol.com
Message-Id: <950201110450_10093643@aol.com>
To: cube-lovers@life.ai.mit.edu
Subject: cubelovers-request@ai.ai.mit.edu

thank you
>greg

From @ibm.co.hennepin.mn.us:WF5435@CO.HENNEPIN.MN.US  Thu Feb  2 13:38:26 1995
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To: cube-lovers@life.ai.mit.edu
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SUBSCRIBE cube-lovers-request Jill Lyons

From @mail.uunet.ca:mark.longridge@canrem.com  Fri Feb  3 03:32:33 1995
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Sender: CRSO.Cube@canrem.com
Subject: More skewb thoughts
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1030.5834.0C1CD83B@canrem.com>
Date: Fri, 3 Feb 1995 00:40:00 -0500
Organization: CRS Online  (Toronto, Ontario)

The following is a follow up to the discussion on the SKEWB
containing quotes from messages of Martin and myself.

>>  The number of positions both David Singmaster and Tony Durham
>>  (the inventor) find for the skewb is 3,149,280.

> Right.  The SKEWB has 75582720 basic states.  Just as with the cube,
> we consider two basic states to be essential equal if the differ only
> by a rotation of the rigid cube.  Since there are 24 rotations of
> the rigid cube, the SKEWB has 3149280 = 75582720/24 essential states.

I just noticed that the number of states of the pyraminx (with vertex
rotations included) also equals 75,582,720. (933,120 * 3^4)

>>Mark continued
>>
>>If we use only one tetrad of the skewb then GAP also finds this
>>  number:
>>
>>  ##    corners                                  centers
>>  ##    (each turn permutes 4)           (each turn permutes 3)
>>  skewb := Group(
>>      ( 1,11,17) ( 2,12,20)( 4,10,18)(22, 6,14) (25,27,29),
>>      ( 2,10,22) ( 1, 9,23)( 3,11,21)(17, 5,15) (25,27,30),
>>      ( 4,14,20) ( 1,15,19)( 3,13,17)( 7,11,23) (25,28,29),
>>      ( 6,12,18) ( 5,11,19)( 7, 9,17)(21, 1,13) (26,27,29)
>>  );;

I'll amend 'each turn permutes 4' to 'rotates one, 3-cycles the
others', although half the corners do move in some way. Also
the operators are RUF, RUB, RDF and lastly LUF.

The corner LDB remains fixed, so just like the 2x2x2 cube we are
fixing a corner.

>Note however, that the corners corresponding to the four generators for
>this subgroups do *not* form a tetrad.  They are the corner RUF and the
>three adjacent corners.

My computer Webster says that a tetrad is 'A group of four'. Perhaps
there is another meaning in geometry or group theory? Certainly I
agree with the 2nd statement.

>Snip< I concur with the Martin's next paragraph (excuse the editing)

>So allow me to use the subgroup H generated by RUF, LUB, RDB, and LDF.
>The corresponding four corners do form a tetrad.

Martin, could you clarify the use of tetrad here?  :-)

>More Snips<

>> Mr. Singmaster had indicated in his last Cubic Circular that we may
>> determine the skewb's orientation if only one of the tetrads are
>> moved.

> I am not certain that I understand what this means.   >Snip<

I'm going to re-read the article and think about this some more.

>> By moving first one tetrad and then the other we can easily change
>> the skewb's orientation in space.

> This comment I don't understand at all.  Could you clarify it a bit?

I shall amend by comment >> above to:

 By moving first one half of the puzzle and then the other we can
easily change the skewb's orientation in space.

 As stated in Douglas Hofstadter's column in the July 1982 issue of
Scientific American, the skewb is a deep-cut puzzle, that is
the part of the puzzle that is being operated on is no
smaller than the stationary portion.

-> Mark <-

From BRYAN@wvnvm.wvnet.edu  Sat Feb  4 12:08:24 1995
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From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Level 11, Whole Cube, Q-turns


 Distance        X  Branching      {m'Xm} Branching Ratio   Local
 from                Factor                Factor    of       Max
 Start                                              Cubes
                                                  to Classes
 0               1                      1                       0
 1              12  12.000              1   1.000    12.000     0
 2             114   9.500              5   5.000    22.800     0
 3           1,068   9.368             25   5.000    42.720     0
 4          10,011   9.374            219   8.760    45.712     0
 5          93,840   9.374          1,978   9.032    47.442     0
 6         878,880   9.366         18,395   9.300    47.778     0
 7       8,221,632   9.355        171,529   9.325    47.931     0
 8      76,843,595   9.347      1,601,725   9.338    47.976     0
 9     717,789,576   9.341     14,956,266   9.338    47.993     0
10   6,701,836,858   9.337    139,629,194   9.336    47.997
11  62,549,615,248   9.333  1,303,138,445   9.333    47.9992

This chart includes a column for local maxima, which my charts
usually do not.  With all the data kept in files instead of memory,
it is not a very natural calculation to determine which positions
are local maxima.  With the data in memory, for any position X
you would calculate the 12 neighbors Xq, and immediately determine
which of the 12 neighbors were one move closer to Start.  It is
easy to identify local maxima in this situation.  With the data
written to files, the neighbors Xq are sorted before determining
which are closer to Start, and there is no (easy) way to relate
a given Xq back to its original X.

However, let me describe the sorting/merging process in a little
more detail.  There is a file containing all cubes X such that
|X|=n.  The neighbors Xq are written to a file.  The file
is sorted, with duplicates deleted.  (Actually, the problem is so
large that there are *many* files containing the neighbors Xq.
Each file is sorted, and then the results are merged).  Finally, the
resulting file is matched against another file containing all
cubes Y such that |Y|=(n-1).  Any matches are deleted, and whatever
is left over becomes the file containing all cubes Z such that
|Z|=(n+1).  The difference between the number of matches deleted
and the number of cubes in the n-1 file is the number of local
maxima of length n-1.  (Remember that all the X's and Y's and Z's
and Xq's are representative elements of M-conjugacy classes.)

The last time through this process, I generated neighbors of level
10 to create level 11, sorted and deleted duplicates, and matched
against level 9 deleting matches.  Hence, the last level for which
I have local maxima information is level 9.

There are not any local maxima through level 9.  I am not really
expecting any until Pons Asinorum at level 12.  However, it would
be nice to verify that Pons Asinorum is the shortest local maximum.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From mschoene@math.rwth-aachen.de  Sun Feb  5 19:07:52 1995
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From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: cube-lovers@life.ai.mit.edu
In-Reply-To: Mark Longridge's message of Fri, 3 Feb 1995 00:40:00 -0500 <60.1030.5834.0C1CD83B@canrem.com>
Subject: Re: More skewb thoughts

Mark Longridge wrote in his e-mail message of 1995/01/29

    If we use only one tetrad of the skewb then GAP also finds this number:

    ... shows how GAP computes the size of this subgroup as 3149280 ...

I replied in my e-mail message of 1995/01/31

    Note however, that the corners corresponding to the four generators for
    this subgroups do *not* form a tetrad.

Mark Longridge replied in his e-mail message of 1995/02/03

    My computer Webster says that a tetrad is 'A group of four'.
    Perhaps there is another meaning in geometry or group theory?

Sorry for the confusion.  There is no special meaning of the word
``tetrad'' that I am aware of, neither in geometry nor in group theory.

I interpreted Mark's ``one tetrad of the skewb'' as ``four corners of the
skewb that are the corners of a regular tetrahedron'', probably because
of the common prefix ``tetra''.

Note that it is problematic to interpret Mark's ``one tetrad of the
skewb'' as ``one group of four corners of the skewb'', since not for all
groups of four corners of the skewb the subgroup generated by the
corresponding generators has size 3149280, for example the subgroup
generated by the generators corresponding to the four corners of the up
face, which I used in my first e-mail message, has size 6298560.

All in all there are 70 different ways to select a 4-tuple of corners
of the cube.  Up to rotation there are 6 (essentially) different types.

      +------*    +------*    *------*    +------*    +------*    +------*
     /|     /|   /|     /|   /|     /|   /|     /|   /|     /|   /|     /|
    *------+ |  *------* |  *------* |  *------* |  *------* |  +------* |
    | |    | |  | |    | |  | |    | |  | |    | |  | |    | |  | |    | |
    | *----|-+  | +----|-+  | +----|-+  | *----|-+  | +----|-+  | *----|-+
    |/     |/   |/     |/   |/     |/   |/     |/   |/     |/   |/     |/
    +------*    +------*    +------+    +------+    *------+    *------+

The first type is what I proposed in my last e-mail message.
The 4 corners are the corners of a regular tetrahedron.
There are 2 such 4-tuples (corresponding to the 2 tetrahedrons).
The subgroup generated by the corresponding generators has size 3149280,
and is a model for the essential SKEWB (i.e. the SKEWB up to rotations).
It leaves the 4 corners RUB, LUF, RDF, and LDB at their home positions,
so it is easy to determine the orientation of an arbitrary state (in the
sense that it is easy to determine how to rotate this state so that the
rotated state is in the subgroup generated by RUB, LUF, RDF, and LDB).

The second type is what Mark proposed in his last e-mail message.  There
are 8 such 4-tuples (corresponding to the 8 possible choices for the
``central'' corner RUF).  The subgroup generated by the corresponding
generators also has size 3149280, and is a model for the essential SKEWB.
It fixes the corner LDB, so it is again easy to determine the orientation
of an arbitrary state.

The third type is what I proposed in my first e-mail message.  There are
6 such 4-tuples (corresponding to the 6 faces).  The subgroup generated
by the corresponding generators has size 6298560, so it is too large by a
factor of 2 to be a model for the essential SKEWB.  It fixes the face
center of the down face.  In this case it is not easy to determine the
orientation of an arbitrary state (in the sense above it is in fact
impossible, because for every state there are *two* rotations such that
the rotated cube is in the subgroup generated by RUB, LUB, RUF, and LUF).

There are 24 4-tuples of the fourth type.  The subgroup generated by
the corresponding generators has size 9447840, so it is too large by
a factor of 3 to be a model for the essential SKEWB.  It fixes nothing.

There are 24 4-tuples of the fifth type, and 6 4-tuples of the sixth
type.  The subgroups generated by the corresponding generators have size
37791360, so it is in fact the group G generated by all 8 generators.

So if we want a model for the essential SKEWB then we have to take one of
the first two types.  My preference is for the first type, which I think
is more special than the second.  Namely there are only 2 such 4-tuples,
whereas there are 8 4-tuples of the second type.  Correspondingly there
are only 2 such subgroups (which are both normal in the group G generated
by 8 generators, though they are conjugated in the group CG generated by
the 8 generators and the rotations of the rigid cube).

Mark Longridge wrote in his e-mail message of 1995/01/29

    By moving first one tetrad and then the other we can easily change
    the skewb's orientation in space.

I replied in my e-mail message of 1995/01/31

    This comment I don't understand at all.  Could you clarify it a bit?

Mark Longridge replied in his e-mail message of 1995/02/03

    I shall amend by comment >> above to:

    By moving first one half of the puzzle and then the other
    we can easily change the skewb's orientation in space.

I interpret that as follows.  By first using an element g1 from the
subgroup H1 generated by RUB, RUF, LUF, and RDF, and then an
element g2 from the subgroup H2 generated by RDB, LDB, LDF, and LUB,
we can acchieve *any* rotation c of the rigid cube.

Now it is true that by first using an element g1 from the
subgroup H1 generated by RUB, RUF, LUF, and RDF, and then an
element g2 from the subgroup H2 generated by RDB, LDB, LDF, and LUB,
we can acchieve any element of the group G generated by all 8 generators
(this follows from the fact that |G| = |H1| |H2| / |H1 <intersect> H2|).

But G contains only one half of the rotations of the rigid cube.  So of
the 24 rotations of the rigid cube we can only achieve 12 (the even ones
if we view the rotations as permutations of the 4 diagonals of the cube).
This becomes obvious if you note that the 8 generators never exchange
the two sets of four corners that form tetrahedrons.

Mark also wrote in his e-mail message of 1995/02/03

    I just noticed that the number of states of the pyraminx (with vertex
    rotations included) also equals 75,582,720. (933,120 * 3^4)

Is this just by chance, or is there some connection between those two
puzzles?  Could you describe the pyraminx?

Have a nice day.

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From @mail.uunet.ca:mark.longridge@canrem.com  Fri Feb 10 11:54:22 1995
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Subject: The Pyraminx Lost
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1035.5834.0C1CF175@canrem.com>
Date: Thu, 9 Feb 1995 23:55:00 -0500
Organization: CRS Online  (Toronto, Ontario)

        Subgroup Sizes of the Pyraminx Octahedron
        ------------------------------------------

8 * 9 = 72 facelets (triangles)

The standard Pyraminx Octahedron has 8 faces, 6 vertices, and 12 edges.
It's vertices rotate. One may imagine a "Master" Pyraminx Octahedron
with edge AND face rotations as well.

Christoph Bandelow has a version of the Pyraminx Octahedron (I call
it "Octa" for short) which has no tips.

Size of Groups without rotating vertex tips:

Name            Subgroup                         # of Elements
----            --------                         -------------

OCT1            <U>                                          4
OCT2            <U, D>                                      16
OCT3            <U, D, F>                          116,121,600
OCT4            <U, D, F, B>                   613,312,204,800
OCT5            <U, D, F, B, L>            502,269,581,721,600
OCT6            <U, D, F, B, L, R>       2,009,078,326,886,400

Size of Groups with rotating vertex tips:

Name            Subgroup                         # of Elements
----            --------                         -------------

OCT1            <U>                                         16
OCT2            <U, D>                                     256
OCT3            <U, D, F>                        7,431,782,400
OCT4            <U, D, F, B>               157,007,924,428,800
OCT5            <U, D, F, B, L>        514,324,051,682,918,400
OCT6            <U, D, F, B, L, R>   8,229,184,826,926,694,400

Approximately  8.2 * 10^18  ..so still less than the 3x3x3 cube

 The number of elements increases by a factor of 4^N for
each successive group if we include the trivial vertex rotations.

        A Skewb Summary
        ---------------

 Without repeating Martin's results on the skewb, (which I concur
with) here is a quick summary on Skewb facts:

It is impossible for any face piece to turn in place 90 degrees.
It is impossible to flip a single face piece 180 degrees.
It is impossible to transpose 2 face pieces.

The Skewb has no non-trivial centre.
The SuperSkewb has non-trivial centre with all 6 face pieces
 rotated 180 degrees.


        The Mystery of the Five Pyraminxi
        ---------------------------------

  Or perhaps that should be Pyraminxes... but I can not resist
comparing the Five Pyraminxes to the Five Wizards of J.R.R Tolkien,
due to their mysterious nature.

  We are probably all familar with the Popular Pyraminx created
by Uwe Meffert. What really confounds me is that Dr. Ronald Turner-
Smith kepts referring to the 5 pyraminxes in ad literature and
his book "The Amazing Pyraminx". The Master Pyraminx I understand,
it has all the basic properties of the standard popular pyraminx
plus all 6 of it's edges can rotate 180 degrees (which flips one
edge, transposes 2 tips, and swaps 2 pairs of interior edge pieces)
giving a total number of permutations of 446,965,972,992,000.

  Then there is the mysterious "Senior Pyraminx" (this is like
Tolkien's Blue Wizards no one knows about). I can only speculate
on the properties of the Senior Pyraminx having never read a
description, and never seen the physical puzzle itself. The only
fact on the Senior Pyraminx I am sure about is that it has less
permutations than the Master Pyraminx. My theory is that the
Senior Pyraminx has all the properties of the standard pyraminx
plus it can rotate SOME of it's edges but not all 6 like the
Master Pyraminx (perhaps one or two?).

 Perhaps Mr. Singmaster, who has seen magic solid variants from
all over the world, can shed some light on the matter!

 -> Mark <-
 Email: mark.longridge@canrem.com

From villa@esaii.upc.es  Fri Feb 10 15:26:58 1995
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help


From ccw@eql12.caltech.edu  Fri Feb 10 19:14:16 1995
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From: ccw@eql12.caltech.edu
Message-Id: <950210161411.25007867@EQL12.Caltech.Edu>
Subject: I have a non-standard Pyraminx. (its magic number is 5)
To: cube-lovers@ai.mit.edu
X-St-Vmsmail-To: ST%"cube-lovers@ai.mit.edu"

It is a shallow cut Dodecahedron.
Could this be one of the "Five Pyraminxes"?
I don't remember what it's official name is, though I thought it was
the Master Pyraminx.  (I will have to check my collection at home)

I only ever sow these once, in J.C Penny's, thankfully I bought one.

I always viewed this as a combining of 2 puzzles, Alexander's Star and
a round one, whose name escapes me at the moment.

My copy of this puzzle has 2 yellow and 2 red faces.  I think they ran
out of colors.
This means that if I am not carefull I can appear to have 2 edges switched.
This is more apparant then real because the stickers for each face have
ridges which can be used to make the proper choice.

There are 12 faces, which can be independantly turned by 72 degrees.
Faces do not move with respect to each other.
There are 20 corners which can only be in Even Permutations.
Corners are like the cube, trios can be spun in the same direction, pairs
can be spun in opposite directions.
There are 30 edges which can only be in Even Permutations.
Edges can flipped in pairs, just like the normal cube.


group size should be
                       20     30
        30!    20!    3      2
        --- *  --- * ---  * ---
        2       2     3      2

        Edge   Corn  Spin   Flip

for the supergroup, increase by a factor of
         12
        5

From mschoene@math.rwth-aachen.de  Sun Feb 12 19:02:40 1995
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Date: Mon, 13 Feb 95 00:59 WET
From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: cube-lovers@life.ai.mit.edu
In-Reply-To: "Martin Schoenert"'s message of Tue, 31 Jan 95 11:43 WET <m0rZG32-00025cC@hobbes.math.rwth-aachen.de>
Subject: Re: Re: Skewb thoughts

I wrote in my e-mail message about the SKEWB of 1995/01/15

    Here is the table for H.  The first column contains the lenght.  The
    second column contains the number of positions of that length in H.
    The third column contains the number of positions of that length that are
    local maxima, i.e., the number of positions <pos> such that for no
    generator <gen> the position <pos>*<gen> is longer.  The fourth column
    contains the number of positions such that for one generator <gen> the
    position <pos>*<gen> is longer.  And so on.  So the eleventh column
    contains the number of positions <pos> such that for all eight generators
    <pos>*<gen> is longer (this happens of course only for the 2 solutions).

    length #pos  #loc max
     0       2       0      0      0      0      0      0      0      0      2
     1      16       0      0      0      0      0      0     16      0      0
     2      96       0      0      0      0      0      0     96      0      0
     3     576       0      0      0      0      0      0    576      0      0
     4    3456       0      0      0      0      0    240   3216      0      0
     5   20496       0      0      0     48    729   2766  16953      0      0
     6  118608      48    161   1231   4228  14779  32993  65168      0      0
     7  630396    8266  33358  76349 121363 153892 137755  99413      0      0
     8 2450966 1025322 621763 381098 234661 128570  47822  11730      0      0
     9 2911712 2768641 126056  15344   1422    199     50      0      0      0
    10  162056  161876    180      0      0      0      0      0      0      0
    11     180     180      0      0      0      0      0      0      0      0

        ... note that this is the H = < RUF, RUB, LUF, LUB > ...

And in my e-mail message of 1995/01/31

    I rerun the computation using the new subgroup H as a model for the
    essential SKEWB.  Here is the output.

     0       1       0      0      0      0      0      0      0  0  1
     1       8       0      0      0      0      0      0      8  0  0
     2      48       0      0      0      0      0      0     48  0  0
     3     288       0      0      0      0      0      0    288  0  0
     4    1728       0      0      0      0      0    120   1608  0  0
     5   10248       0      0      0     36    377   1322   8513  0  0
     6   59304      12     87    662   2217   7561  15698  33067  0  0
     7  315198    4331  16897  37723  61161  76931  66997  51158  0  0
     8 1225483  515249 311594 186221 115830  65096  25012   6481  0  0
     9 1455856 1384909  61839   8280    708     95     25      0  0  0
    10   81028   80938     90      0      0      0      0      0  0  0
    11      90      90      0      0      0      0      0      0  0  0

    As you can see, the numbers in the first column are exactely half of the
    corresponding numbers in my previous message (as was expected).  The
    numbers in the other columns are close to half of the corresponding
    numbers in my previous message but not exactely identical.  I have to
    rethink what those numbers mean and how they relate to the corresponding
    numbers for the basic SKEWB.

        ... note that this is now H = < RUF, LUB, RDB, LDF > ...

The reason that the numbers in the other columns of the second table are
not exactely half of the corresponding numbers in the first table is
rather simple.  They are *both wrong*.

The correct numbers for H = < RUF, LUB, RDB, LDF > are as follows

     0       1       0       0       0       0       0       0       0  0  1
     1       8       0       0       0       0       0       0       8  0  0
     2      48       0       0       0       0       0       0      48  0  0
     3     288       0       0       0       0       0       0     288  0  0
     4    1728       0       0       0       0       0       0    1728  0  0
     5   10248       0       0       0       0     120     240    9888  0  0
     6   59304       0       0      84      96    1740    6024   51360  0  0
     7  315198     198     144    3600    9768   42900   94344  164244  0  0
     8 1225483   15783   73016  199808  316776  343992  208584   67524  0  0
     9 1455856 1001960  365792   74976   11760    1224     144       0  0  0
    10   81028   80308     720       0       0       0       0       0  0  0
    11      90      90       0       0       0       0       0       0  0  0

and the correct numbers for H = < RUF, LUB, RUB, LUF > are exactely twice
as large.

I figured out what those numbers mean.  It is all rather simple.
Everybody who thought about them probably knows everything that follows.
I use the terms from my last few messages about models for the cube.

The basic states of cost 1 are exactely the elements in (F S F), where F
is the subgroup of essentially free elements, and S is the set of simple
elements (the set of generators) of G.  Not all those elements need to be
different.  Assume that there are <n_b> basic states of cost 1.  Each
basic state <g> has <n_b> neighbors, namely the elements <g> (F S F).

The set of neighbors of each state is obviously a union of right cosets
of F.  Furthermore if <g_1> and <g_2> are essentially equal, then there
sets of neighbors are equal.  So we can map the whole concept to the
essential model G/F.

Recall that in the essential model G/F the set of elements of cost 1 was
exactely the set X = { (<x> F) | <x> in F S }.  Assume that there are
<n_e> essential states of coset 1.  Then each essential state (<g> F) has
<n_e> essential neighbors, namely the essential elements (<g> F) X.

We can now count how many of the basic neighbors of a basic state <g>
have smaller cost than <g>.  If all <n_b> basic neighbors have smaller
cost than <g>, then we call <g> a basic local maximum.

Likewise we can count how many of the essential neighbors of the
essential state (<g> F) have smaller cost than (<g> F).  If all <n_e>
essential neighbors have smaller cost than (<g> F), then we call (<g> F)
an essential local maximum.

It is easy to see that a basic element <g> is a basic local maximum if
and only if (<g> F) is an essential maximum.

In fact in most cases the number of basic states that have smaller cost
than <g> is simply (<n_b> / <n_e>) times the number of essential states
that have smaller cost than (<g> F).  One sufficient condition for this
to happen is, that S is invariant under conjugation by S and that all
classes have the same length.

This condition is met for the SKEWB, so the numbers in the first table
*had* to be twice the numbers in the second table.  Sorry about any
confusion I caused.

Have a nice day.

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From mschoene@math.rwth-aachen.de  Sun Feb 12 19:07:52 1995
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Date: Mon, 13 Feb 95 01:05 WET
From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: cube-lovers@life.ai.mit.edu
In-Reply-To: Mark Longridge's message of Thu, 9 Feb 1995 23:55:00 -0500 <60.1035.5834.0C1CF175@canrem.com>
Subject: Re: The Pyraminx Lost

I have never seen the Pyraminx.  I wonder if somebody could tell me
whether the picture I put together from the various comments made about
the Pyraminx is correct?

I am fairly certain that the Pyraminx is a regular tetrahedron.  In the
solved state each of the four faces shows only one of the four colours.

I think the Pyraminx is cut along 8 planes, two planes perpendicular to
each of the four heights (i.e., the four lines that connect a corner with
the center of the opposite face).  I think for the Pyraminx those planes
intersect the height at about 2/5 and 3/5 of the length of the height.

Those planes cut the Pyraminx into 15 pieces (1 central piece, 4 corners,
4 inner pieces, and 6 edes), which are all visible.  Each face is cut
into 10 facelets by them as follows.

                   +
                  / \
                 /   \
                /     \
               /       \
              /         \
             +-----------+
            / \         / \
           /   \       /   \
          +-----+-----+-----+
         / \     \   /     / \
        /   \     \ /     /   \
       /     \     +     /     \
      /       \   / \   /       \
     /         \ /   \ /         \
    +-----------+-----+-----------+

The Pyraminx Star was descibred as a Pyraminx without the centers.
So I guess each face of the Pyraminx Star looks as follows.

                   +
                  / \
                 /   \
                /     \
               /       \
              +---------+
             / \       / \
            /   \     /   \
           /     \   /     \
          /       \ /       \
         +---------*---------+
        / \       / \       / \
       /   \     /   \     /   \
      /     \   /     \   /     \
     /       \ /       \ /       \
    +---------+---------+---------+

The Pyraminx Snub was described as a Pyraminx without the tips.
So I guess each face of the Pyraminx Snub looks as follows.

             +-----------+
            / \         / \
           /   \       /   \
          +-----+-----+-----+
           \     \   /     /
            \     \ /     /
             \     +     /
              \   / \   /
               \ /   \ /
                +-----+

I have no idea what Pyraminx Senior and the Pyraminx Master look like.

Have a nice day.

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From ad@dcs.st-andrews.ac.uk  Mon Feb 13 04:21:08 1995
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To: cube-lovers@ai.mit.edu, ccw@eql12.caltech.edu
From: ad@dcs.st-andrews.ac.uk
Subject: Re: I have a non-standard Pyraminx. (its magic number is 5)

>My copy of this puzzle has 2 yellow and 2 red faces.  I think they ran
>out of colors.

Reminds me of the joke:

"Have you heard about the Irish version of Rubik's Cube?"
"No. Tell me."
"All the fAces were green."


--
Tony Davie                Computer Science            __
Tel: +44 334 463257       St.Andrews University    __/\_\
Fax: +44 334 463278       North Haugh           __/\_\/_/
ad@dcs.st-and.ac.uk       St.Andrews           /\_\/_/\_\
                          Scotland             \/_/\_\/_/
                          KY16 9SS                \/_/\_\
                                                     \/_/

A lottery is a tax on the mathematically challenged



From mschoene@math.rwth-aachen.de  Mon Feb 13 05:57:30 1995
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Date: Mon, 13 Feb 95 11:54 WET
From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: Cube-Lovers@ai.mit.edu
In-Reply-To: "Jerry Bryan"'s message of Sat, 21 Jan 1995 12:44:42 EST <9501212203.AA05451@life.ai.mit.edu>
Subject: Re: Re: superflip in quarter turn metric

Jerry Bryan wrote in his e-mail message of 1995/01/21

    I am taking the liberty of CC'ing Cube-Lovers as well.  A search tree
    giving distances from Start calculates d(I,IY) for all positions IY
    in the domain of inquiry.  With an X-rooted tree, distances are of
    the form d(X,XZ) for all positions XZ in the domain of inquiry.
    In general, it is not the case that d(I,IY)=d(X,XY).  Hence,
    we cannot simply take Z=Y, and elements of the form XY do not produce
    the X-rooted tree.  Therefore, to perform
    two half-depth searches to connect I and X, you really do have to
    construct a standard Start-rooted tree as well as an X-rooted tree.
    You are looking for a position IY=XZ such that |IY|=|XZ|=|X|/2.

First we have to make certain that we agree how to multiply permutations.
If I write <g> <h>, then I mean *first* do <g> and *then* do <h>.
So I compute the image of a point <p> under the permutation (<g> <h>) by
first computing the image of <p> under <g> and then computing the image
of that point under <h>.  For this order of multiplication it is usual
to write <p>^<g> for the image of a point <p> under a permutation <g>
(instead of writing <g>(<p>), which would be better for the other order).
For this order of multiplication we must define conjugation of <g> by <h>
as <g>^<h> := <h>^-1 <g> <h> (instead of <g>^<h> := <h> <g> <h>^-1).

In this notation, it is certainly true that d(<id>,<h>) = d(<g>,<g><h>).
This is because each process that transforms <id> to the state <h>,
will also transform <g> to <g><h>, and likewise each process that
transforms <g> to <g><h> will also transform <id> to <h>.

In a certain sense we don't need this though.  What you are looking
for is a process <p> that effects the state <g>, i.e., <id> <p> = <g>.
If such a process of length 22 exists, then there exist two processes
<p_1> and <p_2> of length 11, such that <id> <p_1> <p_2> = <g>.
We can rewrite this as <id> <p_1> = <g> <p_2>^-1.  Let T be the set of
elements reachable from <id> by a process of length 11.  Note T^-1 = T.
So we see that if there is a process of length 22 effecting <g>,
then the intersection (<id> T) <inter> (<g> T) must be nonempty.
As mentioned above, you can interpret the set (<g> T) as the set
of elements at distance 11 from <g>, but you don't have to.

Now for the superflip <z> you even have d(<id>,<h>) = d(<z>,<h><z>),
since <h><z> = <z><h> because the central <z> commutes with every <h>.
Put differently this means that (<z> T) = (T <z>), i.e., instead
of multiplying each element of T from the left by <z>, you can instead
multiply each element from the right.

Have a nice day.

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From ncramer@bbn.com  Mon Feb 13 06:57:48 1995
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Message-Id: <9502131157.AA28197@life.ai.mit.edu>
Date:     Mon, 13 Feb 95 6:54:06 EST
From: Nichael Cramer <ncramer@bbn.com>
To: ad@dcs.st-andrews.ac.uk
Cc: cube-lovers@ai.mit.edu
Subject:   6:45 am Monday Morning Humor [was: I have a non-standard Pyraminx...]

>Date: Mon, 13 Feb 95 09:17:26 GMT
>From: ad@dcs.st-andrews.ac.uk
>Subject: Re: I have a non-standard Pyraminx. (its magic number is 5)
>
>>My copy of this puzzle has 2 yellow and 2 red faces.  I think they ran
>>out of colors.
>
>Reminds me of the joke:
>
>"Have you heard about the Irish version of Rubik's Cube?"
>"No. Tell me."
>"All the fAces were green."

Well, I have, sitting on the shelf beside me here in my office as I type, a
cube with six blue faces.  What nationality is this I suppose?  (Or is this
because it was -4 [F] when I left home in Vermont this morning?)

>Tony Davie                Computer Science            __
>Tel: +44 334 463257       St.Andrews University    __/\_\
>Fax: +44 334 463278       North Haugh           __/\_\/_/
>ad@dcs.st-and.ac.uk       St.Andrews           /\_\/_/\_\
>                          Scotland             \/_/\_\/_/
>                          KY16 9SS                \/_/\_\
>                                                    \/_/

A scots cube consists wholly, I presume, of alternating red and green
cubies?

N

From mouse@collatz.mcrcim.mcgill.edu  Mon Feb 13 18:30:35 1995
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Date: Mon, 13 Feb 1995 18:30:32 -0500
From: der Mouse <mouse@collatz.mcrcim.mcgill.edu>
Message-Id: <199502132330.SAA11948@Collatz.McRCIM.McGill.EDU>
To: cube-lovers@ai.mit.edu
Subject: Re: Re: superflip in quarter turn metric

In response to a note of mine,

>> A search treegiving distances from Start calculates d(I,IY) for all
>> positions IY in the domain of inquiry.  With an X-rooted tree,
>> distances are of the form d(X,XZ) for all positions XZ in the domain
>> of inquiry.  In general, it is not the case that d(I,IY)=d(X,XY).

whereupon what's-his-name :-) responds

> In this notation, it is certainly true that
> d(<id>,<h>) = d(<g>,<g><h>).  This is because each process that
> transforms <id> to the state <h>, will also transform <g> to <g><h>,
> and likewise each process that transforms <g> to <g><h> will also
> transform <id> to <h>.

This is what I was trying to say in the message that started this: that
one is building a tree of all move sequences no longer than N, which is
to say a certain subset of permutations of the cube.  But these
permutations can be applied to arbitrary positions just as well as as
they can be to Start.  Any Cubist knows this; it's the basis for many
of the common solving macros: that a process that (say) swaps RF and
RB, and TF and TB, can be used to swap whatever cubies happen to be in
those cubicles, even if they aren't the RF/RB/TF/TB cubies.

					der Mouse

			    mouse@collatz.mcrcim.mcgill.edu

From BRYAN@wvnvm.wvnet.edu  Tue Feb 14 13:11:30 1995
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Date:      Tue, 14 Feb 1995 13:10:41 EST
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "der Mouse" <mouse@collatz.mcrcim.mcgill.edu>,
        "Cube Lovers List" <cube-lovers@ai.mit.edu>
Subject:   Re: Re: superflip in quarter turn metric
In-Reply-To: Message of 02/13/95 at 18:30:32 from ,
           mouse@collatz.mcrcim.mcgill.edu

Ok, let's try, try again.  I was incorrect in my response to der Mouse,
and Martin Schoenert's correction is appreciated.

The original issue was as follows:  suppose you have created a data base
for N levels of God's algorithm, beginning with Start at the root of
a tree.  With quarter-turns as generators, there is 1 position at
level zero, 12 positions at level one, 114 positions at level two,
etc.

Now, suppose you want to create a data base for N levels of God's
algorithm, starting with X at the root of the tree.  Can you simply
compose your first tree with X on an element by element basis in order
to obtain the X-rooted data base?  (You do have to be careful about
pre-multiplying vs. post-multiplying as Martin indicated!)

I stated essentially that the superflip was fairly unique in that
you could compose the superflip with the Start-rooted data base
in order to obtain the superflip-rooted data base, but that for
X-rooted data bases in general you would have to perform a complete
search starting with X.  der Mouse (correctly) noted that the
same procedure would work for any position X as for the superflip.
I (incorrectly) took exception with der Mouse, citing my fallacious
distance argument.

However, the way my data bases work, I still think that the
superflip is fairly unique in its ability to be composed with
a Start-rooted data base.  der Mouse was on the right track in
his first post when he questioned whether the issue was the
fact that the data bases only store representative elements of
M-conjugacy classes.  I responded that the storage of representative
elements was not the issue, but in fact it is.

For example, when storing only representative elements of M-conjugacy
classes, consider an F-rooted data base.  Strictly speaking, we would
have to speak of a Repr{m'Fm}-rooted data base, because F might not be
the representative element of {m'Fm}  --  it could be any of the twelve
elements of Q.  When storing only representative elements for
a Start-rooted data base, there is 1 element at level zero, 1 element
at level one, and 5 elements at level two.  For a Repr{m'Fm}-rooted
data base, there is 1 element at level zero, and six (!) elements
at level one -- namely those same elements that are at level zero and
level two of the Start-rooted data base.

Hence, we cannot take the Start-rooted data base and pre-multiply
each element by Repr{m'Fm} to obtain a Repr{m'Fm}-rooted data base.
And in general, we cannot take the Start-rooted data base and
pre-multiply each element by an arbitrary Repr{m'Xm} to obtain
a Repr{m'Xm}-rooted data base.

But for the superflip E, we *can*
take the Start-rooted data base and pre-multiply each element
by Repr{m'Em} to obtain a Repr{m'Em}-rooted data base.  In fact,
note that |{m'Em}|=1, so we must have Repr{m'Em}=E.  However, note
that for each Y in the Start-rooted data base, it is not sufficient
to calculate (EY).  Rather, we must calculate Repr{m'(EY)m}.  That is,
we know by definition that we have Y=Repr{m'Ym}, but we do not
necessarily have (EY)=Repr{m'(EY)m}.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From mbparker@share.netcom.com  Wed Feb 15 20:44:52 1995
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Date: Wed, 15 Feb 95 17:43:19 -0800
From: Michael Benjamin Parker <mbparker@share.mbparker.slip.netcom.com>
Message-Id: <9502160143.AA07457@share.mbparker.slip.netcom.com>
To: Cube-Lovers@ai.mit.edu
Subject: PUZZLE PARTY! in Orange County, CA; 1995 Feb. 18th (Sat) 7:00pm-
Reply-To: mbparker@mit.edu

Dear Cube-Lovers,

One of your members Jerry Slocum mentioned you might be interested in this
event.  If you're in the area this weekend, you're welcome to come to the
puzzle party of the MIT Club of S. California...

				PUZZLE PARTY!

Like to play games?  Need some new challenges?  What some intellectual
stimulation different from your day-to-day routine?  Then gather together your
favorite brain teasers, mental games, IQ tests, & mechanical puzzles, and come
to the PUZZLE PARTY! -- a ``puzzle-potluck'' to bring and share your favorite
puzzles.  We will re-energize our brains as we attempt to decipher the puzzles
of our fellow Southern Cal. friends.  Show off your puzzle collection --or
your puzzle-solving wizardry.  Discover new puzzles and new friends.  Fill
your mind with amusing problems and good conversation as we sit by the
fireside.  Plenty of snacks and refreshments provided.

 WHEN:	Saturday, February 18th, 7pm until...

 WHERE:	506 N. Maplewood, Orange, CA 92666 (Near 55 and 22 freeways), phone
        800-MBPARKER xLIVE.  From 55 fwy, exit west on Chapman Ave., 1st light
        turn right (north) on Tustin, 2nd light turn left (west) on Walnut,
        3rd right is Maplewood: I'm the big yellow house at the corner of
        Walnut and Maplewood.  Use the Maplewood-side entrance.

 COST:  For persons bringing puzzle(s), $4 for each MITCSC member and $6 for
        each non-member.  For ``puzzle-less'' persons, $8 for each member and
        $10 for each non-member.

 RSVP:	You may pay at the door, but please let me know you are coming if
        possible.  Please email, fax, or phone in the following info: 
        your name, address, phone, fax, email, and what you're bringing:

              ___ puzzle-bearing     members at $ 4 each: $___
              ___ puzzle-bearing non-members at $ 6 each: $___
              ___ puzzle-less        members at $ 8 each: $___
              ___ puzzle-less    non-members at $10 each: $___
              ___ <- total persons          total cost -> $___
                 total number of puzzles being brought ___

See you there!

Michael B. Parker, MIT '89

 PERMANENT CONTACT INFO (always forwards to my current address):
  721 E. Walnut Ave., Orange, CA 92667-6833 USA; mbparker@mit.edu (NeXTmail ok)
  800-MBPARKER (800-627-2753) ext: LIVE (5483), MESG (6374), and FAXX (3299)

 CURRENT ADDRESS (modified 2/6/95):
  Orange, CA; live 714-639-6436, fax 714-639-5381, vmail pager 714-413-2090.

From BRYAN@wvnvm.wvnet.edu  Thu Feb 16 03:12:38 1995
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Date:      Wed, 15 Feb 1995 21:47:07 EST
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Start-rooted vs. X-rooted Search Example

It's a fool's errand, I suppose, but in light of our recent discussions,
I created an X-rooted data base up through level 5, where X is the
representative element of {m'Fm}.  Here are the results compared to
a standard Start-rooted data base.  It is most important to realize that
both data bases contain only representative elements, and that the
results with total cubes are derived from the representative elements
by calculating the sizes of the conjugacy classes.

                  Start-                             Repr{m'Fm}-
                  Rooted                             Rooted

                       Representative                   Representative
 Level        Cubes      Elements               Cubes     Elements

 0               1             1                 12         1
 1              12             1                115         6
 2             114             5              1,068        25
 3           1,068            25             10,011       219
 4          10,011           219             93,840     1,978
 5          93,840         1,978            878,880    18,395

Performing the search in this fashion, it seems to me that there are
only four positions for which the search would look the same as for
Start -- Start itself, the Superflip, the Pons Asinorum, and the
composition of the Superflip with Pons Asinorum.

Martin Shoenert and Mark Longridge have convinced me that the Pons
Asinorum and the composition of the Superflip with Pons Asinorum
are not in the center of the cube group.  But I still believe that
the search space for all four position looks essentially the same
because these are the only four positions for which the associated
symmetry group is M.  That is, it is only these four positions for
which X=m'Xm for all m in M.

It was in this sense  --  that the search space structure using
representative elements is the same for Start and for superflip  --
that I meant that two half-depth searches using representative elements
were easy for the superflip, but would be harder for other positions.

Here is a question for Dik Winter and Mike Reid (and my apologies
if I have asked this before): have you tried your Kociemba's
algorithm programs for the composition of Pons Asinorum with
superflip?  I would find the results to be *very* interesting.

Finally, as one last fool's errand, I performed the search for
the first five levels again, this time using cubes instead of
representative elements.  With this last search, the results are
the same whether the root of the search is Start or something
else, which is the point both der Mouse and Martin Schoenert
were making.  In this chart, "level" has to be interpreted as
"distance from root", not "distance from Start".

              Start-            Repr{m'Fm}-
              Rooted              Rooted

 Level        Cubes                 Cubes

 0               1                      1
 1              12                     12
 2             114                    114
 3           1,068                  1,068
 4          10,011                 10,011
 5          93,840                 98,840


 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From mreid@ptc.com  Thu Feb 16 09:48:16 1995
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From: mreid@ptc.com (michael reid)
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To: cube-lovers@ai.mit.edu, bryan@wvnvm.wvnet.edu
Subject: superflip  composed with  pons asinorum
Content-Length: 807

jerry asks

>                               have you tried your Kociemba's
> algorithm programs for the composition of Pons Asinorum with
> superflip?

after superflip, this is probably the second most likely candidate
for an antipode.  there are only 4 positions which are fixed by
all 48 symmetries of the cube.  they are:

start,
superflip,
pons asinorum, and
the composition of superflip and pons asinorum.

obviously start is of no interest and neither is pons asinorum.

my quarter turn version found this in less than one minute:

B3 L1 D1 L1 F3 U3 D3 L1 B3 D3 F3 R1 L3 F3 U1 D1 L2 U1 D1 B2    22q, 20f

so this position is at least as close to start as superflip.

i've tested (but not yet extensively) all the local maxima given by
hoey and saxe.  i'll give a report on these some time soon.

mike

From @mail.uunet.ca:mark.longridge@canrem.com  Thu Feb 16 14:47:44 1995
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Subject: Assorted Pyraminxi
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1049.5834.0C1D048E@canrem.com>
Date: Thu, 16 Feb 1995 00:11:00 -0500
Organization: CRS Online  (Toronto, Ontario)

MS>
I am fairly certain that the Pyraminx is a regular tetrahedron.  In the
solved state each of the four faces shows only one of the four colours.

ML>
This is correct for all of the *tetrahedral* pyraminxes with only
one small exception: the Star Pyraminx has all middle pieces the
same colour. Meffert used the word "pyraminx" as a prefix to just
about all the puzzles he either conceived or planned to market.

MS>
The Pyraminx Star was described as a Pyraminx without the centers.
So I guess each face of the Pyraminx Star looks as follows.

                   +
                  / \
                 /   \
                /  V  \
               /       \
              +---------+
             / \       / \
            /   \  M  /   \
           /  E  \   /  E  \
          /       \ /       \
         +---------*---------+
        / \       / \       / \
       /   \  M  /   \  M  /   \
      /  V  \   /  E  \   /  V  \
     /       \ /       \ /       \
    +---------+---------+---------+

ML>
Actually the above diagram is a good representation of a head-on
view of the popular or standard pyraminx (I've taken the liberty
of embellishing it a little).

There are 4 Vertices (3 colours), 6 Edges (2 colours) and 12 Middle
pieces (single colur) so there are 12 + 12 + 12 = 36 facelets.

The tips (or small vertices) can rotate independently, and the larger
turn includes the rotaion of the adjacent 2 edge pieces and single
middle piece. The small tips each have 3 positions, it's adjacent
middle piece also has 3 positions, and the 6 edges obey the same
basic laws as the cube, so there are:

3^4 * 3^4 * (6!/2) * (2^6 /2) = 75,582,720 combinations or
approximately 75.5 million (993,120 for the snub version)

The math for the pyraminx octahedron is very similar, though
it has 4 positions for the 6 vertices and middle pieces
and 12 edges:

4^6 * 4^6 * (12!/2) * (2^12/2) = 8,229,184,826,926,694,400 or
approximately 8.2 quintillion.

So the snub pyraminx (or if you prefer "The Pyraminx Snub") would
look like:

              +---------+
             / \       / \
            /   \     /   \
           /     \   /     \
          /       \ /       \
         +---------*---------+
          \       / \       /
           \     /   \     /
            \   /     \   /
             \ /       \ /
              +---------+

One could imagine snub octahedrons as well.

MS>
I have no idea what Pyraminx Senior and the Pyraminx Master look like.

ML>
They are visually indistinguishable from the standard pyraminx,
however information on the Senior Pyraminx is exceedingly sketchy.

I've never seen a photograph of a Master Pyraminx in the middle
of an edge turn so I rather doubt a working prototype was ever
made, but you never know...

I'll take a stab at one more calculation...

The pyraminx hexagon has 12 corners and 18 edges and 8 centres.
Each side has 13 facelets so there are 13 * 8 = 104 total facelets.
12!/2 * 3^11 * 16! * 2^15 = 29,087,761,395,446,975,811,708,518,400,000
or approximately 29 nonillion (29^30).

-> Mark <-

From Cube-Lovers-Request@AI.MIT.EDU  Thu Feb 16 15:27:33 1995
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From: mark.longridge@canrem.com (Mark Longridge)
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Date: Thu, 16 Feb 1995 00:49:00 -0500
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Content-Length: 447
Status: R


>The tips (or small vertices) can rotate independently, and the larger
>turn includes the rotation of the adjacent 2 edge pieces and single
>middle piece.

This should read:

The tips (or small vertices) can rotate independently, and the larger
turn includes the rotation of the adjacent 3 edge pieces and 3
middle pieces.

...or approximately 29 nonillion (29^30).

This should read:

 ...or approximately 29 nonillion (29*10^30).

-> Mark <-

[ This message was missing from the mailer archive for some reason.
  I have added it manually.  - Cube-Lovers Moderator ]

From BRYAN@wvnvm.wvnet.edu  Sun Feb 19 23:21:13 1995
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From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Qturn Lengths of M-Symmetric Positions

1. The length of Start is of course 0 qturns.

2. The length of Pons Asinorum is of course 12 qturns.  This result
   has been known since 1980.  However, in the process of testing
   out half-depth searches, I tested Pons Asinorum and encountered
   a minor surprise.  There are five "halfway" positions which
   are unique up to M-conjugacy.  I only expected three.  They
   are:

      a. (RRLL)(FF)          expected -- continue BB etc.
      b. (RRLL)(FB)          expected -- continue FB etc.
      c. (RRLL)(FB')         expected -- continue FB' etc.
      d. (FB')(RRLL)         a surprise to me
      e. (RL')(FB')(RL')     a surprise to me

3. The length of Pons Asinorum composed with Superflip is
   20 qturns.  Half-depth searches through level 9 found nothing.
   A half-depth search at level 10 found ten "halfway" positions which
   are unique up to M-conjugacy.  I have my usual trouble of spinning
   tapes containing representative elements in order to find the
   processes, but I should have them in a couple of days or so.
   I expect we will find that many (or all) of them are really
   closely related, differing only by commuting in fairly trivial
   ways, just as do the five half-way positions for Pons Asinorum.

4. The length of Superflip is 24 qturns.  Half-depth searches
   through level 11 found nothing, so the length is greater than
   22.  Mike Reid has found a Superflip process of length 24.
   Hence, the length is 24.  It would be more satisfying if I
   could perform a half-depth search to level 12, but the
   problem is just too big.  Hence, I have no idea how
   many "halfway" positions there are which are unique up to
   M-conjugacy.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From mouse@collatz.mcrcim.mcgill.edu  Mon Feb 20 15:37:03 1995
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Date: Mon, 20 Feb 1995 15:36:54 -0500
From: der Mouse <mouse@collatz.mcrcim.mcgill.edu>
Message-Id: <199502202036.PAA23249@Collatz.McRCIM.McGill.EDU>
To: cube-lovers@ai.mit.edu
Subject: Re:  Qturn Lengths of M-Symmetric Positions

> 2. The length of Pons Asinorum is of course 12 qturns.  [...]
>       d. (FB')(RRLL)         a surprise to me

Surprising, but explicable.  Write PA as (RRLL)(UUDD)(FFBB).  Since PA
commutes with everything, [PA](FB') = (FB')[PA].  (I am writing X Y to
mean sequence X followed by sequence Y.)  Note also that [PA](F'B) =
(RRLL)(UUDD)(FB').  But then [PA] = [PA](FB')(F'B) = (FB')[PA](F'B) =
(FB')(RRLL)(UUDD)(FB') gives us a length-12 process for PA whose first
half is what you found.

>       e. (RL')(FB')(RL')     a surprise to me

Me too.  By elimination, its second half must be M-equivalent to the
first half, since we can look at these five half-processes as equally
being M-representatives of the reversals of the PA second halves, and
the other four first halves' second halves account for the other four.
(Got that? :-)  In fact, each first half is M-equivalent to the
reversal of the corresponding second half, which is pleasing.

After juggling letters for a while, I've been unable to "justify" this
process the way I did the one above, which leads me to suspect it may
be a fundamentally new process for PA.  Amazing, the things you can
find when you're not looking for them. :-)

> 3. The length of Pons Asinorum composed with Superflip is 20 qturns.
>    [...]  I expect we will find that many (or all) of [the midway
>    positions for this] are really closely related, differing only by
>    commuting in fairly trivial ways, just as do the five half-way
>    positions for Pons Asinorum.

Does this mean you see the fifth process for PA as a trivial
commutation of the known PA process?  How?

					der Mouse

			    mouse@collatz.mcrcim.mcgill.edu

From mreid@ptc.com  Mon Feb 20 16:38:19 1995
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From: mreid@ptc.com (michael reid)
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To: cube-lovers@ai.mit.edu, mouse@collatz.mcrcim.mcgill.edu
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der mouse writes

> > 2. The length of Pons Asinorum is of course 12 qturns.  [...]
> >       d. (FB')(RRLL)         a surprise to me
> 
> Surprising, but explicable.  Write PA as (RRLL)(UUDD)(FFBB).  Since PA
> commutes with everything, [PA](FB') = (FB')[PA].

we've had this discussion before.  pons asinorum does not "commute with
everything".  however, it does commute with "slices" and with cube
symmetries.  (which is all you're really using here.)

> >       e. (RL')(FB')(RL')     a surprise to me
> 
> Me too.  [ ... ]

the only reason i'm not surprised here is that i've read the archives.
dan hoey gave this maneuver on jan 7 1981.  also, it was recently
mentioned by chris worrell (on dec 16 1994).

> > 3. The length of Pons Asinorum composed with Superflip is 20 qturns.
> >    [...]  I expect we will find that many (or all) of [the midway
> >    positions for this] are really closely related, differing only by
> >    commuting in fairly trivial ways, just as do the five half-way
> >    positions for Pons Asinorum.
> 
> Does this mean you see the fifth process for PA as a trivial
> commutation of the known PA process?  How?

er, i think he means that "many" (i.e. 4) of the 5 maneuvers are closely
related.

however, i disagree with jerry's conjecture that the maneuvers for pons
asinorum composed with superflip will be closely related.

i guess we'll know for sure fairly soon.

mike

From BRYAN@wvnvm.wvnet.edu  Tue Feb 21 03:33:17 1995
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From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Pons Asinorum Superflipped Halfway Positions

 1. L' R  D' R  L' F  R' B' U' F'
 2. L' R  D' R  L' F  U' F' R' B'
 3. U' D  L' D  U' F  D' B' R' B'
 4. F' B  U  F' B  D' F' L' D' B'
 5. U' D  F  U' D  R' B  R  U  R
 6. U' D  L' D  U' F  R  B  D  B
 7. F' B  L  F' B  D' F' D' R' D'
 8. F' B  L  F' B  D' R' U' F' D'
 9. U' D  B' U' D  L  D  L  F  R
10. B  B  F  U' D  R  F  D  B  U

This presentation of the data is a little dissatisfying.  These ten
processes yield representative element *positions*.  With a certain
amount of analysis, you could take M-conjugates of the *processes*,
plus maybe a little astute rearranging of commuting moves, and
make the processes look a little more similar, I think.  I actually
did so for the five Pons Asinorum halfway positions, but those
were easy to futz around with compared to the ones above.

Notice, for example, that the first three moves of each of the
first nine processes are M-conjugates.  That is about as far
as I can go in my head.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From mreid@ptc.com  Tue Feb 21 11:23:45 1995
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To: bryan@wvnvm.wvnet.edu, cube-lovers@ai.mit.edu
Subject:   Pons Asinorum Superflipped Halfway Positions
Content-Length: 543

it looks like jerry gave those sequences backwards.  these sequences
pair up (as der mouse points out for pons asinorum) and we get five
essentially different maneuvers:

F3 U3 B3 R3 F1 R1 L3 D3 R1 L3   U1 D3 L3 U1 D3 F1 R1 B1 U1 F1     (1, 8)
B3 R3 F3 U3 F1 R1 L3 D3 R1 L3   U1 D3 L3 U1 D3 F1 U1 F1 R1 B1     (2, 9)
B3 R3 B3 D3 F1 U3 D1 L3 U3 D1   R1 L3 U3 R1 L3 F1 D1 B1 R1 B1     (3, 6)
B3 D3 L3 F3 D3 F3 B1 U1 F3 B1   R2 L1 U1 D3 F1 L1 U1 R1 D1        (4, 10)
R1 U1 R1 B1 R3 U3 D1 F1 U3 D1   F1 B3 D1 F1 B3 R3 B3 R3 U3 R3     (5, 7)

mike

From BRYAN@wvnvm.wvnet.edu  Tue Feb 21 14:31:14 1995
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From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Pons Asinorum Superflipped Halfway Positions (corrected)

As Mike Reid pointed out, the sequences were backwards.  The error was
a human error on my part interpreting the output of the backtracking
program which converted positions to processes.  The error occurred
when I moved the output of the program to E-mail.  Each individual
qturn operator needs to be inverted as in the following corrected table.

 1. L  R' D  R' L  F' R  B  U  F
 2. L  R' D  R' L  F' U  F  R  B
 3. U  D' L  D' U  F' D  B  R  B
 4. F  B' U' F  B' D  F  L  D  B
 5. U  D' F' U  D' R  B' R' U' R'
 6. U  D' L  D' U  F' R' B' D' B'
 7. F  B' L' F  B' D  F  D  R  D
 8. F  B' L' F  B' D  R  U  F  D
 9. U  D' B  U  D' L' D' L' F' R'
10. B' B' F' U  D' R' F' D' B' U'


 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From BRYAN@wvnvm.wvnet.edu  Wed Feb 22 13:06:05 1995
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Date:      Wed, 22 Feb 1995 11:08:56 EST
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Run Times, Storage Requirements, etc.

I have been asked to post some run times, storage requirements, etc.
related to my recent work.

The current model for the whole cube (which I treat as Q[C,E]  --
corners and edges, without storing Face centers explicitly) requires
14 bytes per position.  I know you can identify a position with the
permutation operation which yields that position when applied to Start,
but I certainly think of the data base as consisting of positions
rather than as of operations.

8 corner facelets and 12 edge facelets are stored in
5 bits each, for a total of 100 bits, or 12.5 bytes.  I waste 4 bits,
so 13 bytes are required.  I could get by with 7 corner facelets and
11 edge facelets (saving 10 bits), but it would increase the
processing time.  Most of the time (but not for every single project!)
I only store representative elements of M-conjugate classes.
My representative element function fixes one facelet,
so I could save 5 bits there (and have done so on some previous
versions of the model).  However, I wanted to be able to represent
all cubes in the same format as representative elements (remember that
representative elements of M-conjugate classes are cubes, too!), so
I go ahead and store the fixed facelet.  The 14th byte is the length
of the cube.

I am presently storing cubes of each length in a separate file.  For
short lengths, it is easy to keep the files on disk instead of tape
because they are so small.  This greatly assists the backtracking
process to convert positions to processes.  The files for qturn
lengths 0 through 8 are on disk.  Length 9 is on one tape, length 10
is on nine tapes, and length 11 is on eighty-two tapes.  Each tape
holds about 16 million positions.  All tapes are not exactly the
same length, so some tapes hold a bit more than 16 million and some
a bit less.

These are mainframe cartridge tapes.  Their capacity is not all that
great (225MB) compared to some tapes used for backup on desktop systems,
but their data transfer rate is as fast or faster than disk  --  4.5 Mbyte
per second (36 Mbit per second), vastly faster than most desktop backup
systems I know anything about.

Because the cube positions are segregated into files based on length,
it is not strictly necessary to store the length at all.  Also, the
length could be stored in the four "wasted" bits of byte 13.  Or,
the length could be stored as (length mod 3) to reduce the storage
requirements to 2 bits.  For this model, I have rejected all such
accommodations.  For example, the little project I did to compare
lengths in <U,R> to lengths in <U,R,U2,R2> was greatly assisted by
having the full length stored explicitly.  Also, it is much easier
to deal with the sort program I am using when the sort control
field (which is the cube position) is lined up on byte boundaries.
So I think the various "wasted" space in the model is a good
compromise with some practical processing requirements.

When doing normal qturn searches, generating the neighbors of each
position yields an output file which is initially (before sorting
and eliminating duplicate positions) 12 times larger than the original
file.  With the Pons Asinorum-Superflip project, each X in the data
base is pre-multiplied by PA, Superflip, or their composition rather than
post-multiplied by qturns, and the output file is exactly the same size
as the input file.  The output file has to be sorted anyway, but you know
a priori that there will be no duplicate positions.

For the Pons Asinorum-Superflip project,
it took about 90 minutes to process and sort each input tape.  For
Superflip, I had to go all the way to length 11, so it took about
125 hours to convert 82 input tapes into 82 separate files on 82
separate output tapes.  (Actually, because all tapes aren't the same
length, about half the time the output file extended a few feet of
tape onto a second tape.)  Then, the 82 separate output files had to
be merged into one large output file spanning 82 tapes.  We have
24 tape drives, but we only allow 4 to be used by one job.  Hence,
each merge can only combine 3 files into 1.  (A file can be multiple
tapes,  read one after the other.)  One bunch of merges
reduced the 82 files to 28, a second bunch of merges reduced the 28
files to 10, etc. until only 1 file was left.  Each bunch of merges had
about 82 input tapes total, and about 82 output tapes total, and
each bunch of merges took about 10 hours.  However, I had to spread the
runs over a couple of weeks to keep our operators from shooting me.
Finally, the 82 tapes for positions 11 moves from Superflip were
matched against the 82 tapes for positions 11 moves from Start, again
taking about 10 hours.  (I also checked the shorter lengths along the
way, but it didn't take anywhere near as long for the shorter lengths).

I never found any "halfway" positions for Superflip (would have required
going to length 12).  Finding the "halfway" positions for PA+Superflip
required matching the nine tapes holding positions 10 moves from Start
against the nine tapes holding positions 10 moves from PA+Superflip.
Having found them, I then did mini-searches.

First, the "halfway" positions were the root.  After 3 levels, the
results were matched against the data base for level 7 (which is a disk
file, not tape!).  The matches at that level became the root for a
new mini-search that leaped from level 7 to level 4, etc.

The backtracking search was greatly assisted by the "leaping"
procedure (first suggested to me by Dan Hoey).  The backtracking search
was also assisted by keeping each length segregated, so that the files
for lengths close to Start could be small files, and could be on disk.

The backtracking search is still very much complicated by the fact that
only representative elements are stored, so the backtracking process
rotates and reflects out from under you as you generate it.  I have a
solution for the problem.  It amounts to carrying along both
a cube and the cube's representative element during the backtracking
search, and applying qturns to the cube rather than to the representative
element.  (The representative element is still the one that has to be
matched against the data base.)  In a normal "forward" search where the
data base is being generated, the qturns are applied to the representative
elements and new representative elements are calculated immediately.
The original "unrepresentative" cubes are not used in the forward
search at all as I now do in the backward search.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From mreid@ptc.com  Wed Feb 22 18:14:54 1995
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From: mreid@ptc.com (michael reid)
Message-Id: <9502222328.AA22074@ducie>
To: bryan@wvnvm.wvnet.edu, cube-lovers@ai.mit.edu
Subject: Re: Run Times, Storage Requirements, etc.
Content-Length: 1698

jerry writes about his search for superflip in 22q:

>  [ ... ]                                   However, I had to spread the
> runs over a couple of weeks to keep our operators from shooting me.

isn't it amazing how they never seem to fully understand the importance
of this stuff!

:-)

> Finally, the 82 tapes for positions 11 moves from Superflip were
> matched against the 82 tapes for positions 11 moves from Start, again
> taking about 10 hours.  (I also checked the shorter lengths along the
> way, but it didn't take anywhere near as long for the shorter lengths).

i think this can be done much more efficiently.  well, at least if you
set things up properly in the first place.

suppose that you use an order (for sorting positions) in which the corner
configuration is more significant than the edge configuration.

then, for each position  X  on your huge list, you need to check if
repr(X superflip)  is on the list.  since superflip only affects edges,
the corner configuration of  X superflip  is the same as that of  X.
thus the same holds for  repr(X superflip)  and  repr(X) = X.  therefore,
you only need to look for  repr(X superflip)  nearby in the sorted list.

now, for each corner configuration, load all the positions on your list
with that corner configuration into memory (it shouldn't be too many),
superflip them, compute representative elements, and look for "halfway"
positions.

this way, there's no need to sort or store all the positions  11q  from
superflip (although it wouldn't be hard using this).

of course this also works for pons asinorum and its composition with
superflip.  but it does require that things were set up properly in the
beginning.

mike

From jxs3704@hertz.njit.edu  Thu Feb 23 10:47:47 1995
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From: Codey <jxs3704@hertz.njit.edu>
To: cube-lovers@life.ai.mit.edu
Subject: software for Rubik's cube
Message-Id: <Pine.ULT.3.91.950223104552.12897A-100000@hertz.njit.edu>
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	Does anyone know if there is any software to assist in solving 
the original Rubik's cube?  I just dug my cube up a couple weeks ago and am
still having problems solving it.  Thanks.

-Codey


From miz007@admin.connect.more.net  Thu Feb 23 23:50:22 1995
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To: cube-lovers@life.ai.mit.edu
From: miz007@admin.connect.more.net (Tyler Duncan)
Subject: solving the cube?
Content-Length: 339

I just can't figure out how to solve the rubiks cube. Is there a simple way
to solve it?  I know my cousin had a pattern to follow and it would work
every time, but I haven't seen him in years and can't remember the pattern.

Please reply if you know the answer.


Tyler Duncan
Miz007@admin.connect.more.net
Miz007@mail.connect.more.net



From BECK@vax88a.pica.army.mil  Fri Feb 24 10:22:36 1995
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Date: Fri, 24 Feb 1995 10:22:53 -0500 (EST)
From: BECK@vax88a.pica.army.mil
To: Cube-Lovers@ai.mit.edu
Message-Id: <950224102253.2060111e@VAX88A.PICA.ARMY.MIL>
Subject: new puzzle ???


I was given a puzzle, new to me, whose name I do not know.  It is
magic polyhedra in the shape of a cube.  The surface looks like
Yoshi's puzzle when folded, ie, 12 edge wedges going to the center of
the faces.  It turns on the corners of the cube, in groups of 3 wedges
at a time.  Feels a lot like a skewb but it is not.

The mechanism is analogous to Alexander's Star, ie, on each of the
faces of the core solid there are pyramids fixed to the faces on rods
that are free to turn.  This puzzle has an octahedron (equilateral) as
the core solid and and equilateral pyramids.  {a paper construction is
very easy to do - make an octahedron and 8 pyramids - use a rubber
band through the apex of the pyramid and through the faces of opposing
octahedron faces to attach the pyramids while allowing them to turn.
 the wedges fit between the pyramids.  I do not know how to hold them
down but if you join three and sit them over a pyramid you can get the
idea of the puzzle - i think}

1 - does anybody know what this puzzle is called ??
2 - besides this one and Alexander's Star are there other puzzles that
use this type of mechanism ??
3 - If anybody uses this mechanism with a triangular faced hexahedron
or icosahedron as the core solid please let me know or any other solid
for that matter. 

THE FUTURE IS PUZZLING,
BUT CUBING IS FOREVER !!!

pete

From ronnie@cisco.com  Fri Feb 24 11:23:40 1995
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To: BECK@vax88a.pica.army.mil
Cc: Cube-Lovers@ai.mit.edu
Subject: Re: new puzzle ??? 
In-Reply-To: Your message of "Fri, 24 Feb 1995 10:22:53 EST."
             <950224102253.2060111e@VAX88A.PICA.ARMY.MIL> 
Date: Fri, 24 Feb 1995 08:23:36 -0800
From: "Ronnie B. Kon" <ronnie@cisco.com>

> 1 - does anybody know what this puzzle is called ??
> 2 - besides this one and Alexander's Star are there other puzzles that
> use this type of mechanism ??
> 3 - If anybody uses this mechanism with a triangular faced hexahedron
> or icosahedron as the core solid please let me know or any other solid
> for that matter. 

You left out the most important question:

4 - does anybody know where other people can buy one?

				Ronnie

----------------------------------------------------------------------------
Ronnie B. Kon       | "You couldn't deny that, even if you used both hands"
ronnie@cisco.com    |
(408) 526-4592      |                   -- The Red Queen                
----------------------------------------------------------------------------


From rebdr@mailserv.mta.ca  Sun Feb 26 20:36:47 1995
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To: cube-lovers@life.ai.mit.edu
From: rebdr@mailserv.mta.ca (Chico)
Subject: Rubik's cube
X-Mailer: <Windows Eudora Version 1.4.2b16>

I am a cube lover.  I can solve both the rubik's cube and the rubik's
revenge.  My personal best at the regular cube is 45 seconds, but it usually
takes me about 1min20sec.  

The few cubes I have are getting really beat up, and I would really like to
know if there is still some way to get some new ones.

I would appreciate any help.

cubeless


From ncramer@bbn.com  Tue Feb 28 16:08:33 1995
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Date:     Tue, 28 Feb 95 16:00:25 EST
From: Nichael Cramer <ncramer@bbn.com>
To: Cube-Lovers@ai.mit.edu
Subject:  Custom Cubes

A couple of days back there was brief discussion of some custom-made cubes
(e.g. all six sides the same color, etc).

Just a couple of additional notes on this point:

Some six or eight years ago MIT museum had a major exhibit of puzzles (I
particularly remember the full-sized tangram tables).  In the last room was
a largish display case --the kind of thing your Grandmother displayed her
China in-- filled with an assortment of custom cubes.  For example, one
that I remember was one done for Charles and Diane's wedding; pictures of
HRHes on the various faces.  (I've not been back to the Museum since, so
I'm afraid I don't know the current status of these, or if they are still
on display.)

Likewise a few years back I had occassion to visit the rare books
collection at Widener library.  In the entrance-way there were several
displays (Cotton Mather's Library, a complete set of Jane Austen first
editions, that sort of thing).  One shelf had a handful of cubes as
"objets d'art": e.g. a pair of white cube (they looked like ivory, but
surely not) with words written on the various faces.  Don't recall the
artist.

Cheers
Nichael

From pontius@bartol.udel.edu  Thu Mar  2 11:52:52 1995
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To: cube-lovers@life.ai.mit.edu
From: pontius@bartol.udel.edu (Duane H. Pontius)
Subject: Info?

Greetings,

I was pointed to this group and would like some info.
Please reply to pontius@bartol.udel.edu.

Thanks,
dp



From @mail.uunet.ca:mark.longridge@canrem.com  Fri Mar  3 02:53:46 1995
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Sender: CRSO.Cube@canrem.com
Subject: Centralizers & GAP
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1070.5834.0C1D32BD@canrem.com>
Date: Fri, 3 Mar 1995 02:38:00 -0500
Organization: CRS Online  (Toronto, Ontario)

Martin wrote:

  That is, of the total 980995276800 elements in GE
  only 980995276800/332640 = 2949120 elements centralize P.
  And I used the definition of P from your e-mail of 1995/01/03,
  i.e., P = (F2 B2) (U2 D2) (L2 R2) = (F2 B2) (L2 R2) (U2 D2) = ...
  (one gets the same element independent of the order of the
   three pairs).

Ok.... let's see if I understand this "centralizer" business

(2 ^ 12 / 2 ) * 12! =  980,995,276,800 elements in GE
G is the Group of the cube
GE is the Group of the Edges Only cube
P is the element we call the Pons Asinorum (or 6 X order 2)

Only 2,949,120 elements of GE centralize P,
also only...
     2,949,120 elements of G centralize P

That is, out of all the elements of GE (or G) only 2,949,120 of them
commute with the pons asinorum.

Let's represent the Group of elements of GE that commute with P
as X. Elements of X are represented by x.

Then in all x of X, xP = Px.

But what is really troubling me is:

*  How do you represent a particular cube position (e.g. pons)
   with GAP? *

 If I could do that, then I could verify how many elements of
 the cube group commute with a given cube position:

 Size (Centralizer (cube, pons));

 Should give 2949120 (2,949,120) ... right Martin?

 and

 Size (Centralizer (sq, cube.centre));
 663552

-> Mark <-

From SHV6937@ocvaxa.cc.oberlin.edu  Mon Mar  6 08:41:02 1995
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Sub Cube-Lovers Huy Vo

From @mail.uunet.ca:mark.longridge@canrem.com  Tue Mar  7 00:02:49 1995
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Subject: New GAP insights
From: mark.longridge@canrem.com (Mark Longridge)
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Date: Mon, 6 Mar 1995 23:44:00 -0500
Organization: CRS Online  (Toronto, Ontario)

Okay, I understand the GAP conventions better now.

If we adhere to the following model for the cube:

                         +--------------+
                         |  1    2    3 |
                         |  4  top    5 |
                         |  6    7    8 |
          +--------------+--------------+--------------+--------------+
          |  9   10   11 | 17   18   19 | 25   26   27 | 33   34   35 |
          | 12  left  13 | 20 front  21 | 28 right  29 | 36  rear  37 |
          | 14   15   16 | 22   23   24 | 30   31   32 | 38   39   40 |
          +--------------+--------------+--------------+--------------+
                         | 41   42   43 |
                         | 44 bottom 45 |
                         | 46   47   48 |
                         +--------------+

Then the Pons Asinorum would be:
pons := ( 2,42)( 4,45)( 5,44)( 7,47)(10,31)(12,28)(13,29)(15,26)
        (18,39)(20,36)(21,37)(23,34);;


And the slice group would be:
slice := Group(
    ( 1, 3, 8, 6)( 2, 5, 7, 4)( 9,33,25,17)(10,34,26,18)(11,35,27,19)
    (41,46,48,43)(42,44,47,45)(14,38,30,22)(15,39,31,23)(16,40,32,24),
    ( 9,11,16,14)(10,13,15,12)( 1,17,41,40)( 4,20,44,37)( 6,22,46,35)
    (25,30,32,27)(26,28,31,29)( 3,19,43,38)( 5,21,45,36)( 8,24,48,33),
    (17,19,24,22)(18,21,23,20)( 6,25,43,16)( 7,28,42,13)( 8,30,41,11)
    (33,38,40,35)(34,36,39,37)( 3,32,46, 9)( 2,29,47,12)( 1,27,48,14)
  );;


And the anti-slice group would be:
antisl := Group(
    ( 1, 3, 8, 6)( 2, 5, 7, 4)( 9,33,25,17)(10,34,26,18)(11,35,27,19)
    (41,43,48,46)(42,45,47,44)(14,22,30,38)(15,23,31,39)(16,24,32,40),
    ( 9,11,16,14)(10,13,15,12)( 1,17,41,40)( 4,20,44,37)( 6,22,46,35)
    (25,27,32,30)(26,29,31,28)( 3,38,43,19)( 5,36,45,21)( 8,33,48,24),
    (17,19,24,22)(18,21,23,20)( 6,25,43,16)( 7,28,42,13)( 8,30,41,11)
    (33,35,40,38)(34,37,39,36)( 3, 9,46,32)( 2,12,47,29)( 1,14,48,27)
  );;

Size (antisl) = 6,144
Size (slice)  =   768

These numbers concur with Mr. Singmaster's earlier "Notes".

The following command shows that pons is at the centre of slice group:
Size (Centralizer (slice, pons)) = 768

Once again, I will refer to Martin's earlier statement about
 centralizers:

>  That is, of the total 980995276800 elements in GE
>  only 980995276800/332640 = 2949120 elements centralize P.
>  And I used the definition of P from your e-mail of 1995/01/03,
>  i.e., P = (F2 B2) (U2 D2) (L2 R2) = (F2 B2) (L2 R2) (U2 D2) = ...
>  (one gets the same element independent of the order of the
>   three pairs).

So now that I have the groups and pons element correct:
Size (Centralizer (edge, pons)) = 2,949,120

I wrote some statements before....

> Only 2,949,120 elements of GE centralize P,
> also only...
>     2,949,120 elements of G centralize P

I am only partly correct as....

Size (Centralizer (cube, pons)) = 130,026,464,870,400

As Martin said before:
>     Only one out of 332640 elements of GE (and of G) centralizes P.

Size (cube) / 332640 = 130,026,464,870,400
 or 130 trillion and change.

...the full cube group has many more elements which commute with
pons than the mere edge group!

GAP is a very function-laden beastie:

Size (Intersection (antisl, slice)) = 8
This function gives the number of elements included in both the
anti-slice and slice groups.

Naturally there is a corresponding Union function.

Since I have studied the squares group and the <U, R> group, the
number of elements in the intersection of the two are of
particular interest:

Size (Intersection (ur, sq)) = 72

And now we have a new way to check an old result  :-)
Order (cube, uturn * rturn) = 105

Of course, now that I have answered my old questions, I must
formulate new ones....

A) What is the next most commutative element (the pancentre?)
  after the 12-flip?
B) What is the least commutative element (the anticentre?) of
  the cube group?

-> Mark <-

From mreid@ptc.com  Tue Mar  7 10:11:22 1995
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From: mreid@ptc.com (michael reid)
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To: cube-lovers@ai.mit.edu, mark.longridge@canrem.com
Subject: New GAP insights
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mark writes

[ ... ]

> The following command shows that pons is at the centre of slice group:
> Size (Centralizer (slice, pons)) = 768

this is not hard to see.  the center of the slice group has order 32.
if we hold the corners fixed, then these central elements are exactly
those for which the six face-centers are correct.

[ ... ]

> Of course, now that I have answered my old questions, I must
> formulate new ones....
> 
> A) What is the next most commutative element (the pancentre?)
>   after the 12-flip?
> B) What is the least commutative element (the anticentre?) of
>   the cube group?

i'm sure that GAP can do these.  you're interested in knowing about
the orders of centralizers of various elements.  for an element
g  in a group  G , we have

            |G| / |Z(g)|  =  number of conjugates of  g .

this is because the cosets  G / Z(g)  are in one-to-one correspondence
with the conjugates of  g.  of course, this doesn't help much unless
we know about conjugacy classes in  G.

in the case of the cube group, however, conjugacy classes are easy to
understand.  they are (almost) completely described by cycle structure.
(some cycle structures have two conjugacy classes.)  there are many
different possible cycle structures, but for each it should be easy to
count the number of elements with that cycle structure (and also to
tell whether they comprise a single conjugacy class or split into two).

mike

From @mitvma.mit.edu:JI9316@CMSUVMB.BITNET  Tue Mar  7 12:40:21 1995
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From: "Justin I." <JI9316%CMSUVMB.BITNET@mitvma.mit.edu>
Organization: POOR STUDENTS OF AMERICA
Subject:      UNSUBSCRIBE
To: CUBE-LOVERS@ai.mit.edu
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Please take me off the Cube-Lovers list.  THANK YOU.

JUSTIN INZAURO
FULL TIME STUDENT
JI9316@CMSUVMB.CMSU.EDU
(816)543-4196
                         -IN MEMORY OF BETSY, MY FIRST LOVE-
                          -I WILL NEVER FORGET WHAT HAPPENED-
                           -DON'T BE MAD THAT I WAS UNABLE TO STAY-

From mreid@ptc.com  Tue Mar  7 14:35:00 1995
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From: mreid@ptc.com (michael reid)
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To: cube-lovers@ai.mit.edu, mark.longridge@canrem.com
Subject: New GAP insights
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for what it's worth, i'll make some conjectures about mark's questions.

> A) What is the next most commutative element (the pancentre?)
>   after the 12-flip?

(presumably, start excluded as well)

i'll guess that these four conjugacy classes are tied for next.

corner cycle structure:  (1+)(1+)(1+)(1+)(1+)(1+)(1+)(1-)
edge cycle structure:    (1)(1)(1)(1)(1)(1)(1)(1)(1)(1)(1)(1)

corner cycle structure:  (1+)(1+)(1+)(1+)(1+)(1+)(1+)(1-)
edge cycle structure:    (1+)(1+)(1+)(1+)(1+)(1+)(1+)(1+)(1+)(1+)(1+)(1+)

corner cycle structure:  (1+)(1-)(1-)(1-)(1-)(1-)(1-)(1-)
edge cycle structure:    (1)(1)(1)(1)(1)(1)(1)(1)(1)(1)(1)(1)

corner cycle structure:  (1+)(1-)(1-)(1-)(1-)(1-)(1-)(1-)
edge cycle structure:    (1+)(1+)(1+)(1+)(1+)(1+)(1+)(1+)(1+)(1+)(1+)(1+)

> B) What is the least commutative element (the anticentre?) of
>   the cube group?

i'll guess

corners:  (1)(7)     edges:    (1)(11)
corners:  (1+)(7-)   edges:    (1)(11)
corners:  (1-)(7+)   edges:    (1)(11)
corners:  (1)(7)     edges:    (1+)(11+)
corners:  (1+)(7-)   edges:    (1+)(11+)
corners:  (1-)(7+)   edges:    (1+)(11+)

each of these splits into two conjugacy classes.  i think this is the
example bandelow gives in his book.

mike

From mreid@ptc.com  Wed Mar  8 10:33:00 1995
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From: mreid@ptc.com (michael reid)
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To: cube-lovers@ai.mit.edu
Subject: New GAP insights
Content-Length: 559

i wrote

> > B) What is the least commutative element (the anticentre?) of
> >   the cube group?
> 
> i'll guess
> 
> corners:  (1)(7)     edges:    (1)(11)
> corners:  (1+)(7-)   edges:    (1)(11)
> corners:  (1-)(7+)   edges:    (1)(11)
> corners:  (1)(7)     edges:    (1+)(11+)
> corners:  (1+)(7-)   edges:    (1+)(11+)
> corners:  (1-)(7+)   edges:    (1+)(11+)

after thinking about it, i realized that

corners:  (8)   edges:  (12)

commutes with even fewer elements.  again, elements with
this cycle structure split into two conjugacy classes.

mike

From GPATYK@aol.com  Sat Mar 11 11:23:04 1995
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From: GPATYK@aol.com
Message-Id: <950311103954_46192197@aol.com>
To: cube-lovers@ai.mit.edu
Subject: list

Please take me off list.

thank you 
>>>gregp

From alan@curry.epilogue.com  Sun Mar 12 01:34:35 1995
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Message-Id: <12Mar1995.011352.Alan@LCS.MIT.EDU>
From: Alan Bawden <Cube-Lovers-Request@ai.mit.edu>
Sender: Cube-Lovers-Request@ai.mit.edu
To: GPATYK@aol.com
Cc: cube-lovers@ai.mit.edu
In-Reply-To: GPATYK@aol.com's message of Sat, 11 Mar 1995 10:39:55 -0500 <950311103954_46192197@aol.com>
Subject: list

   Date: Sat, 11 Mar 1995 10:39:55 -0500
   From: GPATYK@aol.com
   Message-Id: <950311103954_46192197@aol.com>
   To: cube-lovers@ai.mit.edu
   Subject: list

   Please take me off list.

   thank you 
   >>>gregp

Looking back through my records, I find that when you asked to be added to
Cube-Lovers back in February, you sent your subscription request to the
entire mailing list.  At the time I explained to you that subscription
requests should be sent to Cube-Lovers-REQUEST -- and that this is in fact
an Internet-wide convention.  In fact, the greeting message I sent you
contained the following text:

   REMEMBER: Our addresses are Cube-Lovers@AI.MIT.EDU for submissions and
   Cube-Lovers-Request@AI.MIT.EDU for administrivia.  Save this message
   somewhere so you don't forget about Cube-Lovers-Request.  If you ever
   want to be -removed- from Cube-Lovers, your request should be sent to
   Cube-Lovers-Request.  If you forget, and send mail concerning your
   subscription to the entire list again, you should expect to be severely
   chastised -- that mistake is considered particularly annoying by
   Internet old-timers.

So here you are, only a month later, screwing up in public -- a
particularly pathetic example of inability to follow simple directions.

To the rest of you on Cube-Lovers: The paragraph quoted above is now sent
to -all- new subscribers (and now all you old-timers have seen it as well),
so don't make the mistake that Greg did -- send your administrative
requests to -me- (Cube-Lovers-Request@AI.MIT.EDU).  If you don't, I have
now given you fair warning that I will feel perfectly free to hold you up
for public ridicule!

			- Alan (Cube-Lovers-Request@AI.MIT.EDU)

From hirsh@cs.rutgers.edu  Mon Mar 13 12:31:57 1995
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Sender: Haym Hirsh <hirsh@cs.rutgers.edu>
Date: Mon, 13 Mar 95 12:31:49 EST
From: Haym Hirsh <hirsh@cs.rutgers.edu>
Reply-To: Haym Hirsh <hirsh@cs.rutgers.edu>
To: cube-lovers@ai.mit.edu
Subject: Announcement - Workshop on Groups and Computation, June 7-9, 1995
Cc: Haym Hirsh <hirsh@cs.rutgers.edu>
Message-Id: <CMM-RU.1.4.795115909.hirsh@pei.rutgers.edu>

Thought this might be of interest to some on this list.
Please do not contact me, I am not involved with the workshop.

Haym

===========================================================================

Date: Mon, 13 Mar 1995 11:34:32 -0500
From: bquigley@dimacs.rutgers.edu

		   WORKSHOP ON GROUPS AND COMPUTATION 

		   Rutgers University, June 7-9, 1995 


Computational group theory is an interdisciplinary field involving the use of
groups to solve problems in computer science and mathematics.  The workshop
will explore the interplay of research which has taken place in a number of
broad areas: 

	Symbolic algebra which has led to the development of algorithms for
	group--theoretic computation and large integrated software packages
	(such as Cayley, Magma and Gap).

	Theoretical computer science which has studied the complexity of
	computation with groups.

	Group theory which has provided new tools for understanding the
	structure of groups, both finite and infinite. 

	Applications of group computation within mathematics or computer
	science, which have dealt with such diverse subjects as simple groups,
	combinatorial search, routing on interconnection networks of
	processors,  the analysis of data, and problems in geometry and
	topology. 

The primary workshop theme is to understand the algorithmic and mathematical
obstructions to efficient computations with groups. This will require an
assessment of algorithms that have had effective implementations and recently
developed algorithms that have improved worst--case asymptotic times.  Many
algorithms of these two types depend heavily on structural properties of
groups (such as properties of simple groups in the finite case), both for
motivation and correctness proofs, while other algorithms have depended more
on novel data structures than on group theory.

The scientific program will consist of a limited number of invited lectures
and short research announcements, as well as informal discussions and software
demonstrations.  Although it is likely that individual talks will have a
theoretical or practical focus, it has become increasingly recognized since
the first DIMACS Workshop on Groups and Computation that there are no clear
dividing lines between theory and practice. Experience has shown that a
thorough discussion of implementation issues produces a deeper understanding
of the mathematical underpinnings for group computations, leading both to new
algorithms and to improvements of existing ones.  Some background for these
discussions will be obtained through software produced by several
participants. 

Organizers are  

	Larry Finkelstein (Northeastern Univ.; {\tt laf@ccs.neu.edu}),

	William M. Kantor (Univ. of Oregon; {\tt kantor@bright.uoregon.edu}) and

	Charles C. Sims (Rutgers Univ.; {\tt sims@math.rutgers.edu}).  

Contact the organizers or DIMACS for information about attending.


From ChadL39788@aol.com  Tue Mar 21 16:04:21 1995
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From: ChadL39788@aol.com
Message-Id: <950321143111_56421409@aol.com>
To: cube-lovers@life.ai.mit.edu
Subject: Rubic's Revenge

Are there any computer programs available to help me solve the Rubic's
Revenge I bought in Japan?  I know the basics for the first three rows, but
not the final side.

From BRYAN@wvnvm.wvnet.edu  Thu Mar 23 01:17:30 1995
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Date:      Thu, 23 Mar 1995 00:29:47 -0500 (EST)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Some Thoughts on Representative Elements

I would like to make some comments on representative elements, but
first there are some preliminaries.

In some of his recent notes, Martin Schoenert has carefully
distinguished between processes, operations, and states. For
the purposes of this note, I wish to distinguish between
processes and operations on the one hand, and states on the
other hand.  For this one note alone, I will always use
upper case letters for processes and operations, and lower
case letters for states.

In functional notation, we often (e.g., calculus) see such things
as y=F(x).  "Function" is synonymous with "operation", and I will
use the two terms somewhat interchangeably.  In calculus, function
composition is often written something like y=FoG(x)=F(G(x)),
where the "o" is the best simulation I can do in E-mail of the
typical function composition symbol.

As has been discussed several times in this forum, the FoG type
of notation is interpreted right-to-left, but in group theory
we more typically write GF for function composition, and it is
interpreted left-to-right.  With the left-to-right notation,
the function argument (if it is written) can be written on the
right as y=GF(x) or on the left as y=(x)GF.  I gather that most
of Cube-Lovers do not like the latter convention, but I am
going to use it anyway.  Indeed, as long as we are utterly
rigorous in our upper-case/lower-case convention, we can
dispense with the parentheses and write y=xGF (or simply
y=xF if the function F is a single function).  Parentheses
can then be used for grouping without any ambiguity that they
might be denoting arguments.

We let G be the cube group and g be the set of cube states.  We
can observe that each X in G is a function on g.  Furthermore,
each X in G is a one-to-one function from g onto g.  Hence,
each inverse X' exists.  These facts are so obvious that they
are virtually never stated.  But we are about to talk about
representative elements, and things are much less well behaved
when we talk about representatives.

One more preliminary:  the idea that a state can be identified
with the operation which effects the state when applied to Start
can be expressed as x=iX.  Hence, there is a one-to-one
correspondence between g and G, and we have |g|=|G|.  Again,
things get much more unruly when we start talking about
representatives.  Also, we can define the composition of the
states x and y as xy=(iX)(iY)=i(XY)=iXY, and this equation
defines an obvious isomorphism between g and G.

We will use * as the symbol for a selection function on the
M-conjugacy classes of g.  We will write * as a postfix operator
so that it reads left-to-right with the rest of our operators.
We have x*=Repr{N'xN} for all N in M, the set of 48 rotations
and reflections.

Note that * is not uniquely determined.  There are
approximately |M| ^ (|G| / |M|) possible choices for *, which is a
truly large number when compared with |G|.  It is conceivable
that we might be able to prove some result for one particular
choice of * that would not be true for every choice of *.  In
general, however, we will assume that * is fixed but arbitrary.

* is a function from g into (but not onto) g.  We let g* be the
set of representative states, and * is a function from g onto g*.
The function * does not have an inverse, but if we consider * to
be a relation, then the inverse relation *' simply maps each
representative x* in g* back to the entire M-conjugacy class
of x*.

Consider the restriction of * to the set g*.  At this point, *
is not very useful, because it is simply the identity.

Define the set Q* as

   Q* = {F*, R*, L*, B*, U*, D*, F'*, R'*, L'*, B'*, U'*, D'*}

This set is near and dear to my heart, because it is how I obtain
nearly all my results.  We will say more about inverses very soon,
but it should be observed immediately that it is not the case that
(F*)'=F'*, nor that (R*)'=R'*, etc.

Each of the twelve elements in Q* are functions from g into g*.
Much more interesting is the restriction of each element of Q*
to g*, so that they are functions from g* into g*.  We are treating
each Q* as composed and not to be decomposed.  You could think of
elements of g that are not also elements of g* appearing briefly
and virtually between the Q and the *, but each Q* is to be
treated as from g* into g* without being decomposed.

Can we do something like define the group G*=<Q*>?  The answer
is no.  I assume it will be totally obvious to many of you why not,
but I have to think about it for a bit.  Martin commented that
the set of representatives could not be a group because the
number of representatives does not divide the size of G.  But
it seems to me that Martin's argument only says that the representatives
are not a subgroup of G.  It seems to me that it does not say that
there could not be a group constructed from the representatives.
But nonetheless, I think it is quite easy to show that there is
no way to make G* into a group.

Before I go on, I suspect that many of you will object to me using
the generator notation G*=<Q*>, even with the caveat that G* is not
a group.  I probably agree, but I am not sure how else to convey
the same idea quite so compactly.  I simply want to define G* as the
set of all compositions of elements of Q*.  The compositions are
all well defined, although they behave terribly.  And G* must be
finite, because g* is finite and there are only finitely many
functions that can be defined on a finite set.

I find the following argument that G* is not a group (and cannot
be made into a group) compelling, although there might well
be simpler arguments.  Do any of the functions Q* have inverses?
In order to do so, they would have to be one-to-one from g* onto
g*.  But i=i*, so i is in g*, and there is only one of the twelve
elements of Q* which maps onto i.  Hence, at least eleven of the
twelve elements of Q* are not onto g*, and therefore do not have
inverses.  It is trivial do choose * in such a way that all twelve
elements of Q* are not onto g*.  However, I have not been able to
decide if there is way to choose * such that one of the elements
of Q* is onto g*.  No doubt, somebody out there will have a
trivial proof one way or the other. In any case, the general lack
of inverses in Q* shows that G* cannot be made into a group.

Even though G* is not a group, it is nonetheless an interesting
structure.  We can think of a "Cayley graph-like" structure where
we connect nodes in g* with arcs from Q*.  I am not sure if such
a structure is properly called a Cayley graph, so just to be
safe I will call it a Cayley* graph.

As we have already seen, the identity state i*=i has only one
neighbor in our Cayley* graph.  Are there any other such states?
Dan Hoey and Jim Saxe's _Symmetry and Local Maxima_ suggests
that there might be 72 such states, namely those states which
they call Q-transitive.  A Q-transitive state has the characteristic
that all its neighbors are M-conjugate.  However, some of the
Q-transitive states are themselves M-conjugate, so the 72 states
collapse somewhat in our Cayley* graph.

There are 4 states which are M-symmetric,
and these states are distinct in our Cayley* graph.  There are
20 states which are H-symmetric but not M-symmetric, but these
states collapse into 10 states in our Cayley* graph.  There are
48 states which are T-symmetric but not M-symmetric, but these
states collapse into 12 states in our Cayley* graph.  Hence, there
are 26=4+10+12 nodes in our Cayley* graph which have only
one neighbor.  These figures are given right at the end of _Symmetry
and Local Maxima_.  They are also given by Martin Schoenert
in _Re: Re: Re: Re: Models of the Cube_ on 1 February 1995.

It has been thoroughly discussed that |X| = |X*| for all X in G.
But of more import for the way I build my data bases is the question
of whether for every x in g, will x* appear in my data base, given
that I generate the data base as i<Q*>.  In other words, can any
x* of length n be decomposed into i(Q_1*)(Q_2*)...(Q_n*)?  I find
this to be one of those things that is obvious yet is hard to
explain.  Hence, I will borrow the following explanation that
was given to me by Dan Hoey.  (Dan's version is much more elegant
than mine.  Any crud in the following is mine, not Dan's.)

First of all, every x* in g* either is the identity or else has
a neighbor in g which is closer to Start.  If it is the identity, it
is in (and indeed is the root of) the data base.  Otherwise, we call
the neighbor y and calculate y*.  Again, y* is either the identity
or else has a neighbor which is closer to Start.  In this manner,
we can backtrack our way to Start from any x*.  However, there is
not (yet) a guarantee that a forward search will find traverse
the same path forwards and find x*, and hence not (yet) a proof
that any of the path to x* except i itself is in the data base.

We note that if y is a neighbor of x in g, then it is immediate
that x is a neighbor of y as well.  We need the same thing in g*
in order to get a forward search that corresponds to the backwards
search.  That is, we need to show that if y* is a neighbor of x* in
g*, then x* is a neighbor of y*.  We now show that if y* is a
neighbor of x* in g*, then x* is a neighbor of y* in g*.

Observe that if v* = w* (i.e., if v and w are M-conjugates), then
the neighbors of v and w are respective M-conjugates as well.
Assume x* is not the identity and let y be a neighbor which is
closer to Start.  Then y* has a neighbor z with z*=x*.  Hence, if
y* is in the data base, then x* is in the data base, and we have
neighbors in both directions.

I think of the preceding result as follows.  Even though the functions
in G* are not individually ("locally") onto, the functions in G* are
collectively ("globally") onto.  Hence, there is an arc _from_ every
node in the Cayley* graph, and there is an arc _to_ every node in the
Cayley* graph, and you can get from anywhere to anywhere in the graph.

The Cayley* graph is not a homomorphism of the Cayley graph (G* is
not a subgroup of G, and is not even a group), but I think of G* as
sort of passing the duck test for a homomorphism.  That is, it
looks like a homomorphism and smells like homomorphism, even though
it isn't.  (Personal opinion is that the duck test often fails in
this manner).  But G* is a collapsing of G which does encode an
enormous amount of information about G.

Roughly speaking, there are two definitions of permutation.  The
more informal definition is simply than a permutation is an
arrangement (or re-arrangement) of objects  ---  e.g., the number
of permutations of n objects taken r at a time.  The more formal
definition is that a permutation is a function on a set which is
one-to-one and onto.  Note that function on a finite set is
one-to-one if and only if it is onto, so in dealing with the
cube we can be a little sloppy at times and speak only of onto
or only of one-to-one.

All the elements in G are permutations and G is finite.  It is
therefore trivial to show that for every X in G, there is
some integer n such that X^n=I.  This means, among other things,
that there exists some integer n such that i(X^n)=i (from Start,
repeat a process enough times and you will return to Start) and
y(X^n)=y (from any position, repeat a process enough times and you
will return to the same position).  The least such n for each
X is the order of X, and considerable discussion has occurred
in this forum concerning the order of various elements of G.

Once again, things become a bit more unruly when we talk about
G* instead of G.  The elements of G* are not permutations because
they are not onto.  So consider what happens when we calculate
something like (X*)^n.  As a specific example, consider
i(R*)^n.

If iR*=r', then i(R*)^n simply oscillates back and forth between
i and r'.  But if * is chosen so that iR* does not equal r', then
all bets are off.  i(R*)^n will enter a loop eventually, but it
will never return to i.  To understand its exact behavior, we would
have to be specific rather than arbitrary about the definition of *.
Similarly, x*(R*)^n might well never return to x*, but it would
loop eventually.  These "eventual loops" rather remind me of
strange attractors.

My data bases are of the form i<Q*>, for example iR*L'*U'*R'*, etc.
As I have discussed before, backtracking through such a structure
is a bit tricky.  Suppose, for example that you have x* in hand
and wish to backtrack to i.  The way to do it that works best for
me is as follows:  begin with x*; find x*(Q_1)* that is closer to
Start;  find x*(Q_1)(Q_2)* that is closer to start (most definitely
NOT x*(Q_1)*(Q_2)*);  find x*(Q_1)(Q_2)(Q_3)* that is closer to
Start (most definitely NOT x*(Q_1)*(Q_2)*(Q_3)*; etc.  It is
doable in the fashion I said NOT to do, but it is extremely
tricky.  You will have the sensation (as I have described before)
that your solution is rotating and reflecting out from under you.
The first way is much easier to keep up with.

Finally, how big is G*?  Remember that g and G are the same size.
Write the restriction of G* to i as iG*.  There is clearly a
one-to-one mapping between g* and iG*, and indeed we can identify
g* with iG* in the same manner in which we identify g with G.
But in iG*, all the elements of Q* are equivalent.  When
we allow the domain of G* to be the entirety of g*, then the elements
of Q* are not equivalent.  Hence, there are at least eleven more elements
in G* than in g*. I don't know how to calculate the size of G*.  But
when Dan Hoey calculated the real size of the cube space, he
calculated the size of g*, not the size of G*.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From itg@athena.compulink.forthnet.gr  Fri Apr  7 03:38:26 1995
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subscribe cube-lovers
Thedore Emmanuel

From patrick@athos.med.auth.gr  Fri Apr  7 11:27:21 1995
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subscribe listname
patrick pfavayi


From ivan@antares.aero.org  Mon Apr 10 19:30:23 1995
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To: cube-lovers@life.ai.mit.edu
Subject: Singmaster's works
Date: 	Mon, 10 Apr 1995 13:16:44 -0700
Message-Id: <10272.797545004@armadillo.aero.org>
From: Ivan Filippenko <ivan@antares.aero.org>

Hello,

Can anybody suggest how I might be able to get copies of David Singmaster's
"Notes on Rubik's Magic Cube" and A. H. Frey and Singmaster's "Handbook of
Cubik Math" ?

How about Christoph Bandelow's "Inside Rubik's Cube and Beyond" ?

I'm also looking for Chris Rowley's "The group of the Hungarian Magic Cube"
(in Algebraic Structures and Applications, Proceedings of the First Western
Australian Conference on Algebra, 1982).


Many thanks,

  -- Ivan
     ivan@aero.org

From @mail.uunet.ca:mark.longridge@canrem.com  Fri Apr 14 03:04:40 1995
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Subject: Slice & Anti-Slice
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1089.5834.0C1DBB1F@canrem.com>
Date: Fri, 14 Apr 1995 03:45:00 -0400
Organization: CRS Online  (Toronto, Ontario)

Some notes on the numeration of the slice and anti-slice
groups....

        Analysis of the 3x3x3 Slice & Anti-Slice Groups
        -----------------------------------------------
                  arrangements           arrangements
Moves Deep     (2q or slice moves)   (4q or double slice moves)

  0                    1                      1
  1                    6                      9
  2                   27                     51
  3                  120                    247
  4                  287                    428
  5                  258                     32
  6                   69                    ---
                     ---                    768
                     768

                  arrangements           arrangements
Moves Deep   (2q or anti-slice moves)   (4q or double anti-slice moves)

    0                   1                     1
    1                   6                     9
    2                  27                    51
    3                 120                   265
    4                 423                   864
    5               1,098                 1,785
    6               1,770                 2,017
    7               1,650                 1,008
    8                 851                   144
    9                 198
                    -----
                    6,144

From BRYAN@wvnvm.wvnet.edu  Fri Apr 14 17:08:12 1995
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Date:      Fri, 14 Apr 1995 17:06:48 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Repetitive Application of Elements of Q*

Recall that Q* has been defined as the set of representatives

   Q* = {F*, R*, L*, B*, U*, D*, F'*, R'*, L'*, B'*, U'*, D'*}

where * has been defined as a function selecting a representative
element of an M-conjugacy class.

I have done a little experimentation with cycles of the form X* ^ n.
As long as the X* are directly in Q*, the sequences are quite short,
and the final cycle is of length 2 in all cases.  I found the latter
surprising initially, but with the wisdom of hindsight, it was
inevitable.  Here is a table of lengths.

     Operation         Length
                         of
                       Sequence


       F*                11
       U*                10
       L*                 7
       R*                 3
       D*                 9
       B*                 2
       F'*                7
       U'*               10
       L'*                4
       R'*                6
       D'*                7
       B'*                2

A couple of points of clarification:

    1.  As an example, for F*, we take i(F*)^n  (that is, apply F*
        to Start repeatedly).  The sequence has 11 elements before it
        repeats, then the 10-th and 11-th element repeat over and
        over again.  In all twelve cases, it is the last two elements
        of the sequence which repeat.

    2.  In order to replicate my results, you would have to define
        a representative element function exactly like mine.  Every
        choice of representative element function can be expected to
        yield different results.


To take a little more interesting case, I tried i(F*D'*) ^ n.  In this
case, there were 63 unique elements in the sequence, and then the
8-th through the 63-rd elements repeated.  Hence, the final cycle had
56 elements rather than the 2 elements of the simpler cases.

I suppose I could try quite a few other cases, but I have no idea how
to predict how long the sequences or the terminal cycles might be.
All I know to expect for sure is for things to be quite ill-behaved.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From mreid@ptc.com  Fri Apr 14 17:30:05 1995
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From: mreid@ptc.com (michael reid)
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To: cube-lovers@ai.mit.edu, mark.longridge@canrem.com, CRSO.Cube@canrem.com
Subject: more on the slice group
Content-Length: 2394

mark's post got me thinking ...  i made a quick hack for the slice
group (which is easy to represent by fixing the corners).  my
figures concur with his.  i wanted to see the number of local maxima.

  90 degree     number of      number of
 slice turns    positions     local maxima

      0              1              0
      1              6              0
      2             27              0
      3            120              0
      4            287              0
      5            258             24
      6             69             69

as i'd hoped, there are local maxima at distance 5.  one such is:

          (FB') (RL') (U'D) (R2L2)       =
          (R2L2) (F'B) (RL') (UD')       =
          (R'L) (FB') (RL') (F'B) (U'D)  =
          (U'D) (F'B) (RL') (U'D) (F'B)  =
          (R'L) (UD') (F'B) (RL') (FB')

(actually i think all are equivalent to this one under symmetries
of the cube.)

this is especially interesting because it is a local maximum in the
full cube group (quarter turn metric) at distance  10q.  according
to jerry bryan's results, there are no local maxima within  9q
of start, so this gives the closest local maximum.  (there may well
be others.)


i also calculated for the other slice metric.  in this metric,
neighbors can have the same distance from start, so a "strong"
local maximum is a position all of whose neighbors are strictly
closer to start.  a "weak" local maximum is a position none of
whose neighbors are further from start.

 90 or 180 degree     number of     number of strong    number of weak
   slice turns        positions       local maxima       local maxima

        0                  1               0                   0
        1                  9               0                   0
        2                 51               0                   0
        3                247               0                   7
        4                428               0                 212
        5                 32               8                  32

the strict local maxima are all equivalent under symmetries of
the cube.  they are the composition of pons asinorum with any
of the eight positions called "six dots".

in the same way, local maxima (within the antislice group) in the
90 degree antislice metric are local maxima in the full cube group
(quarter turn metric).  perhaps mark will tell us more about this.

mike

From BRYAN@wvnvm.wvnet.edu  Fri Apr 14 23:33:20 1995
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Date:      Fri, 14 Apr 1995 21:12:53 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Re: Repetitive Application of Elements of Q*
In-Reply-To: Message of 04/14/95 at 17:06:48 from BRYAN@wvnvm.wvnet.edu

The astute reader will have noticed that there *has* to be an error
in the table as it was originally posted.  It is impossible for there
to be two distinct sequences of length 2, because there is only
one position unique up to M-conjugancy which is only one qturn from
Start.  All of the lengths in the original table except for one were
short by one.  I had two programs  --  one to generate the sequences
to a file, and a second to analyze the sequences.  The first program
did not write Start to the file, resulting in all the sequences
except one being one short.  Here follows the correction.

                       (Incorrect)    (Correct)

>     Operation         Length         Length
>                         of             of
>                       Sequence       Sequence


>       F*                11            12
>       U*                10            11
>       L*                 7             8
>       R*                 3             4
>       D*                 9            10
>       B*                 2             2   !!
>       F'*                7             8
>       U'*               10            11
>       L'*                4             5
>       R'*                6             7
>       D'*                7             8
>       B'*                2             3

>To take a little more interesting case, I tried i(F*D'*) ^ n.  In this
>case, there were 63 unique elements in the sequence, and then the
>8-th through the 63-rd elements repeated.  Hence, the final cycle had
>56 elements rather than the 2 elements of the simpler cases.

Similarly, here there were 64 unique elements in the sequence, with the
9-th through the 64-th elements repeated.  The final cycle still
(correctly) had 56 elements.  The length of the final cycle has to be
even, of course, since qturns are odd.

> = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
>Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
>Associate Director, WVNET                            (304) 293-5540 fax
>837 Chestnut Ridge Road                              BRYAN@WVNVM
>Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From @mail.uunet.ca:mark.longridge@canrem.com  Sun Apr 16 21:14:23 1995
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Sender: CRSO.Cube@canrem.com
Subject: Antislice revisited
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1095.5834.0C1DC195@canrem.com>
Date: Sun, 16 Apr 1995 21:57:00 -0400
Organization: CRS Online  (Toronto, Ontario)

Mike Reid writes:
> 90 or 180 degree     number of     number of strong    number of weak
>   slice turns        positions       local maxima       local maxma
>
>        0                  1               0                   0
>        1                  9               0                   0
>        2                 51               0                   0
>        3                247               0                   7
>        4                428               0                 212
>        5                 32               8                  32
>
> the strict local maxima are all equivalent under symmetries of
> the cube.  they are the composition of pons asinorum with any
> of the eight positions called "six dots".

Very Interesting.

Here's a 12q process for pons asinorum + 6 dots:

 (F2 B2) (T1 D3) (F1 B3) (L3 R1) (T1 D3)


I have some thoughts on the relationship between the groups of the
slice, antislice and squares, although some of it is old news,
discovered by Mr. Singmaster back in 1980. The fact that, at most,
9 anti-slice moves always suffice to restore any position in
this group I think is new.

If we define Combo = < antislice , slice >
I.e. Combo is the group generated by antislice and slice moves.

Then....

Size (Combo) = 15,925,248

The Combo group fully includes the square's group and each group has
some elements in common:

Size (Intersection (antisl, slice, sq)) = 8

Also Size(Combo) / Size Sq) = 24

The Combo group has trivial centre.

I should also note that....
 Size (Intersection (sq, antisl)) = 256.

Mike continues:
> in the same way, local maxima (within the antislice group) in the
> 90 degree antislice metric are local maxima in the full cube group
> (quarter turn metric).  perhaps mark will tell us more about this.

Well, I didn't calculate local maxima for the anti-slice group,
but I will look at it. I did create a file of all the processes
for each anti-slice position, and most of the anti-slice group
antipodes are quite ugly looking!

One of the "not quite so ugly" antipodes:

( F1 B1 L1 R1)^3 + T1 D1 F1 B1 T1 D1 = 18 q

The Centre elements of the Antislice group
------------------------------------------

There are 4 elements which commute with all elements of the
<AS> group, the identity and the three 4 cross order 2 patterns.

(F1 B1) (T1 D1) (L2 R2) (T1 D1) (F1 B1) (T2 D2) = 16 q

      TTT
      TTT
      TTT
RLR   BFB  LRL  FBF
LLL   FFF  RRR  BBB
RLR   BFB  LRR  FBF
      DDD
      DDD
      DDD

Cheers!
 -> Mark <-

From @mail.uunet.ca:mark.longridge@canrem.com  Sun Apr 16 21:39:05 1995
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Subject: DOTC
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1096.5834.0C1DC1A2@canrem.com>
Date: Sun, 16 Apr 1995 22:04:00 -0400
Organization: CRS Online  (Toronto, Ontario)

I am a poor correspondent...

No real excuse... I will send #4 out on Tuesday.

Maybe I'll thrash future issues out in e-mail form.

By the way, I thought your message on symmetric maneuvers
was quite insightful. Did you ever find a symmetric move
for "Cube in a Cube"?

Cheers!
-> Mark <-

From patrick@athos.med.auth.gr  Mon Apr 17 04:22:26 1995
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Sirs ,I am sorry to bother you with junck mail,
but I wish to join Cube-Lovers.
And I have written to you two or three times at the address 
cube-lovers-REQUEST@life.ai.mit.edu and each time after a delay of a No. 
of days I got the response that my mail hadn't been delivered since you 
cuoldn't be found.
Now I know you may not appreciate a new user littering your mail boxes 
with junk mail but I don't know any other way to join in--please excuse me.

Yours sincerely
               Patrick


From alan@curry.epilogue.com  Mon Apr 17 23:54:43 1995
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Date: Mon, 17 Apr 1995 23:50:25 -0400
Message-Id: <17Apr1995.232904.Alan@LCS.MIT.EDU>
From: Alan Bawden <Cube-Lovers-Request@ai.mit.edu>
Sender: Cube-Lovers-Request@ai.mit.edu
To: patrick@athos.med.auth.gr
Cc: Cube-Lovers@ai.mit.edu
In-Reply-To: Patrick's message of Mon, 17 Apr 1995 10:40:17 -0200 (GMT-0200) <Pine.LNX.3.91.950417103117.10111B-100000@athos>
Subject: subscribe

   Date: Mon, 17 Apr 1995 10:40:17 -0200 (GMT-0200)
   From: Patrick <patrick@athos.med.auth.gr>

   Sirs ,I am sorry to bother you with junck mail,
   but I wish to join Cube-Lovers.
   And I have written to you two or three times at the address 
   cube-lovers-REQUEST@life.ai.mit.edu and each time after a delay of a No. 
   of days I got the response that my mail hadn't been delivered since you 
   cuoldn't be found.
   Now I know you may not appreciate a new user littering your mail boxes 
   with junk mail but I don't know any other way to join in--please excuse me.

   Yours sincerely
		  Patrick


I added your subscription a week ago.  You even -responded- to one message
I sent you, so I know that you -can- send mail to Cube-Lovers-Request.

DO NOT SEND ANY FURTHER MESSAGES TO CUBE-LOVERS AS A WHOLE OR YOUR
SUBSCRIPTION WILL BE TERMINATED.

If you are experiencing trouble receiving Cube-Lovers mail, your only
recourse is to complain to -me-, Cube-Lovers-Request -- I will try my best
to help you.  If, for some reason, you are unable to send mail to the
-Request address, then you are simply screwed -- it will -never- do you any
good to send mail to the entire mailing list about list administration.

Everybody please note that the correct administrative address is:

  Cube-Lovers-Request@AI.MIT.EDU

Some mail programs will re-write mail headers to make it look as though the
address is "cube-lovers-request@life.ai.mit.edu", but it isn't.

				- Alan

From BRYAN@wvnvm.wvnet.edu  Thu Apr 20 12:00:31 1995
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Date:      Thu, 20 Apr 1995 11:35:56 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <cube-lovers@ai.mit.edu>
Subject:   Re: Run Times, Storage Requirements, etc.
In-Reply-To: Message of 02/22/95 at 18:28:13 from mreid@ptc.com

On 02/22/95 at 18:28:13 mreid@ptc.com said:

>i think this can be done much more efficiently.  well, at least if you
>set things up properly in the first place.

>suppose that you use an order (for sorting positions) in which the corner
>configuration is more significant than the edge configuration.

Unfortunately, I made the edge more significant for sorting than the
corner.  I plan to change that in the near future.

Strictly speaking, it would not be *too* bad to simply sort the same
data in a different order.  But the problem is worse than that.  I also
made the edge more significant for choosing a representative element
than the corner.  So I need a new representative element function
to make the corner more significant for choosing the representative
element.  Such a change to the program would be trivial, but I would
have to regenerate the data before it was re-sorted.  Otherwise, the
"same" corner configuration would not be the same; equivalent corner
configurations would be M-conjugate rather than equal.

>then, for each position  X  on your huge list, you need to check if
>repr(X superflip)  is on the list.  since superflip only affects edges,
>the corner configuration of  X superflip  is the same as that of  X.
>thus the same holds for  repr(X superflip)  and  repr(X) = X.  therefore,
>you only need to look for  repr(X superflip)  nearby in the sorted list.

Mike's note raises several interesting points.

Suppose we write a cube Z as the disjoint union XY of corners X and
edges Y.  (We could say something like Z[C,E] = Z[C]*Z[E], but let's
keep the notation a little simpler).  A list of cubes in my data
base would then look something like:

    X1 Y3
    X1 Y7
    X1 Y8
    X2 Y1
    X2 Y2
    X2 Y5
     etc.

If we were sufficiently clever, we might be able to save some space
by rewriting the list as something like:

    X1 Y3
       Y7
       Y8
    X2 Y1
       Y2
       Y5
     etc.

In other words, store each corner position only one time.  This is
very similar to some of the indexing schemes that have been described
to store large numbers of cubes very compactly.  I have used some of
these indexing schemes for corners only or edges only, but I have
always rejected them for whole cubes (corners plus edges) because
the representation is so sparse close to Start.  (and you really can't
get very far away from Start with whole cubes!)  But the picture above
just might work.  I'm going to think about it.

Next, notice that Mike's proposal for dealing with Pons Asinorum and
Superflip results in a cube and its neighbor being stored
very close together in the data base.  In this case, a "neighbor"
is a position Z composed with Pons Asinorum or Superflip or both.
More typically, a neighbor is a position Z composed with an element
from Q or from Q+H.

What would be really nice (and which may not be possible) is some
representation for the cube such that a cube Z and its neighbors
Zq or Zh are stored very close together.  Such a representation
would be very helpful in particular for searches accomplished with
massively parallel machines or with farms of workstations.  But I
certainly have never been able to find such a representation.
I have yet to fully understand the Sims table (or FHL table) that
many of you seem to use, so I don't know if it will do the trick
or not.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From mreid@ptc.com  Thu Apr 20 15:46:58 1995
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Date: Thu, 20 Apr 1995 16:03:31 -0400
From: mreid@ptc.com (michael reid)
Message-Id: <9504202003.AA19008@ducie>
To: cube-lovers@ai.mit.edu
Subject: correction and an interesting example
Content-Length: 1221

[ i sent a similar message several days ago, but it appears to
  have gotten lost.  my apologies if anyone already got this. ]

i wrote

> in the same way, local maxima (within the antislice group) in the
> 90 degree antislice metric are local maxima in the full cube group
> (quarter turn metric).

this isn't necessarily true.  one must check that the minimal
maneuvers (within the antislice group) for such positions are
also minimal in the full group.

the position i mentioned

>           (FB') (RL') (U'D) (R2L2)       =   [ ... ]

is quickly checked to require 10 quarter turns, so indeed it
is locally maximal.


here's an example i found of a locally maximal position whose
inverse is not locally maximal:

     A  =   B U2 F2 R U' R' B' R' U F2 U2    (15q)

this position has six symmetries, generated by the cube rotation
C_UFR  and central reflection.  using these symmetries we can
give minimal maneuvers which end with a half turn of any face,
and thus with any of the twelve quarter turns.  the same is not
true of its inverse, and we can easily check that there is no
minimal maneuver for  A  which begins with the quarter turn  B'.
equivalently, the position  A^-1  B'  requires 16 quarter turns.

mike

From mreid@ptc.com  Fri Apr 21 11:20:45 1995
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From: mreid@ptc.com (michael reid)
Message-Id: <9504211537.AA20597@ducie>
To: cube-lovers@ai.mit.edu, bryan@wvnvm.wvnet.edu
Subject:   Re: Run Times, Storage Requirements, etc.
Content-Length: 858

jerry writes

>                   So I need a new representative element function
> to make the corner more significant for choosing the representative
> element.  Such a change to the program would be trivial, but I would
> have to regenerate the data before it was re-sorted.

i don't think you need to regenerate; you just need to put your
old list through your new representative choosing function and
sort the output.  of course, regenerating would give a reasonable
consistency check.

> What would be really nice (and which may not be possible) is some
> representation for the cube such that a cube Z and its neighbors
> Zq or Zh are stored very close together.

remember that the diameter of the group is small.  (my guess is
21 face turns, 24 quarter turns.)  so this isn't possible without
resorting to a liberal definition of "very close".

mike

From @mail.uunet.ca:mark.longridge@canrem.com  Mon Apr 24 04:12:38 1995
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Reply-To: CRSO.Cube@canrem.com
Sender: CRSO.Cube@canrem.com
Subject: Orders of Symmetry
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1104.5834.0C1DD97C@canrem.com>
Date: Sun, 23 Apr 1995 23:29:00 -0400
Organization: CRS Online  (Toronto, Ontario)

>Date: Wed, 8 Dec 93 16:28:29 EST
>From: hoey@AIC.NRL.Navy.Mil (Dan Hoey)
>Message-Id: <9312082128.AA23718@Sun0.AIC.NRL.Navy.Mil>
>To: mark.longridge@canrem.com (Mark Longridge), CRSO.Cube@canrem.com
>Subject: Re: More corrections
>
>> * Hmmm, what are all the possible orders of symmetry? *
>
> M has subgroups of order 48, 24, 16, 12, 8, 6, 4, 3, 2, 1.  Some of
> these subgroups (e.g., A, C) are not symmetry groups of any position,
> so I can't be sure there are positions of all these symmetry orders.

Quite a while ago I asked Dan the question above, and I've thought
a lot about the answer.

So I decided to look at certain cube positions and I wrote a module
to perform
                        C and C + Sm

where
C  = 24 rotations of the cube
Sm = Central Reflection

on any pattern I had in my database, and count how many different
patterns resulted from the 48 operations.

The following are some patterns which I found:

Number of
different       Pattern
patterns        -------
---------

48              R1 U1
24              L2 U2
16              Mark's Pattern 1 (18 q+h, 22 q)
                R2 U3 R1 D1 F1 B1 R3 L3 U1 D1 F3 U1 F3 U2 D3 B2 R2 U1
                (Also 7 clockwise + 1 anticlockwise corner twist)
12              2 dot, 2 T, 2 ARM (sq group antipode, see p108)
 8              6 Dot (a slice pattern)
 6              2 DOT, 4 ARM (sq group antipode, see p99)
 4              ????
 3              4 Dot pattern (slice pattern)
 2              6 H pattern type 2, T2 B2 L2 T2 D2 L2 F2 T2
 1              Pons Asinorum (6 X order 2) or all edges flipped

 It took a while to find a pattern which could be transformed 16
different ways. Still trying to find a pattern which will
result in 4 distinct ways, but I am not optimistic. A random walk
through the cube resulted in a pattern which would transform
48 ways in every case I tried.

>> A) What is the next most commutative element (the pancentre?)
>>   after the 12-flip?
>
> (presumably, start excluded as well)
>
> i'll guess that these four conjugacy classes are tied for next.
>
> corner cycle structure:  (1+)(1+)(1+)(1+)(1+)(1+)(1+)(1-)
> edge cycle structure:    (1)(1)(1)(1)(1)(1)(1)(1)(1)(1)(1)(1)

Here's a small followup to the pancentre question. The reason
why the 7 clockwise + 1 anticlockwise corner twist is the next
most commutative element after the 12-flip & start is because
it has the most number of cube elements (in this case corners)
the same as possible without all the elements being the same,
as with the 12-flip. It must be 7 clockwise + 1 anticlockwise
corner twist because the next most commutative element effecting
edges only would be the 10-flip and that would have 2 elements
not the same as the rest instead of just 1.

-> Mark <-

From ismaia01@bh.bbc.co.uk  Mon Apr 24 23:28:08 1995
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References: <5D6D7C370203370C@-SMF->

subscribe cube-lovers-request@ai.ai.mit.edu

From @mail.uunet.ca:mark.longridge@canrem.com  Tue Apr 25 23:05:02 1995
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Subject: More Cube Orders
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1107.5834.0C1DE11D@canrem.com>
Date: Tue, 25 Apr 1995 23:27:00 -0400
Organization: CRS Online  (Toronto, Ontario)

I said:
>                 Still trying to find a pattern which will
> result in 4 distinct ways, but I am not optimistic.

Jerry adds:
> As one more followup, for each symmetry group order in the above list,
> there exists at least one cube.
> That is, 96 of the 98 subgroups are symmetry groups for at
> least one cube.  The two "missing" subgroups  --  A and C  -- are of
> order 24.  But there is a third subgroup --  H -- of order 24
> (H is the set of 12 even rotations and 12 odd reflections), and there
> are cube positions whose symmetry subgroup is H.  Hence, there are
> cube positions for every symmetry subgroup order.

Well, I figure Jerry is correct and so I kept looking for the magic
pattern which transforms 4 ways...

Number of
different      Pattern
patterns        -------
---------

...
 4              6 flip (UF, UR, FR, DB, DL, BL)
...

So there are 4 types of this 6 flip.

Jerry has said before:
> I believe that Dan and I have solved (sort of independently, and sort
> of working together) the problem you pose (and I give Dan the bulk
> of the credit).  That is, how many cubs are there in each symmetry
> group and each symmetry class?

That sounds harder. Looks like I am specifying only the index of
the symmetry subgroup... perhaps it makes sense to find out
exactly which subgroup of M is the symmetry group of my positions.

It all sounds vaguely familar.... but it will try again tomorrow.

-> Mark <-

From mbparker@share.ai.mit.edu  Thu Apr 27 18:17:58 1995
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	id AA09162; Thu, 27 Apr 95 14:49:35 -0700
Date: Thu, 27 Apr 95 14:49:35 -0700
From: Michael Benjamin Parker <mbparker@share.>
Message-Id: <9504272149.AA09162@share>
To: Cube-Lovers@ai.mit.edu, Jennifer Dubin <JD4090A@american.edu>,
        Stan Isaacs <isaacs@hpcc01.corp.hp.com>, ccw@alumni.caltech.edu,
        Jonathan Haas <positron@starbase.neosoft.com>,
        "Byon Garrabrant" <byon@quicksilver.com>, markc@deltanet.com,
        73613.536@compuserve.com, 70410.1050@compuserve.com,
        ccw@alumni.caltech.edu, mklein@alumni.caltech.edu, mikeh@gordian.com,
        damon@gordian.com, whuang@cco.caltech.edu
Subject: PUZZLE PARTY!  in Orange County, CA; 1995 Apr. 29 (Sat) 7:00pm-
Reply-To: mbparker@mit.edu
Newsgroups: rec.puzzles,geometry.puzzles,rec.games.abstract,oc.general,la.general

Dear puzzle lovers,

     Presenting MIT Club of Southern California's...


                             PUZZLE PARTY 2!

At our first puzzle party in February, Parker's place was packed with
puzzlers of all sorts!  Puzzling continued for hours, with the last
hard-core, half-crazed puzzlers quitting at 5AM.  The party featured the
the puzzles and participation of the Master Puzzler himself, Mr. Jerry
Slocum, a UCLA alumnus who has authored three puzzle books and owns an
unmatched 20,000-piece private collection of puzzles.  [Incidentally, on
May 9th and June 22nd, Jerry and his Puzzle Museum will be featured in
Start to Finish on the Discovery Channel.]

If you didn't have a chance to participate, you'll have another
opportunity at PUZZLE PARTY TWO.  So, dig out your favorite brain teasers,
mental games, IQ tests, and mechanical puzzles.  If you have a
puzzle-freak friend, bring him/her!  Like the first party, this will be a
leisurely puzzling event with amusing problems as well as food, drink, and
conversation by the fireside.  So grab your brain and some puzzles...
and see you there!

 WHEN:  Saturday, April 29th, 7pm until...

 WHERE: In Orange, CA; RSVP as below for directions.

 COST:  For persons bringing puzzle(s), $4 for each MITCSC member and $6
        for each non-member.  For ``puzzle-less'' persons, $8 for each
        member and $10 for each non-member.

 RSVP:  You may pay at the door, but first contact me so I'll know you are
        coming and can give you directions.  Please email, fax, or phone
        in the following info: Your name, address, phone, fax, email,
        and what you're bringing:

              ___ puzzle-bearing     members at $ 4 each: $___
              ___ puzzle-bearing non-members at $ 6 each: $___
              ___ puzzle-less        members at $ 8 each: $___
              ___ puzzle-less    non-members at $10 each: $___
              ___ <- total persons          total cost -> $___
                 total number of puzzles being brought ___


Michael B. Parker, MIT '89

 email mbparker@mit.edu or mbparker@cytex.com
 1-800-MBPARKER(627-2753) xLIVE(5483) MESG(6374) PAGE(7243) FAXX(3299)
 live 714-639-6436, mesg pager 714-413-2090, fax 714-639-5381

From 100343.236@compuserve.com  Sat Apr 29 15:31:10 1995
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From: Dominic Parkinson <100343.236@compuserve.com>
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From mouse@collatz.mcrcim.mcgill.edu  Thu May  4 17:08:31 1995
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Date: Thu, 4 May 1995 17:08:15 -0400
From: der Mouse <mouse@collatz.mcrcim.mcgill.edu>
Message-Id: <199505042108.RAA02560@Collatz.McRCIM.McGill.EDU>
To: cube-lovers@ai.mit.edu
Subject: Re:  SBP "Magic sQ"

Back on

> Date: Fri, 8 Jul 1994 15:24:30 -0400

I quoted and wrote:

>>                                    Sliding Block Puzzle "Magic sQ"
>> Fig.1 is incomplete.                      +---+---+---+
>> Can you complete a magic square           | 2 | 9 | 4 |
>> with minimum sliding steps?               +---+---+---+
>>                                           | 7 | 5 | 3 |
>> You, very easy or not?                    +---+---+---+---+
>>                                           | 1 | 6 | 8 |   |  Fig.1
>>                                           +---+---+---+---+

> [...].  It's then just a straightforward tree search to find a
> solution.  A simple brute-force "meet in the middle" breadth-first
> search finds a solution easily.

If we mark the squares as

        0 1 2
        3 4 5
        6 7 8 9

then I gave a solution which makes the blank space follow the path 8 7
6 3 4 5 8 7 4 1 2 5 8 7 6 3 4 1 0 3 4 7 6 3 0 1 4 7 8 9.  I remarked:

> This solution exhibits curious near-symmetries in portions of the
> path taken by the blank space.  [...]  Perhaps I should modify the
> program so it reports _all_ solutions of this length;

I finally got around to generating all minimal-length solutions.  It
turns out, curiously enough, that each solution is uniquely determined
by the pattern of its middle position.  There are ten solutions, listed
here in terms of the path taken by the blank space, with the middle
position written out:

                              2 3 *
 8 7 4 3 0 1 2 5 4 1 0 3 4 1  4 9 7  5 8 7 6 3 4 1 2 5 8 7 4 5 8
                              1 5 6

                              2 5 *
 8 5 4 3 0 1 2 5 4 1 0 3 4 1  4 9 7  5 8 7 6 3 4 5 8 7 4 1 2 5 8
                              1 6 3

                              7 2 9
 8 7 4 5 2 1 0 3 4 5 8 7 6 3  4 * 6  1 2 5 8 7 6 3 4 1 0 3 4 5 8
                              3 1 5

                              7 4 3
 8 7 4 1 0 3 4 1 2 5 8 7 6 3  2 * 6  1 2 5 8 7 4 1 0 3 6 7 4 5 8
                              9 1 5

                              2 5 7
 8 5 2 1 4 3 0 1 4 5 2 1 0 3  4 * 9  5 8 7 6 3 4 5 8 7 4 1 2 5 8
                              1 6 3

                              2 4 6
 8 7 6 3 4 5 8 7 4 1 2 5 8 7  5 * 1  1 0 3 6 7 4 3 0 1 4 3 6 7 8
                              7 9 3

                              2 4 6
 8 7 4 5 8 7 6 3 4 1 2 5 8 7  3 9 5  3 4 1 0 3 4 7 6 3 0 1 4 5 8
                              * 7 1

                              2 4 6
 8 7 6 3 4 5 8 7 4 1 2 5 8 7  5 9 1  3 4 1 0 3 4 7 6 3 0 1 4 7 8
                              * 7 3

                              2 3 6
 8 7 4 1 2 5 8 7 6 3 4 1 2 5  9 4 5  7 4 1 0 3 6 7 4 3 0 1 4 5 8
                              7 1 *

                              2 9 7
 8 7 4 3 0 1 4 5 2 1 0 3 4 5  3 4 6  7 6 3 4 1 2 5 8 7 6 3 4 5 8
                              1 5 *

Of course, whether this actually matters to anyone is another story :-)

					der Mouse

			    mouse@collatz.mcrcim.mcgill.edu

From Frank=Lindgreen%DATLING%HHASPR@lng.hha.dk  Mon May  8 03:09:54 1995
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From alan@curry.epilogue.com  Mon May  8 04:01:19 1995
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   unsubscribe

For crying out loud, do -not- send administrative requests to the -entire-
mailing list.  You're the third idiot to make this mistake recently, so I
guess I really do have to bother everybody myself with a reminder.

THINK! THINK! THINK!  If you wanted to stop your subscription to a
magazine, would you send postal mail to -all- of the other subscribers?

Send all administrative requests to:

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Everybody got that?

From BRYAN@wvnvm.wvnet.edu  Tue May  9 09:12:52 1995
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Date:      Tue, 9 May 1995 08:48:23 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <cube-lovers@ai.mit.edu>
Subject:   Re: more on the slice group
In-Reply-To: Message of 04/14/95 at 16:03:31 from mreid@ptc.com

On 04/14/95 at 16:03:31 mreid@ptc.com said:
>mark's post got me thinking ...  i made a quick hack for the slice
>group (which is easy to represent by fixing the corners).  my
>figures concur with his.  i wanted to see the number of local maxima.

>  90 degree     number of      number of
> slice turns    positions     local maxima

>      0              1              0
>      1              6              0
>      2             27              0
>      3            120              0
>      4            287              0
>      5            258             24
>      6             69             69

>as i'd hoped, there are local maxima at distance 5.  one such is:

>          (FB') (RL') (U'D) (R2L2)       =
>          (R2L2) (F'B) (RL') (UD')       =
>          (R'L) (FB') (RL') (F'B) (U'D)  =
>          (U'D) (F'B) (RL') (U'D) (F'B)  =
>          (R'L) (UD') (F'B) (RL') (FB')

>(actually i think all are equivalent to this one under symmetries
>of the cube.)

>this is especially interesting because it is a local maximum in the
>full cube group (quarter turn metric) at distance  10q.  according
>to jerry bryan's results, there are no local maxima within  9q
>of start, so this gives the closest local maximum.  (there may well
>be others.)

Results for the slice group under M-conjugacy:

          Level       Number of        Number of
                      Positions       Local Maxima

           0              1               0
           1              1               0
           2              2               0
           3              6               0
           4             16               0
           5             15               1
           6              9               9

Mike's conjecture that all 24 positions which are a local maxima
at level 5 are equivalent under M-conjugation is correct.

I don't yet understand why Mike's position is a local maximum in the
full cube group.  But assuming it is, it is not only the shortest
local maximum, it is the first local maximum which is not
Q-transitive (i.e, we have |{m'Xm}|=24, hence we have |Symm(X)|=2,
and the size of the symmetry groups for Q-transitive positions
must be divisible by 12.).

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From hoey@aic.nrl.navy.mil  Tue May  9 12:11:08 1995
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From: hoey@aic.nrl.navy.mil (Dan Hoey)
Message-Id: <9505091611.AA26374@Sun0.AIC.NRL.Navy.Mil>
To: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
Cc: "Cube Lovers List" <cube-lovers@ai.mit.edu>
Subject: Re: more on the slice group

Jerry Bryan writes:

> I don't yet understand why Mike's position is a local maximum in the
> full cube group.  But assuming it is, it is not only the shortest
> local maximum, it is the first local maximum which is not
> Q-transitive (i.e, we have |{m'Xm}|=24, hence we have |Symm(X)|=2,
> and the size of the symmetry groups for Q-transitive positions
> must be divisible by 12.).

No, the 4-spot pattern is also a local maximum at 12 qtw, although its
symmetry group is of order 16.  Jim Saxe and I reported this on 22
March 1981, in "No short relations and a new local maximum".

Dan
Hoey@AIC.NRL.Navy.Mil

From BRYAN@wvnvm.wvnet.edu  Tue May  9 18:17:13 1995
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Date:      Tue, 9 May 1995 15:31:32 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <cube-lovers@ai.mit.edu>
Subject:   Re: more on the slice group
In-Reply-To: Message of 04/14/95 at 16:03:31 from mreid@ptc.com

On 04/14/95 at 16:03:31 mreid@ptc.com said:

>i also calculated for the other slice metric.  in this metric,
>neighbors can have the same distance from start, so a "strong"
>local maximum is a position all of whose neighbors are strictly
>closer to start.  a "weak" local maximum is a position none of
>whose neighbors are further from start.

I might prefer an alternative set of definitions:  1) a local
maximum is a position none of whose neighbors are further from
Start, 2) a strong local maximum is a position all of whose
neighbors are strictly closer to Start, and 3) a weak local
maximum is a local maximum which is not a strong local maximum.
But in the table which follows, I adopt Mike's definition.

> 90 or 180 degree     number of     number of strong    number of weak
>   slice turns        positions       local maxima       local maxima

>        0                  1               0                   0
>        1                  9               0                   0
>        2                 51               0                   0
>        3                247               0                   7
>        4                428               0                 212
>        5                 32               8                  32

>the strict local maxima are all equivalent under symmetries of
>the cube.  they are the composition of pons asinorum with any
>of the eight positions called "six dots".

Under M-conjugacy, we have

  90 or 180 degree     number of     number of strong    number of weak
    slice turns        positions       local maxima       local maxima

         0                  1               0                   0
         1                  2               0                   0
         2                  4               0                   0
         3                 15               0                   2
         4                 25               0                  16
         5                  3               1                   3

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From BRYAN@wvnvm.wvnet.edu  Tue May  9 19:19:39 1995
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Date:      Tue, 9 May 1995 19:18:18 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@life.ai.mit.edu>
Subject:   Re: Orders of Symmetry
In-Reply-To: Message of 04/23/95 at 23:29:00 from mark.longridge@canrem.com

On 04/23/95 at 23:29:00 mark.longridge@canrem.com said:

> It took a while to find a pattern which could be transformed 16
>different ways. Still trying to find a pattern which will
>result in 4 distinct ways, but I am not optimistic. A random walk
>through the cube resulted in a pattern which would transform
>48 ways in every case I tried.

For well over 99% of the positions we have |{m'Xm}|=48, so it will
be a long time before a random walk finds anything.  If my quick
and dirty calculations are correct, |{m'Xm}|=48 for in excess
of 99.9999986% of the M-conjugate classes.  For positions themselves,
the percentage would be higher still.

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Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From BRYAN@wvnvm.wvnet.edu  Thu May 11 03:58:11 1995
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Date:      Wed, 10 May 1995 22:38:32 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <cube-lovers@ai.mit.edu>
Subject:   Re: more on the slice group
In-Reply-To: Message of 05/09/95 at 12:11:02 from hoey@AIC.NRL.Navy.Mil

On 05/09/95 at 12:11:02 hoey@AIC.NRL.Navy.Mil said:

>No, the 4-spot pattern is also a local maximum at 12 qtw, although its
>symmetry group is of order 16.  Jim Saxe and I reported this on 22
>March 1981, in "No short relations and a new local maximum".

Argh!  After Dan and Mike pointed this out,  I did remember having seen
it in the archives.  Worse still, Dan pointed it out again on
3 August 1992.  But since it has come up, let's take a brief look
at the 22 March 1981 note.

>    With five-qtw searches, it became possible to check another
>conjecture, using an approach that Jim suggested.  The four-spot
>pattern
>
>           U U U
>           U U U
>           U U U
>
>    R R R  B B B  L L L  F F F
>    R L R  B F B  L R L  F B F
>    R R R  B B B  L L L  F F F
>
>           D D D
>           D D D
>           D D D
>
>is solvable in twelve qtw, either by (FFBB)(UD')(LLRR)(UD') or by its
>inverse, (DU')(LLRR)(DU')(FFBB).  It is immediate that a twelve qtw
>path from this pattern to START can begin with a twist of any face in
>either direction.  The program was used to verify that there are no ten
>qtw paths.  (It generated the set of positions attainable at most five
>qtw from START and the set of positions obtainable from the four-spot
>in at most five qtw, and verified that the intersection of the two sets
>is empty.)  Thus the four-spot is exactly twelve qtw from START and all
>its neighbors are exactly eleven qtw from START, proving that the
>four-spot is a local maximum.
>

Call the 4-spot s.  Then, the twelve neighbors form two M-conjugacy
classes:  N1={sL,sL',sF,sF',sR,sR',sB,sB'} and N2={sU,sU',sD,sD'}.
Also, we have s'=s.  Dan and Jim's solution starts in N1 and ends with
a quarter-turn from N2, and since s'=s, we can say "or vice versa".
Hence, we can start a solution with any of the twelve quarter turns,
and therefore s is a local maximum.

There are other positions with the same symmetry characteristics as
the 4-spot.  That is, there are other positions for which the
symmetry group contains sixteen elements.  There are only three subgroups
of M containing sixteen elements, and the three subgroups are M conjugate.
The three M-conjugates of the 4-spot position correspond to the three
conjugate subgroups of M containing sixteen elements.  But what of
other positions with the same symmetry group?  For example, if the
edges of the 4-spot are all flipped, is the position a local maximum?
I don't know, but it is interesting to see how far we can get without
knowing a process.

Call the 4-spot with all edges flipped t.  Then, we certainly have
t'=t.  Is this true of all positions whose symmetry group contains
sixteen elements?  Also, we certainly have the twelve neighbors
forming M-conjugacy classes similar to those for s, N1 with eight
elements and N2 with four.  Is this true of all positions whose symmetry
group contains sixteen elements?  Finally, a solution either starts in
N1 or starts in N2.  If starting in N1 implies ending with a
quarter-turn from N2 or vice versa, then t is a local maximum.
Can we prove such a thing without actually finding a solution?

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From mreid@ptc.com  Thu May 11 17:40:36 1995
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From: mreid@ptc.com (michael reid)
Message-Id: <9505112157.AA02793@ducie>
To: cube-lovers@ai.mit.edu
Subject:   Re: more on the slice group
Content-Length: 1749

jerry writes

> There are other positions with the same symmetry characteristics as
> the 4-spot.  That is, there are other positions for which the
> symmetry group contains sixteen elements.  There are only three subgroups
> of M containing sixteen elements, and the three subgroups are M conjugate.

these subgroups are the 2-sylow subgroups of M.  one of sylow's
theorems states that any two p-sylow subgroups are conjugate.

one of these subgroups is the group of symmetries that preserve
the U-D axis.  call this subgroup  "P".  (this is also the group
of symmetries of the intermediate subgroup of kociemba's algorithm.)

jerry asks about  P-symmetric  positions.  coincidentally, i happened
to investigate these a few weeks back, and here's what i found:
(i calculated by hand, so i'd be grateful for any confirmation.)

there are 128  P-symmetric  positions, 4 of which are  M-symmetric.
they form a subgroup of the cube group (of course) which is
isomorphic to a direct product of 7 copies of  C_2.  in particular,
each such position has order 2 (or 1) as a group element.  thus,
the answer to jerry's question

> Call the 4-spot with all edges flipped t.  Then, we certainly have
> t'=t.  Is this true of all positions whose symmetry group contains
> sixteen elements?

is "yes".  for what it's worth, this group of 128 positions can be
generated by the seven elements

    superflip
    pons asinorum
    four spots
    slice squared               ( U2 D2 )
    eight flip                  ( FB UD RL FB UD RL )
    four pluses                 ( R2 F2 R2 U'D F2 R2 F2 UD' )
    four swapped corner pairs   ( D' B2 U'D F2 U2 L2 B2 L2 B2 U2 L2 F2 U )

however, these positions are not all locally maximal; for instance
U2 D2  is not.

mike

From patrick@athos.med.auth.gr  Fri May 12 09:10:24 1995
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Subject: Re: unsubscrib
In-Reply-To: <8May1995.031910.Alan@LCS.MIT.EDU>
Message-Id: <Pine.LNX.3.91.950512144544.20345B-100000@athos>
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 I have a complaint to make about the way mr. Allan Bawden treats cube 
-lovers.
1)Alot of people are not sure how to join a group when they start out to 
join one. In fact in most cases the people are new to the Internet , and 
in order to learn how to do things in the famous net , they can join a news 
group .
(I am not sugesting that joining a news group teaches them about the 
internet but its one of the things they can do)
2)Now if Mr Allan responds to their mistakes very rudly ,e.g




On Mon, 8 May 1995, Alan Bawden wrote:

>    Date: Mon,  8 May 95 09:11:57 DST
>    From: Frank=Lindgreen%DATLING%HHASPR@lng.hha.dk
>    Subject: unsubscribe
>    To: CUBE-LOVERS@life.ai.mit.edu
>    Cc: 
> 
>    unsubscribe
> 
> For crying out loud, do -not- send administrative requests to the -entire-
> mailing list.  You're the third idiot to make this mistake recently, so I
> guess I really do have to bother everybody myself with a reminder.
> 



3)"you're the third idiot--" etc I mean one gets quite upset about the 
whole thing and one begins to question if it's worth it joining the news 
group if they are so rude.
I can assure you it's a good thing we communicate by computer, because a 
lot of harm might have been caused on the person of Mr. Allan were he 
speaking straight to a person---I'm sure he does'nt speak that way to 
anyone (even to his kids should he be married with children)
4)I sugest Mr. Allan be firm but polite . It's best to keep the 
correspondence bussness - like. In that way we still remain friends ,and 
quite understand what is expected of us to do.




 > THINK! THINK! THINK!  If


5) It sounds like Mr. Allan thinks that the qube member doesn't think. 
For people interested in cubing thats an insult. We think of ourselves as 
people who can think --on an overage better than most people.



 you wanted to stop your 
subscription to a
> magazine, would you send postal mail to -all- of the other subscribers?
> 
> Send all administrative requests to:
> 
>   Cube-Lovers-Request@AI.MIT.EDU
> 
> Everybody got that?
> 

  
Medicine is for every one
Bravo to Doctors without frontiers
Zimbabwe will always be there



From ed@odi.com  Fri May 12 09:47:07 1995
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To: patrick@athos.med.auth.gr
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        CUBE-LOVERS@life.ai.mit.edu
In-Reply-To: Patrick's message of Fri, 12 May 1995 15:33:12 -0200 (GMT-0200) <Pine.LNX.3.91.950512144544.20345B-100000@athos>
Subject: unsubscrib
Organization: Object Design, 25 Mall Rd, Burlington, MA 01803 - 617-674-5337

  Date: Fri, 12 May 1995 15:33:12 -0200 (GMT-0200)
  From: Patrick <patrick@athos.med.auth.gr>

   I have a complaint to make about the way mr. Allan Bawden treats cube 
  -lovers.
  1)Alot of people are not sure how to join a group when they start out to 
  join one.

Alan's complaint is not about people who are trying to join; they might
legitimately be confused about how to go about it.  However, to prevent
the sort of idiocy that he's complaining about, he sends each new member
of the list a "welcome" message that, among other things, says something
like "Please SAVE this message so you will know how to unsubscribe when
you later decide to do so."

If you fail to read his polite-but-firm message and proceed to do exactly
what he warns you not to do, you must expect him to get legitimately angry.

From SCHMIDTG@beast.cle.ab.com  Fri May 12 09:58:46 1995
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Date: Fri, 12 May 1995 9:58:43 -0400 (EDT)
From: SCHMIDTG@beast.cle.ab.com
To: cube-lovers@ai.mit.edu
Message-Id: <950512095843.20201a78@iccgcc.cle.ab.com>
Subject: Rubik's Robot

At the IPC show, a trade show for industrial controls held in Detroit this
week, Kawasaki had a Rubik's cube solving robot.  The robot has two arms and
hands and a video camera.  The robot is handed a scrambled cube, orients the
cube in front of the camera until it has seen enough faces to deduce the
current state of the cube; displays the number of moves required for its
solution; and then solves the cube using both hands (grippers).  The robot
is capable of manipulating the entire cube using only its hands, without
relying on anything else such as a flat surface.  Unfortunately, I missed
the show so this information was obtained second hand.

-- Greg
schmidtg@iccgcc.decnet.ab.com

From mouse@collatz.mcrcim.mcgill.edu  Fri May 12 10:35:18 1995
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Date: Fri, 12 May 1995 10:35:04 -0400
From: der Mouse <mouse@collatz.mcrcim.mcgill.edu>
Message-Id: <199505121435.KAA22790@Collatz.McRCIM.McGill.EDU>
To: cube-lovers@life.ai.mit.edu
Subject: Re: unsubscrib
Cc: patrick@athos.med.auth.gr

Since Patrick <patrick@athos.med.auth.gr> went public with his[%]
complaint, I'll defend publicly, even though neither one is really
on-topic for cube-lovers.  I hope Alan won't get too upset, and I
really hope this doesn't turn into a flamewar; cube-lovers has been
notably free of them in the time I've been on it.

[%] Gender assumed from the given name.

> I have a complaint to make about the way mr. Allan Bawden
(Whose name you consistently misspell, incidentally, thus being rather
rude to him yourself.  You also make many (other) spelling mistakes,
but this one you make consistently.)
> treats cube-lovers.

> 1)Alot of people are not sure how to join a group when they start out
> to join one.  In fact in most cases the people are new to the
> Internet , and in order to learn how to do things in the famous net ,
> they can join a news group .
> (I am not sugesting that joining a news group teaches them about the
> internet but its one of the things they can do)

You seem to be under the impression that cube-lovers is a "news group".
It's not; it's a mailing list.  If you don't know the difference and
can't find anything local to explain it, send me mail privately and
I'll try to explain it.

Yes, it's true, that's one of the things newbies can do.  But that does
not mean that it's a good thing to do, nor that it's reasonable to try
to do it without first learning the correct way to do it.

And when someone stumbles into a milieu without having first bothered
to learn the rudiments of etiquiette as practiced therein, I don't feel
it's unreasonable to meet that someone with a certain degree of
rejection.

> 2)Now if Mr Allan responds to their mistakes very rudly ,e.g

I notice your address is in Greece, so I'll excuse this on grounds of
ignorance.  In English, one does not use titles (such as Mr.) with
given names.  It would be just "Alan" (not "Allan", since he's Alan
Bawden, not Allan Bawden), or "Mr. Bawden".

>>    Date: Mon,  8 May 95 09:11:57 DST
>>    From: Frank=Lindgreen%DATLING%HHASPR@lng.hha.dk
>>    Subject: unsubscribe
>>    To: CUBE-LOVERS@life.ai.mit.edu
>>    Cc: 
>> 
>>    unsubscribe
>> 
>> For crying out loud, do -not- send administrative requests to the
>> -entire- mailing list.  You're the third idiot to make this mistake
>> recently, so I guess I really do have to bother everybody myself
>> with a reminder.

> 3)"you're the third idiot--" etc I mean one gets quite upset about
> the whole thing and one begins to question if it's worth it joining
> the news group if they are so rude.

I don't consider it excessive to be a little harsh to people who are
being that rude.  When Frank=Lindgreen%DATLING%HHASPR@lng.hha.dk was
subscribed to the list to begin with, he[%] was sent a document
explicitly describing how to unsubscribe.

[%] Gender assumed from what looks like a given name in the address.

If a _subscribe_ request goes to the whole list, it's still someone
stumbling in in ignorance.  But at least it can mostly be excused as
ignorance.  An _unsubscribe_ request can't; anyone with cause to send
an unsubscribe request must already have subscribed, and thus been told
the correct way to unsubscribe.

> 4)I sugest Mr. Allan be firm but polite . It's best to keep the
> correspondence bussness - like. In that way we still remain friends
> ,and quite understand what is expected of us to do.

I suggest you cut "Mr. Allan" some slack.  I have participated in and
mailing lists for many years now, and overall, Alan Bawden is one of
the best list admins I've ever seen.  Yes, it would be better if he[%]
could remain firm and polite in the face of such provocation.  But
(apparently unlike you) I don't expect perfection from him.

[%] Again, gender assumed from the given name.  (English needs a good
    gender-neutral animate set of pronouns.)

Perhaps that merely reflects on the poor crop of list admins available.
I don't think so; I certainly doubt I would do much better.  Perhaps
you feel you could do better; I invite you to try.  Start up a list of
your own and see how it goes.

>> THINK! THINK! THINK!  If
> 5) It sounds like Mr. Allan thinks that the qube member doesn't
> think.

I'm inclined to agree.  If Frank=Lindgreen%DATLING%HHASPR@lng.hha.dk
had thought enough to bother rereading the instructions sent when he
subscribed to the list, or even had bothered to learn basic netiquette
for Internet mailing lists in general, he would have known better than
to send the unsubscribe request to the whole list.

> For people interested in cubing thats an insult.  We think of
> ourselves as people who can think --on an overage better than most
> people.

"can think" != "do think".  As is often amply demonstrated on the net.

>> Send all administrative requests to:
>>   Cube-Lovers-Request@AI.MIT.EDU
>> Everybody got that?

This is important enough to bear re-quoting, I feel.

					der Mouse

			    mouse@collatz.mcrcim.mcgill.edu

From news@nntp-server.caltech.edu  Fri May 12 11:45:39 1995
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Path: whuang
From: whuang@cco.caltech.edu (Wei-Hwa Huang)
Newsgroups: mlist.cube-lovers
Subject: Re: Rubik's Robot
Date: 12 May 1995 15:45:33 GMT
Organization: California Institute of Technology, Pasadena
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SCHMIDTG@beast.cle.ab.com writes:

>At the IPC show, a trade show for industrial controls held in Detroit this
>week, Kawasaki had a Rubik's cube solving robot.  The robot has two arms and
>hands and a video camera.  The robot is handed a scrambled cube, orients the
>cube in front of the camera until it has seen enough faces to deduce the
>current state of the cube; displays the number of moves required for its
>solution; and then solves the cube using both hands (grippers).  The robot
>is capable of manipulating the entire cube using only its hands, without
>relying on anything else such as a flat surface.  Unfortunately, I missed
>the show so this information was obtained second hand.

More information can be found by the WWW.  I remember putting a link to
it on one of my pages at http://www.ugcs.caltech.edu/~whuang/.
-- 
   -- Wei-Hwa Huang (whuang@cco.caltech.edu)
http://www.ugcs.caltech.edu/~whuang/
Proponent of rec.games.computer.puzzle

From alan@curry.epilogue.com  Fri May 12 15:03:16 1995
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Date: Fri, 12 May 1995 15:05:36 -0400
Message-Id: <12May1995.144956.Alan@LCS.MIT.EDU>
From: Alan Bawden <Cube-Lovers-Request@ai.mit.edu>
Sender: Cube-Lovers-Request@ai.mit.edu
To: Cube-Lovers@ai.mit.edu
Subject: troublemakers and newbies

Folks, please let's not have any more discussion of mailing list
maintenance on Cube-Lovers.  Let me be the only person to break that rule.
I try to do that as rarely as possible.  That's why I usually wait for more
than one person to make the canonical "unsubscribe" mistake before saying
anything in public.  (Unfortunately, the way people always copy the bad
example makes it necessary to offer an occasional public correction.)

You don't see the mail that goes to Cube-Lovers-Request.  If you did, you
would know that public discussions like this always prompt other people to
drop their subscriptions.  Almost every off-topic message is the last straw
for somebody out there.  Please help Cube-Lovers keep it's subscribers --
let me be the only person who sends messages like this.

				- Alan

From mouse@collatz.mcrcim.mcgill.edu  Mon May 15 16:16:34 1995
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Date: Mon, 15 May 1995 16:16:29 -0400
From: der Mouse <mouse@collatz.mcrcim.mcgill.edu>
Message-Id: <199505152016.QAA02780@Collatz.McRCIM.McGill.EDU>
To: cube-lovers@ai.mit.edu
Subject: Re:  Is there a symbolic cube program?

Back last November, Dave Eaton <devo@vnet.ibm.com> wrote

> Is there a program that allows you to type in Singmaster-style moves
> and then prints out the resultant state, something like this (not
> actual results):

> INPUT:   (R U2 R3 U2)2
> OUTPUT:  (fur,drb,rdf) (fr,dr)

I can't recall whether anyone offered such a thing or not.  I think
there were some related programs, but nothing quite like this.

Anyway, there is now. :-)

Sample run:

% twist
> .set SLICER CUBER R' L
`SLICER' defined
> (SLICER U)4
Cube:
              u b u
              l u u
              u u u
        l u l f f f r r r b u b
        l l l f f f r r r b b b
        l l l f d f r r r b d b
              d f d
              d d d
              d b d
Cycles: (ub)+ (ul)+ (fd)+ (bd)+
Already centered
> .set WRENCH LAST
`WRENCH' defined
> WRENCH U WRENCH U'
Cube:
              u b u
              u u u
              u f u
        l l l f u f r r r b u b
        l l l f f f r r r b b b
        l l l f f f r r r b b b
              d d d
              d d d
              d d d
Cycles: (ub)+ (uf)+
Already centered
> (R U2 R3 U2)2
Cube:
              b u d
              f u u
              f u u
        r r r u f b l l r f b u
        l l l f f u r r r b b b
        l l l f f f l r r b b b
              d d u
              d d d
              d d d
Cycles: (ul,ur,fr) (ulb,urf,flu,frd,bru)
Already centered
> SLICER U
Cube:
              u u u
              f f f
              u u u
        f d f r r r b u b l l l
        l l l f d f r r r b u b
        l l l f d f r r r b u b
              d b d
              d b d
              d b d
Cycles: (u,b,d,f) (ub,bd,df,lu)+ (ur,uf)+ (ulb,ubr,urf,ufl)
Centred: (ul,fu,fr,dr,br,ur,fd,fl,dl,bl) (ulb,fur,lfd,ldb)+ (ubr,frd,drb)+ (ufl)+
> WRENCH CUBEU2 CUBEL WRENCH' (CUBEU2 CUBEL)'
Cube:
              u u u
              l u u
              u u u
        l u l f f f r r r b b b
        l l l f f r f r r b b b
        l l l f f f r r r b b b
              d d d
              d d d
              d d d
Cycles: (ul)+ (fr)+
Already centered
> 

The program is up for anonymous ftp from collatz.mcrcim.mcgill.edu in
/games/cube/twist.c.  You have not a prayer of compiling it under
anything but gcc as it stands, though I think it would be relatively
easy (but perhaps rather ugly) to turn it into vanilla C.  But with gcc
2.6.3, it builds fine for me on a Sun under SunOS 4.1.3 and a NeXT
under NeXT release 2.1, and probably most other machines too.

Read the comment header for a fuller description of its capabilities.

					der Mouse

			    mouse@collatz.mcrcim.mcgill.edu

From bagleyd@source.asset.com  Tue May 16 11:06:02 1995
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From: bagleyd@source.asset.com (David A. Bagley)
Message-Id: <9505161507.AA44189@source.asset.com>
To: cube-lovers@ai.mit.edu
Subject: DINOSAUR RUBIK'S CUBE

Hi
  I just completed coding a new puzzle called xdino.  It is the Rubik's
Dinosaur Puzzle recently mentioned.  (A cube with diagonal X cuts.)
All those who have X should be able to run it.  Also I updated the rest
for the puzzle collection as well.
 
  It is available at ftp.x.org in /contrib/games/puzzles in separate
files.  It will also  be available at sunsite.unc.edu in
/pub/Linux/games/x11 as xpuzzle-4.10.tgz
Any problems with the compilation ... let me know.
It has been tested on Linux, SunOS, Solaris, HP-UX, and VMS.
 
  Here is the README file:
 
 
 
Updates:
  xpuzzles (4.10)
    Random number generator included.
    All puzzles have been put through Sun's cc and lint
    xdino added
    Bug fixed in xmlink. It moved correctly but was hard to turn.
    Bug fixed with control key of xpyraminx.  It turned the whole puzzle
      the wrong way.
    New control key moves for the 2D version of xskewb.
    More freedom in movement in xoct and xpyraminx using control+shift.
    (Later updates to individual puzzles will now be 4.10.1 etc.
     No more different minor version numbers for each puzzle.)
 
  xpuzzles (<4.10)
    Removed lint warnings and added a VMS make.com .
    Conservative guess for random number generator.
    A super Makefile to make all puzzles.
    Puzzles now have undo, save, and recall features.
    xmball and xmlink intitialization bug fixed.
    xmball and xmlink added, both need more efficient methods to draw a sector.
    xrubik only save and undo bug fixed.
      After a save, undo did not work.
      auto-solver - sincere thanks to Michael B. Martin
       <martinm@sps1.phys.vt.edu>
    Some older versions used Motif (3.x), XView (2.x), and SunView (1.x)
 
xrubik is currently the only one in this collection with a auto-solver.
 
The collection includes:
SLIDING BLOCK PUZZLES
xcubes:         expanded 15 puzzle
xtriangles:     same complexity as 15 puzzle
xhexagons:      2 modes: one ridiculously easy, one harder than 15 puzzle
 
ROTATIONAL 3D PUZZLES
  hold down control key to move whole puzzle
  letters that represent colors can be changed in mono-mode
 
xrubik:         a nxnxn Erno Rubik's Cube(tm) (or Magic Cube)
                auto-solves 2x2x2 and 3x3x3 (non-orient mode).
xpyraminx:      a nxnxn Uwe Meffert's Pyraminx(tm) (and Senior Pyraminx),
                a tetrahedron with Period 2, Period 3, and Combined cut modes
                and it also a sticky mode to simulate a Halpern's Tetrahedron
                or a Pyraminx Tetrahedron
xoct:           a nxnxn Uwe Meffert's Magic Octahedron (or Star Puzzler) and
                Trajber's Octahedron with Period 3, Period 4, and Combined
                cut modes and it also includes a sticky mode
xskewb:         a Meffert's Skewb (or Pyraminx Cube), a cube with diagonal
                cuts, each face is cut with a diamond shape
xdino:          a Dinosaur Rubik's Cube, a cube with different diagonal cuts,
                each face is cut with a "X"
xmball:         a variable cut Masterball(tm), variable number of latitudinal
                and longitudinal cuts on a sphere, where the longitudinal cuts
                permit only 180 degree turns.
 
COMBINATION ROTATIONAL AND SLIDING 3D PUZZLES
  hold down shift key to move whole puzzle
  letters that represent colors can be changed in mono-mode
xmlink:         a nxm Erno Rubik's Missing Link(tm)
 
Newbies (especially DOS users 8-) ):
  MS DOS/MS Windows & Mac users, sorry no port currently available.
  What you need:
    80386 or better, or Risc, etc.
    UNIX (or even VMS): Linux and FreeBSD are freely available.
    X: XFree86 is freely available on Linux and FreeBSD distributions.
    gunzip: freely available from GNU and the above distributions.
    tar: freely available from GNU and the above distributions.
  What you do:
    After transfering the PUZZLE file to your machine
      gunzip PUZZLE.tar.gz
      tar xvf PUZZLE.tar
        (tar xvzf PUZZLE.tar.gz may work as a short cut)
    Then read the README generated by the above command.
  Questions about the above or how to find out more about puzzles, no problem,
    my mail address is: bagleyd@source.asset.com

From BRYAN@wvnvm.wvnet.edu  Thu May 18 10:17:25 1995
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Date:      Thu, 18 May 1995 09:29:24 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <mreid@ptc.com>, "Cube Lovers List" <cube-lovers@ai.mit.edu>
Subject:   Re: more on the slice group
In-Reply-To: Message of 05/11/95 at 17:57:14 from mreid@ptc.com

On 05/11/95 at 17:57:14 mreid@ptc.com said:

>jerry asks about  P-symmetric  positions.  coincidentally, i happened
>to investigate these a few weeks back, and here's what i found:
>(i calculated by hand, so i'd be grateful for any confirmation.)

>there are 128  P-symmetric  positions, 4 of which are  M-symmetric.

I can confirm your count.  These positions are called X class positions
in Dan's taxonomy.  That is, there are 124 positions Y unique up to
M-conjugancy for which Symm(Y)=X1 or Symm(Y)=X2 or Symm(Y)=X3, where
X1, X2, and X3 are the three conjugate subgroups of M in Dan's
taxonomy of subgroups which contain sixteen elements. Hence, we
describe the 124 positions as being class X positions.  The 4
M-symmetric positions are also X-symmetric, but we don't count them
as class X.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From BRYAN@wvnvm.wvnet.edu  Thu May 18 11:30:01 1995
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Date:      Thu, 18 May 1995 11:28:36 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <mreid@ptc.com>, "Cube Lovers List" <cube-lovers@ai.mit.edu>
Subject:   Re: more on the slice group
In-Reply-To: Message of 05/11/95 at 17:57:14 from mreid@ptc.com

On 05/11/95 at 17:57:14 mreid@ptc.com said:

>one of these subgroups is the group of symmetries that preserve
>the U-D axis.  call this subgroup  "P".  (this is also the group
>of symmetries of the intermediate subgroup of kociemba's algorithm.)

>there are 128  P-symmetric  positions, 4 of which are  M-symmetric.
>they form a subgroup of the cube group (of course) which is
>isomorphic to a direct product of 7 copies of  C_2.  in particular,
>each such position has order 2 (or 1) as a group element.

If I understand your definition of "P" correctly, the same group is
called X1 in Dan's taxonomy.  X2 similarly preserves the F-B
axis, and X3 similarly preserves the R-L axis.  Hence, there are
three conjugate subgroups of G which preserve a major axis, and
each contains 128 elements:  there are 128 X1-symmetric positions,
128 X2 symmetric positions, and 128 X3 symmetric positions.

I was bothered by your statement that there were 128 P-symmetric
positions at first because I was equating "P-symmetric" with
"X-symmetric" rather than with "X1-symmetric".  There should be
376 X-symmetric positions  --  124 that are X1-symmetric and not
M-symmetric, 124 that are X2-symmetric and not M-symmetric, 124
that are X3-symmetric and not M-symmetric, and 4 that are M-symmetric.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From BRYAN@wvnvm.wvnet.edu  Fri May 19 10:35:47 1995
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Date:      Fri, 19 May 1995 09:00:52 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   AntiSlice Under M-conjugacy

Mark Longridge's antislice results are as follows:

>                  arrangements           arrangements
>Moves Deep   (2q or anti-slice moves)   (4q or double anti-slice moves)
>
>    0                   1                     1
>    1                   6                     9
>    2                  27                    51
>    3                 120                   265
>    4                 423                   864
>    5               1,098                 1,785
>    6               1,770                 2,017
>    7               1,650                 1,008
>    8                 851                   144
>    9                 198
>                    -----                 -----
>                    6,144                 6,144

We now have the following M-conjugacy results (2q moves only,
still working on 4q moves).

  Level              Positions             Local
                                          Maxima

   0                    1                    0
   1                    1                    0
   2                    3                    0
   3                   10                    0
   4                   37                    0
   5                   93                    1
   6                  166                    2
   7                  147                    7
   8                   89                   12
   9                   21                   21
                     ----
                      568

The level 5 local maximum is (U'D')(FB)(FB)(UD)(L'R').  The position is
not its own inverse, but we can use as an inverse (U'D')(FB)(FB)(UD)(LR).
Hence, (U'D')(FB)(FB)(UD) forms a nice "middle" of the sequence.  In
fact, the (U'D')(FB)(FB)(UD) position in some ways seems more
interesting than the local maximum itself.  Does it already have
a name?

I have not verified if the length of the local maximum is 10q in G,
nor if it is a local maximum in G.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU



From alan@curry.epilogue.com  Fri May 19 22:01:05 1995
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Date: Fri, 19 May 1995 22:01:05 -0400
Message-Id: <19May1995.194423.Alan@LCS.MIT.EDU>
From: Alan Bawden <Cube-Lovers-Request@ai.mit.edu>
Sender: Cube-Lovers-Request@ai.mit.edu
Reply-To: Cube-Lovers-Request@ai.mit.edu
To: Cube-Lovers@ai.mit.edu
Subject: The Cube-Lovers Archives expands

Occasionally people have approached me and wondered if there was an FTP
archive of Cube-related material (programs, documents, databases, pictures,
whatever).  I have always replied that the only archive I knew of was the
archive of Cube-Lovers mail that I maintain.  Since this question keeps
coming up, there must be a need to be filled, so I propose to expand our
archives to cover any additional Cube-related material that people might
care to submit.

If you would like to submit a contribution to this archive, please send
mail to Cube-Lovers-Request@AI.MIT.EDU (please do -not- send mail to all of
Cube-Lovers) and include:

  o  The location where I can pick up the files you wish to contribute
     (preferably using anonymous FTP).

  o  A brief description of your contribution, to be included in a master
     index file.

I reserve the right to redescribe, repackage, rename, recompress or totally
reject your contribution.

Periodically I will announce new additions to the archive to all of
Cube-Lovers.  Currently the archives contain nothing other than the
electronic mail archives.  (Although I know of at least one potential
contributor who's been waiting in the wings for a couple of months now...)

Some of you will no doubt have forgotten where the archive is: 

Using FTP, connect to FTP.AI.MIT.EDU, login as "anonymous" (any password),
and go to the directory "pub/cube-lovers".  (From the World Wide Web, you
can use the URL: "ftp://ftp.ai.mit.edu/pub/cube-lovers".)

				- Alan

From BRYAN@wvnvm.wvnet.edu  Sun May 21 15:24:47 1995
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Date:      Sun, 21 May 1995 07:18:49 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   AntiSlice Under M-conjugacy (and a problem with slice)

I have had some posts not get through.  The following will serve
to consolidate several of them. Some of this may be a repeat,
but not all, I think.

Mark Longridge's antislice results are as follows:

>                  arrangements           arrangements
>Moves Deep   (2q or anti-slice moves)   (4q or double anti-slice moves)
>
>    0                   1                     1
>    1                   6                     9
>    2                  27                    51
>    3                 120                   265
>    4                 423                   864
>    5               1,098                 1,785
>    6               1,770                 2,017
>    7               1,650                 1,008
>    8                 851                   144
>    9                 198
>                    -----                 -----
>                    6,144                 6,144

We have the following M-conjugacy results for 2q moves.

  Level              Positions             Local
                                          Maxima

   0                    1                    0
   1                    1                    0
   2                    3                    0
   3                   10                    0
   4                   37                    0
   5                   93                    1
   6                  166                    2
   7                  147                    7
   8                   89                   12
   9                   21                   21
                     ----
                      568

The level 5 local maximum is (U'D')(FB)(FB)(UD)(L'R').  The position is
not its own inverse, but we can use as an inverse (U'D')(FB)(FB)(UD)(LR).
Hence, (U'D')(FB)(FB)(UD) forms a nice "middle" of the sequence.  In
fact, the (U'D')(FB)(FB)(UD) position in some ways seems more
interesting than the local maximum itself.  Does it already have
a name?

I have not verified if the length of the local maximum is 10q in G,
nor if it is a local maximum in G.

We have the following M-conjugacy results for 4q moves.  Strong
and weak local maxima are defined according to my preference.
If you prefer Mike Reid's definition, ignore the "weak"
column and read the "total" column as "weak".

      Level       Positions     Strong      Weak        Total
                               Local Max   Local Max   Local Max

        0              1           0           0           0
        1              2           0           0           0
        2              5           0           0           0
        3             25           0           1           1
        4             75           0           2           2
        5            152           0          19          19
        6            184           1          35          36
        7            108           0          46          46
        8             16           0          16          16
                    ----
                     568

Back on the subject of the slice group, we have the following.

Mark Longridge said:

>By the way GAP gives  NumberConjugacyClasses (slice) = 23

>In your calculations of M-conjugacy classes for the slice group the
>total number of classes is 50, but I think GAP does not use
>M-conjugates but C-conjugates instead. The NumberConjugacyClasses
>function always thrashes with any larger groups unfortunately.
>If you could easily tweak your program perhaps you could
>verify my theory.

Recall that in my work with <U,R>, I had to use W3-conjugacy rather
than M-conjugacy.  The simplest explanation is that all M-conjugates
of a position in <U,R> are not in <U,R>, and in particular the
representative element might not be in <U,R>.  W3 is the largest
subgroup of M such that all conjugates of <U,R> are in <U,R>.

I flirted briefly with the notion that I might have the same problem
with the slice group and the antislice group.  But it seems
immediate that M-conjugacy is appropriate for both slice and
antislice.  For example, think of applying M-conjugacy to
all the individual 2q or 4q moves in a slice or antislice process.
Clearly, the result is still in slice and antislice, respectively.

I doubt that Mark's theory about GAP using C-conjugacy for slice
instead of M-conjugacy is correct.  I already have 50 positions
to 23 for GAP, and using C-conjugacy would just make my results
larger.  For example, RL' and R'L are M-conjugate positions,
but not C-conjugate positions.  I don't have a clue why my
results do not match GAP.  I have double and triple checked
my results, and they seem correct.  For example, I can "expand"
my conjugacy classes, and the results then match Mark's exactly.

How does GAP's NumberConjugacyClasses function work?  By that,
I mean how does it know the subgroup with respect to which
you are taking conjugacy classes (if my terminology is correct)?
For example, how does it know to take C or M or whatever
conjugates?

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From mschoene@math.rwth-aachen.de  Mon May 22 05:14:39 1995
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Date: Mon, 22 May 95 11:13 WET DST
From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: Cube-Lovers@ai.mit.edu
Cc: BRYAN@wvnvm.wvnet.edu
In-Reply-To: "Jerry Bryan"'s message of Sun, 21 May 1995 07:18:49 -0400 (EDT) <wvmail32.1995may19.084306.bryan@wvnvm.wvnet.edu>
Subject: Re: AntiSlice Under M-conjugacy (and a problem with slice)

Sorry, lately I didn't have any time to follow the discussions on
Cube-Lovers (preparing the second upgrade of GAP 3.4 and working long
hours for GAP 4.0).  But Jerry Bryan's message talks about GAP's
'NumberConjugacyClasses' function.

Jerry wrote

    I doubt that Mark's theory about GAP using C-conjugacy for slice
    instead of M-conjugacy is correct.  I already have 50 positions
    to 23 for GAP, and using C-conjugacy would just make my results
    larger.  For example, RL' and R'L are M-conjugate positions,
    but not C-conjugate positions.  I don't have a clue why my
    results do not match GAP.  I have double and triple checked
    my results, and they seem correct.  For example, I can "expand"
    my conjugacy classes, and the results then match Mark's exactly.

GAP's 'NumberConjugacyClasses' follows the general usage in group theory.
The conjugacy class of an element <g> of <G> is the set of elements <h>
that are G-conjugated to <g> (i.e., there exists an element <x> in <G>,
such that <x>^-1 * <g> * <x> = <h>).

Thus GAP is using <slice>-conjugacy classes.  Since GAP is using a
*larger* group, it is not surprising that GAP finds less conjugacy
classes (if M were a subgroup of <slice>, then this had to be so,
because in this case every M-conjugacy class would be a subset of a
<slice>-conjugacy class).

Jerry continued

    How does GAP's NumberConjugacyClasses function work?  By that,
    I mean how does it know the subgroup with respect to which
    you are taking conjugacy classes (if my terminology is correct)?
    For example, how does it know to take C or M or whatever
    conjugates?

It always takes the whole group itself as the acting group.  With some
related functions (e.g., 'ConjugacyClass' itself) you can specify that
you want another group acting, but not with 'NumberConjugacyClasses'.

Mark Longridge wrote (as cited by Jerry)

    In your calculations of M-conjugacy classes for the slice group the
    total number of classes is 50, but I think GAP does not use
    M-conjugates but C-conjugates instead. The NumberConjugacyClasses
    function always thrashes with any larger groups unfortunately.
    If you could easily tweak your program perhaps you could
    verify my theory.

As I wrote above GAP does not use C-conjugates but <slice>-conjugates.
Conjugacy classes in permutation groups are notoriously difficult.
Computing the conjugacy classes of G (the full cube group) for example
is absolutely impossible (without using some theory anyway).

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From mreid@ptc.com  Mon May 22 17:31:26 1995
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From: mreid@ptc.com (michael reid)
Message-Id: <9505222148.AA21637@ducie>
To: cube-lovers@ai.mit.edu
Subject:   AntiSlice Under M-conjugacy (and a problem with slice)
Content-Length: 1713

jerry writes (about the antislice group)

> The level 5 local maximum is (U'D')(FB)(FB)(UD)(L'R').

[ ... ]

> I have not verified if the length of the local maximum is 10q in G,
> nor if it is a local maximum in G.

this is exactly what i tried to explain in my recent posts: the latter
statement follows from the former.  and yes, its length is indeed  10q.
it's pretty easy to find maneuvers which end in each quarter turn:

     (F'B') (UUDD) (FB) (RL)    =
     (U'D') (FFBB) (UD) (R'L')  =
     (R'L') (U'D') (FFBB) (UD)  =
     (R'L') (UD) (FFBB) (U'D')  =
     (RL) (F'B') (UUDD) (FB)    =
     (RL) (FB) (UUDD) (F'B').

> We have the following M-conjugacy results for 4q moves.  Strong
> and weak local maxima are defined according to my preference.

it seems like jerry's terminology is more reasonable, so i'll stop
using mine.  jerry's figures

>       Level       Positions     Strong      Weak        Total
>                                Local Max   Local Max   Local Max
> 
>         0              1           0           0           0
>         1              2           0           0           0
>         2              5           0           0           0
>         3             25           0           1           1
>         4             75           0           2           2
>         5            152           0          19          19
>         6            184           1          35          36
>         7            108           0          46          46
>         8             16           0          16          16
>                     ----
>                      568

beg the obvious question:  what is that strong local maximum,
which is unique up to symmetry?

mike

From hoey@aic.nrl.navy.mil  Tue May 23 13:11:33 1995
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From: hoey@aic.nrl.navy.mil
Message-Id: <9505231711.AA26574@sun13.aic.nrl.navy.mil>
To: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>, cube-lovers@ai.mit.edu
Subject: M-conjugacy vs. C-Conjugacy in the Slice group

Jerry Bryan tried to communicate about this to cube-lovers, but
there's apparently a technical difficulty.

On 20 May, Jerry said:

>I doubt that Mark's theory about GAP using C-conjugacy for slice
>instead of M-conjugacy is correct.  I already have 50 positions
>to 23 for GAP, and using C-conjugacy would just make my results
>larger.  For example, RL' and R'L are M-conjugate positions,
>but not C-conjugate positions.

I emailed him to note to the contrary that RL' and R'L are indeed
C-conjugates, for example under 180 degree rotation around the F-B
axis.  I did wonder, though, whether that meant that there could be
C-conjugate slice positions that were not M-conjugate.

He emailed me:

> We can observe that R and R' are not C-conjugates, nor are L' and L,
> which suckered me into stating that RL' and R'L are not.  But
> rewrite R'L as LR' since opposite face moves commute.  Now, RL'
> and LR' are clearly C-conjugate.

> In fact, I have now verified with a quick search program that
> all M-conjugates in the slice group are also C-conjugates.  Hence,
> there are 50 C-conjugate classes in slice, just as there are
> 50 M-conjugate classes.

> In retrospect, I don't think the search program was necessary....

and continues with an argument that did not convince me, but the
following does:

First, consider the central inversion v, which maps each point of the
cube to its diametric opposite.  Conjugation by v maps each face-turn
(e.g. F) with its diametric opposite in opposite sense (B').  Since
these are the pairs that constitute a slice move, and they commute, we
have:

    v' FB' v = (v' F v)  (v' B' v) = B' F = F B',

and similarly for the other slice moves, showing that each slice move
is its own v-conjugate.  This extends to a proof that each position in
the slice group is its own v-conjugate:

    v' s1 s2 ... sn v = (v' s1 v) (v' s2 v) ... (v' sn v)
                      = s1 s2 ... sn.

Suppose that we have two M-conjugate positions X, Y in the slice
group.  So X = m' Y m for some m in M.  If m is in C, then X and Y are
C-conjugate and we are done.  Otherwise take the central inversion v;
we know that mv is in C.  We also know that

    X = v' X v = v' m' Y m v = (mv)' Y (mv).

So X and Y are C-conjugate in this case as well.  QED.

Note: "Being its own v-conjugate" might as well be called "being
v-symmetric".

Dan
Hoey@AIC.NRL.Navy.Mil

From BRYAN@wvnvm.wvnet.edu  Tue May 23 14:39:03 1995
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Date:      Tue, 23 May 1995 14:04:03 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@lcs.mit.edu>
Subject:   M-conjugacy vs. C-Conjugacy in Slice and Antislice

On 20 May, I said the following:

>I doubt that Mark's theory about GAP using C-conjugacy for slice
>instead of M-conjugacy is correct.  I already have 50 positions
>to 23 for GAP, and using C-conjugacy would just make my results
>larger.  For example, RL' and R'L are M-conjugate positions,
>but not C-conjugate positions.

We already know from Martin Schoenert that GAP is using neither
M-conjugacy nor C-conjugacy, but <slice>-conjugacy.  But my
statement about C-conjugacy vs. M-conjugacy was completely incorrect
in any case.

Dan Hoey pointed out to me that RL' and R'L in fact *are* C-conjugates
under 180 degree rotation around the U-D axis.

We can observe that R and R' are not C-conjugates, nor are L' and L,
which suckered me into stating that RL' and R'L are not.  But
rewrite R'L as LR' since opposite face moves commute.  Now, RL'
and LR' are clearly C-conjugate.

In fact, I have now verified with a quick search program that
all M-conjugates in the slice group are also C-conjugates.  Hence,
there are 50 C-conjugate classes in slice, just as there are
50 M-conjugate classes.

In retrospect, I don't think the search program was necessary.
Suppose X and Y are M-conjugates in the slice group.  Then,
they can be written as M-conjugate sequences.  That is, they
can be written so that the individual slice moves are respective
M-conjugates for some fixed m in M.  (The fact that it might be
possible also to write them so that the individual slice moves
are not respective M-conjugates for some fixed m in M is
irrelevant.)  Furthermore, write the sequence for X so that the
clockwise half of each slice is written prior to the
counter-clockwise half of the slice.  The sequence for
Y with individual slice moves being respective M conjugates of
X may or not have this property.  But if not, then simply reorder
the halves of the slices of Y to put the clockwise half first, and Y
will still be piecewise M-conjugate with X.  Then, the piecewise
M-conjugate slices are also C-conjugate, and therefore the X and
Y positions are C-conjugate.

Antislice is totally different.  For example, RL is M-conjugate to
R'L', but it not C-conjugate.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From BRYAN@wvnvm.wvnet.edu  Tue May 23 14:43:48 1995
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Date:      Tue, 23 May 1995 08:50:40 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   M-conjugacy vs. C-Conjugacy in Slice and Antislice

On 20 May, I said the following:

>I doubt that Mark's theory about GAP using C-conjugacy for slice
>instead of M-conjugacy is correct.  I already have 50 positions
>to 23 for GAP, and using C-conjugacy would just make my results
>larger.  For example, RL' and R'L are M-conjugate positions,
>but not C-conjugate positions.

We already know from Martin Schoenert that GAP is using neither
M-conjugacy nor C-conjugacy, but <slice>-conjugacy.  But my
statement about C-conjugacy vs. M-conjugacy was completely incorrect
in any case.

Dan Hoey pointed out to me that RL' and R'L in fact *are* C-conjugates
under 180 degree rotation around the U-D axis.

We can observe that R and R' are not C-conjugates, nor are L' and L,
which suckered me into stating that RL' and R'L are not.  But
rewrite R'L as LR' since opposite face moves commute.  Now, RL'
and LR' are clearly C-conjugate.

In fact, I have now verified with a quick search program that
all M-conjugates in the slice group are also C-conjugates.  Hence,
there are 50 C-conjugate classes in slice, just as there are
50 M-conjugate classes.

In retrospect, I don't think the search program was necessary.
Suppose X and Y are M-conjugates in the slice group.  Then,
they can be written as M-conjugate sequences.  That is, they
can be written so that the individual slice moves are respective
M-conjugates for some fixed m in M.  (The fact that it might be
possible also to write them so that the individual slice moves
are not respective M-conjugates for some fixed m in M is
irrelevant.)  Furthermore, write the sequence for X so that the
clockwise half of each slice is written prior to the
counter-clockwise half of the slice.  The sequence for
Y with individual slice moves being respective M conjugates of
X may or not have this property.  But if not, then simply reorder
the halves of the slices of Y to put the clockwise half first, and Y
will still be piecewise M-conjugate with X.  Then, the piecewise
M-conjugate slices are also C-conjugate, and therefore the X and
Y positions are C-conjugate.

Antislice is totally different.  For example, RL is M-conjugate to
R'L', but it not C-conjugate.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From BRYAN@wvnvm.wvnet.edu  Wed May 24 15:50:47 1995
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Date:      Wed, 24 May 1995 15:46:45 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <cube-lovers@lcs.mit.edu>
Subject:   Re: AntiSlice Under M-conjugacy (and a problem with slice)
In-Reply-To: Message of 05/22/95 at 17:48:42 from mreid@ptc.com

On 05/22/95 at 17:48:42 mreid@ptc.com said:

>>       Level       Positions     Strong      Weak        Total
>>                                Local Max   Local Max   Local Max
>>
>>         6            184           1          35          36

>beg the obvious question:  what is that strong local maximum,
>which is unique up to symmetry?

I haven't verified in the antislice data base, but it *has* to be
Pons Asinorum, accomplshed as six antislices.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From BRYAN@wvnvm.wvnet.edu  Wed May 24 17:14:08 1995
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Date:      Wed, 24 May 1995 16:36:30 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@lcs.mit.edu>
Subject:   Re: AntiSlice Under M-conjugacy (and a problem with slice)
In-Reply-To: Message of 05/22/95 at 11:13:00 from ,
           Martin.Schoenert@math.rwth-aachen.de

On 05/22/95 at 11:13:00 Martin Schoenert said:

>GAP's 'NumberConjugacyClasses' follows the general usage in group theory.
>The conjugacy class of an element <g> of <G> is the set of elements <h>
>that are G-conjugated to <g> (i.e., there exists an element <x> in <G>,
>such that <x>^-1 * <g> * <x> = <h>).

Just to give an example that I am familiar with, suppose the group
in question were M itself.  Then, NumberConjugacyClasses should yield
10, because the 48 elements in M yield 10 conjugacy classes under
M-conjugation.  If anybody who has GAP also has defined M, you
might give it a try.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From mschoene%math.rwth-aachen.de@samson.math.rwth-aachen.de  Thu May 25 09:39:22 1995
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From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: Cube-Lovers@lcs.mit.edu
Cc: BRYAN@wvnvm.wvnet.edu
In-Reply-To: "Jerry Bryan"'s message of Wed, 24 May 1995 16:36:30 -0400 (EDT) <wvmail32.1995may24.155259.bryan@wvnvm.wvnet.edu>
Subject: Re: Re: AntiSlice Under M-conjugacy (and a problem with slice)

Jerry Bryan wrote

    Just to give an example that I am familiar with, suppose the group
    in question were M itself.  Then, NumberConjugacyClasses should yield
    10, because the 48 elements in M yield 10 conjugacy classes under
    M-conjugation.  If anybody who has GAP also has defined M, you
    might give it a try.

GAP also thinks that M has 10 conjugacy classes (under M-conjugation).

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From BRYAN@wvnvm.wvnet.edu  Thu May 25 12:01:36 1995
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Date:      Thu, 25 May 1995 11:50:55 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Dan Hoey" <hoey@aic.nrl.navy.mil>,
        "Cube Lovers List" <cube-lovers@lcs.mit.edu>
Subject:   Re: M-conjugacy vs. C-Conjugacy in the Slice group
In-Reply-To: Message of 05/23/95 at 13:11:27 from hoey@AIC.NRL.Navy.Mil

I said:

>> In fact, I have now verified with a quick search program that
>> all M-conjugates in the slice group are also C-conjugates.  Hence,
>> there are 50 C-conjugate classes in slice, just as there are
>> 50 M-conjugate classes.

>> In retrospect, I don't think the search program was necessary....

On 05/23/95 at 13:11:27 hoey@AIC.NRL.Navy.Mil said:

>and (Jerry) continues with an argument that did not convince me, but the
>following does:

I think I can both greatly simplify and greatly strengthen the
argument that did not convince Dan.  My argument is based on the idea
(copied from _Symmetry and Local Maxima_) that M-conjugation can be
viewed as a permutation on Q, the set of twelve quarter turns.

Call the six clockwise quarter turns Q+ and the six counter-clockwise
quarter turns Q-.  We can observe that the 24 rotations in M all
map Q+ to Q+ and map Q- to Q-, and that the 24 reflections in M all
map Q+ to Q- and map Q- to Q+.  We also note that in particular,
the central inversion v is a reflection.

Suppose X and Y are M-conjugates in <slice> with Y=m'Xm for some fixed
m in M.  Write X as pairs of quarter turns (each pair is a slice), and
write Y as pairs of quarter turns which are respective M-conjugates
(via the fixed permutation m) of the quarter turns in X.  If the
respective quarter turns have been mapped Q+ to Q+ and Q- to Q-, then
m is a rotation and we are done.

Otherwise, commute the halves of each slice in Y.  We first note that
so commuting is the identity on Y.  We also note that so commuting is
equivalent to performing the permutation operation v on Q, and is
therefore equivalent to performing v-conjugation on Y. (In passing,
we see that this effectively proves Dan's first point, namely that
X=v'Xv for all X in <slice>.  Given that, I would shorten the rest of
Dan's argument by saying Y=m'Xm=v'('m'Xm)v=v'm'Xmv, and noting that
either m or mv is a rotation).

But having started with the "commuting the halves of slices" argument,
I would continue as follows.  Having commuted the halves of the slices,
we still have an M-conjugate (and still the same M-conjugate)
because commuting is equivalent to v-conjugation, v is in M, and
v-conjugation is the identity in <slice>.  Finally, having commuted
the halves of the slices, we are now mapping Q+ to Q+ and Q- to Q-,
so we have a rotation.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
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From @mail.uunet.ca:mark.longridge@canrem.com  Thu May 25 19:56:07 1995
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From: mark.longridge@canrem.com (Mark Longridge)
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Date: Thu, 25 May 1995 18:41:00 -0400
Organization: CRS Online  (Toronto, Ontario)

On 05/22/95 at 11:13:00 Martin Schoenert said:

>GAP's 'NumberConjugacyClasses' follows the general usage in
> group theory.
>The conjugacy class of an element <g> of <G> is the set of elements <h>
>that are G-conjugated to <g> (i.e., there exists an element <x> in <G>,
>such that <x>^-1 * <g> * <x> = <h>).

On  05-24-95 (18:16) Jerry Bryan said:

>Just to give an example that I am familiar with, suppose the group
>in question were M itself.  Then, NumberConjugacyClasses should yield
>10, because the 48 elements in M yield 10 conjugacy classes under
>M-conjugation.  If anybody who has GAP also has defined M, you
>might give it a try.

Ok... let's define C in the context of GAP:

c := Group(
    ( 1, 3, 8, 6)( 2, 5, 7, 4)( 9,33,25,17)(10,34,26,18)(11,35,27,19)
    (20,12,36,28)(21,13,37,29)
    (46,48,43,41)(44,47,45,42)(38,30,22,14)(39,31,23,15)(40,32,24,16),
    ( 9,11,16,14)(10,13,15,12)( 1,17,41,40)( 4,20,44,37)( 6,22,46,35)
    (2,18,42,39)(7,23,47,34)
    (30,32,27,25)(28,31,29,26)(19,43,38,3) (21,45,36,5) (24,48,33, 8)
);;

M is the same as C but with the central reflection:

m := Group(
    ( 1, 3, 8, 6)( 2, 5, 7, 4)( 9,33,25,17)(10,34,26,18)(11,35,27,19)
    (20,12,36,28)(21,13,37,29)
    (46,48,43,41)(44,47,45,42)(38,30,22,14)(39,31,23,15)(40,32,24,16),
    ( 9,11,16,14)(10,13,15,12)( 1,17,41,40)( 4,20,44,37)( 6,22,46,35)
    (2,18,42,39)(7,23,47,34)
    (30,32,27,25)(28,31,29,26)(19,43,38,3) (21,45,36,5) (24,48,33, 8),
    (1,8)(3,6)(2,7)(4,5)
    (17,24)(19,22)(18,23)(20,21)
    (9,16)(11,14)(10,15)(12,13)
    (25,32)(27,30)(26,31)(28,29)
    (33,40)(35,38)(34,39)(36,37)
    (41,48)(43,46)(42,47)(44,45)
);;

Then we have

Size (c) = 24
NumberConjugacyClasses (c) = 5

Size (m) = 48
NumberConjugacyClasses (m) = 10

These results concur with Dan's message from Tue, 28 Dec 93 18:40:52 EST
from the archives.

We can also use GAP to calculate the size of the M-conjugacy class of
a given element:

Size (ConjugacyClass (m, cross4)) = 3

Here we see there are three possible 4 Cross order 2 patterns.

I've tried dabbling in some GAP programming. Say we are looking for an
element in the slice group with 4 variants under M-conjugacy....

a := 0;
x := 0;
z := Elements (slice);

repeat
  a := a+1
  x := Size (ConjugacyClass (m, Random (slice)));
until a = 768 or x = 4

This short program found no elements of size 4 in the slice group.

-> Mark <-

From BRYAN@wvnvm.wvnet.edu  Fri May 26 19:37:17 1995
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Date:      Fri, 26 May 1995 19:33:50 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <mreid@ptc.com>, "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Re: AntiSlice Under M-conjugacy (and a problem with slice)
In-Reply-To: Message of 05/24/95 at 17:14:09 from mreid@ptc.com

On 05/24/95 at 17:14:09 mreid@ptc.com said:
>hi jerry,

>you said

>> On 05/22/95 at 17:48:42 mreid@ptc.com said:
>>
>> >>       Level       Positions     Strong      Weak        Total
>> >>                                Local Max   Local Max   Local Max
>> >>
>> >>         6            184           1          35          36
>>
>> >beg the obvious question:  what is that strong local maximum,
>> >which is unique up to symmetry?
>>
>> I haven't verified in the antislice data base, but it *has* to be
>> Pons Asinorum, accomplshed as six antislices.

>no it hasn't.  this is the equivalent of the face turn metric,
>i.e. your "level" is half the face turn distance.  (RL  and  RRLL
>each count as 1, which is half their face turn count.)

>then pons asinorum is at level 3, and gives the weak local maximum
>at that level.

Mike is correct.

Try instead, (L2R2)(U2D2)(F'B')(U2D2)(L2R2)(F'B').

This is a very pretty pattern which may well have a name, but I
don't know what the name is.  Also, it is its own inverse.

Is the length 12h in <Q,H>?  Is it a local maximum (strong or otherwise)
in <Q,H>?   Is the length 20q in <Q>?  Is it a local maximum in <Q>?

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From mreid@ptc.com  Tue May 30 15:39:31 1995
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From: mreid@ptc.com (michael reid)
Message-Id: <9505301629.AA15624@ducie>
To: cube-lovers@ai.mit.edu
Subject:   Re: AntiSlice Under M-conjugacy (and a problem with slice)
Content-Length: 1649

jerry writes

[ ... ]

> >> >>       Level       Positions     Strong      Weak        Total
> >> >>                                Local Max   Local Max   Local Max
> >> >>
> >> >>         6            184           1          35          36
> >>
> >> >beg the obvious question:  what is that strong local maximum,
> >> >which is unique up to symmetry?

[ ... ]

> Try instead, (L2R2)(U2D2)(F'B')(U2D2)(L2R2)(F'B').
> 
> This is a very pretty pattern which may well have a name, but I
> don't know what the name is.  Also, it is its own inverse.
> 
> Is the length 12h in <Q,H>?  Is it a local maximum (strong or otherwise)
> in <Q,H>?   Is the length 20q in <Q>?  Is it a local maximum in <Q>?

no, yes (otherwise), no, and yes, respectively.

we have seen this pattern several times recently.  this is one of 
those positions with 16 symmetries.  i called it "four pluses" in my 
message of may 11 (although i gave it in a different orientation)

)     four pluses                 ( R2 F2 R2 U'D F2 R2 F2 UD' )

in fact, this maneuver is minimal in both the quarter turn and the
face turn metrics, so its length is  16q, 10f.  it is a weak local
maximum in the face turn metric; one can check that no minimal
maneuver ends with the face turn R.  however, using the 16 symmetries
which preserve the  U-D axis, and inversion, we can give minimal
maneuvers which end with turns of any of the six faces.  this shows
that it's a weak local maximum in the face turn metric.  local
maximality in the quarter turn metric follows in a similar manner.

also, mark pointed out on april 16 that this position lies in the
center of the antislice group.

mike

From @mail.uunet.ca:mark.longridge@canrem.com  Sat Jun  3 03:39:09 1995
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Subject: Super Groups
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1148.5834.0C1E504D@canrem.com>
Date: Sat, 3 Jun 1995 04:14:00 -0400
Organization: CRS Online  (Toronto, Ontario)

Notes on the various Super-Groups
---------------------------------

I have calculated the size of the super-groups for various subgroups
of the cube. I have suffixed the standard group names with the
letter c to show that the centre orientations are significant.

The groups are (ranked smallest to largest):

Size (slice)       =         768
Size (slicec)      =      24,576     Size (slicec) / Size (slice)   = 32

The following reference confirms this calculation and expounds further
on the nature of the slice group...

The Slice Group in Rubik's Cube,
by David Hecker, Ranan Banerji
Mathematics Magazine, Vol. 58 No. 4 Sept 1985

Size (antisl)      =       6,144
Size (antislc)     =      49,152     Size (antislc) / Size (antisl) =  8

Size (sq)          =     663,552
Size (sqc)         =   5,308,416     Size (sqc) / Size (sq)         =  8

Size (ur)          =  73,483,200
Size (urc)         = 587,865,600     Size (urc) / Size (ur)         =  8

Size (cube)  =     43,252,003,274,489,856,000
Size (cubec) = 88,580,102,706,155,225,088,000

Size (cubec) / Size (cube) = 2,048

The case of the super squares group (sqc) is interesting. It is only
possible to rotate opposite centres 180 degrees. There are actually
8 centres in the super square's group:

(1 way)    Identity
(1 way)    All 6 centres rotated 180 degrees
(3 ways)   2 opposite centres rotated 180 degrees
(3 ways)   2 pairs of opposite centres rotated 180 degrees

-> Mark <-

From @mail.uunet.ca:mark.longridge@canrem.com  Sun Jun  4 03:28:58 1995
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Subject: Super Squares Group
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1150.5834.0C1E5238@canrem.com>
Date: Sun, 4 Jun 1995 04:21:00 -0400
Organization: CRS Online  (Toronto, Ontario)

Way back on Thu Aug 20 20:10:13 1992 Mike wrote:

>  R2 F2 B2 L2 U2 L2 F2 B2 R2  ~  D2 ,
>            so that  <U2, F2, R2, L2, B2> = <U2, F2, R2, L2, B2, D2>.

I bet he didn't realize at the time he was finding a minimal
sequence to rotate 2 centres in the super square's group!

If we tack D2 on the end we get the sequence:

R2 F2 B2 L2 U2 L2 F2 B2 R2 D2 =  turn U & D centres 180 degrees
 in 10 face turns.

-> Mark <-

From bagleyd@source.asset.com  Wed Jun  7 17:31:08 1995
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From: bagleyd@source.asset.com (David A. Bagley)
Message-Id: <9506072107.AA36858@source.asset.com>
To: Cube-Lovers@ai.mit.edu
Subject: Dinosaur Rubik's Cube

Hi
  I just just added new modes to my xdino puzzle.  It is the Rubik's
Dinosaur Puzzle recently mentioned.  (A cube with diagonal X cuts.)
In addition to the Period 3 movement it now has a Period 2 movement with
the faces cut up like:
   ___
  |\ /|
  | X |
  |/ \|
   ---
as opposed to just
   \ /
    X
   / \
In the Period 3 movement the cube turns around a corner while in Period 2
movement it turns around the center of an edge.
 
Of course if you want to make it harder there is a "Both" mode where
you can have both turning modes at once.
 
I would like to thank Derek Bosch <bosch@sgi.com> for suggesting the
Period 2 movement -> Bosch's Cube :)
 
I spent a good 10 minutes trying to solve Bosch's Cube and did
not get anywhere.  The Period 3 seems easier.  Be the first in the
Universe to solve it. :)
 
All those who have X should be able to run it.  There are many other
puzzles in the collection as well.
 
Any problems with the compilation or bugs ... let me know.
 
Cheers,
      --__---------------------------------------------------------------
     /  \ \   /           David A. Bagley                                \
    |    \ \ /            bagleyd@source.asset.com                        |
    |     \//\            Some days are better than other days.           |
    |     / \ \                -- A short lived character of Blake's 7    |
     \   /   \_\puzzles   Available at: ftp.x.org/contrib/games/puzzles  /
      -------------------------------------------------------------------

From norgomez@itecs5.telecom-co.net  Mon Jun 12 22:31:21 1995
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From: norgomez@itecs5.telecom-co.net (ORLANDO GOMEZ CAMACHO, BOGOTA-COLOMBIA.)
To: JCCANNEK@ukcc.uky.edu, MTY.ITESM.MX@ukcc.uky.edu,
        commune-list@stealth.acf.nyu.edu, COMP-CEN%UCCVMA.BITNET@vm1.nodak.edu,
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        CRTNET%PSUVM.BITNET@cunyvm.cuny.edu,
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        441495@acadvm1.uottawa.ca, com-priv@psi.com, cjr2@cornell.edu,
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        CPAE@catfish.valdosta.peachnet.edu, CPE-LIST@uncvm1.oit.unc.edu,
        CREA-CPS@nic.surfnet.nl, CREWRT-L@mizzou1.missouri.edu,
        CROMED-L@aearn.bitnet, CSNET-FORUM@sh.cs.net
Subject: Oportunidades de Negocios
X-Vms-To: @LISTA1.DIS

======================================================================

                      OPORTUNIDADES DE NEGOCIOS

         SOLO PARA PERSONAS RESIDENTES EN LOS ESTADOS UNIDOS,
         CON BUENOS CONTACTOS EN EMPRESAS DE ALTA TECNOLOGIA
           EN INFORMATICA, OFIMATICA Y TELECOMUNICACIONES
-----------------------------------------------------------------------

Apreciados amigos de la lista:

Nuestra empresa ha sido contratada por una firma que edita una publicaciOn 
especializada en las Areas de  Computadoras, Telecomunicaciones, OfimAtica 
y Servicios relacionados que  circula  en Venezuela, Ecuador, Peru, Brazil 
y Colombia,  constituyEndose en un medio Unico  de informaciOn sobre estos 
tOpicos en las naciones descritas.

Para comercializar ESPACIOS PUBLICITARIOS de esta publicaciOn, se requiere 
de una compan~Ia o persona que maneje presupuesto de negocios, preferible-
mente  con capacidad econOmica propia e idoneidad para contratar. Es nece-
sario que la persona o empresa resida en los Estados Unidos de AmErica.

A continuaciOn se presentan las caracterIsticas principales de la publica-
ciOn:

    * CirculaciOn efectiva en Brazil, Colombia, Ecuador, Peru y Venezuela.
    * A los anunciadores  se les proveen  en medio magnEtico, las bases de
      datos con informaciOn importante  de estos paises.  A las Empresas y 
      Usuarios de cada naciOn tambiEn se les harA entrega de este material 
      que incluirA informaciOn sobre proveedores de bienes  y servicios en
      los sectores   de Telecomunicaciones, Software, Equipos  de  Oficina,
      Computadores, y Suministros.
    * En  la publicaciOn figuran  empresas  Norteamericanas  que fabrican,
      representan y proveen equipos, programas y servicios en las areas de
      telecomunicaciones, sector electronico, de comunicacion, computacion
      perifericos, componentesm accesorios, suministros, servicios y areas
      afines
    * El tiraje es de Cuarenta Mil (40,000) ejemplares para ser distribui-
      dos en forma gratuita.
    * El perfil de los anunciadores corresponde a compan~Ias Estadouniden-
      ses de estos sectores que se encuentren en ampliar y consolidar su
      operacionalidad en estos paises.

La Empresa OFRECE:

    * Exclusividad
    * Participacion atractiva sobre los negocios
    * Excelente calidad editorial y material informativo y util para generar
      nuevas y mejores oportunidades de negocios en forma proactiva.

==============================================================================
A las personas interesadas, se les ruega contactarnos por esta misma via,
indicando los siguientes datos:
Nombre   :
e-mail   :
Direccion:
Ciudad   :
Telefono :
Fax      :
==============================================================================
Mayores informes:

COMUNICACIONES INTERACTIVAS
Contacto  : Orlando Gomez Camacho
e-mail    : norgomez@itecs5.telecom-co.net
Voice Mail: (+571) 5002072
------------------------------------------------------------------------------ 

From MULL4195@splava.cc.plattsburgh.edu  Wed Jun 14 16:17:44 1995
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Date: Wed, 14 Jun 1995 16:17:58 -0500 (EST)
From: John Mullen <MULL4195@splava.cc.plattsburgh.edu>
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signoff cube-lovers

From comnetlu!georges.helm@eo.net  Thu Jun 15 15:38:16 1995
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Subject: Andras Mezei's book
From: georges.helm@comnet.eo.lu (GEORGES HELM)
Message-Id: <8AB54D5.0063000243.uuout@comnet.eo.lu>
Date: Thu, 15 Jun 95 20:37:00 +1
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Mark Longridge wrote

> Does anyone on Cube-Lovers have that book?

Yes, I do.

The illustrations cover:
Rubik himself, Polytoys, Konsumex, cubes, books, cube-related items
(stickers, pencils, lp's, earrings, t-shirts...)
There is an article on a competition in 1982 in Budapest.
And many more articles I don't understand a word of.

Georges <cubemeister>

From comnetlu!georges.helm@eo.net  Fri Jun 16 03:20:00 1995
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Subject: Andras Mezei's book
From: georges.helm@comnet.eo.lu (GEORGES HELM)
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I have the book. - Georges

From BRYAN@wvnvm.wvnet.edu  Fri Jun 16 14:10:52 1995
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Date:      Fri, 16 Jun 1995 14:11:02 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   10q Local Maxima

There are four of them unique up to M-conjugacy (more
information to come).

 Distance  Cubes  Branch Lcl     {m'Xm}   Branch   Ratio   Local
 from             Factor Max   (M-Conj.   Factor    of       Max
 Start                          Classes)          Cubes to
                                                  Classes
 0              1                       1                     0
 1             12 12.000  0             1   1.000  12.000     0
 2            114  9.500  0             5   5.000  22.800     0
 3          1,068  9.368  0            25   5.000  42.720     0
 4         10,011  9.374  0           219   8.760  45.712     0
 5         93,840  9.374  0         1,978   9.032  47.442     0
 6        878,880  9.366  0        18,395   9.300  47.778     0
 7      8,221,632  9.355  0       171,529   9.325  47.931     0
 8     76,843,595  9.347  0     1,601,725   9.338  47.976     0
 9    717,789,576  9.341  0    14,956,266   9.338  47.993     0
10  6,701,836,858  9.337 42   139,629,194   9.336  47.997     4
11 62,549,615,248  9.333    1,303,138,445   9.333  47.9992

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From BRYAN@wvnvm.wvnet.edu  Fri Jun 16 14:17:43 1995
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Date:      Fri, 16 Jun 1995 14:17:51 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   10q Local Maxima Search Matrix

The individual cells in this chart give numbers of M-conjugacy
classes.  The local maxima are in column 12.  Still more
information to come.

                         Number of Moves Which Go Closer to Start

Level  Total  0         1        2       3    4    5   6  7  8 9 1 1 1
      Classes                                                    0 1 2

 0         1  1         0        0       0     0   0   0  0  0 0 0 0 0
 1         1  0         1        0       0     0   0   0  0  0 0 0 0 0
 2         5  0         2        3       0     0   0   0  0  0 0 0 0 0
 3        25  0        20        4       1     0   0   0  0  0 0 0 0 0
 4       219  0       182       34       2     1   0   0  0  0 0 0 0 0
 5      1978  0      1677      280      20     1   0   0  0  0 0 0 0 0
 6     18395  0     15642     2561     184     8   0   0  0  0 0 0 0 0
 7    171529  0    145974    23773    1721    61   0   0  0  0 0 0 0 0
 8   1601725  0   1362579   222235   16241   663   1   3  0  3 0 0 0 0
 9  14956266  0  12719643  2077549  153026  5954  74  15  2  3 0 0 0 0
10 139629194  0 118711701 19418503 1438825 58862 925 318 11 37 0 8 0 4

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From BRYAN@wvnvm.wvnet.edu  Sat Jun 17 00:35:23 1995
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Date:      Sat, 17 Jun 1995 00:35:31 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   10q Local Maxima Positions

1. (UU)  (F'B')(RL)(RL)(FB)
2. (UD') (F'B')(RL)(RL)(FB)
3. (UD)  (F'B')(RL)(RL)(FB)

4. (UD')(FB')(LR')(FB')(FB')

#4 is the one Mike Reid already found in the slice group.  #3 is the
one I already found in the antislice group.  #1, #2, and #3 are
obviously closely related.  #1 and #2 appear not to be in either
slice or antislice, but I have been fooled before by alternative
sequences which yield the same position.

#1, #2, and #3 all have the property that |{m'Xm}|=6 and
|Symm(X)|=8.  As has already been discussed, #4 has the
property that |{m'Xm}|=24 and |Symm(X)|=2.

The symmetry groups for #1, #2, and #3 are of a type Dan Hoey's
taxonomy calls class P, class S, and class AX, respectively.
These particular classes are hard to describe succinctly without
introducing a lot of notation.  But in all three cases, the symmetry
groups (subgroups of M such that X=m'Xm} consist of four rotations
and four reflections, and have as an axis of symmetry one of
the three major axes of the cube (U-D, F-B, or R-L).  There
are three groups P1, P2, P3 with axis of symmetry U-D, F-B,
and R-L, respectively, and similarly for S1, S2, and S3, and
for AX1, AX2, and AX3.

For #4, we have Symm(X)=HV in Dan's taxonomy, where HV={i,v}, and
where i is the identity in M and v is the central inversion in M.
If proper typography were available, the i and the v would be
upper case script letters to follow Frey and Singmaster.

There are relatively few positions in all of cube space
for which Symm(X)=Pi or Symm(X)=Si or Symm(X)=AXi (i in 1..3).
There are only 10 P positions through level 10 in the search tree (of which
just one is a local maximum).  There is only one S position through
level 10, and only one AX position through level 10, both of
which are of course local maxima.  The positions
are not Q-transitive, but the positions look "symmetric", and they
fulfill the (incorrect) intuition that "symmetric" positions must
be local maxima.  We have no reason to say that other P or S or AX
positions further down the search will be local maxima.

I find position #4 extremely intriguing.  In general, HV is not
very strong symmetry, and there are relatively speaking, quite
a few HV positions in cube space.

We could create an HV position as follows.
Put any edge cubie anywhere (say UF in RD).  Put the "opposite"
cubie in the "opposite" cubicle (DB in LU in this case).  Continue
for the remaining edge cubies, and then do the same thing for the
corners, remembering only to make sure the edges and corners have
the same parity.  You can easily make an HV position that looks
quite "random" to the casual glance, and in fact most HV positions
don't look very "symmetric".

But Mike's position looks very "symmetric" at a casual glance, as
if its symmetry must be much stronger than HV.  I certainly would
not have expected to find an HV position as a local maximum close
to Start.  I think the "look" of Mike's position as "symmetric",
and the fact that it is a local maximum close to Start are
related.  Without getting too long winded, I think the reasons
are two-fold.  First, the corners and edges have much stronger
symmetry separately than they do collectively.  Second, the
symmetry looks much stronger if you ignore the centers (i.e.,
if you ignore the rotational positioning of the cubies), perhaps
in the sense of Dan's CSymm function.  For example, the corners
are properly positioned with respect to each other, even though
they are not properly positioned with respect to the fixed face
centers.

In the next few days, I intend to calculate Symm(X)
for the corners and edges separately for Mike's position, as
well as calculating CSymm(X) for the corners and edges
separately and combined.  I think the results will be
enlightening.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From @mail.uunet.ca:mark.longridge@canrem.com  Sat Jun 17 22:00:06 1995
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Subject: Crazy Corner Pattern
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1161.5834.0C1E7553@canrem.com>
Date: Sat, 17 Jun 1995 05:38:00 -0400
Organization: CRS Online  (Toronto, Ontario)

On Mon, 17 Jan 1994 09:06:59 EST, Jerry wrote:
>
>     Counting M-conjugacy classes of the corners of Rubik's cube
>     -----------------------------------------------------------
>
>                   M-Class     Number     Number
>                     Size        of         of
>                              Classes    Elements
>
>                         1        1  =        1     Start
>                         2        1  =        2     +4 -4 Twist
>                         3        3  =        9
>                         4        1  =        4
>                         6       34  =      204
>                         8       33  =      264
>                        12      301  =     3612
>                        16      104  =     1664
>                        24     9064  =   217536
>                        48  1832428  = 87956544
>
>                     Total  1841970    88179840

I'm trying to find the 1 pattern with M-class size of 4 of the
corners group.
The only pattern that I can find is 4 alternate corners twisted
clockwise which is in the twist orbit.

It does not seem to be any pattern with just corners twisted in
place.

Jerry, if you could *please* identify this pattern before I go
nuts...... !

-> Mark <-

From BRYAN@wvnvm.wvnet.edu  Sun Jun 18 00:35:58 1995
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Date:      Sat, 17 Jun 1995 17:18:00 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   A Third Way to Calculate the Real Size of Cube Space?

We define the real size of cube space to be the number of M-conjugate
classes {m'Ym} for m in M, set of 48 rotations and reflections
of the cube, and for Y in G.

Dan Hoey has calculated the real size of cube space using the
Polya-Burnside theorem.

Dan and I (mostly Dan) have also calculated the same result using
exhaustive computer search.  The computer search is much less
elegant than the Polya-Burnside results, but the search does
provide additional information, such as the number of positions
associated with each symmetry group.  The results from the
computer search have not yet been posted to the list, but a draft
paper is in progress.

In the meantime, it occurs to me that perhaps  --  but only perhaps --
there is a third way to calculate the real size of cube space.
The third way would not require (much) computer searching, but would
provide the same level of detail about number of positions per
symmetry group as does the full blown search.

The idea is based on a posting from Mike Reid.  Mike calculated the
number of positions in G whose symmetry preserves the U-D axis.
Such positions have a symmetry group which is called X1
in Dan's taxonomy.  For these positions, we say Symm(Y)=X1,
where in general for Y in G we have Symm(Y) is the set
(and group) of all m in M such that Y=m'Ym.

X1 contains sixteen elements (eight rotations and eight reflections),
and preserves the U-D axis.  X2 and X3 are conjugate subgroups of X1
and similarly preserve the F-B and the R-L axes, respectively. If Y is
X1-symmetric, then we have {m'Ym}={Y1,Y2,Y3}.  One of the Yi is Y
and is X1-symmetric, one of the Yi is X2-symmetric, and the other
Yi is X3-symmetric.

Mike determined (without computer search) that there are 128
X1-symmetric positions.  Since four of the positions are also
M-symmetric, we have 124 positions Y for which Symm(Y)=X1.
Similar results hold for X2 and X3.  Hence, there are 124
M-conjugacy classes containing cubes for which Symm(Y)=Xi, or
perhaps we might say for which SymmClass(Y)=X.  The important
fact here is that we have determined that there are
124 M-conjugacy classes for symmetry class X without having
to do a computer search.

If we could similarly determine the number of K-symmetric positions
for each of the 98 subgroups K of M without computer search, then
we could calculate the real size of cube space.  You really only
have to determine the size of 33 subgroups.  Just as the solution
for X1 also gave us the solution for X2 and X3, similarly the
solution for any subgroup provides the solution for all conjugate
subgroups, and there are 33 classes of conjugate subgroups.

I usually get myself in trouble when I delve too much into things
I don't understand, but let's try a few examples.  The subgroup
HV={i,v} is easy to understand, where v is the central inversion.
For the edges, the number of HV-symmetric positions should be
24*20*16*12*8*4.  That is, put the first cubie anywhere (24
possibilities) which dictates the location of the respective
"opposite" cubie.  There are then 20 possibilities for the
location of the third cubie which again dictates the position
of the respective "opposite" cubie, and so forth.  In the same
manner, the number of HV-symmetric corner positions is
24*18*12*6.  The number of HV-symmetric positions is then
(24*20*16*12*8*4)*(24*18*12*6)/2 to take parity into account.

Now we have the rub.  In order to calculate the positions for
which Symm(Y)=HV, we must subtract out the HV-symmetric
positions which have stronger symmetry than HV, just as we
subtracted out the M-symmetric positions in Mike's X1 case.
But to do so, we cannot take the subgroups of M in isolation.
We have to do them all, starting with M and working our way
down.  (And HV is pretty far down the food chain.)

Some of the subgroups I can do pretty easily, and for others I have
not a clue.  Recall that A is the subgroup of M consisting of the
24 even rotations and reflections and that C is the subgroup of
M consisting of the 24 rotations (12 even and 12 odd).  As long
ago as _Symmetry and Local Maxima_, Dan Hoey and Jim Saxe
determined that there are only four A-symmetric positions and
only four C-symmetric positions, namely the four that are also
M-symmetric.  Hence, there are no positions for which Symm(Y)=A
nor for which Symm(Y)=C.  But I haven't a clue how they knew,
nor how to go about constructing an A-symmetric or a C-symmetric
position from scratch.  You can't get very far with my proposal
unless you can figure out how to construct K-symmetric positions
for any K.

For one more example, consider H, the set of 12 even rotations and
12 odd reflections.  I know from computer search and also from
_Symmetry and Local Maxima_ that there are 24 H-symmetric positions,
of which 4 are M-symmetric and 20 are H-symmetric without also being
M-symmetric.  The 20 H-symmetric but not M-symmetric positions form
10 M-conjugacy classes for which we would say SymmClass(Y)=H.  It
ought to be easy to derive this result without a computer search, but
again I confess I haven't a clue as to go about constructing the 24
H-symmetric positions from scratch.

Well, I could cheat and look up the Class H positions
in _Symmetry and Local Maxima_, but what about the classes
that haven't been figured out yet?  Also, I could cheat and
use the results from computer search, but that's hardly the
point.

One final point: just as Mike's 128 X1-symmetric positions formed
a group, similarly the set of K-symmetric positions form a group
for all 98 possible values of K.  We have to be a little careful
with our terminology.  The X1-symmetric positions form a group,
as do the X2-symmetric and the X3-symmetric positions.  But if
we want to talk about the X-symmetric positions, we no longer
have a group.  For example, we do not in general have closure
when forming the composition of X1-symmetric positions with
X2-symmetric positions.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From BRYAN@wvnvm.wvnet.edu  Sun Jun 18 08:48:57 1995
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Message-Id: <wvmail32.1995jun18.081858.bryan@wvnvm.wvnet.edu>
Date:      Sun, 18 Jun 1995 08:25:49 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@life.ai.mit.edu>
Subject:   Re: Crazy Corner Pattern
In-Reply-To: Message of 06/17/95 at 05:38:00 from mark.longridge@canrem.com

On 06/17/95 at 05:38:00 mark.longridge@canrem.com said:
>On Mon, 17 Jan 1994 09:06:59 EST, Jerry wrote:
>>
>>     Counting M-conjugacy classes of the corners of Rubik's cube
>>     -----------------------------------------------------------
>>
>>                   M-Class     Number     Number
>>                     Size        of         of
>>                              Classes    Elements
>>
>>                         1        1  =        1     Start
>>                         2        1  =        2     +4 -4 Twist
>>                         3        3  =        9
>>                         4        1  =        4
>>                         6       34  =      204
>>                         8       33  =      264
>>                        12      301  =     3612
>>                        16      104  =     1664
>>                        24     9064  =   217536
>>                        48  1832428  = 87956544
>>
>>                     Total  1841970    88179840

>I'm trying to find the 1 pattern with M-class size of 4 of the
>corners group.
>The only pattern that I can find is 4 alternate corners twisted
>clockwise which is in the twist orbit.

>It does not seem to be any pattern with just corners twisted in
>place.

>Jerry, if you could *please* identify this pattern before I go
>nuts...... !

The position is called T-symmetric in Dan's taxonomy (actually,
there are four T subgroups of M, T1 through T4).  The symmetry
is related to opposite corners, e.g., the UFL-DRB axis,
which is why there are four T subgroups.

Also, the position is Q-transitive, so you can check it out
in _Symmetry and Local Maxima_.  I quote ".... each edge on the
girdle may be swapped with the diametrically opposite edge,
provided that the corners on the girdle are swapped with
their opposites as well."   Here, you would fix the edges and
pay attention only to the swapping of the corners.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From BRYAN@wvnvm.wvnet.edu  Sun Jun 18 15:55:14 1995
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Date:      Sun, 18 Jun 1995 15:55:22 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Re: A Third Way to Calculate the Real Size of Cube Space?
In-Reply-To: Message of 06/17/95 at 17:18:00 from BRYAN@wvnvm.wvnet.edu

I should have said "a fourth way", I think.  Martin Schoernert
performed the same calculation with GAP.  Hence, we have three
ways in hand: 1) Dan's Polya-Burnside method, 2) Martin's GAP
calculations, and 3) brute force computer search.  My new
proposal would then be a fourth way.

Here is a question for Martin: is there any way with GAP to calculate
the number of M-conjugacy classes associated with each symmetry class?
It is this additional information about the "real cube space" which
*is* available via computer search, and for which I am proposing
an alternative which does not involve computer search.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From @mail.uunet.ca:mark.longridge@canrem.com  Sun Jun 18 23:54:16 1995
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Sender: CRSO.Cube@canrem.com
Subject: Re: Crazy Corner Pattern
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1167.5834.0C1E77E1@canrem.com>
Date: Mon, 19 Jun 1995 00:36:00 -0400
Organization: CRS Online  (Toronto, Ontario)

>>I'm trying to find the 1 pattern with M-class size of 4 of the
>>corners group.
>>The only pattern that I can find is 4 alternate corners twisted
>>clockwise which is in the twist orbit.

>>It does not seem to be any pattern with just corners twisted in
>>place.

>>Jerry, if you could *please* identify this pattern before I go
>>nuts...... !

>The position is called T-symmetric in Dan's taxonomy (actually,
>there are four T subgroups of M, T1 through T4).  The symmetry
>is related to opposite corners, e.g., the UFL-DRB axis,
>which is why there are four T subgroups.

>Also, the position is Q-transitive, so you can check it out
>in _Symmetry and Local Maxima_.  I quote ".... each edge on the
>girdle may be swapped with the diametrically opposite edge,
>provided that the corners on the girdle are swapped with
>their opposites as well."   Here, you would fix the edges and
>pay attention only to the swapping of the corners.

Hmmmmm, that's very interesting.

Below is my interpretation of "...corners on the girdle are swapped
with their opposites as well." Let's call it pattern X.

        D U D
        U U U
        D U D
F L B   R F L   B R F   L B R
L L L   F F F   R R R   B B B
F L B   R F L   B R F   L B R
        U D U
        D D D
        U D U

If I understand the terminology correctly, then for this pattern
X = |{m'Xm}|=3 and |Symm(X)|=16, same as the 4 spot.
Also X = |{c'Xc}|=3.

But perhaps this is not the same pattern....

Let's call this next cube arrangement pattern Y.

        F U U
        U U U
        D U L
R L B   R F B   D R R   B B D
L L L   F F F   R R R   B B B
U L L   F F U   L R F   L B F
        D D B
        D D D
        R D U

Y = |{m'Ym}|=4 and |Symm(Y)|=12, and Y=|{c'Yc}| = 4

This pattern was created by swapping 3 pairs of diametrically opposite
corners, which is in the *swap* orbit on a normal cube, but since we
are dealing with corners of Rubik's cube and ignoring edges we can
realize permutations with an odd number of pairs of corners swapped.

-> Mark <-

From BRYAN@wvnvm.wvnet.edu  Mon Jun 19 09:57:45 1995
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Date:      Mon, 19 Jun 1995 09:57:53 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@life.ai.mit.edu>
Subject:   Re: Crazy Corner Pattern
In-Reply-To: Message of 06/19/95 at 00:36:00 from mark.longridge@canrem.com

Perhaps I should have quoted a little more from _Symmetry and
Local Maxima_.  Here is the T-symmetric position given by
Saxe and Hoey.  In their position, the UFL and RBD corners are
in place, and the other three pairs are swapped.  The "girdle"
includes the three pairs that are swapped.  Hence, there
is an axis of symmetry along the UFL-RBD axis.  The odd number of
swaps is compensated by an odd number in the edges.  The compensation
is not required for corners only.

        R D D
        U U D
        U U B

 D L L  F F D  L L F  L F F
 R L L  F F B  L R R  B B F
 R R B  R B B  U R R  B B U

        U U L
        U D D
        F D D

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From BRYAN@wvnvm.wvnet.edu  Mon Jun 19 14:04:49 1995
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Date:      Mon, 19 Jun 1995 13:41:51 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   3x3x3 Cubes for Sale

For the first time in years, I have seen 3x3x3 cubes for sale in a
store.  (For American readers, the store is Hills, which is a regional
chain which competes against WalMart and Kmart.  The price is $8.97 --
U.S. dollars.)  The box has a note signed by Erno Rubik, and gives
the proper size of the cube group.  Also, and I couldn't see inside
the box to verify this, the Face centers seemed to be marked in such
a way as to support the Supergroup.  Rubik's note about the size of
the  problem says it is 4^4 times bigger than the regular problem.
The manufacturer (or maybe distributor?) is Matchbox Toys (or is it
Matchbook Toys?).

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From news@nntp-server.caltech.edu  Mon Jun 19 18:36:29 1995
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From: whuang@cco.caltech.edu (Wei-Hwa Huang)
Newsgroups: mlist.cube-lovers
Subject: Re: 3x3x3 Cubes for Sale
Date: 19 Jun 1995 22:36:17 GMT
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"Jerry Bryan" <BRYAN@wvnvm.wvnet.edu> writes:

>For the first time in years, I have seen 3x3x3 cubes for sale in a
>store.  (For American readers, the store is Hills, which is a regional
>chain which competes against WalMart and Kmart.  The price is $8.97 --
>U.S. dollars.)  The box has a note signed by Erno Rubik, and gives
>the proper size of the cube group.  Also, and I couldn't see inside
>the box to verify this, the Face centers seemed to be marked in such
>a way as to support the Supergroup.  Rubik's note about the size of
>the  problem says it is 4^4 times bigger than the regular problem.
>The manufacturer (or maybe distributor?) is Matchbox Toys (or is it
>Matchbook Toys?).

What you saw was "Rubik's Cube 4th Dimension."  This was released some
time in the early 90s.  The only difference between this and the typical 3x3x3
is that four faces have pictures on their center squares, requiring 
the solver to orient them correctly.  

I belive last year there was a European Rubik Puzzle series that rereleased
the cube, along with other rubik puzzles.

-- 
   -- Wei-Hwa Huang (whuang@cco.caltech.edu)
Homepage (under construction): http://www.ugcs.caltech.edu/~whuang/
I have a proof that NP = P; however, it requires exponential time to write down.

From pbeck@pica.army.mil  Tue Jun 20 08:00:48 1995
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Date:     Tue, 20 Jun 95 8:00:16 EDT
From: Peter Beck (FSAC) <pbeck@pica.army.mil>
To: cube-lovers@ai.mit.edu
Subject:  cube availability
Message-Id:  <9506200800.aa03200@COR6.PICA.ARMY.MIL>


In Feb at the NY international toy show the KOOSH people
told me that they had bought the marketing rights for
RUBIK items from Western Publishing (the Matchbox LABEL).
They had new packaging and were changing the items to
be sold (eg, they said they were going to sell Rubuik's
amgic).

So the items seen in Hills with Matchbox packaging is
old stock that will soon no longer be around.

THE FUTURE IS PUZZLING,
BUT CUBING IS FOREVER !!!

pete beck

From mschoene@math.rwth-aachen.de  Tue Jun 20 08:40:27 1995
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Date: Tue, 20 Jun 95 14:39 WET DST
From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: Cube-Lovers@ai.mit.edu
Cc: BRYAN@wvnvm.wvnet.edu
In-Reply-To: "Jerry Bryan"'s message of Sun, 18 Jun 1995 15:55:22 -0400 (EDT) <wvmail32.1995jun18.154620.bryan@wvnvm.wvnet.edu>
Subject: Re: Re: A Third Way to Calculate the Real Size of Cube Space?

Jerry wrote in his message of 1995/06/17

    We define the real size of cube space to be the number of M-conjugate
    classes {m'Ym} for m in M, set of 48 rotations and reflections
    of the cube, and for Y in G.

    Dan Hoey has calculated the real size of cube space using the
    Polya-Burnside theorem.

    Dan and I (mostly Dan) have also calculated the same result using
    exhaustive computer search.  The computer search is much less
    elegant than the Polya-Burnside results, but the search does
    provide additional information, such as the number of positions
    associated with each symmetry group.  The results from the
    computer search have not yet been posted to the list, but a draft
    paper is in progress.

    In the meantime, it occurs to me that perhaps  --  but only perhaps --
    there is a third way to calculate the real size of cube space.
    The third way would not require (much) computer searching, but would
    provide the same level of detail about number of positions per
    symmetry group as does the full blown search.

    ... detailed description of the method deleted ...

And he continued in his message of 1995/06/18

    I should have said "a fourth way", I think.  Martin Schoernert
    performed the same calculation with GAP.  Hence, we have three
    ways in hand: 1) Dan's Polya-Burnside method, 2) Martin's GAP
    calculations, and 3) brute force computer search.  My new
    proposal would then be a fourth way.

Well, I don't know about *four* ways.

Dan used the Polya-Burnside theorem.  That is, he computed the
number of M-conjugacy classes as the average number of fixed points
of the elements of M w.r.t. to their action on G.  He computed
the number of fixed points of an element m using clever arguments
about the cycle structure of elements of G that m would fix.

I simply observed that the number of fixed points of an element m is
the size of the centralizer of in G, and then used GAP to compute
those.  So I don't think it is correct to call this a way of its own.

The method you propose is indeed different from Dan's method that
uses Polya-Burnside.

I can't figure out how the brute force computer search works.
So I can't tell whether it is really different from the other methods
(and if indeed it is a method to compute the real size of the cube ;-).
Jerry, could you say a little bit more about this computation?

It appears to me that Dan and Jim Saxe must have realized all the
important pieces for your new method when they wrote their seminal
``Symmetry and Local Maxima (long message)'' message of 1980/12/14.
As Jerry points out, they did calculate the important values for
9 of the 33 conjugacy classes of subgroups of M (those whose sizes
are a multiple of 12).  It is neither clear from their message how
they found those 9 classes (in fact they apparently found all 98
subgroups of M), nor how they computed the numbers of elements of
G that have a specific subgroup of M as symmetry group.
Perhaps Dan can say a little bit more about this?

Jerry continued in his message of 1995/06/18

    Here is a question for Martin: is there any way with GAP to calculate
    the number of M-conjugacy classes associated with each symmetry class?
    It is this additional information about the "real cube space" which
    *is* available via computer search, and for which I am proposing
    an alternative which does not involve computer search.

Of course there is ;-).

Given a subgroup H of M, the centralizer of H in G is the set of all
those elements of G that commute with all elements of H.  But this is
of course simply the set of those elements of G that have either H or
a larger group as their symmetry group.

So we can compute the numbers of elements of G with symmetry group H
by computing the size of the centralizer of H in G, and then subtracting
the numbers of those elements that have a symmetry group that is a
proper supergroup of H.  This is easy if we compute those numbers
for all subgroups of M, from the larger subgroups down to the smaller.
Of course it is not neccessarry to do this for all 98 subgroups
of M, but only for one subgroup for each of the 33 conjugacy classes.

Then if we simply divide the number of elements with symmetry group H by
the index of H in M, we obtain the number of M-conjugacy classes into
which those elements fall.

As a GAP program this looks as follows

    # compute the conjugacy classes of subgroups of M
    classes := ConjugacyClassesSubgroups( M );
    numbers := [];

    # for all conjugacy classes of subgroups of M
    for i  in [Length(classes),Length(classes)-1..1]  do

        # select a representative for this conjugacy class
        rep := Representative( classes[i] );

        # compute how many elements have at least this symmetry group
        number := Size( Centralizer( G, rep ) );

        # subtract the number of elements that have a larger symmetry group
        for k  in [Length(classes),Length(classes)-1..i+1]  do
            for sub  in Elements( classes[k] )  do
                if IsSubgroup( sub, rep )  then
                    number := number - numbers[k];
                fi;
            od;
        od;

        # store the number
        numbers[i] := number;

        # print the number of the class
        Print( i, ":\t" );

        # the size of the subgroups in the class
        Print( Size(rep), "\t" );

        # the number of subgroups in the class
        Print( Size(classes[i]), "\t" );

        # the number of elements whose symmetry group lies in the class
        Print( Size(classes[i]), " * ", number, "\t" );

        # and the number of M-conjugacy classes of those elements
        Print( Size(classes[i]), " * ", number, " / ", Index(M,rep), "\n" );

    od;

*Do not try this with GAP 3.4.2 (our latest release).*
GAP 3.4.2 contains several naive functions for permutation groups,
that cause this computation to take a very long time.
But with GAP 3.5 (our current development version),
this produces in about a minute the following table.

CLASS   SIZE   LENGHT NUMBER                    REAL		NAME
33:     48     1      1 * 4                     1 * 4 / 1       (M)
32:     24     1      1 * 0                     1 * 0 / 2       (C)
31:     24     1      1 * 0                     1 * 0 / 2       (AM)
30:     24     1      1 * 20                    1 * 20 / 2      (H)
29:     16     3      3 * 124                   3 * 124 / 3     (X1,X2,X3)
28:     12     4      4 * 12                    4 * 12 / 4      (T1,T2,T3)
27:     12     1      1 * 48                    1 * 48 / 4
26:      8     3      3 * 384                   3 * 384 / 6
25:      8     3      3 * 1408                  3 * 1408 / 6
24:      8     3      3 * 2944                  3 * 2944 / 6
23:      8     3      3 * 1920                  3 * 1920 / 6
22:      8     3      3 * 384                   3 * 384 / 6
21:      8     3      3 * 896                   3 * 896 / 6
20:      8     1      1 * 11892                 1 * 11892 / 6
19:      6     4      4 * 416                   4 * 416 / 8
18:      6     4      4 * 32                    4 * 32 / 8
17:      6     4      4 * 7740                  4 * 7740 / 8
16:      4     6      6 * 96232                 6 * 96232 / 12
15:      4     6      6 * 96256                 6 * 96256 / 12
14:      4     3      3 * 92928                 3 * 92928 / 12
13:      4     3      3 * 437504                3 * 437504 / 12
12:      4     3      3 * 574208                3 * 574208 / 12
11:      4     3      3 * 1163520               3 * 1163520 / 12
10:      4     3      3 * 144640                3 * 144640 / 12
9:       4     3      3 * 62208                 3 * 62208 / 12
8:       4     1      1 * 280272                1 * 280272 / 12
7:       3     4      4 * 3770864               4 * 3770864 / 16
6:       2     6      6 * 424415168             6 * 424415168 / 24
5:       2     6      6 * 2547748032            6 * 2547748032 / 24
4:       2     3      3 * 15285460992           3 * 15285460992 / 24
3:       2     3      3 * 18342768640           3 * 18342768640 / 24
2:       2     1      1 * 45862360944           1 * 45862360944 / 24
1:       1     1      1 * 43252003109885814336  1 * 43252003109885814336 / 48

As expected the numbers in the fourth column add to the size of G.
And the numbers in the fifth column add to 901083404981813616,
the real size of the cube (|M\MG/M|).

For those classes that I could identify I have added their names.
If somebody could describe Dan's taxonomy, I will name the other
classes as well.

Have a nice day.

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From mschoene@math.rwth-aachen.de  Tue Jun 20 09:29:59 1995
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Date: Tue, 20 Jun 95 14:40 WET DST
From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: Cube-Lovers@ai.mit.edu
Cc: BRYAN@wvnvm.wvnet.edu
In-Reply-To: "Jerry Bryan"'s message of Sun, 18 Jun 1995 15:55:22 -0400 (EDT) <wvmail32.1995jun18.154620.bryan@wvnvm.wvnet.edu>
Subject: Re: Re: A Third Way to Calculate the Real Size of Cube Space?

Looking through the old messages about the real size of the cube group,
it appeared to me that no one has shown a proof for the Polya-Burnside
theorem.  Since it is not difficult to prove, I decided to write one up.

In the following I will use TeX notation for formulae, i.e., formulae
are included in '$' signs, '{}' are used to group terms, '^' is used
for superscripts, and '_' for subscripts.

If $g \in G$, then I denote the set of elements that are really
equivalent to $g$ by $g^M$.  Jerry denotes this set by {m'gm},
but $g^M$ is the more common notation in group theory.

The sum $\sum_{g \in h^M}{1/|g^M|}$ is simply 1, since it is the sum over
all elements in one M-conjugacy class (h^M) of 1 over the length of
that M-conjugacy class.  Thus the sum $\sum_{g \in G}{1/|g^M|}$ is the
number of M-conjugacy classes.

Now we need a standard lemma from group theory, which tells us that the
length of a class $g^M$ of an element $g$ under the action of a group $M$
is equal to the size of the group $M$ divided by the size of the subgroup
of those elements of $M$ that fix $g$ (more precisely the index of that
subgroup in $M$, since the lemma is true, even if $M$ is infinite).

So using Jerry's notation this lemma gives $1/|g^M| = |Symm(g)|/|M|$.
Applying that to the above formula we see that the number of
M-conjugacy classes in $G$ is $\sum_{g \in G} {|Symm(g)|/|M|}$
Or, after a trivial change, $1/|M| \sum_{g \in G} {|Symm(g)|}$.

Assume that $(g^m == g)$ is 1 if $g^m$ is equal to $g$ and 0 otherwise.
Then we have $|Symm(g)| = \sum_{m \in M}{(g^m == g)}$.
Thus the number of M-conjugacy classes is
$1/|M| \sum_{g \in G} \sum_{m \in M} {(g^m == g)}$.

Now we can simply change the order of the two summations, so we get
$1/|M| \sum_{m \in M} \sum_{g \in G} {(g^m == g)}$.

But of course $\sum_{g \in G} {(g^m == g)}$ is obviously the number of
fixpoints of $m$.  So we obtain the Polya-Burnside lemma: ``The number of
M-conjugacy classes is the average number of fixpoints of the elements
of $M$ w.r.t. their operation on $G$''.

However, here the operation is special, so we can simplify even further.
$g^m$ here means $m^{-1} g m$, so $(g^m == g)$ means $(m^{-1} g m == g)$,
which is equivalent to $(m == g^{-1} m g)$ (multiply the equation first
by $m$ and then by $g^{-1}$ from the left), which is $(m == m^g)$.

So the number of M-conjugacy classes is
$1/|M| \sum_{m \in M} \sum_{g \in G} {(m == m^g)}$.

But $\sum_{g \in G} {(m == m^g)}$ is simply the size of the subgroup of
those elements in $G$ that fix $m$.  This is the centralizer of
$m$ in $G$.  So the number of M-conjugacy classes is finally
$1/|M| \sum_{m \in M} |Centralizer(G,m)|$.

This is the formulation that I used to compute the real size of the cube
group with GAP.

Have a nice day.

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From BRYAN@wvnvm.wvnet.edu  Tue Jun 20 18:04:35 1995
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Message-Id: <wvmail32.1995jun20.114716.bryan@wvnvm.wvnet.edu>
Date:      Tue, 20 Jun 1995 13:48:51 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Re: Re: A Third Way to Calculate the Real Size of Cube Space?
In-Reply-To: Message of 06/20/95 at 14:39:00 from ,
           Martin.Schoenert@Math.RWTH-Aachen.DE

On 06/20/95 at 14:39:00 Martin Schoenert said:

>I can't figure out how the brute force computer search works.
>So I can't tell whether it is really different from the other methods
>(and if indeed it is a method to compute the real size of the cube ;-).
>Jerry, could you say a little bit more about this computation?

I look at corners and edges separately, and then combine the results.

I can't speak for how Dan did his corner search and his edge search,
but I can describe mine.  Our results do match, which is always nice.

Dan did most of the figuring out of "combining the results" for corners
and edges.

Conceptually, I look at every single corner position X and calculate
Symm(X), and I look at every single edge position Y and calculate
Symm(Y).  In practice, there are some important shortcuts.

I have a data base containing "every" corner position and a second
data base containing "every" edge position with "every" defined in
the following sense.  The set of all positions is partitioned into
equivalence classes of the form {m'Xmc} for the corners and {m'Ymc}
for the edges, for m in M (48 rotations and reflections) and c in C
(24 rotations).  The data bases contain a representative element
from each equivalence class.

For all the cases where |{m'Xmc}|=1152 and |{m'Ymc}|=1152, no searching
is required.  For these cases, we know *a priori* that there are 24
M-conjugacy classes containing 48 elements each, and that for each
of the 1152 positions we have Symm(X)=I and Symm(Y)=I.  Fortunately,
for the vast majority of the cases (over 97% for corners and well
over 99% for edges), the so-called B-class size is 1152.

Suppose the representative for {m'Xmc} is V and for {m'Ymc} is W.
For the remaining cases, the idea is to use Vc and Wc for
each fixed c in C as a base or representative element for an
M-conjugacy class of the form {m'(Vc)m} and {m'(Wc)m}.  The tricky
part here is that while Vc and Vd are distinct when c and d in C
are not equal, nonetheless Vc and Vd may be M-conjugate.  Hence,
we use each Vc and Wc which are distinct up to M-conjugacy as
a representative element for an M-conjugacy class.

Conceptually, we calculate Symm(T) for each T in {m'(Vc)m} for each
fixed c (and for Vc distinct up to M-conjugacy) and summarize the
results.  In practice, it is sufficient to calculate Symm(Vc).

Here comes another tricky part (at least it was until I figured it out).
Initially, I assumed that Symm(T) was the same for all T in {m'(Vc)m}.
However, such is not the case.  Rather, if you calculate each Symm(T), you
will discover that each one will be a group from a class of conjugate
groups, and that each group in the class of conjugate groups will appear
an equal number of times.

Given that you have picked an arbitrary representative from the M-conjugacy
class, you don't know which conjugate group you are going to get when
you calculate Symm(m'(Vc)m).  I got around this issue originally by
calculating a representative group (the 98 subgroups of M then have
33 representative groups, one for each of the 33 symmetry classes).
In the work I am doing now, I map directly from each of the 98 subgroups
to their respective symmetry class.

The exhaustive search then consists of performing the above calculations
for each B-class of the form {m'Xmc} and {m'Ymc} and summarizing the
results (that is, counting all the M-conjugacy classes).  In addition
to an overall total, you summarize by symmetry class, which gives you
the additional information I have talked about, over and above Dan's
Polya-Burnside results.

At this point, the summary results give you the real size of the corners
only problem and the real size of the edges only problem.  While you are
at it, you have to count the M-conjugacy classes for even positions and odd
positions separately in order to put corners and edges together
properly.

Finally, you put the corners and edges together in all possible ways.
The putting together of the corners and edges is described in the
draft I have mentioned, so I will just wait until the draft is
ready before posting the rest.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From hoey@aic.nrl.navy.mil  Thu Jun 22 03:44:23 1995
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Date: Thu, 22 Jun 95 03:43:59 EDT
From: hoey@aic.nrl.navy.mil (Dan Hoey)
Message-Id: <9506220743.AA25013@Sun0.AIC.NRL.Navy.Mil>
To: Cube-Lovers@ai.mit.edu,
        "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
Subject: Ways to Calculate the Real Size of Cube Space?
In-Reply-To: <m0sO2aQ-00026zC@hobbes.math.rwth-aachen.de>

Martin Schoenert says:
> I can't figure out how the brute force computer search works.

and while Jerry Bryan gives one answer, I have another.  For you see,
I also ran a brute force computer search for the symmetry classes,
too.  And while it agrees with Jerry's answers, mine was a
significantly different algorithm, which I discuss a little a the end
of this message.

> It appears to me that Dan and Jim Saxe must have realized all the
> important pieces for your new method when they wrote their seminal
> ``Symmetry and Local Maxima (long message)'' message of 1980/12/14.
> As Jerry points out, they did calculate the important values for
> 9 of the 33 conjugacy classes of subgroups of M (those whose sizes
> are a multiple of 12).  It is neither clear from their message how
> they found those 9 classes (in fact they apparently found all 98
> subgroups of M), nor how they computed the numbers of elements of
> G that have a specific subgroup of M as symmetry group.
> Perhaps Dan can say a little bit more about this?

First, to find all the subgoups of M.  I represented the elements of M
as a list of permutations on faces, found easily enough by finding the
closure of the generators.  Then I did a depth-first search for
subgroups, branching by computing the closure of the current subgroup
with each possible element not in that subgroup, and cutting off the
search on previously-seen subgroups.  It's quick, it's dirty, it
works.  I found the nine subgroups of order a multiple of 12, as shown
in the Hasse subgroup diagram:

    order 48 . . . M_
                  /|\\_ 
                 / | \ \_ 
                /  |  \  \_
    order 24 . A . H . C   \_
                \  |  /      \_
                 \ | /         \_
                  \|/            \
    order 12 . . . E . . . . .T[1..4]

(except we called E "AC" back then).  The trick then was to find all
the E-symmetric positions and all the T1-symmetric positions; the
tasks of finding the full symmetry group of such positions and
counting the positions for T2, T3, and T4 were straightforward.

The best tool we had for figuring out symmetric positions was
essentially the one I wrote about in ``The real size of cube space''
on 4 Nov 94.  For a subgroup J of M, if a position g is J-symmetric,
then g must commute with each operation m in J.  Recall:

The fundamental principle we use in finding whether g commutes with
m can be found by examining the cycles of m.  Suppose m permutes a
cycle (c1,c2,...,c[k-1],ck), so that
    c2=m(c1), c3=m(c2), ..., ck=m(c[k-1]), and c1=m(ck).
For g to commute with m, we have
    g(c2)=m(g(c1)), g(c3)=m(g(c2)), ..., g(ck)=m(g(c[k-1])),
    and g(c1)=m(g(ck)).
So (g(c1),g(c2),...,g(ck)) is also a cycle of m.  Thus g must map each
k-cycle of m to another k-cycle of m, and in the same order.

The orientation question was a lot more difficult, so we ran through a
bunch of little results.  The following is a cleaned-up sample of the
sort of arguments, as I remember them.  In it FRD is an unoriented
corner cubie/cubicle and FRD.D is its down-facing color-tab/facicle.
================================================================
    Lemma 1: Suppose X and Y are corners, and m is in C, m(X)=Y.
             Suppose g(X)=X and g(Y)=Y, and g commutes with m,
             Then g applies the same twist to corners X and Y.
    Proof: Let TX,TY be the clockwise 120-degree rotation of corners
           cubies X and Y, respectively.  Then m(TX(.))=TY(m(.)), as
           can be seen by the fact that we could apply a twist to X in
           place (TX) before moving it to Y with m, or we can perform
           the same twist on Y (TY) after moving it.  So if
           g(X)=TX^k(X), then
           g(Y)=g(m(X))=m(g(X))=m(TX^k(X))=TY^k(m(X))=TY^k(Y).
           performing the same twist on Y as on X,               QED.

    Lemma 2: If g is E-symmetric, then each corner cubie remains in
             its home cubicle (not considering orientation).
    Proof: Supposing otherwise, take a moved cubie (without loss of
           generality) to be FRD, and suppose (w.l.o.g.) it moves to
           one of locations FDL, FLT, or BTL.
        Case 1.  If g(FRD)=FDL, consider operation m to be the
                 120-degree rotation about FRD.
                 m(FRD)=FRD, m(FDL)=FTR.  So
                 g(FRD)=g(m(FRD))=m(g(FRD))=m(FDL)=FTR, contradicting
                 g(FRD)=FDL.
        Case 2.  If g(FRD)=FLT, then take m as in case 1; m(FLT)=BRT,
                 so g(FRD)=g(m(FRD))=m(g(FRD))=m(FLT)=BRT,
                 contradicting g(FRD)=FDL.
        Case 3.  If g(FRD)=BTL, then g(FRD.F) is BTL.B, BTL.T, or
                 BTL.L.
        Case 3a. If g(FRD.F)=BTL.B, then g(FRD.R)=BTL.T, by clockwise
                 adjacency.  But m(FRD.F)=FRD.R and m(BTL.B)=BTL.L,
                 and g(FRD.R)=g(m(FRD.F))=m(g(FRD.F))=m(BTL.B)=BTL.L
                 contradicts G(FRD.F)=BTL.B.
        Cases 3b and 3c work the same way.
           The contradictions establish that FRD does not move,  QED.

    Lemma 3: If g is E-symmetric, then the twists of the four corner
             cubies FRD, FLT, BLD, and BRT agree with each other, and
             and the other four also agree with each other.
    Proof: For any two of the four corners (e.g. FRD, FLT), there is
           a 120-degree rotation in E taking one to the other
           (e.g. the rotation about FTR).  Lemma 1 applies
           immediately to show the twists agree,                 QED.

    Lemma 4. If g is E-symmetric, then the corner cubies are all
             solved, or are rotated alternately in opposite
             directions.
    Proof: From Lemma 2, all the cubies are in their home cubicles.
           If one of the sets from Lemma 3 is twisted, their total
           twist is the twist of a single cubie (since there are four
           of them) and so must be counteracted by having the other
           set twisted in the opposite direction, which is alternate
           corners twisted oppositely; otherwise all corner are
           solved,                                               QED.

    Lemma 5. Let T1 refer to the group that fixes the FRD-BTL axis.
             Then any T1-symmetric position g must keep the FRD and
             BTL cubies in their solved position, and rotated by the
             same amount.
    Proof: Let m be the 120-degree rotation about FRD, which is in T1.
           Since FRD and BTL are the only 1-orbits of m, they are kept
           in place or swapped.  From the proof of Lemma 2 (which uses
           the same m) they cannot be swapped.  Otherwise, the two
           cubies are kept in place, and 180 degree rotation about the
           FL-BR axis, also in T1 fulfills the requirements of Lemma 1
           to show they will both be rotated the same amount,    QED.
================================================================
That's about all I feel like remembering and formalizing right now.
As you can see, it's long, mechanical, and boring.  That's why we
never got around to writing it all down.

Early last year I wrote a computer program to find *all* the symmetry
groups of *all* the positions.  The first part did the corners and
edges separately, counting the number of positions for each symmetry
group and permutation parity.  For each of the 8! or 12! permutations,
I checked to see if the permutation commuted with some nontrivial
operation of M; if not, I just counted the appropriate number of
I-symmetric positions.  Otherwise I applied orientations and counted
up the symmetry groups of each possible orientation.  (It could
probably have been made to go faster, by cutting off partial
permutation or orientation generation as soon as all the non-trivial
operations were ruled out.)

In the above counts, I also kept track each time of whether the
permutation was even or odd.  Then after I had the count of even and
odd permutations for corners and edges in each symmetry group, I had a
program that intersected the symmetry groups, and for each pair of
subgroups J,K, and each parity P, I added
    Corners[J,P] * Edges[K,P]  to  Whole[J intersect K, P], 
with some fancy footwork so I only needed to deal with conjugacy
classes for J and K.  Then for each K, the number of whole K-symmetric
positions was Whole[K,Even]+Whole[K,Odd].

The program finished about the time I realized the application of the
Polya-Burnside theorem.  Then for most of the year, I put off writing
it all up.  I have Jerry to thank for reminding me to get with it, and
for useful comments and discussions on the early drafts.

Dan

From @secyt.gov.ar,@recom.edu.ar:administ@marben.recom.edu.ar  Mon Jun 26 11:47:47 1995
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From BRYAN@wvnvm.wvnet.edu  Tue Jun 27 23:22:41 1995
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Date:      Tue, 27 Jun 1995 23:22:50 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Constructing K-symmetric Cubes

This is a followup on several recent messages concerning the
question of K-symmetric cubes, where K is one of the
98 subgroups of M.

We recall that a permutation is a special kind of function,
namely a one-to-one and onto function on a set.  A very
common technique used with functions is to restrict the
domain to a (usually proper) subset of the original domain.
In the paradigm of a function as a general rule, the general rule
is applied to the subset of the domain to obtain the restriction
of the function.  In the paradigm of a function as a set of ordered
pairs, the restriction is simply a (usually proper) subset of the
set of ordered pairs.  A restriction of a permutation is usually
not a permutation (certainly not a permutation on the original
domain), but it is still a function.

We will treat a permutation on the cube as a set of restrictions
(functions) of the several cubicles.  We take as our first example
the function UFL->UFL.  Unlike cycle notation, it is not
assumed that the other cubicles are fixed;  rather, the other
function values are undefined (e.g., URF->?).

Let X be any function (not necessarily a permutation) whose domain
is some subset of the cubicles.  We define Symm(X) in the standard
fashion  --  Symm(X) is the set of all m in M such that
m'Xm=X.  If X is the function UFL->UFL, we have Symm(X)=AT4, not
Symm(X)=M as you might expect.  AT4 is a subgroup in Dan's taxonomy
containing six elements, and which has an axis of symmetry along
the UFL-DBR axis.

This definition of Symm(X) perhaps requires a minor bit of
justification.  In a function composition such as FG (left-to-
right notation) or G(F(x)) (right-to-left "calculus" notation),
it is sometimes taken as a convention that the range of F must
match the domain of G.  But we can also take the restriction of
G to the intersection of the range of F with the domain of G,
and we do so.  Having done so, Symm(X) is well defined.

We wish to build an M-symmetric permutation containing the
function UFL->UFL, but Symm(UFL->UFL)=AT4 is not a very
auspicious start.  Rather, we define the conditions under which
a function is K-symmetric is follows.  A function X is K-symmetric
if the union of the K-conjugates k'Xk is a function.

Given this definition, the only M-symmetric function on UFL is
in fact UFL->UFL, so we really have made a good start.  Furthermore,
any M-symmetric permutation that contains UFL->UFL must also
contain all the M-conjugates of UFL->UFL, and the union of the
M-conjugates is simply the identity permutation on the corners.

The fact that Symm(UFL->UFL)=AT4 can be of some benefit in our
investigations.  In particular, the fact that |AT4|=6 means that
there are 8 M-conjugates, so taking all the M-conjugates of
UFL->UFL means that all 8 corners are specified.

Our next example will be UFL->LUF (a twist of the corner).  In
this case, we have Symm(X)=ET4.  ET4 is the subgroup of M in
Dan's taxonomy which contains 3 elements including the identity
plus the 1/3 and 2/3 rotations around the UFL-DBR axis.

Of more import, UFL->LUF is not M-symmetric.  However, it is both
C-symmetric (C is the set of 24 rotations) and H-symmetric
(H is the set of 12 even rotations and 12 odd reflections).  As
an aside, we note that it is not A-symmetric, where A is the set
of 24 even rotations and reflections.  But since it is both
C-symmetric and H-symmetric, there is not a unique largest
subgroup K for which we can say it is K-symmetric.

It is easy to see that UFL->LUF is not M-symmetric.  The set of
M-conjugates contains both UFL->LUF and UFL->FLU, so the union
of the M-conjugates is not a function.  We can see the same thing
from the fact that |ET4|=3.  Since |ET4|=3, there are 16 M-conjugates,
but there are only 8 corners to represent the 16 M-conjugates.

Since UFL->LUF is C-symmetric, let's see if we can build a
C-symmetric permutation.  There are 8 C-conjugates (a good
start!), and the 8 C-conjugates twist each of the 8 corners
by 1/3 in the same direction.  Hence, this is a C-symmetric
but not M-symmetric permutation.  Of course, it is an "illegal"
position in the sense that it is not in the same orbit
as Start.

In the same manner, we can build a K-symmetric permutation for
any K.  We start with a K-symmetric function on a single
cubicle. (A function which is K-symmetric is L-symmetric for
any L which is a subgroup of K).  We include all
K-conjugates.  If the cube is completely specified, we stop.
Otherwise, we choose another K-symmetric function for any
previously unspecified cubicle, add in the new K-conjugates,
and so forth, repeating until the entire permutation is
specified.

Needless to say, this construction process suffers from not
preserving orbit.  Additional steps must be taken to assure that
the constructed position is in the desired orbit (usually, the
Start orbit).  And some orbits do not have representatives from
some subgroups, for example it is well known that there are
no C-symmetric but not M-symmetric permutations in the Start
orbit.

To use Dan's adjectives, this process very quickly can become
long, mechanical, and boring.  But I now see how to build a
K-symmetric permutation for any K.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From BRYAN@wvnvm.wvnet.edu  Wed Jun 28 12:52:18 1995
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Date:      Wed, 28 Jun 1995 10:09:51 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Re: Re: A Third Way to Calculate the Real Size of Cube Space?
In-Reply-To: Message of 06/20/95 at 14:39:00 from ,
           Martin.Schoenert@Math.RWTH-Aachen.DE

I guess I am going to have to break down and get a copy of GAP.  It
is truly impressive how much GAP can do so easily.

My interpretation of Martin's GAP program is that it implements the
general outline of the algorithm I described, except that GAP was
able to calculate the number of K-symmetric permutations in a very
simple and direct way, whereas I was going to have to puzzle each one
out by hand.

The heart of Martin's program appears to be the following, and I have
a couple of questions.


>        # compute how many elements have at least this symmetry group
>        number := Size( Centralizer( G, rep ) );

The first question is:  how does the Size function work?  As a simpler
example than the one above, what if you simply say Size(G)?  I am
naively assuming that G is specified to GAP in terms of generators
only, and that it makes no attempt to actually represent each element
of G (too big!).  And I have seen snippets of GAP libraries for G
posted by Mark Longridge, and they look like generators.  I have been
in several group theory books lately, and I don't recall seeing
a general algorithm presented for determining the size of a finite
group based on its generators.

The second question is like unto the first:  how does the
Centralizer function work?  In this particular case, we don't
really need the Centralizer, we only need the Size of the Centralizer,
but the question remains in either case.  Surely, GAP does not
literally try each element of G and each element of rep to see
which elements commute (too big again).  So what is the general
algorithm?

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From rodrigo@lsi.usp.br  Wed Jun 28 17:15:45 1995
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Date: Wed, 28 Jun 1995 14:12:53 +48000
From: Rodrigo de Almeida Siqueira <rodrigo@lsi.usp.br>
To: steinark@ifi.uio.no, bagleyd@source.asset.com,
        Reinaldo Augusto da Costa Bianchi <rbianchi@mozart.lsi.usp.br>
Cc: cube-lovers@life.ai.mit.edu
Subject: Rubik Cube... for Windows ?
In-Reply-To: <199506281753.OAA19377@jaguar.lsi.usp.br>
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On Wed, 28 Jun 1995, steinark@ifi.uio.no wrote:

> id: Email: steinark@ifi.uio.no
> comments:
> Just thought I would mention a Windows version Rubic Cube
> program. It's located at several ftp sites and the file 
> to look for is called 'cubic.zip'. Has a far better
> look-and-feel than XRubic I think. It has its own solving
> algorithm, but this can be rather slow...
> Check it out. If you don't find it I can probably send it
> to you some how.
> 
> Steinar

Hello Steinar,

Would you tell me where (ftp site) can I find cubic.zip ?
I would like to try it and put a link in the "Robot can play with the 
Cube Web homepage" (http://www.lsi.usp.br/~daia/celula/cubo/)

Thank you,

Rodrigo Siqueira
rodrigo@lsi.usp.br
(http://www.lsi.usp.br/usp/rod/rod.html)

From @mail.uunet.ca:mark.longridge@canrem.com  Mon Jul  3 14:15:37 1995
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Subject: Antislice Patterns
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1181.5834.0C1E9C05@canrem.com>
Date: Mon, 3 Jul 1995 14:53:00 -0400
Organization: CRS Online  (Toronto, Ontario)

Patterns in the Anti-Slice Group
--------------------------------

p4   8 flip (Op sides)  (R1 L1 U1 D1 F1 B1) ^2                    (12)
p10a pons asinorum      (L3 R1 U3 D1)^3                           (12)
p16a 4 cross order 2     F1 B1 U1 D1 L2 R2 U1 D1 F1 B1 U2 D2      (12)
p17  4 diagonal         (F1 B1 R1 L1) ^3                          (12)
p18a 4 diagonal,2 cross (F1 B1 R3 L3) ^3                          (12)
p22  2 DOT, 2 Stripe     R1 L1 U2 D2 R3 L3                         (6)
p64a 4 Z                 F1 B1 L3 R3 F1 B1 L1 R1 F3 B3 L1 R1      (12)
p143 Pinwheels           F1 B1 L1 R1 F3 B3 U3 D3 L1 R1 U1 D1      (12)
p175a 6 H order 2        U3 D3 L3 R3 F2 B2 U2 D2 L3 R3 U1 D1      (12)
p198a 2 X, 4 Diag no C   L1 R1 F1 B1 L3 R3 F3 B3 L1 R1 F1 B1      (12)
p201 Pinwheels + Pons    L1 R1 F3 B3 L1 R1 U3 D3 F1 B1 U3 D3      (12)

p201 is a quite interesting position.
The square's group equivalent is no shorter in q turns:

p175 6 H order 2 type 2  U2 B2 L2 U2 D2 L2 F2 U2                   (8)

Note that p201 = |{m'Xm}|=2 and |Symm(X)|=24.

From @mail.uunet.ca:mark.longridge@canrem.com  Mon Jul  3 14:15:29 1995
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Subject: Crazy Corner Pattern Revisited
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1182.5834.0C1E9C06@canrem.com>
Date: Mon, 3 Jul 1995 14:59:00 -0400
Organization: CRS Online  (Toronto, Ontario)

Jerry writes:

>                In their position, the UFL and RBD corners are
>in place, and the other three pairs are swapped.  The "girdle"
>includes the three pairs that are swapped.  Hence, there
>is an axis of symmetry along the UFL-RBD axis.  The odd number of
>swaps is compensated by an odd number in the edges.  The compensation
>is not required for corners only.
>
>        R D D
>        U U D
>        U U B
>
> D L L  F F D  L L F  L F F
> R L L  F F B  L R R  B B F
> R R B  R B B  U R R  B B U
>
>        U U L
>        U D D
>        F D D

Much is explained! (And this is all the way back in file "cube01" in
the archives, Date: 14 December 1980 1916-EST).

What does the 1916- mean in front of EST???

If we put the edges in place in Jerry's (and Dan's) position
then we have a position M-conjugate to the one I posted.
(re-posted below).

        F U U
        U U U
        D U L
R L B   R F B   D R R   B B D
L L L   F F F   R R R   B B B
U L L   F F U   L R F   L B F
        D D B
        D D D
        R D U

Jerry Continues:

>...there is an axis of symmetry along the UFL-RBD axis.  The odd
>number of swaps is compensated by an odd number in the edges.
>The compensation is not required for corners only.

Great! We both concur that this pattern is in the swap orbit.
There is another position with M-class 4 in the twist orbit,
but that one is *utterly* impossible unless one corner twists
on it's own (perhaps due to desperate cube turning?)

There are 2 other types of position with M-class 4, and that is
pinwheels and pinwheels + pons asinorum.

-> Mark <-

(I must be cubing too much, 1916-EST has to be 7:16 pm Eastern
Standard Time).

From kpapado@athena.auth.gr  Wed Jul  5 04:55:00 1995
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From: Kostas Papadopoulos <kpapado@athena.auth.gr>

subscribe listname Kostas Papadopoulos

From hoey@aic.nrl.navy.mil  Sun Jul  9 19:53:12 1995
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From: hoey@aic.nrl.navy.mil (Dan Hoey)
Message-Id: <9507092353.AA03807@Sun0.AIC.NRL.Navy.Mil>
To: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>,
        "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject: Re: 3x3x3 Cubes for Sale

"Jerry Bryan" <BRYAN@wvnvm.wvnet.edu> writes:
> ... I couldn't see inside the box to verify this, the Face centers
> seemed to be marked in such a way as to support the Supergroup.

Just as well, you'd have been disappointed.  As I wrote on 8 Jan 92,

: While most people are content to make each face a solid color, some
: cubes have markings that display whether the face centers are twisted
: with respect to the rest of the cube.
:    [This has recently been done commercially in an spectacularly
:     braindamaged way, in a product known as ``Rubik's cube--the
:     fourth dimension'' or some such nonsense.  The mfrs have marked
:     only four face centers, breaking symmetry while they fail to show
:     the surprising invariant of the Supergroup.  What bagbiters!]

> Rubik's note about the size of
> the  problem says it is 4^4 times bigger than the regular problem.

And it could have been 4^(11/2).

Dan
Hoey@AIC.NRL.Navy.Mil

From phaedrus@future.dreamscape.com  Sun Jul  9 23:39:02 1995
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Subject: Re: 3x3x3 Cubes for Sale
To: hoey@aic.nrl.navy.mil (Dan Hoey)
Date: Sun, 9 Jul 1995 23:38:25 -0400 (EDT)
Cc: BRYAN@wvnvm.wvnet.edu, Cube-Lovers@ai.mit.edu
In-Reply-To: <9507092353.AA03807@Sun0.AIC.NRL.Navy.Mil> from "Dan Hoey" at Jul 9, 95 07:53:10 pm
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> 
> "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu> writes:
> > ... I couldn't see inside the box to verify this, the Face centers
> > seemed to be marked in such a way as to support the Supergroup.
> 
> Just as well, you'd have been disappointed.  As I wrote on 8 Jan 92,
> 
> : While most people are content to make each face a solid color, some
> : cubes have markings that display whether the face centers are twisted
> : with respect to the rest of the cube.
> :    [This has recently been done commercially in an spectacularly
> :     braindamaged way, in a product known as ``Rubik's cube--the
> :     fourth dimension'' or some such nonsense.  The mfrs have marked
> :     only four face centers, breaking symmetry while they fail to show
> :     the surprising invariant of the Supergroup.  What bagbiters!]
> 
> > Rubik's note about the size of
> > the  problem says it is 4^4 times bigger than the regular problem.
> 
> And it could have been 4^(11/2).

And worse, it's not even 4^4, since there are two ways to align the face
centers so that the puzzle looks solved.  The two unmarked faces are
opposite each other, and the four marked faces are marked with symbols
(Rubik's signature, his silouette, "C*4^4", and the "Rubik's Cube"
trademark) that have a definite "up", but don't tie in to the rest of the
cube at all.  If all four are upside down, to "solve" the cube simply
requires turning the cube over.

> 
> Dan
> Hoey@AIC.NRL.Navy.Mil
> 


From BRYAN@wvnvm.wvnet.edu  Mon Jul 10 10:46:59 1995
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Date:      Mon, 10 Jul 1995 10:46:56 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <cube-lovers@ai.mit.edu>
Subject:   Re: Run Times, Storage Requirements, etc.
In-Reply-To: Message of 04/21/95 at 11:37:18 from mreid@ptc.com

On 04/21/95 at 11:37:18 mreid@ptc.com said:

>jerry writes

>> What would be really nice (and which may not be possible) is some
>> representation for the cube such that a cube Z and its neighbors
>> Zq or Zh are stored very close together.

>remember that the diameter of the group is small.  (my guess is
>21 face turns, 24 quarter turns.)  so this isn't possible without
>resorting to a liberal definition of "very close".

This is something I have been thinking about for a long time.  The
idea is that if large searches were run on either massively
parallel machines, or even on farms of workstations, it would be
really nice if most neighbors stayed on the same machine.

Some of my searches get so large that I have to decompose them in
a manner somewhat similar to the manner in which I envision decomposing
searches for parallel processing.  Let me use an example a project
I am working on as we speak.  I am trying to do a complete search
for edges only (with centers).

The data base for level 10 of the search is on four tapes (not so big,
really).  The data base is sorted.  The neighbors will be sorted
according to the same collating sequence.  In order to create level 11,
all neighbors have to be generated and sorted (deleting duplicates),
and then matched against the level 9 data base (again deleting duplicates).

I desire to partition the sorted neighbors using as boundary points for
the partition the first record on each of the four tapes for level 10.
My experience is that boundary points that partition one level of the
data base equally also partitions deeper levels of the data base
equally.  Having partitioned level 11, the sizes of the output files are
rather striking:

                  Neighbors in   Neighbors in   Neighbors in   Neighbors in
                    Tape 1         Tape 2         Tape 3         Tape 4
                   Partition      Partition      Partition      Partition


  Lvl 10 Tape 1    5.2 tapes      2.0 tapes      1.2 tapes      0.2 tapes
  Lvl 10 Tape 2    1.3 tapes      5.0 tapes      1.9 tapes      1.2 tapes
  Lvl 10 Tape 3    1.2 tapes      1.4 tapes      5.1 tapes      2.2 tapes
  Lvl 10 Tape 4    0.2 tapes      1.1 tapes      1.6 tapes      6.0 tapes

In real round numbers, the level 11 data base is going to have about
32 tapes, with about 8 tapes in each partition (the branching factor
is about 8 at this level of the search).  But as the chart above shows,
there is a strong tendency for neighbors to stay in the same partition.
(The chart does not reflect it, but to complete the processing for
level 11, the "Tape 1 partition" will all have to be merged together, as
will the "Tape 2 partition", etc.)

I would emphasize that these "partitions" I am talking about are totally
arbitrary subdivisions of the data into smaller chunks to make the
problem manageable.  But imagine if you will that instead of four tapes,
I had four machines, each with a sizable hard disk.  Each machine would
be assigned one of the four partitions.  As it generated neighbors, each
machine would either store the neighbor locally or would send the
neighbor on to one of the other three machines as required.  Obviously,
the more of the neighbors you can keep locally the better.

With various problems I have worked on, I have had various numbers of
partitions of the data  --  10, 16, 32, 128, etc.  The effect I am
describing is always there, and I am not totally sure why.  I have
some theories, but nothing definitive.  I think the effect is stronger
because I am using representative elements of conjugacy classes or
other equivalence classes than if I were storing all individual cubes.

Also, the effect I am describing can be characterized by the almost
silly statement that "one twist of the cube doesn't change the cube
very much".  But it's true.  For example, the dominant part of the sort
is the Front face, which isn't changed by twists of the Back.  Also,
twists of the Front don't change the Front very much when you
consider that representative elements are being stored.  It is only
twists of the Up, Down, Right, and Left faces which change the Front
very much.  Finally, the sort order for the Front face is the Upper
part, the Right part, the Down part, and the Left part, so that even
a twist of the Left face doesn't change the Front face very much with
respect to its sort order.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From BRYAN@wvnvm.wvnet.edu  Mon Jul 10 19:25:51 1995
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Date:      Mon, 10 Jul 1995 16:20:29 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Partial Results, Edges only (with Face Centers), Qturns

 Level   M-Conjugacy   Branching  Lcl     Positions   Branching  Lcl
            Classes      Factor   Max                   Factor   Max

   0               1               0              1               0
   1               1      1.000    0             12     12.000    0
   2               5      5.000    0            114      9.500    0
   3              25      5.000    0           1068      9.368    0
   4             215      8.600    0           9819      9.194    0
   5            1886      8.772    0          89392      9.104    0
   6           16902      8.962    0         807000      9.028    0
   7          150442      8.900    0        7209384      8.934    0
   8         1326326      8.816    1       63624107      8.825    2
   9        11505339      8.675           552158812      8.678
  10        96755918      8.410          4643963023      8.411

The local maximum unique up to M-conjugacy is the 6-H position
(see Symmetry and Local Maxima).  I do not yet have a process
for the 6-H, but I should be able to have one soon.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From mschoene@math.rwth-aachen.de  Wed Jul 12 06:50:38 1995
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Date: Wed, 12 Jul 95 12:48 WET DST
From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: Cube-Lovers@ai.mit.edu
Cc: BRYAN@wvnvm.wvnet.edu
In-Reply-To: "Jerry Bryan"'s message of Wed, 28 Jun 1995 10:09:51 -0400 (EDT) <wvmail32.1995jun28.094805.bryan@wvnvm.wvnet.edu>
Subject: How to compute the size of a permutation group

Jerry Bryan wrote in his message of 1995/06/28

    I guess I am going to have to break down and get a copy of GAP.  It
    is truly impressive how much GAP can do so easily.

Hah, great, another user for GAP.
Now maybe I can get a raise from my boss ;-)

Jerry continued

    My interpretation of Martin's GAP program is that it implements the
    general outline of the algorithm I described, except that GAP was
    able to calculate the number of K-symmetric permutations in a very
    simple and direct way, whereas I was going to have to puzzle each one
    out by hand.

Let me rephrase that a little bit.  My GAP prgram implements the general
outline of the algorithm you described, except that I am able to tell GAP
to calculate the number of K-symmetric permutations in a very simple and
direct way (but GAP is going to puzzle each one out using an non-simple
and non-direct algorithm I shall outlined in another message).

Jerry continued

    The heart of Martin's program appears to be the following, and I have
    a couple of questions.

    >        # compute how many elements have at least this symmetry group
    >        number := Size( Centralizer( G, rep ) );

    The first question is:  how does the Size function work?  As a simpler
    example than the one above, what if you simply say Size(G)?  I am
    naively assuming that G is specified to GAP in terms of generators
    only, and that it makes no attempt to actually represent each element
    of G (too big!).  And I have seen snippets of GAP libraries for G
    posted by Mark Longridge, and they look like generators.  I have been
    in several group theory books lately, and I don't recall seeing
    a general algorithm presented for determining the size of a finite
    group based on its generators.

For this particular case (asking for the size of a centralizer in a
permutation group), the answer is trivial.  The algorithm that computes
(generators for) the centralizer computes the size of the centralizer as
a byproduct.  I will try to outline this algorithm in another message.

But when you say 'Size(G)', then GAP will indeed use an algorithm that
computes the size of a permutation group given by a set of generators.
The algorithm is called the ``Schreier--Sims'' algorithm.  It was
developed by C.Sims for his investigations of sporadic simple groups, and
is mostly based on a lemma by N.Schreier.  Unfortunately it is *not*
described in any of the generally available textbooks on group theory.

The funny thing is, that this algorithm works very much like the
algorithm puzzlers use to find a method to solve the cube.  Of course it
is lacking the cleverness many puzzlers use to prove that their method is
complete (i.e., solves all possible states), but then the algorithm
(resp. the computer running the algorithm) is a lot more patient.

The first step (for puzzlers and the algorithm) is to find a method to
bring the first facelet to its home place.

The algorithm does this by finding all (24) places to which this facelet
can be moved.  This is done by applying the (6) generators to all the
places found so far, and repeating until no new places are found.  For
each place it also remembers a process that moves the choosen facelet
from its home place to this place, called a representative for the place.

The second step (for puzzlers and the algorithm) is to find enough
processes that leave the first facelet in its home place.  If one has
enough such processes, one can simply start another round of the
algorithm.  I.e., one uses those processes to bring the second facelet to
its home place, finds enough processes that leave the first and the
second facelet in their respective home places, and so on.

In group theoretic terms the set of elements that leave the first facelet
is in its home place is a subgroup (called the stabilizer of that point),
and the problem is to find a set of generators for this subgroup.

So how does the algorithm find such a set of generators?  This is where
Schreier's lemma kicks in.

Let us first take a representative <r1> of a place <p1> (so <r1> moves
the first facelet to the place <p1>).  Then we multiply this by a
generator <g> of the original group.  Say <g> moves the first facelet
from the place <p1> to the place <p2>.  Finally we multiply the product
<r1> * <g> by the inverse of the representative <r2> of the place <p2>.
This obviously moves the first facelet back to its home place, so the
product <r1> * <g> * <r2>^-1 is a process that leaves the first facelet
in its home place, i.e., it is an element of the subgroup.

Now Schreier's lemma says that if we take all 24 times 6 such elements,
then the resulting set is a set of generators for the subgroup.  In fact
Schreier's lemma says that if you select the representatives carefully,
then 24 * (6 - 1) + 1 generators will suffice, and for certain subgroups
(subgroups of free groups) no smaller set of generators will do.

So we now start another round of the algorithm with this (smaller)
subgroup.  How much smaller is this subgroup?  The size of the whole
group is the size of the subgroup times 24 (we get each element of the
whole group exactely once by multiplying each subgroup element by each of
the 24 representatives).  So the subgroup is smaller by a factor of 24.

After at most 48 rounds (for the 48 facelets) we are done.

We can now easily compute the size of the entire group.  It is the size
of the subgroup of those elements that leave the first facelet in its
home place times 24 (the number of possible places for the first
facelet).  The size of this subgroup is the size of the subgroup that
leaves the first and the second facelet in their respective home places
times the number of possible places for the second facelet.  And so on.

We also have a method to solve any state as follows.  We first find the
place <p1> of the first facelet and multiply by the inverse of the
representative of <p1>, to move that facelet back to its home place.
Then we find the place <p2> of the second facelet and multiply by the
inverse of the representative found in the second round of the algorithm,
to move that facelet back to its home place.  Since that representative
was found in the second round it doesn't move the first facelet, so now
the first and the second facelet are in their respective home places.
After at most 48 rounds, we have moved all facelets to their home places.

As has been pointed out before, such a method to solve any state gives us
a membership test for <G>.  Suppose we are given an arbitrary permutation
<x> of the facelets.  Then we simply try to solve <x>.  If we can solve
it, then <x> is an element of <G>.  If we cannot solve it, then <x> is
not an element of <G>.

Now the algorithm as described above has a fatal flaw.  The number of
generators increases with each round.  In the first round we have 6
generators, in the second round we have about 24 * 6 generators, in the
third round we have about 24^2 * 6 generators, and so on.  Most of these
generators will be redundant, i.e., they will lie in the subgroup
generated by the other generators.

The solution is as follows.  Instead of taking all 24 * 6 generators for
the second round, we randomly select some (say 6) of them.  Those will
generate a subgroup <S> of the whole stabilizer (and our hope is that
this subgroup will be equal to the whole stabilizer).  Then we apply the
algorithm to <S>, and compute a method to solve any element of <S>.  As
described above, this gives us a way to test membership in <S>.  We now
compute all 24 * 6 generators for the stabilizer, and for each one test
whether it is an element of <S>.  If they are all elements of <S>, then
<S> is indeed the whole stabilizer, and we are done.  If not, then we
have found a new non-redundant generator for the stabilizer, and we start
anew, this time with 7 generators instead of the originally choosen 6.

The first algorithm is sometimes called the iterative Schreier--Sims,
because it iterates over the stabilizers.  The second algorithm is called
the recursive Schreier--Sims, because it assumes that we have solved the
problem for <S> recursively.  This is the algorithm currently implemented
in GAP.  There is a third variant called Schreier--Todd--Coxeter--Sims,
that uses a Todd--Coxeter algorithm to prove that <S> is the whole
stabilizer without computing all Schreier generators, which is important
when there are many (say 1000000) possible places for the first facelet.

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From BRYAN@wvnvm.wvnet.edu  Wed Jul 12 10:31:12 1995
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Date:      Wed, 12 Jul 1995 10:31:10 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Re: Partial Results, Edges only (with Face Centers), Qturns
In-Reply-To: Message of 07/10/95 at 16:20:29 from BRYAN@wvnvm.wvnet.edu

On 07/10/95 at 16:20:29 Jerry Bryan said:

>The local maximum unique up to M-conjugacy is the 6-H position
>(see Symmetry and Local Maxima).  I do not yet have a process
>for the 6-H, but I should be able to have one soon.

I can now give one way to do it as (UD)(RR)(LL)(UD).  It seems too
simple once you chase it down.  Note that this is *not* the 6-H
position when you apply the process to the whole cube;  you have
to omit the corners for this process to yield the 6-H.  Nonetheless,
the pattern is a nice one when the process is applied to the whole
cube, one that looks familiar, although I cannot place it.  It is
*almost* what I described as the "interesting" part of three of the
10q local maxima on the whole cube, but the "interesting" part of
the 10q local maxima is (U'D')(RR)(LL)(UD) instead.

It might be noted that the length of this position
is also 8q on an edges-only-without-centers cube (see my note
of 8 Dec 1993 22:41:38).  I did not actually provide a process
for the without-centers case, but the same process
works for the 6-H edges-only with or without centers for this position.
Such is not always true.  I have talked about it before, but many
minimal processes for without-center cubes induce an invisible
rotation which becomes visible when the Face centers are included.

This is probably as good a time as any to correct an old error, pointed
out to me by Dan Hoey.  The length of a position without centers is
the minimum taken over C of the length of the same position with
centers  --  that is, the minimum of the respective lengths of the
same position rotated 24 different ways.  For searches without centers
I store representatives of the form Y=Repr{m'Xmc}.  At one point,
I said |Y|=min{|Yc|}.  This is certainly not true.  Y is just one
of the {Yc}, and it is totally arbitrary which one it is.  The difficulty
is really a notational one.  It is the length of Y without centers
which is min{|Yc|}, not the length of Y itself (with centers).  But
I don't have a good way to say "Y without centers" or especially to
say "length of Y without centers".

But in any case, the most interesting cases to me are the ones where
the length without centers matches the length with centers, so that the
minimal process for the without centers case does not induce an
invisible rotation.  The position at hand is such a case.

Finally, the position is in the anti-slice group (i.e.,
(UD)(RL)(RL)(UD)), so the position is a local maximum in the
anti-slice edges only group with a length 4a.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From BRYAN@wvnvm.wvnet.edu  Wed Jul 12 18:04:52 1995
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Date:      Wed, 12 Jul 1995 18:04:53 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Re: Partial Results, Edges only (with Face Centers), Qturns
In-Reply-To: Message of 07/12/95 at 10:31:10 from BRYAN@wvnvm.wvnet.edu

On 07/12/95 at 10:31:10 Jerry Bryan said:

>It might be noted that the length of this position
>is also 8q on an edges-only-without-centers cube (see my note
>of 8 Dec 1993 22:41:38).

Uh, the 8 Dec 1993 22:41:38 note concerns the Superflip composed
with the 6-H.  The plain old 6-H was 8 Dec 1993 23:16:50.

The length of the Superflip composed with the 6-H in edges-without-
centers was 13.  The position is even, so the length must be
at least 14 in edges-with-centers.  My search of edges-with-centers
has not gotten that far yet.  Also, the length 13 Superflip 6-H
process for edges-without-centers induces an invisible rotation,
so it is not as pretty as the length 8 process for 6-H which does
not induce an invisible rotation.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From hoey@aic.nrl.navy.mil  Wed Jul 12 18:02:55 1995
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Date: Wed, 12 Jul 95 18:02:51 EDT
From: hoey@aic.nrl.navy.mil (Dan Hoey)
Message-Id: <9507122202.AA00318@Sun0.AIC.NRL.Navy.Mil>
To: Cube-Lovers@ai.mit.edu
Subject: Happy birthday
References: <12Jul81_134343_DH51@CMU-10A>

(Sorry to have neglected the last thirteen).

             \         \         \         \
             /\        /\        /\        /\
            / /    \  / /    \  / /    \  / /    \
            \/     /\ \/     /\ \/     /\ \/     /\
             \    / /  \ \  / /  \ \  / /  \ \  / /    \
             {}   \/   {}/\ \/   {}/\ \/   {}/\ \/     /\
             {}    \   {/ /  \ \ {/ /  \ \ {/ /  \ \  / /
             {}    {}  {\/   {}/\{\/   {}/\{\/   {}/\ \/
             {}    {}  {}\   {/ /{}\   {/ /{}\   {/ /  \
     +_------{}+_--{}--{}{}--{\/-{}{}--{\/-{}{}  {\/   {}
     | `-_   {}  `-{}  {}{}`-{}\ {}{}`-{}\ {}{}`-{}\   {}
     |S   `+_------{}+_--{}--{}{}--{}--{}{}--{}--{}{}  {}
     |     | `-_   {}  `-{}  {}{}`-{}  {}{}`-{}  {}{}`-{}
     |  S  |A   `+_------{}+_--{}--{}+_--{}--{}+_--{}--{}+_
     |     |    A|M`-_   {}  `-{}  {}  `-{}  {}  `-{}  {}  `-_
     +_   S|     |  M `+_------{}+_------{}+_------{}+_-------`+_
     |L`-_ |A    |M   M|*`-_   {}  `-_   {}  `-_   {}  `-_       `-_
     |  L `+_   A|  M  |* * `+--------`+--------`+--------`+--------`+
     |L   L|O`-_ |M   M|* * *|         |         |         |         |
     |     |O O `+_ M  |*   *|         |         |         |         |
     |L   L|O O O| `-_M|* * *|  Happy  |  Birth  |   day   |   to    |
     +_ L  |O O O|    `+_ * *|         |         |         |         |
     | `-_L|O O O|   Y | `-_*|         |         |         |         |
     |H   `+_ O O|     |D   `+---------+---------+---------+---------+
     |  H  | `-_O| Y   |    D|         |         |         |         |
     |H   H|    `+_    |  D  |         |         |         |         |
     |  H  |     |R`-_ |D    |  Cube   | Lovers  |    @    |   MIT   |
     +_   H|  E  |R   `+_   D|         |         |         |         |
     |W`-_ |     |R R R|E`-_ |         |         |         |         |
     |W   `+_    |R   R|E   `+---------+---------+---------+---------+
     |W W W|A`-_ |R R R|  E E|         |         |         |         |
     |W W W|A   `+_   R|E E E|         |         |         |         |
     |W W W|  A A| `-_R|E E  |   15    |  Years  |  1659   | Messages|
     +_   W|A   A|S   `+_   E|         |         |         |         |
       `-_W|A A  |  S  | `-_E|         |         |         |         |
          `+_   A|S S S|    `+---------+---------+---------+---------+
             `-_A|  S  |     |         |         |         |         |
                `+_   S|     |         |         |         |         |
                   `-_ |     |  Many   |  Happy  |Restores |         |
                      `+_    |         |         |         |         |
                         `-_ |         |         |         |         |
                            `+---------+---------+---------+---------+
--Dan

From vorms@iprolink.ch  Sun Jul 16 07:00:55 1995
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Date: Sun, 16 Jul 1995 12:56:34 +0200
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To: Cube-Lovers@life.ai.mit.edu
From: vorms@iprolink.ch (Beffa Raphael)
X-Mailer: <Windows Eudora Version 1.4.2b16>

Hello my name is Raphael Beffa

        I am a mew user on the web. I am really happy to realise that peaple
interested
on the cube have a space to talk about on the net.
        
        I spent manny hours to develop different prototypes:
An 5^3 cube in 1981 a symetrical one 2^3 cube in 1981 and one more
sophisticated 
5^3 cube 1983.
        I realised a simple simulator on my PC for a 3^3 cube . One
representation
is a classical isometrical perspective , I realised an other representation
I called
a "Planicube" view (6 faces on the same view).

        Is somebody intersted on cube rotating on the summit (4 axes) insted of
rotating on the face (3 axes the original one).
        Or even rotating on the face and on the summit (7 axes).
        Maybe those prototypes allready exist. Please let me now.

        Is anybody able to give me the solution of the Masterball.
                      
        I am happy to meet new frends, Reguards Raphael

                        


From vorms@iprolink.ch  Sun Jul 16 16:16:22 1995
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To: Cube-Lovers@life.ai.mit.edu
From: vorms@iprolink.ch (Beffa Raphael)
X-Mailer: <Windows Eudora Version 1.4.2b16>

Hello my name is Raphael Beffa

        I am a mew user on the web. I am really happy to realise that peaple
interested
on the cube have a space to talk about on the net.
        
        I spent manny hours to develop different prototypes:
An 5^3 cube in 1981 a symetrical one 2^3 cube in 1981 and one more
sophisticated 
5^3 cube 1983.
        I realised a simple simulator on my PC for a 3^3 cube . One
representation
is a classical isometrical perspective , I realised an other representation
I called
a "Planicube" view (6 faces on the same view).

        Is somebody intersted on cube rotating on the summit (4 axes) insted of
rotating on the face (3 axes the original one).
        Or even rotating on the face and on the summit (7 axes).
        Maybe those prototypes allready exist. Please let me now.

        Is anybody able to give me the solution of the Masterball.
                      
        I am happy to meet new frends, Reguards Raphael

                        



From vorms@iprolink.ch  Sun Jul 16 16:16:19 1995
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Date: Sun, 16 Jul 1995 22:11:54 +0200
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Content-Type: text/plain; charset="us-ascii"
To: Cube-Lovers@ai.mit.edu
From: vorms@iprolink.ch (Beffa Raphael)
Subject: Re: 
X-Mailer: <Windows Eudora Version 1.4.2b16>

>Hello my name is Raphael Beffa
>
>        I am a mew user on the web. I am really happy to realise that peaple
>interested on the cube have a space to talk about on the net.
>        
>        I spent manny hours to develop different prototypes:
>An 5^3 cube in 1981 a symetrical one 2^3 cube in 1981 and one more
>sophisticated  5^3 cube 1983.
>        I realised a simple simulator on my PC for a 3^3 cube . One
>representation is a classical perspective , I realised an other
representation I called a "Planicube" view (6 faces on the same view).
>
>        Is somebody intersted on cube rotating on the summit (4 axes) insted of
>rotating on the face (3 axes the original one).
>        Or even rotating on the face and on the summit (7 axes).
>        Maybe those prototypes allready exist. Please let me now.
>
>        Is anybody able to give me the solution of the Masterball.
>                      
>        I am happy to meet new frends, Reguards Raphael
                       


From BRYAN@wvnvm.wvnet.edu  Wed Jul 19 21:47:46 1995
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Date:      Wed, 19 Jul 1995 21:47:47 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Level 11, Edges-with-Face-Centers, Qturns

 Level   M-Conjugacy   Branching  Lcl     Positions   Branching  Lcl
            Classes      Factor   Max                   Factor   Max

   0               1               0              1               0
   1               1      1.000    0             12     12.000    0
   2               5      5.000    0            114      9.500    0
   3              25      5.000    0           1068      9.368    0
   4             215      8.600    0           9819      9.194    0
   5            1886      8.772    0          89392      9.104    0
   6           16902      8.962    0         807000      9.028    0
   7          150442      8.900    0        7209384      8.934    0
   8         1326326      8.816    1       63624107      8.825    2
   9        11505339      8.675    0      552158812      8.678    0
  10        96755918      8.410          4643963023      8.411
  11       750089528      7.752         36003343336      7.753

Note that there are no local maxima at level 9, although one showed
up at level 8.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From FOMEQUE@aol.com  Sat Jul 22 18:25:28 1995
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Date: Sat, 22 Jul 1995 18:21:58 -0400
From: FOMEQUE@aol.com
Message-Id: <950722182158_120701269@aol.com>
To: cube-lovers@life.ai.mit.edu
Subject: Requesting Information

Dear Cube lovers,
   How do I access your service ?
                                                  Sincerely,
                                                      Fomeque

From nichael@sover.net  Sun Jul 23 14:08:25 1995
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Date: Sun, 23 Jul 1995 14:12:48 -0400
To: Cube-Lovers@ai.mit.edu
From: nichael@sover.net (Nichael Lynn Cramer)
Subject: Rubik's Race

[I don't know if this has been discussed before.  Maybe some of the old
timers know about this --the copyright on the box is 1982-- but I don't
recall having seen it mention since I've been on the list...]

Anyway, Saturday we hit a rumage sale at the next town over (Hinsdale VT)
and I picked up an interesting game in the "toys" section.  It's called
"Rubik's Race".  (It was put out by Ideal and seems to be reasonably well
made, so I assume it's "authorized".)

The main "board" is about 18in long by about 7in wide.  The players set at
either end and their section of the board has a recessed area that holds 24
plastic pieces (about 1in square) aranged in a 5X5 grid with one empty
space.

(The square pieces are divided into six groups of four, each four-group
being one of the standard colors on the side of the Cube.)

At the start of a game, one player picks up a small box that contains nine
small "cubies" (about 3/8in on a side), shakes the box and sets it down.
The cubies settle onto the bottom of the box in 3X3 grid (thereby
resembling the face of cube).  Each player then proceeds to slide his
pieces around (using the blank as maneourvering space) until the center 3X3
grid (out of the original 5X5 grid) matches the 3X3 "face" in the small
box.

The board is also divided between the two players' area by a vertical,
hinged piece which can fall towards either end, and which has a "window" in
the center.  When the player finishes his central 3X3 grid, he pulls the
hinged pieces towards himself --the window of which exposes his central
grid-- and declares himself the winner.

An obvious variation is for one player to scramble a cube, set it on the
table and have each player try to match the topmost face.

So, as such the games provides an interesing combination of the Cube, the
15-puzzle and Battleship.   ;-)

(But probably the most amazing part is that, given that I picked this up at
rummage sale, *all* of the nearly sixty small plastic pieces were still
there!  All in all not a bad investment for 10cents.)





Nichael              -  "...did I forget, forget to mention Memphis?
nichael@sover.net           Home of Elvis, and the ancient Greeks."



From munafo@vgi.com  Mon Jul 24 14:59:46 1995
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Date: Mon, 24 Jul 1995 15:00:01 -0400 (EDT)
From: Robert Munafo <munafo@vgi.com>
X-Sender: munafo@frank
To: CUBE-LOVERS List <Cube-Lovers@ai.mit.edu>
Subject: Little keychain cubes
Message-Id: <Pine.HPP.3.90.950721204433.19487B-100000@frank>
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Sort of a strange question -- but at this point I don't know who else to ask. 

Does anyone know how I could get more of those little 25-cm 3x3x3 cubes that
used to be sold on a keychain for about US $3.00? I checked a few toy and
novelty stores in my area, including Games People Play near MIT in Cambridge
(and they weren't very nice about giving me a useful answer either)... 

Barring that, has anyone successfully made their own cubes? Since the 3x3x3
is relatively simple in design I suppose the tolerances wouldn't be all that
bad. 

I'd like to get a bunch of small cubes so I can keep a bunch of cubes in
different configurations. 

Yes, I am reading through the 15-year backlog and am only up to mid-December
1980. I did some grep -i searches for "key\ *chain", "mold" and "cast"  but
found nothing helpful. 

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Davis' principle 36:   Research-then-transfer doesn't work.
- - - - Robert P Munafo - - -  munafo@vgi.com - - - +1.617.276.8960 - - -

From ncramer@bbn.com  Mon Jul 24 16:21:02 1995
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Message-Id: <9507242021.AA21879@life.ai.mit.edu>
Date:     Mon, 24 Jul 95 16:04:53 EDT
From: Nichael Cramer <ncramer@bbn.com>
To: Robert Munafo <munafo@vgi.com>
Cc: CUBE-LOVERS List <Cube-Lovers@ai.mit.edu>
Subject:  Re:  Little keychain cubes

>Date: Mon, 24 Jul 1995 15:00:01 -0400 (EDT)
>From: Robert Munafo <munafo@vgi.com>
>Subject: Little keychain cubes
>
>Does anyone know how I could get more of those little 25-cm 3x3x3 cubes that

25-Cm!?!  Damn son, you must have a helluva lot of keys!     ;-)

>used to be sold on a keychain for about US $3.00? I checked a few toy and
>novelty stores in my area, ...

Right, I know the one's you mean (I've got a two or three of them around
here somewhere) but I haven't seen any of these for sale for years.  If I
remember right I picked mine up at a convience store check-out line.  (Hard
as heck to turn.)

> ... including Games People Play near MIT in Cambridge
>(and they weren't very nice about giving me a useful answer either)... 

Yeah, aren't they a pain to deal with?

>Barring that, has anyone successfully made their own cubes? Since the 3x3x3
>is relatively simple in design I suppose the tolerances wouldn't be all that
>bad. 
>
>I'd like to get a bunch of small cubes so I can keep a bunch of cubes in
>different configurations. 

Probably a dumb idea, but if they're going to stay fixed, couldn't you just
paint some up?

N

From munafo@vgi.com  Mon Jul 24 19:11:44 1995
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Date: Mon, 24 Jul 1995 19:11:50 -0400 (EDT)
From: Robert Munafo <munafo@vgi.com>
X-Sender: munafo@frank
To: CUBE-LOVERS List <Cube-Lovers@ai.mit.edu>,
        Nichael Cramer <ncramer@bbn.com>
Subject: Shallow-cut dodecahedron, and Re: Little keychain cubes
In-Reply-To: <9507242021.AA21879@life.ai.mit.edu>
Message-Id: <Pine.HPP.3.90.950724171843.6874A-100000@frank>
Mime-Version: 1.0
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I have also noticed that the discussion on the "shallow-cut dodecahedron" 
a.k.a. Hungarian "super nova" has come and gone a couple times. What I'm
talking about is a puzzle that looks like a Platonic dodecahedron (12
pentagonal faces) cut like

          +''':'''''':'''+                +'''':'''':''''+
         :   :        :   :              :    :      :    :
        .:...:........:...:.            .'   .'      :    `.
        :   :          :   :            :'''':'''''''':'''':
       :   :            :   :          :    :          :    :
      .'  .'            `.  `.        .'   .'          `.   `.
      :.  :              :  .:   or   :`.  :            :  .':
     :  `:                :'  :      :   `:              :'   :
     +  .'`.            .'`.  +      +   .''.          .'`.   +
      `.:   `.        .'   :.'        `. :   `.      .'   : .'
        `.    `.    .'    .'            `.     `.  .'     .'
          `.    `..'    .'                `.     ::     .'
            `.  .'`.  .'                    `. .'  `. .'
              `:    :'                        `.    .'
                `..'                            `..'

and manipulated by turning a face 72 degrees around its axis of 5-fold
rotational symmetry (an operation that moves 5 edge pieces and 5 corner
pieces).

I'd like to cast the parts for one of these as well. It seems that the only
difficult part would be the 12-pointed "spindle" that holds all the center
pieces. It has to be strong, rigid and perfectly aligned. The easiest thing
would be to screw bolts into a 12-sided die (of the D&D variety) but I think
the alignment would be too poor. 


On Mon, 24 Jul 1995, Nichael Cramer wrote:
> >Does anyone know how I could get more of those little 25-cm 3x3x3 cubes that
> 
> 25-Cm!?!  Damn son, you must have a helluva lot of keys!     ;-)

Oh, yes of course I meant to say 25 mm, not 25 cm. <-:
   Of course, a 25-cm cube would be fun to have, too. I imagine it would be
about a 25x25x25, which would have about 7.3 * 10^2328 combinations (even
without considering face centers). God's Number would be well over 1000, and
even "Pons Asinorum" would be a major effort at 150 QTW. Since it's an
odd-order cube, it could be built by the simplistic "sextapole magnet" 
method (each cubie has a north pole on its inward-pointing faces and a south
pole on its outward-pointing faces, except for the cubiess on the three
central planes which could be hollow steel) [-8

> >I'd like to get a bunch of small cubes so I can keep a bunch of cubes
> >in different configurations. 
> 
> Probably a dumb idea, but if they're going to stay fixed, couldn't you just
> paint some up?

Sort of a minimalist approach... but I have a notebook for recording
patterns. Most of what I want to do involves having a bunch of cubes in the
same pattern, then performing different transformations on each one and
comparing the results.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
More software projects have gone awry for lack of calendar time than for
all other causes combined.
[Fred Brooks, _The Mythical Man-Month_, p. 14 (and again on p. 26)]
- - - - Robert P Munafo - - -  munafo@vgi.com - - - +1.617.276.8960 - - -

From ronnie@cisco.com  Mon Jul 24 20:32:24 1995
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To: Robert Munafo <munafo@vgi.com>
Cc: CUBE-LOVERS List <Cube-Lovers@ai.mit.edu>,
        Nichael Cramer <ncramer@bbn.com>
Subject: Re: Shallow-cut dodecahedron, and Re: Little keychain cubes 
In-Reply-To: Your message of "Mon, 24 Jul 1995 19:11:50 EDT."
             <Pine.HPP.3.90.950724171843.6874A-100000@frank> 
Date: Mon, 24 Jul 1995 17:32:11 -0700
From: "Ronnie B. Kon" <ronnie@cisco.com>

>    Of course, a 25-cm cube would be fun to have, too. I imagine it would be
> about a 25x25x25, which would have about 7.3 * 10^2328 combinations (even

And the the box would probably tout its difficulty as having "more
than 10 billion possibilities."  :-(

When they changed the lottery here in California from pick 6 of 48
numbers to pick 6 of 52 there was a steady stream of people on the
evening news whining about how the lottery used to be hard to win, but
now it's almost impossible.  (I'll save you some trouble: it went from
1 in 12,271,512 to 1 in 20,358,520).

				Ronnie

From Games@puzzles.demon.co.uk  Thu Jul 27 05:38:33 1995
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Date: Tue, 25 Jul 95 10:34:13
Message-Id: <648@puzzles.demon.co.uk>
From: Games@puzzles.demon.co.uk (Yerry Felix)
Organization: Games & Puzzles Magazine
Reply-To: Games@puzzles.demon.co.uk
To: cube-Lovers@ai.mit.edu
Subject: Re: little keychain cubes and IQubes
X-Mailer: Newswin Alpha 0.7
Lines:  19


Hi,

I guess I could keep an eye open for those little keychain cubes when I go 
the next Essen Game Fair. There are myriads of games, puzzles and 
gadgets to be had (100000 people attend) and if I find I supplier
I will post the adress here.

On another note, a while back Games & Puzzles Magazine reviewed a cube 
called IQube. It comprised a cage containing 26 cubelets, there are two 
variants, one with 3 * 2 colours on the individual cubelets and another 
one with 2 * 3 colours. A few days later a friend of mine brought in an 
ancient cube (about 10 years old) which was made to the same principle, 
but with only one colour on one of the faces of the cubelets inside and the other 
5 sides are black. Would anyone know who manufactured these? 

-- Yerry Felix
Games & Puzzles Magazine




From munafo@vgi.com  Fri Jul 28 17:15:16 1995
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From: Robert Munafo <munafo@vgi.com>
X-Sender: munafo@frank
To: CUBE-LOVERS List <Cube-Lovers@ai.mit.edu>
Subject: Re: IQubes
In-Reply-To: <648@puzzles.demon.co.uk>
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Sounds like a 3-d analogue of the "15-puzzle". I have one which is called
"MagicJack", and is imported from Germany.  I bought mine at Games People
Play on Mass Ave. in Cambridge MA, USA (+1.617.492.0711). 

The cubies have varying numbers of sides colored silver, red, and green.
There are three different ways to "solve" it. The silver way is easiest; the
green way requires you to match up patterns on each facelet to make a
continuous loop, conceptually sinilar to Rubik's Tangle. The red solution
also requires a continuous loop, and there are symbols that must be matched
up too, making it harder to find a pattern that is valid. 

I haven't scrambled mine yet, because I haven't had time to write down 
the initial pattern.

On Tue, 25 Jul 1995, Yerry Felix wrote:
> 
> On another note, a while back Games & Puzzles Magazine reviewed a cube 
> called IQube. It comprised a cage containing 26 cubelets, there are two 
> variants, one with 3 * 2 colours on the individual cubelets and another 
> one with 2 * 3 colours. A few days later a friend of mine brought in an 
> ancient cube (about 10 years old) which was made to the same principle, 
> but with only one colour on one of the faces of the cubelets inside and the other 
> 5 sides are black. Would anyone know who manufactured these? 
> 
> -- Yerry Felix
> Games & Puzzles Magazine

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Davis' principle 46:   Avoid design in requirements.
- - - - Robert P Munafo - - -  munafo@vgi.com - - - +1.617.276.8960 - - -

From mbparker@share.ai.mit.edu  Fri Jul 28 19:10:14 1995
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	id AA19733; Fri, 28 Jul 95 16:07:07 -0700
Date: Fri, 28 Jul 95 16:07:07 -0700
From: Michael B. Parker <mbparker@share.cytex.com>
Message-Id: <9507282307.AA19733@share>
To: mitacas@cytex.com, PuzzleParty@cytex.com
Subject: This weekend: the 1995 OC Mensa Annual Regional Gathering
Reply-To: mbparker@cytex.com

[Posted for OC Mensa member Marti L. Hitchcock; please respond to her.  If
you're interested in the local Mensa, this may be a good event to check out.
Over 200 are preregistered already, and they're expecting 100 to 200 more at
the door.]

	 THE 1995 ORANGE COUNTY MENSA ANNUAL REGIONAL GATHERING (RG):

			       ``The M Files''

	      Noon Friday July 28 til 6pm Sunday, July 30, 1995

PROGRAM:
 Friday, July 28
  1100-  Registration opens, Hospitality opens, Construction Workshop
  1200-  New Agent Orientation, Classical Musicale
  1300-  ``Common Problems'', Arts & Crafts (cont), ``What you Know May
         Not be So''
  1500-  ``Magic & Parapsychology'', ``Astrology'', ``Corazon Project''
  1600-  ``UFO lecture, ``An Artist's Perspective'', Psychic lecture
  1700-  ``The FBI Today''
  1800-  Pot luck dinner, Arts & Crafts, Classical Musicale
  1900-  ``Nattering'' by Larry Niven, ``Handwriting Analysis''
  2000-  Space Station Omicron Beta I featuring UFAUX along with psychic
         readings, Sci-fi movies
  2000-	 Rock dance.

 Saturday, July 29
  0900-2400      [About 70% more events than Friday]

 Sunday,   July 30
  0900-1800      [About 20% more events than Friday]

 Saturday & Sunday On-Going
  	Experience the ride of a life-time in David M. Mitchell's lunar
	teleoperations module (VIRTUAL REALITY)

WHERE:	At the Days Inn, 1500 South Raymond Ave., Fullerton (north of the 91
	fwy), CA.  Call 714-635-9000 and mention OC Mensa for the discount
	rate of $54/night.

COST:	AT THE DOOR REG. (after 7/15/95)

	                 ADULT CHILD
	FRI, SAT, SUN    $55   $28
	FRI, SUN         $25   $13
	SAT, SUN         $35   $18

CONTACTS:
	Chuck La Mont, 655 S.  Rosalind Dr., Orange, CA 92669-5124.
	Also Marti L. Hitchcock, 714-750-4333, mlhitch@cytex.com


From dmi@questrel.questrel.com  Sat Jul 29 01:15:47 1995
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	 id WAA19288; Fri, 28 Jul 1995 22:09:57 -0700
From: dmi@questrel.com (Dean Inada)
Message-Id: <9507282209.ZM19286@questrel.questrel.com>
Date: Fri, 28 Jul 1995 22:09:55 -0700
X-Mailer: Z-Mail (3.2.0 26oct94 MediaMail)
To: chris@questrel.questrel.com, mitacas@cytex.com, PuzzleParty@cytex.com
Subject: (Fwd) This weekend: the 1995 OC Mensa Annual Regional Gathering
	This weekend: the 1995 OC Mensa Annual Regional Gathering
Reply-To: mbparker@cytex.com
Mime-Version: 1.0
Content-Type: text/plain; charset=us-ascii

[Posted for OC Mensa member Marti L. Hitchcock; please respond to her.  If
you're interested in the local Mensa, this may be a good event to check out.
Over 200 are preregistered already, and they're expecting 100 to 200 more at
the door.]

	 THE 1995 ORANGE COUNTY MENSA ANNUAL REGIONAL GATHERING (RG):

			       ``The M Files''

	      Noon Friday July 28 til 6pm Sunday, July 30, 1995

PROGRAM:
 Friday, July 28
  1100-  Registration opens, Hospitality opens, Construction Workshop
  1200-  New Agent Orientation, Classical Musicale
  1300-  ``Common Problems'', Arts & Crafts (cont), ``What you Know May
         Not be So''
  1500-  ``Magic & Parapsychology'', ``Astrology'', ``Corazon Project''
  1600-  ``UFO lecture, ``An Artist's Perspective'', Psychic lecture
  1700-  ``The FBI Today''
  1800-  Pot luck dinner, Arts & Crafts, Classical Musicale
  1900-  ``Nattering'' by Larry Niven, ``Handwriting Analysis''
  2000-  Space Station Omicron Beta I featuring UFAUX along with psychic
         readings, Sci-fi movies
  2000-	 Rock dance.

 Saturday, July 29
  0900-2400      [About 70% more events than Friday]

 Sunday,   July 30
  0900-1800      [About 20% more events than Friday]

 Saturday & Sunday On-Going
  	Experience the ride of a life-time in David M. Mitchell's lunar
	teleoperations module (VIRTUAL REALITY)

WHERE:	At the Days Inn, 1500 South Raymond Ave., Fullerton (north of the 91
	fwy), CA.  Call 714-635-9000 and mention OC Mensa for the discount
	rate of $54/night.

COST:	AT THE DOOR REG. (after 7/15/95)

	                 ADULT CHILD
	FRI, SAT, SUN    $55   $28
	FRI, SUN         $25   $13
	SAT, SUN         $35   $18

CONTACTS:
	Chuck La Mont, 655 S.  Rosalind Dr., Orange, CA 92669-5124.
	Also Marti L. Hitchcock, 714-750-4333, mlhitch@cytex.com


---End of forwarded mail from "Michael B. Parker" <mbparker@share>

From alan@curry.epilogue.com  Sat Jul 29 01:43:17 1995
Return-Path: <alan@curry.epilogue.com>
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Date: Sat, 29 Jul 1995 01:43:16 -0400
Message-Id: <29Jul1995.012913.Alan@LCS.MIT.EDU>
From: Alan Bawden <Cube-Lovers-Request@ai.mit.edu>
Sender: Cube-Lovers-Request@ai.mit.edu
To: Cube-Lovers@ai.mit.edu
In-Reply-To: Dean Inada's message of Fri, 28 Jul 1995 22:09:55 -0700 <9507282209.ZM19286@questrel.questrel.com>
Subject: Orange County Mensa?

Many of you are probably wondering why you are getting mail about a Mensa
meeting in Orange County.  It's because some pinheaded loser with only half
a brain thought it would be a good idea to add the entire Cube-Lovers
mailing list to the "PuzzleParty@cytex.com" mailing list.  I have sent a
vigorous complaint to the culprits and to their postmaster.

				- Alan

------- Begin Junk Mail -------

From: Dean Inada <dmi@questrel.com>
Date: Fri, 28 Jul 1995 22:09:55 -0700
To: chris@questrel.questrel.com, mitacas@cytex.com, PuzzleParty@cytex.com
Subject: (Fwd) This weekend: the 1995 OC Mensa Annual Regional Gathering
	This weekend: the 1995 OC Mensa Annual Regional Gathering
Reply-To: mbparker@cytex.com

[Posted for OC Mensa member Marti L. Hitchcock; please respond to her.  If
you're interested in the local Mensa, this may be a good event to check out.
Over 200 are preregistered already, and they're expecting 100 to 200 more at
the door.]

	 THE 1995 ORANGE COUNTY MENSA ANNUAL REGIONAL GATHERING (RG):

			       ``The M Files''

	      Noon Friday July 28 til 6pm Sunday, July 30, 1995

PROGRAM:
 Friday, July 28
  1100-  Registration opens, Hospitality opens, Construction Workshop
  1200-  New Agent Orientation, Classical Musicale
  1300-  ``Common Problems'', Arts & Crafts (cont), ``What you Know May
         Not be So''
  1500-  ``Magic & Parapsychology'', ``Astrology'', ``Corazon Project''
  1600-  ``UFO lecture, ``An Artist's Perspective'', Psychic lecture
  1700-  ``The FBI Today''
  1800-  Pot luck dinner, Arts & Crafts, Classical Musicale
  1900-  ``Nattering'' by Larry Niven, ``Handwriting Analysis''
  2000-  Space Station Omicron Beta I featuring UFAUX along with psychic
         readings, Sci-fi movies
  2000-	 Rock dance.

 Saturday, July 29
  0900-2400      [About 70% more events than Friday]

 Sunday,   July 30
  0900-1800      [About 20% more events than Friday]

 Saturday & Sunday On-Going
  	Experience the ride of a life-time in David M. Mitchell's lunar
	teleoperations module (VIRTUAL REALITY)

WHERE:	At the Days Inn, 1500 South Raymond Ave., Fullerton (north of the 91
	fwy), CA.  Call 714-635-9000 and mention OC Mensa for the discount
	rate of $54/night.

COST:	AT THE DOOR REG. (after 7/15/95)

	                 ADULT CHILD
	FRI, SAT, SUN    $55   $28
	FRI, SUN         $25   $13
	SAT, SUN         $35   $18

CONTACTS:
	Chuck La Mont, 655 S.  Rosalind Dr., Orange, CA 92669-5124.
	Also Marti L. Hitchcock, 714-750-4333, mlhitch@cytex.com


---End of forwarded mail from "Michael B. Parker" <mbparker@share>

------- End Junk Mail -------

From BRYAN@wvnvm.wvnet.edu  Sat Aug 12 20:32:51 1995
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Message-Id: <wvmail32.1995aug12.200305.bryan@wvnvm.wvnet.edu>
Date:      Sat, 12 Aug 1995 20:09:28 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   More on Polya-Burnside

Dan Hoey was the first to write concerning the Polya-Burnside theorem
(he used it in has calculation of the real size of cube space),
and Martin Schoenert recently posted a proof of the theorem.
I would like to add some additional comments.

First, I quote from Martin:

> If $g \in G$, then I denote the set of elements that are really
> equivalent to $g$ by $g^M$.  Jerry denotes this set by {m'gm},
> but $g^M$ is the more common notation in group theory.

I have just copied the {m'gm} notation from others on this list.
I haven't searched the archives to see when it first appeared.
In fact, what I really say is {m'Xm}.  A few points:

    1) {m'Xm} is a shorthand for {m'Xm | m in M}, where X is assumed
       to be a fixed but arbitrary element of G.

    2) m' is used rather than Martin's preferred m^(-1) as a concession
       to the difficulties of using mathematical notation in E-mail.
       Again I haven't searched the archives, but this convention seems to
       go back to the earliest days of Cube-Lovers.

    3) X is used rather than g throughout Cube-Lovers, I think due
       to Singmaster.  Singmaster uses upper case letters for processes
       and lower case letters for the cycles of cubies.
       Hence, we have such things as X=URB'L or Y=RL'UD', etc.  For
       most purposes, we simply identify processes such as X and Y
       with the permutation which is effected by the respective
       process, and there is no loss of generality from such
       identification.  I interpret g in G as being a permutation
       directly, without regard to which process might
       effect the permutation.  But again, for most purposes there
       is no loss of generality in identifying the X's with the g's.

But let's go along with Martin for a moment and write {m'gm}.
We normally interpret this as {m'gm | m in M}, but we could
just as well interpret it as {m'gm | g in G}.  This new interpretation
simply yields G, but in a different order.  To say "in a different
order" is a bit of a corruption because sets don't have order. But
it's useful to think about the "different order" anyway.

M-conjugation for a fixed m in M can be viewed as a permutation on
the set of quarter turns Q.  (See for example Hoey and Saxe's
_Symmetry and Local Maxima_).  But it can also be viewed as a
permutation on G itself.  So for each of the forty-eight m in M,
M-conjugation is a different permutation on G.  This will shortly prove
to be very useful.

What if we interpret {m'gm} as {m'gm | m in M and g in G}?
Again, this is a bit of a corruption, because "m in M and g in G"
will list each element of G forty-eight times, and Set Theory 101
says each element of a set is listed only once.  But
let's ignore that difficulty and picture "m in M and g in G"
as creating a matrix with forty-eight columns indexed by M and
|G| rows indexed by G.  Each column is a different permutation
on G.

Now we detour a minute and note that this matrix contains
|G|*|M| cells.  Given that, how big is G?  Well, it is (|G|*|M|)/|M|.
This may seem tautological, but not quite.  That is, I am not
asking how many rows in a 7 by 3 matrix.  Rather, I am asking
how many rows in a matrix if the matrix has 21 cells and 3 columns.
Trivial though it may be, we have to perform the division to determine
the answer.

I am reminded of an old joke.  A mathematician is asked how many
legs a horse has.  The mathematician observes that the horse has
two front legs, two back legs, two left legs, and two right legs
for a total of eight.  But this procedure counts each leg twice,
so the mathematician divides by two to obtain the correct answer.

Many counting formulas work in a similar fashion.  They overstate
the correct number, and then adjust by dividing out or subtracting
out the excess.  In this manner, the number of cells in the
|G|*|M| matrix overstates the size of G by forty-eight, so we must divide
by forty-eight to get the true size of G.

The Polya-Burnside theorem has to do with counting conjugacy classes.
Martin's proof does not mention the word "matrix", but it effectively
creates a binary matrix with dimension |G|*|M| where each
cell contains the Boolean value (g == m'gm).  In other words, the cell
contains a 1 if g=m'gm and 0 otherwise.  Martin's proof shows that the
number of M-conjugacy classes is the number of 1's divided by forty-eight.

In his note about the real size of cube space,
Dan mentions a book called *Geometry and Symmetry* by Paul Yale.
Yale's book includes a Polya-Burnside proof similar to Martin's.
In an example accompanying his proof, Yale shows a matrix where the
cells are either blank or contain a check mark.  Yale counts the
check marks, and Martin counts the 1's.  Martin's approach has the
advantage that you can count 1's simply by summing them.

To me, the key point in both proofs is the observation that
you get the same answer whether you count the 1's or checkmarks
by row, or whether you count them by column.  This observation
manifests itself in Martin's proof as follows:

> Thus the number of M-conjugacy classes is
> $1/|M| \sum_{g \in G} \sum_{m \in M} {(g^m == g)}$.
>
> Now we can simply change the order of the two summations, so we get
> $1/|M| \sum_{m \in M} \sum_{g \in G} {(g^m == g)}$.

(When I read this last sentence in Martin's proof, the thought that
came to mind was "He transposed the matrix!", even though there is
no matrix there explicitly.)

The essence of Polya-Burnside is that summing by row gives us the
answer we desire (namely the number of M-conjugacy classes),
but summing by column is the calculation which is possible in
practice.  And serendipitously, both sums give us the same answer.

Let us consider each sum in turn.  (Actually, the matrix I am
describing is transposed compared to the one in Yale's book, but
we will continue with |G| rows and |M| columns for the purposes
of this note.)

Martin's proof gives a good explanation why summing by row gives us
the answer we desire.  Let me give a slightly different (but I think
equivalent) explanation.

Suppose |{m'Xm}|=48.  This is the case where Symm(X)=I, so that the
position is "completely asymmetric".  The row indexed by X will
contain a single 1 and forty-seven 0's.  The single 1 will appear
in the column indexed by the identity in M.  The other forty-seven
elements of {m'Xm} will similarly appear in a row containing only
a single 1.  Hence, the number of 1's in these forty-eight rows
will be 48*1, and the number of M-conjugacy classes represented by
these forty-eight rows will be (48*1)/48.  The number of 1's has
overstated the number of conjugacy classes by exactly 48.

Now suppose |{m'Xm}|=24.  We have |Symm(X)|=2.  The row indexed
by X will contain two 1's and forty-six zeros.  Similarly, the
rows indexed by the other twenty-three elements of {m'Xm} will
also contain two 1's and forty-six 0's.  The number of 1's in these
twenty-four rows will be 24*2, and the number of M-conjugacy classes
represented by these twenty-four rows will be (24*2)/48.  Again,
the number of 1's has overstated the number of conjugacy classes
by exactly 48.

The pattern should be clear.  If |{m'Xm}|=16, we will have
(16*3)/48 M-conjugacy classes scattered over three rows.
If |{m'Xm}|=12, we will have (12*4)/48 M-conjugacy classes scattered
over four rows.  Etc.  In all cases, summing the 1's
overstates the number of M-conjugacy classes by exactly forty-eight,
so in all cases we must divide by forty-eight to compensate.
It is therefore clear that to calculate the total number of
conjugacy classes, we simply sum the entire binary matrix and divide
by forty-eight.  It doesn't really matter whether we sum by rows,
sum by columns, some in some other order, or sum in no order at
all.

Polya-Burnside essentially says that we can sum by columns.  The
forty-eight column sums are the number of elements of G which
are fixed by conjugation by the respective elements of M.  Polya-
Burnside is usually stated something like "the number of conjugacy
classes is equal to the average of the number of fixed points  ....",
where there is sufficient language to make sure that the fixed
points in question are the points in G fixed by conjugation by M.
In the matrix at hand, we form the forty-eight column sums,
add them up, and divide by forty-eight.

If that is not an average (adding up forty-eight numbers and dividing
by forty-eight), then I don't know what is.  But I confess this does
not look and feel like an averaging problem to me.  Rather, it looks
and feels like a horse's legs problem where we are overstating the
answer and dividing out the excess.  It feels more comfortable to me
just to add up the entire binary matrix without regard to rows and
columns and then to divide by forty-eight, but that is not the way
Polya-Burnside works.

Forming the forty-eight column sums is no small problem.  Martin's
little GAP program accomplished this task using the Centralizer function.
Unless my E-mail system has lost it, we are still awaiting a description
of GAP's algorithm for calculating the Centralizer.

Dan's method was a "by hand" calculation of the column sums.  He
determined the number of elements of G fixed by m based on an
argument concerning the cycle structure of elements of G.
Dan took one very nice shortcut.  There really is no need to
calculate all forty-eight column sums.  A number of the elements
of M are themselves M-conjugate and there are ten conjugacy classes,
so Dan only had to calculate ten column sums.

One of the ten calculations really wasn't necessary.  The column
indexed by the identity in M contains |G| 1's, so we have
one of the ten required column sums without further ado.  This column
alone shows that the number of M-conjugacy classes is at least
|G|/|M| = |G|/48.  As it turns out, this is very close to the
true value.  The other forty-seven columns of the matrix are extremely
sparse, so relatively speaking, there are not many more
conjugacy classes than |G|/48.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From mbparker@share.ai.mit.edu  Fri Aug 18 19:24:29 1995
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	id AA09843; Fri, 18 Aug 95 16:11:30 -0700
Date: Fri, 18 Aug 95 16:11:30 -0700
From: Michael B. Parker <mbparker@share.cytex.com>
Message-Id: <9508182311.AA09843@share>
To: PuzzleParty@cytex.com, Cube-Lovers@ai.mit.edu, mitacas@cytex.com
Subject: PUZZLE PARTY 3 TOMORROW!, 7pm  (in Orange County, CA)
Reply-To: mlhitch@cytex.com, mbparker@cytex.com


			      PUZZLE PARTY III!

		   "A box without hinges, a lock or a lid,
		     Yet golden treasure inside is hid."

		     Answer this riddle and in you'll be,
		     challenges await at Puzzle Party 3.

		      Arrive at 7, and stay until late,
			August 19th, that's the date!


Bring your brain teasers, mechanical puzzles and mental games, and prepare
yourself to have an incredibly good time with like-minded puzzle gurus.  Only
those seeking a warm and friendly atmosphere, puzzle challenges galore, and an
evening of good times need to enter the door. Home-cooked hamburgers,
hot-dogs, and plenty of snacks and refreshments provided.

This one is sponsored by a member of the Orange County Mensa Chapter, so both
MITCSC and MENSA will be there!

 WHEN:  Saturday, 1995 August 19th, 7pm until...

 WHERE: 11382 Fredrick Drive, Garden Grove, CA (near 22, 57 and 5 freeways).
        FROM 57 or 5 FWY: exit WEST on CHAPMAN; PASS HARBOR; turn SOUTH/LEFT 
             on 9TH then first RIGHT (WEST) onto FREDRICK DRIVE.
        FROM 22 FWY: exit NORTH on HARBOR; turn WEST/RIGHT on CHAPMAN, turn 
             SOUTH/RIGHT on 9TH then first RIGHT (WEST) onto FREDRICK DRIVE.

 COST:	$4      Members & Guests with puzzles
	$6  Non-Members & Guests with puzzles
	$8      Members & Guests w/o  puzzles
	$10 Non-Members & Guests w/o  puzzles

 RSVP:	You may pay at the door, but please try to contact me beforehand so
	I can put you on the list.  Please email, fax, or phone the following
	info: Your NAME, ADDRESS, PHONE, FAX, EMAIL, and what you're bringing:

              ___ puzzle-bearing     members at $ 4 each: $___
              ___ puzzle-bearing non-members at $ 6 each: $___
              ___ puzzle-less        members at $ 8 each: $___
              ___ puzzle-less    non-members at $10 each: $___
              ___ <- total persons          total cost -> $___
                 total number of puzzles being brought ___

 SPONSOR: marti hitchcock, member of Orange County Mensa
          email mlhitch@cytex.com, fax 714-750-4344
          11382 Fredrick Dr., Garden Grove, CA 92640
          day 714-750-4333, eve 714-530-7605

From AirWong@aol.com  Sat Aug 19 15:41:31 1995
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From: AirWong@aol.com
Message-Id: <950819154130_78502965@mail06.mail.aol.com>
To: Cube-Lovers@ai.mit.edu
Subject: 5 X 5 X 5 Rubik's Cubes

Hello All,

I am new to this list, and i would like to ask if anyone knows where I can
get a 5 X 5 X 5 Rubik's Cube. I have already mastered the 3 and 4 cubes, and
I don't think that the 5 will be much different. I would like to try it
anyways.

Aaron Wong
AirWong@AOL.com

From mouse@collatz.mcrcim.mcgill.edu  Sat Aug 19 16:19:27 1995
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Date: Sat, 19 Aug 1995 16:19:15 -0400
From: der Mouse <mouse@collatz.mcrcim.mcgill.edu>
Message-Id: <199508192019.QAA21939@Collatz.McRCIM.McGill.EDU>
To: AirWong@aol.com
Subject: Re:  5 X 5 X 5 Rubik's Cubes
Cc: cube-lovers@ai.mit.edu

> I am new to this list, and i would like to ask if anyone knows where
> I can get a 5 X 5 X 5 Rubik's Cube.

Can't help with this...unless you're in Montreal, in which case you
might give Valet de Coeur a try.  I think that's where I got mine.

> I have already mastered the 3 and 4 cubes, and I don't think that the
> 5 will be much different.

You're right; it won't.  If you can handle the 3-cube and 4-cube,
you've got all the necessary skills.  About the only further challenge
is the supergroup, where (for example) each face has a picture on it,
so that the inner face cubies have to be positioned (and oriented, for
the center cubie on an odd-sized cube) correctly.  I have never seen
such a thing for cubes larger than 3x3x3, but I also have never
bothered really looking for one.

					der Mouse

			    mouse@collatz.mcrcim.mcgill.edu

From ccw@eql12.caltech.edu  Sun Aug 20 18:32:21 1995
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Date:    Sun, 20 Aug 95 15:32:15 PDT
From: ccw@eql12.caltech.edu
Message-Id: <950820153215.26204902@eql12.caltech.edu>
Subject: "Cubism For Fun" Rumor.  (and Dinosaur cube info)
To: cube-lovers@ai.mit.edu, pbeck@pica.army.mil, dik@cwi.nl,
        rik@dutncp8.tn.tudelft.nl, ronnie@cisco.com
X-St-Vmsmail-To: ST%"cube-lovers@ai.mit.edu"

(Note: The second half of this message lists a place to purchase
Rubiks Dinosaur Cube)

Can anyone confirm or refute this rumor?
If true, can enough information be provided to allow ordering?
I heard this rumor last night (or was it this morning :) at a puzzle party.


Rumor:
The Nederlands Kubus Club (NKS, Dutch Cubists Club) has been receiving
a lot of orders for reprints of their newsleter "Cubism For Fun" (CFF).
Because of this, they are going to do a massive (1 time?) reprint of all
of the newsletters, but only enough to fill orders received by September 1.


Questions:
Where do I send my order?
How much for a full set of reprints? (#1 through current)
How much for a full set of the selected English translations from
the early issues?
How much additional would shipping and handling be (to U.S.A.)?
What shipping service will be used?  (in the U.S., only the U.S. Mail
is allowed to deliver to Post Office Boxes)
What are the current membership fees and info?
What is the official name of the Dutch currency?  (so I can tell the bank
what to turn my dollars into.)
Is there any person/address that email inquiries could be sent to?
(in case the one below is no longer valid)


Background (from Cube-Lovers archives):
The first 12 issues of CFF were in Dutch, thereafter in English.
Selected older articles may be available in English translations.
<pbeck@pica.army.mil>
CFF issue number 35 was published in December 1994. <dik@cwi.nl>
At that time, Dik reported that one of the editors was
<rik@dutncp8.tn.tudelft.nl>


Rubiks Dinosaur Cube:
While searching old Cube-Lovers mail to see if any information had been
posted about the above, I noticed that <ronnie@cisco.com> asked where
the Rubiks Dinosaur Cube might be purchased.  I did not see any answers
posted.  I just bought one of these last week.  I got it from
        Gametrends
        36 W. Colorado Bl.      (actually in a side alley)
        Pasadena, CA
        818-577-1882
I believe the price is $13.99.  Before posting this, I confirmed that they
will take phone orders.
If you order one, I suggest you have them test it first.  Make sure all
8 corners turn when in the initial position.  The first one I got had 2
pieces stuck together which I could not free up.  I had to return it.

For those who don't know what this puzzle is...
It is a cube.  All 8 corners turn, rotating 3 edge pieces at a time.
There are 12 visible pieces which are the edges of the cube.
Each face has cuts accross it which look like an "X".  In the solved
position, each face is a solid color.
The name Dinosaur seems to mean nothing, except that there are extra
Dinosaur decorations on the box.  (Can you say Jurassic Park?)


Thanks.
Chris Worrell
ccw@eql.caltech.edu     (current)
ccw@alumni.caltech.edu  (permenant)

From nichael@sover.net  Sun Aug 20 20:22:50 1995
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To: der Mouse <mouse@collatz.mcrcim.mcgill.edu>, AirWong@aol.com
From: nichael@sover.net (Nichael Lynn Cramer)
Subject: Re:  5 X 5 X 5 Rubik's Cubes
Cc: cube-lovers@ai.mit.edu

At 4:19 PM 19/08/95, der Mouse wrote:
>> I am new to this list, and i would like to ask if anyone knows where
>> I can get a 5 X 5 X 5 Rubik's Cube.
>
>Can't help with this...unless you're in Montreal, in which case you
>might give Valet de Coeur a try.  I think that's where I got mine.

I think the last brochure I got from Ishi had some 5bys.  Unfortunately I
can't find a copy.  Anybody else?


Nichael              -  "...did I forget, forget to mention Memphis?
nichael@sover.net           Home of Elvis, and the ancient Greeks."



From diamond@jrdv04.enet.dec-j.co.jp  Sun Aug 20 21:45:11 1995
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From: Norman Diamond  21-Aug-1995 1039 <diamond@jrdv04.enet.dec-j.co.jp>
To: cube-lovers@ai.mit.edu
Apparently-To: cube-lovers@ai.mit.edu
Subject: Re:  5 X 5 X 5 Rubik's Cubes
Mime-Version: 1.0
Content-Type: TEXT/PLAIN; charset=ISO-2022-JP

At 4:19 PM 19/08/95, der Mouse wrote:
>> I am new to this list, and i would like to ask if anyone knows where
>> I can get a 5 X 5 X 5 Rubik's Cube.
>
>Can't help with this...unless you're in Montreal, in which case you
>might give Valet de Coeur a try.  I think that's where I got mine.

Puzzletts in Seattle, USA also has them, I think for $32 each.

It's time I should write to Dr. Bandelow in Germany too; his company used
to be the primary distributor for these.

>> I have already mastered the 3 and 4 cubes, and I don't think that the
>> 5 will be much different.

>You're right; it won't.  If you can handle the 3-cube and 4-cube,
>you've got all the necessary skills.

True, you've got all the necessary skills.  However, there is a pattern
that can occur on the 5x5x5 which needs a new algorithm which can be found
rather easily using the same old skills.

Now, Nob Yoshigahara told me that he has seen a working 6x6x6, and he
received a business card from the person who had manufactured the 6x6x6,
and he was playing with the business card and then lost the business card.
The maker was in Europe.  Anyone know who and where?

-- Norman Diamond        diamond@jrdv04.enet.dec-j.co.jp
[Speaking for Norman Diamond not for Digital.]

From R.F.Hegge@mp.tudelft.nl  Mon Aug 21 05:07:57 1995
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Date: Mon, 21 Aug 1995 11:07:33 +0200
From: R.F.Hegge@mp.tudelft.nl (Rob Hegge)
Subject: Re:  5 X 5 X 5 Rubik's Cubes & buying puzzles
To: cube-lovers@ai.mit.edu
Cc: rob@gf35.mp.tudelft.nl
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Content-Transfer-Encoding: 7BIT
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> At 4:19 PM 19/08/95, der Mouse wrote:
> >> I am new to this list, and i would like to ask if anyone knows where
> >> I can get a 5 X 5 X 5 Rubik's Cube.
> >
> >Can't help with this...unless you're in Montreal, in which case you
> >might give Valet de Coeur a try.  I think that's where I got mine.
> 
> Puzzletts in Seattle, USA also has them, I think for $32 each.
> 

Last year I bought one in San Francisco in a shop located at the 
Trocadero Center (probably misspelled). However I can't really recall the
name of the shop (Game-something), but I did find shops with the same name in
Carmel and Santa Barbera.

> I think the last brochure I got from Ishi had some 5bys.  Unfortunately I
> can't find a copy.  Anybody else?

Having seen mentioned this Ishi so many times I finally take the time to ask some
questions myself:

Does Ishi also sell puzzles by mail order overseas (i.e. in Europe,
especially in the Netherlands) or better: do they a distributor overhere ?
How can I contact them for a brochure ? Is this the same Ishi which sells
GO boards etc ? (I believe I have seen them on a Web-site somewhere).

I am also looking for other company's which sell good puzzles by mail order to the
Netherlands.
Does anyone have the address of Meffert in HongKong ?

Maybe it is good idea to make a list of company's, shops ordered by location etc,
since questions like the above pop up from time to time. I know I would be interested
in a such a list. It would be very useful when I am planning my holidays :).


Rob Hegge.
r.f.hegge@ctg.tudelft.nl










From MOORE@bnlwbc.med.bnl.gov  Mon Aug 21 10:17:25 1995
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Date: Mon, 21 Aug 1995 10:17:24 -0400 (EDT)
From: MOORE@bnlwbc.med.bnl.gov
To: cube-lovers@ai.mit.edu
Message-Id: <950821101724.67e@BNLWBC.MED.BNL.GOV>
Subject: 3x3x3 Rubik's cube

Hello,
	Speaking of cubes, is it possible to still b
	Speaking of cubes, is it possible to still purchase the original
Rubik's cube?  
-RIM

From ishius@ishius.com  Tue Aug 22 13:42:04 1995
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Date: Tue, 22 Aug 1995 10:38:34 -0800
To: Cube-Lovers@ai.mit.edu
From: ishius@ishius.com (ishius@holonet.net)
Subject: Re: 5 X 5 X 5 Rubik's Cubes
Cc: AirWong@aol.com

>Hello All,
>
>I am new to this list, and i would like to ask if anyone knows where I can
>get a 5 X 5 X 5 Rubik's Cube. I have already mastered the 3 and 4 cubes, and
>I don't think that the 5 will be much different. I would like to try it
>anyways.
>
>Aaron Wong
>AirWong@AOL.com

5x5x5 Rubik's Cubes are available for $20 (regularly $30) from Ishi Press
International.  They have two flaws (which is why I am selling them for $20
and not $30).  First, the orange stickers have a tendency to slide.  This can
be fixed by putting a piece of wax paper over the orange stickers and heating
them slightly with a clothes iron, just enough to melt the glue and reset it.
Second, the middle cube faces have a tendency to fall off.  Remove them and
put them back on with a bit of glue.


Always feel free to write me if you have any questions or comments.

Anton Dovydaitis
Customer Support
===========================================================================
Ishi Press International                408/271-0415 vc, 408/271-0416 FAX
1702-H Meridian Avenue, #193            800/859-2086 Toll Free Order Line
San Jose, CA  95125                     ishius@ishius.com (or @holonet.net)



From ccw@eql12.caltech.edu  Tue Aug 22 20:10:37 1995
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Date:    Tue, 22 Aug 95 17:10:31 PDT
From: ccw@eql12.caltech.edu
Message-Id: <950822171031.26205361@eql12.caltech.edu>
Subject: Re: "Cubism For Fun" Rumor
To: cube-lovers@ai.mit.edu
X-St-Vmsmail-To: ST%"cube-lovers@ai.mit.edu"

I wrote
> Rumor:
> The Nederlands Kubus Club (NKS, Dutch Cubists Club) has been receiving
> a lot of orders for reprints of their newsleter "Cubism For Fun" (CFF).
> Because of this, they are going to do a massive (1 time?) reprint of all
> of the newsletters, but only enough to fill orders received by September 1.

Summarizing information I received from Rik van Grol, Editor of CFF.

Reprints are only available to club members.
The next full reprint will probably not be for several years.
Deadline for guaranteed orders is the end of August.  Some small quantities
of extras may also be printed.
Rik estimated the cost as 145 Dutch Guilders, though this is unconfirmed.
Club membership (1995) 30 Guilders.
Orders to Treasurer of the club.  (same address as in archives)

Now all I have to do is join up in the next 8 days. :)
Chris Worrell

From dik@cwi.nl  Tue Aug 22 20:37:47 1995
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From: Dik.Winter@cwi.nl
Message-Id: <9508230037.AA23269=dik@bever.cwi.nl>
To: ccw@eql12.caltech.edu, cube-lovers@ai.mit.edu
Subject: Re: "Cubism For Fun" Rumor
Content-Length: 980

 > Summarizing information I received from Rik van Grol, Editor of CFF.

I add some information:

 > Rik estimated the cost as 145 Dutch Guilders, though this is unconfirmed.
 > Club membership (1995) 30 Guilders.
 > Orders to Treasurer of the club.  (same address as in archives)

# 1-13 NLD 20,- (these are in Dutch, no translation available), the remainder
is in English and sold per year (NLD 5,- for each issues, #25 counts as 5
issues):
#14-16 NLD 15,- (1987)
#17-19 NLD 15,- (1988)
#20-22 NLD 15,- (1989)
#23-25 NLD 35,- (1990)
#27    NLD  5,- (1991)
#28-30 NLD 15,- (1992)
#31-32 NLD 10,- (1993)
#33-35 NLD 15,- (1994)

#26 was never issued (or was it 27?, I disremember).  Total cost English
issues only: NLD 125,-; including Dutch issues NLD 145,-.  So Rik's
estimate is correct.  1995 has upto now #36 and #37.

dik
--
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland, +31205924098
home: bovenover 215, 1025 jn  amsterdam, nederland; e-mail: dik@cwi.nl

From @mail.uunet.ca:mark.longridge@canrem.com  Sat Sep  2 03:19:58 1995
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Reply-To: CRSO.Cube@canrem.com
Sender: CRSO.Cube@canrem.com
Subject: Ranking the Puzzles
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1223.5834.0C1F29E7@canrem.com>
Date: Sat, 2 Sep 1995 03:48:00 -0400
Organization: CRS Online  (Toronto, Ontario)

Ranking the Puzzles by Number of Combinations
---------------------------------------------

    Name                     Combinations   Mechanism
    ----                     ------------   ---------

 1. Rubik's Wahn (5x5x5)     2.8*10^74      Udo Krell
 2. Megaminx                     10^68      Kersten Meier, Ben Halpern
 3. Rubik's Revenge (4x4x4)  7.4*10^45      Unknown
 4. Pyraminx Hexagon (A)     2.9*10^30      No known mechanism
 5. VIP Sphere               4.4*10^26      Unknown
 6. Impossi-ball             2.4*10^25      Wolfgang Kuppers
 7. Picture Cube (3x3x3) (E) 8.8*10^22      Erno Rubik, Dan Hoey
 8. Calendar Cube (3x3x3)(F) 4.4*10^22      Marvin Silbermintz
 9. Rubik's Cube 4th Dim.(D) 1.1*10^22      Erno Rubik
10. Rubik's Cube (3x3x3)     4.3*10^19      Erno Rubik
11. Pyraminx Octahedron      8.2*10^18      Unknown
12. Octagon                  5.4*10^18      Unknown
13. Christoph's Jewel (B)    2.0*10^15      Christoph Bandelow
14. Master Pyraminx (C)      4.5*10^14      Uwe Meffert
15. Barrel                   2.7*10^14      Gumpei Yokoi
16. 15 Puzzle                1.3*10^12      Sam Lloyd
17. Missing Link             8.2*10^10      Marvin Glass & Associates
18. Trillion                 1.0*10^9       Unknown
19. Rubik's Domino (3x3x2)   4.0*10^8       Erno Rubik
20. Picture Skewb            1.0*10^8       Tony Durham, Uwe Meffert
21. Pyraminx                 7.6*10^7       Uwe Meffert
22. Pocket Cube (2x2x2)      3.6*10^6       Enro Rubik
23. Skewb                    3.1*10^6       Tony Durham
24. Snub Pyraminx            9.3*10^5       Uwe Meffert
25. Simple Octahedron        5.0*10^4       No known mechanism

(A) This assumes 90 degree turns for the faces adjacent to the top face
(B) This is a snub Pyraminx Octahedron (Octahedron minus the tips)
(C) This assumes a Pyraminx visually the same as a regular pyraminx
     with rotations about the 4 vertices AND 6 edges.
(D) Yet another picture cube that does not have 4 orientations for
     each of it's 6 centres.
(E) This assumes a cube with centres which can show 4 distinct
     orientations for all 6 centres, and the only example I know
     of is Dan Hoey's Tartan Cube.
(F) Interestingly, due to the 'O' character on one of the centres
     of the Calendar Cube having only 2 distinct orientations,
     this picture cube has only half of the number of combinations
     of the Tartan Cube.

-> Mark <-



From @mail.uunet.ca:mark.longridge@canrem.com  Mon Sep  4 23:11:05 1995
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Subject: Dino Cube
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1225.5834.0C1F3009@canrem.com>
Date: Mon, 4 Sep 1995 23:41:00 -0400
Organization: CRS Online  (Toronto, Ontario)

#   Here are a few Dino cube calculations. The calculations for the
# cube with an X cut on each of the 6 sides, assuming period 3
# rotations of 3 edges (there are 8 of these, one for each corner)

# The Dino cube has 12! /2 = 239,500,800 essential states
# Fixing one edge gives the Dino cube a fixed orientation
#  in space and gives 11! /2 = 19,958,400 combinations

# It has less combinations then the standard pyraminx, but more
#  than the 2x2x2 Rubik's Pocket cube.

# The Dino cube has 12 edges which can not flip, observed by Rubik
#  himself back in 1982 (re: Rubik's Logic & Fantasy in Space.)
# Dino cube has trivial centre

dino := Group(
    (1,24,7)  (2,23,5),
    (2,12,22) (4,11,24),
    (4,19,10) (3,17,12),
    (3,5,20)  (1,6,19),
    (13,21,11) (14,22,9),
    (14,8,23) (16,7,21),
    (16,18,6) (15,20,8),
    (15,9,17) (13,10,18)
  );;

From BRYAN@wvnvm.wvnet.edu  Sat Sep  9 09:52:21 1995
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Date:      Sat, 9 Sep 1995 09:51:56 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   A Proposal for a More General Definition of Symmetry

Our normal definition of symmetry for Rubik's cube is based
ultimately on the 48 symmetries of the standard math book
wire model of a cube, and the 48 symmetries were discovered
long before Rubik's cube was ever dreamed of.  This note is
based on the conviction that these 48 symmetries do not really
capture all that we might think of as "symmetry" when we think
of Rubik's cube.  This note has the further purpose to propose
a more general definition of symmetry for Rubik's cube.

I want to start with a couple of really basic concepts.  I think
every reader of this list knows what a permutation is, namely
it is a one-to-one onto function on a set.  In the case of a finite
set as we have with Rubik, a function on a set is one-to-one
if and only if it onto, so we can sometimes get by with speaking
only of one-to-one or by speaking only of onto.

But what is a symmetry?  A very standard definition is something
like "the set of all rigid motions that transform a given geometric
figure onto itself" (James and James Mathematics Dictionary).
Another way to say it is that the transformation preserves the
figure.

Working with that definition, a symmetry almost inevitably may
be interpreted as a permutation.  With simple geometric figures,
the permutation would usually be described as being a permutation on
Euclidean n-space  --  2-space for a square or circle, etc., and
3-space for a cube or sphere, etc.  Hence, we might think of
a symmetry as being a special kind of permutation, namely one that
preserves a geometric figure in Euclidean n-space.

I have had a difficult time finding books that address the issue
of symmetry vs. permutation to my satisfaction.  It is very hard
to think of a symmetry abstractly enough that it doesn't simply
turn into a permutation right before your eyes.

Paul Yale's "Geometry and Symmetry" doesn't really seem to define
a symmetry (it sort of assumes you know what one is), but it does
describe the relationship between a symmetry and a permutation.
I would paraphrase as follows.  Label your geometric figure in
some fashion  --  e.g., label the edges, label the axes, label
the vertices, or label *something*.  Then, there is a homomorphism
between the set of symmetries and the corresponding set of
permutations on the labels.

But I repeat that it is hard for me to conceive of the set of symmetries
in a sufficiently abstract fashion that the symmetries themselves
aren't already permutations on *some* set or other.  So it seems to me
that Yale could just as well be talking about homomorphisms between
one set of permutations and another set of permutations as talking
about homomorphisms between symmetries and permutations.

A couple of quick additional points, and then I will go on:
1) since we are talking about homomorphisms, it is obvious that
both the set of symmetries and the set of permutations to which
they map are groups, and 2) most homomorphisms between symmetries
and permutations turn out in fact to be isomorphisms.  This latter
observation gives added weight to the notion that symmetries are
just a special kind of permutation.

Given all that has been said so far, we could informally say that
the normal definition of a symmetry is that it is a permutation
that preserves a geometric figure.  Our more general definition
will simply be that a symmetry is a permutation that preserves
some property.  If we were sufficiently liberal in our notion of
"preserving some property", then most any permutation could be
interpreted as a symmetry.  We will not be quite that liberal
by the time we are done, but we will be more liberal than
would be permitted by the standard 48 math book symmetries of
the cube.

But what property of Rubik's cube should we try to preserve if we
want a more general definition of symmetry than the normal one?
I wish to motivate our definition of that property in several
steps.

The standard Rubik's cube definition of symmetry for a position X is
Symm(X) is the set of all m in M such that X=m'Xm, or equivalently
such that mX=Xm.  M is the set of 48 permutations on the Rubik's
cube corresponding to the 48 symmetries of a cube.

Write a position Z as Z=XY, where X is the permutation on the
corners and Y is the permutation on the edges.  We have
Symm(Z)=Symm(XY)=Symm(X) intersect Symm(Y).  For example, we
could have Symm(X)=M, Symm(Y)=I, and Symm(Z)=Symm(XY)=I.
Such a position would look very "symmetrical" because the
corners would be fixed (or "solved"), although the edges
would be scrambled.  Most people would consider such a position
to be more "symmetrical" than one where both the corners and
edges were scrambled, although we would have Symm(Z)=I
in either case.

A couple of points before proceeding:  1)  From an information
theory point of view, Symm(X) and Symm(Y) separately contain
more information than does Symm(XY).  There is an obvious
loss of information when we calculate Symm(X) intersect Symm(Y).
This is a strong indication that Symm(XY) does not tell us
everything we might like to know about the symmetry of a position.
2) The set of positions Z=XY for which Symm(X)=M forms a group
(as does the set of positions for which Symm(Y)=M).  This anticipates
where we are headed, namely that group membership is the property
that we should seek to preserve in a more general definition of
symmetry.

A third (and equivalent) definition for Symm(X) is that Symm(X) is
the set of all m in M such that X'm'Xm=I.  Most readers will
recognize X'm'Xm as a commutator.  Per Dan Hoey, we can generalize
and define CSymm(X) to be the set of all m in M such that X'm'Xm
is in C, the set of 24 rotations of the cube.  For example, if
we have Z=XY as before, then CSymm(X)=M means that the corners
are positioned properly with respect to each other, although they
might be rotated with respect to the fixed face centers.
Such a position would look fairly "symmetrical", even to a
non-cubemeister, even though we might have Symm(Z)=I.

Again, we have the set of all positions for which CSymm(X)=M
forms a group.  Similarly, the set of all positions for which
CSymm(Y)=M forms a group, and the set of all positions
for which CSymm(Z)=CSymm(XY)=M forms a group.

Recall that there are 98 subgroups of M.  For each subgroup K of
M, there is a corresponding subgroup of G consisting of all the
K-symmetric positions.  So would could just as well define
symmetry in terms of these 98 subgroups of G.  But there are far
more than 98 subgroups of G.  (We don't know how many, and I
doubt than even GAP could tell us).  Why not simply define
symmetry in G in terms of subgroup membership in G?  The symmetry
of a position X is then the set of all subgroups of H of G
for which X is in H.  And a symmetry operation (in the sense
that a symmetry is a permutation that preserves something) is
an operation that preserves subgroup membership.

That pretty much completes my proposal, but I have a few closing
remarks.

    1) The proposed general definition of symmetry is
       analogous to the Thistlethwaite algorithm for solving
       the cube.  Typical cube solutions gradually solve more
       and more of the cube.  The "more and more of the cube"
       that gets solved can be characterized as a sequence of
       nested subgroups.  Thistlethwaite reversed the process
       and created a sequence of nested subgroups
       which in turn solves more and more of the cube.  Similarly,
       the standard definition of symmetry implies a set of 98
       subgroups of G.  We reverse the process and let all the
       subgroups of G define symmetry instead.

    2) The proposed general definition of symmetry has the
       virtue that it includes the standard definition as a
       special case, since the 98 K-symmetric subgroups of
       G are in fact subgroups of G.

    3) The proposed general definition of symmetry has the
       virtue that there is only one position that is
       "completely symmetric", namely Start itself (the
       identity permutation).  The standard definition of
       symmetry has four positions which are "completely
       symmetric", which to me is an unsatisfactory state
       of affairs.

       (Recall that we have Symm(X)=M for Start, Pons Asinorum,
       Superflip, and the composition of Pons and Superflip.
       I am still bummed out that this is the case while at the
       same time only Start and Superflip are in the center of
       G.  This suggests that Superflip is "more symmetric"
       than Pons.  I wonder if such a suggestion would be
       supported by my proposed general definition of symmetry?)

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From morabito@omni.voicenet.com  Mon Sep 11 20:48:15 1995
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        Hello. I would like to be part of your mail group.  Please send me 
the letters at morabito@omni.voicenet.com. Thanks.


From @mail.uunet.ca:mark.longridge@canrem.com  Wed Sep 13 02:15:29 1995
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Subject: Alexander's Star
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1229.5834.0C1F4769@canrem.com>
Date: Wed, 13 Sep 1995 02:21:00 -0400
Organization: CRS Online  (Toronto, Ontario)

> From: pbeck@pica.army.mil (Peter Beck)
> Subject: ranking
>
> noticed you did not include
> square-1  or alexanders star
>
> thanks for the compilation


 Very little literature exists on Alexander's Star. Ideal Toy published
a solution booklet, and Adam Alexander (the puzzle's inventor) wrote
a book, and David Singmaster wrote his analysis in one of his
Cubic Circulars. Dr. Singmaster also mentions Alexander's Star in
Rubik's Cubik Compendium in a chapter about variations on Rubik's
cube.

 The Ideal booklet mentions that Alexander's Star has 24 start
positions, and Dr. Singmaster states there are 12 start positions,
each one with it's own mirror reflection, again giving 24 start
positions in total.

 Alexander's Star is akin to an "edge-only" Dodecahedron (Megaminx)
without centres. It has 30 edges, that is 15 pairs of distinct
2 coloured edges.

 To calculate the number of permutations of N objects with N1
objects which are like, N2 objects which are like ...
Nrth objects which are like we use the formula:

           N!
     --------------------
     N1! * N2! * ... Nr!


 In the case of Alexander's Star it is a bit more complicated because
we must contend with edge flips and no centres, however I believe the
correct calculation is....

30! / 2^15 * 2^29 / 60 = 72,431,714,252,715,638,411,621,302,272,000,000
                approx = 7.2 * 10^34

I think the term for this is approximately 72 decillion.
(decillion = 10^33)

Here is a bit of an explanation....

There are 30 total objects, with 15 different pairs of 2 like edges.
29 edges may be flipped in any fashion but the 30th edge is forced.
The Great Dodecahedron has 60 orientations in space.

Here is another way to calculate the same number...

Pick one edge and lock it in position and orientation. We still have
29 edges which can move in position and orientation. There are still
28 edges which can be flipped freely, the 29th edge being forced.
We still have 15 different pairs of 2 like edges.

29! / 2^15 * 2^28 = 7.2 * 10^34 (approx)

 To be completely clear the calculation is really
         29!/2   /   2^15/2 * 2^28

....because only half of the 29! arrangements are possible and only
half of the 2^15 arrangements are possible but both the /2's cancel.

 On the Great Cosmic Ranking List this puzzle has less combinations
than Rubik's Revenge, but more than the VIP sphere.

 I know of no Square 1 calculations whatsoever, but would be
interested in seeing anyone's calculations.

-> Mark <-

From dlitwin@geoworks.com  Thu Sep 14 23:45:03 1995
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Date: Thu, 14 Sep 95 20:40:52 PDT
From: dlitwin@geoworks.com (David Litwin)
Message-Id: <9509150340.AA15388@quark.geoworks.com>
To: cube-lovers@life.ai.mit.edu
Subject: Alexander's Star


	While we are on the subject of the Alexander's star, I have never
been entirely satisfied with the arrangment of its solution.  I've noticed
that most people who look at it don't even know it is solved and I have to
explain that the solution is that all the stickers of pieces laying in a
common plane around the points are the same color.  Hard to say, but I can
point it out to people.  
	To this end I have spent some time trying to find a more satisfying
solution, one more visually clear and simple.  I've only come up with one
alternative that I consider reasonable, and it isn't as pure as I would
like.

	This solution involves grouping colors in the depressions of the
star.  The main problem lies in the fact that the edges come together in
groups of three, but there are 10 stickers of each color so at some point
having all the depressions of the star a single color breaks down.  For
this reason I choose one color (White is my preference) to be an exception
and have it remain in the original configuration, i.e. all in two parallel
planes.  With the rest of the colors, I group them in groups of five: three
in one depression, and two in an adjacent depression with the third color
of this second depression being one of a white.  The result of this is a
set of interlocking "diamonds" that group visually because the are of the
same color.  Unrolled, the star would look like this (this should just fit
on a display of 80 columns):

       /|\             /|\             /|\             /|\             /|\     
      / | \           / | \           / | \           / | \           / | \    
     /  |  \         /  |  \         /  |  \         /  |  \         /  |  \   
    /   |   \       /   |   \       /   |   \       /   |   \       /   |   \  
   /  Y_|_Y  \     /  B_|_B  \     /  O_|_O  \     /  R_|_R  \     /  G_|_G  \ 
  / _-' W `-_ \   / _-' W `-_ \   / _-' W `-_ \   / _-' W `-_ \   / _-' W `-_ \
_/-'_________`-\_/-'_________`-\_/-'_________`-\_/-'_________`-\_/-'_________`-\
|\-,_       _,-/|\-,_       _,-/|\-,_       _,-/|\-,_       _,-/|\-,_       _,-/
| \  `-_Y_-'  / | \  `-_B_-'  / | \  `-_O_-'  / | \  `-_R_-'  / | \  `-_G_-'  / 
|  \  Y " Y  /  |  \  B " B  /  |  \  O " O  /  |  \  R " R  /  |  \  G " G  /  
|   \   |   /   |   \   |   /   |   \   |   /   |   \   |   /   |   \   |   /   
|_O  \  |  /  R_|_R  \  |  /  G_|_G  \  |  /  Y_|_Y  \  |  /  B_|_B  \  |  /  O_
O `-_ \ | / _-' R `-_ \ | / _-' G `-_ \ | / _-' Y `-_ \ | / _-' B `-_ \ | / _-' 
_____`-\|/-'_________`-\|/-'_________`-\|/-'_________`-\|/-'_________`-\|/-'____
    _,-/ \-,_       _,-/ \-,_       _,-/ \-,_       _,-/ \-,_       _,-/ \-,_   
W_-'  /   \  `-_W_-'  /   \  `-_W_-'  /   \  `-_W_-'  /   \  `-_W_-'  /   \  `-_
" O  /     \  R " R  /     \  G " G  /     \  Y " Y  /     \  B " B  /     \  O 
|   /       \   |   /       \   |   /       \   |   /       \   |   /       \   
|  /         \  |  /         \  |  /         \  |  /         \  |  /         \  
| /           \ | /           \ | /           \ | /           \ | /           \ 
|/             \|/             \|/             \|/             \|/             \

 
	I haven't found a nice way of having more solid depressions than this.

	Has anyone else found any nice solutions?

	Dave Litwin

From @mail.uunet.ca:mark.longridge@canrem.com  Thu Sep 21 02:25:16 1995
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To: cube-lovers@life.ai.mit.edu
Reply-To: CRSO.Cube@canrem.com
Sender: CRSO.Cube@canrem.com
Subject: VIP Sphere & Masterball
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1234.5834.0C1F5B3D@canrem.com>
Date: Thu, 21 Sep 1995 02:56:00 -0400
Organization: CRS Online  (Toronto, Ontario)

# Calculations on the VIP sphere
# 32 pieces with all distinct colours
# Sphere where each of the 2 hemispheres rotate 180 degrees
#  and the 4 rows (of 8 pieces each) can slide around
#  the circumference
#
#
# Size (vip) = 437,763,136,697,395,052,544,000,000
# No fixed Orientation
# approx.    = 4.4 * 10^26 or 437 septillion!
# trivial centre

vip := Group(
        (1,2,3,4,5,6,7,8),
        (9,10,11,12,13,14,15,16),
        (17,18,19,20,21,22,23,24),
        (25,26,27,28,29,30,31,32) ,
        (1,28)(2,27)(3,26)(4,25)(9,20)(10,19)(11,18)(12,17),
        (5,32)(6,31)(7,30)(8,29)(13,24)(14,23)(15,22)(16,21)
  );;

Within the 2 orbits of 16 pieces any exchange is possible. One
orbit is "polar" and the other is "equatorial".

(28,1) in vip;
true

 Thus on the VIP Sphere a single 2-cycle is legal, although I
know of no simple process as yet.

  The original calculation by Dr. Singmaster was (16!)^2, and
I have confirmed his result with GAP.

I have also played with the Masterball somewhat. This puzzle is awful!
Just how accurately does this thing have to be lined up to turn it?
It locked up several times on me when I tried to randomize it.

 It is the single most difficult puzzle to turn I have ever
encountered, save the Equator puzzle only.

 My first thoughts on calculating the number of positions on the
Masterball was it was the same as the VIP sphere divided by
2^16 but I'm not sure. I can't use GAP in the case of the
Masterball (rainbow edition in this case) to verify this because
of the identical pieces. The booklet which came with the
Masterball refers to some number like 350 quadrillion but there
are more zeroes than the american quadrillion, and I get a
totally different number anyways.

 Thoughts anyone??

-> Mark <-

From VinylM@aol.com  Fri Sep 22 02:42:39 1995
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From: VinylM@aol.com
Message-Id: <950922024237_105790716@emout05.mail.aol.com>
To: Cube-Lovers@ai.mit.edu
Subject: Rubik's Revenge

I am currently looking for a solution to Rubik's Revenge  (the 4 X 4 cube)

any ideas?

Aron Siegel
404-321-0445

From boland@sci.kun.nl  Sat Sep 23 18:31:59 1995
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To: cube-lovers@ai.mit.edu
Subject: Order problems
Date: Sun, 24 Sep 95 00:31:55 +0200
From: Michiel Boland <boland@sci.kun.nl>

Hello all,

Had a great time reading the archives. What I haven't found
there are order problems: what is the shortest (in terms of
quarter turns or half- and quarter turns, whatever you prefer)
transformation of the cube with a given order?

Here is a list that my good old PC produced this afternoon.
I hope some of you find this interesting. :)

A couple of notes on the list:

"Len" is the length of the transformation in terms of quarter
twists. You will notice that I listed two transforms with order
3: one is minimal wrt quarter-turn metric, the other wrt
half-turn metric.

A notable absentee is number 11. I suspect that (U.R.F2B.D')2 is
the shortest possible with order 11, but my comp just isn't fast
enough to confirm this.

Note that (U.R.F2B.D')2 yields an 11-cycle on the edges (see
also Mark Longridge's mail from 15 Jul 1994.)

(I use dots to maintain readability; personally, I do not like
the U1F2L3 notation, but that's just a matter of taste :)

Order    Len
   1      0
   2      2      U2
   3      6      U.R.U'D'R.D.
   3      8      U2R2U2R2
   4      1      U.
   5      4      U.R.U.R'
   6      4      U2R2
   7      4      U.R.U'F.
   8      4      U.R2D.
   9      4      U.R.F2
  10      4      U'R.U.F.
  11      ?      ????????????
  12      4      U.R.F.D'
  14      6      U'R.U.R'F.D.
  15      6      U.R2U.R2
  16      5      U.R.U'F.D.
  18      5      U.R.U'R'F.
  20      5      U.R.U'L2
  21      6      U2R.U2F.
  22      6      U.R.F2B.D'
  24      4      U.R2D'
  28      4      U.R.U'L.
  30      3      U.R2
  33      4      U.R.F'D'
  35      6      U2R.U2L'
  36      4      U2R'F'
  40      5      U.R.U2L.
  42      6      U.R2U2R'
  44      4      U'R.F'D.
  45      4      U.R.U.L.
  48      5      U2R.U.F.
  55      6      U.R.F'U'B'L.
  56      5      U2R.F'D.
  60      3      U.R'F'
  63      2      U.R'
  66      6      U.R.U.F2L'
  70      6      U.R'U.R.F.R'
  72      4      U.R.U.F'
  77      4      U.R'F'L'
  80      3      U'R'F'
  84      3      U.R.F.
  90      3      U.R.D.
  99      6      U.R2F.L2
 105      2      U.R.
 110      8      U.R.U2R'F.R.L'
 112      6      U.R'U.F'R.D.
 120      4      U.R.F.L'
 126      4      U'R.F'L'
 132      4      U.R.F'L.
 140      4      U.R'U.F'
 144      5      U.R'F'D2
 154      6      U.R.U.F.L.D'
 165      6      U.R'U.F2L'
 168      4      U.R.D2
 180      3      U.R.D'
 198      6      U2R.F.D2
 210      4      U.R'D.L'
 231      4      U.R.F'D.
 240      5      U'R.F'L2
 252      4      U.R.F.L.
 280      5      U'R'U'F.L'
 315      4      U.R.D.L.
 330      6      U2R.F'D'L'
 336      6      U.R.U.F.D2
 360      3      U.R.F'
 420      4      U.R.D.L'
 462      6      U'R.F'D2L'
 495      6      U.R2U.F'L'
 504      5      U.R2F.L'
 630      6      U'R'U'F'L2
 720      6      U'R'U'F'D2
 840      5      U2R'F'D.
 990      6      U'R'U'F'L.D.
1260      6      U.R'U.F'D2
-- 
Michiel Boland <boland@sci.kun.nl>
University of Nijmegen
The Netherlands

From news@nntp-server.caltech.edu  Sun Sep 24 09:54:40 1995
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Path: whuang
From: whuang@cco.caltech.edu (Wei-Hwa Huang)
Newsgroups: mlist.cube-lovers
Subject: Re: Alexander's Star
Date: 24 Sep 1995 13:54:34 GMT
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dlitwin@geoworks.com (David Litwin) writes:


>	While we are on the subject of the Alexander's star, I have never
>been entirely satisfied with the arrangment of its solution.  I've noticed
>that most people who look at it don't even know it is solved and I have to
>explain that the solution is that all the stickers of pieces laying in a
>common plane around the points are the same color.  Hard to say, but I can
>point it out to people.  
>	To this end I have spent some time trying to find a more satisfying
>solution, one more visually clear and simple.  I've only come up with one
>alternative that I consider reasonable, and it isn't as pure as I would
>like.

>	This solution involves grouping colors in the depressions of the
>star.  The main problem lies in the fact that the edges come together in
>groups of three, but there are 10 stickers of each color so at some point
>having all the depressions of the star a single color breaks down.  For
>this reason I choose one color (White is my preference) to be an exception
>and have it remain in the original configuration, i.e. all in two parallel
>planes.  With the rest of the colors, I group them in groups of five: three
>in one depression, and two in an adjacent depression with the third color
>of this second depression being one of a white.  The result of this is a
>set of interlocking "diamonds" that group visually because the are of the
>same color.  Unrolled, the star would look like this (this should just fit
>on a display of 80 columns):

>	I haven't found a nice way of having more solid depressions than this.

>	Has anyone else found any nice solutions?

>	Dave Litwin

I came up with this solution independently.  I also liked on that picked two
opposite depressions on the star and made sure they contained one of each
color.  That allowed me to fill each of the other 18 depressions with
homogenous colors.

-- 
   -- Wei-Hwa Huang (whuang@cco.caltech.edu)
Homepage (under construction): http://www.ugcs.caltech.edu/~whuang/
Microsoft: small and limp.

From boland@sci.kun.nl  Sun Sep 24 12:19:36 1995
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Message-Id: <199509241619.RAA13678@wn1.sci.kun.nl>
To: cube-lovers@ai.mit.edu
Subject: Re: Order problems 
In-Reply-To: Your message of "Sun, 24 Sep 95 00:31:55 +0100."
             <199509232231.AAA29847@wn1.sci.kun.nl> 
Date: Sun, 24 Sep 95 17:19:32 +0100
From: Michiel Boland <boland@sci.kun.nl>

Sorry to follow up on my own post, but I messed up a bit :)

>  12      4      U.R.F.D'

There is also U2R2F' which has less face turns but more quarter turns.

>  70      6      U.R'U.R.F.R'

This transform has in fact order 140!
The correct ones are:

 70   6   U.R.U'R.F.B'
 70   7   U2R'U2F'L'

> 110      8      U.R.U2R'F.R.L'

This should be U'R'U.R.F.D'B'L' (8 qt)

I converted my order-searching program to C and am now running
it on a Sun - expecting results on order 11 soon.
-- 
Michiel Boland <boland@sci.kun.nl>
University of Nijmegen
The Netherlands

From BRYAN@wvnvm.wvnet.edu  Sun Sep 24 18:36:02 1995
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Date:      Sun, 24 Sep 1995 18:35:36 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Michiel Boland" <boland@sci.kun.nl>,
        "Cube Lovers List" <cube-lovers@ai.mit.edu>
Subject:   Re: Order problems
In-Reply-To: Message of 09/24/95 at 00:31:55 from boland@sci.kun.nl

On 09/24/95 at 00:31:55 Michiel Boland said:
>Hello all,

>Had a great time reading the archives. What I haven't found
>there are order problems: what is the shortest (in terms of
>quarter turns or half- and quarter turns, whatever you prefer)
>transformation of the cube with a given order?

I would be curious to hear how you are doing your search.  It is
trivial to see how to calculate the order of a particular
position.  However, it is not obvious to me how to find a
position of a particular order.  I hope it is not the case that
it is in the archives and I just haven't seen it.

I would guess that you are building a search tree of length 0,
length 1, length 2, etc. as has been done many times before,
and calculating the order of each position as you encounter it.
You could then easily build a table of shortest positions of
each order, provided the order appeared in your search.
I would further guess that you have searched down to about
level 6.  However, if that is how you are doing it, I don't
see how you could have proved that the shortest position of
order 110 is of length 8.  I don't see how a PC program could
have searched to level 8 in just a little while.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From boland@sci.kun.nl  Sun Sep 24 19:44:45 1995
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Message-Id: <199509242344.AAA21239@wn1.sci.kun.nl>
To: "Cube Lovers List" <cube-lovers@ai.mit.edu>
Subject: Re: Order problems 
In-Reply-To: Your message of "Sun, 24 Sep 95 18:35:36 -0400."
             <wvmail32.1995sep24.182035.bryan@wvnvm.wvnet.edu> 
Date: Mon, 25 Sep 95 00:44:40 +0100
From: Michiel Boland <boland@sci.kun.nl>

Jerry wrote:
>I would be curious to hear how you are doing your search.  It is
>trivial to see how to calculate the order of a particular
>position.  However, it is not obvious to me how to find a
>position of a particular order.  I hope it is not the case that
>it is in the archives and I just haven't seen it.

I use a simple brute-force method, that is, I compute the order
of each transform and the number of quarter turns. If there is
already a transform with that order & number of qt, I forget all
about it and go to the next transform.

I have the C source available for anyone who wants to peek at it.

Note that (almost) all transforms start with UxRx, since you
have to twist two adjacent faces in order to get something with
order other than 1, 2 or 4. That saves a bit of time.

On my PC, i finished all transforms of length 6 (quarter- and
half turns), and did some of length 7. Fortunately, as I
mentioned earlier, I managed to get it to work on a somewhat
faster machine, and am now waiting for the results of that. I am
searching all transforms of 10 quarter turns or less.

A process with order 110 must have an even number of quarter
turns, since the permutation on the edges has to be even (the
only possibility is to have an 11-cycle on the edges, which is
an even permutation).
Therefore, since there are no processes with order 110 with 6 or
less quarter turns, this proves that the one of length 8 is
indeed the shortest possible.

Cheers,

-- 
Michiel Boland <boland@sci.kun.nl>
University of Nijmegen
The Netherlands

From @mail.uunet.ca:mark.longridge@canrem.com  Sun Sep 24 22:25:13 1995
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Sender: CRSO.Cube@canrem.com
Subject: Least Commutative Element
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1242.5834.0C1F62D8@canrem.com>
Date: Sun, 24 Sep 1995 22:56:00 -0400
Organization: CRS Online  (Toronto, Ontario)

# The 2nd most commutative element of the cube group
# It is the position of 7 clockwise and 1 counter-clockwise twists
# This commutes with 1 out of every 8 elements of the cube

commuter := ( 1,35, 9)( 3,27,33)( 6,11,17)( 8,19,25)(24,43,30)(32,48,38)
            (14,40,46)(16,22,41);;

# The least commutative element of the cube group ( I think! )
# This commutes with 1 out of every 450,541,700,775,936,000
#  or approximately 1 out of every 4.5 * 10^17 patterns

least := ( 1, 6,32,19, 3,41,24,46)( 2, 4,13,42,29, 7,31,15,39,37,26,21)
( 5,28,34,10,20,23,36,18,45,44,47,12)( 8,33,16,30,40, 9,17,38)
(11,48,25,27,22,43,14,35);;

> after thinking about it, i realized that
>
> corners:  (8)   edges:  (12)
>
> commutes with even fewer elements.  again, elements with
> this cycle structure split into two conjugacy classes.
>
> mike

 With GAP we must deal with permutations of cube facelets, and that is
why the permutation 'least' has 3 sets of 8 numbers and 2 sets of
12 numbers. Moreover, as I'm sure Mike will appreciate, the least
commutative element I've found is has a 8-cycle of corners and a
12-cycle of edges.

Size (Centralizer (cube, least));
 96

-> Mark <-

From boland@sci.kun.nl  Mon Sep 25 04:19:19 1995
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To: cube-lovers@ai.mit.edu
Subject: 11-cycle
Date: Mon, 25 Sep 95 09:19:17 +0100
From: Michiel Boland <boland@sci.kun.nl>

U.R.U.F.R'L.U'R'F'D'
This completes the order list. :)
-- 
Michiel Boland <boland@sci.kun.nl>
University of Nijmegen
The Netherlands

From mouse@collatz.mcrcim.mcgill.edu  Mon Sep 25 07:07:11 1995
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Date: Mon, 25 Sep 1995 07:06:43 -0400
From: der Mouse <mouse@collatz.mcrcim.mcgill.edu>
Message-Id: <199509251106.HAA03544@Collatz.McRCIM.McGill.EDU>
To: boland@sci.kun.nl
Subject: Re: Order problems
Cc: cube-lovers@ai.mit.edu

>> I would be curious to hear how you are doing your search.  [...]
> I use a simple brute-force method, that is, I compute the order of
> each transform and the number of quarter turns.  If there is already
> a transform with that order & number of qt, I forget all about it and
> go to the next transform.

This sounds to me as though you're assuming that all transforms with a
given order are equivalent as far as deriving further transforms of
other orders go.  That is, if you find that a given transform X of
length L has order N, it sounds as though you're assuming that there is
no need to store any other transforms of length L and order N.  I'm not
convinced this is justified.  If you've found X of (say) length L and
order N, and then find a different Y of length L and order N, I can't
see any justification for the assumption that you can prune the entire
subtree below Y, because if the cycle decompsition of Y is different
from that of X, they may behave entirely differently when followed by
more twists, even though they have the same order.

					der Mouse

			    mouse@collatz.mcrcim.mcgill.edu

From boland@sci.kun.nl  Mon Sep 25 07:12:23 1995
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Message-Id: <199509251112.MAA05350@wn1.sci.kun.nl>
To: cube-lovers@ai.mit.edu
Subject: Re: Order problems 
In-Reply-To: Your message of "Mon, 25 Sep 95 07:06:43 -0400."
             <199509251106.HAA03544@Collatz.McRCIM.McGill.EDU> 
Date: Mon, 25 Sep 95 12:12:20 +0100
From: Michiel Boland <boland@sci.kun.nl>

>If you've found X of (say) length L and
>order N, and then find a different Y of length L and order N, I can't
>see any justification for the assumption that you can prune the entire
>subtree below Y [...]

I do not prune the search tree. If I say I "forget" about Y,
then I do not mean "forget" all transforms that start with Y.
That would be a bad thing, of course.
-- 
Michiel Boland <boland@sci.kun.nl>
University of Nijmegen
The Netherlands

From mreid@ptc.com  Wed Sep 27 17:28:34 1995
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Date: Wed, 27 Sep 1995 17:50:00 -0400
From: michael reid <mreid@ptc.com>
To: cube-lovers@ai.mit.edu
Subject: number of positions of square1

mark says

>  I know of no Square 1 calculations whatsoever, but would be
> interested in seeing anyone's calculations.

i also haven't seen any figures for square1 (although i guess that
ideal toy corp.'s generic "more than three billion" might apply).

but it's not hard to derive some figures.  i guess there might be
some question about what constitutes a "position".  i think it's
reasonable to consider only configurations where all three axes
are free to turn.

note that up to rotation, there are 29 different shapes for the
top face.  they occur as 19 symmetric shapes and 5 mirror image
pairs.  these are grouped into five different classes according to
the number of 60 degree pieces ("corners"?), which i'll call doubles.

for each shape, we also count the number of rotationally distinct
orientations, as well as the number of orientations where the
locations of the doubles do not block the half-turn axis.


      description   # rotational                       # orientations
name    of shape     symmetries     # orientations    which allow half-turn

type 6-0

A     222222              6               2                  1

type 5-2

B     2222211             1              12                  6
C     2222121             1              12                  4
D     2221221             1              12                  2

type 4-4

E     22221111            1              12                  6
F     22122111            1              12                  4
G     22112211            2               6                  4
H     22212111            1              12                  4
H'    22211121            1              12                  4
I     22211211            1              12                  6
J     22121211            1              12                  6
J'    22112121            1              12                  6
K     22121121            1              12                  4
L     21212121            4               3                  2

type 3-6

M     222111111           1              12                  6
N     221211111           1              12                  6
N'    221111121           1              12                  6
O     221121111           1              12                  8
O'    221111211           1              12                  8
P     221112111           1              12                  6
Q     212121111           1              12                  8
R     212112111           1              12                  6
R'    212111211           1              12                  6
S     211211211           3               4                  2

type 2-8

T     2211111111          1              12                  8
U     2121111111          1              12                  8
V     2112111111          1              12                  8
W     2111211111          1              12                  8
X     2111121111          2               6                  5


the top and bottom faces have complementary type, (i.e. are  6-0 and
2-8, 5-2 and 3-6, 4-4 and 4-4, 3-6 and 5-2, or 2-8 and 6-0).

type     total # valid orientations

6-0                  1
5-2                 12
4-4                 46
3-6                 62
2-8                 37

thus we have  1 * 37 + 12 * 62 + 46 * 46 + 62 * 12 + 37 * 1 = 3678
valid possibilities of the doubles.  each permutation of the doubles
and singles is possible, and the middle layer has two orientations.
any combination of these is possible.  therefore we get a final count
of  3678 * 2 * 8! * 8! = 11958666854400  positions.

mike

From BRYAN@wvnvm.wvnet.edu  Thu Oct  5 21:01:18 1995
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Date:      Wed, 4 Oct 1995 10:26:41 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   <F,U,L,R,D> Question

It is well known that if we define G=<Q> for the twelve quarter turns
q in Q, we can also generate G as G=<F,U,L,R,D>, leaving out B and B'.
Leaving out any other quarter turn would do as well, but I
am going to stick to leaving out B for illustrative purposes.

However, when one of the quarter turns is left out, the length of most
positions will change.  In particular, we will no longer have |B|=1.
My reading of the archives indicates that we do not know what the
length of B would be in this situation, nor what a minimal process
for B would be.

I am going to take a crack at this problem via exhaustive search.
But I like to use representative elements of conjugacy classes in
my searches, and I don't think I can do so in this situation.

For full-blown searches of G, I use M-conjugacy classes.  For subsets
and/or restrictions of G, I use appropriate subsets and/or restrictions
of M.  But I don't think I can use conjugacy classes at all for this
problem.  The group is still G, even though lengths have changed, so
no subset and/or restriction of M is appropriate.  But when G is
generated as <F,U,L,R,D>, we do not necessarily have |X|=|m'Xm|
for all m in M.

Am I missing something obvious?  I don't think so, but in the
meantime I am going to have to start the search without conjugacy
classes.  Bummer.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From bagleyd@source.asset.com  Fri Oct  6 16:34:56 1995
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          id AA51717; Fri, 6 Oct 1995 16:37:09 -0400
Date: Fri, 6 Oct 1995 16:37:09 -0400
From: bagleyd@source.asset.com (David A. Bagley)
Message-Id: <9510062037.AA51717@source.asset.com>
To: cube-lovers@life.ai.mit.edu
Subject: pyraminx-like puzzles

Hi
 
  I have a question, I hope this makes sence. ;)   On a "nxnxn"
tetrahedron with period 2 or period 3 turning or a "nxnxn" octahedron with
period 3 or period 4 turning, can the orientation of any of the center
triangles change when the puzzle is solved?  If so, where does this
start to happen.  I know from "experience" that this is not true on
a pyraminx.
 
A Pyraminx, according to me, has period 3 turning where n = 3.  n is the
number of triangles along an edge not counting corners of triangles.
Center triangles are those triangles that are not on an edge (except for
maybe the corners of the triangle).
 
 
A face of a 2x2x2 has one center triangle
     /\
    /__\
   /\C /\
  /__\/__\
                                                            INSERT MODE
 
A face of a 3x3x3 has 3 center triangles
     /\
    /__\
   /\C /\
  /__\/__\
 /\C /\C /\
/__\/__\/__\
 
 
 
A face of a 4x4x4 has 7 center triangles
 
 
  I would like to know this because it will help me in the design of my
puzzles which I am in the process now of converting to Motif (you will
still be able to run them if you just have X).  I hear that a freely
distributable Lesstif Widget set will be out soon.
(I am also considering a Windows version and am getting ready to convert them).
Estimated completion time for the motif programs: 2 weeks.
 
Cheers,
      --__---------------------------------------------------------------
     /  \ \   /           David A. Bagley                                \
    |    \ \ /            bagleyd@source.asset.com                        |
    |     \//\            Some days are better than other days.           |
    |     / \ \                -- A short lived character of Blake's 7    |
     \   /   \_\puzzles   Available at: ftp.x.org/contrib/games/puzzles  /
      -------------------------------------------------------------------

From BRYAN@wvnvm.wvnet.edu  Fri Oct  6 16:39:08 1995
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Message-Id: <wvmail32.1995oct4.162503.bryan@wvnvm.wvnet.edu>
Date:      Fri, 6 Oct 1995 09:25:11 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Preliminary Search Results for <F,U,L,R,D>

The first step in finding |B| in <F,U,L,R,D> is to build a little
Start-rooted data base using only the ten generators F, F', U, U', etc.
I was able to search eight levels deep with a "quick and dirty" search.
To search deeper would take a good bit longer.  I hope that eight levels
will be enough to calculate |B|.  By constructing a companion
B-rooted data base, I should be able to test up to 15 levels deep.
(B is odd, so eight levels deep for each half-depth search only
gets you 15 levels total instead of 16.)

I haven't started the search for B yet, but I thought the
results so far might be of minor interest in their own right.  As
I indicated before, these are actual cube positions.  I haven't
figured out any way to apply conjugacy classes to this problem.

I find it a little puzzling that the branching factor is not
monotonically decreasing (c.f., level 3 to level 4).


     Level         Number of       Local      Branching
                   Positions        Max        Factor

       0                  1           0
       1                 10           0           10.000
       2                 77           0            7.700
       3                584           0            7.584
       4              4,434           0            7.592
       5             33,664           0            7.592
       6            255,320           0            7.584
       7          1,933,936                        7.575
       8         14,635,503                        7.568

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From alan@curry.epilogue.com  Sat Oct  7 06:36:21 1995
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Message-Id: <7Oct1995.062936.Alan@LCS.MIT.EDU>
From: Alan Bawden <Cube-Lovers-Request@ai.mit.edu>
Sender: Cube-Lovers-Request@ai.mit.edu
To: Cube-Lovers@ai.mit.edu
In-Reply-To: "Samantha Jerrings, President, International Students\
 Association, Eastern Division"'s message of Sat, 7 Oct 1995 01:06:45 -0500 <v01530503ac9bbe6b9a91@[204.141.97.223]>
Subject: ===>> FREE 1 yr. Magazine Sub sent worldwide- 300+ Popular USA Titles

If you're wondering why you got the fllowing piece of mail:

   Date: Sat, 7 Oct 1995 01:06:45 -0500
   From: "Samantha Jerrings, President, International Students\
    Association, Eastern Division" <abacus@abacus.net>

   Hi fellow 'netters,

   My name is Samantha Jerrings and I recently started using a magazine
   subscription club in the USA that has a FREE 1 yr. magazine subscription
   deal with your first paid order- and I have been very pleased with them.
   ...

it's because you're on Cube-Lovers.  Needless to say, I'll be protesting
this misuse of our mailing list.  Please do not respond to -me- about this
message.  And I would certainly encourage you not to order any magazines
from these slime-balls either!

From mark.longridge@canrem.com  Sun Oct  8 00:06:06 1995
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Reply-To: CRSO.Cube@canrem.com
Sender: CRSO.Cube@canrem.com
Subject: Antislice Correction
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1248.5834.0C1F8131@canrem.com>
Date: Sat,  7 Oct 95 23:53:00 -0500
Organization: CRS Online  (Toronto, Ontario)

I'll start with a small correction:

I wrote on Mon, 3 Jul 1995 14:53:00
> Patterns in the Anti-Slice Group
> --------------------------------
>
> p4   8 flip (Op sides)  (R1 L1 U1 D1 F1 B1) ^2                    (12)
> p10a pons asinorum      (L3 R1 U3 D1)^3                           (12)
> p16a 4 cross order 2     F1 B1 U1 D1 L2 R2 U1 D1 F1 B1 U2 D2      (12)
> p17  4 diagonal         (F1 B1 R1 L1) ^3                          (12)
> p18a 4 diagonal,2 cross (F1 B1 R3 L3) ^3                          (12)
> p22  2 DOT, 2 Stripe     R1 L1 U2 D2 R3 L3                         (6)
> p64a 4 Z                 F1 B1 L3 R3 F1 B1 L1 R1 F3 B3 L1 R1      (12)
> p143 Pinwheels           F1 B1 L1 R1 F3 B3 U3 D3 L1 R1 U1 D1      (12)
> p175a 6 H order 2        U3 D3 L3 R3 F2 B2 U2 D2 L3 R3 U1 D1      (12)
> p198a 2 X, 4 Diag no C   L1 R1 F1 B1 L3 R3 F3 B3 L1 R1 F1 B1      (12)
> p201 Pinwheels + Pons    L1 R1 F3 B3 L1 R1 U3 D3 F1 B1 U3 D3      (12)
>
> p201 is a quite interesting position.

The reference to p201 is incorrect. It should read:

  p175a is a quite interesting position.
  ^^^^^

> The square's group equivalent is no shorter in q turns:
>
> p175 6 H order 2 type 2  U2 B2 L2 U2 D2 L2 F2 U2                   (8)
>
> Note that p201 = |{m'Xm}|=2 and |Symm(X)|=24.


Someone also edited my original entry and changed all the T's to U's
which is more consistent and standard.

In addition.....

Dan Hoey wrote on Fri, 28 Oct 94 11:38:15 EDT
> But there's another reason.  Remember the annoying feature that the
> color assignments to faces were never standardized?  The first cube I
> bought had red opposite yellow, blue opposite white, and orange
> opposite green (I think).  Even though in later days most cubes are
> manufactured with opposite faces ``differing by yellow''--red opposite
> orange, blue opposite green, and yellow opposite white--there does not
> seem to be a standard for the handedness of the coloring.  This has
> long been a problem on cube-lovers, where everyone .......

 There was a standard in the cube contests.

 The original Ideal colour arrangement was the tournament standard in
the U.S. and Canada.

Top=White, Down=Blue, Left=Red, Right=Orange, Front=Yellow, Back=Green

-> Mark <-


From mark.longridge@canrem.com  Sun Oct  8 00:27:13 1995
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Subject: Using 5 Generators
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1249.5834.0C1F8132@canrem.com>
Date: Sat,  7 Oct 95 23:54:00 -0500
Organization: CRS Online  (Toronto, Ontario)

From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
Subject:   <F,U,L,R,D> Question

> It is well known that if we define G=<Q> for the twelve quarter turns
> q in Q, we can also generate G as G=<F,U,L,R,D>, leaving out B and B'.
> Leaving out any other quarter turn would do as well, but I
> am going to stick to leaving out B for illustrative purposes.
>
> However, when one of the quarter turns is left out, the length of most
> positions will change.  In particular, we will no longer have |B|=1.
> My reading of the archives indicates that we do not know what the
> length of B would be in this situation, nor what a minimal process
> for B would be.

This problem was solved by David Benson in Oct. 1979, who was one of
the earliest cube pioneers. Dr. Singmaster reports on this in his
2nd Addendum of "Notes".

Let A = R1 L3 F2 B2 R1 L3, then AUA = D1

AUA = R1 L3 F2 B2 R1 L3 U1 R1 L3 F2 B2 R1 L3   (17 q, 13 q+h)

Perhaps Jerry will find something shorter.

-> Mark <-


From mark.longridge@canrem.com  Sun Oct  8 00:27:14 1995
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Reply-To: CRSO.Cube@canrem.com
Sender: CRSO.Cube@canrem.com
Subject: Picture Cubes
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1250.5834.0C1F8133@canrem.com>
Date: Sat,  7 Oct 95 23:55:00 -0500
Organization: CRS Online  (Toronto, Ontario)

David Singmaster (ZINGMAST@VAX.SBU.AC.UK) writes on
Wed Sep 13 11:50:09 1995:
------------------------------------------------
 I think the ranking is not quite fair because the puzzles are of
very different types.  E.g. the 15 puzzle has nearly as many patterns
as the 3^3 but no one would claim it was anywhere near as difficult.
Indeed the Babylon Tower has  36!  = 3.72 x 10^41  basic positions.
One can divide by some small value such as 2 or 6 or perhaps more,
depending on what one considers the same position.  This puts it
between 3 and 4 in your list, but it is not a difficult puzzle, except
that it is hard to see the gradations of the colors!  Indeed, the
commercial  7 x 7  'fifteen puzzles' have  49! =6.08 x 10^62  basic
patterns - again one has to divide by something, in this case 2.
This falls between 2 and 3 in your list, but again it is hardly a
difficult puzzle.

 So I think you are comparing puzzles which are of such different
type that the number of patterns is not a fair comparison of their
difficulties.I would group them in three (or perhaps 2) types.
 Rubik Cube, etc.
 Fifteen Puzzles, etc. in the plane.
 Cylindrical Puzzles - barrels, etc.
------------------------------------------------

 I quite agree. One of my reasons for making that list was to simply
rank all the puzzles by number of combinations only, to show the
feasibility (or lack of) for a brute force search to find
God's Algorithm.

 The major drawback, as you point out, is that difficulty in solving
is not only a function of the number of combinations.

Dr. Singmaster continues:
------------------------
 Re your Case E.  Almost all the picture cubes have all four
orientations distinct on the face centres - both those with nine little
pictures on each face and those with a big picture spread over all
nine facelets.  These are actually pretty common.
-------------------------------------------------

 Case E, that is cases that have only a fraction of the total possible
number of combinations for a Rubik's picture cube, are unfortunately
well represented in my own cube collection.

 The following cubes are all in Case E:

 Rubik's Calendar Cube, Rubik's Cube 4th Dimension, Rubik's World,
Blind Man's Cube (from Germany), Royal Wedding Cube (with Charles
& Di).

 Although I don't doubt that, over all, these cases are exceptional.
In the case of Rubik's World there are 3 blank centre pieces, and
in the Royal Wedding Cube only 2 opposite faces can show all 4
possible orientations.

        Name                  Combinations      Inventor

 8. Picture Cube (3x3x3) (E) 8.8*10^22      Erno Rubik, Dan Hoey
 9. Calendar Cube (3x3x3)(F) 4.4*10^22      Marvin Silbermintz
10. Rubik's Cube 4th Dim.(D) 1.1*10^22      Erno Rubik
11  Rubik's World (G)        2.7*10^21      Erno Rubik
12. Royal Wedding Cube       6.9*10^20      Unknown
13. Rubik's Cube (3x3x3)     4.3*10^19      Erno Rubik


-> Mark <-


From alan@curry.epilogue.com  Sun Oct  8 03:58:52 1995
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From: Alan Bawden <Cube-Lovers-Request@ai.mit.edu>
Sender: Cube-Lovers-Request@ai.mit.edu
To: Cube-Lovers@ai.mit.edu
In-Reply-To: "Samantha Jerrings, President, Association of International\
 Students, Eastern Division"'s message of Sun, 8 Oct 1995 00:00:17 -0500 <v01530509ac9cfcaa0fd7@[198.70.48.43]>
Subject: ===>> FREE 1 yr. Magazine Sub sent worldwide- 300+ Popular USA Titles

   Date: Sun, 8 Oct 1995 00:00:17 -0500
   From: "Samantha Jerrings, President, Association of International\
    Students, Eastern Division" <beef@cow.net>
   ...

Alas, there is little I can currently do about this, beyond complaining to
various postmaster addresses (which I have done).  As the Internet becomes
more of a capitalist free-for-all, even a moderately small mailing list
like Cube-Lovers (less than 200 members) becomes a target for things like
this.  The only real solution, given that it is no longer possible to rely
on people's courtesy, is to go for some kind of moderation.  And in fact, I
am developing some software to moderate Cube-Lovers (and another mailing
list I maintain).  When I'm done, the only changes you all should notice
are: (1) no more junk mail and (2) a slightly longer turnaround time when
you submit a legitimate message.  Until I've got that working all you can
do is ignore these losers.

From tdb@delta1.deltanet.com  Sun Oct  8 12:53:11 1995
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From: tdb@delta1.deltanet.com (Tom D. Baccanti)
Newsgroups: cube-lovers
To: cube-lovers@life.ai.mit.edu
Subject: Intro and question
Date: Sun, 08 Oct 1995 09:33:01 -0700
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Briefly, I am a old time cube fan and I am blind.  I have been blind for
about 9 years now due to complications from diabetes.  I recently
encountered a "Braille Or Tactile Rubik's cube" available from Japan.  I
finally received my cube from there after much expense and effort but it
was worth it.

Describing the cube:  it has designs on each side that can easily be
differentiated from the other.  ie; circles, plus, dotted circle,
smooth, six dotsand textured side.  I am now able to solve all but the
last bottom slice from memory but I would like to ask if anyone has a
xolution that I can read that would make sense to me?  I wish to solve
this and then move on to the other puzzles I have read on this list
mentioned.  I will have to mark them tactilly also.
            
Thanks for the time,

Tom Baccanti

-- 
                       Every crowd has a silver lining.
Tom D. Baccanti       /** Reach me at Internet:  TDB@Deltanet.com
                                                 TDB@crl.com
/* PGP key from: pgp-public-keys@pgp.mit.edu get key id: 8EB942B1

From laz@smartlink.net  Sun Oct  8 15:32:13 1995
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From: Laszlo Takacs <laz@smartlink.net>
To: Cube-Lovers-Request@ai.mit.edu
Cc: Cube-Lovers@ai.mit.edu
In-Reply-To: <8Oct1995.033935.Alan@LCS.MIT.EDU> (message from Alan Bawden on
	Sun, 8 Oct 1995 03:58:50 -0400)
Subject: Re: ===>> FREE 1 yr. Magazine Sub sent worldwide- 300+ Popular USA Titles
Reply-To: laz@smartlink.net

|Until I've got that working all you can do is ignore these losers.

It that really all?  Why not have everyone return the mail 100 fold?

From hazard@niksula.hut.fi  Mon Oct  9 06:19:55 1995
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From: Mikko Haapanen <hazard@niksula.hut.fi>
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On Sun, 8 Oct 1995, Laszlo Takacs wrote:

> |Until I've got that working all you can do is ignore these losers.
> It that really all?  Why not have everyone return the mail 100 fold?

This is dirty but i think it's efficient!

------Mikko Haapanen------hazard@niksula.hut.fi---------
Jos tahto siirt=E4=E4 vuoren, niin tahdon siirt=E4=E4 tahdon
Vuorta siirt=E4m=E4=E4n, ett=E4 vuoren sis=E4=E4n n=E4=E4n -P.Hanhiniemi
--------------------------------------------------------


From ronnie@cisco.com  Mon Oct  9 14:18:38 1995
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To: laz@smartlink.net
Cc: Cube-Lovers-Request@ai.mit.edu, Cube-Lovers@ai.mit.edu
Subject: Re: ===>> FREE 1 yr. Magazine Sub sent worldwide- 300+ Popular USA Titles 
In-Reply-To: Your message of "Sun, 08 Oct 1995 12:32:00 PDT."
             <199510081932.MAA00115@alumina.smartlink.net> 
Date: Mon, 09 Oct 1995 11:18:32 -0700
From: "Ronnie B. Kon" <ronnie@cisco.com>

> |Until I've got that working all you can do is ignore these losers.
> 
> It that really all?  Why not have everyone return the mail 100 fold?

Note that there is an e-mail address buried in the message which is
different from the sending e-mail address.  The sending address is a
forgery, probably an innocent person.

				Ronnie

From alan@curry.epilogue.com  Mon Oct  9 15:07:15 1995
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Date: Mon, 9 Oct 1995 15:07:13 -0400
Message-Id: <9Oct1995.140201.Alan@LCS.MIT.EDU>
From: Alan Bawden <Cube-Lovers-Request@ai.mit.edu>
Sender: Cube-Lovers-Request@ai.mit.edu
To: Cube-Lovers@ai.mit.edu
In-Reply-To: Mikko Haapanen's message of Mon, 9 Oct 1995 12:19:37 +0200 (EET) <Pine.SGI.3.91.951009121643.466A-100000@neppari.cs.hut.fi>
Subject: ===>> FREE 1 yr. Magazine Sub sent worldwide- 300+ ...

   Date: Mon, 9 Oct 1995 12:19:37 +0200 (EET)
   From: Mikko Haapanen <hazard@niksula.hut.fi>
   To: cube-lovers@ai.mit.edu
   On Sun, 8 Oct 1995, Laszlo Takacs wrote:
   > |Until I've got that working all you can do is ignore these losers.
   > It that really all?  Why not have everyone return the mail 100 fold?
   This is dirty but i think it's efficient!

Sigh.  I was hoping I wouldn't have to say this.  PLEASE DO -NOT- CARRY ON
A CONVERSATION ABOUT HOW TO DEAL WITH JUNK MAIL ON CUBE-LOVERS!

From aaweint@io.org  Mon Oct  9 15:35:38 1995
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To: Cube-Lovers@ai.mit.edu
From: aaweint@io.org (Aaron Weintraub)
Subject: Rubik's Revenge, where?

I'm looking fora Rubik's Revenge puzzle.  Do places still sell them?  If
not, is there a place where I can mail order one from?  Thanks for any
information on this.

Aaron
aaweint@io.org


From BRYAN@wvnvm.wvnet.edu  Mon Oct  9 19:48:30 1995
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Date:      Mon, 9 Oct 1995 08:58:39 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@life.ai.mit.edu>
Subject:   Re: Antislice Correction
In-Reply-To: Message of 10/07/95 at 23:53:00 from mark.longridge@canrem.com

On 10/07/95 at 23:53:00 mark.longridge@canrem.com said:

>Someone also edited my original entry and changed all the T's to U's
>which is more consistent and standard.

I agree that U's are more "consistent and standard", but for reasons
that have been discussed several times, T's would probably be better.
However, the effort to switch from U's to T's has not been very
successful.  Personally, I have *tried* to switch to T's, and it just
doesn't feel right.  So I have to share the blame for sabotaging the
switch to T's.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From BRYAN@wvnvm.wvnet.edu  Wed Oct 11 18:10:11 1995
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Date:      Wed, 11 Oct 1995 13:50:51 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@life.ai.mit.edu>
Subject:   Re: Using 5 Generators
In-Reply-To: Message of 10/07/95 at 23:54:00 from mark.longridge@canrem.com

On 10/07/95 at 23:54:00 mark.longridge@canrem.com said:

>This problem was solved by David Benson in Oct. 1979, who was one of
>the earliest cube pioneers. Dr. Singmaster reports on this in his
>2nd Addendum of "Notes".

>Let A = R1 L3 F2 B2 R1 L3, then AUA = D1

>AUA = R1 L3 F2 B2 R1 L3 U1 R1 L3 F2 B2 R1 L3   (17 q, 13 q+h)

>Perhaps Jerry will find something shorter.

Well, without seeing the original article, I am not sure if I would
agree that the problem was solved or not back in 1979.  By that I
mean that the problem I was proposing to solve was finding a minimal
solution.  I don't know if the original article claimed that 17q
was minimal.

I can now confirm that 17q is indeed a minimal solution.  With my
data base of positions through level 8, I was able to perform half-depth
searches to confirm that there are no solutions through 15q.  Given the
existence of a 17q solution, that is sufficient to show that 17q is
minimal.  But just to be sure, I extended my data base of positions
through level 9, and with my extended data base I was able to find
several solutions of length 17q.

I am not quite sure what we should count as a unique solution.  But I can
report that my search found 16 unique (*not* unique up to conjugacy)
half-way positions.  I use the term "half-way" advisedly.  The "half-way"
positions are 9q from Start and 8q from B or vice versa.  I guess you
could say that the vice versa gives you a total of 32=16+16 half-way
positions, but the whole concept of "half-way" is pretty slippery
in this case anyway.

Just because we know that 17q is minimal does not mean that we know
that 13 q+h is minimal.  I have not done any searches of the q+h case.

With my extended data base, the results of the search with five generators
are as follows:

     Level         Number of       Local      Branching
                   Positions        Max        Factor

       0                  1           0
       1                 10           0           10.000
       2                 77           0            7.700
       3                584           0            7.584
       4              4,434           0            7.592
       5             33,664           0            7.592
       6            255,320           0            7.584
       7          1,933,936           0            7.575
       8         14,635,503                        7.568
       9        110,685,344                        7.562


One more thing, and perhaps this particular problem can be put to rest.
I have mentioned several times that I could not figure out how to
use conjugacy for this particular problem.  Well now I have, although
it is too late for doing the search.  You certainly cannot use
M-conjugacy, but you can use a subgroup of M.  The subgroup includes
four rotations and four reflections.

The four rotations are i, b, bb, and bbb, where we use lower case
letters to simulate Frey and Singmaster's script notation for rotations.
For example, b is the whole cube rotation consisting of grasping
the Back face and rotating the whole cube (not just the Back face)
clockwise by 90 degrees.  For the reflections, we use Dan Hoey's
"permutations of face centers" notation.  The four reflections are
(UL)(DR), (UR)(DL), (UD), and (LR).  To tie the two notations
together, we could write the rotations as i=(), b=(ULDR),
bb=(UD)(RL), and bbb=(URDL).  These eight rotations and reflections
form the subgroup of M which is called Q2 in Dan's taxonomy of
the 98 subgroups of M.  Hence, we could have used Q2-conjugacy, which
would have reduced the size of the problem by about eight times.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From hoey@aic.nrl.navy.mil  Wed Oct 11 22:56:23 1995
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Date: Wed, 11 Oct 95 22:56:15 EDT
From: hoey@aic.nrl.navy.mil
Message-Id: <9510120256.AA04061@sun13.aic.nrl.navy.mil>
To: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>, Cube-Lovers@life.ai.mit.edu
Subject: Re: Using 5 Generators

I certainly agree that confirming minimality is an important result.
Thanks, Jerry.

> But I can report that my search found 16 unique (*not* unique up to
> conjugacy) half-way positions.  I use the term "half-way" advisedly.
> The "half-way" positions are 9q from Start and 8q from B or vice
> versa.  I guess you could say that the vice versa gives you a total
> of 32=16+16 half-way positions, but the whole concept of "half-way"
> is pretty slippery in this case anyway.

If I understand this, there are 16 positions at 9q from Start and 8q
from B, and there are 16 other positions at 8q from Start 9q from B.
Is each of the first bunch adjacent to exactly one of the other?  And
vice versa?  It would be good to get them reduced by Q2-conjugacy, as
well.

[in Q2]
> The four rotations are i, b, bb, and bbb, where we use lower case
> letters to simulate Frey and Singmaster's script notation for rotations.
> For example, b is the whole cube rotation consisting of grasping
> the Back face and rotating the whole cube (not just the Back face)
> clockwise by 90 degrees.

The reflections are similarly rrv, rrbv, ttv, and ttbv, where t and r
are the whole-cube rotations by the Top and Right faces, and v is the
central inversion.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From preux@lil.univ-littoral.fr  Thu Oct 12 10:48:47 1995
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Date: Thu, 12 Oct 1995 14:45:45 GMT
From: preux@lil.univ-littoral.fr (Preux Philippe)
To: Cube-Lovers@life.ai.mit.edu
Subject: I am in search of Thistlewaite's algorithm


Hi,

As long as I am a new comer to this mailing list, I will briefly
introduce myself. As a computer science researcher, I am working on
evolutionary algorithms trying to assess their ability to solve
combinatorial optimization problems. One of my old dream has been to
solve the Rubik's cube (for the moment, the very basic 3x3x3 version)
with this kind of algorithms. I have heard about the Thistlewaite's
algorithm which is able to solve the problem in less than 50 or so
moves. I have also heard about a publication of D. Singmaster that, among
other things, describes this algorithm. Thus, I am looking for
information about this algorithm: how does it work precisely? I wonder
whether this algorithm, either a description or an implementation of
it, is available somewhere on the net. I would also be interested in
having a copy of D. Singmaster's report (either via ftp or paper).

Can someone help me? Thanks a lot in any case.

Philippe

-- 

From SHV6937@ocvaxa.cc.oberlin.edu  Thu Oct 12 17:22:08 1995
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Date: Thu, 12 Oct 1995 17:17:21 -0400 (EDT)
From: Huy Vo <SHV6937@ocvaxa.cc.oberlin.edu>
Subject: Unsubscribe me ...
To: Cube-Lovers@life.ai.mit.edu
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Please unsubscribe me from the listserv.  ( shv6937@ocvaxa.oberlin.edu )
I apologize for the occupation of your disk space but the system I am on
is being changed.  

-- huy

From BRYAN@wvnvm.wvnet.edu  Fri Oct 13 15:46:51 1995
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Date:      Fri, 13 Oct 1995 11:47:43 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Dan Hoey" <hoey@aic.nrl.navy.mil>, <Cube-Lovers@life.ai.mit.edu>
Subject:   Re: Using 5 Generators
In-Reply-To: Message of 10/11/95 at 22:56:15 from hoey@aic.nrl.navy.mil

On 10/11/95 at 22:56:15 hoey@aic.nrl.navy.mil said:

>> But I can report that my search found 16 unique (*not* unique up to
>> conjugacy) half-way positions.  I use the term "half-way" advisedly.
>> The "half-way" positions are 9q from Start and 8q from B or vice
>> versa.  I guess you could say that the vice versa gives you a total
>> of 32=16+16 half-way positions, but the whole concept of "half-way"
>> is pretty slippery in this case anyway.

>If I understand this, there are 16 positions at 9q from Start and 8q
>from B, and there are 16 other positions at 8q from Start 9q from B.
>Is each of the first bunch adjacent to exactly one of the other?  And
>vice versa?  It would be good to get them reduced by Q2-conjugacy, as
>well.

I don't think I can answer your questions without further analysis,
and I don't have much time to devote to the problem.  But let me
clarify as follows.  First, the search looked as follows:

        Distance from     Distance from      Total       Number of
          Start                 B           Distance      Matching
                                                          Positions

             0                   1             1               0
             1                   2             3               0
             2                   3             5               0
             3                   4             7               0
             4                   5             9               0
             5                   6            11               0
             6                   7            13               0
             7                   8            15               0
             8                   9            17              16

There is a certain arbitrariness in at least two respects.  For one
example, to test for a total distance of 11, you could just as well use
distances from Start and B respectively of 4 and 7 instead of 5 and 6.
For another example, the Start-rooted tree and the B-rooted tree have
identical structures, so the first two columns could be reversed.
Indeed, you could get the B-rooted tree simply by pre-multiplying the
Start-rooted tree by B.  (This reminds me of one of my most
foolish errors on Cube-Lovers.  For search trees where the nodes are
conjugacy classes (or representatives of conjugacy classes), the
tree looks different depending on which class or representative is at
the root.  But when the nodes are all the positions, then the tree is
essentially the same in all cases, just pre-multiplied by the root.
I once claimed the tree structure depended on the root for trees
containing all positions, confusing that situation with the situation
for trees of conjugacy classes.  Arrg!)

Hence, I am not especially comfortable talking about "half-way" positions.
But continuing anyway, denote the 16 positions which are 8 moves from
Start and 9 moves from B as X_i for i in 1..16.  Then, the 16 positions
which are 8 moves from B and 9 moves from Start are B(X_i) for i
in 1..16.  A solution to the problem would then look something like
(X_j)Y(X_k)', where Y is in Q-{B,B'}.  But I don't think we can say
a priori that there is no Z in Q-{B,B'} and no X_m such that
(X_j)Z(X_m)' is also a solution (Z not equal Y and X_m not equal X_k).

I think to analyze the problem properly you would have to take the
positions X_i for i in 1..16 and the positions B(X_j) for j in 1..16
and match up each X_i with each B(X_j) to see which ones differ
by a quarter turn.  Each X_i is going to match up with at least one
B(X_j) and vice versa, but there might be more than 16 matches
overall.

Reduction by Q2-conjugacy is important, but I don't think it would
tell you how many solutions there are that you would really want to
consider to be unique.  Recall the solution of Pons Asinorum by
half-depth searches.  There are 5 positions unique up to M-conjugacy
which are 6q from Start and 6q from Pons.  But most people would
consider that there are only two minimal solutions to Pons that are
really different.  The trouble is that 4 of the 5 half-way positions
for the Pons are in the middle of a sub-process which commutes.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From BRYAN@wvnvm.wvnet.edu  Fri Oct 13 16:40:29 1995
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Date:      Fri, 13 Oct 1995 15:01:31 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Dan Hoey" <hoey@aic.nrl.navy.mil>, <Cube-Lovers@life.ai.mit.edu>
Subject:   Re: Using 5 Generators
In-Reply-To: Message of 10/11/95 at 22:56:15 from hoey@aic.nrl.navy.mil

On 10/11/95 at 22:56:15 hoey@aic.nrl.navy.mil said:

>[in Q2]
>> The four rotations are i, b, bb, and bbb, where we use lower case
>> letters to simulate Frey and Singmaster's script notation for rotations.
>> For example, b is the whole cube rotation consisting of grasping
>> the Back face and rotating the whole cube (not just the Back face)
>> clockwise by 90 degrees.

>The reflections are similarly rrv, rrbv, ttv, and ttbv, where t and r
>are the whole-cube rotations by the Top and Right faces, and v is the
>central inversion.

I am not quite sure why it took me so long to figure out that
Q2-conjugation was the appropriate symmetry when generating G
without B and B'.  With the wisdom if hindsight, it is perhaps
because the Q2 subgroup of M could be described as fixing the
B and F face centers, and I was thinking only in terms of fixing
the B face center.  Obviously, you cannot fix the B face center
without also fixing the F face center.  Thus, it is clear that
Q2-conjugation is also the appropriate symmetry when generating
G without F and F'.

Q1 and Q3 are subgroups of M which are conjugate with Q2 in Dan's
taxonomy of subgroups of M.  Q1 fixes the T and D face centers, and
would be the appropriate symmetry when generating G without T and T',
or without D and D'.  Q3 fixes the R and L face centers, and would be
the appropriate symmetry when generating G without R and R', or without
L and L'.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From BRYAN@wvnvm.wvnet.edu  Tue Oct 17 12:48:14 1995
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Date:      Mon, 16 Oct 1995 23:08:02 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Correctness of Large Searches

When I was in graduate school in another life many years ago, one of
the big issues was program correctness, and whether programs could
be proven to be correct.  As I recall, the short answer is that
non-trivial programs cannot be proven to be correct.  I haven't
really followed the issue since then, but I doubt that the answer
has changed.

Also, many mathematicians are suspicious or unaccepting of proofs or
other results which involve large computer searches.  For example,
the proof of the Four Color Map Theorem is a famous result involving
a large search which is held in certain suspicion.  For one thing,
the referees for the paper were not able to reproduce the results
of the computer search (not enough computing power).  For another thing,
the referees were neither able to confirm that the programs were error
free (who can?), nor that some sort of computer error (software,
hardware, or procedural) did not occur during the running of the
programs (c.f., the famous Pentium floating point divide error).

Two factors cause me to raise the question of program correctness
at this time.  One is simply that some of the searches we do are
so large, how do we know that the answers are correct?  The other
is that I have found an error in one of my (smaller) searches that
I need to report.

Those of you who actually fight through the details of my longer, more
boring posts will recall that there are essentially three different
models I use from time to time.  For cubes without centers, I usually
use representative elements of sets of the form {m'Xmc}.  For cubes
with centers, I use either representative elements of sets of the
form {m'Xm}, or else I use representatives of the form
Repr{m'Xmc}*C to simulate Repr{m'Xm}.  The latter model is very
compact for cases where complete searches can be accomplished

For the case of the 3x3x3 cube, corners only, with centers, q-turns
only, I recently discovered the following anomaly.  Note in
particular level 8 and below of the search.

          3x3x3 Corners Only -- Qturns


    Distance    Repr{m'Xmc}*C        Repr{m'Xm}
        from          Model             Model
       Start

           0               1                 1
           1               1                 1
           2               5                 5
           3              24                24
           4             149               149
           5             850               850
           6            4257              4257
           7           16937             16937
           8           57800             57848
           9          180639            180787
          10          466052            466220
          11          676790            676786
          12          392558            392342
          13           45744             45600
          14             163               163

        Total        1841970           1841970


The 3x3x3 corners only case has been searched a number of times by
a number of people, all of whom got the same results.  However,
these results all were for every  position, including those which are
M-conjugate.  My searches are for conjugacy classes only, and therefore
I have nobody with whom I can compare results directly.  The only
way I can compare results is to expand the conjugacy classes, and
regrettably I did not do so when I first calculated the corners
only case.

For a variety of reasons to be detailed below, I believe the {m'Xm}
results above are correct and that the {m'Xmc}*C results are incorrect.
But at the same time, I do not believe there is an error per se in
the {m'Xmc}*C model.  Let's see if we can make some sense of this.

When I first discovered the anomaly, my reaction was that
the {m'Xm} model is simpler, and Occum's Razor suggested that
the error was in the {m'Xmc}*C model.  Also, when I expanded
the conjugacy classes for the Repr{m'Xm} model, the results
matched what everybody else had posted.

Hence, I went back to the
{m'Xmc}*C program, which I hadn't looked at in years.  After many
hours of reviewing the model, and many more hours of reviewing
the program itself, I not unsurprisingly found no errors.
Furthermore, I was no longer convinced that Occum's Razor still
applied.  The {m'Xmc}*C model is "more complicated" in some ways,
but on the other hand, every cell of storage for the data base
can be addressed directly.  There is no sorting, merging, and
matching of huge files as there is with the {m'Xm} model.

So I ran the {m'Xmc}*C program again.  Very surprisingly, this time
it produced different answers, and the answers were the same as for
the {m'Xm} model!  So what's going on?  I don't know.

I am presently a bureaucrat (cubing is a hobby), but in previous
lives I have been a technical support person.  In that role, I
have found numerous problems that caused programs to produce incorrect
results, even though the program was "correct" (whatever that means).
I have found hardware errors vaguely similar to the Pentium error,
operating system errors, compiler errors (especially with optimizing
compilers), and subroutine library errors.  So my best guess is that
something of this nature caused the {m'Xmc}*C program to produce
incorrect results one time and correct results another time.
Of course, an uninitialized variable or pointer can cause similar
symptoms, but I have not been able to find anything like that in
my program.

The program is written in Turbo-Pascal for a PC using DOS.  I have
the exact same compiler now I have had for ten years or so.  The
Pascal source code is unchanged from when I ran it before.  The
first time I ran the program, I ran it under DOS, or maybe
in the DOS box in an early version of OS/2 (can't remember for sure),
and it ran on a 286 or an early 386 (can't remember for sure).
This time, it ran in the DOS box under OS/2 Warp on a 486/50.
So lots of things have changed.  Furthermore, both times I ran it,
I used the VDISK facility to cache all the disk files in memory,
and OS/2 Warp surely has a newer VDISK than whatever I was using
before.  I even wondered if the data base was correct the first time,
but maybe I had a bad counting program analyzing the data base.  But
I still had the original data base and counted it again.  It
really was wrong.  So the mystery of the incorrect results is not solved.

In light of the above discussion, I thought it might be appropriate
to summarize some background about my larger searches.  I want to
indicate which ones of them seem pretty well verified, and which ones
of them might benefit from further study.

   <U,R>           -  I believe this one is ok.  I ran the q turn case
                      with and without conjugacy classes.  Upon
                      expanding the conjugacy classes, the results
                      matched the results without conjugacy classes.
                      Also, the results matched results posted by
                      Mark Longridge as far as they went (although
                      my anomaly with corners-only suggests that
                      such matching doesn't prove very much).  I
                      ran the q+h turn case only with conjugacy classes,
                      and expanded then to get the results without
                      conjugacy classes.  The model was Repr{w'Xw}
                      (W-conjugacy instead of M-conjugacy), with
                      much sorting, merging, and matching of external
                      files.

  edges only          This ran for about a year, so there is potential
 (without face        for error.  The model was Repr{m'Xmc}.  The
   centers)           runs to create neighbors were performed primarily
                      on two 486's and a 386, using a mainframe as
                      a file server.  A UNIX station was added for
                      the last few months.  All sorts, merges, and
                      matches were run on the mainframe.  The whole
                      process was driven by REXX scripts on the PC's
                      and UNIX stations (we run REXX on all our UNIX
                      boxes).  There were actually more lines of code
                      for the scripts than for the programs.  The
                      scripts moved files back and forth with FTP,
                      and contained code to detect and correct errors
                      with the FTP's.  The problem has only been
                      run once, so there is no independent verification.

 edges only           This was run twice, once using the {m'Xm} model,
(with face            and once using the Repr{m'Xm}*C model.  The
  centers)            answers matched through level 11.  They did not
                      match at level 12.  I have only posted through
                      level 11 to the list.  I have not had time to
                      find the error.  I do not expect to find the
                      error in any of the programs.  It took about a
                      year to get to level 11, running on the mainframe.
                      I expect the error to be somewhere in a script,
                      probably failing to recover from some error
                      condition like a system failure.  These things
                      run around the clock, and inevitably get caught
                      in system failures from time to time.

                      Although I was not able to do so, this is probably
                      the largest cubing problem where we can conceive
                      of running the problem to completion using today's
                      technology.  Had I been able to do run the problem
                      to completion, the Repr(m'Xm)*C model would have
                      served as verification for the "without face
                      centers" case because the minimum of each
                      row of the solution matrix with centers serves
                      as the solution for the without centers case.

Whole cube            These are the most important runs.  The problem
(edges and corners    has been run twice using the {m'Xm} model, although
 with face centers)   you might not wish to count it as twice.  The
                      difference is that the edges were the major sort
                      in one run, and the corners were the major sort
                      in the other.  The two runs did match, which is
                      a good (but not perfect indication) that there
                      was no error (e.g., such as the
                      scripts failing to include something in
                      in a sort or merge that should have been
                      included).  It would still be nice to have
                      verification by someone else.  I was able to run
                      up through level 11.  I did not try level 12
                      because it was just too big.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From bagleyd@source.asset.com  Tue Oct 17 13:56:03 1995
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          id AA47533; Tue, 17 Oct 1995 14:01:53 -0400
Date: Tue, 17 Oct 1995 14:01:53 -0400
From: bagleyd@source.asset.com (David A. Bagley)
Message-Id: <9510171801.AA47533@source.asset.com>
To: cube-lovers@life.ai.mit.edu
Subject: Motif puzzles

Hi
  I converted my puzzles at ftp.x.org in /contrib/games/puzzles to
use Motif.  By default, the puzzles will compile and link with just the
X-Windows includes and libs.
 
  If there is demand, I could supply the Motif versions statically linked
for SunOS4.1.3.  The Motif versions have no increased functionality but may
be easier to use.
 
Cheers,
      --__---------------------------------------------------------------
     /  \ \   /           David A. Bagley                                \
    |    \ \ /            bagleyd@source.asset.com                        |
    |     \//\            Some days are better than other days.           |
    |     / \ \                -- A short lived character of Blake's 7    |
     \   /   \_\puzzles   Available at: ftp.x.org/contrib/games/puzzles  /
      -------------------------------------------------------------------

From BRYAN@wvnvm.wvnet.edu  Tue Oct 17 14:20:20 1995
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Date:      Tue, 17 Oct 1995 14:19:52 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <Cube-Lovers@life.ai.mit.edu>
Subject:   Re: I am in search of Thistlewaite's algorithm
In-Reply-To: Message of 10/12/95 at 14:45:45 from preux@lil.univ-littoral.fr

On 10/12/95 at 14:45:45 preux@lil.univ-littoral.fr said:

>As long as I am a new comer to this mailing list, I will briefly
>introduce myself. As a computer science researcher, I am working on
>evolutionary algorithms trying to assess their ability to solve
>combinatorial optimization problems. One of my old dream has been to
>solve the Rubik's cube (for the moment, the very basic 3x3x3 version)
>with this kind of algorithms. I have heard about the Thistlethwaite's
>algorithm which is able to solve the problem in less than 50 or so
>moves. I have also heard about a publication of D. Singmaster that, among
>other things, describes this algorithm. Thus, I am looking for
>information about this algorithm: how does it work precisely? I wonder
>whether this algorithm, either a description or an implementation of
>it, is available somewhere on the net. I would also be interested in
>having a copy of D. Singmaster's report (either via ftp or paper).

I don't know if you received any private replies or not, but I
will take a crack at this one.

There are two places I have personally read about Thistlethwaite's
algorithm.  One is Douglas Hofstadter's article in Scientific
American in March of 1981.  The article is reprinted in Hofstadter's
"Metamagical Themas".  The other is Frey and Singmaster's
"Handbook of Cubik Math".  There are probably other sources as well,
but some of them (e.g., Singmaster's Cubik Circulars) may be hard
to come by at your local library.

Any "by hand" solution to the Cube generally involves something like
"corners first, then edges" (or vice versa), or "top layer, then
middle layer, and finally the bottom layer", or (usually) some
combination or variation of these themes.  Any such theme has
states where it is visually obvious that the cube is becoming
more and more solved.

These plateau states generally have the attribute that they define
a subgroup of G.  For example, the set of states where the top layer
is solved is a subgroup of G.  The subgroups defined by the plateau
states form a nested sequence of subgroups as the Cube becomes
more and more solved.  However, progress is not monotonic.  You
almost always have to give up some of your progress temporarily on
the way to the next plateau.

Thistlethwaite's algorithm reverses the role of the plateau states
and the subgroups.  Instead of plateau states defining nested
subgroups, he has nested subgroups defining plateau states.  In fact,
his nested subgroups are arranged in such a way that once a particular
subgroup is achieved, there is no loss of progress on your way to
the next subgroup.  Also, the plateau states achieved by reaching
the next subgroup are generally not visually obvious, and indeed it
is not visually obvious that any progress at all is being made until
the cube is almost solved.

The nested subgroups given by Frey and Singmaster are as follows.
I believe that other nested subgroups have been used as well.

     H0=<L,R,F,B,U,D>=G
     H1=<L,R,F,B,U2,D2>
     H2=<L,R,F2,B2,U2,D2>
     H3=<L2,R2,F2,B2,U2,D2>
     H4=<I>

I do not recall seeing any practitioners of Thistlethwaite's algorithm
posting to Cube-Lovers.  However, there are several practitioner's
of an algorithm called Kociemba's algorithm who contribute to
Cube-Lovers.  Kociemba's algorithm is a close cousin to Thistlethwaite's,
but does not depend on pre-searching as does Thistlethwaite's.  Also,
I believe that the best Kociemba programs now utilize only two
nested subgroups, and are able to achieve results far better than
those achieved by Thistlethwaite.

As I understand it, Kociemba's algorithm differs from Thistlethwaite's
in two primary respects.  Kociemba uses searching, and Thistlethwaite
uses tables (what I called "pre-searching" in the previous paragraph).
Also, Thistlethwaite simply hooks the nested subgroups together and
stops when they are hooked, whereas Kociemba continues searching
to see if improvements are possible by merging the nested subgroups
at their boundaries.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From BRYAN@wvnvm.wvnet.edu  Tue Oct 17 19:45:07 1995
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Date:      Tue, 17 Oct 1995 19:44:40 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Positions 8q from Start, 9q from B, Five Generators

Much analysis is not done, and I have little time.  However, I
am able to print out the positions very easily.

     DDD                FFD                 RRR
     DBD                RBB                 UBR
     DRD                RRD                 RRR

     BUB                UUF                 UUU
     UUB                UUB                 UUU
     BBB                UUR                 UBU

 LLL UUU RRR        BBB LLU FDL         BBB LDL FFF
 LLB DFL BRR        FLB LFU FRR         LLB LFL FRF
 LLL UUU RRR        RRR BLU FFL         BLB LRL FFF

     FFF                DDL                 DBD
     FDF                DDL                 DDD
     FFF                DDB                 DDD

      1.                 2.                  3.

 ----------------------------------------------------

     UUF                LLL                 FFF
     UBF                LBD                 BBL
     RBR                LLL                 LLU

     DUD                UUU                 UUR
     RUF                UUU                 UUR
     FFF                UBU                 UUR

 FBU LLL DDB        FFF RLR BRB         FFF RRD BBB
 LLL DFD RRR        FLF RFR BRR         LLF RFD BRF
 LLL DDB RRR        FFF RUR BBB         RUD BRB ULL

     BBU                DBD                 LDL
     BDU                DDD                 BDD
     BFU                DDD                 DDD

      4.                 5.                  6.

 ----------------------------------------------------

     DBF                ULU                 ULL
     DBF                UBU                 BBL
     LDF                UUU                 UFF

     BFD                FFF                 BUU
     BUD                FUF                 RUU
     BBL                FFF                 RUU

 UUU RRB DRR        LLL DDD RRR         RFF DRB LLL
 LLL UFU RRR        LLB RFU BRR         LLF DFR BRF
 LLL UUB DBR        LLL DDD RRR         RUF DRR BBB

     FFR                BBB                 LDD
     FDL                BDD                 BDD
     FDU                BDB                 FDD

      7.                 8.                  9.

 ----------------------------------------------------

     LBL                FFF                 FFF
     FBD                FBF                 FBF
     FDD                FFF                 FFF

     DFB                DDD                 DDD
     DUB                UUU                 UUU
     DBB                DDD                 DDD

 LLL BUU RRR        LLL BBB RRR         LRL BBB RLR
 LLL UFU RRR        LLL BFB RRR         LLL BFB RRR
 BUU RRR DBF        LLL BBB RRR         LRL BBB RLR

     FFF                UUU                 UUU
     FDL                DDD                 DDD
     UDU                UUU                 UUU

     10.                11.                 12.

 ----------------------------------------------------

     DRR                FFF                 FFF
     RBB                FBF                 FBF
     FFF                FFF                 FFF

     UUU                UUU                 UUU
     UUB                UUU                 DUD
     RUR                UUU                 UUU

 RRD BLB UDL        RLR BBB LRL         RLR BBB LRL
 FLB UFL FRR        RLR BFB LRL         RLR BFB LRL
 BBB ULL FFF        RLR BBB LRL         RLR BBB LRL

     LDD                DDD                 DDD
     LDD                DDD                 UDU
     LDD                DDD                 DDD

     13.                14.                 15.

 ----------------------------------------------------

     FUR
     FBU
     FBU

     DUF
     RUF
     LFF

 LBU BDD RRR
 LLL DFD RRR
 LLU BLL DDD

     RBB
     UDB
     UFB

     16.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From mark.longridge@canrem.com  Wed Oct 18 01:38:43 1995
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Reply-To: CRSO.Cube@canrem.com
Sender: CRSO.Cube@canrem.com
Subject: Dino Cubes Revisited
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1251.5834.0C1F9828@canrem.com>
Date: Wed, 18 Oct 95 01:17:00 -0500
Organization: CRS Online  (Toronto, Ontario)

As I mentioned recently:

> The original Ideal colour arrangement was the tournament standard in
> the U.S. and Canada.
>
> Top  =White,  Down =Blue
> Left =Red,    Right=Orange
> Front=Yellow, Back =Green

Dan Hoey mentions:
> manufactured with opposite faces ``differing by yellow''--red opposite
> orange, blue opposite green, and yellow opposite white--

 The ``differing by yellow'' colouring was pretty common on those
"Wonderful Puzzlers".

 The only difference in original Ideal cubes colouring and the
Wonderful Puzzlers colouring is that Blue and Yellow are transposed.

I just the 3 Dino Cubes I ordered from Gametrends.
More new cubes!

There are 3 different kinds:

A) Each of the 6 faces of the cube is a unique solid colour.
B) Each of the 6 faces of the cube is a unique solid colour with four
 dinosaurs (six different dinos, 1 kind for each face)
C) Each of the 4 tetrads is a unique colour.

The Dino colouring is: (Just like the puzzlers)

TOP  =White, DOWN =Yellow
LEFT =Red,   RIGHT=Orange
FRONT=Blue,  BACK =Green

The packaging is a small white cardboard box devoid of any
trademarks. The name on the box is "Dinosaur Rubik's Cube".

All the regulars on cube-lovers will make mincemeat of this puzzle,
and the 4 colour version is easier still! The mechanism is
quite good (nice and smooth) and it is a good puzzle to have.

The 6 colour version has 19,958,400 combinations.
The 4 colour version has only 42,600 combinations.

-> Mark <-


From BRYAN@wvnvm.wvnet.edu  Wed Oct 18 20:56:25 1995
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Date:      Wed, 18 Oct 1995 20:56:03 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Re: Positions 8q from Start, 9q from B, Five Generators
In-Reply-To: Message of 10/17/95 at 19:44:40 from BRYAN@wvnvm.wvnet.edu

I can add a bit of additional information.  The 16 positions
8q from Start and 9q from B can be reduced to 4 positions
unique up to Q2-conjugacy.  As I have discussed before,
it is still difficult to claim that the 4 positions are really
"different" without further analysis because of the possibility
that the positions are variations within a commuting subsequence
of moves.

I don't really have a Q2-conjugacy program.  It would be easy to
make one, but I don't have time so I used my M-conjugacy program.
Recall that Q2={i,b,bb,bbb,rrv,rrbv,ttv,ttbv}, where b, r, and
t are whole cube rotations of the Back, Right, and Top faces,
respectively, and v is the central inversion.  For 12 of the
16 positions X the program reports Symm(X)={i}, which is to say
m'Xm is not equal X for any m in M except the identity.  Obviously,
the same is true for all m in Q2 since Q2 is a subgroup of M.  We
have |Q2|=8, so |{m'Xm | m in Q2}=6.  Therefore, the 12 positions
for which Symm(X)={i} form two Q2-conjugacy classes.

Using the M-conjugacy program for the other 4 positions is trickier,
but only slightly so.  For the other 4 positions, the M-conjugacy
program reports Symm(X)=HX, where HX={i,bb,rr,tt,v,bbv,rrv,ttv}.
But HX is not a subgroup of Q2, and what we need is sort of
"Symm(X) with respect to Q2", which I will call Symm(X/Q2).
(A better notation is probably available).  It is easy to see
that Symm(X/Q2)=(Symm(X) intersect Q2), and we have
(HX intersect Q2)={i,ttv,bb,rrv}.  This subgroup is called
HQ2 in Dan's taxonomy.

We have |Q2|=8 and |HQ2|=4, so |{m'Xm | m in HQ2}|=2
when Symm(X/Q2)=HQ2.  Therefore, the last 4 positions form
two Q2-conjugacy classes.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From geohelm@pt.lu  Thu Oct 19 03:20:44 1995
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To: cube-lovers@ai.mit.edu
From: Georges Helm <geohelm@pt.lu>
Subject: Thisthlethwaite's algorithm
Message-Id:  <9510190820.aa11995@telinf1.pt.lu>

There are two papers by Morwen Thistlethwaite on this subject, both dating
from 1980.
I got these copies from David Singmaster after having tried unsuccessfully
to get them from Morwen himself. 
I have since then made many copies of these notes for many people all around
the world. 
I will continue to do this as long as I feel comfortable about it.
So if there is still anybody out there who wants them, please e-mail me
directly at geohelm@pt.lu or at georges.helm@comnet.eo.lu
Georges Helm


From boland@sci.kun.nl  Thu Oct 19 10:21:21 1995
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To: cube-lovers@ai.mit.edu
Subject: 3-cycle on edges in group <U,R>
Date: Thu, 19 Oct 95 15:21:18 +0100
From: Michiel Boland <boland@sci.kun.nl>

URURU R'U'R'U'R'
-- 
Michiel Boland <boland@sci.kun.nl>
University of Nijmegen
The Netherlands

From bagleyd@source.asset.com  Thu Oct 19 13:27:59 1995
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From: bagleyd@source.asset.com (David A. Bagley)
Message-Id: <9510191716.AA40695@source.asset.com>
To: cube-lovers@life.ai.mit.edu
Subject: pyraminx-like puzzles (again)

Hi
Recently I asked the question:
>   I have a question, I hope this makes sence. ;)   On a "nxnxn"
> tetrahedron with period 2 or period 3 turning or a "nxnxn" octahedron with
> period 3 or period 4 turning, can the orientation of any of the center
> triangles change when the puzzle is solved?  If so, where does this
> start to happen.  I know from "experience" that this is not true on
> a pyraminx.
 
Well, if you believe proof by example on a simulated puzzle, then
Tetrahedron period 2 turning: never happens
Tetrahedron period 3 turning: starts when n=4 with center triangle
Octahedron period 3 turning: starts when n=4 with center triangle
Octahedron period 4 turning: starts with n=4 with center triangle
 
New versions of my pyraminx and octahedron puzzles will be out soon
to reflect this.
 
Cheers,
      --__---------------------------------------------------------------
     /  \ \   /           David A. Bagley                                \
    |    \ \ /            bagleyd@source.asset.com                        |
    |     \//\            Some days are better than other days.           |
    |     / \ \                -- A short lived character of Blake's 7    |
     \   /   \_\puzzles   Available at: ftp.x.org/contrib/games/puzzles  /
      -------------------------------------------------------------------

From boland@sci.kun.nl  Thu Oct 19 21:41:58 1995
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To: cube-lovers@ai.mit.edu
Subject: Embedding G in a symmetrical group
Date: Fri, 20 Oct 95 02:41:55 +0100
From: Michiel Boland <boland@sci.kun.nl>

It is clear that the group G of the cube (the one with
4.3252x10^19 elements) can be embedded in a
symmetrical group, e.g. S_48, since each move of the cube can be
seen as a permutation of 48 objects. Hence, there is a smallest
number n such that G can be embedded in S_n. I'm curious to find
out what this number is.

It can be shown with some counting arguments that n>=32 (I'm
happy to write these down but it's nicer if you thought about
this first). I would be surprised if n=32 but you never know.
-- 
Michiel Boland <boland@sci.kun.nl>
University of Nijmegen
The Netherlands

From mark.longridge@canrem.com  Fri Oct 20 02:32:34 1995
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Subject: Cube Verification
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1255.5834.0C1F9CFB@canrem.com>
Date: Fri, 20 Oct 95 02:22:00 -0500
Organization: CRS Online  (Toronto, Ontario)

In Jerry's message from
Date:      Mon, 16 Oct 1995 23:08:02 -0400 (EDT)
Subject: The Correctness of Large Seaches

> In light of the above discussion, I thought it might be appropriate
> to summarize some background about my larger searches.  I want to
> indicate which ones of them seem pretty well verified, and which ones
> of them might benefit from further study.
>
>   <U,R>           -  I believe this one is ok.  I ran the q turn case
>                      with and without conjugacy classes.  Upon
>                      expanding the conjugacy classes, the results
>                      matched the results without conjugacy classes.
>                      Also, the results matched results posted by
>                      Mark Longridge as far as they went (although
>                      my anomaly with corners-only suggests that
>                      such matching doesn't prove very much).

Well, I did get up to 12 q turns deep ;-)
Good enough for 2 half deep searches...

 But there is another possible verification method by counting the
number of even positions and odd positions and totalling them.

 Analysis of < U, R > group on 3x3x3 cube by Parity
 --------------------------------------------------

        Even Positions          Odd Positions
        --------------          -------------

        0            1          1           4
        2           10          3          24
        4           58          5         140
        6          338          7         816
        8        1,970          9       4,756
       10       11,448         11      27,448
       12       65,260         13     154,192
       14      360,692         15     827,540
       16    1,851,345         17   3,968,840
       18    7,891,990         19  13,659,821
       20   18,471,682         21  16,586,822
       22    8,039,455         23   1,511,110
       24       47,351         25          87
            ----------             ----------
            36,741,600             36,741,600

This is almost Cube Philosophy... how can we be certain about the true
nature of God's Algorithm? How can we be certain our cube databases
are completely accurate? I suppose it is not really a big problem
as long as the various cube programs all agree, and a human
observer executes such processes on a real cube.

-> Mark <-


From din5w@server.cs.virginia.edu  Fri Oct 20 06:24:11 1995
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From: Dale Newfield <din5w@virginia.edu>
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On Fri, 20 Oct 1995, Michiel Boland wrote:
> It is clear that the group G of the cube (the one with
> 4.3252x10^19 elements) can be embedded in a symmetrical group, e.g. 
> S_48, since each move of the cube can be seen as a permutation of 48 
> objects.

Um...If I were a better net.person, I'd look up which version of the cube 
has that number of elements, but wouldn't it be correct to say that each 
move of the cube is a permutation of the pieces of the cube, i.e. the 26 
cubies?  (Or even, depending on which cube-model you are using(This is 
what I should have looked up), if you ignore center cubie orientation, 
the 20 cubies?)

If that logic holds, then the largest possible S_n would be S_20, much 
less than the 32 that you claim is minimal...

...I think I'm just confused--can you alleviate that problem?

-Dale Newfield


From hazard@niksula.hut.fi  Fri Oct 20 07:00:31 1995
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To: cube-lovers@ai.mit.edu
From: Mikko Haapanen <hazard@niksula.hut.fi>
Subject: Old question about 2 adj edges

Hello!

I want to ask if somebody can tell me how to flip 2 adj. edges (and nothing
else) in 4x4x4 cube? I have my own formula to do this but i cannot write it
down because i have to discover it every time ;) and i turn whole cube so
many times during it. I remember 39 or 40 turns have been the shortest way
i've seen. What i need is that formula in BFUDLR (or something) language.

I have another question about this mailing list. I have seen many lists
similar to cube-lovers in my .nersrc file. Can i find cube-lovers list
somewhere with my newsreader?

Thanks.
------Mikko Haapanen------hazard@niksula.hut.fi---------
Jos tahto siirt   vuoren, niin tahdon siirt   tahdon
Vuorta siirt m  n, ett  vuoren sis  n n  n -P.Hanhiniemi
--------------------------------------------------------



From BRYAN@wvnvm.wvnet.edu  Fri Oct 20 09:02:16 1995
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Date:      Fri, 20 Oct 1995 08:38:41 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@life.ai.mit.edu>
Subject:   Re: Cube Verification
In-Reply-To: Message of 10/20/95 at 02:22:00 from mark.longridge@canrem.com

On 10/20/95 at 02:22:00 mark.longridge@canrem.com said:

> But there is another possible verification method by counting the
>number of even positions and odd positions and totalling them.

But my incorrect results for the 3x3x3 Corners plus face centers
passed this test, and were still incorrect.  Indeed, the incorrect
search found all the positions, the parity of all the lengths
was correct, the maximum length was correct, and the number of
positions at the maximum length was correct.  Yet, some of the
lengths were incorrect.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From BRYAN@wvnvm.wvnet.edu  Fri Oct 20 09:41:53 1995
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Date:      Fri, 20 Oct 1995 09:18:20 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <cube-lovers@ai.mit.edu>
Subject:   Re: Old question about 2 adj edges
In-Reply-To: Message of 10/20/95 at 13:00:30 from hazard@niksula.hut.fi

On 10/20/95 at 13:00:30 Mikko Haapanen said:

>I want to ask if somebody can tell me how to flip 2 adj. edges (and nothing
>else) in 4x4x4 cube?

That reminds me of a question I have been meaning to ask for a long time.
But first here is some context, involving some reminisces about when
I first figured out how to solve the cube.

After buying my first cube, it took me about a week to figure out how
to solve everything except two edges cubies which were flipped.  It took
me about another week to figure out how to unflip them.  Since then, I have
discovered that about half the time, my method of solution yields two
flipped edge cubies which have to be unflipped, and about half the time
it doesn't.  Fair enough  --  50-50 chance, I guess.  (When I am in
practice, I can see the flipped edge cubies coming, and can compensate
as a part of the process of getting the last few edge cubies in place,
but I am seldom really in practice.)

The 4x4x4 was a little tougher.  It took me about a week to figure out
how to solve everything except two edge cubies which were exchanged.
It took me about another year(!) to figure out how to exchange them.

On the 3x3x3, I solve the corners first, then the edges.  I use the
same general method on the 4x4x4, except that there are four face
centers (if you can call them that) on each face to deal with.  So
I solve the corners first, which establishes a frame of reference.
Then, I solve the face centers with respect to the frame of reference
established by the corners.  Finally, I solve the edges.  All the
operators are identical or similar to the ones I use for the 3x3x3.

For the longest time, I thought that a parity argument made one
exchange of the edges impossible, and I was just sure that somebody
was taking my cube apart and putting it back together when I wasn't
looking.  I went through two or three cycles of taking it apart
myself and putting it back together in Start, scrambling it, and
trying to solve it, all in one sitting, before I was convinced that
you really could have just one exchange of the edges.

What I didn't see originally was that there could be invisible
movement of the face centers to compensate for the "bad parity" of
the edges.  I am very poor at solving the 4x4x4, but here is how I
do it.  When I get the "bad parity", I make one slice move (doesn't
disturb the corners), and then solve the face centers again without
simply undoing the once slice move and without worrying about the
edges.  Then, upon solving the edges the second time, there is
"good parity" and all is well.

Finally, here is my question.  On the 4x4x4, I get "bad parity" darn
near every time.  Why isn't it 50-50 like the situation I have with
flips on the 3x3x3?

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From BRYAN@wvnvm.wvnet.edu  Fri Oct 20 11:18:11 1995
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Date:      Fri, 20 Oct 1995 09:37:33 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <cube-lovers@ai.mit.edu>
Subject:   Re: Embedding G in a symmetrical group
In-Reply-To: Message of 10/19/95 at 22:57:15 from din5w@virginia.edu

On 10/19/95 at 22:57:15 Dale Newfield said:


>On Fri, 20 Oct 1995, Michiel Boland wrote:
>> It is clear that the group G of the cube (the one with
>> 4.3252x10^19 elements) can be embedded in a symmetrical group, e.g.
>> S_48, since each move of the cube can be seen as a permutation of 48
>> objects.

>Um...If I were a better net.person, I'd look up which version of the cube
>has that number of elements, but wouldn't it be correct to say that each
>move of the cube is a permutation of the pieces of the cube, i.e. the 26
>cubies?  (Or even, depending on which cube-model you are using(This is
>what I should have looked up), if you ignore center cubie orientation,
>the 20 cubies?)

>If that logic holds, then the largest possible S_n would be S_20, much
>less than the 32 that you claim is minimal...

You are forgetting the twists of the corner cubies and the flips of the
edge cubies.

As an aside, the S_48 upper bound is already based on ignoring the face
centers (i.e., 8 facelets on each of 6 faces of the cube).

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From hazard@niksula.hut.fi  Fri Oct 20 12:29:49 1995
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To: cube-lovers@ai.mit.edu
From: Mikko Haapanen <hazard@niksula.hut.fi>
Subject: Re: Old question about 2 adj edges

>>I want to ask if somebody can tell me how to flip 2 adj. edges (and nothing
>After buying my first cube, it took me about a week to figure out how
>The 4x4x4 was a little tougher.  It took me about a week to figure out
>how to solve everything except two edge cubies which were exchanged.
>It took me about another year(!) to figure out how to exchange them.

It took me about 6 years. That problem was a little background process in my
brains. Then i went to the army. In the evenings i used to 'play' with 4x4x4
until one day. I noticed the same thing as you. 2 flipped edges can be
solved by turning one inner slice 90 degrees. Since then i haven't
discovered anythig new about this parity problem. That's why i asked if
anyone could do it less than 39 turns.

>What I didn't see originally was that there could be invisible
>movement of the face centers to compensate for the "bad parity" of
>the edges.  I am very poor at solving the 4x4x4, but here is how I

I have thought the center piece movements too. It must be very hard to see a
parity from certers, even if they were marked (position and orientation).
There are so many different 'right' positions too.

This reminds me another old question: 3x3x3 are told to have about 4
trillion (or whatever) different positions. How many of these positions are
'solved cube' but with different centerpiece combinations? Once i had 3x3x3
with 6 different pictures (picture/side). Friends asked me to solve it. When
i was completed, they laughed at me and pointed the bottom center piece,
which was out of orientation (i can't remember how many of centers were out
of order).

>Finally, here is my question.  On the 4x4x4, I get "bad parity" darn
>near every time.  Why isn't it 50-50 like the situation I have with

I don't know, but i guess: 1/4 you get it right and 3/4 goes wrong. I think
we must count that there are 2 inner slice states: right and wrong. There
are also 2 outer slice states: right and wrong (2 corners at wrong
position). It makes me think that 1/4.
------Mikko Haapanen------hazard@niksula.hut.fi---------
Jos tahto siirt   vuoren, niin tahdon siirt   tahdon
Vuorta siirt m  n, ett  vuoren sis  n n  n -P.Hanhiniemi
--------------------------------------------------------



From geohelm@pt.lu  Fri Oct 20 12:46:33 1995
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To: cube-lovers@ai.mit.edu
From: Georges Helm <geohelm@pt.lu>
Subject: Re: Old question about 2 adj edges
Message-Id:  <9510201711.aa15691@telinf1.pt.lu>

 how to flip 2 adj. edges (and nothing else) in 4x4x4 cube?

r^2 U^2 r l' U^2 r' U^2 r U^2 r l U^2 l' U^2 r U^2 l r^2 U^2

Georges
geohelm@pt.lu


From news@nntp-server.caltech.edu  Fri Oct 20 14:20:26 1995
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Subject: Re: Old question about 2 adj edges
Date: 20 Oct 1995 18:20:09 GMT
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"Jerry Bryan" <BRYAN@wvnvm.wvnet.edu> writes:
>Finally, here is my question.  On the 4x4x4, I get "bad parity" darn
>near every time.  Why isn't it 50-50 like the situation I have with
>flips on the 3x3x3?

My suspicion is that "bad" or "good" parity is determined by what your
formulae are composed of.  An even number of center-plane moves will
preserve parity, and the opposite is true for odd center-plane moves.

If there really is a reason you're getting bad parity moves, that could
determine it.

However, I ran into a straight of 5 "good parity" situations recently,
and didn't really notice it until later, so perhaps it's a psychological
phenomenon as well (we gloss over good happenings, and dwell on tragedies.)

-- 
int m,u,e=0;float l,_,I;main() {for(;e<1863; putchar((++e>924&&952>
e?60-m:u) ["\n  ude.hcetlac.occ@gnauhw   ]"])) for(u=_=l=0;(m=e%81)
<80&&I*l+_*_ <6&&25>++u;_=2*l*_+e/81*.09-1,l=I)I=l*l-_*_-2+.035*m;}

From news@nntp-server.caltech.edu  Fri Oct 20 17:28:13 1995
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From: whuang@cco.caltech.edu (Wei-Hwa Huang)
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Subject: Re: Old question about 2 adj edges
Date: 20 Oct 1995 21:27:27 GMT
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Mikko Haapanen <hazard@niksula.hut.fi> writes:
>This reminds me another old question: 3x3x3 are told to have about 4
>trillion (or whatever) different positions. How many of these positions are
>'solved cube' but with different centerpiece combinations? Once i had 3x3x3
>with 6 different pictures (picture/side). Friends asked me to solve it. When
>i was completed, they laughed at me and pointed the bottom center piece,
>which was out of orientation (i can't remember how many of centers were out
>of order).

Actually, I think the 4 "trillion" estimate is ignoring the center orientation.
Let's see:
  8! corner positions
x 3^7 corner orientations
x 12!/2 edge positions
x 2^11 edge orientations
= 4.325x10^19
Well, forty-three quadrillion.

Five center orientations force the sixth, so multiply your number by 4^5 to 
get the answer 4.429x10^22 positions, counting center piece 
orientations.  That's 44 quintillion.  Whew.

I remember when I solved the 5x5x5 cube (finally), someone asked me if I
had solved the "invisible" 3x3x3 inside it.  I'm not sure I even want to
think of trying to solve that.  I'll work on the 3x3x3x3 first.  :P

-- 
int m,u,e=0;float l,_,I;main() {for(;e<1863; putchar((++e>924&&952>
e?60-m:u) ["\n  ude.hcetlac.occ@gnauhw   ]"])) for(u=_=l=0;(m=e%81)
<80&&I*l+_*_ <6&&25>++u;_=2*l*_+e/81*.09-1,l=I)I=l*l-_*_-2+.035*m;}

From scotth@ssd.intel.com  Fri Oct 20 19:17:28 1995
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To: Michiel Boland <boland@sci.kun.nl>
Cc: cube-lovers@ai.mit.edu
Subject: Re: Embedding G in a symmetrical group 
In-Reply-To: Your message of "Fri, 20 Oct 95 02:41:55 BST."
             <199510200141.CAA03659@wn1.sci.kun.nl> 
Date: Fri, 20 Oct 95 16:17:04 -0700
From: Scott Huddleston <scotth@ssd.intel.com>


>It is clear that the group G of the cube (the one with
>4.3252x10^19 elements) can be embedded in a
>symmetrical group, e.g. S_48, since each move of the cube can be
>seen as a permutation of 48 objects. Hence, there is a smallest
>number n such that G can be embedded in S_n. I'm curious to find
>out what this number is.
>
>It can be shown with some counting arguments that n>=32 (I'm
>happy to write these down but it's nicer if you thought about
>this first). I would be surprised if n=32 but you never know.
>-- 
>Michiel Boland <boland@sci.kun.nl>
>University of Nijmegen
>The Netherlands

My permutation group memory is rusty.  Is it the "degree" of G you're
asking for?

I also get n = 32 as the smallest |S_n| divisible by |G|.  I suspect
one can argue |G| must divide |A_n| (still requiring only n >= 32),
and if we can argue |G|*12 must divide |S_n| (accounting for all
parities of edges and corners) we get n >= 33.

On the flip side, I'll assert G has degree <= 42 (which is less than
the obvious representation in S_48).  If anyone can prove me either
right or wrong, please do so.  My assertion is based on the following
hand-waving argument:

G is a wreath product (or some kind of product) of the following
subgroups:
A:  S_8   8! corner positions
B:  3^7   3^7 corner orientations
C:  A_12  12!/2 edge positions
D:  2^11  2^11 edge orientation

Clearly subgroup A has degree 8 and C has degree 12.  

I claim (wave hands wildly:-) that BxD has degree at most 22, since it
can be embedded in an S_22.  I use every even factor of 22! for a component
of 2^11, and every third factor of 22! for a component of 3^7.  

Divisibility arguments suggest some smaller S_n might embed 2^11x3^7,
but I don't know whether such embeddings can be realized.  This is an
obvious area to consider for lowering the upper bound on degree(G).  By
divisibility we can conceivably embed 2^11x3^7 in an S_18, but not in
an S_17.

Finally, the degree of a wreath(?) product is bounded by the sum of
the degrees of the multiplicands, hence 
degree(G) <= 8 + 12 + 22 = 42.

If this (alleged) degree 42 construction is valid, is 42 minimal?  I
strongly doubt it.



From mark.longridge@canrem.com  Sat Oct 21 22:24:10 1995
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Subject: Spotty Megaminx
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1256.5834.0C1FA20D@canrem.com>
Date: Sat, 21 Oct 95 22:12:00 -0500
Organization: CRS Online  (Toronto, Ontario)

Observations on the Magic Dodecahedron (Megaminx)
-------------------------------------------------

 I've never seen anything on patterns for the megaminx, with the
sole exception of Kurt Endl's book "Megaminx". Unfortunately there are
no detailed examples, only vague references to "many possible dot
patterns" and "star patterns". A pattern similar to the 6 X order 3
of the cube is on the cover, but only part of the dodecahedron is
visible.

 Using the solving skills I developed myself, I deliberately solved
the megaminx with the centres not matching the surrounding face.
Techniques like mono-twists and mono-flips carried over well from
the cube.

My conclusion: A 10-dot pattern is possible!

 Here is a description....

 One pair of opposite faces is completely solid. The 5 faces adjacent
to solid face A are spotted, also the 5 faces adjacent to solid face
B (opposite to A) are spotted. If we look at one set of 5 faces we
can observe that in this particular 10-spot that the 5 centres
appear rotated to the left, or (since the centres don't really move
in position) that the rest of the face is moved to the right.

 Similarly, in the lower tier of 5 faces, we can observe that 5 centres
appear rotated to the left also.

 Let's try a small thought experiment. Imagine a skeleton, a
disassembled megaminx. Grab the top and bottom with thumb and
forefinger. Now, while keeping the top and bottom centres immobile,
rotate the rest of the puzzle. What happens? The 10 other centres
rotate in the same direction! If we do this on a cube skeleton the
same thing happens, but on a fleshed out cube this would become a
4 cycle of centres, which is in the swap orbit and can't be reached by
face turns. On the megaminx we have 2 five cycles of centres, and this
is legal.

 There are 6 opposite pairs of faces on the megaminx. There are 4 ways
to rotate the centres for each pair to generate a 10 spot. I'll
speculate that there are 6*4 = 24 possible 10-spots.

 I suspect various 12-spots are possible. I have no idea how to
easily permute centre pieces on the megaminx.

-> Mark <-


From dik@cwi.nl  Sat Oct 21 22:52:22 1995
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From: Dik.Winter@cwi.nl
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Subject: Re: Spotty Megaminx
Content-Length: 1116

 >  I've never seen anything on patterns for the megaminx, with the
 > sole exception of Kurt Endl's book "Megaminx".

It is long ago I had it in my hands, and I have no books.  What I say
is from memory; probably correct.  Note that a face turn induces an
even permutation on both the corner and the edge "cubies".  So odd
permutations are not possible.  On the other hand (if I remember well)
*all* combinations of even permutations are possible.

 >  There are 6 opposite pairs of faces on the megaminx. There are 4 ways
 > to rotate the centres for each pair to generate a 10 spot. I'll
 > speculate that there are 6*4 = 24 possible 10-spots.

Right.

 >  I suspect various 12-spots are possible. I have no idea how to
 > easily permute centre pieces on the megaminx.

Indeed.  Every rotation of the center skeleton is possible (if you
consider the remainder fixed...).  So there are 12 centers that can
come out at top; for each center at top you have 5 possible positions
of the remainder leading to 60 configurations.  Of these 24 are
10-spots, 1 is the solved puzzle, so the remainder (35) is 12-spots.

dik

From mreid@ptc.com  Mon Oct 23 11:20:02 1995
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Date: Mon, 23 Oct 1995 11:42:32 -0400
From: michael reid <mreid@ptc.com>
To: boland@sci.kun.nl, cube-lovers@ai.mit.edu
Subject: Re:  Embedding G in a symmetrical group

michiel boland writes

> It is clear that the group G of the cube (the one with
> 4.3252x10^19 elements) can be embedded in a
> symmetrical group, e.g. S_48, since each move of the cube can be
> seen as a permutation of 48 objects. Hence, there is a smallest
> number n such that G can be embedded in S_n. I'm curious to find
> out what this number is.

48.

first note that any homomorphism  G  -->  S_n  can be factored as

           G  -->  S_m_1 x S_m_2 x ... x S_m_k  >-->  S_n

where  m_1, m_2, ... , m_k  are the sizes of the orbits of  G  acting
on  {1, 2, ... , n},  and thus  m_1 + m_2 + ... + m_k = n.
furthermore, the action of  G  on each  {1, 2, ... , m_i}  is
transitive.

transitive  G-sets  are easy to understand.  for any subgroup  H  of  G,
G  acts transitively on the cosets  G/H  by left multiplication.
also, any transitive  G-set  is of this form.  given a homomorphism
G  -->  S_m  with a transitive action, let  H  be the subgroup of  G
that fixes the element  1.  then it's easy to see that the cosets  G/H
are in one-to-one correspondence with elements in the orbit of  1
(which by hypothesis are all of  1, 2, ... , m)  and the action of
G  on  G/H  is isomorphic to the action of  G  on  {1, 2, ... , m}.

the kernel of the homomorphism  G --> sym(G/H)  is the largest normal
subgroup of  G  contained in  H , which is just the intersection of
all  G-conjugates  of  H.

of course, in this case we have  m = (G : H)  (index of  H  in  G).
thus michiel's question can be settled by considering all subgroups of
G  with index less than 48.

unless i've overlooked some, there are exactly 8 such, up to  G-conjugacy.
they are

     G     itself
     G'    = commutator subgroup of  G  =  subgroup of positions an even
             number of quarter turns from start
     C_0   = subgroup where the corner  UFR  is in place, but may be twisted
     C'_0  = commutator subgroup of  C_0  =  intersection of  C_0  and  G'
     E_0   = subgroup where the edge  UR  is in place, but may be flipped
     E'_0  = commutator subgroup of  E_0  =  intersection of  E_0  and  G'
     C_1   = subgroup where the corner  UFR  is in place and is not twisted
     E_1   = subgroup where the edge  UR  is in place and is not flipped.

for each of these, except the last two, the kernel of  G --> sym(G/H)
contains all elements that only flip edges in place and twist corners
in place.
                                                                   number of
  subgroup   index       kernel                                   conjugates

     G         1           G                                           1
     G'        2           G'                                          1
     C_0       8      {all corners in place, may be twisted}           8
     C'_0     16      {all corners in place, may be twisted}           8
     E_0      12      {all edges in place, may be flipped}            12
     E'_0     24      {all edges in place, may be flipped}            12
     C_1      24      {all corners in place, may not be twisted}       8
     E_1      24      {all edges in place, may not be flipped}        12

thus the only way to get and embedding (i.e. injective homomorphism)
G --> S_n  using the subgroups above is

            G --> sym(G/C_1) x sym(G/E_1) >--> S_48

which in fact, is just the action of  G  on the 48 non-center facelets.

i had previously stumbled across this exact same question, so 
now i'm curious: why are you interested in this?  

mike

From bagleyd@source.asset.com  Mon Oct 23 13:57:59 1995
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From: bagleyd@source.asset.com (David A. Bagley)
Message-Id: <9510231733.AA15823@source.asset.com>
To: cube-lovers@life.ai.mit.edu
Subject: pyraminx-like puzzles (yet again)

Hi
> Recently I asked the question:
> >   I have a question, I hope this makes sence. ;)   On a "nxnxn"
> > tetrahedron with period 2 or period 3 turning or a "nxnxn" octahedron with
> > period 3 or period 4 turning, can the orientation of any of the center
> > triangles change when the puzzle is solved?  If so, where does this
> > start to happen.  I know from "experience" that this is not true on
> > a pyraminx.
 
 The reported to answer to * was incorrect
> Well, if you believe proof by example on a simulated puzzle, then
> Tetrahedron period 2 turning: never happens
> Tetrahedron period 3 turning: starts when n=4 with center triangle
> Octahedron period 3 turning: starts when n=4 with center triangle *
> Octahedron period 4 turning: starts with n=4 with center triangle
 
It should be:
Octahedron period 3 turning: starts when n=2 with center triangle
 
The case where n = 3 (here there is no one center triangle) was interesting
because there seemed to be no easy repetition of moves where the colors of
the puzzle would be solved but the orientation of the triangles would
be changed.
 
     /\
    /__\
   /\C /\
  /__\/__\
 /\C /\C /\
/__\/__\/__\
 
After much experimentation, I found a way of rotating
3 center triangles on a face which involved 216 moves, (this can be bettered
by one noting that 2 clockwise rotations = 1 counterclockwise rotation):
 
Repeat 5 times
{
  With reference to the top "C" in diagram, turn center to the right and
    then rotate face clockwise for a total of 42 moves.  One will then get
    a pattern where only 3 center colors are out of place on 3 different faces.
  rotate this face clockwise
}
rotate this face clockwise
 
New versions of my pyraminx and octahedron puzzles are now out.
Cheers,
clockwise
 
New versions of my pyraminx and octahedron puzzles are now out.
Cheers,
      --__---------------------------------------------------------------
     /  \ \   /           David A. Bagley                                \
    |    \ \ /            bagleyd@source.asset.com                        |
    |     \//\            Some days are better than other days.           |
    |     / \ \                -- A short lived character of Blake's 7    |
     \   /   \_\puzzles   Available at: ftp.x.org/contrib/games/puzzles  /
      -------------------------------------------------------------------

From hazard@niksula.hut.fi  Mon Oct 23 14:42:32 1995
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Date: Mon, 23 Oct 1995 20:42:30 +0200
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To: cube-lovers@ai.mit.edu
From: Mikko Haapanen <hazard@niksula.hut.fi>
Subject: pull out the corner?

Hello!

I have a question (yes, again). This subject may be discussed here before,
but i don't understand set theory or other high math, so i ask:

If i had a 3x3x3 cube and i pull out a corner piece. I turn it and push
back. Now the cube cannot be solved. I think the cube is now 'on the other
orbit'. If i pull now an edge piece and flip it, the cube is again on some
other orbit.

Only one of those orbits are legal. How many different illegal orbits there are?

-----Mikko Haapanen------hazard@niksula.hut.fi------
Another toy will help destroy The elder race of man
Forget about your silly whim It doesn't fit the plan
----------------------------------------------------


From BRYAN@wvnvm.wvnet.edu  Mon Oct 23 16:38:39 1995
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Date:      Mon, 23 Oct 1995 16:38:12 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <cube-lovers@ai.mit.edu>
Subject:   Re: pull out the corner?
In-Reply-To: Message of 10/23/95 at 20:42:30 from hazard@niksula.hut.fi

On 10/23/95 at 20:42:30 Mikko Haapanen said:

>I have a question (yes, again). This subject may be discussed here before,
>but i don't understand set theory or other high math, so i ask:

>If i had a 3x3x3 cube and i pull out a corner piece. I turn it and push
>back. Now the cube cannot be solved. I think the cube is now 'on the other
>orbit'. If i pull now an edge piece and flip it, the cube is again on some
>other orbit.

>Only one of those orbits are legal. How many different illegal orbits there are
?

In the terms you are using, there are 12 orbits.  Of these, 1 is
"legal" (contains Start), and 11 are "illegal" (do not contain
Start).

There is a factor of 3 from twisting the corners.  Pull out a corner
piece.  There are 3 ways to put it back in.  You can put it back in
the way it came out, you can twist it right, or you can twist it left.

There is a factor of 2 from flipping the edges.  Pull out an edge
piece.  There are 2 ways to put it back in, flipped or unflipped.

There is a factor of 2 from parity.  The edges can be said to be in
even parity or in odd parity, and the corners can be said to be in
even parity or odd parity.  Normally, the corners and edges are in
the same parity.  A quarter turn changes the parity both for the
edges and for the corners.  But pull out 2 edges pieces (or 2 corner
pieces).  Put them back where they came from, and their parity
remains the same.  Exchange them, and their parity changes.

We therefore have 12=3x2x2.

However (and draw a deep breath), for every expert there is an equal
and opposite expert.  This use of the term "orbit" agrees with some
experts.  However, other experts would say that the corners form an
orbit, that the edges form an orbit, and that the face centers form
an orbit.

I don't know which use of the term orbit is correct (perhaps both are
in the proper context).  But in any case, if you take a cube apart,
there are 12 disjoint sets of positions that you choose from when you
put the cube back together.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

From geohelm@pt.lu  Tue Oct 24 11:31:19 1995
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To: cube-lovers@ai.mit.edu
From: Georges Helm <geohelm@pt.lu>
Subject: availability of cubes and other puzzles
Message-Id:  <9510241349.aa24228@telinf1.pt.lu>

Here is the address of a German friend who is still selling the
following puzzles (among others):
5x5x5, skewb, magic dodecahedron, German calender cube, pyraminx...

Christoph Bandelow
An der Wabeck 37
D-58456 Witten
Germany

Tel.: ++49-2302-71147
Fax : ++49-2302-77001

Books he is selling (among others):
Bandelow: Inside Rubik's Cube and beyond (~$15)
Singmaster: Notes (~$10)

Georges Helm


From boland@sci.kun.nl  Tue Oct 24 17:31:16 1995
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To: Mikko Haapanen <hazard@niksula.hut.fi>
Cc: cube-lovers@ai.mit.edu
Subject: Re: pull out the corner? 
In-Reply-To: Your message of "Mon, 23 Oct 95 20:42:30 +0200."
             <199510231842.UAA28308@nukkekoti.cs.hut.fi> 
Date: Tue, 24 Oct 95 22:31:00 +0100
From: Michiel Boland <boland@sci.kun.nl>

Mikkao Haapanen writes:

>If i had a 3x3x3 cube and i pull out a corner piece. I turn it and push
>back. Now the cube cannot be solved. [...]

This has nothing to do with his question, but one of my old
cubes has become so loose that it has become quite easy to twist
a single corner piece - no doubt other people have expierenced
this phenomenon.

The last stage in my cube-solving algorithm used to be orienting
the corners - this has now become trivial. :)
-- 
Michiel Boland <boland@sci.kun.nl>
University of Nijmegen
The Netherlands

From boland@sci.kun.nl  Mon Oct 30 07:48:01 1995
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To: cube-lovers@ai.mit.edu
Subject: Exchanging just four edges in antislice impossible?
Date: Mon, 30 Oct 95 13:47:59 +0100
From: Michiel Boland <boland@sci.kun.nl>

Hello all,

can anyone provide an easy proof of the fact that it is
impossible to exchange just four edges using just antislice
moves, whilst leaving everything else fixed? (We're talking
about the 3x3 cube of course.)

Another way of putting it: why are the 2xH and 4-dot patterns
not in the antislice group?

I have thought about this a little, but not hard enough to find
an answer. I looked it up in Singmaster's Notes but could not
find a satisfying explanation either.

Here is some more background.

The antislice group is contained in the group of all positions
that are symmetric under `cube half-turns' (the subgroup of M
containing I,(FB)(LR),(FB)(UD) and (UD)(LR)). This group has
(8*4*12*8*4*3*2^2)/2 = 73728 elements.

It can be shown that in the antislice group, the orientation of
the corners is determined by the edge positions [I am willing to
explain this, but it is much easier visualized than written
down], which means that the antislice group contains at most
73728/3=24576 elements.

But apparently the antislice group contains just 6144 elements,
which is a factor 4 below the abovementioned number. This factor
4 is explained by the fact above, which I am trying to prove.
-- 
Michiel Boland <boland@sci.kun.nl>
University of Nijmegen
The Netherlands

From mark.longridge@canrem.com  Tue Oct 31 01:27:02 1995
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Subject: Spotty Megaminx Revisited
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1257.5834.0C1FBDC8@canrem.com>
Date: Tue, 31 Oct 95 01:02:00 -0500
Organization: CRS Online  (Toronto, Ontario)

Notes on the Spot Patterns on the Megaminx
------------------------------------------

Number the faces of the megaminx 1 through 12. Here are all the
possible permutations of the 12 centres:

dod := Group(
   (2,3,4,5,6) (7,8,9,10,11),
   (1,4,10,9,2)(5,11,12,8,6)
  );;

Size (dod) = 60;

NumberConjugacyClasses (dod) = 5;

Elements (dod);

[ (),                                             0   spot
 ( 2, 3, 4, 5, 6)( 7, 8, 9,10,11),                2   5-cycles = 10
 ( 2, 4, 6, 3, 5)( 7, 9,11, 8,10),                2   5-cycles = 10
 ( 2, 5, 3, 6, 4)( 7,10, 8,11, 9),                2   5-cycles = 10
 ( 2, 6, 5, 4, 3)( 7,11,10, 9, 8),                2   5-cycles = 10
 ( 1, 2)( 3, 6)( 4, 8)( 5, 9)( 7,10)(11,12),      6   2-cycles = 12
 ( 1, 2, 3)( 4, 6, 9)( 5, 8,10)( 7,12,11),        4   3-cycles = 12
 ( 1, 2, 6)( 3, 8, 5)( 4, 9, 7)(10,12,11),        4   3-cycles = 12
 ( 1, 2, 8, 7, 5)( 3, 9,12,11, 4),                2   5-cycles = 10
 ( 1, 2, 9,10, 4)( 5, 6, 8,12,11),                2   5-cycles = 10
 ( 1, 3, 2)( 4, 9, 6)( 5,10, 8)( 7,11,12),        4   3-cycles = 12
 ( 1, 3, 9, 8, 6)( 4,10,12, 7, 5),                2   5-cycles = 10
 ( 1, 3)( 2, 4)( 5, 9)( 6,10)( 7,12)( 8,11),      6   2-cycles = 12
 ( 1, 3,10,11, 5)( 2, 9,12, 7, 6),                2   5-cycles = 10
 ( 1, 3, 4)( 2,10, 5)( 6, 9,11)( 7, 8,12),        4   3-cycles = 12
 ( 1, 4,10, 9, 2)( 5,11,12, 8, 6),                2   5-cycles = 10
 ( 1, 4,11, 7, 6)( 2, 3,10,12, 8),                2   5-cycles = 10
 ( 1, 4, 3)( 2, 5,10)( 6,11, 9)( 7,12, 8),        4   3-cycles = 12
 ( 1, 4, 5)( 2,10, 7)( 3,11, 6)( 8, 9,12),        4   3-cycles = 12
 ( 1, 4)( 2,11)( 3, 5)( 6,10)( 7, 9)( 8,12),      6   2-cycles = 12
 ( 1, 5, 7, 8, 2)( 3, 4,11,12, 9),                2   5-cycles = 10
 ( 1, 5, 6)( 2, 4, 7)( 3,11, 8)( 9,10,12),        4   3-cycles = 12
 ( 1, 5,11,10, 3)( 2, 6, 7,12, 9),                2   5-cycles = 10
 ( 1, 5, 4)( 2, 7,10)( 3, 6,11)( 8,12, 9),        4   3-cycles = 12
 ( 1, 5)( 2,11)( 3, 7)( 4, 6)( 8,10)( 9,12),      6   2-cycles = 12
 ( 1, 6, 2)( 3, 5, 8)( 4, 7, 9)(10,11,12),        4   3-cycles = 12
 ( 1, 6, 8, 9, 3)( 4, 5, 7,12,10),                2   5-cycles = 10
 ( 1, 6)( 2, 5)( 3, 7)( 4, 8)( 9,11)(10,12),      6   2-cycles = 12
 ( 1, 6, 5)( 2, 7, 4)( 3, 8,11)( 9,12,10),        4   3-cycles = 12
 ( 1, 6, 7,11, 4)( 2, 8,12,10, 3),                2   5-cycles = 10
 ( 1, 7, 2, 5, 8)( 3,11, 9, 4,12),                2   5-cycles = 10
 ( 1, 7, 9)( 2, 6, 8)( 3, 5,12)( 4,11,10),        4   3-cycles = 12
 ( 1, 7,10)( 2, 8, 9)( 3, 6,12)( 4, 5,11),        4   3-cycles = 12
 ( 1, 7)( 2,11)( 3,12)( 4, 8)( 5, 6)( 9,10),      6   2-cycles = 12
 ( 1, 7, 4, 6,11)( 2,12, 3, 8,10),                2   5-cycles = 10
 ( 1, 8, 3, 6, 9)( 4, 7,10, 5,12),                2   5-cycles = 10
 ( 1, 8)( 2, 6)( 3, 7)( 4,12)( 5, 9)(10,11),      6   2-cycles = 12
 ( 1, 8, 5, 2, 7)( 3,12, 4, 9,11),                2   5-cycles = 10
 ( 1, 8,10)( 2, 9, 3)( 4, 6,12)( 5, 7,11),        4   3-cycles = 12
 ( 1, 8,11)( 2,12, 4)( 3, 9,10)( 5, 6, 7),        4   3-cycles = 12
 ( 1, 9, 6, 3, 8)( 4,12, 5,10, 7),                2   5-cycles = 10
 ( 1, 9)( 2, 3)( 4, 8)( 5,12)( 6,10)( 7,11),      6   2-cycles = 12
 ( 1, 9, 7)( 2, 8, 6)( 3,12, 5)( 4,10,11),        4   3-cycles = 12
 ( 1, 9, 4, 2,10)( 5, 8,11, 6,12),                2   5-cycles = 10
 ( 1, 9,11)( 2,12, 5)( 3,10, 4)( 6, 8, 7),        4   3-cycles = 12
 ( 1,10, 8)( 2, 3, 9)( 4,12, 6)( 5,11, 7),        4   3-cycles = 12
 ( 1,10, 2, 4, 9)( 5,12, 6,11, 8),                2   5-cycles = 10
 ( 1,10, 7)( 2, 9, 8)( 3,12, 6)( 4,11, 5),        4   3-cycles = 12
 ( 1,10)( 2,11)( 3, 4)( 5, 9)( 6,12)( 7, 8),      6   2-cycles = 12
 ( 1,10, 5, 3,11)( 2,12, 6, 9, 7),                2   5-cycles = 10
 ( 1,11, 8)( 2, 4,12)( 3,10, 9)( 5, 7, 6),        4   3-cycles = 12
 ( 1,11, 9)( 2, 5,12)( 3, 4,10)( 6, 7, 8),        4   3-cycles = 12
 ( 1,11, 3, 5,10)( 2, 7, 9, 6,12),                2   5-cycles = 10
 ( 1,11, 6, 4, 7)( 2,10, 8, 3,12),                2   5-cycles = 10
 ( 1,11)( 2,12)( 3, 7)( 4, 5)( 6,10)( 8, 9),      6   2-cycles = 12
 ( 1,12)( 2, 7)( 3,11)( 4,10)( 5, 9)( 6, 8),      6   2-cycles = 12
 ( 1,12)( 2, 8)( 3, 7)( 4,11)( 5,10)( 6, 9),      6   2-cycles = 12
 ( 1,12)( 2, 9)( 3, 8)( 4, 7)( 5,11)( 6,10),      6   2-cycles = 12
 ( 1,12)( 2,10)( 3, 9)( 4, 8)( 5, 7)( 6,11),      6   2-cycles = 12
 ( 1,12)( 2,11)( 3,10)( 4, 9)( 5, 8)( 6, 7)       6   2-cycles = 12

Number    Pattern
------    -------

 1        0 spots
24        2 five-cycles  (10 spot)
15        6 two-cycles   (12 spot)
20        4 three-cycles (12 spot)
--
60   orientations of the dodecahedron, 24 ten-spots, 35 twelve-spots


>>  I suspect various 12-spots are possible. I have no idea how to
>> easily permute centre pieces on the megaminx.
>
> Indeed.  Every rotation of the center skeleton is possible (if you
> consider the remainder fixed...).  So there are 12 centers that can
> come out at top; for each center at top you have 5 possible positions
> of the remainder leading to 60 configurations.  Of these 24 are
> 10-spots, 1 is the solved puzzle, so the remainder (35) is 12-spots.
> dik

 Well, I was confused how there could be 35 twelve-spots (at first),
but I am happy to confirm Dik's memory.

-> Mark <-


From alan@curry.epilogue.com  Fri Nov  3 02:39:27 1995
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From: Alan Bawden <Cube-Lovers-Request@ai.mit.edu>
Sender: Cube-Lovers-Request@ai.mit.edu
To: Cube-Lovers@ai.mit.edu
In-Reply-To: Kathryn Kelly's message of Fri, 3 Nov 1995 01:00:07 -0500 (EST) <Pine.3.89.9511030038.A3828-0100000@ixc>
Subject: Magazine Spam

First off, let me apologize for sending a message that has nothing
whatsoever to do with Rubik's Cube.  But I know from private electronic
mail that many of you were very annoyed by the six copies of the "magazine
club" advertisement that were distributed through Cube-Lovers during the
last month.  I've been urging people to just sit tight and ignore the
messages, because it wasn't entirely clear where the mail was actually
coming from.  All that appeared certain was that the ads were being
gatewayed through an Internet Service Provider in New York named "American
Network, Inc" (domain name IXC.NET).

Recently I've been talking to the responsible folks at American Network.
They are quite apologetic about the whole thing, but apparently the
miscreant is working through accounts that they give away free for the
asking.  So they actually have no way to track the jerk down.  They don't
even know how many of their accounts might all belong to the same person!
This whole experience has apparently woken them up to what a -bad- idea
that is, and they're going to stop.

Besides cleaning up their act about granting free anonymous accounts,
they also tracked down the post office box number that the miscreant uses
for his business.  And they have a request:

   Date: Fri, 3 Nov 1995 01:00:07 -0500 (EST)
   From: Kathryn Kelly <kkelly@ixc.ixc.net>
   ...
   We wish you to do something for us.
   We have the address where the individual sending out this
   material receives his physical mail.
   Please send a letter of complaint to the postmaster at
      Postmaster
      Staten Island NY 10312
   Regarding the person at this address:
      Magazine Club Inquiry Center
      Att. Internet Services Department
      P. O. Box 120990
      Staten Island NY 10312 0990

So those of you who really want to complain to somebody, here's your
chance.  Compose a nice letter to the Postmaster at the address above
explaining that the folks running the business at that P.O.Box are engaging
in anti-social behavior on the Internet.  Unfortunately, I don't believe
there is anything -illegal- about what this jerk is doing (although it is
closely analogous to some things that are illegal to do with a telephone),
so it wont work to demand that the Postmaster actually -do- anything, but
work up a good complaint anyway.

Finally, I urge you all -not- to respond to this message in public.  If you
have further thoughts on Internet advertising, electronic mailing list
administration, or clever acts of revenge, you can send them to -me-, but
don't CC your message to Cube-Lovers as a whole.  The whole point here is
to keep Cube-Lovers relatively free of off-topic mail.  As the list
administrator I get to send out occasional administrivia such as this
message because I do actual -work- to keep the list running.  

And yes, I am still working on some new list management technology that
should eliminate problems like this in the future.  (And yes, I know all
about MajorDomo, listserv, and their relatives -- you don't need to
enlighten me about them.)

			- Alan (Cube-Lovers-Request@AI.MIT.EDU)

(I never imagined when Dave Plummer and I started this list in 1980 that
one day I'd be spending significant time trying to prevent people from
using it to sell magazines.  Amazing.)

From mark.longridge@canrem.com  Sun Nov  5 02:14:26 1995
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Subject: Halpern's Tetrahedron
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1258.5834.0C1FCC4E@canrem.com>
Date: Sun,  5 Nov 95 01:59:00 -0500
Organization: CRS Online  (Toronto, Ontario)

# Ben Halpern's Tetrahedron
# 4 faces rotate

tetra := Group(
  (1,3,5)(2,4,6)(7,13,24)(12,18,23)(11,17,22),
  (7,9,11)(8,10,12)(22,15,3)(2,21,14)(1,20,13),
  (13,15,17)(14,16,18)(11,20,5)(10,19,4)(9,24,3),
  (22,20,24)(21,19,23)(7,15,5)(8,16,6)(9,17,1)
);;

# Size (tetra) = 3,732,480;
#
# Centre (tetra) = (2,12)(4,18)(6,23)(8,21)(10,14)(16,19);
#
# Tetrahedron has 6 edges, 4 corners
#  6! /2 * 2^5 * 4!/2 * 3^3

Just a few more combinations than a 2x2x2 pocket cube...
 -> Mark <-


From rjh@on-ramp.ior.com  Mon Nov  6 22:09:28 1995
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To: Cube-Lovers@ai.mit.edu
From: rjh@on-ramp.ior.com (RonH)
Subject: Rubik's stuff

Saw your message in rec.puzzles.  Please add me to your mailing list 
for Rubik's Cube info.

My address is rjh@on-ramp.ior.com

Thanks in advance!

        RON


From joemcg@catch22.com  Thu Nov  9 13:26:02 1995
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Date: Thu, 9 Nov 1995 10:29:43 -0800 (PST)
From: Joe McGarity <joemcg@catch22.com>
To: "Rubik's Cube Mailing List" <cube-lovers@ai.mit.edu>
Subject: Flowers in you hair
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Hello, 

I just joined this group and I thought that I would throw out a
place here in San Francisco where one can find some interesting things 
(such as the elusive 5x5x5 cube).

Game Gallery
One Embarcadero Center
Street Level
San Francisco, CA  94102

(415) 433-4263

Also, there is a second-hand store on the corner of 17th and Mission 
which seems to have some type of mix-up-and-fix puzzle every time I go 
in.  Let me know what people are looking for and I'll keep an eye out for it.

Later, 


Joe


------------------------------------------------------------------------------

Joe McGarity                      "Mufasa, Mufasa, Mufasa!"   
418 Fair Oaks
San Francisco, CA  94110

joemcg@catch22.com                           

------------------------------------------------------------------------------



From joemcg@catch22.com  Thu Nov  9 14:10:43 1995
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From mark.longridge@canrem.com  Sun Nov 12 01:51:15 1995
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Subject: Magic Platonic Solids
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1260.5834.0C1FDFC0@canrem.com>
Date: Sun, 12 Nov 95 01:33:00 -0500
Organization: CRS Online  (Toronto, Ontario)

First a correction (sorry Dave!)

> # Perhaps David Badley could confirm the following orders:

The above should be "David Bagley".

I have some further comments on the "Magic Platonic Solids".

 One can stretch (abuse?) the concept of the slice and anti-slice
groups of the cube to include the Megaminx (Magic Dodecahedron).
In the case of the Megaminx we can consider one-fifth turns of
opposite faces. Unfortunately my experiments with "slice" turns
on the Megaminx has not generated any spot patterns as yet.

 Ben Halpern was not the only one to make a prototype of a tetrahedron
with rotating faces, as Kersten Meier made one as well.

 Only 3 of the 4 generators of the Halpern-Meier Tetrahedron are
necessary to generate the 3,732,480 possible states. If we use only
2 generators we only get 19,440 possible states.

 It is not possible to swap just 1 pair of corners and 1 pair of
edges, as is possible with the standard Rubik's cube.

 The number of possible states of the Halpern-Meier Tetrahedron break
down like this:

6! /2 * 2^5 * 4!/2 * 3^3 = 3,732,480

The number of pairs of exchanges of the 6 edges must be even.
The number of pairs of exchanges of the 4 corners must be even.
5 of the 6 edges may have any flip, the last edge is forced.
3 of the 4 corners may have any twist, the last corner is forced.

The H-M Tetrahedron is roughly comparable to the 2x2x2 cube and
 the standard Skewb in terms of the number of combinations.

    Halpern's Tetrahedron    3.7*10^6       Ben Halpern, Kersten Meier
    Pocket Cube (2x2x2)      3.6*10^6       Erno Rubik
    Skewb                    3.1*10^6       Tony Durham

-> Mark <-


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From JBRYAN@pstcc.cc.tn.us  Tue Nov 14 09:12:58 1995
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Date: Tue, 14 Nov 1995 09:13:41 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: God's Algorithm for the 1x1x1 Rubik's Cube
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Solving the 1x1x1 Rubik's cube is probably a bit silly and whimsical, but 
let's look at it anyway.  

I was led in this direction by rereading some articles in the archives
from Dan Hoey and others concerning NxNxN Rubik's cubes.  For example,
consider Dan's discussion "Cutism, Slabism, and Eccentric Slabism" from 
1 June 83 19:39:00.  Sometimes degenerate cases are slightly interesting.  
I guess the 1x1x1 case is the most degenerate we have, unless you want to
consider the 0x0x0. 

It seems to me that either cutism or slabism, as Dan calls them, reduce to
whole cube rotations for the 1x1x1 case.  For example, a quarter turn face
turn or a quarter turn slice would be interpreted as a whole cube quarter
turn for the 1x1x1.  Hence, the cube group for the 1x1x1 is simply C, the
group of 24 rotations of the cube. 

By analogy with some of our previous work, I can think of essentially 
three different ways to model the 1x1x1.

   1)  With the 2x2x2, we normally wish to consider the puzzle solved if 
       each face is all of one color.  That is, whole cube rotations are 
       to be considered equivalent.  With the Singmaster fixed face 
       center view of the 3x3x3, the issue of whole cube rotations does 
       not arise.  But with the 2x2x2 we would normally consider 
       (for example) RL' equivalent to I.  The most common way to 
       accomplish this type of equivalence is to fix one of the corners.

       If we fix one of the corners of the 1x1x1, then we have a most 
       remarkable puzzle.  There is only one state, nothing can ever 
       move, and the puzzle is always solved.

   2)  A second way to model the 2x2x2 such that whole cube rotations are 
       considered to be equivalent is to consider the set of states to be 
       the set of cosets of C, that is, the set of all xC.

       If we take this approach with the 1x1x1, then there is only one 
       coset, namely iC (or just C, if you prefer).  The cube can rotate,
       but all 24 states are considered to be equivalent and the puzzle 
       is always solved.

   3)  Finally, if you model the 2x2x2 in such a way that whole cube 
       rotations are considered to be distinct, then you are really
       modelling the corners of the 3x3x3.  Indeed, a naive program 
       that simply modelled the permutations of the 2x2x2 facelets would  
       in fact unwittingly be modelling the corners of the 3x3x3.

       If you take the same approach of modelling the permutations of the 
       1x1x1 facelets, then you in effect are considering whole
       cube rotations to be distinct.  You have a very easy problem,  
       but the problem is not totally trivial as it is with approach #1
       or approach #2.  The rest of this note will therefore consider the
       problem of the 1x1x1 cube where whole cube rotations are considered
       to be distinct. 

Since we need to deal with whole cube rotations, I will use lower case
letters as our standard E-mail simulation of Frey and Singmaster's script
notation for whole cube quarter turns -- t for Top, r for Right, etc.  We
need only three of the six letters because, for example, we have l=r',
d=t', b=f', etc.  I will use t, r, and f. 

We know before we start that there are 24 states.  We also know before we
start that these 24 states form 5 M-conjugacy classes, where M is the set
of 48 rotations and reflections of the cube.  (There are 10 M-conjugacy
classes of M, of which 5 are rotations and 5 are reflections.) Hence, any
discussion of God's algorithm will involve 5 conjugacy classes and 24
states. 

The obvious searches to look at are for qturns only, and for qturns plus 
hturns.  We may generate the qturn case as C=<t,r,f>.  We may generate 
the qturn plus hturn case as C=<t,r,f,t2,r2,f2>.

                    Qturns Only

 Distance Conjugacy Positions  
  from     Classes
  Start

    0          1        1  {i}
    1          1        6  {t,t',r,r',f,f'}       
    2          2       11  {tt,rr,ff},{tr,tr',tf,tf',t'r,t'r',t'f,t'f'}
    3          1        6  {ttf,ttf',ffr,ffr',rrt,rrt'}
  ---        ----     ----
 Total         5       24



                    Qturns Plus Hturns

 Distance Conjugacy Positions  
  from     Classes
  Start

    0        1        1  {i}
    1        2        9  {t,t',r,r',f,f'},{t2,r2,f2}       
    2        2       14  {tr,tr',tf,tf',t'r,t'r',t'f,t'f'},
                         {t2f,t2f',f2r,f2r',r2t,r2t'}
  ---      ----     ----
 Total       5       24


There are some additional problems we can look at.  For an example, an
interesting problem on the 3x3x3 is variously called the stuck axle
problem or the five generator problem.  In the case of the 1x1x1, we have
the "two generator problem" because we certainly can generate C as C=<t,f>
(Proof: r=tft').  But can we generate C with only one generator?  The
answer is no.  (Proof:  Order(i)=1, Order(t)=4, Order(tt)=2, Order(tf)=3,
and Order(ttf)=2.  All the orders are less than 24.  Note that it suffices
to calculate the order for one representative of each conjugacy class.)  I
will leave it as an exercise for the reader to determine the lengths of
each of the 24 positions if we generate C as <t,f>, and to determine the
appropriate conjugacy classes to take into account the symmetry of C
generated as <t,f>.

By the way, do we know the minimum number of generators required to 
generate the 3x3x3?  Here I do not mean the minimum number of quarter 
turns.  I am asking the question if we are permitted to use as generators 
any elements of G.

Here is one final item about the 1x1x1.  We do not know how many subgroups
of G there are for the 3x3x3.  But we do know how many subgroups of C
there are.  There has been much discussion of the 98 subgroups of M which
can be arranged in 33 conjugacy classes.  The subgroups of C are simply 
those subgroups of M which consist entirely of rotations.  There are 30 
such subgroups, and they may be arranged in 11 conjugacy classes. 

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =          
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us                      





From boland@sci.kun.nl  Wed Nov 15 19:34:28 1995
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	id BAA05998 (8.6.10/2.14); Thu, 16 Nov 1995 01:33:07 +0100
Message-Id: <199511160033.BAA05998@wn1.sci.kun.nl>
To: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Cc: Cube-Lovers <cube-lovers@ai.mit.edu>
Subject: Re: God's Algorithm for the 1x1x1 Rubik's Cube 
In-Reply-To: Your message of "Tue, 14 Nov 95 09:13:41 -0400."
             <Pine.PMDF.3.91.951113162306.539286930A-100000@pstcc.cc.tn.us> 
Date: Thu, 16 Nov 95 01:33:05 +0100
From: Michiel Boland <boland@sci.kun.nl>

Jerry wrote:

>But can we generate C with only one generator?  The
>answer is no.  (Proof:  Order(i)=1, Order(t)=4, Order(tt)=2, Order(tf)=3,
>and Order(ttf)=2.  All the orders are less than 24.  Note that it suffices
>to calculate the order for one representative of each conjugacy class.)

Another way to see that C cannot be generated by one generator
is to note that C is not abelian.

Singmaster mentions in his Notes that the cube group G itself
can also be generated by two elements.
-- 
Michiel Boland <boland@sci.kun.nl>
University of Nijmegen
The Netherlands

From hoey@aic.nrl.navy.mil  Wed Nov 29 12:18:52 1995
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Date: Wed, 29 Nov 95 12:14:40 EST
From: hoey@aic.nrl.navy.mil
To: mschoene@math.rwth-aachen.de (Martin Schoenert),
        frb6006@cs.rit.edu (Frank R Bernhart), Cube-Lovers@life.ai.mit.edu
Newsgroups: sci.math
Subject: Generating Rubik's Cube
Message-Id: <9511291210.Hoey@AIC.NRL.Navy.Mil>
References: <DIsoBz.Eu7@rci.ripco.com> <mschoene.817657981@astoria> <1995Nov29.054118.9651@cs.rit.edu>
Distribution: 

About generating the cube's group with arbitrary elements of that
group, mschoene@Math.RWTH-Aachen.DE (Martin Schoenert) writes:

> ... Rubik's cube can be generated by 2 elements.
> Moreover almost any random pair of elements will do the trick....

Actually, I think it's more accurate to say that a random pair of
elements has nearly a 75% probability of generating the cube.  At
least, I'm pretty sure that's an upper bound, and I don't see any
reason why it shouldn't be fairly tight.  That's for the group where
the whole cube's spatial orientation is irrelevant.  I think it's more
like 56% (9/16) if you also need to generate the 24 possible
permutations of face centers.

About the minimal presentation of the cube group on the usual generators,
frb6006@cs.rit.edu (Frank R Bernhart) writes:

> The answers may be in SINGMASTER, et.al.
>    "Handbook of Cubic Math"  or BANDEMEISTER (sp?) "Beyond R. Cube"

I recall Singmaster wanted to know if anyone found a reasonably-sized
presentation; I don't know if any have been found in the intervening
fifteen years.  The best I know of is a few thousand relations, some
of them several thousand letters long.  I've been meaning to try
chopping that down a bit.

Dan                                     posted and e-mailed
Hoey@AIC.NRL.Navy.Mil

From mbparker@share.ai.mit.edu  Fri Dec  1 13:31:02 1995
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From: Michael B. Parker <mbparker@share.cytex.com>
Message-Id: <9512011824.AA00728@share>
To: PuzzleParty@cytex.com, Cube-Lovers@ai.mit.edu,
        506maple-residents@cytex.com, www-designers@cytex.com
Subject: PUZZLE PARTY 4 -- SATURDAY (Dec. 2), 7pm, Orange!
Reply-To: mbparker@cytex.com

			       PUZZLE PARTY IV!

The 3 Puzzle Parties this year have been a big success!  The last party
brought puzzle collectors from as far as Australia, plus the world-famous
Jerry Slocum,... and didn't quit 'til 3:30am!

So if you missed the one before, you absolutely don't want to miss Puzzle
Party 4!...  Bring your brain teasers, mechanical puzzles and mental games,
and prepare yourself to have an incredibly good time.  Join us to dine on a
tasty ``puzzle potluck'' along with drinks and leisurely conversation with
friends by the fireside.  Plenty of snacks & refreshments and good spirits
provided, so grab that brain and some puzzles, and see you there!

 WHEN:	Saturday, December 2nd
	7:00 PM until...

 WHERE: Mike's house, 506 N. Maplewood St., Orange, CA
	From the 5 fwy S, exit 22E, to the end, then 55N, take 2nd exit 
	Chapman West, at 1st light right/north on Tustin, 2nd light left/west 
	onto Walnut, 3rd left is Maplewood.   We are the big yellow house on 
	the NW corner of Walnut and Maplewood.

 COST:	$4   MITCSC     Members & Guests with puzzles
	$6   MITCSC Non-Members & Guests with puzzles
	$8   MITCSC     Members & Guests w/o  puzzles
	$10  MITCSC Non-Members & Guests w/o  puzzles

 RSVP:	You may pay at the door, but please try to contact me beforehand so
	I can put you on the list.  Please email, fax, or phone the following
	info: Your NAME, ADDRESS, PHONE, FAX, EMAIL, and what you're bringing:

              ___ puzzle-bearing     members at $ 4 each: $___
              ___ puzzle-bearing non-members at $ 6 each: $___
              ___ puzzle-less        members at $ 8 each: $___
              ___ puzzle-less    non-members at $10 each: $___
              ___ <- total persons          total cost -> $___
                 total number of puzzles being brought ___

 SPONSOR: Michael B. Parker, MIT '89
	  email mbparker@cytex.com, 1-800-MBPARKER xLIVE, xPAGE, xFAXX

(This info is online!  See http://www.cytex.com/~mitcsc/w96/w96-pzl4.htm)

From hoey@aic.nrl.navy.mil  Sun Dec  3 14:46:32 1995
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From: hoey@aic.nrl.navy.mil
Message-Id: <9512031946.AA24122@sun13.aic.nrl.navy.mil>
To: Cube-Lovers@life.ai.mit.edu
Subject: Re: Generating Rubik's Cube

On the probability that two random elements will generate the entire
cube group, I wrote:

> ... a random pair of elements has nearly a 75% probability of
> generating the cube.  At least, I'm pretty sure that's an upper
> bound, and I don't see any reason why it shouldn't be fairly tight.
> That's for the group where the whole cube's spatial orientation is
> irrelevant.  I think it's more like 56% (9/16) if you also need to
> generate the 24 possible permutations of face centers.

I can now answer the spatial orientation part of the question, and
it's much lower.  We're talking about C, the 24-element group of
proper motions of the whole cube.  If we select two elements at random
with replacement, the probability is only 3/8 that they will generate
the whole group.

The kinds of motions that can take part in a generating pair are a
90-degree rotation about an axis, a 120-degree rotation about a major
diagonal, and a 180-degree rotation about a minor diagonal.  Note that
the last kind fixes two major diagonals and an axis.  Two motions
generate C iff they are

    (48 ways) a 120 and a 180, unless they fix the same major diagonal,
    (48 ways) a 180 and a 90, unless they fix the same axis,
    (24 ways) two 90s at right angles, or
    (96 ways) a 90 and a 120.

The number comes out so even I suspect there's something deeper going
on than the exhaustive analysis I used.

As for generating the (fixed-face) Rubik's group, I still suspect that
two elements almost always generate the entire group unless they are
both even.  Anyone who has a Sims's-algorithm implementation handy
could help out with a Monte-carlo approximation to see if this is
approximately right.  Or, I wonder, is there a way of getting an exact
number, perhaps with the help of GAP?

Dan                                     posted and e-mailed
Hoey@AIC.NRL.Navy.Mil

From mark.longridge@canrem.com  Sun Dec  3 20:32:36 1995
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Sender: CRSO.Cube@canrem.com
Subject: <U,R> & G
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1261.5834.0C201705@canrem.com>
Date: Sun,  3 Dec 95 20:09:00 -0500
Organization: CRS Online  (Toronto, Ontario)

A while back Jerry asked....

>                   Finally, pick any cube X in <U,R>.  We know
> |X| in G <= |X| in <U,R>.  Can anybody find a cube X such that
> |X| in G < |X| in <U,R>?

Well, we basically know the answer is yes. There are elements in
<U,R> which require less moves if we use all the generators of G.

To be more specific, look the 6 twist pattern in <U,R> which
requires 22 q turns:
         ^^^^^^^^^^
>>  Equivalent to (U1 R1)^35= (R1 U1)^35 & Shift Invariant
>>  UR11 = U2 R1 U1 R1 U1 R3 U1 R3 U1 R3 U2 R1 U1 R1 U1 R3 U1 R3 U1 R3

After a bit of computer cubing I found:

p183 6 Twist      R1 U3 R2 U3 R1 D3 U3 R1 U3 R3 D2 R3 U3 R1 D3 U3
                  (18 q, 16 q+h  moves)
                  ^^^^^
I'll spare everyone all the gory details. I'm certain there are
all sorts of other examples, but here is one case where we can
save 4 q turns. It may be of some small interest to see which
of the two processes can be executed more rapidly by the human hand.

-> Mark <-


From mschoene@math.rwth-aachen.de  Mon Dec  4 09:10:17 1995
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Date: Mon, 4 Dec 95 12:49 MET
From: Martin Schoenert <mschoene@math.rwth-aachen.de>
To: hoey@aic.nrl.navy.mil
Cc: Cube-Lovers@life.ai.mit.edu
In-Reply-To: hoey@aic.nrl.navy.mil's message of Sun, 3 Dec 95 14:46:30 EST
	<9512031946.AA24122@sun13.aic.nrl.navy.mil>
Subject: Re: Re: Generating Rubik's Cube

I have used GAP to compute the subgroup generated by 300 random pairs of
elements of G.  151 of those pairs generated the entire group, so the
probability is about 50%.

I don't think we can figure out the exact number, since we don't know the
maximal subgroups of G.  One maximal subgroup we know is the derived
subgroup (on which the upper bound of 75% is based).  Then there are the
8 stabilizers of the corners (of index 8) and the 12 stabilizers of the
edges (of index 12).  Using those it should be possible to push the upper
bound down to something about 60%.

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany


From csstto@alpcom.it  Wed Dec  6 06:16:17 1995
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 How it is possible to access to "Cube lovers at MIT"  ? 
 Do we need a password ?

 Thank You and best regards
    Domenico Inaudi



From alan@curry.epilogue.com  Thu Dec  7 02:44:55 1995
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From: Alan Bawden <Cube-Lovers-Request@ai.mit.edu>
Sender: Cube-Lovers-Request@ai.mit.edu
Subject: Magazine Spam
To: Cube-Lovers@ai.mit.edu

So I thought it was time to send you all an update on unwanted magazine
advertisements that have been broadcast over Cube-Lovers about once a week
for the last few months.  Here's the final story.

First off, there is absolutely -nothing- that I can do IN THE SHORT TERM to
stop these advertisements.  Internet electronic mail was not designed to
prevent unwanted advertising.  As things are set up now, Cube-Lovers is a
simple mailing list, so anybody, anywhere, can send mail to Cube-Lovers and
you all get it.

It turns out that the source of the advertising we've been getting is a
fellow named Kevin Jay Lipsitz <krazykev@kjl.com>.  I've written directly
to Mr. Lipsitz politely asking him to remove Cube-Lovers from his list of
advertising targets (it was hard to be polite, but I was) -- but Mr.
Lipsitz apparently doesn't answer his electronic mail.  Actually, I doubt
he even -reads- his electronic mail, because he is a well-known Spammer,
and probably gets hundreds of complaints a day delivered to his address.
(For those of you new to the Internet, "Spamming" is the technical term for
the kind of advertising Mr. Lipsitz engages in.)

I really doubt that Mr. Lipsitz's technique has sold any magazines to any
of -you-, but I suppose he gets enough suckers to make it pay, and he's got
no motivation to bother removing Cube-Lovers, since MIT is paying for the
resources that he's using to reach you all.  So we're stuck with him.  At
least we're stuck with him until I can get the filtering technology in
place to cut him off.  Which I wanted to avoid, because I have better
things to do with my time, but now I have no choice.  So relief from Mr.
Lipsitz's magazines is on its way eventually, but probably not until
you've seen several more copies of his advertisement -- sorry.

By the way, here's more information on Mr. Lipsitz.  You'll notice that he
has his own domain name: KJL.COM.  They don't give you a domain name unless
you provide a mailing address and a phone number, so the following
information is publicly available from the NIC:

   Kevin Jay Lipsitz (KJL-DOM)
      PO Box 120990
      Staten Island NY  10312-0990

      Domain Name: KJL.COM

      Administrative Contact, Technical Contact, Zone Contact:
	 Lipsitz, Kevin Jay  (KJL2)  krazykev@KJL.COM
	 718-967-1234

      Record last updated on 25-Aug-95.
      Record created on 20-Apr-95.

      Domain servers in listed order:

      NS1.ABS.NET                  206.42.80.130
      NS2.ABS.NET                  206.42.80.131
      NS1.NET99.NET                204.157.3.2

If you want proof that this is the guy, you need only note that the address
given here is the -same- as the address for ordering magazines given in all
those advertisements.  The phone number is in the same area code and
exchange as the Fax number he sometimes gives.  (Although the fact that the
phone number ends in "1234" makes me suspect it is bogus -- I don't think
think the NIC tries to -verify- any of this information.)  Notice that
ABS.NET provides the domain service for KJL.COM.  You will find that
ABS.NET is no more interested in answering your mail than Mr. Lipsitz is.

Finally, I urge you all -not- to respond to this message in public.  If you
have further thoughts on Internet advertising, electronic mailing list
administration, or clever acts of revenge, you can send them to -me-, but
don't CC your message to Cube-Lovers as a whole.  The whole point here is
to keep Cube-Lovers relatively free of off-topic mail.  As the list
administrator I get to send out occasional administrivia such as this
message because I do actual -work- to keep the list running.  

			- Alan (Cube-Lovers-Request@AI.MIT.EDU)

From walts@federal.unisys.com  Thu Dec  7 09:04:57 1995
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Message-Id: <199512071407.JAA05663@homer.MCLN.Federal.Unisys.COM>
Date: Thu, 07 Dec 95 09:06:45 -0800
From: "Walter P. Smith" <walts@federal.unisys.com>
Organization: Installation Services
X-Mailer: Mozilla 1.22 (Windows; I; 16bit)
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To: Cube-Lovers@ai.mit.edu
Subject: Mini Cube & Revenge
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My first message. Oh boy!

It's back. A new version of the 2x2x2 cube previously called the 
Pocket Cube is in production.  The new version is called the Mini 
Cube.  They are available from GameKeepers I got mine in Tysons 
Corner Shopping Mall Virginia, but GameKeepers is a chain and 
should be in most large cities.  If readers can't locate one, let 
me know and I'll get a list of locations.  They also carry a wide 
line of puzzles including Triamid, Snake, regular Rubik's cubes, 
Master Balls etc.


The stickers are glossy paper. I don't think they will be very 
durable.  Also the red and orange sides are very hard to tell 
apart.  How could they be so stupid?  The mechanism seems to work 
better than the old ones.  I can't tell if the inner workings are 
the same.  The original Pocket Cube had a ball in the center with 
six cap like pieces screwed to it (with springs under the screw 
head) to form a series of tracks.  Each piece had a shaft that 
extended down into the grove with a triangular foot on it.  This 
design requires a lot of pieces and drives the price up.  I 
always thought they could be made by making the ball, three of 
the caps and one corner piece, all into one piece.  The Mini Cube 
uses cubies that are solid on all sides.  This may account for 
the smoother action.  They include a complete solution sheet.  
One comes with the Master Ball also.  I personally think 
manufacturers should't do this.  Many people will turn to the 
solution sheet before giving it a good effort and will miss the 
pleasure of solving it for themselves.  My Mini Cube cost over 
six dollars.  A little pricey but a collector should never pass 
up an opportunity.

Walt "The Puzzler"



From serge@nexen.com  Tue Dec 12 17:07:38 1995
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Date: Tue, 12 Dec 1995 17:07:19 -0500
From: Serge Kornfeld <serge@nexen.com>
Message-Id: <199512122207.RAA26101@spank.nexen.com>
To: cube-lovers@ai.mit.edu
Subject: subscribe 



Please subbscribe

Serge     serge@nexen.com







From nichael@sover.net  Wed Dec 13 21:09:22 1995
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Date: Wed, 13 Dec 1995 21:16:38 -0400
To: Cube-Lovers@ai.mit.edu
From: nichael@sover.net (Nichael Lynn Cramer)
Subject: Pocket Stuff

[Possibly minimal relevance; take it in the sense of cool stocking-stuffer
hacks.]

Someone asked a couple of weeks back for pocket/key-ring cubes.  Can't help
with that, but this afternoon in Sandy's and Son's (Inman Square (Cambridge
(Ma))) beside the cash register they had a basket of EtchASketch keyrings.
Seemed pretty solidly built for the the $3.50.

N




From walts@federal.unisys.com  Thu Dec 14 10:23:10 1995
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Message-Id: <199512141525.KAA05516@homer.MCLN.Federal.Unisys.COM>
Date: Thu, 14 Dec 95 10:25:01 -0800
From: "Walter P. Smith" <walts@federal.unisys.com>
Organization: Installation Services
X-Mailer: Mozilla 1.22 (Windows; I; 16bit)
Mime-Version: 1.0
To: Cube-Lovers@ai.mit.edu
Subject: Twist Torus
Content-Transfer-Encoding: 8bit
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In Sept. 1992 Mark Longridge described an idea he had for a 
puzzle.  He called it a Twist Torus.  Well I bought a puzzle that 
fits his description very closely.  I would have sent this in 
sooner except that I am new to Cube Lovers.  I bought mine 
several years ago.  Did he get his design into production or was 
it independently invented or did someone implement Mark's idea?  
Will we ever know?  I bought mine in a department store (cant 
remember which) toy department.  It was not in any packaging and 
cost less than $2 US..  They only had one.  It was quite by 
chance that I determined that it is a puzzle.

It will test my ability to describe it in words but here goes.  
It is torus shaped (dough nut shaped).  At first glance it looks 
like a bracelet.  It has one slice made the same way a bagel is 
sliced.  The puzzle can turn along this cut.  There are eight 
differently colored sections separated by fixed black sections 
around the circumference of the torus.  Each colored section is 
subdivided into 4 sub-segments that can turn at right angles to 
the main circumference.  As a segment is turned, different parts 
of the segment are brought to the other side of the main cut.

It operates smoothly and is brightly colored.  It is fairly easy 
to solve but the geometry is novel and interesting.

Does anyone else have one of these?  Does anyone know who 
manufactured this?  Does anyone know what it is called?

Walt "The Puzzler" Smith





From devo@vnet.ibm.com  Thu Dec 14 12:30:51 1995
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Date: Thu, 14 Dec 95 12:03:41 EST
From: "Dave Eaton" <devo@vnet.ibm.com>
To: Cube-Lovers@ai.mit.edu
Subject: Lumination

Has anyone seen a puzzle called Lumination by Parker Bros.  A
guy at work says he got one from his wife about five years ago.
It is a tetrahedron (like the Pyraminx) except that instead of
having any moving parts, it has lights in the four points that
change color when you rotate the whole puzzle in space.

It sounds really neat.  Does anyone know where I can get one?

Dave Eaton


From serge@nexen.com  Thu Dec 14 13:17:04 1995
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Date: Thu, 14 Dec 1995 13:16:04 -0500
From: Serge Kornfeld <serge@nexen.com>
Message-Id: <199512141816.NAA28270@spank.nexen.com>
To: Cube-Lovers@ai.mit.edu
Subject: Re Twist Torus




>  It will test my ability to describe it in words but here goes.  
>  It is torus shaped (dough nut shaped).  At first glance it looks 
>  like a bracelet.  It has one slice made the same way a bagel is 
>  sliced.  The puzzle can turn along this cut.  There are eight 
>  differently colored sections separated by fixed black sections 
>  around the circumference of the torus.  Each colored section is 
>  subdivided into 4 sub-segments that can turn at right angles to 
>  the main circumference.  As a segment is turned, different parts 
>  of the segment are brought to the other side of the main cut.



I came to US 4 years ago from Russia. Living in Russia I use to collect 
mechanical puzzles. I remember the article in magazine and a picture of the
puzzle you described. I think it was 1987 ????. Article was saying that
there are some problems to actually make this type of puzzle and .....

I cant remember the end of the article and I never saw this toy in real.  

Serge serge@nexen.com

From SCHMIDTG@beast.cle.ab.com  Thu Dec 14 21:36:02 1995
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Date: Thu, 14 Dec 1995 21:35:58 -0500 (EST)
From: SCHMIDTG@beast.cle.ab.com
To: Cube-Lovers@ai.mit.edu
Message-Id: <951214213558.20212e52@iccgcc.cle.ab.com>
Subject: Re: Twist Torus

Walter P. Smith wrote,

>[stuff about a "Twist Torus" puzzle deleted...]
>
>Does anyone else have one of these?  Does anyone know who 
>manufactured this?  Does anyone know what it is called?

I purchased one of these back in 1992 at a toystore in a
Florida mall.  I paid $4.99 for mine.  The tags says:

WrisTwist (tm)
Puzzle & Bracelet

WACO
Riverdale, NJ 07457
Made in Indonesia
Item #20003

The puzzle is still in its pristine state and the color progression is:

Red-Blue-Green-Yellow-OrangeRed-Violet-Orange-LightGreen

-- Greg

From rh@thi.informatik.uni-frankfurt.de  Fri Dec 15 03:39:59 1995
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	id AA07726; Fri, 15 Dec 95 09:40:50 +0100
Date: Fri, 15 Dec 95 09:40:50 +0100
From: rh@thi.informatik.uni-frankfurt.de (Roger Haschke)
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Message-Id: <9512150840.AA07726@riese.informatik.uni-frankfurt.de>
To: cube-lovers@ai.mit.edu
Subject: unsubscribing ...

can anybody please tell me the correct email-adresse
for sending an unsubscribe-command?

thanks - Roger

From alan@curry.epilogue.com  Fri Dec 15 04:07:12 1995
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Date: Fri, 15 Dec 1995 04:06:23 -0500
Message-Id: <15Dec1995.035416.Alan@LCS.MIT.EDU>
From: Alan Bawden <Cube-Lovers-Request@ai.mit.edu>
Sender: Cube-Lovers-Request@ai.mit.edu
To: rh@thi.informatik.uni-frankfurt.de
Cc: cube-lovers@ai.mit.edu
In-Reply-To: Roger Haschke's message of Fri, 15 Dec 95 09:40:50 +0100 <9512150840.AA07726@riese.informatik.uni-frankfurt.de>
Subject: unsubscribing ...

   Date: Fri, 15 Dec 95 09:40:50 +0100
   From: Roger Haschke <rh@thi.informatik.uni-frankfurt.de>

   can anybody please tell me the correct email-adresse
   for sending an unsubscribe-command?

   thanks - Roger

As I'm certain you've been told -multiple- times, the address is:

  CUBE-LOVERS-REQUEST@AI.MIT.EDU

For crying out loud, why can't people remember that?

Let me give everybody a little bit of advice.  For every mailing list you
subscribe to, keep a file that contains the information you will need in
order to cancel or update your subscription.  This isn't hard to do.  I do
it myself.  It's a great way to avoid looking foolish in front of hundreds
of people.  If you don't have such a file for Cube-Lovers already, start
one RIGHT NOW and put this message in it.

		-- Alan Bawden <Cube-Lovers-Request@AI.MIT.EDU>

From walts@federal.unisys.com  Fri Dec 15 08:37:52 1995
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Message-Id: <199512151340.IAA12161@homer.MCLN.Federal.Unisys.COM>
Date: Fri, 15 Dec 95 08:39:46 -0800
From: "Walter P. Smith" <walts@federal.unisys.com>
Organization: Installation Services
X-Mailer: Mozilla 1.22 (Windows; I; 16bit)
Mime-Version: 1.0
To: Cube-Lovers@ai.mit.edu
Subject: WrisTwist
Content-Transfer-Encoding: 7bit
Content-Type: text/plain; charset=us-ascii

I received the following mail from KINSMAN.  I did not reply to 
him or make note of his address.  I am retyping his note to get 
comment from others.  He is refering to the WrisTwist puzzle.


I have just such a puzzle too.  It came from my local toy store. 
 I also have a digital camera sitting next to me.  Should I bring 
mine in and post a low resolution copy in GIF format to the 
group?  -AAK


Does anyone want him to do this? 

Walter P. Smith
walts@federal.unisys.com




From kinsman@ycc.kodak.com  Fri Dec 15 08:51:42 1995
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Date: Fri, 15 Dec 1995 08:50:51 +0500
From: kinsman@ycc.kodak.com (Andy Kinsman 66672)
Message-Id: <9512151350.AA20112@newt.PCD1>
To: cube-lovers@ai.mit.edu
Subject: Re: Twist Torus [small picture]
Content-Length: 61490


Since I have both the puzzle in hand and a camera attached to
my computer... here is a picture of the torus puzzle, slightly
missaligned for effect.

to decode 

save this note into a file, possibly trim stuff before begin
and after end line.  

type 'uudecode the-file'  

find ringpuz.tif in your directory after this operation.  
view it with your favorite tif viewer.  

Get help from another if this doesn't make sense.  Enjoy -AAK

----------------ringpuz.tif.uu included -----------(cut here)-----
begin 644 ringpuz.tif
M34T *@  K$!27E1'4DI17E9(5DT^1T)@:&-475I06U=/6E=06U=17%A,5U%L
M=W)58%I06U906U5)5$],5U%,5U%'4DU37%5065),54Y<9EU?:%] 1S\_1CYD
M:V-47E%,5DE+5DA)4T9?:EQ?:5I+549>:5E67U1375)<:%Y27E1<:%YA:V%%
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end

From walts@federal.unisys.com  Fri Dec 15 09:03:36 1995
Return-Path: <walts@federal.unisys.com>
Received: from www.han.federal.unisys.com by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA17832; Fri, 15 Dec 95 09:03:36 EST
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	id JAA12981; Fri, 15 Dec 1995 09:03:34 -0500
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	id JAA12350; Fri, 15 Dec 1995 09:05:55 -0500
Message-Id: <199512151405.JAA12350@homer.MCLN.Federal.Unisys.COM>
Date: Fri, 15 Dec 95 09:05:31 -0800
From: "Walter P. Smith" <walts@federal.unisys.com>
Organization: Installation Services
X-Mailer: Mozilla 1.22 (Windows; I; 16bit)
Mime-Version: 1.0
To: Cube-Lovers@ai.mit.edu
Subject: Million dollar cube
Content-Transfer-Encoding: 7bit
Content-Type: text/plain; charset=us-ascii

I hope all cube lovers saw the picture and article in USA Today 
newspaper on Wednesday, December 13 in the "Life" section.  It 
reads as follows:

"PRICEY PUZZLE:  Looks like diamonds are a toy's best friend.  
And rubies, sapphires and amethysts, too, in the ultimate 
Rubik's Cube.  To celebrate the 15th anniversary of the brain 
teaser, Diamond Cutters International created an 18-karat gold, 
jewel-encrusted, one-of-a-kind puzzle that'll set you back $1 
million.  Currently on display at DCI's Houston headquarters, the 
fully working replica will hit the road for a European tour 
starting Jan. 20 in London.  And if you buy the cube and can't 
solve it, creator Erno Rubik, who lives in Hungary, will come to 
your home to help out."


It looks like each white cubie has 25 diamonds on it.  The other 
colors are make from different encrusted stones.  It's hard to 
tell how big it is.  If I could afford it, I would buy it and lie 
by saying I can't solve it so I could get a visit from Rubik.  
Watch for tour info.



From nichael@sover.net  Fri Dec 15 09:30:19 1995
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Date: Fri, 15 Dec 1995 09:37:19 -0400
To: Andy Kinsman 66672 <kinsman@ycc.kodak.com>, cube-lovers@ai.mit.edu
From: nichael@sover.net (Nichael Lynn Cramer)
Subject: Re: Twist Torus [small picture]

At 8:50 AM 15/12/95, Andy Kinsman 66672 wrote:
>Since I have both the puzzle in hand and a camera attached to
>my computer... here is a picture of the torus puzzle, slightly
>missaligned for effect.

Sigh...

Please don't mail stuff like this to a list.  If you want to distribute it,
fine; I'm sure there are people who are glad to have it.  In that case
either set up an FTP site or --lacking that-- post an
announcement/invitation and let those who want things like this send you
mail and then you can post to them directly.  But the last thing most of us
need on a dreary friday morning is another 50k bit-bomb in our mailbox.


Nichael                          "... and they opened their thesaurus
nichael@sover.net                      and brought forth gold,
http://www.sover.net/~nichael        and frankincense and myrrh."



From JBRYAN@pstcc.cc.tn.us  Fri Dec 15 10:11:03 1995
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Date: Fri, 15 Dec 1995 10:12:27 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: Million dollar cube
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On Fri, 15 Dec 1995, Walter P. Smith wrote:

> .....          To celebrate the 15th anniversary of the brain 
> teaser, .....

That would make it 1980.  Is that right?  I think Cube-Lovers started in
1980, but I have just been reading some early stuff from Singmaster dated
1978 and 1979. 

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =          
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us                      


From boland@sci.kun.nl  Fri Dec 15 19:22:07 1995
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	id BAA23861 (8.6.10/2.14) for <cube-lovers@ai.mit.edu>; Sat, 16 Dec 1995 01:22:06 +0100
Message-Id: <199512160022.BAA23861@wn1.sci.kun.nl>
To: cube-lovers@ai.mit.edu
Subject: Some notes on the antislice group.
Date: Sat, 16 Dec 95 01:22:05 +0100
From: Michiel Boland <boland@sci.kun.nl>

This is something for the holidays.
I hope someone finds it interesting :)

Some notes on the antislice group.

Notations: Ua = UD, U2a = U2D2, U'a = U'D', Ra = RL, Fa = FB, etc

Consider the three `slices' of edge cubicles (one slice
containing UR, RD, DL, LU, another containing UF, FD, DB and BU,
and another containing FR, RB, BL, LB).

Any operation in the antislice group (the subgroup of G
generated by Ua, Ra and Fa) will map each slice to another
slice. Also, if we restrict ourselves to antislice movements, we
can define an orientation for each slice (choose a fixed cubie
in each slice and define the orientation of the slice to be the
orientation of that cubie).

A fairly obvious subgroup of the antislice group is the one in
which all three slices are in their original position *and* are
oriented correctly. I have been giving this subgroup, which is
in fact a normal subgroup of the antislice group, some study.

To make speaking a little bit easier, I will use the letter T
for this group (don't ask me why :)

If one takes a cube in position START, and applies an operation
in T to it, one finds that, if one looks at a face, each facelet
has either the colour of that face's center, or the colour of
the opposite face's center. Therefore, the patterns generated by
movements in T can be deemed `pretty'.

Some patterns that can be generated from transformations in T
are the Pons Asinorum (U2a R2a F2a), 4 Plusses (Ua Ra U2a Ra Ua F2a),
and 6xH (Ua Ra U2a F2a Ra U'a).
A pattern that cannot be generated from T-movements is the
four-dot pattern (I'm just stating this as a fact; I still
haven't got a proof for it.)

The group T contains 256 elements. It is isomorphic to C_2^8
(the cartesian product of eight copies of C_2). Hence, each
element of T is its own inverse.

The group T is generated by the following elements:

	U2a
	R2a
	F2a
	Fa U2a F'a
	Ua R2a U'a
	Ra F2a R'a
	Ua Ra Ua Ra Ua Ra
	Ra Fa Ra Fa Ra Fa

Note that  Fa Ua Fa Ua Fa Ua  =  Ua Ra Ua Ra Ua Ra  Ra Fa Ra Fa Ra Fa

There are two obvious metrics on the antislice group (and on T):
the `quarter' turn and the `half' turn metric.

It takes at most four `quarter' anti-slice turns to get from any
position in the antislice group to a pattern in T (Ua Ra Fa Ua
is a maximal case in this respect.)

If one groups the members of T by their lengths in either metric
one gets some interesting results.

            length in quarter-turn metric
                0    2    4    6    8
             +---------------------------+
           0 |  1                        |   1
           1 |       3                   |   3
length in  2 |            3              |   3
half-turn  3 |           12    1         |  13
metric     4 |                18         |  18
           5 |                15         |  15
           6 |               192   11    | 203
             +---------------------------+
                1    3   15  226   11      256

The eight elements on the `diagonal' form a subgroup of T
(generated by U2a, R2a and F2a).

If one excludes the 192 elements from row 6 column 6, one also
gets a subgroup of T with 64 elements (generated by the three
elements mentioned above, and FaU2aF'a, UaR2aU'a, and RaF2aR'a).

Each of the 192 elements in the 6th row, 6th column can be
uniquely written in the form

 X Ya Xb Yc Xd Ye

where X and Y are either Ua, Ra or Fa, X and Y are different,
and a,b,c,d,e are either 1 or -1 (these are meant to be
exponents).  (examples: Ua Ra Ua Ra Ua Ra, Ua R'a Ua Ra U'a Ra)

The 11 elements in row 6 column 8 are:

  Ua Ra U2a F2a Ra U'a     (6xH)
  Ua Fa U2a R2a Fa U'a     ( ' )
  Ua Ra U2a Ra Ua F2a      (4x+)
  Ra Fa R2a Fa Ra U2a      ( ' )
  Fa Ua F2a Ua Fa R2a      ( ' )
  Ua Ra U2a F2a Ra Ua      (2xH, 2xDot, 2x+)
  Ua Fa U2a R2a Fa Ua      ( '    '      ' )
  Ra Fa R2a U2a Fa Ra      ( '    '      ' )
  Ra Ua R2a F2a Ua Ra      ( '    '      ' )
  Fa Ua F2a R2a Ua Fa      ( '    '      ' )
  Fa Ra F2a U2a Ra Fa      ( '    '      ' )
-- 
Michiel Boland <boland@sci.kun.nl>
University of Nijmegen
The Netherlands

From dzander@dazzle.sol.net  Sat Dec 16 13:36:18 1995
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From: dzander@dazzle.sol.net (Douglas Zander)
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Organization: The DAZzleman Empire
Subject: Re: WrisTwist
To: Cube-Lovers@ai.mit.edu
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In <199512151340.IAA12161@homer.MCLN.Federal.Unisys.COM>, "Walter P. Smith" writes:
>I received the following mail from KINSMAN.  I did not reply to 
>him or make note of his address.  I am retyping his note to get 
>comment from others.  He is refering to the WrisTwist puzzle.
>
>
>I have just such a puzzle too.  It came from my local toy store. 
> I also have a digital camera sitting next to me.  Should I bring 
>mine in and post a low resolution copy in GIF format to the 
>group?  -AAK
>
>
>Does anyone want him to do this? 
>
>Walter P. Smith
>walts@federal.unisys.com
>

I don't think he should post it to the mailing list but yes, I'd like to
recieve a GIF format picture.  Does he have an ftp site, or www site that
allows lynx download, or could he send it directly to me?
-- 
Douglas Zander          | Editor of GAMES Player's Zine.
dzander@dazzle.sol.net  | An e-zine for subscribers of GAMES Magazine (tm).

From hoey@aic.nrl.navy.mil  Sun Dec 17 02:41:05 1995
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From: hoey@aic.nrl.navy.mil
Message-Id: <9512170741.AA27424@sun13.aic.nrl.navy.mil>
To: Cube-Lovers@life.ai.mit.edu, rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Presenting Rubik's Cube
References: <mschoene.817657981@astoria> <1995Nov29.054118.9651@cs.rit.edu> <9511291210.Hoey@aic.nrl.navy.mil> <49ihhd$j20@muir.math.niu.edu>

A few weeks ago I mentioned the old problem of finding a presentation
of the Rubik's cube group in terms of the usual generators.  This was
posed by Singmaster over 15 years ago, and as far as I know has never
been addressed.  I've made some progress.

I work using a specially selected set of generators, rather than the
usual generators given for the cube.  First I give presentations
separately for the permutation groups of the corners and edges, and
the orientation groups of the corners and edges.  Then I join the
permutation groups with their respective orientation groups to form
the wreath groups, which describe the possible motions of the
respective piece types.  I join the two wreath groups in such a way
that the permutation parity of the two is equal.  Finally, I discuss a
method of converting to the usual generators.

In Coxeter and Moser's _Generators and Relations for Discrete Groups,
2nd ed_ I found Coxeter's presentation 6.271 for the symmetric group
on {1,2,...,n}, n even.  With a modest change of variables, his
presentation is on generators
    v=(1 2) and s=(2 3 ... n)                                ((1))
and relators
    v^2,
    v s^(n-2) (v s^-1)^(n-1),
    (s^-1 v s v)^3, and
    ((s^-1 v)^i (s v)^i)^2,            i=2,...,n/2-1.        ((2))
Here n will be 8 or 12 to present the group of the permutations of
corners or edges, respectively.

The orientation group of the corners (or edges) is the direct product
of n-1 cyclic groups, which can be presented with generators
         r_0=(1)+(2)-
         r_1=(1)+(3)-, ...,
     r_(n-2)=(1)+(n)-,      ((3))
where (k)+ indicates a reorientation of piece k in place and (k)-
indicates the inverse reorientation.  The relators here are
     r_i^d,            (d=3 (corners) or 2 (edges)), and
     r_i r_j r_i^-1 r_j^-1,            0 <= i < j <= n-2.     ((4))

I generate the wreath group with the union of the generators ((1,3)).
The added relators
     v r_0 v r_0
     v r_i v r_0 r_i                   i=1,...,n-2,
     s^-1 r_i s^i r_(i+1)^-1           i=0,...,n-3, and
     s^-1 r_(n-2) s^i r_0^-1                                 ((5))
will permit moving the r_i to the end of a word, after which the
previous relators ((2)) and ((4)) may be used to manipulate the parts
separately, just as a Rubik's cube solvers can perform any needed
permutations before reorientations.

In the wreath group, the r_i are conjugate to each other.  The third
line of ((5)) may be used to define r_k = s^-k r_0 s^k, so I eliminate
r_1,...,r_(n-1) and write r_0 as r.  The last line of ((5)) is then a
consequence of s^(n-1)=e, which is implied by ((2)), according to
Coxeter.  The conjugacy also lets me rewrite ((4)) as
     r^d,            (d=3 (corners) or 2 (edges)), and
     s^-j r s^j r' s^-j r' s^j r,      j=1,...,(n-2)/2.      ((6))

As the discussion turns to working with corners and edges together, I
write cs,cr and es,er for the respective generators.  I use a single
generator v that acts on both corners and edges, to ensure that the
corner permutation has the same parity as the edge permutation.  Since
any identity in {v,cs,cr} must use an even number of v's, the identity
will hold in the when the v operates on edges as well; similarly for
{v,es,er} operating on the corners.

To present the whole cube group, I use all five generators, relators
((2,5,6)) for both corners and edges, and new relators
    es cs es' cs', er cs er cs', es cr es' cr', er cr er cr'
to make the two kinds of generators commute, so they may be separated
in a word.  According to GAP, the complete set of relators is
    er^2,
    v^2,
    cr^3,
    er cr er cr^-1,
    er cs er cs^-1,
    es cr es^-1 cr^-1,
    es cs es^-1 cs^-1,
    (v cr)^2,
    (v er)^2,
    cr cs cr cs^-1 cr^-1 cs cr^-1 cs^-1,
    (er es er es^-1)^2,
    cs cr^-1 cs^-1 v cs cr cs^-1 v cr,
    cs^-1 cr^-1 cs v cs^-1 cr cs v cr,
    (es er es^-1 v)^2 er,
    (es^-1 er es v)^2 er,
    cr cs^2 cr cs^-2 cr^-1 cs^2 cr^-1 cs^-2,
    (cs^-1 v cs v)^3,
    (er es^2 er es^-2)^2,
    (es^-1 v es v)^3,
    cr^-1 cs^-2 v cs^2 cr cs^-2 v cr cs^2,
    cr^-1 cs^2 v cs^-2 cr cs^2 v cr cs^-2,
    er es^-2 v es^2 er es^-2 v er es^2,
    er es^2 v es^-2 er es^2 v er es^-2,
    cr cs^3 cr cs^-3 cr^-1 cs^3 cr^-1 cs^-3,
    ((cs^-1 v)^2 (cs v)^2)^2,
    (er es^3 er es^-3)^2,
    ((es^-1 v)^2 (es v)^2)^2,
    cs^-3 v cs^3 cr cs^-3 v cr cs^3 cr^-1,
    cs^3 v cs^-3 cr cs^3 v cr cs^-3 cr^-1,
    (es^3 er es^-3 v)^2 er,
    (es^-3 er es^3 v)^2 er,
    (cs^-2 v cs^-1 v cs v cs^2 v)^2,
    (es^-2 v es^-1 v es v es^2 v)^2,
    (er es^4 er es^-4)^2,
    (es^-4 er es^4 v)^2 er,
    (es^4 er es^-4 v)^2 er,
    v cs^6 (v cs^-1)^7,
    (es^-3 v es^-1 v es v es^3 v)^2,
    (er es^5 er es^-5)^2,
    (es^5 er es^-5 v)^2 er,
    (es^-5 er es^5 v)^2 er,
    (es^4 v es^-4 v es^-1 v es v)^2,
    v es^10 (v es^-1)^11,                                     ((7))
which has 43 relators of total length 597.  It is apparently beyond
GAP's ability to verify that these relators present the cube group,
though I have verified some smaller wreath groups.

This presentation is of course in terms of generators {v,es,er,cs,cr},
not the generators {F,B,L,R,T,D} natural to the cube. But they can be
translated as follows.  Each quarter-turn Q can be expressed as a word
w(Q) over {v,es,er,cs,cr}, and adding the relators

    F' w(F), B' w(B), L' w(L), R' w(R), T' w(T), D' w(D)     ((8))

will create a presentation on eleven generators
{v,es,er,cs,cr,F,B,L,R,T,D}.  I estimate that the added relators will
be under 70 letters each, and probably less.  If it is desired to
completely eliminate {v,es,er,cs,cr}, that may be done by replacing
each of {v,es,er,cs,cr} with processes in terms of F,B,L,R,T,D,
throughout ((7,8)).  My understanding of the current state of the
software is that the processes will probably be less than 30
quarter-turns each.  This would yield a presentation of 49 relators
and perhaps 2000 letters.  It should be possible to improve this quite
a bit.  I would suggest:

    1. Choosing the corner and edge numbering to reduce the
           rewriting blowup,
    2. Allowing w(Q) to use previously-related Q's as well as
           {v,es,er,cs,cr}.
    3. Adding new relators to abbreviate higher powers, especially
           of es and cs, in the presentation.
    4. Introducing short relators such as F^4=FBF'B'=e to cut down on
           the general verbosity of the relators.

But improvement to the level of actual comprehensibility may require
new ideas.  Perhaps Dave Rusin's "clearer statement" of the question
may help, if I can figure out what it means.

Dan                                        posted and e-mailed
Hoey@AIC.NRL.Navy.Mil

From hoey@aic.nrl.navy.mil  Sun Dec 17 02:47:48 1995
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From: hoey@aic.nrl.navy.mil
Message-Id: <9512170747.AA27428@sun13.aic.nrl.navy.mil>
To: Cube-Lovers@life.ai.mit.edu
Subject: Re: Presenting Rubik's Cube

For the benefit of Cube-Lovers, here is rusin@washington.math.niu.edu
(Dave Rusin)'s remark on finding a presentation of Rubik's cube.

  You have a group  Rubik  generated by the 6 90-degree rotations  g_i.
  Let  F  be the free group on 6 generators  x_i and  f: F --> Rubik the
  obvious homomorphism. There is a big kernel  N  of  f. (It is actually a
  free group: subgroups of free groups are free). You wish to find the
  smallest (free) subgroup  K  of  N  such that  N  is the normal closure
  of  K  in  F. (When you give a presentation of  Rubik  in the form
          Rubik = <g_1, ..., g_6 | word_1=1, word_2=1, ...>,
  you are implicitly describing  K  as the subgroup of  F  generated by the
  corresponding words in the   x_i.)
          To give this process at least a chance of success, you abelianize it:
  Let  N_ab  be the free abelian group  N/[N,N], so that there is a natural
  map from  N  into  N_ab. Since  N  is normal in  F  and  [N,N]  is 
  characteristic in  N, the action of  F  by conjugation on  N  lifts to an
  action of  F  on  N_ab; even better, the subgroup  N < F  acts trivially on
  N_ab, so that  F/N  (i.e., the Rubik group itself) acts on  N_ab.
  We think of  N_ab  as a  Rubik-module (or better, as a  Z[Rubik]-module).
  The subgroup  K < N  also maps to a subgroup K[N,N]/[N,N]  of  N_ab;
  significantly,  N  is the F-closure of  K  iff  N=[K,F]K so that N_ab
  is generated as a  Z[Rubik]-module by  F.
          Thus, the question of what constitutes a minimal set of
  relations is the same as asking for the number of generators needed for
  a certain  Rubik-module. (Of course, while you're at it, you might as well
  ask for a whole presentation or resolution of the Rubik-module. Inevitably,
  you will be led to questions of group cohomology.)

He also included GAP's help file on the cube, which I think has been
posted here already.

Dan
Hoey@AIC.NRL.Navy.Mil

From hoey@aic.nrl.navy.mil  Sun Dec 17 03:12:40 1995
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From: hoey@aic.nrl.navy.mil
Message-Id: <9512170812.AA27433@sun13.aic.nrl.navy.mil>
To: Jerry Bryan <jbryan@pstcc.cc.tn.us>, Cube-Lovers <cube-lovers@ai.mit.edu>
Subject: Re: Million dollar cube

Jerry Bryan wonders about the 15th anniversary celebration:

> That would make it 1980.  Is that right?  I think Cube-Lovers started in
> 1980, but I have just been reading some early stuff from Singmaster dated
> 1978 and 1979. 

In _Rubik's Cubic Compendium_, Erno Rubik remarks that his major
insight occurred in 1974.  He patented the cube in January, 1975 and
it went on sale in Hungary in 1977.

In 1980, one million were sold in Hungary, and U.S. distribution
through Ideal began.  Incidentally, they were always the "Magic Cube"
until Ideal renamed them.

There is some more information in the archives about Bela Szalai
(Logical Games, Inc), who sold the white-faced cubes in the U.S. after
seeing the cube in Hungary in 1978.  I'm not sure whether he actually
beat Ideal to the ship date, or what happened to him after the big
cube bust.

Dan
Hoey@AIC.NRL.Navy.Mil

From diamond@jrdv04.enet.dec-j.co.jp  Sun Dec 17 19:51:58 1995
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From: Norman Diamond  18-Dec-1995 0949 <diamond@jrdv04.enet.dec-j.co.jp>
To: cube-lovers@ai.mit.edu
Cc: hoey@aic.nrl.navy.mil
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Dan Hoey writes:
>There is some more information in the archives about Bela Szalai
>(Logical Games, Inc), who sold the white-faced cubes in the U.S. after
>seeing the cube in Hungary in 1978.  I'm not sure whether he actually
>beat Ideal to the ship date, or what happened to him after the big
>cube bust.

I bought one from him before Ideal's stuff appeared in stores, so I
think he can be considered to have beaten them.  However, when he
re-sized the tabs on the cubies so that the cube wouldn't seem ready
to explode, I think Ideal was shipping.  I wonder what happened to
him during the cube's other explosion (i.e. popularity) let alone
the bust.

-- Norman Diamond            diamond@jrdv04.enet.dec-j.co.jp
[Speaking for Norman Diamond not for Digital.]

From hoey@aic.nrl.navy.mil  Sun Dec 17 21:11:37 1995
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In-Reply-To: hoey@aic.nrl.navy.mil's message of 17 Dec 1995 07:41:01 GMT
Newsgroups: sci.math
Subject: Re: Presenting Rubik's Cube
References: <mschoene.817657981@astoria> <1995Nov29.054118.9651@cs.rit.edu> <9511291210.Hoey@aic.nrl.navy.mil> <49ihhd$j20@muir.math.niu.edu> <HOEY.95Dec17024101@sun13.aic.nrl.navy.mil>
Message-Id: <9512172110.Hoey@AIC.NRL.Navy.Mil>
Distribution: 

In my article on a presentation of the Rubik's cube group last night,
I omitted a relator from list ((7)):

    v es v cs v es^-1 v cs^-1.

This brings the number of relators to 44, with a total length of 605.
Experiments with GAP on some smaller cube-like groups indicate that
with this addition, the presentation is correct.

My apologies for the error.

Dan Hoey                                Posted and e-mailed.
Hoey@AIC.NRL.Navy.Mil

From geohelm@pt.lu  Mon Dec 18 01:59:51 1995
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From: Georges Helm <geohelm@pt.lu>
Subject: Re: Generating Rubik's Cube
Message-Id:  <9512180759.aa10286@mailsvr.pt.lu>

It is Bandelow Christoph: Inside Rubik's cube and beyond

>>  or BANDEMEISTER (sp?) "Beyond R. Cube"
>
Georges Helm           
geohelm@pt.lu


From bagleyd@source.asset.com  Mon Dec 18 12:33:07 1995
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From: bagleyd@source.asset.com (David A. Bagley)
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To: cube-lovers@ai.mit.edu
Subject: neat puzzle web site

Hi
   I was at the Puzzlette's web site and its pretty amazing.

http://www.puzzletts.com/

Cheers, 
  /X\   David A. Bagley
 // \\  bagleyd@perry.njit.edu
((   X  xlockmore, new stuff for xlock @ ftp.x.org//contrib/applications
 \\ //  altris, tetris games for x @ ftp.x.org//contrib/games/altris
  \X/   puzzles, magic cubes for x @ ftp.x.org//contrib/games/puzzles

From serge@nexen.com  Mon Dec 18 13:49:36 1995
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From: Serge Kornfeld <serge@nexen.com>
Message-Id: <199512181847.NAA00173@spank.nexen.com>
To: cube-lovers@ai.mit.edu
Subject: [bagleyd@source.asset.com: neat puzzle web site]

>>Subject: neat puzzle web site

>>Hi
>>   I was at the Puzzlette's web site and its pretty amazing.

>>http://www.puzzletts.com/

>>Cheers, 
>>  /X\   David A. Bagley
>> // \\  bagleyd@perry.njit.edu
>>((   X  xlockmore, new stuff for xlock @ ftp.x.org//contrib/applications
>> \\ //  altris, tetris games for x @ ftp.x.org//contrib/games/altris
>>  \X/   puzzles, magic cubes for x @ ftp.x.org//contrib/games/puzzles



I was at "http://www.puzzletts.com/" also and I like it.
You can also try "http://www.gametrends.com".



Serge serge@nexen.com

From geohelm@pt.lu  Tue Dec 19 02:05:10 1995
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To: cube-lovers@ai.mit.edu
From: Georges Helm <geohelm@pt.lu>
Subject: cube literature
Cc: geohelm@pt.lu
Message-Id:  <9512190804.aa14245@mailsvr.pt.lu>

A list of solutions to Rubik's cube (from 2x2x2 to 5x5x5), pyraminx...
is available now online at my homepage. There are +/- 600 items. The list
is 12k.
http://ourworld.compuserve.com/homepages/Georges_Helm/cubbib.htm

Georges
Georges Helm           
geohelm@pt.lu


From JBRYAN@pstcc.cc.tn.us  Tue Dec 19 17:41:31 1995
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Date: Tue, 19 Dec 1995 17:43:03 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Physical Cubes and Models Thereof
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The general subject of physical cubes and mathematical models thereof
has been discussed many times before, but I have never been totally
satisfied with all of the conclusions.  I'm going to take one more
crack at it.

Let's start with the question of what constitutes a single move and
the argument between the quarter-turners and the half-turners.  There
are good and valid arguments on both sides of the question, and there
is no one "right" answer.  However, the strongest and most succinct
argument in favor of quarter turns is that they are conjugate.  In
the case of the standard 3x3x3 cube, the set Q of twelve quarter
turns is M-conjugate, where M is the set of 48 rotations and
reflections of the cube.

A quarter-turner would normally generate G as G=<Q>.  But given that
Q is M-conjugate, we could say equivalently that G=<{m'Xm | m in M}>
for any X in Q.

     Question: for the 3x3x3, are there any elements X in G other
     than those X in Q itself where we can generate G as G=<{m'Xm |
     m in M}>?  Remember that in most cases we would have 48
     generators available.  

     Clearly, there are X in G such that <{m'Xm | m in M}> does not
     generate G.  For example, the M-conjugates of F2 do not
     generate G.  But I have a feeling that any group that is
     generated by <{m'Xm | m in M}> is an "M-symmetric group" (using
     the term "M-symmetric" very loosely and informally) and is
     therefore a somewhat interesting group.

For the 4x4x4, I will use upper case letters for outer slab moves
(face moves) and lower case letters for inner slab moves.  For
example, L'l'rR would rotate the entire cube away from you by 90
degrees, but the cube would otherwise look unchanged.  If we denote
the set of outer slab moves as Q and the set of inner slap moves as
q, then we can generate a group as G4=<Q,q>.  I am hesitant to say
that G4 is "the" cube group for the 4x4x4, because it is so hard to
agree on what "the" cube group is for higher order cubes.  But in any
case, Q and q are not M-conjugate with each other.  There is in fact
no way to have M-conjugate generators for the 4x4x4 and higher
physical cubes.

For a mathematical model, conjugacy can be repaired.  For example,
there is an operation called Evisceration where inner slabs and
adjacent parallel outer slabs are exchanged.  There is also an
operation called Inflection where inner slabs are exchanged with
their parallel inner slabs, and Exflection where outer slabs are
exchanged with their parallel outer slabs.  We can use Rotations,
Reflections, Evisceration, Inflections, and Exflections to generate a
192 element symmetry group for the 4x4x4 called M4.  We can then show
that Q and q are M4-conjugate, and conjugacy is repaired.  That is,
we can generate G4 as G4=<{m'Xm | m in M4}> for any X in q or Q. 
(See Dan Hoey's article "Eccentric Slabism, Qubic, and S&LM" dated 
1 June 1983.)

     In the previous paragraph, I used the term "symmetry group"
     quite deliberately, although some of you may not agree with the
     way I used it.  I am still struggling to understand how
     narrowly or loosely we should really construe the preservation
     of a geometric property before we declare a permutation to be a
     symmetry.  In the case of M4 above, I think the designation of
     "symmetry" is warranted, although it is a looser interpretation
     than is typical.

But my purpose is to model physical cubes.  Evisceration is not
possible on physical cubes.  Conjugate quarter turn generators are
not possible for physical cubes larger than the 3x3x3 without
Evisceration (or its generalization to the NxNxN case).  Therefore,
we abandon M-conjugation and its generalizations as a criterion for
modeling physical cubes.

Dan's Eccentric Slabism article talked about slab moves (a single
plane of cubies turning together) and cut moves (all the cubies on
each respective side of a plane cut of the cube turning together). 
Evisceration convinced Dan to convert from a Cutist view of the cube
to a Slabist view of the cube.  But Dan fully endorsed the Slabist
view only for even-sided cubes.  His phrase "Eccentric Slabism" refers
to the fact that he still refused to make slab moves for the center
slabs of odd-sided cubes.  The problem is that center slab moves
break M-conjugacy and its generalizations.  But I've already given up
M-conjugacy and its generalizations.  Given that, it seems unnatural
to leave out the center slab moves, so we leave them in.

We next confront the issue that physical cubes are rotated in space
with abandon.  Different rotations of physical cubes are considered
to be equivalent, and/or rotations of physical cubes are considered
to be zero cost operations.  But we desire a mathematical model of a
physical cube to be a group.

My preferred non-computing model of this situation is to treat the
various configurations of the cube as cosets of C, the set of 24
rotations of the cube.  However, this model is awkward for computing. 
For something like the 2x2x2, we more typically do something like
fixing a corner.  We hereby adopt  "fixing a corner" as the solution
for the general NxNxN case.  See below for more details of how we
propose to do so in the general case.

We can note several things about the "fixing the corner" model:

     1.   It breaks M-conjugation.  But we gave up M-conjugation
          anyway.  Consider the 2x2x2 as a good example.  If we
          insist on treating different rotations as equivalent,
          then the 2x2x2 really isn't M-conjugate.  I am simply
          suggesting that the NxNxN physical cube really isn't 
          M-conjugate, no matter the value of N, if we treat
          different rotations as equivalent.

     2.   With the "cosets of C" model, we can make the cosets into
          a group by taking as a representative for each coset the
          unique element which fixes the same corner.  There is
          then an easy isomorphism between the "cosets of C" model
          and the "fixed corner model".  My only trouble with the
          "cosets of C" model is that I keep wanting to call it G/C,
          and you can't call it that.  C is not a normal subgroup
          of G, and we cannot speak of G/C as a factor group of G.

     3.   We can have conjugation and we can have symmetry with a
          "fixed corner" model.  It is just not M-conjugation. 
          Rather, it is the symmetry that preserves the fixed
          corner, and conjugation within that symmetry group.

The 4x4x4 is a good example of how we propose to "fix the corner" for
the general NxNxN case.  Consider our status after L'l'r.  A physical
cubist would say that you were only one move from Start, and would
"solve" the cube simply with R.  But R would yield L'l'rR, which would
leave the cube rotated.  This is fine for our physical cube, but not
so fine for a mathematical model of a physical cube which seeks to
fix a corner.  Hence, we define R as R=Llr', and similarly for the
other slab moves which would otherwise move the fixed corner.  The
generalization to cubes higher than 4x4x4 is obvious.

Actually, I would prefer a slightly different but equivalent
definition for those slab moves which fix a corner.  Frey and
Singmaster use script letters for whole cube moves (those moves in
C).  I would implement R as follows: perform R in the normal sense of
the operation composed with Script-R' (and similarly for other slab
moves that would move the otherwise fixed corner).  So for the 4x4x4,
let's suppose we fixed the TRF corner.  Our generators would be,

    L',l',r,(R)(Script-R'),
    B',b',f,(F)(Script-F'),
    D',d',t,(T)(Script-T')

and their inverses.

Clearly, the same technique works not only for the 4x4x4 and above,
but also for the 2x2x2 and for the 3x3x3.

I am thinking of this in a Slabist interpretation.  However, a case
could be made that the (R)(Script-R') type of moves are really Cutist
moves.

I think all the other problems associated with a mathematical model
of a physical cube can be unified under the heading "Invisible Moves
of Facelets".  The most obvious example is that the Supergroup is
invisible on the 3x3x3 unless the orientations of the face centers
are marked somehow or other.  But with larger cubes (e.g., the face
centers of the 4x4x4), it is not just changes in orientation that are
invisible; there are also invisible changes in location.

In all cases, I would propose initially modeling the "larger group"
(call it L), where invisible changes in location and orientation are
visible.  Number all 16 facelets of each face on the 4x4x4, for
example.  You do have to decide how "large" you wish your larger group
L to be.  For example, to make invisible orientation changes visible,
you have to give a facelet four numbers rather than just one.  The
set of all positions that are equivalent when the "invisible" changes
are ignored  is a subgroup K.  Your final model is then the cosets of
K in L.

The "cosets of K in L" model will always work, but it may be difficult
to deal with computationally.  Ideally, you would be able to find a
subgroup G of L for which you could find an easy isomorphism with the
cosets of K.

As an example, consider the Supergroup of the 3x3x3 and call it L. 
Within L, there is a subgroup K which fixes the corners and edges.  K
is just all the legal face center reorientations.  Therefore, if we
wish to ignore face center orientations our model can be the cosets
of K in L.  There is an easy isomorphism between the cosets of K in
L, and our standard model for the 3x3x3 which we call G.  

In truth, we would never model the 3x3x3 in such a convoluted
fashion.  We would just use G and be done with it.  But for the 4x4x4
and larger cubes, I am not sure there is any choice.

As the cubes get larger, you would generally find that there was
a nested sequence of subgroups  --  K_0, K_1, etc.  -- for which the 
cosets of K_n in the larger group L would produce a useful model.  
For example, on the 4x4x4 one of your K's might be the group of 
permutations that fixed everything but the positions of the center 
facelets within a face (keeping them the proper color, of course).  
But a more stringent K might be the group of permutations that fixed 
everything but the orientations of the center facelets within a face.

I will end by pointing out that Goldilocks would really like the
3x3x3.  Papa Bear's 4x4x4 is too large and Baby Bear's 2x2x2 is too
small.  But Mama Bear's 3x3x3 is just right.  The physical 3x3x3 is
the only physical NxNxN which can be modeled with M-conjugate
generators (assuming we fix the face centers).  And the 3x3x3 is the
NxNxN (physical or mathematical) with the nicest isomorphism between
the cosets of K in L and some reasonable group G.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =          
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us                      


From joemcg@catch22.com  Wed Dec 20 06:16:26 1995
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Date: Wed, 20 Dec 1995 03:20:21 -0800 (PST)
From: Joe McGarity <joemcg@catch22.com>
To: "Rubik's Cube Mailing List" <cube-lovers@ai.mit.edu>
Subject: Luminations
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Yes, I have a Luminations, but no I don't know where to get one.  They 
haven't been on the shelves for about three years.  Someone on this list 
has got to have two of them.  Would the list administrator be opposed to 
some good old-fasioned commerce?  I'll bet for everybody that's looking 
for something there are three people who have an extra one.


------------------------------------------------------------------------------

Joe McGarity                      "Do you expect me to talk?"   
418 Fair Oaks                       
San Francisco, CA  94110          "No, Mr. Bond.  I expect you to die."  

joemcg@catch22.com                           

------------------------------------------------------------------------------



From mouse@collatz.mcrcim.mcgill.edu  Wed Dec 20 06:25:04 1995
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Date: Wed, 20 Dec 1995 06:24:28 -0500
From: der Mouse <mouse@collatz.mcrcim.mcgill.edu>
Message-Id: <199512201124.GAA04566@Collatz.McRCIM.McGill.EDU>
To: cube-lovers@ai.mit.edu
Subject: Re:  Physical Cubes and Models Thereof

> I would propose initially modeling the "larger group", where
> invisible changes in location and orientation are visible.  Number
> all 16 facelets of each face on the 4x4x4, for example.  [...].  For
> example, to make invisible orientation changes visible, you have to
> give a facelet four numbers rather than just one.

The only facelet for which invisible orientation changes are even
possible is the center facelet on an odd-order cube.  Other facelets
always have a fixed orientation with respect to the center of the face
they're on at the moment.  (On the 4x4x4, for example, if you mark
every facelet for orientation, you will find that each center facelets
always has the same corner to the face center, regardless of which face
it's on.)

					der Mouse

			    mouse@collatz.mcrcim.mcgill.edu

From JBRYAN@pstcc.cc.tn.us  Wed Dec 20 08:42:45 1995
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Subject: Re: Physical Cubes and Models Thereof
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On Wed, 20 Dec 1995, der Mouse wrote:

> The only facelet for which invisible orientation changes are even
> possible is the center facelet on an odd-order cube.  Other facelets
> always have a fixed orientation with respect to the center of the face
> they're on at the moment.  (On the 4x4x4, for example, if you mark
> every facelet for orientation, you will find that each center facelets
> always has the same corner to the face center, regardless of which face
> it's on.)

I believe Der Mouse is entirely correct for the physical cube case (which
is the case I was talking about).  Imagine a 99x99x99 or some such large
cube, and for each facelet except the face center itself mark the corner
closest to the face center.  Any slab quarter-turn preserves the fact that
all marked corners remain closest to the face center.  There are two cases
-- a face slab, and any inner slab.  But both cases work.  In the case of
a face slab, the orientations of the facelets on the face of the slab do
change, but the orientations change in lock step with the positions of
the facelets. 

In the case of a mathematical model, Evisceration also preserves facelet
orientation, if I understand correctly how first Singmaster and then Dan
Hoey defined Evisceration.  However, Inflection and Exflection do not
preserve facelet orientation.  Could (or should) the definitions of
Inflection and Exflection be broadened to include and preserve facelet
orientation? 

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =          
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us                      


From SCHMIDTG@beast.cle.ab.com  Wed Dec 20 13:23:50 1995
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Date: Wed, 20 Dec 1995 12:48:47 -0500 (EST)
From: SCHMIDTG@beast.cle.ab.com
To: cube-lovers@ai.mit.edu
Message-Id: <951220124847.202054c5@iccgcc.cle.ab.com>
Subject: luminations


I have some new (unopened) luminations puzzles.
Contact me if you are interested.

-- Greg

From bagleyd@perry.njit.edu  Thu Dec 21 11:10:52 1995
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From: david a bagley <bagleyd@perry.njit.edu>
Message-Id: <199512211617.LAA00923@perry.njit.edu>
Subject: xpuzzles and winpuzz
To: cube-lovers@life.ai.mit.edu
Date: Thu, 21 Dec 1995 11:16:59 -0500 (EST)
X-Mailer: ELM [version 2.4 PL23]
Content-Type: text

Hi
 
My new puzzles are out again.  Here's a brief description:
  5.1
    Mball and Mlink puzzles now draw sectors faster.
    All puzzles have a corrected random number generator for 64 bit
      machines.
    Border color around tiles/pieces makes it look more realistic.
    g (& G) for get of old saved configuration (not e).
    Many other cosmetic changes in the code.  I am getting it in sync
      with MSWindows code (winpuzz).

I hope I don't regret announcing this:  :)
  I am busy porting them to MSWindows.  So far I only ported one, "xcubes".
  I think the rest will be easier now that I have my X-Window-System code
  in sync.  I am looking for anyone with MSWindows AND C experience to Beta
  test and give me some pointers.  (If I could get a small team that would
  be great!)  The executable AND source for MSWindows, when completed will
  be freely redistributable and maintained (as far as I am able to).  So
  far, I am only using windows.h (3.1) to maximize portablity and compiler
  independance.
 
Cheers,
  /X\   David A. Bagley
 // \\  bagleyd@perry.njit.edu
((   X  xlockmore, new stuff for xlock @ ftp.x.org//contrib/applications
 \\ //  altris, tetris games for x @ ftp.x.org//contrib/games/altris
  \X/   puzzles, magic cubes for x @ ftp.x.org//contrib/games/puzzles


From joemcg@catch22.com  Sun Dec 31 11:49:53 1995
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Date: Sun, 31 Dec 1995 08:51:51 -0800 (PST)
From: Joe McGarity <joemcg@catch22.com>
To: "Rubik's Cube Mailing List" <cube-lovers@ai.mit.edu>
Subject: The second challange
Message-Id: <Pine.LNX.3.91.951231084824.31956B-100000@B17.Catch22.COM>
Mime-Version: 1.0
Content-Type: TEXT/PLAIN; charset=US-ASCII


Does anyone have or know where to get Rubik's Magic:  The Second 
Challange?  It is a larger version of the Link the Rings puzzle.  I see 
it advertised in a little flyer that came with Link the Rings, but I have 
never seen a real one.  Also the 4x4x4 Rubik's Revenge eludes capture.  
Happy New Year everyone.

Joe


------------------------------------------------------------------------------

Joe McGarity                      "You'll shoot your eye out."  
P. O. Box 993082                       
Redding, CA  96099-3082             

joemcg@catch22.com                           

------------------------------------------------------------------------------



From S005AXR@desire.wright.edu  Sat Jan  6 17:20:27 1996
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s005axr@discover.wright.edu


From listmast@telegrafix.com  Sun Jan  7 08:17:07 1996
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From: TeleGrafix Information <info@telegrafix.com>
To: customer@telegrafix.com
Subject: Introducing RIP-2 Multimedia Graphics for the Internet
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Happy New Year from TeleGrafix:

     Since you are an important colleague of ours in the online media 
community, TeleGrafix is sending this short news release to brief you 
about an important new Internet technology.  

     Following two years of development, TeleGrafix Communications 
is giving away free communications software that allows you to use 
the new RIPscrip-2 (Remote Imaging Protocol-2 scripting language) 
Internet online multimedia technology.  

     We invite you to sample "RIP-2" multimedia on TeleGrafix's 
Vector Sector BBS at (714) 379-2133.  To fully experience it, please 
download the "shareware" RIPterm v2.2 communications software 
from the BBS.  RIP-2 technical data and RIPterm v2.2 also are 
available for download at http://www.telegrafix.com on the World 
Wide Web.  Browser "plug-ins" to permit viewing of RIP-2 
multimedia on the Web are slated for release in early 1996. 

     RIP-2 enables you to create TV-style multimedia presentations or 
electronic newspapers that fly through the Internet and ordinary 
phone lines at dazzling speeds using regular modems.  RIP-2 encodes 
graphics as hyper-compressed ASCII text files that are as little as 
one-tenth the size of other formats.  It works on any computing 
platform or communications network that uses 7-bit or 8-bit ASCII 
text.  We expect RIP-2 to quickly become an important Internet 
technical standard like HTML, Java or VRML.  

     TeleGrafix is now accepting requests from software developers 
and online system operators who want copies of the RIP-2 Internet 
multimedia language specification when it is published in early 1996.  
The first generation of RIP technology, introduced in 1993, is the 
world's BBS graphics standard. It is used on thousands of BBS 
systems, and is supported by dozens of online software vendors 
including Delrina, Galacticomm, Hayes and Mustang. 

If this message has reached you in error or if you are no longer 
interested in RIPscrip technology,  please tell us via E-mail so you 
won't get additional information. We look forward to helping you, 
and we wish you a Happy New Year. 

     Sincerely,

     Pat Clawson          Mark Hayton         Jeff Reeder
     President/CEO        VP/Technology       Chairman & CyberWizard

                   TeleGrafix Communications Inc.
                  16458 Bolsa Chica Road, Suite 15
                 Huntington Beach, California 92649

          Voice: (714) 379-2131        Fax: (714) 379-2132
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Huntington Beach, CA 92649                  BBS: (714) 379-2133
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FTP: ftp.telegrafix.com



From velucchi@cli.di.unipi.it  Fri Jan 12 13:01:39 1996
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Organization:  Centro di Calcolo - Dip. di Informatica di Pisa - Italy
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From: Mario Velucchi <velucchi@cli.di.unipi.it>
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Subject: CUBE PUZZLE
To: cube-lovers-request@ai.mit.edu
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	Dear FRIENDS,

Is this list only for CUBE PUZZLE or
PUZZLE in general?

THANKS


-- 
Yours Sincerely, MV
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\////////////////////////////////
 Mario Velucchi                              University of PISA
 Via Emilia, 106                 Department of Computer Science
 I-56121 Pisa                   e-mail:velucchi@cli.di.unipi.it
 ITALY                      talk:velucchi@helen.cli.di.unipi.it
                http://www.cli.di.unipi.it/~velucchi/intro.html
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\////////////////////////////////

From mark.longridge@canrem.com  Sun Jan 14 22:25:30 1996
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To: cube-lovers@life.ai.mit.edu
Reply-To: CRSO.Cube@canrem.com
Sender: CRSO.Cube@canrem.com
Subject: Cube Theory
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1278.5834.0C206F74@canrem.com>
Date: Sun, 14 Jan 96 22:03:00 -0500
Organization: CRS Online  (Toronto, Ontario)

Here is some more Cube Theory:

On the standard Rubik's Cube

Using < R, L, F, B, U > to generate D1
Let A = R1 L3 F2 B2 R1 L3, then A U1 A = D1
A U1 A = R1 L3 F2 B2 R1 L3 U1 R1 L3 F2 B2 R1 L3   (17 q, 13 q+h)

Using < R2, L2, F2, B2, U2 > to generate D2
Let A = R2 F2 B2 L2,       then A U2 A = D2
A U2 A = R2 F2 B2 L2 U2 R2 F2 B2 L2               (18 q, 9 q+h)

Slice group pattern, 6 spot, 4 slice moves
p1 = (F1 B3) (L1 R3) (U1 D3) (F1 B3)           (8 q)

Slice group antipode, 6 spot + pons asinorum, 6 slice moves
p2 = (F2 B2) (T1 D3) (F1 B3) (L3 R1) (T1 D3)   (12 q)
p2^6 = I

p7a  Cube in a cube   U2 F2 R2 U3 L2 D1 (B1 R3) ^3 + D3 L2 U1
(15 q+h, 20 q)

   if  A = U3 L2 D1,  then let A' = inverse of A

p7a = U2 F2 R2 + A + (B1 R3) ^3 + A'

Or if we want something more symmetric, there is Mike Reid's...

p7b  Symmetric Maneuver (R3 U1 F2 U3 F3 L1 F2 L3 F1 R1  C_X ) ^ 2
(20 q+h , 24q)

One might even call this maneuver to be "cyclic decomposable".
Even the first half of this sequence generates an interesting pattern.
It would appear that using symmetric maneuvers does not ensure minimal
q or q+h turns.

Perhaps p7a is simpler in terms of notational expression. p7a is
how I actually do "Cube in a cube" in real cubing. Note also that
p7a uses all 6 generators and p7b uses <U, L, R, F> and there may
be a tighter symmetric "Cube in a cube".

Mike, have you tried using the pattern generated by the first half
under your Kociemba algorithm for q turns??
----------------------------------------------------------------------

Megaminx (platonic dodecahedron)  12 faces, 20 corners, 30 edges

tetrahedron  = 4 axis
cube         = 3 axis
dodecahedron = 6 axis

In constructing a dodecahedron, build a bottom, place the 5 adjacent
faces to form a bowl. The top edges now form a skew decagon. Build
another bowl and connect the two bowls together to form a dodecahedron.

-------------------------------------------------------------

Recall that the Halpern-Meier Tetrahedron has 3,732,480 states.
In this count we consider the 4 centre pieces immobile.

The picture H-M tetrahedron has 3,732,480 * 3^3 = 100,776,960 states.
In essence, we can rotate any 3 centres at will, the 4th is forced.
That number may seem familar to some of the more fanatical cubists,
has the picture Skewb has 3,149,280 * 2^5 = 100,776,960 states!

Additionally the possible rotations of the centres of the H-M
tetrahedron are  (), (+ -), (+++) , (---), (-++-), (+--+), (+++-),
(--+-)

#   Order (tetra, r * d) = 45         OR      (r+ d+)^45 = I
Note that (r+ d+)^45 is still the identity, even on the picture
H-M tetrahedron, as 45 is divisible by 3.
-------------------------------------------------------------

-> Mark <-


From mreid@ptc.com  Thu Jan 18 14:13:01 1996
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Date: Thu, 18 Jan 1996 14:38:00 -0500
From: michael reid <mreid@ptc.com>
To: cube-lovers@ai.mit.edu
Subject: Re:  Cube Theory

mark writes

[ ... ]

> p7a  Cube in a cube   U2 F2 R2 U3 L2 D1 (B1 R3) ^3 + D3 L2 U1
> (15 q+h, 20 q)

for what it's worth, on may 19, 1992, i gave the maneuver

     L1 F1 L1 D3 B1 D1 L2 F2 D3 F3 R1 U3 R3 F2 D1  (15f, 18q)

which is minimal in both the face turn and the quarter turn metric.

[ ... ]

> Or if we want something more symmetric, there is Mike Reid's...
> 
> p7b  Symmetric Maneuver (R3 U1 F2 U3 F3 L1 F2 L3 F1 R1  C_X ) ^ 2
> (20 q+h , 24q)

i think this maneuver is well-known, so it shouldn't be attributed
to me.  at least it seems to be a "standard" way of producing the
cube in a cube pattern.

[ ... ]

> Mike, have you tried using the pattern generated by the first half
> under your Kociemba algorithm for q turns??

do you mean the position generated by   R3 U1 F2 U3 F3 L1 F2 L3 F1 R1
(which is just a three cycle of corner-edge pairs) ?  this maneuver
is minimal in both metrics.

back in the days before i did computer-cubing, i found an interesting
maneuver for "cube in a cube" that only turns three faces.  what's the
shortest such maneuver that you know?

mike

From mark.longridge@canrem.com  Tue Jan 23 18:00:31 1996
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Subject: New Rubik's Cube Web Page
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1281.5834.0C2081FF@canrem.com>
Date: Tue, 23 Jan 96 17:19:00 -0500
Organization: CRS Online  (Toronto, Ontario)

I've started my own Web Page for Rubik's Cube.

It has my rubik's chronology and full move list.

Eventually all issues of Domain of the Cube will reside there!

www.dis.on.ca/~cubeman

email: mark.longridge@canrem.com
or     cubeman@admin.dis.on.ca


-> Mark <-


From mreid@ptc.com  Mon Feb  5 16:05:20 1996
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Date: Mon, 5 Feb 1996 16:31:26 -0500
From: michael reid <mreid@ptc.com>
To: cube-lovers@ai.mit.edu
Subject: cube in a cube

recently i wrote

> back in the days before i did computer-cubing, i found an interesting
> maneuver for "cube in a cube" that only turns three faces.  what's the
> shortest such maneuver that you know?

there was never any response to this, but i'll give my solution anyway.

let  X  =  U2 F R' F2 R F2 R F2 R' F U2 .

then  X  produces two two-cycles of corner-edge pairs.  the commutator
[ X , C_UFR ]  produces "cube in a cube" in 22 face / 32 quarter turns
and only turns the faces  U, F  and  R.

the notation  C_UFR  refers to a rotation of the whole cube, and
[ a, b ]  denotes the commutator  a b a^-1 b^-1 .

mike

From rcs@cs.arizona.edu  Mon Feb  5 18:19:06 1996
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Date: Mon, 5 Feb 1996 16:19:03 MST
From: "Richard Schroeppel" <rcs@cs.arizona.edu>
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Received: by leibniz.cs.arizona.edu; Mon, 5 Feb 1996 16:19:03 MST
To: cube-lovers@ai.mit.edu
Subject: Group/graph status?

Has anyone tabulated the number of positions are reachable (from the
initial cube) in one move, two moves, etc.?  Is the diameter of the
graph known?

Rich Schroeppel   rcs@cs.arizona.edu



From news@nntp-server.caltech.edu  Mon Feb  5 21:19:04 1996
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Path: whuang
From: whuang@cco.caltech.edu (Wei-Hwa Huang)
Newsgroups: mlist.cube-lovers
Subject: Re: Group/graph status?
Date: 6 Feb 1996 02:18:21 GMT
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"Richard Schroeppel" <rcs@cs.arizona.edu> writes:
>Has anyone tabulated the number of positions are reachable (from the
>initial cube) in one move, two moves, etc.?  Is the diameter of the
>graph known?

Well, the diameter of the graph would obviously be the upper bound of
God's algorithm.


-- 
Wei-Hwa Huang, whuang@cco.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
Did you know Africa has more sand than the entire Sahara Desert?

From bagleyd@hertz.njit.edu  Tue Feb  6 10:00:56 1996
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Date: Tue, 6 Feb 1996 09:37:50 -0500
From: bagleyd <bagleyd@hertz.njit.edu>
Message-Id: <199602061437.JAA02959@hertz.njit.edu>
To: cube-lovers@life.ai.mit.edu
Subject: xpuzzles and winpuzz

Hi
 
My new X-Windows puzzles are out again.  Here's a brief description:
  5.2
    Puzzles now can change size dynamically.  Puzzles for the most
      part have no maximum size.  For example, you can have a 10x10x10
      Rubik's cube.
    Saved format has changed on most puzzles, should be more
      understandable.
    Drag and drop on a face to move pieces now works on all puzzles.
    Lesstif-0.36 works OK with all puzzles.  Lesstif still has many
      bugs but I have only seen cosmetic and bugs with radio buttons and
      sliders using the puzzles.
    Lots of minor bug fixes and minor improvements.
 
 
  MS Windows, port in progress, E-Mail author if interested
    They are still free and source code is still included. :)
    The best puzzles will be converted last, since they are more complicated.
    (xabacus -> wabacus: port complete (not a puzzle))
    xcubes -> wcubes: port complete
    xtriangles -> wtriangl: port in progress
 
Cheers,
  /X\   David A. Bagley
 // \\  bagleyd@hertz.njit.edu
((   X  xlockmore, new stuff for xlock @ ftp.x.org//contrib/applications
 \\ //  altris, tetris games for x @ ftp.x.org//contrib/games/altris
  \X/   puzzles, magic cubes for x @ ftp.x.org//contrib/games/puzzles

From JBRYAN@pstcc.cc.tn.us  Tue Feb  6 12:24:07 1996
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On Mon, 5 Feb 1996, Richard Schroeppel wrote:

> Has anyone tabulated the number of positions are reachable (from the
> initial cube) in one move, two moves, etc.?  Is the diameter of the
> graph known?

In the quarter turn metric, the number of positions has been tabulated 
out to eleven moves from Start.  The diameter of the graph is known to be 
at least 24 quarter turns because there is one position whose length has 
been verified to be 24q.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7127
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990


From JBRYAN@pstcc.cc.tn.us  Tue Feb  6 13:15:48 1996
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Date: Tue, 06 Feb 1996 13:17:44 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Large Searches with Small Memory
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I have access to less computing resources than I used to have. 
As a result, I have been thinking about ways to conduct large
searches with less memory.  There are no results here, just
some thoughts.

I will assume a quarter turn metric.  It would be easy to
generalize the proposals below to other metrics, but I will
not do so.

We let C[n] be the set of all positions of length n.  C[0] is
just {Start}, and C[1] is just Q, the set of twelve quarter
turns.  We let T[n] be the union of C[k] for k in 0..n.

We assume that any computerized breadth first search for God's
Algorithm would build (in turn) C[0], C[1], etc., and would
store each C[k] in memory in some reasonable indexed and
searchable data structure.  The details of the data structure
do not (I think) matter for the purposes of this note.

We assume that the primary constraint on the search is memory
size rather than time.  These days, most anybody has access to
a machine (or machines) where a problem can be allowed to run
for hundreds or thousands of hours if necessary.  A 486 or
Pentium on your desk would serve nicely, and a UNIX work
station would be even better.  I have used both a 486 (running
OS/2 for multitasking) and a UNIX work station in this manner. 
The machines could be used for other things while the cube
problems were running.

We assume that n is the largest n for which we can store C[n]. 
We assume that if we can store C[n], we can also store T[n]. 
For all practical purposes, T[n] and C[n] are about the same
size close to Start because C[n] is a little more than nine
times larger than C[n-1].

With this structure in hand, we could form all products XY for 
X and Y in T[n].  We would simply form the products; we would 
not try to store them.  But having done so, we would have created 
all positions in T[2n].  In some ways, this is not very useful.  
That is, in general we would not know the length of any 
particular XY, nor would we know the size of C[k] for k in
n+1..2n. 

But as we formed the products, we could check them against any
position of interest, or against a small set of positions of
interest.  We could then determine if any of the interesting
positions were in T[2n], and thus bound their length.  (We
assume that none of the interesting positions are in T[n]. 
Otherwise, we could determine their length directly and none
of these Draconian measures would be necessary.)  Such a 
half-depth search has been discussed quite a few times before in
Cube-Lovers.

But here follows what I think is a new idea.  What if we
formed all products XY for X in C[n] and Y in C[1].  Since
C[1] is Q, this is really just the procedure for a standard
depth first search.  But we can't store C[n+1].  Can we
determine the size of C[n+1] anyway?

Try the following.  The length of each XY is either n-1 or
n+1.  Verify which case we have by reference to C[n-1], and
throw away the products of length n-1.  But if the length of
XY is n+1, do we count it, or do we not?

The problem is that we might have XY=VW for X and V in C[n], Y
and W in C[1], X not equal V, and Y not equal W.  Which do we
count and which do we not?  Actually, there may be up to
twelve such products, so we need a way uniquely to determine
which product to count and which not to count.

Suppose we have XY in hand and wish to know whether to count
it.  Form all products (XY)Z for Z in C[1].  There are twelve
such products, and at least one of them will be in C[n].  We
assume that C[1] can be ordered (probably already is) by our
data structure.  Hence, we count XY only if Y'=Z*, where Z* is
the first Z such that XY(Z) is in C[n].  (Strictly speaking,
we would not have to form all products XY(Z).  We would stop
once we found the first Z such that (XY)Z was in C[n].  We
would then either have Y'=Z*, or we wouldn't.  But this only
reduces the number of products down from twelve to an average
of six.) 

It seems to me that this procedure would work in principle,
but I am not sure how practical it would be.  The problem is
that there would be a lot of products XY(Z) to calculate and
test.  Is there any shorter method to determine whether or not
to count XY?

I normally write my sets C[k] out to a file.  Any analysis I
wish to do is then run against the file after the fact.  With
the new procedure I am describing, any analysis of C[n+1]
would have to be done as the products were being formed.

We can use a similar procedure to determine the size of
C[n+2], C[n+3], etc. up through C[2n], but things get more
complicated and more impractical.

For C[n+2], form all XY for X in C[n] and Y in C[2].  All such
products are in C[n-2], C[n], or C[n+2].  As before, we look
up XY in C[n-2] and in C[n], and throw it away if it is
already there.  We then form all XY(Z) for Z in C[2] (there
will be 114 such products), and count the product only if
Y'=Z*, where Z* is the first Z in C[2] such that XY(Z) is in
C[n].  But this case is even more time consuming than the
C[n+1] case because we will on the average have to look at 57
products (57=114/2).

     As an aside, I have considered creating C[n+1] as the
     product of XY with X in C[n-1] and Y in C[2], rather than
     as the product of XY with X in C[n] and Y in C[1], even
     before running out of memory to store the results.  We
     still have to check for positions whose length is less
     than n+1, and we still have to check for duplicate
     positions of length n+1.  But using C[n-1] and C[2] 
     would automatically eliminate from the search duplicate
     positions such as ZRR'=Z or ZRL=ZLR.  Or better still,
     perhaps to create C[n+1] we should take X from C[n/2] and
     Y from C[(n/2)+1]?  Has anybody tried anything like that?

C[n+3] gets worse still.  If we form all XY for X in C[n] and
Y in C[3], the length of XY may be n-3, n-1, n+1, or n+3.  We
dispose of the n-3 and n-1 cases as before, but then we would
have to have a way to distinguish between the n+1 and n+3
cases.  I think the procedure becomes truly impractical at
this point.

Anyway, that's it.  Has anybody ever tried anything along the
lines I have outlined for a problem too big to store?  If so,
did it work? 

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7127
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990


From pbeck@qa.pica.army.mil  Tue Feb  6 15:25:05 1996
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Date: Tue, 6 Feb 1996 15:19:03 -0500
To: cube-lovers@ai.mit.edu
From: pbeck@qa.pica.army.mil (Peter Beck)
Subject: museum exhibition
Cc: ilanab@wam.umd.edu, jhbeck@gwis2.circ.gwu.edu, KCEBL@aol.com

The "Morris Museum",
6 Normandy Heights Road,
Morristown, NJ 07960,
201-538-0454
is having and exhibit called
"THESE ARE A FEW OF MY FAVORITE THINGS,
A LOOK AT COLLECTORS AND THEIR MAGNIFICENT COLLECTIONS".

I have lent them the following Rubik's cube items to be included.

1.  ephemera
1.1  "Rubik's Cube Solution Poster"
1.2  "The Rubik's Cube Puzzle Poster" -
1.3  "Rubik's Cube Clinic Poster" -
1.4   "Rubik's Cube Button Pin Collection" -
1.5  "We Have The Original Promotion Poster" -
1.6  cardboard dipslay cube

2.  Puzzles
2.1 3x3x3 stuff
2.1.1  57 mm  3x3x3  curiosity cubes
2.1.1.1  Russian box, orange cardboard box with Russian writing
2.1.1.2  Deluxe cube, sold by Ideal Toy Co
2.1.1.3  Blindmans, source unknown
2.1.1.4  Disney Characters , source unknown
2.1.1.5  Pennant, source unknown
2.1.1.5  Calendar Cube,  Ideal Toy Co
2.1.1.6  Video cassette promo Cube,
2.1.2  57 mm 3x3x3 shape adaptations
2.1.2.1  ball in cube, unpackaged, custom non-production
2.1.2.2  "Rubik's World", sold by Ideal Toy Co
2.1.2.3  Japanese "Space Cube"
2.1.2.4  "Magique Cube", common name "Space Shuttle"
2.1.2.5  crystal, custom made by Greg Steven's Seattle WA
2.1.3  3x3x3 size variatations
2.1.3.1 20mm Ideal necklace cube, sold by Ideal Toy Co
2.1.3.2 38mm keychain cube, sold by Dynasty Products
2.2 non-3x3x3 variatations
2.2.1  2x2x2 called Rubik's Pocket Cube", sold by Ideal Toy Co
2.2.2  2x3x3 called "Magic Domino", sold by Politoys with a 1982 copyright,
this version is popularly know as the blind mans version because it has
raised spots for markings instead of paint or stickers.
2.2.3  4x4x4 called "Rubik's Revenge" , sold by Tsakuda in Japan
2.2.4  5x5x5 called "Rubik's Wahn", sold by Arxon, not distributed in USA

3.  Memorabilia
3.1  CUBE in Bottle, custom made by Lee Dian, Malaysia
3.2  Ceramic Lamp Base decorated as a cube, Jessica Beck, USA
3.3 "Le Vrai Magicube" build yourself a cube kit

CHange is IRresistible ...CHAnge is Irresistible ...CHange is IRResistible ...

                         Peter Beck ... aGENT of cHANGe
                         ^^^^^^^^^^
e-mail <pbeck@pica.army.mil> *
 temporary  B-92 x2383    b-62 x5580
 VOICE: 201-724-6684 * DSN:880 * 800-831-2759-1-#-*-4-6684
  FAX: 201-724-4026
   POSTAL  - ARDEC, P. Beck, B-92, Picatinny Arsenal, NJ 07806-5000
           ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
SOFTWARE QUALITY ENGINEERING BRANCH,
... ARDEC, AMSTA-AR-QAT-A, B-62, Picatinny Arsenal, NJ 07806-5000
  --> http://qa.pica.army.mil/qat/sqe/sqe.html * VOICE:201-724-5580 <--

... pb...pb...pb...pb...pb.._ 31 JAN 96 release _..pb...pb...pb...pb...pb...



From mark.longridge@canrem.com  Wed Feb  7 03:00:03 1996
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Subject: < U, F, R > group
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1288.5834.0C20A151@canrem.com>
Date: Wed,  7 Feb 96 02:47:00 -0500
Organization: CRS Online  (Toronto, Ontario)

> there was never any response to this, but i'll give my solution
anyway.

I am working on an engine to search optimal paths for < U, F, R > but
it's not done yet. It's certainly within the bounds of computibility:

Size (u_f_r) = 170,659,735,142,400 (hmmm, 170 trillion maybe not!)

> let  X  =  U2 F R' F2 R F2 R F2 R' F U2 .
>
> then  X  produces two two-cycles of corner-edge pairs.  the commutator
> [ X , C_UFR ]  produces "cube in a cube" in 22 face / 32 quarter turns
> and only turns the faces  U, F  and  R.
>
> the notation  C_UFR  refers to a rotation of the whole cube, and
> [ a, b ]  denotes the commutator  a b a^-1 b^-1 .
>
> mike

Brilliant.

Although harder to remember, X + ( X * C_UFR) will do.

   U2 F1 R3 F2 R1 F2 R1 F2 R3 F1 U2
   F2 R1 U3 R2 U1 R2 U1 R2 U3 R1 F2        (22 f, 32 q)

-> Mark <-


From mark.longridge@canrem.com  Wed Feb  7 03:00:03 1996
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Subject: Cube Musings
From: mark.longridge@canrem.com (Mark Longridge)
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Date: Wed,  7 Feb 96 02:48:00 -0500
Organization: CRS Online  (Toronto, Ontario)

Further Cube Musings
====================

We usually think of positions antipodal to start only, but there are
positions antipodal to any given position.

Given a small enough subgroup of the cube, i.e. one which we can
exhaustively study, it is not hard to determine some examples.

Let's use the square's group and the good ol' pons asinorum.
Pons is antipodal to position X.

        Pons + X = Antipode  (let's use position p135)

p135 2 X, 4 T       L2 B2 D2 F2 T2 F2 T2 L2 T2 D2 F2 T2 L2 D2 F2

Solving for position X is easy enough....

         X = Antipode - Pons

      Position X =  F2 D2 L2 D2 F2 L2 T2 F2 T2 F2 T2 F2 L2

The idea of        (-1) * pons   or  (-pons) is equivalent to
the inverse of pons, since (+pons) + (-pons) = identity.

So Pons and Position X are antipodes of each other. Using this
straightforward method we can find an antipode to any position
in the square's group, or for any other positions in another
small subgroup.

This brings up the idea of a "Rubik's Tour". Such a tour would
touch on a set of interesting patterns within a given subgroup,
or potentially the entire cube group. Of course, "God's Tour"
would not only touch on all the interesting patterns, it would
also sequence all the patterns AND orient them in space such that
the number of q turns would be minimal for the tour! I am currently
working on "God's Tour" for some of the lesser subgroups, touching on
say a dozen patterns for the square's group. If humans and computers
ever resolve "God's Algorithm" there is some solace that there are
problems even more intractible.

Hmmmm, I just had a thought. It would probably be best to group all
the patterns closer to start and work outwards towards the more
antipodal ones.

With the smaller groups a "Total Tour" would be possible! Visit all
elements!

-> Mark <-


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From JBRYAN@pstcc.cc.tn.us  Tue Feb 13 15:45:47 1996
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Subject: Re: Large Searches with Small Memory
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On Tue, 6 Feb 1996, Jerry Bryan wrote:

> But here follows what I think is a new idea.  What if we
> formed all products XY for X in C[n] and Y in C[1].  Since
> C[1] is Q, this is really just the procedure for a standard
> depth first search.  But we can't store C[n+1].  Can we
> determine the size of C[n+1] anyway?

As is often the case, there is nothing new under the sun.  I believe that
the "new" idea I was suggesting is very similar to, or perhaps identical
with, certain aspects (or all) of Shamir's algorithm.  The best references
I have found in the archives are as follows: 

   Alan Bawden     27 May 87  Shamir's talk really was about how to 
                              solve the cube!

   Michael Reid    16 Dec 94  Re: Cyclic Decomposition

   David Moews     23 Jan 95  Shamir's method on the superflip

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7127
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990


From mreid@ptc.com  Tue Feb 13 16:14:27 1996
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From: michael reid <mreid@ptc.com>
To: cube-lovers@ai.mit.edu
Subject: Re:  Group/graph status?
Cc: rcs@cs.arizona.edu

rich schroeppel asks

> Has anyone tabulated the number of positions are reachable (from the
> initial cube) in one move, two moves, etc.?  Is the diameter of the
> graph known?

first note that there are two common ways to define a "move",
any twist of a face (face turn metric), or any 90 degree turn
of a face (quarter turn metric).

jerry bryan was counting (and storing) positions close to start
on magnetic tape.  he gave figures for positions within 7 face
turns on july 19, 1994 and positions within 11 quarter turns on
february 4, 1995.  (jerry, how many reels of tape did this take?)

the diameter isn't known.  the best lower bounds are 20 face turns,
or 24 quarter turns, both from considering the position "superflip".
the best upper bounds are 29 face turns, or 42 quarter turns.

mike

From alan@curry.epilogue.com  Wed Feb 14 18:45:13 1996
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From: Alan Bawden <Cube-Lovers-Request@ai.mit.edu>
Sender: Cube-Lovers-Request@ai.mit.edu
To: Cube-Lovers@ai.mit.edu
Subject: Welcome to the future of the Internet

As those of you who have been reading Cube-Lovers for the last few months
are well aware, we have been suffering through (almost) weekly
advertisements sent by some nutcase trying to sell magazines.  In the past,
I have been able to defeat unwanted advertising on Cube-Lovers by simply
having a chat with the advertiser (or his postmaster).  But the magazine
guy is determined to send a weekly copy of his advertisement to every
mailing list in the world, despite all objections -- there's really no way
to shut him off short of some form of mailing list moderation.
 
So we're forced to make Cube-Lovers a moderated mailing list.  Starting
with this message, every message sent to Cube-Lovers will be screened
before it gets distributed to the rest of you.  For the moment, some of the
steps in the screening process are manual, so there may be a delay before a
message you submit gets out to the rest of the list.  That's the price we
pay to keep Cube-Lovers from becoming a conduit that delivers more
advertising than content.

From your point of view, this change should be completely invisible (beyond
the improved content and increased latency).  It is still the case that you
should address all submissions to Cube-Lovers@AI.MIT.EDU and all
administrative correspondence to Cube-Lovers-Request@AI.MIT.EDU.

There are some other advantages to the new scheme beyond filtering out
advertising:

  1.  Filtering out other inappropriate messages, such as misdirected
      subscription requests.

  2.  Better handling of mailer bounce messages.  This change should almost
      entirely eliminate cases where you submit a message to Cube-Lovers
      and some mailer sends you an error report.  Almost all such messages
      should now be routed correctly to me.  (Of course, mailers will still
      sometimes screw up -- please forward any mailer errors you -do- get
      back to me.)

  3.  Less messages like this one.

  4.  For the moment, the archive at ftp://ftp.ai.mit.edu/pub/cube-lovers
      continues to accumulate the un-filtered messages.  But eventually, I
      hope to keep it free of trash as well.

If this message is delivered smoothly, it should be followed by two more
messages that arrived while I was debugging this stuff.  And you will
-never- see the magazine advertisement that was submitted last weekend!

		- Alan (Cube-Lovers-Request@AI.MIT.EDU)

From tmartin@accucomm.net  Thu Feb 15 07:09:27 1996
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To: Cube-Lovers@ai.mit.edu
From: "Thomas H. Martin" <tmartin@accucomm.net>
Subject: Resolution of the cube

My son has dug out my cube and has a burning interest in it now.  Also, he
has revived my interest in it.  My question is, is there somewhere I can get
the solution for him?  I would take a mailing address, printed copy or even
an old E-mail message.  My son is 13 and quite anxious to work on it and
solve it.

Thanks,

Tommy Martin
Dublin, GA
tmartin@accucomm.net


From JBRYAN@pstcc.cc.tn.us  Thu Feb 15 10:41:04 1996
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Date: Thu, 15 Feb 1996 10:39:54 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Shamir on Breadth First Searches
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Armed with some newfound understanding of Shamir's method,
I would like to revisit the issue of large searches in
small memory.  However, I will be proposing the
application of Shamir's method in a slightly different
form than it has been applied before.

The best discussion of Shamir's method in the archives is
probably Alan Bawden's note of 27 May 87 "Shamir's talk
really was about how to solve the cube!".  The thrust of
Shamir's talk was how to determine the minimal solution
for a given position in a reasonable time and with a
modest (relatively speaking!) amount of memory.

My take on the Shamir's method is two-fold.  First, let
T[n] be the set of positions no greater than n moves from
Start, and let x be the position in question.  Shamir's
method first involves taking the intersection of T[n] with
xT[n].  If the intersection is null, then x is greater
than 2n moves from Start.  Otherwise, the distance from
Start can be determined rather easily.  This is our old
friend which I call the half-depth search.

The efficacy of a half-depth search is dependent on n.  A
reasonable n of say 5 or 6 only permits testing x up to a
distance of about 10 or 12 moves from Start.

The second component of Shamir's method is the clever
part.  You store T[n] and create sort of a virtual T[2n]
which doesn't have to be stored.  You do the same thing
with xT[n] to create a virtual xT[2n].  Forming the
intersection of T[2n] with xT[2n] lets you test positions
up to a distance of 4n from Start while only storing T[n]
and xT[n].  So for n=5, we can test for distances up to 20
moves from Start.

My primary interest lies in creating T[n] for the largest
possible n, rather than testing for particular positions. 
Hence, I want to talk about just the portion of Shamir's
method that lets us get from a real T[n] to a virtual
T[2n].

Let me start be reviewing my understanding of key points
of Shamir's method.  Given two sets of positions S and T,
Shamir tells us how to form all the products st for s in S
and t in T with the products being created in
lexicographic order.  The storage required is order N
rather than order N^2 (provided of course that the
products are only created and are not stored).

For the algorithm to work, T itself has to be in
lexicographic order.  I don't think S has to be in
lexicographic order (see below).  But S and T may well be
the same set, and in any case there is no loss of
generality in requiring that S be in lexicographic order
as well.

Furthermore, T must be stored as a tree, and we might just
as well store S as a tree, too.

Alan gives an excellent description of the required tree
structure.  The structure itself is a very old concept and
is not unique to Shamir's method.  It could be used, for
example, to store a dictionary for a spell-checker.  Such
a tree would branch 26 ways (American alphabet) for each
of the 26 possible first letters.  The tree would branch
again in up to 26 ways for the second letter and for each
subsequent letter, etc.

For Shamir's tree, the "letters" are (usually one byte)
numbers defining the permutations, where the permutation
is simply a vector listing (in order) the values of the
permutation.

Choose a particular s[j] in S and consider all the
products (s[j])(t[k]) for t[k] in T and for k in 1..n. 
There is some ordering of the t[k] values which will put
the {s[j])(t[k]) values in proper lexicographic order. 
The t[k] values themselves are obviously not in
lexicographic order, and may indeed appear to be in a 
"random" order.  But the order is far from random; it is
quite carefully considered.  The genius of Shamir's method
is that it tells us exactly how to accomplish the proper
ordering of the t[k] values to yield lexicographic
ordering of the {s[j])(t[k]) values.

Notice that the required order of the t[k] values is
different for each s[j].  My brief description of the
magic is that (s[j])' is used as a template to tell us how
to traverse the T tree to make the t[k] values come out in
the required order to make the (s[j])(t[k]) values come
out in lexicographic order.  See Alan's note for
additional details.

(Alan doesn't mention (s[j])' explicitly, but that is what
it comes down to.  Shamir reverse engineers s[j], runs it
backwards if you will, to figure out how to make
(s[j])(t[k]) come out right.)

The rest of the algorithm is a little fuzzy to me, but
here is how I think it has to work.  Suppose S contains m
elements and T contains n elements.  What we have done so
far is to create a single sequence of products
(s[j])(t[k]) for some particular, fixed s[j].  The
sequence contains n elements (one for each t in T), and is
in lexicographic order.

We must produce m such sequences, one for each s[j]. 
Then, we must perform an m-way merge of the m sequences. 
The result of our merge is the desired sequence of
products st in lexicographic order.  It is the fact of
this merge that leads me to believe that the s values do
not have to be in lexicographic (or any other particular)
order.

If m is very large (more than a few dozen), such a merge
is not quite so easy as it sounds, and it is the details
of the merge that are most fuzzy to me in Alan's note. 
The merge would normally be accomplished by forming an
ordered queue containing the first element of each
sequence.  The first element would be popped off the
queue, then the next element from that sequence would be
calculated and put into the queue.

The tricky part is that the queue has to be kept ordered. 
It has to be ordered in the first place.  Then, when an
element is popped off and a new element added, the new
element has to be added in the correct place.  Hence, I
would probably implement the queue itself as another tree,
separate from the S and T trees.

Now, we return to our main discussion.  Let Q[k] be the
set of positions which are k quarter turns from Start.  (I
used C[k] in my last note).  Q[1] is just Q, the set of 12
quarter turns.  Store each Q[k] for k in 0..n in its own
"Shamir tree".

Create a virtual Q[n+1] as the lexicographically ordered
set of products st for s in Q[n] and t in Q[1].  Shamir
does not do everything for us.  We have to do some of the
work ourselves at this point.

The first issue is that some of the products will be
duplicate.  But the lexicographical ordering makes the
duplicates easy to detect.  So detect the duplicates and
throw them away.

The second issue is that some of the products are in 
Q[n-1] rather than in Q[n+1].  But since Q[n-1] is also in
lexicographical order, we can keep a finger or toe pointed
to Q[n-1], scanning through it in step with the products
which are generated.  Any product which is found in Q[n-1]
is not counted as being in Q[n+1].

Creating virtual Q[n+2] is like unto creating virtual
Q[n+1].  We form products from Q[n] and Q[2].  An
additional complication is that our fingers and toes must
point to and step along both Q[n] and Q[n-2] looking for
products which are shorter than n+2 quarter turns, and
which therefore are not to be counted.

Now comes the really interesting part  --- creating
virtual Q[n+3].  We form the products from Q[n] and Q[3]. 
As we create the products, we must track along through the
Shamir trees for Q[n-3], Q[n-1], and Q[n+1].  But the
Shamir tree for Q[n+1] is virtual, and isn't really there!

Here is how we do it.  We must create virtual Q[n+1] and
virtual Q[n+3] at the same time, keeping them more or less
in step, with Q[n+1] equal to or one step ahead of Q[n+3]. 
That way, we have the one real element available of the
virtual Q[n+1] that we need to test the virtual Q[n+3]
against.  As a really *old* programmer, I would describe
what we are doing with Q[n+1] and Q[n+3] as a match/merge.

Given the requirement to generate Q[n+1] as we generate
Q[n+3], there is no real reason to generate Q[n+1] by
itself.  If we have enough fingers and toes to point to
and count everything, we might just as well produce Q[n+1]
and Q[n+3] on the same pass of the data.  For that matter,
we might just as well get Q[n+1], Q[n+3], through Q[2n-1]
on the same pass.  Similarly, we might as well get Q[n+2],
Q[n+4], through Q[2n] on the same pass.

This is all very simple in principle.  But in my
experience, keeping track of all those pointers and
counters is a real pain to program.

Can we go again?  That is, can we go from Q[2n] to Q[4n]? 
I think not.  Shamir's method requires that of the S and T
trees, at least the T tree really be there.  We have to
traverse it many times and in all kinds of orders.  Being
there virtually is not enough.

Finally, what about local maxima?  We cannot detect local
maxima by forming Xq for a position X and for all q in Q,
testing to see of all Xq are closer to Start.  (The Xq are
not in lexicographical order.)  I am thinking about the
following as a way to find local maxima, but it may be
bogus.  See what you think.

Suppose a position in one of the virtual Q[n+k]'s that we
are creating is not the product of any st for s in Q[n]
and t in Q[k+2].  For example, suppose there is an element
p in Q[n+1] which is not the product of any st for s in
Q[n] and t in Q[3].  (We could find all such p easily in
our scan of the virtual Q[n+k] trees.)  Could we say that
all such p are local maxima?  I am not sure.

This method works for sure to find local maxima in Q[n-1]
when creating Q[n+1].  In fact, this method is the way I
find local maxima with my large tape searches.  That is,
the method works when you are only going one step ahead. 
If you use Q[n] and Q[1] to create Q[n+1], then all the
products are in Q[n+1] or Q[n-1], and any element of 
Q[n-1] which is not a product of Q[n] and Q[1] is a local
maximum.  But can we say that any element of Q[n+1] that
is not a product of Q[n] and Q[3] is a local maximum?  I
just don't know.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7127
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990


From awechsle@bbn.com  Thu Feb 15 11:20:11 1996
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From: Allan Wechsler <awechsle@bbn.com>
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Date: Thu, 15 Feb 1996 11:20:09 -0500
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To: tmartin@accucomm.net
Cc: Cube-Lovers@ai.mit.edu
In-Reply-To: <199602151206.MAA09438@accucomm.net> (tmartin@accucomm.net)
Subject: Re: Resolution of the cube
Reply-To: awechsle@bbn.com

   Date: Thu, 15 Feb 1996 12:06:18 GMT
   From: "Thomas H. Martin" <tmartin@accucomm.net>

   My son has dug out my cube and has a burning interest in it now.  Also, he
   has revived my interest in it.  My question is, is there somewhere I can get
   the solution for him?  

   Tommy Martin
   Dublin, GA
   tmartin@accucomm.net

Now you've pushed my button.

When the cube first came out, a bunch of us at MIT were wild to solve
it.  There were _no_ published solutions.  At least three or four of
us solved the cube by ourselves, independently.  We twisted and
turned, drew arcane diagrams to show what went where, and although it
sometimes took a couple of weeks, we each managed it.

Then the books started to come out, and as far as I can tell, no one
ever solved it independently again.

The cube is a solvable puzzle.  It is challenging, but it eventually
yields to analysis and experimentation.  Why don't you and your son
_not_ cheat, and actually solve the thing?  You'll be the first to do
it on your own for more than a decade.

What fun is it to read the answer from a book?

-A

From DHUNT1@ukcc.uky.edu  Thu Feb 15 14:35:38 1996
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From: "Andrew F. Hunt" <DHUNT1@ukcc.uky.edu>
Subject:      Subscription
To: CUBE-LOVERS@life.ai.mit.edu
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Please add me to the ube list. Thanks, Andrew F. Hunt

ANDREW F. HUNT
UNIVERSITY OF KENTUCKY






From mark.longridge@canrem.com  Thu Feb 15 15:18:57 1996
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Sender: CRSO.Cube@canrem.com
Subject: Re: Resolution of the cub
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1299.5834.0C20AE8D@canrem.com>
In-Reply-To: <199602151620.LAA03366@chara.BBN.COM>
Date: Thu, 15 Feb 96 15:07:00 -0500
Organization: CRS Online  (Toronto, Ontario)

-> Date: Thu, 15 Feb 1996 12:06:18 GMT
-> From: "Thomas H. Martin" <tmartin@accucomm.net>
->
-> My son has dug out my cube and has a burning interest in it now.
-> Also, he
-> has revived my interest in it.  My question is, is there somewhere I
-> can get
-> the solution for him?
->
-> Tommy Martin
-> Dublin, GA
-> tmartin@accucomm.net
->
-> Now you've pushed my button.
->
-> When the cube first came out, a bunch of us at MIT were wild to solve
-> it.  There were _no_ published solutions.  At least three or four of
-> us solved the cube by ourselves, independently.  We twisted and
-> turned, drew arcane diagrams to show what went where, and although it
-> sometimes took a couple of weeks, we each managed it.
->
-> Then the books started to come out, and as far as I can tell, no one
-> ever solved it independently again.

Bzzzzzz ...... wrong.

I do understand that Allan Wechsler was one of the original solvers,
before the glut of books on the subject, however I solved the cube and
the megaminx independently. Even though there were many cube books,
there was only one megaminx (rubik-type dodecahedron) book, and it was
rather hard to follow.

My advice to anyone who likes puzzles of this type who is already adept
at solving the cube is to solve it using a subgroup like < U, R > or
< U, F, L >.

One of the problems I'm working on is how to get the spot patterns on
the megaminx. To the best of my knowledge this information is not
recorded anywhere on the planet.

(Yes, I can just solve it that way, I'm looking for a short algorithm,
and no one and no program can help!)

So even though we can make mincemeat of the pyraminx, dino cubes,
square 1, Fisher's Cube, Skewbs, The UFO, Rubik's Revenge, Rubik's Wahn,
etc etc there are still a couple of exceptional difficult problems:

Spot Patterns on the Megaminx in a short number of moves.
Dan Hoey's Tartan Cube.
God's Algorithm on the standard cube.

-> Mark <-


From JBRYAN@pstcc.cc.tn.us  Fri Feb 16 12:22:06 1996
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Date: Fri, 16 Feb 1996 12:20:50 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: Shamir on Breadth First Searches
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On Thu, 15 Feb 1996, Don Woods wrote:

> > Can we go again?  That is, can we go from Q[2n] to Q[4n]? 
> > I think not.  Shamir's method requires that of the S and T
> > trees, at least the T tree really be there.  We have to
> > traverse it many times and in all kinds of orders.  Being
> > there virtually is not enough.
> 
> But wait a sec.  Can't we go from Q[2n] to Q[3n]?  We still
> have T sitting there.  As we generate each element of Q[2n],
> we could use it to generate Q[2n] x T, couldn't we?  Or do
> you also need S to "really be there", too?  (I didn't go back
> over your description to try to figure that out; I'm just
> basing my suggestion on the quoted paragraph.)

As I said earlier, this is the part of Shamir's method about which I am
most uncertain.  In my description of forming products st from S and T, I
emphasized the role of T.  But if I understand the method correctly, the
width of what I called an m-way merge is controlled by the size of S
(remember that S contains m elements and T contains n elements).  So the
merge queue itself will also have m elements.  If you try to generate
Q[2n] x T, the width of the merge (and the size of the merge queue) will
have to be as large as Q[2n], which is too large to store.

In answer to your specific question, I would say that the merge queue has 
to really be there.  S might or might not have to really be there, 
depending on exactly how you program the interaction between S and the 
merge queue.  But in any case, the merge queue is as big as S.

(I would gladly accept any and all help discussing these issues with 
anyone who has actually programmed the algorithm.) 

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7127
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990


From AirWong@aol.com  Sun Feb 18 15:41:30 1996
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Date: Sun, 18 Feb 1996 15:41:29 -0500
From: AirWong@aol.com
Message-Id: <960218154129_225194632@mail04.mail.aol.com>
To: Cube-Lovers@ai.mit.edu
Subject: Re  Resolution of the cube

A few messages ago yo was this

-> Now you've pushed my button.
->
-> When the cube first came out, a bunch of us at MIT were wild to solve
-> it.  There were _no_ published solutions.  At least three or four of
-> us solved the cube by ourselves, independently.  We twisted and
-> turned, drew arcane diagrams to show what went where, and although it
-> sometimes took a couple of weeks, we each managed it.
->
-> Then the books started to come out, and as far as I can tell, no one
-> ever solved it independently again.

-Bzzzzzz ...... wrong.

-I do understand that Allan Wechsler was one of the original -solvers,
-before the glut of books on the subject, however I solved the -cube and
-the megaminx independently. Even though there were many cube -books,
-there was only one megaminx (rubik-type dodecahedron) book, and -it was
-rather hard to follow.

I, too, solved the cube independent of a book. However, after I was done, I
found a few books and other algorithms (as well as some friends) and compared
the solutions. There are many ways to solve it, and it was fun trying to
solve the cube combining the different ideas. When I hunt down Thistlewaite's
algorithm it should get even more interesting.

Aaron WOng
AirWong@AOL.com

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On Wed, 7 Feb 1996, Mark Longridge wrote:

> Further Cube Musings
> ====================
> 
> We usually think of positions antipodal to start only, but there are
> positions antipodal to any given position.
> 
> Given a small enough subgroup of the cube, i.e. one which we can
> exhaustively study, it is not hard to determine some examples.

I guess I didn't understand the thrust of this note.  Isn't it a bit simpler?

Let H be some subgroup of G, let A be the set of antipodes of Start within
H, and let h be some element of H.  Then, don't we simply have that the
antipodes of h are the set hA, where hA={ha | a in A}? 

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7127
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990




From JBRYAN@pstcc.cc.tn.us  Tue Feb 20 16:06:52 1996
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On Tue, 13 Feb 1996, michael reid wrote:

> jerry bryan was counting (and storing) positions close to start
> on magnetic tape.  he gave figures for positions within 7 face
> turns on july 19, 1994 and positions within 11 quarter turns on
> february 4, 1995.  (jerry, how many reels of tape did this take?)

It was a little better than 100 tapes.  It was roughly 20GB of data.  I
stored 14 bytes per position (could have done it in 13 bytes, but I stored
the lengths with each permutation).  Each "position" was really a
representative of an equivalence class of M-conjugates (usually)
containing 48 elements.  Hence, it took about 14/48 bytes (about 2.33
bits) to store each position.  This isn't too shabby, but it is nowhere as
compact as the coding scheme discussed in the "How Big is Big?" thread. 

> 
> the diameter isn't known.  the best lower bounds are 20 face turns,
> or 24 quarter turns, both from considering the position "superflip".
> the best upper bounds are 29 face turns, or 42 quarter turns.

A 24q process is known for superflip.  It is known that superflip is
greater than 19f.  Is a 20f process known for superflip?

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7127
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990


From mreid@ptc.com  Wed Feb 21 18:14:30 1996
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From: michael reid <mreid@ptc.com>
To: CRSO.Cube@canrem.com, cube-lovers@ai.mit.edu
Subject: Re:  < U, F, R > group

mark writes

> I am working on an engine to search optimal paths for < U, F, R > but
> it's not done yet. It's certainly within the bounds of computibility:
> 
> Size (u_f_r) = 170,659,735,142,400 (hmmm, 170 trillion maybe not!)

it should be possible to write a good searching program for this
subgroup.  use the filtration

              <U, F, R> ,  <U, F2, R2> ,  <>

(which is just the last two stages of the three stage filtration i
gave on may 22, 1992), and a kociemba-type searching method.
i don't know if it will be feasible to find optimal paths, but
this technique should get pretty close.

let us know what you find out!

mike

From mreid@ptc.com  Thu Feb 22 17:10:57 1996
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From: michael reid <mreid@ptc.com>
To: cube-lovers@ai.mit.edu
Subject: "simplest" solution of the cube?

mark writes

> This brings up the idea of a "Rubik's Tour". Such a tour would
> touch on a set of interesting patterns within a given subgroup,
> or potentially the entire cube group. Of course, "God's Tour"
> would not only touch on all the interesting patterns, it would
> also sequence all the patterns AND orient them in space such that
> the number of q turns would be minimal for the tour! I am currently
> working on "God's Tour" for some of the lesser subgroups, touching on
> say a dozen patterns for the square's group. If humans and computers
> ever resolve "God's Algorithm" there is some solace that there are
> problems even more intractible.

there's a general graph theory conjecture that cayley graphs are
hamiltonian (i.e. have hamiltonian circuits).

if we take the cayley graph formed by generators
{F, F', L, L' U, U', R, R', B, B', D, D'}, the conjecture asserts
that there is a sequence of  N  quarter turns that visits every position
exactly once and returns to START.  (here  N  =  43252003274489856000
is the order of the group.)

so the proposed "simplest" solution to the cube is to apply such a
hamiltonian sequence.  at some point, in the middle of the sequence,
the cube will be solved!  no need to continue with the rest of the
sequence.


i don't think the general conjecture is close to being proved, but
it is known for some special groups and generators.  it would be
interesting to know if anyone can verify the conjecture for the cube
group with quarter turn generators.  (face turn generators would also
be interesting.)

mike

From @uconnvm.uconn.edu:dmoews@xraysgi.ims.uconn.edu  Fri Feb 23 00:39:22 1996
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From: dmoews@xraysgi.ims.uconn.edu (David Moews)
Message-Id: <9602230539.AA10513@xraysgi.ims.uconn.edu>
To: cube-lovers@life.ai.mit.edu, dmoes@xraysgi.ims.uconn.edu
Subject: Implementing Shamir's method


Since there seems to be a surge of interest in Shamir's method, I
thought I would mention a few points about it and my implementation
of it:

1.  How the group must be represented in order to use Shamir's method.

We suppose that elements of our group G are represented by ordered
tuples, which can be ordered lexicographically; we want to generate
the list ST in this lexicographical order.  Suppose that we have an element
s of S, and elements t and u in T which first differ in coordinate i.  For 
Shamir's method to work, we need to be able to order st and su given
only the length i initial segments of t and u.  This is true for
permutation groups if we represent them as acting on {1,...,n}
(st compares to su as s(t(i)) compares to s(u(i)).)
It is also true for the wreath products occurring in the cube group:
suppose G = H wr K, where H is a permutation group acting on {1,...,n},
and K is a product of cyclic groups with index set {1,...,n}.  Then
if we write an element g of G as  ( g(1), ..., g(n), g'(1), ..., g'(n) ),
the g(i)'s being in {1,...,n} and the g'(i)'s in the cyclic groups,
we can write

( h(1), ..., h(n), h'(1), ..., h'(n) ) ( k(1), ..., k(n), k'(1), ..., k'(n) )
= 
( h(k(1)), ..., h(k(n)), h'(k(1)) + k'(1), ..., h'(k(n)) + k'(n) ).

Hence if t and u's first difference is in t(i) != u(i), st and su compare
as s(t(i)) and s(u(i)), and if t and u's first difference is in t'(i) != u'(i),
st and su compare as s'(t(i)) + t'(i) and s'(u(i)) + u'(i).

Since you do a lot of composition in Shamir's method, I felt it best to
leave the permutations unpacked.  I used the wreath product representation
above, with H = S_8 x S_12 and K = (Z/3Z)^8 x (Z/2Z)^12.  Each permutation
then used 8 + 12 + 8 + 12 = 40 bytes.  All members of both S and T must be
stored in memory (see below.)  This used up a lot of memory.  (You could,
of course, also represent the cube group as a permutation group on the 48
facelets.)

2.  The data structure for T.

Jerry Bryan has alluded to this.  I used a tree each of whose leaves
contained a member of T, and each of whose internal nodes contained
an index indicating which tuple coordinate was being branched on, a
value of this coordinate for each son, and pointers to each son.
I also included a pointer to the father to make traversal easier.
The data structure for T does not change during the algorithm; you
can use it with more than one S at once.

3.  The data structure for S.

By traversing the T tree approriately, we can output the sequence
X(s) = (lexicographical sort of {st | t in T}) for each s.  For all elements s
of S, we need to store s itself, and a marker to show our position in X(s)
(for me, this was just a pointer to the T tree.)  

We also need enough structure to make merging the X(s)'s easy.  I used a 
`tree of losers' (cf. Knuth, Chapter 5.)  Since there seems to be
some uncertainty about this, I will go into detail. Let S = {s_0, ..., s_(N-1)}.
The tree will then have 2N nodes: N internal ones, 0 through N-1, and N
leaves, N through 2N-1.  Each internal node i contains a pointer to a leaf.
The leaves contain the actual s_j's, as well as the pointers to T.  Node i 
has nodes 2i and 2i+1 as sons if 0<i<N; node 0 has only node 1 for a son.  
(Since these relations are fixed, no tree traversal pointers need be stored.)  
Leaf N+j always corresponds to s_j.  The tree is initialized by conducting an 
elimination tournament: the leaves are initialized with the first elements of 
the X(s_j)'s, and sons repetitively battle for their fathers, the lesser value 
always winning, i.e., initializing the father's pointer.  After the tournament 
is finished (the overall winner ending up in both nodes 0 and 1) we 
simultaneously, for all i=1,...,N-1, reinitialize node i's pointer from its 
other son, i.e., the loser of the battle that was just played for node i.  
After we do this, each value occurs exactly once in a leaf and once in an 
internal node (everybody in a tournament, except the winner, loses exactly one 
game.)

The advantage of using losers instead of winners is that it makes updating
the tree easy.  Suppose each internal node i points to leaf N + a_i.  To
update, we output the first element of X(s_a_0) and advance X(s_a_0).  We
can then execute the following loop to update the tree:

     i := floor((N + a_0)/2)
     while i > 0 do
         if the next element of X(s_a_0) is greater than the next
            element of X(s_a_i)
             then swap a_i and a_0 (we have a new loser)
         i := floor(i/2)
     od

As you see, we perform many comparisons between the first elements of the 
X(s_i)'s.  It is convenient to store the next element of X(s_i) in
the data structure with s_i.  This uses up much more memory (a comparable 
amount with that taken by S and T themselves) but does speed up the program 
somewhat.

-- 
David Moews                             dmoews@xraysgi.ims.uconn.edu


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From: michael reid <mreid@ptc.com>
To: cube-lovers@ai.mit.edu
Subject: superflip in 20f

jerry asks

> A 24q process is known for superflip.  It is known that superflip is
> greater than 19f.  Is a 20f process known for superflip?

on may 17, 1992, dik winter gave

) Superflip:
) (13+7=20): F B U^2 R F^2 R^2 B^2 U' D F U^2 R' L' U B^2 D R^2 U B^2 U

in my exhaustive search for superflip maneuvers of length <= 19f,
several (but not all) branches of my search found maneuvers of length 20f.
all were equivalent to dik's under the three operations

* conjugation of the sequence by a symmetry of the cube
* cyclic permutation of the sequence
* inversion of the sequence

perhaps it is the case that dik's maneuver is "unique" up to these
operations.

mike

From mark.longridge@canrem.com  Sat Feb 24 01:13:35 1996
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Organization: CRS Online  (Toronto, Ontario)


Mike wrote:
> there's a general graph theory conjecture that cayley graphs are
> hamiltonian (i.e. have hamiltonian circuits).
>
> if we take the cayley graph formed by generators
> {F, F', L, L' U, U', R, R', B, B', D, D'}, the conjecture asserts
> that there is a sequence of  N  quarter turns that visits every
> position exactly once and returns to START.
> (here  N  =  43252003274489856000 is the order of the group.)

Here's an easy example:

Hamiltonian Circuit for < u2, r2 >
12 elements, 12 moves in group to reach each element

         Identity
         /    \
    1.  u2    r2  12.
        |      |
    2.  r2    u2  11.
        |      |
    3.  u2    r2  10.
        |      |
    4.  r2    u2  9.
        |      |
    5.  u2    r2  8.
        |      |
    6.  r2    u2  7.
         \   /
         Antipode

Position at  6. is the antipode
Position at 12. is the identity

Also, I seem to remember that the slice-squared group had 8 elements,
and if you graphed a route through the elements it formed a cube.
After drawing such a graph it is not hard to find a hamiltonian
circuit (using the edges of the cube as a pathway).

This may be true in general for all the platonic solids.
(I need to re-check "Regular Polytopes" by Coxeter).

So we have 2 examples and no counter-examples of the general graph
theory Mike mentions.

-> Mark <-


From JBRYAN@pstcc.cc.tn.us  Thu Mar  7 13:54:55 1996
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Date: Thu, 07 Mar 1996 13:53:22 -0500 (EST)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Shamir and M-Conjugacy
To: Cube-Lovers <cube-lovers@ai.mit.edu>
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With most of my large breadth-first searches of God's Algorithm, I have
used M-conjugacy, where M is the set (and group) of 48 rotations and
reflections of the cube.  Using M-conjugacy reduces the size of the
problem by about 48 times, and allows me to search one or two levels
deeper than would be possible without M-conjugacy. 

I haven't even written a basic program for Shamir's method yet, but it
occurs to me that if Shamir's algoritm could be combined with M-conjugacy,
tremendous benefits would accrue. 

As before, we define Q[n] to be the set of positions which are n quarter
turns from start, and T[n] to be the union of Q[k] for k in 0..n.  I have
been thinking in terms of using Shamir's method on T[5] to get Q[6]
through Q[10].  But T[5] isn't exactly tiny as it contains 105046
elements.  Plus, we already have Q[0] through Q[11] calculated by other
means, so we wouldn't be learning anything new. 

Define Q*[n] to be the set of representative elements of M-conjugacy
classes of length n, and T*[n] to be the union of Q*[k] for k in 0..n. 
T*[5] only contains 2229 elements, which is much more manageable than
105046 elements.  T*[6] contains 20624 elements.  This is still quite
manageable, and might well permit us to calculate Q[12], which would be
something new.  T*[7] contains 192153 elements.  This is right on the bare
edge (maybe past the bare edge) of what could be handled on most machines. 
But if we could handle it, we possibly could calculate Q[13] and Q[14] --
a really major advance in our knowledge of God's algorithm. 

I haven't yet figured out entirely how to marry Shamir's method with
M-conjugacy.  But let me provide a general outline of what would have to
be done, and identify the major problem areas. 

Without repeating all the details, recall that we can in theory modify
Shamir's method to calculate T[2n] (and all the respective Q[k]'s) as the
product (T[n] x T[n]).  Very, very roughly speaking, we seek to calculate
T*[2n] as the product {T*[n] x T*[n]).  But there are many, many
complications along the way. 

Here are some preliminaries.  

First, given a representative X in Q*[n], we can calculate its entire
M-conjugacy class as {m'Xm | m in M}.  I usually just write this set as
{m'Xm}.  In group theory, an element m'Xm is often written as X^m and the
set {m'Xm} is often written as X^M.  I will adopt the group theory
notation to some extent in the remainder of this paper. 

Given Q*[n], we can create Q[n] by simply expanding the M-conjugacy
classes for each X in Q*[n].  In most of my work, the Q[n] which is thus
created is sort of virtualized -- created but not stored.  I will denote
the virtualized version of Q[n] as Q*^M[n] to distinguish it from the real
version.  Notice that we do have Q*^M[n]=Q[n], so Q*^M[n] can serve as a
surrogate for Q[n] most anytime we need it to.  Similarly, we denote the
virtualized version of T[n] as T*^M[n]. 

Second, we define * to be a function (not a permutation) which can be
composed with permutations to calculate a representative element.  We
define X* to be Repr(X), which is really Repr{X^M}.  So we can have such
things as XY* or (X*)(Y*). 

I have generally implemented X* as min{X^M}.  By this, we mean place X^M
in lexicographic order, and choose the first element.  Basing the
representative element of X^M on lexicographic order fits in well with
Shamir's method. 

We now return to the idea of calculating T*[2n] as the product (T*[n] x
T*[n]).  We first note that the product of representatives is not
necessarily a representative, so we would have to calculate (T*[n] x
T*[n])* to assure that all we have is representatives. 

We also note that if we simply calculate all the products st* for s and t
in T*[n], we will have about 48 times too few products.  On the other
hand, if we calculate st* for s and t in T*^M[n], we will have about 48
times more products than we need.  What is required is to calculate st*
for s in T*[n] and t in T*^M[n].  In other words, we expand the
equivalence classes for t but not for s. 

In a sense, this is what I have always done for my non-Shamir searches,
except that I have only advanced by one level of the search at a time. 
That is, I have calculated (Q*[n] x Q*^M[1])* to get to Q*[n+1].  But
remember that Q*^M[1] is just Q[1], which in turn is just Q, the set of
quarter turns. 

That was the sanitized version.  The dirty version is that you have to
calculate (in lexicographic order) o'(s(m'tm))o, for all o in M, for all m
in M, and for all t in T*[n].  The s is a fixed element of T*[n], and the
results for all s in T*[n] are merged in standard Shamir fashion. 
Finally, these products will include representatives and
non-representatives alike, and you have to keep only representatives and
throw away the non-representatives. 

Calculating these products is trivial.  Getting them to come out in
lexicographic order is the hard part.  As I said at the beginning, I am
not sure I know how to do it yet.  But I have some ideas about it that I
will be sharing. 

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7127
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990


From zilla@netcom.com  Sat Mar  9 05:43:22 1996
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From: zilla@netcom.com (Jay Majer)
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Subject: Cube keychains?
To: cube-lovers@ai.mit.edu
Date: Sat, 9 Mar 1996 02:43:15 -0800 (PST)
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Greetings fellow cubers! I hope someone out there can help me. For the 
longest time I've been searching for a Rubik's 3x3 Mini Cube Keychain. 
Are they still being manufactured? If anyone can help me aquire one of 
these, please e-mail.

_______________________________________________________________________
Jay Majer                   "One must avoid all the chewy chunks in
zilla@netcom.com            order to attain pure spiritual creaminess."
jmajer@ucla.edu                                             -A wise man
_______________________________________________________________________

From mouse@collatz.mcrcim.mcgill.edu  Wed Mar 13 07:00:50 1996
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Date: Wed, 13 Mar 1996 07:00:45 -0500
From: der Mouse <mouse@collatz.mcrcim.mcgill.edu>
Message-Id: <199603131200.HAA16027@Collatz.McRCIM.McGill.EDU>
To: cube-lovers@ai.mit.edu
Subject: Re:  Shamir and M-Conjugacy

> T*[7] contains 192153 elements.  This is right on the bare edge
> (maybe past the bare edge) of what could be handled on most machines.

Two hundred thousand elements is on the edge?  Even assuming an
extremely noncompact representation of 20 bytes each (one per
non-center cubie), that's only four megabytes.  The _smallest_ RAM load
I have at home (never mind the machines I have access to at work) is 8
megs, and one machine has 28.  Keeping such a list entirely in-core
would be no problem at all.  Nowhere near the edge.

But Jerry Bryan knows what he's talking about too well to make this
simple a blunder.  Could some kind soul explain what I've obviously
missed?

					der Mouse

			    mouse@collatz.mcrcim.mcgill.edu

From JBRYAN@pstcc.cc.tn.us  Thu Mar 14 08:56:29 1996
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Date: Thu, 14 Mar 1996 08:56:03 -0500 (EST)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: Shamir and M-Conjugacy
To: der Mouse <mouse@collatz.mcrcim.mcgill.edu>
Cc: cube-lovers@ai.mit.edu
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On Wed, 13 Mar 1996, der Mouse wrote:

> > T*[7] contains 192153 elements.  This is right on the bare edge
> > (maybe past the bare edge) of what could be handled on most machines.
> 
> Two hundred thousand elements is on the edge?  Even assuming an
> extremely noncompact representation of 20 bytes each (one per
> non-center cubie), that's only four megabytes.  The _smallest_ RAM load
> I have at home (never mind the machines I have access to at work) is 8
> megs, and one machine has 28.  Keeping such a list entirely in-core
> would be no problem at all.  Nowhere near the edge.
> 
> But Jerry Bryan knows what he's talking about too well to make this
> simple a blunder.  Could some kind soul explain what I've obviously
> missed?
> 

Well, I would argue whether I know what I'm talking about when it comes 
to marrying Shamir with M-conjugacy.  I'm just sort of thinking out loud 
right now, not implementing any code.

But at one point, Bob Moews reported that his implementation of Shamir
required 104 bytes per position.  Of the 104 bytes, 48 bytes were the
position itself.  The rest of the bytes were queues, pointers, and various
overhead of that sort.  I've been guessing that to keep up with all the
pointers and so forth required for M-conjugacy, it might take 200 bytes or
so per position.  But assume optimistically that it could be compressed
down to 100 bytes per position.  Then, we are up to about 20 meg for
T*[7], and about 180 meg for T*[8].  I really think that's a bit too
optimistic, but it's probably not too far off.  But if this guess is off
by even a factor of two, then you would need 40 meg for T*[7]. 

On the other hand, assume these memory estimates are approximately
correct.  At some point, the constraint will become time rather than
memory.  Even on a very fast machine, it might take dozens of years rather
than hundreds of hours to calculate something like T[14] or T[16] as 
(T*[7] x T*[7]) or as (T*[8] x T*[8]). 

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7127
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990



From DNewfield@virginia.edu  Mon Mar 18 14:02:46 1996
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          18 Mar 96 13:19 EST
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Date: Thu, 14 Mar 1996 14:19:04 -0500
From: Dale Newfield <din5w@cs.virginia.edu>
Organization: Computer Science Department
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I just saw a commercial for a puzzle called "Orbix."  It was a sphere 
covered with (I think) 12 colored reflector-like circles, each with a 
button in the center.  They said that there are 4 levels.  It seemed to 
be much like "luminations."  I'm sure that I will buy the first one I 
find. :-)
-Dale Newfield



From isaacs@hpcc01.corp.hp.com  Mon Mar 18 18:46:45 1996
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From: Stan Isaacs <isaacs@hpcc01.corp.hp.com>
Message-Id: <199603182346.AA121182802@hpcc01.corp.hp.com>
Subject: Re: Orbix
To: din5w@cs.virginia.edu (Dale Newfield)
Date: Mon, 18 Mar 96 15:46:41 PST
Cc: cube-lovers@ai.mit.edu
In-Reply-To: <31487128.68F9@cs.virginia.edu>; from "Dale Newfield" at Mar 14, 96 2:19 pm
Mailer: Elm [revision: 70.85.2.1]


> 
> I just saw a commercial for a puzzle called "Orbix."  It was a sphere 
> covered with (I think) 12 colored reflector-like circles, each with a 
> button in the center.  They said that there are 4 levels.  It seemed to 
> be much like "luminations."  I'm sure that I will buy the first one I 
> find. :-)
> -Dale Newfield

No, its more like "Light's Out", on a sphere.  It's a nice puzzle
(although the first one I bought had a broken button, so I had to
exchange it.)  It cost about $20 at Toys-R-Us.  The surface of the
sphere has 12 lights/buttons, dodecahedrally arranged.  When you push
one, in the first puzzle, the 6 surrounding lights change their parity
- if they're on, they go out and vice versa.  The goal is to get all
the lights on, after some random starting pattern.  

The other puzzles have a different combination of which lights go on
when you press a light: I think the second turns out the 6 on the
opposite side from the one pressed, and only do so if the pressed
light was orignally off (or something like that).  Obviously, I
haven't had time to study the details much yet.

The puzzle feels nice, looks good, and should be worth the $20.  

 -- Stan Isaacs

From rodrigo@lsi.usp.br  Mon Mar 25 19:01:33 1996
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Date: Mon, 25 Mar 1996 20:57:25 -0300 (EST)
From: Rodrigo de Almeida Siqueira <rodrigo@lsi.usp.br>
To: Cube-Lovers@ai.mit.edu
Subject: A handling system to fix the Cube
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 ------- Forwarded message --------
 Date: Mon, 25 Mar 1996 19:29:21
 From: jbyland@ln.active.ch

    Name = Byland John
    URL = http://www2.active.ch/~jbyland/
    comments = My Name is Johnny Byland and study Computer-Sience at the 
               Ingenierschule Zuerich. For a work for diploma, we build a 
               Handling-System, who can fix the Rubik-Cube itself.
               This System can detect colors and make moves in all
               relevant axis. We control the system with a PC, programmed 
               in Delphi. If you have good ideas, how we can solve the 
               cube as simple as possible (alogorithm etc), we are 
               recommend, you can send it.
               
               Thanks
                
                Johnny Byland
                Dorfstr. 12
                CH-8330 Pfaeffikon ZH
               
               PS: Have a look at http://www2.active.ch/~jbyland/
               
               E-mail: jbyland@ln.active.ch

From JBRYAN@pstcc.cc.tn.us  Tue Mar 26 08:32:56 1996
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 cube-lovers@ai.mit.edu; Tue, 26 Mar 1996 08:32:33 -0500 (EST)
Date: Tue, 26 Mar 1996 08:32:33 -0500 (EST)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Electronic Citations
To: Cube-Lovers <cube-lovers@ai.mit.edu>
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If I may be permitted a slightly off subject posting ....

Much materal is now online (such as Cube-Lovers) that was never online
before, and much of the online material is being cited in research and
other papers.  I just ran across the best source I have seen for how to
cite such material -- http://www.taft.cc.ca.us/www/tc/tceng/mla.html. 

Since Cube-Lovers is E-mail and archives of E-mail, here is an excerpt 
that addresses citations for E-mail:


E-mail, Listserv, and Newslist Citations

     Give the author's name (if known), the subject line from the posting 
in quotation marks, and the address of the listserv or newslist, along 
with the date.  For personal e-mail  listings, the address may be omitted.

  
Bruckman, Amy S.  "MOOSE Crossing Proposal."  mediamoo@media.
     mit.edu (20 Dec. 1994).

Seabrook, Richard H. C.  "Community and Progress."  cybermind
     @jefferson.village.virginia.edu (22 Jan. 1994) <p>

Thomson, Barry.  "Virtual Reality."  Personal e-mail (25 Jan. 
     1995).


 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7127
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990


From bagleyd@hertz.njit.edu  Thu Mar 28 11:16:46 1996
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Date: Thu, 28 Mar 1996 11:15:55 -0500
From: bagleyd <bagleyd@hertz.njit.edu>
Message-Id: <199603281615.LAA21296@hertz.njit.edu>
To: cube-lovers@life.ai.mit.edu
Subject: simple windows3.1 puzzles

Hi
  I finally got around to making a web page
 
http://hertz.njit.edu/~bagleyd/
 
In it you will find Simple Windows3.1 puzzles.  The more complicated
ones are still in development.  The best of the bunch is wmlink, which
is a Missing Link puzzle.
 
Cheers,
  /X\   David A. Bagley
 // \\  bagleyd@hertz.njit.edu   http://hertz.njit.edu/~bagleyd/
((   X  xlockmore, new stuff for xlock @ ftp.x.org//contrib/applications
 \\ //  altris, tetris games for x @ ftp.x.org//contrib/games/altris
  \X/   puzzles, magic cubes for x @ ftp.x.org//contrib/games/puzzles

From idemoya@mednet.med.miami.edu  Wed Apr  3 20:08:01 1996
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Date: Wed, 03 Apr 96 20:05:22 EST
From: idemoya@mednet.med.miami.edu
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To: cube-lovers@life.ai.mit.edu
Return-Receipt-To: idemoya@mednet.med.miami.edu
Subject: Help me find a RUBIK'S CUBE.............

     
     Hello fellow Rubik's Cube lover:
     
     I must quickly produce a Rubik's cube in the very near future. Please 
     provide me with information that might help me obtain one brand-new unit 
     (or as close to that as possible).
     
     Your copperation is greatly appreciated.
     
     Ivan


From bernier@login.net  Thu Apr  4 21:03:49 1996
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To: cube-lovers@life.ai.mit.edu
From: bernier@login.net (Denise Lemoine)
Subject: list

        bonjour

                j'aimerais ces casse-tetes la!!!

denise lemoine
15 cout
coaticoof p.q.
Madame,
Je suis un peu m=E9lang=E9e. Votre mari m'a dit que vous ne vouliez plus le
compte. En tout cas, j'attends votre cheque. L'adresse est dans ma=
 signature.
Salutations,
At 21:29 96/03/23 -0500, you wrote:
>
>                je ne trouve lpus l'adresse.
>
>                me faire parvenir s.v.p.
>
>denise lemoine
>coaticook
>
>




From aaweint@io.org  Fri Apr  5 16:41:48 1996
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Date: Fri, 05 Apr 1996 16:41:33 -0500
To: cube-lovers@ai.mit.edu
From: Aaron Weintraub <aaweint@io.org>
Subject: Square-1 question

Hi...

I recently got a hold of a Square-1 puzzle and have been trying to solve it.
I can get to the point where it's done, but two edges on one side are
swapped.  How do I swap them back?  Is this a parity problem?  Every move I
have that swaps edges does TWO pairs are a time, so I can't get there with
what I have.  Or can I?  Any help would be appreciated.

-Aaron


From mouse@collatz.mcrcim.mcgill.edu  Sun Apr  7 09:27:42 1996
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Date: Sun, 7 Apr 1996 09:27:39 -0400
From: der Mouse <mouse@collatz.mcrcim.mcgill.edu>
Message-Id: <199604071327.JAA01565@Collatz.McRCIM.McGill.EDU>
To: cube-lovers@ai.mit.edu
Subject: Re:  Square-1 question

> I recently got a hold of a Square-1 puzzle and have been trying to
> solve it.  I can get to the point where it's done, but two edges on
> one side are swapped.  How do I swap them back?  Is this a parity
> problem?  Every move I have that swaps edges does TWO pairs are a
> time, so I can't get there with what I have.  Or can I?

I'm not familiar with Square-1...but if, as seems lkely, it's a square
that you have to get into some arrangement, then consider turning the
whole puzzle 90 or 180 degrees and trying again; that may introduce an
odd permutation and thus make a solution possible.  Or depending on the
definition of "solved" - as I say, I don't know the puzzle - maybe go
for a mirror-reflected state.

					der Mouse

			    mouse@collatz.mcrcim.mcgill.edu

From nichael@sover.net  Sun Apr  7 15:36:20 1996
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Date: Sat, 6 Apr 1996 15:36:52 -0400
To: Aaron Weintraub <aaweint@io.org>, cube-lovers@ai.mit.edu
From: nichael@sover.net (Nichael Lynn Cramer)
Subject: Re: Square-1 question

At 4:41 PM 4/5/96, Aaron Weintraub wrote:
>Hi...
>
>I recently got a hold of a Square-1 puzzle and have been trying to solve it.
>I can get to the point where it's done, but two edges on one side are
>swapped.  How do I swap them back?  Is this a parity problem?  Every move I
>have that swaps edges does TWO pairs are a time, so I can't get there with
>what I have.  Or can I?  Any help would be appreciated.
>
>-Aaron

The quick answer is, yes, in spite of appearances you are actually very far
from finished.

A quick hint is attached below.  (This is only a hint in that it's been two
or three years since I worked with a Square One.  At that time I kept notes
and was going to write up a complete solution but I don't think I ever got
around to doing it [Alan?  Do you remember?]  If I can find those, or can
remember more complete details --time to get the Sq1 back out-- I'll pass
along more details.]


---
---
-0--
---
---
---
--
--
---
---
---
---
--
--
--
--
--
--
--
--
--
--
--
--
--
--
---
--
--
--
--
--
--
--
--
--
--
--
--

Hint:  You're right that you need to swap two pairs together.  The issue
here is that you need to simultaneously swap a pair of edges (the
triangular pieces) and a pair of corners (the quadrilaterals).  And, yes,
you can actually do this. ;-)

In short, in this state the corners only _appear_ to be in the correct
locations.

An analoguous case can occur on the 4X cube where the cube appears to be
_almost_ complete except that two edge peices are flipped.  Again, it looks
like you're close to done, but more accurately you're almost completely
diametrically "across the space of solutions".

Nichael
nichael@sover.net                                               __
http://www.sover.net/~nichael              Be as passersby   -- IC



From nichael@sover.net  Sun Apr  7 15:37:32 1996
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Date: Sat, 6 Apr 1996 15:38:05 -0400
To: Aaron Weintraub <aaweint@io.org>, cube-lovers@ai.mit.edu
From: nichael@sover.net (Nichael Lynn Cramer)
Subject: Re: Square-1 question

At 4:41 PM 4/5/96, Aaron Weintraub wrote:
>Hi...
>
>I recently got a hold of a Square-1 puzzle and have been trying to solve it.
>I can get to the point where it's done, but two edges on one side are
>swapped.  How do I swap them back?  Is this a parity problem?  Every move I
>have that swaps edges does TWO pairs are a time, so I can't get there with
>what I have.  Or can I?  Any help would be appreciated.
>
>-Aaron

The quick answer is, yes, in spite of appearances you are actually very far
from finished.

A quick hint is attached below.  (This is only a hint in that it's been two
or three years since I worked with a Square One.  At that time I kept notes
and was going to write up a complete solution but I don't think I ever got
around to doing it [Alan?  Do you remember?]  If I can find those, or can
remember more complete details --time to get the Sq1 back out-- I'll pass
along more details.]


---
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-0--
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Hint:  You're right that you need to swap two pairs together.  The issue
here is that you need to simultaneously swap a pair of edges (the
triangular pieces) and a pair of corners (the quadrilaterals).  And, yes,
you can actually do this. ;-)

In short, in this state the corners only _appear_ to be in the correct
locations.

An analoguous case can occur on the 4X cube where the cube appears to be
_almost_ complete except that two edge peices are flipped.  Again, it looks
like you're close to done, but more accurately you're almost completely
diametrically "across the space of solutions".

Nichael
nichael@sover.net                                               __
http://www.sover.net/~nichael              Be as passersby   -- IC



From modestr@federal.unisys.com  Mon Apr  8 12:03:25 1996
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Date: Mon, 08 Apr 96 12:03:28 -0700
From: Ron Modest <modestr@federal.unisys.com>
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Regarding swapping two edges on Square-1.

You prompt me to write about something I have been meaning to get around 
to for a long time.

Long ago I found a way to swap two edges using a complicated sequence.  
After considerable unsuccessful effort to improve the solution, I bought 
Richard Snyders amazing book "Turn to 1" from Puzzletts.
( HTTP://WWW.PUZZLETTS.COM/ )  His solution is essentially the same as 
mine.  His method of documentation obscures what is really going on and 
consequently it would be very hard to memorize.  The principal is 
straight forward and follows these steps.

Move all the edge pieces to the same side in an orderly sequence.
Turn the side that has all corner pieces, one position.
Retrace all the moves that brought the edge pieces to the same side.
Fix any thing that got messed up in the process. (this is what I call 
collateral damage)

Snyder's solution optimizes the process to minimize the collateral damage 
but any variation on the steps listed above will work.


On a related subject....
How to get the puzzle into the shape of a cube after initial scrambling.
Snyders book shows pictures of all possible scrambled shapes.  Each has 
instructions for making a few turns and the next diagram to refer to.  
This process may be optimal for getting it into a cube shape but it is 
nearly impossible to memorize.

I am sure everyone who works with the puzzle learns some shapes that are 
close to the cube shape but it may seem nearly impossible to generally 
solve in any orderly way.  Well consider the following strategy:

Collect all the edge pieces on the same side.  They can all be side by 
side in what Snyder calls the Hoofprint pattern or in the Moon pattern 
that has two groups of four edges on the same side.  Then move half of 
the edges to the opposite side.  Then move half of the edges from the top 
to the bottom and half of the edges from the bottom to the top, but do so 
in a way that separates them into groups of two.  You are then with a 
couple of twist of making a cube.

The beauty of the strategy is that to obtain perfect final symmetry, you 
first take it to a position of maximum asymmetry.  Every turn after that 
keeps it symmetric.

This method will not generally be the optimum solution but it is straight 
forward and easily learned.

I said this is related to the previous subject of swapping two edges 
because both require reaching a position will all the edge pieces on one 
side.  I might not have ever found the method for swapping two edges if I 
had not first adopted this method for getting it into a cube shape first.


While I am sending out a message let me recommend that everyone include 
their mail address when the send a message.  Recently a couple of 
messages did not.  I would have send the author a personal message 
answering a simple question but didnt want to bother everyone else.  One 
such question was about obtaining 3x3x3 cubes.  They are available in 
many chain toy stores including "The Game Keeper" and "LearningSmith".  
Most puzzles are available from Puzzletts also.

I also notice that several local stores are carrying Rubiks Magic again. 
 The colors are different than the originals.



Cube Trivia....
In 1982 a Worlds Fair was held in Knoxville Tenn. USA..  At the enterance 
to the Hungarian pavalion was a Rubik's Cube about 4 feet on a side 
mounted on a pedistal.  At that time a Rubik's Cube was a universially 
recognized symbol.



Walter Smith
near Washington D.C.
WALTS@FEDERAL.UNISYS.COM







From aaweint@io.org  Mon Apr  8 21:22:55 1996
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Date: Mon, 08 Apr 1996 21:22:07 -0400
To: cube-lovers@ai.mit.edu
From: Aaron Weintraub <aaweint@io.org>
Subject: Square-1 Question

Hi,

I just wanted to thank everyone who gave me some insight as to how to go
about attacking the parity problem in Square-1.  It was, indeed as I
thought, a parity issue like the double flipped edges in the 4x4x4, and not
just some oversight on my part.  Ron, your tips were particularily helpful.
I didn't have to buy any books or anything, and I can now solve it every
time.  I've got to go find a new puzzle now.

Aaron
aaweint@io.org


From walts@federal.unisys.com  Tue Apr 30 07:42:15 1996
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Date: Tue, 30 Apr 1996 07:42:38 -0700
From: Walt Smith <walts@federal.unisys.com>
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Rubiks Revenge presents several difficult cases when the last few pieces are 
reached.

There has been discussion on this over the years but it does not seem to have 
reached closure.

Mark Longridges list of best known solutions, 
(http://admin.dis.on.ca:80/~cubeman/cmoves.txt) lists some of these but here 
are some improved solutions of my own and additional operators that are 
needed to solve it.

If the center four are done on all sides and the edge pieces are paired up, 
it can be can be treated like a 3x3x3 Rubik Cube.  Several cases come up at 
the end that can not occur on a 3x3x3.  If the corners are done and the last 
few edges are left for last, four cases occur.  Here are my methods that are 
shorter than any others I have seen.

Case 1.  The last three edges can be solved with 3x3x3 techniques.

Case 2.  Two edge pairs are swapped. (swaps RF and RB)

        d2  R2  d2  rR2  d2  r2     (7)

        This is based on combining the following two sequences:
        (d2  R2)2  d2     and    (d2  r2)2

        Each of these is useful in itself.

        This is shorter than Marks Longridges p4.
        (Marks p4 is mislabeled Opp.  It should be Adj and p5 should be 
        labeled Opp)

Case 3.  A single edge pair is flipped.  (flips BL)

        L2
        d1
        R2  d1  R2  d3  L2  u3  B2  u2  B2  u3
        B2  R2  B1  r3  B3  R2  B1  r1  B1     (21)

This is shown in four groups because it proceeds in stages.  The second move 
fixes the parity, the first along with 3rd through 12th fix the faces and the 
last group of moves fix the edges.
This is shorter than Marks p3.


Case 4.  Both Case 2 and Case 3 exist.  The flipped edge pair might be one of 
the swapped edge pairs or it might not be.  Obviously this can be solved by 
using the techniques of Case 2 and 3 applied separately.  I have always 
thought that it should be possible to find an operator that is shorter than 
the sum of these two and possibly shorter that Case 3.  I have not done as 
much study on this case as the others.


If you want to solve the corners last (avoids Case 3 and 4), you may still 
need to solve Case 2 but you may also need to swap two corner pieces.  This 
can be done by applying  Case 2 then fixing the edges then corners.  This 
will take about 40 turns.  The following does it quicker.
(swaps LDB and LDF with the bottom cubie faces remaining bottom faces.)

        R3  D1  L1  D3  R1
        D2  L3  D1  L1  D2
        L1  U3  r2   F2  r2  fF2  r2  f2  U1  L2    (21)

It is listed in three groups to make it easier to memorize as it proceed in 
three stages.
This is shorter than the sequence in Mark Jeays on-line solution at
http://qlink.queensu.ca/~4mj/rr.html  ( or link from Mark Longridges home
http://admin.dis.on.ca:80/~cubeman/ )


I do not have any books on Rubiks Revenge.  If someone out there does, please 
see how these solutions compare.  If anyone has shorter solutions, either 
that they have developed or found in books please submit them so Mark L. can 
improve his list (and so I can improve my technique).  

Also Marks p1 does not work.  Can anyone fix it?



Walter Smith
near Washington D.C.
walts@federal.unisys.com

From a.southern@ic.ac.uk  Tue Apr 30 08:58:46 1996
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To: cube-lovers@ai.mit.edu
From: a.southern@ic.ac.uk (The Official Thermodynamics Fan Club of the UK.)
Subject: Square-1, Super Cubix, Masterball and Rubik's Revenge.
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Hi, I'm on the cube-lovers mailing list.

Bloody Hell! If someone had told me what a square-1 was, I would have been 
able to help with that query. I know the puzzle as "Super Cubix" which was 
written on the side of the 'cube'. I solved this puzzle about six months ago 
and in the same way as Ron Modest did.

Have you tried the Masterball? It is solved in a very similar way, but there 
again it does have very similar properties (having to rotate in one 
direction by pi, but any other multiple in the rest). The Masterball has a 
web site (http://wsd.com/masterball) in which it claims to be unique because 
there are no fixed segments. I once borrowed a 4x4x4 Rubik's Revenge from a 
friend, and it appeared to have no fixed segments. I believe the Masterball 
to be a different puzzle, but with similar internal workings. What do you 
lot think?

p.s. I'm sure this is a common question, but I have many 3x3x3, and one 
5x5x5, can I still get a 4x4x4 anywhere? would anyone consider selling one 
to me?

p.p.s. I'm in London.



From JBRYAN@pstcc.cc.tn.us  Wed May  1 18:33:37 1996
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Date: Wed, 01 May 1996 18:33:30 -0500 (EST)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Shamir and M-Conjugacy Don't Mix
To: Cube-Lovers <cube-lovers@ai.mit.edu>
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I have reluctantly concluded that encoding the nodes of a breadth first
search tree as resentative elements of M-conjugacy classes cannot be
combined with Shamir's method.  The short version of the reason is that
Shamir works only for post-multiplying, and we must both pre-multiply and
post-multiply in order to calculate M-conjugates. 

It is possible, I think, to use a number of what might be called "clever
hacks" involving M-conjugacy to reduce rather considerably the memory
requirements of a program implementing Shamir's method, but the basic
method cannot store just the representative elements.  I will describe the
clever hacks if and when I get a working program.  How "clever" the clever
hacks are may lie in the eye of the beholder.  In all cases, they exchange
reduced storage requirements for increased running time.  It is always a
question as to whether such trade-offs are advantageous or not. 

The longer explanation follows.  I will be starting with some real basics. 

Almost any reasonable introductory or advanced math book will talk about
functions.  There are two basic views of what a function is.  In one view,
a function is a non-empty set X, a non-empty set Y, and some general rule
or correspondence such that for every x in X, there is exactly one y in Y. 
In the other view, you start out with the set of all ordered pairs (x,y)
with x in X and y in Y.  This set is usually called X x Y (X cross Y).  A
function is then a non-empty subset of X x Y such that every x appears
exactly one time as the left hand element of the ordered pair, so that
again for every x in X there is exactly one y in Y. 

The second definition is probably more accurate, but it loses (on purpose,
perhaps) the intuitive feel that there is some sort of "general" rule
relating X and Y.  Indeed, for a finite X and Y, there may be no shorter
way to specify a particular function than simply to list the set of
ordered pairs of which it is comprised. 

A function where X and Y are the same set is said to be a function "on X". 
A function may be one-to-one or onto or both.  There are many (equivalent)
definitions, but my favorites are that a function is onto if there is at
least one x for every y, and a function is one-to-one if there is at most
one x for every y.  Hence, a function is both one-to-one and onto if there
is exactly one x for every y.  This condition is necessary if we wish to
be able to run the function backwards, that is, if we wish to have an
inverse function.  Finally, a function that is on a set and which is
one-to-one and onto is a permutation.  We model the Rubik's cube as a set
of permutations. 

Suppose F and G are functions.  In algebra and calculus, we define the
composition of two functions something like the following: FoG(x)=F(G(x)). 
Proper treatment of this definition would require some care in handling
the domain and range of the respective functions.  But we will dispose of
this issue by simply stipulating that F and G are permutations on the same
set. 

The algebra/calculus notation for function composition yields a
right-to-left evaluation of the functions.  This is especially visible if
we compose more than two functions, e.g., FoGoH(x)=F(G(H(x))).  In group
theory, function composition is more typically written left-to-right such
as HGF for the example at hand, with debates about where the argument
goes.  I prefer in front -- (x)HGF.  In Cube Theory, we almost *always*
write operators left-to-write, following group theory rather than
algebra/calculus. 

This whole left-to-right vs. right-to-left issue is critical for for
proper implementation of Shamir.  It is especially critical to get it
right because essentially all programming languages follow the
algebra/calculus model, whereas Cube Theory follows group theory.  Hence,
everything is always backwards to some extent in a program. 

I'm an *old* programmer, so my first higher level programming language
(after assembler) was FORTRAN.  FORTRAN lets you have statements such as
Y=X(I) or Y=SQRT(X).  FORTRAN has rather obtuse semantics and is hard to
compile.  I can remember at the time I learned FORTRAN being puzzled by
how the compiler could figure out whether parentheses meant function
arguments or whether parentheses meant subscripts.  More "modern"
languages (say, those less than 20 years old) tend to use square brackets
for subscripts, making the compiler's job a bit easier. 

But the vagaries of FORTRAN serve us well at this point.  Suppose we want
to define a permutation (which is after all, just a function) on 1..3.  We
define F as what old FORTRAN programmers called an array with three
elements (more often called a vector these days).  Then, we can assign
values to the array elements, such as F(1)=2, F(2)=3, and F(3)=1. 
Finally, we can write statements such as Y=F(X), which look and act like
functions, although FORTRAN thinks of them as arrays.  (Well, F, X, and Y
would have to be defined as INTEGERS, which is not very FORTRANish, but so
be it). 

What about function composition, say G(F(X))?  It works, but be careful
what you mean.  A very short snippit of code might look something like the
following: 
  
         X=1
         Y=F(X)
         Z=G(F(X))
         PRINT Y, Z

Function composition works as advertized even though these things are
really arrays.  But the composition is calculated only for one particular
value of X, namely X=1.  If we want to calculate the full=blown
composition H=GoF (group theory, H=FG), the code snippit would be

       H(1)=G(F(1))
       H(2)=G(F(2))
       H(3)=G(F(3))

As you can see, this programming way of implementing a permutation as an
array is really the second way in which math books define a function,
namely as a set of ordered pairs.  For example, the function F from above
is simply F={ (1,2), (2,3), 3,1) }.  But the X values were never stored
explicitly.  Rather, they were the array indices.  We would say that the F
array stores the Y values as a vector.  In this case, we would say that
F=[2,3,1]. 

Notice that it is somewhat arbitrary whether X is represented by the
indices and Y is represented by the vector, or vice versa.  The way I have
shown it seems more natural, but the vice versa is certainly tenable. 
Notice also that if we think of the vice versa where X is represented by
the vector and Y is represented by the indices, then we have the inverse
function F'.  Hence, the same vector can represent both F and F'. 

As a practical matter, I really prefer to have indices represent X and to
store the inverse as a separate vector.  Let me switch to a more modern
look and feel, using square brackets.  The code to calculate an inverse
would then look something like. 

     F_inverse[F[1]] := 1;
     F_inverse[F[2]] := 2;
     F_inverse{F[3]] := 3;

You would really do this with a loop, so it would be something like
     
     For i := 1 to 3
        F_inverse[F[i]] := i;

If you translate this loop back into group theory, it more or less states
the identity FF'=I (the looping just makes sure that we touch all our X
values -- the index i is our X value, and the order of F and F' is
backwards between our program and group theory). 

The key component of Shamir's method involves multiplying a permutation t
by each permutation s in a set S, where the set S is in lexicographic
order.  I want to show what happens with both pre-multiplying and
post-multiplying.  In order to deal with representative elements of
M-conjugacy classes, we need both to pre-multiply by m' and to
post-multiply by m, so the issue of pre-multiplying vs. post-multiplying
is critical.  I will use permutations on 1..4 in vector notation for my
examples. 

    t            S                    tS

 [3,1,4,2]    [1,2,3,4]      =     [3,1,4,2]
 [3,1,4,2]    [1,2,4,3]      =     [4,1,3,2]
 [3,1,4,2]    [1,3,4,2]      =     [4,1,2,3]
 [3,1,4,2]    [2,1,3,4]      =     [3,2,4,1]
 [3,1,4,2]    [3,1,2,4]      =     [2,3,4,1]
 [3,1,4,2]    [4,3 2,1]      =     [2,4,1,3]




    S            t                    St

 [1,2,3,4]    [3,1,4,2]      =     [3,1,4,2]
 [1,2,4,3]    [3,1,4,2]      =     [3,1,2,4]
 [1,3,4,2]    [3,1,4,2]      =     [3,4,2,1]
 [2,1,3,4]    [3,1,4,2]      =     [1,3,4,2]
 [3,1,2,4]    [3,1,4,2]      =     [4,3,1,2]
 [4,3,2,1]    [3,1,4,2]      =     [2,4,1,3]
         

Let's take the case of St first.  This is classic Shamir.  S is in
lexicographic order.  Going from S to St, every 1 has been replaced by a
3, every 2 has been replaced by a 1, every 3 has been replaced by a 4, and
every 4 has been replaced by a 2. 

We can get St into lexicographic order by sorting S in the order 2 first,
4 second, 1 third, and 3 fourth.  The vector [2,4,1,3] which controls this
alternative sorting order is simply t'.  Hence, we don't really have to
sort St if S is made into a tree.  Rather, we traverse the S tree using t'
as a template to define an alternative order of traversal, and St
automagically pops out in lexicographic order. 

(By the way, there is a minor error in one of my previous posts.  Allen
Bawden used right-to-left notation in his original Shamir article.  I
copied what he wrote thinking he was using left-to-right notation.  To
properly "copy" what he wrote and also put it in left-to-write notation, I
needed to reverse everything, but I failed to do so.  Hopefully,
everything will be consistent and correct in this article.)

The tS case is much trickier.  Think of S as a matrix.  To get to tS, what
you do is permute the columns.  With the particular t we are using, column
3 becomes column 1, column 1 becomes column 2, column 4 becomes column 3,
and column 2 becomes column 4. 

I really can't think of any Shamir-like tree traversal that would put tS
into lexicographic order.  To see the nature of the problem very clearly,
look at the original S and think of sorting the rows using column 3 as the
major sort.  We can't really move the rows around of course, because we
only have one S and we have to sort it differently for each t.  Just look
down column 3 and think about sorting without actually moving anything. 
Remember that in case of ties, you would then have to look at column 1,
then column 4, etc.  It's a mess, and I don't think you can do it without
adding a data structure much larger than what we already have.  And the
original point of combining Shamir with M-conjugacy was to save memory. 

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7127
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990


From dlitwin@geoworks.com  Wed May  1 20:27:22 1996
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Date: Wed, 1 May 96 17:27:07 PDT
From: dlitwin@geoworks.com (David Litwin)
Message-Id: <9605020027.AA17163@quark.geoworks.com>
To: cube-lovers@ai.mit.edu
In-Reply-To: <9604301252.AA02489@mecmdb.me.ic.ac.uk>
Subject: Where to get the Rubik's Revenge


"The Official Thermodynamics Fan Club of the UK." writes:
> p.s. I'm sure this is a common question, but I have many 3x3x3, and one 
> 5x5x5, can I still get a 4x4x4 anywhere? would anyone consider selling one 
> to me?

	The last time I bought one was in 1993 in Tokyo (at Tokyu Hands
department store).  I'll be visiting again soon and will check to see if
they still have any.  This is the only place I've been able to find them,
outside of those who sell from their private collections.  At this point I
don't think anyone has any available for sale though.  If I find any in
Japan I'll certainly buy as many as possible and let the list know.
	
	The 5x5x5 is more widely available, try:

Peter Beck
pbeck@pica.army.mil

	or

Dr. Christoph Bandelow
An der Wabeck 37
58456 Witten
Germany

	Good luck,

	Dave Litwin

From din5w@dot.cs.virginia.edu  Wed May  1 21:39:44 1996
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Date: Wed, 1 May 1996 21:39:07 -0400 (EDT)
From: Dale Newfield <din5w@virginia.edu>
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To: cube-lovers@ai.mit.edu
Subject: building a Rubik's Revenge
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I know that I have at least some of the parts from at least one broken 
Rubik's Revenge lying around somewhere.  I would suspect that I am not 
alone.  I was wondering if we could satisfy a small portion of the people 
that do not have such a beast merely by pooling our resources, and 
building whole puzzles out of the remains of old ones?  So--I'll check to 
see what I can find, and anyone that likewise has RR carcasses that they 
wouldn't mind donating to such a purpose, mail me.  Let's see what we can 
pull off!

-Dale Newfield
 DNewfield@Virginia.edu

From diamond@jrdv04.enet.dec-j.co.jp  Wed May  1 21:50:39 1996
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From: Norman Diamond  02-May-1996 1049 <diamond@jrdv04.enet.dec-j.co.jp>
To: cube-lovers@ai.mit.edu
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Subject: Re: Where to get the Rubik's Revenge
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David Litwin writes:

>"The Official Thermodynamics Fan Club of the UK." writes:
>> p.s. I'm sure this is a common question, but I have many 3x3x3, and one 
>> 5x5x5, can I still get a 4x4x4 anywhere? would anyone consider selling one 
>> to me?

>	The last time I bought one was in 1993 in Tokyo (at Tokyu Hands
>department store).  I'll be visiting again soon and will check to see if
>they still have any.  This is the only place I've been able to find them,

Seibu and Hakuhinkan used to have them too.  But the distributor took them
off the market last year and they are no longer available.  (This is 4x4x4.)

>	The 5x5x5 is more widely available, try:
>Peter Beck
>pbeck@pica.army.mil
>	or
>Dr. Christoph Bandelow
>An der Wabeck 37
>58456 Witten
>Germany

Indeed Dr. Bandelow is the designer and source of much of this stuff.
Ask for his catalog, which is in English.

The best puzzle store now in Tokyo is Moebius, between Ichigaya and Kudanshita
subway stations.  They have one 5x5x5 in stock at the moment, though it
traversed through Taiwan on its way here.  They also just got in a bunch of
stuff from Jean-Claude Constantin (but unfortunately no more transformations
of Rubik-style shapes).

-- Norman Diamond           diamond@jrdv04.enet.dec-j.co.jp
[Speaking for Norman Diamond not for Digital.]

From nichael@sover.net  Wed May  1 23:23:03 1996
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Date: Wed, 1 May 1996 23:24:16 -0400
To: DNewfield@virginia.edu, cube-lovers@ai.mit.edu
From: nichael@sover.net (Nichael Lynn Cramer)
Subject: Re: building a Rubik's Revenge

At 9:39 PM 5/1/96, Dale Newfield wrote:
>I know that I have at least some of the parts from at least one broken
>Rubik's Revenge lying around somewhere.  I would suspect that I am not
>alone.  I was wondering if we could satisfy a small portion of the people
>that do not have such a beast merely by pooling our resources, and
>building whole puzzles out of the remains of old ones?  So--I'll check to
>see what I can find, and anyone that likewise has RR carcasses that they
>wouldn't mind donating to such a purpose, mail me.  Let's see what we can
>pull off!

An excellent suggestion.

For those who aren't aware of the problem that Dale mentions, the issue is
that --unlike the odd-order cubes that have a central piece which can be
used to anchor its neighbors-- the structural integrity of the RR depends
on a central (internal) plate held in place by a screw whose adjustment is
quite critical.  Of the four RRs that I own, one is very good.  Of the
others, two are so tight as to be almost impossible to turn.  The fourth is
so loose that it simply cannot be made to stay together (I discovered this
for the first time as I was sitting in a movie theater playing with my new
toy; as the lights lowered for the movie to start I felt the sickening
sensation of a cube slowly dissolving in my hands...).

The last time I saw it, it was in a sack in a drawer in my other office.
I'll be glad to contribute that one.

Nichael Cramer
work: ncramer@bbn.com
http://www.sover.net/~nichael



From aaweint@io.org  Thu May  2 00:54:38 1996
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Date: Thu, 02 May 1996 00:54:05 -0400
To: DNewfield@virginia.edu
From: Aaron Weintraub <aaweint@io.org>
Subject: Re: building a Rubik's Revenge

I think this is a good idea.  I don't have any parts to offer, but if anyone
has a spare centre piece with an orange sticker lying around, I could REALLY
use it.   I only have one 4x4x4, and that one piece is broken.  Please
e-mail me if you can help me out on this.

-Aaron
aaweint@io.org


At 09:39 PM 05/01/96 -0400, Dale Newfield wrote:
>
>I know that I have at least some of the parts from at least one broken 
>Rubik's Revenge lying around somewhere.  I would suspect that I am not 
>alone.  I was wondering if we could satisfy a small portion of the people 
>that do not have such a beast merely by pooling our resources, and 
>building whole puzzles out of the remains of old ones?  So--I'll check to 
>see what I can find, and anyone that likewise has RR carcasses that they 
>wouldn't mind donating to such a purpose, mail me.  Let's see what we can 
>pull off!
>
>-Dale Newfield
> DNewfield@Virginia.edu


From phaedrus@dreamscape.com  Thu May  2 01:39:51 1996
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To: Nichael Lynn Cramer <nichael@sover.net>
Cc: DNewfield@virginia.edu, cube-lovers@ai.mit.edu
Reply-To: bmbuck@acsu.buffalo.edu
Subject: Re: building a Rubik's Revenge 
In-Reply-To: Your message of "Wed, 01 May 1996 23:24:16 EDT."
             <v02120d0dadadd60569d5@[204.71.18.82]> 
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Date: Thu, 02 May 1996 01:40:14 -0400
From: Buddha Buck <phaedrus@dreamscape.com>

> At 9:39 PM 5/1/96, Dale Newfield wrote:
> For those who aren't aware of the problem that Dale mentions, the issue is
> that --unlike the odd-order cubes that have a central piece which can be
> used to anchor its neighbors-- the structural integrity of the RR depends
> on a central (internal) plate held in place by a screw whose adjustment is
> quite critical.  Of the four RRs that I own, one is very good.  Of the
> others, two are so tight as to be almost impossible to turn.  The fourth is
> so loose that it simply cannot be made to stay together (I discovered this
> for the first time as I was sitting in a movie theater playing with my new
> toy; as the lights lowered for the movie to start I felt the sickening
> sensation of a cube slowly dissolving in my hands...).

Another problem is that the center pieces are held in place by a foot 
with a rather thin "leg".  My RR died three days after I finally worked 
out a solution.  One of the center pieces lost its foot, and the rest 
of the cube simply dissolved.

I'll have to look to see how much of the cube I can still collect 
together.  Hopefully, I'd be able to donate a central ball in good 
shape, plus perhaps up to 55 pieces (if I can find them all).  Or I'd 
be willing to accept the pieces I'm missing to resurect my RR.
-- 
     Buddha Buck                      bmbuck@acsu.buffalo.edu
"She was infatuated with their male prostitutes, whose members were
like those of donkeys and whose seed came in floods like that of
stallions."  -- Ezekiel 23:20



From din5w@dot.cs.virginia.edu  Thu May  2 11:22:24 1996
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Date: Thu, 2 May 1996 11:22:09 -0400 (EDT)
From: Dale Newfield <din5w@virginia.edu>
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Reply-To: DNewfield@virginia.edu
To: cube-lovers@ai.mit.edu
Subject: Re: building a Rubik's Revenge 
In-Reply-To: <199605020540.BAA17826@zaphod.caz.ny.us>
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OK--how's this for a suggestion:  Since the goal of this project is to
"spread the wealth," let's say that anybody that donates any 4x4x4 parts
and has no other 4x4x4 cube should get one in return, but otherwise, all
reconstructed cubes should go to people that currently have none.  As
well, any excess pieces when we're done can be distributed along with the
cubes (and back to the original donators), so that future catastrophes may
be fixed (a corner, three edges, and three centers, for example).  Rather
than continue this discussion on the mailing list, unless there actually
is anything of general interest in this, I'd suggest we do this through
personal mail from now on. 

So, send me mail regarding what pieces you have been able to dig up and 
stating if you have another 4by cube, or send me mail saying that you've 
been looking for a 4by, and since they seem unavailable elsewhere, would 
like one of these reconstructed ones.

To further save me some typing, I'll include my snail-mail address below, 
to which people that have pieces can send them.

If mail costs turn out to be non-trivial, we could probably ask those 
recieving cubes to pitch in to cover these costs--so I'll try to mark 
down the postage on each package that I recieve, in order to keep a record.

-Dale

BTW--I found and put together what I have last night--an entire cube
except for one broken center piece, and one broken corner piece--this one
should be easily resurrected! (And I have other 4x4x4's so this one is due
for distribution :-)

Dale Newfield
1805 Inglewood Dr. #23A
Charlottesville, VA 22901-2708
U.S.A.

|..|.|..|.||.|..||......||..|.||...|||...|..|.||....|.|..|..||
(No, this barcode is not necessary, but I figured this would be a good 
place to ask: "Has anyone figured out what information is encoded in 
this, or how it is encoded?" :-)

From paul@tarragon.ee.byu.edu  Thu May  2 11:53:34 1996
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Date: Thu, 2 May 1996 09:52:53 -0600 (MDT)
From: Paul Hart <paul@tarragon.ee.byu.edu>
To: David Litwin <dlitwin@geoworks.com>
Cc: cube-lovers@ai.mit.edu
Subject: Re: Where to get the Rubik's Revenge
In-Reply-To: <9605020027.AA17163@quark.geoworks.com>
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On Wed, 1 May 1996, David Litwin wrote:

> "The Official Thermodynamics Fan Club of the UK." writes:
> > p.s. I'm sure this is a common question, but I have many 3x3x3, and one 
> > 5x5x5, can I still get a 4x4x4 anywhere? would anyone consider selling one 
> > to me?
> 
> 	The last time I bought one was in 1993 in Tokyo (at Tokyu Hands
> department store).  I'll be visiting again soon and will check to see if
> they still have any.  This is the only place I've been able to find them,
> outside of those who sell from their private collections.  At this point I
> don't think anyone has any available for sale though.  If I find any in
> Japan I'll certainly buy as many as possible and let the list know.

There are many resources on the World Wide Web for people who enjoy 
puzzles, and in particular cube-style puzzles.

There is a puzzle dealer based in the United States named "Puzzletts" 
that advertises on the WWW at http://www.puzzletts.com.  Among other 
things in their inventory, they carry what I assume are new 3x3x3, 4x4x4, 
and 5x5x5 cubes.  I haven't compared their prices to other distributors 
yet, but they appear to have a *very* large puzzle inventory.

Other cube dealers can likely be located on the page at:

    http://sdg.ncsa.uiuc.edu/~mag/Misc/CubeSources.html

which is where I learned of Puzzletts.  Another good page for cube 
resources and information is:

    http://www.student.informatik.th-darmstadt.de/~schubart/rc_resources.html

I was excited to learn that cubes can still be purchased at at least a 
few retail sources, and with any luck the inventories of these sources will
endure a few more years.

I hope this information has been useful!

-- Paul

--------------------------------------------------------------------------
Paul Hart                                   Computer Systems Administrator 
paul@ee.byu.edu                        Electrical and Computer Engineering
(801) 378-5728                       Brigham Young University, Provo, Utah





From din5w@dot.cs.virginia.edu  Thu May  2 13:28:00 1996
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Date: Thu, 2 May 1996 13:27:30 -0400 (EDT)
From: Dale Newfield <din5w@virginia.edu>
X-Sender: din5w@dot.cs.Virginia.EDU
Reply-To: DNewfield@virginia.edu
To: cube-lovers@ai.mit.edu
Subject: TripleCross by Binary Arts
In-Reply-To: <199605020540.BAA17826@zaphod.caz.ny.us>
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I've been investigating for month or so, now the size and shape of the 
group represented by this puzzle...There are 18! positions, of course, 
although I have yet to do any meaningful investigation in regards to what 
interesting sub-groups exist, or even how many orbits there are in that 
group, as 18! is a very large number (and I've been doing all of this 
work with home-spun tools...I really need to get ahold of GAP and see 
what it can do :-).  Instead I've been doing most of my investigations 
with respect to a fairly arbitrary subgroup--that of distinguishable 
positions (On the actual puzzle there are 9 indestinguishable blank 
tiles, and 3 indestinguishable tiles that each contain one orange dot).
In this group there are just under 3 billion possible positions: 
18!/(9!*3!).  One of the things that I am currently trying to do is, 
using a breadth first search, simply expand the entire graph from start, 
to see how many of these positions are in the same orbit as start, and 
thus to find out experimentally how many orbits there are.  I have not 
really begun the analytical process of looking into these groups yet.  I 
can encode a position (theoretically ~31.5 bits of information) into a 32 
bit unsigned long, along with a 2 bit number representing which direction 
to go to get to start.  I've gone through many iterations, but now I 
believe that I have built a system in which any given position in the 
breadth first search queue takes up 4.5 bytes (on average), and in the 
hashtable that stores the unique positions that popped out of the queue, 
takes up about the same size.  This means that the entire graph can be 
laid out in less than 13 gig of space.  This means that the problem is 
solvable with enough money thrown at it, but is just beyond most of our 
reaches right now (I have this running on a machine w 512M of memory, 
and 1.2 Gig of swap...(Once the job runs into swap it crawls)...so if 
there are 12 orbits I should find out before this run is through.)

As well, I have built a java applet that allows one to play with the 
device.  My goal was to include a "Solve" button that would show a person 
a path toward a solved state.  Actually my goal was to be able to give up 
and "reset" the physical toy back into a start state if I so desired so 
that I can play with it without fear... although the ability to put in 
macros in the applet was quite useful...

Anyway, any comments are welcome.

Binary Arts has a site: http://www.puzzles.com/

the TripleCross can be seen on this page:
http://www.puzzles.com/catalog/b3c1d1.htm

and a link to my TripleCross game is
http://www.cs.virginia.edu/~din5w/tc/
although it is far from complete, and I would appreciate people not 
making links to it yet. (Or telling binary arts people right now--they 
might not like it.)


Sorry this note was so rambling.

-Dale Newfield


From keith@nwwtdi.demon.co.uk  Fri May  3 05:57:07 1996
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Date: Fri, 3 May 1996 10:46:18 +0100
To: Cube-Lovers@ai.mit.edu
From: Keith Gregory <keith@nwwtdi.demon.co.uk>
Subject: Octahedra and tetrahedra
Mime-Version: 1.0
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I have recently been solving a simulation of an octahedral rubiks cube
type puzzle (an octahedron which twists through planes parallel to the
faces). Does anyone know if mechanical versions of such a puzzle exist?
The Skewb is clearly an example of an octahedral puzzle mapped onto a
cube so mechanically it must be possible to manufacture one with 2
triangles on each edge. However I would really like one with 3 or 4 on
each edge. My investigations suggest that, like the cube, if you can
solve a one with 4 on each edge you must have the right techniques for
solving on with 5,6,7,8,9 or 10 along each edge (if you have the
patience). Is there anyone else interested in octahedral puzzles?

Similarly does anyone know if there is a mechanical version of a
tetrahedral puzzle with 4 triangles on each edge rather than the
standard 3?

Thanks
-- 
Keith Gregory

From hoey@aic.nrl.navy.mil  Fri May  3 12:15:04 1996
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From: hoey@aic.nrl.navy.mil
Message-Id: <9605031613.AA08766@sun34.aic.nrl.navy.mil>
To: Dale Newfield <DNewfield@virginia.edu>, cube-lovers@ai.mit.edu
Subject: Re: TripleCross by Binary Arts
In-Reply-To: <Pine.SUN.3.90.960502112226.7956E-100000@dot.cs.Virginia.EDU>

Dale Newfield writes about the Binary Arts puzzle TripleCross.  In
case anyone in cube-lovers hasn't seen it, it's a nice mechanical
puzzle involving the permutation of 18 pieces.  Stripped of the
mechanical parts, it has 18 sliding pieces in an array shaped like:

               1  2
   3  4  5  6  7  8  9
  10 11 12 13 14 15 16
     17 18

Pieces 8, 9, 15, 16, 17, and 18 are distinguishably marked.  Pieces
3, 4, and 5 are marked identically.  The remaining nine pieces are
unmarked.

Pieces are permuted by a plunger-like action.  One kind of plunge is
to move pieces 3-16 left or right one unit.  The other is to move the
second and third columns (4,5,11,12,17,18) up one unit while
simultaneously moving the fifth and sixth columns (1,2,7,8,14,15) down
one unit.  At the end of any process, we it is traditional to restore
both plungers to their original position.  Calling the horizontal
plunger positions Left, Center, and Right and the vertical plunger
positions Up and Down, we may reduce any process to a sequence of ULDC
and URDC and their inverses LUCD and RUCD.  Cycle forms for the first
two are

    ULDC = (1,7,14,15,16,9,2)(4,5,6,13,18,17,11) and
    URDC = (1,2,8,15,14,13,6)(3,10,17,18,12,5,4).

Since these are both even permutations, it's clear the group must be a
subgroup of A18. GAP confirms that all 18!/2 positions are reachable.

Dale writes:

> ... I've been doing most of my investigations 
> with respect to a fairly arbitrary subgroup--that of distinguishable 
> positions (On the actual puzzle there are 9 indestinguishable blank 
> tiles, and 3 indestinguishable tiles that each contain one orange dot).
> In this group there are just under 3 billion possible positions: 
> 18!/(9!*3!)....

Right, though this is not actually a group, but a collection of
cosets.  For instance, you could have two indistinguishable
manipulations that would provide distinguishable outcomes when
preceded by some other manipulation.

Dale writes of his work on a breadth-first search:

> ... I can encode a position (theoretically ~31.5 bits of
> information) into a 32 bit unsigned long, along with a 2 bit number
> representing which direction to go to get to start.

There is a more compressed approach that has proven valuable with some
of the smaller cubes.  The idea is to use the 32-bit number as the
index into a large vector of distances.  Initialize all the distances
to -1, set the distance of SOLVED to zero, and repeatedly scan the
whole array checking neighbors of positions of distance D; if their
distance is -1 set the distance to D+1.

This procedure admits some useful variations.  For one thing, you can
store D mod 3 in two bits, with a fourth value in place of -1.  This
reduces the storage to 735 megabytes.  Also, instead of relying on
your virtual memory system, you can store the vector in a big
random-access file (or several files).  In both cases, it's probably
easiest to use bit-fields of the index to specify which file (if
multiple files), which block of the file (choose a useful power of
two), which byte in the block, and which part of the byte (if packed).
Third, it may be better to generate lists of a few thousand neighbor
indices, sort them, and then scan for neighbors, to reduce thrashing.

If you implement this, I'd be interested in knowing:

    Number of positions at each distance,
    Maximal-distance positions (up to ten or twenty),
    Number of local maxima at each distance (these show up as
        positions none of whose neighbors get set),
    Number of nonstrict local maxima at each distance (i.e. which have
        neighbors at the same distance).

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From cube-lovers-errors@curry.epilogue.com  Tue May  7 15:11:29 1996
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To: cube-lovers@ai.mit.edu
From: Georges Helm <geohelm@pt.lu>
Subject: cube solutions prior to 1980

Some time ago someone asked a question about early cube solutions.
Unfortunately I don't find the corresponding mail, but here is what I have
found.
I have solutions by the following people which were all written in 1979:
Angevine, Beasley, Cairns/Griffiths, Dauphin, Howlett, Jackson (3-D),
Johnson, Maddison, Singmaster, Sweenen and Truran.
At http://ourworld.compuserve.com/homepages/Georges_Helm/cubbib.htm other
info on solutions can be found.
Georges Helm           
geohelm@pt.lu
http://www.geocities.com/Athens/2715
http://ourworld.compuserve.com/homepages/Georges_Helm
-------------------------------------
Phone: ++352-503896 (answer machine)
       ++352-38019 (office)
       ++352-021 19 13 13 (GSM)
Fax:   ++352-38535
-------------------------------------



From cube-lovers-errors@curry.epilogue.com  Wed May  8 02:22:29 1996
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Date: Wed, 08 May 1996 09:19:32 +0300
From: Rubin Shai <s2394459@cs.technion.ac.il>
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Subject: Rubik's cube
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Hi 
I'm looking for solutions to the 2X2X2 cube.
I need solution that put the cubiks ONE AFTER THE OTHER.
Is anyone can help?
Thanks 
Shai Rubin


From cube-lovers-errors@curry.epilogue.com  Fri May 10 16:55:11 1996
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Message-Id: <31932827.7790@cytex.com>
Date: Fri, 10 May 1996 04:27:35 -0700
From: "Michael B. Parker" <mbparker@cytex.com>
Reply-To: mbparker@cytex.com
Organization: CYTEX CORPORATION
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Newsgroups: rec.puzzles,geometry.puzzles,rec.games.abstract,comp.ai.games,rec.games.design,rec.games.misc,oc.general,la.general
To: PuzzleParty@cytex.com, Cube-Lovers@ai.mit.edu, www-designer@cytex.com,
        506maple-residents@cytex.com, mitacas@cytex.com,
        Pierre Wuu <pwuu@aol.com>, "Julie S. Peterson" <jpete@cogent.net>
Cc: Wei-Hwa Huang <whuang@cco.caltech.edu>
Subject: Puzzle Party THIS SATURDAY, 7pm, Cal Tech
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#?????????????????#
                              ? PUZZLE PARTY 5! ?
                              #?????????????????#

HEY, LA!  PUZZLE PARTY 5 IS AT *CALTECH*!, and is hosted by the International
Puzzle Champion (and CalTech Junior) Wei-Wwa Huang!  Wei-Hwa was recently
featured in the LA Times for leading the US Puzzle Team to victory in the
1996 International Puzzle Chapionships.

So show and share the brain teasers and mechanical puzzles you have, and the
mental games you know, and discover dozens of new ones, and new tricks!  (Or,
if you're still puzzleless, *this is the place* to get clued in!)  Plenty of
snacks, refreshments, and good conversation provided.

WHEN:   Saturday, 1996 May 11, 7pm until the wee hours of the morning...

WHERE:  Winnett Student Center, 1200 San Pasqual, Pasadena, CA

  Caltech is located in a rectangle bordered on the north by Del Mar Blvd.,
  south by California Blvd., west by Wilson Avenue, east by Hill Avenue.
  Winnett is the small building right in the middle of campus.  Parking
  near Winnett is limited, so try to find a local parking space and walk
  to Caltech.  There is some parking near the northwest and southeast of
  Caltech, but they can be quite full.  Signs will be posted on and around
  major entrances.

 Directions:
  From 210 fwy: exit south on either:
    (1) Hill Avenue: Caltech will be on your right after 2 to 3 miles; or
    (2) Lake Avenue: Turn left on Del Mar after 2 miles, then right
        on Wilson.  Caltech will be on your left.
  From 110 north: continue until in Pasadena.  Turn right at California.
    Caltech will be on your left after 3 miles.

COST:   $10 Non-MITCSC Members without puzzles
        $ 8     MITCSC Members without puzzles
        $ 6 Non-MITCSC Members with puzzles
        $ 4     MITCSC Members with puzzles
        Free  Caltech Students (with Student ID)
        [Hey, it's a condition of them letting us use the room...]

Please RSVP (with the number of puzzles you'll be bringing) so we know how
many people (and puzzles) to expect.

RSVP:   Mike Parker, MIT '89 - mbparker@cytex.com, 800-MBPARKER xLIVE, xFAXX
        (if you get lost, call 800-MBPARKER xLIVE -- will have cell phone)
HOST:   Wei-Hwa Huang, CIT '97 - whuang@cco.caltech.edu, 818-395-1599

                 PS: for the latest Puzzle Party updates,
                     just tune in to http://www.cytex.com/~mitcsc/

-----------------------------------------------------------------------------
 Michael B. Parker, MIT '89
  CYTEX CORP. President                 http://www.cytex.com/~mbparker/
  email mbparker@cytex.com, direct voice 714-639-6436, fax 714-639-5381

CYTEX CORPORATION,  ** WE PUT YOUR COMPANY ON THE INTERNET **
  506 N. Maplewood St., Orange, CA 92667-6917   *   1-800-33CYTEX (332-9839)
  Dial 800#, then enter extension (pin): SALES(7253), TECH(8324), FAXX(3299)
  World-Wide-Web http://www.cytex.com/  *  email info@cytex.com  (r19960229)


From cube-lovers-errors@curry.epilogue.com  Sat May 11 00:25:22 1996
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Date: Fri, 10 May 1996 18:24:19
From: Sean  Brewer <BREWCREW@gnn.com>
To: cube-lovers@ai.mit.edu
Subject: Rubik's Revenge

  My father has been looking for the sixteen sided Rubik's Revenge  
for years now.  If you have any idea where I can get one for him  
please let me know.  Thanks for any help you can give.




From cube-lovers-errors@curry.epilogue.com  Sat May 11 04:40:17 1996
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Date: Sat, 11 May 96 01:52:57 +0400 (EST)
From: "Robert P. Munafo" <cube%mrob.uucp@ursa-major.spdcc.com>
Reply-To: "Robert P. Munafo" <cube%mrob.uucp@ursa-major.spdcc.com>
Message-Id: <7003%mrob.uucp@ursa-major.spdcc.com>
To: cube-lovers@ai.mit.edu
Subject: The barcode


Dale Newfield wrote:
> [...]
> Charlottesville, VA 22901-2708
> |..|.|..|.||.|..||......||..|.||...|||...|..|.||....|.|..|..||
> (No, this barcode is not necessary, but I figured this would be a good
> place to ask: "Has anyone figured out what information is encoded in
> this, or how it is encoded?" :-)

It's your zip code. The Post Office recognizes the zipcode with some sort of
OCR (Optical Character Recognition) then prints a barcode on the envelope
so that simpler machines can sort the mail later in its delivery path.
Sometimes the sender prints the barcodes (if your mail is part of a large
mailing, like junk mail, magazines, tax forms etc.).

This is from the URL http://www.advanstar.com/autoidnews/barcofaq.txt
> POSTNET symbols are different from other symbologies because the 
> individual bar height alternates rather than the bar width. Each 
> number is represented by a pattern of five bars. A single tall bar is 
> used for the start and stop bars.
> 
> Each symbol includes a check digit defined as the single digit that 
> must be added to the sum of all the digits to make the total the next 
> multiple of 10. For example, 98116's check digit is 5 because:  
> 9+8+1+1+6=25 and 25 + 5 = 30.
> 
> POSTNET can be used for 5-digit, 9-digit ZIP+4, and the new 11-digit 
> Delivery Point Barcode. They are often used in conjunction with one 
> of the three FIM bars (Facing Identification Marks) which are found 
> on the upper right corner of a mail piece like Business Reply Mail.

The encoding is as follows:

  ||...  ...||  ..|.|  ..||.  .|..|  .|.|.  .||..  |...|  |..|.  |.|..
    0      1      2      3      4      5      6      7      8      9

--
Robert P. Munafo                  UUCP: ...!harvard!spdcc!mrob!cube
CUBE-LOVERS Account              Internet: cube%mrob.uucp@spdcc.com


From cube-lovers-errors@curry.epilogue.com  Sat May 11 17:40:28 1996
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From: 100021.1617@compuserve.com
Date: 11 May 96 10:15:57 EDT
To: CUBE-LOVERS@ai.mit.edu, Sean Brewer <BREWCREW@gnn.com>
Subject: Re: Rubik's Revenge
Message-Id: <960511141556_100021.1617_EHV69-2@CompuServe.COM>


> My father has been looking for the sixteen sided Rubik's Revenge  
> for years now.  If you have any idea where I can get one for him  
> please let me know.  Thanks for any help you can give.

Me too. Somebody posted this information some time ago in the cube-lovers 
list...

<http://www.puzzletts.com>

It's a shop with Rubik-related puzzles, among others, including

#130 Rubik's Revenge (4x4x4)
#131 Professor's Cuve (5x5x5)

They sell on the Internet, too. They also have a complete list of 
distribuitors worldwide, including

*Game Preserve
222 D Street, Suite 4
Davis, CA 95616
USA
Tel.: (916) 753 42 63

*Star Magic
4026 24th Street
San Francisco, CA 94114
USA

Greetings,

Alvaro



From cube-lovers-errors@curry.epilogue.com  Sat May 11 17:53:19 1996
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To: cube-lovers@ai.mit.edu
From: The Unofficial Thermodynamics Fan Club of the UK <a.southern@ic.ac.uk>
Subject: Re: Rubik's Revenge
X-Mailer: <PC Eudora Version 1.4>

>  My father has been looking for the sixteen sided Rubik's Revenge  
>for years now.  If you have any idea where I can get one for him  
>please let me know.  Thanks for any help you can give.
>
>

I always thought that the Rubik's cube principle could only be applied to 
Polyhedrons of sides totalling 4, 6, 8, 12 or 20 as these are the only ones 
that could be made from only one shape of side.

If it is a 12 sided Rubik's polyhedron, there is one called the MegaMinx 
made by Uwe Meffret in Hong Kong. This should be available around the world 
as it is still in production.

Meffret's Company is known as:
IDI,
P.O.Box 24455,
Aberdeen,
Hong Kong.

Tel: 852-2518-3080
Fax: 852-2518-3282
(I haven't found anything on the Net for his Company, but I'd be interested 
to learn....)

 
Another possibility is that it could be a 'Cut down' Rubik Cube, but I 
haven't seen any of this type in years.



Andy,

(The Artist Currently Known as the Unofficial Thermodynamics Fan Club of The 
U.K.)



From cube-lovers-errors@curry.epilogue.com  Mon May 27 19:46:08 1996
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Date: Thu, 23 May 1996 12:53:27 -0500 (EST)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Compact Cube Representation for Shamir and Otherwise
To: Cube-Lovers <cube-lovers@ai.mit.edu>
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I said I wasn't going to write again about Shamir's method until
I had a working program.  Well, I don't have a working program
yet but this is only indirectly about Shamir.  Rather, it is
about how we might represent the cube compactly in a way that is
easy to work with.  We would like a compact representation that
is usable by Shamir's method.  But more importantly, we would
like a compact representation that is easily usable for forming
compositions in general.  The compact representation I will
describe is useful in a number of contexts, not just Shamir's
method.

My standard model is an S24 x S24 model, modeling the corner and
edge facelets separately and not modeling the face centers.  At
one byte per facelet, this representation requires 48 bytes per
position without packing.

David Moews has described a wreath product representation (e.g.,
23 Feb 1996) which requires 40 bytes without packing.  There are
8 bytes to describe the position of each corner cubie, and 8 more
bytes to describe the twist of each corner cubie.  Similarly,
there are 12 bytes to describe the position of each edge cubie,
and 12 more bytes to describe the flip of each edge cubie.  This
representation has the virtue of being 8 bytes smaller than the
S24 X S24 representation, while still being easy to work with and
manipulate.

For my very large searches, I always used a supplement
representation for the external files.  That is, I only stored
one facelet from each of the 8 corner cubies and one facelet from
each of the 12 corner cubies for a total of 20 bytes unpacked. 
(I also packed the 20 bytes into 13 bytes to use less tape, but
that is not important for this particular story.)

However, I found the supplement representation awkward to
manipulate, so I always expanded the supplement representation to
a full S24 x S24 representation inside the program.  None of my
programs were more than a few K (not a few Meg, just a few K
because the storage was external), so the extra few bytes were a
non-issue.  But now that I want to implement Shamir, my programs
will be very large.  Therefore, I wanted to figure out how to
manipulate the supplement representation directly.  The
representation itself is not new, but the technique to manipulate
it is.  Here is what I have come up with.  I think it is
applicable to Shamir programs and non-Shamir programs alike.

I will use the corners as an example.  Similar comments would
apply to the edges.  My standard supplement for the corners is
the Front facelets and the Back facelets.  The way I number the
facelets, these are facelets 1 through 4 for the Front and 21
through 24 for the Back.  In the vector notation we have been
talking about in this thread, the supplement of the identity is
[1,2,3,4,21,22,23,24].  1 is mapped to 1, 2 is mapped to 2, 3 is
mapped to 3, and 4 is mapped to 4.  However, 5 is not mapped to
21.  Rather, 21 is mapped to 21, 22 is mapped to 22, etc..  You
have to think of the last 4 indexes as being offset by 16 because
16 of the facelets are left out.  From this vector, we can
reconstruct the fact that 5 is mapped to 5, 6 is mapped to 6,
etc. based on which facelets are part of which cubies.

Composition of these supplement vectors can be hard or easy
depending on what we are trying to do.  Suppose X is a
permutation on the corners represented by an 8 byte supplement
vector and q is a quarter-turn on the corners represented by a 24
byte permutation vector.  Then, the calculation of Xq more or
less "just works", and the composition is an 8 byte supplement
vector.  For some kinds of things you have to worry a little bit
about the offset of the last 4 indexes, but the computer coding
is basically very straightforward.  The code even runs faster
than the code for composing two 24 byte permutation vectors.

But suppose for some reason we need to form qX instead of Xq. 
The q vector will map into values that simply aren't there in the
X vector.  The programming symptom will be an out-of-bounds
subscript.

It doesn't help to use two supplement vectors.  If X and Y are
both supplement vectors, then neither the product XY nor the
product YX can be formed.  The same problem occurs anytime a
supplement vector is pre-multiplied, no matter whether it is 
pre-multiplied by another supplement vector or whether it is 
pre-multiplied by a full-length permutation vector.

With some searches you can probably get by with only post-multiplying 
supplement vectors by full-length permutation
vectors.  I think you could form a breadth first search tree that
way by always post-multiplying by full-length vectors q in Q. 
But I always want to form M-conjugates m'Xm, so I have to be able
to pre-multiply.  Here is how to do it with supplement vectors. 

As I said, my old programs expand an 8 byte supplement vector for
the corners into a 24 byte permutation vectors on the corners
when a position is read from a file into memory.  Two special 24
byte vectors are used in the process.  One of the 24 byte vectors
defines which facelet is to the right of each other facelet on
the corner cubies, and the other of the 24 byte vectors defines
which facelet is to the left of each other facelet on the corner
cubies.  So the supplement is expanded by mapping each of the
8 bytes in the supplement into itself, and in addition by mapping
each of the 8 bytes into its respective right and left.

These "right of" and "left of" vectors can be identified with the
permutations which twist each corner cubie right and left,
respectively.  These permutations are not in the Start orbit. 
But we can nonetheless observe that both of them commute with
every other permutation.

I am focusing this example on the corners, but my old programs
also have to expand a 12 byte supplement vector for the edges
into a 24 byte permutation vector.  The vector which accomplishes
this mapping defines for each edge facelet the other facelet
which is on the same edge cubie.  This permutation can be
identified with Superflip, and Superflip also commutes with every
other permutation.

The center of G consists of the identity plus Superflip.  These
permutations fix the corners and either fix or flip the edges. 
But the center of the constructable group consists of fixing or
flipping the edges composed with fixing or twisting right or
twisting left the corners.  So there are six positions in the
center of the constructable group, and it is precisely these six
permutations which are required to make composition of supplement
vectors work.

I normally write an M-conjugate in E-mail just as m'Xm.  But for
this example, let me write it as (i)m'Xm, where i is the argument
of the permutation and where i runs from 1 to 24 for the corners. 
The trick to make composition of supplements work is going to be
to write the permutation as something like (i)m'k'Xkm, where k is
not really a permutation.  Rather, it is some magic to be defined
below.

The trick is not just for M-conjugation.  It is for pre-multiplication 
in general.  The Xm part of m'Xm is not a problem;
it is the m'X part which is a problem.  Similarly, to multiply
supplement X (or full-length vector X) by supplement Y, the k
trick would be Xk'Yk, which we could group as X(k'Yk) for
emphasis.  As with M-conjugation, I will make the argument
explicit and write (i)Xk'Yk.

But just what is this k?  For the corners, we define k[1] as the
identity, k[2] as twist all corners right, and k[3] as twist all
corners left.  We also define a 24 byte vector j which defines
which corner facelets are in the supplement (a value of 1), right
of the supplement (a value of 2), or left of the supplement (a
value of 3).  j is a function, but is not a permutation.  With my
particular numbering scheme and choice of supplement, j looks
something like [1,1,1,1,2,3,2,3,......3,2,3,2,1,1,1,1].  That is,
only the first four and last four facelets are in the supplement. 
The j vector is used to index k.  For the edges we would define
k[1] as the identity and k[2] as Superflip.

An M-conjugate would then be calculated as 

    (i)m' k[j[t]]' X k[j[t]] m

for i in 1..24 and where t=(i)m'.  Effectively, k'
maps (i)m' into the supplement so that X operates only on the
supplement, and k undoes (untwists and/or unflips) whatever k'
does.  However, the k-conjugation must be applied on a facelet by
facelet basis.  k[1] might be used for one facelet, k[2] for
another facelet, and k[3] for still another.  Nonetheless, since
each of the k's is in the center of the constructable group, we
have X=k'Xk for all X, irrespective of which k is used for a
particular facelet.

It is not strictly necessary, but this procedure would be
slightly simpler if the facelets were renumbered.  That is,
renumber the supplement 1 to 8 for the corners and 1 to 12 for
the edges.

It is easy to see how to construct the tree required by Shamir's 
method using this supplement representation.  The supplement 
representation does not reduce the number of potential branches
out of each node.  But it does reduce the number of levels of nodes.
I plan to have the branching for the first 8 levels of my tree be
controlled by the supplement for the corners, and the branching
for the next 12 levels of my tree be controlled by the supplement
for the edges.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7127
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990



From cube-lovers-errors@curry.epilogue.com  Tue May 28 14:04:51 1996
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Message-Id: <9605281411.AA29962@sarofim-sun.MIT.EDU>
To: cube-lovers@ai.mit.edu
Subject: Square 1 Combinations
Date: Tue, 28 May 1996 10:11:46 EDT
From: "Michael C. Masonjones" <masonjon@mit.edu>

Hi, I'm new to the group, but I have read the entire archive.  I noticed
rather little work done on Square 1.  It seems to me that this puzzle
deserves a closer look for finding God's algorithm.  Mike Reid's
calculations notwithstanding (archive 17), I have found that the problem
can be reduced by at least a factor of 400 if we just get rid of
combinations that result from trivial face turns, and if we note that
the Start position has a degeneracy of 16. (One center slice is assumed
fixed - another factor of 2 is tempting but not possible)

Mike's calcualtion for the number of states would reduce to:
 (2*(1/6)*(9/2)+2*(28/3)*3+(35/4)*(35/4))*2*8!*8!=435891456000
combinations.  Divide by the start degeneracy, multiply by 2 storage
bits per state, and you get a storage requirement of 6.81GB.  This seems
very close to being doable.  Maybe in another 10 years, I can do this
project on my PC, if no one has done it yet. 


On another note, when I signed up, I mentioned to Alan that I must be
crazy enough to join this group since I have a five foot mockup of a
rubik's type puzzle as my coffee table.  He thought its description
might be of general interest. Skip the rest of this paragraph if you
couldn't care less about its origins. I built it for Caltech's ditch day event
Maybe you have heard of it. That's where all the seniors leave for the
day with their room locked only with a puzzle of some sort, and the
object is for the undergraduates to get into the room by solving it
(with a couple of clues, of course).  Anyway, being as it was that I
had a mechanical engineer roommate...  The rest is history, and I now
have a five foot diameter puzzle coffee table.

OK, a description.  The puzzle is a three centered version of the
Puzzler, widely available in the last few years in puzzle/game specialty
stores.  The differences being that it is colored so that the maximum
number of combinatins are possible (including the supergroup of distinguishing
face centers).  For those of you who  have not seen the Puzzler, and thus
have no frame of reference, consider one vertex of a cube and it's 
surrounding faces.  7 vertices, 9 edges.  Faces can undergo 4 quarter
turns.  Extrapolate to the Megaminx and you again get one central vertex
for a total of 10 vertices and 12 edges.  Faces can undergo 1/5 turns.  
Extrapolate again to six sided faces, and you get a flat puzzle with one
central vertex for a total of 13 vertices, and 15 edges.  Faces can
undergo 1/6 turns.

So it is basically the <URF> group for a' cube' with
hexagonal faces.  The extra face over the Puzzler also serves to remove 
the significant parity constraints on the edge pieces.  (compare <UR>
group to <URF> group of regular cube).

You, too, can make a smaller version of the Ditch Day puzzle at home.  
The advantage of the flat puzzle is that it is easily constructed.  I 
built the 6 inch diameter prototype with poster board, lamination, magic
markers, and an easily machined smooth pressboard frame.  You only need
to drill three 3 inch holes.  The rest is trimming.  Oh yeah, you will 
need a plexiglass faceplate to keep the pieces in too.  Cutting out and 
gluing together the poster board to make sufficiently thick pieces was 
the hardest part.

Number of combinations = (13!*15!*3^13*2^15*6^3)/24 = 3.83E33
Difficulty is comparable with Megaminx.

Happy cubing.  This is already too long.

					mikem.
					Mike Masonjones.



From cube-lovers-errors@curry.epilogue.com  Tue May 28 18:56:50 1996
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Date: Tue, 28 May 1996 13:18:51 -0700
From: "Jason K. Werner" <mrhip@neuhelp.corp.sgi.com>
Message-Id: <9605281318.ZM12960@neuhelp.corp.sgi.com>
In-Reply-To: "Michael C. Masonjones" <masonjon@mit.edu>
        "Square 1 Combinations" (May 28, 10:11)
References: <9605281411.AA29962@sarofim-sun.MIT.EDU>
X-Face: 6]L85m[]|?5>dL9qI]8j>PPk/:]fF4Ma`5O&VJU)U.6"lo:gX{D`?bNqWl~),bS~`rrB5+P
                                                                                                                                                                   d=NQ_[sXE*#|;SZ)PanGF^&Q-Ch[[|Q)Pgx%ts.JdPJ,3bwU84qc^s2q"sH{l9+g]$cD&a"?S]PQ)F
                                                                                                                                                                   b~4}Y93=ZOimDi_J^(lR;OLeN^W\]/&!v8S=~8Qw'HJ.ksu:R/!iV:WiExaWEXw!v$&hyp[mC
X-Mailer: Z-Mail-SGI (3.2S.2 10apr95 MediaMail)
To: cube-lovers@ai.mit.edu, "Michael C. Masonjones" <masonjon@mit.edu>
Subject: Re: Square 1 Combinations
Mime-Version: 1.0
Content-Type: text/plain; charset=us-ascii

On May 28, 10:11, Michael C. Masonjones wrote:
> Subject: Square 1 Combinations
.....
> On another note, when I signed up, I mentioned to Alan that I must be
> crazy enough to join this group since I have a five foot mockup of a
> rubik's type puzzle as my coffee table.  He thought its description
> might be of general interest. Skip the rest of this paragraph if you
> couldn't care less about its origins. I built it for Caltech's ditch day
event
> Maybe you have heard of it. That's where all the seniors leave for the
> day with their room locked only with a puzzle of some sort, and the
> object is for the undergraduates to get into the room by solving it
> (with a couple of clues, of course).  Anyway, being as it was that I
> had a mechanical engineer roommate...  The rest is history, and I now
> have a five foot diameter puzzle coffee table.
.....


Speaking of oversized Rubik puzzles...

A good friend of mine built an oversized, fully functional Rubik's Magic about
3 years ago.  She painted all of the artwork that went inbetween the plastic
squares, cut out all the grooves, and used a heavy duty grade of fishing wire
to connect all the pieces.  We kind of thrashed my Rubik's Magic to see how
many wires were used in the Magic and all the paths they took.

I _think_ each square was 1'X1', so that would have made the puzzle 2'X4'.
 It's fun to play with, but only if you have the stamina; it's heavy!  :)

	-Jason

-- 
Jason K. Werner, Silicon Graphics
U.S. Field Operations I/S Sys Admin
mrhip@corp.sgi.com, 415-933-6397
"I will choose free will".....Neil Peart
"These go to eleven".....Nigel Tufnel












*********************** THIS IS A FREE SPEECH ZONE ************************
In defiance of the Communications Decency Act, I refuse to self-censor the
content of my e-mail, my online postings, and my Web pages.  I urge other
Constitutionally-protected Americans to declare their online communications
FREE SPEECH ZONES and to fight any attempts at regulating, censoring and
"dumbing down" the Internet.  The Net is not TV and radio!  Let's keep it
that way.      http://www.eff.org/blueribbon.html       mrhip@corp.sgi.com
**************** I SUPPORT THE EFF'S BLUE RIBBON CAMPAIGN *****************


From cube-lovers-errors@curry.epilogue.com  Tue May 28 23:35:28 1996
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Message-Id: <31ABA343.23B@erols.com>
Date: Tue, 28 May 1996 21:07:15 -0400
From: Charlie Dickman <charlied@erols.com>
Reply-To: charlied@erols.com
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Subject: A 4-dimensional Cube
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I have a document that describes a model of a 4-dimensional Rubik's Cube 
(3x3x3x3) and a program that implements the model. The document is a 
stand-alone that executes on a Macintosh and the implementation of the 
model runs on a Mac as well.

This paper/program is based on an unpublished paper by Harry Kamack and 
Tom Keene that was referenced in Hofstader's '82 SA column.

I'll be more than happy to share them with any interested parties - let 
me know your interest. Anyone with a site on the Web who would like me 
to upload the files please let me know. The document is 305K bytes, the 
program is 197K bytes.

Charlie Dickman
charlied@erols.com


From cube-lovers-errors@curry.epilogue.com  Wed May 29 01:37:34 1996
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Date: Wed, 29 May 1996 00:54:08 -0400
From: AirWong@aol.com
Message-Id: <960529005406_544579496@emout14.mail.aol.com>
To: CUBE-LOVERS@ai.mit.edu
Subject: ULTIMATE Rubik's cube?

I'm not sure if this has been discussed or not, but I was asked the following
question, and I am not sure of thie answer.

Is there an ULTIMATE Rubik's cube that, if an algorithm for it was known, it
would contain an algorthm for ANY Rubik's cube?

I guessed that it was four, but I'm not so sure about it.

Aaron Wong
AirWong@AOL.com


From cube-lovers-errors@curry.epilogue.com  Wed May 29 15:25:56 1996
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To: CUBE-LOVERS@ai.mit.edu
From: "We Love Stress Analysis." <a.southern@ic.ac.uk>
Subject: Ultimate Rubik's Cube. (how to make a 4x4x4)
X-Mailer: <PC Eudora Version 1.4>

I was thinking about something similar to this a while back.

I asked (as did many others) about 4x4x4s, but soon after I sent that E-mail 
I realised that I was holding one with a few extra pieces.

If we can consider a 3x3x3 Rubik's cube as a 2x2x2 with mid-edges and 
centres, in effect we can reverse this and ignore the centres and mid-edges 
of a 3x3x3 thus making it into a 2x2x2. 

Anyone who wanted a 2x2x2 could just pull the stickers from the central 
columns and rows of a 3x3x3. As this makes those pieces indistinguishable, 
they are no longer part of the puzzle.

the relationship between 4x4x4 and 3x3x3 is slightly harder, but the above 
is true for 4x4x4 and 2x2x2 (but if ANYONE tries that with a 4x4x4, I'll hit 
them!!!!!!!).

A more common cube than the 4x4x4 is the 5x5x5 (which is still in 
production, c/o Uwe Meffret). This can be transformed from a 5x5x5 into a 
4x4x4 by removing the central lines of stickers. It can also be transformed 
into a 3x3x3 (why anyone would want to.................) by removing columns 
& rows 2&4 from each side, and 2x2x2 (I won't bother saying it.........) by 
removing columns and rows 2,3,&4 from each side.

Higher orders of cubes aren't in production, but apparently do exist in 
cyberspace, These would display similar properties.

The pattern is simple: smaller cubes with odd number of pieces per side can 
be incorporated with other pieces to form larger cubes with odd number of 
pieces per side. etc. etc. etc. 

2x2x2 + (extras) = 3x3x3
2x2x2 + (extras) = 4x4x4
3x3x3 + 4x4x4 + (extras) = 5x5x5


I'm sure there is some mathematical proof to what I am trying to say, but 
I'm no mathematician.

It might start off: 

                        Pieces          Jump(from last)         used before:    
2x2x2                   8                       8                            
   0
3x3x3                   26                      18                             8
4x4x4                   56                      30                           
   8       
5x5x5                   98                      42                           
   74
6x6x6                   152                     54                           
  ??? 

I would say that the ultimate Rubik cube was in fact the 2x2x2 because it 
features in all the solutions. However, based on this logic, the 8086 is the 
ultimate P.C. as any P.C. can run 8086 software, but an 8086 can't run 80386 
software................





I hope this E-mail hasn't been too scatter brained...................



Andy.







Fact: did you know that British Airways has more Super-Sonic flying time 
than any air force in the world??



From cube-lovers-errors@curry.epilogue.com  Wed May 29 15:27:35 1996
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Date: Wed, 29 May 1996 09:59:23 -0600
To: cube-lovers@ai.mit.edu
From: Tom Magliery <mag@ncsa.uiuc.edu>
Subject: Re: Square 1 Combinations

At 10:11 AM 5/28/96, Michael C. Masonjones wrote:
>On another note, when I signed up, I mentioned to Alan that I must be
>crazy enough to join this group since I have a five foot mockup of a
>rubik's type puzzle as my coffee table.

Neato!

It's long been a carpentry dream of mine to build a giant functioning 3x3x3
Rubik's Cube.  (Just how giant, who knows, but I've usually envisioned
about 1-foot cubies.)  Has anyone ever written up plans or built a giant
cube?

mag

--
.---o  Tom Magliery, Research Programmer                         .---o
`-O-.  NCSA, 605 E. Springfield                  (217) 333-3198  `-O-.
o---'  Champaign, IL 61820          O-        mag@ncsa.uiuc.edu  o---'




From cube-lovers-errors@curry.epilogue.com  Wed May 29 20:41:19 1996
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Date: Wed, 29 May 1996 17:08:37 -0400 (EDT)
From: Nicholas Bodley <nbodley@sunspot.tiac.net>
To: cube-lovers@ai.mit.edu
Subject: Another subscriber
Message-Id: <Pine.SUN.3.91.960529163546.13009J-100000@sunspot.tiac.net>
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 Thanks to Alta Vista and Rubik as a keyword, I did an all-nighter over
the holiday weekend, and discovered you folks. I haven't yet downloaded
the archives, but will do my best to be a good Netizen before posting
"real" queries. 

 The first Cube I saw must have been pre-Ideal; I was riding home on the
West Side IRT local in NYC, and noticed someone sitting at the other end
of the car manipulating a puzzle that looked so unbelievable
(mechanically) that I really wondered whether my perceptions had gone
haywire from too many consecutive late bedtimes and regular mornings. It 
was *some* sleep debt! I have little doubt that it was a Cube. 

 When I got mine, I came quite close to solving it by sheer persistence
and brute force (probably about 4 cubies out of position); beginner's
luck! 

 When Meffert was mentioned in (Martin Gardner's col.?) in Sci. Am., I
wrote away for his catalog, which I'm just about sure I still have. I
bought a "5" from him, and sent another check for more items; never
received them. He said Customs must have confiscated them; Customs never
notified me. At the time, I could afford $112 or so. I consider it lost; 
I hope Meffert used it to good advantage. (This would have been around
1987.) It was fascinating to see that he apparently is active once more.

 I've seen "5"s for sale again within the past year or so, I think at The
Compleat Gamester (?) in Waltham, and also The Games People Play in
Cambridge, which has moved (not too far) about a year ago. 

 Does *everybody* know there's a ball inside Rubik's Revenge? 

 I'm at least as much of a gadget-hound as a puzzle-solver; I have a
decent collection. I get a real bang out of dismantling group-theory
puzzles to see how they're built; almost all can be disassembled, although
(as most people probably know) the "2" (Pocket Cube) is quite hard both to
disassemble and to reassemble. I have the Hungarian Globe, which is truly 
impossible to dismantle, IMO. (I haven't dared to scramble it!) This one 
has printed metal surfaces attached to a plastic structure; the "tiles"
take paths like the grooves in the ball inside the "4" (R.R.).

 I hope I might be forgiven for posting one question that has been paining
me-- I'd dearly love to know the answer! Is it true that a physical 
prototype of the "6" (6 X 6 X 6) has been constructed; if so, could 
anyone tell me the approximate date(s) of messages that discuss it? I 
would not want anyone to do lots of searching on my behalf, but just a 
recollection would be welcome. I'm also very curious about the mechanism 
for a "7"; it seems to me that locking pins (or the equivalent) would be 
necessary. I really wonder whether the mechanical design can be practical.

 I'm also a mechanical calculator (See Erez Kaplan's pages on the Web, in
particular) and also mechanical analog computer enthusiast. Paradise was
being a Navy fire control tech. who correctly diagnosed a loose screw
inside the Mk. 1A main battery computer on a destroyer; it took three
weeks to repair. The Master Technician scheduled things well; it happened
just before the ship went in for its every-3-year yard overhaul. 

 I expect to be enjoying this List!
 
NB   Nicholas Bodley   Autodidact & Polymath  |*| Keep smiling! It makes |
      Waltham, Mass.   Electronic Technician  |*|   people wonder what   |
     nbodley@tiac.net    Amateur musician     |*|  you have been up to.  |
-------------------------------------------------------------------------*



From cube-lovers-errors@curry.epilogue.com  Thu May 30 18:15:45 1996
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Date: Thu, 30 May 1996 15:23:58 -0400
From: der Mouse <mouse@collatz.mcrcim.mcgill.edu>
Message-Id: <199605301923.PAA18428@Collatz.McRCIM.McGill.EDU>
To: cube-lovers@ai.mit.edu
Subject: Re:  Another subscriber

> I hope I might be forgiven for posting one question that has been
> paining me-- I'd dearly love to know the answer!  Is it true that a
> physical prototype of the "6" (6 X 6 X 6) has been constructed; [...]
> I'm also very curious about the mechanism for a "7"; it seems to me
> that locking pins (or the equivalent) would be necessary.  I really
> wonder whether the mechanical design can be practical.

In my opinion mechanical designs for the 7 and above will have to be
fundamentally different from those for the 6 and below, because that's
the point at which the "buried" corner of a corner cubie extends past
the surface of the face during a face turn and thus it's not possible
to build the thing as rigid pieces connected to a central mechanism, at
least not without cutting away part of some face-center cubies.
(Specifically, that buried corner is at sqrt(2)*(.5-1/N) from the
center, taking the cube side as 1 and N as the order of the cube.  The
face is at .5 from the center.  The former becomes greater than the
latter at about N=6.83...not that non-integer N make physical sense.)

This is not to say that a 7 is impossible, just that it will have to be
rather drastically different - somehow, when a turn is started, the
corner cubie will have to be mechnically locked to the rest of the face
that's turning with it.  I can easily enough imagine possible
mechanisms, but coming up with one simple enough to mass-produce at a
price people are likely to be willing to pay would be a major
challenge.

On the other hand, a straightforward locking mechanism could probably
be put together by a good watchmaking shop at no more than the price of
a high-end watch.  A few collectors might go for it, especially since
the result - particularly if made out of metal - would feel much better
than the plastic-on-plastic feel of most cubes.

					der Mouse

			    mouse@collatz.mcrcim.mcgill.edu


From cube-lovers-errors@curry.epilogue.com  Thu May 30 18:15:15 1996
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To: Cube-Lovers@AI.MIT.EDU
From: Wei-Hwa Huang <whuang@cco.caltech.edu>
Subject: Re: Another subscriber
Date: 30 May 1996 05:53:22 GMT
Organization: California Institute of Technology, Pasadena
Lines: 31
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Nntp-Posting-Host: accord.cco.caltech.edu
X-Newsreader: NN version 6.5.0 #12 (NOV)

As a first aside, I'd like to mention that a nice project was to get a
standard 3x3x3 cube into a standard American spaghetti sauce jar.  Makes
a good conversation piece, though hard to scramble quickly...

Nicholas Bodley <nbodley@sunspot.tiac.net> writes:
> I'm at least as much of a gadget-hound as a puzzle-solver; I have a
>decent collection. I get a real bang out of dismantling group-theory
>puzzles to see how they're built; almost all can be disassembled, although
>(as most people probably know) the "2" (Pocket Cube) is quite hard both to
>disassemble and to reassemble. I have the Hungarian Globe, which is truly 
>impossible to dismantle, IMO. (I haven't dared to scramble it!) This one 
>has printed metal surfaces attached to a plastic structure; the "tiles"
>take paths like the grooves in the ball inside the "4" (R.R.).

I feel compelled to mention that there's a small company in Taiwan which
makes two variants of the Hungarian Globe that are harder.  One variant
allows for a move that turns the 9 pieces on one side; the other variant
allows for the 5 pieces at every intersection to be rotated.

Let me dig out the address...

International Puzzles and Games
Fl. 3 No. 192 Chung Ching N. Rd. Sec 2
Taipei Taiwan, Republic of China
Tel: 886-2-5532575
Fax: 886-2-5536757

-- 
Wei-Hwa Huang, whuang@cco.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
Caught Porfiry, Raskolnikov sung his swan Sonia when he went Dounia to Siberia.


From cube-lovers-errors@curry.epilogue.com  Fri May 31 02:25:14 1996
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On the subject of an ultimate cube...
 
There can not possibly be an ultimate cube just like (because) there is
no "ultimate", ie., largest, integer.  But if you can solve the
(N+1)x(N+1)x(N+1) cube then you can surely solve the NxNxN cube. You
would simply reduce the (N+1)x(N+1)x(N+1) permutation to one that is in
the NxNxN group and continue from there like MBW did when looking for
God's algorithm in the 3x3x3 group.
 
The group of an NxNxN cube is a proper subgroup of an (N+1)x(N+1)x(N+1)
cube. For example, the 2x2x2 cube group is the 3x3x3 group minus the
edge moves and the center cubie orientation moves - that is, as
Singmaster pointed out, it is just the corners of the 3x3x3 cube. Adding
the 3rd cut added 2 additional types of cubies to the 2x2x2 cube, the
edges and the centers, and along with them came the edge moves (to form
the group of the 3x3x3 cube) and the center orientations (to form the
3x3x3 super-group). The edge moves alone are a proper subgroup of the
cube group and the cube group is a proper subgroup of the super-group.
 
A similar situation occurs when you go from the 3x3x3 cube to the 4x4x4
cube. If you constrain the cube so that the central 2 slices can not be
moved independently of one another then the 2 central edge pieces act
exactly like the edges of a 3x3x3 cube and the 4 face center pieces act
exactly like the face centers of the 3x3x3 cube. When the central slices
are allowed to move independently of one another permutations are added
to the 3x3x3 group and super-group to make up the 4x4x4 group and
super-group. Thus the 3x3x3 groups are proper subgroups of the 4x4x4
groups.

The pattern continues as the value of N increases with the N+1 group
being larger than the N group and properly containing the N group. So
the answer is no, there is no ultimate cube.


From cube-lovers-errors@curry.epilogue.com  Fri May 31 02:24:27 1996
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Subject: realizing 7x7x7 or larger cubes
In-Reply-To: Your message of "Thu, 30 May 96 15:23:58 EDT."
             <199605301923.PAA18428@Collatz.McRCIM.McGill.EDU> 
Date: Thu, 30 May 96 17:00:59 -0700
From: Scott Huddleston <scotth@ssd.intel.com>


>In my opinion mechanical designs for the 7 and above will have to be
>fundamentally different from those for the 6 and below, because that's
>the point at which the "buried" corner of a corner cubie extends past
>the surface of the face during a face turn and thus it's not possible
>to build the thing as rigid pieces connected to a central mechanism, at
>least not without cutting away part of some face-center cubies.

One solution to this dilemma is to let some of the "cubies" become
"brickies" (i.e., rectangular bricks instead of cubes).  In this approach,
there's no limit in principle on N to how large an NxNxN puzzle you could
build with the standard mechanism.  There is, of course, the lower limit 
you just described to how small the corner cubies could become.


From cube-lovers-errors@curry.epilogue.com  Fri May 31 16:05:37 1996
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To: charlied@erols.com
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Reply-To: bmbuck@acsu.buffalo.edu
Subject: Re: An Ultimate Cube 
In-Reply-To: Your message of "Thu, 30 May 1996 21:19:39 EDT."
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Date: Fri, 31 May 1996 15:41:34 -0400
From: Buddha Buck <phaedrus@dreamscape.com>

> On the subject of an ultimate cube...
>  
> There can not possibly be an ultimate cube just like (because) there is
> no "ultimate", ie., largest, integer.  But if you can solve the
> (N+1)x(N+1)x(N+1) cube then you can surely solve the NxNxN cube. You
> would simply reduce the (N+1)x(N+1)x(N+1) permutation to one that is in
> the NxNxN group and continue from there like MBW did when looking for
> God's algorithm in the 3x3x3 group.

I would not necessarily agree with the assertion that if one can solve 
an N^3 cube, one can solve an (N-1)^3 cube.

Your construction, of solving the N^3 cube into a subgroup that is 
homeomorphic to the (N-1)^3 cube, assumes that one already knows how to 
solve a (N-1)^3 cube. 

>  
> The group of an NxNxN cube is a proper subgroup of an (N+1)x(N+1)x(N+1)
> cube. For example, the 2x2x2 cube group is the 3x3x3 group minus the
> edge moves and the center cubie orientation moves - that is, as
> Singmaster pointed out, it is just the corners of the 3x3x3 cube. Adding
> the 3rd cut added 2 additional types of cubies to the 2x2x2 cube, the
> edges and the centers, and along with them came the edge moves (to form
> the group of the 3x3x3 cube) and the center orientations (to form the
> 3x3x3 super-group). The edge moves alone are a proper subgroup of the
> cube group and the cube group is a proper subgroup of the super-group.

True, and I will conceed that if you know how to solve a 3x3x3, you can 
solve a 2x2x2.
>  
> A similar situation occurs when you go from the 3x3x3 cube to the 4x4x4
> cube. If you constrain the cube so that the central 2 slices can not be
> moved independently of one another then the 2 central edge pieces act
> exactly like the edges of a 3x3x3 cube and the 4 face center pieces act
> exactly like the face centers of the 3x3x3 cube. When the central slices
> are allowed to move independently of one another permutations are added
> to the 3x3x3 group and super-group to make up the 4x4x4 group and
> super-group. Thus the 3x3x3 groups are proper subgroups of the 4x4x4
> groups.

Yes, the 3x3x3 groups are proper subgroups (or, probably more 
accurately, homeomorphic to proper subgroups) of the 4x4x4 groups, but 
that doesn't mean that knowing how to solve the 4x4x4 allows one to 
solve the 3x3x3.

For instance, I can solve a 4x4x4.  However, my solution to the 4x4x4 
involves slice moves that don't exist on a 3x3x3 cube, through all 
stages of my solution, including the final stage.  I cannot directly 
apply my 4x4x4 solution to a 3x3x3 cube.  (I can to the 2x2x2 cube, 
since the techniques for solving the corners are applicable to cubes of 
all order N).  If my solution for solving the 4x4x4 involved reducing 
it to the subgroup of the 4x4x4 generated by face turns only, then yes, 
I could directly solve a 3x3x3 by the methods I use for a 4x4x4, but I 
don't.

> The pattern continues as the value of N increases with the N+1 group
> being larger than the N group and properly containing the N group. So
> the answer is no, there is no ultimate cube.

The question originally asked (by Aaron Wong) was "Is there an ULTIMATE 
Rubik's cube that, if an algorithm for it was known, it would contain 
an algorthm for ANY Rubik's cube?"

There might be no answer to the general question of if -any- algorithm 
was known for the U^3 cube, than an algorithm could be derived for any 
N^3 cube.  For instance, few here would argue the assertion that if you 
can solve a 3x3x3, you can solve a 2x2x2, but from the discriptions 
I've heard of it, I wonder how well Thistlewaite's algorithm would work 
on a 2x2x2 cube.

A Thistlewaite type algorithm for a (2N)^3 cube might very well reduce 
the (2N)^3 cube to the subgroup that is equivilant to a 2^3 cube in its 
final stages.  Such an algorithm would be totally unsuited for solving 
a (2M+1)^3 cube, because there would be no way to reduce that to a 2^3 
cube.  (In general, I would guess that any algorithm for an n^3 cube 
that involved reducing it to an m^3 cube, where n = km, would be 
unsuited for solving a l^3 cube, where l does not have n or m as a 
factor).

However, I think the question can be divided into two parts, if we look 
at it differently (requiring the existance of an algorithm with the 
stated property for order U^3 cubes, rather than requiring that all 
algorithms for order U^3) cubes have the stated property):  First, is 
there a general algorithm that can be used to solve cubes of all 
orders?  I think the answer is "yes".  Second, what is the smallest 
order U^3 cube requiring a complete description of the algorithm?  I 
think the answer is U=5.

My current solution for the 4^3 cube is very closely related to my current solution for the 3^3.  There are only minor changes in one stage, major changes in another, (both to deal with the split edge pieces) and the addition of a completely new stage to handle the centers, which aren't in the 3^3 at all.  Transforming this algorithm to the 5^5 and higher is relatively easy, once I have the 3^3 and 4^4 down.  All the important components of the two lower order solutions are needed for the 5^5, and nothing really new is added.  The same goes for the higher orders.  The tedium of solving increases, but not the real difficulty.

I have been thinking (but haven't done much yet) of writing a collection of web pages describing my general solution (at least, for the 2^3, 3^3, and 4^4 cubes).

-- 
     Buddha Buck                      bmbuck@acsu.buffalo.edu
"She was infatuated with their male prostitutes, whose members were
like those of donkeys and whose seed came in floods like that of
stallions."  -- Ezekiel 23:20




From cube-lovers-errors@curry.epilogue.com  Fri May 31 16:05:07 1996
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From: Wei-Hwa Huang <whuang@cco.caltech.edu>
Subject: Re: Another subscriber
Date: 31 May 1996 07:30:49 GMT
Organization: California Institute of Technology, Pasadena
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der Mouse <mouse@collatz.mcrcim.mcgill.edu> writes:
>In my opinion mechanical designs for the 7 and above will have to be
>fundamentally different from those for the 6 and below, because that's
>the point at which the "buried" corner of a corner cubie extends past
>the surface of the face during a face turn and thus it's not possible
>to build the thing as rigid pieces connected to a central mechanism, at
>least not without cutting away part of some face-center cubies.
>(Specifically, that buried corner is at sqrt(2)*(.5-1/N) from the
>center, taking the cube side as 1 and N as the order of the cube.  The
>face is at .5 from the center.  The former becomes greater than the
>latter at about N=6.83...not that non-integer N make physical sense.)

There's a really simple solution to this.  Just don't make the 7 slices
evenly spaced.

-- 
Wei-Hwa Huang, whuang@cco.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
Caught Porfiry, Raskolnikov sung his swan Sonia when he went Dounia to Siberia.


From cube-lovers-errors@curry.epilogue.com  Sat Jun  1 00:18:13 1996
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Date: Fri, 31 May 1996 19:01:11 -0700
To: Scott Huddleston <scotth@ssd.intel.com>
From: "Ronnie B. Kon" <ronnie@cisco.com>
Subject: Re: realizing 7x7x7 or larger cubes
Cc: cube-lovers@ai.mit.edu

At 05:00 PM 5/30/96 -0700, Scott Huddleston wrote:
>
>>In my opinion mechanical designs for the 7 and above will have to be
>>fundamentally different from those for the 6 and below, because that's
>>the point at which the "buried" corner of a corner cubie extends past
>>the surface of the face during a face turn and thus it's not possible
>>to build the thing as rigid pieces connected to a central mechanism, at
>>least not without cutting away part of some face-center cubies.
>
>One solution to this dilemma is to let some of the "cubies" become
>"brickies" (i.e., rectangular bricks instead of cubes).  In this approach,
>there's no limit in principle on N to how large an NxNxN puzzle you could
>build with the standard mechanism.  There is, of course, the lower limit 
>you just described to how small the corner cubies could become.

I've had this dream of making cubies which attach (via bars or perhaps
electromagnets) to their neighbors, with the smarts to detect the torque of
a turn and release until the turn has been completed.  You could then sell
corner cubies, edge cubies, face cubies, and internal cubies one-at-a-time
and people could build their own puzzles as large as they wanted.

I'll buy enough for an order 10 cube if anyone cares to make this.  :-)

                        Ronnie



From cube-lovers-errors@curry.epilogue.com  Sat Jun  1 00:17:48 1996
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Date: Fri, 31 May 1996 21:27:05 -0400 (EDT)
From: Nicholas Bodley <nbodley@sunspot.tiac.net>
To: Wei-Hwa Huang <whuang@cco.caltech.edu>
Cc: Cube-Lovers@ai.mit.edu
Subject: Cube in a jar
In-Reply-To: <4ojd4i$g2o@gap.cco.caltech.edu>
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On 30 May 1996, Wei-Hwa Huang wrote:

> As a first aside, I'd like to mention that a nice project was to get a
> standard 3x3x3 cube into a standard American spaghetti sauce jar.  Makes
> a good conversation piece, though hard to scramble quickly...

When one does this, is it OK to dismantle the Cube, and then reassemble 
it within the jar? (I assume not).

{Snips}
> Wei-Hwa Huang, whuang@cco.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
> 
Regards to all,

NB   Nicholas Bodley   Autodidact & Polymath  |*| Keep smiling! It makes |
      Waltham, Mass.   Electronic Technician  |*|   people wonder what   |
     nbodley@tiac.net    Amateur musician     |*|  you have been up to.  |
-------------------------------------------------------------------------*



From cube-lovers-errors@curry.epilogue.com  Sat Jun  1 01:40:03 1996
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Date: Sat, 1 Jun 1996 00:34:50 -0400 (EDT)
From: Nicholas Bodley <nbodley@sunspot.tiac.net>
To: der Mouse <mouse@collatz.mcrcim.mcgill.edu>
Cc: cube-lovers@ai.mit.edu
Subject: Locking mechanism for a 7^3 or larger (some thoughts) (fairly long)
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 Some years ago, I think I remember reading (in Douglas Hofstadter's 
second (?) "major" book, as I think of them) that someone had designed a 
mechanism for a 7^3. When I started this message, I thought I'd offer a 
description of such a mechanism to the members of this list, but then 
realized that the problem is even harder than I thought. I hope these 
thoughts are not a waste of bitspace; let me know if so!

 Evidently, the design Hofstadter (probably) alluded to is not generally 
known to this List's members. Perhaps he could enlighten us.

 Also, one wonders whether Erno Rubik has contemplated the mechanical 
design for a 7^3.

 Edge cubies could be retained by schemes such as are now used, but only
until they are moved out "into midair"; then they have one or two rubbing
surfaces completely exposed, so to speak. The mechanism for retaining them
is probably closely related to that for retaining corner cubies when out
of alignment. Once the scheme for retaining corners is worked out, then a
minor variant might be what's needed for the edges. (Or possibly 
conversely...)

 (Ideally, one would not want to prohibit an inner slice from being 
rotated all by itself; less-clever locking schemes might require that 
no more than one plane in the whole Cube be sheared at once.)

 For a first try, the corners would be retained by three locking pins to 
hold them in alignment with their neighbors. However, the problem is to 
retract one of the three pins before shearing an outside layer with 
respect to its neighbor. Until the pin is retracted, you can't shear the 
layer by much! Furthermore, that surface has to be locked once more when 
the corner is realigned, and this must happen automatically, reliably, 
and quickly. IMHO, it takes someone like Mr. Rubik to invent such a 
mechanism, and it might be one of the most difficult to invent yet.

 In the next few paragraphs, I get into speculative electronic/mechanical 
engineering, but try not to presume specialized knowledge any more than 
necessary.

 If one allows an electronic/mechanical scheme, cost balloons, and one has
the disgusting thought of a battery to power things. Sliding contacts to
distribute power from one central battery are bound to be unreliable
unless carefully maintained; they would be a pain. 

 They also would constitute an interesting problem in their own right: Can
power be distributed without short circuits caused 1) by intermediate
physical relative positions, and 2) by any possible configuration of a
Cube? Keep in mind that two paths for current are needed, and they must
never intersect. This is an interesting problem in topology, if I'm
thinking correctly. A related problem is to ask whether every cubie could
always have power connected to it. 

 The thought of a small battery inside each corner and each edge cubie 
(or every other one) is even more painful. 

 But, if there were power inside, here's what could be done. The locking
pins would have a certain amount of "give", to permit limited shear
movement. They could be mounted so they could tilt slightly (say, 10
degrees max.) against spring tension, and their mating sockets could be
narrow "funnels", wide end at the surface. 

 Some sort of sensor would detect misalignment well before the limit of
"give" was reached. Misalignment would cause internal electronics to apply
a pulse to a coil to retract the locking pin. (The pin should have minimal
friction; a polished surface and a Teflon-lined mating hole should do.)
Once the pin was retracted, the electronics would ignore other
misalignments. (Otherwise, one could simply push against a cubie in 
"midair" and detach it!)

 The pin would be kept both extended and retracted magnetically, by a 
remanent alloy that stays magnetized, but which can have its 
magnetization reversed by the flux from a coil. Extending the pin would 
be done by pulsing the coil with the opposite polarity. (The principle is 
closely-related to pulsed magnetic latching relays, which do not require 
continuous coil power in either of their states.)

 As has been implied, once the cubie is realigned, a second 
(reverse-polarity) pulse re-extends the locking pin.

 Pulsed operation should give acceptable battery life; the misalignment
sensor might be somewhat of a challenge to engineer, but not a major
problem. One could consider mechanical contacts; with electronics, they 
wouldn't need to be kept scrupulously clean, only clean enough to permit 
a milliampere or so to flow. (Contamination becomes a factor if you make 
the electronics too sensitive.)

 The electronics seems quite straightforward, and by today's standards,
quite simple and entirely practical. The locking-pin mechanism would be
the most costly, more than likely; it would have to be custom-built, and
might cost $2 US apiece in 10,000 lots, perhaps more. I can imagine many
people-weeks of development to create a decently-reliable locking-pin
design. It's essentially a miniature solenoid. 

 Battery access would be via a screw-threaded cover with a slot that fits 
the edge of a coin; the Cube would look distinctive.

 The clicking sound of the retaining pins would be interesting!

 It's not immediately obvious (to me) how the pins and mating sockets 
should be arrayed; their number must be minimal. Edge cubies would need 
four locks apiece, while corners would need three apiece.

 The mind wants sleep, so I think I'll give this mad message a quick 
proofread and post it.

---------------------
On Thu, 30 May 1996, der Mouse wrote:

{Snips}

> > that locking pins (or the equivalent) would be necessary.  I really

{Snips}

> 
> On the other hand, a straightforward locking mechanism could probably
> be put together by a good watchmaking shop at no more than the price of

> 			    mouse@collatz.mcrcim.mcgill.edu

Best regards to all,

NB   Nicholas Bodley   Autodidact & Polymath  |*| Keep smiling! It makes |
      Waltham, Mass.   Electronic Technician  |*|   people wonder what   |
     nbodley@tiac.net    Amateur musician     |*|  you have been up to.  |
-------------------------------------------------------------------------*





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To: Wei-Hwa Huang <whuang@cco.caltech.edu>
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On 31 May 1996, Wei-Hwa Huang wrote:

{Snips}

> der Mouse <mouse@collatz.mcrcim.mcgill.edu> writes:
> >In my opinion mechanical designs for the 7 and above will have to be
> >fundamentally different from those for the 6 and below, because that's



> There's a really simple solution to this.  Just don't make the 7 slices
> evenly spaced.


> -- 
> Wei-Hwa Huang, whuang@cco.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
_____________________

 Large corner cubies, and rectangular edge pieces to match, when seen 
face-on, right?

 Has anyone worked out the innards? Perhaps a further extension of the
scheme used for the 5^3? (If so, the retaining "foot" on a corner cubie
would have a truly wondrous shape! The "foot" of a 5^3 is quite
impressive.)

 I should have read this earlier; it might have saved bitspace from my 
mad electronic/mechanical scheme! :) It would also be quieter...

My best to all,

NB   Nicholas Bodley   Autodidact & Polymath  |*| Keep smiling! It makes |
      Waltham, Mass.   Electronic Technician  |*|   people wonder what   |
     nbodley@tiac.net    Amateur musician     |*|  you have been up to.  |
-------------------------------------------------------------------------*



From cube-lovers-errors@curry.epilogue.com  Sat Jun  1 16:37:55 1996
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To: Cube-Lovers@AI.MIT.EDU
From: Wei-Hwa Huang <whuang@cco.caltech.edu>
Subject: Re: realizing 7x7x7 or larger cubes
Date: 1 Jun 1996 09:22:10 GMT
Organization: California Institute of Technology, Pasadena
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First, as a comment on the other thread, I think it is safe to say:

If one can solve an (2n+1)^3 cube, then one can solve a (2n)^3 cube.

"Ronnie B. Kon" <ronnie@cisco.com> writes:
>I've had this dream of making cubies which attach (via bars or perhaps
>electromagnets) to their neighbors, with the smarts to detect the torque of
>a turn and release until the turn has been completed.  You could then sell
>corner cubies, edge cubies, face cubies, and internal cubies one-at-a-time
>and people could build their own puzzles as large as they wanted.

It would certainly require a very creative design for the corners; your
description seems to say that in the stable state the corners are not
attached by anything!

Perhaps corner cubies could be equipped with buttons that had to be
depressed before a face would turn?
For instance, imagine a cube with three faces that have a button on
the middle, each one triggering a bar on the opposite side.  When a button
is pressed, the bar retracts into the cube.  This would make a 
workable corner cube, although it would be a bit awkward to press the
face that you wanted to turn!

As another aside, I don't understand the rationale behind the canonical
4x4x4 design.  It would seem to me that it's better to have two rings of
grooves in each dimension, so that the face pieces could have "fatter"
legs and not break off as easily.

-- 
Wei-Hwa Huang, whuang@cco.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
Caught Porfiry, Raskolnikov sung his swan Sonia when he went Dounia to Siberia.


From cube-lovers-errors@curry.epilogue.com  Sat Jun  1 16:43:04 1996
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To: Cube-Lovers@AI.MIT.EDU
From: Wei-Hwa Huang <whuang@cco.caltech.edu>
Subject: Yet another (silly) idea on realizing the 7x7x7
Date: 1 Jun 1996 20:38:23 GMT
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Hey, if the corners are going to fall off, let them!  After all,
anyone who actually bothers to buy a 7x7x7 should know how to solve
the corners...  :)

-- 
Wei-Hwa Huang, whuang@cco.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
Caught Porfiry, Raskolnikov sung his swan Sonia when he went Dounia to Siberia.


From cube-lovers-errors@curry.epilogue.com  Sun Jun  2 04:12:01 1996
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Date: Sun, 2 Jun 96 00:17:03 EDT
From: hoey@aic.nrl.navy.mil
Message-Id: <9606020417.AA04623@sun13.aic.nrl.navy.mil>
To: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Cc: Cube-Lovers <cube-lovers@ai.mit.edu>
Subject: Re: Compact Cube Representation for Shamir and Otherwise

I'm not sure this is so interesting to all of cube-lovers; e-mail me
if you have opinions pro or con.

Jerry writes of the standard S24 x S24 model, which uses 48 bytes per
position without packing.  He also has a "supplement" representation
that uses one facelet from each edge and corner, for 20 bytes.  He
packs them into 13 bytes on tape.

The way I did it the last time I worked on brute force was to
pack eight twelve-bit fields:

     The orientations in two twelve-bit fields (2^11 and 3^7),
     The edge permutation in four twelve-bit fields,
        each of three base-12 digits (12^3), and
     The corner permutation in two twelve-bit fields, each of
        four base-8 digits (8^4).

Unpacking the fields can be done with native arithmetic or table
lookup.  In the latter case, it is better to use 12*11*10 instead of
12^3 and 8*7*6*5 instead of 8^3.

Also, postmultiplying by a fixed permutation can be done with table
lookup without unpacking.  I used this feature for twelve permutations
of particular interest.

I am somewhat rusty on the implications of using this representation
in conjunction with Shamir's algorithm.  I think it provides an
ordering of the permutations that enables at least an approximation to
the random access you need, then you unpack it and do a better job.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil


From cube-lovers-errors@curry.epilogue.com  Tue Jun  4 14:07:52 1996
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Date: Tue, 4 Jun 1996 08:35:17 -0400 (EDT)
From: Nicholas Bodley <nbodley@sunspot.tiac.net>
To: Wei-Hwa Huang <whuang@cco.caltech.edu>
Cc: Cube-Lovers@ai.mit.edu
Subject: Fragile parts in 4^3
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On 1 Jun 1996, Wei-Hwa Huang wrote:

{Mostly snipped}

> As another aside, I don't understand the rationale behind the canonical
> 4x4x4 design.  It would seem to me that it's better to have two rings of
> grooves in each dimension, so that the face pieces could have "fatter"
> legs and not break off as easily.
> 
> Wei-Hwa Huang, whuang@cco.caltech.edu, http://www.ugcs.caltech.edu/~whuang/

  It probably isn't necessary for the legs to be so thin; the mechanical 
engineer probably had optimistic estimates of the likely forces and the 
strength of the particular polymer used. The latter isn't, by any means, 
cheap stuff.

 Wider legs might still meet the constraints that ensure the Cube not fall 
apart.

 (Sorry for a slow reply.)

Regards to all,

NB Nicholas Bodley Autodidact & Polymath |*| Keep smiling! It makes |
      Waltham, Mass.   Electronic Technician  |*|   people wonder what   |
     nbodley@tiac.net    Amateur musician     |*|  you have been up to.  |
-------------------------------------------------------------------------*



From cube-lovers-errors@curry.epilogue.com  Tue Jun  4 18:29:55 1996
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Date: Tue, 4 Jun 1996 18:08:15 -0400
From: Jim Mahoney <mahoney@marlboro.edu>
Message-Id: <199606042208.SAA13307@ marlboro.edu>
To: Cube-Lovers <cube-lovers@ai.mit.edu>
Subject: A essay on the NxNxN Cube : counting positions and solving it


        Thoughts on the NxNxN Cube
        --------------------------

Lately I've seen a few comments here on the NxNxN cube, how to build
one, and how what an algorithm to solve one might look like.  I have
no idea how the inside mechanics should work, nor have I made
significant progress thinking about how to find God's Algorithm for
these guys, but I did work out a recipe for solving the NxNxN, way
back in 1981, and worked out the number of possible positions.
Perhaps given those questions this is the right time to post a synopsis.


My approach is rather pedestrian, and I wave my hands quite a lot,
but I'm confident that an an arbitrarily large cube can be 
unscrambled with the method I describe below.

But first, since this is the first time I've posted to this discussion
group, I guess I should say who I am.  I wrote my undergraduate thesis
on Rubik's Cube fifteen years ago, as part of a math/physics degree.
Ironically, even though I was at MIT and this mailing list is based
there, too, I never knew of it until this year, when I started
(slowly) reading through the archives of this mailing list.  There's a
lot there, and not all of it is exactly quick reading.

These days I teach physics and astronomy at Marlboro College, where I
have occasionally taught an "ideas of group theory" course based on
puzzles like TopSpin and Rubik's Cube, designed to show mostly
non-math majors why group theory is so pretty.  I see from previous
posts that other people have taught similar courses.

Anyway, what follows is one way of thinking about the NxNxN.  Much of
this isn't new, but perhaps it hasn't been said quite this way, and
will therefore be worth saying.  Hope it isn't too long-winded, and
that I don't offend the folks who've posted similar things already by
not referencing them; as I said, I've swallowed some but not all of
the 1500 or so posts to this mailing list.


======================================================================
(I) The Cube Itself  =================================================
======================================================================

First, I will envision the NxNxN Cube (the whole thing) as a solid,
transparent array of N^3 cubies (the pieces its made of), most of
which are 'hidden' on the inside.  (For example, this would imply that
the the 3x3x3 cube has a single hidden cubie at the center.)  All the
real mechanical 3x3x3, 4x4x4, 5x5x5 Cubes that I've seen only have
cubies on the outside, but if you can put back all N^3 cubies in the
one I'm describing then you can certainly do the real ones.  

(In Dan Hoey's notation, I believe that this means I treat the Cube as
the G+C group, where G is generated by the outer slice rotations, and
C is the rotations of the entire thing.  I am not including spatial
inversions, because I'm physicall obtainable positions here.  And
while I agree that G is what I see on the 3x3x3, I think that what I
describe here is a more general and elegant approach to the NxNxN.)

Second, let a 'move' be a rotation of any plane of cubies, including
the interior slices.  There are 3N of these planes, and each has NxN
cubies in it.  Since I'm not going to try to count moves here, it
doesn't matter whether you consider a half turn as one move or two
quarter turns.  The slices are numbered from 1 to N, so that rotating
the 1-slice or N-slice is a rotation of an outside face, while the
(N+1)/2 slice is through the center of an odd Cube.  (I chose this
convention rather than numbering at zero in the center because it lets
me talk about the N'th-layer, as defined below, and the N'th-slice in
the same breath without getting myself confused.)

Third, I imagine that each of face of each cubie has a distinct color,
which I will take as usual to be (Front, Back, Up, Down, Left, Right),
and that the unique 'solved' position has all N^3 cubies in the same
orientation, with their colors all aligned.


======================================================================
(II) Layers, Orbits, and Types =====================================
======================================================================

Now I would like to see how many different kinds of cubies there are,
where they live, and how they behave.  

The first thing to notice is that there are two distinct kinds of
NxNxN Cubes, depending on whether N is even or odd.  For N odd, there
is a single cubie at the center (which I will call the 1-layer),
surrounded by the cubies on the outside of the 3x3x3 (which I will
call the 3-layer), which in turn are surrounded by those cubies on the
outside of the 5x5x5 Cube (the 5-layer), and so on until I reach the
outermost N-layer.  When N is even, the innermost layer is 2x2x2,
which is surrounded by a 4x4x4 "4-layer", and so on.  Thus the entire
Cube is made up of disjoint layers which are either all odd or all
even.  Moreover, it is easy to see that the cubies on a given layer
always stay on that layer; the allowed rotations cycle cubies within a
layer but never between layers.

Next, I will define any complete set of cubies that can move into each
other's position as an "orbit."  (This name is at least suggestive of
the group theory notion of a closed sequence of elements.)  For
example, the 8 corner cubies on the 3x3x3 Cube form one orbit since
any one of those cubies can be put in any of those eight positions.
Likewise, the 12 edge cubies on the 3x3x3 form another orbit.

Finally, distinct orbits which have similar properties will be called
members of the same "type."  For example, the 4x4x4 Cube has an orbit
of eight outer corners on the 4-layer, and a second orbit of eight
corner cubies on the inside, in the 2-layer.  Although these sets
of cubies are in distinct orbits, they are both "Corner" types.
(I know that my notion of what exactly "similar properties" means
is vague here, but I think the general idea is clear.)

One approach to solving the cube, then, is to identify each kind of
type - it turns out there aren't very many - and find some method of
manipulating the cubies in an orbit of that type without disturbing
any other the rest of the Cube.  I'll explain one way to do this
further down, after listing the different types.


======================================================================
(III) The Eight Types ==============================================
======================================================================

Without further ado, here they are.

Name         What:
------       ------

 Central        The unique cubie in the center of Cubes with N odd.

 Corner         Corners in each layer.  In each layer there is 
                1 corner orbit consisting of 8 cubies, each of
                which can be in 3 orientations in each of 8 positions. 
                (8 positions x 3 orientations = 24 total.)
                However, while all 8! position rearrangements are
                permissible, all rotations are not; as is well known, 
                only 1/3 of them are.
                One way to see this is to define "twist" state
                as (0,1,2) for each orientation of a cubie at a 
                corner, and to notice that the sum of all these states
                isn't changed by a single move.  This means that 
                you cannot turn just one corner in place.

 Edge-Single    The ones like the outer edges on the 3x3x3.
                In each odd layer there is 1 of these orbits,
                consisting of 12 cubies, each of which can be in 2 states.
                (12 positions x 2 orientations = 24 total.)
                All 12! placements are accessible, but again only
                some of the flips; you cannot turn just one edge.

 Face-Center    The cubies like the centers of the 3x3x3 face.
                In each odd layer there is on of these orbits,
                which has 6 cubies each of which can be in 4 rotation
                states. (12 places x 4 states each = 24 total.)
                This time all rotations are possible; however, the
                cubies can only move in space as a rigid whole, and
                therefore there are 24 different positions 
                for these cubies, which are completely determined 
                by the orientation of the central cubie.

 Edge-Double,
 Face-Corner,
 Face-Edge,
 Face-Offset
                Each of these orbits consists of exactly 24 cubies, 
                as shown in the pictures below.  There are in general 
                many of each of these orbits in each layer, as given by 
                the formulae (simple geometry and counting - see
                the diagrams) in the table below.  
                *None* of these cubies in these orbits 
                can "flip" or "twist" in place like the Corners 
                and Single Edges do; in every case there are exactly
                24 cubies which implies that there must be only 
                one orientation at each possible position.  Another
                way to see this is to draw in an orientation on 
                each cubie of a given orbit, with arrows, and then
                show that no possible move changes the positions 
                of the arrows.

Here's a summary of the specs for each type.  Note that the "number
of positions" given is for both only one parity, that is, for both an
even or odd number of quarter turns, and ignoring the all other
orbits.  The number of positions of the whole Cube is *not* a simple
product of all these numbers; the parities of different orbits must
agree.  More on this later.

As usual, I use "!" and "^" for factorial and "raise-to-the-power-of",
i.e. 8!=8*7*6*5*4*3*2*1 and 3^7=3*3*3*3*3*3*3.


 - Types --  ( n = which layer ) ------------------------------------ 
 
  Name        # of     # of orbits per layer.   # of positions per orbit
              cubies   (n odd)        (n even)  (both even/odd parity)
  ------      ------   -----------    -------   ---------------------

n=1

  Central        1      1              0               24

n>1

  Corner         8      1              1             (3^7) 8!
  Edge-Single    12     1              0             (2^11) 12!
  Face-Center    6      1              0             (4^6)

n>3

  Edge-Double    24     (n-3)/2        (n-2)/2         24!
  Face-Corner    24     (n-3)/2        (n-2)/2         24!
  Face-Edge      24     (n-3)/2        0               24!

n>5

  Face-Offset    24     (n-3)(n-5)/4   (n-2)(n-4)/4    24!
  
 ---------------------------------------------------------------------
 
It's also convenient to define "h" and "H" such that

  h = n/2       (n even);       H = N/2         (N even)
  h = (n-1)/2   (n odd);        H = (N-1)/2     (N odd)

which makes the counting a bit easier.  "h" stands for "half", and is
the number of the slice just before the center slice, if there is a
center slice.  With this "h", the expressions for the number of orbits
per layer are much simpler, namely

  Name          # of orbits per layer
  -----------    ----------------
  Double Edge    h-1
  Face-Corner    h-1
  Face-Edge      h-1
  Face-Offset    (h-1)(h-2)


And now for the pictures.  This is much easier to visualize in 3D with
real drawings, but I'll do what I can with ASCII.

The smallest layer n that contains all the distinct types (except
the central cubie, of course) is n=7, so I've drawn in one outside
(n=7) plane of a 7x7x7 Cube below and sketched in where they live.
You can either think of this as the outer-most layer of an N=7 Cube,
or part of an inner n=7 slice of a larger Cube.

The slices (rows and columns in the pictures) can be numbered either
left to right or right to left, so when I refer to the "n-slice" I
also mean the "(N+1-n)-slice" where N=(size of entire Cube)=7 here,
and 1<=n<=N is a particular slice.

I also note to the right of each picture which slice rotations can
disturb the cubies in that orbit, and whether a quarter turn of that
kind of move gives an even or odd permutation of the cubies.  ("even"
or "odd" refers to how many pairwise swaps it takes to get that
permutation.  A cycle of 2 cubies is odd, a cycle of 3 cubies is even,
and a cycle of 4 cubies is odd.  For example, it takes an even number
of moves to corner number 1 to corner 2's place, corner 2 to 3's
place, and corner 3 to 1's place.)  These parities will be discussed
further in the next section.

Where there is more than one possible orbit I have used the labels "p"
and "q" to specify which one is shown.  The letter "H" (described
above) is in this case (with N=7) H = (N-1)/2 = 3.



        7  6  5  4  3  2  1
        1  2  3  4  5  6  7
     -----------------------
    | 
 7 1|   C  .  .  .  .  .  C                     n-Corner
    |
 6 2|   .  .  .  .  .  .  .                     ( N >= n >= H )
    |                           
 5 3|   .  .  .  .  .  .  .             Moved By        Parity
    |                                   -------         ------
 4 4|   .  .  .  .  .  .  .     
    |                                   n-slice         odd
 3 5|   .  .  .  .  .  .  .
    |
 2 6|   .  .  .  .  .  .  .
    |
 1 7|   C  .  .  .  .  .  C


        1  2  3  4  5  6  7
     -----------------------
    |
   1|   .  .  .  ES .  .  .                     n-Edge-Single
    |
   2|   .  .  .  .  .  .  .                     ( N >= n >= H )
    |                           
   3|   .  .  .  .  .  .  .             Moved By        Parity  
    |                                   --------        ------  
   4|   ES .  .  .  .  .  ES     
    |                                   n-slice         odd
   5|   .  .  .  .  .  .  .     
    |                                   (H+1)-slice     odd
   6|   .  .  .  .  .  .  .
    |                                           (Note that H+1 is the slice
   7|   .  .  .  ES .  .  .                      through the center.)



        1  2  3  4  5  6  7
     -----------------------
    |
   1|   .  .  .  .  .  .  .                     n-Face-Center
    |
   2|   .  .  .  .  .  .  .             Moved By             Parity
    |                                   --------             ------
   3|   .  .  .  .  .  .  .             center (H+1)-slice     odd
    |
   4|   .  .  .  FC .  .  .
    |                                   Rotated By
   5|   .  .  .  .  .  .  .             ---------
    |                                   n-slice               "odd"
   6|   .  .  .  .  .  .  .
    |
   7|   .  .  .  .  .  .  .


        1  2  3  4  5  6  7
     -----------------------
    |
   1|   .  ED .  .  .  ED .                     n-p-Edge-Double
    |
   2|   ED .  .  .  .  .  ED              ( 1 < p <= H; p=2 shown here.)
    |                           
   3|   .  .  .  .  .  .  .             Moved By       Parity  
    |                                   --------       ------
   4|   .  .  .  .  .  .  .             n-slice        even (2 4-cycles)
    |                                   p-slice        odd  (1 4-cycle)
   5|   .  .  .  .  .  .  .             
    |                                   (The "p-slice" referred to here
   6|   ED .  .  .  .  .  ED            and in the next figures cuts into the
    |                                   Cube in the 3rd dimension not shown
   7|   .  ED .  .  .  ED .             into the paper.  The "n-slice" move
                                        turns this diagram by 90 degrees.)


        1  2  3  4  5  6  7
     -----------------------
    |
   1|   .  .  .  .  .  .  .                     n-p-Face-Corner
    |
   2|   .  FC .  .  .  FC .              ( 1 < p <= H; p=2 shown here.)
    |                           
   3|   .  .  .  .  .  .  .             Moved By     Parity  
    |                                   --------     ------
   4|   .  .  .  .  .  .  .             n-slice      odd   (4 cubies move)
    |                                   p-slice      even  (8 cubies move)
   5|   .  .  .  .  .  .  .                                
    |                                                      
   6|   .  FC .  .  .  FC .           
    |                               
   7|   .  .  .  .  .  .  . 
 

  
        1  2  3  4  5  6  7
     -----------------------
    |
   1|   .  .  .  .  .  .  .                     n-p-Face-Edge
    |
   2|   .  .  .  FE .  .  .              ( 1 < p <= H; p=2 shown here.)
    |                           
   3|   .  .  .  .  .  .  .             Moved By        Parity  
    |                                   --------        ------
   4|   .  FE .  .  .  FE .             n-slice         odd     (4)
    |                                   p-slice         odd     (4)
   5|   .  .  .  .  .  .  .             (H+1)-slice     even    (8)
    |
   6|   .  .  .  FE .  .  .
    | 
   7|   .  .  .  .  .  .  . 



        1  2  3  4  5  6  7
     -----------------------
    |
   1|   .  .  .  .  .  .  .                     n-p-q-Face-Offset
    |
   2|   .  .  FO .  x  .  .               ( 1 < p <= H; p=2 shown here;
    |                                       1 < q <= H; q=3 shown here; 
   3|   .  x  .  .  .  FO .                   p not equal to q. )
    |                       
   4|   .  .  .  .  .  .  .             Moved By        Parity
    |                                   --------        ------
   5|   .  FO .  .  .  x  .             n-slice         odd
    |                                   p-slice         odd
   6|   .  .  x  .  FO .  .             q-slice         odd
    | 
   7|   .  .  .  .  .  .  .     The x's are *not* part of this orbit,
                                but mark a distinct mirror-image orbit.
                                There are no moves which will bring
                                one of the FO's to one of the x's.


Every cubie on any size NxNxN Cube fits one of these patterns.
For large values of N, nearly all the cubies are in Face-Offset orbits.


======================================================================
(IV) Parity and the Total Number of NxNxN Positions ==================
======================================================================

I'd guess most of you who read this will already understand parity
considerations, but let me run through it quickly anyway.

The basic idea is that any permutation of a group of symbols can be
broken into a sequential set of pair exchanges, and the number of
these pair exchanges, even or odd, determines the parity of the
permutation, even or odd.  Thus if ABCDE is re-arranged into BAECD,
then this rearrangement is odd because it requires three pair swaps,
and three is odd: (1) swap AB to BA in the original, (2) swap D and E
to get the D at the end, (3) swap the E and C.  There are other swap
sequences, but all are odd.

Any slice rotation which cycles four cubies ABCD into BCDA puts those
cubies into an odd permutation since it would require three pair swaps
(AB, AC, AD) to change one into the other.

The point all this is that different orbits must have consistent
parities.  The best known Cube example is that on the 3x3x3 cube, a
quarter turn of any outside slice changes the parity of *both* corners
and edges; therefore, positions which have the corners and edges in
different parities are impossible, and therefore one cannot exchange
two corners without exchanging two edges somewhere, too.

On the NxNxN things are a bit messier.  Any given slice rotation will
change the parity of some orbits, and leave many others unchanged.
Most choices of arbitrary placements of cubies won't be a possible
cube position, not only because of the "twist" of the corners and
"flip" of the edges (described above briefly) but also because the
parity of all the orbits must be consistent.

Here's one way to do it.

Since each slice n>1 moves exactly one Corner orbit, and since the n=1
slice moves the Central cubie, I can use the N Corner/Central orbits
to *define* the parity of each slice.  Then the parity of all other
orbits is fixed, and each has available exactly one-half (only one
parity) of its total number of positions as given in the table

Another way of doing the same thing is to count the only one parity,
that is 1/2 of *all* the orbits, and then multiply by 2^N as the
number of ways to choose the parity on each of N slices.

So I can now calculate the total number of available positions T(N)
of the NxNxN cube by multiplying the number of possible positions of
each orbit, taking into account how many orbits there are in each
layer, over all N layers, and keeping the parity in agreement.  (Note
that here I *am* including a rotation of the entire cube as a new
"position".  To take out this factor, divide what follows by 24.  Note
also that I'm distinguishing between different rotations of the face
centers, which is also a bit different from what is usually done.)

The counting is a pretty straightforward.  Using the initials of the
types as abbreviations, the total number of each type of orbit is:

  N odd:
    #1  = 1
    #C  = (N-1)/2
    #ES = (N-1)/2
    #FC = (N-1)/2
    #ED = Sum n= (3, 5, 7, 9, ..., N) of { (n-3)/2  } = (N-1)(N-3)/8
    #FC =  same as #ED
    #FE =  same as #ED
    #FO = Sum n= (3, 5, 7, 9, ..., N) of { (n-3)(n-5)/4 } 
        = (N-1)(N-3)(N-5)/24

  N even:
    #1  = 0
    #C  = N/2
    #ES = 0
    #FC = 0
    #ED = Sum n= (2, 4, 6, 8, ..., N) of { (n-2)/2  } = (N)(N-2)/8
    #FC = same as #ED
    #FE = 0
    #FO = Sum n= (2, 4, 6, 8, ..., N) of { (n-2)(n-4)/4 } 
        = (N)(N-2)(N-4)/24
        
And so

---------
| T(N) = Total Positions of NxNxN cube, all orientations, 
|        all N^3 cubies
| 
| = 2^N (24/2)^#1 (3^7 8!/2)^#C (2^11 12!/2)^#ES (4^6/2)^#FC 
|     * (24!/2)^(#ED+#FC+#FE+#FO)
-----------------------------

which simplifies to either

 T(N odd)= 24 [3^7 8! 2^10 12! 4^6/2]^((N-1)/2) [24!/2]^( (N-1)(N-3)(N+4)/24 )
    =  24 [ 8.85801e22 ]^((N-1)/2) [3.102242e23]^((N-1)(N-3)(N+4)/24)
    =  (44.9) (0.0561)^N (9.52)^(N^3)
 
or

 T(N even) = [3^7 8!]^(N/2) [24!/2]^( N(N-2)(N+2)/24 )
           = (8.817984e7)^N (3.102242e23)^(N(N-2)(N+2)/24)
           = (1.14)^N (9.52)^(N^3)

which is an awful lot of positions no matter how you look at it.

Note that usually the number of 3x3x3 positions is given without the
factor of 24 (spacial rotations) or the 4^6/2 (face center rotations),
which leaves (3^7 8! 2^10 12!) = 4.3e19.

For large N, T(N) is dominated by the (24!/2)^(N^3/24) term, which is
about 9.524^(N^3), which implies that for very big cubes, each of the
N^3 cubies acts as if it has nearly 10 independent places it can be.

(Oh, and the notation 1.23e4 means 1.23 10^4 = 1230.)


=====================================================================
(V) Solving It  =====================================================
=====================================================================

Here's where the handwaving really gets going.  What I describe here
is closer to an outline of a method than a real algorithm, but you can
probably fill in the details yourself.

The basic idea is to use a general-purpose idea for cycling three
cubies of any given orbit without disturbing any other part of the
Cube.  A small variation on this same theme can be used to twist two
corners or single edges in the place, too.

If I can do this for every type, then all I have to do to solve the
whole cube is the following.

 -- An NxNxN Recipe ------------------------------------------

 (A) If N is odd, orient the Central cubie correctly, and 
     at the same time, turn each Face-Center so that it is aligned
     with the Central cubie.  (If you can't see the orientations 
     of the Face-Centers, as is usually the case on the typical
     3x3x3, then just skip that step and move on.)
 
 (B) For each layer, examine the parity of corresponding
     Corner orbit.  If its parity is odd, make one 
     arbitrary 1/4 turn rotation on that layer; otherwise, 
     don't move it.  At this point all the Corner orbits
     have even parity, and therefore *all* the orbits have
     even parity.

 (C) And finally, I have these nested loops:

   (i) Starting at the innermost layer and working outward, 

    (ii)  on each orbit in that layer, 

       (iii) 1. Restore each cubie of that orbit to its proper position 
             with the 3-cycle technique described below.  Since
             all the orbits already have even parity, these even-move
             combinations are enough to restore everything to their
             proper places.

             2. If the current orbit is a Corner or Edge-Single, 
             then once the cubies are in the right places apply 
             the "twist" and "flip" operations described below
             to orient them correctly.

 That's it.  Now all I have to do is describe three tricks, 
  (1) how to cycle three cubies on any given type,
  (2) how to twist two corners in opposite directions, and
  (3) how to flip two Edge-Singles,
 all without disturbing anything else.

 All these tricks are well known, I think.  And there are certainly
 many, many other tricks; however, these are the simplest that 
 I know of that can be generalized to any size Cube.  Moreover, 
 they have the nice property that you can actually "think" your way
 through them without actually needing to memorize a long sequence
 of moves.  


=====================================================================
(VI) How to Cycle Three Cubies  =====================================
=====================================================================

The basic idea is to find a move sequence that will (1) take a chosen
cubie off from its "hot seat" on a chosen slice *without* (here's 
the trick) disturbing any other cubie on that slice.  The rest
of the cube can be completely scrambled by this operation.  Then (2) 
rotate the chosen slice, (3) undo step (1), putting the original
cubie back into its original slice and undo whatever changes were 
made to the other cubies, and (4) undo step 2.

The sequence always of the form 
   A R A' R'
where "A" is step 1, "R" is a rotation of a single slice, and 
the ' mark means, as usual, the inverse operation.

Here's a detailed example, using the Corner orbit of a 3x3x3 cube,
with the top layer as the "chosen slice" and the cubie marked "1" in
the unfolded sketch of a cube below as the focus of attention. In
eight moves the cubies in locations 1, 2, and 3 will trade places.

The starting position:

             U
    a  -  1  - 2 - d -        (a,1,2,d,e,3,g,h) are a Corner orbit.
    |  L  |  F | R | B
    e  -  3  - g - h -        (U, D, L, R, F, B) are the possible
             D                clockwise rotations.


 (1) Get "b" off the chosen slice, without disturbing any other 
     cubie on that slice.  Replace it with the cubie that you 
     want to put in its place.
            
                 e  -  a  - 2 - d -
 ->   L  ->      |     |    |   |   
                 3  -  1  - g - h -
              
             
                 e  -  a  - 2 - d -
 ->   D  ->      |     |    |   |   
                 h  -  3  - 1 - g -


                 a  -  3  - 2 - d -
 ->   L'  ->     |     |    |   |          After L D L'
                 e  -  h  - 1 - g -

      The top layer was (a,b,c,d); now it is (a,f,c,d).  
      "b" has been taken off the top slice, and "f" is in its place.

 (2) Rotate the chosen slice to place a new cubie in the hot seat.

                 3  -  2  - d - a -
 ->   U  ->      |     |    |   |           After (L D L') U 
                 e  -  h  - 1 - g -
 

 (3) Undo step 1, which pops the chosen cubie "b" back to its
      original slice, *and* (here's the key part), restore (nearly) all 
      other cubies to their original locations, since none of the 
      disturbed ones were on the slice that rotated in step (2).

                    3  -  1  - d - a -
 -> L D' L' ->      |     |    |   |           After (L D L') U (L D' L')
                    e  -  2  - g - h -

 
 (4) Undo step 2, restoring the chosen slice back to its original position.

                    a  -  3  - 1 - d -
   ->  U'  ->       |     |    |   |       After (L D L') U (L D' L') U'
                    e  -  2  - g - h -

So the move sequence to cycle corners (1,2,3) is simply 
(L D L') U (L D' L') U'   (reading left to right).

With a few extra moves before this sequence (which should be undone
afterwards) to arrange the cubies which should be moved into the 
places which are actually modified by this operation (or a similar
one), this trick and its variations can be used to put back all 8
corners into their proper places.

And with a bit of exploration, this same idea can be used to cycle
three cubies of any type, in any orbit, on any layer, without
disturbing anything else.  For the Edge-Singles on the 3x3x3, for
example, to bring an edge off the top slice without disturbing
anything else on top, step (1) can be S D S', where "S" vertical is a
rotation of a center slice.  Or it could be F H F', where "H" is a
horizontal rotation of of center slice parallel to the top and bottom;
either works.

I have actually tried this for all eight of the types of orbits, and
it does indeed work.  Yes, I know this is pure handwaving, but this
essay is already long enough, and it really is pretty straightforward
once you get the idea.


=====================================================================
(VII) Turning Corners and Flipping Edges  ===========================
=====================================================================

I don't think I need to say too much about this, because
basically the same tricks that work on the 3x3x3 Cube will
work on the orbits in the NxNxN.  

My usual approach is to find a sequence that will bring 
a corner or single-edge cubie out of its position, and
then back with a turn or flip, *without* changing any cubie on on
slice.  Call that entire operation "A".  Then just like 
before, A R A' R' where "R" is a rotation of the slice which 
was left (nearly) unchanged will restore all the parts of the 
Cube which were messed up by A, and leave only two corners or
two edges turned or flipped.

For example, on the 3x3x3, this sequence will turn two corners.
  [(L DD L') (F' DD F)] U [(F DD F') (L' DD L)] U'
The stuff in the brackets brings a corner cubie off the top (up)
slice, and brings it back with a twist.

If "H" is a clockwise (as viewed from the top) quarter turn 
on the horizontal center slice of the 3x3x3 (the plane parallel
to the top and bottom), then this similar sequence will flip two edges.
  [ (L HH L') U' (F' HH F) U ] U [ U' (F HH F') U (L' HH L) ] U'
Again, the moves in the brackets bring one of the 3x3x3 edge
cubies off the top layer, and bring it back with a twist.

One can also combine several different 3-cycles from section VI
to twist and flip the corners and edges.


=====================================================================
(VII) Comments    ===================================================
=====================================================================

Well, that turned out to be a lot longer than I'd planned.  If anyone
has actually bothered to read down this far, I hope it was worthwile.

I think that when all is said in done, the 3x3x3 is by far the most
interesting of the sizes.  All the new types of orbits on the larger
Cubes are fairly boring, actually, since none of the cubies can be
flipped or turned in place the way that the 3x3x3 corners and edges
can.  And I confess that I like how the only cubie on the 3x3x3 that
you can't see - the one that I like to imagine is hidden in the center
- is the only one that's completely specified by the locations of the
other orbits, namely by the positions of the face centers.

When I was first working out this stuff, back in 1981, I built a 7x7x7
Cube out of colored dice.  None of them were "stuck" to the others; it
was just stack of dice.  Manipulating it was pure hell, but I could
usually squeeze a layer and carefully turn and put it back.  One slip
and I had dice all over the room.  


I have a few other ideas kicking around in the back of my head,
but they'll have to wait for another time, and another note.


Regards,

  Dr. Jim Mahoney                          mahoney@marlboro.edu
  Physics & Astronomy			   <http://www.marlboro.edu/~mahoney>
  Marlboro College, Marlboro, VT  05344



From cube-lovers-errors@curry.epilogue.com  Wed Jun  5 02:48:31 1996
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Date: Wed, 5 Jun 1996 09:33:41 +0300 (IDT)
From: Rubin Shai <s2394459@cs.technion.ac.il>
X-Sender: s2394459@csc
To: Cube-Lovers@ai.mit.edu
Subject: Computer representation to the cube.
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Hi all
I'm looking for an easy to implementation / easy to debug / easy to print 
/ and most important chip to manipulate (in computer time) for the 3X3X3 
cube. Does anyone heard of somthing?
Shai



From cube-lovers-errors@curry.epilogue.com  Wed Jun  5 15:28:47 1996
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Date: Wed, 05 Jun 1996 10:22:31 -0500 (EST)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: A essay on the NxNxN Cube : counting positions and solving it
In-Reply-To: <199606042208.SAA13307@marlboro.edu>
To: Cube-Lovers <cube-lovers@ai.mit.edu>
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On Tue, 4 Jun 1996, Jim Mahoney wrote:

It's going to take a while to absorb your whole note, but I do have a
couple of quick comments/questions. 

> Next, I will define any complete set of cubies that can move into each
> other's position as an "orbit."  (This name is at least suggestive of
> the group theory notion of a closed sequence of elements.)  For
> example, the 8 corner cubies on the 3x3x3 Cube form one orbit since
> any one of those cubies can be put in any of those eight positions.
> Likewise, the 12 edge cubies on the 3x3x3 form another orbit.

This has been discussed before on Cube-Lovers, but I am still puzzled or
curious about the usage of the word "orbit".  Your definition is
consistent with the usage advocated by Martin Schoenert on Cube-Lovers. 
For example, Martin talked about the corner orbit, the edge orbit, and the
face center orbit of the 3x3x3.  (I suppose for completeness, we should
include in this list of orbits the orbit for the invisible center of the
whole 3x3x3 cube.)

David Singmaster, on the other hand, has always talked about the twelve
orbits of the constructable group of the 3x3x3, where orbits are defined
in terms of twists, flips, and parity.  Depending on what you mean by
"closed sequence of elements", your definition may be consistent with
Singmaster's usage as well.  That is, Singmaster's orbits are certainly
closed.  However, Martin says that Singmaster's orbits should be called
cosets. 

Secondly, if my understanding of your model is correct, you are treating
positions as distinct which cannot be distinguished with normal coloring
of a physical cube (even an imaginary physical cube for large N).  The
issue appears as early as the 4x4x4, and persists for larger values of N. 
I don't necessarily disagree with your treatment.  Indeed, it makes the
cube theory tenable.  Otherwise, your model tends to become a coset model
rather than a group model.  But I wondered if my understanding of your
model is correct? 

There are several implications of how you treat visibly indistinguishable
positions.  For example, it impacts your counts of how many positions
there are.  For another example, it impacts your solutions (e.g.,
"invisible" incorrect parity on the 4x4x4.  "Invisible" bad parity can
also occur on the 3x3x3 if you remove the face center color tabs.  A slice
move will give the edges and corners opposite parity that is not visible.)
Perhaps you could discuss these issues with respect to your model. 

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7127
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990




From cube-lovers-errors@curry.epilogue.com  Wed Jun  5 15:27:15 1996
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Date: Wed, 5 Jun 1996 09:30:21 +0100
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From: VANESSA PARADIS WANTS ME <cmaggs@glam.ac.uk>
To: CUBE-LOVERS@ai.mit.edu
X-Vms-To: RUBIKCUBE

To have a bit more challange when doing the cube, complete it so that each 
horizontal slice, is 1 turn (quarter of a full circle) out of place.
Therefore, top and bottom face are one colour, but all side faces contain 3
colours.
Boy, is it hard!

By the way, I can do the cube in 1 minute 26 seconds.
How does that compare with everyone else!
(P.S. Please be fair, I`ve only been doing it for 6 weeks or so!)

Chris

CMAGGS@GLAM.AC.UK


From cube-lovers-errors@curry.epilogue.com  Wed Jun  5 15:27:57 1996
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Date: Wed, 5 Jun 1996 09:53:36 -0400
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From: Allan Wechsler <awechsle@bbn.com>
To: s2394459@cs.technion.ac.il
Cc: Cube-Lovers@ai.mit.edu
In-Reply-To: 
	<Pine.SV4.3.91-heb-2.04.960605092401.2432A-100000@cs.technion.ac.il>
	(message from Rubin Shai on Wed, 5 Jun 1996 09:33:41 +0300 (IDT))
Subject: Re: Computer representation to the cube.
Reply-To: awechsle@bbn.com

   Date: Wed, 5 Jun 1996 09:33:41 +0300 (IDT)
   From: Rubin Shai <s2394459@cs.technion.ac.il>

   Hi all
   I'm looking for an easy to implementation / easy to debug / easy to print 
   / and most important chip to manipulate (in computer time) for the 3X3X3 
   cube. Does anyone heard of somthing?
   Shai

Ani lo y'khol lavin et-ha-anglit shelkha.  Are you looking for a
program?  An algorithm?  A piece of hardware?  A circuit diagram?  And
I apologize in advance if I got your gender wrong in my wretched
Hebrew.

-A



From cube-lovers-errors@curry.epilogue.com  Wed Jun  5 19:50:32 1996
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Date: Wed, 5 Jun 1996 18:02:49 -0500
To: CUBE-LOVERS@ai.mit.edu
From: Kristin Looney <kristin@tsi-telsys.com>
Subject: fastest hands in the midwest...

> By the way, I can do the cube in 1 minute 26 seconds.
> How does that compare with everyone else!

37.72 won me the midwest championship, my best official time was
35.30 seconds which placed me 5th in the country. I think it was 1981.
Now?  I don't get timed very often, but it's still usually under a minute.

I guess it is like riding a bicycle.

Anyone else on this list from those contest days?  Minh Thai - are you
out there?  How about Jeff Verasono? or David P. Conrady?  I've often
wondered what that crazy guy with the bright maroon hair ended up
doing with his life...

Kristin (used to be Wunderlich) Looney
kristin@tsi-telsys.com




From cube-lovers-errors@curry.epilogue.com  Wed Jun  5 19:50:57 1996
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Date: Wed, 5 Jun 96 18:32:29 EDT
From: hoey@aic.nrl.navy.mil
Message-Id: <9606052232.AA22039@sun34.aic.nrl.navy.mil>
To: Nicholas Bodley <nbodley@sunspot.tiac.net>,
        Wei-Hwa Huang <whuang@cco.caltech.edu>
Cc: Cube-Lovers@ai.mit.edu
Subject: Fragile parts in 4^3

On 1 Jun 1996, Wei-Hwa Huang wrote:

{Mostly snipped}

> As another aside, I don't understand the rationale behind the canonical
> 4x4x4 design.  It would seem to me that it's better to have two rings of
> grooves in each dimension, so that the face pieces could have "fatter"
> legs and not break off as easily.

If the center pieces had one leg each (instead of a 1/4-leg) you would
have _one_ groove around each equator (instead of _half_ a groove).
Remember, it's important that the inner sphere stay in sync with at
least one of the sets of face centers so that after you've finished
the turn you will be able to turn in an orthogonal direction.  I don't
know how that would work with the turns of the face.  You might need
a switch that looks kind of like the following where two equators meet:

            I
            I     * * *
            I *           *
============O===============O========
          * I                 *
            I
        *   I                   *
        *   I                   *
        *   I                   *
            I
          * I                 *
            O               O
            I *           *
            I     * * *
            I

where the legs live at the "O" positions when a turn is not in
progress.  But this looks dangerous to me; I think there is a lot of
potential for derailment.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil


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From: hoey@aic.nrl.navy.mil
Message-Id: <9606052254.AA22046@sun34.aic.nrl.navy.mil>
To: Jim Mahoney <mahoney@marlboro.edu>
Cc: Cube-Lovers <cube-lovers@ai.mit.edu>
Subject: Re: A essay on the NxNxN Cube : counting positions and solving it

> All the
> real mechanical 3x3x3, 4x4x4, 5x5x5 Cubes that I've seen only have
> cubies on the outside, but if you can put back all N^3 cubies in the
> one I'm describing then you can certainly do the real ones.  

> (In Dan Hoey's notation, I believe that this means I treat the Cube as
> the G+C group, where G is generated by the outer slice rotations, and
> C is the rotations of the entire thing....

Actually, the distinction between G and G+C is that in the latter we
draw a distinction between cubes that differ by a whole-cube move as
different.

When we take account of the internal cubies I call it the "Theoretical
Invisible cube", described in my Invisible Revenge article 9 August
1982.  A solution method is given in 

	Eidswick, J. A., "Cubelike Puzzles -- What Are They
	and How Do You Solve Them?", 'American Mathematical
	Monthly', Vol. 93, #3, March 1986, pp. 157-176.

that is pretty much like yours, I think.

As for counting the positions, I haven't got around to checking the
numbers in "Groups of the larger cubes", 24 Jun 1987.  You might want
to see how they compare to yours.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil


From cube-lovers-errors@curry.epilogue.com  Wed Jun  5 19:52:19 1996
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From: hoey@aic.nrl.navy.mil
Message-Id: <9606052327.AA22049@sun34.aic.nrl.navy.mil>
To: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Cc: Cube-Lovers <cube-lovers@ai.mit.edu>
Subject: Re: A essay on the NxNxN Cube : counting positions and solving it

Jerry remarks:

> For example, Martin talked about the corner orbit, the edge orbit, and the
> face center orbit of the 3x3x3....

> David Singmaster, on the other hand, has always talked about the twelve
> orbits of the constructable group of the 3x3x3, where orbits are defined
> in terms of twists, flips, and parity....

When a group G has a representation as permutations of a set X, the
orbits are the equivalence classes of X induced by x~y if a (x)g=y for
some g in G.  But these orbits will be different depending on the
representation, and in particular depending on X.

If we represent the Rubik group as the usual permutations on cubies
and facies, the orbits are corners, edges, etc. as Martin uses.  I
agree this is the usual kind of orbit to talk about.

If we represent the Rubik group as permutations on itself (I think
it's called the right regular representation) you get one orbit.  This
is always true of the right regular representation, since for any f, g
in G, let h=f'g, and we have (f)h = g, so f~g.

But consider the constructible group C, the set of positions you can
get by taking the cube apart and putting it back together.  We can
extend the right regular representation to a representation on C.  In
this case, there are twelve orbits of mutually accessible positions.
This is Singmaster's usage.  They are indeed the cosets of C/G, as
with any subgroup of a larger group.

But the fact that we usually do not consider the group structure of C
(as in taking products of reassemblies) militates against calling them
cosets, so I can understand why Singmaster might prefer orbits.  But
we have to remember to disambiguate which kind of orbit we are talking
about.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil


From cube-lovers-errors@curry.epilogue.com  Wed Jun  5 22:17:39 1996
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Date: Wed, 5 Jun 1996 16:47:13 -0400
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From: Allan Wechsler <awechsle@bbn.com>
To: cmaggs@glam.ac.uk
Cc: CUBE-LOVERS@ai.mit.edu
In-Reply-To: <96060509302127@glam.ac.uk> (message from VANESSA PARADIS WANTS
	ME on Wed, 5 Jun 1996 09:30:21 +0100)
Reply-To: awechsle@bbn.com

   Date: Wed, 5 Jun 1996 09:30:21 +0100
   From: VANESSA PARADIS WANTS ME <cmaggs@glam.ac.uk>

   To have a bit more challange when doing the cube, complete it so that each 
   horizontal slice, is 1 turn (quarter of a full circle) out of place.
   Therefore, top and bottom face are one colour, but all side faces contain 3
   colours.

I'm not sure I understand this modified goal.  Isn't this achieved by
solving the cube aas usual, and then giving the top and bottom faces a
clockwise quarter twist?  Then the top and bottom are solid, and the
sides are tricolor horizantal stripes.

Even if I haven't understood the goal position, solving for any
achievable position is not in principle harder than solving for any
other.  There might be perceptual problems, but surely these would go
away after a little practice, no matter what the goal configuration.

-A



From cube-lovers-errors@curry.epilogue.com  Wed Jun  5 22:18:25 1996
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From: Chris and Kori Pelley <ck1@flnet.com>
To: CUBE-LOVERS@ai.mit.edu
Subject: RE: Contest days
Date: Wed, 5 Jun 1996 20:31:12 -0400
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>Anyone else on this list from those contest days?  Minh Thai - are you
>out there?  How about Jeff Verasono? or David P. Conrady?  I've often
>wondered what that crazy guy with the bright maroon hair ended up
>doing with his life...

I was in the contests.  First place in Peoria, IL with 48 seconds or so.
Then I won 5th place in Chicago with 47ish seconds, 3rd in St. Louis
with 46 seconds.  I was only 13 at the time and it was just a thrill to
be there with so many other avid cube-solvers.  I still have my official
Ideal "Cubists Do It Faster" T-shirts.  From the St. Louis contest,
which was held in a large outdoor mall, I managed to come away with
one of the stage props which was a giant cardboard cube about 3 feet
per side.  I used it as a table for a few years, then it finally
collapsed.


Chris Pelley
ck1@flnet.com



From cube-lovers-errors@curry.epilogue.com  Wed Jun  5 22:19:02 1996
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To: Cube-Lovers@AI.MIT.EDU
From: Wei-Hwa Huang <whuang@cco.caltech.edu>
Subject: Re: fastest hands in the midwest...
Date: 6 Jun 1996 02:06:31 GMT
Organization: California Institute of Technology, Pasadena
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Kristin Looney <kristin@tsi-telsys.com> writes:

>Anyone else on this list from those contest days?  Minh Thai - are you
>out there?  How about Jeff Verasono? or David P. Conrady?  I've often
>wondered what that crazy guy with the bright maroon hair ended up
>doing with his life...

I talked with Minh Thai some time last year.  He's currently working with
a firm in Eagle Rock, CA.

-- 
Wei-Hwa Huang, whuang@cco.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
Caught Porfiry, Raskolnikov sung his swan Sonia when he went Dounia to Siberia.


From cube-lovers-errors@curry.epilogue.com  Thu Jun  6 23:32:59 1996
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To: Cube-Lovers@ai.mit.edu
From: "The Official Mech. Eng. Exams Failure Club." <a.southern@ic.ac.uk>
Subject: Re: Fragile parts in 4^3, and limits to a n^3

At 08:35 04/06/96 -0400, you wrote:
>On 1 Jun 1996, Wei-Hwa Huang wrote:
>
>{Mostly snipped}
>
>> As another aside, I don't understand the rationale behind the canonical
>> 4x4x4 design.  It would seem to me that it's better to have two rings of
>> grooves in each dimension, so that the face pieces could have "fatter"
>> legs and not break off as easily.
>> 
>> Wei-Hwa Huang, whuang@cco.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
>
>  It probably isn't necessary for the legs to be so thin; the mechanical 
>engineer probably had optimistic estimates of the likely forces and the 
>strength of the particular polymer used. The latter isn't, by any means, 
>cheap stuff.

It is compared to those injection Moulds that they had made. These will be
the most expensive parts for each puzzle. The more variety of parts, the
more expensive the puzzle.

I reckon that on a 3^3 there must be at least 6 different classes of parts
(the centre pieces on each face have a main part and a cover). I reckon that
the Megaminx has 5 as well. I have not taken apart a 5^3 yet, but there must
be about 8 or 9 if it works on similar lines to the 3^3. This almost doubles
the overheads of the 5^3 production compared to 3^3 production.

This Brings me to two questions (well three really, but the third is
unrelated) :

1) Who has the Injection Moulds for the 4x4x4, and are they still in
functioning order?
2) How many 7^3 (okay, I know there is currently no design) could possibly
be sold, and do you think that this would cover the costs of injection
moulds (optimistically about 15 or so)  and construction (even allowing for
the Retailer to take ~70% of the price)?

my third and final(?) question is:

3)With all this talk of Who is on our group, is there any reason why the
mailing list can't be published? And if not, are there any of the designers
of Rubik's Range or Similiar? Any Company Executives looking for new Ideas?

Tar, Now I've got to go, I've got an exam to fail..................

Andy.



From cube-lovers-errors@curry.epilogue.com  Thu Jun  6 23:33:56 1996
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Date: Thu, 06 Jun 1996 12:23:05 -0500 (EST)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: Compact Cube Representation for Shamir and Otherwise
In-Reply-To: <9606020417.AA04623@sun13.aic.nrl.navy.mil>
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On Sun, 2 Jun 1996 hoey@aic.nrl.navy.mil wrote:

> Jerry writes of the standard S24 x S24 model, which uses 48 bytes per
> position without packing.  He also has a "supplement" representation
> that uses one facelet from each edge and corner, for 20 bytes.  He
> packs them into 13 bytes on tape.
> 
> The way I did it the last time I worked on brute force was to
> pack eight twelve-bit fields:

The way I count it, Dan's result is twelve bytes of eight bit bytes.  So
Dan has bested me by one byte (but see below). 

> 
>      The orientations in two twelve-bit fields (2^11 and 3^7),

It looks like you are only storing the orientation of eleven of the 
twelve edge cubies.  The orientation of the twelfth can be inferred from 
the orientation of the first eleven.  On the other hand, you could store 
all twelve and still fit in twelve bits.  It also looks like you are only 
storing the orientation of seven of the eight corner cubies.  Again, the 
orientation of the eighth can be inferred from the orientation of the 
first seven.  This time it matters.  That is, 3^7 will fit in twelve 
bits, but 3^8 will not, if I am counting it right. 

> Also, postmultiplying by a fixed permutation can be done with table
> lookup without unpacking.  I used this feature for twelve permutations
> of particular interest.
> 
> I am somewhat rusty on the implications of using this representation
> in conjunction with Shamir's algorithm.  I think it provides an
> ordering of the permutations that enables at least an approximation to
> the random access you need, then you unpack it and do a better job.

I think that unpacking would be required to build and traverse the "Shamir
tree", but the unpacking that would be required should be relatively easy. 

I have always been reluctant to form compositions of packed formats
because it seemed both messy and slow to have to unpack both permutations
which are being composed and then to repack the result.  In principle,
what you have to do is about the same as what you have to do in building
and traversing the "Shamir tree" with packed formats.  But in practice it
seems a lot messier to unpack two permutations and to repack their 
composition than to simply unpack one permutation as you traverse a tree. 

Dan seems to have solved the problem for certain specific cases (and for 
post-multiplying only) via table lookup.  It is not clear to me that 
table lookup could be used for the more general case of multiplying any 
permutation by any other permutation.

I have always known that the supplements of the corners and edges could be
reduced by one byte each unpacked, down to seven bytes for the corners and
down to eleven bytes for the edges.  The missing bytes could always be
reconstructed, but I didn't want to bother.  Now, it occurs to me that
working with the supplements alone and never expanding them back to S24 x
S24, there is really no reason ever to reconstruct the missing bytes. 
Hence, we can have an eighteen byte representation unpacked.  The eighteen
byte representation easily packs down to twelve bytes (same as Dan's). 

We could surely do better if we tried.  Log2(|G|) is about 65.22, so we
should be able to represent each position in 66 bits, or in 9 bytes.  But
I don't think the packing required would be worth the trouble. 

66 bits is the theoretical minimum to represent |G| positions if the
positions are independent.  But the positions are not really independent
(e.g., M-conjugacy).  I'm not sure exactly what the best we can do
actually is. 

My Shamir program is going to represent each position as a pair of indices
(i1,i2).  i2 will be a single byte containing 1..48 and indexing M.  i1
will be two or three bytes indexing a table of representatives of
M-conjugacy classes.  The table in turn will consist of eighteen byte
supplements of permutations less the last edge and the last corner cubie. 
A two-byte index will cover quarter turns through level 6 of the tree
(there are 18,395 representatives at level 6).  A three-byte index will
cover quarter turns through level 9 of the tree (there are 14,956,266
representatives at level 9).  I am dubious that I will even get into the
three byte index business.  It will simply take too long.  That is, a two
byte index through level 6 will allow level's 7 through 12 to be
calculated.  It may well take too long to go any further than that. 

Anyway, (i1,i2) means m'[i2]X[i1]m[i2].  The indices (i1,i2) will be
sorted according to the lexicographic order of m'[i2]X[i1]m[i2].  The
"Shamir tree" will be built with the indices as the leaf nodes.  The
effective storage required for each position will be maybe five bytes or
so because you have to count the table of representatives in any honest
accounting of memory requirements.  Thus, the tree structure itself will
be the biggest consumer of memory. 

This approach will not require any packing or unpacking, but it will
require lots of extra multiplying of permutations.  However, at most
points in the processing, it will not be necessary to multiply the entire
permutation.  Multiplying a single cell of the vector will usually suffice
for comparing vectors, traversing the tree, etc. 

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7127
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990









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From: Jim Mahoney <mahoney@marlboro.edu>
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To: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Cc: cube-lovers@ai.mit.edu
In-Reply-To: <Pine.PMDF.3.91.960605084501.9735A-100000@PSTCC6.PSTCC.CC.TN.US> (message from Jerry Bryan on Wed, 05 Jun 1996 10:22:31 -0500 (EST))
Subject: Re: A essay on the NxNxN Cube : counting positions and solving it


>>>>> "Jerry" == Jerry Bryan <jbryan@pstcc.cc.tn.us> writes:
    Jerry>  This has been discussed before on Cube-Lovers, but I am
    Jerry> still puzzled or curious about the usage of the word
    Jerry> "orbit".

So am I, actually.  Dan Hoey just replied with a much more detailed
understanding of the group theory aspects of this word than I have at
present, which I'll have to think about some more.  For myself, I mean
no more and no less than a set of cubies which can move into each
other's positions.  For a 3x3x3 cube which I imagine to be made of of
3^3=27 smaller cubes (cubies), what I call "orbits" are exactly those
cubies at the 8 corners, 12 edges, 6 faces, and 1 (unseen) at the
center.

    Jerry> Secondly, if my understanding of your model is correct, you
    Jerry> are treating positions as distinct which cannot be
    Jerry> distinguished with normal coloring of a physical cube (even
    Jerry> an imaginary physical cube for large N).

Yes, exactly.  As Dan just said, he has discussed this vision of the
cube in earlier notes, and called it the "theoretical invisible cube".

When I started thinking about these larger cubes, I built them by
making piles of dice.  All the inner cubies were there, and all had
definite orientations, and I could see them every time I tried to
rotate a slice - which required carefully seperating out the layers,
turning one, and putting everything back together.  So perhaps that's
why I liked those "invisible" inside pieces.  But it also seemed more
elegant.  The restricted versions (only the outside, only the
orientations of the corners and edges, etc.) are all special cases.

    Jerry> There are several implications of how you treat visibly
    Jerry> indistinguishable positions.  For example, it impacts your
    Jerry> counts of how many positions there are.  For another
    Jerry> example, it impacts your solutions (e.g., "invisible"
    Jerry> incorrect parity on the 4x4x4.  "Invisible" bad parity can
    Jerry> also occur on the 3x3x3 if you remove the face center color
    Jerry> tabs.  A slice move will give the edges and corners
    Jerry> opposite parity that is not visible.)  Perhaps you could
    Jerry> discuss these issues with respect to your model.

I'm not sure what there is to say; you seem to understand the issues.

Yes, I am counting "visibly indistinguishable" positions as different,
especially on the larger cubes, if by "visibly" you mean to only look
at the outside.  I'm assuming that either the whole thing is
transparent, or that you can take it apart, and see the inside cubies
if you like.  There are parity constraints between the different
orbits, including the ones on the inside that are "invisible," but
they turn out to be fairly simple: the parity of each orbit of corners
and the central cubie, from the outer layer all the way down to the
inside, are independent, and can be chosen arbitrarily.  And once
they're fixed, the parity of all the other orbits is given.  

By "bad" parity I assume you mean a case when the edges and corners
have different parities.  Starting from the solved (even parity) 3x3x3
Cube, a slice move definitely does this; four outside edges cycle, and
the corners don't move.  However, on the 3x3x3, this *is* visible,
since the face centers will also have odd parity.  Moreover, the
central cube (which you can't see, of course, and isn't really there
on a real cube) also has odd parity, in a way: it has undergone an odd
number of quarter turns.  On a 4x4x4, a slice move on a solved cube
changes the parity of the inside 2x2x2 corners (which you can't see)
and the edges (which you can).  The parity of the outer corners is
left unchanged, since they didn't move, and the parity of the face
centers is also unchanged, since 8 of them move in two cycles of four
cubies.  Then the fact that the outside edges are odd while the
outside corners are even simply means that the inside 2x2x2 corners
are also odd.  That's all.


Hope that helps,

  Dr. Jim Mahoney                          mahoney@marlboro.edu
  Physics & Astronomy
  Marlboro College, Marlboro, VT  05344




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To: hoey@aic.nrl.navy.mil
Cc: cube-lovers@ai.mit.edu
In-Reply-To: <9606052254.AA22046@sun34.aic.nrl.navy.mil> (hoey@AIC.NRL.Navy.Mil)
Subject: Re: A essay on the NxNxN Cube : counting positions and solving it


>>>>> "hoey" == hoey  <hoey@AIC.NRL.Navy.Mil> writes:
    hoey> When we take account of the internal cubies I call it the
    hoey> "Theoretical Invisible cube", described in my Invisible
    hoey> Revenge article 9 August 1982.  A solution method is given
    hoey> in
    hoey> 	Eidswick, J. A., "Cubelike Puzzles -- What Are They
    hoey>       and How Do You Solve Them?", 'American Mathematical
    hoey>       Monthly', Vol. 93, #3, March 1986, pp. 157-176.

Thanks for the references.  I haven't seen Eidswick's paper yet,
but will check it out.

    hoey> As for counting the positions, I haven't got around to
    hoey> checking the numbers in "Groups of the larger cubes", 24 Jun
    hoey> 1987.  You might want to see how they compare to yours.

I just did, and for the specific case that I describe (which is "s:
Supergroup, i: theoretical invisible group") the formulas are exactly
the same.  I didn't consider nearly the range of alternatives discussed
in that article, but its nevertheless nice to see a confirmation of
the results.

Regards,

  Dr. Jim Mahoney                          mahoney@marlboro.edu
  Physics & Astronomy
  Marlboro College, Marlboro, VT  05344







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Date: Thu, 6 Jun 1996 09:04:04 -0400
From: Rob Hochberg <hochberg@dimacs.rutgers.edu>
Message-Id: <199606061304.JAA20500@dimacs.rutgers.edu>
To: CUBE-LOVERS@ai.mit.edu
Subject: Speed cubing


I've heard about some pretty fast people who've claimed to 
have averages in the low 20's, but I haven't seen them perform.

My buddy from high school, Scott Evans, now living in Austin,
averages about 25 seconds these days.  He's the fastest active
cubist I've seen in the last 10 years.  I'm at about 28 seconds.

Anyone else?

Rob hochberg@dimacs.rutgers.edu



From cube-lovers-errors@curry.epilogue.com  Fri Jun  7 14:55:11 1996
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Date: Wed, 5 Jun 1996 21:43:50 -0700
To: CUBE-LOVERS@ai.mit.edu
From: Lars Petrus <lars@netgate.net>
Subject: Re: fastest hands in the midwest...

>> By the way, I can do the cube in 1 minute 26 seconds.
>> How does that compare with everyone else!

I've teached a lot of people the cube, and after 6 weeks, that is a
really good time. I think it's much better than I did after that time.

To get *really* good times, you need a really good method, averaging
50-60 moves.

>37.72 won me the midwest championship, my best official time was
>35.30 seconds which placed me 5th in the country. I think it was 1981.
>Now?  I don't get timed very often, but it's still usually under a minute.
>
>I guess it is like riding a bicycle.
>
>Anyone else on this list from those contest days?  Minh Thai - are you
>out there?  How about Jeff Verasono? or David P. Conrady?  I've often
>wondered what that crazy guy with the bright maroon hair ended up
>doing with his life...
>
>Kristin (used to be Wunderlich) Looney
>kristin@tsi-telsys.com

I won the swedish championship with 40.48 (*very* hard cubes), and ended
4th in the world championships with 24.57. My personal best is 15.92, and
best average of 10 consecutive solutions about 23.50. Nowadays I'm 2-4
seconds slower, but (fortunately!) I don't do it nearly as much.

Yes, its a lot like riding a bike. Sometimes I haven't done it for years,
get a new cube, and it's just like before. Weird...

  -  -  -  -
For every economist, there exists an equal and opposite economist.

Lars Petrus, Sunnyvale, California  -  lars@netgate.net




From cube-lovers-errors@curry.epilogue.com  Fri Jun  7 14:57:37 1996
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From: Mark Longridge <cubeman@dis.on.ca>
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To: cube-lovers@ai.mit.edu
Subject: Cube Contests
Date: Thu, 6 Jun 96 13:14:17 EDT
Message-Id:  <9606061314.aa26502@admin.dis.on.ca>

Well, I never did manage to contact Minh Thai and Jeff Varasano.

 

I was the sole prize-winning Canadian in the US contests (the 2nd one).

Unfortunately, that was not really allowed by the rules so the feat

was never recorded in the Ideal newsletters, though I still have my

medal from Sept. 28, 1981 from Buffalo New York.

 

My time was around 38 seconds... although I did improve after that a bit.

 

There are other people from the cube contests that are on the internet:

 

Myself, Chris Pelley, Robert Jen....

 

I was never sure if Mike Reid of Cube-Lovers was the same Mike Reid in the

cube contests...

 

I was in the Canadian contests too, although no prizes there. I did get

a yellow certificate for making to the final round (under a minute).

I went to London and Toronto Ontario.

 

Erno Rubik was at the Canadian Championships at the Ontario Science Centre.

Ron Lancaster was also a judge, and Stewart Sims from Ideal Toy was there.



From cube-lovers-errors@curry.epilogue.com  Fri Jun  7 14:58:06 1996
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Date: Thu,  6 Jun 1996 16:49:15 -0400 (EDT)
From: "Jonathan M. Cheyer" <cheyer+@cmu.edu>
To: CUBE-LOVERS@ai.mit.edu
Subject: Re: fastest hands in the midwest...
In-Reply-To: <v02130508addbb30e7cf0@[205.230.130.72]>
References: <v02130508addbb30e7cf0@[205.230.130.72]>


I was the winner of the Massachusetts/New England region with a time
of 48.31 seconds.  I did have the distinction of being the youngest
winner however; I was only 9 years old and it was the summer of 1981.

Kristin, although Adam (my brother) and I did stay in touch with Jeff
Varasano for a few years, we eventually lost contact with him and
everyone else.  The best thing I remember about David Conrady was that
the day before the competition at That's Incredible, he dropped a
Rubik's Cube off of the 15th (or similar) floor and scared the heck
out of an old lady.  I had never seen a cube explode like that
before...


Jon

=============================================
==  Jonathan Cheyer            cheyer@cmu.edu
==  The only thing faster than light is time,
==  so enjoy it while you still can.
=============================================


From cube-lovers-errors@curry.epilogue.com  Sat Jun  8 14:33:05 1996
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Date: Fri, 7 Jun 1996 14:52:51 -0700 (PDT)
From: Darrell Fuhriman <darrell@grumblesmurf.net>
To: CUBE-LOVERS@ai.mit.edu
Subject: verifying correctness of a cube
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So, I was recently given a cube, which once again sparked my interest in 
the whole thing, however it appears that the stickers have been pulled 
off and re-arranged at some point.  (The faces don't match any I've seen 
mentioned in the archives or the various web pages.)

What would be the right way to put it back together to get a "working" cube?

d.



From cube-lovers-errors@curry.epilogue.com  Sat Jun  8 14:32:08 1996
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To: CUBE-LOVERS@ai.mit.edu
Subject: Re: Speed cubing 
In-Reply-To: Your message of "Thu, 06 Jun 1996 09:04:04 CDT."
             <199606061304.JAA20500@dimacs.rutgers.edu> 
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Date: Fri, 07 Jun 1996 15:09:45 -0500
From: Norman Richards <orb@cs.utexas.edu>

> I've heard about some pretty fast people who've claimed to 
> have averages in the low 20's, but I haven't seen them perform.
> 
> My buddy from high school, Scott Evans, now living in Austin,
> averages about 25 seconds these days.  He's the fastest active
> cubist I've seen in the last 10 years.  I'm at about 28 seconds.


  Whenever I see posts like this I have to wonder what methods you guys are
using to solve the cube.  My average casual speed is about 80 seconds and when
I get in blitz mode I usually can average below a minute.  But, I never 
get below say 45 seconds unless I get a really nice starting position and
things just fall into place.  My personal best ever is about 32 seconds, back
when I was in high school.  

  Anyways, I am curious what methods people use.  I've asked around and it 
seems
most people prefer to go top/middle/bottom.  In fact, all the published 
solutions go that way.  But, I learned to go top/bottom/middle because it is 
MUCH easier.  At the same time, I've seen that when solving a single face, 
most solutions suggest doing edge and the corner pieces, but I do corner and 
then edge.  I think this has to do with whether you do the middle or the bottom
next because with the middle/bottom approach you can leave out one corner
piece to do the middle.  But if you do bottom/middle then you can leave out
an edge piece to aid in getting the buttom edge pieces done.


  Here is a quick overview of my solution (because it is different than most
peoples):

  1. do the top face.  4 corners then any 3 edges.  (intuitive)
  2. align bottom corner pieces.  (pattern)
  3. rotate bottom corner pieces so the correct color is showing (pattern)
  4. place the remaining 4 bottom edges and the missing top edge (intuitive)
  5. put the 4 middle edge pieces in the correct positions (intuitive)
  6. rotate edge pieces as needed (pattern)

  
  Anyways, is there one technique that almost all speedy solutions use?  I've
tried the top/middle/bottom solutions but they seem very uninuitive.  Do
most speed people only use patterns? (or at least after the top layer?)  How
many patterns do you use?  When I go for speed, I tend to use two patterns
for step 2, and 4 possible patterns for step 3.  Meaning that I use about
7 patterns max for speed and I use 3 patterns total when doing it casually.  


  


   



From cube-lovers-errors@curry.epilogue.com  Sat Jun  8 14:31:38 1996
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Date: Fri, 7 Jun 1996 15:27:55 -0400 (EDT)
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To: Rob Hochberg <hochberg@dimacs.rutgers.edu>
Cc: CUBE-LOVERS@ai.mit.edu
Subject: Re: Speed cubing
In-Reply-To: <199606061304.JAA20500@dimacs.rutgers.edu>
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My name is Jiri Fridrich and I am the champon of Czechoslovakia from 1982.
I won the championship with time 23.55. When I was at my best, I was able
to solve the cube in 17 seconds on average (average from 10 consecutive
runs). Even today, after all those years, I can solve the cube in 20
seconds on average. I am using about 150 different algorithms and need 60
moves on average. 

Just for the record: The 1983 champion of Czechoslovakia, Robert Pergl,
won with 17.04. There must be *many* guys out there who can solve the cube
consistently below 20 sec :-) 

Jiri

On Thu, 6 Jun 1996, Rob Hochberg wrote:

> 
> I've heard about some pretty fast people who've claimed to 
> have averages in the low 20's, but I haven't seen them perform.
> 
> My buddy from high school, Scott Evans, now living in Austin,
> averages about 25 seconds these days.  He's the fastest active
> cubist I've seen in the last 10 years.  I'm at about 28 seconds.
> 
> Anyone else?
> 
> Rob hochberg@dimacs.rutgers.edu
> 
> 
> 



From cube-lovers-errors@curry.epilogue.com  Sat Jun  8 14:33:28 1996
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Date: Fri, 7 Jun 1996 20:27:52 -0400
From: AirWong@aol.com
Message-Id: <960607202751_212838202@emout15.mail.aol.com>
To: CUBE-LOVERS@ai.mit.edu
Subject: All these fast hands...

All this talk about these old competitions... When was the last one?


From cube-lovers-errors@curry.epilogue.com  Sat Jun  8 14:34:12 1996
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Date: Sat, 8 Jun 96 14:09:59 +0900
From: Norman Diamond 08-Jun-1996 1409 <diamond@jrdv04.enet.dec-j.co.jp>
To: cube-lovers@ai.mit.edu
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Mark Longridge reports:

>I was the sole prize-winning Canadian in the US contests (the 2nd one).
>Unfortunately, that was not really allowed by the rules so the feat
>was never recorded in the Ideal newsletters,

Oh.  Did Ideal's rules constrain the entrants to Hungarians only?

Should puzzlers located outside the US boycott Ideal products?
Unfortunately I didn't know about this in time to do so.

-- Norman Diamond           diamond@jrdv04.enet.dec-j.co.jp
[Speaking for Norman Diamond not for Digital.]


From cube-lovers-errors@curry.epilogue.com  Sun Jun  9 00:34:46 1996
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Date: Sat, 8 Jun 1996 20:48:40 -0400
From: der Mouse <mouse@collatz.mcrcim.mcgill.edu>
Message-Id: <199606090048.UAA04049@Collatz.McRCIM.McGill.EDU>
To: CUBE-LOVERS@ai.mit.edu
Subject: Re:  verifying correctness of a cube

> So, I was recently given a cube, which once again sparked my interest
> in the whole thing, however it appears that the stickers have been
> pulled off and re-arranged at some point.  (The faces don't match any
> I've seen mentioned in the archives or the various web pages.)

Which doesn't necessarily mean it was attacked by a mad sticker-moving
artist; it could just be an off brand. :-)

> What would be the right way to put it back together to get a
> "working" cube?

Well, first I'd try solving it and see if I end up with something like
two edge cubies colored the same, or a corner having colors that belong
to opposite center cubies.  If no such problem is found but a parity
constraint is violated (eg, a single edge cubie flipped), it was just
taken apart, so I'd take it apart and put it back together solved.  If
I _do_ find an "impossible" cubie, then I'd just take the stickers off
and put them back on such that every face is a solid color.  (I'd
probably try to mostly-solve it first, so's to minimize the number of
stickers that need moving, but that's a frill.)  If you take all the
stickers off, all states are the same; there is no mechanical state you
have to match when putting the stickers back on.

					der Mouse

			    mouse@collatz.mcrcim.mcgill.edu


From cube-lovers-errors@curry.epilogue.com  Sun Jun  9 22:45:50 1996
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Date: Sun, 9 Jun 1996 20:49:08 -0500
To: Norman Richards <orb@cs.utexas.edu>, CUBE-LOVERS@ai.mit.edu
From: Kristin Looney <kristin@tsi-telsys.com>
Subject: Methods (Re: Speed cubing)

At 3:09 PM 6/7/96, Norman Richards wrote:
> Here is a quick overview of my solution:
>
> 1. do the top face. 4 corners then any 3 edges. (intuitive)
> 2. align bottom corner pieces. (pattern)
> 3. rotate bottom corner pieces so the correct color is showing (pattern)
> 4. place the remaining 4 bottom edges and the missing top edge (intuitive)
> 5. put the 4 middle edge pieces in the correct positions (intuitive)
> 6. rotate edge pieces as needed (pattern)

my method:

1. do top corners. I always start with white. (intuitive)
2. do bottom corners.
        2a. bring bottom corner color onto bottom face (one of 2 patterns)
        2b. orient bottom corners with each other (one of 2 patterns)
3. fill in all but one edge on top and bottom (intuitive)
4. fill in last edge (pattern)
5. solve middle ring of edges (usually 2 patterns)

The only person I have ever met used this same method was Minh,
the winner of the first U.S. championship. The only differences between
our methods was that he had more patterns memorized for step 5. I
typically do one pattern, which gets me close, and then finish it up
with one more pattern - where Minh could look at that last ring and
instantly know a pattern that would bring the cube into it's final
solved state. His hands were also a lot faster than mine.

The beauty of this method is that there is very little to memorize -
and although it doesn't give me very many 20 second times, it's
always well under a minute. In both steps 2a and 2b, one of the
two pattern options comes up more than 90% of the time - and
you can just do that pattern twice if you want, so you really only
have to know one pattern if you are willing to loose a few seconds
10% of the time. Since the pattern in step 4 is one of the patterns
in step 5, my solution can get you a solved cube reasonably quickly
with only 4 memorized patterns. And with a bit more memorization
it can begin to screeeeeem. For some reason I haven't really had much
need for speed solving these days - under a minute is just fine to keep
up my reputation.

-K.




From cube-lovers-errors@curry.epilogue.com  Sun Jun  9 22:46:11 1996
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Message-Id: <31BB7363.799@erols.com>
Date: Sun, 09 Jun 1996 20:59:15 -0400
From: Charlie Dickman <charlied@erols.com>
Reply-To: charlied@erols.com
X-Mailer: Mozilla 2.01 (Macintosh; U; 68K)
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To: Cube-Lovers <Cube-Lovers@ai.mit.edu>
Subject: Re: A 4-D Rubik's Cube
Content-Type: text/plain; charset=us-ascii
Content-Transfer-Encoding: 7bit

I'm sorry to bother all of you with this again but I have converted the 
document describing the model and implementation of a 4-dimensional 
(3x3x3x3) Rubik's Cube into a word processing document which I can 
transform into several formats, including MS Word 3.0 & 4.0 and RTF.

Anyone interested in a copy in word processing format let me know your 
interest and if you have a preferred format.  Unfortunately, a text only 
document will not be able to include any of the figures and as this does 
a grave injustice to the paper I would rather not ship text only copies.

Regards...
Charlie Dickman
charlied@erols.com


From cube-lovers-errors@curry.epilogue.com  Mon Jun 10 14:52:37 1996
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From: whuang@cco.caltech.edu
Message-Id: <199606101427.HAA07096@accord.cco.caltech.edu>
Subject: Re: "Better" Method?
To: cube-lovers@ai.mit.edu
Date: Mon, 10 Jun 1996 07:27:00 -0700 (PDT)
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> Kristin Looney <kristin@tsi-telsys.com> writes:
> >my method:
> 
> >1. do top corners. I always start with white. (intuitive)
> >2. do bottom corners.
> >        2a. bring bottom corner color onto bottom face (one of 2 patterns)
> >        2b. orient bottom corners with each other (one of 2 patterns)
> >3. fill in all but one edge on top and bottom (intuitive)
> >4. fill in last edge (pattern)
> >5. solve middle ring of edges (usually 2 patterns)
> 
> >The only person I have ever met used this same method was Minh,
> >the winner of the first U.S. championship.
> 
> Odd.  I use the same method, and before this I was almost convinced I was the
> only one.  I guess I never asked Minh.
> 
> > The only differences between
> >our methods was that he had more patterns memorized for step 5. I
> >typically do one pattern, which gets me close, and then finish it up
> >with one more pattern - where Minh could look at that last ring and
> >instantly know a pattern that would bring the cube into it's final
> >solved state. His hands were also a lot faster than mine.
> 
> I only have two patterns for step 5.  One basically permutes three edge
> pieces, the other flips a pair.  It does have the advantage of being
> easy to explain, though.
> 
> Come to think of it, my method is slightly different -- I orient the
> corners BEFORE positioning them.
> 
> >The beauty of this method is that there is very little to memorize -
> >and although it doesn't give me very many 20 second times, it's
> >always well under a minute. 
> 
> Ditto.
> 
> I think the beauty of this "cubie"-oriented method is how easy it
> generalizes to larger cubes.  By just adding 2 patterns, I can solve the
> 4x4x4, and one more pattern gives me the 5x5x5.  I think the "layer"-oriented
> method is much harder to generalize.
> 
> 



From cube-lovers-errors@curry.epilogue.com  Mon Jun 10 14:53:14 1996
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Date: Mon, 10 Jun 1996 09:17:05 -0700
To: CUBE-LOVERS@ai.mit.edu
From: Lars Petrus <lars@netgate.net>
Subject: Re: Speed cubing


>  Anyways, I am curious what methods people use.  I've asked around and it
>seems
>most people prefer to go top/middle/bottom.  In fact, all the published
>solutions go that way.  But, I learned to go top/bottom/middle because it is
>MUCH easier.  At the same time, I've seen that when solving a single face,
>most solutions suggest doing edge and the corner pieces, but I do corner and
>then edge.  I think this has to do with whether you do the middle or the bottom
>next because with the middle/bottom approach you can leave out one corner
>piece to do the middle.  But if you do bottom/middle then you can leave out
>an edge piece to aid in getting the buttom edge pieces done.
>
>
>  Here is a quick overview of my solution (because it is different than most
>peoples):
>
>  1. do the top face.  4 corners then any 3 edges.  (intuitive)
>  2. align bottom corner pieces.  (pattern)
>  3. rotate bottom corner pieces so the correct color is showing (pattern)
>  4. place the remaining 4 bottom edges and the missing top edge (intuitive)
>  5. put the 4 middle edge pieces in the correct positions (intuitive)
>  6. rotate edge pieces as needed (pattern)
>
>
>  Anyways, is there one technique that almost all speedy solutions use?  I've
>tried the top/middle/bottom solutions but they seem very uninuitive.

The layer by layer techniques are useless for speed cubing. Most people
use them, because they are a simple way for the human mind to approach the
problem, but they are not natural "from the cubes point of view".

In the final of the swedish championship, 8 of 11 competitors used the
vanilla layer-by-layer method. The other 3 of us finished 1, 2 and 3!

The basic problem with the layer method is obvious, and very big. When you
have completed the first layer, you can do *nothing* without breaking it up.
So you break it, do something, then restore it, again and again. It's quite
obvious that this layer is in the way of your solution, not a part of it.

My approach was to find something that, once acomplished, did not need to
be broken up. A true step on the way to a solution. What I came up with was
to first solve a 2x2x2 corner. After that, you can move three sides freely,
and not touch what you achieved. Then I expand it to a 2x2x3, which leaves
two sides free. Then you fiddle a bit, and go to 2x3x3 and 3x3x3.

1. do the 2x2x2 (intuitive)
2. expand it to a 2x2x3 (intuitive)

I don't know how to say this in group-babble, but when you move just 2
sides, you can never "truly" rotate an edge or move a corner. This means
by temporarily breaking the 2x2x3 you can very quickly rotate edges and
move corners. On average, you can get the edges correctly rotated in about
5 moves. You could probably move the corners too, in 2-3 more moves, but
it's too hard to see the corner condition to do this while speed cubing.

But if you like, you can do 5 before 4.

3. flip the edges (intuitive/pattern)
4. expand to 2x3x3, using only the 2 free layers (intuitive)
5. move corners (pattern)
6. rotate corners (pattern)
7. move edges (pattern)

Step 4 can be quite hard before you're used to the problems. The others
should be simple for anyine familiar with the cube.

In reality, there are so many special cases in the final layer that 5-7
really is just one phase to solve the final layer.

Using this method, I use on average 60 moves while speed cubing, and I have
done 100 consecutive solutions in an average of 50.53 moves, while taking
time to think about what I'm doing.

>  Do
>most speed people only use patterns? (or at least after the top layer?)  How
>many patterns do you use?  When I go for speed, I tend to use two patterns
>for step 2, and 4 possible patterns for step 3.  Meaning that I use about
>7 patterns max for speed and I use 3 patterns total when doing it casually.

Well, for the final layer I use maybe 20-50 different patterns.


Singmaster asked all the 19 finalists in the Budapest world championship
what method they used. According to him, half used Minhs method, half used
another method, and I was the only one using something else.

  -  -  -  -
For every economist, there exists an equal and opposite economist.

Lars Petrus, Sunnyvale, California  -  lars@netgate.net




From cube-lovers-errors@curry.epilogue.com  Mon Jun 10 14:52:01 1996
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Cc: Norman Richards <orb@cs.utexas.edu>, CUBE-LOVERS@ai.mit.edu
Subject: Re: Methods (Re: Speed cubing)
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I am including my method for solving the cube. It enables me to solve the
cube in 20 seconds on average (since I am not as fast as I used to be 14
years ago :-( ).

1. Do the four edges (white first) (2 sec.)
2. Put the white corner including the corresponding edge from the second
slice. When you put all four white corners, two slices on the cube will be
done. In this stage, almost no algorithms are necessary. Most positions
can be solved with intuition. (4 x 2 sec = 8 sec.)
3. Turn all 8 small cubes from the last slice so that the last face has
the same color. There are only 40 different positions (not counting
symmetrical positions). On average, 10 moves are necessary to do this
phase.(3 sec.).
4. Move the cubes in the last slice so that the cube is solved. There are
only 13 different positions. On average, 10-15 moves are necessary. (4
sec.)

For the whole system, 40+13=53 algorithms are necessary. One also needs
about 8 short algorithms for the second phase. Altogether, 61 algorithms
will enable you to solve the cube in 17 seconds on average, if you can
turn 4 turns per second, and if you can minimize time gaps between
algorithms. 

The handling of the last (3-rd) slice is probably the most efficient
approach ane can come up with. One only needs to carry out two
algorithms to do the 3-rd slice. That is very effective. Breaking the last
slice into four stages (turn edges, turn corners, move edges, move
corners) is less demanding on the algorithmic part, but needs much more
moves and more idle time between algorithms. 

Jiri Fridrich



From cube-lovers-errors@curry.epilogue.com  Tue Jun 11 13:24:32 1996
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From: David Singmaster Computing & Maths South Bank Univ <david.singmaster@sbu.ac.uk>
To: cube-lovers@ai.mit.edu
Message-Id: <009A3B3F.74098380.1922@vax.sbu.ac.uk>
Subject: Orbits, re Jerry Bryan's message of 5 Jun

	First apologies if this has already been discussed - I'm behind on my
email.
	Orbit is used in the context of the action of a group on a set.  The
orbit of a point in the set is the set of all points that the original point
can be carried to by the actions of the group.
	Jerry mentioned that I said there are 12 orbits of the entire cube.
This is correct in that one is thinking of the group of the cube as acting on
patterns or configurations of the entire cube.  Martin Scho"nert is correct
in saying that these are cosets of the cube group in the larger group of
assemblies of the cube, or permutations of the parts.  
	However, it is not always the case that the set being acted upon can be
given a group structure.  E.g. when one considers the action of the cube group
on the individual pieces, then the orbit of a corner piece is the set of
8 corners.
	Perhaps the astronomical imagery can help.  Think of a planet (or
what have you).  There is a group of physical motions of this and the orbit
is the set of positions which these motions can carry the planet to.  The
case of the orbit of a corner piece is quite easy to visualise.  The more
general contexts of the orbits of achievable positions are less easy to
visualize.
	Perhaps another example may help.  Consider the 14-15 or 1 puzzle.
For a given position of the blank, only the even permutations are achievable,
so we can speak of two orbits, each of which has  15!/2  positions.  If we
permit the blank to move about, we again get half the possibilities, i.e.
16!/2  positions, and two orbits.
DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499; 
email:  zingmast  or  David.Singmaster  @vax.sbu.ac.uk


From cube-lovers-errors@curry.epilogue.com  Wed Jun 12 23:26:55 1996
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From: VANESSA PARADIS WANTS ME <cmaggs@glam.ac.uk>
To: CUBE-LOVERS@ai.mit.edu
X-Vms-To: RUBIKCUBE

Off the point completely, anyone got other Rubik puzzles?
My record with the Clock is 8 seconds, but my average is around 15-18 seconds.

Anyone else?


From cube-lovers-errors@curry.epilogue.com  Wed Jun 12 23:25:38 1996
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From: VANESSA PARADIS WANTS ME <cmaggs@glam.ac.uk>
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1 Minute 26!?!?!?!? That's very impressive. I can say that ONCE I got it down
to about 1 minute 10, but that was lucky. What order do you solve the pieces?
Top to Bottom? I'm trying to find the fastest possible method by combining
methods from other people.
=====================================================                                 

I solve top, bottom, middle, rotate.

Pretty quick.

Have tried top, middle bottom, but it is a little slower.

How about you, what metthod do you use


From cube-lovers-errors@curry.epilogue.com  Wed Jun 12 23:26:00 1996
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From: VANESSA PARADIS WANTS ME <cmaggs@glam.ac.uk>
To: CUBE-LOVERS@ai.mit.edu
X-Vms-To: RUBIKCUBE

I have two books.
One (THE SIMPLKE SOLUTION TO THE RUBIK CUBE) which is not very good!

The second, YOU CAN DO THE CUBE, is by Patrick Bossert and was written in 1981.
Does anyone know where Patrick is now?



From cube-lovers-errors@curry.epilogue.com  Wed Jun 12 23:27:11 1996
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From: michael reid <mr@emf.net>
Message-Id: <199606130033.RAA28824@emf.emf.net>
To: cube-lovers@ai.mit.edu
Subject: Re: Cube Contests

mark writes

> I was never sure if Mike Reid of Cube-Lovers was the same Mike Reid in the
> 
> cube contests...

i was only in one contest: october 1982 in some mall in new jersey.
i came in first with a time of 27.3 seconds; very stiff cubes.

but if he really means "contests" (plural), that would mean - shudder -
there are imposters out there!  :-}

mike


From cube-lovers-errors@curry.epilogue.com  Wed Jun 12 23:26:20 1996
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Date: Wed, 12 Jun 1996 09:49:37 +0100
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From:	SMTP%"kristin@tsi-telsys.com" 10-JUN-1996 04:00:41.88
To:	Norman Richards <orb@cs.utexas.edu>, CUBE-LOVERS@ai.mit.edu
CC:	
Subj:	Methods (Re: Speed cubing)

At 3:09 PM 6/7/96, Norman Richards wrote:
> Here is a quick overview of my solution:
>
> 1. do the top face. 4 corners then any 3 edges. (intuitive)
> 2. align bottom corner pieces. (pattern)
> 3. rotate bottom corner pieces so the correct color is showing (pattern)
> 4. place the remaining 4 bottom edges and the missing top edge (intuitive)
> 5. put the 4 middle edge pieces in the correct positions (intuitive)
> 6. rotate edge pieces as needed (pattern)

my method:

1. do top corners. I always start with white. (intuitive)
2. do bottom corners.
        2a. bring bottom corner color onto bottom face (one of 2 patterns)
        2b. orient bottom corners with each other (one of 2 patterns)
3. fill in all but one edge on top and bottom (intuitive)
4. fill in last edge (pattern)
5. solve middle ring of edges (usually 2 patterns)

The only person I have ever met used this same method was Minh,
the winner of the first U.S. championship. The only differences between
our methods was that he had more patterns memorized for step 5. I
typically do one pattern, which gets me close, and then finish it up
with one more pattern - where Minh could look at that last ring and
instantly know a pattern that would bring the cube into it's final
solved state. His hands were also a lot faster than mine.

The beauty of this method is that there is very little to memorize -
and although it doesn't give me very many 20 second times, it's
always well under a minute. In both steps 2a and 2b, one of the
two pattern options comes up more than 90% of the time - and
you can just do that pattern twice if you want, so you really only
have to know one pattern if you are willing to loose a few seconds
10% of the time. Since the pattern in step 4 is one of the patterns
in step 5, my solution can get you a solved cube reasonably quickly
with only 4 memorized patterns. And with a bit more memorization
it can begin to screeeeeem. For some reason I haven't really had much
need for speed solving these days - under a minute is just fine to keep
up my reputation.

-K.



Strange, I always staer with WHITE aswell!!

Anyway, I`ll try that way but I always

1.  Position and orient all white corners and pieces
2.  Position and orient bottom corners.
3.  Position and orient remaing bottom and middle slice pieces.




From cube-lovers-errors@curry.epilogue.com  Thu Jun 13 23:31:29 1996
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On Jun 5, 18:02, Kristin Looney wrote:
> Subject: fastest hands in the midwest...
> > By the way, I can do the cube in 1 minute 26 seconds.
> > How does that compare with everyone else!
>
> 37.72 won me the midwest championship, my best official time was
> 35.30 seconds which placed me 5th in the country. I think it was 1981.
> Now?  I don't get timed very often, but it's still usually under a minute.
>
> I guess it is like riding a bicycle.
>
> Anyone else on this list from those contest days?  Minh Thai - are you
> out there?  How about Jeff Verasono? or David P. Conrady?  I've often
> wondered what that crazy guy with the bright maroon hair ended up
> doing with his life...

I was in two different contests; one was at Six Flags Magic Mountain in
southern California, the other was at some mall in L.A.  I was never fast
enough to get to any of the finalist rounds, but it was fun all the same.

Minh Thai was at the contest at the mall; Ideal had him there to promote it.  I
happened to have a Revenge on me at the time, and he thought he'd demonstrate
to the crowd his solution for it.  He managed not to solve it, but rather
shatter it in about 10 seconds flat.  Not bad, eh?  :)

	-Jason

-- 
Jason K. Werner, Silicon Graphics
U.S. Field Operations I/S Sys Admin
mrhip@corp.sgi.com, 415-933-6397
"I will choose free will".....Neil Peart
"These go to eleven".....Nigel Tufnel


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                  \  /\  /  __/    \  /\  / (_) | | | |_|
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			FREE SPEECH ON THE INTERNET!


From cube-lovers-errors@curry.epilogue.com  Fri Jun 14 13:58:44 1996
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From: David Singmaster Computing & Maths South Bank Univ <david.singmaster@sbu.ac.uk>
To: cube-lovers@ai.mit.edu
Message-Id: <009A3D97.EFAEB000.1938@vax.sbu.ac.uk>
Subject: Storage of cube positions

	Responding to Rubin Shai has made me think of an idea which may have
been discussed already.
	I was wondering if one could reduce the storage required to represent
a position or pattern.  In my notation, one basically records the movement
of each facelet.  However, if  UR -> RB  and we know that the UR piece is moving
to the RB place, then we only need to record the  U -> R  part of the motion.
In order to know which piece is moving, number all the 54 facelets.  Suppose
the facelets of UR are 1,2 in order, and the facelets of RB are  11,12 in order.
Then recording that  1 -> 11  completely describes the movement  UR -> RB.
	So we only need to record the movement of one of the facelets of
each piece - the others have to follow.  Hence we can describe the position
of the cube by a vector of 20 numbers in the range 1 .. 54.
	Of course, one has to pay for this - the composition of movements will
be more complex.  But if storage is your problem rather than time, it may be
worth it.
DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499; 
email:  zingmast  or  David.Singmaster  @vax.sbu.ac.uk


From cube-lovers-errors@curry.epilogue.com  Sat Jun 22 19:28:53 1996
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To: Cube-Lovers@ai.mit.edu
From: Peter Foster <pfoster@pcug.org.au>
Subject: Thistlethwaites algorithm (and others)

Morwen Thistlethwaite devised a solution for Rubik's cube which required at
most 50
turns. I would like to know where I can get a copy of this solution. Can
anyone help?
I have asked David Singmaster about this. He replied that he does have
Thistlethwaite's
solution, but that it has been significantly improved and he does not know
the details.
So, if someone can point me in the right direction I would be most grateful!

While I am interested in Thistlethwaite's solution, it is no use for speed
solving.
There was a recent posting, from Jiri Fridrich, which outlined his speed
solution, as
follows:

>1. Do the four edges (white first) (2 sec.)
>2. Put the white corner including the corresponding edge from the second
>slice. When you put all four white corners, two slices on the cube will be
>done. In this stage, almost no algorithms are necessary. Most positions
>can be solved with intuition. (4 x 2 sec = 8 sec.)
>3. Turn all 8 small cubes from the last slice so that the last face has
>the same color. There are only 40 different positions (not counting
>symmetrical positions). On average, 10 moves are necessary to do this
>phase.(3 sec.).
>4. Move the cubes in the last slice so that the cube is solved. There are
>only 13 different positions. On average, 10-15 moves are necessary. (4
>sec.)
>
>For the whole system, 40+13=53 algorithms are necessary. One also needs
>about 8 short algorithms for the second phase. Altogether, 61 algorithms
>will enable you to solve the cube in 17 seconds on average, if you can
>turn 4 turns per second, and if you can minimize time gaps between
>algorithms. 

Is there any chance of Jiri Fridrich posting these algorithms (or perhaps making
them available via FTP)?

Thanks in advance...

Peter Foster

_______________________________________________________________
Peter Foster                    This sig is dedicated to all those who
616-231-2245                   did not dedicate their sigs to themselves.
pfoster@pcug.org.au



From cube-lovers-errors@curry.epilogue.com  Sat Jun 22 21:43:53 1996
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Date: Sun, 23 Jun 1996 03:41:53 +0200
From: Dik.Winter@cwi.nl
Message-Id: <9606230141.AA09515=dik@hera.cwi.nl>
To: Cube-Lovers@ai.mit.edu, pfoster@pcug.org.au
Subject: Re: Thistlethwaites algorithm (and others)

I think there was in this mail a serious problem of confusion between a
solution by algorithm and a solution by hand.  While Thistlethwaite's
algorithm was (at that time) very fast, nobody would consider to use
the solution by hand.  The current faster algorithms are not even doable
by hand (unless your backtracking is impeccable).

 > There was a recent posting, from Jiri Fridrich, which outlined his speed
 > solution, as follows:

On the other hand Jiri Fridrich's solution is not a solution for computers.
It is only measured in time, not in number of moves.  And time is something
extremely machine-dependant.

 > Is there any chance of Jiri Fridrich posting these algorithms (or perhaps
 > making them available via FTP)?

But the algorithm was posted.  You posted it yourself!


From cube-lovers-errors@curry.epilogue.com  Mon Jun 24 15:34:09 1996
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On average, my algorithm requires about 60 moves. It has been optimized
for speed, not for the number of moves. Once I have little more time, I
may retype all the necessary algorithms into an electronic form and post
them here. I published the algorithms in 1982 in Mlady Svet
(a Czech magazine). But this will probably not help you too much :-(

Jiri Fridrich


On Sat, 22 Jun 1996, Peter Foster wrote:

> Is there any chance of Jiri Fridrich posting these algorithms (or perhaps making
> them available via FTP)?
> 
> Thanks in advance...
> 
> Peter Foster
> 
> _______________________________________________________________
> Peter Foster                    This sig is dedicated to all those who
> 616-231-2245                   did not dedicate their sigs to themselves.
> pfoster@pcug.org.au
> 
> 
> 



From cube-lovers-errors@curry.epilogue.com  Mon Jul  1 00:53:25 1996
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Date: Sun, 30 Jun 1996 21:05:13 -0300
From: FERNANDO VON REICHENBACH <reichen@cnea.edu.ar>
Message-Id: <199607010005.VAA31570@cnea.edu.ar>
To: Cube-Lovers@ai.mit.edu
Subject: Counting moves

  My name is Isidro Costantini, I'm a cube lover since '81.
  I used to have some cube meetings here in Buenos Aires and we have
some interesting formulas. We disserted about how to count cube moves,
and finally decided that any double move (ie: R2) are TWO moves instead
of one. This was because there are a lot of even/odd properties when
you count moves in that way. I'm quite surprised that when I checked
some pages and moves aren't count in that way. For example, to flip two
edges in it's place will always take an even number of moves (14)
(I'll put the shortest formula we have in parenthesis) (always counting
X2 as two moves)  Any 3 edges xchg (12) or Flip 2 corners (14) or
Xchg 3 corners (8) is even.  Any [Xchg 2 corners And Xchg 2 edges] is 
always odd (ie: R1 U3 L1 U2 R3 U1 R1 U2 R3 L3 U1 = 13 counting U2 as two)
  I have a collection of all the combinations of these nonFliping
corner/edges exchange ODD formulas in one face, some of them are of 17
or more movements and I wonder if there are any better than we did.
  ( Where's a place to check for those formulas? )
  Another good example is (xchg 3 edges,noFlip) (12) R2 U1 F1 B3 R2
F3 B1 U1 R2 (9 moves using your way of counting) and another equivalent:
B3 U3 R3 U1 R1 B1 followed by F1 R1 U1 R3 U3 F3 (6+6 moves, same position)
  Another way of counting could be adding the suffix (1,2 or 3) (counting
only clockwise moves) which would preserve parity as well.
  I would be pleased if some one can tell me about this subject.




From cube-lovers-errors@curry.epilogue.com  Tue Jul  2 00:37:39 1996
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Date: Tue, 02 Jul 1996 00:20:44 -0700
From: Mark Longridge <cubeman@dis.on.ca>
Organization: Computer Creations
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 > My name is Isidro Costantini, I'm a cube lover since '81.

   Welcome to cube lovers the mailing list.

>  ( Where's a place to check for those formulas? )

Well, I'm not finished yet, but I do archive all the cube formulas
I get a hold of or compose. Some of the work is with the assistance
of computers and/or mathematical insight.

http://www.dis.on.ca/~cubeman

>  Another good example is (xchg 3 edges,noFlip) (12) R2 U1 F1 B3 R2
> F3 B1 U1 R2 (9 moves using your way of counting) and another
> equivalent: B3 U3 R3 U1 R1 B1 followed by F1 R1 U1 R3 U3 F3
> (6+6 moves, same position).
>  Another way of counting could be adding the suffix (1,2 or 3)
> (counting only clockwise moves) which would preserve parity as well.
> I would be pleased if some one can tell me about this subject.

The sequence  X = (B3 U3 R3 U1 R1 B1 F1 R1 U1 R3 U3 F3) is a very
interesting one.

Note that X = B3 [U3 R3] B1 + F1 [R1 U1] F3

The above makes use of conjugates and commutators.

The following is a top view of a megaminx (magic dodecahedron):
      /\
    /    \
  /        \
  \   U    /
L  \      / R
    \____/
      F

Then the very similar sequence   R+ F+ U+ F- U- R-
                                 L- U- F- T+ F+ L+

...suffices to also 3-cycle the edges  (uf, lf, rf) on the megaminx.

In this case I don't like the U3 = U- or U' notation.
Clearly on the megaminx U3 <> U'

Note that each turn of a face is always turned one way and then back.
The 5-period rotation of a face is never used. In special cases like
these cube moves from the standard 3x3x3 are directly transferable
to the megaminx.

I have found that isoflips and isotwists work very well on the
megaminx.

The shortest flip of 2 adjacent edges uses the same 4 sides
(so I say "this sequence has face-index 4), is the following:

Note use of L-- and L++ etc to denote 2 one-fifth turns of a face!

It is of the form P U1 P' U' which is another commutator.

L-- R++ F+  U-  R+  U+  L++ R++ U+
R-- L-- U-  R-  U+  F-  R-- L++ U-

= 18 face turns or 26 one-fifth turns. Perhaps there is some
improvement to this sequence.

-> Mark <-



From cube-lovers-errors@curry.epilogue.com  Wed Jul  3 04:41:56 1996
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Date: Wed, 3 Jul 1996 05:26:31 -0300
From: FERNANDO VON REICHENBACH <reichen@cnea.edu.ar>
Message-Id: <199607030826.FAA03614@cnea.edu.ar>
To: Cube-Lovers@ai.mit.edu


3/7/96
Hi!, I'm Isidro, yesterday Mark wrote:

> The shortest flip of 2 adjacent edges uses the same 4 sides
> (so I say "this sequence has face-index 4), is the following:
> Note use of L-- and L++ etc to denote 2 one-fifth turns of a face!
> It is of the form P U1 P' U' which is another commutator.
>
> L-- R++ F+  U-  R+  U+  L++ R++ U+
> R-- L-- U-  R-  U+  F-  R-- L++ U-
>
> = 18 face turns or 26 one-fifth turns. Perhaps there is some
> .improvement to this sequence.

  I have a megaminx borrowed and solve it using some of the 3x3x3
knowledge that I have. It took me two years to solve all cases of
the 4x4x4 (borrowed from the same friend), but not trying so hard...
  I suppose you already have this, but just in case I'll send my own
flip edges formulas in the standard cube to see if they could help
with the other (I guess not):
  R'F'L'U B'U B L F R U'B U'B' (14) (flips F & B edges)
the same formula starting in the 4th move (UB'UBLFRU'BU'B' R'F'L')
flips B & L edges.
  A longer, (but faster in my hands) R B R'L U L'B F'D L'D'UB'U'F B'
(16) (flips L & R)
  R' U2 R2 U R' U' R' U2 L F R F' L' (16) (13 counting ^2 as 1) flips F&R
  
  L'B'U R'U'R B L followed by R B U'L U L'B'R' (16) flips B&R and maybe
works on megaminx...

Hope I'm sending something you don't have.

I'm looking for improving this formulas (all of them exchanges (no
fliping) 2 corners and 2 edges on the top face:
  R2 B' R' U' R U R U'B R B'U B R (15/14) xchg BL-FL corn & F-R edg
  F R'U'R F'L'B U'B'L R'U R / L U F U'F'L' (19) xchg BL-BR corn & R-L edg
  F R'F'R U R U'R2U'R U R B'R'B U (17/16) xchg Bl-FR corn & L-B edg
  L'UR'U L U'R U'L F'L' [F F'] U'L'U L F U (17) xchg BL-FR corn & L-F edg
  R U R'U'R'F R [F' F] R U'R'U'R U R'F' (15/14) xchg RF-RB corn & L-R edg
(I've some more but it's enough, besides I must tranlate them from spanish)

  PS: What a coincidence, my first mail were intended to you, Mark.
  


From cube-lovers-errors@curry.epilogue.com  Wed Jul  3 19:36:59 1996
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Date: Wed, 3 Jul 1996 15:17:59 -0300
From: FERNANDO VON REICHENBACH <reichen@cnea.edu.ar>
Message-Id: <199607031817.PAA08199@cnea.edu.ar>
To: Cube-Lovers@ai.mit.edu
Subject: Spanish moves

  Hi! I'm Isidro, (I'm telling so cause I share my mail address)
This is the spanish moving convetion:
  Up = Arriba
  Dwn = Bajo (abajo)
  Lft = Izquierda
  Rgt = Derecha
  Front = Frente (Same letter :)
  Back = Tras (atras)  (This is the one which is confusing, cause the
same letter "B" means differnt things. And I don't have no formulas on
my PC, only some sheets of papers from the '81-'82 and in my head, so
a trnaslating program (very easy to do) it's useless, these last days
I started thinking formulas movements in english, I guess it's the best
choice if we want to exchange things, though I have to recheck everything
to avoid mistakes...
PS: By the way: Which is the preferred convention: U' U3 U- or what?
    Where can I obtain programs for trying to find formulas?
	(I started one myself in Pascal, just interprets moves)

Isidro Costantini       Zappa/Hendrix/King Crimson music lover
  Olivos, Bs.As.        PC hard/soft technician


From cube-lovers-errors@curry.epilogue.com  Wed Jul  3 19:39:39 1996
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Date: Wed, 3 Jul 1996 15:46:44 -0300
From: FERNANDO VON REICHENBACH <reichen@cnea.edu.ar>
Message-Id: <199607031846.PAA08649@cnea.edu.ar>
To: Cube-Lovers@ai.mit.edu
Subject: Moves:

David wrote:
> I chose to count R2 as one move as it takes one hand movement, hence
> it takes about the same time as R, rather than twice as long.  So my
> counting is more appropriate to questions of time or efficiency,
  Although it is may be one hand move, R2 takes longer than R, if we
start thinking that way, I suggest to give different values te each move
(ie: R=1 R2=1.3 R'=1.2 L'=1 L=1.2 ...) depending on how long it would
take to make that move, it could also depend on the previous move...
  We could have an 'efficiency coeficient' of a given formula, but I 
guess that would depend on many subjective factors (ie: if you are right
or left-handed). I disagree with that, in our own cube meetings we used
to have back in '82 (I was 18 then), we accepted the Q method cause it
gave a lot of coherence in ALL formulas, and I'm not a mathemacian or
group theorist, (a program will probably do R2=R+R+ taking exactly twice
the time of one single move), eventually I rather put both counts in
parenthesis, but I definetely choose the Q method. (Look at the samples
in my 30/6 mail, I have a LOT more)

Isidro Costantini       Zappa/Hendrix/King Crimson music lover
  Olivos, Bs.As.        PC hard/soft technician


From cube-lovers-errors@curry.epilogue.com  Fri Jul  5 16:34:40 1996
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Message-Id: <199607051722.AA11168@foxtrot.rahul.net>
To: Cube-Lovers@ai.mit.edu
Subject: Cube Moves
Date: Fri, 05 Jul 96 10:22:26 -0700
From: jmc@rahul.net


  Over the last several days I noticed a number of posts about people
experimenting with moves and different combinations of moves on
the cube, using different notation. I just thought I'd mention
a java applet a wrote for a class. The applet allows you to
enter moves in Singmaster notation and view the results on a cube.
The applet uses fairly standard notation, and I put rather complete
instructions on the page. It was fun to write, and neat to play
with. If you are interesting in finding out new moves, give it
a try. Just playing around, making stuff up, I came up with
(r^b[u,l^f]r^b)^4, which uses commutators, conjugates and exponentiation,
and translaes to (fu,lu,lr) (I think that's the right answer, but I'm
not sure how it's written. Basically the move switches around three
edges). The applet also supports capital letter moves, which is a 
clockwise or counter-clockwise rotation of the whole cube, reorienting
which face is f,l,r.. etc. Read the instructions and enjoy. The 
URL is:
  http://www.reed.edu/~jmc/project/
Tell me what you think and what needs clarification.

   Justin

--
Cthulhu For President, why vote for the lesser of two evils?
http://www.cthulhu.org/jmc/


From cube-lovers-errors@curry.epilogue.com  Fri Jul  5 20:09:33 1996
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Date: Fri, 5 Jul 1996 17:02:18 -0700
To: Cube-Lovers@ai.mit.edu
From: "Larry A. Taylor" <ltaylor@kaiwan.com>
Subject: Rubik's Cube mailing list

I would be interested in descriptions of algorithms and heuristics used
for solutions of the Rubik's Cube problem.  I have a copy of the Rubik's
Math book in which (Freimaster?) the author describes some computer
work done in England or Wales (was it Thistlewaite?)  There seemed to be
no way to contact the author, trace the paper, and even a letter to the
publishers of the book go no clues to the "newsletter" mentioned.

It would be a great benefit to find out more about these computer methods.
A portion of my dissertation work is based on search over the Rubik's cube
domain.

What is the status of the legal dispute?  I was able to buy a cube in a regular
store a short while ago, after apparently being absent for many years.

LAT

Larry A. Taylor, <ltaylor@kaiwan.com>.  UCLA Computer Science Dept., Ph.D.
candidate <ltaylor@cs.ucla.edu>.  DBA North Circle Software, 13104
Philadelphia St, Suite 208, Whitter, CA 90601. Bus. phone, (310) 698-2739.
Fax (310) 698-8164.  <75176.1071@compuserve.com>,  <ltaylo934@aol.com>




From cube-lovers-errors@curry.epilogue.com  Sat Jul 13 04:22:06 1996
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Date: Sat, 13 Jul 1996 01:23:54 -0700
From: Steve Huff <dn00111@durham.net>
Organization: Huff Corp
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Well I was on the right track long ago, but now I have confirmation.

The megaminx has a slice group, analagous to the cube slice group.
All the possible spot patterns are in the megaminx's slice group,
e.g. the 10 spot and the 12 spot patterns. With process M1 we may
easily generate any spot pattern, although there is much room
for improvement.

The slice group of the megaminx is generated by turning the faces
opposite to each other in the same direction (i.e. same direction
looking at a face head-on!)

It is a small enough group to seach from head to tail, although
the exact details are still being worked on.

In the case of process M1, L is opposite to R, not just separated
by a face F, as in processes M2 and M3.

My original diagram is rather limited, but it does illustrate the
idea of L & R separated by F only (as opposed to a real opposites
but I have no satisfactory notation).


      /\
    /    \
  /        \
  \   U    /
L  \      / R
    \____/
      F

Moves for the Magic Dodecahedron (Megaminx)
-------------------------------------------

C_U = Rotate entire dodecahedron clockwise via the U face
suffix notation: f = face turns
                 u = unit turns

M1 10 spot   (L1 R3 C_U)^36    (slice group)                    (72f)

M2 3 cycle of edges (uf, lf, rf)
              R+ F+ U+ F- U- R- L- U- F- U+ F+ L+               (12f)

M3 2 flip     L-- R++ F+  U-  R+  U+  L++ R++ U+                (18f, 26u)
              R-- L-- U-  R-  U+  F-  R-- L++ U-



From cube-lovers-errors@curry.epilogue.com  Fri Jul 26 16:39:05 1996
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Message-Id: <01BB7B0E.AA2487E0@dialup-17.flnet.com>
From: Christopher Pelley <ck1@flnet.com>
To: "'Cube Lovers'" <Cube-Lovers@ai.mit.edu>
Subject: Ray-traced cubes
Date: Fri, 26 Jul 1996 16:22:18 -0400
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For those interested in three-dimensional graphics of Rubik's Cubes, I =
have made a few nice-looking images and put them at:

	http://www.flnet.com/~ck1

The images are 1024x768 JPEG files.  They were inspired by a ray-traced =
image I found on America Online a couple years ago.  There is also a =
scanned image of cube advertisements from the early 80's.  Enjoy!

Chris Pelley
ck1@flnet.com



From cube-lovers-errors@curry.epilogue.com  Mon Aug  5 22:41:01 1996
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From: bagleyd <bagleyd@hertz.njit.edu>
Message-Id: <199608060132.VAA14816@hertz.njit.edu>
Subject: panex puzzle
To: cube-lovers@ai.mit.edu
Date: Mon, 5 Aug 96 21:32:18 EDT
X-Mailer: ELM [version 2.3 PL11]

Hi
  I just made a new puzzle for the X Window System and MS Windows 3.1 or
greater.  The new puzzle is Panex which is very similar to (but a lot harder
than) the Tower of Hanoi.  In fact there is a Hanoi mode in the puzzle.
My wife who is usually indifferent towards puzzles, liked this one.

  I also made updates to my other puzzles; the rubik, dino, and skewb puzzles.
You can pick this stuff up at
http://hertz.njit.edu/~bagleyd/

Source code and README files are also supplied.

Cheers,
  /X\   David A. Bagley
 // \\  bagleyd@hertz.njit.edu   http://hertz.njit.edu/~bagleyd/
((   X  xlockmore, new stuff for xlock @ ftp.x.org//contrib/applications
 \\ //  altris, tetris games for x @ ftp.x.org//contrib/games/altris
  \X/   puzzles, magic cubes for x @ ftp.x.org//contrib/games/puzzles


From cube-lovers-errors@curry.epilogue.com  Tue Aug  6 14:37:19 1996
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Date: Tue, 06 Aug 1996 08:58:23 -0500 (EST)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Commuting Sets
To: Cube-Lovers <cube-lovers@ai.mit.edu>
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If X and Y are sets of permutations, we define XY to be the set
{xy | x in X and y in Y}.  In my various search programs, I have 
encountered a number of cases where we have XY=YX, even though we do not 
in general have xy=yx.  For example, let Q[n] be the set of all positions 
which are n quarter turns from Start.  My standard breadth first search 
is essentially Q[n+1] = Q[n]Q[1] - Q[n-1].  But we could just as well say
Q[n+1] = Q[1]Q[n] - Q[n-1] because Q[n]Q[1] and Q[1]Q[n] are the same set.

I have been wondering, what are the necessary and sufficient conditions 
for XY = YX?  Note that X and Y are not necessarily groups.

I really don't know the answer, and I wondered if anybody out there does. 
I have some suspicions it has something to do with conjugacy.  In all the
cases I have worked with, it it the case that if x in X and y in Y, then
all the K-conjugates of x are also in X and all the K-conjugates of y are
also in Y -- where K is usually M, the set of 48 rotations and reflections
of the cube.  For other searches such as <U,R>, K is the symmetry group
associated with the group being searched. 

It is trivial to make an X and Y that don't "commute" in this matter.  
That is, pick x and y that don't commute and have sets X and Y containing 
only the single elements x and y, respectively.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7127
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990



From cube-lovers-errors@curry.epilogue.com  Fri Aug 30 15:44:49 1996
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Organization:  Centro di Calcolo - Dipartimento di Informatica di Pisa - Italy
From: Mario Velucchi <velucchi@cli.di.unipi.it>
Message-Id: <199608301638.SAA11650@helen.cli.di.unipi.it>
Subject: Chameleon CUBE (I think a NEW <<< from Hungary)
To: cube <Cube-Lovers@ai.mit.edu>
Date: Fri, 30 Aug 1996 18:38:42 +0200 (MET DST)
Cc: Mario VELUCCHI <velucchi@cli.di.unipi.it>
X-Mailer: ELM [version 2.4 PL24]
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	Dear "Cube fans" friends,
in these days I have received from the Hungary the "Chameleon Cube". 
I am not very expert of cubes but I think this is a Magyar new.

For more information/references write to:
---------------
Blazsik ZOLTAN
6701 SZEGED
Pf.:1298
HUNGARY
---------------
I think this a news but ... if this cube is best known ...
i am sorry for the trouble!

Best, Mario VELUCCHI

-- 
Best Regards, MV
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\////////////////////////////////
 Mario Velucchi                              University of PISA
 Via Emilia, 106                 Department of Computer Science
 I-56121 Pisa                   e-mail:velucchi@cli.di.unipi.it
 ITALY                      talk:velucchi@helen.cli.di.unipi.it
                http://www.cli.di.unipi.it/~velucchi/intro.html
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\////////////////////////////////



From cube-lovers-errors@curry.epilogue.com  Sat Aug 31 16:01:10 1996
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Date: Sat, 31 Aug 1996 08:42:30 -0400 (EDT)
From: Assoc Prof W David Joyner <wdj@nadn.navy.mil>
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To: cube-lovers@ai.mit.edu
cc: Assoc Prof W David Joyner <wdj@coronado.nadn.navy.mil>
Subject: cube programs, etc
In-Reply-To: <199608301638.SAA11650@helen.cli.di.unipi.it>
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Hello cube lovers:
 
Several things:
1. I have written computer programs in maple
for (a) the skewb,(b) the rainbow masterball,
(c) the 3x3 Rubik's cube, and (d) the 4x4 Rubik's
cube. You must have maple (http://www,maplesoft.com)
to run it and the program simulates any move of one of 
the above puzzles using maple's 3-d graphics. The idea is
that, using one of these programs, you can
"virtually" make a move, the program draws the
cube in 3-space, and maple allows you to rotate
the cube around with your mouse (assuming you have 
the windows version of maple).
  These programs do not solve the puzzle, only
simulates the moves. It appears to be possible, with
some work, to link these programs with gap to
provide a solution as well, but I don't have the
time to do that.

2. Andrew Southern from London (whom I've lost touch with) 
and I worked out a fairly simple collection of moves to 
help solve the rainbow masterball. These are available.
   Apparently 2-cycle exist on the masterball, unlike
the Rubik's cube. We do not know of a relatively
short expression for one. If anyone out there knows
of one please let me know.

3. This stuff can be found on my www page

http://www.nadn.navy.mil/MathDept/wdj/myhome.html

under "computer programs" and "Rubik's cube like
puzzles". If there are any problems loading them
I'll try to help.
                          - David

PS: FYI, Ishi Press International has moved recently.
They are having a puzzle sale as well (their phone is
(800)859-2086 or (415)323-6996). I think they have
some cheap skewbs. Also, on a recent business
trip I stopped by Puzzletts store in downtown
Seattle - the best puzzle or game store I've ever seen.
Their www address has been posted recently
in this list so I won't repeat it.




From cube-lovers-errors@curry.epilogue.com  Mon Sep  2 19:33:10 1996
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Date: Mon, 2 Sep 1996 22:50:07 +0300 (IDT)
From: Rubin Shai <s2394459@cs.technion.ac.il>
X-Sender: s2394459@csc
To: Assoc Prof W David Joyner <wdj@nadn.navy.mil>
cc: cube-lovers@ai.mit.edu,
        Assoc Prof W David Joyner <wdj@coronado.nadn.navy.mil>
Subject: Re: cube programs, etc
In-Reply-To: <Pine.GSO.3.95.960831081024.20193D-100000@coronado>
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Hi all 
I have a computer program that learn to solve the 2X2X2 cube. I mean that 
after several hours of 'learning' the program knows to solve any legal 
start position of this cube. Before learning the program solve the cube 
after about 15 minuets, after learning it takes about 5 seconds.
The letter from Prof Joyner made me think about the following things:
1. Does anyone have a program (in C) that can take a move (a string or a 
line from a file) and show it on the display.
2. Does anyone know about similar programs to my. Program that 'learn' to 
solve the cube by themselves.
Shai



From cube-lovers-errors@curry.epilogue.com  Fri Sep  6 13:06:11 1996
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To: Rubin Shai <s2394459@cs.technion.ac.il>
cc: cube-lovers@ai.mit.edu
Subject: Re: cube programs, etc
In-Reply-To: <Pine.SV4.3.91-heb-2.04.960902222955.16485A-100000@cs.technion.ac.il>
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On Mon, 2 Sep 1996, Rubin Shai wrote:

> Hi all 
> I have a computer program that learn to solve the 2X2X2 cube. I mean that 
> after several hours of 'learning' the program knows to solve any legal 
> start position of this cube. Before learning the program solve the cube 
> after about 15 minuets, after learning it takes about 5 seconds.
> The letter from Prof Joyner made me think about the following things:
> 1. Does anyone have a program (in C) that can take a move (a string or a 
> line from a file) and show it on the display.

I have no C programs for the cube but MAPLE has a MAPLE-to-C conversion,
but one would have to write their own display.

I don't have a 2x2 program in MAPLE but I'm saving that project for a
student since it is relatively easy, given that I have one for the
3x3 and 4x4 cubes.

> 2. Does anyone know about similar programs to my. Program that 'learn' to 
> solve the cube by themselves.

This is much more serious than anything I have. My programs are simply
"virtual" cubes with no brains. Sounds like your program gives the
cube a brain! Maybe you could post more details. I don't understand
how it works.                  - David Joyner

> Shai
> 
> 
> 




From cube-lovers-errors@curry.epilogue.com  Sat Sep  7 19:51:21 1996
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Date: Fri, 6 Sep 1996 15:08:33 -0700
To: cube-lovers@ai.mit.edu
From: "Larry A. Taylor" <ltaylor@kaiwan.com>
Subject: Re: cube programs, etc
Cc: Rubin Shai <s2394459@cs.technion.ac.il>

>Hi all
>I have a computer program that learn to solve the 2X2X2 cube. I mean that
>after several hours of 'learning' the program knows to solve any legal
>start position of this cube. Before learning the program solve the cube
>after about 15 minuets, after learning it takes about 5 seconds.
>The letter from Prof Joyner made me think about the following things:
>1. Does anyone have a program (in C) that can take a move (a string or a
>line from a file) and show it on the display.
>2. Does anyone know about similar programs to my. Program that 'learn' to
>solve the cube by themselves.
>Shai

Dr. Richard Korf (korf@cs.ucla.edu) included demonstrations of macro learning
on the Rubik's Cube in his dissertation, and in his book on "Learning Macro
Operators."
He may still have his C language code for this available somewhere.

I have used the 2x2x2 and 3x3x3 cube in my work on "Pruning Duplicate Operators
in Depth-First Search." Most available format is Proceedings AAAI-93 (Wash. DC),
or from my web page area.
<http://www.cs.ucla.edu/csd-grads-gs1/ltaylor/WWW/aaai93.ps>
I do not learn to solve the cube, but learn about the cube state space to
speed search.

Neither of our programs produce graphical output.  I may make a Cube page with
a Java applet, unless one of you do it first.

LAT

Larry A. Taylor, <ltaylor@kaiwan.com>.  UCLA Computer Science Dept., Ph.D.
candidate <ltaylor@cs.ucla.edu>.  DBA North Circle Software, 13104
Philadelphia St, Suite 208, Whitter, CA 90601. Bus. phone, (310) 698-2739.
Fax (310) 698-8164.  <75176.1071@compuserve.com>,  <ltaylo934@aol.com>




From cube-lovers-errors@curry.epilogue.com  Wed Sep 11 17:04:33 1996
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Organization:  Centro di Calcolo - Dipartimento di Informatica di Pisa - Italy
From: Mario Velucchi <velucchi@cli.di.unipi.it>
Message-Id: <199609111154.NAA11311@helen.cli.di.unipi.it>
Subject: Chameleon Cube (E-Mail address)
To: cube <Cube-Lovers@ai.mit.edu>
Date: Wed, 11 Sep 1996 13:54:03 +0200 (MET DST)
X-Mailer: ELM [version 2.4 PL24]
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after my precedent e-mail i have received more answers/questions for to
know the E-mail address of my hungarian friend, this is the old
message with the E-mail address:


	Dear "Cube fans" friends,
in these days I have received from the Hungary the "Chameleon Cube". 
I am not very expert of cubes but I think this is a Magyar new.

For more information/references write to:
---------------
Blazsik ZOLTAN
6701 SZEGED
Pf.:1298
HUNGARY
---------------
blazsik@inf.u-szeged.hu
---------------

I think this a news but ... if this cube is best known ...
i am sorry for the trouble!

Best, Mario VELUCCHI


-- 
Best Regards, MV
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\////////////////////////////////
 Mario Velucchi                              University of PISA
 Via Emilia, 106                 Department of Computer Science
 I-56121 Pisa                   e-mail:velucchi@cli.di.unipi.it
 ITALY                      talk:velucchi@helen.cli.di.unipi.it
                http://www.cli.di.unipi.it/~velucchi/intro.html
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\////////////////////////////////



From cube-lovers-errors@curry.epilogue.com  Thu Sep 26 22:37:26 1996
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From: Mario Velucchi <velucchi@cli.di.unipi.it>
Message-Id: <199609261617.SAA20286@helen.cli.di.unipi.it>
Subject: WWW devoted to Recreational Mathematics (CUBE, too ...)
To: cube <Cube-Lovers@ai.mit.edu>
Date: Thu, 26 Sep 1996 18:17:51 +0200 (MET DST)
X-Mailer: ELM [version 2.4 PL24]
MIME-Version: 1.0
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	Dear Friend,
are you interested to Recreational Mathematics and related fields?
If so, see this my new WWW address:

http://www.geocities.com/SiliconValley/9174/material.html

I think you will find a lot of interesting items.
Best Regards, Mario VELUCCHI

-- 
Best Regards, MV
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\////////////////////////////////
 Mario Velucchi                              University of PISA
 Via Emilia, 106                 Department of Computer Science
 I-56121 Pisa                   e-mail:velucchi@cli.di.unipi.it
 ITALY                      talk:velucchi@helen.cli.di.unipi.it
                http://www.cli.di.unipi.it/~velucchi/intro.html
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\////////////////////////////////



From cube-lovers-errors@curry.epilogue.com  Mon Sep 30 23:21:04 1996
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Date: Mon, 30 Sep 1996 22:29:36 -0500 (EST)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Solving One Cubie
To: Cube-Lovers <cube-lovers@ai.mit.edu>
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I've been thinking about a simple little problem I thought I would share. 
Most of the solution is in the archives, but under other guises. 

Suppose you scramble a cube and give it to a cubemeister with instructions
to solve any one cubie.  This is a truly trivial problem, but let's see
what it can teach us. 

The most obvious question is -- what is God's algorithm?  That is, from
any position, what is the minimal solution?  The cubemeister would observe
that for any position, each of the eight corner cubies and each of the
twelve edge cubies has its own individual minimal solution which is easy
to discover.  The cubemeister would then choose the cubie with the
smallest minimal solution and solve it. 

Given this simple technique for God's algorithm, what is the maximal
position?  That is, what is the position where the minimal solution is as
large as possible? 

We start with the edges.  The solution is in the archives in two separate
articles.  On 6 August 1980, David Vanderschel introduced the concept of
Oriented Distance from Home (ODH).  On 7 January 1981, Dan Hoey used the
ODH concept to show that the Pons Asinorum position requires exactly
twelve quarter turns for solution.  But for our purposes, the salient
point is that an edge cubie can be at most four quarter-turns from home. 
There is exactly one such position for each edge cubie.  And the only
position for which each edge cubie is four quarter-turns from home is the
Pons.  So for our trivial little problem, the maximal position for the
edges is the Pons. 

I have found little information in the archives concerning the same
problem for the corners.  (By the way, I have this vision in my mind that
the information for the corners is in there somewhere, but I cannot find
it, neither in the archives nor in Singmaster.  Am I remembering a mirage,
or is it in there somewhere and I can't find it?). 

Vanderschel does not define an Oriented Distance from Home for corners,
but the generalization is obvious.  The following are the ODH values for
the f facelet of the flt cubie. 

        1+2
        +T+
        2+3

   l+2  0+1  1+2  2+3
   +L+  +F+  +R+  +B+
   2+3  1+2  2+3  3+2

        1+2
        +D+
        2+3

The maximum distance from Start for any particular corner cubie is
therefore three quarter-turns.  The question then is whether all eight
corner cubies can be three quarter-turns from Start simultaneously.  There
are probably a number of ways which will work, but the following works
very nicely.  Place each corner cubie in its diametrically opposed corner
cubicle.  For example, place the flt cubie in the bdr cubicle.  The twist
doesn't matter for the individual cubies, except that the overall
configuration for the eight corner cubies must conserve twist.  The reason
that twist doesn't matter is that when a corner cubie is in its
diametrically opposed corner cubicle, all three twists are conjugate (see
below). 

The maximal position for the corners can peacefully co-exist with the Pons
for the edges.  That is, if each corner cubie is in its diametrically
opposed corner cubicle, the parity of the corners is even (as is the
Pons). 

In a certain sense, God's algorithm for a single corner cubie is identical
to God's algorithm for the 1x1x1 cube, which is to say, it is identical to
God's algorithm for the rotation group of the cube (which we normally
denote by C).  (See my note of 14 Nov 1995.)

Here is how it works.  Consider any particular corner cubie such as flt,
and consider any sequence of quarter-turns such as TL where each
quarter-turn moves the cubie in question.  Then, the "same" sequence of
whole cube rotations (tl, in this case) will have the same effect on the
same corner cubie.  Here, we are using the lower case letters t and l to
denote whole cube quarter-turns and the upper case letters T and L to
denote the face quarter-turns. 

The converse is also true if we are careful.  That is, each whole cube
quarter-turn may be denoted in two ways.  For example, t is the same as
d'.  To convert from whole cube rotations back to quarter-turn face turns,
we would convert t to T or to D' depending on whether the cubie in
question were on the Top face or the Down face at the time. 

The same trick does not work for the edges.  The problem is that face
turns and whole cube turns are not fully interchangeable.  For instance, T
and t are interchangeable for the Top edge cubies, as are D and d for the
Down edge cubies.  But there is no equivalent interchange for the
"equator" of edge cubies fl, lb, br, and rf.  (Well, maybe you could do it
if you allowed slice moves, but we are not working with slice moves.)

I am always interested in symmetry, usually as represented by conjugacy. 
For whole cube rotations, there are five conjugacy classes.  (Again, see
my note of 14 November 1995.) For individual cubies, we define conjugacy
as follows.  Let X and Y be functions (not permutations) which are the
restriction of normal permutations to the cubie in question.  Then X and Y
are conjugate if m'Xm=Y for some m in M, the set of 48 rotations and
reflections of the cube.  m' must be restricted to the pre-image of the
domain of X, and m must be restricted to the range of X.  With the various
permutations thus restricted to functions on the single flt cubie, the
conjugacy classes are as follows: 

   1. I
   2. F, F', L, L', T, T'
   3. FF, LL, TT
   4. TL', TB, FT', FR, LF', LD
   5. TL, L'T'
   6. FRR, LDD, TBB
   7. FTT, LFF, TLL

Note that if we treat all the moves as whole cube permutations rather than
as functions on the flt cubie, then #4 and #5 are collapsed down into a
single conjugacy class, as are #6 and #7.  Then, the conjugacy classes are
the same as the ones for the 1x1x1 cube. 

When I first started working on this little problem, I thought the
conjugacy classes for a single cubie might provide a non-arbitrary frame
of reference for defining twist.  They almost do, but not quite. 

       a. When the cubie is in its home cubicle, its twist is
          obvious.  However, we can observe that I, TL, and L'T'
          place the flt cubie in the flt cubicle.  TL and L'T'
          are conjugate, but they are not conjugate to I.  Hence,
          it is natural to take I as the untwisted state.

       b. When the cubie is immediately adjacent to its home
          cubicle (there are three such cubicles), the conjugacy
          classes can be used to define twist.  For example, the
          flt cubie is placed into the ftr cubicle by F, T', and
          by LFF.  F and T' are conjugate, but they are not
          conjugate to LFF.  Hence, we can take LFF as the
          untwisted state.

     c.   When the cubie is immediately adjacent to the
          diametrically opposed cubicle (there are three such
          cubicles), the conjugacy classes can be used to define
          twist.  For example, the flt cubie is placed into the
          frd cubicle by FF, LD, and by LF'.  LD and LF' are
          conjugate, but they are not conjugate to FF.  Hence, we
          can take FF as the untwisted state.

     d.   When the cubie is in the diametrically opposed cubicle
          (there is only one such cubicle), I don't see any way
          to use the conjugacy classes to define twist.  All
          three twists are conjugate, and hence none is
          inherently different from the other two.  For example,
          FRR, LDD, and TBB are all conjugate.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7127
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990






From cube-lovers-errors@curry.epilogue.com  Tue Oct  1 14:32:11 1996
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Date: Tue, 1 Oct 1996 17:19:48 +0100
From: Riccardo Distasi <ric@irsip.na.cnr.it>
Message-Id: <9610011619.AA00774@irsip.na.cnr.it>
To: Cube-Lovers <cube-lovers@ai.mit.edu>
Subject: Intro to cube group theory?

Dear mathematical cubologists,
I am creeping on this list since a few months, but I have to admit
that most of the more advanced mathematical technicalities are beyond
my understanding, mainly because I lack knowledge of the basic
facts and terminology about groups.

All I studied about groups was a part of Birkhoff/McLaine's "Algebra"
some 10 years ago. Is there any good reference on groups where I can
educate myself?  I would prefer freeware papers over costly and
hard-to-find (at least in Italy) books. Does anybody have a hint for
me?  The aim of my training is that of learning about M-conjugacy and
the Shamir algorithm, and to be able to follow the technical
discussions about the Rubik cube that appear on this list.

	Riccardo
--
Riccardo Distasi, ric@irsip.na.cnr.it


From cube-lovers-errors@curry.epilogue.com  Tue Oct  1 19:36:41 1996
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Date: Tue, 1 Oct 1996 18:43:19 -0400 (EDT)
From: Assoc Prof W David Joyner <wdj@nadn.navy.mil>
X-Sender: wdj@coronado
To: Riccardo Distasi <ric@irsip.na.cnr.it>
cc: Cube-Lovers <cube-lovers@ai.mit.edu>
Subject: Re: Intro to cube group theory?
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On Tue, 1 Oct 1996, Riccardo Distasi wrote:

> Dear mathematical cubologists,
> I am creeping on this list since a few months, but I have to admit
> that most of the more advanced mathematical technicalities are beyond
> my understanding, mainly because I lack knowledge of the basic
> facts and terminology about groups.
> 
> All I studied about groups was a part of Birkhoff/McLaine's "Algebra"
> some 10 years ago. Is there any good reference on groups where I can
> educate myself?  I would prefer freeware papers over costly and
> hard-to-find (at least in Italy) books. Does anybody have a hint for
> me?  The aim of my training is that of learning about M-conjugacy and
> the Shamir algorithm, and to be able to follow the technical
> discussions about the Rubik cube that appear on this list.
> 

I think the best book is Bandelow's Inside Rubik's cube and beyond,
which might be in a local library. I have lecture notes for a
course I'm teaching on the Rubik's cube which I can send you
for free. Also, a group-theorist friend of mine has several
hundred copies of an elementary group theory book (printed
by the US government and I think free) available - you can
email me or him (Prof Gaglione, amg@nadn.navy.mil) if you're
interested. Finally, Puzzletts is still selling Singmaster's 
Notes on the Rubik's cube, though they are also out of
print. - David Joyner


> 	Riccardo
> --
> Riccardo Distasi, ric@irsip.na.cnr.it
> 
> 



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From: Timo Berry <taberry@uiah.fi>
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To: Cube-lovers@ai.mit.edu
Subject: An amateur humbly approaches
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Dear Sir(s)!

I'm a student of graphic design in the University of Industrial Arts in
Helsinki. I'm working on a school project and I need some information on
Rubik's Cube. I'm making a piece that takes advantage of the visual
language of the cube, but I need to know more. I'm familiar with the game
from years back but I never really learned how the actual mechanism worked,
nor anything on the history of the game.

I would appreciate a few hints on the vast amount of home pages on the
subject. What I'm mostly looking for is a basic, no-nonsense explanation
of the history and the philosophy and especially the mechanism  of the
cube (I'd hate to take my only cube apart!). Any illustrations of the
mechanism would be great. Are there any working pictures or plans
available?

Sincerely Yours,
Timo Berry

taberry@uiah.fi





From cube-lovers-errors@curry.epilogue.com  Wed Oct  2 14:55:19 1996
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I am located in Austin, Tx. I was wondering where I can purchase a rubik's 
cube. If you could help me out I would greatly appreciate it.





From cube-lovers-errors@curry.epilogue.com  Wed Oct  2 16:52:14 1996
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To: C-Money <C-Money@mail.utexas.edu>
cc: Cube-Lovers@ai.mit.edu
Subject: cubes in austin
In-reply-to: Your message of "Wed, 02 Oct 1996 01:13:22 CDT."
             <199610020614.BAA01730@mail.utexas.edu> 
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Date: Wed, 02 Oct 1996 14:56:54 -0500
From: Norman Richards <orb@cs.utexas.edu>

> I am located in Austin, Tx. I was wondering where I can purchase a rubik's 
> cube. If you could help me out I would greatly appreciate it.

    I just bought a new one a few weeks ago at the Kay-bee toy store in
Highland Mall.  They had about 6 cubes last friday when I was there.  They
also have some triamids and a couple snakes left.  No magic or mini-cube's
though. :(

  The Imaginarium in Higland Mall had the Rubik's C4 cube a while back (the
one where you have to align the centers of 4 faces also), but I was quite
surprised to find the store closed when I went to the mall last week. :(

  All the Rubik's stuff goes for $10 a pop.  If you find a better price
somewhere else, let me know.  Also, if you happen to see any stores here 
that have the mini-cube, please let me know!  


______________________________________________________________________________
orb@cs.utexas.edu                                              soli deo gloria




From cube-lovers-errors@curry.epilogue.com  Wed Oct  2 21:06:02 1996
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From: der Mouse <mouse@holo.rodents.montreal.qc.ca>
Message-Id: <199610030014.UAA03565@Collatz.McRCIM.McGill.EDU>
To: cube-lovers@ai.mit.edu
Subject: Re: cube programs, etc

> Date: Mon, 2 Sep 1996 22:50:07 +0300 (IDT)

Guess who's going through backed-up mail... :-)

> I have a computer program that learn to solve the 2X2X2 cube.  I mean
> that after several hours of 'learning' the program knows to solve any
> legal start position of this cube.  Before learning the program solve
> the cube after about 15 minuets, after learning it takes about 5
> seconds.

Interesting.  Is the program available?

> The letter from Prof Joyner made me think about the following things:
> 1. Does anyone have a program (in C) that can take a move (a string
>    or a line from a file) and show it on the display.

Well, I have something of the sort, though it's for the 3-Cube.  For
example, here's first turning the R face once, then illustrating the
Spratt wrench, first defining a slice turn (and checking it), then
using it to write the wrench more simply than it would be if done
directly with the primitives.

% twist
> R
Cube:
              u u f
              u u f
              u u f
        l l l f f d r r r u b b
        l l l f f d r r r u b b
        l l l f f d r r r u b b
              d d b
              d d b
              d d b
Cycles: (ur,br,dr,fr) (ubr,bdr,dfr,fur) [4]
Already centered
> .set SLICER CUBER R' L
`SLICER' defined
> SLICER
Cube:
              u f u
              u f u
              u f u
        l l l f d f r r r b u b
        l l l f d f r r r b u b
        l l l f d f r r r b u b
              d b d
              d b d
              d b d
Cycles: (u,b,d,f) (ub,bd,df,fu) [4]
Centred: (ul,fl,dl,bl) (ur,fr,dr,br) (ulb,flu,dlf,bld) (ubr,fur,dfr,bdr) [4]
> (SLICER U) 4
Cube:
              u b u
              l u u
              u u u
        l u l f f f r r r b u b
        l l l f f f r r r b b b
        l l l f d f r r r b d b
              d f d
              d d d
              d b d
Cycles: (ub)+ (ul)+ (fd)+ (bd)+ [2]
Already centered
> 

> 2. Does anyone know about similar programs to my. Program that
>    'learn' to solve the cube by themselves.

Someone I know once wrote such a program in Lisp.  (Incidentally, this
was also one of the most stunning examples of hot-spot hand-tuning I
ever saw.  It represented the cube as a bunch of conses pointing to one
another, no leaves at all.  The "apply a rotation" call worked by
juggling links with rplaca and rplacd.  I rewrote this one call in
assembly (approximately the same number of lines of code, incidentally)
and got three orders of magnitude, a factor of a thousand, speed
improvement in the overall program.)

The program was somewhat interesting in that it solved the cube by
experimenting and discovering macros, somewhat akin to the way humans
tend to.

I don't know whether this program still exists anywhere.  If anyone
cares I can try to find out.

					der Mouse

			       mouse@rodents.montreal.qc.ca
		     01 EE 31 F6 BB 0C 34 36  00 F3 7C 5A C1 A0 67 1D


From cube-lovers-errors@curry.epilogue.com  Thu Oct  3 01:05:43 1996
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To: Assoc Prof W David Joyner <wdj@nadn.navy.mil>
From: Georges Helm <geohelm@pt.lu>
Subject: Re: Intro to cube group theory?
Cc: Cube-Lovers <cube-lovers@ai.mit.edu>

At 18:43 01/10/1996 -0400, you wrote:

>I think the best book is Bandelow's Inside Rubik's cube and beyond,
>which might be in a local library. I have lecture notes for a
>course I'm teaching on the Rubik's cube which I can send you
>for free. Also, a group-theorist friend of mine has several
>hundred copies of an elementary group theory book (printed
>by the US government and I think free) available - you can
>email me or him (Prof Gaglione, amg@nadn.navy.mil) if you're
>interested. Finally, Puzzletts is still selling Singmaster's 
>Notes on the Rubik's cube, though they are also out of
>print. - David Joyner
>

I think a very helpful book is

Handbook of Cubic Math by
Alexander H. Frey, Jr. + David Singmaster

Contents:

Preface 
   1.Introduction 
   2.A Cubik Orientation 
   3.Restoring the Cube 
   4.The What, Why, and How of Cube Movements 
   5.Improved Restoration Processes 
   6.The Cube Group and Subgroups 
   7.Permutation Structures and the Order of Groups 
   8.Advanced Restoration Methods 
   9.Epilogue 
A. A Small Catalogue of Processes 
B. Solutions to Exercises 
Index 

It was published by Enslow publishers

Georges
         
geohelm@pt.lu
http://www.geocities.com/Athens/2715
http://ourworld.compuserve.com/homepages/Georges_Helm



From cube-lovers-errors@curry.epilogue.com  Thu Oct  3 14:20:38 1996
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From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: Intro to cube theory?
To: Cube-Lovers <cube-lovers@ai.mit.edu>
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> On Tue, 1 Oct 1996, Riccardo Distasi wrote:

>                                Does anybody have a hint for
> me?  The aim of my training is that of learning about M-conjugacy and
> the Shamir algorithm, and to be able to follow the technical
> discussions about the Rubik cube that appear on this list.

Several good books have already been mentioned here, so I thought I 
would try briefly to answer your specific questions rather than listing 
books again.

I doubt you are going to find any references to M-conjugacy in any Group
Theory Books, nor even in any books that are specific to the Cube.  What
you will find is discussions of conjugacy.  The conjugate of X by Y is
defined either as Y'XY or as YXY'.  Here I am following the E-mail
convention that Y' means Y^(-1) or "Y inverse".  We use Y' because it is
hard to write a proper superscript 1 on E-mail. 

One reason for the different definitions for conjugacy (Y'XY vs. YXY') may
be that some authors use a right-to-left definition for group operators
and some use a left-to-right.  (Cube-Lovers uses left-to-right almost
exclusively).  But I think that even with a consistent left-to-right 
convention, you fill find differences between authors in their definition 
of conjugacy.  I think I remember a discussion in Singmaster about why 
some authors do it one way and others do it the other.  The best I 
recall, both ways of doing it make sense in the proper context.  I will 
try to chase down the reference and post a followup.

I don't think it makes much difference which convention you use as long as
you are consistent.  If Y'XY is a conjugate, then YXY' is also.  That is,
if Y'XY is the conjugate of X by Y, then YXY' is the conjugate of X by Y'. 

Frey and Singmaster use the YXY' convention.  Cube-Lovers (including the
things I have posted) primarily uses the Y'XY convention.  I actually
think the YXY' convention makes more sense.  Roughly speaking, it means to
do one thing, then to do a second thing, and finally to undo the first
thing. The effect is essentially to do the second thing, but to do it
shifted by the first thing. 

For example, suppose you know how to do something to the Top layer of the
cube but you don't know how (or find it awkward) to do the same thing to
the Down layer of the cube.  What you could do is turn the cube upside
down, perform your operation on the Top layer, and then turn the cube
right side up.  You will have performed your operation on the Down layer. 

In Cube-Lovers, we would probably write this as cXc'.  We call the set of
twenty-four rotations of the cube C, and c would be one of the elements of
C that turns the cube upside down.  So c would turn the cube upside down,
X would be your operation, and c' would restore the cube to right side up. 

Except that we would really write it as c'Xc, which in some ways makes no 
sense.  I read it as undo the first thing, then do the second thing, and 
finally do the first thing.  I really do have to chase down Singmaster's 
explanation of why this makes sense.

I confess I struggle with the real geometric significance of Y'XY.  That
is, if we have Z=Y'XY, then what is the relationship between X and Z? 
They have the same cycle structure, but that is about as far as I get in a
geometric interpretation.  Here I am assuming that each of X, Y, and Z are
in the cube group. 

But I find c'Xc or m'Xm easy to interpret.  In Cube-Lovers convention, M
is the set of forty-eight rotations and reflections of the cube to go
along with C as the set of twenty-four rotations of the cube.  So C is a
subset of M and C-conjugacy is a subset of M-conjugacy.  But we nearly
always talk about M-conjugacy. 

But C and M are not really in the cube group G as we usually define it. 
That is, the standard model for G is a fixed face center model where we do
not rotate the whole cube.  To use Group Theory properly with M-conjugacy,
we have to deal with M-conjugacy in terms of a larger group which is
sometimes called MG or G+M.  MG includes all the face turns, rotations,
and reflections of the cube. However, it is the case that if X is in G,
then so too is m'Xm.  So if we want to, we can treat M-conjugacy as a
function on G without having to expand our group to MG. 

Many Group Theory books will talk about symmetry.  A symmetry is just a
special kind of permutation which preserves some kind of property, usually
a geometric property.  For example, there are eight symmetries of a
square.  A square can be rotated in four different ways and still look the
same, and each of the four rotations can be turned inside out.  You can
also think of the "turned inside out" versions as being mirror images, so
they are called reflections.  Similarly, a cube has twenty-four rotations
and twenty-four reflections as symmetries.  This was true long before
Rubik's cube was invented, and you will find discussions of the symmetries
of the cube in books that were written before Rubik's cube was invented. 

Cube-Lovers simply calls the set of forty-eight symmetries of the cube M
on a fairly consistent basis, and so M-conjugacy is born.  It is really
just conjugacy by the symmetries of the cube.  M-conjugacy is important
because it identifies positions which are "really the same", even if they
may look different superficially.  That is, if Y=m'Xm for some m in M,
then X and Y look the same except that they may be rotated or recolored
with respect to each other.  In particular, X and Y may be solved in the
"same way", and each will require the same number of moves for solution. 

My view of Shamir's method is that it really has nothing to do with Group
Theory.  Rather, it has to do with data structures and information theory. 
There are several components of Shamir's method, but the most important is
addressing the following problem.  

Suppose you have a collection of objects in a computer program in some
arbitrary (possibly "random"  order), and suppose you want to eliminate
any duplicate objects to make the collection into a true set in a
mathematical sense.  Almost any algorithm you come up with is equivalent
to sorting the objects to place the duplicate occurences adjacent to each
other, and then scanning the collection front to back to identify the
duplicates.  Now you may not literally sort.  You may build trees, hash 
tables, or any of a number of interesting and efficient structures, but 
they all reduce to sorting at the conceptual level.

A variation on this theme is suppose you have two (or more) such
collections, and you want to eliminate all duplicate objects.  At the
conceptual level, almost any algorithm you come up with is equivalent to
sorting each collection, and then merging and matching the sorted
collections. 

Shamir's method provides a very efficient way to accomplish this
"sorting".  Given a collection of objects which is sorted already, it lets
you create a second collection which is sorted in a totally different 
way, without any of the objects moving in memory  --  by simply 
traversing a search tree in a clever way.

The issue arises in search programs for Rubik's cube because you often 
have a set of cube positions which you need to compose with another 
position or set of positions.  When you are done, you need to "sort and 
match" or "merge and match" the results.  Literally sorting and merging 
can take ridiculous amounts of time and memory.  If the first set of 
positions is already sorted, Shamir's method tells us how to compose the 
first set of positions with other positions in such a way that the newly 
generated sets of positions come out automagically in the right order, 
with no additional sorting required.  Much more detail than this is 
available the the Cube-Lovers archives.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7127
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990



From cube-lovers-errors@curry.epilogue.com  Fri Oct  4 17:33:19 1996
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To: Cube-Lovers-Request@ai.mit.edu
From: Joaquim Folch <estelada@redestb.es>
Subject: Rubik revenge (4x4)
Date: Fri, 4 Oct 1996 12:03:40 +0100

>
>>Dear Sir:
>>
>>I=B4m Joaquin Folch (Barcelona-Spain). I have a blind friend who needs an
>>Rubik cube, large size whith 16 squares per side (Revenge type, 4x4=
 lines),
>>because his actual cube is broken. He played very well and fast whith his
>>marked cube.
>>Please tell me how much I have to pay for it.=20
>>Please answer me. Many thanks to all,   Joaquin.
>>My adress:                      EMail: estelada@redestb.es
>>
>>Joaquin Folch
>>Espigol 6
>>08328 Alella (Barcelona)
>>Spain=20
>>
>
>
>
>
>
>>
>



From cube-lovers-errors@curry.epilogue.com  Wed Oct 16 14:01:16 1996
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Date: Wed, 16 Oct 1996 08:54:11 -0500
To: cube-lovers@ai.mit.edu
From: Peter Beck <pbeck@pica.army.mil>
Subject: largest cube

HI folks,

I would like to revisit the question of
what is the largest (number of slices)
size cube that can be made.  As I
recollect the center spindle mechanism
has been analyzed and the conclusion
was a 5x5x5.

There is a new mechanism used in the
MOZAIKA puzzle (info below).  I wonder
if anybody has analyzed it to determine
what configurations of cubes it could
be used for.

PS  This mechanism also seems to answer
the question of whether or not the cube
is a sliding block puzzle on a spherical
surface or a solid rotating puzzle.


********************************
*  "MOZAIKA" is a spherical sliding block
*  puzzle like Rubik's cube with a new
*  mechanism.  I have only seen the 3x3x3
*  version.  It has 2 types of pieces
*  (the third, a sphere in the center appears to be
*  unnecessary):  a triangular piece analogous
*  to the cubes corner and a rectangular piece
*  analogous to the center piece , using 2 of
*  these to make an edge piece.  The puzzle
*  thus has 3 orthogonal equators made up of
*  the rectangular pieces and the corners.
*
*  These pieces interlock to form a spherical
*  surface - the center is hollow.  The interlock
*  method is that the corner pieces have a rail
*  that the rectangular pieces ride on.  The
*  corner pieces are held in space by the
*  rectangular pieces (sorry for poor description).
********************************
*  FROM:  J&R DESIGNS
*       1126 SOUTH STREET
*        POB 315
*        NILES, MICHIGAN 49120
*  COST:  US $15
*    + $3 POSTAGE USA OR $5 OVERSEAS
*
********************************

   THE FUTURE IS PUZZLING, but CUBING IS FOREVER !!!

Peter Beck,aka, Just Puzzles,  201-625-4191 answering machine

  a cube WEB site;2/27/96 - <http://admin.dis.on.ca:80/~cubeman/>
......................................................
my  career site - updated 5/31/96 <http://qa.pica.army.mil:80/~pbeck/>
......................................................
last modified  31 May 1996




From cube-lovers-errors@curry.epilogue.com  Wed Oct 16 22:51:41 1996
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Date: Thu, 17 Oct 1996 01:49:55 +0200
From: Dik.Winter@cwi.nl
Message-Id: <9610162349.AA04173=dik@bever.cwi.nl>
To: cube-lovers@ai.mit.edu, pbeck@pica.army.mil
Subject: Re: largest cube

 > I would like to revisit the question of
 > what is the largest (number of slices)
 > size cube that can be made.  As I
 > recollect the center spindle mechanism
 > has been analyzed and the conclusion
 > was a 5x5x5.

If I remember well the limit was not order 5 but 6, and not due to
the mechanism but only because during turning the corner cubes will
extend so much outside the cube that they are held by only 2 neighbours.
*But* this holds only if your requirement is that all cubelets have the
same size.  When you allow cubelets to grow when going from the center
you can get larger (although I think even in that case there will be
a limit, right now I am too lazy to think about it even further).

dik
--
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn  amsterdam, nederland; http://www.cwi.nl/~dik/


From cube-lovers-errors@curry.epilogue.com  Wed Oct 23 14:06:47 1996
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Organization:  Centro di Calcolo - Dipartimento di Informatica di Pisa - Italy
From: Mario Velucchi <velucchi@cli.di.unipi.it>
Message-Id: <199610231533.RAA02193@helen.cli.di.unipi.it>
Subject: Re: DEAR TANOFF <<<<<<<<<<<<<<<< (fwd)
To: cube <Cube-Lovers@ai.mit.edu>
Date: Wed, 23 Oct 1996 17:33:50 +0200 (MET DST)
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Forwarded message:
> From velucchi@CLI.DI.Unipi.IT  Wed Oct 23 17:32:34 1996
> Subject: Re: DEAR TANOFF <<<<<<<<<<<<<<<<
> To: TANOFF%SMOOKE@BIOMED.MED.YALE.EDU
> Date: Wed, 23 Oct 1996 17:32:20 +0200 (MET DST)
> 
> > 
> > What is the Siamese Cube?
> > 
> 
> Two (usual/normal) Rubik Cubes in One ...
> 
> 
>    ------
>    |    |
>    |    |
>    -----+-----
>         |    |
>         |    |
>         ------
> 
> The goal is equal to normal cube but the moves are differents ...
> because the two cubes are "uniti" ....
> Do You understand? let me know!
> 
> 
> Sorry for my English and my Picture!
> 
> 
> -- 
> Best Regards, MV
> \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\////////////////////////////////
>  Mario Velucchi                              University of PISA
>  Via Emilia, 106                 Department of Computer Science
>  I-56121 Pisa                   e-mail:velucchi@cli.di.unipi.it
>  ITALY                      talk:velucchi@helen.cli.di.unipi.it
>                 http://www.cli.di.unipi.it/~velucchi/intro.html
> \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\////////////////////////////////
> 


From cube-lovers-errors@curry.epilogue.com  Thu Oct 24 16:27:10 1996
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To: Mario Velucchi <velucchi@cli.di.unipi.it>
From: "The Official Thermo-Fluids Fan Club of the UK. (Andy Southern)" <a.southern@ic.ac.uk>
Subject: Re: DEAR TANOFF <<<<<<<<<<<<<<<< (fwd)
Cc: Cube-Lovers@ai.mit.edu

At 17:33 23/10/96 +0200, you wrote:
>Forwarded message:
>> From velucchi@CLI.DI.Unipi.IT  Wed Oct 23 17:32:34 1996
>> Subject: Re: DEAR TANOFF <<<<<<<<<<<<<<<<
>> To: TANOFF%SMOOKE@BIOMED.MED.YALE.EDU
>> Date: Wed, 23 Oct 1996 17:32:20 +0200 (MET DST)
>> 
>> > 
>> > What is the Siamese Cube?
>> > 
>> 
>> Two (usual/normal) Rubik Cubes in One ...
>> 
>> 
>>    ------
>>    |    |
>>    |    |
>>    -----+-----
>>         |    |
>>         |    |
>>         ------
>> 
>> The goal is equal to normal cube but the moves are differents ...
>> because the two cubes are "uniti" ....
>> Do You understand? let me know!
>> 
>> 
>> Sorry for my English and my Picture!
>> 
>> 
>> -- 
>> Best Regards, MV
>> \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\////////////////////////////////
>>  Mario Velucchi                              University of PISA
>>  Via Emilia, 106                 Department of Computer Science
>>  I-56121 Pisa                   e-mail:velucchi@cli.di.unipi.it
>>  ITALY                      talk:velucchi@helen.cli.di.unipi.it
>>                 http://www.cli.di.unipi.it/~velucchi/intro.html
>> \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\////////////////////////////////
>> 
>
>
>
I think I understand.

There are two cubes, orientated the same, which share a common corner piece.
The shared corner piece has no stickers on it, but is a "Double Inside"
corner piece. The effect is that they share the same line from corner to
corner, passing through the dead centre of the cube. The appearence would be
like a (5x5x5) which had been cut away. 

There would be a cubie at the locations:
(1,1,1),(1,1,2),(1,1,3),(1,2,1),(1,2,2),(1,2,3),(1,3,1),(1,3,2),(1,3,3),
(2,1,1),(2,1,2),(2,1,3),(2,2,1),(2,2,2),(2,2,3),(2,3,1),(2,3,2),(2,3,3),
(3,1,1),(3,1,2),(3,1,3),(3,2,1),(3,2,2),(3,2,3),(3,3,1),(3,3,2),(3,3,3),(3,3
,4),(3,3,5),(3,4,3),(3,4,4),(3,4,5),(3,5,3),(3,5,4),(3,5,5),
(4,3,3),(4,3,4),(4,3,5),(4,4,3),(4,4,4),(4,4,5),(4,5,3),(4,5,4),(4,5,5),
(5,3,3),(5,3,4),(5,3,5),(5,4,3),(5,4,4),(5,4,5),(5,5,3),(5,5,4),(5,5,5),

These cubes would *not* rotate about the apparent centre (3,3,3), but about
the two real centres (4,4,4) and (2,2,2).

I could see there being a few perceptual problems. The conecting cubie at
(3,3,3) would have no colour stickers on it, hence position and rotation
must be determined from the other corners. The cube would also appear to the
operator to turn only the outer slice and middle slice of each cube because
the operator would always use the centre of mass as his/her frame of
referance. That is different to the standard (3x3x3) because the operator
feels the outer slices move.

sorry if this is either wrong or nothing new, I just thought I'd share my
thoughts with you.

Andrew Southern



From cube-lovers-errors@curry.epilogue.com  Fri Oct 25 01:26:18 1996
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Date: Fri, 25 Oct 96 14:00:53 +0900
From: Norman Diamond 25-Oct-1996 1355 <diamond@jrdv04.enet.dec-j.co.jp>
To: cube-lovers@ai.mit.edu
Apparently-To: cube-lovers@ai.mit.edu
Subject: Siamese Rubik's Cubes (was Re: DEAR TANOFF <<<<<<<<<<<<<<<< (fwd))
Mime-Version: 1.0
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A. Southern misinterpreted M. Velucchi's picture:

>>    ------
>>    |    |
>>    |    |
>>    -----+-----
>>         |    |
>>         |    |
>>         ------

Siamese Rubik's cubes share an entire column of cubies, i.e. in the
case of two 3x3x3's they share an edge cubie and two corner cubies.

Cubies cannot move from one cube to the other.  The shared column
of cubies cannot be separated or rearranged.  The effect is like
bandaging an edge column on one 3x3x3 cube and bandaging an edge
column on another 3x3x3 cube and having two identical puzzles.

The idea of bandaging has been extended further by Dieter Gebhardt
(publications in CFF) and others.  Most variations of bandaging
cannot be constructed by joining another cube onto it; they just
have to be done in a simpler and straightforward manner :-)
And even when a collector wants duplicates of some version, there's
no need for two duplicates to be stuck to each other :-)  So there
is no real demand for Siamese cubes any more.  But bandaged cubes,
yeah some variations are really really difficult.

-- Norman Diamond             diamond@jrdv04.enet.dec-j.co.jp
[Speaking for Norman Diamond not for Digital.]


From cube-lovers-errors@curry.epilogue.com  Fri Oct 25 16:08:00 1996
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From: Wei-Hwa Huang <whuang@ugcs.caltech.edu>
Subject: Re: Siamese Rubik's Cubes (was Re: DEAR TANOFF <<<<<<<<<<<<<<<< (fwd))
Date: 25 Oct 1996 14:02:32 GMT
Organization: California Institute of Technology, Pasadena
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Norman Diamond 25-Oct-1996 1355 <diamond@jrdv04.enet.dec-j.co.jp> writes:
>A. Southern misinterpreted M. Velucchi's picture:

>>>    ------
>>>    |    |
>>>    |    |
>>>    -----+-----
>>>         |    |
>>>         |    |
>>>         ------

>Siamese Rubik's cubes share an entire column of cubies, i.e. in the
>case of two 3x3x3's they share an edge cubie and two corner cubies.

>Cubies cannot move from one cube to the other.  The shared column
>of cubies cannot be separated or rearranged.  The effect is like
>bandaging an edge column on one 3x3x3 cube and bandaging an edge
>column on another 3x3x3 cube and having two identical puzzles.

A "creative" question:

Suppose we want to be able to rotate the 17-cubie faces 180 degrees.
Can anyone think of a mechanical structure that could achieve this?



From cube-lovers-errors@curry.epilogue.com  Fri Oct 25 22:41:42 1996
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Date: Fri, 25 Oct 96 18:51:55 EDT
Message-Id: <9610252251.AA14688@sun34.aic.nrl.navy.mil>
From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: cube-lovers@ai.mit.edu
Subject: Re: Siamese Rubik's Cubes

Norman Diamond <diamond@jrdv04.enet.dec-j.co.jp> wrote:
...
> Siamese Rubik's cubes share an entire column of cubies, i.e. in the
> case of two 3x3x3's they share an edge cubie and two corner cubies.
...
> The idea of bandaging has been extended further by Dieter Gebhardt
> (publications in CFF) and others....

Most cases of bandaging create a puzzle whose transition graph is not
the Cayley graph of a group.  For instance, if two disjoint
edge-corner pairs were taped together, you would have some positions
with ten possible QT neighbors and some with eight.

But the corner-edge-corner bandaging does create a group: Fix the
position of the bandaged part, and permute the other 46 facelets (six
corners, eleven edges, and six face centers) with two face moves and
two slice moves.  The resulting group can have at most 5! corner
permutations, as in the two-generator group (see Singmaster or the
archives (21 July 1981, 31 Aug 1994)).  There are at most 11! edge
permutations, and the face center permutations represent the rotation
group of the cube, with 24 elements.  There can be at most 3^5 corner
orientations and 2^10 edge orientations.  Finally, the total
permutation parity (corner, edge, and face center) must be even.  Gap
tells me the group has

    14302911135744000 = 5! 3^5 11! 2^10 24/2 

elements, so all such positions are achievable.  I haven't run the
Supergroup through Gap, so I'm not sure whether it 2048 times as many
positions.

Of course the regular Siamese cube has the square of this many
positions, because there are two cubes.  A different kind of Siamese
cube would be one in which the three 17-cube slabs can rotate 180
degrees with respect to each other.  It would certainly be difficult
to build.  I think the interaction between the slab moves and the
Lucky Six group would make it hard to solve, as well.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil


From cube-lovers-errors@curry.epilogue.com  Sat Oct 26 00:18:08 1996
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Date: Fri, 25 Oct 1996 23:47:50 -0400 (EDT)
From: Nicholas Bodley <nbodley@tiac.net>
To: Wei-Hwa Huang <whuang@ugcs.caltech.edu>
cc: Cube-Lovers@ai.mit.edu
Subject: Re: Siamese Rubik's Cubes (was Re: DEAR TANOFF <(fwd))
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On 25 Oct 1996, Wei-Hwa Huang wrote:

}Norman Diamond 25-Oct-1996 1355 <diamond@jrdv04.enet.dec-j.co.jp> writes:
}>A. Southern misinterpreted M. Velucchi's picture:

{Snips}

}A "creative" question:
}
}Suppose we want to be able to rotate the 17-cubie faces 180 degrees.
}Can anyone think of a mechanical structure that could achieve this?

 Here's hoping that this "stream of consciousness with revisions" style is
acceptable!:

 I can conceive of such a structure, but whether it could be made to work
decently is quite open to doubt. It would have a great many pieces; the
whole top layer would have to consist of cubies with two physical parts,
one that would travel to its new location, and the other which would
remain behind.  Holding the whole works together while rotating it is
difficult enough, but reliably reattaching the two parts of each cubie
once the rotation was complete is borderline crazy!

 Of course, all edge and corner cubies would need to be two-part.

 If someone is ambitious enough to attempt such a design, it would be very
costly and out of the question for mass production.

 It might help if a tool (such as a Torx (TM) wrench) were provided to
insert into both "face-center" cubies (or the common corner cubie) to
unlock the top layer from its underlying parts and to lock the top-layer
cubies together. However, just a clamping frame to hold the top layer
together would make sense, IMO. 

 A strictly-mechanical solution is at least borderline impractical, but
shrewd design with rare-earth magnets might help.

 Dismantle a regular Cube to see what would be involved. An edge cubie has
a "foot" that extends below the top layer, as does a corner cubie. These
"feet" would have to be left behind once a move began.

 It's really nice to have all the unlocking and reattaching taken care of
"automatically" by just the twisting shear force created by gripping the
Siamese Cube, but for such a move as this, that's a formidable luxury.

 If I were an experienced mechanical engineer, I'd say it just isn't
practical. However, it is great fun to think of how it could be done. (If
e-mail had a universal graphics format, illustrations would be nice, but I
honestly don't feel that ambitious!) I also suspect that when it came time
to design in detail, new conceptual problems would arise which might be
extremely difficult to overcome. Consider, for instance, that if you don't
use a clamping frame, the mere act of locking the top layer together has
to hold the corner cubies in place. The locking pieces need to be operated
by sliding members passing through the neighboring edge cubies, and that's
not all, by far.

 My regards to all,


|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*    When the year 2000 begins, we'll celebrate 
|*  Amateur musician  *|*      the 2000th anniversary of the year 0.
--------------------------------------------------------------------------



From cube-lovers-errors@curry.epilogue.com  Mon Oct 28 23:06:18 1996
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Date: Mon, 28 Oct 1996 22:06:45 -0600
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I am looking for anyone who has a Rubik's Revenge they'd like to part
with.

Scott Crawford


From cube-lovers-errors@curry.epilogue.com  Sun Nov  3 21:05:08 1996
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From: Pet Milk <nathank@asms3.dsc.k12.ar.us>
Organization: Arkansas School for Math & Science
To: Cube-Lovers@ai.mit.edu
Date: Sun, 3 Nov 1996 17:41:33 CDT
Subject: Greetings
Priority: normal
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Message-ID: <319C4F363C9@ASMS3.DSC.K12.AR.US>

Hi


My name is Nathan, and, obviously, I'm a newcomer to the list.  I have 
need of some help.  

I'm forced to write a paper concerning some famous individual that 
has contributed to mathematics in some way.  After looking carefully, 
I picked Mr. Rubik.  I searched the Net forever, but only came up 
with an interview that was conducted with Rubik.  However, I need 
more information.  Here's where I need you.  

Does anyone have any further information concerning the work of Mr. 
Rubik in any way?  Anything would be of help:  Net sites, books, 
lists, anything.  I'm not in any bug hurry, however I would like the 
information in time to sort it, etc.  

Thank you for your help...


Nathan

I spy a boy
I spy a girl
I spy a chance to change the world


From cube-lovers-errors@curry.epilogue.com  Sun Nov  3 21:44:47 1996
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From: Stan Isaacs <isaacs@hpcc01.corp.hp.com>
Message-Id: <199611040235.AA111104943@hpcc01.corp.hp.com>
Subject: Book on Bandaged Cubes
To: cube-lovers@ai.mit.edu
Date: Sun, 3 Nov 96 18:35:43 PST
Mailer: Elm [revision: 70.85.2.1]

I just got a book that might be very interesting to cube lovers.  It's 
called "Bandaged Cubes", by Dieter Gebhardt.  Some cube lovers may 
already know Dieter and about bandaged cubes, from articles in the 
CFF magazine; this is all about bandaging in one place.  He presents
notation and classification, and discusses many types, the 
group-theory of them, and how to solve them.  It even has color pictures
of some of the variations.

One type is the C-block cube (also called "Rigit Edge Cube"), which is 
just half of the Siamese cube recently discussed here.

For those who haven't seen articles on this, bandaged cubes are regular
Rubik's cubes with some edges taped together.  If you tape 2 cubies, 
one corner and one edge, that is an "A-block".  If you tape an edge and a
center, that's a "B-block".  2 corners and an edge (3 cubies) is a "C-block.
And so on - he has notation for all the bandage possibilities, and 
discusses (as far as I can tell) all the interesting variations in a 
3x3x3.  (He leaves 4x4x4 and 5x5x5 bandaged cubes for a later time.)

Anyway, if you get tired of Rubik's cube itself, these offer dozens
of variations, each with its own quirks and limitations, and many 
chances for new discoveries.

According to CFF, the booklet can be bought from Dieter for $24 (DM 36)
(including postage) at:
	Dieter Gebhardt
	Norikerstrasse, 23, D-90402
	Nurnberg, GERMANY

Its 100 pages, with 74 figures and 4 color plates.

I highly recommend it.  Every Cube-lover should have a copy.  (Of course, 
now I need a cheap source of blank cubes to tape.)

 -- Stan Isaacs


From cube-lovers-errors@curry.epilogue.com  Mon Nov  4 14:13:43 1996
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From: Wei-Hwa Huang <whuang@cco.caltech.edu>
To: Cube-Lovers@AI.MIT.EDU
Subject: Re: Book on Bandaged Cubes
Date: 4 Nov 1996 16:37:13 GMT
Organization: California Institute of Technology, Pasadena
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Stan Isaacs <isaacs@hpcc01.corp.hp.com> writes:

>For those who haven't seen articles on this, bandaged cubes are regular
>Rubik's cubes with some edges taped together.  If you tape 2 cubies, 
>one corner and one edge, that is an "A-block".  If you tape an edge and a
>center, that's a "B-block".  2 corners and an edge (3 cubies) is a "C-block.
>And so on - he has notation for all the bandage possibilities, and 
>discusses (as far as I can tell) all the interesting variations in a 
>3x3x3.  (He leaves 4x4x4 and 5x5x5 bandaged cubes for a later time.)

Does he cover non-adjacent bandages; for example, two corner cubies?


-- 
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
-------------------------------------------------------------------------------
Not technically an "evil alum".


From cube-lovers-errors@curry.epilogue.com  Mon Nov  4 14:12:38 1996
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On Sun, 3 Nov 1996, Pet Milk wrote:

> Hi
> 
> 
> My name is Nathan, and, obviously, I'm a newcomer to the list.  I have 
> need of some help.  
> 
> I'm forced to write a paper concerning some famous individual that 
> has contributed to mathematics in some way.  After looking carefully, 
> I picked Mr. Rubik.  I searched the Net forever, but only came up 
> with an interview that was conducted with Rubik.  However, I need 
> more information.  Here's where I need you.  
> 
> Does anyone have any further information concerning the work of Mr. 
> Rubik in any way?  Anything would be of help:  Net sites, books, 
> lists, anything.  I'm not in any bug hurry, however I would like the 
> information in time to sort it, etc.  

Have you seen the book "Rubik's cubic compendium", by Rubik, et al?
It has an article by Rubik which is interesting and is still in print
(published by Oxford Univ Press I think). - David Joyner

> 
> Thank you for your help...
> 
> 
> Nathan
> 
> I spy a boy
> I spy a girl
> I spy a chance to change the world
> 
> 




From cube-lovers-errors@curry.epilogue.com  Mon Nov  4 15:29:43 1996
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To: cube-lovers@ai.mit.edu
From: Aben Gentry <gammet@mindspring.com>
Subject: Rubik's Clock...

Hey Guys,

Have any of you figured out how to solve Rubik's clock yet? Also, what is
best source for cubes (and cube-like puzzles) that you know of? ...I
normally shop at Puzzletts.

Aben Gentry
abeng@mindspring.com



From cube-lovers-errors@curry.epilogue.com  Mon Nov  4 21:53:51 1996
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Date: Tue, 5 Nov 1996 13:22:02 +1100 (EST)
From: Leanne Rylands <L.Rylands@st.nepean.uws.edu.au>
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To: Aben Gentry <gammet@mindspring.com>
cc: cube-lovers@ai.mit.edu
Subject: Re: Rubik's Clock...
In-Reply-To: <199611042012.PAA24488@itchy.mindspring.com>
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> 
> Have any of you figured out how to solve Rubik's clock yet? Also, what is
> best source for cubes (and cube-like puzzles) that you know of? ...I
> normally shop at Puzzletts.

Don Taylor and I wrote a book ``Mastering Rubik's Clock''.
Published in 1988 by Simon and Schuster which gives the 
solution.
The clock is very easy to solve (hence the book is very 
thin, only 16 pages).

Leanne Rylands



From cube-lovers-errors@curry.epilogue.com  Tue Nov  5 23:16:36 1996
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From: AirWong@aol.com
Date: Tue, 5 Nov 1996 18:43:57 -0500
Message-ID: <961105184355_222918906@emout11.mail.aol.com>
To: CUBE-LOVERS@ai.mit.edu
Subject: Re: Rubik's Clock...

> Have any of you figured out how to solve Rubik's clock yet?

What exactly is the Rubik's clock? I've only heard of the Rubik's cube, dice,
pyramid, tangle, fifteen... how many puzzles have the Rubik's name on them,
anyway?

Aaron Wong


From cube-lovers-errors@curry.epilogue.com  Wed Nov  6 14:32:49 1996
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To: Cube-Lovers@AI.MIT.EDU
From: Wei-Hwa Huang <whuang@cco.caltech.edu>
Subject: Re: Rubik's Clock...
Date: 6 Nov 1996 16:10:50 GMT
Organization: California Institute of Technology, Pasadena
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AirWong@aol.com writes:

>> Have any of you figured out how to solve Rubik's clock yet?

>What exactly is the Rubik's clock? I've only heard of the Rubik's cube, dice,
>pyramid, tangle, fifteen... how many puzzles have the Rubik's name on them,
>anyway?

Hum de hum...

Cube (Several releases)
Mini Cube
Revenge
4th Dimension (A cube with pictures)
Race
Game
Snake (Many colors, three sizes)
Magic (Link the Rings)
Magic (Make the Cube)
Magic (Unlink the Rings)
Magic Game
Magic Puzzle
Clock
Fifteen
Rabbits
Dice
Triamid
Tangle (4 versions)
Maze

(The Pyraminx, the Octagon, the 5x5x5, and the Missing Link have never
 been labeled with Rubik's name, AFAIK...)

-- 
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
-------------------------------------------------------------------------------
Not technically an "evil alum".


From cube-lovers-errors@curry.epilogue.com  Thu Nov  7 16:07:58 1996
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From: Nicholas Bodley <nbodley@tiac.net>
To: AirWong@aol.com
cc: CUBE-LOVERS@ai.mit.edu
Subject: Re: Rubik's Clock...
In-Reply-To: <961105184355_222918906@emout11.mail.aol.com>
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 Aaron is probably not the only person wondering! By any chance, was this
a temporary lapse, with the calendar cube in mind?

Regards to all, NB

* * *

On Tue, 5 Nov 1996 AirWong@aol.com wrote:

{Snips}

}> Have any of you figured out how to solve Rubik's clock yet?
}
}What exactly is the Rubik's clock? I've only heard of the Rubik's cube, dice,
}pyramid, tangle, fifteen... how many puzzles have the Rubik's name on them,
}anyway?



From cube-lovers-errors@curry.epilogue.com  Thu Nov  7 16:09:46 1996
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Date: Thu, 07 Nov 1996 15:54:02 +0000
To: Cube-Lovers@ai.mit.edu
From: "The Unofficial Thermofluids Fan Club of the UK." <a.southern@ic.ac.uk>
Subject: Re: Rubik's Clock...

I wrote this a few days ago but sent it to the wrong
address......durrrrrrrrrrrrrrrrrrrrrr!

here it is with an amendment.

At 13:22 05/11/96 +1100, you wrote:
>> 
>> Have any of you figured out how to solve Rubik's clock yet? Also, what is
>> best source for cubes (and cube-like puzzles) that you know of? ...I
>> normally shop at Puzzletts.
>
>Don Taylor and I wrote a book ``Mastering Rubik's Clock''.
>Published in 1988 by Simon and Schuster which gives the 
>solution.
>The clock is very easy to solve (hence the book is very 
>thin, only 16 pages).
>
>Leanne Rylands
>
>
>
>

I once set about building a 5x5 rubiks clock, but I never got round to
finishing it as I realised that I already knew how to solve the general NxN
clock.

I did get around to building a 32x2 Rubik's Magic, that's eight Rubiks
Magics built into one array. It required a couple of customised inlays (i.e.
I cut and pasted), and is still prone to misalignment and deligamentation
(i.e. it falls apart a bit), but it works and the solution is just an
extended version of the 4x2. It takes me about half an hour to solve, and is
the equivelent of a good work out, that's why I haven't used it since I was
about 14! 

I only used it to get one above the guys at school that could do the magic
(4x2) in about one second, because I could never get any faster than 2 seconds!

The smallest "Rubik's Magic" I've ever custom built was a 2x1, most people
didn't have any problem with that one!

Has anyone else ever extrapulated a puzzle to form a "Custom Master Edition"?

I'd be interested to hear.



from Wei-Hwa Huang

Cube (Several releases)
Mini Cube
Revenge
4th Dimension (A cube with pictures)
Race
Game
Snake (Many colors, three sizes)
Magic (Link the Rings)
Magic (Make the Cube)
Magic (Unlink the Rings)
Magic Game
Magic Puzzle
Clock
Fifteen
Rabbits
Dice
Triamid
Tangle (4 versions)
Maze

I think there was also Rubik's Illusion, which was a game of chess using
some sort of complex mapping.



Cheers!

Andrew R. Southern,

The unofficial Thermo-Fluids Fan Club of the UK.



From cube-lovers-errors@curry.epilogue.com  Fri Nov  8 17:23:27 1996
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From: Vanessa Paradis WANTS me!! <cmaggs@glam.ac.uk>
To: CUBE-LOVERS@ai.mit.edu
Subject: HELLO AGAIN!
X-VMS-To: CUBE-LOVERS@AI.MIT.EDU

I have solved the Rubiks Clock.
I also have the Rubiks Illusion.

It is like 4-in-a-row game, but you need 5-in-a-row, using a mirror (which is
connected to the back of the board) as another part of the board.

There are 3 types of pieces.  RED, YELLOW and RED/YELLOW.

These can be used to make a row of 5 in any direction


From cube-lovers-errors@curry.epilogue.com  Tue Nov 12 16:10:33 1996
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From: Pete Beck <pbeck@pica.army.mil>
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Subject: Fw: [Dan Galvin:  Thought for Tuesday, Nov 12, 1996]
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----------
> To: pbeck@qa.pica.army.mil
> Subject: [Dan Galvin:  Thought for Tuesday, Nov 12, 1996]
> Date: Tuesday, November 12, 1996 12:40 PM
> 
> 
> ----- Forwarded message # 1:
> 
> Received: from postal.tamu.edu by COR6.PICA.ARMY.MIL id ab27608;
>           12 Nov 96 9:44 EST
> Received: from postal (postal.tamu.edu [128.194.103.24]) by
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> Approved-By:  Dan Galvin <galvin@TAM2000.TAMU.EDU>
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> Date:         Tue, 12 Nov 1996 04:02:02 -0600
> Reply-To: Dan Galvin <galvin@tam2000.tamu.edu>
> Sender: THOUGHT FOR THE DAY <TFTD-L@TAMVM1.TAMU.EDU>
> From: Dan Galvin <galvin@tam2000.tamu.edu>
> Subject:      Thought for Tuesday, Nov 12, 1996
> To: Multiple recipients of list TFTD-L <TFTD-L@TAMVM1.TAMU.EDU>
> 
>                                                                       *
> Easiest Color to Solve on a Rubik's Cube:
>         Black.  Simply remove all the little colored stickers on the
>         cube, and each of side of the cube will now be the original
>         color of the plastic underneath -- black.  According to the
>         instructions, this means the puzzle is solved.
>                 -- Steve Rubenstein
> 
> ----- End of forwarded messages


From cube-lovers-errors@curry.epilogue.com  Tue Nov 12 16:12:10 1996
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From: Carey <g-carey@cfw.com>
To: Cube-Lovers@ai.mit.edu
Subject: Square 1
Date: Tue, 12 Nov 1996 13:59:31 -0500
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Hello,
I'm working on a solution to the Square 1 puzzle.  Does anyone know the
maximum number of moves required?  Also I'm looking for the minimum number
of moves required to solve it if you have three consecutive edge wedges.

Pete Carey
g-carey@cfw.com


From cube-lovers-errors@curry.epilogue.com  Wed Nov 13 16:27:44 1996
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Date: Wed, 13 Nov 1996 16:13:57 -0500 (EST)
From: Michael C Masonjones <mcmj@blazetech.com>
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On Tue, 12 Nov 1996, Carey wrote:

> Hello,
> I'm working on a solution to the Square 1 puzzle.  Does anyone know the
> maximum number of moves required?  Also I'm looking for the minimum number
> of moves required to solve it if you have three consecutive edge wedges.
> 
> Pete Carey
> g-carey@cfw.com
> 
I assume this means a permutation of three edge wedges.
I can do it in 8 flips through the center divisor, the most convenient way
I've found to count moves on Square-1.

Start with the permutation on top (in the square/square configuration,
of course).  Position top and bottom squares so that the left side of the
top edge wedge facing you lies above the central turning slot and the
right side of the bottom edge wedge lies below the same slot.  (If you
flip through the center, you still have square configurations, top and
bottom).

T+n = rotate top n/12 of a turn counterclockwise, as seen from top..
T-n = ..........................clockwise............
B+n, B-n are the same for the bottom face when looking at it from the
bottom.
F = flip through center slot.

Try this:
F T+3 F T-1 B-1 F T-2 B+1 F T-3
F T+3 F T-1 B-1 F T-2 B+1 F T+3

Notice that half of this produces two 2-permutations.

I'm curious if anyone speed cubes Square 1.  My average is about 1:35,
with a best time of 1:15 for partial fluke.  I can always do it in under 2
minutes for the worst parity situation.  How does this compare?  There
have got to be faster people out there, because I can only do the regular
Rubik's cube in 55 seconds on average, which is pretty slow by this
group's standards.

					Mike Masonjones.
					mcmj@blazetech.com



From cube-lovers-errors@curry.epilogue.com  Wed Nov 13 22:28:12 1996
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To: "cube-lovers@ai.mit.edu" <cube-lovers@ai.mit.edu>
Subject: Re: Square 1
Date: Wed, 13 Nov 96 22:37:52 -0500
From: michael <mcmj@world.std.com>
X-Mailer: E-Mail Connection v2.5.03

-- [ From: michael * EMC.Ver #2.5.02 ] --

Whoops! the flip is done with the right hand keeping the left side of the
puzzle stationary, if that was not already clear.  Doing it with a left hand
twist does a double double switch of edge-wedge and corner-wedge pairs.

Sorry for the errata post on this. Maybe I should add something else to
avoid the complete wast of bandwidth.  

Since he asked... The only thing I know about confirmed maximal moves for
Square 1 is that any possible shape can be put back to two squares with at
most 7 flips, and only one configuration requires that many moves.  That's
the one with a square on one side (CECECECE) and the CEECECCE shape (C=
corner wedge, E=edge-wedge) on the other.  Satisfyingly symmetric antipode.

Corrected permutation of 3 edge wedges:

>T+n = rotate top n/12 of a turn counterclockwise, as seen from top..

>T-n = ..........................clockwise............

>B+n, B-n are the same for the bottom face when looking at it from the
bottom. 
>
>F = flip through center slot. **** with right hand.****
>
>Try this:

>F T+3 F T-1 B-1 F T-2 B+1 F T-3

>F T+3 F T-1 B-1 F T-2 B+1 F T+3
>

					Mike Masonjones.
					mcmj@blazetech.com




From cube-lovers-errors@curry.epilogue.com  Wed Nov 13 23:44:48 1996
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Message-Id: <13Nov1996.162951.Alan@LCS.MIT.EDU>
From: Alan Bawden <Cube-Lovers-Request@ai.mit.edu>
Sender: Cube-Lovers-Request@ai.mit.edu
To: Cube-Lovers@ai.mit.edu
In-reply-to: Jerry Slocum's message of 11 Nov 96 18:29:11 EST <961111232911_70410.1050_JHD113-1@CompuServe.COM>
Subject: Directory of Puzzlers

I received the following note from Jerry Slocum, which he asked me to pass
on to Cube-Lovers if I though it was appropriate.  I see nothing wrong with
it, so here it is.

But let me take this opportunity to request that people -not- send things
to Cube-Lovers-Request and ask that I forward them on to Cube-Lovers --
that just makes more work for me.  Please just send what you want to go to
Cube-Lovers to Cube-Lovers.  If you feel the need to make some explanation
to Cube-Lovers-Request, send a separate message to Cube-Lovers-Request.

Thanks!  - Alan

------- Begin Forwarded Message -------

Date: 11 Nov 96 18:29:11 EST
From: Jerry Slocum <70410.1050@compuserve.com>
To: Alan Bawden <Cube-lovers-request@ai.mit.edu>
Subject: Directory of Puzzlers
Message-ID: <961111232911_70410.1050_JHD113-1@CompuServe.COM>

Dear Alan,

In 1994 the Slocum Puzzle Foundation published the Second Edition of the
"Directory of Puzzle Collectors and Sellers". It includes a list of 232 puzzle
collectors and designers and described in detail what puzzles interest them.
More than 120 of them collect, and are interested in, combinatorial Rubik-type
puzzles. It also includes 96 mail order puzzle sellers and 147 retail stores
that sell puzzles (and were recommended by collectors).

I will be sending out letters of inquiry to all collectors & sellers
included in the Directory within the next week for updated information for
a new Third Edition that will be published early in 1997.

Some, but not all, of the Cube Lovers subscribers will receive mailings.

I would be glad to invite Cube Lovers who are puzzle collectors and/or
sellers to email me if they wish to be in the Third Edition and are not in
the current Directory.

I will add a section on puzzle related WWW Internet pages and sites and
expand the coverage of email in the new Directory.  I will send a letter of
inquiry to all that request one and provide their mailing address.  I am
asking for the replies to be returned to me 3 weeks after they are
received.

Let me know if you have any questions or you would like more details.

I would be glad to have this notice posted for Cube Lovers if you think it
is appropriate..

Regards,
Jerry Slocum
257 South Palm Drive, Beverly Hills, CA 90212 USA  Fax 310-274-3644
email:70410.1050@compuserve.com


------- End Forwarded Message -------


From cube-lovers-errors@curry.epilogue.com  Thu Nov 14 14:06:14 1996
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Date: Thu, 14 Nov 1996 08:33:29 -0500 (EST)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: y'xy vs. yxy'
To: Cube-Lovers <cube-lovers@ai.mit.edu>
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As promised, here is my followup on why the conjugate of x by y is y'xy
rather than yxy'.  Recall that y'xy informally means undo y, then do x,
and finally do y.  It seems strange to undo something before you do it,
but nonetheless y'xy is the conventional definition of congugacy rather
than yxy'. 

My first reference is Singmaster, Notes on Rubik's 'Magic Cube', Fifth
Edition, pp. 57-58. We adopt left to right notation so that (a)xy=y(x(a)).
a is the argument, and x and y are permutations which are composed left to
right. 

I paraphrase slightly, but here is what Singmaster says.  We desire the
conjugate of x by y to be x shifted by y.  By "x shifted by y", we mean
the following.  Suppose in cycle notation we have x=(...,a,b,c...).  Then,
x shifted by y is z, where z=(...,(a)y,(b)y,(c)y,...).  I will defer the
presentation of Singmaster's proof, but the final conclusion is that
z=y'xy.  So our definition of the conjugate of x by y becomes y'xy.  By
contrast, we have yxy'=(...,(a)y',(b)y',(c)y',....), or x shifted by y'. 

While I was chasing down this reference in Singmaster, a message arrived
from Dan Hoey giving an alternative justification for the y'xy definition. 
I will quote Dan's message extensively.  Dan first credits Jim Saxe with
the explanation, and then goes on to say the following. 


>                     Suppose we are conjugating elements of a group X
>by elements of a group G.  Congugation by an element g induces a
>permutation on X.  

This is a very old idea in Cube-Lovers.  I believe the first occurrence is
in Symmetry and Local Maxima.  Elements of the standard cube group G were
conjugated by elements of the set M of rotations and reflections of the
cube.  Conjugation of all the elements of G by a fixed element m of M were
viewed as a permutation on G. 

We denote m'gm by g^m for fixed g in G and fixed m in M.  We then denote
{m'gm | m in M} as g^M and {m'gm | g in G} as G^m.  I normally tend to
think of M-conjugation in terms of g^M -- that is, take one fixed element
g and calculate its 48 M-conjugates.  By contrast, G^m means take each g
in G and calculate m'gm using the same fixed m for each g.  It is G^m
which is a permutation on G. 

Dan continues:

>                   It is useful to have the mapping from g to its
>conjugation permutation be a homomorphism into S[X].  Suppose f is the
>mapping
>
>    f: a -> {x -> a' x a}.
>
>To make this a homomorphism, we must have
>
>    f:a.b -> f(a).f(b)
>
>so    {x -> (a.b)' x (a.b)} = {x -> a' x a} . {x -> b' x b}.
>
>The right hand side is the product of two permutations.  

Indeed.  It's probably obvious to everybody else how to form the indicated
composition of the two permutations, but I was bumfuzzled for a while. 
Once I figured it out, I just kicked myself for being so dense.  Let me 
explain.

My day job is as a bureaucrat, but most semesters I am also adjunct
faculty teaching elementary algebra and calculus.  As such, I end up
teaching simple funcions -- e.g., f(x) = x^2 + 1. You teach students to
calculate such things as f(2) or f(3).  Then, you teach them such things
as f(a) and try to explain that "x is a variable" but "a is a constant
that you just don't know the value of".  Finally, you get into such 
things as f(a+b) or f(x^2 + 1).  

The latter is the one that really confuses most of my students.  They can
handle "replace x with 2" or "replace x with a".  But they have great
conceptual difficulty with "replace x with x^2 + 1".  The truth is, it is
a bit of a different concept because it is really function composition in
disguise, although most elementary math books don't teach function
composition for several chapters after introducing functions. 

Anyway, with Dan's equation we really just have a function composition
where in the end we replace x with a'xa.  So x->b'xb becomes
a'xa->b'(a'xa)b.  I kick myself because I couldn't quickly figure out the
same concept that I am forever emphasizing with my students. 

Dan continues:


>                                                         If we are
>writing them left to right, as in f.g (x) = g(f(x)), then it is
>
>              {x->b'(a' x a)b} which corresponds to the left hand side.
>
>>But we write permutation composition from right to left,
>f.g(x)=f(g(x)) we would get
>
>              {x->a'(b' x b)a} ?
>
>for the right hand side, and that is wrong, since a'b' is not (ab)'.
>
>>People who write right to left define conjugation by a as
>f:a->{x->axa'} for this reason.
>

It seems to me that we could rescue the homomorphism and the yxy'
definition, but it would be awkward.  We would have to have the mapping
from g' to its conjugation permutation be the homomorphism, rather than
the mapping from g. 

Now for Singmaster's proof:  given the cycle in our definition of x, we
have x:a->b.  We need y'xy:(a)y->(b)y.  But (a)yy'xy=(a)xy=(b)y.  So y'xy
carries (a)y to (b)y, and we are done. 

Let me finish by talking a little more about the equivalence between
conjugacy and cycle structure.  Again, this is from Singmaster. 

It is the case in Sn that two elements x and z are conjugate if and only
if they have identical cycle structure.  Any finite permutation group may
be viewed as a subgroup of Sn for suitable choice of n.  The theorem may
or may not be true in any particular subgroup of Sn.  The part about
conjugates having identical cycle structure is always true.  But the
converse may or may not be true. 

To say that x and z are conjugate means that there exists some y such that
z=y'xy.  It's easy to see that if x and z have the same cycle structure,
then such a y must exist in Sn (e.g., line up the cycles of x with the
cycles of z, see what goes to what, and that is a y which will work).  The
problem in the general case is that a subgroup of Sn might contain x and z
which have the same cycle structure, without also containing an
appropriate y which would make them conjugate. 

Singmaster shows that the converse of the theorem is true for the
constructable group of the cube, but that it is not true for the standard
cube group G.  The counter-example is as follows.  Let x be a 7-cycle on
the corners and an 11-cycle on the edges -- e.g.,

  x=(C1,C2,C3,C4,C5,C6,C7)(E1,E2,E3,E4,E5,E6,E7,E8,E9,E10,E11).  

Let z be only slightly different (reversing two corners)  --  e.g., 

  z=(C2,C1,C3,C4,C5,C6,C7),(E1,E2,E3,E4,E5,E6,E7,E8,E9,E10,E11)

The obvious conjugating element is y=(C1,C2), which is in the 
constructable group but which is not in G.  There are other conjugating 
elements, but they are all of the form

(C1,C2) (C2,C1,C3,C4,C5,C6,C7)^i (E1,E2,E3,E4,E5,E6,E7,E8,E9,E10,E11)^i,

which also are in the constructable group but not in G.  Hence, x and z 
have the same cycle structure, but are not conjugate in G.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7127
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990



















From cube-lovers-errors@curry.epilogue.com  Sat Nov 16 21:51:20 1996
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Date: Sat, 16 Nov 1996 07:24:20 -0500 (EST)
From: Nicholas Bodley <nbodley@tiac.net>
To: Cube-Lovers@ai.mit.edu
Subject: Non-cubical Rubik cousins; physical realizability
In-Reply-To: <13Nov1996.162951.Alan@LCS.MIT.EDU>
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 (My subject line is a spur-of-the-moment phrase; not deeply considered.)

 I just got to wondering whether some people have considered theoretical
larger analogs of the Magic Domino (btw, would somebody please manufacture
some Magic Dominoes? Binary Arts?). To get back on topic, these would be
Cube-like puzzles with such "cubie counts" ("Dimensions") as 3X3X4, 3X4X4,
etc. Whether these are trivial, I haven't yet thought out; making real,
physical ones might not be simple. If this topic is covered in the
archives, I apologize; in such a case, could someone recommend non-obvious
keywords or names? 

 Is there an agreed-upon concise way of defining the
"size/count/dimensions" of a Cube; i.e., a 2X2X2 is a Pocket Cube, a 4...
is Rubik's Revenge, etc.? How about "order-3" for a regular Rubik's, or
simply (given proper context) "n", so that "2" signifies Pocket, "4"
Revenge, etc.? Perhaps it's just a personal reaction, but I find it
cumbersome to type "5X5X5" more than a few times, for instance.

 Thinking about this brings up another topic, and probably a difficult one
to completely characterize. Given any arbitrary puzzle composed of cubies,
is it always possible to create a mechanism to realize that specific
puzzle physically?

 As far as I know (and here I stick my neck waaaaay out!), there is no
theory of mechanisms in the general case that would, for instance, say
whether an order-2 is realizable (as we know, it can be made, and has
been); the Magic Domino is more of a challenge, imho, in that it isn't as
easy to say whether such a structure can be made. 

 Some matters affecting realizability are relatively easy to anticipate,
such as the matter of holding the corner cubies in place in a "7"  (with
all cubies of equal size) when one plane is rotated with respect to the
other six. Other matters are a question of what's reasonable to design
mechanically; while theoretically possible, some structures might not be
at all practical, because of such problems as cumulative friction, lack of
rigidity, and dimensional tolerances. Such real-world considerations
(unfortunately!) muddy the waters until a really good mind comes along to
settle the mud.

 A preliminary guess at an answer to the question is that probably all
"low-order" collections of cubies are realizable, but we are far from
having a theory of mechanisms that tells us how to design the innards.

 I maintain that the mechanism of the ordinary Rubik's Cube is the most
ingenious simple one ever invented; I have studied mechanisms to a fair
degree. (A good competitor is the programmable pushbutton combination lock
that has five buttons in a row. This is mechanical, digital, programmable,
combinatorial, and sequential.)

Hope and trust this hasn't been a waste of bitspace!

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*    When the year 2000 begins, we'll celebrate 
|*  Amateur musician  *|*      the 2000th anniversary of the year 0.
--------------------------------------------------------------------------



From cube-lovers-errors@curry.epilogue.com  Sun Nov 17 19:42:27 1996
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Date: Mon, 18 Nov 96 09:41:55 +0900
From: Norman Diamond 18-Nov-1996 0937 <diamond@jrdv04.enet.dec-j.co.jp>
To: cube-lovers@ai.mit.edu
Subject: Re: Non-cubical Rubik cousins; physical realizability
Mime-Version: 1.0
Content-Type: TEXT/PLAIN; charset=ISO-2022-JP

By bandaging a 4x4x4, you can make several variations:  3x3x4, 3x4x4, 2x4x4,
and 2x3x3.  The 4x4x4 is no longer made or sold through ordinary distribution
channels any more, but probably still available from Puzzletts at a high
price.  At the IPP a few months ago someone was offering a real 2x3x3 for
around US$40 I think, which is cheaper than a 4x4x4 is now but still rarer.

-- Norman Diamond             diamond@jrdv04.enet.dec-j.co.jp
[Speaking for Norman Diamond not for Digital.]


From cube-lovers-errors@curry.epilogue.com  Tue Nov 26 20:05:10 1996
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To: "Cube-Lovers@AI.MIT.EDU" <cube-lovers@ai.mit.edu>
Subject: Rubic's Revenge
Date: Tue, 26 Nov 96 19:56:00 -0500
From: Bill Edwards <edwards@city-net.com>
X-Mailer: E-Mail Connection v2.5.03

-- [ From: Bill Edwards * EMC.Ver #2.5.02 ] --

Anybody know where I can get some more Rubic's Revenges?  I wore out my last
one more than a year ago.  

Does a 5x5 matrix Rubic's cube exist?  I think I know the general solution,
as an extrapolation from the solution to a 4x4.

Hope to hear from somebody soon.

Bill


From cube-lovers-errors@curry.epilogue.com  Wed Nov 27 14:23:19 1996
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From: Tim Botham <timbo@homemail.com>
To: cube-lovers@ai.mit.edu, Bill Edwards <edwards@city-net.com>
Subject: Re: Rubic's Revenge
Date: Tue, 26 Nov 1996 21:34:10 -0800
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Puzzletts (on the web at http://www.puzzletts.com/) has a good selection of
4x4x4, 5x5x5, and other variations available by mail-order.  Ordering can
be done through their web page.

	Tim

----------
> From: Bill Edwards <edwards@city-net.com>
> To: Cube-Lovers@AI.MIT.EDU <cube-lovers@ai.mit.edu>
> Subject: Rubic's Revenge
> Date: November 26, 1996 4:56 PM
> 
> -- [ From: Bill Edwards * EMC.Ver #2.5.02 ] --
> 
> Anybody know where I can get some more Rubic's Revenges?  I wore out my
last
> one more than a year ago.  
> 
> Does a 5x5 matrix Rubic's cube exist?  I think I know the general
solution,
> as an extrapolation from the solution to a 4x4.
> 
> Hope to hear from somebody soon.
> 
> Bill


From cube-lovers-errors@curry.epilogue.com  Wed Nov 27 14:24:05 1996
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Date: Wed, 27 Nov 1996 10:18:53 -0800
From: Aaron Coles <acoles@fec.gov>
Subject: Rubik's Tangle Puzzles
To: cube-lovers@ai.mit.edu
Reply-to: acoles@fec.gov
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Anyone know where I can purchase Rubik Tangle #1 from?? I already have
2-4.  I lent it to someone and never got it back.  

Also has anyone created the 10x10 grid with these puzzles yet??


From cube-lovers-errors@curry.epilogue.com  Wed Nov 27 16:05:13 1996
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Date: Wed, 27 Nov 1996 12:56:20 -0800
From: Don Woods <don@altum.com>
Message-Id: <199611272056.MAA21078@altum.com>
To: cube-lovers@ai.mit.edu
Subject: Re: Rubik's Tangles

Someone (I've deleted the message) recently asked about the Tangles.
My recollection is that the four puzzles are all the same except for
permutations of the colors.  That is, each Tangle consists of the 24
possible distinct pieces, plus one duplicated piece.  Which piece is
duplicated varies, but the resulting puzzles are the same.  (Oh yeah,
the pattern of the crossing ropes on each tile is also the same for
each puzzle.)  Really disappointing, especially since I think there
were two distinct solutions, and if they'd varied the mix a bit more
they could've had unique solutions as well as having four truly
different puzzles.

Also, another fellow and I independently did some analyses about three
years ago that proved that you cannot make a 10x10 using the four
combined puzzles.  Presumably the marketing blurb that suggests doing
so was written by someone who had no clue whether it was possible or
not.  Again, if they'd varied the puzzles a bit I have no doubt they
could've made the 10x10 achievable as well.

	-- Don.


From cube-lovers-errors@curry.epilogue.com  Thu Nov 28 13:10:28 1996
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Date: Thu, 28 Nov 1996 11:51:48 -0500 (EST)
From: Nicholas Bodley <nbodley@tiac.net>
To: cube-lovers@ai.mit.edu
Subject: Lubricants for puzzles
Message-ID: <Pine.SUN.3.95.961128113717.18706F-100000@sunspot.tiac.net>
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 My apologies in advance if this is a repeated topic; it hasn't appeared
recently, at least.

 Question is: What is a really good lubricant for plastic puzzles (such as
the classic Cube) with moving parts? What's good for metal machinery isn't
necessarily good for plastics; there is even a risk that some additives in
metal lubricants would dissolve or etch some plastics.

 A liquid, probably with a benign solvent to distribute it, would be
desirable. Powdered (or colloidal?) PTFE ("Teflon", a Du Pont TM in the
USA) particles or flakes should help a good bit. A "carrier" grease (which
might as well be a lubricant) would keep any particles in place. 
Molybdenum disulfide might be good, but might also tend to stain hands and
clothing. Some waxes might work. Powdered graphite would probably work
loose and make a mess. Lubricants that stain clothing aren't welcome,
either!

 The lubricant must also be benign toward metal, because the Cube is held
together with metal screws and tensioned by springs.

 I suspect that someone, somewhere, knows about a commercial (proprietary) 
formulation that meets most or all of these criteria. (I would not
recommend WD-40, by the way; I expect it would evaporate after some
months. It has its place, but I don't think it's a good plastic
lubricant.) 

 About 10 years ago, I found such a product, and lubricated some of my
puzzles with it, with good success, but then a major personal crisis came,
and I lost track of what it was...

 I'll try to summarize, if any significant number of replies comes by...

 Thanks in advance, and best regards!

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*    When the year 2000 begins, we'll celebrate 
|*  Amateur musician  *|*    the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------



From cube-lovers-errors@curry.epilogue.com  Fri Nov 29 03:45:13 1996
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Date: Thu, 28 Nov 1996 17:51:53 -0800
To: cube-lovers@ai.mit.edu
From: Lars Petrus <lars@netgate.net>
Subject: Re: Lubricants for puzzles

> My apologies in advance if this is a repeated topic; it hasn't appeared
>recently, at least.
>
> Question is: What is a really good lubricant for plastic puzzles (such as
>the classic Cube) with moving parts?

At first I used ordinary candles. It works fine, but you have to redo
it twice a week if you twist a lot. Later I heard that Silicon Spray
is the best, and I have used it ever since.

The odd thing about it is that for the first few minutes of turning, it
almost GLUES the cube together. It gets VERY hard to turn. The cubes in the
swedish championship were greased with silicon, but not "turned soft", so
it was really hard. Since I had learned the cube with a REALLY bad cube
(there was a shortage of hungarian cubes for months at that time), I had
a big advantage at that competition.


  -  -  -  -
"Madness is the first sign of dandruff"
                                --- Dr Winston O'Boogie

Lars Petrus, Sunnyvale, California  -  lars@netgate.net




From cube-lovers-errors@curry.epilogue.com  Fri Nov 29 03:45:46 1996
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From: AirWong@aol.com
Date: Thu, 28 Nov 1996 22:38:41 -0500
Message-ID: <961128223838_1985119134@emout20.mail.aol.com>
To: cube-lovers@ai.mit.edu
Subject: Dirty Cubes

Hello All!

How do you keep the original Rubik's cube (3X3X3) clean? If you've ever taken
apart the cube, you know what I'm talking about. The dust gets inside and
builds up over time. Is there any way to keep the dust out?

Aaron Wong


From cube-lovers-errors@curry.epilogue.com  Fri Nov 29 18:16:58 1996
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Date: Fri, 29 Nov 1996 10:37:14 -0500 (EST)
From: Michael Swart <mjswart@undergrad.math.uwaterloo.ca>
To: cube-lovers@ai.mit.edu
Subject: Re: Dirty Cubes
In-Reply-To: <961128223838_1985119134@emout20.mail.aol.com>
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Hello

> Hello All!
> 
> How do you keep the original Rubik's cube (3X3X3) clean? If you've ever taken
> apart the cube, you know what I'm talking about. The dust gets inside and
> builds up over time. Is there any way to keep the dust out?

Yep I know what you mean,
	I'm not really sure if there's anyway to keep the dust out.  I
always thought that the dust helps the cube stay 'smooth'.  
	After I think about it a bit, it seems likely that the dust comes
from plastic moving against plastic and so I'm not sure if the dust can be
avoided. But if you do like it clean you can regularly take a damp cloth
to the unassmbled cube.

 Michael Swart
 University of Waterloo
 mjswart@undergrad.math.uwaterloo.ca




From cube-lovers-errors@curry.epilogue.com  Fri Nov 29 18:16:24 1996
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From: Desenvolvimento de Projetos <avelloso@agencia.oesp.com.br>
To: "'AirWong@aol.com'" <AirWong@aol.com>,
        "cube-lovers@ai.mit.edu" <cube-lovers@ai.mit.edu>
Subject: RE: Dirty Cubes
Date: Fri, 29 Nov 1996 09:55:52 -0200
MIME-Version: 1.0
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You must disassemble it sometimes and clean it inside...

-----Original Message-----
From:	AirWong@aol.com [SMTP:AirWong@aol.com]
Sent:	Friday, November 29, 1996 1:39 AM
To:	cube-lovers@ai.mit.edu
Subject:	Dirty Cubes

Hello All!

How do you keep the original Rubik's cube (3X3X3) clean? If you've ever taken
apart the cube, you know what I'm talking about. The dust gets inside and
builds up over time. Is there any way to keep the dust out?

Aaron Wong




From cube-lovers-errors@curry.epilogue.com  Sun Dec  1 00:58:31 1996
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Date: Sat, 30 Nov 1996 12:44:03 -0500 (EST)
From: Nicholas Bodley <nbodley@tiac.net>
To: AirWong@aol.com
cc: cube-lovers@ai.mit.edu
Subject: Re: Dirty Cubes
In-Reply-To: <961128223838_1985119134@emout20.mail.aol.com>
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 Any self-respecting cube is made of plastic that >should< not be affected
by isopropyl alcohol; but don't use it on an unknown plastic, especially
if you don't want to lose that particular Cube. All except the center
"jack" structure should be cleanable with (hand) dishwashing detergent;
don't soak the pieces.

 Rubbing alcohol has other substances in its formulation that could attack
plastic; better to use 91% or 99% grade from a drugstore; shop around for
price. I have seen 91% isopropyl alcohol at maybe 8 times the price it
should be. Figure maybe $3 US per pint tops. It ignites fairly easily, by
the way! Do be careful. 

 Cleaning will remove all the lubricant, so you'll probably want to
restore it. So far, silicone spray (note that it's not spellled "silicon" 
spray; a chemist knows the difference) seems to be the best. If you have
the Cube apart, consider silicone lubricating grease, also, but it should
not contain other substances that would attack the plastic. (If applying
it to a shiny surface removes the gloss when it's wiped off after sitting
there for a short time, don't use it!). I have heard recommendations for
silicone stopcock grease, which would be a scientific lab. item, as a
joint grease for plastic recorders, but for those you want something that
>won't< allow motion with light forces.

 As to keeping dust out: A lubricant that isn't sticky will help; possibly
finely powdered Teflon, or the white powdered lock lubricant that used to
be (and might still be) on the market. Otherwise, it's a matter of
avoiding a dusty environment!

 In the Navy, we called it Preventive Maintenance, seems to me. (It was a
while ago!) Routine disassembly for cleaning maybe once or twice a year
seems reasonable. Fortunately, the "3" (and bigger ones) come apart
gracefully. The Pocket Cube does not! (It also hates to be reassembled as
well. Wonder whether it's assembled with special jigs and tools, or do the
assemblers simply suffer?)

On Thu, 28 Nov 1996 AirWong@aol.com wrote:

}Hello All!
}
}How do you keep the original Rubik's cube (3X3X3) clean? If you've ever taken
}apart the cube, you know what I'm talking about. The dust gets inside and
}builds up over time. Is there any way to keep the dust out?
}
}Aaron Wong

 My best to all,

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*    When the year 2000 begins, we'll celebrate 
|*  Amateur musician  *|*    the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------



From cube-lovers-errors@curry.epilogue.com  Sun Dec  1 00:58:01 1996
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Date: Sat, 30 Nov 1996 09:28:53 -0500 (EST)
From: Nicholas Bodley <nbodley@tiac.net>
Reply-To: Nicholas Bodley <nbodley@tiac.net>
To: Lars Petrus <lars@netgate.net>
cc: cube-lovers@ai.mit.edu
Subject: Re: Lubricants for puzzles
In-Reply-To: <v01530500aec3f0e7f1ed@[205.214.160.57]>
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 Thank you! A second vote for silicone. Another respondent, from Finland,
said one shouldn't spray the outside surfaces, but didn't say why. It's
possible that something in the formulation would react with the adhesive
for the stickers and make it become jelly-like; this happened to my "5",
but I have long ago forgotten what I used on it.

My regards to all,
_  __
|\ |_)
| \|_)





From cube-lovers-errors@curry.epilogue.com  Mon Dec  2 14:17:33 1996
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Date: Mon, 02 Dec 1996 12:03:28 +0000
To: cube-lovers@ai.mit.edu
From: "The Official Thermo-Fluids Fan Club of the UK. (Andy Southern)" <a.southern@ic.ac.uk>
Subject: Dirt?

Hi,

I don't think it's a case of keeping the dirt out as removing it from the cube.
I think the "Dust" is actually parts of the cube that have worn off, and
become entrapped in the cube.

A similar thing happens to roller bearings etc., but with more destructive
results.



Andy Southern.



From cube-lovers-errors@curry.epilogue.com  Tue Dec  3 17:44:19 1996
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Date: Tue, 3 Dec 1996 11:18:42 -0500
From: der Mouse <mouse@holo.rodents.montreal.qc.ca>
Message-Id: <199612031618.LAA12721@Collatz.McRCIM.McGill.EDU>
To: edwards@city-net.com
Subject: Re:  Rubic's Revenge
Cc: cube-lovers@ai.mit.edu

> Does a 5x5 matrix Rubic's [sic] cube exist?  I think I know the
> general solution, as an extrapolation from the solution to a 4x4.

Yes, a 5-Cube exists; I own one.  And yes, if you can solve the 3-Cube
and the 4-Cube, no higher order presents any qualitatively new
challenges to a human.  In theory, the 6-Cube would, because it's the
first one that has one-visible-face cubies that are not on a plane of
symmetry.  The 5-Cube has a 3x3 grid of one-face cubies on each face,
but they are all either (a) face center, (b) non-(a) center slice, or
(c) non-(a) face diagonal.  On the 4-Cube, the four one-face cubies are
all face diagonal, and on the 3-Cube, there's only one (face center)
one-face cubie.  However, at least based on my own experience, I
believe that these new cubies on the 6-Cube will not add any additional
challenge - the one-face cubies are one of the easiest parts of the
cube anyway, and the same basic operations that work for the (c) cubies
on the 4-Cube (and the (b) and (c) cubies on the 5-Cube) will work
equally well for these new cubies.

Incidentally, does anyone know if a physical 6-Cube has ever been made?
If so, and it's not too outrageously priced, I'd be interested in
buying one.

					der Mouse

			       mouse@rodents.montreal.qc.ca
		     7D C8 61 52 5D E7 2D 39  4E F1 31 3E E8 B3 27 4B


From cube-lovers-errors@curry.epilogue.com  Tue Dec  3 19:22:45 1996
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Date: Wed, 4 Dec 96 09:08:55 +0900
From: Norman Diamond 04-Dec-1996 0859 <diamond@jrdv04.enet.dec-j.co.jp>
To: Cube-Lovers@AI.MIT.EDU
Subject: Re:  Rubik's Revenge
Mime-Version: 1.0
Content-Type: TEXT/PLAIN; charset=ISO-2022-JP

Also spracht der Mouse:

>And yes, if you can solve the 3-Cube
>and the 4-Cube, no higher order presents any qualitatively new
>challenges to a human.

Depends on what you mean by a "challenge."  Average puzzlers who found
relatively ordinary algorithms for the 3-cube and 4-cube will discover
that they must find one new algorithm for the 5-cube, but it will be easy.

>In theory, the 6-Cube would, because it's the
>first one that has one-visible-face cubies that are not on a plane of
>symmetry.

Even after reading your explanation, I don't quite believe it, but would
love to own one and find out for sure :-)

Consider that the 4-cube (and 5-cube) can be made harder by forcing the
centre (or inner ring) cubies of each face to be oriented.  If this is
done to a 6-cube then of course the 6-cube becomes harder too.  Otherwise
I think my 4-cube algorithms would solve a 6-cube.

>Incidentally, does anyone know if a physical 6-Cube has ever been made?

Someone told Nob Yoshigahara that his country had solved the problem of
manufacturing the thing.  Then Nob was playing with the guy's business
card and lost it, and has never been able to find the guy again.  Although
Nob is known for a sense of humour at times, he sure wasn't joking when he
admitted this.  The only questions are whether the unknown person was
telling the truth, and who and where he is.

-- Norman Diamond              diamond@jrdv04.enet.dec-j.co.jp
[Speaking for Norman Diamond not for Digital.]


From cube-lovers-errors@curry.epilogue.com  Tue Dec  3 23:34:02 1996
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Date: Tue, 3 Dec 1996 21:54:18 -0500 (EST)
From: Nicholas Bodley <nbodley@tiac.net>
To: Norman Diamond 04-Dec-1996 0859 <diamond@jrdv04.enet.dec-j.co.jp>
cc: Cube-Lovers@ai.mit.edu
Subject: 6-cube and Hofstadter; Meffert
In-Reply-To: <9612040008.AA29913@jrdmax.jrd.dec.com>
Message-ID: <Pine.SUN.3.95.961203214340.27176B-100000@sunspot.tiac.net>


 Around 1985, I think, I was browsing through a book by Douglas
Hofstadter, most likely his _Metamagical Themas_; buried deep within it
was a comment (I think without attribution) that a prototype of the 6-cube
had been built, and a paper design for the 7-cube existed. Digging through
the archives for this mailing list might also yield something.

 He also said that almost all the Magic Dodecahedrons had been melted down
for their plastic, because each one used about $2 worth. This is one of
the sadder pieces of news I have heard.

 Does anyone know whether Uwe M=E8ffert (Meffert, if your char. set isn't
ISO 8859-1 compatible [See Moderator's Note]) is still in business? He's
Swiss, and was (and might still be) primarily an aquaculturist.

 The 6-cube would be fun to manipulate and solve, but what I'd really love
to see is its innards! The insides of the 5-cube are awe-inspiring... 

Regards to all,

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*    When the year 2000 begins, we'll celebrate 
|*  Amateur musician  *|*    the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------


[ Moderator's Note: Actually, I would prefer it if people didn't use
  extended ISO 8859 characters (or MIME, or PGP) in sending messages to
  Cube-Lovers.  This message got to the Cube-Lovers inbox quite mangled
  because of the presence of an 8-bit character in Meffert's name.  I undid
  the damage, and I agree that this -should- just have worked, and perhaps
  someday soon it -will-, but for now, it just doesn't work.  The the
  extent that it is possible, I request that Cube-Lovers subscribers stick
  to submitting messages formatted in plain ASCII.  Thanks!  - Alan ]

From cube-lovers-errors@curry.epilogue.com  Wed Dec  4 01:51:58 1996
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To: cube-lovers@ai.mit.edu
From: Georges Helm <geohelm@pt.lu>
Subject: Mark Longridge's Web Page

I'm looking for the Web page of Mark. (Hello Mark!). His e-mail address
doesn't work any longer. Thank you for helping.

Georges
         
geohelm@pt.lu
http://ourworld.compuserve.com/homepages/Georges_Helm
http://www.geocities.com/Athens/2715


[ Cube-Lovers-Request also wonders what happened to Mark Longridge.  He was
  a frequent contributor to Cube-Lovers.  The electronic mail adress I have
  for him stopped working over a month ago ("User unknown").  - Alan ]

From cube-lovers-errors@curry.epilogue.com  Wed Dec  4 22:36:51 1996
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Date: Wed, 4 Dec 1996 06:56:55 -0500
From: der Mouse <mouse@holo.rodents.montreal.qc.ca>
Message-Id: <199612041156.GAA16203@Collatz.McRCIM.McGill.EDU>
To: cube-lovers@ai.mit.edu
Subject: Re:  Rubik's Revenge

>> And yes, if you can solve the 3-Cube and the 4-Cube, no higher order
>> presents any qualitatively new challenges to a human.

> Depends on what you mean by a "challenge."  Average puzzlers who
> found relatively ordinary algorithms for the 3-cube and 4-cube will
> discover that they must find one new algorithm for the 5-cube, but it
> will be easy.

Are you referring to center-slice non-face-center cubies?  If so, well,
"duh, I missed that".  Dunno why it didn't register on me that the
5-Cube is the first one with such cubies, but it is.

I suppose it's just that to me, all eight non-face-center face cubies
on the 5-Cube "feel the same", so I didn't notice the difference.
(That's also why I think the "new" face cubies on the 6-Cube will not
be a challenge.  What algorithm _are_ you referring to?)

> Consider that the 4-cube (and 5-cube) can be made harder by forcing
> the centre (or inner ring) cubies of each face to be oriented.

True, of course; putting a picture (or equivalent) on the face of the
Cube makes anything above the 2-Cube at least somewhat harder.  But
AFAICT this doesn't explain your first statement, about needing a new
algorithm for the 5-Cube....

					der Mouse

			       mouse@rodents.montreal.qc.ca
		     7D C8 61 52 5D E7 2D 39  4E F1 31 3E E8 B3 27 4B


From cube-lovers-errors@curry.epilogue.com  Wed Dec  4 22:37:15 1996
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Date: Wed, 4 Dec 1996 10:33:02 -0600
To: cube-lovers@ai.mit.edu
From: Tom Magliery <mag@ncsa.uiuc.edu>
Subject: Re: Mark Longridge's Web Page

>I'm looking for the Web page of Mark. (Hello Mark!). His e-mail address
>doesn't work any longer. Thank you for helping.
>
>[ Cube-Lovers-Request also wonders what happened to Mark Longridge.  He was
>  a frequent contributor to Cube-Lovers.  The electronic mail adress I have
>  for him stopped working over a month ago ("User unknown").  - Alan ]

Mark's web page is gone as well.  There is a (now-broken) link to it on my
own cube page:  http://sdg.ncsa.uiuc.edu/~mag/Misc/CubeSoln.html

mag

--
.---o  Tom Magliery, Research Programmer                         .---o
`-O-.  NCSA, 605 E. Springfield                  (217) 333-3198  `-O-.
o---'  Champaign, IL 61820          O-        mag@ncsa.uiuc.edu  o---'




From cube-lovers-errors@curry.epilogue.com  Wed Dec  4 22:36:08 1996
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Date: Wed, 4 Dec 96 17:42:27 +0900
From: Norman Diamond 04-Dec-1996 1742 <diamond@jrdv04.enet.dec-j.co.jp>
To: Cube-Lovers@AI.MIT.EDU
Cc: nbodley@tiac.net
Subject: Re: 6-cube and Hofstadter; Meffert
Mime-Version: 1.0
Content-Type: TEXT/PLAIN; charset=ISO-2022-JP

Nicholas Bodley writes:

> He also said that almost all the Magic Dodecahedrons had been melted down
>for their plastic, because each one used about $2 worth.

I don't think either part of that is true.  I'm nearly certain that the
magic dodecahedron is still in Dr. Bandelow's catalog.  Although to be
pendantic, I believe Meffert's company was in Hong Kong, so the $2 value
of plastic might be accurate in Hong Kong dollars.

> Does anyone know whether Uwe M=E8ffert (Meffert, if your char. set isn't
>ISO 8859-1 compatible [See Moderator's Note]) is still in business?

1.  Maybe around 10 years ago, I read that his company went bankrupt.

2.  The world's largest country uses a language which can't be written
using ISO 8859-1 compatible characters.  So do some neighboring countries.
The world's second-largest country has a small fraction of its population
well educated in a language which can be written using those characters,
but its most commonly used language (and most of its languages) cannot be.

-- Norman Diamond        diamond@jrdv04.enet.dec-j.co.JP
[Speaking for Norman Diamond not for DIGITAL, in Japan or elsewhere.]


From cube-lovers-errors@curry.epilogue.com  Wed Dec  4 22:36:32 1996
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Date: Wed, 04 Dec 1996 10:03:09 +0000
To: Nicholas Bodley <nbodley@tiac.net>, Cube-Lovers@ai.mit.edu
From: "The Official Thermo-Fluids Fan Club of the UK. (Andy Southern)" <a.southern@ic.ac.uk>
Subject: Re: 6-cube and Hofstadter; Meffert

At 21:54 03/12/96 -0500, you wrote:
>
> Does anyone know whether Uwe M=E8ffert (Meffert, if your char. set isn't
>ISO 8859-1 compatible [See Moderator's Note]) is still in business? He's
>Swiss, and was (and might still be) primarily an aquaculturist.
>

Uwe Meffret is in business, in Aberdeen, Hong Kong.
My only question is What the F=56*7c()k (apologies to anyone with a ISO
8859-1)  is a aquaculturist?

> The 6-cube would be fun to manipulate and solve, but what I'd really love
>to see is its innards! The insides of the 5-cube are awe-inspiring... 

I think I could build one from a 4x4x4, but I don't have all that much time,
or a 4x4x4, and I'm not sure which of those is more important!


Andrew Southern



From cube-lovers-errors@curry.epilogue.com  Thu Dec  5 19:07:56 1996
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Date: Thu, 5 Dec 1996 00:23:14 -0500 (EST)
From: Nicholas Bodley <nbodley@tiac.net>
Reply-To: Nicholas Bodley <nbodley@tiac.net>
To: Cube-Lovers@ai.mit.edu
Subject: Thanks, and an apology, too
Message-ID: <Pine.SUN.3.95.961204222023.22514C-100000@sunspot.tiac.net>
MIME-Version: 1.0
Content-Type: TEXT/PLAIN; charset=US-ASCII


 My thanks to the people who replied to my query about Meffert; most said
he's still in business. Thanks also for updates on the Magic Dodecahedron.

 And... BIG apology for tossing in a "High-ASCII" (?) Latin-1 character! 
I only proved my optimism (that it would be OK) and my ignorance (that it
is >not< OK in this List! I'll try to e-mail Alan directly. I would never
have thought that one character would create a mess. Incidentally, and
totally off topic, I found the really-obscure Codepage 819 that makes
MS-DOS totally and directly compatible (no translations to Codepage 850)
with ISO 8859-1, and installed it. Recommended, but e-mail me if you are
interested. 

 Btw, I had problems replying to Norman Diamond; his From: e-mail address
apparently couldn't be found, and my message bounced. Maybe DNS problem?

 [There's just an off-chance that this is double-posted; sorry if so! I
don't think it will be.]

Regards to all, 
_  __
|\ |_)
| \|_)




From cube-lovers-errors@curry.epilogue.com  Thu Dec  5 19:08:19 1996
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Date: Thu, 05 Dec 1996 17:59:05 -0600
From: Scott Crawford <scrawfor@host1.dia.net>
Reply-To: scrawfor@host1.dia.net
X-Mailer: Mozilla 3.0Gold (Win95; I)
MIME-Version: 1.0
To: Cube List <Cube-Lovers@ai.mit.edu>
Subject: Identify this puzzle please!
Content-Type: text/plain; charset=us-ascii
Content-Transfer-Encoding: 7bit

I recently acquired a puzzle through the mail and am not sure what it is
called.  I'll give a shot at describing it:  It has 14 sides - 8
triangles, and 6 squares.  The squares are rotated 45 degrees and each
touch 4 other squares at the corners, while their sides match up with 4
triangles each.  You could achieve the shape of this puzzle by taking a
cube and shaving off the corners to create an equilateral triangle. 
There are no markings or words on it.  If anybody knows, please let me
know, as I am curious if this is the Magic Dodecahedron I've been
reading about.

Thanks,
Scottie


From cube-lovers-errors@curry.epilogue.com  Fri Dec  6 16:12:01 1996
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Message-Id: <6Dec1996.210000.Cube-Lovers@AI.MIT.EDU>
Date: Fri, 6 Dec 1996 21:00:00 GMT
From: Cube Lovers Moderator <Cube-Lovers-Request@AI.MIT.EDU>
To: Cube-Lovers@AI.MIT.EDU
Subject: Cuboctohedron [Digest v21 #77]

Cube-Lovers Digest          Fri, 6 Dec 96         Volume 21 : Issue 77

Today's Topic:
		     Identify this puzzle please!
			    Cuboctohedron

[ Due to a large number of similar messages on a single topic, I have
  gathered them together into digest format.  -Moderator ]

----------------------------------------------------------------------

Date: Thu, 5 Dec 96 16:33:41 PST
Message-Id: <9612060033.AA18270@quark.geoworks.com>
From: David Litwin <dlitwin@geoworks.com>
To: scrawfor@host1.dia.net
Cc: Cube List <Cube-Lovers@ai.mit.edu>
In-Reply-To: <32A761C9.72AD@host1.dia.net>
Subject: Identify this puzzle please!
Reply-To: dlitwin@geoworks.com


	Actually that is a cubo-octahedron and has as its internal
mechanism a normal cube (with different side and corner pieces).
	This has 10 sides, not 12 like the dodecahedron.  The faces of
a dodecahedron are pentagons and the internal mechanism is actually
quite similar in concept to the cube (12 centers on axes instead of 6
centers on axes).

	Dave Litwin

Scott Crawford writes:
> I recently acquired a puzzle through the mail and am not sure what it is
> called.  I'll give a shot at describing it:  It has 14 sides - 8
> triangles, and 6 squares.  The squares are rotated 45 degrees and each
> touch 4 other squares at the corners, while their sides match up with 4
> triangles each.  You could achieve the shape of this puzzle by taking a
> cube and shaving off the corners to create an equilateral triangle. 
> There are no markings or words on it.  If anybody knows, please let me
> know, as I am curious if this is the Magic Dodecahedron I've been
> reading about.

------------------------------

Date: Thu, 5 Dec 1996 19:29:53 -0500 (EST)
From: Dale Newfield <din5w@cs.virginia.edu>
Reply-To: DNewfield@cs.virginia.edu
To: Cube List <Cube-Lovers@ai.mit.edu>
Subject: Re: Identify this puzzle please!
In-Reply-To: <32A761C9.72AD@host1.dia.net>
Message-Id: <Pine.SUN.3.90.961205192726.27532O-100000@dot.cs.Virginia.EDU>


I have at least two of these.  There are semantically equivalent to a 
cube.  I would call them "Truncated cubes" or "truncated hexahedra".  I'm 
not sure what their marketing name was.

-Dale Newfield
 DNewfield@Virginia.edu

------------------------------

Message-Id: <2.2.32.19961206010234.006a133c@uclink4.berkeley.edu>
Date: Thu, 05 Dec 1996 17:02:34 -0800
To: Cube-Lovers@ai.mit.edu
From: Mark Pilloff <mdp1@uclink4.berkeley.edu>
Subject: Re: Identify this puzzle please!

I have one of these things-- got it many years ago and it's not the magic
dodecahedron.  As I recall, you can solve the six square sides like an
ordinary Rubik's cube and the other 8 triangles will automatically fall into
place so it's no harder or more interesting than the regular cube.

Mark

At 05:59 PM 12/5/96 -0600, Scottie wrote:
>I recently acquired a puzzle through the mail and am not sure what it is
>called.  I'll give a shot at describing it:  It has 14 sides - 8
>triangles, and 6 squares.  The squares are rotated 45 degrees and each
>touch 4 other squares at the corners, while their sides match up with 4
>triangles each.  You could achieve the shape of this puzzle by taking a
>cube and shaving off the corners to create an equilateral triangle. 
>There are no markings or words on it.  If anybody knows, please let me
>know, as I am curious if this is the Magic Dodecahedron I've been
>reading about.
>
>Thanks,
>Scottie
>
>
>

************************************
**    Mark D. Pilloff             **
**    mdp1@uclink4.berkeley.edu   **
************************************

------------------------------

Message-ID: <32A7718A.4616@host1.dia.net>
Date: Thu, 05 Dec 1996 19:06:40 -0600
From: Scott Crawford <scrawfor@host1.dia.net>
Reply-To: scrawfor@host1.dia.net
To: Cube List <Cube-Lovers@ai.mit.edu>
Subject: Re: Identify this puzzle please!
References: <9612060033.AA18270@quark.geoworks.com>

David Litwin wrote:
> 
>         Actually that is a cubo-octahedron and has as its internal
> mechanism a normal cube (with different side and corner pieces).
>         This has 10 sides, not 12 like the dodecahedron.  The faces of
> a dodecahedron are pentagons and the internal mechanism is actually
> quite similar in concept to the cube (12 centers on axes instead of 6
> centers on axes).
> 
>         Dave Litwin
> 
> Scott Crawford writes:
> > I recently acquired a puzzle through the mail and am not sure what it is
> > called.  I'll give a shot at describing it:  It has 14 sides - 8
> > triangles, and 6 squares.  The squares are rotated 45 degrees and each
> > touch 4 other squares at the corners, while their sides match up with 4
> > triangles each.  You could achieve the shape of this puzzle by taking a
> > cube and shaving off the corners to create an equilateral triangle.
> > There are no markings or words on it.  If anybody knows, please let me
> > know, as I am curious if this is the Magic Dodecahedron I've been
> > reading about.


Thanks a lot all!!!  Wow! Talk about fast response.  At first glance I
figured I would never be able to solve this one, but upon closer
examination, I realized I already knew how.  It is identical to the
original cube in methods of solving.  Again thanks, you guys are great!!

Scottie

------------------------------

From: Tim Botham <timbo@homemail.com>
To: Cube Lovers Postings <Cube-Lovers@ai.mit.edu>
Subject: Fw: Identify this puzzle please!
Date: Thu, 5 Dec 1996 22:19:08 -0800
Message-Id: <E0vVuBU-0004TR-00@dewey.mindlink.net>

Sounds like you have what I know as a "Truncated Cube", although I'm not
sure if that's the 'official' name.  The Magic Dodecahedron has 12 sides,
each in the shape of a pentagon.  Magic Dodecahedron's are available from
Dr. Christoph Bandelow for 35DM.  He has a variety of "cube puzzles" - for
a catalog, request to: Dr. Christoph Bandelow, An der Wabeck 37, D-58456
Witten, GERMANY.

	Tim

 ---------
From: Scott Crawford <scrawfor@host1.dia.net>
To: Cube List <Cube-Lovers@ai.mit.edu>
Subject: Identify this puzzle please!
Date: December 5, 1996 3:59 PM

I recently acquired a puzzle through the mail and am not sure what it is
called.  I'll give a shot at describing it:  It has 14 sides - 8
triangles, and 6 squares.  The squares are rotated 45 degrees and each
touch 4 other squares at the corners, while their sides match up with 4
triangles each.  You could achieve the shape of this puzzle by taking a
cube and shaving off the corners to create an equilateral triangle. 
There are no markings or words on it.  If anybody knows, please let me
know, as I am curious if this is the Magic Dodecahedron I've been
reading about.

Thanks,
Scottie

------------------------------

Date: Fri, 6 Dec 1996 11:13:10 GMT
Message-Id: <1.5.4.32.19961206120741.002b4114@mentda.me.ic.ac.uk>
To: scrawfor@host1.dia.net, Cube-Lovers@ai.mit.edu
From: "The Official Thermo-Fluids Fan Club of the UK. (Andy Southern)" <a.southern@ic.ac.uk>
Subject: Re: Identify this puzzle please!

At 17:59 05/12/96 -0600, you wrote:
>I recently acquired a puzzle through the mail and am not sure what it is
>called.  I'll give a shot at describing it:  It has 14 sides - 8
>triangles, and 6 squares.  The squares are rotated 45 degrees and each
>touch 4 other squares at the corners, while their sides match up with 4
>triangles each.  You could achieve the shape of this puzzle by taking a
>cube and shaving off the corners to create an equilateral triangle. 
>There are no markings or words on it.  If anybody knows, please let me
>know, as I am curious if this is the Magic Dodecahedron I've been
>reading about.
>
>Thanks,
>Scottie
>
>
>

No, the magic Dodecahedron has 12 regular pentagon faces.

I think you might be talking about Meffret's Creative Challenge, but that is
the Geodesic versions of those shapes on a sphere.

Andy.

------------------------------

Message-ID: <32A84381.753F@host1.dia.net>
Date: Fri, 06 Dec 1996 10:02:09 -0600
From: Scott Crawford <scrawfor@host1.dia.net>
Reply-To: scrawfor@host1.dia.net
To: Cube List <Cube-Lovers@ai.mit.edu>
Subject: Cuboctohedron

There has been quite a few inquiries as to how this puzzle is a puzzle
and not a hunk of cheese as one reader pointed out.

It turns just like a regular cube.  Apparently it has the same mechanism
inside.  It only LOOKS harder, but after taking a few seconds to analyze
it, I noticed that the solution was the same as the cube.  Still a neat
piece in a collection.

Scottie

------------------------------

End of Cube-Lovers Digest
*************************

From cube-lovers-errors@curry.epilogue.com  Mon Dec  9 00:03:51 1996
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From: rmead@ionet.net
Date: Sun, 8 Dec 1996 17:04:40 -0600 (CST)
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To: Cube-Lovers@ai.mit.edu
Subject: rubiks

does anyone know where i could purchase a rubik's cube?  i would really
prefer an original one made in the 1980's. 

rmead@ionet.net



From cube-lovers-errors@curry.epilogue.com  Mon Dec  9 00:04:56 1996
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Date: Sun, 8 Dec 1996 22:08:31 -0500 (EST)
From: Nicholas Bodley <nbodley@tiac.net>
To: Scott Crawford <scrawfor@host1.dia.net>
cc: Cube List <Cube-Lovers@ai.mit.edu>
Subject: Re: Identify this puzzle please!
In-Reply-To: <32A761C9.72AD@host1.dia.net>
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 Sorry, don't think I can help with your puzzle, but the Magic
Dodecahedron has (I surely hope!) twelve identical faces; each face is a
pentagon. ("Do" is "two", and "deca" is "ten", loosely speaking; it's a
prefix for "12", unless I'm 'way off base.)  I'm just about sure that each
face has five corner pieces (you shouldn't, strictly speaking, call them
"cubies")  that it shares with its neighbors; also five middle-edge
pieces. It's been years since I saw a photo (In the Meffert catalog of
around 1986). 

Regards,
_  __
|\ |_)
| \|_)



From cube-lovers-errors@curry.epilogue.com  Mon Dec  9 00:04:32 1996
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To: cube-lovers@ai.mit.edu
From: Dennis Okon <dokon@mit.edu>
Subject: Re: Mark Longridge's Web Page

I just found out that Mark's page is now at: http://web.idirect.com/~cubeman/

- Dennis Okon
dokon@mit.edu



From cube-lovers-errors@curry.epilogue.com  Mon Dec  9 15:41:55 1996
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Date: Mon, 09 Dec 1996 03:21:34 -0600
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Subject: Rubik's 15
References: <199611140236.AA29041@world.std.com>
Content-Type: text/plain; charset=us-ascii
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Does anyone know where I can find information about solving Rubik's 15?

Thanks
Scottie


From cube-lovers-errors@curry.epilogue.com  Mon Dec  9 15:41:26 1996
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Date: Mon, 9 Dec 1996 01:07:30 -0500 (EST)
From: Nicholas Bodley <nbodley@tiac.net>
To: rmead@ionet.net
cc: Cube-Lovers@ai.mit.edu
Subject: Where to get a Cube
In-Reply-To: <199612082304.RAA01952@mail.ionet.net>
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 Have you tried Puzzletts?  As to getting a 1980s Cube, try to get one
with attached colored plastic tiles instead of stickers, made by Ideal. 
Its mechanical design differs in certain details, so that it is to some
degree self-aligning, and much less likely to lock if slightly misaligned.
It is also made of what's probably the best (affordable!) plastic for this
application.  It might have been called a "Deluxe" Cube.

 It's possible that the store in Cambridge, Mass. called "The Games People
Play" might have 1990s Cubes in stock; I tend to think the superior kind
is probably history, but one can always hope. (Sorry! I have just one, and
I'm not selling it!)

I'm reasonably sure that Puzzletts is at http://www.puzzletts.com

Regards,

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*    When the year 2000 begins, we'll celebrate 
|*  Amateur musician  *|*    the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------



From cube-lovers-errors@curry.epilogue.com  Mon Dec  9 19:58:14 1996
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Date: Mon, 9 Dec 1996 18:47:07 -0200 (EDT)
From: Rodrigo de Almeida Siqueira <delirium@ime.usp.br>
X-Sender: delirium@farofa
To: cube-lovers@ai.mit.edu
Subject: Cool Robot Of The Week
Message-ID: <Pine.SUN.3.91.961209183924.2757A-100000@farofa>
MIME-Version: 1.0
Content-Type: TEXT/PLAIN; charset=US-ASCII


Hello Cubers,

The Laboratory of Integrated Systems (at Polytechnic School, Univ. of Sao 
Paulo) has just recieved the "Cool Robot Of The Week" unofficial 
recognition from NASA HQ for "Chip and Dale", a pair of robots that can 
deal and solve the Rubik's Cube!

For more information on the Robots and MPEG, GIF and JPG files with the 
images, check the URL:

   http://www.lsi.usp.br/~daia/celula/cubo/


Rodrigo Siqueira.
rodrigo@lsi.usp.br
delirium@ime.usp.br

Here goes the message:

---------- Forwarded message ----------
Date: Mon, 9 Dec 1996 14:11:56 -0500
From: Dave Lavery <dave.lavery@hq.nasa.gov>
To: delirium@ime.usp.br, rbianchi@lsi.usp.br, fferraz@lsi.usp.br,
    briand@lsi.usp.br
Cc: dave.lavery@hq.nasa.gov
Subject: Cool Robot Of The Week

Congratulations!  The Chip and Dale robots have been selected as "The Cool
Robot Of The Week" for December 9-15, 1996.

The honor of being listed as "Cool Robot Of The Week" is bestowed upon
those robotics-related web sites which portray highly innovative solutions
to robotics problems, describe unique approaches to implementing robotics
system, or present exciting interfaces for the dissemination of
robotics-related information or promoting robotics technology. This award
carries absolutely no monetary value, official recognition, assumed
support, or tangible benefit, other than swamping your web site with a few
dozen extra hits for a week. But everyone else was putting up their "Cool
Site Of The Millenia" lists, so we figured it was our turn too...

Anyway, to find out more about your listing, see the "Cool Robot Of The
Week" web site at:
http://ranier.hq.nasa.gov/telerobotics_page/coolrobots.html

Congratulations once again, and we hope that in some small way this "honor"
will bring some more recognition to your work.

-----------------------------------------------
Dave Lavery
Telerobotics Program Manager
NASA Headquarters, Code SM
e-mail: dave.lavery@hq.nasa.gov
http://ranier.hq.nasa.gov/telerobotics.html
-----------------------------------------------





From cube-lovers-errors@curry.epilogue.com  Tue Dec 10 11:06:20 1996
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Date: Wed, 11 Dec 1996 00:51:55 -0500
From: Mark Longridge <cubeman@idirect.com>
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To: cube-lovers@ai.mit.edu
Subject: Shortest possible Megaminx 2-flip
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I believe I have calculated the shortest possible 2 flip for the
Megaminx (Rubikian Dodecahedron):
 
R- F- U+ L- U- L+ F+ R+
L+ F+ U- R+ U+ R- F- L-
 
16 face turns & 16 unit turns.
 
-> Mark <-


From cube-lovers-errors@curry.epilogue.com  Thu Dec 12 17:46:06 1996
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Date: Thu, 12 Dec 1996 17:18:06 -0500 (EST)
From: Corey Scott <scot3165@ginko.ait.fredonia.edu>
To: Norman Diamond 04-Dec-1996 1742 <diamond@jrdv04.enet.dec-j.co.jp>
cc: Cube-Lovers@ai.mit.edu, nbodley@tiac.net
Subject: Re: 6-cube and Hofstadter; Meffert
In-Reply-To: <9612040842.AA14888@jrdmax.jrd.dec.com>
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i was wondering, i bought a rubic's cube about a month ago and my stickers
are already coming off. what can i do  about it? and why don't they use
something different, like plastic color tabs that are superglued on?
please help because i like my cubeand now i'm afraid to use it, it's
dying.

thank you
corie elizabeth




From cube-lovers-errors@curry.epilogue.com  Thu Dec 12 20:33:29 1996
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From: Tim Botham <timbo@mindlink.net>
To: Corey Scott <scot3165@ginko.ait.fredonia.edu>,
        Norman Diamond 04-Dec-1996 1742 <diamond@jrdv04.enet.dec-j.co.jp>
Cc: Cube-Lovers@AI.MIT.EDU, nbodley@tiac.net
Subject: Re: 6-cube and Hofstadter; Meffert
Date: Thu, 12 Dec 1996 17:28:36 -0800
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An excellent way to re-attach your stickers was taught to me by Mike Green
at Puzzletts.  First clean the blank spot on the cube very well with
rubbing alcohol.  Then put the sticker on piece of two-sided scotch tape,
and trim the tape around the edges of the sticker.  Press it on to the
cube, and presto - it should outlast any of the glued-on stickers.

I tried it, and it worked well.

	Tim
----------
From: Corey Scott <scot3165@ginko.ait.fredonia.edu>
To: Norman Diamond 04-Dec-1996 1742 <diamond@jrdv04.enet.dec-j.co.jp>
Cc: Cube-Lovers@ai.mit.edu; nbodley@tiac.net
Subject: Re: 6-cube and Hofstadter; Meffert
Date: Thursday, December 12, 1996 2:18 PM

i was wondering, i bought a rubic's cube about a month ago and my stickers
are already coming off. what can i do  about it? and why don't they use
something different, like plastic color tabs that are superglued on?
please help because i like my cubeand now i'm afraid to use it, it's
dying.

thank you
corie elizabeth





From cube-lovers-errors@curry.epilogue.com  Fri Dec 13 14:29:29 1996
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Date: Fri, 13 Dec 1996 13:47:01 -0500
From: der Mouse <mouse@holo.rodents.montreal.qc.ca>
Message-Id: <199612131847.NAA17516@Collatz.McRCIM.McGill.EDU>
To: Cube-Lovers@ai.mit.edu
Subject: Stickers falling off [was Re: 6-cube and Hofstadter; Meffert]

> i was wondering, i bought a rubic's cube about a month ago and my
> stickers are already coming off.  what can i do about it?

I have used two solutions.  One is, take the sticker off and then glue
it back on with contact cement; two is, take the stickers off entirely,
clean the glue with rubbing alcohol or some such (careful you don't use
something that dissolves the plastic!), and then paint the cubies.
Pick a paint designed to stick to plastic; model paint probably will
work well, though I used some artist's paints we had around.

> and why don't they use something different, like plastic color tabs
> that are superglued on?

Not being someone who has ever made that decision, I don't know...but
my guess is, cost.. :-(

					der Mouse

			       mouse@rodents.montreal.qc.ca
		     7D C8 61 52 5D E7 2D 39  4E F1 31 3E E8 B3 27 4B


From cube-lovers-errors@curry.epilogue.com  Fri Dec 13 14:54:36 1996
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To: Cube-Lovers@AI.MIT.EDU
From: Wei-Hwa Huang <whuang@ugcs.caltech.edu>
Subject: Re: 6-cube and Hofstadter; Meffert
Date: 13 Dec 1996 19:45:47 GMT
Organization: California Institute of Technology, Pasadena
Lines: 34
Message-ID: <58sbpb$m6p@gap.cco.caltech.edu>
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X-Newsreader: NN version 6.5.0 #2 (NOV)

Corey Scott <scot3165@ginko.ait.fredonia.edu> writes:

>i was wondering, i bought a rubic's cube about a month ago and my stickers
>are already coming off. what can i do  about it? 

Not much.  I find that the only technique that seems to slow down the
sticker-losing is:

1) Don't let anybody else play with it.  Greasy and oily hands damage
 the glue, and every once in a while you'll run into some idiot who 
 will try to impress you by removing the stickers and putting them
 back on in order.

>and why don't they use
>something different, like plastic color tabs that are superglued on?

Probably cost.  Ideal made "Rubik's Game" and the Deluxe "Rubik's Cube"
a while ago, both which had plastic faces and were much more durable.

Unfortunately, no such equivalent was ever produced for the higher-order
cubes; I've already lost three 4x4x4 stickers and one 5x5x5 sticker.

>please help because i like my cubeand now i'm afraid to use it, it's
>dying.

I know the feeling.   Your best bet is not to worry about it, and
to produce your own replacement stickers.



--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
-------------------------------------------------------------------------------
Statistics show that Ster Trek films without Shatner do better at box offices.


From cube-lovers-errors@curry.epilogue.com  Fri Dec 13 16:00:19 1996
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Date: Fri, 13 Dec 1996 15:17:16 -0500 (EST)
From: Dale Newfield <din5w@cs.virginia.edu>
Reply-To: DNewfield@cs.virginia.edu
To: Cube-Lovers@ai.mit.edu
Subject: Re: 6-cube and Hofstadter; Meffert
In-Reply-To: <58sbpb$m6p@gap.cco.caltech.edu>
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I remember that the 5x5x5 cube I got came with a piece of paper that said 
to cover the face with peeling stickers with a sheet of paper, and apply
a warm iron to the paper.

This re-flattened the stickers, and re-strengthened the glue bond.

This would not be a valid solution if the stickers on your cube are 
sliding off rather than curling up.

-Dale



From cube-lovers-errors@curry.epilogue.com  Fri Dec 13 22:55:45 1996
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Date: Sat, 14 Dec 96 11:15:16 +0900
From: Norman Diamond 14-Dec-1996 1039 <diamond@jrdv04.enet.dec-j.co.jp>
To: Cube-Lovers@AI.MIT.EDU
Subject: Stick the non-stickers (was Re: 6-cube and Hofstadter; Meffert)
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Suggestion for lost stickers:

2.  Make a bandaged cube in the style of Dieter Gebhardt or Greg Stevens'
works, and make suitable stickers yourself for the bandaged cube.

1.  Before bandaging the cube in step 2, first you must remove the original
stickers.  Put some of them on the cube that is missing stickers, and save
the others to take care of future losses.

0.  Choose a cube that is already missing the most stickers, and use that
one in making the bandaged cube.

Notice that there is no step 3.  If you do an adequate job of bandaging, it
will be incredibly difficult to solve the cube after messing it up.  So there
is no step 3.  Do not post to this list asking for help in solving it :-)

-- Norman Diamond           diamond@jrdv04.enet.dec-j.co.jp
[Speaking for Norman Diamond not for Digital.]


From cube-lovers-errors@curry.epilogue.com  Mon Dec 16 00:45:32 1996
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Date: Mon, 16 Dec 1996 00:28:02 -0500 (EST)
From: Nicholas Bodley <nbodley@tiac.net>
To: Corey Scott <scot3165@ginko.ait.fredonia.edu>
cc: Norman Diamond 04-Dec-1996 1742 <diamond@jrdv04.enet.dec-j.co.jp>,
        Cube-Lovers@ai.mit.edu
Subject: Stickers coming off
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 It's too easy to say, "buy a better Cube." I hope Ideal (if they're still
the official source) wasn't the source for yours.

 You need to know your solvents fairly well, but you might be able to find
a solvent that would not attack the plastic nor the stickers. Once you use
this to clean off all the adhesive, then use a solvent-cure adhesive like
model-airplane glue or Duco to refasten the stickers. Cyanoacrylate (CA,
"Super Glue") should work, but it requires skill to use; it's easy to use
too much; also skill to keep it off your skin. For many CA formulations,
the surfaces must be clean.

 Good luck!

(Can we persuade Ideal to do a run of their deluxe Cubes, that had plastic
tiles and a self-aligning, low-friction mechanism?)

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*    When the year 2000 begins, we'll celebrate 
|*  Amateur musician  *|*    the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------



From cube-lovers-errors@curry.epilogue.com  Tue Dec 17 15:03:45 1996
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Date: Tue, 17 Dec 1996 14:18:45 BST
From: David Singmaster Computing & Maths South Bank Univ <david.singmaster@sbu.ac.uk>
To: cube-lovers@ai.mit.edu
Message-ID: <009ACFAD.AD4345C0.1124@vax.sbu.ac.uk>
Subject: Cube stickers

	It's distressing to hear that the current cube stickers come off.
This was not a real problem with older cubes, except for the orange stickers
of the 5x5x5.  I found that one can buy plastic sheets from artist's
supply shops.  These have sticky backs, on a plastic backing, like peel-off
labels.  I found one could get quite a selection of colors and they adhered
quite well.
	Regards
DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499; 
email:  zingmast  or  David.Singmaster  @vax.sbu.ac.uk




From cube-lovers-errors@curry.epilogue.com  Tue Dec 17 21:57:04 1996
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Date: Tue, 17 Dec 1996 20:39:33 -0500
To: David Singmaster Computing & Maths South Bank Univ <david.singmaster@sbu.ac.uk>,
        cube-lovers@ai.mit.edu
From: Nichael Cramer <nichael@sover.net>
Subject: Re: Cube stickers

David Singmaster Computing & Maths South Bank Univ wrote:
>This was not a real problem with older cubes, except for the orange stickers
>of the 5x5x5.

Right.  I have 3 5Xs and they're _all_ losing their orange stickers.  (And
only their orange stickers, which I've had trouble with since they were
new.)

What is it with these things??


Nichael Cramer             "...and they opened their thesaurus
nichael@sover.net                  and brought forth gold and frankincense
http://www.sover.net/~nichael          and myrrh."




From cube-lovers-errors@curry.epilogue.com  Wed Dec 18 13:19:59 1996
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From: ba05133@binghamton.edu
Date: Wed, 18 Dec 1996 11:41:04 -0500 (EST)
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To: cube-lovers@ai.mit.edu
Subject: Is there a 15q move for this???
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Is there a 15 quarter move algorithm for this position:

L2 F' L D2 R' B R D2 L B L F L' B'   ?

Maybe, Kociemba's algorithm will help.

Thanks a lot in advance!!

Jiri Fridrich



From cube-lovers-errors@curry.epilogue.com  Wed Dec 18 14:40:42 1996
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Date: Wed, 18 Dec 1996 11:39:51 -0800 (PST)
From: David Barr <davidb@u.washington.edu>
To: cube-lovers@ai.mit.edu
Subject: Quickcam
Message-ID: <Pine.A41.3.95b.961218113401.16826I-100000@homer26.u.washington.edu>
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I have a Color Quickcam camera attached to my PC, and I was wondering
how difficult it would be to write a program that would take two
pictures of a cube (to cover all six sides) and determine the cube
position from the images.  The positions could then be fed into
another program (maybe one of the WWW cube solvers) for further
processing.

Is there any source code out there that does image recognition of cube
positions?



From cube-lovers-errors@curry.epilogue.com  Thu Dec 19 03:26:06 1996
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Date: Thu, 19 Dec 1996 03:24:12 -0500 (EST)
From: Nicholas Bodley <nbodley@tiac.net>
To: Nichael Cramer <nichael@sover.net>
cc: David Singmaster Computing & Maths South Bank Univ <david.singmaster@sbu.ac.uk>,
        cube-lovers@ai.mit.edu
Subject: Re: Cube stickers
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 I've had the same luck, but I thought it was lubricant that had migrated;
I'm reasonably sure I tried lubricating my "5". Lubricant could turn
adhesive into a gel, I think. 

 Why the orange ones only? I think the reason is that they are a different
pigment (mine's fluorescent, pretty sure!) on a different backing, perhaps
plastic, while the others might be a high-quality paper; the plastic might
require a different kind of adhesive.  (Then, again, maybe when the maker
shopped around for sheets of colored adhesive sheet stock, what was
available at a a good price might not have included orange...) 

 It's just possible that the molding compound included a few percent
lubricant, and that interacted with the adhesive to create a gel. Consider
what happens to the adhesive on masking tape when you apply it to
plasticized PVC (vinyl) and leave it there for a few weeks or longer. The
plasticizer (which, I understand, is a actually low-vapor-pressure
solvent)  migrates into the adhesive. Migration of plasticizer is a
problem with plasticized PVC. (Any flexible vinyl is plasticized; rigid
vinyl is quite stiff!))


|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*    When the year 2000 begins, we'll celebrate 
|*  Amateur musician  *|*    the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------



From cube-lovers-errors@curry.epilogue.com  Fri Dec 20 16:50:33 1996
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From: Danny Chamberlin <critter@worldreach.net>
Subject: Re: Cube stickers
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At 02:18 PM 12/17/96 BST, you wrote:
>	It's distressing to hear that the current cube stickers come off.
>This was not a real problem with older cubes, except for the orange stickers
>of the 5x5x5.  I found that one can buy plastic sheets from artist's
>supply shops.  These have sticky backs, on a plastic backing, like peel-off
>labels.  I found one could get quite a selection of colors and they adhered
>quite well.
>	Regards
>DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist

I don't know if that's the case...I still have two original Ideal cubes
(non-Deluxe) and on both of them, shortly after I got them some of the blue
stickers started coming off, so I took the rest off, cleaned the faces off,
and just had 1 blank face on each cube!


From cube-lovers-errors@curry.epilogue.com  Mon Dec 23 17:50:23 1996
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Date: Mon, 23 Dec 1996 15:47:03 -0800
From: Aaron Coles <acoles@fec.gov>
Subject: Rubik's Tangle
To: cube-lovers@ai.mit.edu
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Does anyone know if the solution located at the below address is valid??

	http://www.math.uni-kiel.de/roesler/bruhn/tlsg.htm

[ The moderator has taken the liberty of correcting a small typo here.
  The original submitted message had an underscore ("_") in the URL which 
  I have corrected to be a hyphen ("-").  - Alan ]

From cube-lovers-errors@curry.epilogue.com  Tue Dec 24 15:05:09 1996
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Date: Tue, 24 Dec 1996 12:56:26 -0200 (GMT-0200)
From: Rodrigo de Almeida Siqueira <delirium@ime.usp.br>
X-Sender: delirium@catatau
To: cube-lovers@ai.mit.edu
Subject: Java Rubik's Cube!
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Wow!!!

I've just found and played with a Java version of the Rubik's Cube.
It's great and colorful. You can play and rotate the Cube.

It's available here:

	http://www.zaz.com.br/vestibular/1/Demos/j3.html

have fun.

----------------------------------------
Rodrigo Siqueira
Personal homepage:
http://www.insite.com.br/bio/rodrigo/
----------------------------------------



From cube-lovers-errors@curry.epilogue.com  Tue Dec 24 15:52:02 1996
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Date: Tue, 24 Dec 1996 15:49:47 -0500 (EST)
Message-Id: <199612242049.PAA02532@spork.bbn.com>
From: Allan Wechsler <awechsle@bbn.com>
To: Rodrigo de Almeida Siqueira <delirium@ime.usp.br>
Cc: cube-lovers@ai.mit.edu
Subject: Java Rubik's Cube!
In-Reply-To: <Pine.SUN.3.91.961224124630.9382A-100000@catatau>
References: <Pine.SUN.3.91.961224124630.9382A-100000@catatau>

For those having trouble with
    	http://www.zaz.com.br/vestibular/1/Demos/j3.html
my Portuguese is unreliable, but here is a first cut at a translation:

Magic Cube

The famous magic cube is here in 3D for [desafia]ing.  You can turn
the cube freely (by) dragging with the mouse.  To turn the faces, drag
with the mouse _on the vertices_.  To play: click the [seta] of the
mouse over the cube and type m to mess up all the faces.  With the
[seta] of the mouse over the cube, type i to reinitialize.

(Or you may see:

Your Netscape isn't understanding Java applets!  [Atualize] for a more
recent version.)

Game created by Marcelo Ribeiro.

[I have to confess that I don't understand the heuristic by which it
decides which face to turn.  I actually tried to solve the thing and
it gave me the screaming meemies.]

-A
    
    
    


From cube-lovers-errors@curry.epilogue.com  Thu Dec 26 17:07:53 1996
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From: "joyner.david" <joyner.david@mathnt1.sma.usna.navy.mil>
To: "'acoles@fec.gov'" <acoles@fec.gov>
Cc: "'cube-lovers@ai.mit.edu'" <cube-lovers@ai.mit.edu>
Subject: RE: Rubik's Tangle
Date: Thu, 26 Dec 1996 11:56:11 -0500
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Aaron Coles:
 The Rubik's tangle I bought is 3x3 and in that puzzle there 
are no tiles with a straight yellow. The version of the Rubik's 
tangle which that page you gave describes is not, as far as I 
know, marketed in stores.
 Here is my solution to the store version I bought:
Notation: green=color 1, purple = color 2, red = color 3, 
yellow = color 4. Each tile will be denoted by a 4-tuple
(a,b,c,d), where 
a is the color number of the straight rope,
b is the color number of the quarter circle rope,
c is the color number of the twisted rope (the one going from
 one side to the opposite which is not straight),
d is the color number of the looped rope (the one going from
 one side to an adjacent side which is not in the shape
 of a quarter circle). 
The "orientation" of a tile will be
0 if the straight side is lined up vertically on the left,
1 if it is rotated 90 degrees clockwise from the 
 orientation 0 position,
2 if it is rotated 180 degrees clockwise from the 
 orientation 0 position (the straight side is lined up vertically 
 on the right),
3 if it is rotated 270 degrees clockwise from the 
 orientation 0 position.
I labeled the (2-sided) tiles arbitrarily 1-9, with f for front 
and b for back. They are as follows:

tile 1
f (1,3,4,2), b (1,4,2,3)
tile 2
f (1,4,2,4), b (1,4,3,2)
tile 3 
f (1,2,4,3), b (1,2,3,4)
tile 4 
f (3,1,2,4), b (3,2,4,1)
tile 5
f (3,4,2,1), b (3,4,1,2)
tile 6 
f (3,2,1,4), b (3,1,4,2)
tile 7
f (2,4,3,1), b (2,1,3,4)
tile 8
f (2,1,4,3), b (2,3,1,4)
tile 9 
f (2,3,4,1), b (2,4,1,3)

Mathematics of puzzles:
 In general, let X be a collection of n interlocking puzzle pieces.
Assume that there is a solution to the puzzle for X
which uses every element of X.
Call G a "subpuzzle graph on X" if there is a subpuzzle 
of interlocking pieces constructed from a subset Y of X
such that G is a graph with vertices labeled by the subset Y
of X and two vertices are connected by an edge if and only 
if the corresponding pieces fit or interlock in this subpuzzle.
A "solution" to the puzzle will be a connected subpuzzle 
graph on X having n vertices associated to a solution. 
 For almost every jigsaw puzzle I've seen and for the Rubik's 
tangle puzzle, a solution to a puzzle is a Hamiltonian graph. 

Algorithm for the Rubik's tangle: We shall construct
a subpuzzle graph on the Rubik's tangle pieces as follows:
Notation: Label the positions of the puzzle as 
9-2-3
|  |  |
8-1-4
|  |  |
7-6-5
step 1: Pick a tile (18 possible choices)  and put it in
position number 1 with orientation 0. Draw a corresponding 
vertex and label it with this tile's 4-tuple. 
inductive step: Assume that k tiles have been placed in positions
1 through k and each tile fits with its neighboring tiles, k<n.
Pick a tile which fits with tile k in position number k+1
(in some orientation). Draw a corresponding vertex, label 
it with this tile's 4-tuple and it's orientation. Draw an edge from 
tile k+1 to each neighboring tile.

Solution to Rubik's tangle:
3421,2 - 1342,2 - 3214,3
  |              |           |
1243,2 -  2431   - 3124,1
  |              |           |
2143,2 - 1324,3 - 2413,2

                                                      - David Joyner


>----------
>From: 	Aaron Coles[SMTP:acoles@fec.gov]
>Sent: 	Monday, December 23, 1996 6:47 PM
>To: 	cube-lovers@ai.mit.edu
>Subject: 	Rubik's Tangle
>
>Does anyone know if the solution located at the below address is
>valid??
>
>	http://www.math.uni-kiel.de/roesler/bruhn/tlsg.htm
>
>[ The moderator has taken the liberty of correcting a small typo here.
>  The original submitted message had an underscore ("_") in the URL
>which 
>  I have corrected to be a hyphen ("-").  - Alan ]
>


From cube-lovers-errors@curry.epilogue.com  Sat Dec 28 17:34:33 1996
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From: Jon Teague <jon@jandj.netkonect.co.uk>
To: Cube-Lovers@ai.mit.edu
Subject: Re: Rubik's Tangle
Date: Sat, 28 Dec 1996 14:57:49 -0000
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There are four Rubik's Tangles. Each of them
has 25 pieces and forms a 5x5 square. Once you've
got all four you can mix all of the pieces up and produce
a giant 10x10.

I don't know if Dr Bruhns solution on his page 
<http://www.math.uni-kiel.de/roesler/bruhn/tlsg.htm>
is correct - once I get to print it out in colour at the
office I'll see if it matches the tangle I have (# 4).

Jon Teague
<jon@jandj.netkonect.co.uk>




From cube-lovers-errors@curry.epilogue.com  Mon Dec 30 00:44:06 1996
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Date: Sun, 29 Dec 1996 14:03:06 +0100
From: Rob Hegge <R.F.Hegge@mp.tudelft.nl>
Subject: Re: Rubik's Tangle -> a new one
To: Cube-Lovers@ai.mit.edu
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> There are four Rubik's Tangles. Each of them
> has 25 pieces and forms a 5x5 square. Once you've
> got all four you can mix all of the pieces up and produce
> a giant 10x10.
I have read somewhere that this is not possible.

> I don't know if Dr Bruhns solution on his page 
> <http://www.math.uni-kiel.de/roesler/bruhn/tlsg.htm>
> is correct - once I get to print it out in colour at the
> office I'll see if it matches the tangle I have (# 4).
> 
> Jon Teague
> <jon@jandj.netkonect.co.uk>

Actually there are 5 now, a new one is on sale in the US:
It contains 9 pieces, all of them are double sided with the
ropes in 3D, i.e. the pieces are not flat as compared to the
other 4 sets.

Rob Hegge


From cube-lovers-errors@curry.epilogue.com  Mon Dec 30 17:55:33 1996
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From: Stan Isaacs <isaacs@hpcc01.corp.hp.com>
Message-Id: <199612301945.AA060415116@hpcc01.corp.hp.com>
Subject: Re: 6-cube and Hofstadter; Meffert
To: cube-lovers@ai.mit.edu
Date: Mon, 30 Dec 96 11:45:15 PST
In-Reply-To: <1.5.4.32.19961204100309.002b8e90@mentda.me.ic.ac.uk>; from "The Official Thermo-Fluids Fan Club of the UK." at Dec 04, 96 10:03 am
Mailer: Elm [revision: 70.85.2.1]

> 
> Uwe Meffret is in business, in Aberdeen, Hong Kong.
> My only question is What the F=56*7c()k (apologies to anyone with a ISO
> 8859-1)  is a aquaculturist?

Anybody have his mailing address?

 -- Stan Isaacs
 -- isaacs@corp.hp.com


From cube-lovers-errors@curry.epilogue.com  Mon Jan  6 18:56:19 1997
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Date: Tue, 07 Jan 1997 00:22:25 +0100
From: Manfred Polak <manfred.polak@metronet.de>
X-Mailer: Mozilla 2.02 [de] (Win95; I)
To: Cube-Lovers@ai.mit.edu
Subject: Search

Help! Where can I get a 5*5*5 Rubik's Cube?
I have some problems with my news-server, so if you got some advice, 
please send me a mail!
  Greetings from Munich (Germany)


From cube-lovers-errors@curry.epilogue.com  Wed Jan  8 18:16:23 1997
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From: Manfred Polak <manfred.polak@metronet.de>
X-Mailer: Mozilla 3.01 (Win95; I)
To: Cube-Lovers@ai.mit.edu
Subject: Re:Search

No more mails, please; I've already ordered one!
Many thanks


From cube-lovers-errors@curry.epilogue.com  Thu Jan  9 00:25:55 1997
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Date: Thu, 9 Jan 1997 00:25:05 -0500
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To: Cube-Lovers@ai.mit.edu
From: David Lee Winston Miller <millerd1@sunyit.edu>
Subject: Cube page

Hi,

Just thought you might want to know about my Rubik's Cube page(s).  You can
find it at:

    http://www.sunyit.edu/~millerd1/RUBIK.HTM

It is mainly a demo on solution searching and algorithms.

Thanks,

David Lee Winston Miller

David Lee Winston Miller   Mail: millerd1@sunyit.edu   Homepage:
http://www.sunyit.edu/~millerd1



From cube-lovers-errors@curry.epilogue.com  Tue Jan 28 12:26:26 1997
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To: Cube-Lovers@ai.mit.edu
From: Dennis Okon <dokon@mit.edu>
Subject: Broken Cube

I've got a problem.  My cube (3x3x3) is quite old and very used.
Unfortunately, I was using it the other day and the whole thing fell apart
in my hands - the center cross piece that holds the whole thing together
split in two.  I'm afraid all my attempts with gluing it back together have
failed.  Does anyone know where I can get just the center piece?

- Dennis Okon
dokon@mit.edu



From cube-lovers-errors@curry.epilogue.com  Tue Jan 28 23:59:33 1997
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Date: Wed, 29 Jan 97 08:54:57 +0900
From: Norman Diamond 29-Jan-1997 0854 <diamond@jrdv04.enet.dec-j.co.jp>
To: cube-lovers@ai.mit.edu
Apparently-To: cube-lovers@ai.mit.edu
Subject: Re: Broken Cube
Mime-Version: 1.0
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>the center cross piece that holds the whole thing together split in two.
>Does anyone know where I can get just the center piece?

Sure!  Buy a new cube and remove the 26 visible cubies.

-- Norman Diamond            diamond@jrdv04.enet.dec-j.co.jp
[Speaking for Norman Diamond not for Digital.]


From cube-lovers-errors@curry.epilogue.com  Thu Jan 30 01:31:45 1997
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To: Cube-Lovers@AI.MIT.EDU
From: Wei-Hwa Huang <whuang@ugcs.caltech.edu>
Subject: Taiwan Cube-Solving Invitational
Date: 29 Jan 1997 20:13:35 GMT
Organization: California Institute of Technology, Pasadena
Message-ID: <5cob1f$jiq@gap.cco.caltech.edu>

This is from China World News, 12/17/1996.
The translation is mine.  I can't vouch for any facts in
the article or give any more information.
(I also find a few of the claims a bit exaggerated
and redundant, but what do you expect from the media?)

-------------------------

NEW MAGIC CUBE REVEALED; WELCOMING ALL SPEED SOLVERS
Easy Disassembly Provides a Higher Difficulty of Play;
Taiwan Hosts Invitational Competition; Throws the Gauntlet
to USA, Europe, etc.

[TAIPEI] The craze that mesmerized Taiwan is back.  This new
generation of Rubik's Cube emphasizes its ability to be 
disassembled and assembled.  Play and difficulty are enhanced.
This cube researched and developed by Taiwanese won a gold in the
International Invention Competition (?) in Los Angeles this
past September.  This new Cube is also the focus of 1997's 
International Magic Cube Solving Invitational.  The hosts, intending to
promote basic inventive education, especially welcome youths
to enter the solving competition and challenge their intelligence.

The Chief Organizer of the sponsor of next year's International Cube 
Solving Invitational, the Republic of China's Invention Group,
Chi-Shin (?) Huang [no relation] said that superb inventors always
have three traits: they're capable of prolonged concentration, 
stamina, and will.  Most importantly, that have the trait of 
action: they know what they want to do and they do it.  And this
is a trait they have in common with cube-solvers.

Huang especially emphasizes that the young people now pick up this
toy of the previous generation, since the Rubik's Cube is a
puzzle requiring hand-eye coordination, and in the process of
solving, not only does it help build spatial abilities and 
memory, but also trains the brain in pattern recognition and
organization.  And as far as creativity is concerned, cube solving
is good practice for sustained concentration and patience.

Jein-Chen Zhu (?), a professor at Taiwan University who has
written a book analyzing the mathematical principles of the
cube comments: "This new generation of the Cube is challenging
because it breaks the rules of the original Cube."  Zhu also
said that most speed solvers finish the buttom face first,
and then work towards the upper levels.  Since the new Cube can
be disassembled and the corner cubies rearranged, the new
Cube may have no valid solutions with that method.  "Therefore,
if a solver picks an incorrect starting face, it would be like
building a bridge on top of a riverbed with hidden sand streams:
almost impossible.  The solver must be able to make quick decisions
to finish the solution.

The inventor of the new Cube, Wei-Huang Chang (?) [no relation]
indicates that the breakthrough in his design is that the Cube
can now be easily disassembled, allowing players to observe with
their own eyes the three-dimensional secrets in the structure of
the versatile cube.  The original inventor of the Cube, Rubik,
was himself an expert in the field of construction and originally
invented the cube as a teaching tool.  The new Cube would not
exist without the original design.

The 1997 Competition is projected to be in September.  The 
Committee invites any players from Japan, Korea, Mainland China, Europe,
America, etc. to attend.  No age restriction.  If interested,
please phone (area code 02) 5812314.  [Country code, Taiwan ROC]

----------------------------------

Magic Cube Solved in 19 Seconds
WEN-YUNG KAO (?) IS THE FASTEST SOLVER IN TAIWAN

[TAIPEI]"Mr. Cube" Wen-Yung Kao's fastest solving time is nineteen seconds.
Compared to existing records, he is 2 seconds faster than the 
world record, 22.95 seconds.  

"A Cube solver must make decisions quickly and correctly!" says
Kao, who is a successful industrialist by day.  When he solves 
the cube, all ten of his fingers dextrously spin the cube; in
the last few seconds of his solution, he finishes the solution 
without even having to look at the cube, truly a sight to behold.

Kao notes that actually, there's a trick to solving the cube; he
has had to go through meticulous recordings and observations to
discover it.  He recalls his first encounter with the cube: a
friend of his had started up a factory set to making the cube
and had sent a few to him to play with.  He was immediately 
hooked and spent umpteen hours testing, taking notes, and 
recording the path of every little cubie.  After one month,
he could solve the cube in three minutes.  After half a year
he was even faster, solving in under one minute.  After 20
more months he had cut his time down to 25 seconds.

[TAIPEI]Sixteen years ago the cube-solving craze spread the 
globe.  Due to the complexity of the combinations, a few
turns could make it difficult to restore to its original
state, easily discouraging a would-be solver.  Taiwan inventor
Ji-ren Chang (?) researched the "Gold Key Cube", which 
can be easily disassembled and reassembled, decreasing the
feeling of discouragement.

[PHOTO] CEO of a food company, Wen-Yung Kao is Taiwan's fastest
Cube solver.

[PHOTO] Inventor Ji-Ren Chang's invention "Gold Key" can disassemble
the Cube.

--------------------------



--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
-------------------------------------------------------------------------------
Save the earth.  Then Quit and Shut Down.


From cube-lovers-errors@curry.epilogue.com  Thu Jan 30 17:01:45 1997
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Date: Thu, 30 Jan 1997 15:02:25 -0500 (EST)
From: Dale Newfield <din5w@cs.virginia.edu>
Reply-To: DNewfield@cs.virginia.edu
To: Cube-Lovers@ai.mit.edu
Subject: 2x2x2 at Taco Bell
In-Reply-To: <5cob1f$jiq@gap.cco.caltech.edu>
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I know many of the people on this list are collectors, and are always 
looking for new and unique cubes.  I also figure there may be some people 
on this list that would like to own one, but have yet to acquire a 2x2x2 
cube.  Taco Bell has a number of toys for sale commemerating the Star 
Wars re-release.  One of these toys is a 2x2x2 cube.

FYI,
-Dale Newfield
 Dale@Newfield.org


From cube-lovers-errors@curry.epilogue.com  Thu Jan 30 23:38:19 1997
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Message-Id: <199701310000.TAA11669@orbit.flnet.com>
From: Chris and Kori Pelley <ck1@flnet.com>
To: CUBE-LOVERS@ai.mit.edu
Date: Thu, 30 Jan 1997 19:04:34 -0500
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This is a multi-part message in MIME format.

[ And the moderator really wishes that people would -not- use MIME when
  mailing to Cube-Lovers.  It makes the archives considerably less useful.
				- Alan (Cube-Lovers-Request) ]

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>Taco Bell has a number of toys for sale commemerating the Star 
>Wars re-release.  One of these toys is a 2x2x2 cube.

I checked into this, but it is not a real 2x2x2 magic cube.  It is
one of those folding cube puzzles, where you can turn the
whole thing inside out and it has different graphics on each
assembled exterior.

Too bad though!  An official Star Wars 2x2x2 Rubik's Cube would
really be a find!

ck1@flnet.com
http://www.flnet.com/~ck1/cubes.html

------=_NextPart_000_01BC0EE0.6E7ED400
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Content-Transfer-Encoding: quoted-printable

<html><head></head><BODY bgcolor=3D"#C8E0D8"><p><font size=3D4 =
color=3D"#000000" face=3D"Arial">&gt;Taco Bell has a number of toys for =
sale commemerating the Star <br>&gt;Wars re-release. &nbsp;One of these =
toys is a 2x2x2 cube.<br><br>I checked into this, but it is not a real =
2x2x2 magic cube. &nbsp;It is<br>one of those folding cube puzzles, =
where you can turn the<br>whole thing inside out and it has different =
graphics on each<br>assembled exterior.<br><br>Too bad though! &nbsp;An =
official Star Wars 2x2x2 Rubik's Cube would<br>really be a =
find!<br><br>ck1@flnet.com<br>http://www.flnet.com/~ck1/cubes.html<br><fo=
nt size=3D4><br></p>
</font></font></body></html>
------=_NextPart_000_01BC0EE0.6E7ED400--



From cube-lovers-errors@curry.epilogue.com  Mon Feb  3 01:23:19 1997
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Date: Sun, 2 Feb 1997 23:39:34 -0500 (EST)
From: Nicholas Bodley <nbodley@tiac.net>
To: Dennis Okon <dokon@mit.edu>
cc: Cube-Lovers@ai.mit.edu
Subject: Re: Broken Cube
In-Reply-To: <9701281647.AA05417@MIT.MIT.EDU>
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 Realistically, IMHO, the only practical thing is to buy another. There
are several makes in existence, and they have at least minor detailed
differences in dimensions. I'd love to say something different!  If,
however, you can find someone who's an expert plastic welder (no kidding),
such a person might be able to do a repair.

 In any event, good luck!

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*    When the year 2000 begins, we'll celebrate 
|*  Amateur musician  *|*    the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------



From cube-lovers-errors@curry.epilogue.com  Mon Feb  3 17:42:34 1997
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Message-ID: <32F66942.50AF@ibm.net>
Date: Mon, 03 Feb 1997 14:40:02 -0800
From: Jin "Time Traveler" Kim <chrono@ibm.net>
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CC: Cube-Lovers@ai.mit.edu
Subject: Re: Broken Cube
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As it so happens, I have a spare 'cube' which met with unfortunate
incident a number of years ago.  Let's just say it involved a hacksaw
and a misguided attempt at.... modification.....

Anyway, if you're so inclined, I have no problems mailing you the center
piece if you really want it.. But like someone else said, you'd do just
as well buying a new one.  And besides, the center piece may turn out to
be just a little bit off kilter.  I think it was a knock-off.

-- 
Jin "Time Traveler" Kim
chrono@ibm.net
VGL Costa Mesa


From cube-lovers-errors@curry.epilogue.com  Wed Feb  5 15:45:07 1997
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Date: Wed, 5 Feb 1997 18:19:35 +0200 (IST)
From: Rubin Shai <rubins@cs.technion.ac.il>
To: Cube-Lovers <cube-lovers@ai.mit.edu>
Subject: Magazines about the Rubik cube
In-Reply-To: <Pine.PMDF.3.91.961114075359.20973G-100000@PSTCC6.PSTCC.CC.TN.US>
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Hi all
I'm going to write a report about a program that learn to solve the 2X2X2
Rubik cube by itself.
Does anyone know a magazine that will publish this kind of stuff?

Shai




From cube-lovers-errors@curry.epilogue.com  Thu Feb  6 14:38:30 1997
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Date: Thu, 06 Feb 1997 00:53:16 -0800
From: Jin "Time Traveler" Kim <chrono@ibm.net>
X-Mailer: Mozilla 3.01Gold (Win95; I)
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To: Dennis Okon <dokon@mit.edu>
CC: cube-lovers@ai.mit.edu
Subject: Re: Broken Cube
References: <9702032332.AA09252@MIT.MIT.EDU>
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Dennis Okon wrote:
> 
> It would be wonderful if you could mail it to me.  And, actually I have two
> cubes with broken centers - one of the original type and one of the plastic
> snap in tiles, which apparently (from inspection) both have the same size
> centers.
> 

I will mail you the piece as promptly as possible.  I recently moved
into a new residence days ago, so you will have to bear with me as I
hunt for my puzzle box and hope that I didn't throw away my ravaged cube
in a fit of....confusion?

> Also, I'm interested in hearing more specifically what the modification you
> mentioned was.  The hack sent my mind in all sorts of directions.
> 
> <address omitted>

You want to hear the ugly truth?  Ugh...  Ok..  while thumbing through
Dr. Christoph Bandelow's excellent cube catalog (1992 edition, I think),
I stumbled upon an object called the "Fisher Cube," in which each
handmade cube had a unique "cut" to it.  Basically on a standard cube
the "groove" lines run parallel to the faces of the cube.  On fisher's
cube, two opposite faces had the groove lines running diagonally, which
meant that a scrambled cube looked absolutely jacked up.  I got this
brilliant idea that I could make something similar, so out came the
hacksaw....  After about an hour and a half of cutting I realized that I
had completely ruined a perfectly good cube.  And my bedroom floor (and
desk, and bed, and lamp, and clothes, etc....) was COVERED in black
plastic shavings which, despite their strangely appealing odor, had
managed to make a terrible mess.  So after giving myself disgusted looks
into a mirror (as a form of "self scolding," if you will) I packed all
the pieces away, thinking that maybe someday I would take the Time to
reassemble it or fix it somehow.  After acquiring 4 more cubes over a
period of 5 years, I have lost interest in that project.  

Soooo, that's it.  Nothing particularly drastic, but I think it might be
worth a chuckle or two to those of the mailing list, so what started as
a personal email has become my own public shame.  So I also hope you
won't have a problem with your "private email" being part of this drawn
out response.

-- 
Jin "Time Traveler" Kim
chrono@ibm.net
VGL Costa Mesa

P.S. If I find it, you want the whole mess or just the middle?


From cube-lovers-errors@curry.epilogue.com  Thu Feb  6 14:36:14 1997
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Date: Thu, 6 Feb 97 09:20:10 +0900
From: Norman Diamond 06-Feb-1997 0916 <diamond@jrdv04.enet.dec-j.co.jp>
To: cube-lovers@ai.mit.edu
Subject: Re: Magazines about the Rubik cube
Mime-Version: 1.0
Content-Type: TEXT/PLAIN; charset=ISO-2022-JP

Rubin Shai wrote:

>I'm going to write a report about a program that learn to solve the 2X2X2
>Rubik cube by itself.
>Does anyone know a magazine that will publish this kind of stuff?

Shouldn't the program be smart enough to answer that question for you?  :-)

It would obviously be of interest in Cubism For Fun, the journal of the
Dutch Cube Society (which has been published in English for about the past
10 years or so).  I don't think you'd get paid for it, but the audience
would be right.

I would guess that Scientific American, Journal of Recreational Mathematics,
and Games might also be suitable, depending on possible characteristics of
the program.

-- Norman Diamond            diamond@jrdv04.enet.dec-j.co.jp
[Speaking for Norman Diamond not for Digital.]


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From: Dale Newfield <din5w@cs.virginia.edu>
Reply-To: DNewfield@cs.virginia.edu
To: cube-lovers@ai.mit.edu
Subject: Re: Broken Cube
In-Reply-To: <32F99BFC.7299@ibm.net>
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On Thu, 6 Feb 1997, Jin Time Traveler Kim wrote:
> I got this brilliant idea that I could make something similar, so out 
> came the hacksaw....  After about an hour and a half of cutting I 
> realized that I had completely ruined a perfectly good cube.

I finally got around to building a bandaged pair of cubes 
 XXX
 XXX
 XXXXX
   XXX
   XXX

and a bandaged 5-cube

 XXX XXX
 XXX XXX
 XXXXXXX
   XXX
 XXXXXXX
 XXX XXX
 XXX XXX

I agree--the process was very painful and very full of black shavings, 
but now that I am done, I am very pleased with the final result. :-)

The impetus for this fit of cube-construction came from the book "The 
book of ingenious & diabolical puzzles" by Jerry Slocum and Jack Botermans.

Also resulting from the same fit are a round cube with what look like two 
2x2x2's sticking out of it, and a regular cube in which two opposite 
corners and all the adjacent pieces are round.

All of these came after taking a good look at pages 124 and 125 of this 
book--a page full of a very interesting collection of sequential movement 
puzzles.  Thie is a really neat book that will make any collector drool, 
so keep your eyes out for it :-)

-Dale


From cube-lovers-errors@curry.epilogue.com  Mon Feb 10 15:31:13 1997
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From: Wei-Hwa Huang <whuang@ugcs.caltech.edu>
Subject: Re: Broken Cube
Date: 10 Feb 1997 01:18:36 GMT
Organization: California Institute of Technology, Pasadena
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Dale Newfield <din5w@cs.virginia.edu> writes:
>The impetus for this fit of cube-construction came from the book "The 
>book of ingenious & diabolical puzzles" by Jerry Slocum and Jack Botermans.

>All of these came after taking a good look at pages 124 and 125 of this 
>book--a page full of a very interesting collection of sequential movement 
>puzzles.  Thie is a really neat book that will make any collector drool, 
>so keep your eyes out for it :-)

I believe Jerry is still selling copies of his books.  
Write to him at

257 S. Palm Dr.,
Beverly Hills, CA 90212

(310)273-2270 FAX (310)274-3644
70410.1050@compuserve.com

and ask him to send you a flier of books and prices (nicely).

--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
-------------------------------------------------------------------------------
Save the earth.  Then Quit and Shut Down.


From cube-lovers-errors@curry.epilogue.com  Tue Feb 18 20:39:25 1997
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From: Stan Isaacs <isaacs@hpcc01.corp.hp.com>
Message-Id: <199702190137.AA255196278@hpcc01.corp.hp.com>
Subject: Super-skewb
To: cube-lovers@ai.mit.edu
Date: Tue, 18 Feb 1997 17:37:57 PST
X-Mailer: Elm [revision: 109.19]

Anybody have any good moves for super-skewb centers?  That is, ones
that either twist centers in place, or move them without twisting.
Tony Fisher, in England, makes some wonderful puzzles based on the
Skewb, but in shapes such as an Icosahedron, or Dodecahedron, or
Rhombic Dodecahedron.  These are all actually Super-Skewbs.

 -- Stan Isaacs


From cube-lovers-errors@curry.epilogue.com  Thu Feb 20 01:42:04 1997
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From: Time Traveler <chrono@ibm.net>
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Stan Isaacs wrote:
> 
> Anybody have any good moves for super-skewb centers?  That is, ones
> that either twist centers in place, or move them without twisting.
> Tony Fisher, in England, makes some wonderful puzzles based on the
> Skewb, but in shapes such as an Icosahedron, or Dodecahedron, or
> Rhombic Dodecahedron.  These are all actually Super-Skewbs.
> 
>  -- Stan Isaacs

Tony Fisher still makes those things?  It's HIS fault that I hacked one
of my cubes into fine powder!  Ah, yes.  A GOOD memory. :)

Happen to know how to get a hold of him?

-- 
Jin "Time Traveler" Kim
chrono@ibm.net
VGL Costa Mesa


From cube-lovers-errors@curry.epilogue.com  Thu Feb 20 13:45:33 1997
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To: Cube-Lovers@AI.MIT.EDU
From: Wei-Hwa Huang <whuang@ugcs.caltech.edu>
Subject: IPP17?
Date: 20 Feb 1997 16:11:44 GMT
Organization: California Institute of Technology, Pasadena
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Anyone on the mailing list going to the International Puzzle
Collector's Party in San Francisco? 

It's be nice to meet some more people I know there...

--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
-------------------------------------------------------------------------------
Now surfing the Internet at 24 hours a week.


From cube-lovers-errors@curry.epilogue.com  Thu Feb 20 17:01:07 1997
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From: Stan Isaacs <isaacs@hpcc01.corp.hp.com>
Message-Id: <199702202137.AA029444678@hpcc01.corp.hp.com>
Subject: Re: Super-skewb
To: Cube-Lovers@ai.mit.edu
Date: Thu, 20 Feb 1997 13:37:57 PST
X-Mailer: Elm [revision: 109.19]

Several people asked how to get the so-called "Super-skewbs" (the name
is from an earlier cube-lovers message, not from Tony Fisher.)  They
are made by hand, by Tony Fisher (see below).  He has about 9 puzzles,
and says he plans to have some new ones later this year.  He makes
them by glueing parts onto Skewbs.  The results, I think, are
excellent.  One corner broke off of one of the puzzles I got, but is
easily glued back on.  They move nicely (which means, I think, that
the original Skewbs moved nicely.)  He says he just buys them in
shops, and then makes the transformations at home.  They mostly cost
around $60 - $75 (American), although a triple 5x5x5 is $90, and a
mini-octahedron is $40.  I don't know if these are permanent prices;
they may change if prices of the cubes (or plastic) goes up.  besides
the skewb variations, he as "Siamese" versioin of both the skewb and
the 5x5x5 cube (ie, he combines two into one; I don't have these, so I
don't know exactly how the Siamese Skewb works.)  He also has a Triple
"Triamese 5x5x5", and a mini-octahedron based on the Tetraminx (which
I also want to get a copy of.  His address is:

	Tony Fisher
	9 Cauldwell Hall Road
	Ipswich, Suffolk, IP4 4QD, Great Britain

I guess I ought to put the list of puzzles for clarity:

Skewb variations:
  Icosahedron 	       - $60
  Dodecahedron 	       - $60
  Rhombic-Dodecahedron - $60
  Octahedron	       - $55
  Siamese Skewbs       - $60
  Mental Block	       - $75  (A strange object: looks like a 1x3x3
block, with 9 even squares on the top and bottom, but the edges are
cut by an 'X' in the middle, so there are 2 triangles and 2 pentagons
on each edge.  When twisted, it makes all sorts of strange shapes.)

Siamese 5x5x5		- $75
Triamese 5x5x5		- $90

Mini-Octahedron (based on Tetraminx) - $40

Dave Singmaster pointed out that the instructions for "Sonic's Puzzle
Ball", but Christoph Bandalow, has some super-skewb moves in it.  I've
just glanced at it, so I haven't had a chance to try them yet.  He
seems to have about 3 moves that twist pairs of squares (and some
triangles), of lenght 10, 14 or 18.

  -- Stan isaacs
  -- isaacs@corp.hp.com



From cube-lovers-errors@curry.epilogue.com  Fri Feb 21 13:46:38 1997
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To: Cube-Lovers@AI.MIT.EDU
From: Wei-Hwa Huang <whuang@ugcs.caltech.edu>
Subject: Re: Super-skewb
Date: 21 Feb 1997 16:47:06 GMT
Organization: California Institute of Technology, Pasadena
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Stan Isaacs <isaacs@hpcc01.corp.hp.com> writes:
>Anybody have any good moves for super-skewb centers?  That is, ones
>that either twist centers in place, or move them without twisting.
>Tony Fisher, in England, makes some wonderful puzzles based on the
>Skewb, but in shapes such as an Icosahedron, or Dodecahedron, or
>Rhombic Dodecahedron.  These are all actually Super-Skewbs.

If I remember correctly, all my moves for the Skewb are based on
the "R1L-1R-1L1" move.  Repeating this four-move sequence at
different orientations does everything, including rotating centers
and moving them.

Unfortunately, I don't have any on hand at the moment, so I can't
test them out exactly.

--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
-------------------------------------------------------------------------------
Now surfing the Internet at 24 hours a week.


From cube-lovers-errors@curry.epilogue.com  Fri Feb 28 13:51:54 1997
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From: "joyner.david" <joyner.david@mathnt1.sma.usna.navy.mil>
To: "'cube-lovers@ai.mit.edu'" <cube-lovers@ai.mit.edu>
Subject: impossiballs
Date: Fri, 28 Feb 1997 13:38:46 -0500
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Does anyone know of any impossiballs for sale in a
>store in the US? - David Joyner
>


From cube-lovers-errors@curry.epilogue.com  Sat Mar  1 01:52:20 1997
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Date: Fri, 28 Feb 1997 15:52:23 -0800
From: Time Traveler <chrono@ibm.net>
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To: "joyner.david" <joyner.david@mathnt1.sma.usna.navy.mil>
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joyner.david wrote:
> 
> Does anyone know of any impossiballs for sale in a
> store in the US? - David Joyner
> 

Try Peter Beck of New Jersey.  I don't remember his address (my address
book program was lost along with EVERYTHING else in a hard drive crash)
but I've seen his name on a list or two of puzzle sources.  The last
Time I bought puzzles from him (ohh... about 3 years ago) he had ONE
used Impossiball available.  But who knows, maybe things have improved
since then.

Speaking of which, does anyone know where I can get my hands on a
Rubik's Revenge?  I saw some for sale at www.puzzlett.com, but someone
said that they were an unreliable source (i.e. they ripped him off).

-- 
Jin "Time Traveler" Kim
chrono@ibm.net
VGL Costa Mesa

[ Note from the moderator:
  Peter Beck's electronic mail address is: pbeck@pica.army.mil
			- Alan (aka Cube-Lovers-Request@AI.MIT.EDU) ]

From cube-lovers-errors@curry.epilogue.com  Sun Mar  2 02:53:57 1997
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Date: Sun, 02 Mar 1997 00:50:27 -0500 (EST)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: An Enhancelment for Shamir with M-conjugacy
To: Cube-Lovers <cube-lovers@ai.mit.edu>
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I now have a functioning Shamir program.  I do not consider it quite up to
production quality just yet.  I am still working on a number of
improvements.  Also, all the results that are easily accessible to the
program have already been calculated by other means, so I have no new
search results to report at this time.  I do expect some new results from
time to time, but the runs will probably take weeks or months.  However,
in the process of developing the program, I came up with an enhancement
which I wished to report. 

Recall that the heart of the Shamir method is a mechanism which will for
all s in S and all t in T produce all products st in lexicographic order. 
This basic mechanism can be applied in a number of ways.  For example, it
can be applied to the problem of determining a minimal solution for a
particular position.  It can also be applied to the problem of conducting
a breadth first exhaustive search.  The former is the basic method
developed by Shamir and first reported to this group by Alan Bawden.  The
latter is the problem which I am currently addressing. 

As already reported in several previous messages, my implementation of
Shamir seeks to incorporate M-conjugacy to the maximum extent possible to
reduce memory requirements.  As such, it does not store the sets S and T. 
Rather, it stores the sets A and B, where A and B contain only
representative elements of the M-conjugacy classes which are contained in
S and T, respectively. 

A and B are about 48 times smaller than S and T.  However, we cannot
obtain any meaningful results using only A and B.  Rather, A and B have to
be expanded by M-conjugation to produce S and T.  That is, we represent S
as A^M and T as B^M.  There are no fewer positions, but only three bytes
or so are required to represent each position in A^M and B^M.  So we
produce products st in lexicographic order for s in A^M and t in B^M. 
This model is slow, but it is small. 

At the back end of the algorithm, we determine which st values are
representatives and which are not.  Those which are, we keep.  Those which
are not, we simply discard.  In my earlier messages about combining Shamir
with M-conjugacy, I lamented the difficulty of producing representatives
in lexicographic order.  Simply discarding non-representatives is a crude
but effective way to accomplish the goal.  It is not quite as good as not
producing the non-representatives in the first place, but it is a good bit
better than nothing. 

As an example of the "better than nothing" idea, the Shamir method does
not directly produce ST in lexicographic order.  Rather, it produces St in
lexicographic order for each t in T.  The results then have to be merged. 
The non-representatives are discarded prior to the merge, so that 48 times
fewer positions have to be merged.  Also, the products st are tested byte
by byte as they are produced to determine if they are representatives.  It
is usually possible to determine that a position is not a representative
after no more than two or three bytes, so there are some efficiencies in
the process of discarding non-representatives.  That is, the only products
st which are calculated in their entirety are those for representatives. 

The enhancement I want to report is that it is possible to discard entire
branches of the Shamir tree without examining any of the nodes in the
branch except the root of the branch.  That is, it is possible to show
that entire branches of the tree contain only non-representatives.  Such
branches can be pruned from the search without examining any of the nodes
individually.  Approximately 47/48 of the search tree can be eliminated
from the search tree in this manner.  Unfortunately, the speed up is not
times forty-eight as I hoped, but it is significant nonetheless. 

The model is an S24xS24 model with S24 acting on 0..23.  The corners are
therefore vectors of the form [a,b,c,....], which means 0->a, 1->b, 2->c,
etc.  The identity is [0,1,2,....].  We focus on the corners because we
consider the order of the corners first in our lexicographic order, using
the order of the edges only to break ties on the corners.  We call a
representative element of each M-conjugacy class a canonical form, and all
other elements we call non-canonical. 

The nature of the Shamir search is that it produces successively more
complete partial permutations as a tree is searched.  That is, it produces
[a,?,?,...] at the zeroth level of the tree, [a,b,?,?,...] at the first
level of the tree, [a,b,c,?,?,...] at the second level of the tree, and so
forth until a complete permutation is constructed. 

The enhancement to the method consists of determining which of the partial
permutations are canonical, which are non-canonical, and which are
neither.  A partial permutation is canonical if all daughter nodes are
canonical, a partial permutation is non-canonical if all daughter nodes
are non-canonical, and a partial permutation is neither if there are
daughter nodes of both types. 

>From a theoretical point of view, the type of each node could be
determined by examining each daughter and backing up the results
appropriately, similar to an alpha-beta search.  From a practical point of
view, the whole purpose is to determine the type of node without examining
any of the daughters.  And in practice, we only detect non-canonical nodes
vs. other than non-canonical nodes.  There is no disadvantage to this
procedure because it is only the non-canonical nodes which we wish to
eliminate from the search. 

Determining non-canonical nodes depends both on the particular numbering
scheme which is used for the cube facelets and also upon the particular
representative element function which is chosen.  We number the Front
corner facelets of the cube as follows: 

   0  1

   3  2

The Back corner facelets are then numbered 4..7, with opposite facelets
summing to 7.  All other facelets are numbered by adding 8 to the Front or
Back facelet as you look at the facelets of the cubie in clockwise order. 
For example, the flt cubie is (0,8,16), and the ftr cubie is (1,9,17). 

The representative element function returns the M-conjugate which of all
the elements in the M-conjugacy class is first in lexicographic order. 

Consider the partial permutation [0,?,?,...].  Its M-conjugates are of the
form [?,1,?,?...], [?,?,2,?,?,...], [?,?,?,3,?,?,...], etc.  It is easy to
see that if a representative begins with 0, then there is at least one of
the eight corner cubies somewhere in the cube which is properly
positioned, both with respect to location and with respect to twist. 
Also, it is easy to see that the partial permutation [0,?,?,...] has both
canonical and non-canonical forms as daughters. 

But consider the partial permutations [1,?,?,...] and [3,?,?,...].  They
are conjugate, but the canonical form is [1,?,?,...].  Hence, no canonical
form can begin with 3.  Therefore, we eliminate all permutations which
begin with 3 from the search, and we have eliminated 1/24 of the search
tree.  

I have calculated a table of non-canonical cutoff points for the corners. 
The results are as follows.  Notice that not all cutoffs are at the zeroth
level of the tree as is the cutoff for 3, but nonetheless there are 17
cutoffs at the zeroth level.  That means that there are only 7 (out of 24)
ways to begin a canonical permutation. 


Level       Noncanonical  Positions     
               Nodes      Trimmed        

i= 0    count=        17   62460720          
i= 1    count=        63   11022480          
i= 2    count=       487    4733640          
i= 3    count=      7610    4931280          
i= 4    count=     17830     962820          
i= 5    count=    138978     833868          
i= 6    count=    622745     622745          
Total trimmed              85567553

The positions trimmed figure is based on a corners only search, just to 
give a sense of proportion to the numbers.  The corners only group 
contains about 88 million positions.  For the complete cube, the numbers 
would be larger, but the proportions would be the same.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7127
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990




From cube-lovers-errors@curry.epilogue.com  Thu Mar  6 15:48:05 1997
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From: bagleyd <bagleyd@megahertz.njit.edu>
Message-Id: <199703061759.MAA19236@megahertz.njit.edu>
Subject: winpuzzles and xpuzzles
To: cube-lovers@ai.mit.edu
Date: Thu, 6 Mar 1997 12:59:28 -0500 (EST)
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Content-Type: text

Hi

I made a new release of my puzzles.
They are available for Windows 3.1 and above at
http://megahertz.njit.edu/~bagleyd/
as
winpuz64.zip
source is available also for each puzzle.

For X
ftp://sunsite.unc.edu//pub/Linux/games/x11/strategy/xpuzzles-5.4.1.tgz
Also available at the site below, you may need a mirror though to get it from
ftp.x.org since they only allow 50 users.

They puzzles include:

I updated all the txt/manual files. So they now contain a little history
of the origins of each with references.

Fixes for the Masterball

Auto solve capability for Panex thanks to Rene Jansen <RENE.R.J.JANSEN@RCC.nl>
(based on the algorithm in Quantum Jan/Feb 96).
Panex with 3 tiles on each side is known to be solvable in 42 moves
The version here solves it in 45.  Rene and I would both like to improve this.

For the X versions I added: Username, concurrency check, configure

-- 
Cheers, 
  /X\   David A. Bagley
 // \\  bagleyd@bigfoot.com   http://megahertz.njit.edu/~bagleyd/
((   X  xlockmore, new stuff for xlock @ ftp.x.org//contrib/applications
 \\ //  altris, tetris games for x @ ftp.x.org//contrib/games/altris
  \X/   puzzles, magic cubes for x @ ftp.x.org//contrib/games/puzzles


From cube-lovers-errors@curry.epilogue.com  Sat Mar  8 21:56:13 1997
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From: Mitch269@aol.com
Date: Sat, 8 Mar 1997 21:37:53 -0500 (EST)
Message-ID: <970308213750_1383812770@emout11.mail.aol.com>
To: Cube-Lovers@ai.mit.edu
Subject: RUBIK'S CUBE/RUBIK'S REVENGE

Does anyone know of an address I could send to order Rubik's Puzzles?  I
can't seem to find either one anywhere.  Thanks.

Mitch Brewer
mitch269@aol.com


From cube-lovers-errors@curry.epilogue.com  Sun Mar 23 14:19:14 1997
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From: bandecbv@rz.ruhr-uni-bochum.de
To: cube-lovers@ai.mit.edu
Date: Sun, 23 Mar 1997 17:47:14 +0000
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Subject: Super-skewb
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> Anybody have any good moves ... (Stan Isaacs, 19 Feb 1997) 
> Several people asked ... (Stan Isaacs, 20 Feb 1997)
>
Hi, Cube-Lovers!
My name is Chistoph Bandelow, and I am new in this exclusive club.
Having finally aquired a modem and the necessary software a few weeks 
ago, I'm still overwhelmed and confused by everything. I have read 
all the contributions from July 1996 till now and the complete "Index 
by Subject". 

To begin with something, I would like to make some remarks about the
Skewb, especially the wonderful Skewb variations made by Tony Fisher.  I
suggest to call them "Fisher's Icosahedron", "Fisher's Dodecahedron", and
so on..Owing all of these puzzles, I am very enthusiastic because of their
originality and first-class quality. Just in case you are considering to
acquire one or the other from him: Tony Fisher is a very reliable, modest
and decent person.

THE SKEWB has been first described to a larger public by Douglas R. 
Hofstadter in the July 1982 edition of Scientific American. This 
treatise is included in Hofstadter's book "Metamagical Themas" (Basic 
Books 1985 and Bantam Books 1986). It was here where Hofstadter 
suggested the name 'skewb' as a reminder of skew and cube. 
One of the wonderful features of the Skewb is that we don't have to 
quarrel how to count single moves: no trouble with outer layer moves 
versus slice moves or with 90=B0 moves versus 180=B0 moves, there is just 
one type of move rotating one half of the Skewb by 120=B0 against the 
other half. 
>From 1985  to 1988 I had an intensive correspondence with Ronald 
Fletterman where I used the following 

NOTATION: Hold the (ordinary cubical) Skewb such that there is a 
unique Right upper corner, Left upper corner, Front upper corner and 
Back upper corner. R, L, F and B respectively denote a clockwise 120=B0 
rotation  of the associated half cube (as seen from the outside). R', 
L', F' and B' denote counterclockwise turns. Small letters r, l and 
so on are used for rotations around the bottom corners. Though this 
notation is short and handy, it is probably not as good as the one 
used by D. J. Joyner in his Skewb page (see 
http://www.nadn.navy.mil/MathDept/wdj/rubik.html) 
But I stick on my old notation, especially after Ronald Fletterman, 
used this notation in two papers in CFF (Cubism For Fun, the 
newsletter  of the Dutch Cubist Club) in which he provides an 
enormous collection of Skewb maneuvers, see CFF 17 (May 1988) and  
CFF 18 (September 1988). The square center pieces of the Skewb can 
only rotate pairwise and only by 180=B0, not by 90=B0. Ronald 
Fletterman's collection covers these "invisibles". He, the 
perfectionist,  gives maneuvers for all 5 possible cases: 2 
neighboring squares, 2 opposing squares, 4 squares all but 2 
neighborings, 4 squares all but 2 opposing, all 6 squares. However, 
all his other maneuvers (those for the corner pieces of the Skewb) 
pay absolutely no heed to the orientation of the center pieces. Every 
short and elegant solution method for the new Skewb variations of 
Tony Fisher or for the beautiful round versions of the Skewb like 
Mickey's Challenge or Sonic's Puzzle Ball or the Mach Balls require 
maneuvers for the corner pieces which do not twist the center pieces. 
Some of those maneuvers are given in my 80 page booklet about 
Mickey's Challenge or about Sonic's Puzzle Ball or on the leaflets 
accompanying some other puzzle balls from Meffert. Here is a 
selection:

SOME SUPER SKEWB MANEUVERS. 

1. (FL'R)^6 (18)  twists the 2 neighboring squares on the left side. 

2. R'BLF'L'FRLB'R'FRF'L' (14) achieves the same thing. 

3. FfRr'f'FfF'R'f'rRF'R' (14) twists the top and bottom square. 

4. (RF')^2 (R'F)^2 (8) twists the four top corner pieces: 

... the right and front one clockwise, the left and back one counter-

... clockwise, in short: (+R) (+F) (-L) (-B). 

5. R'FR' (F'R)^2 fF'f (Ff')^2 (14)  twists 2 corner pieces: (+R)(-L)

6. rF'rfR'F'rfR'r'F (11) twists the top front corner piece clockwise 

... and exchanges (3-cycles) the 3 neighboring corner pieces, 

... in short: (+F) (L,R,f). The very last notation does not precisely 

... describe the effect of the maneuver since the orientation of the

... three corner pieces is not given. A remarkable and fundamental 

... difference between Rubik's Cube and the Skewb is that the Skewb 

... does not allow pure corner-3-cycles: It is impossible to achieve 

... (L,R,f) without any other corner piece change! 

CALL TO CUBE-LOVERS: I'm convinced that the maneuvers given above, 
especially number 2, 3 and 5, may be improved. Who can give shorter 
maneuvers or a good maneuver for (+L) (+R) (+f) ?

PROPAGANDA. One last remark: I don't want to offend the good rules of 
netiquette by doing any kind of advertising here. But to avoid 
unnecessary questions and loss of time, allow me to say that I will 
send my free mail order cube catalog (1996 edition, this is still the 
latest) to everybody who requests one and provides his postal 
address. 

Christoph 
Christoph Bandelow
mailto:Christoph.Bandelow@rz.ruhr-uni-bochum.de


From cube-lovers-errors@curry.epilogue.com  Thu Mar 27 11:43:20 1997
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Date: Thu, 27 Mar 1997 04:56:51 -0500 (EST)
From: Nicholas Bodley <nbodley@tiac.net>
To: bandecbv@rz.ruhr-uni-bochum.de
cc: cube-lovers@ai.mit.edu
Subject: "Propaganda": OK with me!
In-Reply-To: <199703231549.QAA15923@sun529.rz.ruhr-uni-bochum.de>
Message-ID: <Pine.SUN.3.95.970327045127.914C-100000@sunspot.tiac.net>
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 Here's one list recipient who thinks Christoph Bandelow was quite
courteous in his "propaganda"; I was quite glad to finally have a
convenient way to request his catalog. I hope he's not overwhelmed by
requests! As some (perhaps most) of you know, he has quite a reputation
(and a good one) already.

 Christoph: It's fine with me! Next message (private) will be a
catalog request to you. Welcome to the 'Net!

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*    When the year 2000 begins, we'll celebrate 
|*  Amateur musician  *|*    the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------



From cube-lovers-errors@curry.epilogue.com  Thu Mar 27 14:21:52 1997
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Date: Thu, 27 Mar 1997 11:18:35 -0800
From: Time Traveler <chrono@ibm.net>
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Subject: Re: Super-skewb
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bandecbv@rz.ruhr-uni-bochum.de wrote:

> PROPAGANDA. One last remark: I don't want to offend the good rules of
> netiquette by doing any kind of advertising here. But to avoid
> unnecessary questions and loss of time, allow me to say that I will
> send my free mail order cube catalog (1996 edition, this is still the
> latest) to everybody who requests one and provides his postal
> address.
> 
> Christoph
> Christoph Bandelow
> mailto:Christoph.Bandelow@rz.ruhr-uni-bochum.de

I must vouch for Dr. Bandelow, as I had an opportunity to purchase a
number of pieces from him some years ago, specifically an intriguing
variant of the spherical skewb called the Moody Ball by Gerd Braun.  

-- 
Jin "Time Traveler" Kim
chrono@ibm.net
VGL Costa Mesa


From cube-lovers-errors@curry.epilogue.com  Tue Apr  1 15:56:31 1997
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X-Originating-IP: [194.239.24.235]
From: Philip  Knudsen <philipknudsen@hotmail.com>
To: cube-lovers@ai.mit.edu
Subject: Greg's Cubes
Content-Type: text/plain

Hi everybody!

I'm Philip Knudsen, and i live in Copenhagen, Denmark. Last year i found out
about this list, and read most of the old messages. The past few months i have
subscribed to the list using a free email account at my library.
Now for my 1st message: Since the talk is of Fisher's Skewb variants (which
sound very interesting indeed), i would like to mention to those whom it might
interest, that Greg Stevens, Seattle, is making various types of cube variants.
I have one of his catalogues, and have also received from him an "Off Center
Cube".
His designs, about 12 different, seem to be in the following categories:
1) Bandaged Cubes, made from standard 3x3x3
2) Shape modifications, made from standard 3x3x3
3) 1) and 2) combined
4) Bandaged Square-1
5) Bandaged 4x4x4 (where did he get the raw material???)
6) Shape variations of Pyraminx, i.e. "Pyraminx Star"

The "Off Center-Cube" is a 3. It is well made, and very tough to solve. In fact
i still need to find out how to turn the center pieces, which by the way are
disguised as edge pieces.
Last but not least - Greg also is a very reliable and decent person.
If anyone's interested i shall forward his postal address to the list. 




---------------------------------------------------------
Get Your *Web-Based* Free Email at http://www.hotmail.com
---------------------------------------------------------


From cube-lovers-errors@curry.epilogue.com  Wed Apr  2 15:54:57 1997
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From: Philip  Knudsen <philipknudsen@hotmail.com>
To: cube-lovers@ai.mit.edu
Subject: Greg's Cubes
Content-Type: text/plain

Greg is not online. His postal address is:
Greg Stevens,
313 N.E. 151st,
Seattle, WA 98155,
U.S.A.
I'll try and find his phone number too.

Good Luck!           Philip K



---------------------------------------------------------
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From cube-lovers-errors@curry.epilogue.com  Wed Apr  2 15:54:25 1997
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From: "joyner.david" <joyner.david@mathnt1.sma.usna.navy.mil>
To: "'cube-lovers@ai.mit.edu'" <cube-lovers@ai.mit.edu>
Subject: RE: Greg's Cubes
Date: Wed, 2 Apr 1997 07:14:54 -0500
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I have bought cubes from Greg Stevens and can second Philip Knutsen's 
opinion about his reliability. - David Joyner

>----------
>From: 	Philip  Knudsen[SMTP:philipknudsen@hotmail.com]
>Sent: 	Tuesday, April 01, 1997 9:36 AM
>To: 	cube-lovers@ai.mit.edu
>Subject: 	Greg's Cubes
>
>Hi everybody!
>
>I'm Philip Knudsen, and i live in Copenhagen, Denmark. Last year i
>found out
>about this list, and read most of the old messages. The past few months
>i have
>subscribed to the list using a free email account at my library.
>Now for my 1st message: Since the talk is of Fisher's Skewb variants
>(which
>sound very interesting indeed), i would like to mention to those whom
>it might
>interest, that Greg Stevens, Seattle, is making various types of cube
>variants.
>I have one of his catalogues, and have also received from him an "Off
>Center
>Cube".
>His designs, about 12 different, seem to be in the following
>categories:
>1) Bandaged Cubes, made from standard 3x3x3
>2) Shape modifications, made from standard 3x3x3
>3) 1) and 2) combined
>4) Bandaged Square-1
>5) Bandaged 4x4x4 (where did he get the raw material???)
>6) Shape variations of Pyraminx, i.e. "Pyraminx Star"
>
>The "Off Center-Cube" is a 3. It is well made, and very tough to solve.
>In fact
>i still need to find out how to turn the center pieces, which by the
>way are
>disguised as edge pieces.
>Last but not least - Greg also is a very reliable and decent person.
>If anyone's interested i shall forward his postal address to the list. 
>
>
>
>
>---------------------------------------------------------
>Get Your *Web-Based* Free Email at http://www.hotmail.com
>---------------------------------------------------------
>
>


From cube-lovers-errors@curry.epilogue.com  Thu Apr  3 12:16:11 1997
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From: Karen Angelli <kangelli@tiac.net>
To: "'Cube-Lovers@AI.MIT.EDU'" <Cube-Lovers@ai.mit.edu>
Subject: Cube History: Triumph and Defeat
Date: Thu, 3 Apr 1997 07:58:25 -0500

A recent addition to the Cube-Lovers list, I would like to share one of
the least known events in the history of Cube entertainment, and its
repercussions in the mass media and ultimate disgrace of one of the most
powerful men in entertainment.

It was the early '80s and Rubik's-Cube-Mania was all the rage.  Although
not nearly as accomplished as some of this list's members were, I could
solve the cube in about one minute.  I was also a lifeguard at a public
pool, and a locally renowned under-water swimmer (with a personal best
of 75 meters).  With such amazing and narrowly acclaimed accomplishments
in such diverse fields of endeavor, it was only natural that I would
feel public pressure to combine the two.  Thus was born underwater cube
solutions.

I took my best lubricated cube, a dive mask and a weight belt, and
started solving the cube in 10 feet of water.  After several practice
attempts, to ensure that I could hold my breath long enough to complete
the cube, I volunteered my services to the local synchronized swimming
club which was looking for an opening act for their show.

The show took place before a not so overflowing crowd during the busiest
season of the year, the local Nordic Fest celebration of Scandinavian
heritage.  The international crowd of aquatic enthusiasts was stunned
when I was introduced.  My bikini clad assistant handed the pristine
cube to one of the audience members to randomize and returned to me.
Then, in four and a half feet of water, I submerged and started solving
the cube.

After about 10 seconds of hurried twisting, I dropped the cube and lost
my place - I had to start over.  In practice, it had never taken me more
than about 1 min, 15 seconds to solve the cube, and I had practiced
holding my breath comfortably for about 1 min, 25 seconds.  I wasn't
sure whether I would complete the task.  After about 1 min 30 seconds,
my sister started yelling for someone to help me.  However, at the 1 min
38 second mark, I surfaced - to thunderous applause.  Certainly one of
the greatest moments of cube history.  How sad that this would lead to
infamy and no greater laurels.

After hearing the story of how I wowed a normally reserved Iowan crowd,
my classmates in college in Pennsylvania encouraged me to find a larger
audience, on a national stage.  Naturally, I wrote to NBC's Late Night
with David Letterman to pitch my idea for a stupid human trick.
Uncharacteristically, however, Dave turned down a good idea.  I haven't
forgiven him since.

I hope I haven't disappointed any Dave fans out there, but the truth had
to surface some day.  Thanks for keeping the flame alive Cube-lovers.

Pete Reitan



From cube-lovers-errors@curry.epilogue.com  Fri Apr  4 11:35:52 1997
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Date: Fri, 4 Apr 1997 09:28:08 -0500 (EST)
From: Jiri Fridrich <fridrich@binghamton.edu>
X-Sender: fridrich@bingsun2
To: Cube-Lovers@ai.mit.edu
Subject: Pretty patterns with Kociemba (help)
Message-ID: <Pine.SOL.L3.93.970404092400.8134A-100000@bingsun2>
MIME-Version: 1.0
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I would like to ask for your help to find short algorithms for some
pretty patterns below. The algorithms are just working algorithms and
are probably too long. Can anybody apply Kociemba's algorithm to 
those positions? 

The patterns form a small portion of a very large collection of
pretty patterns found by Mirek Goljan and Peter Nanasy from Czech
Republic. The algorithms below are the awkward "outliers" for which
we were unable to find a reasonably short "logical" algorithm. It
looks like Kociemba's algorithm is the only chance. 

Thanks in advance for your help!

Jiri Fridrich

P.S.: Visit my speed cubing page at
http://ssie.binghamton.edu/~jirif. The complete collection of
pretty patterns will be there soon.

F'R2D'RsF RsF D'R2D F'L2F D'R2D2B LsU'F'      28,23,20  DLV
U'R B'R'U R'D2F U F U'F'D'U2BsLsD F           22,20,18  U4V
R2F'D'F2L F'R F R'aF'D B LsDsF'U2             22,19,17  U5V
F'R2D'RsF RsF D'R2D F'L2F D'R2D2F UsL'U'      28,23,20  LV
D B'L'B2L B D2B'L'B2L B D'L2B2D2              22,16     L'2
D2B2D2B2U2F2U R2U F2U2R2D R2D		      26,15	L'8
FaR2F'aD'aR2U'F2D R2D2F2U F2U                 24,17     L'9
D2F2L2U2D B R B2R'B'D2R'B'R2B R D             24,17     L'10
LsF2R2D'FsU'F2sDFsDR2F2Rs                     24,18,13  [SS']
U L D R2U R U'R B'D B'D'B2D'L'U'              18,16     [VH]
R'D2R B'U2B R'D2R B'U2B . U B2L BsL2R'FsU2D L'B2U'      [VHH']
R U'R B'D B'D'B L B'U R'U R U2B2R2L'U         22,19     [VSS']  
LsF2R2D'FsU'F2sDFsDR2F'L'FsU F'U'BsR                    [SS'H]
U'R F'R'F L F'DsBsL U'R'D R D'F               18,18,16  [DOO']
U L D R2U R U'L DsBsR'B L'B'L2B'D'F'          22,20,18  [DVH]
LsF2R2D'FsU'F2sDFsDR2F'R'DsBR'F'LsU                     [DSS'H]
B'L'D L'U'BsLsU'R DsR'D R U R'D L             20,20,17  [DHOO']
B'L F'L2FsR'B R F'LsB'R B2L F L'              20,18,16  [VO]
LsU FsD F2sU'FsU'F L'FsU F'U'BsR              24,22,16  [SHH']
R'F2L'D'L F2B D B'D'FsL B L2D L F'R           22,19,18  [SS'O]
D'L F R U2L U2R2U2L'U R2U R'F'L'D             22,17     [VHO]
R'F'L BsD'F'D B'L'B L F R'DsBsR U             20,20,17  [DSO']
LsU FsD F2sU'FsU'F R'DsB R'F'LsU              24,22,16  [DSHH']
R'D B D R'DsB L B'U'aR B'D'R2B'D'R            20,19,18  [DVSO]
D F'UsL'DsF UsL2F'L DsF'LsD BsR'D B'R'        26,25,19  [DU3U4]
U L F'L DsF'LsD BsR'D LsU'L'U'F'aU'BsL F2U    28,27,22  [DU2U3]
R2sF2R2F2sR2F2.FD'L'DLD'L'DLD'L'DLF'
FD'L'DLD'L'DLD'L'DLF'. F2sR2sD F2sU2R2sD
D'F'D F DsB L'D L UsB'D'B R'F'D'F'D           20,20,18  [WORKS(14)]
U'B2RaU2R'B'U L'B UsR'B U'R B'DsB2R2U         26,22,20  [WS(14)S'(23)]
R D'F2D F'R F UsF DsF'R2D F D'R D F'D'        23,21,19  [ORKK'S(14)(23)]
R U'F R'B'D2R'U'BsLsD B L U'R'U'F U'Ls        23,22,19  [DK]
D'R'BULsB'UB'U'BL'BLB'UB'U'BRsDB'DsF'UsLB     30,30,26  giant meson 1
L D R2D'L2U B'D'B D'R'D R DsL D'B2D           22,19,18  giant meson 2
R2L'DBR'D'BLBL'D'B'D2LDL'U'FD'F'RFL'FLF2R'D2U           
DFRsU'B'D'R'aD'LsBDFD2F2DF'R'B'L'DLBF2RD'R2D        

Notation:

Ra = antislice RL, Fa = FB, etc.
Fs = slice move FB', Rs = RL', etc.
F2s = 180 deg. slice move, etc.

The three-tuple in the second column means the number of quarter,
face, and slice moves. You can Ignore the cryptic notation in square
brackets.



**********************************************************************
|  Jiri FRIDRICH, Research Associate, Dept. of Systems Science and   |
|  Industrial Engineering, Center for Intelligent Systems, SUNY      |
|  Binghamton, Binghamton, NY 13902-6000, Tel.: (607) 797-4660,      |
|  Fax: (607) 777-2577, E-mail: fridrich@binghamton.edu              |
|  http://ssie.binghamton.edu/~jirif/jiri.html                       |
**********************************************************************

......................................................................

Remember, the less insight into a problem, the simpler it seems to be!
----------------------------------------------------------------------



From cube-lovers-errors@curry.epilogue.com  Tue Apr 15 13:20:28 1997
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From: Goyra <Goyra@cheerful.com>
To: Cube-Lovers@ai.mit.edu
Subject: Java cubes
Date: Tue, 15 Apr 1997 10:53:27 +0100
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	Yo!

	The world's largest collection of Java
Rubik puzzles is now under construction at

	http://www.iol.ie/~goyra/Rubik.html

All comments are appreciated. An up-to-date Java 
browser is required.

				Goyra



From cube-lovers-errors@curry.epilogue.com  Fri May  9 15:02:56 1997
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Date: Fri, 09 May 1997 18:03:26 BST
From: David Singmaster Computing & Maths South Bank Univ <david.singmaster@sbu.ac.uk>
To: cube-lovers@ai.mit.edu
Message-ID: <009B402B.C196AB40.297@vax.sbu.ac.uk>
Subject: Underwater cube solving

	I found the story in my Cubic Circular 1, p. 16.  Pete promised to
marry his lady when she got down to 60 seconds, which wasn't too hard as she 
was already at 70 seconds.  Her name was Chris Clark.  The TV recorded only
showed a lot of bubbles!  One would need a underwater TV camera to get 
antything interesting.
DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499; 
email:  zingmast  or  David.Singmaster  @sbu.ac.uk


From cube-lovers-errors@oolong.camellia.org  Fri May  9 15:21:07 1997
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Date: Fri, 09 May 1997 17:16:09 BST
From: David Singmaster Computing & Maths South Bank Univ <david.singmaster@sbu.ac.uk>
To: cube-lovers@ai.mit.edu
Message-ID: <009B4025.26B4C360.894@vax.sbu.ac.uk>
Subject: Underwater cube solving

	I've been away from my email for some time and have just seen the
message of 3 Apr 1997 from Pete Reitan via Karen Angelli.
	In England, we also had some underwater cube solving.  This is probably
in my Cubic Circular somewhere, but I can't find it.
	A lecturer in mathematics at the open University, Pete Strain, got
interested in the Cube when I brought early examples to the Open University in
early 1979.  He got married about that time and took on the name Strain-Clark,
and his wife was also interested.  They performed underwater for Anglia
Television, probably in 1980.  This is a regional television and I've never
seen the program.  They had similar problems to Pete - in particular, her
face mask leaked and she couldn't see for the last 15 seconds and had to
solve the cube by memory!
	I've looked through my Cubic Circular again and can't find that
I ever included the above.
	Some day I may assemble another issue, mostly of anecdotes.
DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499; 
email:  zingmast  or  David.Singmaster  @sbu.ac.uk


From cube-lovers-errors@oolong.camellia.org  Sat May 10 00:03:41 1997
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From: Goyra <Goyra@cheerful.com>
To: cube-lovers@ai.mit.edu
Subject: Strange new Rubik puzzle in Java
Date: Sat, 10 May 1997 02:28:38 +0100
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	Those of you with Java browsers may want to
see a new Rubik puzzle I've just posted at

	http://www.iol.ie/~goyra/Rubik.html

It's a dodecahedron sliced on 8 axes. I would not have 
believed that it was possible that you could take the
20  evenly spaced corners of a dedecahedron and find 
8 of them that are ALSO evenly spaced - but there
it is.

					David Byrden



From cube-lovers-errors@oolong.camellia.org  Sat May 10 15:47:13 1997
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Date: Sat, 10 May 1997 15:48:18 -0400
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From: karen angelli <kangelli@tiac.net>
Subject: Extra-long Rubik's magic rings
Mime-Version: 1.0
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I just ran across a store selling a double long rubik's magic rings (the
flat folding puzzle) that instead 2x4 squares with three rings was 2x8 (I
think)with about six rings.  They were made by matchbox in 1987.  I would
have purchased one, but the price seemed like it might be steep.  Does
anyone know if these are common or can be found elsewhere? Or should I rush
back and buy one before they disappear?  Please let me know.
Pete Reitan



From cube-lovers-errors@oolong.camellia.org  Sun May 11 16:56:07 1997
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Date: Sun, 11 May 1997 10:49:22 -0400
From: "Richard W. Pearson, Jr." <rwpearso@ipass.net>
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Subject: Re: Extra-long Rubik's magic rings
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> I just ran across a store selling a double long rubik's magic rings (the
> flat folding puzzle) that instead 2x4 squares with three rings was 2x8 (I
> think)with about six rings.  They were made by matchbox in 1987.  I would
> have purchased one, but the price seemed like it might be steep.  Does
> anyone know if these are common or can be found elsewhere? Or should I rush
> back and buy one before they disappear?  Please let me know.
> Pete Reitan

I'd say they're pretty uncommon.  I've been looking for one for about 5
years now.  Would you mind sending me the address and phone number of
the company that is selling them.  I've also been looking for a "Rubik's
Magic Create the Cube."  It was originally described as a 'Level 2'
Rubik's Magic.  I snapped the strings on mine and haven't seen one
since.

Thanks,
Ricky Pearson
rwpearso@ipass.net


From cube-lovers-errors@oolong.camellia.org  Sun May 11 17:46:25 1997
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Date: Sun, 11 May 1997 17:44:53 -0400
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From: Aaron Weintraub <aaweint@interlog.com>
Subject: Re: Extra-long Rubik's magic rings
Mime-Version: 1.0
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>I'd say they're pretty uncommon.  I've been looking for one for about 5
>years now.  Would you mind sending me the address and phone number of
>the company that is selling them.  I've also been looking for a "Rubik's
>Magic Create the Cube."  It was originally described as a 'Level 2'
>Rubik's Magic.  I snapped the strings on mine and haven't seen one
>since.

I have one of these that I bought when they originally came out.  I
wouldn't really call it a "level 2" puzzle, as the mechanics are identical
to that of the original Rubik's Magic.  The goal is different, however.
Each plate is divided into different coloured quarters.  The object is to
get both the proper shape - a cube resting on one of it's edges, centred
atop a "platform" of the two other plates - and the proper colour
orientation - the three faces that join in each corner of the cube must
have identical colours in the quadrant at that corner.

-Aaron



From cube-lovers-errors@oolong.camellia.org  Mon May 12 13:39:03 1997
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From: Pete Beck <pbeck@pica.army.mil>
To: cube-lovers@ai.mit.edu
Subject: Re: Extra-long Rubik's magic rings
Date: Mon, 12 May 1997 07:53:38 -0400
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As I recollect the 12 panel Magic Rings was called the Master's Edition. 
Since they have been out of manufacture for quite awhile suggest you buy it
when you see it.

The 8 panel edition has been released by ODDZON and is in some toy stores
along with some other Rubik's items.

As some of you might remember there also was a 4 panel magic sold.  If you
have ever taken one apart you know that any multiple of 4 is possible. 
Awhile back I posted intstructions on CUBE LOVERS for restringing and or
making your own variant.  The panels are just 2 pieces of plastic with the
design sandwiched in bewteen.  They are not glued but held together by the
strings.


Does anybody know where to get I BET YOU CAN"T ?? a variant made with
hexagon panels.



THE FUTURE IS PUZZLING,
BUT CUBING IS FOREVER !!!

Pete, aka JUST PUZZLES


From cube-lovers-errors@oolong.camellia.org  Mon May 12 13:40:54 1997
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To: Cube-Lovers@AI.MIT.EDU
From: Wei-Hwa Huang <whuang@ugcs.caltech.edu>
Subject: Re: Extra-long Rubik's magic rings
Date: 12 May 1997 15:47:57 GMT
Organization: California Institute of Technology, Pasadena
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X-Newsreader: NN version 6.5.0 #2 (NOV)

"Richard W. Pearson, Jr." <rwpearso@ipass.net> writes:
>> I just ran across a store selling a double long rubik's magic rings (the
>> flat folding puzzle) that instead 2x4 squares with three rings was 2x8 (I
>> think)with about six rings.  They were made by matchbox in 1987.  I would
>> have purchased one, but the price seemed like it might be steep.  Does
>> anyone know if these are common or can be found elsewhere? Or should I rush
>> back and buy one before they disappear?  Please let me know.
>> Pete Reitan

>I'd say they're pretty uncommon.  I've been looking for one for about 5
>years now.  Would you mind sending me the address and phone number of
>the company that is selling them.  I've also been looking for a "Rubik's
>Magic Create the Cube."  It was originally described as a 'Level 2'
>Rubik's Magic.  I snapped the strings on mine and haven't seen one
>since.

It has 12 panels and 5 rings, and was marketed as "Unlink the Rings".
I have one with the original packaging.  They are hard to find;
I only found mine through a very lucky occurence (and paid only $1
for it!)

I am still searching for a "Make the Cube."  I also have a broken one.
Perhaps I should just take the string from one of my normal Magics
and transfer it.


--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
-------------------------------------------------------------------------------
The cynical poet -- is worst of the worst.
What others are thinking -- he says out loud first.


From cube-lovers-errors@oolong.camellia.org  Mon May 12 21:52:48 1997
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Date: Tue, 13 May 1997 02:50:14 BST
From: David Singmaster Computing & Maths South Bank Univ <david.singmaster@sbu.ac.uk>
To: Goyra@cheerful.com
CC: cube-lovers@ai.mit.edu
Message-ID: <009B42D0.D8685A60.2@vax.sbu.ac.uk>
Subject: RE: Strange new Rubik puzzle in Java

	What Goyra is describing as 'Strange new Rubik puzzle in Java' is
clearly based on the fact that one can inscribe a cube in a dodecahedron
using 8 of the dodecahedron's vertices.  This is pretty well known and one
can probably see versions of it in some of the books on polyhedra.
Adapting a cube to having a dodecahedral appearance was certainly considered
in the 1980s, thoguh I can't remember if anybody ever made them for sale
- e.g Greg Stevens or Jean-Claude Constantin.  I don't have any in my 
collection, so I'd be grateful to hear hpw to get one.
	Incidentally, a 2x2x2 version is being made in Spain in the shape
of Mickey Mouse's head (and perhaps Donald Duck's head).  These are
supposed to be coming on the market here soon.
DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499; 
email:  zingmast  or  David.Singmaster  @sbu.ac.uk


From cube-lovers-errors@oolong.camellia.org  Mon May 12 23:04:42 1997
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Date: Tue, 13 May 1997 03:38:27 BST
From: David Singmaster Computing & Maths South Bank Univ <david.singmaster@sbu.ac.uk>
To: cube-lovers@ai.mit.edu
Message-ID: <009B42D7.94F5F2E0.2@vax.sbu.ac.uk>
Subject: Rubik's Magic

	Peter Beck says one can make these with any number of panels which is
a multiple of four.  Actual any even number is possible.  I have several
example of  2 x 3.  Actually, there are two examples - the third is the
one with six hexagons.
	Both the  2 x 3 examples were promotional items for magazines, one
in Italy and one in France.  I happened to be in France when the French 
example was attached to a magazine called Super in Jun 1988.  I bought about
a dozen example, but I have no spares left!
	The  2 x 2  was marketed in four forms and one could assemble all
four into a bigger pattern.  However, I've also got three Hungarian versions.
One has religious symbols and comes in a folder with a picture of the Pope
on the front, apparently commemorating the Pope's visit to Hungary or
Austria.
DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499; 
email:  zingmast  or  David.Singmaster  @sbu.ac.uk


From cube-lovers-errors@oolong.camellia.org  Mon May 12 23:04:11 1997
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Date: Tue, 13 May 1997 03:27:02 BST
From: David Singmaster Computing & Maths South Bank Univ <david.singmaster@sbu.ac.uk>
To: whuang@ugcs.caltech.edu
CC: cube-lovers@ai.mit.edu
Message-ID: <009B42D5.FC6B2820.2@vax.sbu.ac.uk>
Subject: Re: Extra-long Rubik's magic rings

	Several people have remarked about having broken strings on a
version of Rubik's Magic.  When one of mine first came apart, I thought
it would be impossible to get it back together correctly as I thought
there would be some critical weaving of the strings.  However, once
one examines it closely, one sees that no magic is needed.  Simply rethread
the strings along the correct paths and it works.  All one needs is a small
screwdriver to stretch the end of a loop over the corner of a panel.  So
it is much easier to assemble or to reassemble than one initially thinks
and I'd encourage anyone who has a little manual dexterity to remake a
broken one using strings from other ones.
DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499; 
email:  zingmast  or  David.Singmaster  @sbu.ac.uk


From cube-lovers-errors@oolong.camellia.org  Tue May 13 12:38:16 1997
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To: cube-lovers@ai.mit.edu
From: "Andrew R. Southern" <a.southern@ic.ac.uk>
Subject: Re: Extra-long Rubik's magic rings

At 07:53 12/05/97 -0400, you wrote:
>As I recollect the 12 panel Magic Rings was called the Master's Edition. 
>Since they have been out of manufacture for quite awhile suggest you buy it
>when you see it.
>
>The 8 panel edition has been released by ODDZON and is in some toy stores
>along with some other Rubik's items.
>
>As some of you might remember there also was a 4 panel magic sold.  If you
>have ever taken one apart you know that any multiple of 4 is possible. 
>Awhile back I posted intstructions on CUBE LOVERS for restringing and or
>making your own variant.  The panels are just 2 pieces of plastic with the
>design sandwiched in bewteen.  They are not glued but held together by the
>strings.
>
>
>Does anybody know where to get I BET YOU CAN"T ?? a variant made with
>hexagon panels.
>

I must point out at this point that the magic's stringing pattern actually
revovles around just two patterns, which are equivalent to each other in
some positions.

These patterns actually only require three tiles. Three tiles will not make
a circuit, but they will make a nice line or L-shape. 

For a bit of real(?) excitment(??) you could try to make one out of just two
panels. This is highly possible, as I did it back when I was 12 or 13 in
Senior school. The secret is that you loop one of the "Ligaments" as I used
to call them back around one panel twice. I cannot remember whether or not
this could go around for ever or whether it had a definiate start and end
point, but it certainly confused my classmates.

I also know that it is possible to add just two panels onto an ordinary
Rubik's Magic, creating a 2x5 array, which I don't have to say, is not
exactly divisible by four. This was fully functional, but when it was folded
in half, it did not go into "Loop" or "Tie Fighter", it went into "Fish"
from either side, unless the last two squares were folded in before folding
along the centre line. I called this the "Diabolical Edition" because it was
harder than the Master Edition, and I made about four of them for members of
my old school (Blue Coat, Liverpool).

loop:
!--!
!__!

tie fighter:
\/\/
/\/\

Fish:
\/-\
/\_/

I also created a "master Edition" from Two ordinary magics when someone drew
freehand the "Sandwich Filler" pieces of paper. 

But my Piece de Resistance is the 64-panelled monstrousity that used to take
me an hour to solve, and was a complete work-out as there was so much
to-and-fro ing of the entire magic. I had to use the pictures from 12
magics, and is just an extension of the master editions solution. When it is
in a 32x2 state, it was higher than my (then) best friend, and you require
quite substantial floor space to solve it.

-Andy Southern
(The artist formerly known as the Unofficail Thermo-Fluids Fan Club of the UK).





From cube-lovers-errors@oolong.camellia.org  Tue May 13 15:14:04 1997
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MMDF-Warning:  Parse error in original version of preceding line at moag.epilogue.com
Subject: Re: Strange new Rubik puzzle in Java
Date: Tue, 13 May 1997 20:03:49 +0100
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> From: David Singmaster Computing & Maths South Bank Univ
<david.singmaster@sbu.ac.uk>


> Adapting a cube to having a dodecahedral appearance was certainly
considered
> in the 1980s, thoguh I can't remember if anybody ever made them for sale

	This puzzle is depicted on page 335 of "Metamagical Themas"
by Douglas Hofstadter, which has just come back into print. He says that
it was in Meffert's catalogue way back then, called the "Pyraminx Ball"


							David Byrden



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Date: Tue, 13 May 1997 22:26:22 -0400
To: cube-lovers@ai.mit.edu
From: karen angelli <kangelli@tiac.net>
Subject: 2x6 Rubik's Magics
Mime-Version: 1.0
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Wow, what a response.  
	Yes, you are all correct, the ones I saw are 6x2, not 8x2.  However, as I
now know, almost any number are possible.  I saw the puzzles for sale at
The Games People Play at 1100 Massachusetts Ave, Cambridge, Ma 02138
617-492-0711 (not far from the cube-lovers administrators at MIT) (They are
also available from Puzzletts in Seattle).  The owner of Games People Play
had about five for sale, and also several of the original black background
8panel Magics.  She sells the 12 panel for $20 and the 8 panel for $15.
However, the buyer must beware.  Each of the 12 panel ones I looked at in
Games People play had one broken string each.  I bought one despite the
flaw, because she gave me the phone number for someone who knows how to fix
them, and the $20 price was substantially below the $35 plus s.h. from
Puzzletts.  I had lunch with my Rubiks Magic repair-man today, got a lesson
in Rubiks magicology and now have a fully functional toy.
	According to my sources, the Magics can function with as few as half of
the original strings.  Each string set is double strung for extra strength.
 Accordingly, the thing will still work properly if any one breaks, or if
any number break, so long as both strings on any particular loop both
break.  He suggests removing any broken string as soon as possible so that
it does not get in the way of the other ones and create a cascading loose
loop catastrophe.  
	He happened to have a number of extra loops of fishing line because he
fixes the puzzles for a puzzle mail-order company.  The company told him
that there was a large batch of puzzles made with defective strings.  They
send him the puzzles, and he canabalizes them to make n-1 Magics from n
defective magics.
	He fixes the Magics by wrapping the loop around the same path as the one
path that has only one loop around it.  There are only a couple of rules to
follow: the string paths on one side cross the string paths on the other
side at 90degree angles, and when the two strings from one side pass the
strings from the other side between panels, the two from one side must be
either both outside of the two from the other side, or both inside the ones
from the other side.
	Personally, I've always been too paranoid about my puzzles to abuse them
enough to actually find out how to take them apart.  I also usually
hesitated at spending the money to get several puzzles when one seemed
sufficient.  I guess I need to learn how to live life on the wild side.

'e-ya later, 
Peter Reitan



From cube-lovers-errors@oolong.camellia.org  Tue May 20 13:57:16 1997
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From: Philip Knudsen <philipknudsen@hotmail.com>
To: cube-lovers@ai.mit.edu
Subject: Triamid
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Date: Tue, 20 May 1997 07:47:33 PDT

Hi everybody!

I recently purchased a Rubik's Triamid, Matchbox version.
Do any of you know whether that version differs from the new Oddzon release? The 
Oddzon Tangle certainly is different from the Matchbox Tangle.
The leaflet that comes with the new Oddzon Tangle says the following puzzles are 
in the new series:
Tangle, 2x2x2 Mini Cube, Rubik's Cube, Snake, and Triamid.
I'd like to know if they have re-released even more of the old stuff.
None of them are available where i live (Copenhagen, Denmark)
Also, if anyone has a Rubik's Hat they's like to part with, i'd be real 
interested.

Philip K





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Get Your *Web-Based* Free Email at http://www.hotmail.com
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From cube-lovers-errors@oolong.camellia.org  Thu May 22 16:39:15 1997
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Subject: Re: Triamid
Date: 22 May 1997 18:49:46 GMT
Organization: California Institute of Technology, Pasadena
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Philip Knudsen <philipknudsen@hotmail.com> writes:
>I recently purchased a Rubik's Triamid, Matchbox version.
>Do any of you know whether that version differs from the new Oddzon release? The 
>Oddzon Tangle certainly is different from the Matchbox Tangle.

As far as I can tell, the only difference is the Tangle.
(Well, and the colors of the Snake and the Magic.)

>The leaflet that comes with the new Oddzon Tangle says the following puzzles are 
>in the new series:
>Tangle, 2x2x2 Mini Cube, Rubik's Cube, Snake, and Triamid.

I believe the Magic was released in this release as well.

>I'd like to know if they have re-released even more of the old stuff.
>None of them are available where i live (Copenhagen, Denmark)
>Also, if anyone has a Rubik's Hat they's like to part with, i'd be real 
>interested.

Heh heh.  (No, I don't have one.  I do have a Rubik's Maze, though.)


>Philip K





>---------------------------------------------------------
>Get Your *Web-Based* Free Email at http://www.hotmail.com
>---------------------------------------------------------

--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
-------------------------------------------------------------------------------
The cynical poet -- is worst of the worst.
What others are thinking -- he says out loud first.


From cube-lovers-errors@oolong.camellia.org  Tue May 27 15:51:48 1997
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From: Philip Knudsen <philipknudsen@hotmail.com>
To: cube-lovers@ai.mit.edu
Subject: Tangle & Triamid
Content-Type: text/plain
Date: Tue, 27 May 1997 04:34:47 PDT

Thanks alot for the info on Triamid (no major difference).

Pete Beck asked:

  How is tangle different.  I know it is plastic and the strings are    raised 
but is the puzzle different?

The Oddzon version has only nine pieces, which are double-sided. For these 18 
sides, all pieces from the original tangle, except the six with a straight 
yellow line are used. Of the 9 puzzle pieces, 3 have straight red on both sides, 
3 have a straight green, and 3 a straight purple. The result is a somewhat 
easier puzzle, at least it's solveable without computer aid.

Back to the Triamid: Has anyone ever tried to build a large one out of several 
small?  To build a 4x4 one would need three Triamids to get enough connector 
pieces. The question is, would the puzzle work, or would it be too hard to 
disconnect a 3x3 top from a 4x4 base?
Me, i'm seriously considering buying 2 extra. 

Philip K



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Get Your *Web-Based* Free Email at http://www.hotmail.com
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From cube-lovers-errors@oolong.camellia.org  Tue May 27 15:52:28 1997
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From: Goyra <Goyra@cheerful.com>
To: cube-lovers@ai.mit.edu
Subject: Java Rubik Gallery now open
Date: Tue, 27 May 1997 19:46:27 +0100
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	Those of you with Java browsers, powerful
PCs and fast Internet connections may want to 
revisit my Rubik Gallery at

	http://www.iol.ie/~goyra/Rubik.html

I have just installed Rubik cubes of all sizes up to 11.
I intend to maintain the Gallery as the world's
largest collection of "Rubik" puzzles, a place where 
you can play with that obscure puzzle you were never
able to find in the shops. Any suggestions as to what 
should go in it next, are welcome.


					David Byrden




From cube-lovers-errors@oolong.camellia.org  Tue May 27 17:37:53 1997
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Date:     Tue, 27 May 97 17:18:18 EDT
From: Nichael Cramer <ncramer@bbn.com>
To: Goyra <Goyra@cheerful.com>
cc: cube-lovers@ai.mit.edu
Subject:  Re:  Java Rubik Gallery now open

>From: Goyra <Goyra@cheerful.com>
>Subject: Java Rubik Gallery now open
>Date: Tue, 27 May 1997 19:46:27 +0100
>
>
>	Those of you with Java browsers, powerful
>PCs and fast Internet connections may want to 
>revisit my Rubik Gallery at
>
>	http://www.iol.ie/~goyra/Rubik.html
> [...]

I freely admit that I tuned in fully expecting to be disappointed.  But this
is really slick.

N


From cube-lovers-errors@oolong.camellia.org  Wed May 28 16:35:19 1997
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Date:     Wed, 28 May 97 8:30:53 EDT
From: Nichael Cramer <ncramer@bbn.com>
To: cube-lovers@ai.mit.edu
Subject:  [gknauth:  Professor cracks Rubik's cube mystery]

[Simply passing along bits --N]


----- Forwarded message # 1:

Date: Wed, 28 May 97 08:13:32 EDT
From: gknauth@BBN.COM
Subject: Professor cracks Rubik's cube mystery

> From: Andy Lee <alee@worldstreet.com>

Professor cracks Rubik's cube mystery

LOS ANGELES (May 27, 1997 5:43 p.m. EDT) - A University of California
computer science professor has solved the long-standing mystery of Rubik's
cube, university officials said Tuesday.

Richard Korf found a way to line up the colored squares of the cube in an
average 18 moves and a maximum of 20, officials said without explaining
exactly how it is done.

Rubik's cube, launched in the 1970s by the Hungarian Erno Rubik, became a
worldwide phenomenon, with people spending hours trying to manipulate it
into color-coordinated rows.

Korf is due to reveal his method at a national conference on artifical
intelligence July 28 in Providence, Rhode Island.

Copyright 1997 Nando.net, Agence France-Presse

----- End of forwarded messages


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Date: Wed, 28 May 1997 14:33:20 -0700
From: Joe McGarity <joemcg3@snowcrest.net>
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Forgive me for saying so, but, "Bulls**t"


From cube-lovers-errors@oolong.camellia.org  Wed May 28 17:40:08 1997
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To: cube-lovers@ai.mit.edu
Cc: Nichael Cramer <ncramer@bbn.com>
Subject: Re: [gknauth:  Professor cracks Rubik's cube mystery]
References: <199705281233.IAA05016@life.ai.mit.edu>
From: Mark Atwood <zot@ampersand.com>
Date: 28 May 1997 17:38:41 -0400
In-Reply-To: Nichael Cramer's message of Wed, 28 May 97 8:30:53 EDT
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Nichael Cramer <ncramer@bbn.com> writes:
> 
> ----- Forwarded message # 1:
> 
> Date: Wed, 28 May 97 08:13:32 EDT
> From: gknauth@BBN.COM
> Subject: Professor cracks Rubik's cube mystery
> 
> ...
> 
> Richard Korf found a way to line up the colored squares of the cube in an
> average 18 moves and a maximum of 20, officials said without explaining
> exactly how it is done.
> ...
> 

Is this a new upper bound?

-- 
Mark Atwood       | We must not remind them
zot@ampersand.com | that Giants walk the Earth.


From cube-lovers-errors@oolong.camellia.org  Thu May 29 00:28:22 1997
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Date: Wed, 28 May 1997 15:18:10 -0700
From: Richard E Korf <korf@cs.ucla.edu>
Message-Id: <199705282218.PAA17014@denali.cs.ucla.edu>
To: Cube-Lovers@ai.mit.edu
Subject: rumor control

Dear Cube-Lovers,
   Apparently some work I did recently has gotten badly mangled by the press.  I
have NOT resolved the question of whether or not 20 face turns is the maximum
distance one can get from a scrambled cube.
   What I did is to write a heuristic search program that finds optimal
solutions to arbitrary scrambled cubes.  The algorithm is very different from
the method of Fiat, Moses, Shamir, et al, and seems to be competitive with their
algorithm in terms of time and space.  The current version of my program is
practical for cubes up to 18 moves away from solved.
   Out of 10 randomly generated cubes, one was solved in 16 moves, 3 required 17
moves, and 6 required 18 moves, suggesting that the median optimal solution
length is probably 18 moves.
   A paper on this work will be presented at the National Conference on
Artificial Intelligence (AAAI-97) in Providence, RI in July. I'd be happy to
send a postscript copy of the paper to anyone who is interested, unless there
are a lot of requests, in which case I'll just post it on my web site and put a
pointer here.  In addition, if there is enough interest, I could write a short
summary of the paper for this list. Thanks for your attention.
                            -rich korf



From cube-lovers-errors@oolong.camellia.org  Thu May 29 00:28:49 1997
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Nichael Cramer wrote:
> 
> [Simply passing along bits --N]
> 
> ----- Forwarded message # 1:
> 
> Date: Wed, 28 May 97 08:13:32 EDT
> From: gknauth@BBN.COM
> Subject: Professor cracks Rubik's cube mystery
> 
> > From: Andy Lee <alee@worldstreet.com>
> 
> Professor cracks Rubik's cube mystery
> 
> LOS ANGELES (May 27, 1997 5:43 p.m. EDT) - A University of California
> computer science professor has solved the long-standing mystery of Rubik's
> cube, university officials said Tuesday.
> 
> Richard Korf found a way to line up the colored squares of the cube in an
> average 18 moves and a maximum of 20, officials said without explaining
> exactly how it is done.
> 
> Rubik's cube, launched in the 1970s by the Hungarian Erno Rubik, became a
> worldwide phenomenon, with people spending hours trying to manipulate it
> into color-coordinated rows.
> 
> Korf is due to reveal his method at a national conference on artifical
> intelligence July 28 in Providence, Rhode Island.
> 
> Copyright 1997 Nando.net, Agence France-Presse
> 
> ----- End of forwarded messages 
Let's say the cube-lovers of the world are skeptical...

Are we talking about q turns or q+h turns?
My own conjecture (which I have kept to myself until now) was that Mike Reid's
12-flip pattern in 24 q turns was the antipode in q turns only. I have no
proof of this fact.
 
It is possible Professor Korf has found a totally new approach to the rubik
problem. Dik Winter (months and months before) never did find any position on
the 3x3x3  which required more than 20 moves in the q+h metric. 
 
Conventional wisdom (using Kociemba type algorithms) was that the god's algorithm
for the standard 3x3x3 cube was intractible.
 
In case case, without more evidence, this news message does not add to the
existing level of cube knowledge. 
 
I'm still waiting and watching for any optimal solutions to the Megaminx 
spot patterns!
 
Perhaps Professor Korf has a mathematical proof. It does seem unlikely that he
sifted through all the possible positions.
 
-> Mark


From cube-lovers-errors@oolong.camellia.org  Thu May 29 00:29:17 1997
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Subject: Professor cracks Rubik's cube mystery
Message-ID: <19970528.222126.9494.2.shaggy34@juno.com>
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From: Josh D Weaver <shaggy34@juno.com>
Date: Wed, 28 May 1997 23:22:16 EDT


What do the skeptics have to say about the 20 move solve?  It was said
that in theory it could be done; but is it really possible?  

Does anyone know how long time wise it took Richard Korf to solve it
using the 20 move method?

I'm just a 15 year old who can solve the cube so I don't know much about
theories or mathematical patterns.  So if someone can explain it to me
that would be great.

If it all that is claims to be, then I can't wait to know how to use the
method. 


From cube-lovers-errors@oolong.camellia.org  Thu May 29 00:30:06 1997
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Message-Id: <970528233918.21401a3a@iccgcc.cle.ab.com>
Subject: Re: Solved in 20 moves?

From:	SMTP%"joemcg3@snowcrest.net" 28-MAY-1997 22:02:17.57
To:	SCHMIDTG
CC:	
Subj:	Solved in 20 moves?

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On the subject of a new upper bound for Rubik's cube solution, Joe
McGarity wrote:

>Forgive me for saying so, but, "Bulls**t"

While it's certainly possible that the news release may have distorted the
original message, I certainly would not in any way discount the work of
Dr. Korf as he is a recognized authority in the area of computer search
(you'll find his name within the index of any decent book on the subject).

As far as I know he is also the first, and only, person to have developed
a computer program capable of discovering a general solution to the cube
starting only from basic knowledge of cube states and operators.

Forgive me for keeping an open mind on this one.

-- Greg


From cube-lovers-errors@oolong.camellia.org  Thu May 29 15:24:53 1997
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From: Benjamin Wong <chi@cse.unsw.edu.au>
To: Josh D Weaver <shaggy34@juno.com>
Date: Thu, 29 May 1997 17:34:55 +1000 (EST)
X-Sender: chi@pipe13.orchestra.cse.unsw.EDU.AU
cc: cube-lovers@ai.mit.edu
Subject: Re: Professor cracks Rubik's cube mystery
In-Reply-To: <19970528.222126.9494.2.shaggy34@juno.com>
Message-ID: <Pine.OSF.3.95.970529172749.8049A-100000@pipe13.orchestra.cse.unsw.EDU.AU>

On Wed, 28 May 1997, Josh D Weaver wrote:

._@_.
._@_.What do the skeptics have to say about the 20 move solve?  It was said
._@_.that in theory it could be done; but is it really possible?  
._@_.

There was a prove that within 16-18 face move, the number of possible
pattern exceed the number of possible pattern of the cube,
hence they deduce that any pattern can be acheive within 16-18 move,
ie: it can be done, 

if u can move forward 18 move, 
by backtracking, it can be move backward within 18 move.


._@_.Does anyone know how long time wise it took Richard Korf to solve it
._@_.using the 20 move method?
._@_.
._@_.I'm just a 15 year old who can solve the cube so I don't know much about
._@_.theories or mathematical patterns.  So if someone can explain it to me
._@_.that would be great.
._@_.
._@_.If it all that is claims to be, then I can't wait to know how to use the
._@_.method. 

I don't think "the method",(if it actually exist,) can be applied to human,
i think it involved a lot of AI research, where computer will think
faster and can plan say 5 to 6 move ahead, and see if it recognise any pattern


I will not put much hope on trying to solve in 20 move 
with a real cube by hand.


o------------------------------------------------------o
|Error: Reality.sys Corrupt?  Reboot Universe [Y,N,Q]  |
+---------------o--------------------------------------o
| Benjamin Wong |  E-mail: chi@cse.unsw.edu.au         |
|  		|  or       benjaminwong@hotmail.com   |
| 		|  http://www.cse.unsw.edu.au/~chi     |
o---------------o--------------------------------------o
|=A1u=C2=E5=A5=CD=A1I=BD=D0=B0=DD=A1y=BA=B5=BF=DF=B2=B4=A1z=AA=BA=A6=A8=A6]=ACO=AC=C6=BB=F2=A1H=A1v                |
|Quick Quiz:  Describe Universe ? Give Three Example.  |
o------------------------------------------------------o



From cube-lovers-errors@oolong.camellia.org  Thu May 29 15:26:00 1997
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I have been reprimanded.  

And justly so.  Of course the first step to science is to retain an open
mind.  My comment was not to be taken 100% seriously, but to hopefully
point out with a bit of humor that the burden of proof here is on those
making the claim.  In retrospect the post should have read, "Bulls**t
:-)"  With the additional challange of proving me wrong invited and
openly reviewed by myself.  I would be delighted to be proved wrong in
this matter, but I maintain that the evidence should be carefully
examined and discussed as a group before we jump to a concensus on the
subject.

Having climbed out of the hole I dug for myself, I will now remove my
feet from my mouth and check out what Prof. Korf has to say.

By the way, I will do PR work for interested corporations and
politicians for a fee.

Joe McGarity




From cube-lovers-errors@oolong.camellia.org  Thu May 29 16:30:21 1997
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To: Cube-Lovers@AI.MIT.EDU
From: Wei-Hwa Huang <whuang@ugcs.caltech.edu>
Subject: Re: rumor control
Date: 29 May 1997 19:55:39 GMT
Organization: California Institute of Technology, Pasadena
Message-ID: <5mkmvr$b2v@gap.cco.caltech.edu>
References: <cube-lovers.199705282218.PAA17014@denali.cs.ucla.edu>
NNTP-Posting-Host: pyro.ugcs.caltech.edu
X-Newsreader: NN version 6.5.0 #3 (NOV)

Richard E Korf <korf@cs.ucla.edu> writes:
>Dear Cube-Lovers,
>   Apparently some work I did recently has gotten badly mangled by the press.

My symphathies.  I remember a news article a few years ago when the
largest Mersenne prime was discovered; it went something like:

"Mathematicians have discovered a number that has 224,375 digits and
is divisible by 1."

>   A paper on this work will be presented at the National Conference on
>Artificial Intelligence (AAAI-97) in Providence, RI in July. I'd be happy to
>send a postscript copy of the paper to anyone who is interested, unless there
>are a lot of requests, in which case I'll just post it on my web site and put a
>pointer here.  In addition, if there is enough interest, I could write a short
>summary of the paper for this list. Thanks for your attention.

Please do so.  I am sure many people on this list would be interested in
seeing it.


--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
-------------------------------------------------------------------------------
Question everything.  Learn something.  Answer nothing.  -- Engineer's Motto


From cube-lovers-errors@oolong.camellia.org  Thu May 29 21:58:15 1997
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From: Richard E Korf <korf@cs.ucla.edu>
Message-Id: <199705300024.RAA18247@denali.cs.ucla.edu>
To: Cube-Lovers@ai.mit.edu
Subject: Description of algorithm for finding minimal-move solutions to Rubik's Cube

Dear Cube-Lovers,
   Here is the promised short description of my algorithm for finding optimal
solutions to Rubik's Cube.  I use the face-turn metric, meaning any twist of a
face, even 180 degrees, counts as a single move.  A twist of a center slice
can only be accomplished by two twists of the outside faces.
   The algorithm is a heuristic search, called Iterative-Deepening-A*, or IDA*,
for any artificial intelligence (AI) folks in the group.  Given a scrambled
cube, it first looks for solutions one move long, then solutions two moves long,
then three moves, etc.  Each iteration searches for solutions of successively
greater length, until a solution is found.  At that point it quits, returning
what must be an optimal solution, barring program bugs.  This is a completely
brute-force approach to the problem.  At a million twists per second, searches
to depth 10 would take almost 3 days.
    To make this approach practical, we need a function that given a cube state
will efficiently calculate a lower bound on the number of moves needed to solve
it.  This is called a heuristic evaluation function.  For example, we can
precompute the number of moves needed to solve each edge cubie individually from
each possible position and orientation. Then given a state of the cube, we sum
the number of moves needed to solve all 12 edge cubies individually, and divide
by 4, since each move moves 4 edge cubies.  This heuristic, called 3D Manhattan
Distance, has an average value of 5.5. The important thing is that this function
always return a lower bound on the number of moves needed to solve a state.
    During our search we compute the Manhattan Distance of each state.  If we
are looking for solutions of length 10, for example, and we have a state that is
5 moves from the initial state, and its Manhattan Distance from the solved state
is 6 moves, we don't have to search that path any deeper, since it will take at
least 11 moves to get to the goal along that path, since 6 is a lower bound on
the number of moves needed to solve the state.  Adding the Manhattan Distance
heuristic to our search algorithm lets us search to depth 14 in about 3 days.
We could do the same thing with the corner cubies, and take the maximum of the
two values, but that doesn't help much.
    To do better, we need a more accurate heuristic function.  For that, we use
an idea call "Pattern Databases" due to Culberson and Schaeffer. See Culberson,
J.C., and J. Schaeffer, Searching with pattern databases, Proceedings of the
11th Conference of the Canadian Society for the Computational Study of
Intelligence, published in Advances in Artificial Intelligence, Gordon McCalla
(Ed.), Springer Verlag, 1996.
   For example, if we consider just the corner cubies, there are only about 88
million possible states they could be in (8!x3^7). We exhaustively build and
store a table, using breadth-first search, that tells us the minimum number of
moves needed to solve just the corner cubies from every possible combination,
ignoring the edge cubies.  This value ranges from 0 to 11 moves, averages 8.764
moves, and requires only 4 bits per state. (We could reduce this further using
an idea of Dan Hoey's published in this list awhile ago.)  This table only has
to be computed once, taking about a half hour, and requires about 42 megabytes
of memory to store (a megabyte is 1048576 bytes).
    Then, during the search, we compute the heuristic lower bound on the number
of moves to the goal by looking up the configuration of the corner cubies, and
using the number of moves stored in the table. 8.764 is a lot better than 5.5.
    Finally, we divide the edge cubies into two groups of six, and compute a
similar table for each group.  There are too many combinations of all 12 edge
cubies to build a single table.  The final heuristic function we use is the
maximum of 3 different values, the moves needed to solve the corner cubies, and
the moves needed to solve each group of six edge cubies. The total memory for
all three tables is 82 megabytes.
    Given more memory, we could built larger tables, for example, considering 7
edge cubies at a time.  This would give a more accurate heuristic value, and
reduce the running time of the search algorithm.  In fact, an informal analysis
of the performance of the algorithm suggests that its speed will increase
linearly with the amount of available memory.  Thus, given twice as much memory,
the algorithm should run in roughly half the time.  Disks and other secondary
storage are of no use, since the access time is much too slow to be worthwhile.
   The current version of the program is written in C on a Sun Ultra-Sparc
Model-1 workstation with 128 megabytes of memory.  It generates about 700,000
states per second.  Depth 16 searches typically take less than 4 hours, depth 17
searches take about 2 days, and complete depth 18 searches take about 27 days. A
complete depth 19 search would take about a year.  Each depth takes roughly
13.34847 times longer than the previous, which is the branching factor of the
problem space.
   The algorithm is easily parallelized. Given 18 processors, for example, we
make all 18 possible first moves, and hand each of the resulting states to a
different processor to solve. This will give roughly linear speedup with the
number of processors, since the amount of time needed to search to the deeper
levels is very consistent from one state to the next. 
   Sorry for the length of this message, but I hope it will of interest to some
of you. If you'd like the full paper, just send me a message. Thanks very much.
                                  -rich




From cube-lovers-errors@oolong.camellia.org  Thu May 29 22:04:04 1997
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From: Ed Schwalenberg <ed@odi.com>
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To: joemcg3@snowcrest.net
Cc: cube-lovers@ai.mit.edu
In-Reply-To: Joe McGarity's message of Thu, 29 May 1997 00:04:46 -0700 <338D2A8E.56E4@snowcrest.net>
Subject: Okay, I give.
Organization: Object Design, 25 Mall Rd, Burlington, MA 01803 - 617-674-5337

  Date: Thu, 29 May 1997 00:04:46 -0700
  From: Joe McGarity <joemcg3@snowcrest.net>

  I have been reprimanded.  

  And justly so.  ...

This all reminds me of a time a few years ago when I read an article in
the Boston Globe about a physics professor from Iowa who claimed to have
discovered a "black box" way to remove radioactivity.  I sneered at the
idea, until a friend of mine pointed out that the professor didn't say how
long you had to leave the radioactive material in the box....
  

[ The moderator will allow no further messages on the topic of 
  how the press screws up explaining science, mathematics and
  technology to the public.  ]

From cube-lovers-errors@oolong.camellia.org  Fri May 30 14:28:22 1997
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I hope that I am not alone in that much of Prof. Korf's description is a
little over my head.  Perhaps the Professor or others can recomend books
on AI or group theory to those of us whose education only goes as far as
trigonometry.  This is a very interesting subject and I find myself
wanting to understand it better.  Is there anything out there aimed at
the beginner?  If so I would very much like to see it.




From cube-lovers-errors@oolong.camellia.org  Fri May 30 18:38:34 1997
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Date: Fri, 30 May 1997 13:44:48 -0700
From: Jin "Time Traveler" Kim <chrono@ibm.net>
Organization: The Fourth Dimension
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Joe McGarity wrote:
> 
> I hope that I am not alone in that much of Prof. Korf's description is a
> little over my head.  Perhaps the Professor or others can recomend books
> on AI or group theory to those of us whose education only goes as far as
> trigonometry.  This is a very interesting subject and I find myself
> wanting to understand it better.  Is there anything out there aimed at
> the beginner?  If so I would very much like to see it.

Prof. Korf's solution to the Cube sounds like it basically maps all
possible iterations within a given number of steps.  Once you know all
the possible combinations given the maximum number of turns, you can
then just compare a scrambled cube to the map and see if it falls within 
one of the available templates.  And of course, the more moves you
calculate out to, the longer it's going to take due to the geometrically
increasing number of possible movements.  

Yes, the solution, as the good Professor explains it, is definately over
my head, but I think I know how he is going about solving it.  Brute
force.  Of course, the DETAILS of how it's done are over my head as
well.  Ultimately I think the cube is more satisfying if I solve it
myself, even if it takes me 3 minutes and dozens of twists.

-- 
Jin "Time Traveler" Kim
chrono@ibm.net
VGL Costa Mesa


From cube-lovers-errors@oolong.camellia.org  Fri May 30 19:43:03 1997
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Date: Fri, 30 May 97 19:41:15 EDT
Message-Id: <9705302341.AA06714@sun34.aic.nrl.navy.mil>
From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: chrono@ibm.net
Cc: cube-lovers@ai.mit.edu
Reply-To: hoey@aic.nrl.navy.mil
Subject: Re: The rest of us

> Prof. Korf's solution to the Cube sounds like it basically maps all
> possible iterations within a given number of steps.  Once you know all
> the possible combinations given the maximum number of turns, you can
> then just compare a scrambled cube to the map and see if it falls within 
> one of the available templates.

No, you've misunderstood.  Rich doesn't attempt to "map all possible
iterations" (by which you seem to imply some sort of preprocessing so
as to be able to recognize any position at a given depth).  After all,
according to his estimate of the median, there should be over 10^18 of
them at depth 18f, and finding them all would take his program many
thousands of years.

Instead, he takes advantage of knowing what the target cube is by
using a measure called a "heuristic".  A heuristic estimates how far a
given process is from solving the cube, such as the "oriented distance
from home" (ODH) that appears in the archives.  Then if you have tried
7f turns and you know it will take at least 12f more, you know you're
on the wrong track, and look elsewhere.

But there are lots and lots of wrong tracks, and you need to recognize
and discard them quickly.  The ODH isn't that good an estimate, so it
doesn't discard enough of them--it would still take 250 years to
search for one position.  Finding better ways of estimating how far
you are from the goal is what the research is about.

So just because he says it's "brute force" doesn't mean you list all
the positions in advance.  You definitely need to know what the goal
position is for Rich's approach to work.  In particular, his method
does not seem to be applicable to finding what the furthest position
is.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil


From cube-lovers-errors@oolong.camellia.org  Fri May 30 20:33:43 1997
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Date: Sat, 31 May 1997 02:30:28 +0200
From: Herbert Kociemba <kociemba@hrz1.hrz.th-darmstadt.de>
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Subject: Re: Description of algorithm for finding minimal-move solutions to Rubik's Cube
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Richard E Korf wrote:
> 
> Dear Cube-Lovers,
>    Here is the promised short description of my algorithm for finding optimal
> solutions to Rubik's Cube.

>From the description it is evident, that the algorithm Richard E Korf
uses is basically identical to the the sub-algorithm which is used in
each stage of my two stage algorithm to solve the cube. What he calls
"heuristic functions" are the "pruning tables" of Dik Winter and Michal
Reid and the "filters" in the original description of the algorithm in
CFF 28 (April 1992) of the Dutch Cubist Club. Here is a short summary of
this algorithm in the version I implemented in a windows95 program
requiring 16Mbyte of Ram. When I will have included a help-function
within the next weeks, I will offer it to all interested cubists for
free: 

In phase 1, the cube is transformed to an element of the subgroup
generated by <U,D,R2,L2,F2,B2>. This is equivalent to restore the
orientation of the 8 corners and 12 edges and to put the 4 edges of the
UD-slice in that slice. There are 3^7=2187 possible states for the
corner orientations, 2^11=2048 possible edge orientations and
12*11*10*9/(1*2*3*4)=495 possible positions for the 4 edges of the
UD-slice. The "heuristic functions" consist of three tables, using 4
bits for each entry. The first table stores the minimum numbers to solve
the 2187*2048 possible states to restore the orientation of both edges
and corners, the other tables have 2187*495 and 2048*495 entries and
store the corresponding minimum numbers.
Dik Winter proved, that 12 moves always suffice to get to this subgroup.

In phase 2, the cube is solved in this subgroup, using only
U,D,R2,L2,F2, and B2. Now we have do restore the permutations of the
corners, edges and middle slice. There are 8! states for the corner
permutations, 8! states for the edge permutations and 4! states for the
permutations of the UD-slice. The "heuristic functions" consist of only
two tables, storing the minimum numbers to restore both edges and
UD-Slice and both corners and UD-Slice, having 8!*4! entries each. 
The table for the minimum numbers to restore both edges and corners
would have 8!*8! entries and is not possible with the current hardware.
Michael Reid proved, that 18 moves always suffice in this subgroup.

Having found a solution in stage1 and stage2 the algorithm does not
stop, but generates other solutions in stage1. So if for example we have
10 moves in stage1 and 12 moves in stage2, there might be a solution
with 11 moves in stage1 but only 10 moves in stage2. Running long
enough, the algorithm will find the overall optimal solution, having no
moves in stage2 then.

Due to the smaller size of the subgroups a first solution usually will
be
found within seconds. This first solution  is optimal for phase1, but
indeed (usually) not optimal for the overall solution. Typically you
will have solutions with less then 20 moves within minutes and the
optimal solution for states with lets say less then 16 moves will be
found within a reaseonable time.

Herbert



From cube-lovers-errors@oolong.camellia.org  Fri May 30 21:24:38 1997
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Sender: Haym Hirsh <hirsh@cs.rutgers.edu>
Date: Fri, 30 May 97 21:10:39 EDT
From: Haym Hirsh <hirsh@cs.rutgers.edu>
Reply-To: Haym Hirsh <hirsh@cs.rutgers.edu>
To: joemcg3@snowcrest.net
Subject: Re: The rest of us
In-Reply-To: Your message of Fri, 30 May 1997 11:14:19 -0700
Cc: cube-lovers@ai.mit.edu
Message-ID: <CMM-RU.1.5.865041039.hirsh@pei.rutgers.edu>

Here's a brief attempt at a "layman's" description of Professor Korf's
work:

Imagine you have a function that takes as input a messed-up Rubik's
cube, and outputs a guess of how many moves it will take to get it to
the solved state.  Further, assume this guess is never greater than
the correct number of moves -- sometimes your solution-length guesser
may make a correct guess, but sometimes (or even perhaps always) it
may underestimate the number.

There is an algorithm called A* that is guaranteed to find a shortest
solution sequence for any Rubik's cube it is given, as long as it is
given a solution-length guesser that has this never-overestimates-the-
number-of-moves-to-solved guarantee.

The problem is that A*'s guarantee is only that it will return a
shortest solution to any cube, with no guarantee on how long it will
take to find it.  Due to this run-time issue A* is only applicable to
the most trivial of problems.

However, in the mid-80s Professor Korf presented a tractable variant
of A*, called IDA* (Iterative Deepening A*) that has the same
guarantee as A* on finding shortest solutions, but is much faster.

The problem now, though, is that even IDA* can also take a long time.
Its salvation, however, is that, loosely speaking, the better the
solution-length-guessing-function is, the faster IDA* will run.  Thus,
for example, you could use a function that always returned 0 as the
guess for how many moves you are from the start.  It's not a
particularly clever guess, but it obeys the rule that it never
overestimates the solution length.  Therefore, you could use it with
IDA* (or, for that matter, A*) to find shortest solutions to any cube.
Except that it would run too slow, because the solution-length guesser
is so dumb.  A better solution-length guesser would help IDA* run
faster.

Professor Korf came up with a way to more intelligently guess what the
solution length will be for arbitrary cubes -- it gives something much
closer to the true value, but still without overestimating.  A
simplified form of this would be to figure out how many moves at
minimum it will take to get the corners in place, and use this
corners-only solution length as a guess.  This will never overestimate
the solution length, since to get everything in place you certainly
have to get at least the corners into their proper positions, and it
is better than a guesser that always returns 0.

Professor Korf also had to figure out how to compute these guesses in
an efficient fashion, since guesses will be requested many many times
by IDA* as it explores possible intermediate cubes in its search for
the solution.  To do this he enumerated all 88 million configurations
of corners (different cubes with different arrangements of edges but
with identical corners are considered identical configurations).  For
each he figured out the minimum number of moves that would be
necessary to get them into their correct position in a solved cube if
edges were ignored (taking a non-trivial, but non-infinite, amount of
time to do this for each of the 88 million configurations).  Finally,
he generated a table with 88 million entries, with each entry
corresponding to a corner configuration and containing the solution
length for that configuration.  This created a way to quickly compute
his more accurate corner-centric solution-length guesser, via table
lookups.

In truth Professor Korf improves on this even further by developing a
better solution-length guesser that does similar things with edges as
I just described with corners, also using tables for efficient guess
calculation.  The result is a solution-length guesser that is accurate
enough to allow IDA* to solve the 10 random cubes that he generated.

More specifically, Professor Korf generated random cubes by taking a
solved cube and making 100 random turns to it.  He did this 10
separate times, and got 10 messed-up cubes.  He then ran IDA* using
his table-based solution-length guesser, and solved all 10, one in 16
moves, three in 17 moves, and the rest in 18 moves.  Because he used
IDA*, and because his solution-length guesser never overestimates
solution lengths, his solutions are guaranteed to be optimal (due to
IDA*'s mathematical guarantees).

This does not argue that 18 is the longest solution possible for any
cube.  Just that for the 10 he generated randomly, none required more
than 18.  Perhaps some cubes are more than 18 moves away from start.
None simply happened to arise amongst his 10 cases.



From cube-lovers-errors@oolong.camellia.org  Sat May 31 00:19:54 1997
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Date: Fri, 30 May 97 23:15:17 EDT
Message-Id: <9705310315.AA19052@sun13.aic.nrl.navy.mil>
From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: Haym Hirsh <hirsh@cs.rutgers.edu>, cube-lovers@ai.mit.edu
Subject: More on Korf's method

Herbert Kociemba wrote:

> From the description it is evident, that the algorithm Richard E Korf
> uses is basically identical to the the sub-algorithm which is used in
> each stage of my two stage algorithm to solve the cube. What he calls
> "heuristic functions" are the "pruning tables" of Dik Winter and Michal
> Reid and the "filters" in the original description of the algorithm in
> CFF 28 (April 1992) of the Dutch Cubist Club.

First, the term "heuristic function" is not Rich's invention for the
lower-bound function of A*.  I learned that term in 1970 from
Nillson's textbook on Artificial Intelligence.  And second, even if
"pruning tables" and "filters" are essentially nothing but heuristic
functions, that still does not make the algorithms "basically
identical".  From the description, I think Rich's heuristic functions
are quite a different type from what you use (though I do not
understand either exactly yet).  I also suspect that the difference
between A* and IDA* (which has not really been explained here yet) may
play a larger role than you give it credit for.

But thanks for the description of your algorithm (some of which has
previously filtered into the archives), and the offer of a program
(what language?).  My guess is that your heuristics have a good chance
of being more effective at finding optimal solutions for random cubes
than Rich's, though perhaps some ideas from Rich need to be
incorporated.

Haym Hirsh <hirsh@cs.rutgers.edu> wrote:

> Here's a brief attempt at a "layman's" description of Professor
> Korf's work:

[ which I omit ]

Thanks very much for the explanation.  It agrees with my understanding
of the paper, as far as that goes.  But do you have a succinct
explanation of what makes IDA* more tractable than A*?  That's the
part I missed.

Now when Rich found his first ten random cubes (well, he doesn't _say_
they're the _first_ he tried, but they had better be) took under 18f
moves each, you mention

> This does not argue that 18 is the longest solution possible for any
> cube.  Just that for the 10 he generated randomly, none required more
> than 18.  Perhaps some cubes are more than 18 moves away from start.
> None simply happened to arise amongst his 10 cases.

First, we know 18f is not optimal, because the 12-flip is proven to
require 20f moves exactly (unless Mike Reid made a mistake, or I
misunderstood).

But we _can_ say there's at most one chance in 1024 that the first ten
random cubes you pick will all be closer than the median to solved.
So this demonstrates Rich's claim that the median optimal solution is
very likely 18f.

Dan


From cube-lovers-errors@oolong.camellia.org  Sat May 31 16:03:06 1997
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Date: Sat, 31 May 1997 10:26:59 -0400
To: Cube-Lovers@ai.mit.edu
From: Nichael Lynn Cramer <nichael@sover.net>
Subject: Website [was: Description of algorithm for ...]
Cc: Richard E Korf <korf@cs.ucla.edu>, joemcg3@snowcrest.net, chrono@ibm.net,
        Hoey@aic.nrl.navy.mil, kociemba@hrz1.hrz.th-darmstadt.de,
        hirsh@cs.rutgers.edu

Because of interest I've run into from friends elsewhere, I'm planning to
collect the (useful) messages from this thread and hang them on a (possibly
short-term) page on my website.

Of course, this all assumes that there are no objections from the
"participants"; in particular if you are in the CC list above, I'm
currently planning to include one of your messages.  (Similarly, I'll be
including any future "relevant" messages, again assuming the poster doesn't
object.)

Any other (relevant) matter is welcome.


Nichael                                 "Pull down...
nichael@sover.net                           ...tear up."
http://www.sover.net/~nichael/                    -D. Martin




From cube-lovers-errors@oolong.camellia.org  Sat May 31 16:02:29 1997
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From: Tiffp1@aol.com
Date: Sat, 31 May 1997 09:15:49 -0400 (EDT)
Message-ID: <970531091548_-1732449048@emout01.mail.aol.com>
To: hirsh@cs.rutgers.edu, joemcg3@snowcrest.net
cc: cube-lovers@ai.mit.edu
Subject: Re: The rest of us

DOES ANYONE KNOW ANY STORES NEAR HENDERSON,DURHAM,OR RALEIGH WHERE I CAN BUY
A RUBIK'S CUBE AND A RUBIK' S TRIAMID



From cube-lovers-errors@oolong.camellia.org  Sat May 31 16:03:43 1997
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From: SCHMIDTG@iccgcc.cle.ab.com
Date: Sat, 31 May 1997 14:15:25 -0400 (EDT)
To: cube-lovers@ai.mit.edu
Message-Id: <970531141525.2140f541@iccgcc.cle.ab.com>
Subject: A* versus IDA*

On Haym Hirsh's description of Professor Korf's work, Dan Hoey wrote:

>Thanks very much for the explanation.  It agrees with my understanding
>of the paper, as far as that goes.  But do you have a succinct
>explanation of what makes IDA* more tractable than A*?  That's the
>part I missed.

Sorry, perhaps not so "succinct", but here goes:

For problems with constant or near constant branching factors, such
as the cube, both A* and IDA* exhibit exponential time complexity.
In "big O" complexity notation this would be O(b^d) where b is the
branching factor and d is the depth of the search.

The major difference between the two algorithms is in respect to
the space complexity.  A* minimally requires that all frontier nodes
be stored in memory.  This is true of all breadth-first-search (BFS)
algorithms and thus requires O(b^d) space complexity (i.e. exponential
storage -- very bad!).  BFS may also incur some additional time complexity
that depends upon the implementation details of how the stored nodes are
searched and managed.

On the other hand, IDA* is a depth-first-search (DFS) algorithm.  DFS
algorithms require only a linear amount of storage with respect to search
depth (i.e. it has O(d) storage requirements) since it only needs to store
the current path it is exploring.  It uses a cost threshold to determine
when it has gone deep enough and should backtrack to the next unexplored
node (as determined by the current path).  Since the cost threshold is based
on a heuristic estimate (really just an informed "guess"), a solution
may not be immediately found if the guesss was too low, and the search may
have to be repeated with an increased cost threshold, in order to find a
solution.  At first glance, this may seem inefficient, however when one
considers the branching factor (e.g. somewhere in the neighborhood of 13
for the cube) only a small percentage of the search time may be taken up
by the earlier searches.

The bottom line is that A*'s exponential memory requirements limit
its usefulness to small, one might even say "toy", problems.  So an
even bigger issue is that one is likely not to have the memory capacity
to solve the problem at hand using A*.  Note that secondary mass storage
devices do not typically help, since they drastically reduce the number
of node evaluations per second.

Having said that, I've neglected the effect of some other factors such as
duplicate node detection.  BFS can detect duplicate nodes if it stores all
of them and searches through its list of nodes. IDA* implicitly avoids many
of them because their high cost.  IDA* can also be augmented in other ways
(e.g. hash tables) to account for duplicate node checking if this is a
signficant issue with the search space at hand.

There are also some problem dependent factors such as the nature of the
search space and the quality of the cost heuristic.  Consider the limiting
case where we have a "perfect cost heuristic" capable of always leading us
down the optimal path.  If we had such as thing, the time complexity of
these algorithms would be O(b*d) (i.e. linear with respect to depth).
In that case, it would be overkill to use either of these search methods,
but the notion of a "perfect cost heuristic" helps demonstrate the
importance of good heuristics and corresponding reduction in search
exploration.

Professor Korf has consistently broken new ground with respect to solving
previously unsolved problems.  During the mid 80's he was the first to solve
random instances of the 15 puzzle using IDA*.  Since he has used so called
"admissible" heuristics, (heuristics which never overestimate the cost
to the goal state) the solutions are guaranteed optimal.  I have been
writing search programs for over twelve years and consider IDA* to be a
real "gem".  As an aside, I've applied IDA* (augmented with hashing for
duplicate node detection) to solve all but a few hundred of the 32000
instances of Microsoft's "FreeCell" puzzle game that comes packaged
with Win95 and NT.

So to summarize, neglecting details, both A* and IDA* have similar time
complexity requirements, namely exponential.  A* also has exponential
storage requirements whereas IDA* has linear memory requirements.  The
space advantage of IDA* therefore greatly increases the scope of problems
that can be attacked by this method.

Hope I've served to clarify rather than to further obfuscate.

-- Greg


From cube-lovers-errors@oolong.camellia.org  Sat May 31 17:19:14 1997
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Date: Sat, 31 May 97 16:55:01 EDT
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From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: cube-lovers@ai.mit.edu
Subject: Searching and Metrics in (Korf 1997)

INTRODUCTION

I've read through Rich Korf's paper now, and I have a few ideas on the
paper and how the method might be improved.  This is fairly long, so
I've broken it up into two parts.  Part 1 has a bit about searching
methods (answering the question I asked in my last message) and some
concerns about the face-turn vs quarter-turn metric.  Part 2 covers
some ideas I have on the heuristics he uses.  Eventually I hope to air
my concerns over how realistic the memory-based analysis is, but I'm
not sure I understand it well enough yet.

SEARCHING

I asked yesterday what made IDA* a more tractable method than A*.  I
think I've got it now.  Both use a heuristic function h(p) that is
guaranteed not to underestimate the number of moves to solve position
p.  And both may have to check every position p (encountered at depth
g(p)) for which g(p)+h(p) is less than the optimum.

But A* is essentially a breadth-first algorithm.  You have to make a
list of all the nodes for which g(p)+h(p) is minimum before you try a
higher value.  For this problem, there are too many positions to store
conveniently.  IDA* is a variant that allows depth-first search.  If
you have a lower bound L, you search depth-first for all positions
that have g(p)+h(p)<=L.  You will find a solution if and only if the
optimal solution is at depth L; if you fail you try again with L+1.

The big advantage of IDA* is that you don't need to represent a
database of all the frontier positions at once, you try them one at a
time.  IDA* has two disadvantages, though.  First, whenever you fail a
search, you lose all the information from previous searches with
smaller values of L, except that they failed.  But if the number of
positions at each depth is much larger than the previous (ten or
thirteen times larger, in this case) this loss is small.  Second, your
depth-first search may visit the same position more than once, if it's
reachable by more than one near-optimal path.  This seems to occur for
only a few percent of positions as far as we've seen, but it
eventually gets to be all of them near the global maximum.  The issue
of duplication leads to my question about metrics.

METRICS

Rich uses the face-turn metric, which has been discussed here earlier.
But the justification he gives is one I haven't seen before.  He
claims the face metric

    ... leads to a search tree with fewer duplicate nodes.  For
    example, two consecutive clockwise twists of the same face leads
    to the same state as two counterclockwise twists.

But this is a bad example of duplication.  No one who is familiar with
the cube-lovers archives (e.g. my message of 9 January 1981) would
generate both of the above nodes, any more than they would generate
the duplicate nodes caused by composing two commuting moves like F, B
in both possible orders.  Rich knows not to do the latter, as he
discusses in the paper.

In case I haven't been sufficiently explicit about this, the way to
avoid this kind of duplication in the quarter-turn metric is to
require:

    1. The move after F        must not be F',
    2. The move after F' or FF must not be F or F',
    3. The move after B        must not be F, F', or B',
    4. The move after B' or BB must not be F, F', B, or B',
    5. The same as rules 1-4 with F,B replaced by R,L, respectively, and
    6. The same as rules 1-4 with F,B replaced by T,D, respectively.

So two questions remain. First, is there really a difference in the
duplication of positions in the two metrics?  I think Jerry Bryan's
table shows that only about 1.74% of the 63 billion positions are
duplicated at 11q.  Do we have statistics on duplication for the
face-turn metric?  Second, is there any technical justification for
using the face-turn metric?  I'm aware that most of the published
literature uses it, and that small numbers of moves sound more
impressive than large ones, but these aren't very satisfying reasons.
As far as I know, the problem of finding optimal solutions can be
fruitfully approached in either of the metrics, or in any of several
other metrics.

[ End of part 1. ]


From cube-lovers-errors@oolong.camellia.org  Sat May 31 17:56:04 1997
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Date: Sat, 31 May 97 17:26:17 EDT
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From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: cube-lovers@ai.mit.edu
Subject: Heuristics in (Korf 1997)

[ As I mentioned, before, this is part 2. Just after I sent part 1, I
  saw Greg Schmidt's explanation of IDA*.  I hope someone finds the
  parts of our messages that don't duplicate each other to be
  instructive enough to justify the parts that do. ]

HEURISTIC FUNCTIONS

Herbert Kociemba notes three interesting heuristics based on the
number of moves to reach the subgroup <T,D,R2,L2,F2,B2>.  In fact,
Mike Reid calculated (and Dik Winter verified) the exact distances in
this 2.2-billion element coset space (see archives at 7 Jan 1995 and
following).  Mike shows how you could look up this distance in a
64-megabyte table, and Dik suggests how this could be made into a
database half the size (though I think the performance penalty might
be too high).

These coset differences form an admissible heuristic.  There are a lot
of other interesting subgroups, and some of their coset spaces may
yield useful heuristics.

But the coset spaces Rich uses are those relative to the subgroup that
fixes a certain number of pieces: The corners, or either of two
subsets of six edges.  It's unfortunate he didn't notice that the
latter two tables could use the same database.  The way you do this is
to choose your two sets S, T of 6 edges such that there is a whole
cube move m in M for which m(S)=T.

Here's a formalism of how the database for fixing set S works.  Make a
database that maps a position p to the length of the shortest sequence
x for which px fixes each piece s in S.  Thus the distance h(p) from
position p to the goal position q is the length of pq', for which we
get a lower bound by looking up pq' in the database.

To find the heuristics based on fixing the pieces in T, we could make
a new database.  But px(s) = s exactly when (mpm' mxm')(m'(s))=m'(s).
That is to say, when the m-conjugate position (mpm' mxm') fixes the
piece m'(s).  So if we look up mpm' in our database, it will give the
length of the shortest sequence mxm' that fixes each m'(s) -- i.e.,
that fixes each t in T.

This also gives us 94 more admissible heuristics for free, at least in
terms of table space.  Of course we can use the other 46 elements of
M.  What might not be obvious is that the lower bound we get by
looking up x in the database is probably not the same as the lower
bound we get by looking up x'.  But the length of x is the length of
x', so we could get 48 more heuristics by looking up the inverse and
it's M-conjugates.  By taking the maximum of the 96 values formed by
looking up mpq'm' and mqp'm' in the data base, we may get a much
better lower bound for the solution length.

Of course, we could take any database of lower bounds and use this
process to get up to 96 times as many bounds.  The distance to
Kociemba's subgroup is such a lower bound, but it unfortunately is so
symmetric that I think we only get a 6-fold improvement (or perhaps
3-fold; I'm losing my intuition on inverses in those cosets).  Perhaps
just fixing an asymmetric subset of edges and corners might be the
best solution.

[ End of part 2. ]


From cube-lovers-errors@oolong.camellia.org  Sat May 31 18:34:15 1997
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Date: Sun, 1 Jun 1997 00:20:40 +0200
From: Dik.Winter@cwi.nl
Message-Id: <9705312220.AA22245=dik@hera.cwi.nl>
To: cube-lovers@ai.mit.edu
Subject: Re: More on Korf's method

Herbert Kociemba:
 > From the description it is evident, that the algorithm Richard E Korf
 > uses is basically identical to the the sub-algorithm which is used in
 > each stage of my two stage algorithm to solve the cube.

This is right.

Dan Hoey:
 >              From the description, I think Rich's heuristic functions
 > are quite a different type from what you use (though I do not
 > understand either exactly yet).

Not really.  Rich's heuristic functions are (precomputed) distances 
along some coordinates of a multidimensional space.  His best apparently
are the corner positions and twice one half of the edge positions.
Similarly in both phases of Herbert's algorithm similar heuristic
functions (pruning tables, filters, ...) are used.

Of course the choice of heuristic function plays an essential role.
For instance, Herbert's original algorithm uses in the first phases
three heuristic functions all three based on a single coordinate
in a three dimensional space.  I modified it to use three heuristic
functions based on two dimensional coordinate planes in that same
space.  Depending on the problem to solve, this may be better or not,
in this case it is (much) better.  A similar modification did I do
in the second phase.

 >                    My guess is that your heuristics have a good chance
 > of being more effective at finding optimal solutions for random cubes
 > than Rich's, though perhaps some ideas from Rich need to be
 > incorporated.

As far as the first is concerned, I think so too.  When Herbert's
algorithm is run through to the end it will find an optimal solution
indeed, and in the search for that optimal solution it will use a
new heuristic function for the total solution: the result of previous
suboptimal solutions that come in pretty fast, which is used to prune
the second phase rigorously.  I have been able to proof (with my
modification of Herbert's algorithm) some pretty large (18-20 turn)
solutions optimal.  I do not think Rich's algorithm will be able to do
that in reasonable time.

 > First, we know 18f is not optimal, because the 12-flip is proven to
 > require 20f moves exactly (unless Mike Reid made a mistake, or I
 > misunderstood).

No, this is right indeed.

 > But we _can_ say there's at most one chance in 1024 that the first ten
 > random cubes you pick will all be closer than the median to solved.
 > So this demonstrates Rich's claim that the median optimal solution is
 > very likely 18f.

Something I did estimate already a long time ago.  I have done a few
hundred random cubes (a few thousand?  I do no longer remember) back
so many years ago.  As I remember, I let the program look for optimal
solutions upto 18f (longer is a bit time consuming).  As I remember,
there were only very few that could *not* be solved in 18f.  There must
be a discussion about this in the archives.

dik
--
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn  amsterdam, nederland; http://www.cwi.nl/~dik/


From cube-lovers-errors@oolong.camellia.org  Sat May 31 19:10:39 1997
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Date: Sun, 1 Jun 1997 00:57:00 +0200
From: Dik.Winter@cwi.nl
Message-Id: <9705312257.AA23367=dik@hera.cwi.nl>
To: cube-lovers@ai.mit.edu
Subject: Re: Searching and Metrics in (Korf 1997)

 > So two questions remain. First, is there really a difference in the
 > duplication of positions in the two metrics?  I think Jerry Bryan's
 > table shows that only about 1.74% of the 63 billion positions are
 > duplicated at 11q.  Do we have statistics on duplication for the
 > face-turn metric?

I do not think there are really statistics.  But I have done a complete
analysis on similar puzzles (domino, 2x2x2) where the number of positions
from start increases roughly by the factor you would expect if you
eliminate elementary duplicates as you listed.  This both for face
turns and quarter turns.  This increase goes on until shortly before
the maximum of turns when the number of new configurations drops
dramatically.  I think the figures are in the archives.  I have no
reason to expect the case to be different for the cube, rather all my
experiments lead me to predict that the same is the case with the cube.

 >                    Second, is there any technical justification for
 > using the face-turn metric?

None, except that the diameter of the group will be larger.  (But not
much larger.)  And that makes it in Rich's algorithm only computationally
more expensive.

 >                              I'm aware that most of the published
 > literature uses it, and that small numbers of moves sound more
 > impressive than large ones, but these aren't very satisfying reasons.
 > As far as I know, the problem of finding optimal solutions can be
 > fruitfully approached in either of the metrics, or in any of several
 > other metrics.

The last is almost certainly not true.  In the archives there must be an
article by Michael Reid where he made an attempt to generalize Kociemba's
algorithm.  I.e. find a subgroup of the total group, phase 1 is to
find a path to that subgroup and phase 2 is to find a path to the
solution.  I disremember the entire contents, but as far as I remember
the optimal partitions all used a subgroup which needed only half turns
for at least one face pair.  Looking for quarter turn optimization is
not really feasable with such a partition.

dik
--
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn  amsterdam, nederland; http://www.cwi.nl/~dik/


From cube-lovers-errors@oolong.camellia.org  Sat May 31 19:11:01 1997
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Sender: Haym Hirsh <hirsh@cs.rutgers.edu>
Date: Sat, 31 May 97 19:07:01 EDT
From: Haym Hirsh <hirsh@cs.rutgers.edu>
Reply-To: Haym Hirsh <hirsh@cs.rutgers.edu>
To: Dan Hoey <Hoey@aic.nrl.navy.mil>
Cc: cube-lovers@ai.mit.edu
Subject: Re: More on Korf's method
In-Reply-To: Your message of Fri, 30 May 97 23:15:17 EDT
Message-ID: <CMM-RU.1.5.865120021.hirsh@athos.rutgers.edu>

> Thanks very much for the explanation.  It agrees with my understanding
> of the paper, as far as that goes.  But do you have a succinct
> explanation of what makes IDA* more tractable than A*?  That's the
> part I missed.

Here are a couple of attempts to explain why and how IDA* is a win over A*.
In my attempt to generate a description for the layman I tried to err
on the side of saying too much rather than too little -- my apologies
if I belabor the obvious for anyone.

The highest-level explanation is that A* may need to store a number of
intermediate results whose number is some exponential function of the
length of the solution -- e.g., b^l: l is the solution length, b,
roughly speaking, is the number of new cubes you can get from a given
cube, aka the "branching factor", and "^" means exponentiation --
whereas IDA* will only store a number of intermediate results whose
number is a linear function of the solution length -- e.g., b*l.  The
apparent paradox is that, to do this, IDA* does redundant work --
exploring some intermediate results many times because of its inferior
"bookkeeping".  However, it is usually the case that the extra work is
more than paid off by memory savings.  This is particularly true for
Rubik's cube, since the higher the branching factor (i.e., roughly 18
for the cube if you count crudely), the less the redundant work.

In a bit more detail, the difference between A* and IDA* is similar to
the difference between breadth-first search and depth-first iterative
deepening.  Imagine you want to generate all cubes that are reachable
in d steps from the start.  What you can do is generate all cubes that
are one step away, then generate all that are one step away from those
that are one step away (resulting in a list of all that are two steps
away), then all that are one step away from those that are two steps
away (resulting a list of all that are three steps away), etc.  At the
final step in this process you have a list of all cubes that are d-1
steps away, and you generate all cubes that are one step away from any
item in this list.  This generates all cubes that are d steps away.
The process is known as breadth-first search.  It's main problem is
that the list of all cubes that are d-1 steps away will have size
roughly b^(d-1).

Depth-first search, on the other hand, generates all cubes that are
one step away from start, puts all but one of them (i.e., b-1 of them)
on a list, and takes the one that wasn't placed on the list and
(recursively) generates all cubes that are d-1 steps away from it.
When you are done with this first depth-one cube, take one of the
other cubes that are one step from start (which is one of those
stashed away in the afore-mentioned list) and do the same thing,
generating, in turn, all cubes that are d-1 steps away from it.  This
continues until all the items that were put on the list have been
explored -- i.e., they have had all cubes at depth d-1 from them
returned.  This is depth-first search.  Because at each recursion
level it saves only b-1 things, at worst it winds up saving roughly
(b-1)*d cubes in its search.

Now imagine you have a cube that you know is at most (but not
necessarily exactly) d steps from the start, and you want to know what
the shortest solution to it is.  One approach would be to do a
breadth-first search to depth 1 and see if you have it, continue to
depth 2 and see if you have it, etc., until you reach depth d.  A
second possible approach would appear to be to use depth-first search
to depth d, but this is not guaranteed to give a shortest solution.
To see this, imagine that the cube is two moves from start, but it is
also four moves away if you make the wrong first move.  If the result
of that wrong first move is the cube that depth-first search chooses
to "expand" first (with the "correct" one waiting its turn on the list
of cubes to be seen later if you haven't found your cube), you will
find your desired cube via the depth-four solution.  You didn't find
the depth-two solution.

This problem with depth-first search leads to the idea of depth-first
iterative deepening.  The basic idea of iterative deepening is simple.
First do a depth-first search to depth 1.  If you haven't found it,
throw away all your work and start over, doing a depth-first search to
depth 2.  Again, if you haven't found it, throw away all your work and
start over, doing a depth-first search to depth 3.  This continues,
until you hit the right depth for finding it.  This process is
guaranteed to find the shortest solution, but seems silly,
regenerating everything you did in the previous depth-first search
when you add one to the depth.  The interesting observation that makes
this a win is that the percent of overall effort spent on previous
depths is only a small fraction of the effort spent on the final
depth-first search.  So you penalize yourself a little redundancy, but
are rewarded with much more modest and realistic memory requirements.

The step from this to A* vs IDA* is actually not too large.  The basic
idea is to use depth-first search, but instead of using a depth bound
d, instead don't go any farther from a cube if the sum of the number
of steps to get to it plus the guess on how many more steps are needed
to get to a solved cube exceeds some threshold.  You start with a
small threshold, and slowly keep increasing it, each time starting
over again from scratch, until the threshold is just barely high
enough to find the solution.  If you do this in the correct way (for
example, upping the threshold each time in the appropriate fashion
based on the values you observed in your previous iteration), you can
prove that the solution IDA* finds is the shortest possible (as long
as the solution-length guesser never over-estimates the correct value).



From cube-lovers-errors@oolong.camellia.org  Sun Jun  1 00:58:07 1997
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From: SCHMIDTG@iccgcc.cle.ab.com
Date: Sun, 1 Jun 1997 0:19:28 -0400 (EDT)
To: cube-lovers@ai.mit.edu
Message-Id: <970601001928.2140f226@iccgcc.cle.ab.com>
Subject: Re: Searching and Metrics in (Korf 1997)

Dan Hoey wrote:

>I've read through Rich Korf's paper now, and I have a few ideas on the
>paper and how the method might be improved...

I feel that there is a fundamental point regarding cube solving approaches
that is worth making.  First there is what I shall call the "Kociemba approach".
I would loosely paraphrase this approach as follows:

"If one examines the behavior of the cube in great detail and applies group
theoretic concepts to the spaces generated by the cube, one can apply this
analysis to great effect towards producing an algorithm capable of solving
the cube."

Then there is what I shall refer to as the "Korf approach".  Loosely
paraphrased as:

"The cube is a difficult combinatorial puzzle that provides a fertile
ground for advancing the state of the art of search methods."

It should be apparent that the goal of these two approaches is quite
different.  The "Kociemba approach" is focused only on solving the
cube.  All domain specific knowledge about the cube problem, such as
specific group theoretic properties of the cube, can and should be
applied.  Whereas the "Korf approach" attempts to be a general
approach, applicable to other problems, not just the cube (i.e. the
cube problem is merely incidental).  This approach should not rely
on methods that are specific to a single (or at least very narrow)
class of problems.  This is close to the definition of so called
"weak" methods in AI.  Weak methods are those that do not rely
heavily on human provided domain specific knowledge.  For example,
an expert system does not classify as a weak method, whereas a
search method that is given little more than a description of
the problem space and a desired goal does.

Stepping back a bit, I would say that Korf's method applies to
a broad class of problems, much larger than the cube alone.  In
effect, his method says:

"If one can partition the desired end goal of a problem into subgoals,
and for each subgoal provide a cost to achieve that subgoal, then one
can use this information to produce a higher quality (i.e. "more informed"
in search parlance) heuristic.  An effective way to achieve this is by
firt precomputing the cost information associated with each subgoal and
storing it in a table.  This table can be reused across multiple problem
instances and represents an effective way to utilize available memory
to improve the effectiveness of the search."

And so I feel it is necessary to say that any suggestion for improving
Korf's method, should take this distinction into account since solving
the cube is only incidental to the emphasis of the research work that
professor Korf is involved in.

In no way am I intending to imply that methods which are not consistent
with the "Korf approach" should not be considered.  In fact, I find this
a fascinating topic.  I'm just pointing out it is both a virtue and a
goal of his approach that he hasn't leveraged extended knowledge of
"cube principles" in order to devise an effective algorithm.

I hope I haven't misrepresented Mr. Kociemba or Mr. Korf in this
discussion.  If so, please speak up.

-- Greg


From cube-lovers-errors@oolong.camellia.org  Sun Jun  1 05:20:49 1997
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Date: Sun, 01 Jun 1997 11:17:59 +0200
From: Herbert Kociemba <kociemba@hrz1.hrz.th-darmstadt.de>
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To: cube-lovers@ai.mit.edu
Subject: Generalisation of Korf's method?
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Greg wrote:

> It should be apparent that the goal of these two approaches is quite
> different.  The "Kociemba approach" is focused only on solving the
> cube.  All domain specific knowledge about the cube problem, such as
> specific group theoretic properties of the cube, can and should be
> applied.  Whereas the "Korf approach" attempts to be a general
> approach, applicable to other problems, not just the cube (i.e. the
> cube problem is merely incidental).

Maybe. But I'm not sure about that. I am no specialist concerning IDA*
search, but I think it is worth while to examine, for which general
class of problems a two phase algorithm is profitable, that means doing
IDA* search to some subgoal which consists of an appropriate subset
(including the end goal) of the problem space (phase1) and appropriate
heuristic functions, then doing IDA* search from here to the end goal
(phase2). Then continuing with IDA* search in phase1, creating more
solutions and then doing IDA* search in phase2 with a maximal length of
L=N-n1-1, where N is the length of the already found (usually not
optimal) last solution and n1 is the lenght in phase1.
L decreases for two reasons when the alogrithm is running: Every new
solution found will have an length N at least one smaller then the
previous solution and nl will increase also.
If you have enough time, you wait until nl=N, then you have the
guaranteed optimal overall solution.

This approach could be valuable, if the problem space is very large, and
in this case you get a sequence of solutions with decreasing length
which might be better than waiting for the optimal solution for 100
years with single IDA* search.
In the case of Rubik's cube the sequences of solutions seem to converge
very quick to a solution with minimal overall length in many cases,
though it might be difficult to prove this rigorously.
I would be interesting to see, how the two phase algorithm handles the
10 cubes, Rich Korf solved.

Dik.Winter@cwi.nl wrote:
 
> Of course the choice of heuristic function plays an essential role.
> For instance, Herbert's original algorithm uses in the first phases
> three heuristic functions all three based on a single coordinate
> in a three dimensional space. 

This is not quite true. I don't think that the algorithm would have
worked then satisfactory. I did not add the details in the description
of the algorithm in CFF28, because I did not want to hide the
principles.
Because of the limited memory (1MB!! at this time) I worked in four
dimensional space and also used heuristic functions based on
two-dimensional coordinate-planes. To get 4 dimensions, I did not use
the turns of the 6 faces, but I fixed the DLB-corner. Instead of the L,
B, and D turns I did R, F and U turns together with a slice move then,
which is of course identical with L, B and D turns and then turning the
whole cube. The additional coordinate is the position of the
center-cubies(24 states). In this way you have only 7! corner
permutations (instead of 8!) and 3^6 possible corner orientations
(instead of 3^7). Only this fact made it possible for me use two
dimensional coordinate planes for the heuristic functions.
     
Herbert



From cube-lovers-errors@oolong.camellia.org  Sun Jun  1 17:21:28 1997
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Date: Sun, 1 Jun 1997 14:09:08 -0800
To: cube-lovers@ai.mit.edu
From: Dave Dyer <ddyer@netcom.com>
Subject: new search heuristics


I wonder about the local "shape" of the distance heuristics accuracy.  How inaccurate are
they typically, and how inaccurate can the worst cases be?  Is it typically the case
that all the estimates for nearby positions are similar, or are there outliers?
Are the good and bad guesses uniformly distributed, or are they clustered? Depending
on the shape of the space, I can imagine strategies specifically designed to improve,
by looking a little harder for a good estimate.

Are there useable heuristics which are guaranteed to overestimate
distance?  If so, then they could be used in concert with positions
known to be distant from solved to improve the bounds for A*.  For example,
with an overestimating heuristic (n) for a position known to be more than 18
moves from start, we could use 18-n as an independant underestimate of distance
from start.

---
My home page:  http://www.andromeda.com/people/ddyer/home.html





From cube-lovers-errors@oolong.camellia.org  Sun Jun  1 18:00:42 1997
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Date: Sun, 01 Jun 1997 17:59:49 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: Description of algorithm for finding minimal-move solutions to
 Rubik's Cube
In-reply-to: <338F7124.73A6@hrz1.hrz.th-darmstadt.de>
To: Cube-Lovers <cube-lovers@ai.mit.edu>
Message-id: <Pine.PMDF.3.95.970601174941.90493E-100000@PSTCC6.PSTCC.CC.TN.US>
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On Sat, 31 May 1997, Herbert Kociemba wrote:

> Having found a solution in stage1 and stage2 the algorithm does not
> stop, but generates other solutions in stage1. So if for example we have
> 10 moves in stage1 and 12 moves in stage2, there might be a solution
> with 11 moves in stage1 but only 10 moves in stage2. Running long
> enough, the algorithm will find the overall optimal solution, having no
> moves in stage2 then.

I have always been curious about the termination criteria for your
algorithm -- that is, how long is "long enough"?.  It appears that you are
effectively moving moves from stage2 to stage1 until stage2 is empty.  I
wonder if you could describe this process a little further.  I have always
rather naively assumed that you were (for example) combining an R at the
end of stage1 with an R' at the end of beginning of stage2, simply
cancelling out the moves.  But you certainly appear to be doing some a
little more elaborate than simple cancellation.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990



From cube-lovers-errors@oolong.camellia.org  Sun Jun  1 21:39:09 1997
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Date: Sun, 1 Jun 97 19:45:11 EDT
Message-Id: <9706012345.AA19309@sun34.aic.nrl.navy.mil>
From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: SCHMIDTG@iccgcc.cle.ab.com, cube-lovers@ai.mit.edu
Subject: Re: Searching and Metrics in (Korf 1997)

I wrote:

>I've read through Rich Korf's paper now, and I have a few ideas on the
>paper and how the method might be improved...

Greg Schmidt replies:

> ...And so I feel it is necessary to say that any suggestion for improving
> Korf's method, should take this distinction into account since solving
> the cube is only incidental to the emphasis of the research work that
> professor Korf is involved in....

The method I suggested for reusing tables for new heuristics should be
applicable to any group-theoretic puzzle for which there are
symmetries mapping generators to generators.  For instance, the N^2-1
puzzles have the 8-fold symmetry D4, and so could have one set of
tables used for 16 heuristics.

Given the central nature the memory-performance tradeoff plays in the
paper, I imagine this is quite relevant to Rich's research.

Of course, I also discussed a number of other topics in that message,
but I don't think they were the ones you were addressing in your remarks.

Dan


From cube-lovers-errors@oolong.camellia.org  Sun Jun  1 21:38:37 1997
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Date: Sun, 01 Jun 1997 18:20:28 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: Description of algorithm for finding minimal-move solutions to
 Rubik's Cube
In-reply-to: <338F7124.73A6@hrz1.hrz.th-darmstadt.de>
To: Cube-Lovers <cube-lovers@ai.mit.edu>
Reply-to: Jerry Bryan <jbryan@pstcc.cc.tn.us>
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On Sat, 31 May 1997, Herbert Kociemba wrote:

> Dik Winter proved, that 12 moves always suffice to get to this subgroup.
> 
> Michael Reid proved, that 18 moves always suffice in this subgroup.

If I interpret this correctly, what you have at this point is a
Thistlethwaite algorithm with G -> <U,D,R2,L2,F2,B2> -> I, proving that
any position can be solved in no more than 30f moves.  Is this the correct
interpretation?  But more importantly, is there anything in the part of
your algorithm where you combine stage1 and stage2 which would establish
an upper bound which is less than 30f? 

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990




From cube-lovers-errors@oolong.camellia.org  Sun Jun  1 21:38:55 1997
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Date: Sun, 01 Jun 1997 18:35:50 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: Description of algorithm for finding minimal-move solutions to
 Rubik's Cube
In-reply-to: <338F7124.73A6@hrz1.hrz.th-darmstadt.de>
To: Cube-Lovers <cube-lovers@ai.mit.edu>
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On Sat, 31 May 1997, Herbert Kociemba wrote:

> 
> In phase 1, the cube is transformed to an element of the subgroup
> generated by <U,D,R2,L2,F2,B2>. 

.....

>           The "heuristic functions" consist of three tables, using 4
> bits for each entry. The first table stores the minimum numbers to solve
> the 2187*2048 possible states to restore the orientation of both edges
> and corners, the other tables have 2187*495 and 2048*495 entries and
> store the corresponding minimum numbers.

Obviously, any conjugate of <U,D,R2,L2,F2,B2> would do as well as any
other.  Have you looked for conjugate positions in your table?  For
example, you might have a position x which is 10 moves from
<U,D,R2,L2,F2,B2>, but position y which is a conjugate of x might be only
be 9 moves from <U,D,R2,L2,F2,B2>.  In this case, you might as well solve
y, knowing that the number of moves to solve x is the same as the number
of moves to solve y, and knowing that the solutions are conjugate.  Also,
by subjecting your entire table to this kind of analysis (if you haven't
already), you might find an upper bound for stage1 which is less than 12f.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990



From cube-lovers-errors@oolong.camellia.org  Sun Jun  1 21:39:26 1997
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Date: Mon, 2 Jun 1997 01:45:38 +0200
From: Dik.Winter@cwi.nl
Message-Id: <9706012345.AA07277=dik@hera.cwi.nl>
To: cube-lovers@ai.mit.edu
Subject: Re: Description of algorithm for finding minimal-move solutions to Rubik's Cube

 > I have always been curious about the termination criteria for your
 > algorithm -- that is, how long is "long enough"?.

"Long enough" means that it is certain that no shorter solutions can
be found.

 >                                                    It appears that you are
 > effectively moving moves from stage2 to stage1 until stage2 is empty.

Not really.  What is happening is a breadth first search in phase1.  The
search is continued although phase1 solutions are found (and a lot are
found on the way).  This is continued until you find that deepening the
search in phase1 will not give shorter solutions.  For each phase1
solution a phase2 solution is looked for, also in a breadth first manner,
and also as deep as is needed to find shorter total solutions.

This method finds very quickly solutions for phase1 in about 10-12 turns
with matching solutions in phase2 with 12-14 turns for a total of about
22-26 (very rarely the first solution found has over 26 turns).  When
for instance a solution is found with 10 turns in phase1 and 12 in phase2
(breadth first so this is optimal with the particular path in phase1),
search in phase1 continues and for each new solution in phase1 solutions
are searched for in phase2 with limited depth.  So first 10 turns in
phase1 and at most 11 in phase2, followed by 11 in phase1 and at most
10 in phase2.


From cube-lovers-errors@oolong.camellia.org  Sun Jun  1 21:39:51 1997
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Date: Sun, 01 Jun 1997 21:06:18 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: Description of algorithm for finding minimal-move solutions to
 Rubik's Cube
In-reply-to: <199705300024.RAA18247@denali.cs.ucla.edu>
To: Richard E Korf <korf@cs.ucla.edu>
Cc: Cube-Lovers@ai.mit.edu
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On Thu, 29 May 1997, Richard E Korf wrote:

>    For example, if we consider just the corner cubies, there are only about 88
> million possible states they could be in (8!x3^7). We exhaustively build and
> store a table, using breadth-first search, that tells us the minimum number of
> moves needed to solve just the corner cubies from every possible combination,
> ignoring the edge cubies.  This value ranges from 0 to 11 moves, averages 8.764
> moves, and requires only 4 bits per state. (We could reduce this further using
> an idea of Dan Hoey's published in this list awhile ago.)  This table only has
> to be computed once, taking about a half hour, and requires about 42 megabytes
> of memory to store (a megabyte is 1048576 bytes).

I have an old program, developed on a 286 PC with a 10MB harddisk, which
stores the entire solution for the corners in about 2.5MB.  Details are in
the archives, but it uses representatives of the form Repr{m'Xmc}.  The
representatives consitute the solution of the 2x2x2 and take about .625MB.
The remaining storage is in a format I call Repr{m'Xmc}*C to take care of
the corners of the 3x3x3.  However, I would guess that even though this
format would save a great deal of memory, it would also very much slow
your program down, rather than speeding it up, because of the rather
arcane indexing required. 

This brings up a point which I think has been taken for granted in the
archives but which I think has never been spelled out in detail. In its
most simple-minded form, a search involves storing both permutations and
distances from Start.  But sometimes you can get by with storing only the
permutations, and sometimes you can get by with storing only the
distances.

In this case, you are storing the entire corner group, and therefore you
can get by with storing only the distances.  That is, you have obviously
developed an easy-to-calculate function and an inverse to map between the
corner permutations and an index set, say 1..|GC| or maybe 0..|GC-1|. 
Hence, you don't need to store the permutations themselves.  My
Repr{m'Xmc} technique stores only a subset of the permutations.  There is
a one-to-one correspondence between the subset which is stored and an
index set, but it is not very easy to calculate (actually, it involves
some binary searching).

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990



From cube-lovers-errors@oolong.camellia.org  Sun Jun  1 21:40:09 1997
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From: SCHMIDTG@iccgcc.cle.ab.com
Date: Sun, 1 Jun 1997 21:22:20 -0400 (EDT)
To: cube-lovers@ai.mit.edu
Message-Id: <970601212220.2140c63e@iccgcc.cle.ab.com>
Subject: Re: Generalisation of Korf's method?

Herbert Kociemba wrote:

>Greg wrote:
>
>> It should be apparent that the goal of these two approaches is quite
>> different.  The "Kociemba approach" is focused only on solving the
>> cube.  All domain specific knowledge about the cube problem, such as
>> specific group theoretic properties of the cube, can and should be
>> applied.  Whereas the "Korf approach" attempts to be a general
>> approach, applicable to other problems, not just the cube (i.e. the
>> cube problem is merely incidental).
>
>Maybe. But I'm not sure about that. I am no specialist concerning IDA*
>search, but I think it is worth while to examine, for which general
>class of problems a two phase algorithm is profitable, that means doing
>IDA* search to some subgoal which consists of an appropriate subset
>(including the end goal) of the problem space (phase1) and appropriate
>heuristic functions, then doing IDA* search from here to the end goal
>(phase2). Then continuing with IDA* search in phase1, creating more
>solutions and then doing IDA* search in phase2 with a maximal length of
>L=N-n1-1, where N is the length of the already found (usually not
>optimal) last solution and n1 is the lenght in phase1.
>L decreases for two reasons when the alogrithm is running: Every new
>solution found will have an length N at least one smaller then the
>previous solution and nl will increase also.
>If you have enough time, you wait until nl=N, then you have the
>guaranteed optimal overall solution.
>
>This approach could be valuable, if the problem space is very large, and
>in this case you get a sequence of solutions with decreasing length
>which might be better than waiting for the optimal solution for 100
>years with single IDA* search.

I think the two phase approach is relevant to a larger class of problems
than just the cube, or for that matter, combinatorial puzzles in general.
In fact, I think the two phase aspect is what is commonly referred to
as "bidirectional search".  When it can be applied, it has the potential
for reducing search complexity from O(b^d) to O(b^(d/2)).  That is to
say that it can cut the exponent of an exponential search in half.

My earlier comment was more in regards to leveraging specific knowledge
of cube groups.  I'm not sure that specific knowledge of cube groups is
highly applicable to Korf's work as this seems in conflict with the
desire to develop general search methods that are domain independent.
I suspect that he might consider this "cheating" with respect to the
thrust of his work having greater applicability outside combinatorial
puzzle problems.  But if the goal is to develop a highly optimized
Rubik's cube solver, then inputting specific knowledge of the cube
problem is reasonable.

Although I don't yet fully understand the pruning tables used in Kociemba's
algorithm, I suspect they are different than Korf's tables (which I do
understand), especially with respect to how they are applied to pruning
the search.

I have a hunch that combining Korf's heuristic methods with a Kociemba
style two phase approach (a.ka. "bidirectional search) could result in 
a more effective cube solver.

-- Greg


From cube-lovers-errors@oolong.camellia.org  Mon Jun  2 13:10:30 1997
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From: SCHMIDTG@iccgcc.cle.ab.com
Date: Sun, 1 Jun 1997 22:39:43 -0400 (EDT)
To: cube-lovers@ai.mit.edu
Message-Id: <970601223943.2140c63e@iccgcc.cle.ab.com>
Subject: Re: A* versus IDA*

Jerry Bryan wrote:

>On Sat, 31 May 1997 SCHMIDTG@iccgcc.cle.ab.com wrote:
>
>>   As an aside, I've applied IDA* (augmented with hashing for
>> duplicate node detection) to solve all but a few hundred of the 32000
>> instances of Microsoft's "FreeCell" puzzle game that comes packaged
>> with Win95 and NT.
>> 
>
>Interesting.  I once upon a time found some notes on the Web that
>indicated that all but one of the prepackaged instances of FreeCell had
>been solved by hand (by humans).  I gather that many people worked on the
>project in a parallel processing mode.  Only one of the games was said to
>be unsolvable, and this was claimed to have been proven by computer
>search.

The unsolvable instance is FreeCell game #11982.  And you are correct
to point out that a computer program has proven the unsolvability of
this particular instance.  It is a curious fact that only one of the
32000 instances turned out to be unsolvable.

[...some discussion of Baker's game deleted...]
>Nevertheless, my program demonstrated that the game can be won about 60%
>of the time.  I am sure that your IDA* FreeCell program, were it to be
>adapted to Baker's Game, would be vastly more effective than my rather
>primitive program was.  Would you have any interest in investigating
>Baker's Game with your program.?

I am familiar with Baker's game and as you mentioned, it is harder
to win at than FreeCell.  It would be fairly simple to adapt my
program to Baker's game and it would be interesting to compare performance.
As I mentioned, my program uses a sort of "hash cache" for duplicate
node checking (as the search space for FreeCell is a graph, not a tree).
There is a flaw in this mechanism as some false hits can occur and effectively
over prune the search tree.  I believe that was why I was unable to solve
approximately 300 of the 32000 instances.  I don't know if that flaw would
be amplified in Baker's game where there are fewer paths to a solution as
compared to FreeCell.  Unfortunately, eliminating this behavior would
require a complete overhaul of the duplicate node checking mechanism
as opposed to a simple "quick fix".

At any rate, I'm willing to make this program available to interested
parties (BTW, its written in C++).  I would prefer to upload it to an
FTP site if possible.  Is anyone aware of an FTP site available for use
by cube-lovers?

-- Greg


From cube-lovers-errors@oolong.camellia.org  Mon Jun  2 13:10:15 1997
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Date: Sun, 1 Jun 1997 22:03:44 -0400 (EDT)
To: cube-lovers@ai.mit.edu
Message-Id: <970601220344.2140c63e@iccgcc.cle.ab.com>
Subject: Re: Searching and Metrics in (Korf 1997)

Dan Hoey wrote:

>The method I suggested for reusing tables for new heuristics should be
>applicable to any group-theoretic puzzle for which there are
>symmetries mapping generators to generators.  For instance, the N^2-1
>puzzles have the 8-fold symmetry D4, and so could have one set of
>tables used for 16 heuristics.
>
>Given the central nature the memory-performance tradeoff plays in the
>paper, I imagine this is quite relevant to Rich's research.

I suppose it depends upon where one is willing to draw the line with
respect to so called "weak methods" (i.e. search techniques that don't
rely on information about the problem domain).  If the methods are specific
only to group theoretic puzzles, I think they are interesting and useful,
but somewhat less relevant to advancing the state of the art of weak methods
as applied to larger classes of problems.  I'm under the impression that
Rich's research is focused on the latter, as opposed to the goals of the
typical hard-core cube enthusiast.

By the way, if someone is aware of a paying full-time research position
focusing only on solving the cube, and related, puzzles, please let me
know and I'll sign up tomorrow! :)

Having said that, I think the table optimization you described is a very
clever way to take advantage of symmetries for these types of problems.
Eventually, I hope that the knowledge gained by this, and related, threads
can be synthesized into a new algorithm that surpasses all previous cube
solving program.  Now that would be exciting to cube-lovers!

-- Greg


From cube-lovers-errors@oolong.camellia.org  Mon Jun  2 13:10:47 1997
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Date: Sun, 01 Jun 1997 23:57:43 -0400
To: cube-lovers@ai.mit.edu
From: Hersch Pilloff <pilloff@radix.net>
Subject: 5x5x5 practical Q's
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Hello,

I'm proud to say that after significant quantities of blood, sweat, and
tears, I have finally solved the 5x5x5 cube.  I used some techniques from
the good old 3x3x3 cube as well as some general techniques I've found
useful over the past (often of the form A R A' R' where A is a set of
rotations preserving one face and R is a rotation of that face).  One
problem I faced along the way and have not been able to solve to my
satisfaction is an issue of parity.  I often put the big cube in a state
where exactly two "equivalent" off-center edge pieces on the upper face
were interchanged with one another.  They had the correct orientation, I
simply needed to switch the two pieces.  I would like to know if anyone has
an effective means of dealing with this situation.

It was immediately apparent that the ARA'R' technique would not work
because interchanging two pieces requires a change in parity which the
ARA'R' won't produce.  I tried interchanging "identical" pieces from the
interior of the cube and then returning to the top face to see if any
change had resulted.  This was met with only marginal sucess-- after enough
fiddling I was able to produce two pairs of interchanged, properly
oriented, off-center edge pieces on the upper face which I could, after
some further manipulation, handle with an ARA'R' scheme.  Still, this isn't
very satisfactory to me because I don't much like the idea of having to
randomly interchange pieces until I produce a workable situation without
any more definite strategy.  Undoubtedly, most of you have been more
successful at this endeavor than I have, so I'd appreciate any available
wisdom.

Thanks,
Mark Pilloff

P.S.  I'm not using my usual account.  If you email a reply to me, please
send it to mdp1@uclink4.berkeley.edu and not whatever return address is
listed above or below.  Or just mail the list-- I'm on that as well.


From cube-lovers-errors@oolong.camellia.org  Mon Jun  2 13:11:24 1997
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Date: Mon, 02 Jun 1997 09:50:34 -0400 (Eastern Daylight Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: A* versus IDA*
In-reply-to: <970531141525.2140f541@iccgcc.cle.ab.com>
To: cube-lovers@ai.mit.edu
Reply-to: Jerry Bryan <jbryan@pstcc.cc.tn.us>
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On Sat, 31 May 1997 SCHMIDTG@iccgcc.cle.ab.com wrote:

> On the other hand, IDA* is a depth-first-search (DFS) algorithm.  DFS
> algorithms require only a linear amount of storage with respect to search
> depth (i.e. it has O(d) storage requirements) since it only needs to store
> the current path it is exploring.  It uses a cost threshold to determine
> when it has gone deep enough and should backtrack to the next unexplored
> node (as determined by the current path).  

Like everybody else, I am still trying to grasp IDA*.  The way it
backtracks reminds me a little bit of alpha-beta pruning.  There are
major differences, of course, because alpha-beta works with two person
games such as chess.  But the pruning idea seems very similar anyway. 

This raises a question.  It has been twenty years or so since I have
worked on a problem with alpha-beta, but my best recollection is that it
basically reduces the effective branching factor by the square root. For
example, chess is usually considered to have a branching factor in the
high 30's or low 40's, and alpha-beta gets the effective branching
factor down to about 6 or 7.  (The efficacy of this effect is dependant
on how well the nodes are ordered at each level of the tree.) 

So can we say something similar about IDA*?  That is, is there an
effective branching factor which is less than the actual branching
factor?  Is a little bigger than 13 for face turns effectively reduced
to some j, and is a little bigger than 9 for quarter turns effectively
reduced to some k?  It would seem so to me, and your algorithm then
becomes O(j^d) or O(k^d) instead of O(b^d). 

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990




From cube-lovers-errors@oolong.camellia.org  Mon Jun  2 13:11:40 1997
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Date: Mon, 02 Jun 1997 11:23:05 -0400 (Eastern Daylight Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: Searching and Metrics in (Korf 1997)
In-reply-to: <9705312055.AA16150@sun34.aic.nrl.navy.mil>
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Reply-to: Jerry Bryan <jbryan@pstcc.cc.tn.us>
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On Sat, 31 May 1997, Dan Hoey wrote:

> 
> So two questions remain. First, is there really a difference in the
> duplication of positions in the two metrics?  I think Jerry Bryan's
> table shows that only about 1.74% of the 63 billion positions are
> duplicated at 11q.  Do we have statistics on duplication for the
> face-turn metric?  Second, is there any technical justification for
> using the face-turn metric?  I'm aware that most of the published
> literature uses it, and that small numbers of moves sound more
> impressive than large ones, but these aren't very satisfying reasons.
> As far as I know, the problem of finding optimal solutions can be
> fruitfully approached in either of the metrics, or in any of several
> other metrics.
> 

I do not think there is really any difference of the sort that Prof. 
Korf was talking about.  I have done extensive breadth-search searches
with both the quarter-turn and face-turn metrics (although I have done
more work on the quarter-turn). Dan has posted what I will call
"syllable analyses" for both metrics.  Dan's "syllable analyses" (e.g.,
FB=BF, etc., where FB and BF are our syllables) provide an upper bound
on possible branching factors. The observed branching factors are
extremely close to Dan's upper bounds for both metrics out to the levels
I have searched. Hence, I see no advantage either way. 

Dan's syllable analyses are essentially based on short relations. 
Observed branching factors deviate from Dan's tight bounds only to the
extent that there exist longer relations which are not taken into
account by Dan's formulas.  The fact that the bounds are tight simply
reflects that there are not very many longer relations, at least out to
the level that has been searched.

Far enough from Start, there will be a major break in branching factors,
and the break will occur closer to Start for the face-turn metric than
for the quarter-turn metric.  This break in branching factors will be
reflective of longer relations which do not simply contain shorter
relations. And again, these longer relations will be shorter in
face-turns than in quarter-turns.  But that's just the way the metrics
work.

We seem to be degenerating into one of our periodic face-turn vs. 
quarter-turn arguments (I am a confirmed quarter-turner because
quarter-turns are conjugate).  But this reminds me of something I ran
across in Singmaster which I found curious. Singmaster is a face-turner,
but he nevertheless defined the length of a position as the number of
quarter-turns from Start for the position.  In other words, he did not
identify the number of moves to solve a position with the length of the
position.  Is his definition of "length" standard in group theory?  If
so, we would have only one length for each position, although we could
have several metrics for "number of moves from Start" (face-turns and
quarter-turns are by no means the only possible metrics).  In the same
vein, is the term "diameter" reserved to mean the maximum length for any
position so that the diameter is unique, or do we have one diameter for
quarter-turns, another diameter for face-turns, etc.? 

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990




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Date: Mon, 02 Jun 1997 09:29:19 -0400 (Eastern Daylight Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Size of God's Algorithm
In-reply-to: <970531141525.2140f541@iccgcc.cle.ab.com>
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Reply-to: Jerry Bryan <jbryan@pstcc.cc.tn.us>
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On Sat, 31 May 1997 SCHMIDTG@iccgcc.cle.ab.com wrote:

> On the other hand, IDA* is a depth-first-search (DFS) algorithm. 

This reminds me of an note I have been meaning to post to the list for a
long time.  The issue is, how big is God's algorithm?  By that, I do not
mean how large is cube space, neither the size of G nor the number of
M-conjugacy classes.  What I mean is, what is the size of the smallest
program which can calculate God's algorithm? In the size, we have to
include not only the executable code, but also the size of any tables.

To be more specific, suppose our task is to write a totally
self-contained function called cubelen which given a position x would
return the number of moves from Start for that position (e.g., 
L := cubelen(x); ).  We are specifically not worried about running
time, which might be longer than the age of the universe.  For example, 
given the existence of cubelen, we could call it once for each x in G, 
and thereby determine the complete frequency distribution of distances,
including the group diameter. 

Under these rules, the answer is actually very silly (or it may be the
rules which are silly).  With tight assembler coding, it can be done in
about 10^4 bits, certainly no more than 10^5 (about 10^3 to 10^4 bytes). 
All we have to do is calculate all processes containing (in turn) 0
moves, 1 move, 2 moves, etc., and comparing each generated position with
x until it matches.  We would make no attempt to eliminate sequences
such as RR', nor would we make any attempt to recognize that RL is the
same position as LR.  We only have to store two positions, x and the
current product, and there is no real tree structure.

This is very roughly speaking a depth search first, except that the
bookkeeping requirements (and attendant memory requirements) are a bit
less than with a typical depth search search.  That is, all you have to
keep track of is an index set from 1 to the number of moves (12 for
quarter-turns, 18 for face-turns), with one index for each level of the
search.  The branching factor is a constant, equal to the size of the
index set. 

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990




From cube-lovers-errors@oolong.camellia.org  Mon Jun  2 13:16:04 1997
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Date: Mon, 2 Jun 1997 10:10:29 -0700
From: Richard E Korf <korf@cs.ucla.edu>
Message-Id: <199706021710.KAA21887@denali.cs.ucla.edu>
To: cube-lovers@ai.mit.edu
Subject: miscellaneous comments

Dear Cube-Lovers,
   Here's a few comments on the recent flurry of messages. I guess I owe you all
an apology for being partly responsible for filling up your mailboxes lately,
yet here I go, sinning again.
   Perhaps it is no longer necessary due to the excellent messages by Haym
Hirsh, Greg Schmidt, and Dan Hoey, but I've written a 20 page survey article on
artificial intelligence search algorithms that I would be happy to send to
anyone on request. The first 10 pages cover things like breadth-first search,
depth-first search, depth-first iterative deepening, A*, and
Iterative-Deepening-A*. The rest talks about two-player game search and
constraint satisfaction. When ordering, please specify if you are interested in
the Rubik's Cube article or the search article, and allow 6 to 8 hours for
delivery.
   Regarding the quarter-turn metric, as long as one is careful to eliminate the
obvious duplicate states as Dan points out, it shouldn't matter much whether you
use the quarter-turn or face-turn metric.  While solutions are longer in the
quarter-turn metric, the branching factor, which is the average number of
operators that apply to a given state, is correspondingly lower. The branching
factor for the face-turn metric is about 13.34847, and the branching factor for
the quarter-turn metric should be about 9.
   Jerry Bryan is right on when he talks about the memory savings from storing
an entire subgroup, and the importance of efficient indexing.  For my heuristic
tables, no states are actually stored, just the number of moves to solve them.
The states are "respresented" by the indexes in the table.
   Here's the indexing problem.  Write out all the permutations of say 4
elements, 24 in all, in lexicographic, or any other, order. Now number the
permutations from 0 to 23.  The problem then is given a permutation of N
elements, compute its sequential number in your ordering scheme.  The obvious
algorithms do this in roughly N^2 time, but it would be nice to able to do it
faster.
   To put all this in perspective, there are two obvious but impractical
implementations of "God's algorithm". One is brute-force depth-first
iterative-deepening search, with no heuristic functions. At a million twists per
second, this would take about 700,000 years on average, but almost no
memory. The other is a complete lookup table storing every state. This would be
very fast once the table was built but would take a few bits for each one of the
4x10^19 states.  We don't have the time for the former approach, nor the storage
for the latter.  But by using both a lot of time, and a lot of memory, we can
find optimal solutions.  Most of the different design choices presented by these
types of approaches amount to a tradeoff between time and space. It remains to
be seen what choices lead to the best algorithms.

                                -rich

  





From cube-lovers-errors@oolong.camellia.org  Mon Jun  2 13:55:08 1997
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Date: Mon, 2 Jun 1997 13:39:44 -0400
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From: Alan Bawden <Cube-Lovers-Request@ai.mit.edu>
Sender: Cube-Lovers-Request@ai.mit.edu
To: Cube-Lovers@ai.mit.edu
In-reply-to: SCHMIDTG@iccgcc.cle.ab.com's message of Sun, 1 Jun 1997 22:39:43 -0400 (EDT) <970601223943.2140c63e@iccgcc.cle.ab.com>
Subject: ftp://ftp.ai.mit.edu/pub/cube-lovers/contrib

   From: SCHMIDTG@iccgcc.cle.ab.com
   Date: Sun, 1 Jun 1997 22:39:43 -0400 (EDT)
   ...
   At any rate, I'm willing to make this program available to interested
   parties (BTW, its written in C++).  I would prefer to upload it to an
   FTP site if possible.  Is anyone aware of an FTP site available for use
   by cube-lovers?

This seems like a good time to remind everyone that there -is- a
Cube-Lovers FTP archive for situations such as this:

  ftp://ftp.ai.mit.edu/pub/cube-lovers/contrib

I'm always happy to put more things into this directory, provided you don't
make it too much work for me.  The ideal contribution consists of a single
file (preferable a gzipped tar file) plus a -brief- description for the
README file.  You've got to figure out how to get the file to me in some
reasonable way -- letting me pick it up from some other anonymous FTP site
is easiest.

If you make me think too hard, for example by expecting me to compose the
description for the README file, or by making me pick out the Cube related
files from some larger collection, then I'm liable to never get around to
doing it.

I also reserve the right to redescribe, repackage, rename, recompress or
totally reject your contribution.


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From: Jin "Time Traveler" Kim <chrono@ibm.net>
Organization: The Fourth Dimension
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To: cube-lovers@ai.mit.edu
Subject: Re: 5x5x5 practical Q's
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Hersch Pilloff wrote:
> 
> Hello,
> 
> I'm proud to say that after significant quantities of blood, sweat, and
> tears, I have finally solved the 5x5x5 cube.  I used some techniques from
> the good old 3x3x3 cube as well as some general techniques I've found
> useful over the past (often of the form A R A' R' where A is a set of
> rotations preserving one face and R is a rotation of that face).  One
> problem I faced along the way and have not been able to solve to my
> satisfaction is an issue of parity.  I often put the big cube in a state
> where exactly two "equivalent" off-center edge pieces on the upper face
> were interchanged with one another.  They had the correct orientation, I
> simply needed to switch the two pieces.  I would like to know if anyone has
> an effective means of dealing with this situation.
> 
> It was immediately apparent that the ARA'R' technique would not work
> because interchanging two pieces requires a change in parity which the
> ARA'R' won't produce.  I tried interchanging "identical" pieces from the
> interior of the cube and then returning to the top face to see if any
> change had resulted.  This was met with only marginal sucess-- after enough
> fiddling I was able to produce two pairs of interchanged, properly
> oriented, off-center edge pieces on the upper face which I could, after
> some further manipulation, handle with an ARA'R' scheme.  Still, this isn't
> very satisfactory to me because I don't much like the idea of having to
> randomly interchange pieces until I produce a workable situation without
> any more definite strategy.  Undoubtedly, most of you have been more
> successful at this endeavor than I have, so I'd appreciate any available
> wisdom.
> 
> Thanks,
> Mark Pilloff
> 
> P.S.  I'm not using my usual account.  If you email a reply to me, please
> send it to mdp1@uclink4.berkeley.edu and not whatever return address is
> listed above or below.  Or just mail the list-- I'm on that as well.

I found that the solution book for the Rubik's Revenge was also a very
useful tool to solving the 5x5x5. The only pieces I had to figure out
myself through trial and error were the interior face pieces that have
edge contact with the middle piece (i.e. four pieces that including the
center piece would make a Plus "+" sign).

-- 
Jin "Time Traveler" Kim
chrono@ibm.net
VGL Costa Mesa


From cube-lovers-errors@oolong.camellia.org  Mon Jun  2 17:32:10 1997
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Date: Mon, 02 Jun 1997 23:03:11 +0200
From: Herbert Kociemba <kociemba@hrz1.hrz.th-darmstadt.de>
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Reading the many contributions in the mailing list in the last days, I
state, that the insight im my two phase algorithm solving the cube
ranges from misunderstood to partly understood, so I will add some more
really detailed explanation here.

The memory requirements for the search algorithm are of the order
O(d*log b), where b is the branching factor and d is the solution depth,
so it definitely is not breadth-first search with O(b^d) nor is it
bidirectional search with O(b^d/2).
The "log b" is no misprint, it is due to the special situation when
dealing rotational puzzles.

The orientations of the corners, the edges and the position of the four
middleslice edges are mapped to {0,1,...,3^7-1},{0,1,...,2^11-1} and
{0,1,...,12*11*10*9/4! -1} by appropriate functions in phase1.
Every state of the cube is represented by a triple (x,y,z) in stage1,
and a face turn maps this triple to another triple (x',y',z').

Let us denote (x0,y0,z0) for the triple, when arriving in the subgroup
<U,D,R2,L2,F2,B2>. All elements of this subgroup have this same triple
(x0,y0,z0), because neither edge nor corner orientations can be changed
here and the four middleslice edges stay in their position too (only
their permutation changes, but the mapping function for z ignores the
permutation).

Before applying the search algorithm we use the inverses of the mapping
functions to create lookup-tables for each coordinate, so that a face
turn can be performed with three table lookups, which is very effective.

The three heuristic functions in phase1 also are table-based. From a
pair (x,y) we compute the index i=3^7*y + x which will be a unique
number out of {0,1,...,3^7*2^11-1}. At tableposition i we store the
minimum number of moves we need to get from (x,y) to (x0,y0), ignoring
the z coordinate. Of course this minimum number never is greater than
the number of moves to go from (x,y,z) to (x0,y0,z0), so it is accurate
for the use in an IDA* type search. The other two tables for (x,z) and
(y,z) are constructed in a corresponding way.
Because these minimum-number never exceed 9 in phase 1, 4 bits will do
per tableentry.

Now I *try* to describe the search algorithm for phase1. The
implementation in my program has slight modifications, but they would
not improve the readability of the description. For example I omit the
part how to reduce the branching factor forbidding  the move sequences
RR2 or UDU etc.
During the search algorithm, we only store the current state (x,y,z).
Instead of storing the node path we store the applied move sequence,
which is equivalent but more adopted for our problem. We use 1 Byte for
every move. Let denote the list for the move sequence with A, A[i] then
is the i's element of the list. The sequence is stored in reverse order,
A[0] holds the last move of the solving sequence when a solution is
found. The iteration depth is denoted with L1.

1. On initialization set L1=1, i=0, A[0]=0.

2. Apply a face turn to (x,y,z) using the generated lookup tables, the
face turn according to the number A[i]: If A[i]==0, apply U. When we
write 0:U for that the following table shows what faceturn(s) to apply: 

O:U, 1:U, 2:U, 3:UR, 4:R, 5:R, 6:RF, 7:F, 8:F, 9:FD, 10:D, 11:D, 12:DL,
13:L, 14:L, 15:LB, 16:B, 17:B, 18:B.

In the case  A[i]=18 all branches  had been handled and this last B move
resets the cube to the state of the node where it came from at the
current depth -1. We reset A[i] to 0, increment i and goto 3. then.

If A[i]<18 increment A[i].Then compute the indices for the heuristic
tables using the triple (x,y,z) and check, if the current depth (which
is L1-i) plus the tablevalue v (which is a heuristic for the minimum
length to solve the cube from this state) exceeds L1, which is
equivalent to v>i. If that happens for any of the three tables, we prune
that branch and goto 2., to generate the next node of the same depth.

If v<=i, we first check, if i=0. In this case the current depth is the
iteration depth L1 and we have found a solution for phase1, because v=0
only can happen for all three heuristic tables, when we are in state
(x0,y0,z0). Goto phase2 then.
But if i>0, we have to generate the node at the current length + 1. We
decrement i and goto 2.  

3. If i==L1 now, we have searched the complete tree with lenght L1. In
this case we increment L1, set A[i]=0 and goto 2. to build again our
first depth-one node. 
If i<L1 we also goto 2. to build our next node at depth L1-i.

If you analyze the preceeding phase1 algorithm you will see that it is
indeed just an IDA* with lowerbound heuristic functions based on tables.

When we have found a solution in phase1, from the list A we construct
the maneuver sequence, apply it to the cube and are in <U,D,R2,L2,F2,B2)
now. With the help of new mapping functions we construct a triple
(a,b,c), where a,b, and c represent now the permutation of the 8
corners, the 8 edges outside the UD-slice and the four edges in the
UD-slice. Almost everything is analog to phase1, most important that we
have a maximum L2MAX for the iteration lenght L2 (except for the first
search in phase2), because we can assume L1+L2<N when we already have
found a solution with length N=L1'+L2'. Because of the maximum iteration
lenght the search can fail, in this case we go directly to 2. of phase1,
else we print out the solution and go to 2. of phase1.
Because N decreases and L1 increases, L2MAX decreases, and when we have
L2MAX=0 (we will not have time to wait for that in most cases) we have
found an optimal solution.

It seems very difficult to establish an upper bound for the maneuver
length for the algorithm except for the first solution, because the
maximum length in phase1=12 and in phase2=18.
If you still have any question about the algorithm, please ask. If
myself would be interested if you have any idea how to improve the
performance of the algorithm, using for examples symmetries of the cube
or the asymmetry of phase2.

Herbert



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From: David Barr <davidb@u.washington.edu>
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Subject: Re: 5x5x5 practical Q's
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On Sun, 1 Jun 1997, Hersch Pilloff wrote:

> Hello,
> 
> I'm proud to say that after significant quantities of blood, sweat, and
> tears, I have finally solved the 5x5x5 cube.  I used some techniques from
> the good old 3x3x3 cube as well as some general techniques I've found
> useful over the past (often of the form A R A' R' where A is a set of
> rotations preserving one face and R is a rotation of that face).  One
> problem I faced along the way and have not been able to solve to my
> satisfaction is an issue of parity.  I often put the big cube in a state
> where exactly two "equivalent" off-center edge pieces on the upper face
> were interchanged with one another.  They had the correct orientation, I
> simply needed to switch the two pieces.  I would like to know if anyone has
> an effective means of dealing with this situation.

Take a look at some of the web pages about the 4x4x4 or 5x5x5 cubes;
they have a sequence of about 20 moves for swapping two edge pieces.
I used to have it memorized.

Here's what I would do now (when I don't have the sequence memorized).
Turn either the 4th or 2nd horizontal slice one quarter turn, then use
your normal moves to put this slice's edge pieces back into place.
You will end up with a solvable number of pieces out of place on the
top.

I don't understand what ARA'R' means.  When I solve the 5x5x5 cube, I
end up using mostly one type of sequence that moves three pieces.  I'm
not very familiar with 5x5x5 notation, so I'll describe a notation
that can describe the sequence.  I learned this notation from a book
on the 4x4x4 that I bought in the early 80s.

A number (1-5) refers to a slice move that causes a vertical column of
pieces on the front face to move to the bottom face.  1'-5' are the
inverses.  L or R refers to a move that will move the bottom row of
pieces on the front face to either the left or right side.  Here's a
sequence that moves three corner pieces:

  1 L 5 R 1' L 5' R

The shorthand for this move is 1L5, because the rest of the moves can
be predicted from the first three moves.  If you replace 1 and 5 with
any two other columns (and possible swap the L and R moves), you can
design a sequence that will move 3 pieces of any type.

1L5 and 5R1 will move three corner pieces.
1L3 and 5R3 will move three edge pieces.
1L2, 1L4, 5R2 and 5R4 will move three off-center edge pieces.
2L3 and 4R3 will move three center near-edge pieces.
2L4 and 4R2 will move three center near-corner pieces.

A lot of the time, the three pieces you want to move aren't in
locations that correspond to one of these sequences.  In that case,
make a move that puts the three pieces in the right place for one of
the sequences, do the sequence, then undo your original move.

For all I know, this is just another way of describing the moves that
you are making.

David



From cube-lovers-errors@oolong.camellia.org  Mon Jun  2 17:30:49 1997
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Sender: Haym Hirsh <hirsh@cs.rutgers.edu>
Date: Mon, 2 Jun 97 16:13:55 EDT
From: Haym Hirsh <hirsh@cs.rutgers.edu>
Reply-To: Haym Hirsh <hirsh@cs.rutgers.edu>
To: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Cc: cube-lovers@ai.mit.edu
Subject: Re: A* versus IDA*
In-Reply-To: Your message of Mon, 02 Jun 1997 09:50:34 -0400 (Eastern
        Daylight Time)
Message-ID: <CMM-RU.1.5.865282435.hirsh@pei.rutgers.edu>

> Like everybody else, I am still trying to grasp IDA*.  The way it
> backtracks reminds me a little bit of alpha-beta pruning.  There are
> major differences, of course, because alpha-beta works with two person
> games such as chess.  But the pruning idea seems very similar anyway. 

Actually, in many respects IDA* and alpha-beta pruning are opposites.

Alpha-beta pruning is a way to avoid exploring possible moves in
game-tree search that are guaranteed to have no influence on what the
final move will be.  It lessens the work that traditional minimax
search would have to do by eliminating from consideration some of the
possibilities it would ordinarily consider.  Alpha-beta results in a
smaller search tree than minimax would ordinarily have.

In contrast, IDA* =adds= additional work that, in theory, would not
be necessary if the search was done optimally.  It basically keeps
A*'s search tree, but searches parts of it multiple times.  Although
this seems wrong-headed, it winds up being a win because, although it
does some redundant work, the search method has much more realistic
space requirements.

Alpha-beta is a search-space-reduction method for the minimax search
procedure.  IDA* is a search-space-maintaining method that replaces
A*'s search of this space with a new search algorithm that explores
some of A*'s possibilities multiple times, but at a tremendous savings
of internal "bookkeeping" memory requirements.

Finally, there is another analogy between alpha-beta and IDA* that is
potentially misleading: both use evaluation functions to evaluate the
merit of a given state (e.g., cube or board position), and, moreover,
the more accurate the evaluation function, the better the performance
of the search method.  However, in the case of alpha-beta, you expect
a better evaluation function to yield better moves, i.e., it changes
the output of the search method, returning something that looks
better.  (Indeed, this seems to be a major reason for Deep Thought's
success in its recent match against Kasparov.)  In contrast, as long
as the evaluation function given to A* or IDA* never overestimates the
cost of solving a given state (i.e., number of moves to solving a
given cube), they are guaranteed to return an optimal solution.  Here
the improved performance refers to run-time -- a better evaluation
function typically means A* or IDA* will explore fewer states on its
way to finding an optimal solution.  In the case of A* this means it
reduces its run-time from very very unreasonable to very unreasonable,
whereas for IDA*, in at least this problem, it reduces it from
unreasonable to feasible.


At the risk of having him bombarded with lots of email requests, I
strongly encourage those who are interested in understanding this
further to take up Professor Korf's offer of his survey of search
methods in AI.  He is an excellent writer and has made many major
contributions to this area.  Alternatively, most introductory
textbooks on artificial intelligence cover search methods in a fairly
early chapter (although not all cover IDA* in adequate depth).
However, both options will probably assume some familiarity with basic
concepts in computer science, so may not be accessible to all.

Haym

---------------------------------------------------------------------------
Haym Hirsh                               office: Busch Campus, CoRE 317
Associate Professor                      email:  hirsh@cs.rutgers.edu
Department of Computer Science           phone:  +1 732-445-4176
Rutgers University                       fax:    +1 732-445-0537
New Brunswick, NJ 08903                  http://www.cs.rutgers.edu/~hirsh
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From cube-lovers-errors@oolong.camellia.org  Mon Jun  2 17:31:24 1997
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From: Goyra <Goyra@cheerful.com>
To: cube-lovers@ai.mit.edu
Subject: Re: 5x5x5 practical Q's
Date: Mon, 2 Jun 1997 21:30:19 +0100
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> From: Hersch Pilloff <pilloff@radix.net>


> I'm proud to say that after significant quantities of blood, sweat, and
> tears, I have finally solved the 5x5x5 cube.  I used some techniques from
> the good old 3x3x3 cube 


	Congratulations! I remember going through this about 8
years ago, when I got my 5X5. My approach for the 3X3 was to 
solve the corners, then the middle edges; so those techniques carried 
over without a change. The rest was simpler, as far as I recall. Trouble 
was, even knowing the finished procedure, it took about an hour each 
time.


							David



From cube-lovers-errors@oolong.camellia.org  Mon Jun  2 22:09:15 1997
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From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: cube-lovers@ai.mit.edu
Subject: Memory-Performance tradeoffs in (Korf 1997)

This is the third part of my series on Rich Korf's paper.  It covers
what I think is the most interesting part of the research, but
(intentionally) also the least rigorous.  Rich makes an attempt to
estimate how many positions will be examined by IDA* as a function of
the memory used by the heuristic.  I have to admit I may have missed
something here, but this is my take at understanding, explaining, and
a few queries about the result.

I plan at least one other message to clarify some of the points in the
previous parts.  But right now I will note the most glaring error,
which is that heuristic functions are actually guaranteed NOT TO
OVERESTIMATE the true distance to of a solution.  Thanks to Clive
Dawson for letting me know I said exactly the opposite.  Urk!

DEFINITIONS

Search will be undertaken on a problems in G, with |G|=N.  For x in G,
Depth(x) is the length of the shortest process to solve x.

A heuristic is a function h on G satisfying h(x) <= Depth(x) for every
position x in G.  The special heuristic h0(x)=0 is the "trivial
heuristic".

Work(h,Depth(x)) is roughly the total number of nodes visited in
searching for x using IDA* with heuristic h.  Roughly, because we
average over all the positions at that depth.

The average number of operators/generators applicable to a position is
called the branching factor b.  This is a constant over the positions
we will consider, and in the following I will write logb(x) for the
logarithm to the base b.

For most heuristics h, we partition the space G into a certain number
of parts, such that h(x) is a constant over each part; we write
Part(h,x) for the part containing x and extend h to the parts by
writing h(Part(h,x))=h(x). We can use any partition to define such a
heuristic h by defining

    h(Part(h,SOLVED))=0, and for x not in Part(h,SOLVED),

    h(x)=1 + max over all y in Part(h,x),
                min over all neighbors z of y,
                   h(z).

The number of parts of a partition defining h is called Size(h).  We
make a table of size Size(h) once containing the heuristic values of
the parts, and look up h(x)=h(Part(h,x)) over the course of the
search.

If each primitive operator maps parts to parts, then the "max" in the
definition of h(x) is over only one value.  This occurs, for instance
if "primitive"="group generator" and Part(h,x)="Coset of a subgroup
with respect to x".  Size(h) is then the order of the subgroup.

ESTIMATES

These are the rough estimates that Rich uses (as I understand them).

Most of these exponential-growth spaces have one depth
       Mode(Depth) = Mode({Depth(x) : x in G})
at which most of the nodes appear, and almost all of the nodes appear
very close to that depth (so the answer doesn't change much if we take
Mean or Median instead of Mode.  Rich uses Mean).  If the branching
factor stays nearly constant to the end, we should find that

       Mode(Depth) ~ logb(N).                   (#1)

When heuristics are defined on parts, and the branching factor
of the partition space is the same as the branching factor of the
whole space,

       Mode(h) ~ logb(Size(h))                  (#2)

since there are Size(h) partitions.

If we examine all positions up to depth d, there are about b^d of
them, so

       Work(h0,d) ~ b^d.                        (#3)

Finally, we might hope that in searching with a consistently
underestimating heuristic, we might be doing something like examining
all the positions up to the amount of underestimation, followed by a
non-branching search to the end:

       Work(h,d) ~ Work(h0,d-Mode(h)).          (#4)

THE RESULT

Using these estimates we can calculate

      Mode(Work(h,Depth(x))) ~ Work(h,logb(N))                 by #1
                             ~ Work(h0,logb(N)-Mode(h))        by #4
                             ~ Work(h0,logb(N)-logb(Size(h)))  by #2
                             ~ b^(logb(N)-logb(Size(h)))       by #3
                             = N / Size(h).

This is the really fundamental result of Rich's paper.

ERRORS

There are some ways in which this model is known to be flawed.  Rich
notes that actually

#4   Work(h,d) > Work(h0,d-Mode(h)),

by over two orders of magnitude.  He conjectures that a "locality of
understimation" effect causes most of the search to be concentrated in
the parts of the space for which h is worst.

He hopes this will be balanced out by

#2    Mode(h) > logb(Size(h))

because the branching is not perfect.  This effect is stronger on the
branching on the parts of h than on G, because there are fewer of the
former.  He finds that under the effects of these two errors, the
answer is off by a factor of 1.4 for his experiments.

I am wondering about a few other effects.  For one thing, I am not at
all sure how well the heuristics model the exponential behavior of the
search space, with a strongly defined mode.  I think that if
Mode!=Mean, you find the entire argument falls apart (but I may be
missing something).  I would like to know something more like the
curve for the heuristics, rather than just the mean.

Second, Rich is combining heuristics based on partition search to form
a different kind of heuristic.  Say we form h=max(h1,h2), where h1 and
h2 have about the same size.  Estimate #2 would say

       Mode(h) = logb(Size(h))
               = logb(2 Size(h1))
               = Mode(h1) + logb(2).

But I think the strongly modal behavior of these heuristics may not
allow the mode to be increased this easily.  We might find that
Mode(h)=Mode(h1), but with a more pronounced peak.

My third quibble is on whether the branching factor b is the same for
the coset spaces as for the whole space G.  I'm concerned that some
generators might lie in the subgroup used to form a heuristic, so they
would map a coset to itself, lowering the effective branching factor
for heuristics.  But I'm not sure about this--mapping how close this
estimate holds is a ripe direction for research.

Dan


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To: cube-lovers@ai.mit.edu
From: Nichael Lynn Cramer <nichael@sover.net>
Subject: Re: 5x5x5 practical Q's

At 9:30 PM +0100 6/2/97, Goyra wrote:
>> From: Hersch Pilloff <pilloff@radix.net>
>> I'm proud to say that after significant quantities of blood, sweat, and
>> tears, I have finally solved the 5x5x5 cube.  I used some techniques from
>> the good old 3x3x3 cube
>	Congratulations! I remember going through this about 8
>years ago, when I got my 5X5. My approach for the 3X3 was to
>solve the corners, then the middle edges; so those techniques carried
>over without a change. The rest was simpler, as far as I recall. Trouble
>was, even knowing the finished procedure, it took about an hour each
>time.

Ditto congrats.

As far as solving, I find it useful (from a "finger-exercise" point of
view) to give myself "stunt" solutions to work towards.  For instance with
the 5X, I like to start with a single center face (say, blue).  Then I
solve the remaining "ring" of inner blue faces in order.  Then the ring of
blue edge and corner pieces (again, in circluar order).  Then each
successively higher "horizontal" slice (again, in order around the cube)...
and so on until the cube is finished.

Needless to say, some backup is occassionally necessary.  But this can be
a pleasant way to pass the time.

Nichael
nichael@sover.net               "Did I forget, forget to mention Memphis,
http://www.sover.net/~nichael/      Home of Elvis and the ancient Greeks..."




From cube-lovers-errors@oolong.camellia.org  Mon Jun  2 22:08:35 1997
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From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: cube-lovers@ai.mit.edu
Subject: Indexing (was Re: miscellaneous comments)
In-Reply-To: <199706021710.KAA21887@denali.cs.ucla.edu>

Rich Korf wrote:

>    Here's the indexing problem.  Write out all the permutations of
> say 4 elements, 24 in all, in lexicographic, or any other,
> order. Now number the permutations from 0 to 23.  The problem then
> is given a permutation of N elements, compute its sequential number
> in your ordering scheme.  The obvious algorithms do this in roughly
> N^2 time, but it would be nice to able to do it faster.

I thought everyone knew this, but it seems not.  The procedure is
this: Make a fresh copy of P and its inverse Pinv, represented as
arrays on [0..N-1].  For k from N-1 down to 1, do

            i = Pinv[k];
            Pinv[P[k]] = i;
            P[i] = P[k].

The loop invariant is that P[0..k] is a permutation on [0..k] and
Pinv[0..k] is its inverse.

Conceptually, you are exchanging P[k-1] with P[Pinv[k-1]] to turn P
into the identity permutation.  But instead, you leave stuff in the
part of the P and Pinv arrays that you don't need to use because you
decrement k.  That stuff you leave records what exchanges you (would
have) made, so it encodes the index in a variable base: 0<=P[k]<=k and
you take the sum (P[k] k!) to get the index.  The permutation parity
is |{k : P[k]==k}| mod 2.

This requires O(N) operations on integers of size O(N log N), so the
time is O(N^2 log N).  But if we don't charge extra for the integer
size, it's an O(N) algorithm.  If you're using the index to lookup
something in a table that exceeds the integer size you usually need to
split the index into integer-sized subindices anyway (one tells you
which byte in the file, another tells you which file on the disk,
another tells you which disk...).

Oh, and you can run the algorithm in reverse to convert the
variable-base index back into a permutation.  (This part doesn't need
Pinv).  If you fill the P[k] with Random[0..k] and do this, you get a
fair shuffle.  (I wish programs would randomize their cubes this way.
Somehow I never trust the 100 random turns.)

I think the only reason people don't think of this balking at the
initial overhead of making a copy of P and calculating its inverse.
But then we go and spend quadratic time searching for the bits and
pieces we need.

Dan
[ Still working on part 3...]


From cube-lovers-errors@oolong.camellia.org  Mon Jun  2 22:08:57 1997
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From: Richard E Korf <korf@cs.ucla.edu>
Message-Id: <199706022314.QAA22320@denali.cs.ucla.edu>
To: jbryan@pstcc.cc.tn.us
CC: cube-lovers@ai.mit.edu
In-reply-to: <Pine.WNT.3.96.970602093611.-408329C-100000@ER123A.PSTCC.CC.TN.US> (message from Jerry Bryan on Mon, 02 Jun 1997 09:50:34 -0400 (Eastern Daylight Time))
Subject: Re: A* versus IDA*

Dear Jerry and fellow Cube-Lovers,
   Unfortunately, the analysis of IDA* and A* is much harder than the
corresponding analysis of alpha-beta minimax, and is still an open research
problem.  The reason is that the running time of IDA* or A* depends on the
accuracy of the heuristic function. If we use zero for the heuristic everywhere,
which is still a lower bound, these algorithms degenerate into brute-force
searches, which take b^d time, where b is the branching factor and d is the
optimal solution length. On the other hand, if our heuristic function is
perfect, and always gives us the exact number of moves needed to solve a state,
then these algorithms run in time proportional to d, the length of the optimal
solution. This would be practically instantaneous in this case.
   My experience with my program is that the ratio of the number of nodes
generated in searching from one level to the next is approximately the same as
brute-force branching factor of about 13.35. The effect of the heuristic tables
is to start this geometric progression at 8 or 9 instead of 0. This is greatly
oversimplified, but conveys the gist of what I've been able to observe.
                          -rich

                                   -rich



From cube-lovers-errors@oolong.camellia.org  Tue Jun  3 01:54:34 1997
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From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: cube-lovers@ai.mit.edu
Subject: Re: Searching and Metrics in (Korf 1997)

My earlier remarks on searching, outside of saying "underestimate" for
"overestimate", are pretty much redundant given the availability of
Rich's survey article.  Here are some remarks on what I said about the
quarter-turn metric.

I said "No one who is familiar with the cube-lovers archives (e.g. my
message of 9 January 1981) would generate both [ FF and F'F' ] any
more than they would generate [ both FB and BF]".  This is both too
strong and too weak a statement.

First, there are search techniques that are unable to determine what
the last move was, so they must go ahead and generate such moves
(hopefully discarding them later).  I should have replaced "any more
than they would" with "if they could avoid" or something.

But second, I seem to have given the impression this is "a cube thing"
rather than "a search thing".  But counting unit turns instead of
multiple turns is easily generalized, moreso than the problem of
commutativity.

Suppose you have generators g1, g2, g3, ..., gk of a group, so that
their inverses g1', ..., gk' are also generators.  The order of a
generator g, o(g), is the minimum positive integer for which g^o(g) is
the identity.  (I try to avoid using the asymptotic little-o when I'm
talking about this.)

What is the appropriate rule for the set of possible next generators?

The generalization of the face-turn metric is the "power-turn" metric.
We count gi, gi^2, gi^3, ..., gi^(o(gi)-1) as generators.  Then after
each gi^k we refuse to allow the next move to involve gi.  We deal
with commutativity by also refusing to allow the next move to involve
gj if j<i and gi commutes with gj.  We need only keep a k-valued state
variable around to know what moves are available next (though this
doesn't solve the commutativity problem in general, see below).

For the "unit-turn" metric, we wish to count gi^2 as twice as
expensive as gi if o(gi)>3.  We still assume gi' is the same cost as
gi, so if o(gi)=3 we have gi^2=gi'.  For each gi, let the half-order
h(gi)=Floor((o(gi)-1)/2).  The state variable assumes 2 h(gi) values
1,2,...,h(gi) and -1,-2,...,-h(gi).  If the previous state did not
involve gi, then gi enters state 1 and gi' enters state -1.
Thereafter, positive states can accept gi and increment, and negative
states can accept gi' and decrement, to the maximum of their range.
There is one more case: if o(gi) is even, then state h(gi) can accept
one more gi and go to state -h(gi).  Otherwise gi and gi' are
prohibited.  This prevents backtracking (gi gi' or gi' gi),
overturning (gi^x or gi'^x where 2x>o(gi)) and halfway-duplication
(gi'^(o(gi)/2)).  So the total number of states is
2(h(g1)+h(g2)+...+h(gK)).  Commutative duplication is prevented by the
same prohibition as for the power-turn metric.

So for the unit-turn cube metric, we need 12 states (two per face).
The megaminx requires 48 states (12 per face) because the generators
have order 5.

This completely solves the problem about there being more duplication
in the unit-turn metric than the power-turn metric.  But the problem
of commuting generators is more complicated, as I remarked with
respect to the Megaminx on 23 September 1982.  We can find commuting
pairs {A,B} and {B,C} such that {A,C} do not commute.  Remember that
when we are ordering generators, we require that commuting generators
appear in order.  But suppose A<B<C.  This rule would allow duplicate
processes BCA and CAB.  I did some work on trying to avoid this, but
I'm not sure I got a satisfactory solution.

Dan
Hoey@AIC.NRL.Navy.Mil


From cube-lovers-errors@oolong.camellia.org  Tue Jun  3 01:54:51 1997
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Date: Sun, 01 Jun 1997 10:35:40 -0400
From: "Richard W. Pearson, Jr." <rwpearso@ipass.net>
Reply-To: rwpearso@ipass.net
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To: Tiffp1@aol.com
CC: cube-lovers@ai.mit.edu
Subject: Re: The rest of us
References: <970531091548_-1732449048@emout01.mail.aol.com>
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Tiffp1@aol.com wrote:
> 
> DOES ANYONE KNOW ANY STORES NEAR HENDERSON,DURHAM,OR RALEIGH WHERE I CAN BUY
> A RUBIK'S CUBE AND A RUBIK' S TRIAMID

I've seen them in quite a few toy stores and children's educational
stores.  Offhand, I can think of Zany Brainy in Pleasant Valley Shopping
Center in Raleigh.  

Ricky Pearson


From cube-lovers-errors@oolong.camellia.org  Tue Jun  3 01:55:07 1997
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To: cube-lovers@ai.mit.edu
Message-ID: <19970602.222516.3174.1.shaggy34@juno.com>
X-Mailer: Juno 1.15
X-Juno-Line-Breaks: 1-3
From: Josh D Weaver <shaggy34@juno.com>
Date: Mon, 02 Jun 1997 23:26:10 EDT

How do you solve the cube? I  solve it by layers.  How are some other
ways to solve it?  And what does "<U,D,R2,L2,F2,B2>" this mean?

Josh


From cube-lovers-errors@oolong.camellia.org  Tue Jun  3 14:53:11 1997
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From: Pete Beck <pbeck@pica.army.mil>
To: cube-lovers@ai.mit.edu
Subject: orbix
Date: Tue, 3 Jun 1997 06:53:21 -0400
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If anybody is looking for an ORBIX I saw them last night at my local
Kay Bee (rockaway townsquare mall, NJ) reduced from $25 to $10.

Pete


From cube-lovers-errors@oolong.camellia.org  Tue Jun  3 14:53:33 1997
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Date: Tue, 3 Jun 1997 16:52:46 +0100
To: cube-lovers@ai.mit.edu
From: Tony Davie <ad@dcs.st-and.ac.uk>
Subject: FreeCell

Could someone describe the FreeCell puzzle for us non-windows people?


                                                     _____
                                                    /    /\
Tony Davie                Computer Science         /    /  \
Tel: +44 1334 463257      St.Andrews University   /    /    \
Fax: +44 1334 463278      North Haugh            /    /  /\  \
ad@dcs.st-and.ac.uk       St.Andrews            /    /  /  \  \
                          Scotland             /    /  /\   \  \
                          KY16 9SS            /    /  /  \   \  \
                                             /    /__/____\   \  \
                                            /              \   \  \
http://www.dcs.st-and.ac.uk/~ad/Home.html  /________________\   \  \
                                           \                     \  \
                                            \_____________________\ /

In theory, there is no difference between theory and practice, but
in practice there is a great deal of difference.




From cube-lovers-errors@oolong.camellia.org  Tue Jun  3 14:52:20 1997
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Date: Tue, 03 Jun 1997 00:23:43 -0700
From: Jin "Time Traveler" Kim <chrono@ibm.net>
Organization: The Fourth Dimension
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Josh D Weaver wrote:
> 
> How do you solve the cube? I  solve it by layers.  How are some other
> ways to solve it?  And what does "<U,D,R2,L2,F2,B2>" this mean?
> 
> Josh

Corners first, middle edges interconnecting edges, and finally centers.
<U,D,R2,L2,F2,B2> that is read as Up turn, Down turn, Left turn twice,
Front turn twice, Back turn twice.  It's standard notation referring to
the faces of a cube, assuming you're using one face as a reference
point.  The notation typically uses 90 degree clockwise turns.

-- 
Jin "Time Traveler" Kim
chrono@ibm.net
VGL Costa Mesa


From cube-lovers-errors@oolong.camellia.org  Tue Jun  3 14:52:43 1997
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Date: Tue, 03 Jun 1997 06:13:22 EDT
From: Richard M Morton                             RMM      - ICOMSOLS <richard_morton@icom-solutions.com>
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Subject: Number of States for 2x2x2 cube

-------------------------------------------------------------------------------

  Reading about the number of possible states (88million) for the corners
  of the 3x3x3 cube (equiv. to 2x2x2) made me try working this out for
  myself. My logic was :

  Cube has 8 corners, each of which can have 3 orientations. Number of
  possible states is

  (8*3)*(7*3)*(6*3)*(5*3)*(4*3)*(3*3)*(2*3)*(1*3) = 8!*3**8 = 264,539,520

  This figure of course includes some states only possible by disassembling
  the cube (or maybe by twisting it in a fourth dimension ?). Without this
  the last corner can only have one orientation so the number of states
  achievable by twisting only in 3d is 8!*3**7 = 88179840

  I assume that this is the correct figure but what I would like to know is
  whether my logic is correct ie is the assumption about the last corner
  being fixed in orientation the only requirement (I am not a mathematician
  so please excuse me if this is obvious).

  Richard Morton  "I'm Brian and so's my wife"


From cube-lovers-errors@oolong.camellia.org  Tue Jun  3 22:32:12 1997
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From: Don Woods <don@clari.net>
Message-Id: <199706031957.MAA14492@madrigal.clari.net>
To: cube-lovers@ai.mit.edu
Subject: Re: FreeCell
Cc: ad@dcs.st-and.ac.uk

> Could someone describe the FreeCell puzzle for us non-windows people?

This is perhaps getting a bit off-topic for cube-lovers, since FreeCell
is a card game, but it can also be thought of as a puzzle in which you
are to unscramble a randomised bunch of cards, so maybe it's not so far
off-topic at that.  Anyway, since you asked:

Shuffle a standard 52-card deck and deal them out to form 8 columns of
cards; 4 columns of 7 and 4 columns of 6.  All the cards are face up,
and you should spread the columns (while keeping the cards overlapping
within each column) so you can see all the cards.

Above the columns are 8 initially-empty spaces.  4 of these are reserved
for collecting the 4 suits in order: the object of the game is to move
Ace-deuce-3-4-...-queen-king of each suit onto these spaces.  The other 4
are "free cells": each free cell can be used to hold any SINGLE card.

You move only one card at a time.  The only cards you can move are the
last card of a column (i.e., cards that don't have other cards on top
of them) or a card in a free cell.  A card can move to (1) an empty free
cell, (2) an empty column (i.e., a column where you've moved out all of
the cards), or (3) a column whose last card is the opposite color and
one rank higher than the card being moved (i.e., you can place a red 6
on a black 7, etc.).  A card can also be moved to the suit-collecting
piles if it's the next card needed in that suit, i.e., an Ace to empty
suit spot, a deuce onto the Ace of the same suit, etc.

The Windows version of the game lets you specify a number and then
generates a deck based on that number, so you can get the same initial
layout by giving the same number again.  Thus it offers only a small
fraction of the possible layouts.  I wasn't part of the effort that
solved the Windows layouts, but I did write a program to solve FreeCell
layouts and fed it a million random layouts, and it solved all but 14.
So I don't find it surprising that all but 1 of the 32000 Windows
layouts is solvable.

There are several other puzzle-type solitaire card games with a similar
theme.  E.g., Seahaven Towers uses 10 columns instead of 8, and two of
the free cells start with cards in them (each column has 5 cards).  In
Seahaven Towers, moves onto a column must be matching suit instead of
opposite color, i.e., 6 of clubs onto 7 of clubs.  Also, only a King
can be moved into an empty column.  Thus the moves in Seahaven Towers
are much more restricted; my program for that game runs a lot faster.
About 90% of Seahaven Towers layouts can be solved.

	-- Don.

-------------------------------------------------------------------------------
-- 
-- Don Woods (don@clari.net)		        ClariNet provides on-line news.
-- http://www.clari.net/~don		        I provide personal opinions.
--


From cube-lovers-errors@oolong.camellia.org  Tue Jun  3 22:32:23 1997
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Message-Id: <9706040106.AA26157@sun34.aic.nrl.navy.mil>
From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: Herbert Kociemba <kociemba@hrz1.hrz.th-darmstadt.de>,
        cube-lovers@ai.mit.edu
Subject: Re: Detailed explanation of two phase algorithm

Herbert,

Thank you very much for the description of your algorithm.  This fills
many of the gaps that have been left by the fragmentary discussions on
cube-lovers over the years.  Though perhaps it's all my fault--I would
not have felt so left out if I had taken the effort to subscribe to
CFF.

> The memory requirements for the search algorithm are of the order
> O(d*log b), where b is the branching factor and d is the solution
> depth, so it definitely is not breadth-first search with O(b^d) nor
> is it bidirectional search with O(b^d/2).

I like the extremely compact stack size.  It might make it possible to
take advantage of very large (disk-based) heuristic tables.  The idea
is that if you can collect enough positions at once, you can sort them
and look up all the heuristics in one pass through the table, so you
get sequential access instead of random access.  The usual problem
with this is you need massive parallelism--or simulate it with massive
multiprogramming.  I'm hoping the tiny stack size makes
multiprogramming cheap enough.

I hope this is not doing too much violence to your idea, which is
remarkably parsimonious of memory usage.  I guess I'm looking for ways
that increased memory could make it faster.

I would be careful in classifying search algorithms by memory size,
though.  With Shamir's modification of bidirectional search, for
instance, you can get much less than O(b^(d/2)) memory use.  Still
nothing like the very low usage of your idea, though.

> The "log b" is no misprint, it is due to the special situation when
> dealing rotational puzzles.

Well, not only rotational puzzles--any grouplike puzzle could achieve
this--any puzzle where you can undo your moves.  For instance, on a
fifteen puzzle, you could use an encoding like
    0: R 1:LD 2:TL 3:RT 4:D
to explore the four possiblities for the direction to move the blank
square at any node.  So knowing the position of the puzzle and the
current index would tell you how to modify the position for the next
index.

> ...If you analyze the preceeding phase1 algorithm you will see that
> it is indeed just an IDA* with lowerbound heuristic functions based
> on tables.

It sure is.  Now here's a question.  Should it be combined with phase2
in such a way that the entire search is IDA*?

The way I would see this happening is that whenever you get to phase 2
(at depth i with a cutoff of L1), instead of allowing the phase2
search to iteratitively deepen, you run a depth-first A* search with a
fixed depth cutoff of L1-i.  It would take longer to get the first
answer, but when you got an answer it would be optimal, and you would
never spend time searching longer processes.  I'm just hoping this
might find optimal solutions faster.  Possibly it could be combined
with global heuristics like Rich used.

One final thing, which I'm not sure ever got asked, much less
answered, is that Mike Reid did an exhaustive search of the subgroup
<T,D,F2,R2,B2,L2> (7 Jan 1995).  Did this verify that the optimal
face-turn process for each element of the subgroup is a word on those
generators?  Or are there shortcuts that use forbidden quarter-turns?

Dan


From cube-lovers-errors@oolong.camellia.org  Wed Jun  4 01:26:56 1997
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Date: Tue, 03 Jun 1997 23:57:22 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: Number of States for 2x2x2 cube
In-reply-to: <199706031013.GAA17406@life.ai.mit.edu>
To: Richard M Morton RMM - ICOMSOLS <richard_morton@icom-solutions.com>
Cc: cube-lovers@ai.mit.edu
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On Tue, 3 Jun 1997, Richard M Morton RMM - ICOMSOLS wrote:

> 
>   Reading about the number of possible states (88million) for the corners
>   of the 3x3x3 cube (equiv. to 2x2x2) made me try working this out for
>   myself. My logic was :

Actually, the corners of the 3x3x3 are not usually considered to be
equivalent to the 2x2x2.  There are 24 times fewer states in the 2x2x2
because any of the 24 rotations in space of the 2x2x2 are considered to be
equivalent.  Another way (and the typical computer way) to model this
effect is to fix one of the corners of the 2x2x2.

> 
>   Cube has 8 corners, each of which can have 3 orientations. Number of
>   possible states is
> 
>   (8*3)*(7*3)*(6*3)*(5*3)*(4*3)*(3*3)*(2*3)*(1*3) = 8!*3**8 = 264,539,520
> 
>   This figure of course includes some states only possible by disassembling
>   the cube (or maybe by twisting it in a fourth dimension ?). Without this
>   the last corner can only have one orientation so the number of states
>   achievable by twisting only in 3d is 8!*3**7 = 88179840
> 
>   I assume that this is the correct figure but what I would like to know is
>   whether my logic is correct ie is the assumption about the last corner
>   being fixed in orientation the only requirement (I am not a mathematician
>   so please excuse me if this is obvious).
> 

Your logic is correct.  The larger size of 264,539,520 is the number of
corner configurations in the so-called constructable group.  The
constraint on the twist of the last corner cubie gets you down to
88179840, and is the only other constraint on the corners alone.  When you
add in the edge cubies, there are two additional constraints  --  one for
the edge cubies alone, and one for the corner and edge cubies combined
(they have to have the same parity).

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990



From cube-lovers-errors@oolong.camellia.org  Wed Jun  4 17:30:05 1997
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From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: FreeCell
In-reply-to: <v01520d01afb9ee177570@[138.251.192.26]>
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Cc: cube-lovers@ai.mit.edu
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On Tue, 3 Jun 1997, Tony Davie wrote:

> Could someone describe the FreeCell puzzle for us non-windows people?
> 

The game has already been described, so I won't do it again.  But I will
make a couple of comments.

The Windows version of the game has a certain charm that does not exist
when you play it with real cards.  The rules require that you move one
card at a time.  In actual practice you often move a group of cards as
if they were a single entity, knowing full well that you could move them
one at a time and still stay within the rules of the game.  The Windows
version of the game understands this concept, and will move an entire
group of cards for you with one mouse click.  It's almost as if the
program is defining a macro operator for you on the fly.  Also, there
are totally obvious moves, such as always playing aces to their payoff
cells without further ado as soon as the aces are uncovered.  The
program plays these totally obvious moves for you. At the end of the
game there may be several dozen such obvious moves in a row, and it
makes a nice visual effect that you cannot get with real cards.  (By the
way, the program fails to make some moves which are totally obvious (to
me, at least), but any discussion of such things should probably be
taken off line.  I am curious if our moderator is going to let through
such an off topic message, anyway.)

But by contrast, my personal experience is that graphics cube
manipulation programs have less charm than the real thing.  There is
just something nice about the feel of the thing in your hands, and in
its obvious 3-D solidness. 

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990




From cube-lovers-errors@oolong.camellia.org  Wed Jun  4 17:30:26 1997
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Date: Wed, 04 Jun 1997 18:28:06 +0200
From: Herbert Kociemba <kociemba@hrz1.hrz.th-darmstadt.de>
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To: Dan Hoey <Hoey@aic.nrl.navy.mil>, cube-lovers@ai.mit.edu
Subject: Re: Detailed explanation of two phase algorithm
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Dan Hoey wrote:
> 
> > ...If you analyze the preceeding phase1 algorithm you will see that
> > it is indeed just an IDA* with lowerbound heuristic functions based
> > on tables.
> 
> It sure is.  Now here's a question.  Should it be combined with phase2
> in such a way that the entire search is IDA*?
> 
> The way I would see this happening is that whenever you get to phase 2
> (at depth i with a cutoff of L1), instead of allowing the phase2
> search to iteratitively deepen, you run a depth-first A* search with a
> fixed depth cutoff of L1-i.  It would take longer to get the first
> answer, but when you got an answer it would be optimal, and you would
> never spend time searching longer processes.
>
I try to understand but I can't grasp the idea. Maybe I did not describe
clearly enough, how phase1 and phase2 work together.Doing the entire
search in one single IDA* will result inevitably in a program, which
applys IDA* to the whole cube group in my opinion. Rich Korf did this
now. I could not do this (though I would have liked to do) due to a lack
of the appropriate hardware. So I had to apply it to the much smaller
groups G2=<U,D,R2,L2,F2,B2> in phase2 and G1=G/G2 in phase1.
Is the "i" you use above the same "i" I used in my explanation? There we
have i=0 every time we enter phase2. Beyond that I can't see how to
guarantee optimal solutions whith a phase2 at all, because we only allow
R2,L2,F2 and B2 moves there.
Why does the algorithm work so surprisingly well? Let us assume, that
every time we enter phase2 while the algorithm is running, the state is
a random sample of G2. From the archives (7 Jan 95), where Michael Reid
computed the distances of the elements of G2 from SOLVED we compute for
example, that the chance to get a state with distance less than 9 in
phase2 is about 2.6*10^-4. But in phase1 even with a very modest
iteration depth we already produce very very many solutions for G2.
Today I applied my algorithm to Rich Korfs Random Cube Nr.10 with
minimum length 18 for example, and within a minute I had a solution with
20 moves, 12 in phase1 and 8 in phase2. The next solution took about 15
minutes and gave a 19 move solution with 14 in phase1 and 5 in phase2. I
did not try for more than an hour, in this time no 18 move solution
appeared.

> One final thing, which I'm not sure ever got asked, much less
> answered, is that Mike Reid did an exhaustive search of the subgroup
> <T,D,F2,R2,B2,L2> (7 Jan 1995).  Did this verify that the optimal
> face-turn process for each element of the subgroup is a word on those
> generators?  Or are there shortcuts that use forbidden quarter-turns?
> 
There definitely are shortcuts with quarter turns. I just tried the
first of the antipodes of phase2 Mike Reid gave (7 Jan 1995) with 18
moves. They usually are hard to solve with the algorithm, but because of
the asymmetrie of stage2, conjugation with moves, that turn the whole
cube lead to a much easier to solve state. Within less than a minute a
had  the generator B R2 U2 L2 R2 B2 F' . U' R2 U F' D2 R2 B' D F' D' 
(17).
It would be interesting, if it is possible to reduce all of the
antipodes of phase2 to 17 moves, which would reduce algorithms upper
bound for the maneuver length.

Herbert



From cube-lovers-errors@oolong.camellia.org  Thu Jun  5 00:34:59 1997
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Date: Thu, 5 Jun 1997 00:28:40 -0400
From: michael reid <reid@math.brown.edu>
To: Hoey@aic.nrl.navy.mil, cube-lovers@ai.mit.edu
Subject: correction of priority

dan writes

> Herbert Kociemba notes three interesting heuristics based on the
> number of moves to reach the subgroup <T,D,R2,L2,F2,B2>.  In fact,
> Mike Reid calculated (and Dik Winter verified) the exact distances in
> this 2.2-billion element coset space (see archives at 7 Jan 1995 and
> following).

these distances (in the face turn metric) were originally calculated
by dik winter.  see his message of may 28, 1992, "Corrected calculations
are now done."  i mentioned this in my message of january 7, 1995.

mike


From cube-lovers-errors@oolong.camellia.org  Thu Jun  5 01:18:40 1997
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Date: Thu, 5 Jun 1997 00:55:43 -0400
From: michael reid <reid@math.brown.edu>
To: Dik.Winter@cwi.nl, cube-lovers@ai.mit.edu
Subject: Re: More on Korf's method

dik writes

>  > But we _can_ say there's at most one chance in 1024 that the first ten
>  > random cubes you pick will all be closer than the median to solved.
>  > So this demonstrates Rich's claim that the median optimal solution is
>  > very likely 18f.
> 
> Something I did estimate already a long time ago.  I have done a few
> hundred random cubes (a few thousand?  I do no longer remember) back
> so many years ago.  As I remember, I let the program look for optimal
> solutions upto 18f (longer is a bit time consuming).  As I remember,
> there were only very few that could *not* be solved in 18f.  There must
> be a discussion about this in the archives.

this is not quite right.  i consulted the archives and found dik's
message of august 3, 1993 "Diameter of cube group?"  the details are
as follows:
he tested 9000 random cube positions using kociemba's algorithm and
found that they were all within 20 face turns of start.  (this took
two months of computer time.)  however, he was not so interested in
finding _optimal_ solutions, but instead was satisfied with a maneuver
of length <= 20f.

this seems to be the most fundamental difference between kociemba's
algorithm and korf's method: kociemba is interested in sub-optimal
solutions (optimal solutions are ok, too), whereas korf has no interest
in sub-optimal solutions.

dik says in that message that he generated random cubes by taking
random sequences of 100 face turns, which is the same as korf did.
this is probably adequate randomness; however, i would do things
differently: first generate a random permutation of the edges, then
a random permutation of the corners, with the same parity, then
random flips for the edges and random twists for the corners.
in reality, it probably doesn't make any difference.  but i would
choose the latter method as a matter of principle.  this is just
my own philosophy.  (i think dan hoey recently expressed similar
sentiments.)

mike


From cube-lovers-errors@oolong.camellia.org  Thu Jun  5 01:28:40 1997
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From: michael reid <reid@math.brown.edu>
To: Hoey@aic.nrl.navy.mil, cube-lovers@ai.mit.edu,
        kociemba@hrz1.hrz.th-darmstadt.de
Subject: Re: Detailed explanation of two phase algorithm

dan writes

> One final thing, which I'm not sure ever got asked, much less
> answered, is that Mike Reid did an exhaustive search of the subgroup
> <T,D,F2,R2,B2,L2> (7 Jan 1995).  Did this verify that the optimal
> face-turn process for each element of the subgroup is a word on those
> generators?  Or are there shortcuts that use forbidden quarter-turns?

i guess i didn't ask this explicitly, but i certainly thought about it.
there's no reason to preclude the existence of such shortcuts.
i posted the antipodes for this subgroup, so anyone who wants to search
for shortcuts can do so.  in fact, herbert gives a shortcut for the
first position.

herbert replies

) There definitely are shortcuts with quarter turns. I just tried the
) first of the antipodes of phase2 Mike Reid gave (7 Jan 1995) with 18
) moves. They usually are hard to solve with the algorithm, but because of
) the asymmetrie of stage2, conjugation with moves, that turn the whole
) cube lead to a much easier to solve state. Within less than a minute a
) had  the generator B R2 U2 L2 R2 B2 F' . U' R2 U F' D2 R2 B' D F' D'
) (17).
) It would be interesting, if it is possible to reduce all of the
) antipodes of phase2 to 17 moves, which would reduce algorithms upper
) bound for the maneuver length.

yes, it would be interesting, but it would not improve the upper bound.
recall that i already gave the upper bound of 29 face turns.  to do
this, one must verify that the antipodes in stage 2 can be avoided by
choosing the last turn in stage 1 properly.  specifically, if
sequence R'  reduces the position to the intermediate subgroup
<U,D,F2,R2,B2,L2> , then so does  sequence R .  the last face turn
in stage 1 is always a quarter turn of either F, L, B or R.  so we
can always change the direction of this quarter turn, if necessary.
my program verified that by making the proper choice, we can avoid the
positions at distance 18f.

however, the same approach can probably improve the upper bound in
the quarter turn metric (currently 42q).  here the antipodes in stage 2
are at distance 30q.  in fact, the diameter of the whole cube group
is probably less than that (24q ???), so most of the positions at large
distance have shortcuts.  there are a lot of these positions to test,
and finding shortcuts isn't always easy.

this is all discussed in my message "new upper bounds" of jan 7 1995.

mike


From cube-lovers-errors@oolong.camellia.org  Thu Jun  5 16:09:12 1997
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From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: More on Korf's method
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On Thu, 5 Jun 1997, michael reid wrote:

> this seems to be the most fundamental difference between kociemba's
> algorithm and korf's method: kociemba is interested in sub-optimal
> solutions (optimal solutions are ok, too), whereas korf has no interest
> in sub-optimal solutions.

Good explanation, but I guess I still am unclear on one point.  It seems
that Kociemba's algorithm finds sub-optimal solutions which are either
very close to optimal (or may actually be optimal -- by accident, as it
were), and finds them very quickly.  It also seems that Kociemba's
algorithm will eventually find optimal solutions if it runs long enough,
but "long enough" may be a long time.  I think that I am hearing that
"long enough" means that phase1 has essentially subsumed phase2 to the
point that phase2 contains no moves.  Is this correct -- that is, does
the Kociemba algorithm guarantee us an optimal solution only after the
solution is derived entirely in phase1?  If not, at what point does the
algorithm itself guarantee an optimal solution? 

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990



From cube-lovers-errors@oolong.camellia.org  Thu Jun  5 16:29:27 1997
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From: Wei-Hwa Huang <whuang@ugcs.caltech.edu>
Subject: Re: 5x5x5 practical Q's
Date: 5 Jun 1997 19:48:39 GMT
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Yes, I was part of the Freecell Project, and it was indeed achieved
by many people working in parallel.

The rest of this mail will probably be uninteresting to the
advanced Cube-lovers on this list.  "Cubism for Dummies"?  

David Barr <davidb@u.washington.edu> writes:
>I don't understand what ARA'R' means.

This is basically the commutator.

A and A' represent a sequence of moves and the inverse of said sequence.
R and R' represent a sequence of moves and the inverse of said sequence.

The general technique is to discover sequences A and R that
intersect in as few pieces as possible, then the sequence ARA'R'
will only involve those pieces.  I.e., suppose sequence A does:
  (1) an action J on set X and
  (2) an action K on set Y.
Suppose sequence R does:
  (3) an action L on set Y and
  (4) an action M on set Z.
X, Y, and Z are disjoint; so, Y is defined as the intersection
of the two sets that A and R operate on.
Ao, the sequence ARA'R' becomes
  (1) actions J and J' on set X.
  (2) actions K, L, K', L' on set Y.
  (3) actions M and M' on set Z.
The actions on X and Z cancel out, and you end up with KLK'L' on set Y.
If set Y is small, K and L are necessarily simple.

Here's an example (take out your cube now):

I use this following sequence to flip two edge cubies on the same slice:

1.  Position the cube so that the edge cubies are at FL and FR.
2.  RF'UR'F
3.  Move the center slice to the right.
4.  F'RU'FR'
5.  Move the center slice to the left.

The two edge cubies are now flipped.

In this example, A = RF'UR'F, and R (sorry for the redundant notation,
hope it doesn't confuse anyone) = Moving the center slice.

If you watch the center slice closely during RF'UR'F, you'll 
notice that all that sequence does is to flip the FR cubie
while maintaining the rest of the center slice.

It also creates a lot of junk on the top and bottom faces, 
but since the R sequence only involves the center slice,
that junk gets restored in step 4, when we reverse RF'UR'F.

To reiterate, our steps are:
A  : flip an edge cubie, creating some junk
R  : move the center slice so that another edge is there
A' : remove that junk, flipping another edge cubie you just positioned
R' : move that center slice back.


It is my feeling that a algorithm consisting only of a few basic
moves and the ARA'R' technique is the most elegant algorithm of all.
(Sorry, God.)

Parity of the face pieces aside, the algorithms I use to solve the
3x3x3, the 4x4x4, and the 5x5x5 are almost completely based on
this method.  They appear very impressive to an onlooker; because
of the complexity of A and R, it often appears as if I am messing
up the cube, then bringing it back to a slightly more ordered
state, then bringing it back to chaos, then ordering it again,
etc., etc,. until the cube is solved.  The order in which I
solve the cube (corners, paired edges, corner faces, edge faces,
center edges, corner units, face centers) contributes to 
this effect.

(Similarly, the most impressive way of assembling a jigsaw puzzle
is to look at each piece one by one and place each one where it
belongs, instead of isolating the edge pieces first, etc.)
--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
-------------------------------------------------------------------------------
Question everything.  Learn something.  Answer nothing.  -- Engineer's Motto


From cube-lovers-errors@oolong.camellia.org  Thu Jun  5 18:43:26 1997
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From: Josh D Weaver <shaggy34@juno.com>
Date: Thu, 05 Jun 1997 18:25:04 EDT

Does anyone know more about the following method?

	1. do top corners. I always start with white. (intuitive)
	2. do bottom corners.
	        2a. bring bottom corner color onto bottom face (one of 2
patterns)
	        2b. orient bottom corners with each other (one of 2
patterns)
	3. fill in all but one edge on top and bottom (intuitive)
	4. fill in last edge (pattern)
	5. solve middle ring of edges (usually 2 patterns)

I received it from someone off the mailing list but they didn't know
where I could find more info on this pattern.  If anyone can point me to
a url that would be helpful, or just explain it directly to me.

Josh


From cube-lovers-errors@oolong.camellia.org  Thu Jun  5 22:52:37 1997
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From: SCHMIDTG@iccgcc.cle.ab.com
Date: Thu, 5 Jun 1997 19:32:06 -0400 (EDT)
To: cube-lovers@ai.mit.edu
Message-Id: <970605193206.214149bd@iccgcc.cle.ab.com>
Subject: Re: Detailed explanation of two phase algorithm

Herbert Kociemba wrote:

>Reading the many contributions in the mailing list in the last days, I
>state, that the insight im my two phase algorithm solving the cube
>ranges from misunderstood to partly understood, so I will add some more
>really detailed explanation here.

Well I must account for most, if not at least some, of the misunderstanding.
Although I wasn't quite certain, I was previously under the impression
that since Korf's tables were much larger and took longer to generate that
they were somehow "better".  I see now that the nature of the tables depends
almost entirely upon the particular subgroup that is being searched and so
Korf's tables may not be totally applicable to the specific subgroups
utilized in your approach.

>The memory requirements for the search algorithm are of the order
>O(d*log b), where b is the branching factor and d is the solution depth,
>so it definitely is not breadth-first search with O(b^d) nor is it
>bidirectional search with O(b^d/2).

Actually, when I mentioned O(b^d/2) for bidirectional search I was
referring to time complexity as opposed to space complexity.  It still
strikes me that the two phase search is a form of bidirectional search
where one searches from both ends and the two solutions must "meet"
in the middle.  I suppose it depends upon how strict one wants to
interpret the definition of bidirectional search.

>If you analyze the preceeding phase1 algorithm you will see that it is
>indeed just an IDA* with lowerbound heuristic functions based on tables.

I do not believe your phase1 is *exactly* IDA* as I think there is a
subtle difference.  IDA* limits search depth based on reaching a
cost threshold whereas phase1 simply iterates uniformaly at depth
1, 2, ... N pruning nodes within the bounds of the current search depth.
At the start of a search, IDA* consults its heuristic function to determine
the initial threshold.  As IDA* examines nodes, it keeps track of the
minimum cost of any node that exceeded the current threshold.  It uses
that as the cost threshold for the next repeated search.  So IDA* could
conceivably choose 2, 4, 5, 7, .. N for a sequence of cost threshold's
(as opposed to 1, 2, ... N) during the search.

It is this aspect of IDA* combined with the notion of an admissible
heuristic that guarantees that the first solution IDA* finds is optimal.
(Granted, you are doing a two phase search and optimality of phase1 does
not guarantee optimality of the combined solution).  I agree that your
overall search is guaranteed to eventually find the combined optimal
since it iterates through all possible depth combinations.

I don't know to what extent, if at all, this difference is signficant.  Of
course if it turns out that in phase1, one always happens to exceed the
current threshold by 1 for the cube problem, then I think the two algorithms
would effectively behave identically for this problem.  I don't know if this
is in fact the case.  But I would say that an initial depth limit computed
from your pruning tables from the initial cube state would start you off
searching at a depth greater than one with no loss of optimality.


>Herbert

Thank you for the detailed explanation of your algorithm.

-- Greg


From cube-lovers-errors@oolong.camellia.org  Thu Jun  5 22:52:52 1997
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Date: Thu, 05 Jun 1997 22:50:09 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Some Face Turn Numbers
To: Cube-Lovers <cube-lovers@ai.mit.edu>
Reply-to: Jerry Bryan <jbryan@pstcc.cc.tn.us>
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The fact that 10% of Rich Korf's random sample was 16f from Start seemed a
little funny, so I did some calculations.  At most, about 2.9% of the
positions in G are 16f from Start.  I am sure that the problem is no more
than that the sample size is very small.
 
Here is the calculation of the 2.9% figure.  The following table gives the
best known results for face turns.  The results through depth 7 have been
calculated (my message of 19 July 1994).  The rest are based on Dan Hoey's
recursion formula PH[n] = 6*2*PH[n-1] + 9*2*PH[n-2] for n>2, where PH[n]
is the number of face turns which are n moves from Start (Dan's note of 16
Sep 1981).  I think it would have been ok to use a branching factor of
13.231 from depth 8 on, but just to be safe I used Dan's formula (the
results are essentially the same either way).  Hence, we have PH[16] <=
1.47E+18, which is just about 2.9% of |G|. 


  d     #       b     Sigma #

  0         1                1
  1        18 18            19
  2       243 13.500       262  
  3      3240 13.333      3502 
  4     43239 13.345     46741
  5    574908 13.296    621649
  6   7618438 13.252   8240087
  7 100803036 13.231 109043123
  8  1.35E+09 13.360  1.46E+09
  9  1.80E+10 13.347  1.94E+10
 10  2.40E+11 13.349  2.59E+11
 11  3.20E+12 13.348  3.46E+12
 12  4.28E+13 13.348  4.62E+13
 13  5.71E+14 13.348  6.17E+14
 14  7.62E+15 13.348  8.24E+15
 15  1.02E+17 13.348  1.10E+17
 16  1.36E+18 13.348  1.47E+18
 17  1.81E+19 13.348  1.96E+19
 18  2.42E+20 13.348  2.61E+20

Now, I am going to do something strange.  I am going to assume as per
Rich's results that the branching factor from 16f to 17f is 3, and from
17f to 18f is 2 (Rich found 1 position at 16f, 3 at 17f, and 6 at 18f).
Doing so yields the following unhappy result (the total positions are less
than |G| which is about 4.3E+19).

 17  4.07E+18  3.000  5.54E+18
 18  8.14E+18  2.000  1.37E+19

If we assume the 16f position was an accident (occurred more than three
times too often, which is not surprising with the small sample size), we
can suppose the branching factor does not break until going from 17f to
18f, and we get the following. 
				        	
 17  1.81E+19 13.348  1.96E+19
 18  3.62E+19  2.000  4.23E+20

It's just totally a wild guess, but I would suspect that the correct
numbers are closer to the following because I don't think the branching
factor will collapse all at one depth.

 17  6.79E+18  5.000  8.25E+18
 18  3.39E+19  5.000  4.22E+19

I am a little shy of |G| with this guess, but I am not too far off.  What
we need is a larger sample. 

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990



From cube-lovers-errors@oolong.camellia.org  Thu Jun  5 22:58:32 1997
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Date: Thu, 5 Jun 1997 22:56:56 -0400 (EDT)
To: cube-lovers@ai.mit.edu
Message-Id: <970605225656.21412b24@iccgcc.cle.ab.com>
Subject: Categorization of cube solving programs

Cube Lovers,

Since I'm interested in such things, I came up with the following
categories of cube solving programs in general order of increasing
sophistication:

Class 1:	Simply provide a simulation of the cube and allow the
		user to manipulate the cube model via some cube
		operator notation (e.g. Singmaster).  May also allow
		the user to save/restore/replay lists of moves and
		possibly to build up sequences of "macro operators".
		Might allow the user to input a given cube state.
		Might also allow the user to 'solve' the cube by
		replaying all user made moves in reverse.  Often
		these programs have very nice 3D graphics.

Class 2:	A program which solves the cube by implementing a
		canned algorithm (or 'book procedure').  Basically,
		a straight forward implementation of a known
		cube solving process.  These programs generally
		work through a series of stages to solve the cube
		and do not generally produce optimal solutions.

Class 3:	A program that when given a specific instance of the
		cube, attempts to 'discover' or learn a sequence which
		will solve that particular instance.  That sequence is
		not usually considered general purpose since it will only
		apply to solving from the given position.  It may be
		possible to find a minimal sequence tailored to the
		given position.  The Kociemba and the recent Korf
		program fall into this category.  These programs
		are capable of discovering optimal or near optimal
		solutions to a given cube instance.

Class 4:	A program which attempts to discover an ALGORITHM to
		solve ALL randomized cubes.  The program starts off only
		with a model of the cube and attempts to discover a general
		procedure which solves all permutations of the cube.
		Korf wrote a program to do this in the mid 1980s.
		The program was able to learn a complete set of
		sequences (a.k.a. 'macros') sufficient to solve any
		scrambled cube.  The resulting algorithm is very
		much like a Class 2 algorithm as it works in stages
		to solve the cube and does not generally produce
		optimal solutions.  I believe Korf's program is
		the only program ever achieved that can be placed
		in this category.

Needless to say, I find Class 3 and Class 4 cube programs the most
interesting.  Someday, I would like to develop a Class 4 program
of my own that works on principles that differ from the Korf
program.

-- Greg


From cube-lovers-errors@oolong.camellia.org  Fri Jun  6 01:56:45 1997
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Date: Fri, 06 Jun 1997 01:13:57 +0200
From: Herbert Kociemba <kociemba@hrz1.hrz.th-darmstadt.de>
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To: cube-lovers@ai.mit.edu, Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: More on Korf's method
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Jerry Bryan wrote:
> Is this correct -- that is, does
> the Kociemba algorithm guarantee us an optimal solution only after the
> solution is derived entirely in phase1? 

It's like you supposed. I see no way, how to *guarantee* an optimal
solution before that, though in many cases you *find* an optimal
solution much earlier. You may have more than one minimum maneuver, and
the chance is good, that one of these maneuvers has a tail with
maneuvers only out of <U,D,R2,L2,F2,B2>.

Herbert



From cube-lovers-errors@oolong.camellia.org  Fri Jun  6 11:47:18 1997
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Where are you able to find 5x5x5 cubes that don't instantly fall apart? 
I have had nothing but trouble with mine.  Is there more than one on the
market?  I am only aware of the one manufactured in Germany.  Not only
have I never been able to solve it, but I have never been able to
scramble it.  It simply crumbles away in my hands.  The orange stickers
seem to have a habit of fleeing the cube in terror.  (It's always the
orange ones on any cube that fall off first.  Has anyone else noticed
this?)  At any rate, if there is some other source for these cubes or if
I simply got a bad one, someone please let me know.  Wei-Hwa Huang's
post has made me want to pick it up and try it out, but so far, mine is
only for looks.

Joe



From cube-lovers-errors@oolong.camellia.org  Fri Jun  6 11:48:04 1997
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On Thu, 5 Jun 1997 SCHMIDTG@iccgcc.cle.ab.com wrote:

> Cube Lovers,
> Class 4:	A program which attempts to discover an ALGORITHM to
> 		solve ALL randomized cubes.  The program starts off only
> 		with a model of the cube and attempts to discover a general
> 		procedure which solves all permutations of the cube.
> 		Korf wrote a program to do this in the mid 1980s.
> 		The program was able to learn a complete set of
> 		sequences (a.k.a. 'macros') sufficient to solve any
> 		scrambled cube.  The resulting algorithm is very
> 		much like a Class 2 algorithm as it works in stages
> 		to solve the cube and does not generally produce
> 		optimal solutions.  I believe Korf's program is
> 		the only program ever achieved that can be placed
> 		in this category.
> 
Hello Greg
I read Korf's work about macro learning and in particular his work about
the cube. Also I don't have a program that learn to solve the 3X3X3 cube I
have succeeded to write a program that learn to solve the 2X2X2 cube.
I have used the Micro-Hillary algorithm:
reference can be found in http://www.cs.technion.ac.il/~shaulm/
The basic problem in order to finish this work for the 3X3X3 cube was the
lack of a 'good' heuristic function. 
Maybe after the last achievements in the heuristic field this algorithm can
be used to solve the 3X3X3 cube.
Regards 
Shai
 




From cube-lovers-errors@oolong.camellia.org  Fri Jun  6 11:48:24 1997
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From: Allan Wechsler <awechsle@bbn.com>
To: SCHMIDTG@iccgcc.cle.ab.com
Cc: cube-lovers@ai.mit.edu
Subject: Categorization of cube solving programs
In-Reply-To: <970605225656.21412b24@iccgcc.cle.ab.com>
References: <970605225656.21412b24@iccgcc.cle.ab.com>

    [SCHMIDTG@iccgcc.cle.ab.com:]
    Class 4:	A program which attempts to discover an ALGORITHM to
    		solve ALL randomized cubes.  The program starts off only
    		with a model of the cube and attempts to discover a general
    		procedure which solves all permutations of the cube.

There's a classic procedure, the Furst-Hopcroft-Luks algorithm, that
can solve any permutation puzzle in polynomial time.  Here's a
citation I snarfed from the web.

Merrick Furst, John Hopcroft, and Eugene Luks. Polynomial-time 
    algorithms for permutation groups. In 21st Annual Symposium 
    on Foundations of Computer Science, pages 36-41, Syracuse, 
    New York, 13-15 October 1980. IEEE.

The solutions it finds are typically very long -- for the cube, it's
typically thousands of moves.  But that's really not too shabby -- the
first solution _I_ found was thousands of moves long too.

The algorithm is extremely mechanical.  It involves building a library
of tools that leave more and more of the pieces fixed, permuting fewer
and fewer of them.

-A


From cube-lovers-errors@oolong.camellia.org  Fri Jun  6 11:48:49 1997
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To: Cube-Lovers@AI.MIT.EDU
From: Wei-Hwa Huang <whuang@ugcs.caltech.edu>
Subject: Re: (none)
Date: 6 Jun 1997 14:01:26 GMT
Organization: California Institute of Technology, Pasadena
Message-ID: <5n957m$6dj@gap.cco.caltech.edu>
References: <cube-lovers.19970605.172331.12518.1.shaggy34@juno.com>
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Josh D Weaver <shaggy34@juno.com> writes:
>Does anyone know more about the following method?

My main method is a variant of this.

I'll switch your top and bottom because I like looking at the top:

>1. do bottom corners. I always start with white. (intuitive)
>2. do top corners.
>    2a. bring top corner color onto top face (one of 2 patterns)

I only use one pattern:
  RUR'URU2R'
It's very easy to remember.  (I only use easy-to-remember patterns.)
This pattern rotates three corners.  Technically, I also use the
inverse, which may count as another pattern:
  RU2R'U'RU'R'
Step 2a can be done with at most two applications of the pattern.

>    2b. orient top corners with each other (one of 2 patterns)
I only need one pattern:
  F2DF2D2R2DR2
This switches two adjacent pairs of corners on both top and bottom.
Step 2b can be done with at most two applications.

I sometimes use this pattern as a short cut:
  F2R2F2

(Technically, these patterns actually involve moving 
slices as well when I generalize them to 4x4x4 and up.)

>3. fill in all but one edge on top and bottom (intuitive)
>4. fill in last edge (pattern)

Actually, I fill in all but one edge on the top AND all
but one edge on the bottom.  Filling in the two
remaining ones at once is more intuitive than pattern.

>5. solve middle ring of edges (usually 2 patterns)

I used to do this with a pattern that permuted three edge pieces
while flipping two of them.  Now I do something more elegant, IMHO.

Let A denote shifting the middle "ring" slice to the right.
Then my first pattern is:
  F2AF2A'
which permutes three edge pieces.  Eventually they're
all positioned correctly, and I use this pattern to flip
two edge cubies:
  RF'UR'FA*F'RU'FR'A*
(* means to repeat as many times as is appropriate).

Sometimes in step 3  I don't bother with the orientation of
the last two edges.  This is because I can use this pattern:
  RARARARA
that is *really* easy to do and flips four edge pieces, three
on the middle slice and one not.

--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
-------------------------------------------------------------------------------
Question everything.  Learn something.  Answer nothing.  -- Engineer's Motto


From cube-lovers-errors@oolong.camellia.org  Fri Jun  6 21:25:13 1997
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In-Reply-To: Joe McGarity <joemcg3@snowcrest.net>
        "5x5x5 Stuctural Integrtity" (Jun  6,  1:16)
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On Jun 6,  1:16, Joe McGarity wrote:
> Subject: 5x5x5 Stuctural Integrtity
> Where are you able to find 5x5x5 cubes that don't instantly fall apart?
> I have had nothing but trouble with mine.  Is there more than one on the
> market?  I am only aware of the one manufactured in Germany.  Not only
> have I never been able to solve it, but I have never been able to
> scramble it.  It simply crumbles away in my hands.  The orange stickers
> seem to have a habit of fleeing the cube in terror.  (It's always the
> orange ones on any cube that fall off first.  Has anyone else noticed
> this?)  At any rate, if there is some other source for these cubes or if
> I simply got a bad one, someone please let me know.  Wei-Hwa Huang's
> post has made me want to pick it up and try it out, but so far, mine is
> only for looks.

I live in the Bay Area (California), and there's a mall with a toy store that
sells them from time to time.  That's were I got mine, and it's in excellent
shape.  I've abused and thrashed mine over and over again, and it's still all
"stuck" together; never had a piece break or fall out on me.

Let me know if you want more details on location.

	-Jason


--
Jason K. Werner                 Email: mrhip@sgi.com
Systems Administrator           Phone: 415-933-6397
USFO I/S Technical Services     Fax:   415-932-6397
Silicon Graphics, Inc.          Pager: 415-317-4084, mrhip_p@sgi.com

"Winning is a habit"-Vince Lombardi;"These go to eleven"-Nigel Tufnel


From cube-lovers-errors@oolong.camellia.org  Fri Jun  6 21:24:21 1997
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Subject: 5x5x5 Stuctural Integrtity


	The orange sticker problem seems to be with all of them.  Mine had
some small documentation with it mentioning that putting a piece of paper
on the orange side and ironing it a bit will help fix the stickers on the
cube.  It worked well for me and I've not had any problems with them
anymore.
	As to keeping it in one piece, pop off the caps of the center
pieces and tighten the screws underneath.  Adjust this tension to tight and
your cube won't move well, too loose and it will fall apart.  I had trouble
getting the caps off one of my 5x5x5s, but on the other they often fall off
on their own, good luck.

	Dave Litwin

Joe McGarity writes:
> Where are you able to find 5x5x5 cubes that don't instantly fall apart? 
> I have had nothing but trouble with mine.  Is there more than one on the
> market?  I am only aware of the one manufactured in Germany.  Not only
> have I never been able to solve it, but I have never been able to
> scramble it.  It simply crumbles away in my hands.  The orange stickers
> seem to have a habit of fleeing the cube in terror.  (It's always the
> orange ones on any cube that fall off first.  Has anyone else noticed
> this?)  At any rate, if there is some other source for these cubes or if
> I simply got a bad one, someone please let me know.  Wei-Hwa Huang's
> post has made me want to pick it up and try it out, but so far, mine is
> only for looks.


From cube-lovers-errors@oolong.camellia.org  Fri Jun  6 21:24:01 1997
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From: Herbert Kociemba <kociemba@hrz1.hrz.th-darmstadt.de>
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Subject: Re: Detailed explanation of two phase algorithm
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Greg wrote:
> 
> I do not believe your phase1 is *exactly* IDA* as I think there is a
> subtle difference.  IDA* limits search depth based on reaching a
> cost threshold whereas phase1 simply iterates uniformaly at depth
> 1, 2, ... N pruning nodes within the bounds of the current search depth.
> At the start of a search, IDA* consults its heuristic function to determine
> the initial threshold.  As IDA* examines nodes, it keeps track of the
> minimum cost of any node that exceeded the current threshold.  It uses
> that as the cost threshold for the next repeated search.  So IDA* could
> conceivably choose 2, 4, 5, 7, .. N for a sequence of cost threshold's
> (as opposed to 1, 2, ... N) during the search.
> .....
> 
> I don't know to what extent, if at all, this difference is signficant.  Of
> course if it turns out that in phase1, one always happens to exceed the
> current threshold by 1 for the cube problem, then I think the two algorithms
> would effectively behave identically for this problem.  I don't know if this
> is in fact the case.  But I would say that an initial depth limit computed
> from your pruning tables from the initial cube state would start you off
> searching at a depth greater than one with no loss of optimality.

I hardly can imagine a problem, where it makes any practical difference,
if you start with an initial treshhold determined by the heuristic
function for the initial cube state (let's denote it h0) or just start
with a treshhold of 1: In the latter case all depth 1 nodes will be
pruned immediately, and you generate exactly b*(h0-1) nodes, before you
start the search with treshhold h0, b denoting the branching factor.
Because h0<=9 in phase1 and b=18 for the first node, you generate at
most 162 nodes too much, which from a practical point of view is
nothing. In the general case, h0<=N, where N is the minimal solution
length, and you generate at most (N-1)*b nodes too much - so it really
makes no difference.

Does it make a difference if you increase the treshhold to the cost of
the lowest-cost node, that was pruned during the iteration or just
increase the treshhold by 1, if you start the next iteration? In case of
the cube, this question seems a bit academical. I can't believe, that it
is possible to omit a certain iteration depth >h0, though I must admit
that I found no obvious proof for that using the properties of the
heuristic functions in phase1 or phase2 (and it only depends on these
functions).

Herbert



From cube-lovers-errors@oolong.camellia.org  Sat Jun  7 15:57:59 1997
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From: SCHMIDTG@iccgcc.cle.ab.com
Date: Sat, 7 Jun 1997 1:50:04 -0400 (EDT)
To: cube-lovers@ai.mit.edu
Message-Id: <970607015004.21414d85@iccgcc.cle.ab.com>
Subject: Re: Detailed explanation of two phase algorithm

On the subject of differences between IDA* and the phase1 algorithm,
Herbert Kociemba wrote:

>Because h0<=9 in phase1 and b=18 for the first node, you generate at
>most 162 nodes too much, which from a practical point of view is
>nothing. In the general case, h0<=N, where N is the minimal solution
>length, and you generate at most (N-1)*b nodes too much - so it really
>makes no difference.

If one can show that the cost values have a property such that they will
prune all nodes at levels less then h0, then I would agree with your
assessment.  However, I do not see why that should necessarily be
the case as the costs examined at lower levels could be overly optimistic
with respect to h0.

For example.  Let's say that we are iterating through the search and
our current depth limit happens to be 5.  Now we examine the first node
at level one.  Your analysis above assumes that this node (and all others
at this level for that matter) will be pruned.  Now let's also say that
I now consult the pruning tables and compute an overly optimistic lower
bound cost of 3.  We now add the 3 to our current depth of 1 and since
3+1<5 we would not prune that node.  So if this can occur, I think one
is actually looking at evaluating b^(h0-1) additional nodes in the worst
case.

>Does it make a difference if you increase the treshhold to the cost of
>the lowest-cost node, that was pruned during the iteration or just
>increase the treshhold by 1, if you start the next iteration? In case of
>the cube, this question seems a bit academical. I can't believe, that it
>is possible to omit a certain iteration depth >h0, though I must admit
>that I found no obvious proof for that using the properties of the
>heuristic functions in phase1 or phase2 (and it only depends on these
>functions).

While I believe your phase1 algorithm is certainly ID (iterative deepening),
I do not believe it is A* since the depth limit is not based purely on
the cost.  Given an A* search algorithm, certain hard conclusions can
be proven (such as the guarantee that the first solution found is optimal
if an admissible heuristic is employed).  I don't know if these same
conclusions can be proven with the phase1 algorithm.  However, I agree
that from a purely practical standpoint and considering this particular
application of the algorithm to the cube problem this *may* not be an
important distinction.  But I don't think that conclusion has yet
been fully established.

-- Greg


From cube-lovers-errors@oolong.camellia.org  Sat Jun  7 16:01:19 1997
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From: Goyra <Goyra@cheerful.com>
To: cube-lovers@ai.mit.edu
Subject: Generalised notations?
Date: Sat, 7 Jun 1997 10:53:21 +0100
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	Hi.

	I've got a bunch of Java cubes and puzzles of
other shapes (deodecahedron, icosahedron, octahedron)
located at

	http://www.iol.ie/~goyra/Rubik.html

At the moment they've got only a visual interface. I
am looking at the possibility of giving you a way
to trace and replay movements using standard 
Cube notation.


	Trouble is, many of these shapes have never been
given a standard notation and I don't want to invent a 
crude one. Can anyone tell me where to find the
definitive work that has been done in the area of generalising 
the notation?

					David Byrden



From cube-lovers-errors@oolong.camellia.org  Sat Jun  7 16:13:35 1997
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Date: Sat, 07 Jun 1997 16:14:04 -0400
To: cube-lovers@ai.mit.edu
From: karen angelli <kangelli@tiac.net>
Subject: 5x5x5 practical Q's
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>From Peter Reitan, not Karen Angelli:

The 'parity' problem Mark Pilloff discussed in his 5x5x5 practical Q's
posting is a common frustrator when it comes to solving either the 4x4x4 or
5x5x5.  If you've ever read any of the on-line descriptions of Rubik's
Revenge solutions in various web pages, some of them suggest mixing the
cube back up and starting over, hoping to get the correct orientation 50%
of the time.  However, there is help available.

I happen to have a nifty algorithm that solves your problem.  However, I am
not fluent in cube-algebra notation, so I will try to explain it as well as
possible.  This algorithm is equally applicable to either the 4x4x4 or 5x5x5.

When I solve either cube, I solve opposite sides, corners, edges and center
pieces first.  My second goal is to correctly orient the edge cubies.  On
the 5x5x5, the center edge cubies is equivalent to solving the 3x3x3
edgies.  The 4x4x4 edgies is the same as solving the 5x5x5 non-center
edgies.  I have, and it sounds like Mark Pilloff has, moves that will pair
up the appropriate edgies together.  As he notes, half of the time, one
pair of the edgies are flipped into the wrong orientation.  To flip the
wayward edgie pair (w-edgies) back into line, I use the following method,
which I 'discovered' through brute force trial and error, and my faith in
cube symmetry.

Hold the cube so that the offending w-edgies are oriented horizontally, on
the top of the cube, and at the back of the cube.  Each individual edgie
cublet of the offending w-edgie pair belongs to its own vertical slice -
one to the left of center running vertically around the cube (L-slice) and
the slice to the right of center (R-slice) (these are the interior vertical
slices, not the right or left face slices).  The move involves only 180
twists of the right Face (T-Face) and 90 twists of the R-Slice, either
toward you or away from you.

1. 	a. 180 T-Face
	b. 90  R-Slice away from you
	c. 180 T-Face
	d. 90  R-Slice toward you
	e. Rotate the entire cube toward you 90(move the top front edge to the
bottom front edge.)

2.	Repeat 1.
3.	Repeat 1.
4.	Repeat 1, steps 1-3.

The parity problem is now more or less solved, but you have a couple more
steps to put the edgies back to their final resting place.

You will notice that two things have happened - 1. the center slices are
now rotated 90 degrees from the orientation of the right and left face
slices (you can fix that simply), and 2. the edgie pair whose parity
orientation you have corrected has now swapped places with one of the other
edge pairs, (the edge pair that started on the front top edge).  This can
be quickly corrected with the following algorithm (which you probably
already know):

Orient the cube so that the swapped edge pairs are on the top face, and one
of the pairs is at the front top edge and the other pair is at the back top
edge.

1. 	a. 180 T-Face
	b. 180 R-Slice
2. repeat 1
3. repeat 1
4. 	a. 180 T-Face

Since this method of arranging the outer, interior edge cubies on the 5x5x5
does not preserve the center edge cubies, I wait solve the center edge
cubies afterwards.  This is done in the same way as the 3x3x3.  Although
correct orientation of the center cube edgies can corrupt the two opposite
faces that I solved first, as a matter of style, I always like to have the
two opposite sides solved first.  That way, I solve a greater percentage of
the cube using intuitive moves, rather than with the more constrained
end-game moves where you need to preserve a greater portion of the cube
with each new solution.

Having written these moves down, I am struck by their similarity to each
other, and their similarity to the ARA'R' method Wei-Hua Huang describes
(with apologies to God) as the most elegant algorithm of all.  The core of
both of the algorithms I describe can be represented with ARA'R'.  

1. 	a. 180 T-Face				TF
	b. 90  R-Slice away			RS
	c. 180 T-Face				TF'
	d. 90  R-Slice toward		RS'
	e. Rotate cube 90			90

1. 	a. 180 T-Face				TF
	b. 180 R-Slice			RS
2. repeat 1					TF' 
						RS'

In addition, the first algorithm involves three iterations of one
algorithm, and then an incomplete core algorithm.  The second algorithm,
similarly, involves three repetitions of a core algorithm, followed by an
unfinished core algorithm.  

Step 1
Repeat 1
Repeat 1
Repeat most of 1

Not being a mathematician, I don't know what else to say.  Perhaps I'll
think about it a bit longer.  Any comments?

'e-ya later,
Pete




From cube-lovers-errors@oolong.camellia.org  Sat Jun  7 16:26:58 1997
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Date: Sat, 07 Jun 1997 16:28:33 -0400
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From: karen angelli <kangelli@tiac.net>
Subject: 5x5x5 Structural Integrity
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>From Peter Reitan, not karen Angelli

Joe McGarity has had some trouble with the structural integrity of his
5x5x5 cube (is there a more poetic name out there, in the same vein as
rubik's pocket, rubik's cube or rubik's revenge? 5x5x5 seems cumbersome to
me.  If not, let me suggest the "Big Cube").  I have had similar problems,
but I have managed to get past them, and can now freely twist my cube (at
least so far).

I bought my "Big Cube" from Puzzletts' on-line puzzle store.  I think that
their supplier is from Germany.  You can also purchase them directly from
Christoph Bandelow in Germany.  Like Joe's, my cube had several initial
problems - some of the orange stickers, paradoxically, did not want to
stick, and three of the center-piece caps fell off.  After several moments
of panic, I ran to my super glue supplier and took a few hits (no, I did
not sniff it).  Since then, the orange stickers have lived up to their
name, and the center-piece caps have stayed in place.  

I hope that these suggestions can get you on the right track.

'e-ya later,
Pete



From cube-lovers-errors@oolong.camellia.org  Sat Jun  7 22:28:39 1997
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Date: Sun, 08 Jun 1997 04:18:23 +0200
From: Herbert Kociemba <kociemba@hrz1.hrz.th-darmstadt.de>
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To: SCHMIDTG@iccgcc.cle.ab.com, cube-lovers@ai.mit.edu
Subject: Re: Detailed explanation of two phase algorithm
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SCHMIDTG@iccgcc.cle.ab.com wrote:
> 
> On the subject of differences between IDA* and the phase1 algorithm,
> Herbert Kociemba wrote:
> 
> >Because h0<=9 in phase1 and b=18 for the first node, you generate at
> >most 162 nodes too much, which from a practical point of view is
> >nothing. In the general case, h0<=N, where N is the minimal solution
> >length, and you generate at most (N-1)*b nodes too much - so it really
> >makes no difference.
> 
> If one can show that the cost values have a property such that they will
> prune all nodes at levels less then h0, then I would agree with your
> assessment.  However, I do not see why that should necessarily be
> the case as the costs examined at lower levels could be overly optimistic
> with respect to h0.

In phase1, the heuristic function is
h(x,y,z):=max{h1(x,y),h2(x,z),h3(y,z)}, where for example
h1(x,y):=min {lenght of maneuvers m with m(x,y,z)=(x0,y0,z0)|z<495},
and (x0,y0,z0) denotes the goal state.
>From this it follows, that |h(x,y,z)-h(x',y',z')| <=1, if (x',y',z') is
a state which is the result of a single move applied on (x,y,z).
In particular this is true for the initial state (x,y,z) and any
depth-one node (x',y',z'). The cost f of the depth-one node  is f=1 +
h(x',y',z'), and from the above we have h0:=h(x,y,z)<=1 + h(x',y',z')=f
When we now make an iteration with an iteration depth d and d<h0, we
have
d<h0<=f, but d<f means, that the depth-one node will be pruned, q.e.d.

> Given an A* search algorithm, certain hard conclusions can
> be proven (such as the guarantee that the first solution found is optimal
> if an admissible heuristic is employed).  I don't know if these same
> conclusions can be proven with the phase1 algorithm.

Obviously the first solution is optimal for phase1, because the
maneuverlength of a later solution cannot be smaller.

Herbert



From cube-lovers-errors@oolong.camellia.org  Sat Jun  7 22:38:09 1997
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Date: Sat, 07 Jun 1997 19:49:45 -0400
To: cube-lovers@ai.mit.edu
From: karen angelli <kangelli@tiac.net>
Subject: 5x5x5 practical Q's
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WARNING - My original posting contained a typo.  I said "right face" where
I meant to say "Top Face".
Corrected version is below.


Date: Sat, 07 Jun 1997 16:14:04 -0400
To: cube-lovers@ai.mit.edu
From: karen angelli <kangelli@tiac.net>
Subject: 5x5x5 practical Q's

>From Peter Reitan, not Karen Angelli:

The 'parity' problem Mark Pilloff discussed in his 5x5x5 practical Q's
posting is a common frustrator when it comes to solving either the 4x4x4 or
5x5x5.  If you've ever read any of the on-line descriptions of Rubik's
Revenge solutions in various web pages, some of them suggest mixing the
cube back up and starting over, hoping to get the correct orientation 50%
of the time.  However, there is help available.

I happen to have a nifty algorithm that solves your problem.  However, I
am not fluent in cube-algebra notation, so I will try to explain it as well
as possible.  This algorithm is equally applicable to either the 4x4x4 or
5x5x5.

When I solve either cube, I solve opposite sides, corners, edges and
center pieces first.  My second goal is to correctly orient the edge
cubies.  On the 5x5x5, the center edge cubies is equivalent to solving the
3x3x3 edgies.  The 4x4x4 edgies is the same as solving the 5x5x5 non-center
edgies.  I have, and it sounds like Mark Pilloff has, moves that will pair
up the appropriate edgies together.  As he notes, half of the time, one
pair of the edgies are flipped into the wrong orientation.  To flip the
wayward edgie pair (w-edgies) back into line, I use the following method,
which I 'discovered' through brute force trial and error, and my faith in
cube symmetry.

Hold the cube so that the offending w-edgies are oriented horizontally, on
the top of the cube, and at the back of the cube.  Each individual edgie
cublet of the offending w-edgie pair belongs to its own vertical slice -
one to the left of center running vertically around the cube (L-slice) and
the slice to the right of center (R-slice) (these are the interior vertical
slices, not the right or left face slices).  The move involves only 180
twists of the Top Face (T-Face) and 90 twists of the R-Slice, either toward
you or away from you.

1. 	a. 180 T-Face
	b. 90  R-Slice away from you
	c. 180 T-Face
	d. 90  R-Slice toward you
	e. Rotate the entire cube toward you 90(move the top front edge to the
bottom front edge.)

2.	Repeat 1.
3.	Repeat 1.
4.	Repeat 1, steps 1-3.

The parity problem is now more or less solved, but you have a couple more
steps to put the edgies back to their final resting place.

You will notice that two things have happened - 1. the center slices are
now rotated 90 degrees from the orientation of the right and left face
slices (you can fix that simply), and 2. the edgie pair whose parity
orientation you have corrected has now swapped places with one of the other
edge pairs, (the edge pair that started on the front top edge).  This can
be quickly corrected with the following algorithm (which you probably
already know):

Orient the cube so that the swapped edge pairs are on the top face, and
one of the pairs is at the front top edge and the other pair is at the back
top edge.

1. 	a. 180 T-Face
	b. 180 R-Slice
2. repeat 1
3. repeat 1
4. 	a. 180 T-Face

Since this method of arranging the outer, interior edge cubies on the
5x5x5 does not preserve the center edge cubies, I wait solve the center
edge cubies afterwards.  This is done in the same way as the 3x3x3.
Although correct orientation of the center cube edgies can corrupt the two
opposite faces that I solved first, as a matter of style, I always like to
have the two opposite sides solved first.  That way, I solve a greater
percentage of the cube using intuitive moves, rather than with the more
constrained end-game moves where you need to preserve a greater portion of
the cube with each new solution.

Having written these moves down, I am struck by their similarity to each
other, and their similarity to the ARA'R' method Wei-Hua Huang describes
(with apologies to God) as the most elegant algorithm of all.  The core of
both of the algorithms I describe can be represented with ARA'R'.  

1. 	a. 180 T-Face				TF
	b. 90  R-Slice away			RS
	c. 180 T-Face				TF'
	d. 90  R-Slice toward		RS'
	e. Rotate cube 90			90

1. 	a. 180 T-Face				TF
	b. 180 R-Slice			RS
2. repeat 1					TF' 
						RS'

In addition, the first algorithm involves three iterations of one
algorithm, and then an incomplete core algorithm.  The second algorithm,
similarly, involves three repetitions of a core algorithm, followed by an
unfinished core algorithm.  

Step 1
Repeat 1
Repeat 1
Repeat most of 1

Not being a mathematician, I don't know what else to say.  Perhaps I'll
think about it a bit longer.  Any comments?

'e-ya later,
Pete



[ The moderator lightly edited this message to make it easier to read.  ]

From cube-lovers-errors@oolong.camellia.org  Sun Jun  8 17:54:41 1997
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Date: Sun, 8 Jun 1997 10:31:06 -0400 (EDT)
From: Nicholas Bodley <nbodley@tiac.net>
To: Jerry Bryan <jbryan@pstcc.cc.tn.us>
cc: Tony Davie <ad@dcs.st-and.ac.uk>, cube-lovers@ai.mit.edu
Subject: Virtual cubes that you can feel
In-Reply-To: <Pine.WNT.3.96.970604085505.-409527D-100000@ER123A.PSTCC.CC.TN.US>
Message-ID: <Pine.SUN.3.95.970608101917.11440J-100000@sunspot.tiac.net>
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On Wed, 4 Jun 1997, Jerry Bryan wrote:

{Snips}
}
}But by contrast, my personal experience is that graphics cube
}manipulation programs have less charm than the real thing.  There is
}just something nice about the feel of the thing in your hands, and in
}its obvious 3-D solidness. 

+ + +

 In a recent issue of EE Times, the newspaper for EEs, was a short article
that points out that we are getting closer to having force-feedback
interfaces for our (personal) computers; these already exist in research
machines.

 In such interfaces, you can feel the virtual objects; there are electric
motors (or the equivalent) to generate computer-controlled forces
according to the positions of the various parts of the mechanism; combined
with data about the position and the location of the virtual object, the
motors turn on when you "make contact" with the virtual object.

 I see no reason why we couldn't define a virtual cube with on-screen
graphics to let us see it; we would then have the sensation of
manipulating it. There would be no need for a mechanism (sad to say!) to
hold the cubies together. (I love the innards...) 

 Of course, there would be a reset button, and provisions to preset any
arbitrary configuration. This would take some of the fun out of cube
manipulation, naturally. 

My best regards,

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*    When the year 2000 begins, we'll celebrate 
|*  Amateur musician  *|*    the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------



From cube-lovers-errors@oolong.camellia.org  Sun Jun  8 17:54:59 1997
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Date: Sun, 8 Jun 1997 11:31:51 -0400 (EDT)
From: Nicholas Bodley <nbodley@tiac.net>
To: Cube Mailing List <Cube-Lovers@ai.mit.edu>
Subject: Designations for the cubes (proposal)
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 Peter (Reitan, I think; sorry) (who is not Karen) brought up the
clumsiness of such designations as "5X5X5". I find these downright clumsy
to type (although the Caps Lock key helps). In my private world, I simply
refer to the Pocket Cube as "[the] two", the original Rubik's as "[the]
three", Revenge as "[the] four", and the biggest available as "[the]
five". 

 I think that provided we understand that we are referring to the
well-known family of true cubes, it should be OK simply to refer to "the
three", for instance. Granted, these names require more keystrokes, but
numerals should be OK, as in "the 3".

 There is some risk of being obscure; I feel that there are ways to deal
with that. We don't seem to have oodles of newcomers to the List every
day, who would need to be directed to FAQ.

 By any chance, is there more info. about a 6 or a 7? I last heard that a
prototype 6 had been built; I'd really love to know what the mechanism is
like. (I wrote about this at some length, a good number of months ago.)

My best regards,

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*    When the year 2000 begins, we'll celebrate 
|*  Amateur musician  *|*    the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------



From cube-lovers-errors@oolong.camellia.org  Sun Jun  8 17:54:20 1997
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Date: Sun, 8 Jun 1997 09:27:57 -0400 (EDT)
From: Nicholas Bodley <nbodley@tiac.net>
To: Cube Mailing List <Cube-Lovers@ai.mit.edu>
Subject: Special-purpose hardware for solving cubes, etc.
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 If I keep my facts straight, Deep Blue used quite a good supply of
special-purpose ICs ("chips"), designed specifically for searching ahead
in chess. Scanning the recent messages in the List about searches, as a
relatively-uninformed amateur, I wonder whether some special-purpose ICs
could be designed to help with some of the algorithms you are discussing.

 These days, there are some excellent software tools for designing
practical Application-Specific ICs (ASICs); while I have no specific
details to offer, it seems likely that part of the computational task
could be offloaded onto hardware. The recent emergence of single chips
that contain lots of memory and some processing should be considered, for
it is likely that such chips would be much more useful/powerful than those
with little or no memory. 

 By any chance, is IBM looking for new sub-worlds to conquer?

My best regards to all,

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*    When the year 2000 begins, we'll celebrate 
|*  Amateur musician  *|*    the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------



From cube-lovers-errors@oolong.camellia.org  Sun Jun  8 22:01:08 1997
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Date: Sun, 08 Jun 1997 18:25:09 -0400
From: Mathieu Girard <mgirard@videotron.ca>
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I am in a quest to buy both 4x4x4 Rubik's Revenge and also 5x5x5 cube..=20
but i just can't find any!=20

I live in Qu=E9bec, Canada, but i could order it by mail... even in
europe..

If u could please send me the complete geographical address or the Email
of a store where to buy those cubes... i would really appreciate... u
are my last resource...

By the way.. if it is not asking too much... could u please give me an
aproximate of the prices of thoses cubes... in canadian or us dollar...

Thank u very much...


From cube-lovers-errors@oolong.camellia.org  Sun Jun  8 22:01:29 1997
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From: SCHMIDTG@iccgcc.cle.ab.com
Date: Sun, 8 Jun 1997 19:31:31 -0400 (EDT)
To: cube-lovers@ai.mit.edu
Message-Id: <970608193131.21411978@iccgcc.cle.ab.com>
Subject: Re: Detailed explanation of two phas algorithm

Herbert Kociemba wrote:

>> If one can show that the cost values have a property such that they will
>> prune all nodes at levels less then h0, then I would agree with your
>> assessment.  However, I do not see why that should necessarily be
>> the case as the costs examined at lower levels could be overly optimistic
>> with respect to h0.
>
>In phase1, the heuristic function is
>h(x,y,z):=max{h1(x,y),h2(x,z),h3(y,z)}, where for example
>h1(x,y):=min {lenght of maneuvers m with m(x,y,z)=(x0,y0,z0)|z<495},
>and (x0,y0,z0) denotes the goal state.
>From this it follows, that |h(x,y,z)-h(x',y',z')| <=1, if (x',y',z') is
>a state which is the result of a single move applied on (x,y,z).

Let me try to add a bit more detail that I find helpful in this analysis.
Consider the following heuristic cost formula:

1.1	f[n] = g[n] + h[n]

Where:

f[n] is an estimate of the total path length (i.e. cost) for some
node n.
g[n] is the actual cost of the path to get to node n.
h[n] is an estimate of the cost of the path to get from node n to the
goal node (x0,y0,z0).

Let d be the current iteration depth and let D be the depth limit.  Since
g[n] = d for this problem we can slightly simplify 1.1 to:

1.2	f[n] = d + h[n]

Also since we're not interested in specific nodes, but rather all nodes
at a specific depth, let n in 1.2 represent "some node at depth n".

>In particular this is true for the initial state (x,y,z) and any
>depth-one node (x',y',z'). The cost f of the depth-one node  is f=1 +
>h(x',y',z'), and from the above we have h0:=h(x,y,z)<=1 + h(x',y',z')=f

The cost of the depth one node is:

1.3	f[1] = 1 + h[1]		where h[1] = h(x',y',z')

and our initial estimate is:

1.4	f[0] = 0 + h[0]

or simply

1.5	f[0] = h[0]

>When we now make an iteration with an iteration depth d and d<h0, we
>have

1.6	d<h[0]

and from 1.5

1.7	d<f[0]

>d<h0<=f, but d<f means, that the depth-one node will be pruned, q.e.d.

I don't see that as 1.7 does not seem sufficient to prune since we only
prune when:

1.8	d + h[d] > D

or

1.9	f[d] > D

but I don't think we have shown that.

And if we want to show that all depth one nodes will be pruned when
we are at some search depth d where 1 < d < h[0] we would need to show
that:

1.9	1 + h[1] > h[0]

for all nodes at depth 1 and I don't think we have shown that either.

It seems to me that the validity of 1.9 can only be determined by taking
into consideration properties of the particular heuristic function h[n]
that is used.  For any given admissible heuristic h[n], 1.9 will either
be true or false for all depth 1 nodes and I think one has to show this
based on properties of the heuristic function alone.  Have I completely
missed some important point? (please be patient with me if I have :) )

>> Given an A* search algorithm, certain hard conclusions can
>> be proven (such as the guarantee that the first solution found is optimal
>> if an admissible heuristic is employed).  I don't know if these same
>> conclusions can be proven with the phase1 algorithm.
>
>Obviously the first solution is optimal for phase1, because the
>maneuverlength of a later solution cannot be smaller.

Very true and for the same reason non-heuristic breadth first search always
finds optimal solutions first.  I should have clarified that my interest
in this thread is to point out differences between phase1 and IDA* whether
or not these differences are important to the cube problem.  I do agree
that phase1 is optimal in this particular application.  However, I do
not believe that it would necessarily be optimal for problems where there
is an indirect relationship between path cost and search depth whereas IDA*
would be optimal in that case.

graphically, this could be illustrated by the following trivial search
tree:

    (1)
    / \
  (2) (3*) cost = .9
  /
(4*) cost = .7

Suppose nodes 3 and nodes 4 were both solutions.  Even though node 4
has a lower cost, phase1 would find node 3 to be our first solution
whereas IDA* wouldn't.

With respect to your program, this is completely academic, but I think
it does point out a subtle, but important difference between the two
algorithms.  However, it might be important to someone wanting to apply
these algorithms to some other problem.  Would you grant me that?

Best regards,

-- Greg


From cube-lovers-errors@oolong.camellia.org  Mon Jun  9 12:25:03 1997
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From: Goyra <Goyra@cheerful.com>
To: cube-lovers@ai.mit.edu
Subject: Want Java cubes?
Date: Mon, 9 Jun 1997 10:42:52 +0100
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	I got an email from a list member
indicating that he would like to try my Java cubes and
polyhedra but has no suitable Internet access.

	If anyone else is in this position, I could email
the Java class files directly to them, and they could use them by
installing Sun's free Java Development Kit on their machine.
It's available for a few operating systems. Of course, without
Internet access, this begs the question of how you would get the 
JDK in the first place.

					David Byrden





From cube-lovers-errors@oolong.camellia.org  Mon Jun  9 19:04:20 1997
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Date: Mon, 09 Jun 1997 16:01:06 -0500
To: cube-lovers@ai.mit.edu
From: Tom Magliery <mag@ncsa.uiuc.edu>
Subject: Re: Designations for the cubes (proposal)
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At 11:31 AM 6/8/97 -0400, Nicholas Bodley unabashedly said:
>
> Peter (Reitan, I think; sorry) (who is not Karen) brought up the
>clumsiness of such designations as "5X5X5". I find these downright clumsy
>to type (although the Caps Lock key helps). In my private world, I simply
>refer to the Pocket Cube as "[the] two", the original Rubik's as "[the]
>three", Revenge as "[the] four", and the biggest available as "[the]
>five". 
>
> I think that provided we understand that we are referring to the
>well-known family of true cubes, it should be OK simply to refer to "the
>three", for instance. Granted, these names require more keystrokes, but
>numerals should be OK, as in "the 3".

I have another suggestion, which might be slightly less likely to require
explanation to a newcomer.  I know how to *pronounce* it, but I'm not sure
how I would recommend *spelling* it.  (Considerations include terseness,
ease of typing -- which is of course not the same thing!, and likeliness to
be mispronounced by a reader.)

The pronunciation is three-bye, four-bye, five-bye, ...

Possible spellings include:
	3by, 4by, 5by, ...
	3-by, 4-by, 5-by, ...
	3x, 4x, 5x, ...
	three-by, four-by, five-by, ...

mag
-- 
.---o  Tom Magliery, Research Programmer         (217) 333-3198  .---o
`-O-.  NCSA, 605 E. Springfield      O-       mag@ncsa.uiuc.edu  `-O-.
o---'  Champaign, IL 61820       http://sdg.ncsa.uiuc.edu/~mag/  o---'


From cube-lovers-errors@oolong.camellia.org  Mon Jun  9 19:03:57 1997
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Date: Mon, 09 Jun 1997 21:17:47 +0200
From: Herbert Kociemba <kociemba@hrz1.hrz.th-darmstadt.de>
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SCHMIDTG@iccgcc.cle.ab.com wrote:
>
> And if we want to show that all depth one nodes will be pruned when
> we are at some search depth d where 1 < d < h[0] we would need to show
> that:
> 
> 1.9     1 + h[1] > h[0]
> 

Why do you say 1 < d < h[0] and not  d = 1? What I wanted to show is,
that unter the assumption

2.0        D < h(0)

all depth-one nodes will be pruned. As you correctly stated before, for
pruning we need

1.8        d + h(d) > D

and in the case d=1 this means

1.9a       1 + h(1) > D, 

which is different from 1.9, because of 2.0 . 
But 1.9a can be shown easily:

In my last message, I tried to explane that

2.1        |h(n-1)-h(n)| <=1,

I try to explain it once more in other words. A node at depth n is
generated from a node at depth n-1 by applying a single face-turn on it.
And as I told, h is defined by

            h(x,y,z):=max{h1(x,y),h2(x,z),h3(y,z)},

where for example h1(x,y) is the length of the shortest maneuver
sequence which transforms (x,y,z) to (x0,y0,z') for any z' (this means
the z-coordinate is ignored). And this length can maximal change by one
when applying a single move. The same holds for h2(x,z) and h3(y,z). For
this reason, h(x,y,z) also can change maximal by one, which implies 2.1
.

In the case n=1, from 2.1 follows 

       h(0) <= 1 + h(1), and because of 2.0 we have

       D < h(0) <= 1 + h(1),

which proves 1.9a .


>     (1)
>     / \
>   (2) (3*) cost = .9
>   /
> (4*) cost = .7
> 
> Suppose nodes 3 and nodes 4 were both solutions.  Even though node 4
> has a lower cost, phase1 would find node 3 to be our first solution
> whereas IDA* wouldn't.

I don't think we are far away from each other. Of course, the phase1 (or
phase2) algorithm does not claim to be an universal IDA* for any sort of
problem. But for a special problem like the cube you can simplify the
general IDA* and the simplified algorithm will be equivalent to the one
I developed for phase1.

Best regards,

Herbert



From cube-lovers-errors@oolong.camellia.org  Mon Jun  9 21:30:42 1997
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Date: Tue, 10 Jun 1997 02:38:52 +0000
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Mathieu Girard wrote:

>I am in a quest to buy both 4x4x4 Rubik's Revenge and also 5x5x5 
>cube.. but i just can't find any!

Regrettably the 4x4x4 cubes seem to be sold out everywhere in the 
world.  With the 5x5x5 Magic Cubes (in Japan once sold under the name 
Professor's Cube), the situation is much better: They are still 
available from me (and as far as I know nowhere else now). Since they 
will probably never be produced again (the production cost is too 
high and the general interest too low), they will also be sold out 
soon. I shall inform the cube-lovers when this is the case.
My price of the 5x5x5 cube is 40 DM or 24 USD plus postage. 
I send my free mail order catalog (containing also many other 
twisting puzzles like the Magic Dodecahedron , the Skewb, the 
Pyraminx, Mickey's Chellenge and several books and details how to 
order) to  every cube-lover requesting it and providing a postal address. 

A few days ago, Joe McGarity complained bitterly about his 5x5x5 cube 
which fell apart. Fortunately, I did not encounter this problem 
before, and Joe  is not in my files so he has probably not bought his cube 
from me.  On the other hand you will destroy every twisting puzzle by 
twisting it with force without sufficient aligning the layers before 
every single move. Since  the 5x5x5 cube contains 98 visible little cubies 
compared to the 26 of the 3x3x3 (and 92 compared to 20 if we only 
count the freely floating ones), one should accept that it requires a 
little bit more care. 

Joe McGarity also mentioned that some orange stickers sometimes do 
not  behave according to there name. I have to admit that this 
sometimes also  happens with my 5x5x5 cubes. Furtunately 
it happens only to the orange stickers and it can be repaired easily by 
warm pressure or - better - some glue. 
 
Christoph
Christoph Bandelow
mailto:Christoph.Bandelow@rz.ruhr-uni-bochum.de


From cube-lovers-errors@oolong.camellia.org  Tue Jun 10 15:15:53 1997
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Date: Tue, 10 Jun 1997 08:16:53 -0400 (EDT)
From: Nicholas Bodley <nbodley@tiac.net>
To: Tom Magliery <mag@ncsa.uiuc.edu>
cc: cube-lovers@ai.mit.edu
Subject: Re: Designations for the cubes (proposal)
In-Reply-To: <3.0.1.32.19970609160106.00ade4a0@sdgmail.ncsa.uiuc.edu>
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 I like Tom's version; it is concise, distinctive enough not to be
confused (however, I haven't studied all the various math. notations going
around), and easy to say and type. Of the ones you gave, I prefer such a
form as "3-by". Thanks! 


|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*    When the year 2000 begins, we'll celebrate 
|*  Amateur musician  *|*    the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------



From cube-lovers-errors@oolong.camellia.org  Tue Jun 10 15:16:08 1997
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Date: Tue, 10 Jun 1997 13:07:15 -0400 (Eastern Daylight Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: Some Face Turn Numbers
In-reply-to: <Pine.PMDF.3.95.970605215908.177208C-100000@PSTCC6.PSTCC.CC.TN.US>
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On Thu, 5 Jun 1997, Jerry Bryan wrote:

>                                         The following table gives the
> best known results for face turns.  The results through depth 7 have been
> calculated (my message of 19 July 1994).  The rest are based on Dan Hoey's
> recursion formula PH[n] = 6*2*PH[n-1] + 9*2*PH[n-2] for n>2, where PH[n]
> is the number of face turns which are n moves from Start 

Rats, here is a little correction.  I think my meaning was clear from
the overall context of the note, but Dan's formula is an upper bound, so
it should read PH[n] <= 6*2*PH[n-1] + 9*2*PH[n-2] for n>2.  For depth 0
through 7, my table provided exact values.  As I hope was clear from the
context, my table included upper bounds rather than exact values for
depths greater than 7.  My apologies. 

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990



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To: "Mailing List, Rubik's Cube" <cube-lovers@ai.mit.edu>
Subject: 5x cubes
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Yes, the cube I purchased was not from Cristoph Bandelow.  He is off the
hook.  It was purchased in San Francisco at the Game Gallery.  Although
I am considering buying my next one from him.  And no, I don't force it
either.  I treated it with kindness and love, yet it betrayed me.




From cube-lovers-errors@oolong.camellia.org  Wed Jun 11 00:38:42 1997
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From: SCHMIDTG@iccgcc.cle.ab.com
Date: Wed, 11 Jun 1997 0:35:55 -0400 (EDT)
To: cube-lovers@ai.mit.edu
Message-Id: <970611003555.21417ec3@iccgcc.cle.ab.com>
Subject: Re: Detailed explanation of two phase algorithm

Herbert Kociemba wrote:

>SCHMIDTG@iccgcc.cle.ab.com wrote:
>>
>> And if we want to show that all depth one nodes will be pruned when
>> we are at some search depth d where 1 < d < h[0] we would need to show
>> that:
>> 
>> 1.9     1 + h[1] > h[0]
>> 
>
>Why do you say 1 < d < h[0] and not  d = 1?

Oops, I think that should have been 'D' and not 'd'.

>[...slightly different restatement of earlier proof omitted...]

After examining this once again, I have now satisfied myself
that it is correct.  It's just that for some reason, I seem to
find the result rather counter-intuitive.  But that makes the
result all the more interesting.

So I think this may yet me another case where the phase1 algorithm
differs slightly from IDA*, but the difference is not significant
since, in this case, one is able to prove a special property of the
heuristic that demonstrates that the number of nodes explored by the
two algorithms is comparable.  At this point, I think we can wind down
this thread, (I do hope others on this list have found it interesting)
and I will still continue to think of possible ideas for improving
the algorithm.

I do have one last question regarding the pruning tables.  While
the three tables used in phase1 are clear, I do not recall reading
a description of the tables that are used in phase2.

I examined Dik Winter's program and he seems to have a few more
"maximum move" (i.e. "mm" tables) than I expected, namely:

phase1
------
mm_twists[]
mm_flips[]
mm_choices[]
/* and the following "mixed" tables */
mm_tf[][]	/* twist & flip */
mm_tc[][]	/* twist & choice */
mm_fc[][]	/* flip & choice */

phase2
------
mm_eperms[]	/* edge perms */
mm_cperms[]	/* corner perms */
mm_sperms[]	/* slice orderliness */
/* "mixed" tables follow */
mm_cs[][]	/* corner perms & slice orderliness */
mm_es[][]	/* edge perms and slice orderliness */

Are you using the same tables?  Or are the "mixed" tables ones that Dik
added to the algorithm?  It appears that Dik was able to use them because
he had a machine with more memory at his disposal than your 1MB Atari ST.
His program can be built with or without the "mixed" tables and is 11MB
with them.  He also mentions that the small program finds a reasonable
solution in 30 minutes whereas the large program finds it in only a
few seconds.

I have also been studying his code to try to understand how he generates
these tables.  He does not seem to be using breadth-first-search to
fill in these tables as Korf does.

I will be interested in looking at your new program when it becomes
available.

Thanks again for your patience.

Best regards,

-- Greg


From cube-lovers-errors@oolong.camellia.org  Wed Jun 11 14:05:23 1997
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Date: Wed, 11 Jun 1997 08:39:17 -0700
From: "Jason K. Werner" <mrhip@corp.sgi.com>
Message-Id: <9706110839.ZM926@isdn-rubik.corp.sgi.com>
In-Reply-To: Joe McGarity <joemcg3@snowcrest.net>
        "5x cubes" (Jun  9, 19:15)
References: <339CB8B7.1B8D@snowcrest.net>
Reply-to: "Jason K. Werner" <mrhip@sgi.com>
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To: cube-lovers@ai.mit.edu, Mark Pilloff <mdp1@uclink4.berkeley.edu>,
        Mathieu Girard <mgirard@videotron.ca>, joemcg3@snowcrest.net
Subject: Re: 5x cubes
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Here's the place where I got my 5x cube, which has been extremely solid and
nearly indestructible:

	Game Gallery
	2855 Stevens Creek Blvd.
	Santa Clara, CA  95050
	USA
	408-241-4263
	408-241-5945 FAX
	http://www.gamegallery.com


From cube-lovers-errors@oolong.camellia.org  Wed Jun 11 14:04:50 1997
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Date: Wed, 11 Jun 1997 08:32:27 -0400 (EDT)
From: der Mouse <mouse@rodents.montreal.qc.ca>
Message-Id: <199706111232.IAA00315@Twig.Rodents.Montreal.QC.CA>
To: cube-lovers@ai.mit.edu
Cc: Mathieu Girard <mgirard@videotron.ca>
Subject: Re: ...
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> I am in a quest to buy both 4x4x4 Rubik's Revenge and also 5x5x5
> cube.. but i just can't find any!

> I live in Qu=E9bec, Canada,

Where?  Montr=E9al, or Qu=E9bec City, or what?  I'm in Montr=E9al; I got =
my
5-Cube at Valet de Coeur, on the west side of St-Denis, somewhere a bit
south of Mont-Royal.  I don't know whether they still have them; this
_was_ back in 1993 (December 15, according to my records).

> By the way.. if it is not asking too much... could u please give me
> an aproximate of the prices of thoses cubes... in canadian or us
> dollar...

I paid $45.01, including tax, for my 5-Cube.  But as I remarked above,
that _was_ three and a half years ago.

					der Mouse

			       mouse@rodents.montreal.qc.ca
		     7D C8 61 52 5D E7 2D 39  4E F1 31 3E E8 B3 27 4B


From cube-lovers-errors@oolong.camellia.org  Wed Jun 11 16:11:56 1997
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To: Cube-Lovers@AI.MIT.EDU
From: Wei-Hwa Huang <whuang@ugcs.caltech.edu>
Subject: Re: 5x5x5 Stuctural Integrtity
Date: 11 Jun 1997 19:01:30 GMT
Organization: California Institute of Technology, Pasadena
Message-ID: <5nmsma$t63@gap.cco.caltech.edu>
References: <cube-lovers.199706060927.FAA00399@dlitwinHome.geoworks.com>
NNTP-Posting-Host: blend.ugcs.caltech.edu
X-Newsreader: NN version 6.5.0 #2 (NOV)

David Litwin <dlitwin@emf.net> writes:
>	The orange sticker problem seems to be with all of them.  Mine had
>some small documentation with it mentioning that putting a piece of paper
>on the orange side and ironing it a bit will help fix the stickers on the
>cube.  It worked well for me and I've not had any problems with them
>anymore.

I must say that the first sticker I lost on my 5x5x5 was red.

(And I don't know where it went!  ARGH!!!)


--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
-------------------------------------------------------------------------------
Inspiration strikes suddenly, so be prepared to defend yourself.


From cube-lovers-errors@oolong.camellia.org  Wed Jun 11 16:11:24 1997
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Date: Wed, 11 Jun 1997 20:51:44 +0200
From: Herbert Kociemba <kociemba@hrz1.hrz.th-darmstadt.de>
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To: cube-lovers@ai.mit.edu
Subject: Re: Detailed explanation of two phase algorithm
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SCHMIDTG@iccgcc.cle.ab.com wrote:
> I do have one last question regarding the pruning tables.  While
> the three tables used in phase1 are clear, I do not recall reading
> a description of the tables that are used in phase2.

In phase2, the state of the cube also is described by a triple (x,y,z),
in this case 0<=x<8! describes a permutation of the 8 corners, 0<=y<8!
describes a permutation of the 8 UD-slice edges and 0<=z<4! describes a
permutation of the middleslice edges. Because the overall permutation
must be even, only half of the triples belong to physical cubes. We
could correct this, by defining the z coordinate to describe one of the
12 possibilities for the locations of two middleslice edges - the other
two edges will then be corrected automatic. But there are good reasons
not to do so (which I think is not necessary to explain here).

> I have also been studying his code to try to understand how he generates
> these tables.  He does not seem to be using breadth-first-search to
> fill in these tables as Korf does.
> 
I only use the "mixed" tables. How to generate the tables is quite
obvious and though I don't know how Dik does it I'm sure it is similar:

1. On initialisation set all elements of the table to 0xf (we use four
bits per entry), only the element belonging to (x0,y0,z0) is set to 0.
Set  L=0, n_done=1, n_old=1 (n_done denotes the number of valid
tableentries).
2. Check all elements of the table one after the other. If an entry is
0xf, do nothing. If the entry is L, compute the the 18 possible "child
nodes" and check, if the corresponding tableentry is 0xf. Only in this
case set it to L+1 and increment n_done.
3. Check if n_done=n_old. In this case we are ready. Else increment L,
set n_old=n_done and goto 2.

> I will be interested in looking at your new program when it becomes
> available.

I'm writing too much to this mailing list and do not work at  my
windows-help! The program itself is ready. I did a two hours run on each
of Rich Korfs 10 random cubes on a Pentium133 with 16MB RAM and the
result were really pleasing: The generated maneuver lenghts were on the
average less than 1 move away from Rich Korfs optimal solutions
(exactly: 9 moves more for the 10 cubes).

Best regards,

Herbert



From cube-lovers-errors@oolong.camellia.org  Wed Jun 11 20:44:06 1997
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From: SCHMIDTG@iccgcc.cle.ab.com
Date: Wed, 11 Jun 1997 20:42:11 -0400 (EDT)
To: cube-lovers@ai.mit.edu
Message-Id: <970611204211.2141971d@iccgcc.cle.ab.com>
Subject: Re: Detailed explanation of two phase algorithm

Herber Kociemba wrote:

[...much additional detailed explanation deleted...]

Thank again, I found this information helpful, especially when my only
other option is to examine code in great detail in order to extract out
the general principles.

>> I will be interested in looking at your new program when it becomes
>> available.
>
>I'm writing too much to this mailing list and do not work at  my
>windows-help!

Sorry about that.  I'll stop with my questions.  In fact, no need
to even answer this response!  I'm sure your program will be well
worth the wait :).

> The program itself is ready. I did a two hours run on each
>of Rich Korfs 10 random cubes on a Pentium133 with 16MB RAM and the
>result were really pleasing: The generated maneuver lenghts were on the
>average less than 1 move away from Rich Korfs optimal solutions
>(exactly: 9 moves more for the 10 cubes).

Very impressive.  And if you perform some longer runs and find optimal
solutions, please be sure to let us know the run times.

Best regards,

-- Greg


From cube-lovers-errors@oolong.camellia.org  Thu Jun 12 13:23:55 1997
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From: Nicholas Bodley <nbodley@tiac.net>
To: Wei-Hwa Huang <whuang@ugcs.caltech.edu>
cc: Cube-Lovers@ai.mit.edu
Subject: Re: 5x5x5 Structural Integrtity (Stickers)
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 One cure might be to take a solved cube (more convenient...) and remove
all the red stickers; carefully clean off the adhesive, perhaps with 99%
isopropyl alcohol, and paint the surfaces neatly with the type of paint
used for plastic model kits. I have also thought of removing the stickers,
cleaning all the adhesive off both the stickers and the cubies, and then
reattaching the stickers with a different type of adhesive.

 These are just ideas, and I hope no source of trouble.

 My best to all,

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*    When the year 2000 begins, we'll celebrate 
|*  Amateur musician  *|*    the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------



From cube-lovers-errors@oolong.camellia.org  Wed Jun 18 16:19:15 1997
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To: cube-lovers@ai.mit.edu
Date: Wed, 18 Jun 1997 15:05:54 -0500
Subject: Square One
Message-ID: <19970618.150557.11350.0.shaggy34@juno.com>
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From: Josh D Weaver <shaggy34@juno.com>

Does anyone know how to solve one of those "Square One" puzzles?

Josh


From cube-lovers-errors@oolong.camellia.org  Wed Jun 18 18:43:38 1997
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From: Herbert Kociemba <kociemba@hrz1.hrz.th-darmstadt.de>
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The Windows95 program which implements my algorithm to solve Rubik's
Cube is now availabe at

http://home.t-online.de/home/kociemba/cube.htm

It not only solves Rubik's cube, but also does a few other nice
things...


Herbert


[ Moderator's note: This program is also available in the Cube-Lovers Archive.
  See: ftp://ftp.ai.mit.edu/pub/cube-lovers/contrib/cubexp10.zip
				- Alan ]

From cube-lovers-errors@oolong.camellia.org  Thu Jun 19 01:16:47 1997
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From: Benjamin Wong <chi@cse.unsw.edu.au>
To: Josh D Weaver <shaggy34@juno.com>
Date: Thu, 19 Jun 1997 11:37:43 +1000 (EST)
X-Sender: chi@pipe02.orchestra.cse.unsw.EDU.AU
cc: cube-lovers@ai.mit.edu
Subject: Re: Square One
In-Reply-To: <19970618.150557.11350.0.shaggy34@juno.com>
Message-ID: <Pine.OSF.3.95.970619113450.27424A-100000@pipe02.orchestra.cse.unsw.EDU.AU>

On Wed, 18 Jun 1997, Josh D Weaver wrote:

._@_.Does anyone know how to solve one of those "Square One" puzzles?
._@_.


http://www.cfar.umd.edu/~arensb/Square1/
is the only page on the net (that i can find)
which describe how to solve square 1
however, either i can not follow instruction,
or error in it's instructuion
i just can not solve it with their algorimthm
I bought square 1 mess them up,
only manage to solve them 2 times. 
(beginner luck)
but the page does not help very much.


._@_.Josh
._@_.

o------------------------------------------------------o
|Error: Reality.sys Corrupt?  Reboot Universe [Y,N,Q]  |
+---------------o--------------------------------------o
| Benjamin Wong |  E-mail: chi@cse.unsw.edu.au         |
|  		|  or       benjaminwong@hotmail.com   |
| 		|  http://www.cse.unsw.edu.au/~chi     |
o---------------o--------------------------------------o
|=A1u=C2=E5=A5=CD=A1I=BD=D0=B0=DD=A1y=BA=B5=BF=DF=B2=B4=A1z=AA=BA=A6=A8=A6]=ACO=AC=C6=BB=F2=A1H=A1v                |
|Quick Quiz:  Describe Universe ? Give Three Example.  |
o------------------------------------------------------o



From cube-lovers-errors@oolong.camellia.org  Thu Jun 19 12:31:32 1997
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After three months of agony and wondering if Square One was actually the
puzzle box from Clive Barker's Hellraiser, I managed to come up with a
solution that covered all the bases, i.e. worked every time.  I have
never tried to write it out in a step by step form however.  I will try
to cover the basics.

First the puzzle looks like a Rubik's Cube when solved in that it is a
cube with a solid color on each face, but the similarity ends there. 
Square One more closely resembles the Orb, Masterball and Smart Alex in
the ways that it moves.  If you can solve any of those you will be a
step closer to the Square One.  I see the square one as nearly identicle
to the Smart Alex.  The shapes of the pieces are different, but they
move as a disk divided into sectors (exactly like the Masterball). 
There are six pieces on each face if you count the small sectors as half
pieces.  Count the pieces and you will see what I mean.  The little ones
are half the the size of the angle of the big ones.  The idea then for
me was to get the little ones paired up like the picture in the
instruction booklet.  Once they were paired correctly I could solve it
just like the Masterball or Smart Alex making it look like it did when
it was new in the package (remember it came in a slightly scrambled
state with instructions on how to solve it from there in about six
moves).  Then I could just follow the booklet for the final part.  Like
I said it took three months and scores of note paper to finally get it. 
When I did, the walls opened up and the Cenobites took me away, but it
was worth it.

I hope I haven't caused more confusion.  It is difficult to describe
without having one in my hands to show you.  This is just a sketchy
overview of how I solve it.  If I get a chance to document this solution
I will send you a copy, but it probably won't be for a while.  I'm sure
that someone has a better solution and I'd be interested in seeing what
others have come up with.  My solution takes about twenty minutes to do
and there must be a faster way.  The ones I have trouble with are the
Sqewb and the Alexander's Star.  Anybody got a good solution for any of
these?

Joe McGarity




From cube-lovers-errors@oolong.camellia.org  Thu Jun 19 12:32:15 1997
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From: "joyner.david" <joyner.david@mathnt1.sma.usna.navy.mil>
To: "'Josh D Weaver'" <shaggy34@juno.com>
Cc: "'cube-lovers@ai.mit.edu'" <cube-lovers@ai.mit.edu>
Subject: RE: Square One
Date: Thu, 19 Jun 1997 08:06:58 -0400
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>----------
>From: 	Josh D Weaver[SMTP:shaggy34@juno.com]
>Sent: 	Wednesday, June 18, 1997 4:05 PM
>To: 	cube-lovers@ai.mit.edu
>Subject: 	Square One
>
>Does anyone know how to solve one of those "Square One" puzzles?

There's a paper on my web page which indirectly
explains how. (It's actually a math paper written with
a student of mine explaining the group theory of the puzzle.)
What's useful are some of the moves which we give.
If you can't print it out (It's a dvi file) I'll mail it to you
if you give me your postal address. 
http://www.nadn.navy.mil/MathDept/wdj/rubik.html

The idea, if I remember, is
1. get into a square form,
2. use the special moves we give (moves which permute 3
pieces only and leave the others alone, for example)
to solve the puzzle as one solves the Rubik's cube.
          
                                        - David Joyner


>
>Josh
>
>


From cube-lovers-errors@oolong.camellia.org  Thu Jun 19 12:32:39 1997
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From: Carl Woolf <woolf@newvision.com>
To: chi@cse.unsw.edu.au
CC: shaggy34@juno.com, cube-lovers@ai.mit.edu
In-reply-to: 
	<Pine.OSF.3.95.970619113450.27424A-100000@pipe02.orchestra.cse.unsw.EDU.AU>
	(message from Benjamin Wong on Thu, 19 Jun 1997 11:37:43 +1000 (EST))
Subject: Re: Square One

Square One is a great puzzle!

I think there is an instruction booklet, published in Massachusetts or
thereabouts, and available from Puzzlets (mgreen@puzzletts.com). I
developed a set of techniques that let me solve the thing, but I
haven't worked my notes into a form intelligible by other humans (or
by me on a bad day).

-- 
-- Carl 
-----------------------------------------------------
Business:           woolf@newvision.com

Personal:           woolf@ccs.neu.edu
                    http://www.ccs.neu.edu/home/woolf



From cube-lovers-errors@oolong.camellia.org  Thu Jun 19 21:55:51 1997
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To: "cube-lovers@ai.mit.edu" <cube-lovers@ai.mit.edu>
Subject: Square One Solution
Date: Thu, 19 Jun 97 21:53:07 -0500
From: Mike Masonjones <mcmj@world.std.com>
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Cube Lovers,

This is long, but complete solutions are long.  I even left quite a bit of
the boring and obvious stuff out, and it still took me 3 hours to write it. 
Any errors, let me know.  I hope this satisfies.

Apparently I have the least extracurricular life of all of you, since I know
how to solve Square One much quicker than any reports I've seen here.  Quite
a dubious honor, I suspect.  Anyway, here goes my solution, which with a
little practice, guarantees a solution within 75-100 seconds for someone who
can do The Cube in about 55-60 seconds (assuming there's a correlation in
hand speed from puzzle to puzzle).

The solution is completely my own, as is the notation.  Sorry if it offends
anyone.

1. Start by getting to the cube state for the top and bottom faces.  Ignore
the middle slice til the very end.  This can be most efficiently done by
memorizing a table. A pretty good (but not error-free) one is on the only
square-one web site in existence that I know of. Sorry, you'll have to find
the site yourself with a browser, since I can't look it up right now.  I
have a scheme written down somewhere that is quite a bit easier to memorize,
but why should I take the fun away from any of you looking for the solution
yourselves.  If there is a big response to this letter, then I will dig out
my easy table.  

Tables are difficult to memorize, so I usually just try to get to a six
pointed star on one side and all the little wedges plus the remaining 2 big
wedges on the other.  It is easy to get to one of the five possible states
that result, thus requiring memorization of only 5 solutions to get back to
a cube.  This method takes about 5-10 seconds longer, an average, than the
table technique.  

Using a notation where L represents a large wedgie, and S a small wedgie,
the five possible states can be written as:

1) bottom = LLLLLL, top = LLSSSSSSSS
2) bottom = LLLLLL, top = LSLSSSSSSS
3) bottom = LLLLLL, top = LSSLSSSSSS
4) bottom = LLLLLL, top = LSSSLSSSSS
5) bottom = LLLLLL, top = LSSSSLSSSS

For cases 1),3),5), rotate the top face so that it will be sliced
symmetrically between the two L pieces when the center is flipped.  The next
move in each of these cases involves moving the top face one way and the
bottom face the other, when looking from the front.  (Front will be the term
used from now on to denote the end nearest you of the central cut through
which flipping occurs (a 180 turn of one half of the cube)).  After a flip,
cases 1) and 5) should give two barrel shapes (LLSSLLSS), top and bottom. 
You should aim in case 3) for two tomahawk (LLSSLSLS) shapes.  Any self-
respecting cubist should be able to get home from here.

Cases 2 and 4 are a little more complicated.  For both cases align the top
so that the left half of the top face reads, going clockwise, SLSSS.  Flip
right side.  Now rotate the bottom so that when you flip with the right hand
, the top will read SlSSSSSLL starting from the front and going clockwise. 
Now rotate the top 1/12 turn counterclockwise and the bottom so it reads
LLLLSSL going clockwise from the front and flip again.  Now you're in an
easy state to get home from (LLLLSSSS on top and LLSLSSLS on bottom).

2. Now that you're in a cube state top and bottom, get all the wedgies on
their correct side (top and bottom face all the same color, respectively). 
This is very straightforward and intuitive.  I usually start with one large
wedgie, and sequentially put in one at a time next to it going around a face
until you get down to one S wedgie stuck on the wrong side.  Sometimes it is
easier to do LSL on one half of the top, and then do LSL on the other, and
then putting in the second to last S between the groups.

Now position the top face so that the Odd S wedgie (O) is positioned as
LSLOLSLS going clockwise from the front.  Put the bottom odd wedgie in front
with the bottom square skewed from the top (bottom should read LSLSLSLO
going clockwise from the front).  Now do FT4B1FT-4B-1FT4B1FT-3F, where F =
flip with right hand, Bx = turn bottom face clockwise x/12 of a turn, B-x =
same thing counterclockwise, Tx, T-x mean similar things.

3. Now get L's positioned.

Case 1. No L's are correctly adjacent to each other.  Position top and
bottom (top = LSLSLSLS, bottom = SLSLSLSL, each going clockwise from front).
 Now go
FB3FT-3B-3FT3F, turn the whole puzzle 180 degrees so that the back of the
central cut is now the front, and repeat the move.

Case 2. Two sets of adjacent pairs are out of whack, one on top, and one on
bottom.
Do the move for case 1 once, with the components of the pairs in question
all nearest the front.

Case 3. Only one adjacent pair correct.  Position the top so that the
correctly adjacent pair (denoted as A) is positioned as ASASLSLS, and the
bottom reads LSLSLSLS (same conventions as before).  Now do FB-3FB3FB-3FB3F.

Case 4. Only one pair incorrect.  Position the top (with the incorrect pair)
so that the correctly adjacent pair is positioned as LSASASLS, and the
bottom reads LSLSLSLS.  Do the move in Case 3 twice with a T3 between
instances.

Case 5. One side is OK, the other has no correct adjacent pairs.  Bad side =
top.  top = LSLSLSLS, bottom = LSLSLSLS.  Do the move in Case 3 twice with
T6 between instances.

4. Now check for parity.

With the L's in place it is easy to identify whether you need to change the
parity of the system.  It should take an even number of switches to right
the S's at this stage.  A cycle of three is even, since it would take two
switches to fix it.  A cycle of two or four is odd.  If the overall parity
is odd, do the following:
starting with top = LSLSLSLS, bottom = LSLSLSLS, go FT3B3FT1B2FT2B2FT-
2FT2B2FT3B2FT-3B-3T-3B-1FT-2B-2F
This may not be the optimum way, but it preserves the corners, and it's easy
to remember the path.  (Try it)

5. Place the S's (they should already be on their correct face). The most
useful moves are the below: All permutations of S's can be solved with
application of a maximum of three of these short moves in sequence, combined
with the appropriate turns in between to set things up.  

Move 1. Start with top = LOLSLSLO, bottom = LOLSLSLO, where O = pairs that
will be switched on a given side.  Do FT-3FT1B1FT2B-1F. Repeated twice with
a T3 between makes a three-cycle on the top side.

Move 2. top = LSLOLSLO, bottom = LSLOLSLO, O definition same as Move 1.  Do
FT1B1FT6FT-1B-1F.

There may be quicker solutions than applying these moves for a 4-cycle/2-
cycle combination or a 4-cycle/4-cycle combination, where you have to apply
3 moves in succession. I'd like to hear about suggestions.  I haven't
investigated it too much since these modes come up so rarely.

6. Fix middle slice.  If square shaped but wrong, do FT6B6F. Otherwise,
position the bad half on the right, and do FB6FB6F.

Congratulations, you have a solved Square One.

Happy cubing.  

Mike Masonjones



From cube-lovers-errors@oolong.camellia.org  Thu Jun 19 21:55:37 1997
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Date: Thu, 19 Jun 1997 11:00:50 -0700
To: cube-lovers@ai.mit.edu
From: Mark Pilloff <mdp1@uclink4.berkeley.edu>
Subject: Re: Square One

At 12:12 AM 6/19/97 -0700, J.M. wrote:
>The ones I have trouble with are the
>Sqewb and the Alexander's Star.  Anybody got a good solution for any of
>these?
>
>Joe McGarity

I finally came up with a solution to the Alexander Star last year.  I
haven't ever written out all of the details, but here are some helpful
hints.  First of all, the star is almost identical to the Megaminx (aka,
magic dodecahedron, etc.) with all of the corners pieces removed.  The only
reason I say "almost" is that on the star, every individual piece is doubly
degenerate.  This sometimes leads to a problem wherein using the Megaminx
moves seems to leads to an insoluble position.  The trick in this case is
two swap two of the degenerate pieces while disturbing as little of the rest
of the star as possible.  This has always worked perfectly for me.  As for
the rest of the star, I usually find that I can solve most of the star
(except for the uppermost regions) just by inspection.  From there, very
slight modifications of Rubik's cube manipulations are useful.  It's
worthwhile to note that locally, the star (and the megaminx) are identical
to the cube (except, perhaps, for the missing corners on the star).  Thus,
cube moves which only affect small portions of the cube will often be
successful on the star or megaminx.  In any event, I'm not going to write
out explicit moves because I believe solutions to the megaminx are floating
on the net as well, but I hope this is somewhat helpful.  If there is really
strong demand for explicit solutions, I'll see what I can do about that.

Good luck,
Mark

************************************
**    Mark D. Pilloff             **
**    mdp1@uclink4.berkeley.edu   **
************************************




From cube-lovers-errors@oolong.camellia.org  Fri Jun 20 11:36:32 1997
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To: Carl Woolf <woolf@newvision.com>
From: Georges Helm <geohelm@pt.lu>
Subject: Re: Square One
Cc: cube-lovers@ai.mit.edu

At 09:19 19/06/1997 -0400, you wrote:
>Square One is a great puzzle!
>
>I think there is an instruction booklet, published in Massachusetts or
>thereabouts, and available from Puzzlets (mgreen@puzzletts.com). I
>developed a set of techniques that let me solve the thing, but I
>haven't worked my notes into a form intelligible by other humans (or
>by me on a bad day).
>

There is a solution in Puzzle World. Check out details at
http://ourworld.compuserve.com/homepages/Georges_Helm/cubeold.htm

Georges
         
geohelm@pt.lu
http://ourworld.compuserve.com/homepages/Georges_Helm
http://www.geocities.com/Athens/2715



From cube-lovers-errors@oolong.camellia.org  Fri Jun 20 11:36:18 1997
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To: joemcg3@snowcrest.net
From: Georges Helm <geohelm@pt.lu>
Subject: Re: Square One
Cc: cube-lovers@ai.mit.edu

At 00:12 19/06/1997 -0700, you wrote:

>The ones I have trouble with are the
>Sqewb and the Alexander's Star.  Anybody got a good solution for any of
>these?

I have solutions for both.
Check out my list of solutions at
http://ourworld.compuserve.com/homepages/Georges_Helm/cubeold.htm
Alexander's book on his star
Flettermann has a good solution for the skewb (but I realize I don't have
him listed)
I can send copies, though
Georges
         
geohelm@pt.lu
http://ourworld.compuserve.com/homepages/Georges_Helm
http://www.geocities.com/Athens/2715



From cube-lovers-errors@oolong.camellia.org  Sat Jun 21 14:03:33 1997
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From: Goyra <Goyra@cheerful.com>
To: cube-lovers@ai.mit.edu
Subject: Re: Square One
Date: Thu, 19 Jun 1997 19:00:58 +0100
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> >Does anyone know how to solve one of those "Square One" puzzles?


	Can anyone tell me what this looks like so I can 
put up a Java version?


						David Byrden



From cube-lovers-errors@oolong.camellia.org  Mon Jun 23 21:05:29 1997
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Date: Mon, 23 Jun 1997 18:33:40 -0400
From: "Jonathan R. Ferro" <jferro@knave.ece.cmu.edu>
Message-Id: <199706232233.SAA51996@knave.ece.cmu.edu>
Organization: Electrical and Computer Engineering, CMU
X-Disclaimer: This disclaimer is not required by Leader Kibo.
    This article does not necessarily represent the opinions of Leader Kibo.
    Have a nice day!
X-Exclaimer: Yow!
To: cube-lovers@ai.mit.edu
Subject: An art project...

http://www.wunderland.com/EBooks/Window/Pages/SUTW-JD.html

-- Jon


From cube-lovers-errors@oolong.camellia.org  Tue Jun 24 20:45:27 1997
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Date: Tue, 24 Jun 1997 17:30:04 -0700
From: Jin "Time Traveler" Kim <chrono@ibm.net>
Organization: The Fourth Dimension
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Jonathan R. Ferro wrote:
> 
> http://www.wunderland.com/EBooks/Window/Pages/SUTW-JD.html
> 
> -- Jon

Very impressive.  How many cubes and were they altered in any way except
turning them?

-- 
Jin "Time Traveler" Kim
chrono@ibm.net
VGL Costa Mesa


From cube-lovers-errors@oolong.camellia.org  Wed Jun 25 13:29:10 1997
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Date: Wed, 25 Jun 1997 18:11:31 BST
From: David Singmaster Computing & Maths South Bank Univ <david.singmaster@sbu.ac.uk>
To: cube-lovers@ai.mit.edu
Message-ID: <009B651B.AC285B40.328@vax.sbu.ac.uk>
Subject: Pilloff's query about 5^3;  Korf's method.

	I've been away and have just seen email for late May and early June.
	Pilloff asks about getting the parity of the edges correct on the 5^3.
As he notes, the commutators cannot solve this.  Examining the basic moves,
one sees that the rotation of a inner face changes the parity of the edges
while conserving the parity of some pieces, but not of others.  However, the
pieces whose parity is not conserved are the pieces next to the centre and 
there are four apparently identical copies of these, so one can simulate
an exchange by a 3-cycle with two pieces the same.  
	Hence one wants to apply a rotation of an inner face.  What one does
is to move the two edges into the same inner face.  Then rotate the face.  Then
make a 3-cycle of the edges.  This produces an exchange of edges - and rather
messes up the centre pieces.  Then put things back.
	The edges go   A B C D  to  D A B C  to  B A C D.
	Because this is relatively easy to do, but messes up the centres, I
normally do the edges and corners first and then put centres in place.
	
	Re: Korf.  Someone has said it reminds them of alpha-beta pruning.
It reminds me of branch and bound search.  Both are older names for the general
process of using information about the remainder of the problem to estimate
the number of steps for a solution via a partial solution.

	Going back to Pilloff's query, I have several methods in my files for
exchanging two edges without moving anything else (I think, but that seems to
contradict what I said about parities??)
DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499; 
email:  zingmast  or  David.Singmaster  @sbu.ac.uk


From cube-lovers-errors@oolong.camellia.org  Wed Jun 25 13:28:50 1997
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Date: Wed, 25 Jun 1997 09:53:21 -0500
To: cube-lovers@ai.mit.edu
From: Kristin Looney <kristin@tsi-telsys.com>
Subject: Re: An art project...
Cc: jake@wunderland.com

>> http://www.wunderland.com/EBooks/Window/Pages/SUTW-JD.html
>
> Very impressive.  How many cubes and were they altered in any
> way except turning them?

100 cubes, scrambled - not altered in any other way.

This is, I believe, the sixth cube sculpture that Jacob has done in
the window of my gameroom.  A seventh is currently in progress.
Previous sculptures include: "Rubik's Cube", "Merry Xmas" with a
picture of a Christmas tree, a bizarre (and not too successful)
abstract thingamabob, a pacman with several ghosts, and the Apple
logo.  I think the pacman is probably my favorite, it's a hard choice.

They are all very much worth a look...  I have pictures of them, and
I will encourage Jake to put them on his web page for all to see.

The one he is currently working on is, believe it or not, TWO SIDED.
Unlike most which he just sorta fiddles with while playing games
at my gameroom table, this one was carefully planned out in advance
with graph paper.

-K.
5th fastest hands in the nation (at least back in 1981)


kristin@wunderland.com
www.wunderland.com/kristin
-------------------------------------------------------------
"I'm really angry that I, a superior human being in every way, have
less money than my neighbor, who's wife I would love to nail, if only
I weren't so busy sleeping and eating pork chops."

                                        -- George "Cannonball" Carlin,
                                           on the 7 deadly sins




From cube-lovers-errors@oolong.camellia.org  Wed Jun 25 17:18:25 1997
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Date: Wed, 25 Jun 1997 22:14:32 BST
From: David Singmaster Computing & Maths South Bank Univ <david.singmaster@sbu.ac.uk>
To: cube-lovers@ai.mit.edu
CC: notation@vax.sbu.ac.uk, for@vax.sbu.ac.uk, 4^3@vax.sbu.ac.uk,
        and@vax.sbu.ac.uk, 5^3@vax.sbu.ac.uk
Message-ID: <009B653D.9EE57540.331@vax.sbu.ac.uk>
Subject: notation for 4^3 and 5^3

	David Barr has described some moves for the 5^3 using his own notation.
In the early 1980s, I extended my 3^3 notation to the 4^3 and this can be used
on the 5^3 as well.  I described this in my Cubic Circular but perhaps it would
be useful to give it here.
	I will describe the situation for the 4^3.  On the 5^3, we don't ever
have to make a slice move of just a central layer.  Further, a combination of
4^3 and 3^3 processes will solve the 5^3, so we don't really need to label the 
central layers.
	Consider the four layers from  L  to  R.  I denote the inner layers
by  l  and  r.  So the four layers are:  LlrR.  Similarly we have four layers:
FfbB  and  UudD.  To describe a piece requires more effort than before, but
each piec lies in three layers and we can describe the piece by these layers.
E.g.  FUR  is a corner piece;  FUr is an edge piece, lying on the  FU  edge -
but there are two of these and they are distinguished as being  FUr  and  FUl
(I usually give the layers in clockwise order, but it is not essential and
there are times when it is more informative to use the other order.);  Fur  is
a face-centre piece, the one in the upper right corner of the inner four cells
of the  F  face.  If you have been paying attention, you will ask about  fur.
This is one of the body-centre cells, invisible to you unless you make a
transparent cube!
	Using the standard notation of  [F, R] = FRF'R', we find a number
of easy 3-cycles.
	[[F,R],L] = (FLU,ULB,RFU)
	[[F,R],l] = (FlU,UlB,DfR)  (I've copied this from my Circular, but
I wonder if it's right as I thought there'd be some symmetry with the
preceding??)
	[[F,r],l] = (Flu,Ulb,Drb)
In theory, these and a careful consideration of parity are sufficient to
solve the 4^3 and the 5^3.  However, the parity problem is a bit awkward.
	In my original approach, I got the corners in place and then all edges
except leaving the four edges along the  FU  and  BU  edges.  Examine the 
parity of these carefully.  If they are in an odd permutation, apply
r^2 D^2 l' D^2 r^2  which 4-cycles these four edges and moves some centres. 
Once the parity is corrected, there is little difficulty restoring the
rest of the cube.
	For the 5^3, once you have paired up the edges, one can solve the
central edges by treating the 5^3 as a 3^3 with fat slices.
	To correct a single pair of edges, one can use the following.
rrDDl'DDrr rrD'RR [[R,U],l'] RRDrr  =  rrDDl'DR'UR'U'l'URU'lRDrr  =
(UBl,UBr) (Ful,Ubl,Bdl,Ufr) (Fdl,Ufl,Bul,Ubr).  This messes up some centres,
but they are not too hard to restore.  Indeed applying  
rrUUr (uurrll)^2 r'UUrr  wil correct the  F  and  B  centres disturbed by the
above, leaving a  180 degree  rotation of the four  U  centres.
	After I had developed the notation and solution, a Peter Lees pointed
out an unexpected feature.  The exchange of upper and lower case letters is
a duality.  The dual of  URF  is  urf,  while the dual of  URf  is  urF.
This gives us a Pricniple of Duality:  The dual of a sequence of moves is
the same process on the dual pieces.  E.g., we had
[[F,R],l] = (FUl,UBl,DRf),  so  [[f,r],L] = (fuL,ubL,drF).
	This duality allows one to transfer a number of 3^3 processes to
4^3 processes and to solve the invisible interior part of the cube!
	By always moving an outer layer with its inner layer, one is
obviously simulating the 2^3.  However, if one always moves, say  R  and  l
together, one is also simulating the 2^3 in eight copies!  Ooops, one
wants to move  R  and  l'  together.  If one moves  R  and  l  together,
I think you get eight versions of the 2^3, but each is a reflection of
its neighbours!
	If you are tired of thinking about God's Algorithm on the 3^3,
try the 4^3.  I'm not even sure how to count moves.  E.g., to do  r,  one
normally does  Rr  and then  R',  so does  r  count as one move or two?
Likewise, does  Rr  count as one move or two?
	Enough for now.
DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499; 
email:  zingmast  or  David.Singmaster  @sbu.ac.uk


From cube-lovers-errors@oolong.camellia.org  Wed Jun 25 18:29:12 1997
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Date: Wed, 25 Jun 1997 23:28:16 BST
From: David Singmaster Computing & Maths South Bank Univ <david.singmaster@sbu.ac.uk>
To: cube-lovers@ai.mit.edu
Message-ID: <009B6547.EBD77100.328@vax.sbu.ac.uk>
Subject: 4^3 and 5^3

	I've just seen a comment about the 5^3 saying the writer had problems
with the four pieces at distance 1 from the center.  My approach treated
both these and the pieces ate distance 2 from the center in the same way.
We know that the commutator of two slice moves on the 3^3 produces two
3-cycles of the centres.  Applying this idea to the 4^3, we find that
[f,r] gives two 3-cycles of central pieces, one on each face.  By turning
one face and then inverting, we get a 3-cycle of central pieces, two being
in one face.  E.g.  [[r,b],U] = (Fur,Ubr,Ubl).  A similar result holds if
we combine any two inner moves, so we can replace the  b  above by a central
slice on the 5^3 to obtain a 3-cycle of the pieces at distance 1 from the
central piece, while the process directly gives us a 3-cycle of the pieces at
distance 2 from the central piece.  Although tedious, these moves mean that
once one has the corners and edges in place, the rest of the problem is
easy - though very tedious - it generally took me about an hour to do the
5^3, assuming I could get the corners and edges correct without making too
many mistakes.
DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499; 
email:  zingmast  or  David.Singmaster  @sbu.ac.uk


From cube-lovers-errors@oolong.camellia.org  Wed Jun 25 23:51:13 1997
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Date: Thu, 26 Jun 1997 04:44:18 BST
From: David Singmaster Computing & Maths South Bank Univ <david.singmaster@sbu.ac.uk>
To: cube-lovers@ai.mit.edu
Message-ID: <009B6574.123497C0.321@vax.sbu.ac.uk>
Subject: names for cubes

	What's wrong with  2^3, 3^3, 4^3, 5^3,  pronounced  2 cube,
3 cube, 4 cube, 5 cube.  If you have superscripts available, you can use them
instead of the uparrows.
DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499; 
email:  zingmast  or  David.Singmaster  @sbu.ac.uk


From cube-lovers-errors@oolong.camellia.org  Thu Jun 26 15:40:18 1997
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Date: Thu, 26 Jun 1997 07:44:43 -0400 (EDT)
From: Nicholas Bodley <nbodley@tiac.net>
To: David Singmaster Computing & Maths South Bank Univ <david.singmaster@sbu.ac.uk>
cc: cube-lovers@ai.mit.edu, notation@vax.sbu.ac.uk, for@vax.sbu.ac.uk,
        4^3@vax.sbu.ac.uk, and@vax.sbu.ac.uk, 5^3@vax.sbu.ac.uk
Subject: Hidden cubies; Spaceball
In-Reply-To: <009B653D.9EE57540.331@vax.sbu.ac.uk>
Message-ID: <Pine.SUN.3.95.970626073036.5876A-100000@sunspot.tiac.net>
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 Reading David Singmaster's recent posts, it seems almost obvious to me
that (for the ambitious, which I'm definitely not!), computer simulations
that are unhindered by real-world mechanical constraints and opacity can
permit manipulation of the normally-hidden (and physically-nonexistent)
internal cubies. It's very likely that a new collection of maneuvers would
need to be developed for this.

 I haven't thought much about coloring the internal faces of the outer
cubies...

 Incidentally, if this weren't the cube-lovers' List, I would have split
that first sentence into a few shorter ones.

 Sorry if "cubie" is not the most-preferred term; should be no great
problem.

 On another topic, it seems to me that an ideal device for controlling a
computer-simulated Cube (or other similar puzzle) would be the Spaceball,
a ball that you can grip. It senses torque around all three mutually-
orthogonal axes, as well as "translational" force along those axes. It's
not a consumer item; not sure it's still being made. I'm reasonably sure
of the tradename. It was/is used with workstations. "Spaceball" sounds
much like the name of a puzzle.

 (I expect some astute reader to tell me that the MIT Media Lab did just
this thing 5 years ago!)

My best to all,

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*    When the year 2000 begins, we'll celebrate 
|*  Amateur musician  *|*    the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------



From cube-lovers-errors@oolong.camellia.org  Thu Jun 26 16:35:00 1997
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Date: Thu, 26 Jun 1997 13:33:57 -0700
From: Jin "Time Traveler" Kim <chrono@ibm.net>
Organization: The Fourth Dimension
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To: Nicholas Bodley <nbodley@tiac.net>
CC: cube-lovers@ai.mit.edu
Subject: Re: Hidden cubies; Spaceball
References: <Pine.SUN.3.95.970626073036.5876A-100000@sunspot.tiac.net>
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Nicholas Bodley wrote:
> 

>  On another topic, it seems to me that an ideal device for controlling a
> computer-simulated Cube (or other similar puzzle) would be the Spaceball,
> a ball that you can grip. It senses torque around all three mutually-
> orthogonal axes, as well as "translational" force along those axes. It's
> not a consumer item; not sure it's still being made. I'm reasonably sure
> of the tradename. It was/is used with workstations. "Spaceball" sounds
> much like the name of a puzzle.
> 
>  (I expect some astute reader to tell me that the MIT Media Lab did just
> this thing 5 years ago!)
> 
> My best to all,
> 

Actually, the Spaceball that you talk about is still in existence of
sorts.  I have three Spaceballs.  Actually, they were known as the
Spacetec Spaceball Avengers.  Those are no longer produced.  They've
been replaced by the newer model, the SpaceOrb 360.  Not to get too far
off subject, but the SpaceOrb is used by some people to play Quake.  You
can't beat a good Mouse and Keyboard for Quake, but the SpaceOrb's
multiple axes of movement does allow for some interesting
possibilities.  Due to its 3D nature, I think the SpaceOrb would be a
natural extension for the solving of 3d puzzles in graphical
environments.

-- 
Jin "Time Traveler" Kim
chrono@ibm.net
VGL Costa Mesa


From cube-lovers-errors@oolong.camellia.org  Thu Jun 26 16:55:53 1997
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Date: Thu, 26 Jun 1997 13:53:47 -0700
From: "Jason K. Werner" <mrhip@corp.sgi.com>
Message-Id: <9706261353.ZM3850@neuhelp.corp.sgi.com>
In-Reply-To: Nicholas Bodley <nbodley@tiac.net>
        "Hidden cubies; Spaceball" (Jun 26,  7:44)
References: <Pine.SUN.3.95.970626073036.5876A-100000@sunspot.tiac.net>
Reply-to: "Jason K. Werner" <mrhip@sgi.com>
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To: Nicholas Bodley <nbodley@tiac.net>, cube-lovers@ai.mit.edu
Subject: Re: Hidden cubies; Spaceball
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On Jun 26,  7:44, Nicholas Bodley wrote:
> Subject: Hidden cubies; Spaceball

>  On another topic, it seems to me that an ideal device for controlling a
> computer-simulated Cube (or other similar puzzle) would be the Spaceball,
> a ball that you can grip. It senses torque around all three mutually-
> orthogonal axes, as well as "translational" force along those axes. It's
> not a consumer item; not sure it's still being made. I'm reasonably sure
> of the tradename. It was/is used with workstations. "Spaceball" sounds
> much like the name of a puzzle.

In case anyone is interested:

	http://www.spacetec.com/
	http://www.spaceorb.com/


	-Jason


--
Jason K. Werner                 Email: mrhip@sgi.com
Systems Administrator           Phone: 415-933-6397
USFO I/S Technical Services     Fax:   415-932-6397
Silicon Graphics, Inc.          Pager: 415-317-4084, mrhip_p@sgi.com

"Winning is a habit"-Vince Lombardi;"These go to eleven"-Nigel Tufnel


From cube-lovers-errors@oolong.camellia.org  Fri Jun 27 16:39:01 1997
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Date: Fri, 27 Jun 1997 16:28:04 -0400
From: Corey Folkerts <CFolkerts@compuserve.com>
Subject: First corners, then sides
To: Cube-Lovers@ai.mit.edu
Message-ID: <199706271628_MC2-1961-C7B8@compuserve.com>



        During the last 5 months or so I have been fiddling around with a
Rubik's Cube that I got for Toys R Us. I have become pretty good at solving
it with the layers method (90 seconds is my record so far.) My question is
this : Is the method of solving the cube corners first and then sides any
faster (once one becomes good at it) then solving it by layers? If so,
could someone please reply with a description of that method. I don't know
the names of any fancy moves, so if possible please  use F B U D L R for
the faces during specific moves and expain the other parts of the strategy
clearly. I realize this is asking alot. Thanks in advance to anyone who
replies!

        Corey Folkerts


From cube-lovers-errors@oolong.camellia.org  Sat Jun 28 14:03:46 1997
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Date: Sat, 28 Jun 1997 05:50:29 -0400 (EDT)
From: Jiri Fridrich <fridrich@binghamton.edu>
X-Sender: fridrich@bingsun2
To: Corey Folkerts <CFolkerts@compuserve.com>
cc: Cube-Lovers@ai.mit.edu
Subject: Re: First corners, then sides
In-Reply-To: <199706271628_MC2-1961-C7B8@compuserve.com>
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It appears that both systems are good if worked out into sufficient
detail. As far as I know, the fastest speed cubists can achieve an average
of 16-17 seconds irrespective! of whether they use corners-edges or
by-slices systems. You can look at a system which I have developed some
time ago and which enables me to solve the cube in 17 sec. on average. It
is described in detail here: http://ssie.binghamton.edu/~jirif/. I also
recommend the section on speed cubing tips.

Good luck!

Jiri

On Fri, 27 Jun 1997, Corey Folkerts wrote:

> 
> 
>         During the last 5 months or so I have been fiddling around with a
> Rubik's Cube that I got for Toys R Us. I have become pretty good at solving
> it with the layers method (90 seconds is my record so far.) My question is
> this : Is the method of solving the cube corners first and then sides any
> faster (once one becomes good at it) then solving it by layers? If so,
> could someone please reply with a description of that method. I don't know
> the names of any fancy moves, so if possible please  use F B U D L R for
> the faces during specific moves and expain the other parts of the strategy
> clearly. I realize this is asking alot. Thanks in advance to anyone who
> replies!
> 
>         Corey Folkerts
> 
> 

**********************************************************************
|  Jiri FRIDRICH, Research Associate, Dept. of Systems Science and   |
|  Industrial Engineering, Center for Intelligent Systems, SUNY      |
|  Binghamton, Binghamton, NY 13902-6000, Tel.: (607) 797-4660,      |
|  Fax: (607) 777-2577, E-mail: fridrich@binghamton.edu              |
|  http://ssie.binghamton.edu/~jirif/jiri.html                       |
**********************************************************************

......................................................................

Remember, the less insight into a problem, the simpler it seems to be!
----------------------------------------------------------------------



From cube-lovers-errors@oolong.camellia.org  Sun Jun 29 22:09:51 1997
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From: Alex Joukov <lllykob@telecom.lek.ru>
To: Cube-Lovers@ai.mit.edu
Subject: Newcomer: Algoritms. How to solv U-slice if D-slice and UD-slice have been solvd yet? 
Date: Mon, 30 Jun 1997 03:11:05 +0400
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Dear Cube-Lovers,
How to solv U-slice if D-slice and UD-slice have been solvd yet? I remember
that in about 1994 when I was a child I used 3 or 5 algoritms (with mirrow
variants). But now I don't remember the way. 
I read cube-lovers arhives (from #0) and step by step find out some
algoritms. But before I will be able to solve Cube I lose a lot of
interesting in
archive messages. I just can't try a lot not having solved Cube!
Please help me. A need just "1. U2DFTU'RD'   2. UR'D2L'D2 3..." without
comments.
Or, may be somebody have created such type FAQ which is accesible for ftp? 

lllykob@telecom.lek.ru

Sasha


From cube-lovers-errors@oolong.camellia.org  Mon Jun 30 15:09:03 1997
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Date: Mon, 30 Jun 1997 14:08:40 -0400 (Eastern Daylight Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Inverses of Local Maxima
To: cube-lovers@ai.mit.edu
Reply-to: Jerry Bryan <jbryan@pstcc.cc.tn.us>
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One of the oldest unsolved problems on Cube-Lovers (aside from God's
Algorithm itself) has to do with inverses of local maxima.  It seems
obvious that the inverse of a local maximum also ought to be a local
maximum.  But is it necessarily so? 

In Symmetry and Local Maxima, Jim Saxe and Dan Hoey suggest that it may
not be.  Their example is UFF, which can end with F or F' because we can
write it as UF'F'. But the inverse is F'F'U', which can only end with U'
Hence, there are very simple positions where the number of q-turns with
which the position can end is different than the number of q-turns with
with the inverse of the position can end.  If the same thing should
happen with a local maximum, then the inverse would not be a local
maximum.

On the other hand, for all known local maxima in G, the inverse is also
a local maximum.  What are we to think? 

I have some small progress.  I can report that for the corners-only
group, there are local maxima for which the inverse is not a local
maximum.  The results were obtained with my new Shamir program. 

For each position x, we define E(x) to be the set of all quarter-turns
with which a minimal process for the position can end.  As an example,
if x=UFF, then E(x)={F,F'}.  

E(x) is a subset of Q, the set of twelve quarter-turns, or equivalently
it is an element of P(Q), the power set of Q.  As such, it is
conveniently represented in my program as a bit-string of twelve bits. 
In this notation, we would say that a position x is a local maximum if
E(x)=Q or if |E(x)|=12. 

We also define S(x) to be the set of all quarter-turns with which a
minimal process for a position can start.  In this notation, for x=UFF
we would say that |S(x)|=1 and |E(x)|=2.  So the general question for
local maxima becomes the following:  if |E(x)|=12, does it necessarily
follow that |S(x)|=12?

My program calculates S(x) and E(x) as follows.  Any breadth-first
search may be characterized as calculating products of the form z=xy for
suitable choices of x and y.  Most typically, x comes from Q[n], the set
of all quarter-turns of length n, and y comes from Q[1], the set of all
quarter-turns of length 1.  But in my more general Shamir program, x
comes from Q[m] and y comes from Q[n] to form products of length m+m. In
any case, S(z) is the union of S(x) over all x which can be a part of a
product which produces z, and E(z) is the union of E(y) over all y which
can be a part of a product which produce z.  For each q in Q, we
initialize with S(q)=E(q)={q} and go from there. 

Here is a portion of a printout from my program.


 |x|  |E(x)| |S(x)|  M-Conjugacy  Positions
                        Classes  

  0      0      0          1          1
  1      1      1          1         12
  2      1      1          2         96
  2      2      2          3         18
  3      1      1         12        576
  3      1      2          3         96
  3      2      1          3         96
  3      2      2          4         96
  3      3      3          2         60

As you can see, the effect pointed out by Saxe and Hoey first shows up
three moves from Start, where there are six positions unique up to
M-conjugacy where |S(x)| is not equal to |E(x)|. (Actually, three of the
six positions are just the inverses of the other three.)

The first local maxima are six moves from Start in the corners-only
group.

 |x|  |E(x)| |S(x)|  M-Conjugacy  Positions
                        Classes  

  6     12     12          8        114

As you can see, there are 114 local maxima of which 8 are unique up to
M-conjugacy.  However, for all 8 of them, the inverse is also a local
maximum so we discover nothing new. 

The new discovery occurs 7 moves from Start.

 |x|  |E(x)| |S(x)|  M-Conjugacy  Positions
                        Classes  

  7     12      8          4        120
  7     12     10          3        144
  7     12     12         14        336

As you can see, there are 21 local maximu unique up to M-conjugacy.  For
14 of them, the inverse is also a local maximum.  But for the other 7,
the inverse is not a local maximum.  In 4 cases, we have |S(x)|=8, and
in 3 cases we have |S(x)|=10. 

Here follow summaries for local maximum up to a distance of 11 moves
from Start.

 |x|  |E(x)| |S(x)|  M-Conjugacy  Positions
                        Classes  


  8     12      6         14        576
  8     12      8         12        576
  8     12     10         86       4128
  8     12     11         13        624
  8     12     12        272      12012

  9     12      4         26       1152
  9     12      6         31       1344
  9     12      8         24       1152
  9     12     10         14        576
  9     12     12        131       5976

 10     12      2         14        576
 10     12      4         88       4032
 10     12      6        218      10368
 10     12      8        144       6336
 10     12     10        168       8064
 10     12     12        140       5664

 11     12      4        384      18432
 11     12      6       2687     128688
 11     12      8       5550     264192
 11     12     10       5014     240576
 11     12     12       3617     166224


 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990







From cube-lovers-errors@oolong.camellia.org  Mon Jun 30 19:21:04 1997
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Date: Mon, 30 Jun 1997 16:05:31 -0700 (PDT)
From: Bill Webster <captainhaddock@rocketmail.com>
Subject: Hi
To: cube-lovers@ai.mit.edu
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Hi,

I first encountered the cube sometime in the early 80s when the fad
hit Australia. Solving it absorbed all my spare time for two weeks
along with several wads of A4 and $15 for a second cube which I used
to develop what I called 'sequences'. My experiments on the second
cube were conducted with great care, as I had not yet discovered that
Satan's Algorithm could be had for the price of a small screwdriver. I
never became a speed-freak, or even employed operators outside my own
meagre discoveries, but could solve the cube comfortably inside
three minutes. I solved corners, then edges because the first
reasonable sequences I discovered were edge-disrupting corner
operators. My operators were short and scant so my method incurred a
lot of short term memory overhead, manipulating faces into susceptible
positions, applying the sequence, then inverting the prior
manipulation.

A friend of mine aquired a cube at about the same time and much to my
chagrin, solved it in less than a day, without paper, without
explicitly developing any operators. In fact, he couldn't give a
satisfactory account of exactly how he'd solved it. He may have just
got *extremely* lucky and stumbled on something close to START, but I
don't think so. I handed him a scrambled cube a couple of weeks later
and he was quite taken aback - he wasn't going through all that again,
he'd done it hadn't he?. I always felt that my own solution was
somewhat contrived after witnessing this feat. Does anyone else have
examples of GestaltCube?

I have coded a C++ class which represents cube states and operators
and which includes methods to manipulate the cube. I would like to
implement overloaded C++ operators in a manner which is consistent
with (and perhaps extends) the appropriate mathematical notation *if
this is feasible*. Is there anyone out there familiar with both
grammars and willing to make suggestions? I am aware and prepared to
accept that the use of some (C++) operators may introduce
inefficiencies in the form of temporary objects created during
expression evaluation - such operators will not be used in time
critical code.

I have been using a freeware ray-tracer, POV-Ray to produce
'photo-realist' cube images. I intend to extend my software to export
animation scripts, so that I can produce (externally rendered)
animated solutions. These take forever to trace, so their value is
aesthetic rather than practical. I have been experimenting with solid
gold cubes inlaid with coloured marble etc., but still prefer the
platonic form. I have the POV source for a static cube if anyone is
interested.

The POV team are true heroes - details...

 "The internet home of POV-Ray is reachable on the World Wide Web
  via the address http://www.povray.org and via ftp as ftp.povray.org."

 "POV-Ray can be used under MS-DOS, Windows 3.x, Windows for
   Workgroups 3.11, Windows 95, Windows NT, Apple Macintosh 68k,
   Power PC, Commodore Amiga, Linux, UNIX and other platforms."

Regards, Bill Webster

_____________________________________________________________________
Sent by RocketMail. Get your free e-mail at http://www.rocketmail.com



From cube-lovers-errors@oolong.camellia.org  Mon Jun 30 21:31:40 1997
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Date: Mon, 30 Jun 1997 20:01:16 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: Inverses of Local Maxima
In-reply-to: <Pine.WNT.3.96.970630120803.-397377S-100000@ER123A.PSTCC.CC.TN.US>
To: cube-lovers@ai.mit.edu
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On Mon, 30 Jun 1997, Jerry Bryan wrote:

> 
> I have some small progress.  I can report that for the corners-only
> group, there are local maxima for which the inverse is not a local
> maximum.
> 

There is a minor interesting point that might be added.  When we find a
local maximum x for which |S(x)|<12, we can form a new, longer local
maximum qx for suitable q in Q.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990



From cube-lovers-errors@oolong.camellia.org  Mon Jun 30 21:57:52 1997
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Date: Mon, 30 Jun 1997 21:37:39 -0400
From: michael reid <reid@math.brown.edu>
To: cube-lovers@ai.mit.edu, jbryan@pstcc.cc.tn.us
Subject: example of a local maximum whose inverse is not a local maximum

jerry bryan asks if the inverse of a local maximum is necessarily a local
maximum.  the following example shows that this need not be the case.
the interesting "six-two-one" pattern is produced by the sequence

     B U2 F2 R U' R' B' R' U F2 U2  (15q)

this position has six symmetries, generated by the cube rotation  C_UFR
and central reflection.  therefore we also have the maneuvers

     L F2 R2 U F' U' L' U' F R2 F2
     D R2 U2 F R' F' D' F' R U2 R2
     F' D2 B2 L' D L F L D' B2 D2
     R' B2 L2 D' B D R D B' L2 B2
     U' L2 D2 B' L B U B L' D2 L2

for the same position.  it is not hard to check (by computer) that
these are minimal maneuvers.  note that for each quarter turn, we
have a maneuver that ends with that quarter turn.  thus, from this
position, any quarter turn brings us closer to start, so our position
is a local maximum.

consider now the inverse position; it is produced by


     U2 F2 U' R B R U R' F2 U2 B'  (15q)

it is not hard to check (by computer) that applying the quarter turn  B'
to this moves us further from start (16q), so this position is not locally
maximal.

note that this is already in the archives; i first reported it on april 20,
1995 in my message "correction and an interesting example"

mike


From cube-lovers-errors@oolong.camellia.org  Sat Jul  5 20:21:49 1997
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Date: Sat, 5 Jul 1997 16:58:20 -0400
From: michael reid <reid@math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: optimal cube solver

the recent work by rich korf on finding optimal solutions has prompted
me to try my hand at writing an optimal cube solving program.  so far,
i've done this for the face turn metric.  a description of my program
follows.

let  H  denote the intermediate subgroup  <U, D, F2, R2, B2, L2>  which
we've seen before.  we'll use distances to this intermediate subgroup
for our pruning tables (or "pattern databases").  calculating these
distances involves doing a breadth first search on the coset space
H \ G , and storing these distances in memory.  (i've written this as
a right coset space, rather than a left coset space.)  this search has
been done several times, by dik winter and by myself.

some review.  positions in  H  are characterized by the following.
corners cannot change orientation; their  U  or  D  facelet remains on
the  U  or  D  face.  similar, edges cannot change orientation.
furthermore, the four U-D slice edges remain in the U-D slice.
therefore, cosets in  H \ G  are described by triples  (c, e, l),
where  c  denotes corner orientation,
       e  denotes edge orientation
and    l  denotes the location of the four U-D slice edges.

there are  3^7 = 2187 possible corner orientations,
          2^11 = 2048 possible edge orientations
         / 12 \
and      \  4 / = 495 possible U-D slice edge locations.

all combinations are possible, so there are  2187 * 2048 * 495 =
2217093120  cosets.  since this is too many configurations to store
in memory, we use symmetry to to reduce this number.

there are 16 symmetries of the cube that preserve the U-D axis, and
therefore the intermediate subgroup H.  rather than store all the
cosets, we'll just store one of each up to symmetry.  actually, this
is slightly more complicated than necessary; instead, we could just
divide the corner coordinate by symmetry.  this is what i did in my
message of january 7, 1995.  however, i encountered a pitfall along
the way.  i discovered (very late in the development stage) the need
for very large transformation tables.  although i continued with the
same approach at that time, i gave two options for overcoming this
problem:

>       i) only use the 8 symmetries that preserve my choice of
>          12 edge facelets.
> 
>      ii) combine the two coordinates edge and location into a single
>          coordinate and divide this coordinate by the 16 symmetries.

of these, clearly the second is the better choice, since it utilizes
more symmetry.  this new edge coordinate has  2048 * 495 = 1013760
possibilities.  up to symmetry, there are 64430 possibilities.  we
need room for  64430 * 2187 = 140908410  cosets in memory.  for each
of these, we store its distance to the identity coset.  this is an
integer between 0 and 12 (inclusive), so each is stored in half a
byte.  thus the whole table requires 67 megabytes.

essentially, what we're doing here is changing coordinates from
(c, e, l)  to  (c, e', s),  where  e'  is our new edge coordinate,
and  s  is a symmetry coordinate.  some cosets have multiple
coordinates in this new system, but that causes no harm.

a breadth first search of this space takes under 11 minutes.  the
increase in speed is partially due to a more powerful computer, and
partially due to switching to "backward searching" (or "bidirectional
search") at the optimal time.

we'll also use distances to the intermediate subgroups
<F, B, U2, R2, D2, L2>  and  <R, L, F2, U2, B2, D2>.  we don't need to
store additional coset spaces, since we can derive that information
from our first coset space.  note that the cube rotation C_UFR takes
the subgroup  <U, D, F2, R2, B2, L2>  to the subgroup
<F, B, U2, R2, D2, L2>.  therefore it transforms the first coset space
into the second coset space.  furthermore, it preserves distances, so
the one pattern database suffices for all three applications.

an attractive feature of this approach is that it uses the 16
symmetries to reduce the size of the pattern database, and then uses
the remaining symmetry of the cube in applying it in different
orientations.

these are the only pruning tables my program currently uses.  note that
they cannot "see the entire group".  specifically, let
H_0 = <U, D, F2, R2, B2, L2>,  H_1 = <F, B, U2, R2, D2, L2>,
H_2 = <R, L, F2, U2, B2, D2>,  and let  T  denote the intersection of
these three subgroups.  for a given position, the three distances to
these subgroups depend only upon the corresponding coset in  T \ G .
thus  T  might be thought of as a "target subgroup".

this target subgroup  T  is interesting.  it consists of those positions
that "look like" they're in the "square group"  <F2, L2, U2, B2, R2, D2>,
i.e.  F  and  B  colors mix only with each other, and similarly for
R  and  L , and also for  U  and  D.  however, this is strictly larger
than the square group; it contains the square group as a subgroup of
index 6.

the searching is done in the way that korf describes as "IDA*" (or at
least the "ID" part of that terminology).  we traverse the tree of all
sequences of length 1, hoping to find a solution.  that generally fails,
so we continue to sequences of length 2, and so forth, until a solution
is found.  the "A*" part of the algorithm is to use the pruning tables
to avoid searching large parts of the tree that are guaranteed not to
bear fruit.

in his paper, korf uses the expected value of his heuristic functions to
get an estimate of how effective they are at pruning the search tree.
actually, he should subtract 1 from this expected value, since we must
generate (at least partially) the top node of a subtree that gets pruned.
this is only a rough estimate; getting a more precise figure is a delicate
matter which i won't address here.  korf reports an expected value of 8.878.
i generated 10 million random cubes (i did not use the long sequence of
random twists method) and got an expected value of 9.941.

my program generates slightly more than 500000 nodes per second.  korf
generates them at 700000 per second, so i've got more overhead per
node.  however, it generates many fewer nodes, since it prunes the
search tree more efficiently.

i solved korf's ten random cubes, and found all minimal solutions,
rather than stopping at the first.  this entailed one complete search
through length 16f, three through length 17f and six through length 18f.
the position at distance 16f has a unique minimal solution, as do the
three positions at distance 17f.  of the six positions at distance 18f,
one has a unique minimal solution, one has 3 minimal solutions, two
have 4 minimal solutions and two have 6 minimal solutions.  the total
run time for these was just under 198 hours.  korf estimates 4000 hours
for the same search, so on these positions, my program is twenty times
as fast.  my computer has a 200 MHz pentium pro processor, and is
configured with 128 megabytes of RAM.

i'd expect a similar increase in performance for most positions, but
not all.  for example, positions inside the target subgroup  T  run
very slowly, as do positions very close to it.  hopefully, most of
these are close enough to start, so that searches don't have to go very
deep.  i suspect that there are probably also positions that give korf's
program difficulty.

as you can see, i've made only minor modifications to korf's method.
the only differences are:
1. use different pattern databases that allow more efficient pruning.
2. apply the same pattern database in multiple orientations.
3. allow a target subgroup larger than just the identity.

it's clear that more experimentation is needed with different pattern
databases.  for any subgroup  K  of  G , we could consider distances
to that subgroup.  it seems likely that we want small subgroups, so
that the average distance is large.  for this reason, using symmetry
to reduce the size of the database is an important tool.  i encourage
others to experiment with different subgroups.

more results to come ...

mike



From cube-lovers-errors@oolong.camellia.org  Mon Jul  7 02:22:55 1997
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Date: Mon, 7 Jul 1997 01:04:35 -0400
From: michael reid <reid@math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: symmetry reductions for superflip

a simple counting argument shows that some cube positions are at least
18 face turns from start, and thus the diameter of the cube group is
at least 18f.  in january 1995, i showed, by exhaustive search,
that the position "superflip" is exactly 20 face turns from start.
therefore the diameter is at least 20f.  this gave the first
improvement to the lower bound obtained by the counting argument.

the searching method i used at the time was my version of kociemba's
algorithm.  although my symmetry reductions fit together quite well
with kociemba's algorithm, this might not be the most appropriate
searching method to use for this purpose.  (i guess i could have
hacked it not to bother looking for solutions longer than 19f.
i don't remember why i didn't do this.)

my new optimal solving program can do an exhaustive search in much
less time.  the symmetry reductions are similar, but much simpler.
i will try be more coherent this time with my explanation, hopefully
without being overbearing.

the first thing to note is that dik winter found a maneuver for
superflip in 20f:

  F B U2 R F2 R2 B2 U' D F U2 R' L' U B2 D R2 U B2 U   (20f)

therefore our concern is with searching for maneuvers of length
at most 19f.

there are three ways to transform a maneuver for superflip to get
another such maneuver, which do not change its length:

1. we may conjugate the maneuver by any symmetry of the cube.
2. we may cyclically shift the maneuver; i.e. replace

      sequence_1 sequence_2     by     sequence_2 sequence_1

3. we may replace the maneuver by its inverse.

(in fact, we won't use 3 here, but it might be helpful elsewhere.)
our first result is

proposition 1.  any maneuver for superflip in 19f contains a 180 degree
                face turn.
proof.  if the proposition were false, then superflip would be an
        odd number of quarter turns from start, contradiction.  qed.

the relevance of this proposition is

proposition 2.  suppose that a maneuver for superflip contains a 180 degree
                face turn.  then it can be transformed, using the above
                tranformations, into a maneuver that begins with  U R2.
proof.  we first claim that the maneuver has two consecutive "syllables"
        such that the first contains a 90 degree face turn and the second
        contains a 180 degree face turn.  a "syllable" is a sequence of
        one or two face turns along the same axis; e.g. U D2.  by
        hypothesis, the maneuver has a syllable that contains a half turn.
        if the claim is false, then the preceding syllable contains no
        90 degree turns, and therefore consists only of half turns.  but
        then the syllable before that contains only half turns, by the
        same reasoning.  continuing in this way, we see that every syllable
        consists only of half turns.  therefore we have a maneuver for
        superflip consisting only of half turns.  this is a contradiction,
        so the claim is true.
             now, since the individual face turns within a syllable
        commute, we may suppose that the maneuver has a 90 degree face
        turn followed by a 180 degree face turn, which are along
        different axes, and thus are adjacent faces.  now we may
        conjugate by an appropriate symmetry of the cube to suppose that
        these turns are  U R2.  finally, we may cyclically shift the
        maneuver so that these are the first two turns.  qed.

proposition 3.  suppose that superflip is exactly 19 face turns from
                start.  then applying the sequence  U R2  to it brings
                us 2 face turns closer to start, i.e. 17f  from start.
proof.  apply proposition 1 and proposition 2.  qed.

we now know how to handle the case that superflip's distance from start
is exactly 19f.  if the distance is less than 19f, we use the following

proposition 4.  under any circumstances, applying the sequence  U R2
                to superflip brings us at least  1f  closer to start.
proof.  a minimal maneuver for superflip must contain a 90 degree twist,
        and we may suppose that the next face turned is an adjacent one.
        by cyclically shifting the maneuver, we may bring these two
        turns to the beginning.  furthermore, by symmetry, we may
        suppose that the first turn is  U  and the second is some twist
        of the  R  face.  now by applying  U  to superflip, we've moved
        1f  closer to start, and applying  R2  to this doesn't move us
        any further from start, since it either combines with, or cancels
        the next turn in the minimal maneuver.  qed.

putting this all together, we get our desired result.

proposition 5.  suppose that superflip is within 19f of start.  then the
                position  superflip U R2  is within 17f of start.
proof.  this is just combining props 3 and 4.  qed.

i don't claim that these are the best reductions possible.  they
suffice for our purposes.

i tested the position  superflip U R2  (i.e. the position obtained by
first doing superflip, and then doing the sequence  U R2) with my
optimal solver.  my program took 2 hours and 40 minutes to exhaustively
search this position through 17 face turns (not including about 11
minutes to generate all the lookup tables).  there were no solutions.
thus superflip is exactly 20 face turns from start.  when i did the
search in january 1995, the run time was 6 days.  so we see quite a bit
of improvement.

mike


From cube-lovers-errors@oolong.camellia.org  Tue Jul  8 00:16:07 1997
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Date: Tue, 8 Jul 1997 00:18:18 -0400
From: michael reid <reid@math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: superfliptwist requires 20 face turns

i can now show that the pattern "superfliptwist" is exactly 20 face
turns from start.  this position was proposed as a likely antipode
of start by cubologist christoph bandelow.  in the german edition
of his book "einfuerung in die cubologie" he offered a prize for the
shortest maneuver for this pattern.  the prize was collected by rainer
aus dem spring, who found a maneuver in 22 face turns.  much later,
a maneuver of length 20f was found by herbert kociemba:

     D F2 U' B2 R2 B2 R2 L B' D' F D2 F B2 U F' L R U2 F'  (20f)

as one of the first applications of his ingenious searching algorithm.

i'll try not to be so verbose with my symmetry reductions this time.

first note that "superfliptwist" does not describe a unique position
of the cube; there are two possible orientations.  in this context,
i use the term "position" to refer to one of the 43252003274489856000
possible configurations, and the term "pattern" to refer to an
equivalence class of positions under symmetries of the cube.
(this concept has been discussed by dan hoey and jerry bryan as
the "real size of cube space" i.e. the number of patterns.)

the following two facts are easily verified:

* superfliptwist commutes with the square of each face turn.
* it does not commute with 90 degree slices (e.g.  U D') or 90 degree
  antislices (e.g.  U D), however, if  A  is a 90 degree slice or
  antislice, then
                    A  superfliptwist  A^(-1)

  is also superfliptwist, but in the other orientation.

these facts lead to the importance of the following

proposition.  superfliptwist is not in the subgroup generated by slices
              and antislices.  (note that this group contains all squares
              of face turns.)
proof.  we may ignore the corners and just show that all edges cannot be
        flipped in this subgroup.  to do this, we choose dominant facelets
        on the 12 edges as follows: choose the U or D facelet of the edges
        in the R-L slice, the R or L facelet of the edges in the F-B slice
        and the F or B facelet of the edges in the U-D slice.  now we may
        define the flip of an edge that is not in its correct location.

        all edges start in the correct orientation.  a 90 degree slice or
        antislice along the U-D axis changes the orientation of all eight
        edges in the F-B slices and R-L slices.  similarly, a 90 degree
        slice or antislice along the F-B or R-L axis flips all edges in
        two different slices.  within this subgroup, either all edges in
        a given slice are flipped, or none are flipped, and furthermore,
        the number of the three slices with flipped edges is even, i.e.
        0 or 2.  however, superfliptwist has all three slices with
        flipped edges, so it is not in this subgroup.  qed.

now consider the first syllable of a minimal maneuver for superfliptwist.
("syllable" was defined in my previous message.)  if this is a single
180 degree turn, then we may cyclically shift this to the end of the
maneuver.  similarly, a slice squared may also be shifted to the end
of the maneuver.  furthermore, 90 degree slices and or antislices may
also be shifted to the end of the maneuver, with only the mild effect
of changing which orientation of superfliptwist we're doing.  from the
proposition, we eventually find a syllable which is not of these types,
and is therefore of type  U  or  D2 U.  in the case of  D2 U , we may
shift the  D2  to the back of the maneuver, so we may suppose that the
first face turn is  U .  furthermore, by conjugating by the cube rotation
C_U, if necessary, we may suppose that our maneuver solves our preferred
orientation of superflip.  the second face turn is in a different syllable,
so it is an adjacent face.  conjugating by C_U2, if necessary, brings this
face to either  R  or  F.  therefore we may suppose that the first two
face turns are one of the six sequences

     U R ,   U R2 ,   U R',   U F ,   U F2   or  U F' .

to show that superfliptwist is not within 19f of start, i tested the
six patterns obtained by applying these sequences to it.  it took my
program 7.5 hours to exhaustively search all of these through 17f.
(these positions ran a bit faster than most of the others i've tested.
this is partly because superfliptwist is 15 face turns from my "target"
subgroup, so larger parts of the search tree are pruned.)  no solutions
were found, so superfliptwist requires 20 face turns.

i also let the first situation run partially through depth 18f.  in about
4 and a half hours, it found a solution which yields

  U R F' B U' D' F U' D F L F' L' U R D F U R L  (20f, 20q)

this is automatically minimal in the quarter turn metric!

mike


From cube-lovers-errors@oolong.camellia.org  Tue Jul  8 17:33:24 1997
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Date: Tue, 8 Jul 1997 16:43:50 -0400
From: michael reid <reid@math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: composition of superflip and pons asinorum

the position which is the composition of superflip and pons asinorum is
exactly 19 face turns from start.  some time ago, jerry bryan found that
this position is exactly 20 quarter turns from start, and he gave all
minimal maneuvers, up to symmetry.  one of these is 19 face turns long:

     B' D' L' F' D' F' B U F' B R2 L U D' F L U R D   (19f, 20q)

symmetry reductions for this position are much simpler (but not nearly
as good) as for superflip and/or superfliptwist.  if the first face turn
is a 90 degree turn, then by symmetry, we may suppose it is  U .  if the
first face turn is a 180 degree turn, then we may suppose it is  U2 .
i tested the two positions obtained by applying these possible initial
turns.  my program took about 6 and a half hours to exhaustively search
these through 17 face turns.  no solutions were found, and therefore the
original position is more than 18 face turns from start.

i realize that this is not nearly as satisfying as obtaining all minimal
maneuvers.  that will take about 13 times as long, but is feasible with
my current program.

mike


From cube-lovers-errors@oolong.camellia.org  Thu Jul 10 00:56:16 1997
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Date: Thu, 10 Jul 1997 00:54:58 -0400
From: michael reid <reid@math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: all minimal maneuvers for superflip (face turn metric)

i can now give all minimal maneuvers for superflip in the face turn
metric.  recall that there are three operations we may apply to any
maneuver for superflip which give another maneuver of the same length:

1. we may conjugate the maneuver by any symmetry of the cube.
2. we may cyclically shift the maneuver; i.e. replace

      sequence_1 sequence_2     by     sequence_2 sequence_1

3. we may replace the maneuver by its inverse.

my original search (january 1995) for superflip in 19 face turns was
divided into 16 cases.  since i used my (unhacked) version of kociemba's
algorithm, the search through each case produced maneuvers for superflip,
and 8 of these cases found maneuvers of length 20f.  i previously reported
that these were each equivalent to dik winter's maneuver, using the three
operations above.  however, i was mistaken about this; there were two
different maneuvers which differ only very slightly.

to facilitate an exhaustive search through 20f, i'll use a result of a
previous search.

proposition.  any maneuver for superflip in 20f contains a 180 degree
              face turn.
proof.  otherwise the maneuver would be 20 quarter turns long.  however, i
        did an exhaustive search through 20q and found no maneuvers.  qed.

(in fact, this quarter turn result was later improved by jerry bryan, who
showed that superflip is not within 22q of start, and therefore is exactly
24q from start.)

now the symmetry reductions show that we may take the first two face turns
to be  U R2 .  my program exhaustively searched the position   superflip U R2
through 18f.  it took 35 hours, and found 30 maneuvers, which came in two
different types:

     U R2 F B R B2 R U2 L B2 R U' D' R2 F   R' L  B2 U2 F2   (20f)
     U R2 F B R B2 R U2 L B2 R U' D' R2 F   D2 B2 U2 R' L    (20f)

note that these are the same except for the last 5 face turns.  (this gives
the relation   R' L B2 U2 F2 R L' U2 B2 D2  = identity; alternatively, the
same sequence produces  (f++)(d++)  in the supergroup.)

from this, we can count the exact number of 20f sequences for superflip.
both of the above may be cyclically shifted in 23 different ways.  we get
23 different ways, instead of 20, because there are three separate pairs
of consecutive twists of opposite faces.  we'd consider

     sequence_1  U D  sequence_2     and     sequence_1  D U  sequence_2

to be the same, but we wouldn't consider

     U  sequence  D     and     D  sequence  U

to be the same.  yet cyclic shifting of these last two produces the same
maneuver.

we can also conjugate by any of the 48 symmetries of the cube, and we can
also invert any of the maneuvers.  all these operations produce different
maneuvers, so we get a total of  2 * 23 * 48 * 2 = 4416  different maneuvers.

by counting, the number of different sequences of length  <= 19f  is about
82 times as many positions the cube has.  thus a position has, on average,
82 maneuvers of length  <= 19f, although superflip has 0.  the number of
different sequences of length  20f  is about 1016 times the number of
positions, so a position has, on average, 1016 different maneuvers of
length 20f.  superflip has more than 4 times that many.

here are the 30 solutions my program found for  superflip U R2.  hopefully
i haven't made any mistakes this time.  they should all be equivalent to
one of the two listed above.

     U R2    F  U2 F2 D2 R' L  U  R2 F' B' R  D2 L  F2 R  D2 R  D    (20f)
     U R2    F  B  R  B2 R  U2 L  B2 R  U' D' R2 F  R' L  B2 U2 F2   (20f)
     U R2    F  B  R  B2 R  U2 L  B2 R  U' D' R2 F  D2 B2 U2 R' L    (20f)
     U R2    F  R' L  F2 D2 B2 U  R2 F' B' R  D2 L  F2 R  D2 R  D    (20f)
     U R2    F2 L2 F  D2 R  L  D  R2 D  F2 U  R2 D  F' B' D2 L  D'   (20f)
     U R2    F2 L2 F' U2 R' L' U' R2 U' F2 D' R2 U' F  B  U2 L' D'   (20f)
     U R2    F2 L2 B  D2 R' L' D  F2 U  R2 D  F2 D  F  B  D2 R  D'   (20f)
     U R2    F2 L2 B' U2 R  L  U' F2 D' R2 U' F2 U' F' B' U2 R' D'   (20f)
     U R2    F' U2 B2 D2 R  L' D' R2 F' B' R' B2 R' D2 L' B2 R' D    (20f)
     U R2    F' B' R  D2 L  F2 R  D2 R  U  D  R2 F  U2 F2 D2 R' L    (20f)
     U R2    F' B' R  D2 L  F2 R  D2 R  U  D  R2 F  R' L  F2 D2 B2   (20f)
     U R2    F' R  L' B2 D2 F2 D' R2 F' B' R' B2 R' D2 L' B2 R' D    (20f)
     U R2    U  B2 D  R2 U  F' B' U2 L  F2 R2 B2 U' D  B  U2 R  L    (20f)
     U R2    U  B2 D  R2 U  F' B' U2 L  U' D  R2 B2 L2 B  U2 R  L    (20f)
     U R2    U  R  L  U2 F  L2 F2 R2 U' D  L  U2 F' B' U  R2 D  F2   (20f)
     U R2    U  R  L  U2 F  U' D  F2 R2 B2 L  U2 F' B' U  R2 D  F2   (20f)
     U R2    U2 L2 F' B  R  F2 U' D' F  L2 B  U2 F  L2 F  R  L  F2   (20f)
     U R2    B  R' L  B2 U2 F2 D  R2 F' B' R  U2 L  B2 R  U2 R  D    (20f)
     U R2    B  D2 B2 U2 R' L  D  R2 F' B' R  U2 L  B2 R  U2 R  D    (20f)
     U R2    B' R  L' F2 U2 B2 U' R2 F' B' R' F2 R' U2 L' F2 R' D    (20f)
     U R2    B' D2 F2 U2 R  L' U' R2 F' B' R' F2 R' U2 L' F2 R' D    (20f)
     U R2    D  F2 U  R2 U  R  L  U2 F  L2 F2 R2 U' D  L  U2 F' B'   (20f)
     U R2    D  F2 U  R2 U  R  L  U2 F  U' D  F2 R2 B2 L  U2 F' B'   (20f)
     U R2    D  F2 U  R' L' U2 B  L2 F2 R2 U' D  R  U2 F  B  U  F2   (20f)
     U R2    D  F2 U  R' L' U2 B  U' D  F2 R2 B2 R  U2 F  B  U  F2   (20f)
     U R2    D  F2 D  F  B  D2 R  U  D' R2 F2 L2 B  D2 R' L' D  F2   (20f)
     U R2    D  F2 D  F  B  D2 R  B2 R2 F2 U  D' B  D2 R' L' D  F2   (20f)
     U R2    D  F' B' D2 L  U  D' R2 F2 L2 F  D2 R  L  D  R2 D  F2   (20f)
     U R2    D  F' B' D2 L  B2 R2 F2 U  D' F  D2 R  L  D  R2 D  F2   (20f)
     U R2    D2 L2 F  B' L  B2 U  D  B  D2 B  L2 F  D2 B  R' L' B2   (20f)

mike


From cube-lovers-errors@oolong.camellia.org  Sun Jul 13 19:35:33 1997
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Date: Sun, 13 Jul 1997 19:37:58 -0400
From: michael reid <reid@math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: minimal maneuvers for composition of superflip and pons asinorum

i've finished calculating all minimal maneuvers (in the face turn metric)
for the composition of superflip and pons asinorum.  in my search for
maneuvers of length  <= 18f  for this position, i used symmetry to show
that we may suppose that the first face turn is either  U  or  U2 .
in fact, there is more symmetry available, and this time i will use it.

in the case beginning with  U , there are four symmetries, generated
by the cube rotation  C_U .  using these, we may suppose that the second
face turn is one of  D , D2 , D' , R , R2  or  R' .

in the case beginning with  U2 , there are eight symmetries.  these are
generated by the cube rotation  C_U  and reflection through the left-right
plane.  using these, we may assume that the second face turn is one of
D , D2 , R  or  R2 .

we can reduce these cases somewhat further.  the cases beginning with
U D2  and with  U2 D  are equivalent, so only one needs to be seacrhed.

the cases beginning with  U D'  and with  U2 D2  can also be eliminated.
both  U D'  and  U2 D2  commute with both pons asinorum and with superflip,
so we may cyclically shift these turns to the end of the maneuver.  our
position cannot be achieved only using these "slice" turns, so we'll
always be able to cyclically shift until we do not begin with a slice turn.
(alternatively, note that any maneuver of length  19f , or any odd length
cannot consist only of slice turns!)

that leaves seven cases to search.  my program took just less than one
day to search all through 17f.  it found 26 maneuvers, 16 for the case
beginning with  U D .  however, this case has 8 symmetries, so there are
just 2 different maneuvers, each in 8 different orientations.

this leaves 12 different maneuvers, which come in 6 pairs of inverses.
they are:

     U  R    F  D  R  U' D  L' U' D  F' B2 R  L' D' F' L' B' R'   (19f)
     U  D    F  R  L' F  B' L  D2 R  L  F' B' U' L2 F  B' U2 L'   (19f)
     U  D    F' B' L' U2 F' B  L2 U' R' L' F' U' D  F' B  D' L2   (19f)
     U2 R    F  U  F  B' L' D' F  B' L  B  R  L' U  D2 B' R' U2   (19f)
     U2 R    F  U2 D' R' L  F' L' F  B' U  L  F  B' D' B' R' U2   (19f)
     U2 R    U2 D2 R  U' L' U  B  R  F2 U' D  B' R' F' D  B' L2   (19f)

and their inverses.  the first of these is the maneuver found by
jerry bryan.  it's also the only of these that is  20 quarter turns
long, which is consistent with his findings.

mike


From cube-lovers-errors@oolong.camellia.org  Mon Jul 14 12:37:17 1997
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Date: Mon, 14 Jul 1997 11:39:21 -0400
To: cube-lovers@ai.mit.edu
From: karen angelli <kangelli@tiac.net>
Subject: hockey puck puzzle
Mime-Version: 1.0
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I recently purchased a hockey puck puzzle from Whole Systems Design, who
also brought us the Masterball.  Since I received it, I've played with for
only a couple of hours without beginning any systematic attempt at solving
it.  My initial impression is that, although it looks like is should be
very easy, it may actually be pretty hare.  Or, it's one of those puzzles
that is so easy that I'm thinking too hard and can't see the answer.  I've
never heard anyone in the cube-lovers discuss the puzzle, and I wondered if
any of you had any impressions of it.

The puzzle has the shape, size and feel of a regulation hockey puck, and is
divided into twelve wedges.  It's basically a flattened masterball in which
only the lines of longitude twist, and the lines of latitude do not.  There
are several designs, with various degrees of difficulty and redundancy.
For example, on some, there are printed hockey players, Maple leafs, or
American flags.  On my version, the wedges are numbered consecutively from
one through twelve.  The pristine version is with the numbers lined up in
consecutive order.

I'm not really interested in learning a solution from anybody, but I would
be interested in comments about whether you think the puzzle is harder or
easier than it looks.

YOu can see pictures of the hockey puck puzzle or order them at
www.wsd.com/HockeyPuck/home.

'e-ya later,

Pete.





From cube-lovers-errors@oolong.camellia.org  Mon Jul 14 14:02:22 1997
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Date: Mon, 14 Jul 1997 13:59:16 -0400 (EDT)
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From: Allan Wechsler <awechsle@bbn.com>
To: karen angelli <kangelli@tiac.net>
Cc: cube-lovers@ai.mit.edu
Subject: hockey puck puzzle
In-Reply-To: <3.0.1.32.19970714113921.006aa4a4@pop.tiac.net>
References: <3.0.1.32.19970714113921.006aa4a4@pop.tiac.net>

Please give a clearer description of the puck.  The photo gives only a
slight clue about how many pieces there are, and how they are
arranged.  Are there twelve pieces, or twenty-four, or more?

-A


From cube-lovers-errors@oolong.camellia.org  Tue Jul 15 13:18:50 1997
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Date: Tue, 15 Jul 1997 17:57:11 +0200
From: Rob Hegge <R.F.Hegge@MP.TUDelft.NL>
Subject: Description of hockey puck puzzle
To: cube-lovers@ai.mit.edu
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The hockey puck puzzle is a flat disk with a diameter of about 9 cm
or 3.5 inches and a thickness of about 2.5 cm or 1 inch. It basically
consists of a circle (the hart) and a "ring" surrounding the circle.
The circle is cut into two equal halves like "(|)". The two halves are
connected so that you can turn one half upside down, while holding the
other half. The ring is cut (from front to back) into 12 equal wedges,
each of which is attached to the circle by a dovetail so that the ring
with the wedges can be moved around the circle. One can also flip six
wedges including one half of the circle around so that afterwards those
6 wedges and the half circle face backwards. Thus the puzzle is similar
to a puzzle called saturn (which has only 8 wedges ?). The type of moves
reminds me of moves possible on square-1.

In the puzzle I own the 12 wedges on the front are numbered from 1 to 12
and on the back with the letters of "hockeypuzzle", while the left half
circle contains the letters "pu" and right half circle the letters "ck"
as shown below. I do not have it here so this was straight from memory.

    front:        back:

     12 1             c k
   11     2         o     e
 10    |    3     h    |    y
       |             pu|ck
  9    |    4     p    |    e
    8     5         u     l
      7 6             z z

The three "|" denote the cut through the circle.
A flip as described above would give for instance

     12 k             c 1      
   11     e         o     2   
 10    |    y     h    |    3   
       |ck           pu|    
  9    |    e     p    |    4  
    8     l         u     5       
      7 z             z 6


while then a clockwise turn of the ring for one wedge would give:

     11 12            1 2
   10     k         c     3
  9    |    e     o    |    4
       |ck           pu|    
  8    |    y     h    |    5
    7     e         p     6
      z l             u z

For a rotational puzzle it is not that difficult.

Rob



From cube-lovers-errors@oolong.camellia.org  Tue Jul 15 13:18:23 1997
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Date: Tue, 15 Jul 1997 09:41:28 -0400 (EDT)
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From: Allan Wechsler <awechsle@bbn.com>
To: jandr@xirion.nl
Cc: Allan Wechsler <awechsle@bbn.com>, cube-lovers@ai.mit.edu
Subject: Re: hockey puck puzzle
In-Reply-To: <9707151143.AA27610@la1.apd.dec.com>
References: <199707141759.NAA00043@spork.bbn.com>
	<9707151143.AA27610@la1.apd.dec.com>

    [Jan de Ruiter:]
    The puzzle contains edge pieces and center pieces, all with the same
    thickness as the puck.
    There are always two center pieces which together form the inner
    circle.
    In this case there are 12 edge pieces which together form the outer
    ring. Simpeler pucks might contain less than 12 edge pieces, but always
    an even number.
    The possible moves are:
    1. rotate one center piece with half of the edge pieces 180 degrees
       relative to the other center piece and the other half of the edge
       pieces.
    2. rotate the edge pieces around the center pieces, always multiples of
       30 degrees, or 1/12 of a circle (or more if there are less edge
       pieces)
    
    I know the puck with 6 edge pieces is near trivial to solve.
    I haven't tried the other ones yet.
    
    Jan de Ruiter
    
Thanks for the description -- Pete ("Karen Angelli") provided an
identical one in a private reply.  But this is still incomplete.  Are
the obverse and reverse of the individual pieces distinguishable?
Suppose I manage to flip every other edge piece over in place (not
sure this is possible).  Does it then look solved?  Or do the two
sides have different colors or a distinguishing mark or something?

I haven't tried it, but I can't imagine that this puzzle would be very
difficult.

-A




From cube-lovers-errors@oolong.camellia.org  Tue Jul 15 13:17:34 1997
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Subject: Re: hockey puck puzzle
To: Allan Wechsler <awechsle@bbn.com>
Date: Tue, 15 Jul 1997 13:42:54 +0200 (CETS)
From: Jan de Ruiter <jandr@apd.dec.com>
Cc: cube-lovers@ai.mit.edu
In-Reply-To: <199707141759.NAA00043@spork.bbn.com> from "Allan Wechsler" at Jul 14, 97 01:59:16 pm
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> 
> Please give a clearer description of the puck.  The photo gives only a
> slight clue about how many pieces there are, and how they are
> arranged.  Are there twelve pieces, or twenty-four, or more?
> 
> -A
> 
The puzzle contains edge pieces and center pieces, all with the same
thickness as the puck.
There are always two center pieces which together form the inner
circle.
In this case there are 12 edge pieces which together form the outer
ring. Simpeler pucks might contain less than 12 edge pieces, but always
an even number.
The possible moves are:
1. rotate one center piece with half of the edge pieces 180 degrees
   relative to the other center piece and the other half of the edge
   pieces.
2. rotate the edge pieces around the center pieces, always multiples of
   30 degrees, or 1/12 of a circle (or more if there are less edge
   pieces)

I know the puck with 6 edge pieces is near trivial to solve.
I haven't tried the other ones yet.

Jan de Ruiter


From cube-lovers-errors@oolong.camellia.org  Wed Jul 16 09:37:39 1997
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From: "joyner.david" <joyner.david@mathnt1.sma.usna.navy.mil>
To: "'cube-lovers@ai.mit.edu'" <cube-lovers@ai.mit.edu>
Subject: RE: hockey puck puzzle
Date: Wed, 16 Jul 1997 07:53:21 -0400
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>----------
>From: 	Jan de Ruiter[SMTP:jandr@apd.dec.com]
>Sent: 	Tuesday, July 15, 1997 7:42 AM
>To: 	Allan Wechsler
>Cc: 	cube-lovers@ai.mit.edu
>Subject: 	Re: hockey puck puzzle
>
>> 
>> Please give a clearer description of the puck.  The photo gives only a
>> slight clue about how many pieces there are, and how they are
>> arranged.  Are there twelve pieces, or twenty-four, or more?

Spencer Robinson, a former student, and I have written a WWW 
page which sketches an easy but rather inefficient solution of the 12
piece 
hockey puck puzzle:
http://www.nadn.navy.mil/MathDept/wdj/mball/puck.htm
Have fun!                             - David Joyner

>> 
>> -A
>> 
>


From cube-lovers-errors@oolong.camellia.org  Wed Jul 16 10:18:34 1997
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Date: Wed, 16 Jul 1997 09:44:21 -0400 (Eastern Daylight Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: No Local Maxima 11q from Start
To: cube-lovers@ai.mit.edu
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My new Shamir program has now generated the entire search tree for the
standard cube group G up to 11q from Start.  This search was
accomplished once before using my old tape spinning programs, so there
is limited new information. 

One good result is that all the numbers match between the two programs. 
The matching results were obtained using different programs,
implementing different algorithms, in different programming languages,
on different hardware platforms, and under a different operating
systems.  So I feel pretty good about the numbers.  They have been
posted before, so I won't post them again.  With problems this big, it
is always good to have some sort of independent verification because it
is impossible to verify anything by hand.

Another interesting result is in fact new.  The old program was only
able to determine local maxima up to 10q from Start while calculating
the 11q tree.  The new program is able to determine local maxima up to
the same distance from Start it is searching.  There are no local maxima
11q from Start.  I find this result somewhat surprising, since there are
four local maxima (unique up to M-conjugacy) 10q from Start.  The new
program did confirm the previously known 10q local maxima, but failed to
find any 11q local maxima.

In its search for local maxima, for each position x the program
calculates the set E(x) of quarter turns with which a minimal process
for the position can end.  We call |E(x)| the maximality of x, and a
position is a local maximum if its maximality is 12.  At a distance of
11q from Start, there exist positions with maximality values for every
number in 1..11.  This is the first time we have found any positions
with a maximality of 9 or 11.  (See my note of 16 June 1995, "10q Local
Maxima Search Matrix".)  There seem to be more positions with even
maximality values than odd, and a maximality of 11 is especially
interesting because such a position is "almost" a local maximum. 

I am disappointed in the speed of my program.  For this run, it
identified about 1100 patterns (representative elements of M-conjugacy
classes) per second.  This corresponds to about 50,000 positions per
second (about 48 times 1100).  The program is running on a Pentium P166
with 16MB memory under Windows/95.  My concern is that I have worked so
hard to make the program run in small amounts of memory that it is
running too slow.  I am now going to take out a few of the memory saving
techniques to see if I can speed it up a bit. 

The program is actually about 20MB, and runs successfully on a 16MB
machine due to the good graces of virtual memory.  In fact, I can
calculate out to 11q from Start even on an 8MB machine.  But trying to
calculate out to 12q from Start fails on the 16MB machine (the program
is the same size for 11q from Start and for 12q from Start because I am
storing all positions up to 6q from Start.  The program would only have
to be made larger if I were to try calculating 13q from Start or 14q
from Start.)

When I say the program fails at 12q from Start, I mean that the virtual
memory thrashes unmercifully, and therein lies an interesting tale.  Why
should the program be able to calculate 11q without thrashing, but
thrash so badly at 12q?  It has to do with the Shamir algorithm itself. 

Recall that we are producing products of the form ST in lexicographic
order.  To be specific, we are producing products of the from St in
lexicographic order for all t in T and merging the results.  S itself is
already in lexicographic order.  Think of processing a dictionary, and
thing of processing S in lexicographic order.  We essentially process
all the A's, followed by all the B's, then all the C's, etc.  There is
very good locality of reference as far as the virtual memory goes. 

Moving up to St, we might first process all the N's, then all the E's,
then all the Z's, etc, but there is still very good locality of
reference. There is an occasional big jump in where we are referencing
memory, but most of the time we reference elements of the set S which
are very close together in memory. 

When we calculate 11q from Start, S is the set Q[6] of positions which
are 6q from Start, and T is the set Q[5].  Because Q[5] is only about
1/9 as big as Q[6], the real memory working set to calculate Q[6]Q[5] is
only about 10% of the total virtual memory of the program, maybe about
2MB.  But when we move up to calculating 12q, we move up to Q[6]Q[6] and
the real memory working set becomes the whole 20M program.  This simply
doesn't work on a 16MB machine.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990



From cube-lovers-errors@oolong.camellia.org  Thu Jul 17 14:17:15 1997
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Date: Wed, 16 Jul 1997 22:41:08 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: No Local Maxima 11q from Start
In-reply-to: <Pine.WNT.3.96.970716085137.-650973C-100000@ER123A.PSTCC.CC.TN.US>
To: cube-lovers@ai.mit.edu
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On Wed, 16 Jul 1997, Jerry Bryan wrote:

>   (See my note of 16 June 1995, "10q Local
> Maxima Search Matrix".)  

>                      There seem to be more positions with even
> maximality values than odd, ...

This statement is bogus, which is clear if you look at my chart from 1995. 
Most of the positions have a maximality of 1 this close to Start.  I was
thinking of another situation.  That is, when I looked at local maxima in
the corners only group, the inverses of the local maxima tended to have
even maximality.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990







From cube-lovers-errors@oolong.camellia.org  Wed Jul 23 16:50:10 1997
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Date: Wed, 23 Jul 1997 06:42:55 -0400
From: Edwin Saesen <ERCO@compuserve.com>
Subject: Where can I get...?
To: CUBE <CUBE-LOVERS@ai.mit.edu>
Message-ID: <199707230643_MC2-1B6E-D8FD@compuserve.com>
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Hi everyone,

sorry if this is a boring question for most of you (and if it is, please
reply by private mail instead of to the list), but I'm new here, and my
most important question at the moment is where to get any of the followin=
g
puzzles:

Rubik's Wahn (5x5x5) (maybe also called Professor's cube, Ultimate or
Master Revenge)
Rubik's Revenge (4x4x4)
Rubik's Domino (3x3x2)
Rubik's Pocket Cube (2x2x2)
Pyraminx
Pyraminx Octahedron
Megaminx

I'd prefer sources in germany, although I doubt it that I can find any of=

them here...

Michael


From cube-lovers-errors@oolong.camellia.org  Wed Jul 23 17:35:00 1997
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Date: Wed, 23 Jul 1997 14:24:27 -0700
From: Jin "Time Traveler" Kim <chrono@ibm.net>
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Organization: The Fourth Dimension
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Edwin Saesen wrote:
> 

> Rubik's Wahn (5x5x5) (maybe also called Professor's cube, Ultimate or
> Master Revenge)
> Rubik's Revenge (4x4x4)
> Rubik's Domino (3x3x2)
> Rubik's Pocket Cube (2x2x2)
> Pyraminx
> Pyraminx Octahedron
> Megaminx
> 
> I'd prefer sources in germany, although I doubt it that I can find any of
> them here...
> 
> Michael

You are quite in luck.  There is a great source of puzzles in Germany.

Christoph.Bandelow@rz.ruhr-uni-bochum.de
bandecbv@rz.ruhr-uni-bochum.de

I'm not sure which is correct, or maybe they both work, but Dr.
Christoph Bandelow has a catalog available for people to buy various
puzzles.

He should have all of the above puzzles available for purchase, except
the 4x4x4, 3x3x2, and the Megaminx.  All of them are rather hard to
locate.  Especially tough is the 4x4x4 because of its high demand.  I've
been looking for one as well for a number of years.  By many accounts,
there are no more for sale anywhere except maybe private collections. 
And if you own one, you're not likely to part with it anyway.  So good
luck.

-- 
Jin "Time Traveler" Kim
chrono@ibm.net
VGL Costa Mesa
http://www.geocities.com/timessquare/alley/9895
http://www.slamsite.com


From cube-lovers-errors@oolong.camellia.org  Wed Jul 23 22:49:31 1997
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From: Christoph Bandelow <bandecbv@mailhost.rz.ruhr-uni-bochum.de>
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Edwin Saesen asked for a source of "Rubik's Wahn" (5x5x5 magic cube) 
and some other rotational puzzles. 

As our busy "Time Traveler"  Jin Kim already kindly remarked, the 
5x5x5 Magic Cube and many other rotational puzzles are still 
available from me. This refers also to the Megaminx, both the one 
made in Hong Kong - sold as Megaminx - and the slightly better one 
made in Hungary - sold as Super Nova or as Magic Dodecahedron.
By the way, I do now sell the 5x5x5 cube under the name "Giant Magic 
Cube" (In Germany: "Riesen-Zauberwuerfel", in France: "Cube Magique 
Geant") , and I hope this sounds nice and doesn't create too much new 
confusion. At least the price is still the same: 40 DM or 24 USD. 
Just email me your postal address to receive your free copy of my 
mail order catalog.

Cube-Lovers: Please notice my new slightly simplified email address.

Christoph 
Christoph Bandelow
mailto:Christoph.Bandelow@ruhr-uni-bochum.de


From cube-lovers-errors@oolong.camellia.org  Wed Jul 23 22:49:09 1997
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From: Tim Mirabile <tim@mail.htp.com>
To: CUBE <CUBE-LOVERS@ai.mit.edu>
Subject: Re: Where can I get...?
Date: Wed, 23 Jul 1997 23:16:49 GMT
Organization: http://www.webcom.com/timm/
Message-ID: <33d68e99.988371@mail.htp.com>
References: <199707230643_MC2-1B6E-D8FD@compuserve.com> <33D6768B.707E@ibm.net>
In-Reply-To: <33D6768B.707E@ibm.net>

On Wed, 23 Jul 1997 14:24:27 -0700, Jin "Time Traveler" Kim <chrono@ibm.net>
wrote:

>He should have all of the above puzzles available for purchase, except
>the 4x4x4, 3x3x2, and the Megaminx.  All of them are rather hard to
>locate.  Especially tough is the 4x4x4 because of its high demand.  I've
>been looking for one as well for a number of years.  By many accounts,
>there are no more for sale anywhere except maybe private collections. 
>And if you own one, you're not likely to part with it anyway.  So good
>luck.

Hmmm.  I would imagine a large number of the 4x4x4 cubes sold in the U.S. as
"Rubik's Revenge" were sold to non-cube-lovers and are now tucked away in attics
and basements with all the other toys the kids have outgrown.  For this reason
I've considered checking out local garage and yard sales.  (Before it became
widely known what the value of old baseball cards could be, these sales were an
excellent way for collectors to pick up large boxes of old cards for only a few
dollars).

Anyway, I hope Christoph still has the Megaminx because I just sent him a check
for some items including this. :)  There was no indication he was out of them in
the catalog I just got.

--
For USCF & FIDE rated chess on Long Island -> http://www.webcom.com/timm/
TimM on the Free Internet Chess Server - telnet://fics.onenet.net:5000/
Webmaster, tech support - ICD/Your Move Chess & Games: http://www.icdchess.com/
The opinions of my employers are not necessarily mine, and vice versa.


From cube-lovers-errors@mc.lcs.mit.edu  Fri Jul 25 12:21:37 1997
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Mail-from: From R.F.Hegge@MP.TUDelft.NL Thu Jul 24 04:40:13 1997
Date: Thu, 24 Jul 1997 10:36:30 +0200
From: R.F.Hegge@MP.TUDelft.NL (Rob Hegge)
Subject: Re: Where can I get...?
To: CUBE-LOVERS@ai.mit.edu
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> Rubik's Wahn (5x5x5) (maybe also called Professor's cube, Ultimate or
> Master Revenge)
> Rubik's Revenge (4x4x4)
> Rubik's Domino (3x3x2)
> Rubik's Pocket Cube (2x2x2)
> Pyraminx
> Pyraminx Octahedron
> Megaminx
> 
> I'd prefer sources in germany, although I doubt it that I can find any of
> them here...

Most of them can indeed be bought from C.Bandelow.

The only source for a 4x4x4 that I know of is Puzzletts, where I bought mine
a year ago for about 50 US $. You can see their online mailorder catalog at
www.puzzletts.com. They are based in Seattle, USA.

I myself am still looking for Rubik's Domino (3x3x2).

Rob, 
r.f.hegge@ctg.tudelft.nl

PS
If you or anyone else is interested in puzzles like this. There still exist
a club called "Nederlandse Kubus Club" (NKC) or Dutch Cubist Club.
About half of its members are not Dutch, despite its name, but American,
German, Japanese etc. They publish a magazine called Cubism For Fun (CFF)
three times a year, which deals with all kinds of (mechanical) puzzles, like
polyform puzzles, sliding puzzles, rotational puzzles, burrs etc. They also
organise a Cube Day once a year in the Netherlands for members with talks
about puzzles and where new and SECOND HAND puzzles can be traded, bought,
admired etc.

From cube-lovers-errors@mc.lcs.mit.edu  Fri Jul 25 12:39:18 1997
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Mail-from: From whuang@ugcs.caltech.edu Thu Jul 24 15:30:43 1997
To: mlist-cube-lovers@nntp-server.caltech.edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Re: Where can I get...?
Date: 24 Jul 1997 19:27:15 GMT
Organization: California Institute of Technology, Pasadena
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Tim Mirabile <tim@mail.htp.com> writes:

>Hmmm.  I would imagine a large number of the 4x4x4 cubes sold in the
>U.S. as "Rubik's Revenge" were sold to non-cube-lovers and are now
>tucked away in attics and basements with all the other toys the kids
>have outgrown.  For this reason I've considered checking out local
>garage and yard sales.

I suggest any cube-lover who is considering going to local garage
and yard sales to not do so!  This way, *I* can get all of the 
old cubes and other puzzles!!  In fact, I've acquired TWO 4x4x4's
by this method in the last five years, as well as one of Nob
Yoshigahara's Pineapple Puzzles for only fifty cents!  The old 
lady who sold it to me even cautioned me not to eat it!

(Hmm... I'm not doing a very good job of convincing you guys to
avoid these sales, am I?  ;-)  )

--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
finger me for /etc/passwd

From cube-lovers-errors@mc.lcs.mit.edu  Fri Jul 25 13:54:12 1997
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Mail-from: From pdwyer@ecn.net.au Fri Jul 25 05:16:15 1997
Message-Id: <199707250905.AA10662@ecn.net.au>
From: "Peter Dwyer" <pdwyer@ecn.net.au>
To: "CUBE" <CUBE-LOVERS@ai.mit.edu>
Subject: Re: Where can I get...?
Date: Fri, 25 Jul 1997 18:57:16 +1000

> Hmmm.  I would imagine a large number of the 4x4x4 cubes sold in the
> U.S. as "Rubik's Revenge" were sold to non-cube-lovers and are now
> tucked away in attics and basements with all the other toys the kids
> have outgrown.  For this reason I've considered checking out local
> garage and yard sales.  (Before it became widely known what the
> value of old baseball cards could be, these sales were an excellent
> way for collectors to pick up large boxes of old cards for only a
> few dollars).

I got my 4x4x4 from a fleamarket for $1.  I was so exicited when I saw
it I told the lady I would pay anything and she said $1 :-)

Donna
 



From cube-lovers-errors@mc.lcs.mit.edu  Fri Jul 25 20:58:24 1997
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Mail-from: From chrono@ibm.net Fri Jul 25 17:00:49 1997
Message-Id: <33D8EF4F.69C5@ibm.net>
Date: Fri, 25 Jul 1997 11:24:15 -0700
From: "Jin \"Time Traveler\" Kim" <chrono@ibm.net>
Reply-To: chrono@ibm.net
Organization: The Fourth Dimension
To: CUBE <CUBE-LOVERS@ai.mit.edu>
Subject: Re: Where can I get...?
References: <199707250905.AA10662@ecn.net.au>

Peter Dwyer wrote:
> 
> > Hmmm.  I would imagine a large number of the 4x4x4 cubes sold in the
> > U.S. as "Rubik's Revenge" were sold to non-cube-lovers and are now
> > tucked away in attics and basements with all the other toys the kids
> > have outgrown.  For this reason I've considered checking out local
> > garage and yard sales.  (Before it became widely known what the
> > value of old baseball cards could be, these sales were an excellent
> > way for collectors to pick up large boxes of old cards for only a
> > few dollars).
> 
> I got my 4x4x4 from a fleamarket for $1.  I was so exicited when I saw
> it I told the lady I would pay anything and she said $1 :-)
> 
> Donna
> 

Indeed, while we are sharing 4x4x4 stories, I got mine (on an "extended"
borrow) while digging around in a friend's garage.  (Just hunting for
junk) I found the Rubik's Revenge in a plastic bucket behind some old
paint cans with all teh stickers peeled off.  The cube is in great
shape, but the stickers are pretty trashed (had to Krazy Glue them back
on).  But at least it works well.

-- 
Jin "Time Traveler" Kim
chrono@ibm.net
VGL Costa Mesa
http://www.geocities.com/timessquare/alley/9895
http://www.slamsite.com




From cube-lovers-errors@mc.lcs.mit.edu  Fri Jul 25 21:02:50 1997
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Mail-from: From tim@mail.htp.com Fri Jul 25 19:32:56 1997
From: tim@mail.htp.com (Tim Mirabile)
To: CUBE-LOVERS@ai.mit.edu
Subject: Re: Where can I get...?
Date: Fri, 25 Jul 1997 23:29:20 GMT
Organization: http://www.webcom.com/timm/
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On Thu, 24 Jul 1997 10:36:30 +0200, R.F.Hegge@MP.TUDelft.NL
 (Rob Hegge) wrote:


>The only source for a 4x4x4 that I know of is Puzzletts, where I
>bought mine a year ago for about 50 US $. You can see their online
>mailorder catalog at www.puzzletts.com. They are based in Seattle,
>USA.

I tried to order from them first, in the beginning of June.  There was
no response to my order or the followup email I sent.

--
TimM on the Free Internet Chess Server - telnet://fics.onenet.net:5000/
The opinions of my employers are not necessarily mine, and vice versa.

From cube-lovers-errors@mc.lcs.mit.edu  Sat Jul 26 12:33:57 1997
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Mail-from: From ERCO@compuserve.com Sat Jul 26 05:11:01 1997
Date: Sat, 26 Jul 1997 05:06:24 -0400
From: Edwin Saesen <ERCO@compuserve.com>
Subject: Re: Re: Where Can I get...
To: CUBE <CUBE-LOVERS@ai.mit.edu>
Message-Id: <199707260506_MC2-1B9B-F7B3@compuserve.com>

Hi everyone,

thanks for all of the responses for my original inquiry about those various
cubes.

Rob wrote:
>There still exist
>a club called "Nederlandse Kubus Club" (NKC) or Dutch Cubist Club.
This sounds interesting. Can you tell me how to contact them? And, does
anyone know if there's a similar organization in germany?

Wei-Hwa Huang wrote:
>I suggest any cube-lover who is considering going to local garage
>and yard sales to not do so!  This way, *I* can get all of the
>old cubes and other puzzles!!
Well, it seem s to me that all those people in the USA are much more lucky
than I am in germany. I've been going to flea markets for years (not only
for cubes, mainly for buying records), and I've never ever seen a 4x4x4 for
sale there, only lots of 3x3x3s.

Concerning Puzzletts:
>The only source for a 4x4x4 that I know of is Puzzletts,

>I tried to order from them first, in the beginning of June.  There was
>no response to my order or the followup email I sent.

I asked a friend of mine in the USA to order one for me, so I simply *HOPE*
that the above described situation isn't their regular way of doing
business. What might help is the fact that one of their retailers is in the
city where my friend lives, so he might have a chance of getting one there.

Concerning my old 4x4x4: Is there any way to fix broken center pieces?
That's the reason why I need a new one, I still have all the pieces but two
of them are broken and I have no idea if there's a way to fix them (I doubt
it, though).

Furthermore, does anyone of you maybe have a broken 4x4x4 and is willing to
sell that one? Maybe I can fix my old one with some of those pieces then.
(But probably they wouldn't fit together if they were done by different
companies).
I also could do with someone willing to sell a working 4x4x4 by the way :-)

Sorry for the length of this

Michael


From cube-lovers-errors@mc.lcs.mit.edu  Sun Jul 27 21:30:37 1997
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Mail-from: From nbodley@tiac.net Sun Jul 27 11:09:16 1997
Date: Sun, 27 Jul 1997 11:05:58 -0400 (EDT)
From: Nicholas Bodley <nbodley@tiac.net>
To: Edwin Saesen <ERCO@compuserve.com>
Cc: CUBE <CUBE-LOVERS@ai.mit.edu>
Subject: Broken 4^3s; advice on repairs to plastic (medium length)
In-Reply-To: <199707260506_MC2-1B9B-F7B3@compuserve.com>
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 Edwin's mention of broken center pieces reminds me of a period roughly 15
years ago when I was quite active disassembling various cubes and their
relatives; I'm fairly sure that it was these center pieces that impressed
me as being fragile. Even the wondrous innards of the 5^3 are not as
delicate, although fascinating. 

 I'm trying to remember without much success whether I actually broke a
piece. I think I did. Some plastics, I'm just about sure, can't be
dissolved by any readily-available liquids, and i think the 4^3s were made
of such material. (Acrylics can be "solvent-welded" very successfully.)
Plastic welding is possible with probably any thermoplastic (i.e., one
that can be melted after being molded), but is a skill just as is metal
welding, and needs expensive hot-air tools (others also?) designed for the
purpose.  I would not recommend it for something so precious as a 4^3,
unless you can find an expert.

 Finally, adhesives are worth considering. After all, the colored stickers
are retained by adhesive. Not at all sure, but I think I did have success
with cyanoacrylate (CA) (famous in the USA for its tradename "Krazy
Glue"). This is a strange substance that seems not to be well understood,
and it might be a good idea to learn more about it (even if you already
think you know) before trying a critical repair. At least, if your repair
is unsuccessful it will come apart, with probably little damage (no
promises!) to the surfaces. The adhesive can then be scraped off. (Of
course, you'll let it cure before reassembly...)

 As with almost any adhesive, clean surfaces are quite important. 99%
isopropyl alcohol (no longer costly; try a good drugstore) is a worthy
cleaner; rather few, uncommon contemporary plastics are attacked by it
(but the clear printhead drive rack in an Oki printer disintegrated in a
very few minutes, maybe 8 years ago). I doubt that any plastics used for
these sorts of puzzles would be sensitive to alcohol.

 Not sure of my information, but I believe that curing of CA is triggered
by absence of oxygen combined with an imperceptibly thin film of water on
the surfaces (don't try to wet them!). Very low humidity might inhibit
curing.

 For those who are really serious, the model-builders' magazines have ads
for different versions of CA adhesives; after all, models have been made
of plastics since WW II. Look into (i.e., catalog pages (Web?)) the
products of the Loctite Corp., which makes a variety of industrial
adhesives. There are other companies like Loctite, but Loctite has the
most-successful marketers IMHO. (Note that there's no "k" in "Loctite"!)

 By the way, the Pocket Cube (2^3) is a bear to disassemble and even worse
to reassemble. If it weren't for the really-good-quality polymer chosen
for it, it (more than likely) could not be manufactured. The difficulty is
in that the cubies have to be distorted ("sprung") to disassemble it. 
Whether this plastic retains its ability over many years to be bent out of
shape but not crack, I don't know! 

 All of this is offered with the best intentions; if I'm wrong about some
particular, and you're reasonably sure that I am, by all means please let
me know!  

 The brittle plastics used before WW II, in general, are a different
matter; alcohol is probably a bad idea in general. The modest amount I do
know is really off-topic. 

 I hope that this isn't too lengthy; this List seems to be willing to
carry long messages at times.

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*    When the year 2000 begins, we'll celebrate 
|*  Amateur musician  *|*    the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Sun Jul 27 21:32:43 1997
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Mail-from: From nbodley@tiac.net Sun Jul 27 11:32:09 1997
Date: Sun, 27 Jul 1997 11:28:40 -0400 (EDT)
From: Nicholas Bodley <nbodley@tiac.net>
Reply-To: Nicholas Bodley <nbodley@tiac.net>
To: Edwin Saesen <ERCO@compuserve.com>
Cc: Cube Mailing List <Cube-Lovers@ai.mit.edu>,
        "Dr. Christoph Bandelow" <Christoph.Bandelow@ruhr-uni-bochum.de>
Subject: Fit of 4^3 pieces; 5^3 query
In-Reply-To: <199707260506_MC2-1B9B-F7B3@compuserve.com>
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 (Sorry to have forgotten to include this in the previous message about
repairing plastic.) 

 Others (such as Drs. Christoph Bandelow or David Singmaster) might know
better than I, but I strongly suspect that there was only one specific
maker, and probably one set of molding dies, for the 4^3. If so, all
pieces regardless of by what path they reached the owner, should fit. A
good, close look, perhaps under a magnifier, would give an initial
judgment about whether a given piece should be trial-assembled. If it has
about the same amount of friction as the others, it really ought to be OK.

 The 3^3 most definitely has been made by several different companies from
their own molding dies; interchangeability is by no means assured!

 Could someone enlighten me about the 5^3? I got mine from Dr. Uwe
Meffert; I wonder whether Dr. Bandelow's came from the same set of dies?

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*    When the year 2000 begins, we'll celebrate 
|*  Amateur musician  *|*    the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------




From cube-lovers-errors@mc.lcs.mit.edu  Sun Jul 27 21:34:49 1997
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Mail-from: From nbodley@tiac.net Sun Jul 27 12:09:36 1997
Date: Sun, 27 Jul 1997 12:06:18 -0400 (EDT)
From: Nicholas Bodley <nbodley@tiac.net>
To: Cube Mailing List <Cube-Lovers@ai.mit.edu>
Subject: 4^3 innards: There's a ball in there, but which way does it point?
Message-Id: <Pine.SUN.3.95.970727112852.20916C-100000@sunspot.tiac.net>
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 Uniquely to all cubes from the Pocket Cube to the 5^3, only the 4^3
(Rubik's Revenge) has a ball inside it. (I do hope that I don't cause
curious people to break their fragile center pieces trying to open up
theirs!) This is really a repetition of some old information, but with the
current interest in 4^3s, it seems sufficiently interesting to repeat it.
However, afaik, the questions about the ball's orientation are probably
new.

 The ball (made of at least 9 pieces plus eight screws, as I recall) 
consists of a center piece, essentially a sphere (possibly two
hemispheres) with "octants"  fastened to it; these have the geometry of
90-90-90-degree right spherical triangles, although in practical detail
they differ. Gaps between these octants define three circumferential
grooves that correspond to the Earth's equator and two orthogonal
meridians of longitude. (All 3 grooves are orthogonal.)  (Sorry for the
redundancies;  trying to be clear to everybody). 

 These grooves are "undercut" on one side, so they have an inverted
L-shaped cross-section. The cubies (center ones only, as I recall) have
feet that tuck under the extended edges of the octants' grooves; this
keeps them in place. Machinists know well of the T-slots that work with
clamps to hold work in place on machine tools; these are similar, but the
cubies are free to slide in the L-slots. 

 If I'm thinking clearly (not too sure!), the ball has a 120-degree
rotational symmetry about an unique diameter. One octant has no undercuts;
its opposite, I think, has all three edges undercut. Other octants have
some edges undercut.

 Practical details dictate that when one half of the 4^3 is rotated, one
half is definitely locked to the internal ball. However, you probably
don't know which half it is! (The mathematical folk here might find it fun
to predict which half is the locked half for any given configuration; this
might even not be a trivial problem. (As well, does the locked part always
end up in the same place when the Cube is solved? I suspect so, but am not
sure.)  (Anybody for a translucent-cubie 4^3?) 

 Is it possible to maneuver the internal ball so that it has
effectively revolved by a half turn (or quarter turn?) about any given
axis, while preserving the exterior configuration? 

 My best regards to all,

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*    When the year 2000 begins, we'll celebrate 
|*  Amateur musician  *|*    the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------



From cube-lovers-errors@mc.lcs.mit.edu  Sun Jul 27 21:38:21 1997
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Mail-from: From nbodley@tiac.net Sun Jul 27 20:25:05 1997
Date: Sun, 27 Jul 1997 20:21:46 -0400 (EDT)
From: Nicholas Bodley <nbodley@tiac.net>
Reply-To: Nicholas Bodley <nbodley@tiac.net>
To: Cube Mailing List <Cube-Lovers@ai.mit.edu>
Cc: Javier Susaeta <javier.susaeta@nexo.es>,
        Mark Glusker <glusk@mechcad3.engr.sgi.com>
Subject: Making parts for puzzles (somewhat off-topic) (medium length)
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 For reasons I'm not certain of, but probably having to do with a case of
brain fade caused by not eating a proper breakfast (I live alone!), I
posted a short article to the wrong mailing list, no less. It was to an
unmoderated list that goes to people interested in mechanical calculators,
so the few replies I have had so far found it interesting. 

 I do hope this is not too far off-topic; I have included replies that
contain material that would be helpful to anyone who has any ideas for
making new or replacement parts, and has access to the technologies.

 The length is not hopeless...


Subject:  Rapid Prototyping and conceivable Cube reissues

 (Posted 09:53 AM 27/07/97 -0700 with follow-up (by gracious permission)
from a couple of people who >do< know what they're talking about.)

 This is close to being peripheral to the topics of the List, but it might
be worth pointing out that it is now possible by several methods to
"print" solid objects in 3 dimensions from digital data in the proper
format. I'm fairly sure it's even possible to make molds.

 These processes are sometimes known as Rapid Prototyping, and (betraying
linguistic ignorance or lack of concern) something like 3-D Lithography.
("Lithography" has roots meaning writing on stone.) (I suggest Solid
Object Synthesis...)  The field is still evolving and improving.

 The data files for these processes are ordinarily created by CAD
(Computer-Aided Design/Drafting) programs. The programs create the data
for numerous thin slices of an object.

 The data for a slice directs the machine to make a solid counterpart, and
the object is created progressively by creating a bonded stack of the
requisite number of slices. (I shouldn't go into how, but one early method
uses a pool of photopolymer and a flat "stage" that progressively moves
deeper into the pool. A bright UV light source positioned by the data
triggers polymerization (solidifies).)

 My point is that it is now considerably less expensive and much quicker
than it used to be, to make a prototype of a puzzle design, although the
materials used for this process (afaik) don't have the requisite
durability yet. With copyrights taken care of, it might be possible to
make limited numbers of given designs. It would (at present) be too costly
to make individual puzzles by these processes, but the economics could
change in a decade.

 Whether it is possible to pour melted plastic into molds to make the
pieces of a Cube-like puzzle is at present very doubtful, but it might be,
in the future. (Rattlebacks, also known as celts, look just like
solidified poured plastic.)

 It's unfortunate that the economics that make possible the affordable
production of large numbers of such puzzles as the 4^3 also mean that
starting another production run has to be economically justified.

 Let's hope that the tooling for manufacturing such miracles as the 5^3
and the Magic Dodecahedron is preserved! 

 Failing that, at least any CAD files for doing numerical machining of the
molding dies really ought to be kept and backed up.

 Hope I'm not too far off-topic.

{"afaik" = "as far as I know", pointed out as a courtesy to non-native
users of English}

{END of the text I posted to the wrong List -nb}

* * *

{The following replies are posted with the gracious permission of their
respective authors. }


Here's a reply from Javier Susaeta<javier.susaeta@nexo.es>:

Yes, I have read a bit about rapid prototyping and believe that it has a
tremendous potential. I cannot tell you where I read it, but I remember a
case where a complete intake manifold was designed by CAD/CAM and then a
single copy was (slowly) built, layer by layer, with one of the just-born
rapid-prototyping machines. The material used was aluminium powder plus a
plastic agglomerant.
 Once the manifold was "finished" it was baked in an oven in order to
eliminate the agglomerant and weld together the metal particles. The result
was not so solid as a cast aluminium manifold, but nevertheless perfectly
usable.

 (Referring to more details, Javier said:)

 I remember having read it in internet, a few months ago.  The intake
manifold was for a specially-built big diesel of a bulldozer or a similar
machine. I am not sure, but perhaps it was from Caterpillar or a similar
US company. They resorted to rapid prototyping because of costs.  It was a
very special engine, and they needed 1 (yes, one) intake manifold. The
fixed costs for a single casting were enormous, and a manifold is so
convoluted that machining it out of a block of metal was impossible. So
they resorted to this new technique. 

Regards

Javier Susaeta

   Here's some more on the topic, from Mark Glusker:

Date: Sun, 27 Jul 1997 14:33:36 -0700 (PDT)
From: Mark Glusker <glusk@mechcad3.engr.sgi.com>

This process has improved dramatically over the past few years.  The 
materials are now quite durable.  It is not necessarily cheaper to use
stereo lithography: it is best for small parts with lots of detail-per-
cubic-inch. Simple parts or large parts are best fabricated using 
conventional machining methods. If you are making a part to be used as a
master for a mold, don't forget to enlarge the original by several percent
(depending on the final material) to account for molding shrinkage.

Despite the improvements in this field, the best stereo lithography part
is still not nearly as good as a well machined part.  It's very much
like the difference between a well printed photograph and a scanned image
printed on a good laserwriter.  Used appropriately, it is a great tool
but Bridgeport is not about to go the way of Friden or Monroe.

{Bridgeport is a very-famous maker of milling machines; Friden and Monroe
were once-vital makers of mechanical desktop calculators; they are now
history. -nb}

 [Mark also wrote:]

--
Mark Glusker, glusk@sgi.com
>From glusk@mechcad3.engr.sgi.com Sun Jul 27 19:13:05 1997
Date: Sun, 27 Jul 1997 16:04:04 -0700 (PDT)
From: Mark Glusker <glusk@mechcad3.engr.sgi.com>

 I just received some stereo lithography parts last week that were made of
polycarbonate.  It is finally possible to use that process for more than
just verifying the shape of objects.  However, if you wanted to replicate
a cube (I assume you mean a Rubik-like mechanism)  you would need to do
lots of post-finishing to remove the "raster" ridges from the parts so
they will move smoothly against one another.  This post-finishing is done
by hand and will certainly affect your final tolerances.

 I use ProEngineer (on a Silicon Graphics workstation, naturally) which
can automatically generate an STL file, the standard data format used by
the stereo lithography vendors.  There are lots of vendors around here,
and several have relationships with prototype die casters for limited
production runs of parts (5 to 100 pieces, typically).

  {At this point, Mark offered help; however, since the help was a
personal offer to me, I don't want to post his comments directly.  -nb}

{snip}

 ... I could go on for quite some time on this subject! 

 There are similar digital processes for replicating flat metal parts from
a CAD file, with similar economic tradeoffs, in this case related to
perimeter-per-area of part.  That would be a great way to replicate a
missing piece of a mechanical calculator,... {Here, Mark's helpful
comments were welcome, but off-topic for this List. -nb}
Regards,

Mark

 

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*    When the year 2000 begins, we'll celebrate 
|*  Amateur musician  *|*    the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------




From cube-lovers-errors@mc.lcs.mit.edu  Sun Jul 27 22:51:19 1997
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Mail-from: From Hoey@AIC.NRL.Navy.Mil Sun Jul 27 22:45:48 1997
Date: Sun, 27 Jul 1997 22:45:37 -0400
Message-Id: <199707280245.WAA12048@sun30.aic.nrl.navy.mil>
From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: cube-lovers@ai.mit.edu
Cc: Nicholas Bodley <nbodley@tiac.net>
Subject: Administrata and a reference

The administrata is that Alan Bawden has asked me to take over
cube-lovers-request for a few weeks, while he is recovering from
surgery.  After seventeen years of running cube-lovers, he deserves a
vacation, but I would rather he find someplace more pleasant to spend
it than a hospital.  I'm sure we all wish him a speedy recovery.

I will be adding and removing addresses and filtering out abuse, but I
will not be updating the archives or the collection of reader
contributions.  As always, send to cube-lovers-request@ai.mit.edu for
administrative services.

The reference is for Nicholas Bodley, who in one of his very
informative messages on Rubik's Revenge raised questions of the
possible orientations achievable by the internal sphere without
changing the exterior.  I answered that question in my message on
"Invisible Revenge" on 9 August 1982.  The sphere can be placed in any
of 24 orientations, and I showed how to do so.  If we consider the
sphere modulo its functional symmetry (fixing one corner of the cube)
we will distinguish only 8 of these orientations.  I also mentioned
how to determine which of these 8 orientations the interior sphere is
in on a physical cube, without disassembly.  See
    ftp://ftp.ai.mit.edu/pub/cube-lovers/cube-mail-4
for details.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From cube-lovers-errors@mc.lcs.mit.edu  Mon Jul 28 12:11:59 1997
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Mail-from: From CFolkerts@compuserve.com Mon Jul 28 05:22:17 1997
Date: Mon, 28 Jul 1997 05:18:01 -0400
From: Corey Folkerts <CFolkerts@compuserve.com>
Subject: 2^3 Reassembly
To: <Cube-Lovers@ai.mit.edu>
Message-Id: <199707280518_MC2-1BB6-48C8@compuserve.com>

My 2^3 burst into pieces while I was playing around with it a while
back. I was amazed and intrigued by the number of internal pieces it
contained; many more than the 3^3.  Anyway, after a couple minutes I
got it all put back together, and started playing with it again. One
problem: when I attempted to rotate the cube on one of the axes, it
gave me a lot of resistance. If I continued to force it, the whole
thing burst and was reduced once again to a pile of little black
plastic pieces. After a few more random tests, I examined the pieces
and noticed, as I'm sure many have, that some of the small internal
pieces are slightly different than the others.  This fact leads me to
believe that the 'special' pieces need to be oriented correctly with
respect to each other in order for the cube to work correctly.  I
would be most appreciative if someone could please inform the manner
in which they need to be placed.

        Thanks in advance,
        Corey Folkerts

From cube-lovers-errors@mc.lcs.mit.edu  Mon Jul 28 12:14:45 1997
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Mail-from: From R.F.Hegge@MP.TUDelft.NL Mon Jul 28 05:30:18 1997
Date: Mon, 28 Jul 1997 11:26:33 +0200
From: R.F.Hegge@MP.TUDelft.NL (Rob Hegge)
Subject: Re: Where Can I get...
To: CUBE-LOVERS@ai.mit.edu
Message-Id: <9707280926.AA23236@sumatra.mp.tudelft.nl>

Sorry,
but I have just been informed that Puzzletts has
been out of 4x4x4's for several years now. I probably
received one of their last.

>I tried to order from them first, in the beginning of June.  There was
>no response to my order or the followup email I sent.

Maybe they put your order on their 'wish list' and will contact you
after they obtained one. But given the fact the 4x4x4 are quite rare
now it does not seem likely.

> >There still exist
> >a club called "Nederlandse Kubus Club" (NKC) or Dutch Cubist Club.
> This sounds interesting. Can you tell me how to contact them? And, does
> anyone know if there's a similar organization in germany?
I will look up the address etc and send the info to the list asap. 
For germany I would not know.

Rob


From cube-lovers-errors@mc.lcs.mit.edu  Mon Jul 28 12:56:42 1997
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Mail-from: From hart@netprofitlc.com Mon Jul 28 11:16:15 1997
Date: Mon, 28 Jul 1997 09:01:14 -0600 (MDT)
From: Paul Hart <hart@netprofitlc.com>
To: "Jin \"Time Traveler\" Kim" <chrono@ibm.net>
Cc: CUBE <CUBE-LOVERS@ai.mit.edu>
Subject: Re: Where can I get...?
In-Reply-To: <33D8EF4F.69C5@ibm.net>
Message-Id: <Pine.GSO.3.96.970728085652.25201B-100000@socrates.netprofitlc.com>

On Fri, 25 Jul 1997, Jin "Time Traveler" Kim wrote:

> Indeed, while we are sharing 4x4x4 stories, I got mine (on an "extended"
> borrow) while digging around in a friend's garage.

I've got another interesting story, too, about my two 4x4x4 cubes.
Earlier this year, by sheer coincidence, I happened across two Rubik's
Revenge cubes at a local hobby store.  Each of the cubes was authentic,
and still in the (unopened) original box.  I easily slapped down my $12.95
(USD) for each cube, needlessly to say.  :-)

Paul Hart

From cube-lovers-errors@mc.lcs.mit.edu  Mon Jul 28 15:20:07 1997
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Mail-from: From chrono@ibm.net Mon Jul 28 14:58:51 1997
Message-Id: <33DCE54F.5463@ibm.net>
Date: Mon, 28 Jul 1997 11:30:39 -0700
From: "Jin \"Time Traveler\" Kim" <chrono@ibm.net>
Reply-To: chrono@ibm.net
Organization: The Fourth Dimension
To: Cube-Lovers@ai.mit.edu
Subject: Re: 2^3 Reassembly
References: <199707280518_MC2-1BB6-48C8@compuserve.com>

Corey Folkerts wrote:
> 
> My 2^3 burst into pieces while I was playing around with it a while
> back. I was amazed and intrigued by the number of internal pieces it
> contained; many more than the 3^3.  Anyway, after a couple minutes I
> got it all put back together, and started playing with it again. One
> problem: when I attempted to rotate the cube on one of the axes, it
> gave me a lot of resistance. If I continued to force it, the whole
> thing burst and was reduced once again to a pile of little black
> plastic pieces. After a few more random tests, I examined the pieces
> and noticed, as I'm sure many have, that some of the small internal
> pieces are slightly different than the others.  This fact leads me to
> believe that the 'special' pieces need to be oriented correctly with
> respect to each other in order for the cube to work correctly.  I
> would be most appreciative if someone could please inform the manner
> in which they need to be placed.
> 
>         Thanks in advance,
>         Corey Folkerts

You have experienced a problem which has led me to purchase a total of
FOUR 2x2x2 cubes.  Not even of my own undoing either.  In two cases,
friends attempted to play with the cube and disassembled them, and were
unable to properly reassemble them.  I have pieces of each in separate
boxes, minus several pieces each.  In one case I dropped it, it flew
open, and I DID manage to reassemble it properly.  But it ALSO fell
victim to a careless reassembly by a friend who also carelessly
disassembled it.  That one I reassembled AGAIN and gave it to someone. 
They later told me they "broke" it, which means it's in pieces and since
they live 450 miles away, I can't exactly help them.  I have a fourth,
still in its bag, untouched by human hands.  Oh yes, and I bought a
fifth one (actually, it was the fourth, so the untouched one is
technically the fifth) but I had to return it and get another one (the
fifth) because it had been disassembled before and reassembled
incorrectly (the pieces only turned on one axis).

-- 
Jin "Time Traveler" Kim
chrono@ibm.net
VGL Costa Mesa
http://www.geocities.com/timessquare/alley/9895
http://www.slamsite.com

From cube-lovers-errors@mc.lcs.mit.edu  Tue Jul 29 10:14:22 1997
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Mail-from: From ERCO@compuserve.com Tue Jul 29 02:39:59 1997
Date: Tue, 29 Jul 1997 02:22:43 -0400
From: Edwin Saesen <ERCO@compuserve.com>
Subject: Re: Where can I get...?
To: CUBE <CUBE-LOVERS@ai.mit.edu>
Message-Id: <199707290223_MC2-1BD0-779D@compuserve.com>

Paul wrote:
>I've got another interesting story, too, about my two 4x4x4 cubes.
>I easily slapped down my $12.95
>(USD) for each cube, needlessly to say.  :-)

>From all of these stories of almost everyone in the USA either finding two
copies or a single one for around US$1, I take it that they were MUCH more
common in the USA than they were in germany :-(

Rob wrote:
>but I have just been informed that Puzzletts has
>been out of 4x4x4's for several years now. I probably
>received one of their last.

:-((( Ok then, I'm desperate now. Is there ANYONE willing to sell a spare
4x4x4 one? I'll be willing to pay the $50 puzzletts were charging for them
(as I tried to order one from them...).
(Hey Paul, this is *THE* chance for you to get your money back, and have
your own copy of the 4x4x4 for free plus having about $24 leftover to buy
24 copies of the 4x4x4 on your local flea markets...).
But honest, if anyone ever sees one of those for sale somewhere, please get
it for me. I think my chances of ever finding one here in germany are
virtually zero, as I haven't been able to replace my own copy for six years
or so...

Michael


From cube-lovers-errors@mc.lcs.mit.edu  Tue Jul 29 10:52:34 1997
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Mail-from: From nbodley@tiac.net Tue Jul 29 03:22:14 1997
Date: Tue, 29 Jul 1997 03:10:32 -0400 (EDT)
From: Nicholas Bodley <nbodley@tiac.net>
To: Corey Folkerts <CFolkerts@compuserve.com>
Cc: Cube-Lovers@ai.mit.edu
Subject: Re: 2^3 Reassembly
In-Reply-To: <199707280518_MC2-1BB6-48C8@compuserve.com>
Message-Id: <Pine.SUN.3.95.970729023932.13860A-100000@sunspot.tiac.net>

 My, it's been a while since I opened up mine. I hope I remember. What
perplexes me (I was about to say, "puzzles me") is that I don't recall how
it could be assembled wrong; surely there aren't two internal mechanisms
that differ in some details? 

 Perhaps it would help if I describe the internal structure (from memory).

 The basic structure that I remember is built on a "jack"; it's basically
mutually-orthogonal extensions from a center, so to speak; like a physical
embodiment of the axes of 3-D Cartesian coordinates. What I recall is that
three of these, all adjacent, have a square cross-section; the other three
have a circular cross-section, with a diameter significantly smaller than
a side of the square.

 Each of three round projections fits into a hole through a rotating
long, thin square prism. (In the vernacular, square sticks with round
holes through their centers.) In mine, I am just about certain that all of
these had the same length.

 Each cubie (all were identical internally) is hollow, but cut away with
concave arcs that allow them to turn with respect to their neighbors.

 The cubies are kept together by 12 "clips". These fit into the cutout
arcs; when you assemble the Cube, you put two cubies next to each other
(they touch) and fit this "clip" so that it keep s them together. To
install it, you move the clip away from the imaginary geometrical center
of the whole puzzle.

 As I recall them, the "clips" are essentially quadrants (1/4 circles).
They consist of two parallel planes with a gap between them; the sides of
the cubies fit into this gap. The parallel planes are joined at the inner
edges.

 When the whole Cube is assembled, the square extensions of the "jack", as
well as the square sticks that turn on the other ends of the jack serve to
keep the clips from moving toward the center of the whole Cube.

 This is a structure that could not be either assembled or disassembled if
it were made of rigid materials. It's only because the cubies (at least!)
are made of a strong plastic that has good mechanical spring properties
and can be harmlessly deformed (within limits), that it is possible to
make this structure.

 >>> It's conceivable that the cube was misassembled so that one or more
"clips" didn't actually straddle both of its cubies. When assembled, there
should not be any gaps between the cubies, and all movement should be
reasonably free of friction.

 I'd love to know how these were assembled in the first place; did the
mfr. have special tools to temporarily deform the parts? Did the
assemblers develop very strong hand muscles?

 Btw, it's a challenge to describe the innards in words.

 I hope this helps!

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*    When the year 2000 begins, we'll celebrate 
|*  Amateur musician  *|*    the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------



From cube-lovers-errors@mc.lcs.mit.edu  Tue Jul 29 14:33:28 1997
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Mail-from: From chrono@ibm.net Tue Jul 29 13:43:28 1997
Message-Id: <33DE2A54.62F2@ibm.net>
Date: Tue, 29 Jul 1997 10:37:24 -0700
From: "Jin \"Time Traveler\" Kim" <chrono@ibm.net>
Reply-To: chrono@ibm.net
Organization: The Fourth Dimension
To: Cube-Lovers@ai.mit.edu
Subject: Re: 2^3 Reassembly
References: <Pine.SUN.3.95.970729023932.13860A-100000@sunspot.tiac.net>

Nicholas Bodley wrote:

>  The cubies are kept together by 12 "clips". These fit into the cutout
> arcs; when you assemble the Cube, you put two cubies next to each other
> (they touch) and fit this "clip" so that it keep s them together. To
> install it, you move the clip away from the imaginary geometrical center
> of the whole puzzle.
> 
>  As I recall them, the "clips" are essentially quadrants (1/4 circles).
> They consist of two parallel planes with a gap between them; the sides of
> the cubies fit into this gap. The parallel planes are joined at the inner
> edges.

Part of the problem is that the clips weren't all identically shaped. 
If all of the clips were shaped the same, then reassembly wouldn't be a
problem.  But because they ARE shape differently, there is a question of
whether the position of one is important relative to the position of the
others.

-- 
Jin "Time Traveler" Kim
chrono@ibm.net
VGL Costa Mesa
http://www.geocities.com/timessquare/alley/9895
http://www.slamsite.com



From cube-lovers-errors@mc.lcs.mit.edu  Tue Jul 29 15:10:58 1997
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Mail-from: From CFolkerts@compuserve.com Tue Jul 29 14:09:01 1997
Date: Tue, 29 Jul 1997 13:58:09 -0400
From: Corey Folkerts <CFolkerts@compuserve.com>
Subject: Re: 2^3 Reassembly
To: Cube-Lovers <Cube-Lovers@ai.mit.edu>
Message-Id: <199707291358_MC2-1BDB-4183@compuserve.com>

Nicholas Bodley writes:

> The cubies are kept together by 12 "clips". These fit into the cutout
>arcs; when you assemble the Cube, you put two cubies next to each other
>(they touch) and fit this "clip" so that it keep s them together. To
>install it, you move the clip away from the imaginary geometrical center
>of the whole puzzle. 

My 2^3 has these 12 clips as I discovered when it first burst. However, I
would like to confirm something. Nine of my clips are identical, the 1/4
circle shape. However, its the other three that are causing me trouble. One
of them is identical to the other nine except that on one of the two planes
it has a very small notch cut out of it. The notch is an arc and I'm
guessing it is probably about 1 mm deep. The other two have one of the 1/4
planes identical to the first nine, but the second plane extends far
beyond, doubling the "height" of the clip. If viewed from the side which
has the extended plane it is a diamond instead of a 1/4 circle.  All of the
other internal pieces are identical to your description

        I would like to know if everyone else has these altered clips in
their 2^3s.

        Corey Folkerts


From cube-lovers-errors@mc.lcs.mit.edu  Tue Jul 29 16:24:18 1997
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Mail-from: From whuang@ugcs.caltech.edu Tue Jul 29 14:32:02 1997
To: Cube-Lovers@AI.MIT.Edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Re: Broken 4^3s; advice on repairs to plastic (medium length)
Date: 29 Jul 1997 18:28:24 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <5rlco8$em3@gap.cco.caltech.edu>
References: <cube-lovers.Pine.SUN.3.95.970727101355.20916A-100000@sunspot.tiac.net>

Nicholas Bodley <nbodley@tiac.net> writes:
> By the way, the Pocket Cube (2^3) is a bear to disassemble and even worse
>to reassemble. If it weren't for the really-good-quality polymer chosen
>for it, it (more than likely) could not be manufactured. The difficulty is
>in that the cubies have to be distorted ("sprung") to disassemble it. 
>Whether this plastic retains its ability over many years to be bent out of
>shape but not crack, I don't know! 

Does anyone know whether the mechanisms for the old 2^3's are the same
as the recent new 2^3 releases?


--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
I hate formication.  It should be abolished entirely.

From cube-lovers-errors@mc.lcs.mit.edu  Wed Jul 30 10:35:07 1997
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Mail-from: From nbodley@tiac.net Wed Jul 30 08:15:13 1997
Date: Wed, 30 Jul 1997 08:11:44 -0400 (EDT)
From: Nicholas Bodley <nbodley@tiac.net>
To: "Jin \"Time Traveler\" Kim" <chrono@ibm.net>
Cc: Cube-Lovers@ai.mit.edu
Subject: Re: 2^3 Reassembly
In-Reply-To: <33DE2A54.62F2@ibm.net>
Message-Id: <Pine.SUN.3.95.970730080230.2403A-100000@sunspot.tiac.net>

On Tue, 29 Jul 1997, Jin "Time Traveler" Kim wrote:

}Nicholas Bodley wrote:
}
}>  The cubies are kept together by 12 "clips". These fit into the cutout

{Snips}

}Part of the problem is that the clips weren't all identically shaped. 
}If all of the clips were shaped the same, then reassembly wouldn't be a
}problem.  But because they ARE shaped differently, there is a question of
}whether the position of one is important relative to the position of the
}others.
}
}-- 
}Jin "Time Traveler" Kim
}chrono@ibm.net

 It's not likely that there are two or more designs of the 2^3; sorry if I
misled anyone. I might have been lucky; however, I don't recall needing to
sort the clips. 

 This >is< something to look out for if you disassemble a 2^3.

 It's likely that the clips for the swiveling long blocks would be
different from those for the rigid extensions of the "jack". Less likely
is that the different cavities used to mold several clips at once had
different shapes where the differences had no effect on operation, but
anyone familiar with such mechanisms would be able to tell.  The
mechanical engineer on this project has the answers!

 Sure hope this isn't memory fade; I'm going on 62...

 My best regards to all,

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*    When the year 2000 begins, we'll celebrate 
|*  Amateur musician  *|*    the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------



From cube-lovers-errors@mc.lcs.mit.edu  Wed Jul 30 12:26:38 1997
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Mail-from: From nbodley@tiac.net Wed Jul 30 08:21:49 1997
Date: Wed, 30 Jul 1997 08:18:30 -0400 (EDT)
From: Nicholas Bodley <nbodley@tiac.net>
To: Corey Folkerts <CFolkerts@compuserve.com>
Cc: Cube-Lovers <Cube-Lovers@ai.mit.edu>
Subject: Re: 2^3 Reassembly
In-Reply-To: <199707291358_MC2-1BDB-4183@compuserve.com>
Message-Id: <Pine.SUN.3.95.970730081345.2403B-100000@sunspot.tiac.net>

 As to the notches Corey mentions, they might not matter, but all three
pieces I would guess might of a later design that would simplify assembly;
they are probably the last three pieces to be assembled.

 It sounds as though one needs to very careful and make sketches when
disassembling a 2^3! All of mine were bought probably around 1985 or so,
or earlier; they are old. 

 Please don't think that simply because I'm partly informed that I'm an
expert! I'm not.

 Good luck to all,

NB

On Tue, 29 Jul 1997, Corey Folkerts wrote:

}Nicholas Bodley writes:
}
}> The cubies are kept together by 12 "clips". These fit into the cutout
}>arcs; when you assemble the Cube, you put two cubies next to each other
}>(they touch) and fit this "clip" so that it keep s them together. To
}>install it, you move the clip away from the imaginary geometrical center
}>of the whole puzzle. 
}
}My 2^3 has these 12 clips as I discovered when it first burst. However, I
}would like to confirm something. Nine of my clips are identical, the 1/4
}circle shape. However, its the other three that are causing me trouble. One
}of them is identical to the other nine except that on one of the two planes
}it has a very small notch cut out of it. The notch is an arc and I'm
}guessing it is probably about 1 mm deep. The other two have one of the 1/4
}planes identical to the first nine, but the second plane extends far
}beyond, doubling the "height" of the clip. If viewed from the side which
}has the extended plane it is a diamond instead of a 1/4 circle.  All of the
}other internal pieces are identical to your description
}
}        I would like to know if everyone else has these altered clips in
}their 2^3s.
}
}        Corey Folkerts
}
}

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*    When the year 2000 begins, we'll celebrate 
|*  Amateur musician  *|*    the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Wed Jul 30 15:23:06 1997
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Mail-from: From whuang@ugcs.caltech.edu Wed Jul 30 14:13:18 1997
To: Cube-Lovers@AI.MIT.Edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Re: 2^3 Reassembly
Date: 30 Jul 1997 18:09:49 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <5ro01d$as6@gap.cco.caltech.edu>
References: <cube-lovers.199707291358_MC2-1BDB-4183@compuserve.com>

Corey Folkerts <CFolkerts@compuserve.com> writes:
>Nicholas Bodley writes:
>> The cubies are kept together by 12 "clips". These fit into the cutout
>>arcs; when you assemble the Cube, you put two cubies next to each other
>>(they touch) and fit this "clip" so that it keep s them together. To
>>install it, you move the clip away from the imaginary geometrical center
>>of the whole puzzle. 
>My 2^3 has these 12 clips as I discovered when it first burst. However, I
>would like to confirm something. Nine of my clips are identical, the 1/4
>circle shape. However, its the other three that are causing me trouble. One
>of them is identical to the other nine except that on one of the two planes
>it has a very small notch cut out of it. The notch is an arc and I'm
>guessing it is probably about 1 mm deep. The other two have one of the 1/4
>planes identical to the first nine, but the second plane extends far
>beyond, doubling the "height" of the clip. If viewed from the side which
>has the extended plane it is a diamond instead of a 1/4 circle.  All of the
>other internal pieces are identical to your description

>        I would like to know if everyone else has these altered clips in
>their 2^3s.

I believe so.  Those "special" serve the same purpose as the protruding
octants on the 4^3 internal ball -- to anchor one of the blocks in
each plane.  Otherwise, one of the planes of "clips" may be offset 
45 degrees (not obvious from the outside), and the other planes
become unturnable.  Make sure that one weird clip is in each
plane before assembly.


--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
I hate formication.  It should be abolished entirely.

From cube-lovers-errors@mc.lcs.mit.edu  Thu Jul 31 17:38:54 1997
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Mail-from: From ponder@austin.ibm.com Thu Jul 31 16:55:58 1997
Date: Thu, 31 Jul 1997 15:52:23 -0500
From: ponder@austin.ibm.com (Ponder)
Message-Id: <9707312052.AA16660@roosevelt.austin.ibm.com>
To: Cube-Lovers@ai.mit.edu
Subject: Rubik's octahedron's etc.
Cc: ponder@austin.ibm.com


I have a Rubik's Cube and a Megaminx Dodecahedron. There
are some tetrahedral puzzles available but they do not
correspond precisely to the Rubik's cube, in that they do
not have well-defined center pieces and the corners are
freely rotating. As far as I can tell, nobody has an octa-
hedron or an icosahedron that works on these principles
either.

Its hard to expect the puzzle companies to come out with
anything like these since they're in it for a profit. I
heard that Meffert's company closed down before they could
produce most of the puzzles they intended. Does anyone
have designs for puzzles like these that could be built in
a machine-shop? (Preferably that you've already patented,
to eliminate any legal concerns!!). I imagine I could try
to hack something together, but it would take an awful lot
of trial-and-error especially since some internal designs
would hold together better than others.

I'm publishing a paper in the Journal of Recreational
Mathematics on solving these other puzzles, but it would
be real nice to have demo models, even if it takes some
work. The Octahedron is particularly interesting because
it forbids edge-flips and it would be more convincing if
I do more than show it on paper.

Thanks,

	Carl Ponder
	ponder@austin.ibm.com



From cube-lovers-errors@mc.lcs.mit.edu  Thu Jul 31 18:52:22 1997
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Mail-from: From ponder@austin.ibm.com Thu Jul 31 17:16:55 1997
Date: Thu, 31 Jul 1997 16:13:34 -0500
From: ponder@austin.ibm.com (Ponder)
Message-Id: <9707312113.AA33486@roosevelt.austin.ibm.com>
To: Cube-Lovers@ai.mit.edu
Subject: puzzle to be simulated
Cc: ponder@austin.ibm.com


I've seen a number of Rubik's Cube simulations on the web,
and was wondering if any of you would be interested in
implementing the following puzzle that I call "the hell-hole".
It has 16 faces that each work like the faces of a rubik's
cube, but:

	1] The faces are layed out on a 4x4 grid. Each
	   "corner" joins four surrounding faces instead
	   of three.

	2] The grid is rolled into a cylinder and then
	   joined at both ends to forma torus.

However, the torus is given a "twist" when you join the two
ends together, as follows:

	  1 2 3 4
	  _ _ _ _
	a|_|_|_|_|b
	b|_|_|_|_|c
	c|_|_|_|_|d
	d|_|_|_|_|a
	  1 2 3 4

First join the 1-2-3-4 sides together to form the cylinder,
then the a-b-c-d ends together to get the torus. Each of the
squares is a 3x3 face like a Rubik's Cube.

The combinatorics get a *lot* messier because of the twist.
Without it, you can't "flip" the edge pieces. With it, you
can, but only by moving the edge-piece in a full-circle around
the torus.

No way to build it, either, since the pieces would need to
flex between convex and concave. It could be simulated on a
computer, however. I have a paper coming out in the Journal of
Recreational Mathematics on how to solve these kinds of things,
and it is pretty messy.

Thanks,

	Carl Ponder
	ponder@austin.ibm.com



From cube-lovers-errors@mc.lcs.mit.edu  Thu Jul 31 21:36:12 1997
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Date: Thu, 31 Jul 1997 21:38:18 -0400
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From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: Cube-Lovers@ai.mit.edu, ponder@austin.ibm.com
In-Reply-To: <9707312113.AA33486@roosevelt.austin.ibm.com>
Subject: Re: puzzle to be simulated

Carl Ponder describes his "hell-hole" puzzle that (if I undestand
correctly) has 36 facets laid out like

   m : n n n : p p p : r r r : s s s : a
   ..:.......:.......:.......:.......:..
   s : a a a : b b b : c c c : d d d : e
   s : a a a : b b b : c c c : d d d : e
   s : a a a : b b b : c c c : d d d : e
   ..:.......:.......:.......:.......:..
   d : e e e : f f f : g g g : h h h : j
   d : e e e : f f f : g g g : h h h : j
   d : e e e : f f f : g g g : h h h : j
   ..:.......:.......:.......:.......:..
   h : j j j : k k k : l l l : m m m : n
   h : j j j : k k k : l l l : m m m : n
   h : j j j : k k k : l l l : m m m : n
   ..:.......:.......:.......:.......:..
   m : n n n : p p p : r r r : s s s : a
   m : n n n : p p p : r r r : s s s : a
   m : n n n : p p p : r r r : s s s : a
   ..:.......:.......:.......:.......:..
   s : a a a : b b b : c c c : d d d : e

with opposite boundaries identified so that the letters match up.  A
turn rotates 25 facets--one of the 3x3 "faces" marked with dots--and
the sixteen neighboring facets from the neighboring faces.

> The combinatorics get a *lot* messier because of the twist.
> Without it, you can't "flip" the edge pieces. With it, you
> can, but only by moving the edge-piece in a full-circle around
> the torus.

If I've got the puzzle right, you could get edge flippability just by
using a 3x3 array of faces instead of 4x4.  Or perhaps 3x5.  Another
nice idea that uses a "square" torus is

   k : l l l : m m m : n n n : d
   ..:.......:.......:.......:..
   h : a a a : b b b : c c c : j
   h : a a a : b b b : c c c : j
   h : a a a : b b b : c c c : j j j : k k k : l
   ..:.......:.......:.......:.......:.......:..
   n : d d d : e e e : f f f : g g g : h h h : a
   n : d d d : e e e : f f f : g g g : h h h : a
   n : d d d : e e e : f f f : g g g : h h h : a
   ..:.......:.......:.......:.......:.......:..
   c : j j j : k k k : l l l : m m m : n n n : d
   c : j j j : k k k : l l l : m m m : n n n : d
   c : j j j : k k k : l l l : m m m : n n n : d
   ..:.......:.......:.......:.......:.......:..
   f : g g g : h h h : a a a : b b b : c c c : j

This corresponds to a square of side sqrt(13) with opposite edges
identified, cut on the bias into 13 square faces.
    ............................
    :.'   a  .' `.. m .'     `.:
    : `..   .'     `..'   n  .':
    :    `..'   b  .' `..   .' :
    : d  .' `..   .'     `..' d:
    :   .'     `..'   c  .' `..:
    :`..'   e  .' `..   .'     :
    :.' `..   .'     `..'   j  :
    :'     `..'   f  .' `..   .:
    :   k  .' `..   .'     `..':
    :..   .'     `..'   g  .' `:
    :  `..'   l  .' `..   .'   :
    :h .' `..   .'     `..'   h:
    : .'  a  `..'   m   .'`..  :
    :.'.......'b`.......'.n..`.:

after which each face is cut up into nine facets.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil


From cube-lovers-errors@mc.lcs.mit.edu  Fri Aug  1 09:34:41 1997
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Mail-from: From nbodley@tiac.net Thu Jul 31 22:03:44 1997
Date: Thu, 31 Jul 1997 22:00:20 -0400 (EDT)
From: Nicholas Bodley <nbodley@tiac.net>
To: Ponder <ponder@austin.ibm.com>
Cc: Cube-Lovers@ai.mit.edu
Subject: Re: Rubik's octahedrons etc.
In-Reply-To: <9707312052.AA16660@roosevelt.austin.ibm.com>
Message-Id: <Pine.SUN.3.95.970731215713.21913D-100000@sunspot.tiac.net>

 If you were lucky, you could use a good CAD program to define the shapes,
and NC machine tools to produce them; also possible is Rapid Prototyping.

 A graphic computer simulation would also be a substitute for a physical
puzzle, although holding one in your hand beats just about any graphics.

 Good luck!

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*    When the year 2000 begins, we'll celebrate 
|*  Amateur musician  *|*    the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------



From cube-lovers-errors@mc.lcs.mit.edu  Fri Aug  1 10:47:57 1997
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Mail-from: From Goyra@iol.ie Fri Aug  1 09:14:33 1997
Message-Id: <199708011310.OAA00278@mail.iol.ie>
From: "David Byrden" <Goyra@iol.ie>
To: <Cube-Lovers@ai.mit.edu>
Subject: Re: Rubik's octahedrons, etc.
Date: Fri, 1 Aug 1997 14:09:19 +0100

> From: Ponder <ponder@austin.ibm.com>

> I'm publishing a paper in the Journal of Recreational
> Mathematics on solving these other puzzles, but it would
> be real nice to have demo models, even if it takes some
> work. The Octahedron is particularly interesting because
> it forbids edge-flips 


	Just for you, I have put up a new Java Octahedron at the
Rubik Gallery

	http://www.iol.ie/~goyra/Rubik.html

The new one is in the Cousteau Collection and has corner-centred
slices.There is another one in the Plato Collection with face-centred 
slices. If the one you are thinking about is deeper or it twists in a 
different way, drop me a line and I can probably brew it up for you.


						David



From cube-lovers-errors@mc.lcs.mit.edu  Fri Aug  1 20:34:49 1997
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Mail-from: From Hoey@AIC.NRL.Navy.Mil Fri Aug  1 20:32:48 1997
Date: Fri, 1 Aug 1997 20:31:56 -0400
Message-Id: <199708020031.UAA28142@sun30.aic.nrl.navy.mil>
From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: Cube-Lovers@ai.mit.edu, ponder@austin.ibm.com
In-Reply-To: <199708010138.VAA24279@sun30.aic.nrl.navy.mil> (message from Dan
	Hoey on Thu, 31 Jul 1997 21:38:18 -0400)
Subject: Re: puzzle to be simulated

Continuing on my quest for Ponderesque cube-planes made of a square
torus, possibly cut on the bias, I notice that any square torus will
have k=a^2+b^2 faces for some a>b; if b=0 there is no bias.  Carl
Ponder suggests (if I understand it correctly) cutting each face into
nine facets, and permuting the puzzle by turning a face together with
the 16 neighboring facets.  As with Rubik's cube, the edge facets move
in pairs (called edge cubies).  The corner facets move in quadruplets
(called corner cubies).

There are k corner cubies that can apparently achieve any of four
orientations (twist) each, and 2k edge cubies that can apparently
achieve any of two orientations (flip) each.  The face center cubies
can achieve any of four orientations (twist) each but never move.

So the "constructible" group size--the size of the group before we
consider parity and orientation constraints--is k! (2k)! 16^k, or
k! (2k)! 64^k for the supergroup.

But if a+b is even, we can shade the faces in a checkerboard, and the
shades never change when we turn the faces.  So in this case, the
edges never flip, and the corners have only two orientations.  The
checkerboard-constructible group size is then k! (2k)! 2^k, or 
k! (2k)! 8^k for the supergroup.

Everyone who knows Rubik's cube will suspect (and everyone who
understands orientation theory will know!) that the corner
orientations must sum to zero (mod 4) and the edge orientations must
sum to zero (mod 2).  If a+b is even, there is only a corner
orientation constraint (mod 2).  [See my article of 23 September 1982
for a sketch of orientation theory.  Essentially, if the orientation
group of a kind of piece is Abelian then there is an orientation
constraint of the order of that orientation group.]

The permutation parity constraint is also familiar to anyone who knows
the cube.  The edge permutation parity must equal the corner
permutation parity, and in the supergroup the parity of the face
center twist must also be equal (mod 180 degrees).

So we should find groups of size k! (2k)! 2^f(k), where

    f(k)=4k - 4     a+b odd, or
    f(k)= k - 2     a+b even

for the permutation group, and

    f(k)=6k - 5     a+b odd, or
    f(k)=3k - 3     a+b even

for the supergroup.

I've used GAP to find the group sizes for
     (a,b) = (2,0), (2,1), (3,0), (3,1), (3,2), (4,0), (4,1),
and the group sizes agree except for (2,0) and (2,1).  The group
is smaller than expected by a factor of

      5040 for the (2,0) permutation group,
     20160 for the (2,0) supergroup,
         6 for the (2,1) permutation group, and
        12 for the (2,1) supergroup.

I'm not too surprised about the (2,0) groups (for instance, all four
corner cubies move cyclically on every turn!) but I don't see why
(2,1) is pathological.  Maybe it's one of those special group things
that happen for just one permutation group.

By the way, I suggest that a simulation of these should not try to
map them to a curved torus, but to a toroidal tesselation of the
plane.  Then when you turn one piece, you see a lattice of other
pieces turning in synchrony.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From cube-lovers-errors@mc.lcs.mit.edu  Mon Aug  4 10:22:51 1997
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Mail-from: From R.F.Hegge@MP.TUDelft.NL Mon Aug  4 09:38:48 1997
Date: Mon, 04 Aug 1997 15:34:47 +0200
From: R.F.Hegge@MP.TUDelft.NL (Rob Hegge)
Subject: CFF was Re: Where Can I get...
To: Cube-Lovers@ai.mit.edu
Message-Id: <9708041334.AA09170@sumatra.mp.tudelft.nl>

Edwin wrote:
> Rob wrote:
> >There still exist
> >a club called "Nederlandse Kubus Club" (NKC) or Dutch Cubist Club.
> This sounds interesting. Can you tell me how to contact them? And, does
> anyone know if there's a similar organization in germany?

Sorry I do not know of anything similar to NKC in Germany.

I put most of the information about CFF/NKC including some tables of
contents on
	http://wwwtg.mp.tudelft.nl/~rob/cff.html
Please don't hesitate to email me if you still have questions.

After finally finding an original Rubik's Cube I have some 3x3x3's left.
What are the most interesting bandaged cubes or other puzzles one can
make of them ?

Rob

From cube-lovers-errors@mc.lcs.mit.edu  Mon Aug  4 19:43:16 1997
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Mail-from: From Hoey@AIC.NRL.Navy.Mil Mon Aug  4 19:41:34 1997
Date: Mon, 4 Aug 1997 19:41:24 -0400
Message-Id: <199708042341.TAA09295@sun30.aic.nrl.navy.mil>
From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: Cube-Lovers@ai.mit.edu, ponder@austin.ibm.com
In-Reply-To: <199708020031.UAA28142@sun30.aic.nrl.navy.mil> (message from Dan
	Hoey on Fri, 1 Aug 1997 20:31:56 -0400)
Subject: Re: puzzle to be simulated

I've found out why the cube-plane groups related to the 1^2+2^2 square
torus are 1/6 the size we would expect.  It's the corners.  The group
has five corners {1,2,3,4,5} and five generators {A,B,C,D,E} that
operate on corners as

  5..CC/DDD`5    A: (1,2,4,3)
  EEE`1..DD/E    B: (2,3,5,4)
  .EE/AAA`2..    C: (3,4,1,5)
  B`3..AA/BBB    D: (4,5,2,1)
  B/CCC`4..BB    E: (5,1,3,2)
  5..CC/DDD`5

These generators do not generate the 120-element group S5, rather they
generate a 20-element subgroup known to GAP as

    5:4       = A split extension of C5 by C4      or equivalently
    H(2^2,5)  = <a,b ; a^5 = b^4 = a b a^2 b = 1 >.

Neither of these tells me a lot, except that the fact that this group
has index 6 in S5 means that there are six "orbits" of corner
permutations.

Dan
Hoey@AIC.NRL.Navy.Mil


From cube-lovers-errors@mc.lcs.mit.edu  Mon Aug  4 21:26:36 1997
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Mail-from: From cubeman@idirect.com Mon Aug  4 20:14:53 1997
Message-Id: <33E6709C.4277@idirect.com>
Date: Mon, 04 Aug 1997 20:15:25 -0400
From: Mark Longridge <cubeman@idirect.com>
To: cube lovers <cube-lovers@ai.mit.edu>
Subject: Megaminx a.k.a. Supernova

Has anyone ever written a simulation of the Megaminx for the PC?
 
I'm thinking about Megaminx moves and I'm on the verge of writing
a simulation from scratch (starting with my file for GAP), to
help myself to compose sequences for patterns.
 
I am particularly interested in processes for the 10-spot and 
12-spot.
 
Ultimately I should write one, something with coarse face movement,
algebraic move entries like  ( F+ B- )^12 with whole megaminx
rotations.

This is interesting to me as it lies outside of recorded cube
literature. 
 
Mark
Web Page At: http://web.idirect.com/~cubeman
 
I've also managed to compile Mike Reid's ANSI C cube solver for
MS DOS using DJGPP. It makes good use of DPMI.

From cube-lovers-errors@mc.lcs.mit.edu  Tue Aug  5 10:16:49 1997
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Mail-from: From Goyra@iol.ie Tue Aug  5 06:34:57 1997
Message-Id: <199708051031.LAA16445@mail.iol.ie>
From: "David Byrden" <Goyra@iol.ie>
To: "<cube-lovers@ai.mit.edu"@life.ai.mit.edu
Subject: Re: Megaminx a.k.a. Supernova
Date: Tue, 5 Aug 1997 11:30:44 +0100

> From: Mark Longridge <cubeman@idirect.com>


> Has anyone ever written a simulation of the Megaminx for the PC?

	Mark:

	There is a virtual Megaminx at my Rubik Gallery

	http://www.iol.ie/~goyra/Rubik.html

It's not written for the PC: it's in Java, so it works on alll the major
kinds of computer. In fact, you don't need to download or install
anything, just use a Java browser and you can play with the 
puzzles immediately.

	Control is via the mouse. The puzzles are not self-solving
but if anyone wants to write a solver and make use of my graphical 
representation of the puzzles, get in touch.

> I am particularly interested in processes for the 10-spot and 
> 12-spot.

	What exactly are these, I may want to put them in the 
Gallery.


						David

From cube-lovers-errors@mc.lcs.mit.edu  Tue Aug  5 11:42:18 1997
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Mail-from: From Goyra@iol.ie Tue Aug  5 06:34:57 1997
Message-Id: <Resent.199708051031.LAA16445@mail.iol.ie>
From: "David Byrden" <Goyra@iol.ie>
To: Cube-Lovers@ai.mit.edu
Subject: Re: Megaminx a.k.a. Supernova
Date: Tue, 5 Aug 1997 11:30:44 +0100

> From: Mark Longridge <cubeman@idirect.com>


> Has anyone ever written a simulation of the Megaminx for the PC?

	Mark:

	There is a virtual Megaminx at my Rubik Gallery

	http://www.iol.ie/~goyra/Rubik.html

It's not written for the PC: it's in Java, so it works on alll the major
kinds of computer. In fact, you don't need to download or install
anything, just use a Java browser and you can play with the
puzzles immediately.

	Control is via the mouse. The puzzles are not self-solving
but if anyone wants to write a solver and make use of my graphical
representation of the puzzles, get in touch.

> I am particularly interested in processes for the 10-spot and
> 12-spot.

	What exactly are these, I may want to put them in the
Gallery.


						David

[ Moderator's note:  Sorry for those of you who get this Cube-Lovers
  message twice.  I accidentally sent a version with mangled headers,
  which several mailer daemons refused to process. -- Dan]

From cube-lovers-errors@mc.lcs.mit.edu  Tue Aug  5 12:36:40 1997
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Mail-from: From Hoey@AIC.NRL.Navy.Mil Tue Aug  5 12:04:25 1997
Date: Tue, 5 Aug 1997 12:04:08 -0400
Message-Id: <199708051604.MAA13056@sun30.aic.nrl.navy.mil>
From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: Goyra@iol.ie
Cc: cube-lovers@ai.mit.edu
In-Reply-To: <199708051031.LAA16445@mail.iol.ie> (Goyra@iol.ie)
Subject: Re: Megaminx a.k.a. Supernova

Mark Longridge <cubeman@idirect.com> wrote:

> I am particularly interested in processes for the 10-spot and 
> 12-spot.

"David Byrden" <Goyra@iol.ie> asked for clarification.

Spot patterns are those in which all the corner and edge cubies agree
with each other, but not necessarily with all the face centers.  They
are so named because the non-matching face centers show up as
contrasting spots.

Mark reported some analysis on them on 31 Oct 95, apparently from GAP.
As he reported, there are five conjugacy classes:

    0. The identity,
    1. The  72-degree twelve-spot,
    2. The 144-degree twelve-spot,
    3. The 120-degree ten-spot,
    4. The 180-degree ten-spot.

The angle given is the displacement of the corners-and-edges ensemble
from the face-centers ensemble.  In cases 1 and 2, the rotation is
about an axis through two opposite face centers; in case 3, through
opposite corners; in case 4, through opposite edges.

Of course, there's no reason to expect optimal processes for these
patterns to the same length.  Interestingly, while the square of the
72-degree is the 144-degree, it is also the case that the square of
the 144-degree is the 72-degree (up to conjugacy).

It's also interesting to consider star patterns, in which the edges
agree with the face centers, and the corners agree with each other.
These come in the same classes as the spots.

A third type of pattern is a distorted checkerboard, in which the
corners and face centers agree with each other, and the edges agree
with each other.  These come in the same classes as well.  I had hoped
to find some in which the edges were apparently reflected with respect
to the face centers (as in the Pons Asinorum and the order 6 6-X
patterns on the cube) but they seem to be in the wrong orbit for the
Megaminx.

Dan
Hoey@AIC.NRL.Navy.Mil

From cube-lovers-errors@mc.lcs.mit.edu  Tue Aug  5 15:30:18 1997
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Mail-from: From reid@math.brown.edu Tue Aug  5 14:31:16 1997
Message-Id: <199708051827.OAA07000@life.ai.mit.edu>
Date: Tue, 5 Aug 1997 14:33:36 -0400
From: michael reid <reid@math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: 20f maneuvers for superfliptwist

because of the historical interest in the pattern superfliptwist, i
decided to find all  20f  maneuvers for it.  this took about 114 hours
of searching, using the symmetry reductions i described in an earlier
message.

by "all" maneuvers, i mean that any 20 face turn sequence for
superfliptwist can be transformed in to one on my list by conjugating
by one of the 24 symmetries that fix superfliptwist, or by inverting
the sequence and conjugating by the cube rotation  C_U , and perhaps
again by one of the 24 symmetries.  some extra processing by hand
was required to eliminate inverse pairs.  hopefully i haven't made
any errors here.  the list of maneuvers is


   U  R  F' B  U' D' F  U' D  F  L  F' L' U  R  D  F  U  R  L   (20f, 20q)
   U  F  D  L  U  R' F' R  F  U' D  F  U' D' F' B  L  U  R  L   (20f, 20q)

   U  R' F' U' F' R' B' D2 R' D  R  L' B  R  F  B2 R' U' B  D'  (20f, 22q)
   U  F  L  D  L  F  R  U2 F  U' F' B  R' F' R2 L' F  D  R' D'  (20f, 22q)
   U  R' U' D  F' U' F2 B' U  L  F' R  L' U  B  L  B  U  F  R2  (20f, 22q)
   U  R' U' D  F' U' B' R' F' R  U  F2 R  U  B  L  B  U  F  R2  (20f, 22q)
   U  F  L  D  L  F  R  U2 F  U' F' B  R' F' L' U' R' U  F  R2  (20f, 22q)
   U  F  R' F' L' U' R' U' D  F' U  F2 R  U  L  B  L  U  F  R2  (20f, 22q)
   U  F  D  L  D  F  R  L' U' L  F  U2 D' F' U' F' B  R' F  R2  (20f, 22q)
   U  F  D  L  D  F  R  U2 F  R  U' R' D' F' U' F' B  R' F  R2  (20f, 22q)

   U  R  B  D2 L  B  R' D' R' B  L' D2 L  B' D2 R' F  B2 D' R   (20f, 24q)
   U  R' B  L2 U' L2 U' B  U' L2 D  R  B  D  F  U2 R' L' B' R'  (20f, 24q)
   U  F  B' R' U2 L  U' R2 B' L' F2 U' R' D' L2 U  D  B  D' B   (20f, 24q)
   U  R  L2 F  U  F  U  F  L  D2 L' D' L  U' D  F2 B  L' F  R2  (20f, 24q)
   U  R  L' B' R' F  R' F' B' D2 F  U  B  L2 D  R  U2 B  D' B2  (20f, 24q)
   U  R' U2 D  F  B' R  F' R' F2 R  U  B  U  B  U  R2 L  B  L2  (20f, 24q)
   U  R' B  R  U' L' U2 B' R2 L' D  B2 L  U' B  R  F  U  B  L2  (20f, 24q)
   U  R' D2 B' U' F2 R' D' L' U2 R  L  B  L' B  R  F  B' D' R2  (20f, 24q)
   U  F  D  L  U  F' R  U2 B  R' L2 U' F2 R' F' L  U  L' F  R2  (20f, 24q)
   U  F2 R' U' F' R' L2 U  B  U  L' F  B' U2 D  L  U' D' B  R2  (20f, 24q)
   U  F' B' L  F  B2 U' D  L' B  U  B  R' L2 D' B' R' D2 B  R2  (20f, 24q)

   U  R2 B  L' U2 B' R' L  F2 D  F  L2 D  R' F2 D' R  L' U2 B   (20f, 26q)
   U  F  R' L  D  B  R2 U2 L2 D' R' D' R  L2 U' F  L  D2 B  R2  (20f, 26q)
   U  F2 L  D  B' R  L2 F' R' F' L2 B2 R2 U  F  R' L  D  B  R2  (20f, 26q)

   U  F2 R' L' U  F' U' D2 B2 U' B  D  R' L2 D2 L2 D2 F  U2 B'  (20f, 28q)


the maneuver that herbert kociemba found is equivalent to the  28q
maneuver.

the two maneuvers that are  20q  long can be obtained from one another
by inverting the first, then cyclically shifting the antislice to the
end of the maneuver, and then reorienting.

the first of the  26q  maneuvers is quite interesting.  it can also be
written as

      (U  R2 B  L' U2 B' R' L  F2 D   C_UF)^2

where  C_UF  is a cube rotation about the UF - DB edge axis (as in
bandelow's book).

mike

From cube-lovers-errors@mc.lcs.mit.edu  Tue Aug  5 19:12:26 1997
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Mail-from: From Hoey@AIC.NRL.Navy.Mil Tue Aug  5 19:11:22 1997
Date: Tue, 5 Aug 1997 19:11:12 -0400
Message-Id: <199708052311.TAA13218@sun30.aic.nrl.navy.mil>
From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: Cube-lovers@ai.mit.edu
Subject: Glyph patterns

Years ago I thought for a while on a taxonomy of some of the pretty
patterns.  Mark's bringing some of them up on the Megaminx has
reminded me of them.

My favorite class of pretty patterns is the "glyph" patterns.  These
are the patterns on which each face of the cube has facets of at most
two colors.  They include most of the pretty patterns we've discussed
on the list.  The "glyphs" here are the partition of the nine facets
into two colors, where we aren't concerned with which colors but with
the partition.  I call the part of the glyph that includes the face
center the "figure" and its complement the "ground".

There are only 51 glyphs up to the symmetry of the square, or 70 if we
distinguish chiral pairs.  Some of the common ones we have discussed
are blank, X, plus, dot, bar, T, slash, and H.  I recall seeing a
cubing book that assigns 26 of the glyphs to letters of the alphabet,
where you try to place all the letters of your favorite six-letter
word on the cube, or something like that.

Classification and analysis of glyph patterns is often simplified by
separating out the corner-glyph from the edge-glyph.  There are only 6
each of these subglyphs (up to symmetry), mostly determined by how
many "figure" facets of each type there are.

Name       0     1     2     D     3     4    
        +-----+-----+-----+-----+-----+-----+
        |.   .|X   .|X   X|X   .|X   X|X   X|
Corner  |  .  |  .  |  .  |  .  |  .  |  .  |
        |.   .|.   .|.   .|.   X|.   X|X   X|
        +-----+-----+-----+-----+-----+-----+
        |  .  |  X  |  X  |  X  |  X  |  X  |
Edge    |. . .|. . .|X . .|. . .|X . X|X . X|
        |  .  |  .  |  .  |  X  |  .  |  X  |
        +-----+-----+-----+-----+-----+-----+

So a type-2D glyph would have the corner-glyph 2 and the edge-glyph D.
There are two type-2D glyphs, called T and U.

An important subclass of the glyph patterns are the "isoglyphs", which
have the same glyph on all six faces.  We've talked about the 6-T,
6-plus, 6-X, 6-H, and 6-spot patterns.  Recall that you can twist just
two opposite corners of the cube--I think Hofstadter called this a
boson or something.  I was amused to find that there is just one other
6-corner isoglyph of the cube.

Another subclass are the "continuous" glyph patterns, in which the
glyphs on neighboring faces match along the edge.  That is to say, a
facet of an edge cubie and an adjacent facet of a corner cubie have
the same color if and only if the other facet of the edge cubie and
the adjacent facet of the corner cubie have the same color.  This
matching condition gives the 6-plus patterns much of their charm.

When every cubie of a continuous glyph pattern has either all "figure"
facets or all "ground" facets, we call the pattern a "reassembled"
glyph pattern.  In this case, we can envision the cube having been cut
apart into figure and ground cubies and put back together in a
different orientation.  Note that the reorientation may include a
reflection, as we see in the Pons Asinorum.

Some of the prettiest reassembled glyph patterns have corner type 4 on
all faces--I call them "path patterns", because you can consider them
a road map going around the cube.  In 1981 Dave Ackley found one he
called the "four-way street", which is the unique continuous type-41
isoglyph.  If you can find it, you know what I'm talking about.

I've been considering writing a program (or perhaps sparking someone
else's interest in writing a program) to count and classify all the
glyph patterns, possibly by using corner-edge reduction.  It might be
interesting to see if there is a set of nine cubes that exhibits all
51 glyphs, or if not what the smallest panglyphic set is.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil


From cube-lovers-errors@mc.lcs.mit.edu  Wed Aug  6 11:16:57 1997
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Mail-from: From reid@math.brown.edu Tue Aug  5 22:42:50 1997
Message-Id: <199708060239.WAA23888@life.ai.mit.edu>
Date: Tue, 5 Aug 1997 22:45:08 -0400
From: michael reid <reid@math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: more patterns at distance 20f

i can also give three more patterns that are exactly 20 face turns from
start.  all three are very symmetric; they have 24 symmetries.  the
symmetry group is "H" in dan hoey's taxonomy.  in particular, all are
local maxima (in the quarter turn metric).  see hoey and saxe's
"symmetry and local maxima" (december 14, 1980) for more info about this.

the first pattern is the composition of superfliptwist with pons asinorum.
you may recall that i suggested this pattern to dik winter when he was
looking for positions that couldn't be solved in 20 face turns or less.
he did succeed in solving it in 20f, using kociemba's algorithm, but it
took much longer than most other positions did.

the other two are inverses of one another.  they can both be described
as the composition of superfliptwist with 6 H's.  however, the patterns
"6 H's" and "superfliptwist" each come in two orientations.  therefore,
fix your favorite orientation of 6 H's; now there are two different
orientations of superfliptwist which may be composed.  this gives two
distinct patterns, and the positions are inverses.

by symmetry, we may assume that the first face turn of a maneuver for
any of these positions is either  U  or  U2.  to confirm that the pattern
superfliptwist . pons asinorum   is not within 19f of start, we need to
search the positions

          superfliptwist . pons asinorum . U
and       superfliptwist . pons asinorum . U2

completely through depth 18f.  similarly, for the second pattern,
two complete searches through depth 18f were required.  the third
pattern is the same distance from start as is its inverse, so this one
doesn't require further testing.

my optimal solver did not find the minimal maneuvers for these, although
it certainly would have, if i'd let it continue searching long enough.
however, one can find 20f maneuvers using kociemba's algorithm:

   superfliptwist . pons asinorum:
   D' R' B' L2 U' L  B' D' R' D' B2 D2 B' U  D2 R2 F2 D' L' B'   (20f)

   superfliptwist . 6 H's:
   B' L2 D  B2 R' D2 F' L2 U' L' F' B  U' R  D' R2 F2 R2 U' D2   (20f)

it would be nice to find a position that was not within 20f of start.
of course, we don't know if any such positions exist.  my guess is that
they do, but that's only a hunch.  dik winter examined 9000 random
positions and found that they were all within 20f of start.  therefore
the positions we're looking for are extremely scarce.  i think that
looking at positions with a lot of symmetry seems to be the right way
to approach this.  i've tested some of the most symmetric positions,
but each that i examined was solved in 20f or less.

mike

From cube-lovers-errors@mc.lcs.mit.edu  Wed Aug  6 12:37:59 1997
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Mail-from: From saxonia@imn.htwk-leipzig.de Wed Aug  6 11:44:11 1997
Date: Wed, 6 Aug 1997 17:40:38 +0200
From: saxonia@imn.htwk-leipzig.de (Ralf Laue)
Message-Id: <9708061540.AA24642@imn.htwk-leipzig.de>
To: Cube-Lovers@ai.mit.edu
Subject: Rubik's Cube World Records

RUBIK'S CUBE WORLD RECORD LIST
------------------------------

I have created a list of Rubik's Cube world records in the WWW with the URL:
http://www.imn.htwk-leipzig.de/~saxonia/list/rubik.html

It is about speed cubing world records and other funny stuff (cube marathon
record etc.)

I would be glad about comments and corrections to this list. (New record
categories are very welcome: Particularly I am interested in the 4x4x4 
cube speed solving world record!)
If you do not have access to the WWW, just send me an e-mail, and I will send
you the list by e-mail.

If you have your own WWW site about Rubik's Cube, I would be glad if you
would create a link to my URL. 
(The list is a part of my WWW information about unusual world records at 
http://www.imn.htwk-leipzig.de/~saxonia/homepage.html            )

Very Best Wishes,

Ralf Laue

___________________________________________________________________________
Please excuse me for a delay in replying to your e-mail.
I cannot read my incoming mail daily.
-----------------------------------------------------------------------------
Ralf Laue              e-mail: saxonia@imn.htwk-leipzig.de
P. O. Box 80           Read my Homepage about unusual world records:
04181 Leipzig          http://www.imn.htwk-leipzig.de/~saxonia/homepage.html
Germany
-----------------------------------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Wed Aug  6 19:25:21 1997
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Mail-from: From Hoey@AIC.NRL.Navy.Mil Wed Aug  6 19:23:53 1997
Date: Wed, 6 Aug 1997 19:10:30 -0400
Message-Id: <199708062310.TAA17135@sun30.aic.nrl.navy.mil>
From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: Cube-lovers@ai.mit.edu
Subject: Reassembled patterns (was Glyph patterns)

I wrote:

> When every cubie of a continuous glyph pattern has either all "figure"
> facets or all "ground" facets, we call the pattern a "reassembled"
> glyph pattern.  In this case, we can envision the cube having been cut
> apart into figure and ground cubies and put back together in a
> different orientation....

On second thought, I prefer the definition that a (2-part) reassembled
pattern is one that can be partitioned into two sets of cubies, where
the cubies of each set are in agreement with each other.  This
definition differs from the previous in two ways.  Reassembled
patterns need not be continuous--"laughter" is a noncontinuous glyph
pattern.  And not all continous glyph patterns with figure/ground
cubies meet this definition--e.g. flip the LF and RD edge cubies.

We may also speak of 3-part reassembled patterns, though they are not
necessarily glyph patterns.  Are there any particularly nice ones?
Cube-in-a-cube-in-a-cube comes to mind.

Call an "N-part" pattern one that requires cutting into at least N
parts for reassembly.  Surely every position can be reassembled from
at most 21 parts, since that's all the pieces there are.  Is this
achievable?  We could restrict the reorientation of the parts to C,
but in some cases (e.g. pons asinorum) we can manage with fewer parts
if we allow reorienting some of the edges by M.  Is there a 20-part
pattern that would require 21 parts if the orientations were
restricted to C?

In the supergroup, can we manage a 24-part position?  A 23-part
position that requires 24 parts for C-reorientation?

Dan Hoey
Hoey@AIC.NRL.Navy.Mil


From cube-lovers-errors@mc.lcs.mit.edu  Wed Aug  6 22:13:35 1997
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Mail-from: From reid@math.brown.edu Wed Aug  6 22:12:07 1997
Message-Id: <199708070208.WAA08375@life.ai.mit.edu>
Date: Wed, 6 Aug 1997 22:14:21 -0400
From: michael reid <reid@math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: optimal solver for quarter turns

i have my optimal cube solver working for quarter turns.  it seems to
be as effective as the face turn version.  some minimal maneuvers
it has found are

   cube in a cube in a cube
   U' L' U' F' R2 B' R  F  U  B2 U  B' L  U' F  U  R  F'  (20q)

   six X's, order 6
   F  U' L2 F' L' D  R  U' D  L' B  U2 F' L' D' F  D  R   (20q)

   ron's cube within the cube
   F  D' F' R  D  F' R' D  R  D  L' F  L  D  R' F  D'  (17q)

and it has also confirmed minimality of known maneuvers for several
other patterns.

mike

From cube-lovers-errors@mc.lcs.mit.edu  Thu Aug  7 10:59:18 1997
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Mail-from: From reid@math.brown.edu Wed Aug  6 23:51:22 1997
Message-Id: <199708070348.XAA10924@life.ai.mit.edu>
Date: Wed, 6 Aug 1997 23:53:38 -0400
From: michael reid <reid@math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: superflip requires 24 quarter turns

with my optimal solver, i can show that superflip is exactly
24 quarter turns from start.  this was already shown by jerry
bryan, so this confirms his result.

first some history.  david plummer gave a 28q maneuver for
superflip on december 10, 1980.  apparently there was no
improvement to this until january 1995, when i implemented
kociemba's algorithm for quarter turns.  after a lot of searching,
where i specified the initial sequence  R' U2 , it found

   R' U2 B  L' F  U' B  D  F  U  D' L  D2 F' R  B' D  F' U' B' U  D' (24q)

mark longridge notes that this sequence has a remarkable symmetry,
namely that it may be written as

   (R' U2 B  L' F  U' B  D  F  U  D'  C_-1)^2 ,

where  C_-1  denotes central reflection.

later in january 1995, i completed an exhaustive search for superflip
in 20 quarter turns, without finding any maneuvers.  i used my quarter
turn version of kociemba's algorithm, which took 29 cpu hours.  this
improved the lower bound of the diameter of the cube group to 22q.
the previous lower bound was 21q, obtained by a counting argument.

in february 1995, jerry bryan improved this result to show that superflip
is not within 22 quarter turns, and thus is exactly 24 quarter turns
from start.  this also improved the lower bound for the diameter to 24q.

we'll use symmetry to reduce the size of the search space dramatically.
consider three cases for a minimal maneuver for superflip.

1)  it contains a half turn (i.e. two consecutive quarter turns of the
    same face).
2)  it does not contain a half turn, but contains two consecutive turns
    of opposite faces.
3)  otherwise.

in case 1, as in the face turn situation, we may suppose that the first
three quarter turns are  U R2 .

in case 2, by cyclically shifting, we may suppose these two turns are
the first two.  if they form a slice (U D') then we may take the
first three quarter turns to be  U D' R .  if they form an antislice,
then we may take the first three quarter turns to be either  U D R  or
U D R' .

in case 3, i claim that we may find three consecutive turns of mutually
adjacent faces.  otherwise, if the first two faces turned were  U  and  R,
then we'd only be turning  U , R , D  and  L .  however, edges cannot
change orientation when only these faces are turned.  thus the claim holds,
and by cyclically shifting, we may suppose that these three faces are
U , R  and  F .  by symmetry, we may suppose that they're turned in that
order.  now we have eight cases:

     U  R  F      U  R  F'     U  R' F      U  R' F'
     U' R  F      U' R  F'     U' R' F      U' R' F'

we can eliminate two of these by using inversion.  inverting the case
U' R  F  gives  F' R' U .  conjugating this by the appropriate cube
reflection gives  U  R  F' , and these three turns can be cyclically
shifted to the beginning of the maneuver.  similarly, the case  U' R' F
can be transformed to the case  U' R  F .

thus ten cases remain.  to show that superflip is not within 22q of start,
these cases must be searched through 19q.  my program took 22 hours to
searched these completely, and no maneuvers were found.

iw would be nice to know all the minimal maneuvers for superflip.  the
branching factor is about 9.37, so an exhaustive search would take about
22 * (9.37)^2  hours, which is about 80 days.  this is feasible, but is
definitely a long term project.  i've already searched the first case,
(beginning with  U R2) which would seem to be the most likely, through 21q.
this took about 147 hours.  i expected it to find a lot of maneuvers, but
it only found 4, in two inverse pairs.  the first is equivalent to the
maneuver above, and the new one is

   U  R2 F' R  D' L  B' R  U' R  U' D  F' U  F' U' D' B  L' F' B' D' L' (24q)

mike

From cube-lovers-errors@mc.lcs.mit.edu  Thu Aug  7 15:46:48 1997
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Mail-from: From reid@math.brown.edu Thu Aug  7 15:46:42 1997
Message-Id: <199708071943.PAA07912@life.ai.mit.edu>
Date: Thu, 7 Aug 1997 15:48:49 -0400
From: michael reid <reid@math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: composition of superflip and pons asinorum

my optimal cube solver has also found all minimal maneuvers for the
composition of superflip and pons asinorum.  this was previously done
by jerry bryan, so the purpose here is to confirm his results.

up to symmetry, there are 10 maneuvers of length 20q, which occur in
5 inverse pairs.  they are

     U  R  F  D  R  U' D  L' U' D  F' B2 R  L' D' F' L' B' R'     (20q)
     U  R  U  F  U  F  B' L' F  B' R  L' B' R  L' U  F' U' R' U'  (20q)
     U  R  U  F  D  R  L' B' R  L' F  B' L' F  B' D  F' U' R' U'  (20q)
     U  R  D  B  U  R  L' F' R  L' F' B  L' F' B  U  B' D' R' U'  (20q)
     U  R  D  B  D  F' B  L' F' B  R  L' F' R  L' D  B' D' R' U'  (20q)

this agrees exactly with jerry bryan's results.

mike


From cube-lovers-errors@mc.lcs.mit.edu  Fri Aug  8 11:22:01 1997
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Mail-from: From kociemba@hrz1.hrz.th-darmstadt.de Fri Aug  8 09:15:32 1997
Message-Id: <33EB1ABE.1DC8@hrz1.hrz.th-darmstadt.de>
Date: Fri, 08 Aug 1997 15:10:22 +0200
From: Herbert Kociemba <kociemba@hrz1.hrz.th-darmstadt.de>
To: Dan Hoey <Hoey@aic.nrl.navy.mil>, cube-lovers@ai.mit.edu
Subject: Continuous isoglyph patterns
References: <199708052311.TAA13218@sun30.aic.nrl.navy.mil>

Dan Hoey made some proposals concerning 2-colored cube patterns. The
"coninuous"-condition i find especially interesting. I added this
feature to my Cube Explorer program and found exactly 34 continuous
isoglyphs (plus the trivial solution). I don't know if there are any
among them, which have not been described somewhere else before.
Here are generators for the patterns (I searched only about 1 minute for
most generators of the patterns, so there is no claim for the maneuvers
to be optimal):

D F2 D' . R B2 R' D F2 D' R B2 R'  (12)
U . L' D' U B2 D U' L' U'  (9)
U2 L' B2 . F R' D' R2 D B2 F' U L  (12)
D' U . L' R B' F D' U  (8)
R2 D L2 U' B2 D' U2 . R' F' U R B' L' D' F L2 B2 R U'  (19)
F2 L2 U L2 U' F2 D . B L R' D' U' F' D2 F' R2 F R  (18)
B2 U' B2 D' B2 D . L' B2 F' U R D2 R' D' U' F U  (17)
L2 D2 B D2 B' D2 B L2 . D B R D2 F L2 D F' R'  (17)
B2 U' B2 L2 B2 U2 B2 U' D2 . R U R' D' L U F U' D' L  (19)
U R2 D . F' L D2 U2 R' D2 U2 F D' R2 U'  (14)
U B2 . L B F' L2 R' B' F D U2 L' B2 U'  (14)
D' U . F' U L' R B' U F D' U R'  (12)
U' B2 F2 L2 U B2 U' L2 F2 . B' U R' F D' R2 D2 R' F'  (18)
F2 R2 D R2 D U F2 D' . R' D' F L2 F' D R U'  (16)
D U2 R2 D' U' . R D B2 R2 B2 R2 D B2 D2 R U'  (16)
D U2 L2 U R2 U' L2 U . R' B2 L2 F' L2 B' R' F' L D U'  (19)
D2 U F2 D' L2 U R2 B2 . R B2 R2 U2 B' L2 D2 R2 D R' U'  (19)
D2 R B2 R . F L B' F U' R L' U' F' D2 F' L2  (16)
D' B2 F2 D' U L2 . F' L R' F' D U' R D B2 R  (16)
U' R2 F2 U2 . L' D2 B' L2 U' L2 D2 L U2 F' U2  (15)
L2 U2 R' . B' D U' B2 D' R' D L D2 F D U2 L2  (16)
U' F2 U . R U2 R2 U2 R' F' R2 F U' F2 U  (14)
U2 R2 F2 U B2 D' . L' F L' F L' F D B2 U  (15)
D2 R F2 L' D2 R . B D2 F' L2 U' R' D L F D L' D L'  (19)
D' L2 F2 L2 B2 R2 U F2 U2 . L' F R B D R U' L F' U2 F  (20)
B2 L2 R2 U B2 R2 D F2 U' . B F U2 R' B2 L2 D U' B' L' R'  (20)
U B2 U2 L2 U F2 R2 B2 U' L2 D2 F2 U' . B L2 R2 D2 U2 F'  (19)
L2 . R' B2 F2 D2 B2 F2 L2 R2 U2 R'  (11)
D U L2 B2 D U' . F' U F' R F2 R' F D' B2 L2 D' U'  (18)
L2 U' B2 F2 D . R D F' U' R2 B2 U' B D2 B' F' L U'  (18)
D F2 R2 F2 R2 U F2 . R F2 R D2 U' F L' F' L D  (17)
B2 R2 F' U2 D2 L2 R2 B . U' L R B' F U D B2 F2 R' F2  (19)
D' L2 R2 D2 B2 F2 U' . R' B' F D' U L R' F2 D2 U2 F'  (18)
B2 F2 L2 R2 D2 U2  (6)

If you copy and paste the maneuvers from this message to a text file,
you can load them into Cube Explorer and directly watch the results.

The response to my Cube Explorer 1.0 program showed me, that the
userinterface and the terminology of the program are confusing (if not
to say bad) and some features are missing which should be there.
I almost completed Version 1.5 now. When I put it to my homepage in a
few days, I will send another message.

--Herbert

From cube-lovers-errors@mc.lcs.mit.edu  Fri Aug  8 12:47:55 1997
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Mail-from: From ERCO@compuserve.com Fri Aug  8 03:45:46 1997
Date: Fri, 8 Aug 1997 00:52:32 -0400
From: Edwin Saesen <ERCO@compuserve.com>
Subject: Alexander's star etc.
To: CUBE <CUBE-LOVERS@ai.mit.edu>
Message-Id: <199708080052_MC2-1CA0-3F67@compuserve.com>

Hi everyone,

can anyone tell me a way to disassemble an Alexander's Star without
breaking it? I've had this one for YEARS (about 10 or so), and never
really found a way to do it. It's very difficult to use, and I think a
little lubricationg will do wonders :-)

On a different note, I visited Christoph Bandelow on this wednesday,
and it was absolutely incredible to see the range of puzzles he has
(although he said he's not collectiong anymore). Anyway, I got quite a
few nice things from him, including a 5x5x5 and two replacement center
pieces for my 4x4x4 (YES!!! YES!!!!!!) :-)

Michael

PS: This doesn't mean I'm not interested in a second 4^3 one anymore -
I would feel more comfortable with a second 4^3 to have a chance to
replace broken pieces myself, but at least now it's working again.

From cube-lovers-errors@mc.lcs.mit.edu  Sat Aug  9 15:13:50 1997
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Mail-from: From Hoey@AIC.NRL.Navy.Mil Sat Aug  9 15:09:29 1997
Date: Sat, 9 Aug 1997 15:09:01 -0400
Message-Id: <199708091909.PAA07964@sun30.aic.nrl.navy.mil>
From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: kociemba@hrz1.hrz.th-darmstadt.de
Cc: cube-lovers@ai.mit.edu
Subject: Re: Continuous isoglyph patterns

Bravo, Herbert!  A very nice list.  It's surprising how many of them
are reassembled patterns, too.  Only the second and tenth are not
reassembled, and both fail by using a reassembled pattern to
camouflage a small distortion.  Pattern #2 is pattern #3 composed with
a two-flip, and pattern #10 is pattern #9 composed with a three-cycle
of edges.

There are four elements of M used to perform the reorientation of the
reassembled patterns.  Over half of them use the order-3
major-diagonal rotation, of Plummer's cross: patterns 1, 3, 4, 6, 7,
9, 11, 12, 13, 15, 16, 17, 18, 19, 21, 22, 23, 24, 25, 27, 30, 31, and
32.  Several use the order-2 major-diagonal rotation, of Christman's
cross: patterns 5, 8, 14, 20, 26, 28, and 29.  The only pattern
reassembled by the major diagonal reflected rotation is the order-6
6-X, pattern 33, And the only pattern reassembled by central
reflection is Pons Asinorum, pattern 34.

I've also classified these patterns by the glyph that appears on the
faces (modulo my clerical errors).  Patterns I know traditional names
for are given with an asterisk; I've made up temporary descriptive
names otherwise.

 Glyph  Type  Patterns

 X . X        2. Girdle 3-cycle, distorted
 X X X   01   3. Girdle 3-cycle
 X X X

 X . X        8. Christman's girdle
 . X X   02   9. Off-girdle 3-cycles
 X X X       10. Off-girdle 3-cycles, distorted
             11. Girdle 3-cycles

 X . X
 . X X   03  21. Plummer's C's
 X . X

 X . X       32. Plummer's X
 . X .   04  33. Order-6 X
 X . X       34. * Pons Asinorum


 . X X
 X X X   10   1. * Meson
 X X X
 
 . X X
 X X .   11   7. Meson & girdle 3-cycle
 X X X

 . . X       13. Plummer's cluster
 . X X   12  14. Christman's cluster
 X X X

 . X X
 X X .   12  16. Meson & girdle 3-cycles
 X . X

 X . X
 X X .   13  24. Plummer's Y's
 X . .

 . . X       25. Plummer's cluster & girdle 3-cycles
 . X .   14  26. Christman's cluster & girdle
 X . X       

 . X .
 X X X   30  15. Plummer's rabbits
 . X X

 . X X
 . X X   31  22. Plummer's P's
 . X .

 X X .
 X X .   32  23.  * Cube in a cube
 . . .

 . X .       29. Christman's arrow
 X X .   32  30. Plummers's arrow
 . . X

 X . .
 . X X   33  16. Plummer's bend
 . . .

 X . .        5. Christman's comma
 . X .   34   6. Plummer's comma  
 . . .

 . X .       27. * Plummer's Cross
 X X X   40  28. * Christman's Cross
 . X .

 . X .
 X X X   41  31. * Four-way street
 . . .

 . X .       18. Plummer's cube out of cube in a cube
 X X .   42  19. * Worm
 . . .       20. Christman's cube out of cube in a cube

 . X .
 . X .   43  12. Plummer's U's
 . . .

 . . .
 . X .   44   4. * Six-spot
 . . .

Dan
Hoey@AIC.NRL.Navy.Mil

From cube-lovers-errors@mc.lcs.mit.edu  Thu Aug 14 18:29:06 1997
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Mail-from: From reid@math.brown.edu Thu Aug 14 18:19:32 1997
Message-Id: <199708142216.SAA16395@life.ai.mit.edu>
Date: Thu, 14 Aug 1997 18:21:24 -0400
From: michael reid <reid@math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: patterns with 24-fold symmetry

i've finished computing minimal maneuvers for those positions with
24-fold symmetry.  these positions were classified by dan hoey and
jim saxe in their note "symmetry and local maxima."  there are 24
such positions; they form an abelian subgroup of type 6, 2, 2.
we may take as generators  superfliptwist, pons asinorum, and 6 H's.
of these 24 positions, 4 have 48-fold symmetry; i'll include these
here as well.  the other 20 positions occur in 10 pairs which differ
only in orientation; i.e. there are 10 "patterns".

some of these maneuvers were found earlier by others; i'll acknowledge
this to the extent that i'm aware of it.  in addition, most maneuvers
were found by kociemba's algorithm, and a few by my optimal solver,
which is based on the same ideas.


positions with 48-fold symmetry

start  (0q, 0f)  no turns needed

superflip
 R' U2 B  L' F  U' B  D  F  U  D' L  D2 F' R  B' D  F' U' B' U  D'  (24q)
 U  R2 F' R  D' L  B' R  U' R  U' D  F' U  F' U' D' B  L' F' B' D' L' (24q)

 U  R2 F  B  R  B2 R  U2 L  B2 R  U' D' R2 F  R' L  B2 U2 F2   (20f)
 U  R2 F  B  R  B2 R  U2 L  B2 R  U' D' R2 F  D2 B2 U2 R' L    (20f)
dik winter was the first to find a 20f maneuver.


pons asinorum
 U2 D2 F2 B2 R2 L2  (12q, 6f)

 F  B' U2 D2 R2 L2 F  B'  (12q)
 F  B' U' D  F  B' U' D  F  B' U' D   (12q)
this last maneuver is due to dan hoey.


pons asinorum composed with superflip
 B' D' L' F' D' F' B  U  F' B  R2 L  U  D' F  L  U  R  D   (20q, 19f)

 F' U' B' R' F  R  L' D' R  L' U  D' L' U  D' F  R  B  U  F   (20q)
 B' R' F' U' F  R  L' D' R  L' U  D' L' U  D' F  U  F  R  B   (20q)
 B' R' B' D' F  U' D  L' U' D  R  L' U' R  L' F  D  B  R  B   (20q)
 R  U  R  B  R' U' D  F  U' D  F  B' D  F  B' R' B' R' U' R'  (20q)

 U  D  F  R  L' F  B' L  D2 R  L  F' B' U' L2 F  B' U2 L'   (19f)
 U  D  F' B' L' U2 F' B  L2 U' R' L' F' U' D  F' B  D' L2   (19f)
 U2 R  F  U  F  B' L' D' F  B' L  B  R  L' U  D2 B' R' U2   (19f)
 U2 R  F  U2 D' R' L  F' L' F  B' U  L  F  B' D' B' R' U2   (19f)
 U2 R  U2 D2 R  U' L' U  B  R  F2 U' D  B' R' F' D  B' L2   (19f)
jerry bryan found the 20q maneuvers.


positions with 24-fold symmetry

superfliptwist
 U  R  F' B  U' D' F  U' D  F  L  F' L' U  R  D  F  U  R  L   (20q, 20f)
herbert kociemba was the first to find a 20f maneuver.

supertwist
 U  R' B  D  B  U  L  D  B' D2 R  U' F  L  F  R  D  L  F' L2  (22q)
 B' L2 U  D  R2 B' D2 F2 D' R2 F  B  L2 D' B2 U2  (16f)
dik winter first found the 16f maneuver.

6 H's
 D2 L2 B2 U2 D2 B2 R2 D2  (16q, 8f)
jim saxe found this maneuver.

superflip composed with 6 H's
 U  F' L' F' B  U  R  F' B  U' B' U  D' R2 L' B  U'  (18q, 17f)

superfliptwist composed with pons asinorum
 U  F  B  D  R  L  U' F2 B2 R  L  D' F  B  D  R  L  D  F' B'  (22q, 20f)
dik winter was the first to find a 20f maneuver.

supertwist composed with pons asinorum
 F  L  D  F  U' B2 R  F  R' F' R  F  L2 U' R  D  B  R   (20q)
 B2 L  U2 F' B' U2 R' F2 L2 F' U2 R' L' U2 B  R2  (16f)

superfliptwist composed with 6 H's  (type 1)
 U  F  B  U' R  L  U  F  B  R2 L2 D' F  B  U' R  L  D' R' L'  (22q, 20f)
superfliptwist composed with 6 H's  (type 2)
 inverse of type 1

supertwist composed with 6 H's   (type 1)
 U2 L  U  B  D  L  U  B' R' L' F' D  R  U  F  D  L' U2  (20q)
 L' B2 U' D' B2 R' U2 L2 U  B2 R  L  F2 U  R2 U2  (16f)
supertwist composed with 6 H's   (type 2)
 inverse of type 1


some of these maneuvers have some symmetry.  i find the maneuver for
superfliptwist composed with pons asinorum especially interesting.
it is composed of: twists of the  U  or  D  face, and antislices
along the R-L and F-B axes:

     U (FB) D (RL) U' (F2B2) (RL) D' (FB) D (RL) D (F'B')

therefore, when we conjugate this maneuver by the cube rotation  C_U2,
we get the same maneuver!  the maneuver for superfliptwist composed with
6 H's has the same type of symmetry.

mike

From cube-lovers-errors@mc.lcs.mit.edu  Fri Aug 15 11:12:17 1997
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Mail-from: From reid@math.brown.edu Thu Aug 14 21:48:30 1997
Message-Id: <199708150145.VAA22661@life.ai.mit.edu>
Date: Thu, 14 Aug 1997 21:50:27 -0400
From: michael reid <reid@math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: isoglyphs

dan's idea of glyph patterns, especially isoglyphs is quite interesting.
herbert has given all the "continuous" isoglyphs.  i spent some time
looking for other isoglyphs, and was surprised at how many exist.
herbert, how did you find those?  is that part of your pattern generator?
if your program can also find all "discontinuous" isoglyphs, then i guess
there's not much point in trying to do it by hand.

mike


From cube-lovers-errors@mc.lcs.mit.edu  Fri Aug 15 13:38:15 1997
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Mail-from: From goyra@iol.ie Fri Aug 15 12:42:42 1997
Message-Id: <199708151639.RAA24296@GPO.iol.ie>
From: "Goyra" <goyra@iol.ie>
To: <cube-lovers@ai.mit.edu>
Subject: Got a new shape for a Rubik puzzle?
Date: Fri, 15 Aug 1997 17:38:13 +0100

The Rubik Gallery is a website with Java
Rubik puzzles. There are dodecahedrons,
icosahedrons, cubes, etc etc. The software 
can support any shape at all, twisting in any
way imaginable, provided the axes all meet 
at one point.

I'd like to add some more intricate and
strange shapes of puzzle, and before I sit down 
to bust my brain over the geometry, I want to 
ask the list members for ideas.

You guys are mathematical geniuses and 
I'm sure some of you already have ideas for
wierd puzzles that will never see the light of day
in physical form; perhaps ones with Penrose 
tiles as the facelets, perhaps ones that twist
in non-intuitive ways to create surprising 
shapes.If you'd like us all to see, twist and 
solve your puzzle, please tell me about it.

David

http://www.iol.ie/~goyra/Rubik.html

From cube-lovers-errors@mc.lcs.mit.edu  Fri Aug 15 16:56:00 1997
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Mail-from: From Hoey@AIC.NRL.Navy.Mil Fri Aug 15 15:27:45 1997
Date: Fri, 15 Aug 1997 15:27:33 -0400
Message-Id: <199708151927.PAA03768@sun30.aic.nrl.navy.mil>
From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: reid@math.brown.edu
Cc: cube-lovers@ai.mit.edu
In-Reply-To: <199708150145.VAA22661@life.ai.mit.edu> (message from michael
	reid on Thu, 14 Aug 1997 21:50:27 -0400)
Subject: Re: isoglyphs

Mike,

I'm glad you like glyphs, and I'd also like to know about the other
isoglyphs (by which I don't want to minimize my interest in the wealth
of optimal processes you've been producing!).  In particular, we've
seen isoglyphs with all corner types except D (which we know is
impossible) and with all edge types.  So we might wonder if all the
glyph types not involving corner type D are achievable.  But I know
there is no isoglyph of type 4D (stripe).  Are there others?

One superset of the isoglyphs that might be worth looking is the
partial isoglyphs, in which all faces are either the same glyph or
blank (type 00).  This allows corner type D (laughter is
4 type D4 + 2 type 00).  These even come in continuous varieties
(slice is 4 type 4D + 2 type 00).  Is there a partial isoglyph pattern
for every glyph?

And what about chiral partial isoglyphs, for which isoglyphicity is
redefined to require the same handedness for orientable patterns?  I'm
pretty sure all the isoglyphs we've seen so far are chiral, but are
there isoglyphs achievable only non-chirally?

Dan
Hoey@AIC.NRL.Navy.Mil

From cube-lovers-errors@mc.lcs.mit.edu  Fri Aug 15 18:58:41 1997
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Mail-from: From Hoey@AIC.NRL.Navy.Mil Fri Aug 15 19:00:35 1997
Date: Fri, 15 Aug 1997 19:00:26 -0400
Message-Id: <199708152300.TAA04077@sun30.aic.nrl.navy.mil>
From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: reid@math.brown.edu
Cc: cube-lovers@ai.mit.edu
In-Reply-To: <199708142216.SAA16395@life.ai.mit.edu> (message from michael
	reid on Thu, 14 Aug 1997 18:21:24 -0400)
Subject: Re: patterns with 24-fold symmetry

Mike Reid writes:

> i've finished computing minimal maneuvers for those positions with
> 24-fold symmetry.  these positions were classified by dan hoey and
> jim saxe in their note "symmetry and local maxima."  there are 24
> such positions; they form an abelian subgroup of type 6, 2, 2.

It took me a while to understand that.  For the benefit of other
cube-lovers, since any finite Abelian group can be decomposed into a
direct product of cyclic groups, it can be typified by listing the
orders of its factors.

> we may take as generators  superfliptwist, pons asinorum, and 6 H's.
> of these 24 positions, 4 have 48-fold symmetry; i'll include these
> here as well.  the other 20 positions occur in 10 pairs which differ
> only in orientation; i.e. there are 10 "patterns".

It may be better to take the order-6 generator to be one of the
6-H-supertwists.  Then you can tell the M-symmetric positions because
they project to the identity of the 6-factor.  Writing p, f, t, h for
pons, superflip, supertwist, and 6-H, I get the following table of
positions (suffixed with optimal qtw:ftw).

        i          p          f            fp
      ...............................................
    i : i    0:0   p   12:6   f    24:20   fp   20:19
      :
   th : th  20:16  th' 20:16  fth  22:20   fth' 22:20
      :
    t : t   22:16  pt  20:16  ft   20:20   fpt  22:20
      :
    h : h   16:8   h   16:8   fh   18:17   fh   18:17
      :
    t : t   22:16  pt  20:16  ft   20:20   fpt  22:20
      :
   th': th' 20:16  th  20:16  fth' 22:20   fth  22:20

The last two rows could be omitted, just as the last column could be
with your decomposition:

        i          h            p          h
      ...............................................
    i : i   0:0    h    16:8    p   12:6   h    16:8
      :
   ft : ft 20:20   fth  22:20   ftp 22:20  fth' 22:20
      :
    t : t  22:16   th   20:16   tp  20:16  th'  20:16
      :
    f : f  24:20   fh   18:17   fp  20:19  fh   18:17
      :
    t : t  22:16   th'  20:16   tp  20:16  th   20:16
      :
   ft : ft 20:20   fth' 22:20   ftp 22:20  fth  22:20

This has the advantage of having patterns on each row nearer each
other.

By the way, this isn't a complete list of optimal maneuvers, is it?
Are you looking to find such a list?  Or would it be too difficult (or
too voluminous)?

And I'm looking forward to seeing optimal maneuvers for the
T-symmetric positions (if I'm not being too presumptuous).

Dan
Hoey@AIC.NRL.Navy.Mil

From cube-lovers-errors@mc.lcs.mit.edu  Fri Aug 15 20:32:05 1997
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Mail-from: From reid@math.brown.edu Fri Aug 15 19:48:02 1997
Message-Id: <199708152344.TAA05912@life.ai.mit.edu>
Date: Fri, 15 Aug 1997 19:49:54 -0400
From: michael reid <reid@math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: Re: isoglyphs

perhaps i am missing something, but doesn't

     D2 R2 U2 D2 R2 U D   (12q, 7f)

produce an isoglyph of type 4D ?

are there any isoglyphs of type 21 or 23 ?  i haven't found any.
each has three possible patterns:

...     .*.         .*.
***     .**         ***
***  ,  ***   and   *.* ,

.*.     ...         ...
.*.     **.         .*.
*.*  ,  *.*   and   *** .

i hadn't even considered chiral versus achiral isoglyphs.  indeed,
all the "continuous" isoglyphs given by herbert are chiral.
achiral isoglyphs certainly exist, for example

     D2 R2 U' B' L  B  U  B  L  F2 R  D' L2 U2 B2 D    (22q, 16f)

of type 11; pattern

*..
***
***

and others can be derived from this.  i suspect that there is no
chiral form of this isoglyph, but i'm not absolutely certain.

another interesting note is that the inverses of the "continuous"
isoglyphs are also isoglyphs; in fact the same pattern, perhaps in
a different orientation.  however, there is at least one (probably
more) isoglyph whose inverse is not an isoglyph.  instead of giving
it right here, i'll challenge other readers to find it/them.

mike


From cube-lovers-errors@mc.lcs.mit.edu  Sat Aug 16 20:08:01 1997
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Mail-from: From kociemba@hrz1.hrz.th-darmstadt.de Sat Aug 16 05:28:59 1997
Message-Id: <33F571EE.764B@hrz1.hrz.th-darmstadt.de>
Date: Sat, 16 Aug 1997 11:25:02 +0200
From: Herbert Kociemba <kociemba@hrz1.hrz.th-darmstadt.de>
To: cube-lovers@ai.mit.edu
Subject: Re: isoglyphs
References: <199708152344.TAA05912@life.ai.mit.edu>

michael reid wrote:

> perhaps i am missing something, but doesn't
> 
>      D2 R2 U2 D2 R2 U D   (12q, 7f)
> 
> produce an isoglyph of type 4D ?

That's right, and it is the only one which exists of this type. But it
is no continuous isoglyph.
> 
> are there any isoglyphs of type 21 or 23 ?  i haven't found any.
> each has three possible patterns:
> 
> ...     .*.         .*.
> ***     .**         ***
> ***  ,  ***   and   *.* ,
> 
> .*.     ...         ...
> .*.     **.         .*.
> *.*  ,  *.*   and   *** .

Here are generators for all isoglyphs of your second pattern:
D2 R2 B2 L2 U2 F2 U2 B2 R2 U2 R2 U2  (12)
U2 R2 F2 R2 U2 B2 D2 B2 L2 U2 R2 U2  (12)
U2 L2 F2 R2 U2 B2 D2 B2 R2 U2 R2 U2  (12)
D2 L2 B2 L2 U2 F2 U2 B2 L2 U2 R2 U2  (12)

and here for the fifth:
D2 R2 F2 L2 U2 B2 D2 F2 L2 U2 L2 U2  (12)
U2 R2 B2 R2 U2 F2 U2 F2 R2 U2 L2 U2  (12)
U2 L2 B2 R2 U2 F2 U2 F2 L2 U2 L2 U2  (12)
D2 L2 F2 L2 U2 B2 D2 F2 R2 U2 L2 U2  (12)

For the others, no isoglyphs exist.

--Herbert

From cube-lovers-errors@mc.lcs.mit.edu  Sat Aug 16 20:49:30 1997
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Mail-from: From kociemba@hrz1.hrz.th-darmstadt.de Sat Aug 16 05:29:55 1997
Message-Id: <33F571F4.2C84@hrz1.hrz.th-darmstadt.de>
Date: Sat, 16 Aug 1997 11:25:08 +0200
From: Herbert Kociemba <kociemba@hrz1.hrz.th-darmstadt.de>
To: cube-lovers@ai.mit.edu
Subject: Re: isoglyphs
References: <199708150145.VAA22661@life.ai.mit.edu>

michael reid wrote:
> 
> dan's idea of glyph patterns, especially isoglyphs is quite interesting.
> herbert has given all the "continuous" isoglyphs.  i spent some time
> looking for other isoglyphs, and was surprised at how many exist.
> herbert, how did you find those?  is that part of your pattern generator?
> if your program can also find all "discontinuous" isoglyphs, then i guess
> there's not much point in trying to do it by hand.

With the pattern generator it's indeed very easy to find the isoglyphs. 
I restricted myself to continuous isoglyphs, because I had the most
interest in them an because the number is quite limited. There are many,
if you do not use the "continuous" condition. By the way, Mike, it would
be nice to complete the chapter "continous isoglyphs" by computing the
shortest generators for them with your program. This hopefully should
not take too long, because most of the generators seem to be rather
short.

Here are for example the (not necessarily continuous) isoglyphs, which
built the "snake patterns". There are 13 of them:

R2 B' U2 B' . D' F' U D' L B' F L D R2 D F2  (16)
D' B2 F2 D' U L2 . F' L R' F' D U' R D B2 R  (16)
R L2 B2 R' . D' L' D B2 F L2 U' L U' F'  (14)
B2 U2 F2 D2 F2 U . R' F' L2 U2 L R U' L2 F2 L' F  (17)
D B2 L2 D2 . F' D2 L B2 D F2 U2 F U2 R' D2  (15)
D2 R D2 F2 L U2 . B R' D R2 D' R B D2 L U2 L'  (17)
U' B2 R2 U2 . F' D2 L' F2 U' F2 D2 F U2 R' U2  (15)
D2 B2 L B2 F2 L' . U B F' L F2 L' B' F D2 U'  (16)
D2 U' B2 D2 U2 L2 U2 . B' U2 L F2 U B2 D2 B D2 R' U2  (18)
U2 R' F2 R' D2 F2 . B U' L' U' B D L' F2 D'  (15)
U' F2 D R2 U L2 U2 . B' U B' U R D' L F' R D2  (17)
U2 F2 D F2 L2 U2 . F' U2 L F2 D B2 D2 F D2 R' F2  (17)
D2 F2 D2 U2 R D2 U2 R' . U R L' B D2 B' R' L D2 U'  (18)


P.S.: I could not hold what I promised, the Cube Explorer 1.5 version
(which for example has the "continuous" feature) still is not ready....

--Herbert

From cube-lovers-errors@mc.lcs.mit.edu  Sat Aug 16 21:30:20 1997
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Mail-from: From reid@math.brown.edu Sat Aug 16 17:12:29 1997
Message-Id: <199708162109.RAA29706@life.ai.mit.edu>
Date: Sat, 16 Aug 1997 17:14:10 -0400
From: michael reid <reid@math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: Re: patterns with 24-fold symmetry

dan asks

> By the way, this isn't a complete list of optimal maneuvers, is it?
> Are you looking to find such a list?  Or would it be too difficult (or
> too voluminous)?

for most, it isn't a complete list.  the main reason is that it's too time
consuming to calculate all of them.  (perhaps it is also too voluminous,
but i'm not sure.)

on an individual basis, here's what i gave:

superflip: all 20f maneuvers, up to inversion, cyclic shifting, and
           conjugation by cube symmetries.
           all _known_ 24f maneuvers, up to the same transformations.

pons asinorum: all 12q maneuvers, up to conjugation by cube symmetries.
               all 6f maneuvers, up to conjugation by cube symmetries.
               (the inverse of each maneuver is the same as some
               conjugate by a cube symmetry.)

superflip composed with pons asinorum:  all 20q and 19f maneuvers, up to
                                        inversion and conjugation by cube
                                        symmetries.

for the H-symmetric patterns (24-fold symmetry), i was less ambitious.
for each, i gave a single minimal maneuver in each metric.  also, in the
cases where there is a maneuver that is minimal in both metrics, i gave
such a maneuver.

> And I'm looking forward to seeing optimal maneuvers for the
> T-symmetric positions (if I'm not being too presumptuous).

how did you know what i'm working on next?  ;-)
i also plan to examine AC-symmetric positions and X-symmetric positions
(if i understand your terminology correctly).  however, there are so
many of these (124) in the X-symmetric case, that i probably will have
to settle for sub-optimal maneuvers.

mike

From cube-lovers-errors@mc.lcs.mit.edu  Sat Aug 16 22:34:17 1997
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Mail-from: From Hoey@AIC.NRL.Navy.Mil Sat Aug 16 22:35:27 1997
Date: Sat, 16 Aug 1997 22:35:13 -0400
Message-Id: <199708170235.WAA07984@sun30.aic.nrl.navy.mil>
From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: reid@math.brown.edu
Cc: cube-lovers@ai.mit.edu
In-Reply-To: <199708152344.TAA05912@life.ai.mit.edu> (message from michael
	reid on Fri, 15 Aug 1997 19:49:54 -0400)
Subject: Re: isoglyphs

> perhaps i am missing something, but doesn't

>      D2 R2 U2 D2 R2 U D   (12q, 7f)

> produce an isoglyph of type 4D ?

Oops, you're right.  I goofed because I half-remembered a different
result, that there are no 6-bar patterns of the nice symmetric sort,
with three mutually perpendicular pairs of parallel bars.

>  ...
> i hadn't even considered chiral versus achiral isoglyphs.  indeed,
> all the "continuous" isoglyphs given by herbert are chiral.
> achiral isoglyphs certainly exist, for example

>      D2 R2 U' B' L  B  U  B  L  F2 R  D' L2 U2 B2 D    (22q, 16f)

> of type 11; pattern

> *..
> ***
> ***

> and others can be derived from this.  i suspect that there is no
> chiral form of this isoglyph, but i'm not absolutely certain.

Modulo some oversight, I think this is true, and not hard to
demonstrate.

Recall that a "ground" facet is one that is not on its home face.
First note that a corner cubie will have 0, 2, or 3 ground facets.  So
on any isoglyph of corner type 1, there are a total of 6 ground corner
facets, and these ground facets must appear on two corner cubies
(three ground facets each) or three corner cubies (two ground facets
each).  If two corner cubies, those cubies must be antipodes, and they
are either rotated (forming a meson, FTR+ BLD- or equivalent) or
exchanged ((FTR,BLD) or equivalent, implying odd edge permutation
parity).  If ground facets appear on three corner cubies, the cubies
must be a three-cycle of cubies on nonadjacent corners ((FTR,FDL,BTL)
or equivalent).

I've done some analysis by facets on these three cases, which is too
messy to describe, but which leads me to the conclusion that the above
position is the only isoglyph of its pattern, implying the conclusion
that there is no chiral form.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From cube-lovers-errors@mc.lcs.mit.edu  Mon Aug 18 13:05:48 1997
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Mail-from: From jbryan@pstcc.cc.tn.us Sun Aug 17 00:42:09 1997
Date: Sun, 17 Aug 1997 00:38:39 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Calculating Local Maxima in Face Turn Metric
To: Cube-Lovers <cube-lovers@ai.mit.edu>
Reply-To: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Message-Id: <Pine.PMDF.3.95.970816234247.40618B-100000@PSTCC6.PSTCC.CC.TN.US>

While everybody else has been doing isoglyphs and minimal maneuvers for
highly "symmetric" positions, I have been trying to figure out how to
calculate local maxima for the face turn metric with my Shamir program.
It's a bit trickier than with the quarter turn metric, but I think I have
it.  If anybody sees any holes in the following, please let me know. 

Recall that I have been creating Start-rooted search trees by forming
products of the form z=xy for |x|=n and |y|=m, for 1<=m<=n.  Prior to my
Shamir program, I would fix n (it gets larger iteratively), and just let
m=1 to advance one level of the tree at a time.  But the Shamir method
lends itself to jumping forward several levels at one fell swoop.

If E(w) is the set of all moves with which a minimal maneuver for w can
end, then E(z) is the union of E(y) over all the y which can be used to
create z.  We now introduce an alternative interpretation for E(w).  E(w) 
is the set of all moves whose inverses carry w one move closer to Start. 
The alternative interpretation works for both quarter turns and face
turns. 

My program is all set up to calculate E(w) for either the quarter turn
metric or the face turn metric.  The only difference is that the
representation of E(w) for the quarter turn metric is a bit string of 12
bits and for the face turn metric is a bit string of 18 bits. 

But using E(w) to calculate local maxima for the face turn metric will
yield only what we have agreed to call strong local maxima, namely those
local maxima where every face turn moves one move closer to Start.  We
desire also to calculate weak local maxima, where one or more face turns
may leave the distance from Start unchanged.  To this end, we define E2(w) 
to be the set of all moves whose inverses leave w the same distance from
Start.

For quarter turns, the required initializations are E(q)={q} for all q in
Q, the set of twelve quarter turns.  E2(q) is of course null in all cases.

For face turns, the required initializations are:

  E(q)   = {q} for all q in Q
  E(q2)  = {q2} for all q2 in H, the set of six half turns
  E2{q}  = {q',q2} for all q in Q
  E2{q2} = {q,q'} for all q2 in H

To be pedantically complete, we could define E3(w) to be the set of all
moves whose inverse leaves w one move further from Start.  Note that E(w),
E2(w), and E3(w) are disjoint, and their union is Q for quarter turns and
Q+H for face turns. 

For quarter turns, we have defined the maximality of w to be |E(w)|,
wherein we have a local maximum if |E(w)|=12.  The corresponding
definition of maximality for face turns is an ordered pair
(|E(w)|,|E2(w)|), where w is a local maximum if |E(w)|+|E2(w)|=18 and
where w is a strong local maximum if |E(w)|=18. (A local maximum which is
not a strong local maximum is a weak local maximum.)

The only thing I am worried about is the following.  Given the proposed
initializations and calculations for E(w) and E2(w) for face turns, will
E(w) and E2(w) be disjoint automagically, or is their disjointedness
something which will have to be tested? 

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990

From cube-lovers-errors@mc.lcs.mit.edu  Mon Aug 18 13:49:17 1997
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Mail-from: From kociemba@hrz1.hrz.th-darmstadt.de Sun Aug 17 03:43:15 1997
Message-Id: <33F6AA41.3C98@hrz1.hrz.th-darmstadt.de>
Date: Sun, 17 Aug 1997 09:37:37 +0200
From: Herbert Kociemba <kociemba@hrz1.hrz.th-darmstadt.de>
To: cube-lovers@ai.mit.edu
Subject: Re: isoglyphs
References: <199708170235.WAA07984@sun30.aic.nrl.navy.mil>

[Mike Reid wrote:]
> >  ...
> > i hadn't even considered chiral versus achiral isoglyphs.  indeed,
> > all the "continuous" isoglyphs given by herbert are chiral.
> > achiral isoglyphs certainly exist, for example
> 
> >      D2 R2 U' B' L  B  U  B  L  F2 R  D' L2 U2 B2 D    (22q, 16f)
> 
> > of type 11; pattern
> 
> > *..
> > ***
> > ***
> 
> > and others can be derived from this.  i suspect that there is no
> > chiral form of this isoglyph, but i'm not absolutely certain.
> 

Dan Hoey wrote:

>...
> I've done some analysis by facets on these three cases, which is too
> messy to describe, but which leads me to the conclusion that the above
> position is the only isoglyph of its pattern, implying the conclusion
> that there is no chiral form.

There are two more isoglyphs of this pattern,

B2 D . R D' B2 F' R2 U' B' R' U F R'  (13)
D2 R D2 L B2 R' B2 . D' R F D' L' F' D2 L'  (15)

Could someone tell me, what chiral and achiral exactly mean?

--Herbert

From cube-lovers-errors@mc.lcs.mit.edu  Mon Aug 18 14:10:15 1997
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Mail-from: From jbryan@pstcc.cc.tn.us Sun Aug 17 14:08:40 1997
Date: Sun, 17 Aug 1997 14:05:08 -0400 (Eastern Daylight Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: Calculating Local Maxima in Face Turn Metric
In-Reply-To: <Pine.PMDF.3.95.970816234247.40618B-100000@PSTCC6.PSTCC.CC.TN.US>
To: Cube-Lovers <cube-lovers@ai.mit.edu>
Message-Id: <Pine.WNT.3.96.970817140206.-250917Q-100000@GN209A.PSTCC.CC.TN.US>

On Sun, 17 Aug 1997, Jerry Bryan wrote:

> The only thing I am worried about is the following.  Given the proposed
> initializations and calculations for E(w) and E2(w) for face turns, will
> E(w) and E2(w) be disjoint automagically, or is their disjointedness
> something which will have to be tested? 
> 

Well, I see that I never actually gave the calculation of E2(w), just
the initialization.  But it works the same as for E(w), namely if z=xy,
then E2(z) is the union of E2(y) over all the y which can be used to
make minimal maneuvers for z.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990

From cube-lovers-errors@mc.lcs.mit.edu  Mon Aug 18 16:16:56 1997
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Mail-from: From nbodley@tiac.net Sun Aug 17 14:33:24 1997
Date: Sun, 17 Aug 1997 09:23:04 -0400 (EDT)
From: Nicholas Bodley <nbodley@tiac.net>
To: Dan Hoey <Hoey@aic.nrl.navy.mil>
Cc: Cube Mailing List <Cube-Lovers@ai.mit.edu>
Subject: Patterns on larger cubes (Was Re: isoglyphs)
In-Reply-To: <199708170235.WAA07984@sun30.aic.nrl.navy.mil>
Message-Id: <Pine.SUN.3.95.970817091106.22282A-100000@sunspot.tiac.net>

 Perhaps many List subscribers have realized that (unless I'm very
confused!) while the current highly-evolved discussion of patterns
(isoglyphs, etc.) pertains only to 3^3s, there must be enormous "worlds to
conquer"  when one considers [maneuvers] to create patterns on the 4^3
(Rubik's Revenge) and the 5^3.

 I understand little of the current discussion about isoglyphs (even if
the term itself makes sense); nevertheless, it's delightful to see such
discussions going on, and I have great respect for those who do understand
and can contribute. (I wouldn't have it any other way!)

 It's perhaps of interest to consider a Theory of Mechanisms, in which it
would be possible (eventually) to design an optimum set of innards for,
say, a 5^3, or to rigorously prove that what exists is optimal.
Connections with topology and kinematics would be not at all unexpected. 
Close connections with CAD would also make sense.

My best regards to all,

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*    When the year 2000 begins, we'll celebrate 
|*  Amateur musician  *|*    the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Mon Aug 18 16:49:39 1997
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Mail-from: From kociemba@hrz1.hrz.th-darmstadt.de Sun Aug 17 22:56:57 1997
Message-Id: <33F7B8DA.4C23@hrz1.hrz.th-darmstadt.de>
Date: Mon, 18 Aug 1997 04:52:11 +0200
From: Herbert Kociemba <kociemba@hrz1.hrz.th-darmstadt.de>
To: cube-lovers@ai.mit.edu
Subject: Cube Explorer 1.5 now available!

I finally succeeded in updating my Cube Explorer program. You can
download it from

http://home.t-online.de/home/kociemba/cube.htm

The current discussion in this mailing list concerning isoglyphs shows
that is pretty error prone to classify these patterns by hand. The
program is a good help here.

--Herbert

From cube-lovers-errors@mc.lcs.mit.edu  Mon Aug 18 17:45:09 1997
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Mail-from: From scotth@ichips.intel.com Mon Aug 18 17:36:26 1997
Message-Id: <199708182132.OAA19914@ichips.intel.com>
To: Cube Mailing List <Cube-Lovers@ai.mit.edu>
Subject: d-dimensional cube mechanisms
Date: Mon, 18 Aug 1997 14:32:59 -0700
From: Scott Huddleston <scotth@ichips.intel.com>

Several years ago I worked out a solution to the d-cube 3^d, for d>3.
This is most interesting combinatorially if you assume you're
restricted to only rotating entire (d-1)-faces at a time, so that's
what I assumed in my solution.

But when I thought about building a mechanism for the d-cube, I
came to the surprising (to me) conclusion that any natural extension 
of the 3^3 mechanism to d dimensions would allow you to rotate any 
2-face.  I concluded that any mechanism that would restrict you to 
only rotating entire (d-1)-faces would require some sort of complex 
interlocking mechanism that would have to engage and disengage 
whenever a (d-1)-face was to be rotated.

Has anyone else thought about this problem (d-cube mechanisms)
enough to confirm or refute my conclusions?

Best,
- Scott Huddleston
scotth@ichips.intel.com

From cube-lovers-errors@mc.lcs.mit.edu  Mon Aug 18 18:37:28 1997
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Mail-from: From Hoey@AIC.NRL.Navy.Mil Mon Aug 18 18:16:31 1997
Date: Mon, 18 Aug 1997 18:16:18 -0400
Message-Id: <199708182216.SAA00604@sun30.aic.nrl.navy.mil>
From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: kociemba@hrz1.hrz.th-darmstadt.de
Cc: cube-lovers@ai.mit.edu
In-Reply-To: <33F6AA41.3C98@hrz1.hrz.th-darmstadt.de> (message from Herbert
	Kociemba on Sun, 17 Aug 1997 09:37:37 +0200)
Subject: Re: isoglyphs

Herbert Kociemba asks:

> Could someone tell me, what chiral and achiral exactly mean?

"Chirality" is Lord Kelvin's word for "handedness" as in "appearing in
two mirror-image varieties."

A "chiral isoglyph" is one in which the handedness of the glyph is
taken into account in testing for isoglyphy,* so that the glyph
appears only in one variety.  Neither Mike's original isoglyph nor the
                                                       *..     ..*
two you found are chiral isoglyphs--they all have both *** and *** .
                                                       ***     ***

Mike used "achiral" for an isoglyph that fails to be a chiral
isoglyph, though I would tend to use "non-chiral".  I would rather use
"achiral" for a situation that lacked chirality, as in an isoglyph of
a mirror-symmetric glyph.

[* Thanks to Allan Wechsler for inventing the word "isoglyphy".  His
   alternate term, "isoglyphism", is still looking for a good use. ]

Dan
Hoey@AIC.NRL.Navy.Mil

From cube-lovers-errors@mc.lcs.mit.edu  Tue Aug 19 10:53:09 1997
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Message-Id: <199708190550.BAA21896@life.ai.mit.edu>
Date: Tue, 19 Aug 1997 01:55:51 -0400
From: michael reid <reid@math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: Re: isoglyphs

herbert writes

> With the pattern generator it's indeed very easy to find the isoglyphs. 

i'm still unclear about what your pattern generator does.  could you
describe what it does, for the benefit of those who haven't seen your
program?

>                                               By the way, Mike, it would
> be nice to complete the chapter "continous isoglyphs" by computing the
> shortest generators for them with your program.

i will do this soon.  right now the program is busy with T-symmetric
positions.  after that (or if there's a break) i'll give it the
continuous isoglyphs to think about.

there's one last pattern for which i could not find any isoglyph.
it's the 32 pattern of type

..*
*..
***

all others, except those previously mentioned as impossible (patterns of
corner type D, and the 21 and 23 types which we previously discussed)
have isoglyphs.  can your program find isoglyphs of this type, or show
that none exist?

mike

From cube-lovers-errors@mc.lcs.mit.edu  Tue Aug 19 13:29:02 1997
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Mail-from: From jbryan@pstcc.cc.tn.us Tue Aug 19 13:17:24 1997
Date: Tue, 19 Aug 1997 13:13:34 -0400 (Eastern Daylight Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: isoglyphs
In-Reply-To: <199708190550.BAA21896@life.ai.mit.edu>
To: michael reid <reid@math.brown.edu>
Cc: cube-lovers@ai.mit.edu
Message-Id: <Pine.WNT.3.96.970819130121.-192999I-100000@GN209A.PSTCC.CC.TN.US>

On Tue, 19 Aug 1997, michael reid wrote:

> 
> > With the pattern generator it's indeed very easy to find the isoglyphs. 
> 
> i'm still unclear about what your pattern generator does.  could you
> describe what it does, for the benefit of those who haven't seen your
> program?
> 


I'll take a crack at this one.  (The program is great, by the way.)

The basic mode of the pattern generator allows you to specify a pattern
for one of the 3x3 faces of the cube, and the program finds all the
positions (unique up to symmetry) where each of the six 3x3 faces has
this same pattern.  It doesn't really matter which colors you specify in
your one face, since you are really only specifying a pattern.  For
example, I have played with corner facelets and center facelet all one
color and edge facelets all another color, or center facelet one color
and all the edge and corner facelets another color (yields the 6-spot),
etc.

The patterns I have played with have very few (or sometimes, no) 
solutions.  I don't know what happens if you choose a pattern with many,
many solutions (maybe there really aren't all that many such positions,
given that all six 3x3 faces have to have the same pattern). 

There is an expanded mode which I haven't played with much yet where you
can give up to four 3x3 patterns.  Each of the six faces on the cube
then has to have a pattern that matches any one of the (up to) four
which you specified.

The so-called pattern editor I have described seems to operate
essentially instantaneously.  But having generated the position, you can
then ask the program to find a near-optimal solution using the Kociemba
algorithm.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990

From cube-lovers-errors@mc.lcs.mit.edu  Wed Aug 20 08:33:01 1997
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Mail-from: From cubeman@idirect.com Wed Aug 20 00:53:03 1997
Message-Id: <33FA7810.3446@idirect.com>
Date: Wed, 20 Aug 1997 00:52:32 -0400
From: Mark Longridge <cubeman@idirect.com>
To: cube lovers <cube-lovers@ai.mit.edu>
Subject: Mike Reid's Cube Program 

I've updated my web page to include Mike Reid's cube program.
 
I've also updated my own cube program to save arrangements the way
Mike's program requires. The MS-DOS source and executables are 
available at:
 
http://web.idirect.com/~cubeman/rubik.zip
http://web.idirect.com/~cubeman/miker.zip
 
I guess cubing is back in fashion.
 
P.S. To Herbert Kociemba, your 1.5 version is great!

From cube-lovers-errors@mc.lcs.mit.edu  Wed Aug 20 09:33:07 1997
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Mail-from: From CFolkerts@compuserve.com Wed Aug 20 03:55:06 1997
Date: Wed, 20 Aug 1997 03:50:41 -0400
From: Corey Folkerts <CFolkerts@compuserve.com>
Subject: 5x5x5 Solution
To: Cube-Lovers <Cube-Lovers@ai.mit.edu>
Message-Id: <199708200350_MC2-1DA6-9307@compuserve.com>

        I recently got my hands on a 5x5x5 from Dr. Christoph Bandelow,
however, I'm am at an almost complete loss as to how to solve it.  I think
that if I ignore the center row in all axes then it is pretty much a 4x4x4,
but if I then treat it as a 4x4x4 the center axes get all screwed up. 
        I would really like to know if there is a site that has a
description of the solution to a 5x5x5 or if someone could describe it to
me in a message. Thanks in advance.

        Corey Folkerts

From cube-lovers-errors@mc.lcs.mit.edu  Wed Aug 20 13:02:50 1997
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Mail-from: From Goyra@iol.ie Wed Aug 20 12:36:22 1997
Message-Id: <199708201632.RAA08938@mail.iol.ie>
From: "David Byrden" <Goyra@iol.ie>
To: "<Cube-Lovers@ai.mit.edu"@life.ai.mit.edu
Subject: Re: 5x5x5 Solution
Date: Wed, 20 Aug 1997 17:28:52 +0100

> From: Corey Folkerts <CFolkerts@compuserve.com>


>         I recently got my hands on a 5x5x5 from Dr. Christoph Bandelow,
> however, I'm am at an almost complete loss as to how to solve it. 

	I just extended the technique that had worked for me on the 
smaller cubes.

	Solve the corners, then solve the inner edges, then eventually
the faces in the interior of the sides. By choosing these 5 subsets
of the faces, which of course do not exchange faces with each other, 
you break the cube into a sequence of 5 smaller problems. Working 
inwards from the outermost faces is best because you can easily find 
operators (combinations of moves) that affect the inner faces in some 
way but preserve the outer ones that you have already solved.

						David

From cube-lovers-errors@mc.lcs.mit.edu  Wed Aug 20 14:23:23 1997
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Mail-from: From Goyra@iol.ie Wed Aug 20 12:36:22 1997
Message-Id: <Resent.199708201632.RAA08938@mail.iol.ie>
From: "David Byrden" <Goyra@iol.ie>
To: Cube-Lovers@ai.mit.edu
Subject: Re: 5x5x5 Solution
Date: Wed, 20 Aug 1997 17:28:52 +0100

> From: Corey Folkerts <CFolkerts@compuserve.com>


>         I recently got my hands on a 5x5x5 from Dr. Christoph Bandelow,
> however, I'm am at an almost complete loss as to how to solve it. 

	I just extended the technique that had worked for me on the 
smaller cubes.

	Solve the corners, then solve the inner edges, then eventually
the faces in the interior of the sides. By choosing these 5 subsets
of the faces, which of course do not exchange faces with each other, 
you break the cube into a sequence of 5 smaller problems. Working 
inwards from the outermost faces is best because you can easily find 
operators (combinations of moves) that affect the inner faces in some 
way but preserve the outer ones that you have already solved.

						David

[ Moderator's note-- The previous copy of this message had bad headers.
  Sorry.  --Dan]

From cube-lovers-errors@mc.lcs.mit.edu  Wed Aug 20 15:06:25 1997
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Mail-from: From cubeman@idirect.com Wed Aug 20 14:21:03 1997
Message-Id: <33FB355F.3A5D@idirect.com>
Date: Wed, 20 Aug 1997 14:20:15 -0400
From: Mark Longridge <cubeman@idirect.com>
To: cube lovers <cube-lovers@ai.mit.edu>
Subject: Corrections

Oops.. those were the wrong URLs
 
Here are the correct ones:

http://web.idirect.com/~cubeman/rubik/rubik.zip
http://web.idirect.com/~cubeman/rubik/miker.zip
 
Or just go to http://web.idirect.com/~cubeman and click on
the appropriate link.
 
-> Mark <-

From cube-lovers-errors@mc.lcs.mit.edu  Wed Aug 20 16:12:54 1997
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Mail-from: From reid@math.brown.edu Wed Aug 20 14:55:47 1997
Message-Id: <199708201852.OAA00834@life.ai.mit.edu>
Date: Wed, 20 Aug 1997 14:57:28 -0400
From: michael reid <reid@math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: Re:  Mike Reid's Cube Program

mark longridge writes

> I've updated my web page to include Mike Reid's cube program.

this is my sub-optimal solver, i.e. kociemba's algorithm, which
can handle either quarter turns or face turns.  (it's not yet
in its final form, but that may take a while.)

to those who are interested in my optimal solver, it will be
available soon.  it requires about 85 megabytes, so your computer
will need at least that much RAM.

mike

From cube-lovers-errors@mc.lcs.mit.edu  Wed Aug 20 17:27:39 1997
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Mail-from: From Hoey@AIC.NRL.Navy.Mil Wed Aug 20 17:22:17 1997
Date: Wed, 20 Aug 1997 17:22:07 -0400
Message-Id: <199708202122.RAA08061@sun30.aic.nrl.navy.mil>
From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: scotth@ichips.intel.com
Cc: Cube-Lovers@ai.mit.edu
In-Reply-To: <199708182132.OAA19914@ichips.intel.com>
Subject: Re: d-dimensional cube mechanisms

Scott Huddleston <scotth@ichips.intel.com> writes:

> Several years ago I worked out a solution to the d-cube 3^d, for d>3.
> This is most interesting combinatorially if you assume you're
> restricted to only rotating entire (d-1)-faces at a time, so that's
> what I assumed in my solution.

I thought that was most natural, but there are certainly at least two
others.  In my article of 22 Dec 1993 in
ftp://ftp.ai.mit.edu/pub/cube-lovers/cube-mail-11 I mentioned how to
show they are different, at least in the 3^4 hypercube.  I didn't go
into enough detail to determine "most interesting"--can you expand on
what you mean?  By the way, do you know if this is the Kamack and
Keane model?  I've never seen their paper, and I never figured out how
to read Charlie Dickman's document.)

> But when I thought about building a mechanism for the d-cube, I
> came to the surprising (to me) conclusion that any natural extension 
> of the 3^3 mechanism to d dimensions would allow you to rotate any 
> 2-face.

Wow! I can verify it for the most obvious natural extension I can
describe.  We could form the 3^d Rubik's ball by taking the unit ball
B_d={z in R^d : |z|<=1} and some sufficiently small constant c, and
cutting the ball with the hyperplanes

     P_i,s = {z in B_d : z_i = s c} for each index i in {1,...,d},
                                    for each sign s in {-1,1}.

We arrange that all rotations are allowed that keep these pieces
inside the unit ball (perhaps enforcing this by attaching extensions
in the shape of a cube). It's then easy to see that a representative
2-face

     {z in B_d : z_3 > c, z_4 > c, ..., z_d > c}

can be rotated by mapping

    (z_1,z_2,z_3,...,zn) -> (z_1 cos th + z_2 sin th,
                             z_1 sin th + z_2 cos th,
                             z_3, z_4, ..., z_d).

That is a very surprising observation to me, too!  Hands up anyone who
_isn't_ surprised!!

In fact, I think it may be worse (geometrically, if not
combinatorically).  For instance, I expected that a cubical hyperface
of the 3^4 could only be turned by rotating the cube about an
orthogonal axis.  But it looks to me like you could rotate the cube
around any axis you like.  Maybe you have to eventually rotate it so
it occupies it's original space before you rotate a perpendicular
hyperface, but I'm still somewhat annoyed that you can put it in weird
orientations in between.

> I concluded that any mechanism that would restrict you to 
> only rotating entire (d-1)-faces would require some sort of complex 
> interlocking mechanism that would have to engage and disengage 
> whenever a (d-1)-face was to be rotated.

> Has anyone else thought about this problem (d-cube mechanisms)
> enough to confirm or refute my conclusions?

What I haven't proven is that say, any decomposition of the ball into
3^d pieces that admit (d-1)-face rotations will also admit 2-face
rotations.  If you've got that, it would pretty much support your
conclusion, but I don't know how to verify it off-hand.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From cube-lovers-errors@mc.lcs.mit.edu  Wed Aug 20 19:26:37 1997
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Mail-from: From kociemba@hrz1.hrz.th-darmstadt.de Wed Aug 20 19:12:10 1997
Message-Id: <33FB67F0.3D1C@hrz1.hrz.th-darmstadt.de>
Date: Wed, 20 Aug 1997 23:56:00 +0200
From: Herbert Kociemba <kociemba@hrz1.hrz.th-darmstadt.de>
To: cube-lovers@ai.mit.edu
Subject: Re: isoglyphs
References: <199708190550.BAA21896@life.ai.mit.edu>

michael reid wrote:

> i'm still unclear about what your pattern generator does.  could you
> describe what it does, for the benefit of those who haven't seen your
> program?


Though I used the word "pattern generator" myself I would like to ban it
now, because the word generator is already uses for maneuvers (solvers
versus generators). Let's talk about "pattern search".

The pattern search is implemented in principle in the same way as you
could try to built a pattern manually: First you "remove" all cubies,
then again you add one after the other to the next free position and
check if there is any contradiction with the pattern(s) in the Pattern
Editor. If yes, the cubie is removed and added again in a different
orientation or location. This is done recursivly, until al positions are
filled.
If there is no bug in the code, the pattern search should find *all*
cubes which are possible with the patterns given in the Pattern Editor.
> 
> there's one last pattern for which i could not find any isoglyph.
> it's the 32 pattern of type
> 
> ..*
> *..
> ***
> 
> all others, except those previously mentioned as impossible (patterns of
> corner type D, and the 21 and 23 types which we previously discussed)
> have isoglyphs.  can your program find isoglyphs of this type, or show
> that none exist?

My program finds no solution, so there also should not exist any.

--Herbert

From cube-lovers-errors@mc.lcs.mit.edu  Wed Aug 20 21:46:19 1997
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Mail-from: From tim@mail.htp.com Wed Aug 20 21:13:21 1997
From: tim@mail.htp.com (Tim Mirabile)
To: Cube-Lovers@ai.mit.edu
Subject: Re: 5x5x5 Solution
Date: Thu, 21 Aug 1997 01:09:43 GMT
Organization: http://www.webcom.com/timm/
Message-Id: <33fc8852.2330703@mail.htp.com>
References: <Resent.199708201632.RAA08938@mail.iol.ie>
In-Reply-To: <Resent.199708201632.RAA08938@mail.iol.ie>

The method I use is inefficient I'm sure, but I was able to solve the 5x5x5
right away only learning one specialized move.

First I solve the 3x3x3 centers on all six sides using mostly
intuitive methods, with an occasional 3x3x3 move (turning the two
outer slices together on each twist to simulate a 3x3x3) thrown in.  I
usually work on the "corner centers" first, then the "edge centers",
just like I would on a 3x3x3.

Then I solve the "outer" corners (holding the three center slices together),
followed by the "center" edge pieces (holding the two outer slices together)
also using purely 3x3x3 methods.

Then I work on the "off center" edge pieces using moves like r1 U2 r3 U2
or l3 U2 l1 U2.  This messes up a row of center pieces but if you do
it three times total they are restored, and 5 of the off center edges
are permuted.  I usually improvise by making a twist or two to get the
edges I want to permute in the right place, followed by reversing
these afterward.  I also work with the slightly messed up centers at
times making sure that I restore them later as I permute other sets of
edges.

Finally I use one 4x4x4 specific move from Mark Longridge's page at
http://web.idirect.com/~cubeman/revenge.txt which also explains the extended
notation.  (http://web.idirect.com/~cubeman/ is his main page of course).

The move I use is this:

p3:  Flip UF edge pair:   r2 (D2 l1)^2 D1 l3 r3 d2 l1 r1 D3 l3 r3 d2 B2 r1
B2 l3 B2 l1 B2 r2

On the 5x5x5, this not only flips (and swaps) the edges, but swaps two
center pieces.  But if you hold the edges you want to flip at UB
instead of UF, and do (p3) U2 (p3) U2 (p3), the edges will be flipped
and the centers restored.  You can also improvise here if you need to
flip more than one pair - messing up centers with the first flip and
restoring them with the second.  Since these edges here are swapped as
well as flipped, you can also use these moves to swap a single pair of
edges.

--
Webmaster, tech support - ICD/Your Move Chess & Games: http://www.icdchess.com/
The opinions of my employers are not necessarily mine, and vice versa.

From cube-lovers-errors@mc.lcs.mit.edu  Thu Aug 21 10:17:18 1997
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Mail-from: From reid@math.brown.edu Thu Aug 21 00:10:16 1997
Message-Id: <199708210406.AAA21524@life.ai.mit.edu>
Date: Thu, 21 Aug 1997 00:11:54 -0400
From: michael reid <reid@math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: Re: isoglyphs

thanks for the description of your program, herbert.  i was not so much
concerned with the algorithm the pattern finder uses (although this is
nice to know) as i was with how the user may specify patterns, etc.
jerry bryan gave a good description of this.

here are several other applications that your pattern finder should
be able to handle easily:

1. classify all "snakes".  you've already done the part where they
   consist only of faces of type 42; there's also face type 4D to
   consider.
2. confirm my results on the "czech check" problem, i.e. classify all
   patterns that have exactly eight squares correct on each face.
3. find all "partial isoglyphs".

and they are certainly many other applications.  these three come
to mind immediately.  it looks like i'll have to find a windows machine
to try out your program; it sure sounds excellent.

mike

From cube-lovers-errors@mc.lcs.mit.edu  Thu Aug 21 12:54:21 1997
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Mail-from: From reid@math.brown.edu Thu Aug 21 00:45:19 1997
Message-Id: <199708210441.AAA22489@life.ai.mit.edu>
Date: Thu, 21 Aug 1997 00:47:02 -0400
From: michael reid <reid@math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: partial isoglyphs

dan recently introduced the concept of "partial isoglyphs", in which
some faces are solid, and the others are glyphs of the same pattern.
i looked into this a little and didn't find much.  only the case
of two opposite solid faces seems to have many possible glyph types,
although some of these possible types may have many solutions.

here's what i found

6 solid faces: start
5 solid faces: no possibilities
4 solid faces:
   other two faces opposite:  types 02, 0D and 04 are possible
   other two faces adjacent:  type 0D is possible
3 solid faces:
   mutually adjacent:  type 02 is possible
   not mutually adjacent:  types 01 and 0D are possible
2 solid faces:
   adjacent:  types 01, 02, 0D and 03 are possible
   opposite:  many possible types 
1 solid face:  types 01, 02 and 0D are possible

but it seems like herbert's cube explorer program can settle this
matter completely.  maybe he, or someone else wants to investigate
this.

mike

From cube-lovers-errors@mc.lcs.mit.edu  Thu Aug 21 14:21:46 1997
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Mail-from: From Hoey@AIC.NRL.Navy.Mil Thu Aug 21 12:51:15 1997
Date: Thu, 21 Aug 1997 12:20:58 -0400
Message-Id: <199708211620.MAA00539@sun30.aic.nrl.navy.mil>
From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: Goyra@iol.ie
Cc: cube-lovers@ai.mit.edu
Subject: Re: Megaminx a.k.a. Supernova
In-Reply-To: <199708051604.MAA13056@sun30.aic.nrl.navy.mil>

Glyph-lovers may recall I was led to that discussion by analyzing the
five conjucacy classes of spot patterns on the Megaminx.  I called
them

    0. The identity,
    1. The  72-degree twelve-spot,
    2. The 144-degree twelve-spot,
    3. The 120-degree ten-spot,
    4. The 180-degree ten-spot.

Thanks to David Singmaster for noticing this is wrong.  It should have
been

    0. The identity,
    1. The  72-degree ten-spot,
    2. The 144-degree ten-spot,
    3. The 120-degree twelve-spot,
    4. The 180-degree twelve-spot.

I'm glad to see someone's paying attention around here.

Dan
Hoey@AIC.NRL.Navy.Mil

From cube-lovers-errors@mc.lcs.mit.edu  Thu Aug 21 16:12:37 1997
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Mail-from: From jbryan@pstcc.cc.tn.us Thu Aug 21 16:11:04 1997
Date: Thu, 21 Aug 1997 16:07:30 -0400 (Eastern Daylight Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: isoglyphs
In-Reply-To: <199708210406.AAA21524@life.ai.mit.edu>
To: michael reid <reid@math.brown.edu>
Cc: cube-lovers@ai.mit.edu
Message-Id: <Pine.WNT.3.96.970821151359.-226699B-100000@GN209A.PSTCC.CC.TN.US>

On Thu, 21 Aug 1997, michael reid wrote: 

> here are several other applications that your pattern finder should
> be able to handle easily:
> 


> 2. confirm my results on the "czech check" problem, i.e. classify all
>    patterns that have exactly eight squares correct on each face.

If I understand your question correctly, and if I am using Herbert's
program correctly, there are 54 such isoglyphs unique up to M-conjugacy.
3 of them involve only corners as the incorrect facelet, and 51 of them
involve only edges as the incorrect facelet.

(I am assuming that by definition the center facelet is always correct,
thus eliminating the 6-spot from consideration.  If you count the
6-spot, then there are of course 2 such isoglyphs unique up to
M-conjugacy.) 

If you tell Herbert's program to consider only continuous isoglyphs with
exactly eight squares correct on each face, there are 3 such isoglyphs
unique up to M-conjugacy. 1 of them involves only corners as the
incorrect facelet, and 2 of them involve only edges as the incorrect
facelet.  (I suppose you would say that the other 51 are discontinuous.) 

Herbert's program only lists positions which are unique up to
M-conjugacy. Here's a modest suggestion.  It might be nice for the
program to list the size of the M-conjugacy class for each such
position.  That way, you could count both "positions" and "patterns",
where I am using "position" to mean any element of G and "pattern" to
mean an M-conjugacy class (or a representative thereof). 

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990

From cube-lovers-errors@mc.lcs.mit.edu  Thu Aug 21 18:46:52 1997
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Mail-from: From kociemba@hrz1.hrz.th-darmstadt.de Thu Aug 21 18:12:32 1997
Message-Id: <33FCB0F1.3E08@hrz1.hrz.th-darmstadt.de>
Date: Thu, 21 Aug 1997 23:19:45 +0200
From: Herbert Kociemba <kociemba@hrz1.hrz.th-darmstadt.de>
To: cube-lovers@ai.mit.edu
Subject: Re: isoglyphs

michael reid wrote:

> here are several other applications that your pattern finder should
> be able to handle easily:
> 
> 1. classify all "snakes".  you've already done the part where they
>    consist only of faces of type 42; there's also face type 4D to
>    consider.

There are 57 different patterns, which have face type 42 or 4D. I don't
think I should list generators for them here. The pattern-computation
with my program took only about 3 minutes on a PC with 486 processor, so
anybody who wants could repeat the computation and then create
generators for the patterns he/she(?) is interested in.

> 2. confirm my results on the "czech check" problem, i.e. classify all
>    patterns that have exactly eight squares correct on each face.

Cube Explorer finds 56 patterns, which confirms your result.

> and they are certainly many other applications.  these three come
> to mind immediately.  it looks like i'll have to find a windows machine
> to try out your program; it sure sounds excellent.

And I thought, window machines are very popular....

--Herbert

From cube-lovers-errors@mc.lcs.mit.edu  Fri Aug 22 21:50:40 1997
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Mail-from: From reid@math.brown.edu Fri Aug 22 21:15:50 1997
Message-Id: <199708222345.TAA12112@life.ai.mit.edu>
Date: Fri, 22 Aug 1997 19:51:01 -0400
From: michael reid <reid@math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: minimal maneuvers for T-symmetric positions

dan hoey denotes by  T  a subgroup of size 12 of the symmetries of the cube
that preserves one of the long diagonals.  there are 4 conjugates of  T ,
each preserves a different long diagonal.  i'll choose the  UFR - DLB
diagonal.  (note that "preserving" allows the diagonal to be reversed
in direction.)

in "symmetry and local maxima," hoey and saxe classify all cube positions
with T-symmetry.  for each choice of a  T  subgroup, there are 16 such
positions.  they form a commutative subgroup of the cube group of type
2, 2, 2, 2.  (this means that it is isomorphic to a product 
C_2 x C_2 x C_2 x C_2  of cyclic groups of order 2.)  we may take
generators of these cyclic factors to be  superflip, pons asinorum, and
the two positions  (UB+) (UL+) (FL+) (FD+) (RD+) (RB+)  (number 1 below)
and  (DRF, UBL) (FLD, BUR) (LUF, RBD) (UB, DF) (UL, DR) (FL, BR)
(number 5 below).

4 of these positions have more symmetry, namely the subgroup generated
by  superflip  and  pons asinorum.  for the other 12 positions, minimal
maneuvers are given below.  i've also given a maneuver that is minimal
in both metrics, whenever such a maneuver exists.

 1.  B  U  L' F' U  R  U2 D2 F' L  U' B' L  D  R2 L2 B2             (22q, 17f)
 2.  F  U  D' R2 U2 R' B' U' F  R' D  R  L' F  U' F  U' R'          (20q, 18f)
 3.  U  R  U' F  D  R  L' B' L' F  R  F  B' U' L' D  B' D'          (18q, 18f)
 4.  D' R' U  B' D' R' L  F  L  B' R' F  B' U  L  D' F  U' D2       (20q, 19f)
 5.  D' B' D' R' B  L  B  U' B  U  R  D  R  L  D  R' L2 D           (19q)
     B2 L  U' L  D  R' L' D2 R  U  L' B2 U  R2 U2 F2 U              (17f)
 6.  U  L  U  D  F  B' U' D  L2 F  U  D  B' R  L  B' U' D' F  U     (21q)
     D' L  F' B' L  F2 B2 U  R  L' U  D' L  F' R2 L2 F2 U' D2       (19f)
 7.  F  U  D  L2 F2 L2 U' F  B  D  F' B' D' F'                      (17q, 14f)
 8.  U  F  B' L  U  F  B' L  D  F  D' R2 L  F  U' B' L  F2 R        (21q)
     U  B  R' F2 U' D' L2 D2 R2 B2 L' F' B  R2 F2 R' D  F           (18f)
 9.  U  B  U2 L  F' B2 U' F' B  L  U' B' D' F  R  U  B  R  L'       (21q)
     U  F2 D  B' U' B2 R  B2 D' F2 U' D2 B2 L' U2 B  D2             (17f)
10.  U  F  B  D' L' U' B' L' F  R' L' D  L  U  F  U  B  D  B  R' D' (21q)
     U  F  U  D' B' D2 R  U  D  R2 D2 B  R  L2 F2 B' D  R2 L'       (19f)
11.  U  R  F' B  D  B' U' F  B' D' F  R  L' D2 R  D' B' U' D  F'    (21q)
     D' F2 U2 B2 R  F' L  U' F2 B  R' F' D  L2 D  R2 F2 U' F2       (19f)
12.  U  B' D' R  F' R' F' R2 B' D  F' B2 D  B' L' U  F  B  R'       (21q)
     U  F2 B  R2 F2 B' U  L  U  B' U2 R2 L' U  B  U' L' U'          (18f)

mike

From cube-lovers-errors@mc.lcs.mit.edu  Mon Aug 25 11:47:21 1997
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Mail-from: From jbryan@pstcc.cc.tn.us Wed Aug 20 12:41:14 1997
Date: Wed, 20 Aug 1997 12:37:38 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Open and Closed Subgroups of G
To: Cube-Lovers <cube-lovers@ai.mit.edu>
Message-Id: <Pine.PMDF.3.95.970820122826.19610A-100000@PSTCC6.PSTCC.CC.TN.US>

I am going to use the terms open and closed with a certain trepidation. 
These terms may already have a conventional meaning in group theory of
which I am not aware.  If so, my apologies in advance.  Anyway,... 

The Hofstadter articles about the cube in Scientific American many years
ago made one point very strongly.  Namely, that in solving the cube by
hand you often temporarily have to give up hard won progress.  For
example, suppose you solve the cube bottom layer first, middle layer (the
middle between the top and bottom) second, and the top layer last. (This
isn't the way I do it, and I doubt that anybody does it in quite this way,
but it makes for a good example.) 

Using this method, it is almost certain that you will have to disturb the
bottom layer to solve the middle layer, and it is almost certain that you
will have disturb the bottom and middle layers to solve the top layer.
(It's an interesting exercise to characterize those few positions where
you wouldn't have to disturb previously solved layers to complete your
task.) 

The set of positions where the bottom layer is fixed constitute a group,
as do those positions where the two bottom layers are fixed.  Hence, the
series of plateaus involved in this particular solution define a sequence
of nested subgroups. 

The Thistlethwaite algorithm reverses the roles of the solution algorithm
and the sequence of nested subgroups.  Instead of a solution algorithm
defining a sequence of nested subgroups, a sequence of nested subgroups
defines a solution algorithm.  In some ways, I think this is a distinction
without a difference;  it is more like two sides of the same coin. 

However, I think there is one really fundamental difference with the
Thistlethwaite algorithm.  Namely, you never have to give up any of your
"hard won progress". That is, after you make it to a particular subgroup
in the nested sequence, no subsequent move takes you out of that subgroup.
So let's define any subgroup of G having this property as closed, and any
subgroup which is not closed as open.  To be a little more specific, a
closed subgroup is a subgroup such that for every position in the
subgroup, there is a maneuver back to Start which never leaves the
subgroup. 

I am a quarter-turner, but I think that perhaps closed subgroups are an
argument which has not yet been specifically articulated for counting
half-turns as one move. That is, many of the subgroups which are used in
Thistlethwaite and Thistlethwaite-like algorithms forbid quarter-turns
along one or more axes at some point in the process.  Restricting
quarter-turns assists your sequence of nested subgroups all to be closed
subgroups.  It seems to me that such an approach leads very naturally to
counting half-turns as one move.

Finally, when I first read about Thistlethwaite's algorithm, I naively
assumed that the algorithm's maneuvers from the next to last subgroup in
the sequence to the last subgroup in the sequence (namely to Start itself)
were minimal in G. This is clearly not true in general.  The first example
I remember where this began becoming clear to me was the <U,R> group. We
could solve a cube as G -> <U,R> -> I, although the jump from G to <U,R>
is quite a big jump.  But a minimal maneuver in <U,R> is certainly not
necessarily minimal in G. 

Let's call a subgroup H of G a closed minimal subgroup if every minimal
maneuver in H is also minimal in G.  <U,R> is closed, but it is not closed
minimal.  I have tried to think about which subgroups of G are clearly
closed minimal.  I can't think of many.  In fact the only examples I can
think of are subgroups generated as <s>, where s is a syllable or
syllables along the same axis. So <F>, <F2>, <F,B>, <F2,B>, and <F2,B2>
and their conjugates are closed minimal subgroups.  (Well, I and G are
closed minimal, but it hardly seems fair to count them.)

This whole area was discussed rather thoroughly in the recent spate of
messages about Korf's paper and Kociemba's algorithm, but without using
the terms closed or closed minimal. I do not believe any of the subgroups
which were discussed were claimed to be closed minimal, although I think
essentially all of them were closed. The most interesting subgroup which
was discussed was Mike Reid's co-called T, which is the intersection of
<F,B,U2,D2,R2,L2> with its conjugates.  I wonder if Mike's T subgroup is
closed minimal? 

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990

[Moderator's note: This message has been delayed by a clerical error
 on my part.--Dan]

From cube-lovers-errors@mc.lcs.mit.edu  Tue Aug 26 12:05:11 1997
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Mail-from: From cubeman@idirect.com Tue Aug 26 01:20:45 1997
Message-Id: <3402678A.7E62@idirect.com>
Date: Tue, 26 Aug 1997 01:20:10 -0400
From: Mark Longridge <cubeman@idirect.com>
To: cube lovers <cube-lovers@ai.mit.edu>
Subject: Old URL no longer works!

Sorry to trouble everyone with this...
 
I just discovered my defaults on my page are different.
 
That is, the URL http://web.idirect.com/~cubeman no longer works!

Use http://web.idirect.com/~cubeman/index.html instead!
 
More interesting stuff to follow...

From cube-lovers-errors@mc.lcs.mit.edu  Tue Aug 26 21:10:39 1997
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Mail-from: From jbryan@pstcc.cc.tn.us Tue Aug 26 16:33:22 1997
Date: Tue, 26 Aug 1997 16:29:44 -0400 (Eastern Daylight Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: Open and Closed Subgroups of G
In-Reply-To: <199708252135.RAA25741@cyber1.servtech.com>
To: cube-lovers@ai.mit.edu
Reply-To: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Message-Id: <Pine.WNT.3.96.970826120155.-226699M-100000@GN209A.PSTCC.CC.TN.US>

On Mon, 25 Aug 1997, christopher f. chiesa wrote:

> 
> Jerry, judging from the fact that you were ABLE to go into a lot of
> details, it strikes (humble, little ol', un-group-theory-educated) me that
> perhaps you "read too much into" Hofstadter's remark about "giving up
> hard-earned progress" -- either that, or _I_ have for years been reading
> TOO LITTLE into the remark!
> 
> To wit, Jerry, although I can't really follow all the stuff about groups
> and subgroups and closure and all that, I think it suffices at the
> layman's level to say that "you certainly make it SOUND" as though there
> are a multitude of non-trivial cases in which one DOESN'T "have to give up
> hard-earned progress" en route to a solution of the (3^3) Cube.
> 
> Me, I don't get it.  Seems to me that Hofstadter was speaking from
> the simple perspective of the uneducated layman who, by hook or by
> crook, manages after hours/days/weeks of effort to solve, say, the
> TOP LAYER of the cube.  To the layman, at this point, ANY turn of ANY face 
> other than U itself, or D which does not directly interact with it, is
> going to constitute "giving up hard-earned progress!"  Now, maybe,
> from the perspective of all that math I can't follow, a simple "quarter
> turn of any face other than U or D" at this point does NOT constitute
> "giving up progress," but I assure you, the poor fellow who's just solved
> the top layer for the first time is going to have a HEART ATTACK if you so
> much as walk up and give the R face a quarter turn.  And I'm pretty sure
> it was from THAT simplistic perspective that Hofstadter was speaking.
> 
> Does this conflict with YOUR interpretation, Jerry?  I'd hate to think I
> blew all this hot air into the mailinglist over NOTHING. :-)
> 
> Is there a layman-comprehensible description of "the Thistlethwaite
> algorithm" available anywhere (preferably for free, preferably online)?
> I've been hearing about it for years but have never seen any details.
> 

I'll take a crack at a number of your questions. 

There are probably a lot of people on the list who are not conversant
with group theory, but I'll bet essentially everybody knows at least a
little bit about basic set theory.  I'm old enough that when I was
taking algebra in high school in the early 60's, "new math" was all the
rage. It turns out that "new math" was really just set theory. 

Many traditionalists were aghast that this wierd "new math" stuff was
being taught.  Never mind that it was hundreds of years old.  And never
mind that set theory is the foundation of nearly all modern mathematics. 
For example, most formal treatments of the concept of a function view a
function as a set of ordered pairs (which has to satisfy certain rules). 

Anyway, sets are just collections of objects where the basic rule is
that for any object you can unequivically determine that the object
either is or is not in the set.  We might write a set as something like
{a,b,c} where the braces denote the set and the elements a, b, and c are
listed within the braces.  We often give names to sets, as in A={a,b,c}. 
And finally, we have subsets, where for example {a,b}, {a,c}, {a}, etc.
are subsets of {a,b,c}.  Subsets can have names as well.  For example,
we might say B={a,b} and then we would say that B is a subset of A.

If you know what sets are, then it's easy to talk about groups. 
Oversimplifying slightly, a group is just a set, an operation on that
set, and a short list of rules.  As a simple example, the real numbers
and addition form a group.  The real numbers are the set and addition is
the operation.  Exactly what the short list of rules is does not matter
for now, but be assured that real numbers and addition do comply with
the required rules for a group.

Just as there are subsets of sets, there are subgroups of groups.  For
example, the rational numbers form a subset of the real numbers.
Similarly, the rational numbers and addition form a subgroup of real
numbers and addition.  The integers form a subset of the rational
numbers.  The integers and addition form a subgroup of the rational
numbers and addition.

It is very common to be a little sloppy and simply identify a group as
being the set if the operation is well understood.  So we might say that
the integers form a group if it is well understood that we are talking
about addition as being the group operation.  So I will be a little
sloppy myself to make my sentences a little shorter. 

Not every subset is a subgroup.  For example, the set of even integers
is a subgroup of the integers, but the set of odd integers is not a
subgroup of the integers.  If you add two even integers together, the
result is an even integer.  But if you add two odd integers together,
the result is not an odd integer.  One of the group rules is that if you
combine two elements from the set together, then the result must also be
in the set or else you don't have a group.

In the case of the cube, the set is the collection of all positions
which can be reached by scrambling the cube in all possible ways, and
the operation is "followed by".  For real numbers and addition, we might
write x+y to indicate adding x and y together.  For cube positions, we
might write XY to mean "X followed by Y".  Even though it is not
multiplication in the every day sense of real numbers, XY is often
called a product and the "followed by" operation is often called
multiplication.

Basic operations on the cube consist of twisting one face.  These
operations are called F, B, U, D, L, and R if you twist the Front, Back,
Up, Down, Left, and Right faces clockwise by 90 degrees.  The respective
counterclockwise twists are called F', B', U', D', L', and R'.  The
respective 180 twists are called F2, B2, U2, D2, L2, and R2.  For 180
degree twists, it doesn't matter whether your twist is clockwise or
counterclockwise. 

Notice, for example, that FF=F2.  That is "F followed by F" is the same
thing as turning the Front face by 180 degrees.  Also, F'F'=FF=F2, etc.,
and FFF=F' (90+90+90 degrees clockwise is the same thing as 90 degrees
counterclockwise), etc.

There are many ways to define a group or a subgroup.  One of the more
common ways is in terms of generators.  The generator notation is <S>,
where S is some set or list of elements.  For example, with the integers
and addition, we might define a subgroup as <3>.  This means {3, 3+3,
3+3+3, ...} so <3> is the group of all integers which are divisible by
3.

One of the rules for groups is that every element in the set must have
an opposite, usually called an inverse.  For example, with integers and
addition the opposite of 3 is -3, and the opposite of -3 is 3.  The
generator notation automatically includes inverses.  So if we write <3>
for integers and addition, it is the same as writing <3,-3>.  So we
could write <3> or <-3> or <3,-3> and it would all mean the same thing,
namely {..., -6, -3, 0, 3, 6, ...}.  (To simplify things, I lied
slightly in the previous paragraph when I left out the negative numbers. 
Groups require inverses, and with addition the way you get inverses is
to include the negative numbers.) 

With the cube, the way you get inverses is that F' is the inverse of F
and F is the inverse of F', etc.  F2 is its own inverse, so we would
write (F2)'=F2. 

Given all that, the way we would write generators for the cube group
would be as <F,B,U,D,L,R>.  Remember that we do not have to include F',
B', etc. because they are included automatically.  On the other hand, if
you are left handed you might want to write the generators as
<F',B',U',D',L',R'> and you would not have to include F, B, etc. because
they would be included automatically.  Also, you do not have to include
F2, B2, etc. because we can get F2 as FF, we can get B2 as BB, etc. 

The <F,B,U,D,L,R> notation essentially says the following.  Beginning
with a cube which is solved (which is at Start), we get the cube group
by combining together the F, B, U, D, L, and R operations in all
possible ways.  This is just another way of saying that we would
scramble the cube in all possible ways by turning all six of the faces
in all possible ways. 

We now have enough definitions and notations in place to start talking
about Thistlethwaite's algorithm.  Consider what it means to say <U>. 
This means that you can twist the Up face any way you want, but you
can't twist any of the other faces.  This also means that the group is
{U,UU,UUU,UUUU}. 

If you are new at this, you ought to have a few questions.  For example,
where is U'?  Well, U' is the same thing as UUU, so U' is included (270
clockwise is the same thing as 90 degrees counterclockwise).  Where is
U2?  Well, U2 is the same thing as UU, so U2 is included.  What about
UUUUU and UUUUUU etc.?  Well, UUUU is the same thing as not twisting at
all (360 degrees clockwise is the same thing as not twisting), so
UUUUU=U, UUUUUU=UU, etc.  No matter how you twist, as long as you
confine yourself to the Up face, there are only four possible positions.

UUUU is normally written as I (for the identity).  Every group must have
an identity.  For addition, the identity is zero.  For the cube, we
normally just write I.  It should be obvious, for example, that UU'=I
and that U'U=I.  That is, if you twist the Up face 90 degrees clockwise
and immediately twist the Up face 90 degrees counterclockwise, you are
back where you started. 

Now, let's go back to the idea of solving the bottom two layers of the
cube first, then solving the Up layer.  It is very likely that after
solving the bottom two layers, the Up layer would look very scrambled. 
But we might get very lucky and discover that Up layer was already in
<U>.  That is, it might already be in one of the four positions, U, UU,
U'=UUU, or I.  If the Up layer were already at I by accident, then the
whole cube would already be solved.  If it were in one of the other
three positions, then we could finish solving the cube by simply
twisting the Up face and there would be no need to disturb either of the
bottom two layers.

The Thistlethwaite algorithm accomplishes the same sort of thing, except
that it is by design rather than by luck. 

We have already considered the group <F,B,U,D,L,R> where you scramble
the cube in any way you want using any twist of any face.  Now, consider
the group <F2,B2,U,D,L,R>.  What this means is that starting with a
solved cube, we scramble it any way we want by making any twists we want
of the U, D, L, and R faces, but for the F and B faces we can only make
180 degree twists.  It turns out that by so doing, we cannot reach as
many positions as we can if we allow 90 degree twists of all six faces. 
Hence, we would say that <F2,B2,U,D,L,R> is a subgroup of <F,B,U,D,L,R>. 

A key point of the <F2,B2,U,D,L,R> subgroup is that if we can create a
position in it by using only the indicated moves, than we can also
reverse the process and solve any position in it by using only the
indicated moves.

A position in the <F2,B2,U,D,L,R> subgroup is "somewhat solved" in much
the same sense that a cube with the bottom layer solved is "somewhat
solved", but a position in <F2,B2,U,D,L,R> still looks pretty scrambled. 
There is some disagreement among Cube-Lovers as to whether you can look
at a scrambled cube and determine easily whether it is in
<F2,B2,U,D,L,R> or not.  I will leave that question unaddressed for the
purposes of this note.

The real point is that suppose that you were in <F,B,U,D,L,R> and by
some clever strategy or other managed to get your cube into
<F2,B2,U,D,L,R>.  You would have made some hard won progress. 
Furthermore, you could solve the cube without giving up any of your hard
won progress because you could solve the cube without making any more 90
degree F and B moves.  This is very much unlike the situation of solving
by layer, where inevitably you must give up some hard won progress.  It
was thinking along these lines that led me to think in terms of closed
and open groups, namely those where you can or cannot proceed without
giving up any of your hard won progress.

The Thistlethwaite algorithm tells you how to get from <F,B,U,D,L,R> to
<F2,B2,U,D,L,R>.  It continues by a progression of subgroups that goes
something like <F2,B2,U2,D2,L,R> and then <F2,B2,U2,D2,L2,R2> on its way
to Start.  So Thistelthwaite is trying to get into a position where 90
degree turns are no longer necessary and the solution can be completed
using only 180 degree moves.

Let's go back to the <U> subgroup just for a minute, where
<U>={I,U,U2,U'}.  <U2> is a subgroup of <U> where <U2>={I,U2}.  As a
silly example of the Thistlethwaite technique, we could go from <U> to
<U2> and then on to I.  For example, suppose we were at U, which is in
<U>.  Since we are bound and determined to get into <U2>, we could make
the move U which takes us into <U2> and we could complete the solution
by making the move U2.  Hence, we have solved the position U with two
moves (namely U U2) when one would have sufficed (namely U').  As silly
as this example is, it is illustrative of the way in which
Thistlethwaite's method is suboptimal, and how Thistlethwaite's method
can be improved. 

Finally, I think the most elegant sequence of closed subgroups is more
in the vein of starting with a corner and working your way out from the
corner by layer. For example, you might first solve a 2x2x2 subcorner of
a 3x3x3, then solve the other three faces.  This approach does not
inherently involve any preference for 180 degree turns.  I like it
because I do not like counting 180 degree turns.  The trouble with this
approach is that, for example, <U,R,F> is an awfully big group to solve
all at one go. Mike Reid has suggested breaking down <U,R,F> into
<U2,R,F> etc. to make the problem manageable, but then we are back into
using 180 degree turns. 

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990



From cube-lovers-errors@mc.lcs.mit.edu  Tue Aug 26 21:22:54 1997
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Mail-from: From walsmith@erols.com Tue Aug 26 21:03:14 1997
Message-Id: <34037BAB.7923@erols.com>
Date: Tue, 26 Aug 1997 20:58:19 -0400
From: Walter Smith <walsmith@erols.com>
Reply-To: walsmith@erols.com
To: cube-lovers@ai.mit.edu
Subject: Got a new shape...?

On 8/15/97 David Goyra asked for ideas for simulated puzzles.

Obviously there are infinite possibilities.  If you want a source of
inspiration for simulated or real puzzles, I recommend the following
book:

Shapes, Space and Symmetry
by Alan Holden
Dover Publications, Inc.

I got mine at Boarders Bookstores.  It is a book about three dimensional
shapes.  It discusses symmetry and other properties with a minimum of
mathematical terms.  It gives instructions (and pictures) on
constructing many shapes from cardboard or wire.

Any solid shape could be cut (or cuts) parallel to the sides, between
opposite corners, between opposite edges, along edges or any combination
of the foregoing.  You will see the shapes of the common puzzles and
ideas for hundreds more.

Walt Smith
WALSMITH@EROLS.COM

From cube-lovers-errors@mc.lcs.mit.edu  Wed Aug 27 14:39:14 1997
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Mail-from: From reid@math.brown.edu Wed Aug 27 14:33:56 1997
Message-Id: <199708271830.OAA29527@life.ai.mit.edu>
Date: Wed, 27 Aug 1997 14:36:47 -0400
From: michael reid <reid@math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: minimal maneuvers for "continuous" isoglyphs

i finished computing minimal maneuvers for the "continuous" isoglyphs.
some may be in a different orientation than herbert gave them.
i also give a maneuver that is simultaneously minimal in both the
quarter turn metric and the face turn metric when there is such a
maneuver.


*.*
***  (type 01)
***

 1. (girdle 3-cycle)
 F  R' L  U' R' U  R  L' B' R  F' B   (12q, 12f)

 2. (distorted girdle 3-cycle)
 U  R  U  D' F2 U' D  R  U'  (10q, 9f)


*.*
.**  (type 02)
***

 3. edge hexagon of order 2
 U  B2 U' F' U' D  L' D2 L  U  D' F  D' L2 B2 D'  (20q, 16f)

 4. edge hexagon of order 3
 U' D  L' B  D  B' U2 D' B' R' B  R  U' L  D'  (16q)
 F  L  B  U  F2 B2 R  F2 B2 L' U' B' L' F'  (14f)

 5. (off-girdle 3-cycles)
 B' U  F2 L' F2 U' F' B  L  B2 U  B2 L' F   (18q, 14f)

 6. (distorted off-girdle 3-cycles)
 F  L  B  R  D' F  B2 L' F' B  L' F' D  R  F  R' F'  (18q)
 U  R2 D  F' L  U2 D2 R' U2 D2 F  D' R2 U'  (14f)


*.*
**.  (type 03)
*.*

 7. (plummer's C's)
 F  U' F  B' D2 B' U' D  R  B2 R  L' B  R' F  U' D  R'  (20q)
 L2 U2 R' B' U' D  B2 D' R' D  L  D2 F  U2 D  L2  (16f)


*.*
.*.  (type 04)
*.*

 8. pons asinorum
 U2 D2 F2 B2 R2 L2  (12q, 6f)

 9. checkerboards of order 3
 F  B2 R' D2 B  R  U  D' R  L' D' F' R2 D  F2 B'  (20q, 16f)

10. checkerboards of order 6
 R' D' F' D  L  F  U2 B' L  U  D' R' D' L  F  L2 U  F'  (20q)
 R2 L2 U  B  L2 D' F  B2 R  L' F' B  R  D  F2 L' U'  (17f)


***
***  (type 10)
**.

11. meson
 U  F' D  F  U' F' L' U' L  D' L' U  L  F   (14q)
 D  F2 D' R  B2 R' D  F2 D' R  B2 R'  (12f)


*.*
***  (type 11)
**.

12. (meson & girdle 3-cycle)
 F' L' B' D2 B' D' B  D' R  F' R  F  R2 B  L  F   (18q, 16f)


***
**.  (type 12)
*..

13. two twisted peaks
 F  B' U  F  U  F  U  L  B  L2 B' U  F' L  U  L' B   (18q)
 F  D2 B  R  B' L' F  D' L2 F2 R  F' R' F2 L' F'  (16f)

14. exchanged peaks
 F  U2 L  F  L' B  L  U  B' R' L' U  R' D' F' B  R2  (19q)
 F2 R2 D  R2 U  D  F2 D' R' D' F  L2 F' D  R  U'  (16f)


*.*
.**  (type 12)
**.

15. (meson & girdle 3-cycles)
 F  B' R  F' U  L  U' F  B' D' B  D  L' B  D' R' D  F'  (18q, 18f)


*.*
**.  (type 13)
*..

16. (plummer's Y's)
 R  U' R  B' R  F  R' U  D' R  L' B' L  F  L' F' R  F'  (18q)
 L  F  B' U' R' B' R' L' U2 L  D' R  F  R2 B2 L' F   (17f)


*.*
.*.  (type 14)
*..

17. (plummer's cluster & girdle 3-cycles)
 R  U' F  U  F' D' R  F  D' R  L' F  B' D' R  F' L  F'  (18q)
 F  B2 U  R  L2 B' L  F  D' L' B  L  B' U  L' U' D2  (17f)

18. (christman's cluster & girdle)
DL DB DR DF UL UB UR UF LB LF RB RF DLB URB UBL ULF DRF DFL UFR DBR
 F  U  R' U' R  U2 R' B' R' F  R' D  R' L  U  F  D' F  B' R'  (21q)
 F2 U  R2 L' U2 D' F2 U' B  R  L' B' U2 B  U' R  B' L   (18f)


.*.
***  (type 30)
.**

19. (plummer's rabbits)
 F  L' F  R' U  R  U' F' L  U  R' U' R  F'  (14q, 14f)


.*.
.**  (type 31)
.**

20. twisted cube edges, orthogonal bars
 F  L' U  L  U' R' U  F' L  F  L' U' R  F'  (14q, 14f)


...
.**  (type 32)
.**

21. cube in a cube
 F  L  F  U' R  U  F2 L2 U' L' B  D' B' L2 U   (18q, 15f)


.*.
**.  (type 32)
..*

22. twisted duck feet
 U2 F' B  D  B' U  D2 L  U2 F  L  F  U' R' B' R  F'  (20q, 17f)

23. exchanged duck feet
 U  F  R2 F' D' R  U  B2 U2 F' R2 F  D  B2 R  B'  (21q, 16f)


...
**.  (type 33)
..*

24. (plummer's bend)
 F  R  B' R  U  R  F  D' L' F2 R  U' F  R' B' R  F'  (18q)
 F' R  U2 L' F  B  U  B2 R' U  R2 D' R2 U' L' U   (16f)


...
.*.  (type 34)
..*

25. twisted chicken feet
 D2 R  U  L' F2 R  F' U  F' U' B' U  F  D' L  F'  (18q, 16f)

26. exchanged chicken feet, cherries
 F  L' D' B' L  F  U  F' D' F  L2 B' R' U  L2 D' F   (19q, 17f)


.*.
***  (type 40)
.*.

27. christman's cross
 U  R  L' F2 U2 F2 R' L  U2 F2 U   (16q, 11f)

28. plummer's cross
 U' D2 R  B2 D' R' U  D' R  L' D  R  F2 D' R2 L    (20q, 16f)


...
***  (type 41)
.*.

29. four way street
 L  U2 F' U  F  L2 U  L  F' D' F2 L' D' L  D2 F'  (20q, 16f)


...
.**  (type 42)
.*.

30. exchanged rings
 B' U' B' L' D  B  U  D2 B  U  L  D' L' U' L2 D    (18q)
 F  U  D' L' B2 L  U' D  F  U  R2 L2 U' L2 F2  (15f)

31. twisted rings
 F  D  F' D2 L' B' U  L  D  R  U  L' F' U  L  U2  (18q, 16f)

32. anaconda, worm
 L  U  B' U' R  L' B  R' F  B' D  R  D' F'  (14q, 14f)


.*.
.*.  (type 43)
...

33. six U's type 6
 U  D' F' U  R  L' B' U  F  U  D' R'  (12q, 12f)


...
.*.  (type 44)
...

34. six spot, six O's
 U  D' R  L' F  B' U  D' (8q, 8f)


mike

From cube-lovers-errors@mc.lcs.mit.edu  Mon Sep  1 22:33:57 1997
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Mail-from: From SCHMIDTG@iccgcc.cle.ab.com Mon Sep  1 16:35:03 1997
From: SCHMIDTG@iccgcc.cle.ab.com
Date: Mon, 1 Sep 1997 16:32:10 -0400 (EDT)
To: cube-lovers@ai.mit.edu
Message-Id: <970901163210.20217b13@iccgcc.cle.ab.com>
Subject: Re: Open and Closed Subgroups of G

I'd like to thank Jerry for taking the time to put together his 
message discussing basic group theory as it applies to the cube as
well as the basics of Thistlewaite's algorithm.  Although I consider
myself somewhat beyond the "layman" level in this area, I'm not
always able to follow the various posts to this group.  Besides,
it's also helpful to read a little "refresher" every now and then
to help reinforce and clarify previously digested concepts.

It might also be helpful for someone to cover the basics of cube
parity.  Although I think I understand the basic group theoretic
concepts of permutation parity, the asymmetry of the marked faces
of the cube have never quite left me feeling comfortable about
how this concept is applied to the cube.  Hofstadter, covers this,
but does not discuss it in enough detail for one to fully grasp
the concept.

Regards,

-- Greg Schmidt

From cube-lovers-errors@mc.lcs.mit.edu  Mon Sep  1 23:26:56 1997
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Mail-from: From SCHMIDTG@iccgcc.cle.ab.com Mon Sep  1 16:50:08 1997
From: SCHMIDTG@iccgcc.cle.ab.com
Date: Mon, 1 Sep 1997 16:46:33 -0400 (EDT)
To: cube-lovers@ai.mit.edu
Message-Id: <970901164633.20217b13@iccgcc.cle.ab.com>
Subject: Re[2]: Open and Closed Subgroups of G

Oh, and I forgot to mention...

My ultimate goal of understanding parity would be such that someone could
hand me an arbitrary permutation puzzle and I'd be able to examine it and
determine from the set of legal moves both the parity constraints and also
be able to construct a parity test valid from any given puzzle state.

I find it interesting that the method seems to differ across puzzles.
For example, 15 puzzle parity can be determined by the number of pairwise
exchanges required to solve the puzzle, whereas with the cube, it seems
a more direct approach is possible by examining cubie orientations with
respect to marked cubicles.

Still, I'm somewhat mystified.

Regards,

-- Greg Schmidt

From cube-lovers-errors@mc.lcs.mit.edu  Tue Sep  2 11:08:15 1997
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Mail-from: From nbodley@tiac.net Tue Sep  2 09:46:52 1997
Date: Tue, 2 Sep 1997 08:44:30 -0400 (EDT)
From: Nicholas Bodley <nbodley@tiac.net>
To: SCHMIDTG@iccgcc.cle.ab.com
Cc: cube-lovers@ai.mit.edu
Subject: Parity (Was Re: Re[2]: Open and Closed Subgroups of G)
In-Reply-To: <970901164633.20217b13@iccgcc.cle.ab.com>
Message-Id: <Pine.SUN.3.95.970902084236.25019C-100000@sunspot.tiac.net>


If I understand parity, Greg's examination would reveal whether someone
had reassembled a Cube (or other mathematically-related puzzle) into a
state that can't be solved.

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*   Waltham is now in the new 781 area code.
|*  Amateur musician  *|*   617 will be recognized until the end of 1997.
--------------------------------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Wed Sep  3 18:01:12 1997
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Mail-from: From lvt-cfc@servtech.com Wed Sep  3 13:05:50 1997
From: "christopher f. chiesa" <lvt-cfc@servtech.com>
Message-Id: <199709031702.NAA20567@cyber1.servtech.com>
Subject: Re: Open and Closed Subgroups of G
To: cube-lovers@ai.mit.edu
Date: Wed, 3 Sep 1997 13:02:11 -0400 (EDT)

Greg Schmidt (SCHMIDTG@iccgcc.cle.ab.com) mentions discomfort about 
how concepts of "parity" are applied to the Cube.  I second the
notion! :-)  

I assume that by "parity" we mean that which is conserved as the "twist"
of corner cubies or the "flip" of edge cubies.  I myself have a HELL of a 
time determining a particular corner cubie's precise amount (N/3, N an 
integer) of "twist," or a particular edge cubie's precise amount (N/2, N an
integer) of "flip," other than in the case of an observable change in ONLY
that particular cubie -- and moreover, ONLY in its ORIENTATION.  Any change
in a cubie's POSITION, relative OR absolute, renders my notions of "twist"
and "flip" rather fuzzy.

F'rinstance, start with a Cube in the "solved" state and perform the sequence
(generator?): 

   R' D2 R F D2 F' U2 F D2 F' R' D2 R U2

You will find that "FRU has been twisted -1/3 ("one 'notch' CCW"), and BLU
has been twisted +1/3 ("one 'notch' CW")," relative to their previous
orientations (i.e., relative to "solved") -- and that this is easy to assess
largely because the "solved" state of the rest of the Cube makes it very
clear how the corner cubies' orientations have changed (and their positions
have NOT).  The sequence/generator would produce the same net effect
(twisting FRU -1/3, and BLU +1/3) when performed on the Cube in ANY state; it
would merely be more difficult for the casual observer to identify against
the background of a "scrambled" Cube state.

But, back to the start-from-"solved" example.  If I now make the single turn

   B'

I no longer find it so easy to characterize the corner-twist parity state of
the Cube, because (all of) the corner-cubies affected by this particular
Cube-state-change have left their previous positions, leaving me to wonder,
"RELATIVE TO WHAT" their twist is to be assessed.  How is it done?  What can 
now be said about the "twist state" of, say, the former BLU (now BRU) cubie?
What about the former BLD (now BLU) cubie?

My efforts to "reason it out," within the limitations of my group-theory
background (which is now infinitely broader thanks to Jerry Bryan!), lead to
what almost seems a paradox.  For what it's worth, I present it for your
discussion, and will be very interested to hear what you Cubemeisters are
able to contribute!

Observe that the orientations of all corners in the F layer remain unchanged
by the B' operation last performed.  In particular, the FRU cubie retains its
-1/3 twist relative to (what's left of) the "solved" state.  Assuming that
the "twist" of a cubie which "hasn't moved" REMAINS THE SAME, as opposed to
being, say, "implicitly redefined" by the movement of OTHER cubies, I can
still say a few things -- though not as many things as I would like! -- about
the twist-states of the corner-cubies in the "B layer" after that B' face
turn. 

Invoking twist-parity-conservation (let's just say "twist-conservation,"
okay?), I assert that "the TOTAL twist of all corner cubies in the B layer
must still be 'some integer plus 1/3,'" so as to "cancel out" the -1/3 twist
remaining on FRU.  The B' turn thus imparted "some integer" TOTAL twist,
which is to say, a total of 0 "net" twist, to the corner cubies in the B
layer -- but was it e.g. "0, 0, 0, 0" or "+1/3, +1/3, -1/3, -1/3?"  (I 
believe all other combinations reduce to these.)  Note that this boils down
to asking, "does a face turn, if it twists corner-cubies AT ALL, twist ALL
FOUR the SAME WAY (i.e. apply the same "net twist" to all four), or NOT?"
Is there a definitive answer?  A standard assumption?  Proof or disproof of
either?  It seems there would _have_ to be, in order to have "meaningful"
discussions of "twist" at all.

For a while I thought I could prove that it was the "0, 0, 0, 0" case, but it
turned out that one of my working assumptions was equivalent to STATING that
it was the "0, 0, 0, 0" case.  I was only "proving" my own ASSUMPTION.  Glad
I didn't post THAT. :-)

Naturally, analogous issues and questions will arise when discussing
edge-cubie "flip" and the conservation thereof. :-)

All in all, I'd be VERY interested in seeing the professional theoretical
dissection of this issue!

...

That's all I have today on the subjects of "twist," "flip," and "parity/
conservation thereof."  But before I go, I'll leave you with two more
demented, blue-sky thoughts.  Beware; this is what I get for reading Star
Trek novels before bed, and again at breakfast...

1) At the edge of my intuition, beyond my ability to formalize, I fancy
   I sense that there might be a way of looking at the Cube, perhaps 
   through the use of additional spatial dimensions or their mathemati-
   cal equivalents, in which the Cube is in some sense "always" in the 
   "solved" state, or at least in which it is trivially obvious where lies
   the "direct path" back TO the "solved" state.  I'm visualizing some 
   sort of extra-spatial "rubber bands," or "strings" (in those higher
   spatial dimensions specifically so as to avoid "tangling" issues)
   that "trace" the route (or "net" route) taken by each cubie, or arbi-
   trary collection of cubies, from its/their position(s)-and-orienta-
   tion(s) in the "solved" state, to its/their p(s)-and-o(s) in a "scram-
   bled" Cube.  In such a perception, one could simply "tug on the 
   strings" and "pull" the Cube back to "solved."  Does this make ANY 
   kind of sense to ANYBODY else here?  I feel as though I can "almost
   see it."

2) Is there a notion, has anybody done any work, on Cube states which
   are each other's "duals?"  I define the "dual" of a Cube state X as
   that Cube state reached by performing, on a "solved" Cube, the same
   sequence of turns/moves which "solve" Cube state X.  In other words,
   define a sequence of turns which transforms the Cube from state X
   to "solved," then apply that sequence again to the "solved" cube to 
   arrive at state Y.  State Y is then the "dual" of state X.  Ques-
   tions abound:

      - does each state have EXACTLY ONE dual?  Or many, depending on
        the specific sequence (as we know, there are many) of moves
        performed in solving state X ?  (My gut feeling is that each
        state has exactly one dual.  This would seem to be pretty easy
        to prove using the group-theory math at the disposal of many
        readers here.)

      - are there states which are their OWN duals?  (Yes, clearly; 
        the trivial "checkerboard" pattern arising from a single 180-
        degree turn of each face, is its own dual)

      - a state which is its own dual, is a "two-cycle" with the 
        "solved" state: perform the generating sequence on either and
        get to the other.  Are there "three-cycles?"  "Four-cycles?"
        etc.?

Looking forward to the followups,

Chris Chiesa
  lvt-cfc@servtech.com


From cube-lovers-errors@mc.lcs.mit.edu  Wed Sep  3 18:42:04 1997
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Mail-from: From jbryan@pstcc.cc.tn.us Wed Sep  3 13:55:04 1997
Date: Wed, 03 Sep 1997 13:51:02 -0400 (Eastern Daylight Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: Open and Closed Subgroups of G
In-Reply-To: <970901163210.20217b13@iccgcc.cle.ab.com>
To: SCHMIDTG@iccgcc.cle.ab.com
Cc: cube-lovers@ai.mit.edu
Message-Id: <Pine.WNT.3.96.970903133533.-229605I-100000@GN209A.PSTCC.CC.TN.US>

On Mon, 1 Sep 1997 SCHMIDTG@iccgcc.cle.ab.com wrote:

> It might also be helpful for someone to cover the basics of cube
> parity.  Although I think I understand the basic group theoretic
> concepts of permutation parity, the asymmetry of the marked faces
> of the cube have never quite left me feeling comfortable about
> how this concept is applied to the cube.  Hofstadter, covers this,
> but does not discuss it in enough detail for one to fully grasp
> the concept.
> 

I'll take your question as literal, assuming you mean just parity and
not twist and flip, and assuming you know the basic group theoretic
concepts of permutation parity.

Parity of the cube is best described (I think) as applying to whole
cubies rather than to facelets.  As such, a quarter turn of any face is
a 4-cycle on the corner cubies and a 4-cycle on the edge cubies.  A
4-cycle is odd, which is to say that it can be decomposed into an odd
number of 2-cycles.  The "obvious" way to decompose a 4-cycle is into
three 2-cycles.  Although decomposition of a 4-cycle into 2-cycles is
not unique, any such decomposition will contain an odd number of
2-cycles.

Start is even for both the edges and the corners (the identity consists
of zero 2-cycles).  If you any quarter turn from Start, both edges and
corners become odd.  Make another quarter turn, both edges and corners
become even.  Make another quarter turn, both edges and corners become
odd.  Etc.  Edges and corners are either both even or both odd. 

In the constructable group, you can have odd corners with even edges or
vice versa.  For example, remove two edge cubies from a cube and
exchange them without moving any of the other cubies around.  You will
be changing the parity of the edges without changing the parity of the
corners.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990

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Date: Thu, 04 Sep 1997 12:50:11 -0400 (Eastern Daylight Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: Open and Closed Subgroups of G
In-Reply-To: <199709031702.NAA20567@cyber1.servtech.com>
To: "christopher f. chiesa" <lvt-cfc@servtech.com>
Cc: cube-lovers@ai.mit.edu
Message-Id: <Pine.WNT.3.96.970904120838.-229605B-100000@GN209A.PSTCC.CC.TN.US>

On Wed, 3 Sep 1997, christopher f. chiesa wrote:

> 2) Is there a notion, has anybody done any work, on Cube states which
>    are each other's "duals?"  I define the "dual" of a Cube state X as
>    that Cube state reached by performing, on a "solved" Cube, the same
>    sequence of turns/moves which "solve" Cube state X.  In other words,
>    define a sequence of turns which transforms the Cube from state X
>    to "solved," then apply that sequence again to the "solved" cube to 
>    arrive at state Y.  State Y is then the "dual" of state X.  Ques-
>    tions abound:

The concept of "dual" which you are describing is standard in group
theory (and be extension, in cube theory).  A "dual" is properly called
an inverse.  If you have a sequence of turns which creates a position,
the inverse sequence consists of writing the turns in reverse order, and
converting clockwise turns to counterclockwise turns and vice versa.  So
the inverse of FRU' is UR'F'.  If there are multiple sequences for a
position (and most typically there are), you can do the same thing for
any such sequence. 

Also, a position can be described in terms of which cubies have gone
where.  For example, you might have something like
 
              flu  -->  fur
              fur  -->  frd
              frd  -->  fdl
              fdl  -->  flu

(flu is the front-left-up cubie etc.  Standard Singmaster notation uses
lower case letters for cubies and upper case letters for the moves
themselves.)

You could get the inverse by reversing the arrows like so.

 
              flu  <--  fur
              fur  <--  frd
              frd  <--  fdl
              fdl  <--  flu

More commonly, you would write the inverse by swapping the cubie
designations between the left and right side of the arrows like so.

 
              fur  -->  flu
              frd  -->  fur
              fdl  -->  frd
              flu  -->  fdl


I don't know what you mean by "any work", but here are some standard
information about inverses.  The length of a position X is the same as
the length of its inverse X', where length is the minimum number of
moves to create the position.  If X' is the inverse of X, then X is the
inverse of X'.  The symmetry of an inverse X' is the same as the
symmetry of a position X (see Symmetry and Local Maxima in the archives
for a discussion of symmetry).  A local maximum is a position such that
no matter which move you make, you will be one move closer to Start.  It
is not necessarily the case that the inverse of a local maximum is also 
a local maximum. 


> 
>       - does each state have EXACTLY ONE dual?  Or many, depending on
>         the specific sequence (as we know, there are many) of moves
>         performed in solving state X ?  

Yes, inverses are unique, both for groups in general, and for cubes in
particular.

> 
>       - are there states which are their OWN duals?  (Yes, clearly; 
>         the trivial "checkerboard" pattern arising from a single 180-
>         degree turn of each face, is its own dual)

You have answered your own question.  Many positions are their own
inverse.  Some of them are much more complicated than the one which you
describe.

> 
>       - a state which is its own dual, is a "two-cycle" with the 
>         "solved" state: perform the generating sequence on either and
>         get to the other.  Are there "three-cycles?"  "Four-cycles?"
>         etc.?
> 

The proper term for the concept you are describing is order.  If you
repeat a maneuver n times from Start and return to Start, then the
position is of order n.  (Strictly speaking, the order of a position is
the smallest n which will work.  Obviously, if n will work then so too
will 2n, 3n, etc.)  There are many different orders for which there are
cube positions of that order. One of David Singmaster's early Cubic
Circulars (I don't have the reference handy) had a table of possible
cube orders and how many positions there were of each order.

The term cycle is also very important in group theory (and by extension
in cube theory).  Suppose you look at a scrambled cube and determine
that cubie a has gone to cubie b's place, cubie b has gone to cubie c's
place, and cubie c has gone to cubie a's place, then a, b, and c form a
3-cycle.  The way I have defined this particular 3-cycle, you could
write it as (a,b,c), as (b,c,a), or as (c,a,b).  This so-called cycle
notation is circular, so it does't really matter which you write first. 
However, (a,c,b) is a different cycle than (a,b,c).  In fact, (a,c,b) is
the inverse of (a,b,c).  Just for emphasis, (a,b,c) is not like an
ordered pair (or really an ordered triple in this case).  (a,b,c) means
a goes to b, b goes to c, c goes to a.

As an example of a cycle in purely cube terms, the cycle for the example
I gave earlier would be (flu,fur,frd,fdl), so it is a 4-cycle. 

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990

From cube-lovers-errors@mc.lcs.mit.edu  Fri Sep  5 21:03:58 1997
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Mail-from: From Hoey@AIC.NRL.Navy.Mil Fri Sep  5 21:08:00 1997
Date: Fri, 5 Sep 1997 21:07:48 -0400
Message-Id: <199709060107.VAA04503@sun30.aic.nrl.navy.mil>
From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: lvt-cfc@servtech.com
Cc: cube-lovers@ai.mit.edu
In-Reply-To: <199709031702.NAA20567@cyber1.servtech.com> (lvt-cfc@servtech.com)
Subject: Re: Open and Closed Subgroups of G (fwd)

Chris Chiesa <lvt-cfc@servtech.com>, among other things, writes

> If I now make the single turn

>    B'

> I no longer find it so easy to characterize the corner-twist parity state of
> the Cube, because (all of) the corner-cubies affected by this particular
> Cube-state-change have left their previous positions, leaving me to wonder,
> "RELATIVE TO WHAT" their twist is to be assessed.

At the risk of being repetitious, the answer is, "relative to the home
orientation of the position they find themselves in".  You choose a
special facelet for each corner cubie.  When the cubie is in its home
position, its twist is the position of its special facelet relative to
the home of the special facelet.  When cubie X is in cubie Y's home
position, the twist of cubie X is the position of X's special facelet
relative to the home of Y's special facelet.  The edges are done the
same way, except mod 2.

Cube-lovers can find this in Vanderschel's article (6 Aug 1980) and
the extension by Saxe (3 September 1980).  I mentioned (23 September
1982) that the choice of special facelets is arbitrary, and that a
conservation of twist occurs for a set of pieces of any puzzle that

    1. have an Abelian orientation group, and
    2. are moved in untwisted cycles by the generators.

This is true even if not all the cycles have the same length.  For
instance, we could have a Rubik's cube in which generators move
corners in permutations like (FTR,FRD,FDL,FLT)(BRT,BTL,BLD), and twist
would be preserved.  The key is that for each piece, the minimum power
of the generator that returns that piece to its home position must
also return it to its home orientation.

I'm quite uncertain about what orientation constraints can arise in
puzzles with non-Abelian orientation groups.  For instance, the
hypercorners of a Rubik's tesseract have the symmetry group A4, and
any orientation is achievable up to a constraint imposed by an Abelian
quotient of A4 of type 3 (See 22 Oct 1982).  Does every group have a
unique maximal Abelian quotient?  Is that the only orientation
constraint that can occur?

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

[ Moderator's Note: Cube-lovers will be down Saturday and Sunday due to
  major electrical work at MIT. ]

From cube-lovers-errors@mc.lcs.mit.edu  Mon Sep  8 09:47:22 1997
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Mail-from: From kociemba@hrz1.hrz.th-darmstadt.de Sun Sep  7 17:51:08 1997
Message-Id: <3411D734.6471@hrz1.hrz.th-darmstadt.de>
Date: Sun, 07 Sep 1997 00:20:36 +0200
From: Herbert Kociemba <kociemba@hrz1.hrz.th-darmstadt.de>
To: cube-lovers@ai.mit.edu
Subject: Number of maneuvers with n face turns

The number of maneuvers with 1, 2, 3,.. face turns for Rubik's cube are
of course well known and are 18, 243, 3240... But I did not see a closed
formula for these numbers before, so maybe you find the following
formula interesting:

Let r:= sqrt(6), then you have with n face turns

P(n) = [(3+r)*(6+3r)^n + (3-r)*(6-3r)^n]/4

maneuvers.  Because the second part in brackets is much smaller than the
first, asymptotically you have

(3+r)*(6+3r)^n /4 maneuvers.

Even for small n, this approximation is very good. So for n=3 you get
3240.33 instead of 3240. The asymptotic branching factor P(n+1)/P(n) is
therefore (6+3r), which is about 13.348469 .

Herbert

From cube-lovers-errors@mc.lcs.mit.edu  Tue Sep  9 11:01:44 1997
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Mail-from: From reid@math.brown.edu Tue Sep  9 00:17:33 1997
Message-Id: <199709090413.AAA00748@life.ai.mit.edu>
Date: Tue, 9 Sep 1997 00:20:27 -0400
From: michael reid <reid@math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: maximal abelian quotients

dan asks

>                                              Does every group have a
> unique maximal Abelian quotient?

yes.  let  G  be a group.  it's not difficult to show that

1)  the commutator subgroup  G'  is normal,
2)  the quotient group  G / G'  is abelian, and
3)  if  G --> A  is a homomorphism to any abelian group  A ,
    then  G'  is in the kernel, so there is a unique  homomorphism
    G / G' --> A  such that the original homomorphism is the composite
    G --> G / G' --> A .

this last one is kind of technical, but in the special case where
A = G / N  for some normal subgroup  N , it says that if  G / N  is
abelian, then  N  contains the commutator subgroup.  thus,  G / G'
is the maximal abelian quotient of  G .

the quotient  G / G'  is sometimes written  G^ab  (the "abelianization" of G).
as you might guess, this is an important construction in group theory,
and it's one of the reasons why commutator subgroups are important.

mike

From cube-lovers-errors@mc.lcs.mit.edu  Tue Sep  9 14:56:02 1997
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Mail-from: From jbryan@pstcc.cc.tn.us Tue Sep  9 11:06:36 1997
Date: Tue, 09 Sep 1997 11:02:32 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: Open and Closed Subgroups of G
In-Reply-To: <199709060107.VAA04503@sun30.aic.nrl.navy.mil>
To: Cube-Lovers <cube-lovers@ai.mit.edu>
Cc: lvt-cfc@servtech.com
Message-Id: <Pine.PMDF.3.95.970909103952.3344N-100000@PSTCC6.PSTCC.CC.TN.US>

On Fri, 5 Sep 1997, Dan Hoey wrote:

> Chris Chiesa <lvt-cfc@servtech.com>, among other things, writes
> 
> > If I now make the single turn
> 
> >    B'
> 
> > I no longer find it so easy to characterize the corner-twist
> > parity state of the Cube, because (all of) the corner-cubies
> > affected by this particular Cube-state-change have left their
> > previous positions, leaving me to wonder, "RELATIVE TO WHAT" their
> > twist is to be assessed.
> 
> At the risk of being repetitious, the answer is, "relative to the home
> orientation of the position they find themselves in".  You choose a
> special facelet for each corner cubie.  When the cubie is in its home
> position, its twist is the position of its special facelet relative to
> the home of the special facelet.  When cubie X is in cubie Y's home
> position, the twist of cubie X is the position of X's special facelet
> relative to the home of Y's special facelet.  The edges are done the
> same way, except mod 2.

Dan's response (plus his references in the Cube-Lovers archives) pretty
well covers it.  I would just like to add a couple of points.

  1. There is a reference in the archives to a way of demonstrating
     conservation of twist without first establishing a frame of
     reference, but I can't find the reference.  The best I can
     recall, the same technique did not work for edges.  But I prefer
     the frame of reference technique anyway because it is closely
     tied to some of the more usual ways of representing the cube in a
     computer.

  2. For example, number the corner facelets from 1 to 24.  Each
     facelet has two companion facelets which are bound to it on
     the same cubie.  By knowing where one of the three facelets
     of a cubie is in a computer program, you automatically know
     where the other two facelets are, so you only have to store
     one of the three facelets.  The one that you store can be the
     "special" facelet that Dan described for the purposes of
     determining conservation of twist.

     The collection of eight "special" facelets for the corners
     have been described in the archives as constituting a
     supplement for the group, but I have yet to find a discussion
     group supplements in any group theory book.

     As Dan says, your choice of "special" facelet is totally 
     arbitrary for each cubie, but most typically you choose
     the Front and Back facelets, or the Right and Left facelets,
     or something equally well organized.

  3. For another example, number the corner cubies from 1 to 8, and 
     for each of the cubies describe the twist with a number from 0
     to 2.  This is essentially a wreath product representation of
     the cube.  The numbers from 0 to 2 which describe the twist
     can be used to describe whether a cubie is twisted when it is
     not home, and can therefore be used to prove conservation of
     twist.

     Without knowing any more than I do about supplements, it seems
     very likely that it should be easy to represent any group 
     which can be representated as a supplement as a wreath product
     and vice versa.  The isomorphism seems obvious.  I wonder if
     anybody out there can shed any light on this issue?

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990

From cube-lovers-errors@mc.lcs.mit.edu  Fri Sep 12 17:53:00 1997
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Date: Fri, 12 Sep 1997 17:07:47 -0400 (Eastern Daylight Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: isoglyphs
In-Reply-To: <199708182216.SAA00604@sun30.aic.nrl.navy.mil>
To: cube-lovers@ai.mit.edu
Reply-To: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Message-Id: <Pine.WNT.3.96.970912155335.-953687V-100000@GN209A.PSTCC.CC.TN.US>

On Mon, 18 Aug 1997, Dan Hoey wrote:

> A "chiral isoglyph" is one in which the handedness of the glyph is
> taken into account in testing for isoglyphy,* so that the glyph
> appears only in one variety.

> 
> Mike used "achiral" for an isoglyph that fails to be a chiral
> isoglyph, though I would tend to use "non-chiral".  I would rather use
> "achiral" for a situation that lacked chirality, as in an isoglyph of
> a mirror-symmetric glyph.

Let me see what I can do to muddy these waters. 

It seems to me that we might ought to consider the chirality of an
isoglyph as being a different issue than the chirality of a glyph.  I
think the two are clearly related, but I am not sure that the one
necessarily derives from the other. 

As to a glyph, it seems to me that a glyph is chiral only if conjugating
the position by each of the four reflections of the square yields a
different set of positions than does conjugating the position by each of
the four rotations of the square.  Hence, you can have a glyph which
occurs in right-handed or left-handed forms, or one that doesn't.  This
is the simple part. 

I think the situation with isoglyphs is a little more complicated.  For
example, form an isoglyph using both the right-handed and the
left-handed forms of a chiral glyph.  You might have 6 right-handed
glyphs and 0 left-handed glyphs, 5 right-handed glyphs and 1 left-handed
glyph, etc.  If there are unequal numbers of right-handed and
left-handed glyphs, then it seems natural to define the handedness of
the isoglyph as being that of the dominate glyph.  But what if there are
three right-handed glyphs and three left-handed glyphs?

Up to symmetry, there are only two ways to partition the six faces of a
cube into two sets of three faces.  For example, the F, U, and B faces
can be of the same chirality, or the F, U, and R faces can be of the
same chirality (or any conjugates of these choice of faces).  In the
first case, the cube is partitioned like a universal joint, or maybe
like a cubic baseball.  Such a position seems to me to lack chirality. 
In the second case, three faces with the same chirality cluster around a
common corner.  Again, such a position seems to me to lack chirality. 
So an isoglyph which lacks chirality can contain chiral glyphs.

On the other hand, even on an isoglyph consisting of three right-handed
and three left-handed glyphs, you still might be able to find a
distinguishing characteristic of the right hand part that was different
from the left-handed part.  For example, the glyph boundaries which were
internal to the right-handed part of the isoglyph might be continuous
whereas the glyph boundaries which were internal to the left-handed part
of the isoglyph might not be continuous.  Or for another example, the
rotations of the three right-handed faces relative to each other might
be different than the rotations of the left-handed faces relative to
each other. 

(By the way, I have not verified that any of these positions I have
described are actually in G.  I guess I am thinking in terms of the
constructible group of the facelets -- conceptually, peeling all the
facelets off and reattaching them.) 

On the other hand, two glyphs which lack chirality when placed side by
side can be chiral.  For example,

       XOOXXX   (the base glyph is XXX
       XXXOXO                      OXO
       XOOOXO                      OXO )

I really haven't thought through the implications of using six glyphs
instead of two, but it seems to me quite likely that an isoglyph could
be constructed using six glyphs which lack chirality and which are the
same pattern, and where the we could attribute chirality to the isoglyph
as a whole. 

I have thought about this in terms of Herbert's Cube Explorer 1.5
program.  The pattern editor has a check box for continuous.  If you
don't check the box, the program finds both continuous and
non-continuous isoglyphs.  If you do check the box, it finds only
continuous ones.  So I have considered what would happen if the program
had a check box for chiral.  What should it do?  The obvious thing would
be that in normal operation, it would consider conjugates of both
rotations and reflections of the square when building an isoglyph from a
glyph, but that if the chiral box were checked it would consider only
conjugates of rotations of the square.  But is that sufficient to
satisfy our various definitions of chiral, achiral, and/or non-chiral?
I'm not sure.  Maybe Dan or Mike would be kind enough to clarify further
their thoughts on this issue. 

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990

From cube-lovers-errors@mc.lcs.mit.edu  Sun Sep 14 22:54:52 1997
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Mail-from: From reid@math.brown.edu Sat Sep 13 21:32:35 1997
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Date: Sat, 13 Sep 1997 21:33:59 -0400
From: michael reid <reid@math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: optimal solver is available

for those who are interested in my optimal cube solver, you can now
get it from the web page

     http://www.math.brown.edu/~reid/rubik/optimal_solver.html

i've reduced the size of my transformation tables, so now i think there's
a reasonable chance that it will run within 80Mb of RAM.

enjoy the program.  if you make any new exciting discoveries, please share
them with the entire mailing list.

mike

From cube-lovers-errors@mc.lcs.mit.edu  Mon Sep 29 13:08:14 1997
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Mail-from: From reid@math.brown.edu Sun Sep 28 14:45:54 1997
Message-Id: <199709281845.OAA08068@life.ai.mit.edu>
Date: Sun, 28 Sep 1997 14:46:39 -0400
From: michael reid <reid@math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: idea for smaller optimal solver

a number of people have told me that they don't have 80Mb of RAM on their
computers, so that my optimal solver won't work on their machine.  here's
an idea for an optimal solver that uses much less memory; it should fit
within 16Mb, or 20Mb at the most.  of course, it's a space/time tradeoff,
but perhaps will still be fairly good.

in my current program, i use distances to the subgroup

     H  =  <U, D, F2, R2, B2, L2>

as my "heuristic" function.  there is another subgroup,  H'  , which
contains  H  as a subgroup of index 8.  H'  is the subgroup of all
elements of  H  composed with all (valid) flips of U-D slice edges.

another way to describe  H'  is the subgroup of all elements where
the U face has only the colors U and D, and the same for the D face.
from this latter description, we see that if  H1' , H2'  and  H3'
are the three orientations of this subgroup, then their intersection
is the subgroup of elements that "look like" they're in the square
group.  this is the same target subgroup that my current program has.

the subgroup  H'  also has 16 symmetries.  using this to reduce the
size of the pattern database, and storing each entry with 4 bits,
it should take about 8.5Mb.  my current program also has about 8.5Mb
of transformation tables (but 3Mb of these are not used while searching).
the transformation tables will probably be slightly smaller (certainly
no larger), so it seems plausible that this could run with 16Mb of RAM.

what about running time?  in his paper, rich korf hypothesizes that
the number of nodes generated should be roughly proportional to the
inverse of the size of the pattern databases.  this suggests that
using the smaller tables above would result in about 8 times as many
nodes as my current program.  this isn't bad, especially given that
the branching factor (6 + 3 * sqrt(6) = 13.348469  for face turns,
9.3736596  for quarter turns) is larger than this.  so this approach
would be within 1 turn of my current program.

i don't foresee having enough spare time anytime soon to program this,
so i'll just post it here and maybe someone who is interested will
program this.

mike

From cube-lovers-errors@mc.lcs.mit.edu  Tue Sep 30 12:12:19 1997
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Mail-from: From C.McCaig@Queens-Belfast.AC.UK Tue Sep 30 05:39:44 1997
From: C.McCaig@queens-belfast.ac.uk
Date: Tue, 30 Sep 1997 10:27:55 GMT
To: cube-lovers@ai.mit.edu
Message-Id: <009BB113.FAA1EEF6.44@a1.qub.ac.uk>
Subject: 4x4x4 solution

i recently borrowed a friends 4x4x4, and i know the basic method for
solving it.  ie get the 6 centres, pair up all the edges, and then
solve for the normal cube.  however, about half the time i end up
with a single edge pair inverted and cant figure out a move for
reorientating the single edge pair.  usually i break a few pairs
and try and reorientate them this way, but this seems rather longwinded...
does anyone have a move for this?.  for example, say the green edge
is on the blue face, and the blue edge is on the green face...

thanks.

clive 
---
clive mccaig
queens university belfast
northern ireland

From cube-lovers-errors@mc.lcs.mit.edu  Tue Sep 30 17:44:19 1997
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Mail-from: From Cube-Lovers-Request@ai.mit.edu Tue Sep 30 15:39:10 1997
Date: Tue, 30 Sep 1997 17:43:57 -0400
Message-Id: <30Sep1997.174357.Cube-Lovers@AI.MIT.EDU>
From: Cube Lovers Moderator <Cube-Lovers-Request@AI.MIT.EDU>
Sender: Cube-Lovers-Request@AI.MIT.EDU
To: Cube-Lovers@AI.MIT.EDU
Subject: 4x4x4 solution -- [Digest v23 #159]

Cube-Lovers Digest         Tue, 30 Sep 1997      Volume 23 : Issue 159

Today's Topic:
		            4x4x4 solution

[ I have gathered together several similar messages on a single topic,
  putting them in digest format.  It would be nice to get an explicit
  process for this problem, though.  --Moderator. ]

----------------------------------------------------------------------

Date: Tue, 30 Sep 1997 13:39:46 -0400 (EDT)
From: der Mouse  <mouse@Rodents.Montreal.QC.CA>
To: C.McCaig@queens-belfast.ac.uk
Cc: cube-lovers@ai.mit.edu
Subject: Re: 4x4x4 solution

> i recently borrowed a friends 4x4x4, and i know the basic method for
> solving it.  [...]  however, about half the time i end up with a
> single edge pair inverted and cant figure out a move for
> reorientating the single edge pair.

Make a single 90-degree inner-slice turn, then solve as before.  This
introduces an odd permutation on the edge pairs, which gets you back
into easily solvable space.  (It's usually easiest if you make sure
that the two swapped edge cubies are part of the slice turn, by placing
on the same slice beforehand if necessary.)

I'm not sure quite what the parity constraint here is.  There is some
kind of even-parity constraint on the edge cubies, it appears, with a
linked constraint on the face centres, but it's not as simple as the
parity of the edge and face permutations being both even or both odd,
because the single slice turn introduces two nonoverlapping 4-cycles on
the face centre cubies - which is, overall, an even permutation on
them.

I do notice, though, that a slice turn produces a 4-cycle on the edges
and two 4-cycles on the face centres; a face turn produces a 4-cycle on
the face centres and two 4-cycles on the edges (and a 4-cycle on the
corners, which may or may not be relevant).  I wonder if there's a
multiple-of-three constraint lurking.

Doubtless some group theorist has long ago worked out exactly what the
constraints are, but I haven't heard.  (I tried to work through a
group-theory text recently, got stalled along about the time it got to
cosets, quotient groups, normal subgroups, etc.)

					der Mouse

			       mouse@rodents.montreal.qc.ca
		     7D C8 61 52 5D E7 2D 39  4E F1 31 3E E8 B3 27 4B

------------------------------

Date: Tue, 30 Sep 1997 14:11:28 -0400 (EDT)
From: Allan Wechsler <awechsle@bbn.com>
To: C.McCaig@queens-belfast.ac.uk
Cc: cube-lovers@ai.mit.edu
Subject: 4x4x4 solution

    [C. McCaig:]
    i recently borrowed a friends 4x4x4, and i know the basic method for
    solving it.  ie get the 6 centres, pair up all the edges, and then
    solve for the normal cube.  however, about half the time i end up
    with a single edge pair inverted and cant figure out a move for
    reorientating the single edge pair.  usually i break a few pairs
    and try and reorientate them this way, but this seems rather longwinded...
    does anyone have a move for this?.  for example, say the green edge
    is on the blue face, and the blue edge is on the green face...
    
What's happened here is that you've got those two edge-cubies
_exchanged_.  Here's how it works.  Look at any face.  You will see
eight edge stickers, arranged around the face like eight square
dancers (or Irish set dancers, if you prefer).  Now I hope you are
familiar with one or the other of these kinds of folk-dancing, because
otherwise what I am going to say won't make sense.  Those eight decals
are either men or women, and no matter how they dance around the cube,
they will never change sex.  Every edge-cubie on the 444 has one
permanently male and one permanently female sticker.  If I haven't
clarified things, at least I've spiced them up a bit.

House around to home.

- -A

------------------------------

Date: Tue, 30 Sep 1997 14:18:17 -0400 (Eastern Daylight Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: 4x4x4 solution
To: C.McCaig@Queens-Belfast.AC.UK
Cc: cube-lovers@ai.mit.edu

On Tue, 30 Sep 1997 C.McCaig@Queens-Belfast.AC.UK wrote:

> i recently borrowed a friends 4x4x4, and i know the basic method for
> solving it.  ie get the 6 centres, pair up all the edges, and then
> solve for the normal cube.  however, about half the time i end up
> with a single edge pair inverted and cant figure out a move for
> reorientating the single edge pair.  usually i break a few pairs
> and try and reorientate them this way, but this seems rather longwinded...
> does anyone have a move for this?.  for example, say the green edge
> is on the blue face, and the blue edge is on the green face...
> 

Your problem is one of parity.  You have two edges cubies swapped (this
swap is visible) and two face center (centre) cubies of the same color
swapped (this swap is invisible).  You have to have an even number of
swaps in the total cube.  If you want an even number in the edges (and
you do), then you also have to have an even number in the face centers,
even if swaps in the face centers are invisible.

There is probably a more elegant solution, but the following will work. 
If you encounter the situation you describe, make any middle slice
quarter turn.  This will disturb the centers.  The centers will now have
an even numbers of swaps. Solve the centers again without simply undoing
the middle slice you just made.  The parity of the edges will then be
ok.  (I'm assuming that your solution for the face centers will maintain
their parity after you correct it as described.)

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990

------------------------------

End of Cube-Lovers Digest
*************************

From cube-lovers-errors@mc.lcs.mit.edu  Tue Sep 30 18:10:34 1997
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Mail-from: From roger.broadie@iclweb.com Tue Sep 30 18:05:38 1997
From: roger.broadie@iclweb.com (Roger Broadie)
To: <cube-lovers@ai.mit.edu>
Subject: Re: 4x4x4 solution
Date: Tue, 30 Sep 1997 23:02:48 +0100
Message-Id: <19970930230037.AAA21244@home>

C.McCaig@queens-belfast.ac.uk wrote:

> ...I recently borrowed a friends 4x4x4, and I ... can't figure out a
> move for reorientating the single edge pair....

It is possible to solve the problem with a sequence based on a quarter
turn of a central slice, since that, like a swap of two edge pieces,
involves an odd-parity cycle of the edge pieces.  Thus

	r2 U2 r U2 r2

(where r is the turn of the inner slice next to R in the direction
parallel to R)

puts a 4-cycle of edges onto the top face, but leaves you with the task
of restoring the centres.
It was the desire to find something less cumbersome that first lead me
to investigate the archives of this list, and there the answer was:

	Date: Fri, 20 Oct 95 12:46:32 -0400 (EDT)
	From: Georges Helm <geohelm@pt.lu>
	Subject: Re: Old question about 2 adj edges


	how to flip 2 adj. edges (and nothing else) in 4x4x4 cube?


	r^2 U^2 r l' U^2 r' U^2 r U^2 r l U^2 l' U^2 r U^2 l r^2 U^2

	Georges
	geohelm@pt.lu

It does indeed contain an odd number of turns of the central slices to
give the desired parity.

Roger Broadie 

From cube-lovers-errors@mc.lcs.mit.edu  Wed Oct  1 11:53:15 1997
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Mail-from: From dokon@MIT.EDU Tue Sep 30 19:30:29 1997
Message-Id: <3.0.32.19970930192820.006ce8ac@po9.mit.edu>
Date: Tue, 30 Sep 1997 19:28:21 -0400
To: cube-lovers@ai.mit.edu
From: Dennis Okon <dokon@mit.edu>
Subject: God's Number

I just found out that Keith Randall for the theory group of LCS (Lab for
Computer Science) at MIT gave a talk Monday about God's number for the
rubik's cube.  He upped the lower bound 24 and gave "evidence" that it is
24.  I don't know what moves he was counting (e.g. slice, quarter).
Unfortunately, I missed it.  Does anyone have any information on this?
I'll see what I can find out.

-Dennis Okon
dokon@mit.edu

From cube-lovers-errors@mc.lcs.mit.edu  Wed Oct  1 13:18:18 1997
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Mail-from: From ERCO@compuserve.com Wed Oct  1 01:56:54 1997
Date: Wed, 1 Oct 1997 01:52:24 -0400
From: Edwin Saesen <ERCO@compuserve.com>
Subject: Re: 4x4x4 solution -- [Digest v23 #159]
Sender: Edwin Saesen <ERCO@compuserve.com>
To: CUBE <CUBE-LOVERS@ai.mit.edu>
Message-Id: <199710010152_MC2-225C-6120@compuserve.com>

jbryan@pstcc.cc.tn.us wrote:
>Your problem is one of parity.  You have two edges cubies swapped
>(this swap is visible) and two face center (centre) cubies of the
>same color swapped (this swap is invisible).  You have to have an
>even number of swaps in the total cube.  If you want an even number
>in the edges (and you do), then you also have to have an even number
>in the face centers, even if swaps in the face centers are invisible.

I've had this problem as well. If I understand you correctly, this
problem simply doesn't occur anymore as soon as you number (or mark in
any other way) the center pieces which
a) makes solving the cube a bit more difficult
b) makes sure that you'll always get back to the original
configuration of center pieces.

I've had a similar problem on my 5x5x5 as well, and I assume that
marking the nine center pieces might solve the problem as well.
On my 4x4x4 I also had a problem of having two pairs of edges
exchanged which simply can't happen on a 3x3x3. By experimenting with
3x3x3 moves I found a 24move solution to this, and I wonder if that's
also sort of automatically solved by marking center pieces.

Can anyone confirm this?

Michael Ehrt
 ---------------------------------------------
ERCO@compuserve.com

From cube-lovers-errors@mc.lcs.mit.edu  Wed Oct  1 13:55:36 1997
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Mail-from: From darinh@ldr.com Wed Oct  1 13:22:36 1997
Message-Id: <34328707.3792@ldr.com>
Date: Wed, 01 Oct 1997 10:23:24 -0700
From: Darin Haines <darinh@ldr.com>
Organization: Litho Development & Research <http://www.ldr.com>
To: Cube <Cube-Lovers@ai.mit.edu>
Subject: Piece for a Rubik's Revenge

Hi Everyone,

Does anyone know of someone wanting to sell a BROKEN Rubik's Revenge?
or maybe a center piece from the same?

Did anyone else have problems with the center pieces breaking on their
RR?  or am I the only one?

My RR has been sitting useless on the shelf since '84.  Hey, I was young
and careless. ;-)

-Darin

[Moderator's note: I'll be away from cube-lovers from 2 Oct to 5 Oct.
 Messages received during that time will be distributed on the 6th. ]

From cube-lovers-errors@mc.lcs.mit.edu  Wed Oct  1 15:49:18 1997
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Mail-from: From bagleyd@americas.sun.sed.monmouth.army.mil Wed Oct  1 14:41:30 1997
From: bagleyd@americas.sun.sed.monmouth.army.mil (David Bagley x21081)
Message-Id: <199710011842.OAA21977@asia.sed.monmouth.army.mil>
Subject: Piece for Alexander's Star
To: Cube-Lovers@ai.mit.edu
Date: Wed, 1 Oct 1997 14:42:15 -0400 (EDT)
In-Reply-To: <34328707.3792@ldr.com> from "Darin Haines" at Oct 1, 97 10:23:24 am

> 
> Hi Everyone,
> 
> Does anyone know of someone wanting to sell a BROKEN Rubik's Revenge?
> or maybe a center piece from the same?
> 
That reminds me:
If anyone needs a piece or 2 for the Alexander's Star let me know.
They seem to break pretty easily IMHO.  Please specify colors.
I have a center piece too.

Mine broke a while back and I have since gotten another one.
-- 
Cheers,
 /X\  David A. Bagley
(( X  bagleyd@bigfoot.com  http://wauug.erols.com/~bagleyd/
 \X/  xlockmore  ftp://wauug.erols.com/pub/X-Windows/xlockmore/index.html

From cube-lovers-errors@mc.lcs.mit.edu  Wed Oct  1 16:57:31 1997
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Mail-from: From Cube-Lovers-Request@ai.mit.edu Wed Oct 1 16:53:46 1997
Date: Wed, 1 Oct 1997 16:53:46 -0400 (EDT)
Message-Id: <01Oct1997.165346.Cube-Lovers@AI.MIT.EDU>
From: Cube Lovers Moderator <Cube-Lovers-Request@AI.MIT.EDU>
To: Cube-Lovers@AI.MIT.EDU
Subject: 4x4x4 solution -- [Digest v23 #165]

Cube-Lovers Digest         Wed, 1 Oct 1997      Volume 23 : Issue 165

Today's Topic:
                           4x4x4 solution

----------------------------------------------------------------------

Date: Wed, 1 Oct 1997 08:20:34 -0400 (EDT)
From: Assoc Prof W David Joyner <wdj@nadn.navy.mil>
To: C.McCaig@queens-belfast.ac.uk
Cc: cube-lovers@ai.mit.edu
Subject: Re: 4x4x4 solution

On Tue, 30 Sep 1997 C.McCaig@queens-belfast.ac.uk wrote:

> i recently borrowed a friends 4x4x4, and i know the basic method for
> solving it.  ie get the 6 centres, pair up all the edges, and then
> solve for the normal cube.  however, about half the time i end up
> with a single edge pair inverted and cant figure out a move for
> reorientating the single edge pair.  usually i break a few pairs
> and try and reorientate them this way, but this seems rather longwinded...
> does anyone have a move for this?.  for example, say the green edge
> is on the blue face, and the blue edge is on the green face...

The idea is on the www page
http://www.nadn.navy.mil/MathDept/wdj/solve4.txt
Try
L2^2*D1^2*U2*F1^3*U2^3*F1*D1^2*L2^2*L1*U1*L1^3*U2^3*L1*U1^3*L1^3
(due to Jeff Adams). - David Joyner

------------------------------

Date: Wed, 1 Oct 1997 14:13:50 -0400 (EDT)
From: Nichael Cramer <nichael@sover.net>
To: Edwin Saesen <ERCO@compuserve.com>
Cc: CUBE <CUBE-LOVERS@ai.mit.edu>
Subject: Re: 4x4x4 solution -- [Digest v23 #159]

On Wed, 1 Oct 1997, Edwin Saesen wrote:

> On my 4x4x4 I also had a problem of having two pairs of edges
> exchanged which simply can't happen on a 3x3x3.

I find it most convienent to think of this situation as the following:

One of the "center-slices" containing one of the "swapped" edge pieces is
rotated by 90 degrees.

(This is roughly analogous to the the 3X case where the whole cube is
solved except for two corners and two edge pieces being --respectively--
swapped.  The problem is that the unfinished face is 90dg out of phase.)

Rotate that center-slice by 90 degrees and re-solve from there.

This is surely not the most efficient (i.e. shortest) solution; but it is
conceptually straight forward.


Nichael Cramer
work: ncramer@bbn.com
home: nichael@sover.net
http://www.sover.net/~nichael/

------------------------------

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Date: Wed, 1 Oct 1997 16:23:14 -0400
From: Jim Mahoney <mahoney@marlboro.edu>
To: ERCO@compuserve.com
Cc: CUBE-LOVERS@ai.mit.edu
Subject: Re: 4x4x4 solution -- [Digest v23 #159]


      >Your problem is one of parity.  You have two edges cubies swapped
      >(this swap is visible) and two face center (centre) cubies of the
      >same color swapped (this swap is invisible).  You have to have an
    Edwin>  I've had this problem as well.... If I understand you
    Edwin> correctly, this problem simply doesn't occur anymore as
    Edwin> soon as you number (or mark in any other way) the center
    Edwin> pieces which a) makes solving the cube a bit more difficult
    Edwin> b) makes sure that you'll always get back to the original
    Edwin> configuration of center pieces.

This isn't quite true, at least not on the 4x4x4.

While it is true that parity is the question at hand, and also that on
the 4x4x4 cube a quarter of a central slice performs an odd permutation
on the edges which is otherwise "invisible", it is *not* true that
marking the centers will help.  The reason is that a quarter turn on a
center slice of the 4x4x4 performs a cyclic rearrangement of 4 edges -
an odd permutation - while at the same time rearranges *two* sets of 4
central pieces - an even permutation of the centers.  Thus parity does
not prohibit swapping two edges while leaving the centers untouched.

Moreover, in fact there are move sequences which will exchange two
edges without disturbing the position of any other piece, corner or
center - though I don't have any on hand which are short.  If there's
interest, though, I can produce a move sequence to exchange two 4x4x4
edges while leaving all corners and centers in their original
positions.

A cross-section looks like this.  A quarter turn cycles the four E's,
the four C1's, and the four C2's.  This is an odd permutation of the
E's but an even permutation of the C's.  (All the C's are corners, and
can be put into each other's positions with a combination of face and
center turns.)

   E  C1  C2  E

   C2         C1

   C1         C2

   E  C2  C1  E


A full account of parity and possible 4x4x4 moves gives

  4x4x4
  type     ,  how many  ,  parity after:  1/4 face turn , 1/4 center turn
- ---------------------------------------------------------------------
  corners  |   8                         |  odd           |  even (untouched)
  edges    |   24=2x(12 edges)           |  even (8 move) |  odd
  centers  |   24=4x(6 faces)            |  odd           |  even (8 move)


Thus to solve a 4x4x4 cube you must have made both (1) an even
total number of moves on the faces (to restore the corners and centers to
even parity), as well as (2) an even total number of moves on the center
slices, to restore the edges to even parity.

The parity constraints on the 5x5x5 are a bit different.  In that case
there are two types of edges (the one in the middle of an edge vs the
ones next to the corners) and three types of centers.  Each has its
own parity change under each different slice.  A bit of playing around
shows that any central slice move which cycles 4 edges must also cycle
several kinds of centers.  At least one of those center cycles is odd.
Therefore on the 5x5x5 you cannot exchange a pair of edges without
also exchanging two centers somewhere.  So marking where the centers
go will help on the 5x5x5.

Regards,

  Jim Mahoney                          mahoney@marlboro.edu
  Physics & Astronomy
  Marlboro College, Marlboro, VT  05344

------------------------------

End of Cube-Lovers Digest
*************************

From cube-lovers-errors@mc.lcs.mit.edu  Wed Oct  1 17:46:05 1997
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Mail-from: From SCHMIDTG@iccgcc.cle.ab.com Wed Oct  1 16:48:58 1997
From: SCHMIDTG@iccgcc.cle.ab.com
Date: Wed, 1 Oct 1997 16:48:44 -0400 (EDT)
To: cube-lovers@ai.mit.edu
Cc: ljl@basmark.com
Message-Id: <971001164844.2023493c@iccgcc.cle.ab.com>
Subject: New cube program available

I've recently finished my implementation of Kociemba's algorithm and it
is now available from the cube-lovers ftp site at:

ftp.ai.mit.edu /pub/cube-lovers/contrib/kcube1_0.zip

The .zip files contains a README.TXT file, commented C++ source, and
an executable program that runs on Win95/NT.  Here's a brief description
of the program that appears in the README file within the "contrib"
directory:

File: kcube1_0.zip
Author: Greg Schmidt <Greg.Schmidt@ab.com>
Description:

    A cube solver that implements Kociemba's algorithm.  This program was
    written for the express purpose of understanding the algorithm in
    sufficient detail for me to implement it.  The source code is included
    and commented with the hope of providing others with a similar
    understanding.


I welcome feedback concerning any aspects of this program.  Many thanks to
Dik Winter and especially Herbert Kociemba for answering some of my
detailed questions as well as allowing me to use their ideas and offer
them to cube-lovers in the form of this program.

Regards,

-- Greg

From cube-lovers-errors@mc.lcs.mit.edu  Wed Oct  1 19:01:21 1997
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Mail-from: From joemcg3@snowcrest.net Wed Oct  1 18:42:12 1997
Message-Id: <3432D0DB.4B19@snowcrest.net>
Date: Wed, 01 Oct 1997 15:38:19 -0700
From: Joe McGarity <joemcg3@snowcrest.net>
Reply-To: joemcg3@snowcrest.net
To: "Mailing List, Rubik's Cube" <cube-lovers@ai.mit.edu>
Subject: My Revenge is Complete

How strange that I both have a broken Rubik's Revenge and need a piece
to an Alexander's Star.  What are the odds?

I haven't looked at it for quite some time, but I think my Revenge is
complete.  The problem is the ball in the center.  One of the corners
is broken (if you have seen a dissasmbled RR it makes sense for a ball
to have corners) and has resisted all attempts at being glued.  So it
may be that my broken cube will not be of any help to Darin Haines, but
the rest of the cubies are intact if anyone needs any of them.  Which
brings up that I have a fairly large collection of broken or otherwise
incomplete puzzles.  I suspect that this is true for many of us. I would
be more than willing to do some trading if anyone has any particular
needs.  Let me know.

Joe McGarity

From cube-lovers-errors@mc.lcs.mit.edu  Wed Oct  1 20:02:37 1997
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Mail-from: From randall@theory.lcs.mit.edu Wed Oct  1 19:49:18 1997
Date: Wed, 1 Oct 1997 19:46:08 -0400
Message-Id: <199710012346.TAA06162@hemp>
From: Keith H Randall <randall@theory.lcs.mit.edu>
To: reid@math.brown.edu
Cc: cube-lovers@ai.mit.edu
In-Reply-To: <199710012120.AA25636@theory.lcs.mit.edu> (message from michael
	reid on Wed, 1 Oct 1997 17:19:02 -0400)
Subject: Re: God's Number

   Don Dailey, Aske Plaat, and myself have a program that will do a
complete 22-ply search in about 24 hours on an 8 processor Sun
machine.  The program measures distance in the QT (quarter-turn)
metric.

I've run some experiments on random cubes, summarized as follows:

112 random odd cubes:
20   depth 19
92   depth 21

57 random even cubes:
41   depth 20
16   depth 22

>From this random sample, it seems as if less than 1% of cubes are
depth 23, let alone more than depth 24.  In fact, the only depth 23
cubes I know of so far are the twelve cubes 1 move away from the
superflip.  This fact gives some evidence that God's number is
probably 24.

By the way, below are solutions and depths for all of the symmetric
cubes enumerated by Hoey and Saxe in their message of Sun, 14 Dec 80.
These are obvious cubes to try because they are local maxima, and they
are all depth 22 or less except for the superflip.  Only one
representative from each of the 26 conjugacy classes is given.  All
solutions were obtained from the program, except for the superflip
solution which is absconded from a post from Reid on Tue, 10 Jan 95.
All depths are exact minimal depths, i.e. no shorter solutions exist.

M-symmetric cubes
0	solved
		--
12	pons asinorum
		F F B B L L R R U U D D
24	superflip
		R' U U B L' F U' B D F U D' L D D F' R B' D F' U' B' U D'
20	pons asinorum * superflip
		F' U' B' R' F R L' D' R L' U D' L' U D' F R B U F

T-symmetric cubes
22	girdleflip
		F F U F F B' U R' L B U F D' F F B D' R L' B' D' F
19	girdleswap
		F U F R U' L' U' B U' B' R' F' R' L' F' R L L F'
21	girdleflip * girdleswap
		F U' L U F' U' B B D B U B' D' R D' R' B' R' D R B'
22	girdleflip * pons asinorum
		F F U L F L' D' R L' U' L L U U R F' B D' F' U R' D'
17	girdleswap * pons asinorum
		F R F B R' F' B' L D D F F D D R' L' F'
21	girdleflip * girdleswap * pons asinorum
		F R' L B R U' R R U' D F' R F' B L B R' F B' U' L'
20	girdleflip * superflip
		F U U F' R' U' L F' D F B' L U' L U' F' L U D' F
21	girdleswap * superflip
		F R F B U D' F' B R R U F B D' R L D' F' B' U F
21	girdleflip * girdleswap * superflip
		F U D B' R' F' D' R' U R' L' B R F U F D B D L' B'
20	girdleflip * pons asinorum * superflip
		F F B R' F U' B' R' L D L U' R' U' D F L B' D F
21	girdleswap * pons asinorum * superflip
		F U U B D' L' U F F B R' U R B U D' L B U D' L
21	girdleflip * girdleswap * pons asinorum * superflip
		F B U F' U' F R B' R' F' U R' U F B U' F' B' U R U'

H-symmetric cubes
22	plummer
		F F R B' U L U R F L U' L L B R' D F D B L F D'
16	six-H
		F F R R F B' R R L L F' B R R B B
20	plummer * six-H
		F F U F' R' B' D' F' R U D L B' U' F' L' B' U' F F
20	plummer^2 * six-H
		F F U F R B U F R' U' D' L' B D F L B U' F F
20	plummer^2 * pons asinorum
		F R U F D' B B L F L' F' L F R R D' L U B L
20	plummer^2 * superflip
		F B U F R L' U' D' L U' D L B L' B' U F D L U
18	six-H * superflip
		F R' U D D F' B R F R' L D' F' R' L U L F'
22	plummer * six-H * superflip
		F U D F' R L F U D R R L L B' U D F' R L B' R' L
22	plummer^2 * six-H * superflip
		F B U F' B' D R' L' U F F B B R' L' D' F' B' D R' L' D'
22	plummer * pons asinorum * superflip
		F B R' U' D' R' F' B' R U' D' F F B B L U' D' R' F' B' L'

reference for cube names:
pons asinorum
        W B W
        B W B
        W B W

O R O   G Y G   R O R   Y G Y
R O R   Y G Y   O R O   G Y G
O R O   G Y G   R O R   Y G Y

        B W B
        W B W
        B W B

superflip
        W Y W
        O W R
        W G W

O W O   G W G   R W R   Y W Y
Y O G   O G R   G R Y   R Y O
O B O   G B G   R B R   Y B Y

        B G B
        O B R
        B Y B

plummer
        Y W Y
        W W W
        G W G

W O W   O G R   W R W   R Y O
O O O   G G G   R R R   Y Y Y
B O B   O G R   B R B   R Y O

        G B G
        B B B
        Y B Y

six-H
        W W W
        B W B
        W W W

O O O   G Y G   R R R   Y G Y
R O R   G G G   O R O   Y Y Y
O O O   G Y G   R R R   Y G Y

        B B B
        W B W
        B B B

girdle flip (about ULF-DRB axis)
        W Y W
        W W R
        W W W

O O O   G G G   R W R   Y W Y
Y O O   G G R   G R R   Y Y O
O B O   G B G   R R R   Y Y Y

        B G B
        O B B
        B B B

girdle swap (about ULF-DRB axis)
        R B B
        W W B
        W W Y

B O O   G G B   O O G   O G G
R O O   G G Y   O R R   Y Y G
R R Y   R Y Y   W R R   Y Y W

        W W O
        W B B
        G B B


                                        -Keith

From cube-lovers-errors@mc.lcs.mit.edu  Wed Oct  1 20:49:17 1997
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Mail-from: From roger.broadie@iclweb.com Wed Oct  1 19:12:26 1997
From: roger.broadie@iclweb.com (Roger Broadie)
To: <Cube-Lovers@ai.mit.edu>
Subject: Re: 4x4x4 solution
Date: Thu, 2 Oct 1997 00:09:47 +0100
Message-Id: <19971002000735.AAA23683@home>

I'm tempted to try a little more analysis of the parity constraints on
the 4x4x4 cube, though no doubt it's all been done before.  As der
Mouse said,

A slice turn produces a 4-cycle on the edges and two 4-cycles on the
face centres; a face turn produces a 4-cycle on the face centres and
two 4-cycles on the edges (and a 4-cycle on the corners, which may or
may not be relevant).

I think it is very relevant. We can set the effects out as follows:

Turn		Piece		Cycle(s)	Parity
------          -------         --------        -------

Slice		edge		1x4		odd
		centre		2x4		even

Face		edge		2x4		even
		centre		1x4		odd
		corner		1x4		odd

The consequence is that the parity of the centre pieces depends
entirely on the number of face turns - any slice turns do not affect
the parity of these pieces since the changes they introduce will be of
even parity.  For face turns, the changes to the parity of the corner
pieces and the centre pieces are the same.  Hence if the corner pieces
are in place, the centres will be in an even permutation, and that will
not be changed even if the edge pieces are in an odd permutation, which
was the essence of Clive McCaig's original question. Nor will that be
changed by any turn of a central slice to bring them back to an even
permutation.

I the corners are correct (which I guess is the normal situation when
the problem with the swapped edge pieces shows up) then, though I say
so with some hesitation, I do not think Jerry Bryan is right in saying
that the pair of swapped edge pieces will be matched by a pair of
swapped centre pieces.  For example, the process I quoted switches edge
pieces, and though it has no visible effect on the centre pieces, it
does in fact change the positions of the centre pieces on the front
face (if I have correctly identified the results of a bit of hasty work
with little Post-it stickers).  However, the whole block of four
rotates through 180 degrees, which is two 2-cycles and thus of even
parity.  Edwin Saesen could mark the centre pieces, get them back to
their original position and still find the edge pieces swapped, but
that will not prevent his correcting the edge pieces, and then, if he
wants to, correcting the centre pieces with even-parity processes.

Luckily, for the 4x4x4, we do not have to worry about twists for the
edge pieces or the centre pieces, since that is fixed geometrically for
each position they can occupy.  When an edge piece is in its home
position it must be the right way round.  When it moves to its
next-door position it must flip.  I imagine this is the point behind
Allan Wechsler's charming square-dancing analogy.  The centre pieces
always present the same corner to the central intersection of the face.

Roger Broadie

From cube-lovers-errors@mc.lcs.mit.edu  Mon Oct  6 19:55:23 1997
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Mail-from: From Goyra@iol.ie Wed Oct  1 20:07:28 1997
Message-Id: <199710020007.BAA10545@GPO.iol.ie>
From: "Goyra (David Byrden)" <Goyra@iol.ie>
To: <cube-lovers@ai.mit.edu>
Subject: For all cube programmers
Date: Thu, 2 Oct 1997 01:02:23 +0100

	When writing a program to manipulate
the Cube, you're interested in your algorithm.
The output usually looks like RLULRURL
because you won't waste time programming
any graphics.

	I will shortly release a freeware
software component that displays a standard
Rubik's Cube. You can incorporate it into
your software and manipulate the cube
directly. See your cube solutions executed in
front of your eyes.

	For an idea of what this component
will look like, take a Java browser to my
pages at

http://www.iol.ie/~goyra/Rubik.html

	The component will be a Java Bean,
meaning you can use it in Java, and also
in any Activex environment such as Visual
C++ or Visual Basic.

	Anyone with suggestions about how
the programmatic interface to the component
should look, please mail me.

				David



From cube-lovers-errors@mc.lcs.mit.edu  Mon Oct  6 21:04:33 1997
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Mail-from: From Cube-Lovers-Request@ai.mit.edu Mon Oct 6 21:02:21 1997
Date: Mon, 6 Oct 1997 21:02:21 -0400 (EDT)
Message-Id: <06Oct1997.210221.Cube-Lovers@AI.MIT.EDU>
From: Cube Lovers Moderator <Cube-Lovers-Request@AI.MIT.EDU>
To: Cube-Lovers@AI.MIT.EDU
Subject: 4x4x4 solution -- [Digest v23 #170]

Cube-Lovers Digest         Mon, 6 Oct 1997      Volume 23 : Issue 170

Today's Topic:
                           4x4x4 solution

----------------------------------------------------------------------

Date: Wed, 1 Oct 1997 23:26:34 -0400 (EDT)
From: Nichael Cramer <nichael@sover.net>
To: Roger Broadie <roger.broadie@iclweb.com>
Cc: Cube-Lovers@ai.mit.edu
Subject: Re: 4x4x4 solution
Message-Id: <Pine.BSI.3.91.971001225256.25718B-100000@granite.sover.net>

Roger Broadie wrote:
> A slice turn produces a 4-cycle on the edges and two 4-cycles on the
> face centres; a face turn produces a 4-cycle on the face centres and
> two 4-cycles on the edges (and a 4-cycle on the corners, which may or
> may not be relevant).
>
> I think it is very relevant. We can set the effects out as follows:
>
> Turn		Piece		Cycle(s)	Parity
> ------        -------         --------        -------
>
> Slice         edge            1x4                odd
>               centre          2x4                even
>
> Face          edge            2x4                even
>               centre          1x4                odd
>               corner          1x4                odd
>
> The consequence is that the parity of the centre pieces depends
> entirely on the number of face turns - any slice turns do not affect
> the parity of these pieces since the changes they introduce will be of
> even parity.  For face turns, the changes to the parity of the corner
> pieces and the centre pieces are the same.  Hence if the corner pieces
> are in place, the centres will be in an even permutation, and that will
> not be changed even if the edge pieces are in an odd permutation, which
> was the essence of Clive McCaig's original question. Nor will that be
> changed by any turn of a central slice to bring them back to an even
> permutation.


As one of the folks who advocated rotating a center slice, let me explain
my (admittedly non-optimal) process for getting out of this fix and perhaps
you can explain where my reasoning is wrong.

1] Imagine a 4X which is completely solved except for two flipped (i.e.
swapped) edge-pieces.

2] For simplicity's sake --and without loss of generality--, assume the 2
flipped/swapped pieces are adjacent and in the top front location.  So the
top of the cube will look like this:

   X X X X
   X X X X
   X X X X
   X 1 2 X

(Here the numbers are meant to indicate only where the cubies are located,
having nothing to do with their colors.)

3] I now rotate one of the center slices (say, the one on the right, i.e.
the one containing the cubie "2") 90dg away from me.

4] The top of the cube now looks like:

   X X 2 X
   X X O X
   X X O X
   X 1 3 X

5] I can now perform the 3-cycle 1->3->2 (i.e. without affecting any of the
rest of the cube).

The top of the cube now looks like:

   X X 3 X
   X X O X
   X X O X
   X 2 1 X

6] In particular, note that "2" and "1" are now in their correct positions
(and, of course, necessarily in their proper "flip" orientation).

7] Moreover, note that I now have exactly three edge cubes in the wrong
place (i.e. "3" from above and the other two edge cubes which were
misplaced during my original 90dg rotation of the center slice).

I can now perform a 3-cycle on these edges pieces (similar to the one used
in step 5 above) again without affecting any of the other locations on the
cube.

8] My cube now has all the edge pieces in their correct location.

9] I now have only to "fix" the 8 central-face cubes which were misplaced
during my initial 90dg twist.  I can now do this is short order.

QED[?]

Nichael Cramer
work: ncramer@bbn.com
home: nichael@sover.net
http://www.sover.net/~nichael/

------------------------------

From: roger.broadie@iclweb.com (Roger Broadie)
To: "Nichael Cramer" <nichael@sover.net>
Cc: <Cube-Lovers@ai.mit.edu>
Subject: Re: 4x4x4 solution
Date: Thu, 2 Oct 1997 23:12:39 +0100

> From: Nichael Cramer <nichael@sover.net>
> To: Roger Broadie <roger.broadie@iclweb.com>
> Cc: Cube-Lovers@ai.mit.edu
> Subject: Re: 4x4x4 solution
> Date: 2 October 1997 4:26
>
> As one of the folks who advocated rotating a center slice, let me
> explain my (admittedly non-optimal) process for getting out of this
> fix and perhaps you can explain where my reasoning is wrong.
>[followed by a procedure in which a quarter turn of a centre slice is
followed, first, by a 3-cycle of edges on the top to restore the two
swapped pieces, second, by a 3-cycle of edges to restore the other
displaced edges, and, third, by restoring the displaced centres]

I absolutely agree with your reasoning. A quarter turn of a central
slice must be at the heart of any procedure to perform an edge swap,
because it is the only way to change the parity of the edges.  That was
what I said in my first post on 1 October 1997.

In my second post I was trying to look at the effect of that quarter
turn of the central slice on the centre pieces, and show that, as they
had been subjected to an even permutation by reason of the centre-slice
turn, the centre pieces could not have undergone an invisible swap of a
single pair of centre pieces.  Having made a single quarter turn of the
central slice, all the other edge and centre pieces can be restored
with processes of even parity, like your two 3-cycles.

Roger Broadie

------------------------------

End of Cube-Lovers Digest
*************************

From cube-lovers-errors@mc.lcs.mit.edu  Mon Oct  6 22:28:22 1997
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Mail-from: From Cube-Lovers-Request@ai.mit.edu Mon Oct 6 22:27:32 1997
Date: Mon, 6 Oct 1997 22:27:32 -0400 (EDT)
Message-Id: <06Oct1997.222732.Cube-Lovers@AI.MIT.EDU>
From: Cube Lovers Moderator <Cube-Lovers-Request@AI.MIT.EDU>
To: Cube-Lovers@AI.MIT.EDU
Subject: 4x4x4 solution -- [Digest v23 #171]

Cube-Lovers Digest         Mon, 6 Oct 1997      Volume 23 : Issue 171

Today's Topic:        Pieces of broken cubes:

                    Rubik's Revenge (Clarified)
                      My Revenge is Complete
                    Piece for a Rubik's Revenge
                     Piece for Alexander's Star

----------------------------------------------------------------------

Date: Thu, 02 Oct 1997 08:01:03 -0700
From: Darin Haines <darinh@ldr.com>
To: Cube <Cube-Lovers@ai.mit.edu>
Subject: Rubik's Revenge (Clarified)

I guess my terminology was incorrect.  The parts I need are actually the
center cubies of which there are 4 on each side (for a total of 24).  It
sounds like Joe McGarity's broken RR will help me out just fine (as will
a couple of other responses I've received).

I got to looking last night and found that I actually need 3 (not just
1) of these center cubies.

- -Darin

------------------------------

To: cube-lovers@ai.mit.edu
From: "Bryan Main" <bmain@caddscan.com>
Subject: Re: My Revenge is Complete
Date: Thu, 02 Oct 1997 14:32:15 Eastern Daylight Time

At 03:38 PM 10/1/97 -0700, you wrote:

>I haven't looked at it for quite some time, but I think my Revenge is
>complete.

How stable are the 4x4x4 and 5x5x5?  I was thinking on getting one but they
cost quite a lot of money and was wondering how easy it is to break them.
Also what kind of paint should I use to paint my cubes as I have 4 normal
ones and would like to make different patterns on them to make them more
intersting.  Also any patterns would be helpful.

bryan
__________________________________________________________________
Bryan Main
Cartographic Specialist
http://caddscan.com

------------------------------

Date: Thu, 2 Oct 1997 22:42:11 -0400 (EDT)
From: Nicholas Bodley <nbodley@tiac.net>
To: Darin Haines <darinh@ldr.com>
Cc: Cube <Cube-Lovers@ai.mit.edu>
Subject: Re: Piece for a Rubik's Revenge

On Wed, 1 Oct 1997, Darin Haines wrote:

{Snips}
}Did anyone else have problems with the center pieces breaking on their
}RR?  or am I the only one?

 These pieces >are< rather fragile, as I remember.

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*    Waltham is now in the new 781 area code.
|*  Amateur musician  *|*   617 will be recognized until 1 Dec. 1997.

------------------------------

Date: Sun, 5 Oct 1997 18:40:17 -0400 (EDT)
From: Nicholas Bodley <nbodley@tiac.net>
To: David Bagley x21081 <bagleyd@americas.sun.sed.monmouth.army.mil>
Cc: Cube-Lovers@ai.mit.edu
Subject: Re: Piece for Alexander's Star

They do break easily. I haven't had mine out of storage for some time, but
I well remember that it needed conscious care when manipulating; nothing
like a properly-lubricated deluxe Ideal 3^3 (the one with plastic color
tiles, and changes to the shapes of the pieces that tend to make it
self-align).

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*    Waltham is now in the new 781 area code.
|*  Amateur musician  *|*   617 will be recognized until 1 Dec. 1997.

------------------------------

End of Cube-Lovers Digest
*************************

From cube-lovers-errors@mc.lcs.mit.edu  Mon Oct  6 23:26:59 1997
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Date: Mon, 6 Oct 1997 23:26:19 -0400 (EDT)
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From: Cube Lovers Moderator <Cube-Lovers-Request@AI.MIT.EDU>
To: Cube-Lovers@AI.MIT.EDU
Subject: God's number -- [Digest v23 #172]

Cube-Lovers Digest         Mon, 6 Oct 1997      Volume 23 : Issue 172

Today's Topic:

                           God's Number

----------------------------------------------------------------------

Date: Thu, 02 Oct 1997 17:04:33 -0400 (Eastern Daylight Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: God's Number
To: Keith H Randall <randall@theory.lcs.mit.edu>
Cc: reid@math.brown.edu, cube-lovers@ai.mit.edu
Message-Id: <Pine.WNT.3.96.971002113239.-1012347D-100000@GN209A.PSTCC.CC.TN.US>

On Wed, 1 Oct 1997, Keith H Randall wrote:

>    Don Dailey, Aske Plaat, and myself have a program that will do a
> complete 22-ply search in about 24 hours on an 8 processor Sun
> machine.  The program measures distance in the QT (quarter-turn)
> metric.
>
> I've run some experiments on random cubes, summarized as follows:
>
> 112 random odd cubes:
> 20   depth 19
> 92   depth 21
>
> 57 random even cubes:
> 41   depth 20
> 16   depth 22

Wow.  I am impressed with how much data you have.  For the case of
random cubes and guaranteed optimal solutions, I believe this is the
most data which has been posted to Cube-Lovers.

It would be nice to examine enough cases to raise the probability that a
few positions of length 17q would show up for odd cubes and of length
18q for even cubes.  At this distance from Start, the branching factor
for one level is about 9.3, so the branching factor for two levels
(e.g., between level 17 and level 19) would be about 85 or so.  So you
are just at the edge of the sample size where you would expect the
shorter lengths to show up.

Notwithstanding that, I decided to play with the numbers to see if I
could make any reasonable projection about the overall distribution of
lengths in the quarter-turn metric.  Here is what I have come up with.

Consider the 19q case.  Your results suggest that about 17.8% of odd
positions, and hence about 8.6% or 8.7% of all positions are exactly 19q
from Start.  (The sample size does not support an estimate of that
precision, of course, but let's continue anyway).  It's easy to
calculate that no more than about 8.4% of positions can be 19q from
Start.  From this, I would conclude two things.  First, your results
seem right on, well within the bounds of sampling error. Second, your
results suggest that it is very unlikely that the branching factor drops
below about 9.3 until you pass 19q from Start.  Using the best available
known results, plus using your results as an estimate, plus some other
guessing, I would propose that the actual search tree for the q-turn
case looks something like the following.

Distance Number   Branching    Cumulative
 from     of        Factor      Number of
 Start   Positions              Positions


   0            1                       1
   1           12   12.000             13
   2          114    9.500            127
   3         1068    9.368           1195
   4        10011    9.374          11206
   5        93840    9.374         105046
   6       878880    9.366         983926
   7      8221632    9.355        9205558
   8     76843595    9.347       86049153
   9    717789576    9.341      803838729
  10   6701836858    9.337     7505675587
  11  62549615248    9.333    70055290835
  12    5.838E+11    9.333      6.538E+11
  13    5.449E+12    9.333      6.102E+12
  14    5.085E+13    9.333      5.696E+13
  15    4.746E+14    9.333      5.316E+14
  16    4.430E+15    9.333      4.961E+15
  17    4.134E+16    9.333      4.631E+16
  18    3.859E+17    9.333      4.322E+17
  19    3.601E+18    9.333      4.034E+18
  20    1.546E+19    4.294      1.950E+19
  21    1.657E+19    1.071      3.606E+19
  22    6.035E+18    0.364      4.210E+19
  23           12    0.000      4.210E+19
  24            1    0.083      4.210E+19

Notice that my table does not quite reach |G|, so there are probably a
few more positions than this at 20q, 21q, and 22q from Start (there
can't be more any closer to Start than that). Also, the branching factor
probably does not remain constant at 9.333 all the way out to 19q from
Start; it probably declines slightly, maybe to 9.300 or so.  Finally,
the distribution is probably bimodal, with modes at 20q and 21q (it
almost has to be bimodal because of odd/even parity considerations).

(By the way, I am making no claim whatsover that the diameter of the
cube group is 24q.  This is only an educated guess based on the evidence
at hand.  In fact I tend to doubt it.  I think the branching factor in
the chart just drops off too sharply at levels 21q, 22q, and 23q for
the chart to be real.)

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990

------------------------------

Date: Sun, 5 Oct 1997 18:54:32 -0400
From: michael reid <reid@math.brown.edu>
To: cube-lovers@ai.mit.edu, randall@theory.lcs.mit.edu
Subject: Re: God's Number

keith randall writes

>    Don Dailey, Aske Plaat, and myself have a program that will do a
> complete 22-ply search in about 24 hours on an 8 processor Sun
> machine.  The program measures distance in the QT (quarter-turn)
> metric.

wow, that's quite a bit faster than my optimal solver!  how about searches
through other depths (20q, 21q, 23q, ... )?  does the run time depend upon
the input position?  could you describe your searching algorithm?  i'm
sure that this would be of interest to many people on the cube-lovers
mailing list.

> By the way, below are solutions and depths for all of the symmetric
> cubes enumerated by Hoey and Saxe in their message of Sun, 14 Dec 80.

i already posted data for these positions, but it's always nice to have
confirmation.  however, ...

> 22    girdleflip * pons asinorum
>               F F U L F L' D' R L' U' L L U U R F' B D' F' U R' D'

this is solvable in 18q:

) 3.  U  R  U' F  D  R  L' B' L' F  R  F  B' U' L' D  B' D'          (18q, 18f)

although i gave it in a different orientation.

> 22    plummer * six-H * superflip
>               F U D F' R L F U D R R L L B' U D F' R L B' R' L

there's a slight typo here; the last twist should be  L' .

mike

------------------------------

End of Cube-Lovers Digest
*************************

From cube-lovers-errors@mc.lcs.mit.edu  Tue Oct  7 13:06:59 1997
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Mail-from: From nichael@sover.net Tue Oct  7 00:45:56 1997
Message-Id: <v03010d00b05f54faf661@[204.71.18.82]>
In-Reply-To: <06Oct1997.222732.Cube-Lovers@AI.MIT.EDU>
Date: Mon, 6 Oct 1997 23:03:54 -0400
To: Cube-Lovers@ai.mit.edu
From: Nichael Lynn Cramer <nichael@sover.net>
Subject: Re: 4x4x4 solution -- [Digest v23 #171]
Cc: bmain@caddscan.com

[ Moderator's note--The subject is misleading, because I erroneously 
  titled Digest v23 #171 as "4x4x4 solutions".  It was actually about
  "Pieces of broken cubes"--the discussion of breakability and the idea
  of trading pieces of cubes.  I regret the error. ]

>To: cube-lovers@ai.mit.edu
>From: "Bryan Main" <bmain@caddscan.com>
>Subject: Re: My Revenge is Complete

>How stable are the 4x4x4 and 5x5x5?  I was thinking on getting one but they
>cost quite a lot of money and was wondering how easy it is to break them.

5Xs are pretty stable; each side has a fixed center piece (i.e. like a 3X).
I've had three and never had any problem with any of them.

4Xs are another ballgame altogether.  Since they don't have a fixed center,
they depend on an internal configuration, consisting of a cluster of four
plates, to hold the faces on.

Each of these plates is held on with a screw and this adjustment is
_critical_.  Too  tight and it can be all but impossible to twist the
faces; too loose and the cube tends to dissolve in your hands.

I've owned four; one was fine, one was OK/usable, one was too stiff to use
and one couldn't be kept together.  So, the "usability" rate was approx
1/3.  (OTOH I picked them all up for $2/ea at a ToysRU clearance...)

Nichael
nichael@sover.net                      6.501
http://www.sover.net/~nichael/            -- the ln of the Beast

From cube-lovers-errors@mc.lcs.mit.edu  Tue Oct  7 17:04:52 1997
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Mail-from: From jbryan@pstcc.cc.tn.us Tue Oct  7 16:59:58 1997
Date: Tue, 07 Oct 1997 16:59:25 -0400 (Eastern Daylight Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Maximality Analysis Through 11q
To: cube-lovers@ai.mit.edu
Message-Id: <Pine.WNT.3.96.971007165003.-773443A-100000@GN209A.PSTCC.CC.TN.US>

Not too long ago, I reported that my Shamir program had completed
searching through 11q from Start, that the results did confirm my
previous results using tape spinning programs, that no local maxima were
found 11q from Start, and that otherwise nothing new was found.  I have
come to realize that there is a small bit of new information.  I really
should post the maximality analysis in its entirety, because the whole
row 11q from Start is new.  The row 11q from Start does include the
failure to find any new local maxima.  As always, the local maxima are
in the right-most column, where all 12 moves go closer to Start.


                      Maximalility Analysis
               In Terms of Patterns (M-conjugacy classes)


                Number of Moves which go Closer to Start
                                                                  1 1 1
    0          1         2        3      4     5    6   7   8  9  0 1 2
|x|
 0  1          0         0        0      0     0    0   0   0  0  0 0 0
 1  0          1         0        0      0     0    0   0   0  0  0 0 0
 2  0          2         3        0      0     0    0   0   0  0  0 0 0
 3  0         20         4        1      0     0    0   0   0  0  0 0 0
 4  0        182        34        2      1     0    0   0   0  0  0 0 0
 5  0       1677       280       20      1     0    0   0   0  0  0 0 0
 6  0      15642      2561      184      8     0    0   0   0  0  0 0 0
 7  0     145974     23773     1721     61     0    0   0   0  0  0 0 0
 8  0    1362579    222235    16241    663     1    3   0   3  0  0 0 0
 9  0   12719643   2077549   153026   5954    74   15   2   3  0  0 0 0
10  0  118711701  19418503  1438825  58862   925  318  11  37  0  8 0 4
11  0 1107594690 181433604 13517370 576891 11843 3442 251 321 10 21 2 0





                      Maximalility Analysis
                      In Terms of Positions


                Number of Moves which go Closer to Start

    0           1          2         3        4      5     6     7
|x|
 0  1           0          0         0        0      0      0     0
 1  0          12          0         0        0      0      0     0
 2  0          96         18         0        0      0      0     0
 3  0         912        144        12        0      0      0     0
 4  0        8544       1368        96        3      0      0     0
 5  0       80088      12816       912       24      0      0     0
 6  0      749376     120612      8640      252      0      0     0
 7  0     7001712    1135104     82152     2664      0      0     0
 8  0    65391504   10645824    777936    28200     48     56     0
 9  0   610499652   99666528   7338720   280800   3048    624    96
10  0  5698027296  931905180  69049264  2796978  43800  12336   528
11  0 53164171632 8708296416 648777868 27618360 563880 159024 11904



                   1   1   1
         8    9    0   1   2
|x|
 0       0    0    0   0   0
 1       0    0    0   0   0
 2       0    0    0   0   0
 3       0    0    0   0   0
 4       0    0    0   0   0
 5       0    0    0   0   0
 6       0    0    0   0   0
 7       0    0    0   0   0
 8      27    0    0   0   0
 9     108    0    0   0   0
10    1296    0  138   0  42
11   14856  408  828  72   0


 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990

From cube-lovers-errors@mc.lcs.mit.edu  Wed Oct  8 12:48:49 1997
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Mail-from: From dzander@solaria.sol.net Tue Oct  7 19:21:35 1997
From: Douglas Zander <dzander@solaria.sol.net>
Message-Id: <199710072320.SAA09034@solaria.sol.net>
Subject: broken rubik's cube: help!
To: cube-lovers@ai.mit.edu (cube)
Date: Tue, 7 Oct 97 18:20:57 CDT

Hello,
  I wonder if someone can suggest a way to fix my 3x3x3 cube.  The screw
  that holds a center cubie to the spindle has stripped out of the spindle.
  I thought of just super-glueing it back in; would this work?  Also, I
  wonder if there is to be any tension (compression) on the spring inside
  the center cubie when the screw is set in?  I'm afraid that I will have to
  open up the center cubie and use a driver to screw the cube together
  again.  I don't want to pry open my center cubie.
  Thanks for any suggestions.

-- 
 Douglas Zander                |
 dzander@solaria.sol.net       |
 Milwaukee, Wisconsin, USA     |


From cube-lovers-errors@mc.lcs.mit.edu  Tue Oct 14 12:51:26 1997
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Mail-from: From jbryan@pstcc.cc.tn.us Mon Oct 13 16:19:54 1997
Date: Mon, 13 Oct 1997 16:18:30 -0400 (Eastern Daylight Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: God's Number
In-Reply-To: <3.0.32.19970930192820.006ce8ac@po9.mit.edu>
To: Dennis Okon <dokon@mit.edu>
Cc: cube-lovers@ai.mit.edu
Message-Id: <Pine.WNT.3.96.971013161347.-780495Z-100000@GN209A.PSTCC.CC.TN.US>

On Tue, 30 Sep 1997, Dennis Okon wrote:

> I just found out that Keith Randall for the theory group of LCS (Lab for
> Computer Science) at MIT gave a talk Monday about God's number for the
> rubik's cube.  He upped the lower bound 24 and gave "evidence" that it is
> 24.  I don't know what moves he was counting (e.g. slice, quarter).
> Unfortunately, I missed it.  Does anyone have any information on this?
> I'll see what I can find out.

Was there ever any more information on this?

The lower bound for the diameter of the cube group was raised to 24q on
19 February 1995.  I would be very surprised if Keith Randall presented
a position requiring 24f.  I don't know of any published results in
metrics which include both slice and face turns.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990

From cube-lovers-errors@mc.lcs.mit.edu  Tue Oct 14 18:44:16 1997
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Mail-from: From fb91@dial.pipex.com Tue Oct 14 08:00:33 1997
Message-Id: <199710141200.IAA10374@life.ai.mit.edu>
From: "Richard Armitage" <fb91@dial.pipex.com>
To: <Cube-Lovers@ai.mit.edu>
Subject: VRML puzzles/newsletter
Date: Tue, 14 Oct 1997 12:59:59 +0100

We are shortly to create a full VRML site of cubes, and other similar
puzzles a la Rubiks and spacecubes.  It will contain both free and for sale
items and will evolve as demand requests from people like you!!

I am going to be publishing a monthly newsletter from November 1997,
covering 3D puzzles (real and virtual) and SpaceCubes news.  You can sign
up from the first page of our website or by sending an e-mail to
info@spacecubes.com with the subject newsletter.  You will receive a
SpaceCubes info standard letter for now until we set up all the right
autoresponses but I wiil  happily deal with all feedback.

Thankyou and looking forward to giving you good challenges

Richard
Richard Armitage (SpaceCubes Marketing)
tel:44 191 281 6011 US fax 2125048016
autorespond:<mailto:info@spacecubes.com>
<mailto:Richard.Armitage@dial.pipex.com> or <Query@spacecubes.com>
<http://www.spacecubes.com>

From cube-lovers-errors@mc.lcs.mit.edu  Thu Oct 23 12:00:22 1997
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Mail-from: From whuang@ugcs.caltech.edu Thu Oct 23 07:05:58 1997
To: Cube-Lovers@AI.MIT.Edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Magic Make-a-cube
Date: 23 Oct 1997 11:05:18 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <62nb1e$q4a@gap.cco.caltech.edu>

I have just acquired what appear to be the components to Rubik's Magic
Make-a-cube.  Unfortunately, two color paper pieces are missing.  Can 
anyone tell me what the color arrangements are, and in what order? Thanks
for anything you can dig up.


--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
"[Lucy's eyes] look like little round dots of India ink..." -- Charlie Brown

From cube-lovers-errors@mc.lcs.mit.edu  Fri Oct 31 11:43:29 1997
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Mail-from: From MO374@cnsvax.albany.edu Wed Oct 29 11:06:48 1997
Date: Wed, 29 Oct 1997 11:02:53 -0500 (EST)
From: Mary Osielski <MO374@cnsvax.albany.edu>
Subject: Where to buy one???
To: cube-lovers@ai.mit.edu
Message-Id: <01IPDKFAISLE90NIU9@cnsvax.albany.edu>

     I'm trying to buy a regular, standard, run-of-the-mill Rubik's cube
which I now realize is not so easy.  Can you please direct me to a
source?  Are they no longer produced?  I got your address from the
mountains of material on the Internet about Rubik.  Is there a store,
a phone number, a person from whom I can buy one.  I'm in Albany, NY
but mail-order is fine.  Thanks in advance for the help!
       Mary Osielski
       mo374@cnsvax.albany.edu

From cube-lovers-errors@mc.lcs.mit.edu  Fri Oct 31 12:09:35 1997
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Mail-from: From mouse@Rodents.Montreal.QC.CA Thu Oct 30 12:30:33 1997
Date: Thu, 30 Oct 1997 12:29:32 -0500 (EST)
From: der Mouse  <mouse@rodents.montreal.qc.ca>
Message-Id: <199710301729.MAA08700@Twig.Rodents.Montreal.QC.CA>
To: cube-lovers@ai.mit.edu
Subject: Re: A* versus IDA*

It's a little off-topic and rather old (June 1st) anyway, so I'll make
this quick:

> [...discussion of FreeCell and "Baker's game"...]

Could someone interested in these contact me?  I'd like to learn more
about Baker's game, whatever that is, and discuss some empirical
results with with Seahaven.

					der Mouse

			       mouse@rodents.montreal.qc.ca
		     7D C8 61 52 5D E7 2D 39  4E F1 31 3E E8 B3 27 4B

From cube-lovers-errors@mc.lcs.mit.edu  Fri Oct 31 12:45:42 1997
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Mail-from: From mouse@Rodents.Montreal.QC.CA Thu Oct 30 19:12:43 1997
Date: Thu, 30 Oct 1997 19:11:51 -0500 (EST)
From: der Mouse  <mouse@rodents.montreal.qc.ca>
Message-Id: <199710310011.TAA10960@Twig.Rodents.Montreal.QC.CA>
Mime-Version: 1.0
Content-Type: text/plain; charset="iso-8859-1"
Content-Transfer-Encoding: 8bit
To: cube-lovers@ai.mit.edu
Subject: Re: Categorization of cube solving programs

This is a response to a pretty old message:

> Date: Thu, 5 Jun 1997 22:56:56 -0400 (EDT)

However, I kept the message around, which usually means I never did
anything with it.  If I already did, my apologies to the list for
duplication.

> Since I'm interested in such things, I came up with the following
> categories of cube solving programs in general order of increasing
> sophistication:

> Class 1:	Simply provide a simulation of the cube and allow the
> 		user to manipulate the cube model [...].  Often these
> 		programs have very nice 3D graphics.
> Class 2:	A program which solves the cube by implementing a
> 		canned algorithm (or 'book procedure').  [...]
> Class 3:	A program that when given a specific instance of the
> 		cube, attempts to 'discover' or learn a sequence which
> 		will solve that particular instance.  [eg, Kociemba]
> Class 4:	A program which attempts to discover an ALGORITHM to
> 		solve ALL randomized cubes.  [...]  Korf wrote a
> 		program to do this in the mid 1980s.  [Such programs
> 		generally produce Class-2-ish solutions.]  I believe
> 		Korf's program is the only program ever achieved that
> 		can be placed in this category.

I wish to speak to the last sentence of the Class 4 description.  Back
in my larval stage (mid-'80s), someone at a lab I worked for build a
Class 4 program in Franz Lisp.  It wasn't fast, but that was probably
because it had nothing more than a VAX-11/780 to run on.  (I remember
it particularly as it was one of the most impressive pieces of hot-spot
optimization I ever did; replacing about 20 lines of Lisp with about 20
lines of assembly got a speedup of between two and three orders of
magnitude overall.)

I have no idea whether the program still exists in any form.  I do
believe I can still reach its author, if anyone would like me to
inquire.

					der Mouse

			       mouse@rodents.montreal.qc.ca
		     7D C8 61 52 5D E7 2D 39  4E F1 31 3E E8 B3 27 4B

From cube-lovers-errors@mc.lcs.mit.edu  Fri Oct 31 21:19:56 1997
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Mail-from: From mouse@Rodents.Montreal.QC.CA Thu Oct 30 19:28:57 1997
Date: Thu, 30 Oct 1997 19:28:16 -0500 (EST)
From: der Mouse  <mouse@rodents.montreal.qc.ca>
Message-Id: <199710310028.TAA11074@Twig.Rodents.Montreal.QC.CA>
To: cube-lovers@ai.mit.edu
Subject: Re: 5x5x5 Stuctural Integrtity

> Where are you able to find 5x5x5 cubes that don't instantly fall
> apart?

I've owned only one 5-Cube and have had no mechanical problem with it
at all.  I bought it in mid-December 1993, but unfortunately I don't
know where it came from.  I probably got it at a retail toy/game store
here in the city called Valet de Coeur ("Jack of Hearts" in French),
but (a) am not sure of even that by now (though I have trouble
imagining where else might have had it) and (b) I have no idea where it
was made or what distributor they got it from.

> The orange stickers seem to have a habit of fleeing the cube in
> terror.  (It's always the orange ones on any cube that fall off
> first.  Has anyone else noticed this?)

I sure have, with my 5-Cube.  Three of the 25 have come off, and one
has been completely lost (the other two are attached to the cube with a
piece of masking tape, pending my doing something more permanent).  I
may do to it what I did to one of my 3-Cubes recently: take all the
stickers off and use plastic-model paint to color the cubies.  (I
actually may do this to just the orange face, since that's the only
problematic one.)

					der Mouse

			       mouse@rodents.montreal.qc.ca
		     7D C8 61 52 5D E7 2D 39  4E F1 31 3E E8 B3 27 4B

From cube-lovers-errors@mc.lcs.mit.edu  Fri Oct 31 22:06:26 1997
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Mail-from: From jferro@knave.ece.cmu.edu Fri Oct 31 13:50:54 1997
Date: Fri, 31 Oct 1997 13:50:10 -0500
From: "Jonathan R. Ferro" <jferro@knave.ece.cmu.edu>
Message-Id: <199710311850.NAA26736@knave.ece.cmu.edu>
Organization: Electrical and Computer Engineering, CMU
To: cube-lovers@ai.mit.edu
In-Reply-To: <01IPDKFAISLE90NIU9@cnsvax.albany.edu> (message from Mary Osielski on Wed, 29 Oct 1997 11:02:53 -0500 (EST))
Subject: Re: Where to buy one???

"Mary" == Mary Osielski <MO374@cnsvax.albany.edu> writes:
Mary>  I'm trying to buy a regular, standard, run-of-the-mill Rubik's
Mary> cube which I now realize is not so easy.  Can you please direct me
Mary> to a source?  Are they no longer produced?

There has been a new run (I'm not sure if it's by Ideal or not), and I
saw two on the shelf under the Lego Brand Construction Blocks (tm) (Note
to self: kill the lawyers) at K-Mart just last week.

-- Jon

From cube-lovers-errors@mc.lcs.mit.edu  Sat Nov  1 22:17:34 1997
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Mail-from: From SCHMIDTG@iccgcc.cle.ab.com Sat Nov  1 20:01:52 1997
From: SCHMIDTG@iccgcc.cle.ab.com
Date: Sat, 1 Nov 1997 20:01:05 -0500 (EST)
To: cube-lovers@ai.mit.edu
Message-Id: <971101200105.20201302@iccgcc.cle.ab.com>
Subject: Re: Categorization of cube solving programs

"der Mouse" wrote:

>This is a response to a pretty old message:

>> Since I'm interested in such things, I came up with the following
>> categories of cube solving programs in general order of increasing
>> sophistication:

>[...Class 1 through Class 2...]

> Class 3:	A program that when given a specific instance of the
> 		cube, attempts to 'discover' or learn a sequence which
> 		will solve that particular instance.  [eg, Kociemba]
> Class 4:	A program which attempts to discover an ALGORITHM to
> 		solve ALL randomized cubes.  [...]  Korf wrote a
> 		program to do this in the mid 1980s.  [Such programs
> 		generally produce Class-2-ish solutions.]  I believe
> 		Korf's program is the only program ever achieved that
> 		can be placed in this category.

In retrospect, Class 4 programs are not necessarily more sophisticated
than Class 3 programs especially when one considers that the latter
should be be able to produce a macro-table solution by solving for
a sufficient set of specific sequences.  Perhaps, I'm overly fascinated
by a learning program which, in essence, outputs a solving program but
I don't want to discount the fact that there are some very interesting
and sophisticated Class 3 programs out there.

Richard Korf points out a suggestion by Jon Bently that the learning
program can be be interleaved with the solving program, as co-routines,
and only running the learning program when a new macro is needed to
solve a particular problem instance.  Thus, the specific entries
required in the macro-table do not have to be planned out in advance.

>I wish to speak to the last sentence of the Class 4 description.  Back
>in my larval stage (mid-'80s), someone at a lab I worked for build a
>Class 4 program in Franz Lisp.  It wasn't fast, but that was probably
>because it had nothing more than a VAX-11/780 to run on.  (I remember
>it particularly as it was one of the most impressive pieces of hot-spot
>optimization I ever did; replacing about 20 lines of Lisp with about 20
>lines of assembly got a speedup of between two and three orders of
>magnitude overall.)

>I have no idea whether the program still exists in any form.  I do
>believe I can still reach its author, if anyone would like me to
>inquire.

It would be interesting to compare the approach of this program to
Korf's learning program.  If the program is still available I suggest
it would make a quite excellent addition to the cube lovers archive.

Regards,

-- Greg Schmidt

From cube-lovers-errors@mc.lcs.mit.edu  Mon Nov  3 12:42:36 1997
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Mail-from: From mouse@Rodents.Montreal.QC.CA Sun Nov  2 06:52:25 1997
Date: Sun, 2 Nov 1997 06:51:36 -0500 (EST)
From: der Mouse  <mouse@rodents.montreal.qc.ca>
Message-Id: <199711021151.GAA27954@Twig.Rodents.Montreal.QC.CA>
To: cube-lovers@ai.mit.edu
Subject: Re: Categorization of cube solving programs

>> Class 3:	A program that when given a specific instance of the
>> 		cube, attempts to [solve it]  [eg, Kociemba]
>> Class 4:	A program which attempts to [find an algorithm to solve
>> 		arbitrary cubes].

> In retrospect, Class 4 programs are not necessarily more
> sophisticated than Class 3 programs especially when one considers
> that the latter should be be able to produce a macro-table solution
> by solving for a sufficient set of specific sequences.

Sure...but who picks the specific instances for them?

> Richard Korf points out a suggestion by Jon Bently that the learning
> program can be be interleaved with the solving program, as
> co-routines, and only running the learning program when a new macro
> is needed to solve a particular problem instance.

This means that the solving program has to imagine macros, try to
choose a useful one, determine whether it's actually possible (you
gotta keep the program from trying to produce, for example, a single
edge flipper).  You also have to decide when it's worth trying for a
macro and when it's better to just hit the (sub)problem with brute
force.  I would expect all these problems to be quite hard.

>> I wish to speak to the last sentence of the Class 4 description.
>> Back in my larval stage (mid-'80s), someone at a lab I worked for
>> build a Class 4 program in Franz Lisp.  [...]

>> I have no idea whether the program still exists in any form.  I do
>> believe I can still reach its author, if anyone would like me to
>> inquire.

> It would be interesting to compare the approach of this program to
> Korf's learning program.  If the program is still available I suggest
> it would make a quite excellent addition to the cube lovers archive.

I'll send off a missive to the author.

					der Mouse

			       mouse@rodents.montreal.qc.ca
		     7D C8 61 52 5D E7 2D 39  4E F1 31 3E E8 B3 27 4B

From cube-lovers-errors@mc.lcs.mit.edu  Mon Nov  3 13:18:15 1997
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Mail-from: From nbodley@tiac.net Sun Nov  2 20:32:15 1997
Date: Sun, 2 Nov 1997 20:31:16 -0500 (EST)
From: Nicholas Bodley <nbodley@tiac.net>
To: der Mouse <mouse@rodents.montreal.qc.ca>
Cc: cube-lovers@ai.mit.edu
Subject: 5^3 orange stickers
In-Reply-To: <199710310028.TAA11074@Twig.Rodents.Montreal.QC.CA>
Message-Id: <Pine.BSF.3.96.971102202540.20000H-100000@shell2.tiac.net>

 I think these might be made of plastic instead of paper, and they seem to
have a different adhesive. I thought mine were loose because I had tried
several lubricants on my cube, and the lube. had interacted with the
adhesive; apparently not. Someone who's smart with solvents might be able
to remove all the adhesive, and reattach them with a better adhesive.
CA ("Krazy Glue"; cyanoacrylate) might be good, as might plastic-model
cement. However, one should be careful; a 5^3 is not something to
mistreat!

My regards to all,

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*    Waltham is now in the new 781 area code.
|*  Amateur musician  *|*   617 will be recognized until 1 Dec. 1997.
--------------------------------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Mon Nov  3 14:03:20 1997
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Mail-from: From bmain@caddscan.com Mon Nov  3 10:09:08 1997
To: cube-lovers@ai.mit.edu
From: "Bryan Main" <bmain@caddscan.com>
Subject: Re: Where to buy one???
Date: Mon, 03 Nov 1997 10:07:11 EST
Message-Id: <19971103100711.0054df7a.in@caddscan.com>

At 01:50 PM 10/31/97 -0500, you wrote:
>"Mary" == Mary Osielski <MO374@cnsvax.albany.edu> writes:
>Mary>  I'm trying to buy a regular, standard, run-of-the-mill Rubik's
>Mary> cube which I now realize is not so easy.  Can you please direct me
>Mary> to a source?  Are they no longer produced?
>
>There has been a new run (I'm not sure if it's by Ideal or not), and I
>saw two on the shelf under the Lego Brand Construction Blocks (tm) (Note
>to self: kill the lawyers) at K-Mart just last week.

The new ones, at least the ones that I've gotten in the last year or less,
are made by Oddz-on (sp?).  I think that they still make them but I haven't
looked in a few months.  I called them a few months ago to see if they had
plans to make a 4x4x4 but they said no.  Also they did make 2x2x2's for
awhile but I don't think they do anymore, plus the 2's were hard to rotate
and fell apart eaisly.

bryan
__________________________________________________________________
Bryan Main
Cartographic Specialist
http://caddscan.com
CADDScan Engineering Inc.
NOAA Site Number: 301-713-0388 X 110

From cube-lovers-errors@mc.lcs.mit.edu  Mon Nov  3 19:04:36 1997
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Date: Mon, 3 Nov 1997 13:48:19 -0700 (MST)
From: cube-lovers-request@ai.mit.edu
To: cube-lovers@ai.mit.edu
Reply-To: Paul Hart <hart@iserver.com>
Subject: Auction on Rubik's Revenge (4x4x4) cubes

Paul Hart <hart@iserver.com> has announced he has 6 unopened Rubik's
Revenge cubes for sale to the highest bidder.  The Cube-lovers list
will not include details of the offer; I am passing this information
on only because a number of persons on this list have asked about
finding Rubik's Revenge cubes, apparently without success.  Contact
hart@iserver.com for any further information.  As always, beware of
fraud.

                Dan Hoey, Interim moderator
                Cube-Lovers-Request@AI.MIT.Edu

From cube-lovers-errors@mc.lcs.mit.edu  Mon Nov  3 19:40:01 1997
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Mail-from: From dzander@solaria.sol.net Mon Nov  3 18:52:31 1997
From: Douglas Zander <dzander@solaria.sol.net>
Message-Id: <199711032351.RAA14876@solaria.sol.net>
Subject: Re: Where to buy one???
To: bmain@caddscan.com (Bryan Main)
Date: Mon, 3 Nov 97 17:51:42 CST
Cc: cube-lovers@ai.mit.edu (cube)
In-Reply-To: <19971103100711.0054df7a.in@caddscan.com> from "Bryan Main" at Nov 3, 97 10:07:11 am

 Can you comment how good the new cubes from Oddz-on rotate?  Are they
 smooth and slick like the original Rubik's Cubes were or hard to turn like
 the knock-offs were?
--
 Douglas Zander                |
 dzander@solaria.sol.net       |
 Milwaukee, Wisconsin, USA     |


From cube-lovers-errors@mc.lcs.mit.edu  Tue Nov  4 14:24:17 1997
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Mail-from: From SCHMIDTG@iccgcc.cle.ab.com Tue Nov  4 01:38:07 1997
From: SCHMIDTG@iccgcc.cle.ab.com
Date: Tue, 4 Nov 1997 1:37:40 -0500 (EST)
To: cube-lovers@ai.mit.edu
Message-Id: <971104013740.202034d6@iccgcc.cle.ab.com>
Subject: Re: Categorization of cube solving programs

"der Mouse" wrote:

>>> Class 3:	A program that when given a specific instance of the
>>> 		cube, attempts to [solve it]  [eg, Kociemba]
>>> Class 4:	A program which attempts to [find an algorithm to solve
>>> 		arbitrary cubes].
>
>> In retrospect, Class 4 programs are not necessarily more
>> sophisticated than Class 3 programs especially when one considers
>> that the latter should be be able to produce a macro-table solution
>> by solving for a sufficient set of specific sequences.
>
>Sure...but who picks the specific instances for them?

See below...

>> Richard Korf points out a suggestion by Jon Bently that the learning
>> program can be be interleaved with the solving program, as
>> co-routines, and only running the learning program when a new macro
>> is needed to solve a particular problem instance.
>
>This means that the solving program has to imagine macros, try to
>choose a useful one, determine whether it's actually possible (you
>gotta keep the program from trying to produce, for example, a single
>edge flipper).  You also have to decide when it's worth trying for a
>macro and when it's better to just hit the (sub)problem with brute
>force.  I would expect all these problems to be quite hard.

Although I haven't verified this with Richard Korf, I think there
is a very simple approach to this.  Consider each cubie to have one
of two states, either "fixed" or "don't care".  Initially, all cubies
are in the "don't care" state.  If a cubie state is "don't care" then
that means we disregard it's position (i.e. location and orientation)
in the target state for a particular macro.

Number all 20 of the corner and edge cubies.  Now perform the following
"Pidgin C" algorithm:

Mark all cubies[1 through 20] as "don't care" in current_cube_state

for (i = 1 to 20)
{
  target_state = cubies 1 through i in proper home cubicle position
    and marked as "fixed", all other cubies are in a "don't care" state

  Construct a unique macro index =
    f(IN = current_cubie_position[i], IN = desired_cubie_position[i])

  if (the macro at "index" doesn't exist)
  {
    Class_3_Solve(IN = current_cube_state, IN = target_cube_state, OUT = macro)
      add the new "macro" to the macro table at "index"
  }

  Apply the macro to our current_cube_state

  Mark cubie[i] as "fixed" in current_cube_state
}

Note: Class_3_solve must be able to accept an initial and goal state
augmented with the "fixed" and "don't care" markings and should honor
the constraints implied by them.  To put it another way, if a cubicle
is marked as "don't care" then a valid target state allows this
cubie to be placed in any other cubicle not currently occupied by a
"fixed" cubie.  Not really a big deal for any search procedure as we
are simply relaxing the goal state condition to a partial match rather
than requiring an exact match.

So we start out by solving for one cubie only and ignore the effect
this has on the remaining 19 cubies.  We continue doing this, each
time successively fixing another cubie and ignoring the rest, until
all cubies are finally in place.  For any valid cube configuration,
we are always guaranteed to find a macro that can solve this subproblem.
Actually, we will never fully iterate to all 20 cubies since it
is impossible to move just a single cubie.  For example, the very last
subproblem for cubie #19 might be an edge flip.  Once we've discovered
and applied the appropriate macro for this particular edge flip
we will have also flipped the #20 cubie and placed the cube in its
solved configuration.

Initially, the macros are very easy to find since most cubies
can be relocated.  At the very end, we can only move very few
cubies, and the macros are more difficult.  But a class 3 program
can solve any cube and thus can find even the most difficult
macros (e.g. an edge flip).

Eventually, once we've solved enough cube instances, our macro table
will be complete and all future cubes can be solved via macro table
lookup without the aid of the solving portion of the program.

Regards,

-- Greg

From cube-lovers-errors@mc.lcs.mit.edu  Thu Nov  6 18:53:55 1997
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Mail-from: From bmain@caddscan.com Thu Nov  6 10:15:24 1997
To: Richard E Korf <korf@cs.ucla.edu>
From: "Bryan Main" <bmain@caddscan.com>
Subject: Re: Where to buy one???
Cc: cube-lovers@ai.mit.edu
Date: Thu, 06 Nov 1997 10:13:37 Eastern Standard Time
Message-Id: <19971106101337.000d3578.in@caddscan.com>

At 11:26 AM 11/5/97 -0800, you wrote:
>Douglas,
> I bought an Oddz-on Cube the other day, and although I don't have a
> large basis for comparison, it seems to work pretty well.
>                         -rich

This got sent to me and I think that it was for the list so I'm forwarding
it.  On the same note I have three of these cubes and they work well but the
stickers become old fast.  They begin to come off around the edges and the
protective cover sometimes comes off.
__________________________________________________________________
Bryan Main
Cartographic Specialist
http://caddscan.com

From cube-lovers-errors@mc.lcs.mit.edu  Thu Nov  6 19:32:59 1997
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Mail-from: From tenie1@juno.com Thu Nov  6 18:13:12 1997
To: Cube-Lovers@ai.mit.edu
Subject: Better way to flip a middle edge?
Message-Id: <19971106.151149.11046.0.tenie1@juno.com>
From: tenie1@juno.com (Tenie Remmel)
Date: Thu, 06 Nov 1997 18:10:26 EST

Is there a short way to flip a middle edge cubie without disturbing the
top layer or the other middle edges?  I mean, better than replacing it
with one from the bottom and then putting it back in the correct
orientation, which takes 15 moves.

--Tenie Remmel (tenie1@juno.com)

From cube-lovers-errors@mc.lcs.mit.edu  Fri Nov  7 13:52:09 1997
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Mail-from: From cubeman@idirect.com Thu Nov  6 23:15:29 1997
Message-Id: <34629664.2B5E@idirect.com>
Date: Thu, 06 Nov 1997 23:17:40 -0500
From: Mark Longridge <cubeman@idirect.com>
To: cube lovers <cube-lovers@ai.mit.edu>
Subject: Availablility of Rubik's Cube

I've followed the thread about the availibility of Rubik's Cubes.
 
Ideal Toy is once again manufacturing Rubik's Cube for the mass market. 
New packaging (with the correct number of permutations) has been 
purchased by myself in a Canadian Toys R Us store just recently.

I know nothing of the Oddz-On cubes, but the new Ideal Toy cubes are
wonderful. The new cubes sport a new logo and brighter colours, but
they use the same colour arrangement as the Ideal Cubes of old.

There is also an official site for Rubik's Cube at http://www.rubiks.com
Unfortunately they are not answering their mail and are attracting a
mostly younger crowd. They are also using Karl Hornell's rubik's cube
java applet (sporting the incorrect colour arrangement I might add) 
without giving any mention of Karl's name. It is an exact byte for 
byte copy. Although Karl does give out the Rubik's Cube java applet
as freeware, I think he deserves credit from the Ideal web site.

As for my own web site (which does sport Karl's java applet with the
correct standard colouration, and also BEFUDDLER support!) I intend
to record the entire chronology of all the cube contests from every
country, including all the records from the World Championships.

My Rubik's Cube web page is currently http://web.idirect.com/~cubeman

If anyone has any information about the cube contests I have missed,
please email me. Thanks!

-> Mark Longridge <-
The Cubeman of the Internet

From cube-lovers-errors@mc.lcs.mit.edu  Fri Nov  7 14:22:06 1997
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Mail-from: From C.McCaig@Queens-Belfast.AC.UK Fri Nov  7 05:33:32 1997
From: C.McCaig@queens-belfast.ac.uk
Date: Fri, 07 Nov 1997 10:28:08 GMT
To: cube-lovers@ai.mit.edu
Message-Id: <009BCEF0.4DC8B4D9.41@a1.qub.ac.uk>
Subject: Re: Where to buy one???

i've noticed that here in northern ireland, there are a couple of
places selling cubes.  one is a standard copy of the rubik's cube,
and the other is called "magic cube" which has holographic stickers
on it, and the cubies are much squarer making it very difficult to
take apart.. it comes with a locking key which allows you to remove
one of the faces..  the turning mechanism is _really_ loose, too
loose in fact, but mine hasnt fallen apart.   

as an aside, i have an original cube, that my grandmother bought me
16 or 17 years ago, and it's still got all it's original stickers!

clive
---
Clive McCaig
Dept. Applied Mathematics
Queens University Belfast
Northern Ireland

From cube-lovers-errors@mc.lcs.mit.edu  Fri Nov  7 14:46:43 1997
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Mail-from: From davidbarr@iname.com Fri Nov  7 02:09:32 1997
Sender: davidb@davidb.concentric.net
Message-Id: <3462BE64.3F20493A@iname.com>
Date: Thu, 06 Nov 1997 23:08:20 -0800
From: David Barr <davidbarr@iname.com>
Organization: Medweb
To: Tenie Remmel <tenie1@juno.com>, Cube-Lovers@ai.mit.edu
Subject: Re: Better way to flip a middle edge?
References: <19971106.151149.11046.0.tenie1@juno.com>

Tenie Remmel wrote:

> Is there a short way to flip a middle edge cubie without disturbing the
> top layer or the other middle edges?  I mean, better than replacing it
> with one from the bottom and then putting it back in the correct
> orientation, which takes 15 moves.
>
> --Tenie Remmel (tenie1@juno.com)

 Take a look at http://ssie.binghamton.edu/~jirif/Mike/middle.html.  I
think this is the sequence you want:

2) R2 D2 F' R2 F D2 R D' R

This sequence flips the cubie on the front-right edge without disturbing
the upper face or the other middle edges.

--
mailto:davidbarr@iname.com
http://www.concentric.net/~Davebarr/

From cube-lovers-errors@mc.lcs.mit.edu  Fri Nov  7 18:43:53 1997
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Mail-from: From Michael.Swart@switchview.com Fri Nov  7 09:33:48 1997
Message-Id: <199711071431.JAA05030@support.switchview.com>
From: "Michael Swart" <Michael.Swart@switchview.com>
To: <Cube-Lovers@ai.mit.edu>, "Tenie Remmel" <tenie1@juno.com>
Subject: Re: Better way to flip a middle edge?
Date: Fri, 7 Nov 1997 09:26:36 -0500

> Is there a short way to flip a middle edge cubie without disturbing the
> top layer or the other middle edges?  I mean, better than replacing it
> with one from the bottom and then putting it back in the correct
> orientation, which takes 15 moves.

F' L' R F L' R T' B2 T L R' F L R' D2 F'

18 q turns 16 h turns. Guess it wasn't any better but you may notice
that this sequence also leaves the bottom intact except for one flipped
cubie at  the back down edge.

Michael Swart

From cube-lovers-errors@mc.lcs.mit.edu  Sun Nov  9 14:40:21 1997
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Mail-from: From charlied@erols.com Sun Nov  9 01:06:22 1997
Message-Id: <l03110702b08b02c04251@[207.172.128.13]>
Date: Sun, 9 Nov 1997 01:05:27 -0500
To: Cube-Lovers@ai.mit.edu
From: Charlie Dickman <charlied@erols.com>
Subject: A 4 Dimensional Rubik's Cube

About 18 months ago I sent this group an email about a Mac based program I
have written that simulates a 4 dimensional (3x3x3x3) Rubik's Cube based on
an unpublished paper by Harry Kamack and Tom Keene.

Some of you who were interested in the paper that describes the model and
the program had difficulty with the copies I sent you and, I suspect, were
unable to read it after you received it. Someone suggested that I translate
the document into HTML and this email is to let you know that I have done
that and will send either a ZIP or STUFFIT archive of the document to
anyone interested.

I know that maybe I should get a web site and put the paper there but I'm
not up for designing a web page or maintaining it. If you would like a copy
of the document and would also like to put it on your web site, let me know
that too.

The HTML version of the document consists of 36 fairly small GIFs that
illustrate the words. The STUFFIT archive is 328K and the ZIP file is 320K.
The documentation for the Mac based ZIP program claims that the file can be
successfully unZIPped on non-Mac platforms. The STUFFIT archive is
self-extracting if you're Mac enabled.

Send me an email if you are interested in either the program or the HTML
document or both. If you just want the document, tell me which format you
want. If you ask for the program I will assume you have a Mac and will send
everything in a STUFFIT sea.

Regards to all...

Charlie Dickman
charlied@erols.com

From cube-lovers-errors@mc.lcs.mit.edu  Sun Nov  9 15:59:22 1997
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Mail-from: From whuang@ugcs.caltech.edu Sun Nov  9 12:26:11 1997
To: cube-lovers@ai.mit.edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Re: Where to buy one???
Date: 9 Nov 1997 17:25:11 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <644rln$45r@gap.cco.caltech.edu>
References: <cube-lovers.009BCEF0.4DC8B4D9.41@a1.qub.ac.uk>

C.McCaig@queens-belfast.ac.uk writes:
>and the other is called "magic cube" which has holographic stickers
>on it, and the cubies are much squarer making it very difficult to
>take apart.. it comes with a locking key which allows you to remove
>one of the faces..  the turning mechanism is _really_ loose, too
>loose in fact, but mine hasnt fallen apart.   

Ah... this is a recent Taiwanese invention.

--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
"[Lucy's eyes] look like little round dots of India ink..." -- Charlie Brown

From cube-lovers-errors@mc.lcs.mit.edu  Mon Nov 10 10:58:12 1997
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Mail-from: From whuang@ugcs.caltech.edu Sun Nov  9 12:36:08 1997
To: cube-lovers@ai.mit.edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Re: Better way to flip a middle edge?
Date: 9 Nov 1997 17:35:21 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <644s8p$4er@gap.cco.caltech.edu>
References: <cube-lovers.199711071431.JAA05030@support.switchview.com>

"Michael Swart" <Michael.Swart@switchview.com> writes:
>> Is there a short way to flip a middle edge cubie without disturbing the
>> top layer or the other middle edges?  I mean, better than replacing it
>> with one from the bottom and then putting it back in the correct
>> orientation, which takes 15 moves.

>F' L' R F L' R T' B2 T L R' F L R' D2 F'

>18 q turns 16 h turns. Guess it wasn't any better but you may notice
>that this sequence also leaves the bottom intact except for one flipped
>cubie at  the back down edge.

This can be slightly improved:

D R' F D' R' L B D' R B' D R L' F'

14 quarter turns; does exactly the same thing.

--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
"[Lucy's eyes] look like little round dots of India ink..." -- Charlie Brown

From cube-lovers-errors@mc.lcs.mit.edu  Mon Nov 10 11:32:49 1997
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Mail-from: From cubeman@idirect.com Sun Nov  9 23:05:52 1997
Message-Id: <34668899.248A@idirect.com>
Date: Sun, 09 Nov 1997 23:07:53 -0500
From: Mark Longridge <cubeman@idirect.com>
To: cube lovers <cube-lovers@ai.mit.edu>
Cc: joyner.david@mathnt1.sma.usna.navy.mil
Subject: Megaminx, the 10-spot and GAP

First of all, the STANDARD colour arrangement used by Ideal Toy is as
follows:

UP      = White
DOWN    = Blue
FRONT   = Yellow
BACK    = Green
LEFT    = Red
RIGHT   = Orange

All the official Ideal Toy 2x2x2, 3x3x3 & 4x4x4 cubes used this
arrangement.
Even my 5x5x5 cube is the same.


Secondly, I have at last resolved the 10-spot pattern for the megaminx
in GAP. I created the process m1a which is the sequence of operators
to generate the 10-spot. I had no C_U operator, so it was more
difficult than I thought it would be.

To see all the gory details surf to the following URLs
(These are all GAP text files)

http://web.idirect.com/~cubeman/dodeca.txt   describes the megaminx
http://web.idirect.com/~cubeman/megaop.txt   describes operators
http://web.idirect.com/~cubeman/spot.txt     generates the 10-spot

Note that after executing spot.txt (which loads the other necessary
files) in gives the order of process m1a correctly as 5.

This generator uses all of the megaminx operators except the top and
bottom faces, so it is a pretty good test of the correctness of the
all of dodeca.txt, megaop.txt, and spot.txt

I believe this is the first simulation of the megaminx generating
the 10-spot although Dr. David Joyner is very close! His work is
more graphically interesting (using Maple to generate 3d pics
of the megaminx) but his operators to rotate the whole megaminx
are cooked. However, we have both verified that processes
m2, m3 and m3a are correct and have been graphed correctly using
Maple.

-> Mark <-

From cube-lovers-errors@mc.lcs.mit.edu  Tue Nov 11 20:07:56 1997
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Mail-from: From tim@mail.htp.net Tue Nov 11 00:07:26 1997
From: tim@mail.htp.net (Tim Mirabile)
To: cube lovers <cube-lovers@ai.mit.edu>
Subject: Re: Megaminx, the 10-spot and GAP
Date: Tue, 11 Nov 1997 05:06:29 GMT
Organization: http://www.webcom.com/timm/
Message-Id: <3467e58b.881450@mail.htp.net>
References: <34668899.248A@idirect.com>
In-Reply-To: <34668899.248A@idirect.com>

On Sun, 09 Nov 1997 23:07:53 -0500, Mark Longridge <cubeman@idirect.com> wrote:

>First of all, the STANDARD colour arrangement used by Ideal Toy is as
>follows:
>
>UP      = White
>DOWN    = Blue
>FRONT   = Yellow
>BACK    = Green
>LEFT    = Red
>RIGHT   = Orange
>...

I remember having one if the early Ideal cubes (at least I think it
was), and green was opposite blue.

Recently I bought one of those "odds-on" cubes, and within a week I
wore the plastic coating off the faces, so I decided to peel all the
stickers off and paint it using model paint.  I decided to keep the
most similar colors opposite each other (blue-green, red-orange,
yellow-white).  I find this arrangement makes things easier when
cubing under dim lighting. :)

--
Long Island chess -> http://www.webcom.com/timm/     TimM on ICC and A-FICS
The opinions of my employers are not necessarily mine and vice versa.

From cube-lovers-errors@mc.lcs.mit.edu  Tue Nov 11 20:43:51 1997
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Mail-from: From Anders.Larsson@hvi.uu.se Tue Nov 11 02:25:24 1997
Message-Id: <346807C1.ACFE4D68@hvi.uu.se>
Date: 	Tue, 11 Nov 1997 08:22:41 +0100
From: Anders Larsson <Anders.Larsson@hvi.uu.se>
To: cube lovers <cube-lovers@ai.mit.edu>
Subject: Colour arrangements (Was: Re: Megaminx, the 10-spot and GAP)
References: <34668899.248A@idirect.com>

Mark Longridge wrote:

> First of all, the STANDARD colour arrangement used by Ideal Toy is as
> follows:
>
> UP      = White
> DOWN    = Blue
> FRONT   = Yellow
> BACK    = Green
> LEFT    = Red
> RIGHT   = Orange

In front of me I hold a cube from one of the first batches from Hungary
with the following colour arrangement:

Up = white
Down = yellow
Front = blue
Back = green
Left = red
Right = orange

Does anybody know the history why this colour arrangement was changed?

BTW: Even if Ideal Toys has their own local standard, it doesn't change
the original ("correct") colour arrangement.

/Anders

--
Anders Larsson, PhD
Institute of High Voltage Research   Tel.: +46 (0)18 532702
Uppsala University                   Fax.: +46 (0)18 502619
Husbyborg                            E-mail: Anders.Larsson@hvi.uu.se
S-752 28 Uppsala, Sweden
http://www.hvi.uu.se/IFH/staff/Anders/Anders.html

From cube-lovers-errors@mc.lcs.mit.edu  Wed Nov 12 21:40:03 1997
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Mail-from: From ck1@home.com Tue Nov 11 22:26:29 1997
From: "Chris and Kori Pelley" <ck1@home.com>
To: <CUBE-LOVERS@ai.mit.edu>
Subject: Colors and other variations between brands
Date: Tue, 11 Nov 1997 22:26:17 -0500
Message-Id: <01bcef1a$bcbe2f60$da460318@CC623255-A.srst1.fl.home.com>

Most of the early "clone" cubes had the Blue/Green arrangement instead of
the Blue/White.  Most Ideal cubes seemed to have the Blue/White.  There were
exceptions, though...

I remember there were several factories where Ideal had their cubes made.
Some factories were better than others in terms of their quality.  My
favorites were the ones that said "Made in Korea" on a little peel-off gold
sticker.  Back in those days I would refer to "my Korean cube."  Believe it
or not, these all had the Blue/Green arrangement but they were genuine Ideal
cubes!  Their cubes were also the smoothest.  I still have one of them that
is in near perfect condition.  It was the cube I used in the competitions.

The other factories included Japan and Hong Kong.  The Japanese cubes seemed
more prevalent and I still have at least three of those-- all featuring the
Blue/White arrangement.

The earliest Rubik's Cube I ever saw had strange colors-- grey instead of
white and the shades of green and blue were very different from later cubes.
I don't think it was an Ideal cube.

The Blue/White arrangement definitely won out as Ideal's "standard"
arrangement since their 4x4x4 Revenge and 2x2x2 Pocket Cubes featured the
identical coloring.  Some Ideal 3x3x3 cubes were Blue/White but
"non-standard" because the Yellow/Green would be reversed (mirror image).
Who knows why these variations existed-- probably something as simple as
some factory tech switching the sticker feeds accidentally?

The new "Rubik's Cubes" made by Oddz-On are not all that great, in my
opinion.  They look shiny and great in the box, but after mild use the
stickers get ruined.  The Square-1 puzzles suffer the same fate.  Also their
turning mechanism is nowhere near the quality of the "Korean cubes."  Their
2x2x2 "Mini-Cube" as it is now called also lacks in quality compared to the
old Ideal Pocket Cubes.  Still, it warms my heart to see them back in toy
stores again!

Much better are the "Magic Cube" clones that appeared last year.  I have
purchased several of these (only $3.99 at Walgreen's!) and they turn very
smoothly.  The holographic stickers are different, but they don't wear out
like the Oddz-On cubes.  Also, mine feature the Blue/White arrangement!

I recently saw a post that Ideal is now making cubes again.  This seems
strange since I thought they went out of business, but I could be wrong.
Anybody know the real scoop?

Finally, Square-1 seems to have made a reappearance.  I thought they only
made one batch of these, but maybe they've made another lately?

Chris Pelley
ck1@home.com
http://members.home.net/ck1




From cube-lovers-errors@mc.lcs.mit.edu  Wed Nov 12 22:10:31 1997
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Mail-from: From chrono@ibm.net Wed Nov 12 00:37:48 1997
Message-Id: <34694088.5E3AC959@ibm.net>
Date: Tue, 11 Nov 1997 21:37:12 -0800
From: "Jin 'Time Traveler' Kim" <chrono@ibm.net>
Organization: The Fourth Dimension
To: cube lovers <cube-lovers@ai.mit.edu>
Subject: Rubik's Cube Color arrangements
References: <34668899.248A@idirect.com> <346807C1.ACFE4D68@hvi.uu.se>

I have serveral cubes spanning over a decade and a half in front of me.
Here's the quick color arrangements:

Rubik's Cube (circa 1982)
Up = White
Down = Yellow
Front = Blue
Back = Green
Left = Orange
Right = Red

I've owned this cube for what feels like forever.  I'm not even sure how I
even found it again, because I dug it out of some old junk after not having
a puzzle for about 5 years.  This is slightly different from the one you
described as being from Hungary.  I suspect mine is from there too, so
maybe production values weren't as high as they could be.

Rubik's Cube "4th Dimension" (Golden Toys, circa 1988) - Poor quality in my
opinion, as the stickers are paper with a clear plastic laminate, but
despite only being taken out of the box only 4 Times ever, the plastic
laminate is already peeling in spots.
Rubik's Mini Cube (OddzOn, circa 1996)
Rubik's Cube (OddzOn, circa 1997)

All of the above have the same color arrangement as what you described
below as being the "Ideal" solution, which I believe isn't the best.  I
still think that the opposite pairing of red/orange, white/yellow, and
blue/green makes for the best balanced color combination.  Not to mention
it's also the best quality with plastic stickers instead of paper.

Anders Larsson wrote:

> Mark Longridge wrote:
>
> > First of all, the STANDARD colour arrangement used by Ideal Toy is as
> > follows:
> >
> > UP      = White
> > DOWN    = Blue
> > FRONT   = Yellow
> > BACK    = Green
> > LEFT    = Red
> > RIGHT   = Orange
>
> In front of me I hold a cube from one of the first batches from Hungary
> with the following colour arrangement:
>
> Up = white
> Down = yellow
> Front = blue
> Back = green
> Left = red
> Right = orange
>
> Does anybody know the history why this colour arrangement was changed?
>
> BTW: Even if Ideal Toys has their own local standard, it doesn't change
> the original ("correct") colour arrangement.

--
Jin "Time Traveler" Kim
chrono@ibm.net
VGL Costa Mesa
http://www.geocities.com/timessquare/alley/9895
http://www.slamsite.com

From cube-lovers-errors@mc.lcs.mit.edu  Wed Nov 12 22:54:25 1997
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Mail-from: From richard_morton@icom-solutions.com Wed Nov 12 04:56:12 1997
Date: Wed, 12 Nov 1997 17:31:37 GMT
From: David Singmaster <david.singmaster@sbu.ac.uk>
To: Anders.Larsson@hvi.uu.se
Cc: cube-lovers@ai.mit.edu
Message-Id: <009BD319.4B14FD66.61@ice.sbu.ac.uk>
Subject: RE: Colour arrangements (Was: Re: Megaminx, the 10-spot and GAP)

	The colour arrangement on the early Hungarian cubes was quite random!!
I even have two examples where two faces have the same colour!!  It was not
until about 1980 that the idea of having a standardised colour pattern was
adopted and the most common was to have the opposite faces differ by yellow.
That is the opposite faces were  White - Yellow,  Blue - Green,  Red - Orange.
Rubik went to some effort to select six colours that would be maximally
distinct, but I think the yellow, red and orange tended to be too close in the
sense that either the orange was too close to the red or too close to the
yellow! 
	However, this does not completely determine the colour pattern.  Just
as with a die, there are two possible arrangements.  Conway and Guy etc.
observed that Blue, Orange and Yellow meet at a corner and they can occur
clockwise or counterclockwise, spelling  BOY  or  YOB.  Some people have
expressly asked me for one form rather than the other!
	An early anecdote, from about 1979.  A friend's son was trying to help
another friend solve his cube over the telephone.  This is a pretty formidable
task at the best of times, but their two cubes had different colour patterns,
so the son was making statements like: turn the red face, that's blue on your
cube, ....
 
DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499; 
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Thu Nov 13 13:15:59 1997
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Mail-from: From richard_morton@icom-solutions.com Wed Nov 12 04:56:12 1997
Message-Id: <199711120955.EAA01537@life.ai.mit.edu>
Date: Wed, 12 Nov 1997 04:55:28 EST
From: "Richard M Morton" <richard_morton@icom-solutions.com>
To: cube-lovers@ai.mit.edu
Subject:  Cube Colours

  Mark Longridge wrote:

  > First of all, the STANDARD colour arrangement used by Ideal Toy is as
  > follows:
  >
  > UP      = White
  > DOWN    = Blue
  > FRONT   = Yellow
  > BACK    = Green
  > LEFT    = Red
  > RIGHT   = Orange


  Is the orientation of the above fixed in some way or is it arbitrary ?


  My second cube (can't remember what happened to the first one) is a later
  edition (not sure if it is Ideal) with the same arrangement to above
  except, the orientation is different (UP is either RED or ORANGE) The
  reason I say this is that the LEFT,RIGHT,DOWN and FRONT faces have
  symbols printed in the centre as follows :


  YELLOW - signature of Erno Rubik
  WHITE - Rubik's CUBE tm
  GREEN - C*4**4 (actually uses superscript 4 for the power)
  BLUE - silhouette (of Erno Rubik)


  The symbols are designed to make the cube harder to solve - the challenge
  is to solve the cube with the centre cubes all in the correct
  orientation. I recall that there are sequences of moves that rotate pairs
  of centre cubes.


  This cube is definitely a lot stiffer than my original cube but the
  novelty of speed cubing has worn off anyway.


  Richard Morton
  (If my employers views are not necessarily those of my own, why am I
  still working here ?)
  Icom Solutions
  http://www.icom-solutions.com/offprods/default.htm

From cube-lovers-errors@mc.lcs.mit.edu  Thu Nov 13 13:57:08 1997
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Mail-from: From Geoffroy.VanLerberghe@ping.be Wed Nov 12 15:37:45 1997
Message-Id: <346A13C7.7C38@ping.be>
Date: Wed, 12 Nov 1997 21:38:32 +0100
From: Geoffroy Van Lerberghe <Geoffroy.VanLerberghe@ping.be>
To: Cube-Lovers <Cube-Lovers@ai.mit.edu>
Subject: Cubes in London

In one month I am going to London for a few days and I would like to
know where I can buy brainteasers there (mainly Rubik's cubes and
related puzzles).
Could you help me and send me all the information you have?

In Brussels, Belgium, you can (sometimes) find Magic Dodecahedron,
Pyraminx, Skewb and "555" cube at

   Dedale
   Galerie du Cinquantenaire
   Avenue de Tervuren 32
   1040 Brussels

Thank you for your help.

Geoffroy <Geoffroy.VanLerberghe@ping.be>

From cube-lovers-errors@mc.lcs.mit.edu  Thu Nov 13 14:26:35 1997
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Mail-from: From tenie1@juno.com Thu Nov 13 13:02:44 1997
To: Cube-Lovers@ai.mit.edu
Date: Thu, 13 Nov 1997 10:01:41 -0800
Subject: 6x6x6 cube design
Message-Id: <19971113.100159.5094.0.tenie1@juno.com>
From: tenie1@juno.com (Tenie Remmel)

I am attempting to design a 6x6x6 cube.  My idea to make it structurally
sound is to attach both the center cubies and the middle edge cubies to
a ball in the center.  Then all other pieces are wedged behind those.  I
think that extending from the 5x5x5 design the same way the 4x4x4 was
extended from the 3x3x3 design would be way too flimsy, mainly because
the centers would have to be attached via long, thin struts which are
apt to break easily unless made out of metal, which would make the thing
way too heavy.  The width of the cubies probably could not be more than
14 or 15 mm; if they were larger, the cube would be quite big and so it
would be difficult to manipulate.

Unfortunately the ball would be quite complicated, with six or even nine
tracks in it instead of just three as in the 4x4x4 cube.  It might have
to be made of metal instead of plastic (it shouldn't be too heavy if it
is hollow).  Also the 152 pieces will be a real pain to put together...

Of course, even if it can be built, does anyone know how to solve it?

Here is a rather crude diagram of a cross section through the center of
the cube.  Actually it is just a quarter of a cube.
------------------------------------------------------------------------

 aaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbcccccccccccccccccccc
 aaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbcccccccccccccccccccc
 aaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbbbbbbbcccccccccccccccccccc
 aaaa......aaaabbbbbbbbbbbbbbbbbbbbbbbbbbbbbbcccccccccccccccc
 aaaa..............bbbbbbbbbbbbbbbbbbbbbbbbbbcccccccccccccccc
 aaaaaaaa................bbbbbbbbbbbbbbbbbbbbcccccccccccccccc
 aaaaaaaa....................bbbbbbbbbbbbbbbbcccccccccccccccc
 ................................bbbbbbbbbbcccccccccccccccccc
 ..................................bbbbbbccccccbbbbbbbbcccccc
 ....................................bbccccccbbbbbbbbbbcccccc
 ..............................cccc..ccccccbbbbbbbbbbbbbbbbbb
 ..............................ccccccccccbbbbbbbbbbbbbbbbbbbb
 ................................cccccc....bbbbbbbbbbbbbbbbbb
 ..................................cccccc....bbbbbbbbbbbbbbbb
 ....................................cccc......bbbbbbbbbbbbbb
 ..............................................bbbbbbbbbbbbbb
 ................................................bbbbbbbbbbbb
 ................................................bbbbbbbbbbbb
 ..................................................bbbbbbbbbb
 ..................................................bbbbbbbbbb
 ..................................................bbbbbbaaaa
 ....................................................bbbbaaaa
 ....................................................bbbbaaaa
 ....................................................aaaaaaaa
 ....................................................aaaaaaaa
 ......................................................aaaaaa
 ..............................................aaaa....aaaaaa
 ..............................................aaaa....aaaaaa
 ..............................................aaaaaaaaaaaaaa
 ..............................................aaaaaaaaaaaaaa

------------------------------------------------------------------------
BTW, Does anyone have experience with TurboCAD?  Can it be used to
design this type of thing?  It sure would be easier to use a computer
program than to use graph paper.

I believe that the 6x6x6 is the largest mechanically possible, because
with the 7x7x7 and higher cubes, the corner cubies aren't attached to
anything at all!  Is this correct?

Also what is the mechanism for a 2x2x2 cube?  Could it be extended to
make a more stable 4x4x4 and/or 6x6x6 cubes...

And how about a GigaMinx, a 5x5 version of the MegaMinx magic pentagonal
dodecahedron, with five pieces on each edge, 31 pieces on each face
(5 corners, 11 edges and 11 central pieces), 242 pieces total.  I would
draw a diagram if it wasn't so hard to make a pentagon out of chars...

--Tenie Remmel (tenie1@juno.com)

[ Moderator's note:  The purported impossibility of a Rubik's 7^3 has been
  discussed and refuted repeatedly on this list, and several mechanisms
  have been proposed for it; see the archives.  It is not true that the
  corner cubies "aren't attached to anything".  Each corner will be attached
  to at least two edge cubies, though not always the same two edge cubies.
  You should also look in the archives to find descriptions of the 2^3, some
  as recently as 28 July.  Unfortunately, I haven't been able to understand
  it.  I'd like to see a clear description, as I haven't got a 2^3 handy to
  try myself.  -Dan ]

From cube-lovers-errors@mc.lcs.mit.edu  Fri Nov 14 10:37:37 1997
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Mail-from: From cubeman@idirect.com Thu Nov 13 21:03:37 1997
Date: Thu, 13 Nov 1997 19:06:53 -0500 (EST)
From: Mark Longridge <cubeman@idirect.com>
To: Richard M Morton <richard_morton@icom-solutions.com>
Cc: cube-lovers@ai.mit.edu
Subject: Re: Cube Colours
In-Reply-To: <199711120955.EAA01537@life.ai.mit.edu>
Message-Id: <Pine.BSI.3.94.971113184945.11986A-100000@hometown.idirect.com>

On Wed, 12 Nov 1997, Richard M Morton wrote:

>   Mark Longridge wrote:
> 
>   > First of all, the STANDARD colour arrangement used by Ideal Toy is as
>   > follows:
>   >
>   > UP      = White
>   > DOWN    = Blue
>   > FRONT   = Yellow
>   > BACK    = Green
>   > LEFT    = Red
>   > RIGHT   = Orange
> 
> ...
Ok folks, one last bit of info about the cube colour controversy

The colouring "standard" I was referring to was used by Canadian and
the USA cube contests. Having said that there were probably contests
where people brought there own cubes, and that would make it potpourri.

Moreover, this was stipulated in the rules of the contest. I still have
the form. The only difference between differ by yellow and the standard
Ideal cube was the transposition of yellow and blue.

There isn't really a standard orientation, save for the orientation 
I use in my own cube programs. All the Ideal cubes I have conform
to White/Blue, Yellow/Green, Red/Orange for Top/Down, Front/BACK,
Left/Right.

So I suppose it is open to interpretation. I thought David Singmaster 
might mention what colour arrangement was used in the World Championship.

So a case may be made for both "Differ by Yellow" and Ideal Contest
Colours.

Would someone like to pick one?? :-)

-> Mark <-
The Colourist

From cube-lovers-errors@mc.lcs.mit.edu  Fri Nov 14 11:12:41 1997
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Mail-from: From chrono@ibm.net Thu Nov 13 22:03:44 1997
Message-Id: <346BBF7F.ADFDE0D4@ibm.net>
Date: Thu, 13 Nov 1997 19:03:27 -0800
From: "Jin 'Time Traveler' Kim" <chrono@ibm.net>
Organization: The Fourth Dimension
To: Cube-Lovers@ai.mit.edu
Subject: Re: 6x6x6 cube design
References: <19971113.100159.5094.0.tenie1@juno.com>

Tenie Remmel wrote:

> I am attempting to design a 6x6x6 cube.  My idea to make it structurally
> sound is to attach both the center cubies and the middle edge cubies to
> a ball in the center.  Then all other pieces are wedged behind those.  I
> think that extending from the 5x5x5 design the same way the 4x4x4 was
> extended from the 3x3x3 design would be way too flimsy, mainly because
> the centers would have to be attached via long, thin struts which are
> apt to break easily unless made out of metal, which would make the thing
> way too heavy.  The width of the cubies probably could not be more than
> 14 or 15 mm; if they were larger, the cube would be quite big and so it
> would be difficult to manipulate.

> Of course, even if it can be built, does anyone know how to solve it?

If it can be built and scrambled, it can be solved.  In fact, it could
make for a very interesting puzzle since it could behave identically
to a 3x3x3 if one wanted it to, just like a 4x4x4 can be manipulated
like a 2x2x2.  Heck, the 6x6x6 could also behave like a 2x2x2...  One
puzzle could take the place of two others.  Sort of a "mix and match"
difficulty setting.  Regardless, I suspect that many would applaud the
ingenuity of a 6x6x6 if it was executed elegantly and worked well,
like the 5x5x5.

> I believe that the 6x6x6 is the largest mechanically possible, because
> with the 7x7x7 and higher cubes, the corner cubies aren't attached to
> anything at all!  Is this correct?

The moderator of the mailing list stated that a 7x7x7 cube could be
built, but I counter that it would require "cubes" of dissimilar size
or some kind of groove type scheme, which actually isn't quite in the
spirit of a cube.  Even a 6x6x6 would require some careful engineering
since the corner cubes just barely overlap.

> Also what is the mechanism for a 2x2x2 cube?  Could it be extended to
> make a more stable 4x4x4 and/or 6x6x6 cubes...

The mechanism of the 2x2x2 is similar to the 4x4x4, which makes both
of them rather stiff.

> And how about a GigaMinx, a 5x5 version of the MegaMinx magic pentagonal
> dodecahedron, with five pieces on each edge, 31 pieces on each face
> (5 corners, 11 edges and 11 central pieces), 242 pieces total.  I would
> draw a diagram if it wasn't so hard to make a pentagon out of chars...

I'm sure supersets of many existing puzzles have been considered.  I
myself spent some hours contemplating and drafting the possibility of
a pyraminx to the next level.  I called it Tut's Curse as a sort of
'project' name, despite the fact that Tut was never buried in a
pyramid.  Maybe that's why I never completed the project.  Oh well.
The best laid plans of mice and men...

--
Jin "Time Traveler" Kim
chrono@ibm.net
VGL Costa Mesa
http://www.geocities.com/timessquare/alley/9895
http://www.slamsite.com

From cube-lovers-errors@mc.lcs.mit.edu  Fri Nov 14 11:46:24 1997
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Mail-from: From cube-lovers-request@life.ai.mit.edu Fri Nov 14 10:27:59 1997
Date: Fri, 14 Nov 1997 15:25:06 GMT
From: David Singmaster <david.singmaster@sbu.ac.uk>
To: tenie1@juno.com
Cc: cube-lovers@ai.mit.edu
Message-Id: <009BD499.F2FD74E5.202@ice.sbu.ac.uk>
Subject: RE: 6x6x6 cube design

	First, regarding the 6^3 and 7^3.  As noted, when you get to
these sizes, the connection of the corners while turning becomes
problematic.  For the 6^3, the overlap is about 15% of the edge length
of the cubie, probably too small to be practicable.  One can imagine
some clever mechanism to hold onto the corners, but it would be tricky
and I've never seen one clearly described.
	However, if you think about it, there's no reason for all the
levels of the cube to be the same size.  That it, the parallel cutting
planes of the entire cube do not have to be equally spaced.  One can
thus have the corner cubies be very large with much smaller centre
cubies.  The edge cubies will be cuboids, rather than cubes.  Using
this idea, one can make arbitrarily large cubes, but the interior
pieces become impossible to manipulate.

	Now let me try my hand at describing three versions of the 2^3.
	I'll start with the simplest which was sent to me from Japan
about 1980.  This had a steel sphere in the middle and each cubie had
a magnet in it.  Although the sphere and the cubies were carefully
machined, when one moved it quickly, a piece would catch against
another piece and lift off and then fall off.  Not very successful.
	The second version was patented by Ishige in Japan about 1977?
and several versions were made.  I received a batch of seven with
different colouring patterns made by a German sports firm - three or
four had broken just in the post!  This version has a central sphere
and six of what I call 'umbrellas' sticking out toward each face
centre.  Each of the pieces has a notch around the part that rest
against the inner sphere.  The umbrellas catch into these notches.
One can also think of the cubies as having their own umbrellas, but of
triangular form and concave.  This is the same mechanism used in the
Impossiball.
	The third version is the most common and is shown in Rubik's
Hungarian patent, but is hard to interpret as I've never had the text
translated.  Basically, his 2^3 is a 3^3 with the edge and centre
pieces concealed.  I gather from earlier messages that there were
several versions of this, but I only recall one, but I only ever took
a few apart.
	At the very centre was a cube.  On each face was a square rod
extending almost to the face center.  The ends of these had a +
groove.  Between the rods were pieces in the form of a quadrant with a
groove on the outer, curved, edge.  When all these pieces are in
place, each of the midplanes of the cube is seen to contain a circle
with a groove on its outer edge.  The corner pieces are basically
hollow, but each interior face is a layer ending in a quarter-circular
curve, which fits as a tongue into the groove just mentioned.  Where
two of these meet, at the interior edge of the piece, a section is cut
away to allow the piece to slide past the projections of the end of
the square rods.
 	In theory, one might be able to avoid the quadrant pieces, but
I think they give the structure stability.
	A more serious problem is that the inner, concealed, pieces
can get out of synch with the visible pieces, The early patent of
Gustafson left gaps so one could see the inner pieces and move them.
The method used by Rubik and in some similar puzzles is to fix one
corner piece to the inner structure by some method.  Rubik's 2^3 did
this by making some of the rod ends solid rather than grooved (or
perhaps they were fixed to the central cube so they couldn't rotate).
One could also not notch one of the corner pieces.  Whatever one does,
it must have the effect of preventing one corner from moving in
relation to the inner structure.  I seem to recall that the 4^3 uses
this idea also.
	Don't know how much this helps, but that's the best I can do
off-hand.

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499; 
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Sat Nov 15 22:37:51 1997
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sat Nov 15 14:55:00 1997
Date: Sat, 15 Nov 1997 14:54:07 -0500 (EST)
From: Nicholas Bodley <nbodley@tiac.net>
Reply-To: Nicholas Bodley <nbodley@tiac.net>
To: David Singmaster <david.singmaster@sbu.ac.uk>
Cc: tenie1@juno.com, cube-lovers@ai.mit.edu
Subject: RE: 6x6x6 cube design; also notes about the 2^3 innards. (Fairly long)
In-Reply-To: <009BD499.F2FD74E5.202@ice.sbu.ac.uk>
Message-Id: <Pine.BSF.3.96.971115130333.25685A-100000@shell2.tiac.net>


 There's a short mention in passing in Douglas Hofstadter's (second
major?) book (Metamagical Themas?) to the effect that a physical
prototype exists for the 6^3, and a paper design for the 7^3. This was
ca. 1982, iirc.

On Fri, 14 Nov 1997, David Singmaster wrote:

{Snips}

}	Now let me try my hand at describing three versions of the 2^3.

}	The third version is the most common and is shown in Rubik's
}Hungarian patent, but is hard to interpret as I've never had the text
}translated.  Basically, his 2^3 is a 3^3 with the edge and centre
}pieces concealed.

 The ones I had were quite difficult to take apart and reassemble; if
they weren't made of a strong, resilient engineering plastic, they would
not have been possible to make, I would say.

}	At the very centre was a cube.

 In mine, this cube was almost tiny; perhaps 15% (along an edge) of the
size of a cubie as seen from the outside.

}  On each face was a square rod
}extending almost to the face center.

 In mine, just about sure that three adjacent faces of this inner cube
each had thin cylindrical rods extending toward the face centers. These
were surrounded by square rods of the same width as the other three
which were part of the center cube.

 The thin cylindrical rods served as pivots for the square rods of the
same width. When you rotated one half of the Cube, these pivots allowed
one half to rotate with respect to the other without prying anything
apart.

 The three fixed square rods, which are extensions and "part of" the
center cube, stayed fixed within their half of the Cube when the other
half was rotated, much as the ball inside a 4^3 stays fixed.

} The ends of these [rods -nb] had a +
}groove.  Between the rods were pieces in the form of a quadrant with a
}groove on the outer, curved, edge.  When all these pieces are in
}place, each of the midplanes of the cube is seen to contain a circle
}with a groove on its outer edge.

 The aforementioned rods are required to keep the quadrants from moving
inward and therefore out of engagement with the inner, "cut-away" edges
of the cubies. If that were to happen, the Cube would fall apart.
(Please see the next paragraph.) (When I tried to describe the innards
of a 2^3 a while back, I called these quadrants "clips". My hat's off to
Mr. Singmaster for his fluency!)

}  The corner pieces are basically
}hollow, but each interior face is a layer ending in a quarter-circular
}curve, which fits as a tongue into the groove just mentioned.  Where
}two of these meet, at the interior edge of the piece, a section is cut
}away to allow the piece to slide past the projections of the end of
}the square rods.
} 	In theory, one might be able to avoid the quadrant pieces, but
}I think they give the structure stability.

 With all due respect, without them, the Cube would instantly fall
apart! They are essential.

}	A more serious problem is that the inner, concealed, pieces
}can get out of synch with the visible pieces.

 The natural tendency is to squeeze the cubies of each half together
when maneuvering.

 Because the thin square rods molded along with the center cube are
"attached" to adjacent faces of that cube, the other three faces of that
cube carry the swiveling rods. No matter how you pick up the Cube, one
half will contain a fixed rod. Squeezing the cubies together around that
rod will make the center cube stay aligned with those four cubies that
are squeezing one of its rods. (Actually, the cubies squeeze the
quadrants, and the quadrants squeeze the rods.)

 Keeping that center cube aligned also means it will keep aligned the
four rods that have their axes in the current shear plane. These rods
will then keep the quadrants aligned with the half of the cubie that is
squeezing the fixed rod.

 The four quadrants in the swiveling half will squeeze the hollow,
swiveling rod, which will rotate around the thin cylindrical [rod] that
extends from that face of the center cube.

 I'm indebted to Mr. Singmaster for his clarifying description.

 This mechanism seems to be a real challenge to describe solely in
words! Here, a few images equal many kB of ASCII...

}DAVID SINGMASTER, Professor of Mathematics and Metagrobologist
}  email:
}zingmast or David.Singmaster @sbu.ac.uk

* * *

 Here's another go, for those who have the patience:

 Imagine that each cubie is hollow. (They really are.) Imagine that they
are separated from each other in 3-D space by moderate and equal
distances, but still not tilted with respect to each other. In other
words, there's a large gap between any two.

 Now, imagine a spherical rotary cutter, spinning in the center of the
3-D array of 8 cubies. Move the cubies toward the cutter, along radii of
the cutter passing through their outermost corners. Don't tilt or
rotate, just translate radially inward toward the cutter along a
[45-degree] axis.

 Let the cutter machine a curved outline in each of the three inner
faces. (The diameter of the cutter is maybe 80% of the edge of a
complete Cube.) Make the cutter disappear, and you have a spherical
cutaway inside the whole cluster of eight. (This is real, in essence.)

 The cubies are hollow, and they really have this curved "cut" in each
of their concealed inside faces. Of course, this was molded in, not
machined by a cutter.

 Now, you need something to hold the cubies together.

 If you've seen a radar corner reflector used by small boat owners,
think of one made of three intersecting, mutually-orthogonal circles.
They intersect at a common, center point. Make this corner reflector
tiny, maybe 3,5 (3.5) cm (?) in diameter.

 Cut this apart into eight quadrants. Make them thick, if they aren't.
Make a rectangular groove in each curved edge. Remove some material from
the straight edges; line up the curved edges with a circle (same size as
the original structure before you cut it apart) on your workbench).
Space them equally apart. The gaps form a cross (or an "X", if you like
45-degree angles).  The rods will go into those gaps.

 OK: These are now positioned the way they will be in one of the three
shear planes in a Cube.

 [The radius of the corner reflector is somewhat bigger than that of the
ball cutter.)

 Thinking back to the corner reflector, if you replace all 12 quadrants
where they used to be in 3-D space, with gaps between them as they were
on the bench, that is how they are positioned in an aligned Cube.

 To start assembling the Cube, you take four cubies, lay them down next
to each other (touching) with colors properly aligned, but with their
inside surfaces facing upward. Pick up four quadrants, placing the
grooves you made (in the curved edges) onto the curved "cutaways" in the
adjacent inside edges of the hollow cubies, where the cubies touch. As
long as these quadrants don't move toward each other, they will keep the
cubies together.

 This, in two dimensions, is what holds the Cube together. The next four
quadrants fit into the remaining cutouts. They lie flat, and form a
circle, the way they did when you laid them down on the workbench.

 To keep the quadrants away from the center, you now insert the center
cube and its rods. However, assembling the remaining four cubies (and
their four quadrants) to what you have so far, is, in the real world, a
major struggle. It involves some worrisome distortions of the pieces!
This "geometric interference" is also what makes it so hard to
disassemble.

 Wonder how these are assembled at the factory?

My best regards to all,

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|* 'T was the night before Xmas, and all through
|*  Amateur musician  *|*  the coffeehouse, not a creature was stirring.
--------------------------------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Sat Nov 15 23:21:58 1997
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Mail-from: From roger.broadie@iclweb.com Thu Nov 13 19:09:04 1997
From: roger.broadie@iclweb.com (Roger Broadie)
To: "Geoffroy Van Lerberghe" <Geoffroy.VanLerberghe@ping.be>,
        "Cube-Lovers" <Cube-Lovers@ai.mit.edu>
Subject: Re: Cubes in London, plus OddzOn, clones and colours
Date: Fri, 14 Nov 1997 00:07:33 -0000
Message-Id: <19971114000519.AAA4398@home>

London's most famous toy-shop is Hamley's in Regent St.  On the fourth
floor they have a wall of OddzOn cubes at 39.00 pounds each, together
with snakes and magics.

I say OddzOn because that is the name given in the copyright notice at the
back of the instruction booklet, although in this country that name appears
nowhere else. The packaging (a rectangular cardboard casing with a clear
central panel holding the cube at an angle) gives the distributors as
Toybrokers Ltd.  This cube sometimes appears in the British Toys 'R' Us,
and my family bought one in Jenners, the big Edinburgh department store,
this summer.  I  wouldn't rate it as highly as the Ideal cubes.

I had a quick look for clones in the sort of shop in London that I have
seen them in in the past, but found none.  I bought a couple of Taiwanese
clones in Dublin a few weeks ago - they came in a cube-sized cardboard box
with a picture of a cube on the front with two yellow centre pieces, five
green edge pieces and five red corner pieces.  I did not complain that the
cube inside did not match this picture.  I tried both the sample on display
and one of the cubes which I later bought.  They turned quite well.  I did
not try the other one until I got it home, when to my annoyance I found it
much stiffer.  The colours are rather dull, but yellow and white are
opposite, which I prefer, because then the colours of the opposite faces
seem to have a sensible connection helping recognition of a piece that is
in the opposite face to its home face. They cost 35.00 Irish pounds each .

Among various other puzzle, Hamleys also has those from Meffert, including
the skewb and an annoying dodecahedron - the colours are duplicated at
opposite poles.  So does Toys 'R' Us.  What I have not been able to find is
the 5x5x5 that is shown on the Meffert packaging.

The OddzOn cube is the one that is associated with the www.rubiks.com site,
which reproduces the instruction booklet and uses the same logo in chubby
capitals.  Rubik himself is clearly involved - he is quoted on the site.

I had concluded that Ideal Toys (the US company) had gone out of business,
having failed to find any reference to it currently.  There is a British
company Ideal Toys (UK) Limited, but that is a subsidiary of Triumph Adler
AG.  There was a company The Ideal Toy Company Limited,  but that was
dissolved, as was CBS Toys Limited, which may have been connected.

OddzOn Products Inc appears to be a subsidiary of Hasbro Inc.

I have a suspicion that Ideal may have deliberately adopted the
yellow-opposite-green configuration to create a new colour arrangement that
would help them expunge clones by relying on their trade dress.

Roger Broadie

From cube-lovers-errors@mc.lcs.mit.edu  Sun Nov 16 14:26:45 1997
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sun Nov 16 06:12:15 1997
From: roger.broadie@iclweb.com (Roger Broadie)
To: "Cube-Lovers" <Cube-Lovers@ai.mit.edu>
Subject: Re: Cubes in London
Date: Sun, 16 Nov 1997 11:07:06 -0000
Message-Id: <19971116110836.AAA18652@home>

I apparently wrote

> .. at 39.00 pounds .. 

Before people get the wrong idea about the price of cubes on this side of
the Atlantic, I'd better say that the OddZon cube was 9 British pounds in
Hamleys and the Dublin clone was 5 Irish pounds.  I used the pound symbol
and the conversion - I suspect both machine and human - went awry.

I can add to my slightly meandering note on the Dublin clone that the
central spider has now bust.  

Life is full of new hazards.

Roger Broadie

[ Sorry, you're a victim of moderator error.  While replacing the
  pound symbol with the word "pounds", and I left in an extra 3.
  Thanks for the information.  --Dan ]

From cube-lovers-errors@mc.lcs.mit.edu  Sun Nov 16 14:57:58 1997
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sun Nov 16 07:24:33 1997
Message-Id: <199711161045.KAA30105@GPO.iol.ie>
From: "Goyra (David Byrden)" <Goyra@iol.ie>
To: Cube-Lovers@ai.mit.edu
Subject: Re: Cubes in London, plus OddzOn, clones and colours
Date: Sun, 16 Nov 1997 10:43:38 -0000

> From: Roger Broadie <roger.broadie@iclweb.com>


> What I have not been able to find is
> the 5x5x5 that is shown on the Meffert packaging.

	Try writing to Dr. Christophe Banelow
	An Der Wabeck 37
	D-58456   Witten
	Germany
	tel: 49 2302 71147
	fax: 49 2302 77001

	I have his catalogue here and he lists the 5^3, Skewb,
Dedecahedron, Pyraminx, Octahedron, Magic Jewel, among
others.


					David

From cube-lovers-errors@mc.lcs.mit.edu  Mon Nov 17 00:07:22 1997
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sun Nov 16 17:31:22 1997
Date: Sun, 16 Nov 1997 17:30:08 -0500 (EST)
From: Nicholas Bodley <nbodley@tiac.net>
To: "Goyra (David Byrden)" <Goyra@iol.ie>
Cc: Cube-Lovers@ai.mit.edu
Subject: Re: Cubes in London, plus OddzOn, clones and colours
In-Reply-To: <199711161045.KAA30105@GPO.iol.ie>
Message-Id: <Pine.BSF.3.96.971116172625.27404H-100000@shell2.tiac.net>

On Sun, 16 Nov 1997, Goyra (David Byrden) wrote:

{Snips}

}
}	Try writing to Dr. Christophe Banelow
}	An Der Wabeck 37
}	D-58456   Witten
}	Germany
}	tel: 49 2302 71147
}	fax: 49 2302 77001

I hope I'm not being rude to point out minor typos; his name should be
"Christoph Bandelow". It's an easy slip to make.

Regards,

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|* 'T was the night before Xmas, and all through
|*  Amateur musician  *|*  the coffeehouse, not a creature was stirring.
--------------------------------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Mon Nov 17 21:35:42 1997
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Mail-from: From cube-lovers-request@life.ai.mit.edu Mon Nov 17 04:26:22 1997
Message-Id: <34700D89.24426818@ibm.net>
Date: Mon, 17 Nov 1997 01:25:29 -0800
From: "Jin 'Time Traveler' Kim" <chrono@ibm.net>
Organization: The Fourth Dimension
To: Cube-Lovers@ai.mit.edu
Subject: Color schemes revisited
References: <cube-lovers.Pine.SUN.3.90.970206150433.26015M-100000@dot.cs.Virginia.EDU> <5dlt1c$baq@gap.cco.caltech.edu>

An interesting thing to note regarding the color patterns on cubes...
on the Rubik's Cube home page (http://www.rubiks.com) a picture is
displayed showing Dr. Rubik himself holding a mixed cube in his hand.
On a whim I decided to figure out what the color scheme of the cube
was.

If you wish to figure out yourself without being told, or if you just
want to try to refute my guess (I'm no stranger to being wrong) then
don't read the "answer" to the puzzle that's below my .sig.
Otherwise, if you can't be bothered with minor trivialities like this
one (it's really not that difficult to figure out the colors anyway)
then read on.

--
Jin "Time Traveler" Kim
chrono@ibm.net
VGL Costa Mesa
http://www.geocities.com/timessquare/alley/9895
http://www.slamsite.com



I determined that the color scheme of the cube held in Erno Rubik's hand is:
Front: Red
Back: Orange
Left: Green
Right: Blue
Top: White
Bottom: Yellow

From cube-lovers-errors@mc.lcs.mit.edu  Mon Nov 17 22:05:10 1997
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Mail-from: From cube-lovers-request@life.ai.mit.edu Mon Nov 17 10:56:56 1997
Date: Mon, 17 Nov 1997 15:53:23 GMT
From: David Singmaster <david.singmaster@sbu.ac.uk>
To: chrono@ibm.net
Cc: cube-lovers@ai.mit.edu
Message-Id: <009BD6F9.65F6A7C1.425@ice.sbu.ac.uk>
Subject: Re: 6x6x6 cube design

	I'm sure that this has been mentioned before, but the 6^3
etc. actually introduce no further complications than present on the
4^3 (and 5^3).  There are just more types of center pieces, but they
all behave in much the same way.  In my message on notation and
solution of the 4^3, I gave a method of producing a 3-cycle of center
pieces and it can be used for each class of centre pieces - the puzzle
doesn't get any more interesting, just longer!
	The 6^3 introduces a slightly interesting feature
theoretically in that the center pieces break up into more classes
than one might initially expect because the piece at the (1,2)
location is not in the same class as the piece at the (2,1) location.
(Taking a corner as (0,0).)

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499; 
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Mon Nov 17 22:44:50 1997
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Mail-from: From cube-lovers-request@life.ai.mit.edu Mon Nov 17 10:58:38 1997
Date: Mon, 17 Nov 1997 15:57:19 GMT
From: David Singmaster <david.singmaster@sbu.ac.uk>
To: cubeman@idirect.com
Cc: cube-lovers@ai.mit.edu
Message-Id: <009BD6F9.F2D7ACBC.209@ice.sbu.ac.uk>
Subject: Re: Cube Colours

	According to what I wrote in my Cubic Circular 3/4 of Summer 1982, the
cubes used in the World Championship were of the +/- yellow BOY pattern.

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499; 
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Tue Nov 18 23:50:41 1997
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Mail-from: From cube-lovers-request@life.ai.mit.edu Tue Nov 18 23:43:40 1997
To: Cube-Lovers@ai.mit.edu
Date: Tue, 18 Nov 1997 20:42:55 -0800
Subject: Rubiks Revenge moves
Message-Id: <19971118.204255.7126.1.tenie1@juno.com>
From: tenie1@juno.com (Tenie Remmel)

Is there an easy way to cycle three adjacent top edges on the
Rubiks Revenge?  I can't find one shorter than 62 moves, but if
there was a short one I could simplify my solution greatly.

. b c .    . a b .
a . . . => c . . .
. . . .    . . . .
. . . .    . . . .

Hopefully it won't mess up the corners, but it's ok if it does.

I'd also like to see some short moves for the following 3-cycles:

. * * .    . . * .    . . . .    . . * .
. . . .    * . . *    * . . *    . . . *
* . . .    . . . .    . . . *    * . . .
. . . .    . . . .    . . . .    . . . .

Is there a good source anywhere for moves, pretty patterns, etc. for
the Rubiks Revenge?  It's quite difficult to find information about
it.  Also is there an automatic move generating program for the higher
order cubes like 'Cube Explorer' is for the 3x3x3?

--Tenie Remmel (tjr19@juno.com)

From cube-lovers-errors@mc.lcs.mit.edu  Thu Nov 20 11:46:33 1997
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Nov 19 08:29:34 1997
From: bagleyd@americas.sun.sed.monmouth.army.mil (David Bagley x21081)
Message-Id: <199711191329.IAA26271@java.sed.monmouth.army.mil>
Subject: Re: A 4 Dimensional Rubik's Cube
To: charlied@erols.com (Charlie Dickman), Cube-Lovers@ai.mit.edu
Date: Wed, 19 Nov 1997 08:29:09 -0500 (EST)
In-Reply-To: <l03110700b0969495b904@[207.172.130.170]> from "Charlie Dickman" at Nov 17, 97 08:01:03 pm

Hi All


I added Charlie Dickman's Tesseract (A 4 Dimensional Rubik's Cube) to my
web pages ( http://www.tux.org/~bagleyd/ ).  Its in two parts, the docs
(mind twisting stuff) and the Mac Program.

Charlie Dickman:  If you make any updates I'll be happy to update the pages.

By the way, I recently reorganized my web pages.... same old junk but its
presented better.  :)

-- 
Cheers,
 /X\  David A. Bagley
(( X  bagleyd@bigfoot.com  http://www.tux.org/~bagleyd/
 \X/  xlockmore and more   ftp://ftp.tux.org/pub/people/david-bagley

From cube-lovers-errors@mc.lcs.mit.edu  Thu Nov 20 12:17:58 1997
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Sender: davidb@davidb.concentric.net
Message-Id: <34735538.113A5129@iname.com>
Date: Wed, 19 Nov 1997 13:08:08 -0800
From: David Barr <davidbarr@iname.com>
Organization: Medweb
To: Tenie Remmel <tenie1@juno.com>, Cube-Lovers <Cube-Lovers@ai.mit.edu>
Subject: Re: Rubiks Revenge moves
References: <19971118.204255.7126.1.tenie1@juno.com>

Tenie Remmel wrote:
>
> Is there an easy way to cycle three adjacent top edges on the
> Rubiks Revenge?  I can't find one shorter than 62 moves, but if
> there was a short one I could simplify my solution greatly.
>
> . b c .    . a b .
> a . . . => c . . .
> . . . .    . . . .
> . . . .    . . . .

I hold the cube so the bottom looks like this:

. . a .
. . . b
. . . c
. . . .

and do this sequence:

  F' b2 L2 / R' D r' D' R D r D' / L2 b2 F

Capital letters are outer slices.  Small letters are inner slices.
The slashes are just to show the different parts of the sequence.  The
middle part, if performed alone, will cycle three edges.  The first
part of the sequence positions the cubies we want to move into the
positions of the cubies that are cycled by the middle sequence.  The
last part of the sequence simply reverses the first part.

Left view of cube:

  . . . .
  . . . a
  . . . .
  . c b .

  b2 R2 / L D' l D L' D' l' D / R2 b2

Bottom view of cube:

  . . a .
  c . . b
  . . . .
  . . . .

  F' / R' D r' D' R D r D' / F

Bottom view of cube:

  . . . .
  a . . c
  . . . b
  . . . .

  b2 U' F / R' D r' D' R D r D' / F' U b2

Bottom view of cube:

  . . a .
  . . . b
  c . . .
  . . . .

  F' b' L2 / R' D r' D' R D r D' / L2 b F

Here are some other three cycles you may find useful:

  R' D l D' R D l' D'
  R' D L D' R D L' D'
  r' D l D' r D l' D'

-- 
mailto:davidbarr@iname.com
http://www.concentric.net/~Davebarr/

From cube-lovers-errors@mc.lcs.mit.edu  Thu Nov 20 12:50:09 1997
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Mail-from: From cube-lovers-request@life.ai.mit.edu Thu Nov 20 12:06:06 1997
Date: Thu, 20 Nov 1997 12:05:48 -0500
Message-Id: <20Nov1997.115617.Hoey@AIC.NRL.Navy.Mil>
From: Dan Hoey <Cube-Lovers-Request@ai.mit.edu>
Sender: Cube-Lovers-Request@ai.mit.edu
To: cube-lovers@ai.mit.edu
Subject: Auction on Rubik's Revenge (4x4x4) cubes (REPOST)
Reply-To: Paul Hart <hart@iserver.com>

whuang@ugcs.caltech.edu (Wei-Hwa Huang) has passed on a Usenet
announcement of Paul Hart's auction of 6 unopened Rubik's Revenges, as
mentioned previously in Cube-Lovers.  The auction ends November 22.
For details on the offer read http://www.enol.com/~hart, check a
Usenet search engine, or inquire by e-mail to Paul Hart
<hart@iserver.com>.

                                - Dan Hoey
                                  Interim Cube-Lovers-Request operator

From cube-lovers-errors@mc.lcs.mit.edu  Thu Nov 20 13:16:42 1997
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Nov 19 09:32:05 1997
Date: Wed, 19 Nov 1997 14:30:34 GMT
From: David Singmaster <david.singmaster@sbu.ac.uk>
To: chrono@ibm.net
Cc: cube-lovers@ai.mit.edu
Message-Id: <009BD880.292EDB5C.279@ice.sbu.ac.uk>
Subject: RE: Color schemes revisited

	In 1979(?) when I had my company David Singmaster Ltd which dealt in
Cubes and cube-related items, we had a tee-shirt designed showing a jumbled
cube with the caption  Rubik's Cube Cures Sanity.  Only one person ever wrote
in pointing out that the cube was impossible!  From the colouring of various
visible pieces, one could tell that the white face was adjacent to all five
other colours!!

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499;
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Thu Nov 20 13:47:50 1997
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Nov 19 08:55:52 1997
Message-Id: <v03010d00b09896ddf366@[204.71.18.82]>
In-Reply-To: <19971118.204255.7126.1.tenie1@juno.com>
Date: Wed, 19 Nov 1997 08:56:30 -0500
To: Tenie Remmel <tenie1@juno.com>, Cube-Lovers@ai.mit.edu
From: Nichael Lynn Cramer <nichael@sover.net>
Subject: Re: Rubiks Revenge moves

Tenie Remmel wrote:
>Is there an easy way to cycle three adjacent top edges on the
>Rubiks Revenge?  I can't find one shorter than 62 moves, but if
>there was a short one I could simplify my solution greatly.
>
>. b c .    . a b .
>a . . . => c . . .
>. . . .    . . . .
>. . . .    . . . .
>
>Hopefully it won't mess up the corners, but it's ok if it does.

The way I approach this is to begin with the following simple 3-cycle for
edge cubies (note that this cycles only the cubies and leave the rest of
the cube unaltered):

1] Imagine the involved cubies in the following configuration:

Top face :  . . c .      Left face: . . . .
            . . . .                 a . . .
            . . . .                 . . . .
            . b . .                 . . . .

2] Perform the following sequence:

 - Rotate Front Face by 1/4 turn clockwise.
 - Rotate the slice just below the Top Layer by 180 dgs.
 - Rotate the Front Face by 1/4 turn counter-clockwise.
 - Rotate the Top Face by 180 dg.
 - Rotate Front Face by 1/4 turn clockwise.
 - Rotate the slice just below the Top Layer by 180 dgs.
 - Rotate the Front Face by 1/4 turn counter-clockwise.
 - Rotate the Top Face by 180 dg.

This will result in:

Top face :  . . b .      Left face: . . . .
            . . . .                 c . . .
            . . . .                 . . . .
            . a . .                 . . . .

with all other cubies in their original locations.

3] Once this step is mastered, it is now only a question of moving the
cubies that you want to swap into the approriate location for this operator
to do its work.

For example, in your example above this can be accomplished by (this
assumes that the figure you have drawn above is your Top Face):

 - Rotating the Left-most two slices 1/4 turn clockwise (i.e. towards you)
 - Rotating the Top Face 1/4 turn counter-clockwise.

If you now rotate the entire cube by 90dgs clockwise, you will see your
three cubies are now in the proper location to use the above operator.
(When you're done with the operator, repeat the steps just above in the
reverse order to finish.)

>I'd also like to see some short moves for the following 3-cycles:
>
>. * * .    . . * .    . . . .    . . * .
>. . . .    * . . *    * . . *    . . . *
>* . . .    . . . .    . . . *    * . . .
>. . . .    . . . .    . . . .    . . . .

These are just variations on the above.  They will be left an exercise for
the reader.  ;-)

Hope this helps.
Nichael



Nichael
nichael@sover.net               deep autumn   my neighbor what does she do
http://www.sover.net/~nichael/                                 --Basho

From cube-lovers-errors@mc.lcs.mit.edu  Thu Nov 20 20:37:44 1997
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Sender: davidb@davidb.concentric.net
Message-Id: <3474866B.D1136871@iname.com>
Date: Thu, 20 Nov 1997 10:50:19 -0800
From: David Barr <davidbarr@iname.com>
Organization: Medweb
To: Cube-Lovers <Cube-Lovers@ai.mit.edu>
Subject: Re: Rubiks Revenge moves
References: <19971118.204255.7126.1.tenie1@juno.com> <34735538.113A5129@iname.com>

David Barr wrote:

> I hold the cube so the bottom looks like this:
>
> . . a .
> . . . b
> . . . c
> . . . .
>
> and do this sequence:
>
>   F' b2 L2 / R' D r' D' R D r D' / L2 b2 F

Actually, you can save a couple moves by doing

d2 L' R' D r' D' R D r D' L d2

but the pieces affected will be on the right side instead of the bottom.
-- 
mailto:davidbarr@iname.com
http://www.concentric.net/~Davebarr/

From cube-lovers-errors@mc.lcs.mit.edu  Thu Nov 20 20:54:43 1997
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Message-Id: <34736B2E.F9F4962@marlboro.edu>
Date: Wed, 19 Nov 1997 17:41:50 -0500
From: Jim Mahoney <mahoney@marlboro.edu>
Organization: Marlboro College
To: Tenie Remmel <tenie1@juno.com>
Cc: Cube-Lovers@ai.mit.edu
Subject: Re: Rubiks Revenge moves
References: <19971118.204255.7126.1.tenie1@juno.com>

Tenie Remmel wrote:

> Is there an easy way to cycle three adjacent top edges on the
> Rubiks Revenge?  I can't find one shorter than 62 moves, but if
> there was a short one I could simplify my solution greatly.
>
> . b c .    . a b .
> a . . . => c . . .
> . . . .    . . . .
> . . . .    . . . .

You can cycle these three edges on the 4x4x4 in 14 quarter turns
without disturbing the corners.

With the "up" and "front" faces like this (in a kind of projection view;
the corners are given by "*"; the "right" face is not shown),

       *  .  .  *
      .  .  .  .
     .  .  .  C
    *  A  B  *
    .  .  .  .
    .  .  .  .
    *  .  .  *

a procedure to cyle A,B,C is as follows:
 (1) 2 preparation moves which put C on "down" slice and B on "up/back"
 (2) 3 moves to get A off top slice and replace with C.
 (3) 2 moves (1/2 rotate) the top slice to put B where C (orginally A) was.
 (4) undo (2), restoring bottom layers and bring A back to top, in new spot.
 (5) undo (3)
 (6) undo (1), the prep moves.

(In each case "undo" means to do the inverse of the same moves in the
opposite order; that is,  "undo" ABC means C'B'A' where C' is the inverse
of move C.)

The hardest part I have of describing the specifics of each of
these is the notation; each of the 6 steps is only a couple of moves.

Let me define U,R,F as the Up, Right, and Front faces, and
number the slices by integers 1 to 4, so for example (F1,F2,F3,F4)
are clockwise quarter turns on the 4 slices (front to back)
parallel to the front face.  Counterclockwise turns are indicated
with either a ' (indicating "inverse") or lower case, so
F1' = a counterclockwise quarter turn of the front face.

In pictures,

        *  -  -  *                      *  -  -  *
      .  1  2  .                      .  3  1  .
     .  3  4  .                      .  4  2  .
    *  -  -  *     ==>  U1 ==>      *  -  -  *
    |  a  b  |                      |  a  b  |
    |  c  d  |                      |  c  d  |
    *  -  -  *                      *  -  -  *


        *  -  -  *                      *  -  -  *
      .  1  2  .                      .  1  2  .
     .  3  4  .                      .  3  4  .
    *  -  -  *     ==>  F1' ==>     *  -  -  *
    |  a  b  |                      |  b  d  |
    |  c  d  |                      |  a  c  |
    *  -  -  *                      *  -  -  *

where the letters are on the "front" face
and the numbers are on "up" face.

Then with this notation, steps (1) through (6) of this procedure are
  (1) F2  R2
  (2) F3' U4' F3
  (3) U1  U1
  (4) F3' U4  F3'
  (5) U1'  U1'
  (6) R2' F2'
which is 14 moves.

[Moderator's note: Certainly (4) should be F3' U4 F3, but that still
 cycles the wrong edges.  With (2)=R2 U4' R2', (4)=R2 U4 U2' we cycle
 the correct triple of edges, but in inverse order. ]

I confess that I don't have a 4x4x4 anymore and so can't try this -
I may have visualized one of the details wrong.  Hope not.

> I'd also like to see some short moves for the following 3-cycles:
>
> . * * .    . . * .    . . . .    . . * .
> . . . .    * . . *    * . . *    . . . *
> * . . .    . . . .    . . . *    * . . .
> . . . .    . . . .    . . . .    . . . .

You can do any of these as variations of the method I give above
with different "preperation" moves to get the edges into the
proper positions, namely two on the same slice which can be turned
into one another (the top slice in these cases),
and the third on another slice (usually the bottom slice) which
can replace one of the edges from the top slice.

I have a discussion of the NxNxN cube which includes in section (VI)
this recipe for 3-cycles of any kind of edge, corner, or face
piece; you can read it at
http://www.marlboro.edu/~mahoney/cube/NxN.txt if you're interested.


Regards,

 Jim Mahoney   (mahoney@marlboro.edu)
 Marlboro College

From cube-lovers-errors@mc.lcs.mit.edu  Thu Nov 20 21:52:54 1997
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Nov 19 11:58:12 1997
Date: Wed, 19 Nov 1997 11:53:14 -0500
Message-Id: <000E9C40.001706@scudder.com>
From: jdavenport@scudder.com (Jacob Davenport)
Subject: Re: Rubiks Revenge moves
To: Cube-Lovers@ai.mit.edu

Forget three adjacent top edges, I just want to cycle two of them.  I've
been solving a 5x5x5, and finally figured out how to make it look like a
3x3x3 so that I could solve nearly all of it.  However, the one place where
I cannot do that is solving the second and fourth edges from any side, and
have been using a short move that cycles three of them:

.axx.      .bxx.
y....      z....
y....  =>  z....
b....      a....
.czz.      .cyy.

The move is 2L F' L F 2L' (where 2L means the second layer from the
left)

This works great for getting nearly all the edges in place, but I have
two edges that are switched, and every time I use this move to put
them in place, I either leave two other edges out of place or leave
four edges out of place.  That is, I have the following:

.baa.                              .aaa.
b....                              c....
b....  which I can only make into  c....
a....                              b....
.ccc.                              .cbb.

which does not help.

I believe that my move works on 4x4x4 edges, and any move that helps a 4x4x4
cube will probably help me.

From cube-lovers-errors@mc.lcs.mit.edu  Sat Nov 22 22:56:58 1997
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sat Nov 22 17:53:28 1997
From: roger.broadie@iclweb.com (Roger Broadie)
To: <Cube-Lovers@ai.mit.edu>
Cc: "Tenie Remmel" <tenie1@juno.com>
Subject: Re: Rubiks Revenge moves
Date: Sat, 22 Nov 1997 22:51:33 -0000
Message-Id: <19971122224927.AAA7296@home>

Tenie Remmel <tenie1@juno.com> wrote (19 November 1997 )

> Is there an easy way to cycle three adjacent top edges on the
> Rubiks Revenge?  I can't find one shorter than 62 moves, but if
> there was a short one I could simplify my solution greatly.
>
> . b c .    . a b .
> a . . . => c . . .
> . . . .    . . . .
> . . . .    . . . .

Rather than just throw a few more solutions into the pot, I'd like to start
with some comments on the sort of process everyone, including me, seems to
use to deliver 3-cycles of edge pieces in the 4x4x4.  It is of the general
form

    [P, TQT']

where the square brackets are used to show a commutator, that is, [A,B]
means ABA'B'.

In this process P and Q are turns of layers that are parallel to one
another, and T is a turn of a layer transverse to P and Q.  For instance, P
and Q could be L and r and T could be U (capitals for the outer layers,
lower case for the neighbouring inner layers, with sense parallel to the
corresponding outer layer).  That gives

   [L, UrU'] == L UrU'.L' Ur'U'

which is a (not especially appealing) 3-cycle of edges.

In fact any process of this form is a 3-cycle provided it takes one piece
from the layer Q into the layer P.  That will happen if T is a quarter turn
in either sense - I haven't found anything useful with T as a half turn.
But P and Q can be any power.  The reason that processes of this form are
3-cycles is simple.  If two permutations intersect at only one element,
then their commutator is a 3-cycle.  Thus if

   A = (...a1, x, a2...) and B = (...b1, x, b2...)

then

   [A,B] -> (a1, x, b1)

If you do just UrU' you will find there is a line of displaced pieces along
the intersection Ub, but no other displaced piece in any of the layers
parallel to r.  Any of these pieces can be picked out to form part of a
3-cycle by selecting the layer that is parallel to r and contains the piece
and using a turn of that layer as the component P of the commutator, with
UrU' forming the component TQT'.

In general, if all of P, Q and T are outer layers we will have a 3-cycle of
corner pieces, if two are outer layers and one an inner layer we will have
a 3-cycle of edge pieces, if one is an outer layer and the other two inner
layers we will have a 3-cycle of centre pieces, and if all three are inner
layers we will have done nothing visible to the cube, but in fact there
will have been an invisible 3-cycle of the pieces of the imaginary internal
2x2x2.

We can derive the last of these cases from the first
quite neatly applying a fascinating concept called evisceration, which I
recently met trawling through the archives. It was first quoted from David
Singmaster's Cubic Circular by Stan Isaacs on 26 May 1983 and our present
acting moderator also discussed it on 1 June 1983. If you turn a cube
inside out by changing each outer layer in a process into an inner and
vice-versa (i.e. capitals to lower case), then, in the effect of the
process, you will interchange corner pieces with the pieces of the internal
2x2x2, and edge pieces with centre pieces. Making P, Q and T all to be
outer layers gives just a 3-cycle of corner pieces; therefore applying
evisceration takes that cycle into one on the pieces of the internal 2x2x2.

Singmaster's Notes on Rubik's Magic Cube, the fifth edition, interprets
processes of the type

   [P, TQT']

as

   [P,[T,Q]].

This expands to P TQT'Q'.P' QTQ'T', but the sequence Q'P'Q in the centre
reduces to just P', giving the same expansion as before. Of course, the two
components of the commutators TQT' and [T,Q] have different total effects,
but what they have in common is that they put the same single piece into P.
 We can look at them both as being sort of like a mono-operation.  Let's
call it a "monopop": each process pops a piece into P; you then turn P,
then reverse the pop operation, which extracts a different piece, and
finally restore P.

It's relatively straightforward to use this form to design specific
processes.  Say we want to move an edge piece from ULf to FLd and keep the
third member of the 3-cycle in the top layer.  Then we can take P to be L
to achieve the required part of the cycle.  We now know that Q must be in r
or l.  Let's take l.  The transverse move in T has to take a piece from l
into the point of intersection of the two components of the commutator,
FLd.  So it must be in F. Playing with F and l shows that the following
does the job.

   [L, FlF'] == L FlF'. L' Fl'F' -> (ULf, FLd, UBl)

If we'd taken Q to be in r we'd have needed a bit more care to keep the
third piece of the 3-cycle in the top layer, but [L, F'r^2F] does, putting
it at UBr.

If we want to move a piece along a diagonal - from ULf to FLu, say - we
need to use the other component of the commutator, TQT'.  Thus we can build
3-cycles which include ULf to FLu around the component U'FU.  For instance

   [f', U'FU] -> (ULf, LFu, RUf)

With a clear head and a good following wind it's possible to work out these
processes on the fly.

They also transform nicely into another process of the same type by cycling
the elements, which has the effect of conjugating the original process.
Thus the last process can be dealt with as follows

   U[f', U'FU]U' = [Uf'U', F] -> (UFr, LFu, BUr)

This cycling procedure comes from Singmaster.

Let's now think about top-layer edge processes. I'll denote the pieces like
this.


   X a1 a2 X

   d2 o o b1

   d1 o o b2

   X c2 c1 X

The purpose of the numbering in pairs is to emphasize that the processes
come in pairs.  Each process has a twin created by changing each inner
layer turn into its next-door inner neighbour. Thus the simplest U process
of the general type we're using is of this form:

   [l, F'LF] -> (a1, c2, d1)

Its twin is

   [r', F'LF] -> (a2, c1, d2)

In the twin process, each edge piece is changed into its next door
neighbour.

We want to capture this regularity.  I will therefore represent this pair
by

   [M', F'LF] -> (a*, c', d)

In this representation, M is either r or l', the asterisked piece defines
the layer that contains M and primed letters denote a piece with the
opposite suffix number to the asterisked piece.  Obviously, these suffixes
are closely related to flip in a 3x3x3 and the assignment of the numbers is
arbitrary.  Some assignments are more helpful than others in a particular
context, and the method used in the diagram above is the obvious one of
giving the same number ("flipperty"?) to the pieces in the positions that a
single piece moves into during a complete U turn.

Here, then is a complete set of top layer 3-cycles of edge pieces, to
within a reflection.  It comes from a fairly systematic search I did for
processes of the type [P, TQT'] that can be conjugated by at most one turn
into a top-layer process. They are oriented in a way I find easy to do.

I will leave them as commutators, because it is very easy to perform the
full set of turns from them.  The T/T' sequence remains constant for both
halves and the only adjustment needed is to invert P and Q the second time
around.  Inverses are also easy to perform, since all one has to do is read
off the second component first.

   (a*, c', d) [M', F'LF]

   (a*, b, c)    F2 [D R2 D', M2] F2

   (c*, a, b')    R2 [M' D' M, U2] R2

   (d1, b2, b1)   (Bb)' [U l U', R2] (Bb)
   (d2, b1, b2)   (Ff) [U r' U', R2] (Ff)'

   (d2, c2, d1)   (Bb) [L2, D l D'] (Bb)'  (Tenie's question)
   (d1, c1, d2)   f' [L2, D r' D'] f

The last two pairs could have been left in the M form if say N was
introduced to represent either f or b'. But keeping them separate lets us
save a wrist movement for the first three by combining the inner and
neighbouring outer layers for a turn relative to the central cut.  That
won't work for the final process, since the F layer is already included in
the 3-cycle and can't be amalgamated with the f layer.

All these process can be directly transferred to the 3x3x3, using the one
single central slice as M (or l' or r).  The primes then correspond to
flipping the edge piece relative to the top surface.

Roger Broadie
22 November 1997

From cube-lovers-errors@mc.lcs.mit.edu  Sat Nov 22 23:38:50 1997
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sat Nov 22 08:31:49 1997
Message-Id: <3.0.32.19971122082850.007b25c0@po9.mit.edu>
Date: Sat, 22 Nov 1997 08:28:51 -0500
To: cube-lovers@ai.mit.edu
From: Dennis Okon <dokon@mit.edu>
Subject: Large Cube

I realize this is impossible to make, at least in the traditional way, but...

I just saw an add for some computer conference which displayed a 9x9x9 cube
(with  some rather strange colors - at least 7 different ones too!).  The
text read something like (really paraphrased):

"When you were young you could work a cube in 27 seconds, but now you're
older and only have a week to solve this."

So, my question is: Assuming the cube is solvable, can it be done?  What
kind of order of growth to the current algorithms, for human algorithms and
computer algorithms, have in relation to the size of the cube?

-Dennis

[ Moderator's note: For solvability, see J.A. Eidswick's article
"Cubelike Puzzles -- What Are They and How Do You Solve Them?"
(American Mathematical Monthly', 93:3 (March 1986), pp. 157-176).
There are some loose bounds in my two articles of 24 Jun 1987 in the
archives at <ftp://ftp.ai.mit.edu/pub/cube-lovers/cube-mail-5.gz>. ]

From cube-lovers-errors@mc.lcs.mit.edu  Mon Nov 24 20:30:40 1997
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sun Nov 23 06:10:45 1997
Message-Id: <199711231109.LAA02433@GPO.iol.ie>
From: "Goyra (David Byrden)" <Goyra@iol.ie>
To: <cube-lovers@ai.mit.edu>
Subject: Re: Large Cube
Date: Sun, 23 Nov 1997 11:01:33 -0000

> From: Dennis Okon <dokon@mit.edu>


> I just saw an add for some computer conference which displayed a 9x9x9
cube


	Java version can be played with at 

	http://www.iol.ie/~goyra/Rubik.html


a "back" button is under development, too.

					David

From cube-lovers-errors@mc.lcs.mit.edu  Mon Nov 24 21:00:19 1997
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Mail-from: From cube-lovers-request@life.ai.mit.edu Mon Nov 24 19:45:36 1997
To: Cube-Lovers@AI.MIT.Edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Re: Rubiks Revenge moves
Date: 25 Nov 1997 00:44:22 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <65d716$29p@gap.cco.caltech.edu>
References: <cube-lovers.19971122224927.AAA7296@home.SOMEWHERE>

roger.broadie@iclweb.com (Roger Broadie) writes:
>Tenie Remmel <tenie1@juno.com> wrote (19 November 1997 )
>> Is there an easy way to cycle three adjacent top edges on the
>> Rubiks Revenge?  I can't find one shorter than 62 moves, but if
>> there was a short one I could simplify my solution greatly.
>>
>> . b c .    . a b .
>> a . . . => c . . .
>> . . . .    . . . .
>> . . . .    . . . .

>Rather than just throw a few more solutions into the pot, I'd like to start
>with some comments on the sort of process everyone, including me, seems to
>use to deliver 3-cycles of edge pieces in the 4x4x4.  It is of the general
>form

>    [P, TQT']

>where the square brackets are used to show a commutator, that is, [A,B]
>means ABA'B'.

>In this process P and Q are turns of layers that are parallel to one
>another, and T is a turn of a layer transverse to P and Q.  

Count me among the few "self-taught" solvers who don't actually
use this, then.  The one I worked out for myself a long time ago
turns out to be:

[r, FUF']

which is of a similar form, but P and Q are not parallel.

As a consequence of this, the permutation is not "clean": i.e.,
some other cubies get disturbed.  As these are all face cubies
anyway, I just modified my solution so that I do the face cubies
last.  :-)

--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
"[Lucy's eyes] look like little round dots of India ink..." -- Charlie Brown

From cube-lovers-errors@mc.lcs.mit.edu  Thu Dec  4 21:21:56 1997
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Mail-from: From cube-lovers-request@life.ai.mit.edu Thu Dec  4 11:42:31 1997
Message-Id: <19971204164104.18805.qmail@hotmail.com>
From: "John Coffey" <john2001@hotmail.com>
To: Cube-Lovers@ai.mit.edu
Date: Thu, 04 Dec 1997 09:41:03 MST

I have made a DOS program that solves the square-1 rubik's cube
variant.  If you would like to have this program then please 
contact me.  Source code is also available.

john2001@hotmail.com
John Coffey.
http://www.xmission.com/~jrcoffey/chess.htm
http://www.xmission.com/~jrcoffey/play.htm

[ Moderator's note: Soon to appear in
 ftp://ftp.ai.mit.edu/pub/cube-lovers/contrib/ ]

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Mail-from: From cube-lovers-request@life.ai.mit.edu Mon Dec 22 19:05:02 1997
Message-Id: <199712230003.AAA21827@GPO.iol.ie>
From: "David Byrden" <David@Byrden.com>
To: <Cube-Lovers@ai.mit.edu>
Subject: Return of the Cube?
Date: Tue, 23 Dec 1997 00:01:20 -0000

	I have just seen a toy expert
on UK tv say that the Rubik Cube is 
making a comeback this year. Any 
comment?

			David

From cube-lovers-errors@mc.lcs.mit.edu  Tue Dec 23 20:21:04 1997
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Mail-from: From cube-lovers-request@life.ai.mit.edu Tue Dec 23 18:19:28 1997
Message-Id: <199712232318.BAA01255@mail2.dial-up.net>
From: "Frederik Strauss" <fstrauss@icon.co.za>
To: <Cube-Lovers@ai.mit.edu>
Subject: Re: Return of the cube?
Date: Wed, 24 Dec 1997 01:18:17 +0200

 Here in South Africa the Rubiks cube is now for sale in one of the 
biggest newsagents, where previously it was hard to find anywhere.

From cube-lovers-errors@mc.lcs.mit.edu  Tue Dec 23 21:04:46 1997
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Mail-from: From cube-lovers-request@life.ai.mit.edu Tue Dec 23 03:03:42 1997
Message-Id: <349F637D.4DA23A12@ibm.net>
Date: Mon, 22 Dec 1997 23:08:45 -0800
From: "Jin 'Time Traveler' Kim" <chrono@ibm.net>
Organization: The Fourth Dimension
To: Cube-Lovers@ai.mit.edu
Subject: Re: Return of the Cube?
References: <199712230003.AAA21827@GPO.iol.ie>

David Byrden wrote:

>         I have just seen a toy expert
> on UK tv say that the Rubik Cube is
> making a comeback this year. Any
> comment?
>
>                         David

How about, "yay?"  I hope that by having the cube re-emerge and become
once again a fairly popular item, many other unique pieces will make a
return, like the RUBIKS REVENGE, as an example.  And perhaps then they
will become a stable commodity instead of just burning out after just an
intense period of a couple of years.  A Hula Hoop is a fad.  A piece
like the Cube is Timeless.

--
Jin "Time Traveler" Kim
chrono@ibm.net
VGL Costa Mesa / Puente Hills
http://www.geocities.com/timessquare/alley/9895
http://www.slamsite.com

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Message-Id: <199712242237.AAA11999@mail2.dial-up.net>
From: "Frederik Strauss" <fstrauss@icon.co.za>
To: "Cube Lovers" <Cube-Lovers@ai.mit.edu>
Subject: 5x5x5
Date: Wed, 24 Dec 1997 00:20:56 +0200

 Hi...

I have the 5x5x5 cube and can solve it, but it takes me about 15 minutes.
I've looked on the net but can find nothing on this cube. Does anyone where
I can get a bit more info on it, like moves or patterns? If anyone else can
solve it, I'd like to know how you do it.

Cu
Fred

fstrauss@icon.co.za

From cube-lovers-errors@mc.lcs.mit.edu  Thu Dec 25 23:43:21 1997
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Mail-from: From cube-lovers-request@life.ai.mit.edu Thu Dec 25 19:23:57 1997
Date: Thu, 25 Dec 1997 19:22:44 -0500 (EST)
From: Nicholas Bodley <nbodley@tiac.net>
To: "Jin 'Time Traveler' Kim" <chrono@ibm.net>
Cc: Cube Mailing List <Cube-lovers@ai.mit.edu>
Subject: Re: Return of the Cube?
In-Reply-To: <349F637D.4DA23A12@ibm.net>
Message-Id: <Pine.BSF.3.96.971225190234.4737G-100000@shell2.tiac.net>

 Jin Kim's comment, with which I heartily agree, leads me to wonder
whether some entrepreneur might support small-scale manufacture of a
very well-made 3^3, on the order of the Ideal deluxe Cube (I have
forgotten what they called it).

 That Cube had a redesigned mechanism (it differed only in the details)
that would tolerate much more misalignment, before making a maneuver,
than the typical Cubes. It had attached plastic colored tiles instead of
stickers, and was made of an excellent "engineering plastic".

 There are commercially-available materials that are self-lubricating,
and these could be used for the wearing surfaces. The pivots could be
true bearings, that is to say, like those in a typical piece
of machinery. (The Ideal Cube might have had such.} Tests to 250,000
revolutions might be reasonable.

 Whether it makes sense to try to improve upon that already very-good
design, I can't say.

 Consider chess, go, checkers or dominoes. I think the Cube quite
rightly should take its place with them.

 Imho, the Cube and its direct derivatives are the most ingenious
mechanisms ever invented.

On Mon, 22 Dec 1997, Jin 'Time Traveler' Kim wrote:

{Lots snipped}

} A piece like the Cube is Timeless.
--
}Jin "Time Traveler" Kim

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  ----------------Amateur musician--------------
|*  nbodley@tiac.net <<<-- Possible change to nbodley@shore.net; will let
oodles of folks know if I do. I'd try to overlap by a month or so to give
time to change over. Apologies in advance if so! Hope I don't have to.

From cube-lovers-errors@mc.lcs.mit.edu  Fri Dec 26 22:50:16 1997
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Date: Fri, 26 Dec 1997 10:03:05 +0100 (MET)
Message-Id: <1.5.4.16.19971226100243.1137cba4@mailsvr.pt.lu>
X-Sender: geohelm@mailsvr.pt.lu
To: "Frederik Strauss" <fstrauss@icon.co.za>
From: Georges Helm <geohelm@pt.lu>
Subject: Re: 5x5x5
Cc: Cube-Lovers@ai.mit.edu

I have a German book by Kurt ENDL on how to solve the whole bunch of 2x2x2,
3x3x3, 4x4x4 and 5x5x5 cubes.
I have a xeroxed copy of a solution by myself.
I do upper middles, edges, corners.
Then 2d, 3d and 4th layer edges.
Then 2d, 3d and 4th layer middles.
Then last layer corners and finally last layer edges.

Sometimes parity is uneven, i,e, there remain 2 edges to swap, and
there is a move I use to resolve that problem without disturbing the
rest of the cube by Helmut GEMBITZKY.

Georges Helm

geohelm@pt.lu
http://ourworld.compuserve.com/homepages/Georges_Helm/
http://www.geocities.com/Athens/2715

[ Moderator's note: As has been mentioned previously, there is a
general solution method in Eidswick, J. A., "Cubelike Puzzles -- What
Are They and How Do You Solve Them?", 'American Mathematical Monthly',
Vol. 93, #3, March 1986, pp. 157-176, though it's not optimized. ]

From cube-lovers-errors@mc.lcs.mit.edu  Sat Dec 27 20:24:59 1997
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sat Dec 27 11:45:01 1997
Message-Id: <l03110700b0cadfe5ce36@[207.172.128.129]>
Date: Sat, 27 Dec 1997 11:43:58 -0500
To: cube-lovers <cube-lovers@ai.mit.edu>
From: Charlie Dickman <charlied@erols.com>
Subject: Eidswick Reference

In a recent message regarding a question about solving the 5x5x5 cube our
moderator mentioned the reference

Eidswick, J.A., "Cubelike Puzzles -- What Are They and How Do You Solve
Them?", 'American Mathematical Monthly', Vol. 93, #3, March 1986, pp.
157-176

Does anyone know if this article/document is available on the web?

Charlie Dickman
charlied@erols.com

From cube-lovers-errors@mc.lcs.mit.edu  Tue Dec 30 18:20:01 1997
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Mail-from: From cube-lovers-request@life.ai.mit.edu Mon Dec 29 08:34:36 1997
Message-Id: <3.0.3.32.19971229083148.0055d700@caddscan.com>
Date: Mon, 29 Dec 1997 08:31:48 -0500
To: Cube-Lovers@ai.mit.edu
From: "Bryan Main" <bmain@caddscan.com>
Subject: Re: 5x5x5
In-Reply-To: <1.5.4.16.19971226100243.1137cba4@mailsvr.pt.lu>

At 10:03 AM 12/26/97 +0100, Georges Helm wrote:
>I have a German book by Kurt ENDL on how to solve the whole bunch of 2x2x2,
>3x3x3, 4x4x4 and 5x5x5 cubes.
>I have a xeroxed copy of a solution by myself.
>I do upper middles, edges, corners.
>Then 2d, 3d and 4th layer edges.
>Then 2d, 3d and 4th layer middles.
>Then last layer corners and finally last layer edges.
>
>Sometimes parity is uneven, i,e, there remain 2 edges to swap, and
>there is a move I use to resolve that problem without disturbing the
>rest of the cube by Helmut GEMBITZKY.
>
>Georges Helm

I just got one of these for christmas and had a question or two.  First is
there a cube program so I can play with it and not destroy all the work I
have done?  And I have solved one side, and all of the edges without much
problems.  However, can I solve the middle pieces without destroying the
edges?  As of yet I haven't found a way to keep the one side I have
finished and move one of the center pieces on another side.  I don't want
moves, I just want to know if it is possible to solve this way or if I
need to start looking at another way to solve it.

bryan

From cube-lovers-errors@mc.lcs.mit.edu  Tue Dec 30 18:50:07 1997
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Mail-from: From cube-lovers-request@life.ai.mit.edu Mon Dec 29 10:22:12 1997
Message-Id: <v02140b01b0cd629638a2@[205.230.130.11]>
Date: Mon, 29 Dec 1997 10:21:01 -0500
To: cube-lovers@ai.mit.edu
From: kristin@tsi-telsys.com (Kristin Looney)
Subject: a Rubiks Xmas

not only did I get 24 new cubes (as a gift to fill out my
gameroom window which now holds 120 cubes poised
and ready for new cube art) but my 9 year old niece
brought me a cube to mix up...  which she then solved!

anyone know any kids younger than 9 that can solve the
cube?  I'm sure there are younger cubists out there...  but
I sure was impressed.

-Kristin
kristin@wunderland.com
http://www.wunderland.com

_________________________________________________________
Kristin Looney / Manager, Information Systems / TSI TelSys Inc.
7100 Columbia Gateway Drive * Columbia, MD 21046 * 410.872.3939
klooney@tsi-telsys.com  * www.tsi-telsys.com * fax 410.872.3901

From cube-lovers-errors@mc.lcs.mit.edu  Tue Dec 30 19:54:57 1997
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Mail-from: From cube-lovers-request@life.ai.mit.edu Mon Dec 29 10:42:38 1997
Date: Mon, 29 Dec 1997 10:27:04 -0500
Message-Id: <00115225.001706@scudder.com>
From: jdavenport@scudder.com (Jacob Davenport)
Subject: Re: 5x5x5
To: "Cube Lovers" <Cube-Lovers@ai.mit.edu>

My way of solving the 5x5x5 has been to think about the cube in 3x3x3
terms.  When I solve a 3x3x3, I do top corners, bottom corners, top
and bottom edges at the same time, and then middle edges.

When I do a 5x5x5, I think of the middle corners (those cubies
directly diagonal from the center) as corners to the 3x3x3, ignoring
completely the outside edges, and I solve them so that all the middle
corners are aligned like a 3x3x3 would be aligned.  Then I solve the
middle edges (those cubies directly next to the center) like I would
solve the edges from 3x3x3 cubes.  This leaves the nine cubies in the
center of each face sovled.

I then use a move which many people use when solving a 4x4x4 to get
all the edge pieces together without disturbing the center squares.
This finally leaves me with messed up corners, solved center squares,
and the three edges on each side together.  I then view this as a
3x3x3 and solve that using my normal method.  The only drawback is
that I sometimes cannot get the edges to all work together, and the
reason is that I have inadvertantly switched two middle corners, and
it takes a long time for me to fix them.

If anyone wants more information on this solution, I'll spell it out
in detail on my web page.

-Jacob Davenport

http://wunderland.com/wts/jake

From cube-lovers-errors@mc.lcs.mit.edu  Thu Jan  1 22:43:23 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Dec 31 19:31:26 1997
To: Cube-Lovers@ai.mit.edu
Date: Wed, 31 Dec 1997 14:31:51 -0800
Subject: Where to get Dino Cube?
Message-Id: <19971231.143151.14726.0.tenie1@juno.com>
From: tenie1@juno.com (Tenie Remmel)

Is there a current source for the Dino Cube?  Both the
vertex turning kind (with 12 pieces one for each edge)
and the edge turning kind (with 24 pieces four on each
face).

Is Gametrends still around?  The phone number given in
a message in 1995 does not work.

--Tenie Remmel (tenie1@juno.com)

From cube-lovers-errors@mc.lcs.mit.edu  Thu Jan  1 23:48:05 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Dec 31 20:13:48 1997
Date: Wed, 31 Dec 1997 17:12:53 -0800 (PST)
Message-Id: <v01530500b0d02e83f283@[207.71.218.5]>
To: Cube-Lovers@ai.mit.edu
From: lowfrqcy@west.net (Ryan Blum)
Subject: Getting a 4x4x4 or a 5x5x5

As a relative newbie to the cube world, I only have a 3x3x3.  Would there
be any chance that I would find a used 4x4x4 or 5x5x5 cube around for less
than an arm and a leg? The new 4's went for around $100 in that auction a
little while ago, and that scared me....

Any Info would be greatly appreciated!
Thanks,
Ryan

[Moderator's note: The 5x5x5 cubes may be more available.  You can use
 one to practice 4x4x4 moves by ignoring the central slices (except in
 that you  have to keep them aligned with an adjacent slice).]

From cube-lovers-errors@mc.lcs.mit.edu  Fri Jan  2 00:51:34 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Thu Jan  1 08:00:23 1998
Message-Id: <199801011258.OAA17338@mail2.dial-up.net>
From: "Frederik Strauss" <fstrauss@icon.co.za>
To: "Cube Lovers" <Cube-Lovers@ai.mit.edu>
Subject: Re: 5x5x5
Date: Wed, 31 Dec 1997 04:17:14 +0200

>From: Bryan Main <bmain@caddscan.com>
>Subject: Re: 5x5x5

>problems.  However, can I solve the middle pieces without destroying the
>edges?  As of yet I haven't found a way to keep the one side I have

Yes, as a matter of fact that is the current method I use to solve it,
e-mail me if you want the moves.

Cu
Fred

fstrauss@icon.co.za

From cube-lovers-errors@mc.lcs.mit.edu  Fri Jan  2 21:37:09 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Fri Jan  2 12:33:26 1998
Date:     Fri, 2 Jan 1998 09:31:22 PST
From: ccw@eql12.caltech.edu (Chris Worrell)
Message-Id: <980102092825.23006d20@eql12.caltech.edu>
Subject:  Re: Where to get Dino Cube?
In-Reply-To: Your message <19971231.143151.14726.0.tenie1@juno.com> dated 31-Dec-1997
To: tenie1@juno.com
Cc: cube-lovers@ai.mit.edu

> Is Gametrends still around?  The phone number given in
> a message in 1995 does not work.

I think I was the one who posted that.
Gametrends (in Pasadena, CA) has been gone for more than a year.

Chris Worrell (ccw@EQL12.caltech.edu)

From cube-lovers-errors@mc.lcs.mit.edu  Sat Jan  3 02:01:09 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Fri Jan  2 19:50:46 1998
Message-Id: <v03010d00b0d2f87a2ed3@[204.71.18.82]>
In-Reply-To: <199712230003.AAA21827@GPO.iol.ie>
Content-Type: text/plain; charset="us-ascii"
Date: Fri, 2 Jan 1998 15:15:49 -0500
To: Cube-Lovers@ai.mit.edu
From: Nichael Lynn Cramer <nichael@sover.net>
Subject: Re: Return of the Cube?
Cc: David Byrden <David@byrden.com>

David Byrden wrote:
>	I have just seen a toy expert
>on UK tv say that the Rubik Cube is
>making a comeback this year. Any
>comment?

NWell, this is more a "cube-sighting" than an answer to the question whether
cubes will soon become more easily available, but I notice that in the new
(i.e. 4Jan98) New York Times Book Review, on the  inside of the front cover
is a small ad consisting of a (b&w) photo of a (3X) cube with the
over-laying copy:

          NEVER
            A
           DULL
         WEEKEND

   Our enhanced, two-part
     Weekend section is
   full of ideas about ways
      to broaden your
         horizons.


I couldn't quite figure out, though, whether the ad intended the cube to
symbolize the "broaden[ed] horizons", or the "dull weekend"...

From cube-lovers-errors@mc.lcs.mit.edu  Sat Jan  3 02:35:28 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sat Jan  3 01:40:29 1998
Date: Sat, 3 Jan 1998 01:39:21 -0500
Message-Id: <3Jan1998.011449.Alan@LCS.MIT.EDU>
From: Alan Bawden <Alan@lcs.mit.edu>
To: Cube-Lovers@ai.mit.edu
Subject: "A Message from Professor Erno Rubik"

I just accidentally tripped across http://www.rubiks.com/.  Much to my
amusement the home page has a little message from Erno Rubik that begins:

  It is 23 years since I created the Cube, some 17 years since this simple
  little six-coloured object attained its great world wide appeal. I often
  wondered what impact the Internet would have had if it had been around at
  the time.  Cube awareness, for one thing, would have spread even faster,
  aggravating the already severe Cube shortages in the market place.
  Suggestions and disputes about different approaches to solving it would
  surely have filled the screens. Amusing, unusual, interesting tales to do
  with the Cube would have criss-crossed the globe on the Net and intrigued
  mathematicians would have proposed and discussed Cube related theories
  on-line.

Those of you who have been on this mailing list for the last 17 years will
recognize this as an uncannily accurate description of exactly what -did-
happen!  (Except, of course, the Net was so much smaller then that we had
little effect on the market place.)

So accurate is Prof. Rubik's description that I'd be surprised if he (or his
ghost writer) hasn't actually read through some of our earliest archives.

				- Alan

From cube-lovers-errors@mc.lcs.mit.edu  Sat Jan  3 21:37:36 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sat Jan  3 05:56:11 1998
From: roger.broadie@iclweb.com (Roger Broadie)
To: <Cube-Lovers@ai.mit.edu>, "Bryan Main" <bmain@caddscan.com>
Subject: Re: 5x5x5
Date: Sat, 3 Jan 1998 10:55:50 -0000
Message-Id: <19980103105500.AAA11342@home>

> From: Bryan Main <bmain@caddscan.com>
> To: Cube-Lovers@ai.mit.edu
> Subject: Re: 5x5x5
> Date: 29 December 1997 13:31

> I just got one of these for christmas and had a question or two.
> First is there a cube program so I can play with it and not destroy
> all the work I have done?  And I have solved one side, and all of
> the edges without much problems.  However, can I solve the middle
> pieces without destroying the edges?  As of yet I haven't found a
> way to keep the one side I have finished and move one of the center
> pieces on another side.  I don't want moves, I just want to know if
> it is possible to solve this way or if I need to start looking at
> another way to solve it.

> bryan

Yes, if the corners of the top layer are also in the right place.  You can
move them around by normal 3x3x3 moves, but in doing so you may find that
the parity of the edge pieces is changed.  If you can swap a pair of edge
pieces on a 4x4x4, all will be well, and all the pieces in the ring of
eight around the piece at the centre of each face can be dealt with by
3-cycles to move these pieces to a different face or around on the same
face.

There is a hidden complication.  The new type of pieces introduced by the
5x5x5 are those at N, S, E and W in the central block of nine in each face.
 If the corner pieces of the cube are correctly placed, the parity of these
new pieces is tied to that of the edge pieces introduced by the 4x4x4, i.e.
those next to the corner pieces of the cube.  So if a pair of these edge
pieces is swapped, so will be a pair of the new 5x5x5 central pieces.  But
the swap of the edge pieces will cure them at the same time.  Often the
change to the centre pieces will not even show, because it will take place
within the same face.  Thus the sequence Georges Helm gave some time ago to
swap the 4x4x4 edges also cures the 5x5x5 centre pieces.  If it is applied
to a cube in the start position, it swaps Bl and Br visibly, and
interchanges FHl and FHr (where H is the central slice parallel to U)
invisibly.  It also makes an even-parity change to the 4x4x4 centre pieces
on the front face - in fact it rotates by 180 degrees the (l, u+H+d) and
(r, u+H+d) strips on this face.

Roger Broadie

From cube-lovers-errors@mc.lcs.mit.edu  Mon Jan  5 21:48:30 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sun Jan  4 10:49:11 1998
Message-Id: <Version.32.19980104102220.00f2a460@mail.interlog.com>
Date: Sun, 04 Jan 1998 10:23:51 -0500
To: cube-lovers@ai.mit.edu
From: Aaron Weintraub <aaweint@interlog.com>
Subject: Re: a Rubiks Xmas
In-Reply-To: <v02140b01b0cd629638a2@[205.230.130.11]>

Kristin,

Not to brag or anything, but I first solved the cube when I was 7 years
old.  I'm 22 now and still haven't lost interest.

-Aaron

At 10:21 AM 12/29/97, Kristin wrote:
>anyone know any kids younger than 9 that can solve the
>cube?  I'm sure there are younger cubists out there...  but
>I sure was impressed.
>
>-Kristin
>kristin@wunderland.com

From cube-lovers-errors@mc.lcs.mit.edu  Tue Jan 13 13:12:03 1998
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Date: Mon, 12 Jan 1998 23:25:18 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Face Turns Nine Moves from Start
To: Cube-Lovers <cube-lovers@ai.mit.edu>
Message-Id: <Pine.PMDF.3.95.980112230236.123792A-100000@PSTCC6.PSTCC.CC.TN.US>

I have some new search results for the face turn metric.  Here is a
summary of the new search.

Face Turns     Patterns  Positions    Branching    Positions/
from Start                               Factor     Patterns

	0             1           1                    1.000
	1             2          18      18.000        9.000
	2             9         243      13.500       27.000
	3            75        3240      13.333       43.200
	4           934       43239      13.345       46.294
	5         12077      574908      13.296       47.604
	6        159131     7618438      13.252       47.875
	7       2101575   100803036      13.231       47.965
	8      27762103  1332343288      13.217       47.991
	9     366611212 17596479795      13.207       47.998


The results at 8f and 9f from Start are new.  Previously, the face turn
metric had only been searched through 7f from Start.  All the results in
terms of patterns (M-conjugacy classes) are new.  Previously, the face
turn metric had been searched only in terms of positions.

Note that the branching factor does not change very much.  We already know
(or strongly suspect by statistical arguments based on the results of
Kociemba, Winter, Reid, and Korf) that it cannot change much this close to
Start.  Otherwise, the mode of the distribution would be greater than the
18f which is strongly suspected to be the case.

I have not yet installed the logic to detect weak local maxima.  The logic
to detect strong local maxima is installed with an interesting result. Two
patterns were detected at 9f from Start which are strong local maxima.
Regrettably, I have no idea what they are.  I will have to add something
to the program to print out strong local maxima when they are detected.
All I know is that the patterns are at least "somewhat symmetric" in that
they collectively represent only 32 positions.

I have begun to suspect that strong local maxima are fairly rare in the
face turn metric.  Recall that a strong local maximum is one where all 18
face turns carry the cube closer to Start.  A weak local maximum, by
contrast, is a local maximum where at least one face turn leaves the cube
the same distance from Start.  If I have not made a mistake in analyzing
them (which is entirely possible), the only one of Mike Reid's "highly
symmetric" positions which is a strong local maximum is superflip.  Even
Pons Asinorum is not a strong local maximum.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990

From cube-lovers-errors@mc.lcs.mit.edu  Wed Jan 14 19:00:40 1998
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Date: Wed, 14 Jan 1998 13:42:20 -0500 (Eastern Standard Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Performance Analyzers for Cube (and other) Programs
To: cube-lovers@ai.mit.edu
Reply-To: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Message-Id: <Pine.WNT.3.96.980114133501.-824401J-100000@GN209A.PSTCC.CC.TN.US>

[Moderator's note:  Please reply directly to Jerry.]

This is a little off topic, but many cube searching programs run for
dozens or hundreds of hours and we are always interested in speeding
them up.  The best speed ups usually come from algorithm improvements,
but I am also interested in more mundane program improvements.

Through the years, I have used various tools, usually for FORTRAN,
usually on mainframes, which will analyze a running program, telling you
where (which routines, which lines of source code) the program is
spending its time.  I am now running mostly C programs, mostly on a PC.
I confess I am clueless as to what performance analysis tools might be
available in this environment.  (I use Borland C++ if it matters.)  Any
suggestions would be gratefully accepted.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990

From cube-lovers-errors@mc.lcs.mit.edu  Thu Jan 15 12:55:44 1998
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To: Cube-Lovers@ai.mit.edu
From: Dave Dyer <ddyer@netcom.com>
Reply-To: Dave Dyer <ddyer@netcom.com>
Subject: save a cube for the price of a stamp

My trusty 4x4x4 has died.  I urgently need 2 replacement "1-sided"
cubelets, preferably white and yellow.  I'm hoping someone has a
similarly defunct 4x4x4 that they saved for sentimental reasons (or to
admire the amazing internal mechanism) and will send me some spare
parts.

From cube-lovers-errors@mc.lcs.mit.edu  Fri Jan 16 16:29:01 1998
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Date: Fri, 16 Jan 1998 17:58:10 +0000 (GMT)
From: Jonathan Tuliani <jont@dcs.rhbnc.ac.uk>
To: Cube-Lovers <Cube-Lovers@ai.mit.edu>
Subject: MEGAMINX
Message-Id: <Pine.OSF.3.96.980116175549.13054B-100000@platon>

The following is based on an email I sent recently to Kurt Endl.  He says that
he does not have time to work on this at present, and, with a thesis to write,
nor do I!  I expect that someone has heard of this and the answer is
known--I am new to this discussion group.  Otherwise, hopefully
someone will find it sufficiently interesting to think it through.

I was delighted to be given a MEGAMINX this Christmas, together with
Kurt Endl's instruction booklet.  I resolved to attempt the puzzle
without looking in the booklet, at least at first.

My approach was similar--I built a solution in layers, starting
at the bottom (which I shall call the north pole, consistent with the
notation in the latter part of the book) and proceeding, layer by
later towards the south pole.  I was successful in my efforts until I
reached the south cap, at which point I became stuck.  My problem was to
position and orient the south pole edges.  Try as I might, the best I
could do was to reach a position where two south pole edges needed to be
exchanged.  And try as I might, I could not find a way to do this.

After a week, I gave up and turned to the instructions.  I was delighted
to see that their approach was similar to mine, and fascinated by the
simple moves L_{**}, L^{**}, R_{**} and R^{**} used.  My
methods were, of course, far less elegant.

I was able to start at section 8, `Setting the South Pole
edges'.  The procedure for setting the edges affects the southern
equatorial corners, which you then arrange later.  My layer-by-layer
approach had, of course, already set these corners correctly before
attempting the south pole edges (and indeed before setting the southern
equatorial edges).  Perhaps this was the key?  Having the southern
equatorial corners set should not affect the validity of the book's method,
which should work with any arrangement of these corners.  But following
the instructions, I was unable to position the last two south pole edges
correctly.  The statement ``The remaining two South Pole edges will be
correctly placed again at the same time'' on page 21, section 8 of the
instructions must be incorrect--here after all was a counterexample!  My
last two south pole edges needed to be exchanged.

I ignored the problem for the time being.  I oriented the two edges
correctly, with them still in the wrong positions.  I was then able to
complete the MEGAMINX, positioning and orienting the southern equatorial
and south pole corners as in the instructions.  The result was a complete
MEGAMINX, except that it appeared that two little triangular stickers,
each on the border of the southern cap, had been exchanged.

After some thought, I have found a way out of this problem.  I believe
that this is a detail that may be required for solution in some
circumstances that is not in this instruction booklet.  I will try to
describe what I think went wrong.

The twelve faces of the MEGAMINX are coloured using only six colours, with
opposite faces bearing the same colour.  Thus, each edge piece has an
`identical twin' on the opposite side of the completed MEGAMINX.  When
solving the MEGAMINX from a totally jumbled position, these twins are
indistinguishable and may therefore be assigned either to their own
original position, or to their twin's position at random.  In this sense
the solution of MEGAMINX is not unique.  (Strictly, the solution may still
be unique but we have not proved conclusively that it is so, and have
demonstrated reason to believe that it may not be.)

Now, suppose we return to the two edges I wished to exchange.  Imagine
them as south pole edges also in adjacent faces.  Turn the MEGAMINX so
that the two faces concerned are in the left and right positions (using
your terminology).  We need to exchange these edges, but the problem we
have is that every sequence of moves seems to rotate 3 edges cyclicly,
rather than exchange a pair!  For example, the book uses L_{**} or R_{**} to
bring one of the two edges concerned down into the front southern
equatorial edge (``...the edge we have misused so brutally...''), and the
same moves to move it back up again.  But these moves, together with any I
can find, cycle 3 edges.  How can moves cycling 3 edges be used to
exchange just 2 edges?  I was stuck.

The solution comes from the `identical twins' I talked about before.
Suppose one of the two edges I'm interested in is, say, yellow/blue, and
the other yellow/orange, so the southern pole is yellow.  Tucked away on
the opposite side of MEGAMINX is *another* yellow/blue edge.  By a simple
sequence of moves, this may be brought into the postion of the front
southern equatorial edge.  Now consider cycling these three edges.  As two
of the edges are identical, cycing these three edges *looks* like a swapping
of just two edges!  Now we return to the other side of MEGAMINX the twin
of the piece that was originaly there.  (Some fixing of the MEGAMINX is
required to repair the damage caused by bringing an edge from the opposite
side of the MEGAMINX to the front and sending it's twin back, but this
isn't too hard.) Now, having apparently `swapped' two adjacent edges, we
can proceed as per the instructions and complete the MEGAMINX.  This I
have done.

Have any other people encountered this problem, or was I just
extremely unlucky?  The question arising is, of course, just how many
solutions of MEGAMINX are there?  There are 10 of these `identical
twin' edge pairs.  I reckon swapping just one twin pair is not possible in
a complete solution, but that swapping any even number of twins may be
(unproven), and so there are 512 solutions, each of which would be
distinct if 12 different colours had been used on the original puzzle.
Does anyone fancy having a stab at this conjecture?

Jonathan Tuliani

Mathematics Department
Royal Holloway, University of London
Egham
Surrey TW20 0EX
U.K.

jont@dcs.rhbnc.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Sun Jan 18 15:03:51 1998
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References: <l03110702b0e413fd88f6@[207.172.131.170]> <l03020905b0e2b255bd83@[192.0.2.1]>
Date: Sat, 17 Jan 1998 00:04:00 -0500
To: cube-lovers <cube-lovers@ai.mit.edu>
From: Charlie Dickman <charlied@erols.com>
Subject: Re: save a cube for the price of a stamp

Dave...

The other day you wrote...

>>>My trusty 4x4x4 has died.  I urgently need 2 replacement "1-sided"
>>>cubelets, preferably white and yellow.  I'm hoping someone has a
>>>similarly defunct 4x4x4 that they saved for sentimental reasons (or to
>>>admire the amazing internal mechanism) and will send me some spare
>>>parts.

I took the liberty of forwarding your message to Mike Green at Puzzletts.
Here is his response.

>I still have parts for the 4x4x4 @ $2.50 each plus postage.  I believe I
>can handle your request.  ( - :
>
>MG

Hope this helps.

Charlie Dickman
charlied@erols.com

From cube-lovers-errors@mc.lcs.mit.edu  Sun Jan 18 15:40:29 1998
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Date: Sun, 18 Jan 1998 15:39:22 -0500 (EST)
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From: Cube Lovers Moderator <Cube-Lovers-Request@AI.MIT.EDU>
To: Cube-Lovers@AI.MIT.EDU
Subject: Megaminx -- [Digest v23 #257]

Cube-Lovers Digest         Sun, 18 Jan 1998      Volume 23 : Issue 257

Today's Topic:

                           Megaminx [5 messages]

----------------------------------------------------------------------

Date: Fri, 16 Jan 1998 21:59:11 -0500
To: Cube-Lovers <Cube-Lovers@ai.mit.edu>
From: Charlie Dickman <charlied@erols.com>
Subject: Re: MEGAMINX

Where can one obtain Kurt Endl's instruction booklet for the megaminx?

Charlie Dickman
charlied@erols.com

------------------------------

From: roger.broadie@iclweb.com (Roger Broadie)
To: "Cube-Lovers" <Cube-Lovers@ai.mit.edu>
Cc: "Jonathan Tuliani" <jont@dcs.rhbnc.ac.uk>
Subject: Re: MEGAMINX
Date: Sat, 17 Jan 1998 22:54:40 -0000

In my opinion Jonathan is quite right in his analysis of the problem and
its solution.  When cubes or similar puzzles are coloured ambiguously, it
is always possible that the puzzle will be in an apparently impossible
configuration which must be cured by a move which changes identically
coloured pieces in an invisible way.  I am lucky enough to have a
dodecahedron puzzle from the 80s called the Supernova, which was made in
Hungary and sold in the UK by Pentangle.  That used twelve different
colours, and the problem did not arise.  Perhaps the instructions for the
Megaminx were originally written for this form.

Other puzzles can show similar effects - the variant of the cube with the
vertical edges bevelled so that it is octagonal in horizontal cross-section
has edge-pieces in the middle horizontal slice that have only one colour,
so their orientation is ambiguous and the top and bottom edge-pieces can
appear to have impossible flip states, such a single edge-piece flipped.

Jonathan remarks that every sequence of moves seems to rotate 3 edges
cyclically.  The reason (as usual) is to be found in parity considerations.
A single turn of  a face of the dodecahedron yields two 5-cycles, one of
the edge-pieces and one of the corners.  Both these are even permutations,
and it is therefore impossible to create a permutation of odd parity by any
combination of turns.  Hence a 3-cycle is the minimum possible, and a
single 2-cycle is impossible.

Obviously any 3-cycle of edges can in principle be conjugated into a form
to solve Jonathan's problem, but I thought I'd look for a process for a
3-cycle of edge-pieces which would take one piece from the bottom into the
top and preserve orientation (in the sense that the faces in the top and
bottom remain so).  I don't have the instruction booklet that goes with the
Megaminx, so I don't know what notation it used, or indeed what anyone else
may have used for this puzzle, so with the usual apologies if I'm ignoring
a standard notation:

Position the puzzle on a table, so there is a horizontal top plane and
bottom plane, and turn it so that one of the faces in the top band directly
faces you.  Call that face F, the top T, the two faces on either side of F
respectively Ru on the right and Lu on the left, and the two faces in the
lower band that join in the centre Rl on the right and Ll on the left (i.e.
u=upper and l=lower).  The arrangement of faces you see is thus (nb use a
non-proportionally spaced typeface);

                 T

          Lu     F      Ru

             Ll      Rl

                 D
Then

Rl Ll' T Ru Lu' F^2 Ru' Lu  T' Ru Lu' F^-2 Ru' Lu  Ll Rl'

does (TRu, TF, Drl).

Yes, you can swap pairs of edges - this is an even permutation which can be
composed out of 3-cycles, and in general any even number of swaps is
possible.

Roger Broadie

------------------------------

Date: Sun, 18 Jan 1998 02:28:59 +0100 (MET)
From: Dik.Winter@cwi.nl
To: Cube-Lovers@ai.mit.edu
Subject: Re: MEGAMINX

 >                    I reckon swapping just one twin pair is not possible in
 > a complete solution, but that swapping any even number of twins may be
 > (unproven), and so there are 512 solutions, each of which would be
 > distinct if 12 different colours had been used on the original puzzle.

It is not so difficult to prove.  Just as with the cube, also for the
dodecahedron it is easy to see that whenever you turn a face, the
parity of the edge and corner permutations remain the same.  So a
single swap of two edges is not possible, that is an odd permutation
and would also require an odd permutation of the corner.  However,
interchanging two pairs is possible.  Actually any even permutation
of the edges is possible with the corners in place.  This is because
there are simple procedures that rotate a triple of edges, leaving the
corners in place.  Actually these procedures can be extremely similar
to those used for the cube.  Anyhow, this proves it.

dik

[ Moderator's note: Lest a reader misunderstand, let me note that the
  parity situation is different between the cube and megaminx.  On the
  cube an odd permutation of edges is achievable provided the corner
  permutation is also odd.  On the megaminx, neither the corner
  permutation nor the edge permutation can ever be odd. ]

------------------------------

From: "Philip Knudsen" <philipknudsen@hotmail.com>
To: cube-lovers@ai.mit.edu
Subject: Re: MEGAMINX
Date: Sun, 18 Jan 1998 00:36:43 PST

Jonathan Tuliani wrote:

>I was delighted to be given a MEGAMINX this Christmas, together with
>Kurt Endl's instruction booklet.

This is a recently re-issued version of the Megaminx, using 6 colors
instead of 12. This puzzle is available at
Spielkiste <http://www.twfg.de/puzzle/default.htm>
The original Meffert Megaminx, however, used 12 colors, and the same
goes for a slightly different hungarian version called Supernova.
Christoph Bandelow <Christoph.Bandelow@ruhr-uni-bochum.de> still has a
few of the latter for sale.

>The result was a complete MEGAMINX, except that it appeared that two
>little triangular stickers, each on the border of the southern cap,
>had been exchanged.

This indeed is caused by the fact that in this version of the puzzle
there exist 15 (not 10) pairs of edges that are alike. Thus solving the
puzzle is somewhat similar to solving Alexander's Star and Impossiball
at the same time!

____________________________________
Philip Knudsen
Recording Artist
Vendersgade 15, 3th
DK - 1363 Copenhagen K
Denmark
Phone  : +45 3393 2787
E-mail : philipknudsen@hotmail.com

------------------------------

Date: Sun, 18 Jan 1998 14:08:10 -0500
From: Walter Smith <walsmith@erols.com>
To: cube-lovers@ai.mit.edu
Subject: Megaminx

On 1/6/98 Jonathan Tuliani described a Megaminx with 6 colors.  I have
a Megaminx purchased when they first came out.  It has 10 colors.
They are positioned so that there is only one "twin" pair on the
puzzle.  There are two red/yellow edge pieces.  It was a major
personal discovery when I found that the puzzle was only solvable if
the red/yellow pieces were put in the "right" places.  Only having one
twin pair made the puzzle particularly difficult because it took so
long before I noticed their existence.

I am sure that many readers will want to know if Megaminx is back in
production.  Jonathan, were did yours come from?

Walt Smith
walsmith@erols.com
Germantown Md.

------------------------------

End of Cube-Lovers Digest
*************************

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Date: Sun, 18 Jan 1998 23:59:30 -0500
From: michael reid <reid@math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: Re:  Face Turns Nine Moves from Start
Cc: jbryan@pstcc.cc.tn.us

jerry writes about strong local maxima in the face turn metric.
he says that superflip is such a position, but pons asinorum is not.
there are some other positions with a high degree of symmetry that are
also strong local maxima, for example

     pons asinorum  composed with  superflip
     superfliptwist
     supertwist

and some of the  T-symmetric positions that may not have standard names

     #1.  B  U  L' F' U  R  U2 D2 F' L  U' B' L  D  R2 L2 B2
     #4.  D' R' U  B' D' R' L  F  L  B' R' F  B' U  L  D' F  U' D2
     #5.  B2 L  U' L  D  R' L' D2 R  U  L' B2 U  R2 U2 F2 U
     #6.  D' L  F' B' L  F2 B2 U  R  L' U  D' L  F' R2 L2 F2 U' D2
     #9.  U  F2 D  B' U' B2 R  B2 D' F2 U' D2 B2 L' U2 B  D2
    #11.  D' F2 U2 B2 R  F' L  U' F2 B  R' F' D  L2 D  R2 F2 U' F2

and there might be more among the  H-symmetric  and  T-symmetric
positions (i can't tell right now without doing more searching).

yes, strong local maxima in the face turn metric are probably quite rare.
in the quarter turn metric, any global maximum is necessarily a local
maximum, because of parity considerations.  however, in the face turn
metric, a global maximum is a local maximum, but it may not be a strong one!
so the only strong local maxima we have here are found as a result of
(lots of) computer searching.

i look forward to seeing your 2 strong local maxima at distance 9f from
start.

mike

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Date: Mon, 19 Jan 1998 10:25:32 +0000 (GMT)
From: Jonathan Tuliani <jont@dcs.rhbnc.ac.uk>
To: Cube-Lovers <Cube-Lovers@ai.mit.edu>
Subject: MEGAMINX
Message-Id: <Pine.OSF.3.96.980119102221.31083B-100000@platon>

I'm glad my comments regarding MEGAMINX seemed to have sparked interest.
As someone points out, and as I realised over the weekend, there are of
course 15 pairs of `twin' edges, not 10 as I stated before.  Of course the
conjecture I made (suggested by one correspondent to be true, but well
beyond my group theory) would mean there are then 2^{15-1} = 16384
`distinct' complete solutions to this version of the puzzle!

I'll ask my friend where he got the MEGAMINX and the instruction booklet
from.  He said there were a fair few other similar puzzles there as well.

Jonathan Tuliani

Date: Mon, 19 Jan 1998 18:14:56 +0000 (GMT)
From: Jonathan Tuliani <jont@dcs.rhbnc.ac.uk>
To: Cube-Lovers <Cube-Lovers@ai.mit.edu>
Subject: Source for my Megaminx

Since somebody asked...apparently, my Megaminx came from Toys'R'Us, a
large toy retailer here in the UK.  The instruction booklet came with it. 

Jonathan

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Sender: hainesd@ai.mit.edu
Message-Id: <34C37E50.41C67EA6@kentrox.com>
Date: Mon, 19 Jan 1998 08:24:48 -0800
From: Darin Haines <hainesd@kentrox.com>
To: Cube-Lovers@ai.mit.edu
Subject: re: save a cube for the price of a stamp

My situation was similar to Dave's.  I sent him a response, and forgot
to cc it to the list.  I thought it would be beneficial [for me to send
it] for everyone on the list who may be in the same, or similar boat.
Here's what I told Dave...

The guy to talk to is Christoph Bandelow.  His email address is
Christoph.Bandelow@ruhr-uni-bochum.de.  (I'm pretty sure he monitors the
list.)  He is located in Germany.  He also has a bunch of other puzzles
that you will be interested in.  He is very prompt with delivering
orders, and is very easy to work with.

Hope this helps.

-Darin

Dave Dyer wrote:
>
> My trusty 4x4x4 has died.  I urgently need 2 replacement "1-sided"
> cubelets, preferably white and yellow.  I'm hoping someone has a
> similarly defunct 4x4x4 that they saved for sentimental reasons (or to
> admire the amazing internal mechanism) and will send me some spare
> parts.

From cube-lovers-errors@mc.lcs.mit.edu  Tue Jan 20 01:11:57 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Mon Jan 19 17:29:03 1998
Message-Id: <19980119222713.19893.qmail@hotmail.com>
From: "HADER MESA" <hamepa@hotmail.com>
To: cube-lovers@ai.mit.edu
Subject: deseo cubos
Date: Mon, 19 Jan 1998 14:27:11 PST

[Moderator's note: As Cube-lovers is conducted in ASCII, the ISO
 accent characters in the original message are replaced by
 two-character sequences here.  E.g., "informaci'on" refers to an
 acute accent over the "o".]

Hola, yo soy un aficionado al cubo de rubik, y me gustar'ia tener en mi
poder el cubo de rubik 3x3x3 y sus variantes (4x4x4, 5x5x5,2x2x2, etc)
ya que el que yo ten'ia, lo perd'i, y en mi pa'is no lo he podido
conseguir, adem'as yo recibo muchas noticias de los cube-lovers porque
estoy inscrito en su grupo de noticias, que con un gran esfuerzo logro
traducir.  El ingl'es se me dificulta y por eso escribo en espa~nol.
Si me pueden dar alguna informaci'on sobre donde los puedo adquirir, o
comprar a trav'es de Internet, estar'ia muy agradecido.

Cordialmente:
                      Hader Mesa Pareja
                      hamepa@hotmail.com

From cube-lovers-errors@mc.lcs.mit.edu  Tue Jan 20 12:14:46 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Tue Jan 20 11:57:15 1998
Date: Tue, 20 Jan 98 10:55:25 CST
Message-Id: <9801201655.AA23859@dvorak.amd.com>
Sender: clive1@dvorak.amd.com
From: "HADER MESA" <hamepa@hotmail.com>
To: cube-lovers@ai.mit.edu
Subject: Translation: [hamepa@hotmail.com: deseo cubos]
Reply-To: "HADER MESA" <hamepa@hotmail.com>

Here's a translation of Hader's message for those who are interested:

Hi, I am a fan of Rubik's cube, and I would like to have in my
posession the 3x3x3 cube and its variants (4x4x4, 5,x,5, 2x2x2, etc.)
since the one I had, I lost, and in my country I haven't been able to
obtain it.  Also, I receive many items from the cube-lovers because I
am already subscribed to your news group, that I manage to translate
with great effort.  English is hard for me, and that is why I write
in Spanish.  If you can give me any information about where I can
obtain them, or buy them over the Internet, I would be very grateful.

Cordially,

		Hader Mesa Pareja
		hamepa@hotmail.com

------
Original message:

[Moderator's note: As Cube-lovers is conducted in ASCII, the ISO
 accent characters in the original message are replaced by
 two-character sequences here.  E.g., "informaci'on" refers to an
 acute accent over the "o".]

Hola, yo soy un aficionado al cubo de rubik, y me gustar'ia tener en mi
poder el cubo de rubik 3x3x3 y sus variantes (4x4x4, 5x5x5,2x2x2, etc)
ya que el que yo ten'ia, lo perd'i, y en mi pa'is no lo he podido
conseguir, adem'as yo recibo muchas noticias de los cube-lovers porque
estoy inscrito en su grupo de noticias, que con un gran esfuerzo logro
traducir.  El ingl'es se me dificulta y por eso escribo en espa~nol.
Si me pueden dar alguna informaci'on sobre donde los puedo adquirir, o
comprar a trav'es de Internet, estar'ia muy agradecido.

Cordialmente:
                      Hader Mesa Pareja
                      hamepa@hotmail.com

From cube-lovers-errors@mc.lcs.mit.edu  Thu Jan 22 12:46:32 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Thu Jan 22 00:44:32 1998
Date: Thu, 22 Jan 1998 00:43:14 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: Face Turns Nine Moves from Start
In-Reply-To: <Pine.PMDF.3.95.980112230236.123792A-100000@PSTCC6.PSTCC.CC.TN.US>
To: Cube-Lovers <cube-lovers@ai.mit.edu>
Message-Id: <Pine.PMDF.3.95.980122001920.61579B-100000@PSTCC6.PSTCC.CC.TN.US>

On Mon, 12 Jan 1998, Jerry Bryan wrote:

> I have not yet installed the logic to detect weak local maxima.  The logic
> to detect strong local maxima is installed with an interesting result. Two
> patterns were detected at 9f from Start which are strong local maxima.
> Regrettably, I have no idea what they are.  I will have to add something
> to the program to print out strong local maxima when they are detected.
> All I know is that the patterns are at least "somewhat symmetric" in that
> they collectively represent only 32 positions.

I have not yet added the logic for weak local maxima, but further perusing
of my printout from the run which has already been made does yield a bit
of confirmation to some previous results reported by others.

At each distance from Start, my program summarizes the number of patterns
and positions by the symmetry class (one of the 33 symmetry classes of M,
the group of 48 symmetries of the cube).  Hence, I can easily look for
"highly symmetric" positions based on the symmetry class.  Of the 72
positions defined as q-transitive by Jim Saxe and Dan Hoey in Symmetry and
Local Maxima, only 4 of them show up in the search through 9f.  One of
them is at 0f (Start), one of them is at 6f (Pons Asinorum, a weak local
maximum, only the six half turns move closer to Start), and two of them
are at 8f (the two conjugate 6-H positions, weak local maxima with only
the six half turns moving closer to Start).

We therefore know that the two patterns at 9f which are strong local
maxima are not q-transitive, and cannot be shown to be local maxima by
symmetry considerations alone.

Strictly speaking, we already knew all this based on Dan's study of Pons
Asinorum from many years ago, and based on Mike Reid's recent studies of
highly symmetric positions with his optimal cube solver.  Jim Saxe found
the 8f processes for the 6-H positions many years ago, but I do not
believe that they were shown to be minimal until Mike's recent work.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990

From cube-lovers-errors@mc.lcs.mit.edu  Wed Jan 28 13:49:36 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Jan 28 13:18:22 1998
Date: Wed, 28 Jan 1998 18:14:52 GMT
From: David Singmaster Computing <david.singmaster@sbu.ac.uk>
To: cube-lovers@ai.mit.edu
Message-Id: <009C0FA1.17365270.46@ice.sbu.ac.uk>
Subject: Megaminx

	The UK stores of Toys R Us have been selling a number of
Meffert's products, including the Megaminx, Skewb, Impossiball, etc.
I had assumed these were also available in the USA.  Is this not the
case??

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499;
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Wed Jan 28 17:37:14 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Jan 28 13:32:58 1998
Date: Wed, 28 Jan 1998 10:31:35 -0800
From: mrhip@sgi.com (Jason Werner)
Message-Id: <9801281031.ZM27368@neuhelp.corp.sgi.com>
To: Cube-Lovers@ai.mit.edu
Subject: 9X9X9 cube (fictional)

If you can get your hands on the February 1998 edition of "Sys Admin: The
Journal For UNIX Systems Administrators", check out the ad on page 56.

Or, catch a glimpse at:

	http://www.sd98.com/

Enjoy!

	-Jason

--
Jason K. Werner                         Email: mrhip@sgi.com
Systems Administrator                   Phone: 650-933-9393
USFO I/S Technical Support              Fax:   650-932-9393
Silicon Graphics, Inc./Cray Research    Pager: 888-491-2906, mrhip_p@sgi.com

  "Winning is a habit"-Vince Lombardi "These go to eleven"-Nigel Tufnel

From cube-lovers-errors@mc.lcs.mit.edu  Thu Jan 29 13:56:49 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Jan 28 15:39:47 1998
Message-Id: <cUSapUSNAlMATHNT1-980128203809Z-449@mathnt1.sma.usna.navy.mil>
From: "joyner.david" <joyner.david@mathnt1.sma.usna.navy.mil>
To: "'David Singmaster Computing'" <david.singmaster@sbu.ac.uk>,
        "'cube-lovers@ai.mit.edu'" <cube-lovers@ai.mit.edu>
Subject: RE: Megaminx
Date: Wed, 28 Jan 1998 15:38:09 -0500

>From: 	David Singmaster Computing[SMTP:david.singmaster@sbu.ac.uk]
>Sent: 	Wednesday, January 28, 1998 1:14 PM
>To: 	cube-lovers@ai.mit.edu
>
>	The UK stores of Toys R Us have been selling a number of
>Meffert's products, including the Megaminx, Skewb, Impossiball, etc.
>I had assumed these were also available in the USA.  Is this not the
>case??

No this is not the case. The Toys R Us don't even sell Rubik's cubes
around were I live (in the Washington DC area). - David Joyner

[ Moderator's note: Michael Swart <Michael.Swart@switchview.com>
  notes that the Toys Ya us stores in Kitchener-Waterloo, Ontario
  Canada don't have them either. ]

From cube-lovers-errors@mc.lcs.mit.edu  Thu Jan 29 15:45:06 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Thu Jan 29 02:24:42 1998
Date: Thu, 29 Jan 1998 02:22:17 -0500
From: Edwin Saesen <ERCO@compuserve.com>
Subject: Re: 9X9X9 cube (fictional)
To: CUBE <CUBE-LOVERS@ai.mit.edu>
Message-Id: <199801290222_MC2-3110-7043@compuserve.com>

>If you can get your hands on the February 1998 edition of "Sys Admin:
>The Journal For UNIX Systems Administrators", check out the ad on
>page 56.

That will probably be similar to the one I saw in the February edition
of Visual Basic Programmer's Journal on page 90. I wanted to post
about this one, but somehow forgot...

Although, I think a deadline of one week to solve it should be enough
time...after I worked out the 5x5x5, the 9x9x9 should be just more of
the same :-)

Michael Ehrt

From cube-lovers-errors@mc.lcs.mit.edu  Fri Jan 30 13:08:04 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Thu Jan 29 05:57:13 1998
Message-Id: <34D05FEC.B6FE6B14@ibm.net>
Date: Thu, 29 Jan 1998 02:54:37 -0800
From: "Jin 'Time Traveler' Kim" <chrono@ibm.net>
To: Cube-Lovers@ai.mit.edu
Subject: Rubik's Cube FAQ
References: <9801281031.ZM27368@neuhelp.corp.sgi.com>

Is there such a thing as a Rubik's-type puzzle FAQ?  There is interest among
several people who wish to create one (specifically for www.rubiks.com) but if
there's already another that exists, there's no reason to duplicate effort if
it's not necessary.  If anybody knows of such a FAQ, please let me know.

--
Jin "Time Traveler" Kim
chrono@ibm.net
VGL Costa Mesa / Puente Hills
http://www.geocities.com/timessquare/alley/9895
http://www.slamsite.com

From cube-lovers-errors@mc.lcs.mit.edu  Fri Jan 30 13:43:05 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Thu Jan 29 15:22:57 1998
Message-Id: <3.0.3.32.19980129151831.005569e0@caddscan.com>
Date: Thu, 29 Jan 1998 15:18:31 -0500
To: cube-lovers@ai.mit.edu
From: "Bryan Main" <bmain@caddscan.com>
Subject: RE: Megaminx
In-Reply-To: <cUSapUSNAlMATHNT1-980128203809Z-449@mathnt1.sma.usna.navy.mil>

>No this is not the case. The Toys R Us don't even sell Rubik's cubes
>around were I live (in the Washington DC area). - David Joyner

It's kind of difficult to find the Cubes in Toy's R Us but they are there.
They also have some new game out for two people, but I don't remember what
it's called.  I know that the cubes are not where you would expect them to
be, with other games and puzzles, they are normally on an end cap near the
front of the stores on the bottom shelf.  Also a lot of the other toy
stores have them and they all have them in strange places around the store.

I don't think that I have seen anything new besides the new game.  They
normally sell the cube, snake, magic rings, and a pyramid that comes apart
and you put back together.

bryan

From cube-lovers-errors@mc.lcs.mit.edu  Mon Feb  2 14:45:11 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sun Feb  1 04:11:00 1998
From: peter@mold.demon.co.uk (Pete Thomas)
To: Cube-Lovers@ai.mit.edu
Subject: Centre Turns - A simple solution?
Date: Sun, 01 Feb 1998 09:07:10 GMT
Organization: Virtual Mold
Reply-To: peter@mold.demon.co.uk
Message-Id: <34db3a3e.3154932@post.eng.demon.net>

I've a  3 x 3 cube with a rather abstract pattern made up of three
colours. It does require the centered to be orientated correctly.

If I solve the cube, but fail to get the centers correct; is there an
easy solution to rotating them about their axis (I would guess in
opposite pairs?   Singmaster Notation if poss..

Regards

Pete

(Cube dabbler over the past 20 years).

Pete
---------------------------------------------

Virtual Mold.... Better than the real thing!
http://www.mold.demon.co.uk

From cube-lovers-errors@mc.lcs.mit.edu  Mon Feb  2 15:25:57 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sun Feb  1 12:38:55 1998
Message-Id: <199802011737.MAA08347@life.ai.mit.edu>
From: "John Coffey" <john2001@hotmail.com>
To: <cube-lovers@ai.mit.edu>, "Bryan Main" <bmain@caddscan.com>
Subject: Re: Megaminx
Date: Sun, 1 Feb 1998 10:30:04 -0700

Just as a side note, I was able to find a Square 1 puzzle at 
KayBee toys on closeout.   I had been looking for one for about
2 months.

Thanks,

John coffey

From cube-lovers-errors@mc.lcs.mit.edu  Mon Feb  2 18:07:51 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Fri Jan 30 17:00:36 1998
Date: Fri, 30 Jan 1998 22:59:06 +0100 (MET)
Message-Id: <199801302159.WAA02024@relay.euronet.nl>
To: Cube-Lovers@ai.mit.edu
From: Sytse de Maat <4xs2fs@euronet.nl>
Subject: Re: Rubik's Cube FAQ

At 02:54 29-1-98 -0800, chrono@ibm.net (Jin "Time Traveler" Kim) wrote:
>Is there such a thing as a Rubik's-type puzzle FAQ?...

At least a universal notation for 4x4x4 and 5x5x5 would be very
welcome to me.

Sytse de Maat <4xs2fs@euronet.nl>
Designer of a 5x5x5 cube in 1982

From cube-lovers-errors@mc.lcs.mit.edu  Tue Feb  3 14:15:18 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Tue Feb  3 04:22:43 1998
Message-Id: <3.0.5.16.19980203101216.2f57e01a@vip.cybercity.dk>
Date: Tue, 03 Feb 1998 10:12:16
To: cube-lovers@ai.mit.edu
From: Maria Skou & Philip Knudsen <skouknudsen@email.dk>
Subject: Lights Out Cube

I just heard there's a new puzzle out in the U.S. called "Lights Out Cube".
My first thought was that it's probably from the same company (Tiger
Electronics) who introduced the "Lights Out" and "Deluxe Lights Out".
Anybody who can confirm or knows some more??
It would also be nice if anyone knew a store in the San Francisco Area
where these puzzles are available (i have a friend who can buy them for
me). I haven't seen them in Europe.

Thanks in advance,

Philip K
recording and performing artist
Vendersgade 15, 3th
DK - 1363 Copenhagen K
Phone:  +45 33932787
Mobile: +45 21706731
E-mail: skouknudsen@email.dk
E-mail: philipknudsen@hotmail.com
E-mail: 4521706731@sms.tdk.dk (leave subject blank!)

From cube-lovers-errors@mc.lcs.mit.edu  Tue Feb  3 16:10:34 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Tue Feb  3 15:49:29 1998
Message-Id: <01a501bd30e5$5e06f440$da460318@CC623255-A.srst1.fl.home.com>
From: "Chris and Kori Pelley" <ck1@home.com>
To: <cube-lovers@ai.mit.edu>
Subject: Re: Lights Out Cube
Date: Tue, 3 Feb 1998 15:50:28 -0500

>I just heard there's a new puzzle out in the U.S. called "Lights Out Cube".

Yes the cube is by the makers of Lights Out.  It is available here in the
U.S. at places like Target and Wal-Mart.  It's quite a conversation piece
because people often ask if it's an electronic Rubik's Cube!  Of course it's
just a 3-D version of the Lights Out game.  It's fun because the puzzles get
progressively more difficult, plus there is a multi-player mode.

Chris Pelley
ck1@home.com

From cube-lovers-errors@mc.lcs.mit.edu  Thu Feb  5 13:30:56 1998
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Mail-from: From hoey@AIC.NRL.Navy.Mil Thu Feb  5 12:17:12 1998
Date: Thu, 5 Feb 98 12:16:55 EST
Message-Id: <9802051716.AA02692@sun33.aic.nrl.navy.mil>
From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: cube-lovers@mc.lcs.mit.edu
Subject: Test

This is a test of cube-lovers forwarding.  It shouldn't go to anyone
but the administrator, but if it does, that's a mistake.

    Dan Hoey
    Interim Cube-lovers administrator

From cube-lovers-errors@mc.lcs.mit.edu  Mon Feb  9 17:53:53 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Mon Feb  9 15:36:15 1998
Message-Id: <v02140b0ab10517d31577@[205.230.130.11]>
Date: Mon, 9 Feb 1998 15:36:13 -0500
To: Cube-Lovers@ai.mit.edu
From: kristin@wunderland.com (Kristin Looney)
Subject: looking for a phone number...
Reply-To: kristin@wunderland.com (Kristin Looney)

Does anyone on this list have contact information
for Oddz-On?

-K.
kristin@wunderland.com
http://www.wunderland.com/wts/kristin

From cube-lovers-errors@mc.lcs.mit.edu  Sat Feb 14 15:02:57 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sat Feb 14 09:38:36 1998
Message-Id: <199802141438.JAA02445@life.ai.mit.edu>
From: "David Byrden" <David@Byrden.com>
To: <Cube-Lovers@ai.mit.edu>
Subject: Rubik lawyers up in arms over website
Date: Sat, 14 Feb 1998 14:37:00 -0000

        I have a website at

        http://Byrden.com/puzzles/

where I keep playable versions of many
varieties of cube, pyramid, dodecahedron,
etc etc. All of them are built in Java
and have an "undo" button allowing you
to explore your moves.

        The site was titled "The Rubik Gallery"
in honour of Erno Rubik. It explicitly said
that there was no further connection with him.

        I got a letter from a Washington
firm of lawyers about a week ago, saying
that they were the advisers to Seven Towns
Limited, holders of the 'Rubik' trademark.
The site was "diluting" the value of the
trademark and causing "customer confusion".
I was engaging in "unfair competition"
(despite not selling or distributing anything
or taking any money or having any advertising
on the site).

        Not only did they want the word 'Rubik'
removed from the website, they wanted one of
the Java puzzles removed as well. They called it
an "electronic version of the RUBIK'S CUBE".
Fair enough, being a hexahedron sliced into 26
equal parts it bore a certian visual resemblance,
but obviously there was none of their mechanism
involved. It was all brand-new software.

        Anyway, I took it off, and they seem happy
enough now. But...does anyone know what company
owns the rights to the 4x4 cube?


                                David Byrden

From cube-lovers-errors@mc.lcs.mit.edu  Sun Feb 15 23:45:18 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sun Feb 15 19:06:46 1998
Date: Sun, 15 Feb 1998 19:06:35 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Strong Local Maxima 9f from Start
To: Cube-Lovers <cube-lovers@ai.mit.edu>
Message-Id: <Pine.PMDF.3.95.980215185935.291516B-100000@PSTCC6.PSTCC.CC.TN.US>

#1. D2 F2 L2 D' U  L2 F2 D' U'
#2. U  D  B2 R2 U  D' L2 B2 U2

These positions "look" very symmetric, especially #2, but I have not yet
examined their symmetry characteristics in detail.  They are certainly not
Q-transitive.  I do not know if either position has been reported before,
has a name, etc.  The corners are identical between the two positions, but
the edges are a good bit different.  I find both positions to be very
pretty.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990

From cube-lovers-errors@mc.lcs.mit.edu  Mon Feb 16 00:16:58 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sat Feb 14 16:39:52 1998
Message-Id: <15Feb1988.235225.Cube-Lovers@AI.MIT.EDU>
To: Cube-Lovers@AI.MIT.EDU
Subject: Rubik lawyers up in arms over website -- Digest v23 #279]
Date: Sun, 15 Feb 1998 23:52:25 EST

Cube-Lovers Digest         Sun, 15 Feb 1998      Volume 23 : Issue 279

Today's Topic:

         Rubik lawyers up in arms over website [3 messages]

----------------------------------------------------------------------

From: "Philip Knudsen" <philipknudsen@hotmail.com>
To: cube-lovers@ai.mit.edu
Subject: Re: Rubik lawyers up in arms over website
Date: Sat, 14 Feb 1998 13:38:35 PST

This confirms a common (european?) prejudice about the U.S. and their
tendency to file lawsuits over just about anything.
I feel certain Rubik himself would have nothing against Byrden's
online cube. On the other hand it's good to know his business is well
taken care of...
Don't know about the 4x4x4 copyright, but it's pretty well known Rubik
did not design the actual mechanism for it. Ideal just used his name
to market the puzzle.

By the way, I did visit the www.Byrden.com site some time ago and really
liked it, not the least the "special octahedron".
____________________________________
Philip K

[ Moderator's note: I don't think the archive has anything about the
  origin of the 4^3 (or any other) design.  Can you give a source for
  this well-known information? ]

------------------------------

Date: Mon, 16 Feb 98 08:59:30 +0900
From: Norman Diamond  16-Feb-1998 0859 <diamond@jrdv04.enet.dec-j.co.jp>
To: cube-lovers@ai.mit.edu
Subject: Re: Rubik lawyers up in arms over website

David Byrden wrote:

>I got a letter from a Washington
>firm of lawyers about a week ago, saying
>that they were the advisers to Seven Towns
>Limited, holders of the 'Rubik' trademark.
>The site was "diluting" the value of the
>trademark and causing "customer confusion".

Fine.  Please restore your web site, but say:

"Please enjoy these puzzles yourself.  However, we
cannot honor the famous Dr. Rubik because his lawyers
won't let us honor him."

>I was engaging in "unfair competition"
>(despite not selling or distributing anything
>or taking any money or having any advertising
>on the site).

It doesn't matter if you take money or not.  I thought it
didn't hurt if you had advertisements honoring Dr. Rubik,
but ...

Well, I guess everyone had better remove Rubik's signature
from our instances of his merchandise, because whenever we
play with one of his products, we're advertising his name
illegally.  [Note:  This is the only sarcastic sentence
in this message.  Please take the rest seriously.]

>        Not only did they want the word 'Rubik'
>removed from the website, they wanted one of
>the Java puzzles removed as well. They called it
>an "electronic version of the RUBIK'S CUBE".
>Fair enough, being a hexahedron sliced into 26
>equal parts it bore a certian visual resemblance,
>but obviously there was none of their mechanism
>involved. It was all brand-new software.

It is true that none of their mechanism is involved.
Therefore I believe their patent doesn't apply.  That is,
if they actually still have a pattent, after Ishige and
some American who preceded all of them (whose name I've
forgotten) ...  but wait, it's been more than 20 years
(or 17 in the US), so ALL their patents have expired.

So don't honor Dr. Rubik.  Please restore all of your
mathematical puzzles.

-- Norman Diamond               diamond@jrdv04.enet.dec-j.co.jp
[Speaking for Norman Diamond not for Digital.]

------------------------------

Date: Sun, 15 Feb 1998 20:23:38 -0500
From: Alan Bawden <Alan@LCS.MIT.EDU>
To: David@Byrden.com
Cc: Cube-Lovers@ai.mit.edu

   From: "David Byrden" <David@Byrden.com>
   Date: Sat, 14 Feb 1998 14:37:00 -0000
	   ...
	   Not only did they want the word 'Rubik'
   removed from the website, they wanted one of
   the Java puzzles removed as well. They called it
   an "electronic version of the RUBIK'S CUBE".
   Fair enough, being a hexahedron sliced into 26
   equal parts it bore a certian visual resemblance,
   but obviously there was none of their mechanism
   involved. It was all brand-new software.

This latter seems totally outrageous to me.  If I were in your shoes, I
would consider contacting the EFF to see if they were interested in making
a case out of this.  The request that you remove Rubik's name from your
site is the kind of petty stupidity we're seeing all to often these days,
and is probably pretty mundane to the cyberlawyers at EFF, but the notion
that they can torpedo your software if it merely duplicates the user
interface (the "look-and-feel") of their physical puzzle might be something
genuinely new.  Heck, do these guys claim that they own the underlying
mathematical group?

------------------------------

End of Cube-Lovers Digest
*************************

From cube-lovers-errors@mc.lcs.mit.edu  Tue Feb 17 18:08:36 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Mon Feb 16 19:02:35 1998
Date: Mon, 16 Feb 1998 19:02:17 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: Strong Local Maxima 9f from Start
In-Reply-To: <Pine.PMDF.3.95.980215185935.291516B-100000@PSTCC6.PSTCC.CC.TN.US>
To: Cube-Lovers <cube-lovers@ai.mit.edu>
Message-Id: <Pine.PMDF.3.95.980216185414.311707A-100000@PSTCC6.PSTCC.CC.TN.US>

On Sun, 15 Feb 1998, Jerry Bryan wrote:

> 
> #1. D2 F2 L2 D' U  L2 F2 D' U'
> #2. U  D  B2 R2 U  D' L2 B2 U2
> 

It ocurred to me that because these positions are strong local maxima (and
the shortest ones, at that), maybe I should show maneuvers of length 9f
ending with each of the 18 possible face turns.  Here they are.  No
uniqueness is claimed for the maneuvers.

#1

F2 L2 U2 B  F' D2 L2 B' F' 
F2 L2 U2 B  F' D2 L2 F' B' 
R2 D2 F2 L  R' F2 D2 L' R' 
R2 D2 F2 L  R' F2 D2 R' L' 
D2 F2 L2 D' U  L2 F2 D' U' 
D2 F2 L2 D' U  L2 F2 U' D' 
B2 R2 D2 B  F' U2 R2 B  F  
B2 R2 D2 B  F' U2 R2 F  B
L2 U2 B2 L  R' B2 U2 L  R
L2 U2 B2 L  R' B2 U2 R  L
U2 B2 R2 D' U  R2 B2 D  U
U2 B2 R2 D' U  R2 B2 U  D
B' F' U2 R2 B  F' L2 U2 F2
B  F  D2 L2 B  F' R2 D2 B2
L' R' F2 U2 L  R' U2 F2 R2
L  R  B2 D2 L  R' D2 B2 L2
D  U  L2 B2 D' U  B2 L2 U2
D' U' R2 F2 D' U  F2 R2 D2



#2

B2 U2 R2 B' F  L2 U2 B' F'
B2 U2 R2 B' F  L2 U2 F' B'
R2 F2 D2 L  R' U2 F2 L' R'
R2 F2 D2 L  R' U2 F2 R' L'
D2 L2 B2 D' U  F2 L2 D' U'
D2 L2 B2 D' U  F2 L2 U' D'
F2 D2 L2 B' F  R2 D2 B  F
F2 D2 L2 B' F  R2 D2 F  B
L2 B2 U2 L  R' D2 B2 L  R
L2 B2 U2 L  R' D2 B2 R  L
U2 R2 F2 D' U  B2 R2 D  U
U2 R2 F2 D' U  B2 R2 U  D
B  F  R2 D2 B' F  U2 R2 F2
B' F' L2 U2 B' F  D2 L2 B2
L' R' U2 F2 L  R' B2 U2 R2
L  R  D2 B2 L  R' F2 D2 L2
D  U  B2 R2 D' U  L2 B2 U2
D' U' F2 L2 D' U  R2 F2 D2

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990

From cube-lovers-errors@mc.lcs.mit.edu  Tue Feb 17 19:44:21 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Mon Feb 16 07:57:12 1998
Message-Id: <17Feb1998.181625.Cube-Lovers@AI.MIT.EDU>
Date: Tue, 17 Feb 1998 08:16:25 -0500
Subject: Rubik lawyers up in arms over website -- Digest v23 #281]
To: Cube-Lovers@ai.mit.edu
From: Cube-Lovers@ai.mit.edu

Cube-Lovers Digest         Tue, 17 Feb 1998      Volume 23 : Issue 281

Today's Topic:

         Rubik lawyers up in arms over website [3 messages]

----------------------------------------------------------------------

Date: Mon, 16 Feb 1998 07:58:28 -0500
To: Cube-Lovers@ai.mit.edu
From: Nichael Lynn Cramer <nichael@sover.net>
Subject: Laches
Message-Id: <v03010d04b10de3f62631@[204.71.18.82]>

>From: "Philip Knudsen" <philipknudsen@hotmail.com>
>
>This confirms a common (european?) prejudice about the U.S. and their
>tendency to file lawsuits over just about anything.
>I feel certain Rubik himself would have nothing against Byrden's
>online cube. On the other hand it's good to know his business is well
>taken care of...

[Apologies for taking this even farther off topic.  But in an attempt to
clear up one point...]

Indeed it's quite possible that Rubik knows nothing about it.

The issue here is the legal concept of "Laches", which says --effectively--
that if it can be shown that you, the copyright owner, did not pursue all
incidents of copyright infringement of which you were aware, then the
copyrighted item is in real danger of being declared as in the public
domain.

(This is also what's behind those silly stories of, say, some vet in the
wilds outside Buccolia, Maine with a picture of Snoopy painted on his barn
who one day gets a letter from Charles Schulz's lawyers requesting that he
either remove the picture or else sign a license arrangement at a zillion
dollars a year.)

So, in short, a copyright owner is legally "required" to go after _any_
known or perceived abuse of the copyright or face the very real danger of
losing it.

Nichael Cramer
nichael@sover.net                     Gather the folks, tell the stories,
http://www.sover.net/~nichael/             break the bread.   -- John Shea

------------------------------

To: cube-lovers@ai.mit.edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Date: 16 Feb 1998 17:38:31 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <6c9tin$2ni@gap.cco.caltech.edu>

"David Byrden" <David@Byrden.com> writes:
>        I got a letter from a Washington
>firm of lawyers about a week ago, saying
>that they were the advisers to Seven Towns
>Limited, holders of the 'Rubik' trademark.
>The site was "diluting" the value of the
>trademark and causing "customer confusion".
>I was engaging in "unfair competition"
>(despite not selling or distributing anything
>or taking any money or having any advertising
>on the site).

They have to say stuff like this to demonstrate that they've protected
their trademark.  Apparently the word "Rubik", when applied to puzzles,
is trademarked.  In US law, if one doesn't protect a trademark by this
manner, one may lose it.

Of course, since Rubik is also the name of a person, you should be able to
use "Rubik" when referring to the person.  The specific thing they're
worried about is phrases like "The Rubik Page" or "Rubik Puzzles."
Change the wording to "Puzzles based on those invented by Erno Rubik,"
and I don't think they can touch you.

>        Not only did they want the word 'Rubik'
>removed from the website, they wanted one of
>the Java puzzles removed as well. They called it
>an "electronic version of the RUBIK'S CUBE".
>Fair enough, being a hexahedron sliced into 26
>equal parts it bore a certian visual resemblance,
>but obviously there was none of their mechanism
>involved. It was all brand-new software.

This is bunk.  No one can trademark that stuff.  At most, there's a patent
(which you can't have violated).  The lawyers are just asking for extra.

- --
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
- ---------------------------------------------------------------------------
"...he put a wire in his cap and called himself Marconi."

------------------------------

Date: Mon, 16 Feb 1998 00:45:33 -0500
From: mark longridge <cubeman@idirect.com>
Subject: Re: Rubik lawyers up in arms over website -- Digest v23 #279]
Message-Id: <34E7D27D.6083@idirect.com>

> From: "Philip Knudsen" <philipknudsen@hotmail.com>

> ...Don't know about the 4x4x4 copyright, but it's pretty well known
> Rubik did not design the actual mechanism for it. Ideal just used
> his name to market the puzzle.

Probably (I'm not totally certain) it was Udo Krell, an inventor whose
design was used by Uwe Meffert to make the 5x5x5.

Norman Diamond  <diamond@jrdv04.enet.dec-j.co.jp> wrote:
> ...
> It is true that none of their mechanism is involved.
> Therefore I believe their patent doesn't apply.  That is,
> if they actually still have a pattent, after Ishige and
> some American who preceded all of them (whose name I've
> forgotten) ...  but wait, it's been more than 20 years
> (or 17 in the US), so ALL their patents have expired....

What about Karl Hornell's Java Applet "Rubik Unbound"?? It's
all over the internet on hundreds of sites including my
own!!

I don't think the name Rubik itself can expire since that
is his name... so the name of the product is always
"Rubik's Cube"... ummmm right? :-)

[Moderator's note: _Patents_ expire.  _Trademarks_ don't necessarily
 expire.  _Names_ are not protected by law. ]

Alan Bawden <Alan@LCS.MIT.EDU> wrote:
> ...  The request that you remove Rubik's name from your
> site is the kind of petty stupidity we're seeing all to often these days,
> and is probably pretty mundane to the cyberlawyers at EFF, but the notion
> that they can torpedo your software if it merely duplicates the user
> interface (the "look-and-feel") of their physical puzzle might be something
> genuinely new.  Heck, do these guys claim that they own the underlying
> mathematical group?

I don't think you can prevent people from making java applets and the
like of cubes... but I think they (the lawyers that be) can protect
Rubik's name. I don't think anyone can say "Don't show a rubik's
cube-like construction on your web page".

The other guy who history has forgotten was Larry Nichols who made
a 2x2x2 cube called twizzle which routinely came apart and was
rejected by Ideal Toy!

I'd restore the 3x3x3 java applet if it was my web page.

- -> Mark <-

------------------------------

End of Cube-Lovers Digest
*************************

From cube-lovers-errors@mc.lcs.mit.edu  Wed Feb 18 13:29:22 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Feb 18 11:08:58 1998
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Date: Wed, 18 Feb 1998 16:56:04
To: cube-lovers@ai.mit.edu
From: Philip Knudsen <skouknudsen@email.dk>
Subject: Game designers [was Re: Rubik lawyers...]

Dan wrote:

>I don't think the archive has anything about the origin of the 4^3 (or any
>other) design.  Can you give a source for this well-known information?

Maybe i was a little fast to state that it is "generally known Rubik did
not design the 4x4x4 mechanism". I've also digged through the entire list
archives as well as my own stuff, and have found nothing which directly
indicated that Rubik DID design it. Now the earliest mention of 4x4x4 is in
Hofstadter's article in S.A. from march '81, page 26. Quote: "Rest assured,
it's being developed in the Netherlands, and it may be ready soon..."

>From: mark longridge -
>Probably (I'm not totally certain) it was Udo Krell, an inventor whose
>design was used by Uwe Meffert to make the 5x5x5.

Don't you think it would be known that Udo Krell also invented the 4x4x4 if
this was indeed the case?
All of this is getting a little vague, it could be nice to have this matter
cleared up by someone who has some REAL info! Maybe someone who worked for
Ideal at the time.

Yours Truly,

Philip K

From cube-lovers-errors@mc.lcs.mit.edu  Wed Feb 18 16:13:49 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Feb 18 15:12:55 1998
Message-Id: <v02140b06b110e8ce5e6a@[205.230.130.11]>
Date: Wed, 18 Feb 1998 15:13:03 -0500
To: cube-lovers@ai.mit.edu
From: kristin@wunderland.com (Kristin Looney)
Subject: working for Ideal...  

At 4:56 PM 2/18/98, Philip Knudsen wrote:

> All of this is getting a little vague, it could be nice to have this
> matter cleared up by someone who has some REAL info! Maybe
> someone who worked for Ideal at the time.

I worked for Ideal at the time...   but only as a 16 year old kid
demonstrating the cube and giving away free T-Shirts and posters
in shopping malls in the Chicago area.  Sorry, I have no idea who
invented the 4x4 mechanism.  I do remember anxiously waiting
for the mail every day for a couple of weeks when they had said
they were sending me one hot off the assembly line...

-K.
kristin@wunderland.com
http://www.wunderland.com/wts/kristin
http://www.wunderland.com/Home/Rubik.html

From cube-lovers-errors@mc.lcs.mit.edu  Wed Feb 18 19:38:23 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Feb 18 15:29:06 1998
Message-Id: <v02140b07b110ebcc12af@[205.230.130.11]>
Date: Wed, 18 Feb 1998 15:29:15 -0500
To: cube-lovers@ai.mit.edu
From: kristin@wunderland.com (Kristin Looney)
Subject: custom cubes / awesome cube art

Two other URL's to point the members of this list to...


First, I have perfected custom cube sticker technology!   Check out:
      http://www.wunderland.com/WTS/Kristin/CustomCubes.html

If anyone has a custom cube they have been itching to make
for years...  as long as you provide the cube sans stickers and
artwork in the size and format that I need it in...  and probably
postage to send it back to you if very many people take me up
on this...   talk to me...  together we can make you the cube of
your dreams.   A word of warning:  Odz On has done an excellent
job of solving the stickers-always-falling-off problem on the
new cubes.  You will have a MUCH easier time getting the stickers
off an old cube than you will one of the new ones.  Zarf designed
a REALLY cool cube last week...  a picture of it should go up on this
page with this weeks update on Thursday.


Also, Jake has been continuing to evolve his cube art...  Check out:
      http://www.wunderland.com/WTS/Jake/CubeInfo/

I have a window in my gameroom with 120 cubes arranged within
it...  and every couple of weeks or so Jake comes over and solves
them into some sort of a cool 36 x 30 pixel picture.  Jake and Zarf
and Andy and myself have all designed images, and we are not out
of ideas yet...  but I'm guessing Jake would take outside design
submissions if you asked him.

-K.
kristin@wunderland.com
http://www.wunderland.com/wts/kristin

To all the fishies in the deep blue sea, Joy.

From cube-lovers-errors@mc.lcs.mit.edu  Thu Feb 19 15:32:08 1998
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Mail-from: From zingmast@sbu.ac.uk Wed Feb 18 13:38:54 1998
Sender: zingmast@sbu.ac.uk
Date: Wed, 18 Feb 1998 18:36:07 +0000
From: David Singmaster <zingmast@sbu.ac.uk>
To: cube-lovers@ai.mit.edu
Message-Id: <009C2024.8A2B6B63.8@ice.sbu.ac.uk>
Subject: RE: Rubik lawyers up in arms over website -- Digest v23 #279]

	Norman Diamond refers to the patents expiring.  However, Rubik
only had a Hungarian patent.  As a result, the various Rubik companies
took legal action under copyright law, and copyright lasts much
longer.
	Perhaps I should hassle Seven Towns about the use of my notation!

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499; 
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Thu Feb 19 16:53:14 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Feb 18 21:36:39 1998
Date: Thu, 19 Feb 1998 02:03:04 +0000
From: David Singmaster <zingmast@sbu.ac.uk>
To: skouknudsen@email.dk
Cc: cube-lovers@ai.mit.edu
Message-Id: <009C2062.FA899020.3@ice.sbu.ac.uk>
Subject: RE: Game designers [was Re: Rubik lawyers...]

	In my Cubic Circular 1 (Autumn 1981), I recorded that Wim
Osterholt, of the Netherlands, had made and patented a 4^3 which he
showed me.  I don't remember it and I'm not sure when he brought it to
London - perhaps Summer 1981?  I also recorded that Rainier Seitz
(product manager of Arxon which was Ideal's German agent) showed me
some German patents and applications for the 4^3 and 5^3.  In Cubic
Circular 2 (Spring 1982), I record talking with another person who had
devised a 4^3 mechanism.  In Cubic Circular 3/4 (Spring/Summer 1982),
I describe playing with examples.  However, I don't recall ever
knowing who devised the mechanism that was produced for Ideal.  It was
common knowledge that it was not Rubik's mechanism.  One may be able
to get details from the web site that Oddz On (sp??) has set up.  Tom
Kremer (of Seven Towns, who is Rubik's agent) is supervising this site
and he would be one of the most likely people to know.  I have a huge
file of material comprising all US puzzle patents and I'll look there,
but I think I would have known about the patent for the 4^3 already.

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499; 
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Mon Feb 23 17:21:17 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Mon Feb 23 11:00:45 1998
Date: Mon, 23 Feb 1998 10:59:54 -0500 (Eastern Standard Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: Strong Local Maxima 9f from Start
In-Reply-To: <Pine.PMDF.3.95.980215185935.291516B-100000@PSTCC6.PSTCC.CC.TN.US>
To: Cube-Lovers <cube-lovers@ai.mit.edu>
Message-Id: <Pine.WNT.3.96.980223104308.-220451A-100000@GN209A.PSTCC.CC.TN.US>

On Sun, 15 Feb 1998, Jerry Bryan wrote:

> #1. D2 F2 L2 D' U  L2 F2 D' U'
> #2. U  D  B2 R2 U  D' L2 B2 U2
>
> These positions "look" very symmetric, especially #2, ....

The reason #2 "looks" so symmetric is that it is an isoglyph.  That is,
each face contains only two colors and the pattern of colors is the same
on all six faces.  If I am reading Dan Hoey's glyph table correctly, the
glyph is of type 20.  The glyph looks like the following, and the glyph
itself is fairly symmetric.

  XOX
  OXO
  OOO

On a lark, I asked Herbert Kociemba's Cube Explorer 1.5 program to find
all isoglyphs which can be built with this glyph.  Any such isoglyph is
likely to be pretty.  Up to symmetry, it found four isoglyphs (one of
which is #2, which is a strong local maximum in the face turn metric).
The other three are as follows:

F2 U2 L' R  D2 F2 L' R                       8f
F2 U2 B2 L2 U' B2 U' B2 L2 D2 L2 U  R2 U'   14f
U' L2 D' L2 D  B2 F2 L2 R2 D  F2 U' F2 U'   14f

Cube Explorer 1.5 was able to show that the 8f maneuver is minimal.
This position is not a strong local maximum, because the shortest strong
local maxima are 9f.

Cube Explorer 1.5 was not able to show that the 14f maneuvers are
minimal in the time I gave it (six hours each on a Pentium 166).  But I
suspect that 14f is in fact minimal.  Also, I do not know if the 14f
maneuvers are strong local maxima because my search for strong local
maxima extended only through 9f from Start.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990

From cube-lovers-errors@mc.lcs.mit.edu  Mon Feb 23 18:59:06 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Mon Feb 23 18:53:55 1998
Date: Mon, 23 Feb 1998 15:53:21 -0800 (PST)
Message-Id: <199802232353.PAA05251@denali.cs.ucla.edu>
From: Richard E Korf <korf@cs.ucla.edu>
To: jbryan@pstcc.cc.tn.us
Cc: cube-lovers@ai.mit.edu
In-Reply-To: <Pine.WNT.3.96.980223104308.-220451A-100000@GN209A.PSTCC.CC.TN.US>
	     (message from Jerry Bryan on Mon, 23 Feb 1998 10:59:54 -0500
	     (Eastern Standard Time))
Subject: Re: Strong Local Maxima 9f from Start

The two 14f move isoglyphs reported by Jerry Bryan in his last message
do indeed require 14f moves.

                        -rich

From cube-lovers-errors@mc.lcs.mit.edu  Tue Feb 24 17:38:27 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Tue Feb 24 05:49:14 1998
Date: Tue, 24 Feb 1998 11:48:55 +0100 (MET)
From: Christian <C.Eggermont@inter.nl.net>
X-Sender: ceggermo@hengstdal.nijmegen.inter.nl.net
Reply-To: Christian <C.Eggermont@inter.nl.net>
To: cube-lovers@ai.mit.edu
Subject: RE: Rubik lawyers up in arms over website -- Digest v23 #279]
In-Reply-To: <009C2024.8A2B6B63.8@ice.sbu.ac.uk>
Message-Id: <Pine.SUN.3.91.980224114558.20791B-100000@hengstdal.nijmegen.inter.nl.net>

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist wrote:

> 	Norman Diamond refers to the patents expiring.  However, Rubik
> only had a Hungarian patent.  As a result, the various Rubik companies
> took legal action under copyright law, and copyright lasts much
> longer.

I think the following URL will answer most questions and resolves the
issue:

http://www.csun.edu/~hcmth014/comicfiles/copyright.html

> 	Perhaps I should hassle Seven Towns about the use of my notation!

Mmm interesting Idea... (-:

Christian

---------------------------------------------------
E-Mail: C.Eggermont@inter.NL.net
Homepage: http://www.inter.nl.net/users/C.Eggermont
---------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Wed Feb 25 12:25:52 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Feb 25 00:39:17 1998
To: cube-lovers@ai.mit.edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Taiwanese Invention of the Cube?
Date: 25 Feb 1998 05:38:19 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <6d0aob$6vq@gap.cco.caltech.edu>

I have heard through multi-generation, unreliable sources that the Cube
was first invented and patented by a Taiwanese person.  This story
strikes me as strongly false, but perhaps may have some basis 
somewhere.  Any guesses?  Perhaps a particularly different sort of
mechanism was patented?  A trademark?  

--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
"...he put a wire in his cap and called himself Marconi."

From cube-lovers-errors@mc.lcs.mit.edu  Wed Feb 25 19:33:53 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Feb 25 16:47:43 1998
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Message-Id: <199802252147.NAA14815@liquefy.ugcs.caltech.edu>
Subject: RE: Taiwanese Invention of the Cube? (fwd)
To: zingmast@sbu.ac.uk, cube-lovers@ai.mit.edu
Date: Wed, 25 Feb 1998 13:47:07 -0800 (PST)
Reply-To: whuang@ugcs.caltech.edu

David Singmaster Computing & Maths South Bank Univ typed something like
 this in a previous message:

>From zingmast@sbu.ac.uk  Wed Feb 25 13:17:35 1998
Sender: zingmast@sbu.ac.uk
Date: Wed, 25 Feb 1998 21:14:57 +0000
From: David Singmaster Computing & Maths South Bank Univ
      <zingmast@sbu.ac.uk>
To: whuang@ugcs.caltech.edu
Message-ID: <009C25BA.E326A9B3.4@ice.sbu.ac.uk>
Subject: RE: Taiwanese Invention of the Cube?

	The earliest idea was due to someone in California, named
William O.  Gustafson (US patent 3,081,089 of 12 Mar 1963).  He had a
2^3 in the shape of a sphere, but he had the problem of keeping the
interior parts in synch with the outer parts and so he left gaps
between the pieces.  Basically he had an interior sphere with grooves
and the pieces had lips.  There are two versions - the first seems
like it wouldn't work well, if at all, but the second seems fairly
feasible.
	Unfortunately, Gustafson let his patent lapse, so the patent
of Larry Nichols was the next, with US patent 3,655,201 (applied 4 Mar
1970, issued 11 Apr 1972).  This had only the idea of a cubical puzzle
and no practical mechanism, so I don't consider it very significant,
but Nichols sued Rubik, more or less successfully - I never heard the
conclusion of the story.
	Frank Fox (UK patent 1,344,259, applied for on 9 Apr 1970 and
issued on 16 Jan 1974) seems to be next.  He had a 3^3 sphere with
tongue and grooves holding the pieces together, with a hollow center.
He had let his patent lapse also.
	In 1976-1980, Terutoshi Ishige devised and patented two
mechanisms for a 2^3., similar to Rubik's.  This may be the source of
the story you were asking about.
	However, there is another odd story.  The first French writers
on the Cube record that a old friend said he had played with such a
cube (in wood) in Istanbul in 1920 and in Marseilles about 1935.
However, no further evidence of such an early version has appeared.
	I forgot to set this to send myself a copy.  Could you forward
it back to me.  Also you might like to send it to
cube-lovers@ai.mit.edu
	Regards

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499;
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

-- 
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
Have you heard the one about the guy Jean who visited Japan?

From cube-lovers-errors@mc.lcs.mit.edu  Wed Feb 25 21:29:52 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Feb 25 18:29:45 1998
Message-Id: <9802252330.AA19883@jrdmax.jrd.dec.com>
Date: Thu, 26 Feb 98 08:30:27 +0900
From: Norman Diamond  26-Feb-1998 0830 <diamond@jrdv04.enet.dec-j.co.jp>
To: cube-lovers@ai.mit.edu
Subject: Re: Taiwanese Invention of the Cube?

Wei-Hwa Huang wote:
>I have heard through multi-generation, unreliable sources that the Cube
>was first invented and patented by a Taiwanese person.

Invention is more or less possible.  Surely someone like the famous
Mr. Wu (whose given names I've forgotten) would be able to invent it.
But if it happened, surely it would be hard to say who came first.

As for patenting, somehow the mixture of "patent" and "Taiwan" in the
same sentence strikes me as an oxymoron.

>A trademark?

Somehow the mixture of "trademark" and "Taiwan" strikes me as an
oxymoron too, even though they're not in the same sentence.
Want to try "copyright" next?  :-)

-- Norman Diamond             diamond@jrdv04.enet.dec-j.co.jp
[Speaking for Norman Diamond not for Digital.]

From cube-lovers-errors@mc.lcs.mit.edu  Fri Mar  6 19:16:00 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Mar  4 16:17:05 1998
To: cube-lovers@ai.mit.edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Re: Taiwanese Invention of the Cube?
Date: 4 Mar 1998 21:16:09 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <6dkgap$rsv@gap.cco.caltech.edu>
References: <cube-lovers.9802252330.AA19883@jrdmax.jrd.dec.com>

Norman Diamond  26-Feb-1998 0830 <diamond@jrdv04.enet.dec-j.co.jp> writes:
>As for patenting, somehow the mixture of "patent" and "Taiwan" in the
>same sentence strikes me as an oxymoron.
>Somehow the mixture of "trademark" and "Taiwan" strikes me as an
>oxymoron too, even though they're not in the same sentence.
>Want to try "copyright" next?  :-)

Is it possible to copyright the Cube?  That's why I didn't try it.

In any case, stop sneering -- Taiwan has local copyright, trademark,
and patent laws, and has had them for decades.  Sure, they haven't
honored international copyright laws, but then again, most
other countries don't think Taiwan exists as an independent
country.  When it became economically viable to honor international
copyright, they did so -- such legislation was passed in 1994.

Perhaps you are getting a biased view from living in Japan?
--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
Have you heard the one about the guy Jean who visited Japan?



From cube-lovers-errors@mc.lcs.mit.edu  Sun Mar  8 19:35:17 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sun Mar  8 03:41:58 1998
Date: Sun, 8 Mar 1998 09:41:21 +0100 (MET)
Message-Id: <1.5.4.16.19980308094102.437739b8@mailsvr.pt.lu>
To: rshep@simplex.nl
From: Georges Helm <geohelm@pt.lu>
Cc: geohelm@pt.lu, schubart@best.com, Cube-Lovers@ai.mit.edu

Hi,
You once asked a question about early rubik's cube solutions
(on Schubart's web page)
I have solution from 1979 by
ANGEVINE James
BEASLEY J.D.
CAIRNS Colin / GRIFFITHS Dave
CLAXTON Mike
DAUPHIN Michel (Mathematique et Pedadogie 24/79)
EASTER Bob
HOWLETT G.S.
JACKSON William Bradley
JOHNSON K.C.
MADDISON Richard
NELSON Roy
RODDEWIG Ulrich
SWEENEN John
TAYLOR Don (1978)
TRURAN Trevor (Computer Talk 7.11.1979)

Regards
Georges
Georges Helm

geohelm@pt.lu
http://ourworld.compuserve.com/homepages/Georges_Helm/
http://www.geocities.com/Athens/2715

From cube-lovers-errors@mc.lcs.mit.edu  Mon Mar  9 10:21:54 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sun Mar  8 18:58:07 1998
Message-Id: <9803082359.AA00210@jrdmax.jrd.dec.com>
Date: Mon, 9 Mar 98 08:59:07 +0900
From: Norman Diamond  09-Mar-1998 0859 <diamond@jrdv04.enet.dec-j.co.jp>
To: cube-lovers@ai.mit.edu
Subject: Re: Taiwanese Invention of the Cube?
Reply-To: diamond@jrdv04.enet.dec-j.co.jp, whuang@ugcs.caltech.edu

Wei-Hwa Huang replied to me:

>>As for patenting, somehow the mixture of "patent" and "Taiwan" in the
>>same sentence strikes me as an oxymoron.
>>Somehow the mixture of "trademark" and "Taiwan" strikes me as an
>>oxymoron too, even though they're not in the same sentence.
>>Want to try "copyright" next?  :-)

>Is it possible to copyright the Cube?  That's why I didn't try it.

Some puzzle designers do copyright their designs.

When one compares patents with copyrights, copyright makes sense.
Patents are intended for inventions that improve the quality of life
and will become important in industry after the patents expire, so
that the inventors starve.  Copyrights are for frivolous entertainment
like puzzles and photos, so they bring royalties for the lifetime of
the creator plus 50 years to the heirs.  One can only wonder why
patents were ever granted for puzzles.

>In any case, stop sneering -- Taiwan has local copyright, trademark,
>and patent laws, and has had them for decades.  Sure, they haven't
>honored international copyright laws,

Guess which part of that I was sneering at.

>but then again, most other countries don't think Taiwan exists as an
>independent country.

The Republic of China also thinks Taiwan doesn't exist as an independent
country.

>When it became economically viable to honor international
>copyright, they did so -- such legislation was passed in 1994.
>Perhaps you are getting a biased view from living in Japan?

No, my unbiased view was based on observations that I had made for decades.

=====

Mr. Huang and I had this discussion in private e-mail already.  I didn't
know that he was going public with it too.  Anyway if I understand
correctly, Mr. Huang agreed with my point after that, so there's no
need to repeat the rest of the discussion unless I misunderstood.

[Moderator's note: In any event, further discussion on this topic
 should be sent to Wei-Hwa Huang and Norman Diamond, rather than to
 cube-lovers.  I somewhat regret passing _any_ of it on.  The topic of
 intellectual property and its legal status is vast, and has eaten
 bigger lists than this. ]

=====

-- Norman Diamond                 diamond@jrdv04.enet.dec-j.co.jp
[Speaking for Norman Diamond not for Digital.]

From cube-lovers-errors@mc.lcs.mit.edu  Mon Mar  9 11:40:13 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sun Mar  8 20:10:06 1998
Message-Id: <l03110700b128ef688d6e@[207.172.133.11]>
Date: Sun, 8 Mar 1998 20:09:26 -0500
To: tomkeane@mail.del.net, cube-lovers <cube-lovers@ai.mit.edu>
From: Charlie Dickman <charlied@erols.com>
Subject: Rubik's Tesseract Solution

Tom and other cube-lovers,

I have completed a solution to the Rubik Tesseract and have included it in
the program and it's associated documentation but neither is ready for
prime time just yet.

I was wondering if there was anyone who would be kind enough to review the
documentation and see if the write-up of the solution is reasonably
intelligible and provide me some feedback before I make it and the program
generally available. It is an HTML document (332K self-extracting-archive)
that you can read with your browser.

Thanks,

Charlie Dickman
charlied@erols.com

From cube-lovers-errors@mc.lcs.mit.edu  Wed Mar 11 13:07:59 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Mar 11 07:40:13 1998
To: cube-lovers@ai.mit.edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Blindfold Cube-solving
Date: 11 Mar 1998 12:39:04 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <6e60l8$2bc@gap.cco.caltech.edu>

Is there anyone who knows some good techniques for blindfold cube-solving?

I can solve the cube in about 7 "peeks" or so, but that's still quite
a ways from looking at the cube once and solving it behind one's back.

--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
Smoking cigarettes are bad for you, so smoking cigarettes is bad for you.

From cube-lovers-errors@mc.lcs.mit.edu  Wed Mar 11 14:44:29 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Mar 11 13:58:59 1998
Date: Wed, 11 Mar 1998 13:58:48 -0500 (EST)
From: Jiri Fridrich <fridrich@binghamton.edu>
To: Wei-Hwa Huang <whuang@ugcs.caltech.edu>
Cc: cube-lovers@ai.mit.edu
Subject: Re: Blindfold Cube-solving
In-Reply-To: <6e60l8$2bc@gap.cco.caltech.edu>
Message-Id: <Pine.SOL.L3.93.980311134754.22079A-100000@bingsun1>

I believe that solving the cube blindfolded in one shot is very difficult
if not impossible. One could memorize the orientation of all cubies and
their permutation. Then use algorithms for turning the cubes without
moving them, and then algorithms for permuting them. One would need to
define orintation of cubies on the cube and then the permutation
algorithms would have to preserve that orientation. This system would
presume one really long "peek" and excellent memory, of course :)

Using my system (http://ssie.binghamton.edu/~jirif), I could probably
bring down the number of peeks to four with some practice ... Of course,
seven is no sweat.

Jiri

*********************************************
 Jiri FRIDRICH, Research Scientist
 Center for Intelligent Systems
 SUNY Binghamton
 Binghamton, NY 13902-6000
 Ph/Fax: (607) 777-2577
 E-mail: fridrich@binghamton.edu
 http://ssie.binghamton.edu/~jirif/jiri.html
*********************************************

From cube-lovers-errors@mc.lcs.mit.edu  Fri Mar 13 12:20:32 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Thu Mar 12 13:47:19 1998
Sender: mark@ampersand.com
To: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Cc: cube-lovers@ai.mit.edu
Subject: Re: Blindfold Cube-solving
References: <6e60l8$2bc@gap.cco.caltech.edu>
From: Mark Atwood <zot@ampersand.com>
Date: 12 Mar 1998 13:47:11 -0500
In-Reply-To: whuang@ugcs.caltech.edu's message of 11 Mar 1998 12:39:04 GMT
Message-Id: <v667ljbvbk.fsf@colon.dev.ampersand.com>

whuang@ugcs.caltech.edu (Wei-Hwa Huang) writes:
> 
> Is there anyone who knows some good techniques for blindfold cube-solving?
> 
> I can solve the cube in about 7 "peeks" or so, but that's still quite
> a ways from looking at the cube once and solving it behind one's back.

I have heard of something like "cubes for the blind". Probably either
have a different textured material attached to each cubie face, or a
Braille glyph embossed into each cubie face.

(Never tried to solve one blind, but I could probably solve on in about
a dozen or so glances. But for a while I worked on solving them with
my feet, after seeing someone do it on TV.)

-- 
Mark Atwood       | Thank you gentlemen, you are everything we have come to
zot@ampersand.com | expect from years of government training. -- MIB Zed

[ Moderator's note: You'll notice this is a different topic.  Perhaps
  Wei-Hwa Huang should consider his problem "memory solving" rather
  than "blindfold solving".  I've heard that John Conway has a good
  memory method, I think requiring five peeks (cf Roger Frye, 20 Oct
  1981).  There are also several mentions of tactile cubes in the
  archives. ]

From cube-lovers-errors@mc.lcs.mit.edu  Fri Mar 13 15:11:08 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Thu Mar 12 17:13:30 1998
Message-Id: <v03007801b12cbee0df07@[10.10.10.2]>
In-Reply-To: <6e60l8$2bc@gap.cco.caltech.edu>
Date: Wed, 11 Mar 1998 17:33:58 -0500
To: cube-lovers@ai.mit.edu
From: Kristin Looney <kristin@wunderland.com>
Subject: Re: Blindfold Cube-solving

Wei-Hwa Huang wrote:

> Is there anyone who knows some good techniques for blindfold
> cube-solving?
>
> I can solve the cube in about 7 "peeks" or so, but that's still quite
> a ways from looking at the cube once and solving it behind one's back.

This brings back fond memories of the trip to CA for the first National
Cube contest back in '81...  us nine finalists were taken on a day trip to
Disney Land and we had a race to see who could solve the cube the fastest
in the line to space mountain.  As the line winds inside the building,
it is really quite dark, and we were on our hands and knees trying to
get whatever light we could from the running lights on the floor.

I don't remember who won...  but it was a huge amount of fun.

-K.
kristin@wunderland.com
http://www.wunderland.com/wts/kristin

To all the fishies in the deep blue sea, Joy.

From cube-lovers-errors@mc.lcs.mit.edu  Fri Mar 13 16:02:47 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Thu Mar 12 16:09:47 1998
Date: Thu, 12 Mar 1998 22:09:22 +0100
Message-Id: <199803122109.WAA06383@dataway.ch>
To: Cube-Lovers@ai.mit.edu
From: Geir Ugelstad <ugelstad@dataway.ch>
Subject: Rules for speed-cubing

Hello,

What are the exact rules for speed cubeing?

I have seen that in the World-campionship it was legal to look at
the cube 15 seconds and then put it back on the table.  How long time
did it take from puting it back on the table (after looking) and the
real start???

Ys Geir Ugelstad

From cube-lovers-errors@mc.lcs.mit.edu  Fri Mar 13 17:08:16 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Thu Mar 12 23:37:59 1998
Date: Thu, 12 Mar 1998 22:34:54 -0600 (CST)
From: "J. David Blackstone" <jxb9451@utarlg.uta.edu>
Subject: Oddz On website
In-Reply-To: <009C2062.FA899020.3@ice.sbu.ac.uk>
To: David Singmaster <zingmast@sbu.ac.uk>
Cc: skouknudsen@email.dk, cube-lovers@ai.mit.edu
Message-Id: <Pine.PMDF.3.95.980312223127.539020398A-100000@UTARLG.UTA.EDU>
Mime-Version: 1.0
Content-Type: TEXT/PLAIN; charset=US-ASCII

On Thu, 19 Feb 1998, David Singmaster wrote:
> common knowledge that it was not Rubik's mechanism.  One may be able
> to get details from the web site that Oddz On (sp??) has set up.  Tom

  I may have missed it, but could someone provide the URL of this website?

-----------------------------------------
J. David Blackstone
jxb9451@utarlg.uta.edu
http://www.geocities.com/Athens/Acropolis/1341
-----------------------------------------



From cube-lovers-errors@mc.lcs.mit.edu  Tue Mar 17 10:14:50 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Fri Mar 13 14:48:44 1998
From: Phil Servita <meister@epilogue.com>
Sender: meister@khitomer.epilogue.com
To: cube-lovers@ai.mit.edu
Subject: not quite blind cubing
Date: Fri, 13 Mar 98 14:48:43 -0500
Message-Id:  <9803131448.aa12167@khitomer.epilogue.com>


whuang@ugcs.caltech.edu (Wei-Hwa Huang) writes:
> 
> Is there anyone who knows some good techniques for blindfold cube-solving?
> 
> I can solve the cube in about 7 "peeks" or so, but that's still quite
> a ways from looking at the cube once and solving it behind one's back.

Back when i was still in college, myself and a friend would occasionally
perform our "geek party trick", which was that we would sit on the floor, 
back-to-back, and someone would toss one of us a scrambled cube. Whoever 
caught it would look at it, make a single quarter-turn on it, and pass it 
over their shoulder to the other person, who would look at it and make another
quarter turn, pass it back, and so on. We could solve it in this fashion in 
just under 2 minutes. 

                                              -phil

From cube-lovers-errors@mc.lcs.mit.edu  Tue Mar 17 10:46:50 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sat Mar 14 03:49:32 1998
Message-Id: <3.0.3.32.19980313231431.00835810@netcom13.netcom.com>
Date: Fri, 13 Mar 1998 23:14:31 -0800
To: Mark Atwood <zot@ampersand.com>
From: Ray Tayek <rtayek@netcom.com>
Subject: Re: Blindfold Cube-solving
Cc: cube-lovers@ai.mit.edu
In-Reply-To: <v667ljbvbk.fsf@colon.dev.ampersand.com>

At 01:47 PM 3/12/98 -0500, Mark Atwood wrote:
>...
>I have heard of something like "cubes for the blind". Probably either
>have a different textured material attached to each cubie face, or a
>Braille glyph embossed into each cubie face.
>...

my wife teaches blind kids. do you know where i could get some braile cubes?

thanks

Ray (will hack java for food) http://home.pacbell.net/rtayek/
hate Spam? http://www.compulink.co.uk/~net-services/spam/

[ Moderator's note:  There are quite a few notes in the archives about
  adding tactile labels to cubes.  Adding characters in Braille should
  be about the easiest thing to do--I'm sure she has a DYMO embosser. ]

From cube-lovers-errors@mc.lcs.mit.edu  Tue Mar 17 11:02:50 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sun Mar 15 15:27:20 1998
From: roger.broadie@iclweb.com (Roger Broadie)
To: "Cube Lovers Submissions" <Cube-Lovers@ai.mit.edu>
Subject: Ideal's patent for 4^3
Date: Sun, 15 Mar 1998 20:29:20 -0000
Message-Id: <19980315202713.AAA21006@home>

On 19 Feb 1998 David Singmaster wrote:

        In my Cubic Circular 1 (Autumn 1981), I recorded that Wim
        Osterholt, of the Netherlands, had made and patented a 4^3
        which he showed me.  I don't remember it and I'm not sure when
        he brought it to London - perhaps Summer 1981? I also recorded
        that Rainier Seitz (product manager of Arxon which was Ideal's
        German agent) showed me some German patents and applications
        for the 4^3 and 5^3.  In Cubic Circular 2 (Spring 1982), I
        record talking with another person who had devised a 4^3
        mechanism.  In Cubic Circular 3/4 (Spring/Summer 1982), I
        describe playing with examples.  However, I don't recall ever
        knowing who devised the mechanism that was produced for Ideal.

        It was common knowledge that it was not Rubik's mechanism 

I have just come across Ideal's patent for its 4^3. It is US Patent No
4,421,311.  The inventor was Peter Sebesteny, and the original application
was made in Germany on 8 Feb 1981, so it may have been one of the patents
David Singmaster was shown.  It can be viewed at the IBM patent site from 

        http://www.patents.ibm.com/details?patent_number=4421311

One of the references cited by the US Patent Examiner was to page 29 of
David Singmaster's "Notes on Rubik's Magic Cube" - undoubtedly the remark
"One can imagine the 4x4x4 cube or the 3x3x3x3 hypercube.  The first might
be makeable but its group seems to be much more complicated.  The second is
unmakeable, but its group structure may be determinable."

The corresponding European patent application was taken through to the
point where it was ready for grant, but then allowed to lapse.  The next
stage would have been quite expensive and have required Ideal to translate
the specification into the languages of the European countries in which it
was to be in force.  And the US was not renewed when the first renewal fees
became due in 1986.  Presumably by then Ideal had lost interest in the
patent - they may have calculated there was zero chance of anyone launching
an imitation, given the number of 4^3s that had been left unsold.

I don't have ready access to information about the German application, but
I suspect it was applied for by Sebesteny on his own behalf, and he then
interested Ideal in it.  

Roger Broadie

From cube-lovers-errors@mc.lcs.mit.edu  Tue Mar 17 11:43:37 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sun Mar 15 06:19:03 1998
To: cube-lovers@ai.mit.edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Re: Blindfold Cube-solving
Date: 15 Mar 1998 11:17:40 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <6egdck$cvj@gap.cco.caltech.edu>
References: <cube-lovers.v667ljbvbk.fsf@colon.dev.ampersand.com>

The Moderator wrote:
>[ Moderator's note: You'll notice this is a different topic.  Perhaps
>  Wei-Hwa Huang should consider his problem "memory solving" rather
>  than "blindfold solving".  I've heard that John Conway has a good
>  memory method, I think requiring five peeks (cf Roger Frye, 20 Oct
>  1981).  There are also several mentions of tactile cubes in the
>  archives. ]

I used the term "blindfold solving" patterned after "blindfold chess",
where two players merely recite moves to each other, using no actual
pieces or board.  

As far as "solving in the dark" goes, it reminds me that I have a cube
in which under certain lamps, the yellow and white colors are 
indistinguishable.  Solving such a cube can occasionally give a few 
trip-ups!

--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
Smoking cigarettes are bad for you, so smoking cigarettes is bad for you.

From cube-lovers-errors@mc.lcs.mit.edu  Tue Mar 17 14:28:15 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Tue Mar 17 13:30:17 1998
Date: Tue, 17 Mar 1998 13:30:10 -0500 (Eastern Standard Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: Blindfold Cube-solving
In-Reply-To: <6egdck$cvj@gap.cco.caltech.edu>
To: Wei-Hwa Huang <whuang@ugcs.caltech.edu>
Cc: cube-lovers@ai.mit.edu
Message-Id: <Pine.WNT.3.96.980317132801.-960339T-100000@GN209A.PSTCC.CC.TN.US>

On Sun, 15 Mar 1998, Wei-Hwa Huang wrote:

> As far as "solving in the dark" goes, it reminds me that I have a cube
> in which under certain lamps, the yellow and white colors are
> indistinguishable.  Solving such a cube can occasionally give a few
> trip-ups!

I have had the same problem with orange and red, especially on my 2x2x2.
I have a "latter day" 2x2x2 (my kids lost my first one), and the colors
in general do not seem quite true to the colors on my 3x3x3 and 4x4x4
cubes.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990

From cube-lovers-errors@mc.lcs.mit.edu  Tue Mar 24 12:51:28 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Tue Mar 24 11:52:07 1998
Message-Id: <3517E49F.DF5B21BF@mail.retina.ar>
Date: Tue, 24 Mar 1998 13:51:44 -0300
From: Isidro <reichen@mail.retina.ar>
Reply-To: isidroc@usa.net
Organization: Frank Zappa's Fan Club
To: Cube Lovers Submissions <Cube-Lovers@ai.mit.edu>
Subject: 5^3 quiz

I need to know the answers for these questions:
Who invented 5^3?
What is the commercial name?
How many cubies it has?

--
Isidro: isidroc@usa.net

[ Moderator's note: There was a note last July mentioning "Rubik's
  Wahn (5x5x5) (maybe also called Professor's cube, Ultimate or Master
  Revenge)"--any other names?
  The number of cubies is obviously 98--why didn't you just count them? ]

From cube-lovers-errors@mc.lcs.mit.edu  Wed Mar 25 10:09:36 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Tue Mar 24 16:14:21 1998
Message-Id: <3.0.5.16.19980324220550.0bd76334@vip.cybercity.dk>
Date: Tue, 24 Mar 1998 22:05:50
To: cube-lovers@ai.mit.edu
From: Philip Knudsen <skouknudsen@email.dk>
Subject: RE: 5^3 quiz

To my knowledge, the 5x5x5 was invented by Udo Krell. It was produced by
Uwe Meffert in 1983. I read somewhere that Dr. Chr. Bandelow had the Hong
Kong factory finish extra puzzles from previously manufactured parts around
1990, don't know if this is true. Bandelow is still selling this puzzle,
under the name "Giant Magic Cube". It also seems Meffert reissued the 5x5x5
one or two years ago, under the name "Professor's Cube". This new version
might have other colors than the original. I have seen the puzzle under the
name "Ultimate Cube" several times, the name "Master Revenge" however is
new to me.
Since Meffert is the manufacturer, the "most" official name for the 5x5x5
is probably "Professor's Cube".

Philip K
recording and performing artist
Vendersgade 15, 3th
DK - 1363 Copenhagen K
Phone:  +45 33932787
Mobile: +45 21706731
E-mail: skouknudsen@email.dk
E-mail: philipknudsen@hotmail.com
Sms: 4521706731@sms.tdk.dk (short message, no subject)

From cube-lovers-errors@mc.lcs.mit.edu  Wed Mar 25 12:56:17 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Tue Mar 24 16:46:37 1998
Date: Tue, 24 Mar 1998 22:46:49 +0100 (MET)
Message-Id: <199803242146.WAA06298@relay.euronet.nl>
To: Cube-Lovers@ai.mit.edu
From: Sytse <4xs2fs@euronet.nl>
Subject: Re: 5^3 quiz

Isidro,

Who invented 5^3?

At least I did. In 1982 I designed and built a 5^3 cube, all in plywood.
Although I did not aplly for a patent or other registration, as I was only a
schoolboy by then, the local newspaper recorded this event. As the wooden
prototype was not as speedy as necessary, I later designed a simulator for
the Sinclair ZX Spectrum (a then so called 'personal computer' with an
amazing 48K RAM memory). This simulator also included a 6^3 cube. 7^3 was
not possible as this did not fit in the screen, which was my parents
television set. Oh, those were the days!

Nowadays I am an architect.

Kind regards,

Sytse de Maat

P.S. If you happen to know other designers of 5^3, please mail me.

[ Moderator's note: Can you describe the design that held the plywood
  model together while allowing it to turn? ]

From cube-lovers-errors@mc.lcs.mit.edu  Wed Mar 25 15:19:48 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Mar 25 02:37:53 1998
Date: Wed, 25 Mar 1998 08:37:42 +0100
Message-Id: <199803250737.IAA30286@dataway.ch>
To: Cube-Lovers@ai.mit.edu
From: Geir Ugelstad <ugelstad@dataway.ch>
Subject: Jiri's system for solving Rubiks's cube

hello cube-lovers

For all of you that haven't been into Jiri's home page at 
http://ssie.binghamton.edu/~jirif, you should realy look into it!
Bouth the method and presentation is of very high standard!
I bought myself a system in 1982 but I was so dissapointed that I
trow it In the garbage just after.  With the system I bought in 1982 
it was not possible to make it faster than 2-3 minutes.  With Jiri's
system it should be possible in about 17 sec.!

Ys Geir Ugelstad

PS: Question to Jiri.  How far are you able to do the foreplanning
the 15 sec.  before the time start to run?  Hopefully longer than
"Place the four edges from the first layer"?

From cube-lovers-errors@mc.lcs.mit.edu  Thu Mar 26 11:46:40 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Mar 25 16:14:40 1998
Date: Wed, 25 Mar 1998 16:10:54 -0500 (EST)
From: Jiri Fridrich <fridrich@binghamton.edu>
To: Geir Ugelstad <ugelstad@dataway.ch>
Cc: Cube-Lovers@ai.mit.edu
Subject: Re: Jiri's system for solving Rubiks's cube
In-Reply-To: <199803250737.IAA30286@dataway.ch>
Message-Id: <Pine.SOL.L3.93.980325160352.11481B-100000@bingsun1>

On Wed, 25 Mar 1998, Geir Ugelstad wrote:

> it was not possible to make it faster than 2-3 minutes.  With Jiri's
> system it should be possible in about 17 sec.!

Yes, you are right - with my system AND a lot of time on your hands :) I
am pretty sure that the systems of other top speed cubists are at
least as as good as mine. The system is only half of the secret.

> PS: Question to Jiri.  How far are you able to do the foreplanning
> the 15 sec.  before the time start to run?  Hopefully longer than
> "Place the four edges from the first layer"?

Nope. 15 seconds is not a long time to plan more than the four edges. Of
course, as you proceed, you will usually be able to spot the corners with
their appropriate cubies from the second layer in some nice position and
continue without delays ...

Jiri

*********************************************
 Jiri FRIDRICH, Research Scientist
 Center for Intelligent Systems
 SUNY Binghamton
 Binghamton, NY 13902-6000
 Ph/Fax: (607) 777-2577
 E-mail: fridrich@binghamton.edu
 http://ssie.binghamton.edu/~jirif/jiri.html
*********************************************

From cube-lovers-errors@mc.lcs.mit.edu  Thu Mar 26 12:46:25 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Mar 25 18:23:23 1998
Message-Id: <9803252324.AA16745@jrdmax.jrd.dec.com>
Date: Thu, 26 Mar 98 08:24:24 +0900
From: Norman Diamond  26-Mar-1998 0817 <diamond@jrdv04.enet.dec-j.co.jp>
To: cube-lovers@ai.mit.edu
Subject: RE: 5^3 quiz

I bought my first 5^3 from a department store in Japan in 1985, while it
was alongside the 3^3 and 4^3 on the mass market.  Bought my second one
from Dr. Bandelow some time later.  In Japan it was called "Professor Cube"
which could be taken as "Professor's Cube" because it would be a bit too
awkward to pedantically insert the syllable for possessive form (in
Japanese grammar) between two polysyllabic foreign words.

(Tangential details:
pu-ro-fue-so-ru kyu-u-bu is 5 + 3 syllables, while
pu-ro-fue-so-ru no kyu-u-bu would be 5 + 1 + 3 syllables.)

The magic dodecahedron reached the mass market around 1989 or so.

Those were the days.  Some time around 1993, the mass market shifted to
computer games.

-- Norman Diamond                diamond@jrdv04.enet.dec-j.co.jp
[Speaking for Norman Diamond not for Digital]

From cube-lovers-errors@mc.lcs.mit.edu  Thu Mar 26 15:10:34 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Thu Mar 26 10:47:00 1998
Date: Thu, 26 Mar 1998 15:36:25 +0000
From: David Singmaster <david.singmaster@sbu.ac.uk>
To: skouknudsen@email.dk
Cc: cube-lovers@ai.mit.edu
Message-Id: <009C3C55.665587E6.39@ice.sbu.ac.uk>
Subject: RE: 5^3 quiz

Bandelow's leaflet, which he encloses with the 5^3, states that the
mechanism was invented by Udo Krell, of Hamburg(?).  I haven't seen
the patent but perhaps Bandelow has details.

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499; 
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Thu Mar 26 15:58:05 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Mar 25 19:10:32 1998
Message-Id: <01BD5821.7C9449E0@jburkhardt.ne.mediaone.net>
From: John Burkhardt <jburkhardt@mediaone.net>
To: <cube-lovers@ai.mit.edu>
Subject: new to list
Date: Wed, 25 Mar 1998 19:09:07 -0500

Hi, I just found and joined this list.

So I am looking for any and all oddball cube variations I can find.
Does anyone have anything to sell or trade.  I can trade for a "Magic
Dodecahedron" which is the start shaped Hungarian version of the
Megaminx and I might be willing to part with a 5x5x5 cube for anything
really interesting.  I'm looking for an original Tomy Megaminx.  Also
the octahedron puzzle which is like two Pyraminx's glued together
(there might be an official name).  I am also searching for a 4x4x4
but I know they are really hard to find these days (mostly because
they tend to break).

The Dodecahedron puzzle is really amazing.  It was actually harder
than the 5x5x5 cube.  IT took me about 3 hours to work it out!  I
think once you know the 3x3x3 then all the same moves do similar
things and you can easily solve 4x4x4 or 5x5x5 with variations.  Of
course there are some cool things you can do with these.

I must say that I was disappointed with one web page that listed a
bunch of moves for the 3x3x3 cube.  I was trying some of them out and
thinking, my god, how did anyone figure this out, only to then
discover that a computer had figured them out.  OK, that's certainly
an interesting problem, but I have much more fun discovering them on
my own.  Interstingly enough, solving the dodecahedron led me to some
neat new moves for the original cube!

So where can we go from here? Have we made all the regular polyhedra
into puzzles?  Is there hope of actually building 6x6x6 and beyond
cubes?  Is there really any point to doing it?  I suppose they would
allow for some nice patterns. Does anyone know of any puzzles that are
not in George Helm's collection?  I just bought a Magic Cube puzzle at
Walgreens for $3.  It's a 3x3x3 with psychedelic stickers on it...

From cube-lovers-errors@mc.lcs.mit.edu  Fri Mar 27 09:48:24 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Fri Mar 27 06:50:30 1998
Message-Id: <01BD594B.F9EFBF20@jburkhardt.ne.mediaone.net>
From: John Burkhardt <jburkhardt@mediaone.net>
To: <cube-lovers@ai.mit.edu>
Subject: Stickers
Date: Fri, 27 Mar 1998 06:45:47 -0500

Does anyone know where to find cube stickers?  They must come from
somewhere!  I found some vinyl lettering once and the periods were
exactly the right size for a 5x5x5 cube.  But they don't come in
orange.  There must be a way to buy sheets of the stuff.  Any ideas?

From cube-lovers-errors@mc.lcs.mit.edu  Fri Mar 27 13:44:27 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Fri Mar 27 11:01:41 1998
Date: Fri, 27 Mar 1998 11:02:04 -0500 (EST)
From: Nichael Cramer <nichael@sover.net>
To: John Burkhardt <jburkhardt@mediaone.net>
Cc: cube-lovers@ai.mit.edu
Subject: Re: Stickers
In-Reply-To: <01BD594B.F9EFBF20@jburkhardt.ne.mediaone.net>
Message-Id: <Pine.BSI.3.91.980327105221.2759A-100000@granite.sover.net>

John Burkhardt wrote:

> Does anyone know where to find cube stickers?  They must come from
> somewhere!  I found some vinyl lettering once and the periods were
> exactly the right size for a 5x5x5 cube.  But they don't come in
> orange.  There must be a way to buy sheets of the stuff.  Any ideas?

Ah, yes, the orange stickers on the 5X ....    ;-)


Anyway, don't they have sticker sets in any colors other than in the
standard cube-pallette?  Black or grey come to mind.  Not quite the
optimal solution, of course, but it would still give you a useable cube. 

Nichael

--
Nichael Cramer
work: ncramer@bbn.com
home: nichael@sover.net
http://www.sover.net/~nichael/


(The cool bit about letters, of course, is that on the 5X5 face in
question, you could, say, put almost all the letters of the alphabet --or
some other personalized message(s) of your choice-- and give yourself a
little something extra to shoot for as you solve the cube.)

From cube-lovers-errors@mc.lcs.mit.edu  Mon Mar 30 14:54:27 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Fri Mar 27 20:28:38 1998
Date: Fri, 27 Mar 1998 20:28:57 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: All the Isoglyphs [long]
To: Cube-Lovers <cube-lovers@ai.mit.edu>
Message-Id: <Pine.PMDF.3.95.980327202039.282847A-100000@PSTCC6.PSTCC.CC.TN.US>


Dan Hoey introduced glyphs and isoglyphs on 5 August 1997.  A glyph is a
cube face containing no more than two colors, and an isoglyph is a cube
position where every face contains the same glyph.  Isoglyphs tend to be
very striking and pretty patterns. 

Each corner and edge facelet of a glyph can be the same or a different
color than the center facelet, so there are 2^8 or 256 possible glyphs. 
Dan reported that there are 51 glyphs unique up to symmetry (70 if chiral
pairs are distinguished). 

On 8 August 1997, Herbert Kociemba reported that there are 35 continuous
isoglyphs unique up to symmetry (including Start).  A continuous isoglyph
is one for which each glyph matches the neighboring glyph along the edge.
Herbert did not include the non-continuous glyphs because there are so
many, and because non-continuous glyphs are sometimes not so striking and
pretty as the continuous glyphs. 

On 9 August, Dan Hoey classified Herbert's isoglyphs according the their
respective glyphs, and provided the usual name for the isoglyphs where a
usual name existed.  Where a usual name did not exist, Dan provided a
reasonable name based on the names of closely related isoglyphs. 

On 27 August, Mike Reid gave minimal maneuvers for all the continuous
isoglyphs in both the quarter-turn and face-turn metrics. 

I have now calculated all the isoglyphs, using Herbert's Cube Explorer 1.5
program.  All I really did was to put each of the 51 glyphs into the
program in turn.  I can only guess, but this has to be more or less what
Herbert did to obtain his results.  The only difference is that I asked
the program to calculate both continuous and non-continuous isoglyphs, so
the task was a bit bigger.  My report is much in the spirit of Herbert's
original report.  I have made no effort to calculate minimal maneuvers,
nor have I made any attempt to associate names with the maneuvers. 

However, my report does include all the glyphs along with their associated
isoglyphs.  In fact, for each glyph I have included the entire equivalence
class of glyphs under the rotations and reflections of the square (either
1, 2, 4, or 8 glyphs in each equivalence class).  There is, of course, no
necessary relationship between the number of glyphs in the equivalence
class and the number of isoglyphs.  You only need to put one glyph from
the equivalence class into Cube Explorer 1.5 to create the isoglyph, and
any one glyph from the equivalence class will do as well as any other.

I can report that of the 51 glyphs unique up to symmetry, 8 of them
produce only continuous isoglyphs, 17 of them produce only non-continuous
isoglyphs, 14 of them produce both continuous and non-continuous
isoglyphs, and 12 of them produce no isoglyphs. 

In addition to confirming Herbert's figure of 35 continuous isoglyphs, I
can report that there are 249 non-continuous isoglyphs.  In the category
of "most isoglyphs", one glyph has 2 continuous and 49 non-continuous
isoglyphs, and another has 4 continuous and 46 non-continuous isoglyphs. 

The only other thing that probably requires explanation about the chart
that follows is that there is a two character code below each glyph.  This
is a hexadecimal representation of a binary number based on the following
pattern,

  765
  4X3
  210

where the number includes 2^k if facelet k is the same color as the center
facelet.  This is not intended as a new classification to replace Dan's. 
It is just a bookkeeping technique I used (a 16x16 matrix) to keep track
of the 256 glyphs. 


000
0X0
000

 00

D' U  L' R B' F D' U  (8) *    continuous






000 000 00X X00
0X0 0X0 0X0 0X0
00X X00 000 000

 01  04  20  80

R2 D L2 U' B2 D' U2  R' F' U R B' L' D' F L2 B2 R U'  (19) continuous
B2 D F2 U' L2  B' D2 B U B' D2 F L R' D U F'  (17)         continuous





000 000 000 0X0
0X0 0XX XX0 0X0
0X0 000 000 000

 02  08  10  40

D' U  B D' L' R F D' B' D' U L  (12) *               continuous
L2 D'  B' F L' D U' F L' R U B' F'  (13)             not
F2 D' L2  B' D' U' R B L F L F U' F'  (14)           not
F2 D L2 R2 U'  B' U L D L D2 R' F' D' B' D  (16)     not
U R2 D B2 D  F D' B' L' B' D2 F' L' F U F' R'  (17)  not
R2 U2  B' F D B2 L' R D2 F' R2 F2 D' U  (14)         not
B2 D2 R2  B' F D' F2 L' R U2 F' L2 D U  (14)         not
D' U' L2  F' D2 L R' B2 D' B F' R2 D2 B2  (14)       not
B2 D U' L2  F' D2 L R' B2 D' B F' R2  (13)           not
R2 D2  R' B' L' B D' R2 B' R B' D' R  (13)           not




000 000 000 000 00X 0XX X00 XX0
0X0 0X0 0XX XX0 0XX 0X0 XX0 0X0
0XX XX0 00X X00 000 000 000 000

 03  06  09  14  28  60  90  C0

F2 D F2 D B2 L2 U L2 D'  L D L' B' L U' F' U R' U'  (19)       not
F2 U L2 U L2 U'  F U' F' D2 B L R U' B' D' R F'  (18)          not
D' R2 D2 B2 U' F2 U' L2  B D R D' U F U' B' U2 B' R' U'  (20)  not





000 00X X00 X0X
0X0 0X0 0X0 0X0
X0X 00X X00 000

 05  21  84  A0

F2 U2  L' R D2 F2 L' R  (8)                      not
F2 U2 B2 L2 U' B2 U' B2 L2 D2 L2 U R2 U'  (14)    not
U' L2 D' L2 D B2 F2 L2 R2 D F2 U' F2 U'  (14)     not
U2 L2 F2 D U' B2 L2 D' U'  (9)                    not





000 00X X00 XXX
0X0 0XX XX0 0X0
XXX 00X X00 000

 07  29  94  E0

(none)






000 000 0X0 0X0
0XX XX0 0XX XX0
0X0 0X0 000 000

 0A  12  48  50

F2 D' R2 D'  L' U' L' R B D' U B L F2 L U2  (16)            continuous
U B2  L D B' F L' D U' L' R F' D2 R'  (14)                  continuous
U' B2 R2 U2  F' D2 L' F2 U' F2 D2 F U2 R' U2  (15)          continuous
D2 U B2 D U'  R' D2 B' R2 D2 L' R' D' B2 L B  (16)          not
B2 U2 F2 D2 F2 U  R' F' L2 U2 L R U' L2 F2 L' F  (17)       not
U F2 L2 U2  B' U2 L F2 U B2 D2 B D2 R' U2  (15)             not
F2 D2 B2 D' B2 L2 U2  B D2 R F2 D F2 D2 B' U2 L F2  (18)    not
U2 B2 F2 D' F2 R2 D2  B U2 L' U2 B' D2 F2 D F2 R' U2  (18)  not
D2 U' B2 U2  F D2 L D2 F' D2 L2 F2 U B2 R' U2  (16)         not
F2 U R2 D U  B' D' B' D' F L' F D' U2 L U2  (16)            not
D2 U' B2 U2  F D2 L D2 F U2 R2 B2 U' B2 R' U2  (16)         not
U2 F2 D F2 L2 U2  F' U2 L F2 D B2 D2 F D2 R' F2  (17)       not
F2 D L2 D2 B2 R2 B2 L2 F2 U2  R B U L U B U' L' U  (19)     not





000 000 0XX XX0
0XX XX0 0XX XX0
0XX XX0 000 000

 0B  16  68  D0

U2 F2 R2 U' L2 D  B R' B R' B R' D' L2 U'  (15)    continuous





000 000 00X 00X 0X0 0X0 X00 X00
0XX XX0 0X0 XX0 0X0 0X0 0X0 0XX
X00 00X 0X0 000 00X X00 0X0 000

 0C  11  22  30  41  44  82  88

D U2 L2 U R2 U' L2 U  R' B2 L2 F' L2 B' R' F' L D U'  (19)   continuous
U' F2 L2 D2 U F2 U2  F' L' D2 B2 R' D' B R' U L2 B2 F'  (19) not





000 000 00X 0XX X00 X0X X0X XX0
0XX XX0 0X0 0X0 0X0 0XX XX0 0X0
X0X X0X 0XX 00X XX0 000 000 X00

 0D  15  23  61  86  A8  B0  C4

D2 L2 F2 R2 U2 B2 D2 F2 R2 U2 R2 U2  (12)   not
U2 L2 B2 L2 U2 F2 U2 F2 L2 U2 R2 U2  (12)   not
U2 R2 B2 L2 U2 F2 U2 F2 R2 U2 R2 U2  (12)   not
D2 R2 F2 R2 U2 B2 D2 F2 L2 U2 R2 U2  (12)   not





000 000 00X 0X0 0X0 0XX X00 XX0
0XX XX0 0XX 0XX XX0 XX0 XX0 0XX
XX0 0XX 0X0 00X X00 000 0X0 000

 0E  13  2A  49  54  70  92  C8

(none)





000 000 00X 0XX X00 XX0 XXX XXX
0XX XX0 0XX 0XX XX0 XX0 0XX XX0
XXX XXX 0XX 00X XX0 X00 000 000

 0F  17  2B  69  96  D4  E8  F0

D2 R2 F2 U2 F2 U2 F2 U2 R2 B2              (10)     not
F2 L2 D2 B  F  R2 B  F' R2                 (9)      not
F2 U2 L2 F2 D  U  R2 F2 D  U' B2           (11)     not
U2 L2 B2 D2 F2 U2 F2 U2 R2 B2              (10)     not
U2 L2 B2 U2 B2 D2 F2 U2 R2 B2              (10)     not
U2 F2 L2 B2 U2 B2 D2 F2 U2 R2              (10)     not
L2 D2 F2 L2 U' B2 L2 R2 F2 D' R2           (11)     not
D2 R2 F2 D2 B2 D2 F2 U2 R2 B2              (10)     not
U2 L2 R2 D  F2 U' R2 F2 U2 F2 D' U2 F2 U'  (14)     not
D' R2 D' B2 U2 B2 F2 R2 B2 F2 U' F2 U'     (13)     not
U' B2 U' F2 D2 B2 F2 R2 B2 F2 U' F2 U'     (13)     not
B2 U  B2 U' L2 D2 F2 U' R2 U  F2           (11)     not
D' R2 D' F2 D2 L2 R2 U' B2 F2 D  R2 U' F2 U'  (15)  not
L2 D  F2 U' R2 F2 U2 F2 D  B2 U' B2 U2     (13)     not
F2 D  F2 U' R2 U2 F2 U' R2 D B2            (11)     not
D2 B2 U' L2 U  B2 U  B2 D' R2 D' R2 U'     (13)     not
D' B2 D  L2 D2 B2 U  B2 U'                 (9)      not
D' L2 B2 D  L2 D2 B2 U  B2 R2 U'           (11)     not
D' U' L2 D2 U2 B2 D' U'                    (8)      not
L2 U2 B2 L2 D  B2 L2 R2 F2 U'              (10)     not
L2 U2 R2 D' U' B2 R2 B2 D' U'              (10)     not
L2 D2 L2 B2 U2 F2 D2 F2 R2 F2 R2 U2        (12)     not
L2 D2 B2 D2 F2 R2 B2 R2 F2 U2 R2 U2        (12)     not
B  F  D2 L2 B  F  (6) *                             not





000 0X0
XXX 0X0
000 0X0

 18  42

L2 U2 L2 R2 U2 L' R'  (7) *      not





000 000 00X 0X0 0X0 0XX X00 XX0
XXX XXX XXX 0X0 0X0 0X0 XXX 0X0
00X X00 000 0XX XX0 0X0 000 0X0

 19  1C  38  43  46  62  98  C2

D2 L2 D R2 U B2 U2  B R' B' D B2 R' F R2 F' U R'  (18)  not





000 0X0 0X0 0X0
XXX 0XX XX0 XXX
0X0 0X0 0X0 000

 1A  4A  52  58

D  F2 R2 F2 R2 U  F2 R  F2 R  D2 U' F  L' F' L  D         (17)   continuous
B2 L2 U' B2 F2 D2 B2 R  B' F2 U' B' D2 L' B' U  L2 D' U'  (19)   not
D' B2 U' B2 F2 D  F2 D2 F  L2 U  L  F' D' F2 L' U'        (17)   not
L2 F2 U  B2 U2 F2 R2 B2 R2 F  R2 D  F2 R' D' B' D' B' R' U  (20) not
F2 D2 R2 B2 D2 F2 D' F2 L' B2 U' L  B' L  D  L  B' R2 U' F2  (20) not
L2 D  R2 U' L2 F2 L2 D' B2 F' L  R  B  D2 R' B  F  L  F2 U'  (20) not
L2 D2 U2 L' U2 L' R2 D2 U2 R' U2 R'                          (12) not
U2 R2 B2 F2 D2 U2 L' B2 F2 R' D2 L' R                        (13) not





000 000 0X0 0X0 0XX 0XX XX0 XX0
XXX XXX 0XX XX0 0XX XXX XX0 XXX
0XX XX0 0XX XX0 0X0 000 0X0 000

 1B  1E  4B  56  6A  78  D2  D8

U  R2 U' F' U2 F2 U2 F  R  F2 R' U  R2 U'  (14)             continuous
B2 D  U2 R2 D  F2 B' L2 D2 F  L  F2 L' R2 F' U' F' U  (18)  not





000 0XX X0X XX0
XXX 0X0 XXX 0X0
X0X 0XX 000 XX0

 1D  63  B8  C6

F2 L' R  B2 U2 L R' D2     (8)   not
D' U' B2 L2 D' U R2 F2 U2  (9)   not





000 0XX XX0 XXX
XXX 0XX XX0 XXX
XXX 0XX XX0 000

 1F  6B  D6  F8

 (none)





00X X00
0X0 0X0
X00 00X

 24  81

 (none)





00X X00 X0X X0X
0X0 0X0 0X0 0X0
X0X X0X 00X X00

 25  85  A1  A4

D' B2 L2 F2 R2 F2 U R2 U2 F' R  B  L  D  B  U' F R' U2 R   (20) continuous
B2 L2 R2 U  R2 B2 U L2 U' B  F  D2 L' B2 R2 D' U B' L' R'  (20) continuous
F2 R2 U2 B2 D' R2 D L2 D2 R2 F2 U' R2 U'  (14)                  not





00X 00X 00X 0XX X00 X00 X00 XX0
0X0 0XX XX0 0X0 0X0 0XX XX0 0X0
XX0 X00 X00 X00 0XX 00X 00X 00X

 26  2C  34  64  83  89  91  C1

 (none)





00X 00X X00 X00 X0X X0X XXX XXX
0X0 0XX 0X0 XX0 0XX XX0 0X0 0X0
XXX X0X XXX X0X 00X X00 00X X00

 27  2D  87  95  A9  B4  E1  E4

L2 U' R2 D U2  L' B2 F' D' R' B2 D L2 R2 U2 F' L U  (18) continuous
B R2 B' F2 L2 B'  L' D2 R D' L2 U F' R2 B' L B U  (18)   not





00X 0XX X00 XX0
0XX XX0 XX0 0XX
XX0 X00 0XX 00X

 2E  74  93  C9

 (none)





00X X00 XXX XXX
0XX XX0 0XX XX0
XXX XXX 00X X00

  2F  97  E9  F4

U' L2 R2 F2 U L2 U' F2 R2 L' U B' R  D' B2 D2 B' R'  (18)  continuous
R2 B2 D  B2 D U  R2 D' B' D' R F2 R' D  B  U'  (16)        continuous
D' L2 U' F2 U F2 U2 F2 D' L2 U B2 U'  (13)                 not





00X 0X0 X00 X0X
XX0 0X0 0XX 0X0
00X X0X X00 0X0

  31  45  8C  A2

 (none)





00X 0X0 0X0 X00
XX0 0XX XX0 0XX
0X0 X00 00X 0X0

 32  4C  51  8A

D  U  L2 B2 D U' F' U F' R  F2 R' F  D' B2 L2 D' U'  (18)    continuous
R2 B2 D2 L2 U L2 D  B L2 U2 B2 L' R2 F2 D' U  R  U2 R'  (19) continuous





00X 0X0 0X0 0XX X00 X0X X0X XX0
XX0 0XX XX0 XX0 0XX 0XX XX0 0XX
0XX X0X X0X 00X XX0 0X0 0X0 X00

 33  4D  55  71  8E  AA  B2  CC

R2 B2 D  U  L' R' D2 L' R  D  U  (11)               not
B2 L2 D2 L2 R2 B' F' R2 B  F' R2  (11)              not
B2 R2 B2 R2 F2 U2 B2 R2 U2 R2  (10)                 not
R2 F2 D  U  L' R' U2 L' R  D' U'  (11)              not
R2 F2 D' U' L' R' U2 L  R' D  U  (11)               not
L2 D' U  F2 L  R  B2 L  R  D  U  (11)               not
F2 L2 B2 R2 F2 U2 B2 R2 D2 R2  (10)                 not
R2 B2 D' U' L' R' D2 L  R' D' U'  (11)              not
B2 L2 B2 U  R2 U' B2 R2 U2 R2 U B2 U'  (13)         not
R2 F2 U' B2 D' R2 U2 F2 U' F2 D' B2 L2  (13)        not
B2 F2 R2 D  F2 D' L2 U2 B2 U' B2 U R2  (13)         not
D' R2 D2 B2 R2 B2 U  B2 D  U  B2 U' F2 U'  (14)     not
B2 R2 U2 L2 D' F2 R2 U' L2 D2 B2 U' R2 F2 U'  (15)  not
B2 R2 F2 R2 F2 D  F2 D' L2 U2 B2 U' B2 U R2  (15)   not
F2 U  R2 U  F2 D2 L2 U  B2 U  L2 R2 F2  (13)        not
B2 F2 L2 R2 D' B2 D  B2 D2 R2 U F2 U'  (13)         not
B2 R2 D' R2 U  F2 D2 L2 U  B2 D' L2 F2  (13)        not
U  F2 R2 U2 F2 D' R2 U2 B2 D  F2 R2 U'  (13)        not
B2 F2 L2 R2 D  U  R2 F2 D' U'  (10)                 not
F2 L2 F2 R2 F2 U2 F2 D2 U2 R2  (10)                 not
L2 D2 R2 U2 B2 F2 R2 F2 R2 U2  (10)                 not
L2 D2 R2 F2 U2 B2 D2 F2 R2 F2 R2 U2  (12)           not
L2 D2 B2 U2 F2 L2 B2 R2 F2 U2 R2 U2  (12)           not
B2 L2 F2 L2 F2 D2 F2 R2 D2 U2  (10)                 not





00X 0XX X00 X0X X0X X0X X0X XX0
XX0 0X0 0XX 0X0 0X0 0XX XX0 0X0
X0X X0X X0X 0XX XX0 X00 00X X0X

 35  65  8D  A3  A6  AC  B1  C5

B2 L' D2 L' B2 L  U2 F2 U2 R' F' L2 D' L  F2 U  F' L  B  R  (20) not
D2 L2 U2 F' L2 F  R2 F  U  B' D2 F2 R  U' F2 L' U2 B2 D  F  (20) not
L2 U' L2 B2 U' R2 F  R  B' D  L  U  B' U' R2 D  R2 F  L  F  (20) not





00X 0XX X00 XX0
XX0 0XX 0XX XX0
XX0 X00 0XX 00X

 36  6C  8B  D1

 (none)





00X 0XX X00 X0X X0X XX0 XXX XXX
XX0 0XX 0XX 0XX XX0 XX0 0XX XX0
XXX X0X XXX 0XX XX0 X0X X00 00X

 37  6D  8F  AB  B6  D5  EC  F1

D2 R2 F2 L2 F2 D  R2 D' R2 U2 F2 U' R2 U'  (14)   not
D2 B2 D' L2 D  F2 U2 R2 U  R2 U  F2  (12)         not
U' L2 U' L2 D2 F2 U' F2 U' F2 R2 B2 L2  (13)      not
R2 D  F2 U  R2 D2 L2 B2 D' B2 U  L2 F2 U2  (14)   not
F2 U  R2 D' F2 R2 U2 L2 D  B2 U'  (11)            not
F2 R2 D' B2 U  F2 D2 F2 R2 D' F2 U  F2  (13)      not
R2 B2 L2 U  B2 U  R2 D2 F2 U  L2 U' B2 U2  (14)   not
U2 L2 F2 L2 F2 U  F2 U' F2 U2 L2 U' L2 U'  (14)   not





00X 0X0 X00 XXX
XXX 0X0 XXX 0X0
00X XXX X00 0X0

 39  47  9C  E2

B2 D2 L  R' D2 B2 L  R'  (8)       not
U2 R2 F2 D' U  B2 L2 D' U'  (9)    not





00X 0X0 0X0 0X0 0X0 0XX X00 XX0
XXX 0XX XX0 XXX XXX XX0 XXX 0XX
0X0 XX0 0XX 00X X00 0X0 0X0 0X0

 3A  4E  53  59  5C  72  9A  CA

D' U2 B2 U2 L2 U  B  U' L2 B2 R' B2 R  F2 D2 F2 D  F'  (18)      not
F2 D' B2 U2 F2 U  R2 D  B  L2 B' R  B2 U2 F  D2 L' U' F  (19)    not
D  F2 D  U2 F2 L2 D' B2 F  D  L' B2 L2 F  D  F  D2 U2 F2 R  (20) not





00X 0X0 0X0 0XX X00 XX0 XXX XXX
XXX 0XX XX0 XXX XXX XXX 0XX XX0
0XX XXX XXX 00X XX0 X00 0X0 0X0

 3B  4F  57  79  9E  DC  EA  F2

D2 R2 B2 L2 U2 F2 U2 B2 R2 U2 R2 U2  (12)  not
U2 R2 F2 R2 U2 B2 D2 B2 L2 U2 R2 U2  (12)  not
U2 L2 F2 R2 U2 B2 D2 B2 R2 U2 R2 U2  (12)  not
D2 L2 B2 L2 U2 F2 U2 B2 L2 U2 R2 U2  (12)  not





00X 0XX X00 XX0
XXX 0X0 XXX 0X0
X00 XX0 00X 0XX

 3C  66  99  C3

 (none)





00X 0XX X00 X0X X0X XX0 XXX XXX
XXX 0X0 XXX XXX XXX 0X0 0X0 0X0
X0X XXX X0X 00X X00 XXX 0XX XX0

 3D  67  9D  B9  BC  C7  E3  E6

L2 F2 L2 U  R2 D' F2 U' R2 D  R2 U  R2 U'  (14)  not
D2 R2 B2 D  B2 U  R2 B2 D2 F2 D' B2 U  B2  (14)  not
F2 L2 U2 F2 D' R2 D  L2 D2 R2 F2 U' R2 U'  (14)  not
D2 B2 R2 U  B2 U  F2 R2 D2 F2 R2 U  L2 U'  (14)  not
B2 L2 R2 D' F2 L2 U' B' F' L  D2 F2 R  D' U' F' L' R'  (18)  not





00X 0XX 0XX 0XX X00 XX0 XX0 XX0
XXX 0XX XX0 XXX XXX 0XX XX0 XXX
XX0 XX0 XX0 X00 0XX 0XX 0XX 00X

 3E  6E  76  7C  9B  CB  D3  D9

 (none)





00X 0XX X00 XX0 XXX XXX XXX XXX
XXX 0XX XXX XX0 0XX XX0 XXX XXX
XXX XXX XXX XXX 0XX XX0 00X X00

 3F  6F  9F  D7  EB  F6  F9  FC

L2 D' L  B2 U' B' L  B U  B2 L' B  D  (13)             not
U' F2 D' L' U' F  U2 L U2 F  D  R2 F' R' U2  (15)      not
U  R2 D2 B2 U' F  L2 B R  D  R  B  R' D' F2 U2  (16)   not





0X0
XXX
0X0

 5A

U  B2 U2 L2 U  F2 R2 B2 U' L2 D2 F2 U' B  L2 R2 D2 U2 F'  (19) continuous
L2 R' B2 F2 D2 B2 F2 L2 R2 U2 R'  (11)   continuous





0X0 0X0 0XX XX0
XXX XXX XXX XXX
0XX XX0 0X0 0X0

 5B  5E  7A  DA

D  U2 R2 D' U' R  D  B2 R2 B2 R2 D  B2 D2 R  U'  (16)  continuous





0X0 0XX X0X XX0
XXX XX0 XXX 0XX
X0X 0XX 0X0 XX0

 5D  73  BA  CE

 (none)





0X0 0XX XX0 XXX
XXX XXX XXX XXX
XXX 0XX XX0 0X0

 5F  7B  DE  FA

B2 D2 B2 R2 F2 L2 U2 L2 F2 R2  (10)              not
B2 D2 B2 L2 U' F2 U' F2 R2 U2 L2 U  R2 U'  (14)  not
B2 L2 D2 L2 U' F2 U  B2 U2 F2 R2 U' R2 U'  (14)  not
D2 L2 D  U' L2 F2 D  U' F2 U2  (10)              not





0XX X0X X0X XX0
XX0 0XX XX0 0XX
X0X XX0 0XX X0X

 75  AE  B3  CD

D2 U R2 D' F2 U F2 R2  B R2 F2 U2 L' F2 D2 B2 D B' U'  (19) continuous
U2 L2 F2 R2 F2 U B2 U' B2 D2 L2 U' L2 U'  (14)               not





0XX 0XX X0X X0X XX0 XX0 XXX XXX
XX0 XXX XXX XXX 0XX XXX 0XX XX0
XXX X0X 0XX XX0 XXX X0X XX0 0XX

 77  7D  BB  BE  CF  DD  EE  F3

L2 U' L2 D' L2 D  F' L2 R' U  B D2 B' D' U' R  U  (17)       continuous
L2 B2 F2 D2 L2 B2 U  R2 U  B2 F D' B2 U2 L  F2 L D' B'  (19) not





0XX XX0
XXX XXX
XX0 0XX

 7E  DB

 (none)





0XX XX0 XXX XXX
XXX XXX XXX XXX
XXX XXX 0XX XX0

 7F  DF  FB  FE

U' L2 U  F' R2 F  U' L2 U  F' R2 F  (12)    continuous
R2 D  B2 D' B2 U' B2 U  B2 U  R2 U'  (12)   not





X0X
0X0
X0X

 A5

D2 F2 U' B2 F2 L2 R2 D R' B  F  D' U  L  R  D2 U2 F' U2  (19) continuous
R' D2 U2 L2 B2 F2 L' F D' U  R2 B2 F2 R' L' B  F' U'     (18) continuous
B2 F2 L2 R2 D2 U2  (6) *   continuous





X0X X0X X0X XXX
0X0 0XX XX0 0X0
XXX X0X X0X X0X

 A7  AD  B5  E5

F2 U2 B2 F2 L2 U' B2  L D2 F' R2 B L2 R  U' R' D' F' R  (19) continuous
L2 R2 D2 L2 D' U  F L' R U2 B2 U B F' R' D2 R' U2     (18) not
B2 U2 F2 U2 L2 D  U' B' U2 L' R  B2 D' R2 B' F' L' R   (18) not
B2 R2 D  U' F2 U2 R2 B  D  U  B' F' L' R' F' D' U  R2  (18) not
B2 L2 D' U2 L2 B2 R' B  U2 R  B  L' D' F2 R            (15) not
R2 D  L2 B2 F2 R2 U' R2 D2 U2                          (10) not
R2 F2 D  B2 L2 B  F  L  B2 L2 D  U' F' L' R  D'        (16) not
R2 U2 L2 R2 U' R2 D  B  F' L' B2 R2 D' U  B  L' R'     (17) not
U  L2 U2 F2 U  F2 R2 F' L  F  U  B' D' R' D  R2 D2 F' D  (19) not
U  B2 F2 R2 D2 U2 R  D2 U2 R2 B  D  U' L  R2 B' D'     (17) not
D2 U  B2 U2 B2 R2 U2 B' R2 D  U  F  U' L  R' B  L  F  U'  (19) not
U2 L2 B2 R2 U2 B2 R2 D  U  B  U  B' L' F2 U' L' U  R' D  U'  (20) not
U  F2 L2 R2 F2 L2 R' B  D  B2 F2 U' B' L' U'           (15) not
D' B2 D' L2 F2 D' L' R  F  D2 L2 F  D' U' R  B  F'     (17) not
R2 U' B2 L2 F2 U' L2 D  F2 L2 F  D  F  U  L  U  B' U' L'  (19) not
D' U' B2 F2 L' R  F  R2 D' U  F2 L' B' F  U2           (15) not
R2 D  L2 F2 U  F2 D  R2 B  R' D  R  D2 R2 B' F2 D  F  R2  (19) not
F2 D' B2 D  F2 L2 D2 U  L2 U  R  D  R2 D2 L' F' U2 B2 U  F  (20) not
U2 B  U2 R2 D2 B  L2 U' B' U  L  F2 U  R  F' D' R2 B' R'  (19) not
U2 L2 R2 D' F2 U' B2 R  F' D  R' B' D  F  R' U  L' D2 F  U  (20) not
D2 B2 U2 R2 B2 R2 U2 F2 R2 U2                              (10) not





X0X X0X XXX XXX
0XX XX0 0XX XX0
XXX XXX X0X X0X

 AF  B7  ED  F5

F2 L2 D' R2 B2 L2 R2 F  U2 L2 D' L' D' R2 F' D' L' F2    (18) continuous
D2 L2 D' F2 D  U  F' R2 D' L' R  F' L' R' B' U' R2 U2    (18) continuous
D  U  F2 R' B  D2 U2 F' D2 U2 R  F2 D' U'                (14) continuous
U  B2 L  B  F' L2 R' B' F  D  U2 L' B2 U'                (14) continuous
B2 L2 U' L2 U2 B2 R2 U' B  U  R' D' L' D2 L  D  B  D  U' (19) not
B2 F2 L2 D2 R2 U  L  F  D' L2 R2 D2 U' F' R' D  L2 U2    (18) not
R2 U' L2 R2 D  B2 D  R  F' R' B  L' R  U  L' U' F  R'    (18) not
F2 D2 B2 U  L2 B2 L2 D' R' B  R' D' L2 B' D' B  L2 R2 U' (19) not
B2 D' R2 D  R2 D' B  R' F  R' D  L2 F' U  L  B' L  U'    (18) not
R2 F2 L2 D' B2 U' B2 L2 F2 L  B' L' U' B' D' B  D  B' D' (19) not
B2 L2 F2 D' L2 B2 D' L  B  L  D  F2 D  B' D  U2 F' U2    (18) not
D  L2 F2 D  U2 B2 R2 F' D  R  U2 L2 F' L2 U' R' U  B     (18) not
U2 R2 F2 D  U' B2 R2 B2 R2 D' U'                         (11) not
D2 L2 B2 D' U  B2 R2 B2 R2 D' U'                         (11) not
D2 B2 D2 L2 D2 L2 U2 F2 R2 F2 R2 U2                      (12) not
D2 B2 U2 R2 U2 L2 U2 F2 R2 F2 R2 U2                      (12) not
B2 F2 D2 U  L2 D2 R  D' L2 R  D' B2 D  F  D  F' U'       (17) not
B2 F2 D' F2 D  U  R  D' L2 R  U' L2 U  F  D  F' U'       (17) not
R2 U2 B2 L2 F2 R2 D2 F2 U2 F2 R2 U2                      (12) not
L2 F2 U2 B2 U2 R2 B2 R2 F2 U2 R2 U2                      (12) not
L2 U2 F2 R2 F2 R2 U2 B2 U2 F2 R2 U2                      (12) not
R2 F2 D2 F2 D2 R2 B2 L2 F2 U2 R2 U2                      (12) not
D2 B2 L2 U2 R2 U2 B2 L2 U2 F2 R2 U2                      (12) not
D2 B2 L2 D2 L2 D2 B2 L2 U2 F2 R2 U2                      (12) not
U2 R2 F2 D  R2 F2 R2 F  L  D2 L' D' F' L' U2 B2 R'       (17) not
B2 L2 U  R2 D  U' L2 B  L  R2 D' L' B  D  L' R2 B  L'    (18) not
F2 L2 B2 R2 D2 F2 D2 F2 R2 U2 R2 U2                      (12) not
F2 R2 B2 R2 U2 B2 U2 F2 L2 U2 R2 U2                      (12) not
L2 U2 R2 F2 U2 B2 U2 R2 F2 R2 F2 U2                      (12) not
L2 D2 L2 B2 D2 B2 U2 R2 F2 R2 F2 U2                      (12) not
F2 R2 B2 R2 D2 F2 D2 F2 L2 U2 R2 U2                      (12) not
D' R2 B2 R2 D' R2 B' D' F' L' U' B' U  L  F  D  R2       (17) not
L2 U2 R2 D2 R2 U2 B  D' U2 R  F  D2 U2 B' L' D' B'       (17) not
L2 B2 D' B2 L2 B2 F  R  B  R2 U  F2 R  U  B  U  F'       (17) not
R2 D  F2 D  U' R2 L  B  D  L2 R2 D' L' B  F2 D' R2       (17) not
B2 D  U' L2 D2 F' D  U' R  F  D  U' R' D' U'             (15) not
L2 D2 L2 D' U' F2 L' D' U  B  L  D' U  B' D' U'          (16) not
U' L2 F2 L2 D  F2 L2 D' U  L  B' L' D' L2 D  B  D  L     (18) not
R2 B2 R2 D' F2 L2 U  L' D' L2 R  F' R' D  R  F  R' U'    (18) not
U' F2 D' F2 L2 D2 U  B' L' B  D  L' U' L' F2 L' U        (17) not
D' L2 D' L2 B2 D' B' L' B  D  L' U' L' F2 L' U           (16) not
R2 U  F2 L2 B2 U  L2 D' R2 F2 R2 U'                      (12) not
D  L2 D' F2 D' R2 U2 R2 U2 B  D2 F' R' U  R' D2 U  B     (18) not
R2 B2 D2 F2 D2 R2 F2 L2 F2 U2 R2 U2                      (12) not
U  F2 R2 F2 D' L2 D  U2 B2 L2 F  D' R' F  R' D  R  F  R' (19) not
L2 B2 L2 B2 D2 R2 U2 R2 B2 U2 F2 U2                      (12) not
B2 R2 U2 R2 D2 R2 B  F' R2 B' F'                         (11) not
B2 R2 D2 L2 U2 R2 B  F' R2 B' F'                         (11) not
U' B2 D' L2 U' L2 R2 D  L  D  U' F  D  F  D' U  L  D     (18) not
B2 D' B2 F2 D' L2 U2 B2 R' B  R  U2 L' F' L' U' F2 L2 U  (19) not





X0X XXX
XXX 0X0
X0X XXX

 BD  E7

L2 B  F' L2 R2 B  F' R2             (8)  not
F2 L2 U2 L2 R2 B' D2 U2 F U2 R2 F2  (12) not
R2 B2 F2 R2 U' B2 F2 D2 L2 R2 U'    (11) not
L2 R2 D  B2 F2 R2 B2 F2 R2 U'       (10) not
U2 B2 R2 D2 U2 R2 F2 U2             (8)  not





X0X XXX XXX XXX
XXX 0XX XX0 XXX
XXX XXX XXX X0X

 BF  EF  F7  FD

U  R' D' U  F2 D  U' R' U'                    (9) *    continuous
L2 U' F2 B  D' R' D2 R  B' F2 L  U            (12) *   continuous
D' B' R2 B' D  U' L  B2 D2 U2 R  D2 U'        (13)     not
D2 U  R' D2 U2 B2 L' D' U  B  R2 B  D         (13)     not
F2 D' L2 R2 B' L' R  D  B2 D  L' R  F' D' F2  (15)     not
U  F2 U2 F  L' U' B' U2 B  L  U  F' U         (13)     not
U' F2 U  F2 R  B' U' R' U  R  B  U2 R' F2     (14)     not
U' R  U  L' R  B' R' B  U' L  R' F            (12)     not
U  L2 B2 D' F2 R2 B2 U' F2 U'                 (10)     not
L2 R2 U' L2 R2 D' L' R  F2 L  R'              (11)     not
L2 R2 U  L2 B2 R2 D' R2 B' L2 U2 R2 F  L2     (14)     not
F2 D  U2 R2 B  R2 U2 R2 B  R2 D' F2           (12)     not
D2 R2 B2 R2 D' R2 B2 R2 D                     (9)      not
F2 R2 U  L2 U' R2 F2 L  R  F' U2 F  L' R'     (14)     not
R2 D  B  R' D' R' B' R' D  B  R'              (11) *   not
R2 D2 B  D2 R2 B2 L  B2 U2 F2 R  F' U2        (13)     not
D2 B2 U2 L' U2 B' D2 R2 U2 F  U2 R            (12)     not
L2 D2 F2 D2 R  D2 F' R2 D2 L2 B  U2 R         (13)     not
F2 U' F2 D2 B2 D2 F2 U' F2                    (9)      not
R2 D  R2 B  R2 D  R2 B' R2 D' R2 B'           (12)     not
D2 B2 R2 D2 R2 B  R2 D2 R2 B' D2              (11)     not
U2 F2 U2 L2 U2 F  U2 L2 U2 F' U2              (11)     not
R2 D2 B2 D2 U2 F  D2 L2 D2 F' U2              (11)     not
D2 R2 B2 L2 U' R2 F2 R2 U  L2 R2              (11)     not
U  F' U2 L  D2 B2 U2 R  D2 F' U'              (11) *   not
D' U' B  D  B  U  R2 D' L  R2 B' L'           (12)     not
R2 D  F  D' L' B  F  L' B' U  L  F2 R2        (13)     not
F2 D' L2 R2 U  L  R' U2 L  R'                 (10)     not
D' R2 B2 R2 D  R2 B2 R2 D2                    (9)      not
D2 B2 R2 D2 R2 D2 R2 U2 F2 U2                 (10)     not
U  R' F  L  R' D' R  D' L' R  F  U'           (12)     not
F2 L2 F' L2 R2 B  L2 B  L2 R2 F               (11)     not
D2 R2 U2 F2 D2 U2 B' U2 L2 U2 B               (11)     not
R2 D' L2 B2 L2 D  R2 U2 R' F2 D2 F2 R  U2     (14)     not
D2 R2 U' R2 F2 L2 D  R2 B  L2 U2 R2 F' D2     (14)     not
D  U  B' R' D2 B' D' R  B  D' U' B  D'        (13)     not
R2 F2 L2 D' B2 L  B2 D2 F2 R' B2 U            (12)     not
U' R  B  U' R' U' B' U' R  B  U2              (11) *   not
L2 U  L2 F2 R2 D' L2 B' L2 D2 R2 F            (12)     not
U2 R2 B' L' R' B2 L' R' F' L2                 (10) *   not
D2 B  D2 U2 F  D  U' R2 D' U'                 (10) *   not
R2 U' L' U' B' L' B  U  L  U  B  R2           (12) *   not
D2 L  D2 F' R2 D2 L2 B  D2 R' U2 F2           (12) *   not
U' L  D' U  F  L  R' U' L' R  F' D            (12) *   not
R2 F2 L' F' L' R  U  L  U  L  R' F  R2        (13)     not
U' L2 D' L' D' U  B  D  B  D  U' L  U         (13)     not
D  U  R' D' U  F' D  U' R  D' U  F  U2        (13)     not
U' L2 B  L' R  D' L  D' L  R' B  L  U         (13)     not
D  U  F2 U2 B2 D' U' F  D2 U2 B' R2           (12)     not
D  U2 B2 U  B  U' B' U' B2 F  L' D  L  F' D' U2        (16)  not
B2 F2 R2 D' F2 R2 F2 D' L' U2 F' L2 U2 L2 F' D2 R  F2  (18)  not





XXX
XXX
XXX

 FF

      (0)  continuous   (this is Start)

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990

From cube-lovers-errors@mc.lcs.mit.edu  Mon Mar 30 15:45:45 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Fri Mar 27 21:26:20 1998
Message-Id: <l03110703b1420addbe67@[207.172.129.171]>
In-Reply-To: <01BD594B.F9EFBF20@jburkhardt.ne.mediaone.net>
Date: Fri, 27 Mar 1998 21:26:52 -0500
To: cube-lovers@ai.mit.edu
From: Charlie Dickman <charlied@erols.com>
Subject: Re: Stickers

>Does anyone know where to find cube stickers?  They must come from
>somewhere!  I found some vinyl lettering once and the periods were
>exactly the right size for a 5x5x5 cube.  But they don't come in
>orange.  There must be a way to buy sheets of the stuff.  Any ideas?

I have found some adhesive backed vinyl sheets at a local Art Emporium but
they are mostly irridescent shades and you have to cut the pieces to size
yourself. I seem to recall that there was an orange color but I'm not sure.

Charlie Dickman
charlied@erols.com

From cube-lovers-errors@mc.lcs.mit.edu  Mon Mar 30 16:23:46 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sun Mar 29 04:29:11 1998
Message-Id: <3.0.5.16.19980329094205.097f34b6@vip.cybercity.dk>
Date: Sun, 29 Mar 1998 09:42:05
To: cube-lovers@ai.mit.edu
From: Philip Knudsen <skouknudsen@email.dk>
Subject: Eclipse and Pyramorphix

There are two new puzzles out,
by the two most prominent veterans respectively:

 1) Rubik's Eclipse, which is some sort of two-player game and,
    according to the people who have it, a real gem.
 2) Pyramorphix, by Meffert.
    David Byrden's Twisty Puzzles page shows a picture of
    a 2x2x2 Pyraminx together with the text
    "A solid version of this amazing puzzle is now available
    from Uwe Meffert, called the Pyramorphix".
    Now the 2x2x2 pyraminx looks like an old east german puzzle,
    which was a 2x2x2 cube in tetrahedral shape. The shape changed
    when the puzzle was scrambled, so the name Pyramorphix would
    apply. However the east german puzzle was not by Meffert.

Now if anyone knows more about these new puzzles,
or where to get them, please reply.


Philip K
recording and performing artist
Vendersgade 15, 3th
DK - 1363 Copenhagen K
Phone:  +45 33932787
Mobile: +45 21706731
E-mail: skouknudsen@email.dk
E-mail: philipknudsen@hotmail.com
Sms: 4521706731@sms.tdk.dk (short message, no subject)

From cube-lovers-errors@mc.lcs.mit.edu  Tue Mar 31 10:02:23 1998
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	id KAA07294; Tue, 31 Mar 1998 10:02:22 -0500 (EST)
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Mail-from: From cube-lovers-request@life.ai.mit.edu Mon Mar 30 21:08:16 1998
Message-Id: <19980331020806.13788.qmail@hotmail.com>
X-Originating-Ip: [206.114.5.101]
From: "HADER MESA" <hamepa@hotmail.com>
To: zot@ampersand.com, rtayek@netcom.com
Cc: cube-lovers@ai.mit.edu
Subject: i need information!!!
Date: Mon, 30 Mar 1998 18:08:05 PST

Hello, I am a fond of the cube of Rubik, but in my country it is very 
difficult to get it.  
She/he would want to know if you can give me information about where I 
can get the cube and their different variants.  
For the information that you can to give, I thank him a lot.  
  
Cordially:  Hader Mesa 

From cube-lovers-errors@mc.lcs.mit.edu  Wed Apr  1 10:55:58 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sun Mar 29 18:20:22 1998
Date: Sun, 29 Mar 1998 18:20:42 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: All the Partial Isoglyphs
To: Cube-Lovers <cube-lovers@ai.mit.edu>
Message-Id: <Pine.PMDF.3.95.980329181553.328319A-100000@PSTCC6.PSTCC.CC.TN.US>

I have been able to calculate all the partial isoglyphs a little more
quickly than expected.  I can report that there are 10 continuous partial
isoglyphs and 130 non-continuous partial isoglyphs, unique up to symmetry.

Here is a breakdown of how the solid faces can be arranged.

      97 - two solid faces, opposite to each other
      11 - two solid faces, adjacent to each other
      25 - one solid face
       1 - three solid faces, mutually adjacent to each other
       2 - three solid faces, not mutually adjacent to each other
       3 - four solid faces, other two opposite to each other
       1 - four solid faces, other two adjacent to each other
     ---
     140

The partial isoglyphs are all included in the chart which follows.  If
nothing is listed with respect to the manner in which the solid faces are
arranged, then there are two solid faces opposite to each other.
Otherwise, the arrangement of the solid faces is listed explicitly.

This chart follows the same format as the previous one I posted for all
the isoglyphs, except that this time I included only a single
representative glyph for each partial isoglyph, rather than the complete
equivalence class of glyphs.


000
0X0
000

 00

D B2 F2 D U' L2 R2 U'  (8)  continuous




000
0X0
00X

 01

 (none)




000
0X0
0X0

 02

D B2 L2 R2 F2 U' L2 B2 F2 R2  (10)               not
D' B2 F2 D U L2 R2 U'  (8)                       not
F2 D2 B2 D U L2 D' U'  (8)                       not
B2 D U' L2 D U'  (6) *                           not
D B2 D L2 . B F' D2 R' B2 R2 D' U F L R'  (15)   not




000
0X0
0XX

 03

L2 U2 B2 D' B2 R2 D2 B2 R2 U' B2 L2 U2 R2 U'  (15)  not




000
0X0
X0X

 05

D U F2 D' L2 . B' F U2 L' D U' F2 R2 F' L R'  (16) not
D B2 L2 R2 F2 U' L2 B2 F2 R2 U2  (11)              not
L2 F2 U B2 F2 U2 B2 F2 U' F2 R2  (11)              not




000
0X0
XXX

 07

D' B2 F2 D' U L2 R2 U'  (8)  continuous




000
0XX
0X0

 0A

L2 D2 R2 F2 U2 R2 F2 U2 F2 U2  (10)  not
D B2 D' U' L2 R2 U2 R2 U'  (9)       not




000
0XX
0XX

 0B

B2 L2 U2 L2 U' L2 B2 D2 F2 U F2 R2 U2 R2 U'  (15)  not




000
0XX
X00

 0C

F2 L2 U2 L2 U L2 F2 D2 B2 U' B2 R2 U2 R2 U'  (15) not




000
0XX
X0X

 0D

L2 D2 R2 F2 U2 R2 F2 U2 F2  (9)            not
B2 R2 F2 D U B2 L2 F2 L2 D U  (11)         not
F2 R2 B2 D' U' B2 R2 F2 R2 D' U'  (11)     not




000
0XX
XX0

 0E

L2 U2 F2 D F2 R2 D2 F2 R2 U F2 L2 U2 R2 U'  (15)   not
R2 U2 F2 D' F2 L2 U2 B2 R2 U' B2 R2 U2 R2 U'  (15) not
F2 U2 R2 U F2 D2 F2 R2 D' F2 R2 D2 F2 R2 U'  (15)  not




000
0XX
XXX

 0F

B2 D U' L2 D U  (6) *   not




000
XXX
000

 18

D2 U2  (2) *               continuous
D' U  (2) *                continuous
L2 F2 L2 R2 F2 R2  (6) *   not
F2 U2 B2 F2 U2 F2  (6) *   not




000
XXX
00X

 19

 (none)




000
XXX
0X0

 1A

D F2 R2 B2 F2 R2 D' U R2 U'  (10)       not
D' B2 U2 B2 L2 R2 D2 F2 L2 R2 D'  (11)  not
L2 B2 F2 R2 D' U2 L2 B2 F2 R2 U'  (11)  not
U' L2 R2 U2 L2 R2 U'  (7) *             not





000
XXX
0XX

 1B

B2 D2 L2 U' F2 D2 F2 L2 D F2 L2 U2 B2 R2 U'  (15)  not




000
XXX
X0X

 1D

L2 B2 F2 R2 D U2 L2 B2 F2 R2 U'  (11) not
F2 U L2 R2 D2 L2 R2 U' F2  (9)        not




000
XXX
XXX

 1F

D  (1) *    continuous
D2  (1) *   continuous




00X
0X0
X00

 24

D' L2 F2 U2 B2 R2 U2 L2 D U' R2 U'  (12)  not




00X
0X0
X0X

 25

 (none)




00X
0X0
XX0

 26

B2 L2 D U F2 L2 D U F2 R2  (10)      not
L2 D' U' F2 D U . L R' U2 L R  (11)  not
D' U' F2 D' U . L R' U2 L R  (10)    not
U' B2 L2 D2 R2 F2 U2 F2 R2 U'  (10)  not




00X
0X0
XXX

 27

L2 D2 R2 B2 U R2 B2 D2 L2 F2 U' F2 D2 R2 U'  (15)  not





00X
0XX
XX0

 2E

 (none)




00X
0XX
XXX

 2F

L2 D2 B2 D B2 L2 U2 B2 L2 D' B2 R2 U2 R2 U'  (15)   not




00X
XX0
00X

 31

L2 F2 L2 R2 F2 R2 U2  (7) *  not




00X
XX0
0X0

 32

F2 D2 L2 U B2 D2 B2 L2 D' B2 L2 U2 F2 R2 U'  (15)  not




00X
XX0
0XX

 33

D F2 R2 B2 F2 R2 D' U R2 U  (10)    not




00X
XX0
X0X

 35

F2 L2 D2 R2 D' R2 F2 D2 B2 D' B2 R2 U2 R2 U'  (15)  not




00X
XX0
XX0

 36


 (none)




00X
XX0
XXX

 37

L2 U2 F2 D F2 R2 U2 F2 R2 D' F2 L2 D2 R2 U'  (15)  not
B2 U2 F2 L2 D L2 D2 R2 B2 D B2 D2 F2 R2 U'  (15)   not
F2 R2 D2 L2 D L2 F2 U2 F2 D F2 R2 U2 R2 U'  (15)   not





00X
XXX
00X

 39

U2 F2 U L2 . B' F U2 R' F2 R2 D U' B L R'  (15) not
B2 L2 R2 F2 D' L2 B2 F2 R2 U'  (10)             not
B2 U' B2 L2 R2 F2 D' F2  (8)                    not





00X
XXX
0X0

 3A

B2 U2 R2 U' B2 D2 B2 R2 D B2 R2 D2 B2 R2 U'  (15) not





00X
XXX
0XX

 3B

L2 D2 L2 F2 U2 R2 B2 U2 F2  (9)        not
U' R2 B2 R2 D F2 D' R2 B2 R2 U  (11)   not
B2 R2 F2 D' U' B2 L2 F2 R2 D' U'  (11) not





00X
XXX
X00

 3C

D' L2 B2 U2 F2 R2 U2 L2 D' U R2 U'  (12) not
F2 R2 D U L2 B2 R2 B2 D' U' F2 R2  (12)  not
D' R2 B2 U2 B2 R2 U2 R2 D' U R2 U'  (12) not




00X
XXX
X0X

 3D

 (none)




00X
XXX
XX0

 3E

B2 L2 D2 B2 R2 F2 L2 U2 F2 R2  (10)  not
D' U' L2 D' U . B F' D2 B F  (10)    not
U2 L2 . B' L2 D2 U2 R2 F' R2  (9)    not
D2 L2 . B' D2 L2 R2 U2 F' R2  (9)    not





00X
XXX
XXX

 3F

R2 U2 F2 D' F2 L2 D2 B2 R2 D B2 L2 D2 R2 U'  (15)  not





0X0
XXX
0X0

 5A

D F2 R2 F2 D' U R2 F2 R2 U'  (10)         not
B2 F2 D2 L2 R2 D B2 F2 U2 L2 R2 U'  (12)  not




0X0
XXX
0XX

 5B

 (none)




0X0
XXX
X0X

 5D

B2 F2 L2 R2 D B2 F2 L2 R2  (9)   not
B2 F2 L2 R2 D2 B2 F2 L2 R2  (9)  not





0X0
XXX
XXX

 5F

D' L2 B2 F2 R2 U' L2 B2 F2 R2  (10)  not
L2 B2 D' B2 L2 R2 F2 U' F2 R2  (10)  not





0XX
XX0
X0X

 75

F2 D2 B2 L2 D L2 U2 L2 F2 D F2 U2 F2 R2 U'  (15)  not





0XX
XX0
XXX

 77

R2 D2 B2 D' B2 R2 D2 F2 L2 D F2 R2 U2 R2 U'  (15) not





0XX
XXX
XX0

 7E

D' R2 F2 U2 F2 R2 U2 R2 D U' R2 U'  (12)  not




0XX
XXX
XXX

 7F

 (none)




X0X
0X0
X0X

 A5

D2 R2 U2 L2 R2 U2 R2 U2  (8)  not  (four solid, other two opposite)
D2 L2 F2 L2 R2 F2 R2 U2  (8)  not





X0X
0X0
XXX

A7

D' B2 U' L2 . B' F U2 R' D U' B2 L2 B L R'  (15)                not
B2 L2 R2 F2 D' U2 L2 B2 F2 R2 U'  (11)                          not
U2 F2 D' U' R2 U2 . R B2 F L' R D L' B2 F2 R B  (17)            not *1
D2 L2 B2 R2 U' F2 L2 D U' . R B2 U' F' D2 U' R F2 D L' R2  (20) not *2
B2 R2 F2 U' L2 U . R B D2 B' R' D' R' F2 L R2 B2 U'  (18)       not *2
D F2 L2 F2 D' U' R2 D' R2 . B' D2 B' D' L' U L2 R' U' R'  (19)  not *2
D' L2 R2 D' U' B2 F2 U'  (8)                                    not
F2 L2 D2 B2 U2 B2 F2 R2 F2 U2  (10)                             not
F2 D' F2 D B2 U B2 F2 U2 L2 U F2 . R B U' B2 U B' R'  (19)      not
D' F2 R2 B2 F2 R2 D' U R2 U  (10)                               not

      *1  two solid faces, adjacent
      *2  one solid face






X0X
0XX
XXX

 AF

D . F' D2 U2 B R B' D2 U2 F L' D'  (12) *                  continuous *1
R2 U2 . L B L U R' U R' D' F' D'  (12) *                   not *2
L2 U2 R2 D2 R2 U2  (6) *                                   not *4
U F2 D U2 L2 U' F2 . L' U' F D2 U L' F2 U2 B' R  (17)      not *3
D B2 D' U2 . F D F U' R' U F' U' R' U' B2  (15)            not *2
F2 U' L2 U L2 U' B2 . L' U' B U2 B' U L' F2 D' R2  (17)    not
R2 D' R2 U' R2 U . R U L F2 D2 L' U' B' D B D  (17)        not *2
L2 U2 R2 D2 R2 . B' L' B' U' F U' F D R D F2  (16)         not *2
B2 D2 B2 U2 F2 L2 D' . F' D' L' U L' U R B R  (16)         not *2
L2 U2 R2 F2 D2 R2 F2 D2 F2 U2  (10)                        not
F2 L2 U2 B2 D R2 . B' L' D' L' D' B' U' B' F U' B U2  (18) not
R2 D2 B2 D2 F2 U . F' D' L' D' B D L D F' R2  (16)         not *3
U F2 L2 U2 L2 F2 R2 D' U' . B' D' L' B' R' B' R B L  (18)  not *2
B2 L2 . B' D' L' U B' R2 U' L F' D' F  (13)                not *2
B2 R2 F2 L2 U2 F2 R2 U2 F2 U2  (10)                        not

          *1 three solid faces mutually adjacent
          *2 one solid face
          *3 two solid faces adjacent
          *4 four solid, other faces opposite






X0X
XXX
X0X

 BD

D B2 F2 D' U L2 R2 U'  (8)                  continuous
D2 B2 D2 U2 F2 U2  (6) *                    not *4
R2 . F' U2 L2 D2 B2 L2 U2 R2 F' R2  (11) *  not *2
F2 D2 L2 . F D2 U2 B' R2 U2 F2  (10) *      not *3
F2 R2 U2 . B' D2 U2 F D2 R2 F2  (10) *      not *1
D2 U2 B2 U2 R2 . F' D2 U2 B L2 U2 F2  (12)  not *5
L2 D' U' B2 F2 D' U' R2  (8)                not
L2 D2 R2 . B' U2 F U2 L2 U2 F R2 F'  (12)   not *5
B2 F2 L2 R2 U' B2 F2 D2 L2 R2 U'  (11)      not
L2 R2 U2 B2 F2 U B2 F2 U2 L2 R2 U'  (12)    not
L2 R2 D2 L2 B2 U B2 F2 D' F2 R2 U2  (12)    not
D B2 L2 B2 D U' R2 F2 R2 U'  (10)           not

      *1 - three solid faces, not mutually adjacent
      *2 - four solid faces, other two faces adjacent
      *3 - two solid faces, adjacent
      *4 - four solid faces, other two faces opposite
      *5 - one solid face





X0X
XXX
XXX

 BF

D' U R2 F2 D U' . R' D U' B' L2 B D' U R'  (15) continuous
L2 D U . B D' B' U' L2 D L D L' D2  (13)        continuous *1
B2 D U' L2 D' U  (6) *                          not
D B2 U2 . L' U2 B2 D2 R' D  (9) *               not *2
R2 D U' . B D' B' D' U R D R  (11) *            not *2
F L R' D2 L' R F  (7) *                         not *2
L2 U2 . B U2 L2 D2 F D2  (8) *                  not *1
L2 . F L R' D2 L' R F L2  (9) *                 not *2
U2 L2 D2 . B' L2 U2 R2 F'  (8) *                not *1
L2 D U' . F' L F D' U L' B' L'  (11) *          not *2
F2 U2 L2 D2 . B' L2 U2 R2 F  (9) *              not *2
D2 . B' L' R D2 L R' B' D2  (9) *               not *3
R2 U2 . B D2 L2 U2 F D2  (8) *                  not *1
U2 B2 U2 L2 U2 . B D2 R2 U2 F' D2  (11)         not *2
D . R B2 F2 L' U' L B2 F2 R'  (10) *            not *2
D . R' B F' U R' U' B' F R  (10) *              not *2
D . F' R' B' L' D' L B R F  (10) *              not *1
B2 D L2 U . R U R' F U2 L D' L B'  (13)         not *2
R2 D . F D' F' R2 D' B' D B  (10) *             not *1
D F2 D R2 . F R2 D2 R2 F R2 D F2 D'  (13)       not *2
D' F2 U2 B2 U2 F2 D'  (7) *                     not
F2 R2 U2 . B' U2 R2 U2 B' U2 F2  (10) *         not *2
F2 D2 . F D2 R2 D2 F D2 R2 F2  (10) *           not *2
B2 R2 U' L2 U R2 B2 R2 U F2 U' R2  (12)         not
D L2 B2 F2 R2 U' L2 B2 F2 R2  (10)              not
F' L2 R2 B2 L2 R2 F'  (7) *                     not
L2 . B L' B' D2 R' B' R B D2 L'  (11) *         not *1
D U' . B F' U' B' F R2 D' U F'  (11) *          not

       *1 - 2 solid, adjacent
       *2 - 1 solid
       *3 - 3 solid, not mutually adjacent




XXX
XXX
XXX

 FF

 (none)

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990

From cube-lovers-errors@mc.lcs.mit.edu  Fri Apr  3 17:18:01 1998
Return-Path: <cube-lovers-errors@mc.lcs.mit.edu>
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Apr  1 06:05:52 1998
To: cube-lovers@ai.mit.edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Re: new to list
Date: 1 Apr 1998 09:19:33 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <6ft0r5$6kj@gap.cco.caltech.edu>
References: <cube-lovers.01BD5821.7C9449E0@jburkhardt.ne.mediaone.net>

John Burkhardt <jburkhardt@mediaone.net> writes:
>The Dodecahedron puzzle is really amazing.  It was actually harder
>than the 5x5x5 cube.  IT took me about 3 hours to work it out!  I
>think once you know the 3x3x3 then all the same moves do similar
>things and you can easily solve 4x4x4 or 5x5x5 with variations.  Of
>course there are some cool things you can do with these.

Really??  I found the Dodecahedron significantly easier than the 4x4x4.
The Dodecahedron gives more "space" for moves...

--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
Smoking cigarettes are bad for you, so smoking cigarettes is bad for you.

From cube-lovers-errors@mc.lcs.mit.edu  Fri Apr  3 18:45:49 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sun Mar 29 18:56:43 1998
Date: Sun, 29 Mar 1998 18:57:08 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: All the Partial Isoglyphs
In-Reply-To: <Pine.PMDF.3.95.980329181553.328319A-100000@PSTCC6.PSTCC.CC.TN.US>
To: Cube-Lovers <cube-lovers@ai.mit.edu>
Message-Id: <Pine.PMDF.3.95.980329182138.328319B-100000@PSTCC6.PSTCC.CC.TN.US>

On Sun, 29 Mar 1998, Jerry Bryan wrote:

> Here is a breakdown of how the solid faces can be arranged.
>
>       97 - two solid faces, opposite to each other
>       11 - two solid faces, adjacent to each other
>       25 - one solid face
>        1 - three solid faces, mutually adjacent to each other
>        2 - three solid faces, not mutually adjacent to each other
>        3 - four solid faces, other two opposite to each other
>        1 - four solid faces, other two adjacent to each other
>      ---
>      140
>

As this table shows, the vast majority of partial isoglyphs involve two
solid faces opposite to each other.  The basic reason for this is the
corners.  If the corners are not fixed, then the only partial isoglyphs
which are possible have two solid faces opposite to each other.
Conversely, the 43 partial isoglyphs which do not have two solid faces
opposite to each other do fix the corners.

In fact, 67 of the partial isoglyphs derive from just 5 of the glyphs,
namely those which fix the corners.  If the corners of the partial
isoglyph are fixed, you can think of the edges as consisting of a set of
strongly constrained edge flips and swaps.  (Be careful -- if the corners
are fixed, then *any* resultant position can be thought of as just a bunch
of edge flips and swaps.  But for partial isoglyphs, the possible edge
flips and swaps are strongly constrained.)

The glyph which yields the most partial isoglyphs is the one my charts
call BF, whick looks like the following.

X0X
XXX
XXX

With this glyph, each face of a partial isoglyph can have at most one edge
cubie which is swapped or flipped, but on a cube-wide basis there are
quite a few different ways to arrange for this to happen.

Another interesting glyph which fixes the corners is called BD on my
charts, and which appears as follows.

X0X
XXX
X0X

As an isoglyph, this glyph yields five different patterns on the 6-H
theme.  As a partial isoglyph, this glyph yields a number of pretty 2-H,
3-H, 4-H, and 5-H patterns.  You may also think of the H patterns as
complicated edge swappers/flippers, with exactly zero or two edges
swapped/flipped on each face, and with the coloring requirements for
partial isoglyphs being maintained.

The following two glyphs (A7 and AF in my charts) are in the same spirit
as the H, except that the configuration of the edges on each face which
are swapped/flipped is slightly different than for the H.

X0X X0X
0X0 0XX
XXX XXX

Finally, for completeness in the list of glyphs which fix the corners, the
glyph called A5 on my charts appears as follows.

X0X
0X0
X0X

However, this glyph only yields two partial isoglyphs.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990

From cube-lovers-errors@mc.lcs.mit.edu  Fri Apr  3 19:32:23 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sun Mar 29 19:35:16 1998
Date: Sun, 29 Mar 1998 19:35:43 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: partial isoglyphs
In-Reply-To: <199708210441.AAA22489@life.ai.mit.edu>
To: Cube-Lovers <cube-lovers@ai.mit.edu>
Message-Id: <Pine.PMDF.3.95.980329185822.328319C-100000@PSTCC6.PSTCC.CC.TN.US>

On Thu, 21 Aug 1997, michael reid wrote:

> dan recently introduced the concept of "partial isoglyphs", in which
> some faces are solid, and the others are glyphs of the same pattern.
> i looked into this a little and didn't find much.  only the case
> of two opposite solid faces seems to have many possible glyph types,
> although some of these possible types may have many solutions.
>
> here's what i found

Note that all the glyph types which Mike lists (01, 02, 0D, 04, and 03 in
Dan's taxonomy) fix the corners.  Thus, his note below points out that in
order to have anything other than two solid faces opposite to each other,
you must fix the corners.

The correspondence between Dan's taxonomy and my charts is 01=BF, 02=AF,
03=A7, 04=A5, and 0D=BD.  As I said earlier, the identfication numbers on
my charts are not a taxonomy.  Rather, they provide a unique
identification for each of the 2^8 glyphs.

>
> 6 solid faces: start
> 5 solid faces: no possibilities

> 4 solid faces:
>    other two faces opposite:  types 02, 0D and 04 are possible

All three possibilities do occur in my chart.


>    other two faces adjacent:  type 0D is possible

This possibility does occur in my chart.


> 3 solid faces:
>    mutually adjacent:  type 02 is possible

This possibility does occur in my chart.


>    not mutually adjacent:  types 01 and 0D are possible

Both possibilities do occur in my chart.


> 2 solid faces:
>    adjacent:  types 01, 02, 0D and 03 are possible

All  four possibilities do occur in my chart.

>    opposite:  many possible types

Indeed!


> 1 solid face:  types 01, 02 and 0D are possible
>

All three possibilities do occur in my chart.  In addition, I found three
partial isoglyphs of type 03 with one solid face.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990



From cube-lovers-errors@mc.lcs.mit.edu  Sun Apr  5 16:13:15 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Mon Mar 30 20:40:51 1998
Date: Mon, 30 Mar 1998 20:41:13 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Pretty vs. Not-So-Pretty Isoglyphs
To: Cube-Lovers <cube-lovers@ai.mit.edu>
Message-Id: <Pine.PMDF.3.95.980330202452.358663A-100000@PSTCC6.PSTCC.CC.TN.US>

After looking at a lot of isoglyphs and partial isoglyphs in the last
little while, I wonder if it's not the case that some of the
non-continuous isoglyphs are prettier than some of the continuous ones,
and that some of the partial isoglyphs are prettier than some of the
isoglyphs?

Continuous isoglyphs do *in general* seem prettier than non-continuous
ones, and isoglyphs do *in general* seem prettier than partial isoglyphs.
But consider the following two (counter?) examples.

The glyph

   000
   XXX
   000

yields (among other things) L2 F2 L2 R2 F2 R2, which is a non-continuous
partial isoglyph.  It looks about as follows (quite pretty and striking,
in my opinion):

      XXX
      XXX
      XXX

  0X0 0X0 0X0 0X0
  0X0 0X0 0X0 0X0
  0X0 0X0 0X0 0X0

      XXX
      XXX
      XXX



On the other hand,

   U B2 R2 F2 L2 U L2 F2 U2 R' B' R F' L' U2 B2 R2 B' D' U'

is a real mess in my opinion, even though it is a continuous isoglyph. It
looks something like the following.

      X00
      0X0
      XXX

  XOX XXX X00 00X
  XX0 0X0 0X0 0XX
  X00 00X XXX X0X

      00X
      0XX
      X0X


Notice that the partial isoglyph which was my first example "looks" fairly
continuous, even though it really isn't.  The reason it looks that way is
that it is continuous along all the edges where the non-solid glyphs come
together.  Call such a non-continuous partial isoglyph quasi-continuous.
I think your eye tends to ignore the solid faces anyway, so that a
quasi-continuous partial isoglyph tends to be very striking and very
pretty.  For example, there are a number of 4-H and 4-T patterns among the
partial isoglyphs which are quasi-continuous and which are very pretty.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990

From cube-lovers-errors@mc.lcs.mit.edu  Sun Apr  5 23:28:33 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sun Apr  5 18:06:04 1998
Date: Sun, 5 Apr 1998 18:05:59 -0400 (EDT)
From: der Mouse  <mouse@rodents.montreal.qc.ca>
Message-Id: <199804052205.SAA03822@Twig.Rodents.Montreal.QC.CA>
To: cube-lovers@ai.mit.edu
Subject: Re: Pretty vs. Not-So-Pretty Isoglyphs

> On the other hand,
>    U B2 R2 F2 L2 U L2 F2 U2 R' B' R F' L' U2 B2 R2 B' D' U'
> is a real mess in my opinion, even though it is a continuous
> isoglyph.

I think this (the pattern, not the operator to produce it) is actually
rather striking and pretty - provided you look at the cube along the
URB-LDF corner-to-corner axis.

					der Mouse

			       mouse@rodents.montreal.qc.ca
		     7D C8 61 52 5D E7 2D 39  4E F1 31 3E E8 B3 27 4B

From cube-lovers-errors@mc.lcs.mit.edu  Wed Apr  8 12:17:06 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Apr  8 11:04:08 1998
To: Cube-Lovers@ai.mit.edu
Date: Wed, 8 Apr 1998 07:55:06 -0700
Subject: A workable 6x6x6 cube design (probably)
Message-Id: <19980408.075506.7150.0.tenie1@juno.com>
From: tenie1@juno.com (Tenie Remmel)

I have found that the 6x6x6 cube can only be made practical if the
outer rows of cubies are slightly larger (about 3mm or 1/8 inch).
If the rows are all the same size then some cross-sections of pieces
(e.g. the corner pieces) are less than 3 sq-mm, and other pieces are
extremely thin (0.6mm in some places).

If the plastic is black (or white) and the stickers are all the same
size then the inequality in the size of the cubies will be effectively
masked.  The stickers would have to be spaced evenly.  The cube will
look as if it has a small 'border' but the perception will be that the
cubies are the same size.

This design is actually almost as strong as the 4x4x4 cube.  It contains
an internal frame plus 256 movable pieces of ten different types.  No
cross section of a piece is smaller than 7 sq-mm (the 4x4x4 has center
pieces with 9 sq-mm cross section).  Two of the types of piece
(FACE EDGE PIECE, SPACER PIECE 2) come in two mirror image forms, so the
number of molds that would be needed to produce this is 14 (counting two
for the internal frame).  The internal mechanism would need to be greased
to allow it to turn smoothly, but it should be no worse than the 5x5x5.

The following is an exact geometric description of each piece.  To be
able to understand this you need to know how to use Cartesian and Polar
coordinates.  All pieces are intersections of planes, spheres, and
hyperboloids (which can probably be approximated as cones).

The SPACER PIECE 2 could probably be replaced by some sort of rectangular
but rounded blob-like thing, it does not need to be an exact shape and
the cube might turn more smoothly if it is rounded.  It also might then
be possible to make it symmetrical so they could be produced with a
single mold, which would slightly reduce production cost.

Comments, suggestions and quibbles are welcome.



LEGEND - x,y,z are Cartesian coordinates, r is distance from origin
         Dx, Dy, Dz is distance from x, y, z axis respectively

NO TOLERANCES - pieces must be shrunk away from all sides a little bit

DIMENSIONS assume that the size of an inner CUBIE is 100 and the size of
an outer CUBIE is 125, this allows the pieces to be much stronger than if
the cubies were all the same size.  The puzzle occupies the space such
that
-325<x<325, -325<y<325, -325<z<325.

Maximum vertical play needed (i.e. strain in the radial direction, mostly
on the inside) is about 12 units, or 1.5 mm on a 78mm cube.  If all parts
are designed with a tolerance of 2 units (0.24 mm) (e.g. by shrinking
each
part 1 unit (0.12 mm) away from all sides) then the maximum vertical play
needed is about 8 units (1.0 mm).  Extra space could be provided (shrink
inner face of frame away from center by another couple of units), if this
is too much.


PUZZLE CONTAINS:
    08 CORNER PIECE
    24 INNER EDGE PIECE
    24 OUTER EDGE PIECE
    48 FACE EDGE PIECE
    24 FACE CORNER PIECE
    24 CENTER PIECE
    24 INTERNAL EDGE PIECE
    08 INTERNAL CORNER PIECE
    24 SPACER PIECE 1
    48 SPACER PIECE 2
    01 INTERNAL FRAME



----------------------------------------------------------------------

CORNER PIECE consists of:

        all points such that 200<x<325, 200<y<325, 200<z<325
    AND all points such that x>175, y>175, z>175, 280<r<360

INNER EDGE PIECE consists of:

        all points such that 0<x<100, 200<y<325, 200<z<325, r>320
    AND all points such that 0<x<70, y>175, z>175, 280<r<360

OUTER EDGE PIECE consists of:

        all points such that 100<x<200, 200<y<325, 200<z<325, r>360
    AND all points such that 100<x<175, y>175, z>175, 320<r<360
    AND all points such that 70<x<175, y>175, z>175, 280<r<320

FACE EDGE PIECE consists of:

        all points such that 0<x<100, 100<y<200, 200<z<325, r>360
    AND all points such that 0<x<100, 100<y<175, z>175, 320<r<360
    AND all points such that 0<x<70, 70<y<175, z>175, 280<r<320

FACE CORNER PIECE consists of:

        all points such that 100<x<200, 100<y<200, 200<z<325, r>360
        all points such that 100<x<175, 100<y<175, z>175, 320<r<360
    AND all points such that 70<x<175, 70<y<175, z>175, 280<r<320

CENTER PIECE consists of:

        all points such that 0<x<100, 0<y<100, 200<z<325, r>360
    AND all points such that 0<x<70, 0<y<70, z>175, 280<r<360
    AND all points such that 0<x<120, 0<y<120, z>0, 240<r<280
    AND all points such that x>60, y>60, z>0, 200<r<240,
Dz<sqrt(z^2+60^2)
    AND all points such that x>30, y>30, z>0, 100<r<200,
Dz<sqrt(z^2+30^2)

INTERNAL EDGE PIECE consists of:

        all points such that 60<x<z, -60<y<60, z>0, 200<r<240
    AND all points such that 30<x<z, -30<y<30, z>0, 100<r<200

INTERNAL CORNER PIECE consists of:

        all points such that Dx>sqrt(x^2+60^2), Dy>sqrt(y^2+60^2),
                             Dz>sqrt(z^2+60^2), x>0, y>0, z>0,
                             200<r<240
    AND all points such that Dx>sqrt(x^2+30^2), Dy>sqrt(y^2+30^2),
                             Dz>sqrt(z^2+30^2), x>0, y>0, z>0,
                             100<r<200
    AND all points such that x>120, y>120, z>120, 240<r<280

SPACER PIECE 1 consists of:

        all points such that x>120, y>120, 0<z<120, 240<r<280

SPACER PIECE 2 consists of:

        all points such that 70<x<100, 175<y<200, z>175, 320<r<360

INTERNAL FRAME consists of

        all points such that -60<x<60, -60<y<60, 200<r<240
    AND all points such that -60<x<60, -60<z<60, 200<r<240
    AND all points such that -60<y<60, -60<z<60, 200<r<240
    AND all points such that Dx<30, r<200
    AND all points such that Dy<30, r<200
    AND all points such that Dz<30, r<200
    AND all points such that r<100

--Tenie Remmel (tenie1@juno.com)

From cube-lovers-errors@mc.lcs.mit.edu  Thu Apr  9 16:30:17 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Apr  8 18:05:08 1998
To: Cube-Lovers@ai.mit.edu
Date: Wed, 8 Apr 1998 13:45:07 -0700
Subject: A workable 6x6x6 cube design (probably) - correction
Message-Id: <19980408.144131.8926.2.tenie1@juno.com>
From: tenie1@juno.com (Tenie Remmel)

Yikes, there were errors in my geometric description.
Here is a (hopefully) correct version:




CORNER PIECE consists of:

        all points such that 200<x<325, 200<y<325, 200<z<325
    AND all points such that 175<x<325, 175<y<325, 175<z<325, 280<r<360

INNER EDGE PIECE consists of:

        all points such that 0<x<100, 200<y<325, 200<z<325, r>320
    AND all points such that 0<x<70, 175<y<325, 175<z<325, 280<r<360

OUTER EDGE PIECE consists of:

        all points such that 100<x<200, 200<y<325, 200<z<325, r>360
    AND all points such that 100<x<175, 175<y<325, 175<z<325, 320<r<360
    AND all points such that 70<x<175, y>175, z>175, 280<r<320

FACE EDGE PIECE consists of:

        all points such that 0<x<100, 100<y<200, 200<z<325, r>360
    AND all points such that 0<x<100, 100<y<175, 175<z<325, 320<r<360
    AND all points such that 0<x<70, 70<y<175, z>175, 280<r<320

FACE CORNER PIECE consists of:

        all points such that 100<x<200, 100<y<200, 200<z<325, r>360
        all points such that 100<x<175, 100<y<175, 175<z<325, 320<r<360
    AND all points such that 70<x<175, 70<y<175, z>175, 280<r<320

CENTER PIECE consists of:

        all points such that 0<x<100, 0<y<100, 200<z<325, r>360
    AND all points such that 0<x<70, 0<y<70, 175<z<325, 280<r<360
    AND all points such that 0<x<120, 0<y<120, z>0, 240<r<280
    AND all points such that x>60, y>60, z>0, 200<r<240,
                             Dz<sqrt(z^2+60^2)
    AND all points such that x>30, y>30, z>0, 100<r<200,
                             Dz<sqrt(z^2+30^2)

INTERNAL EDGE PIECE consists of:

        all points such that 60<x<z, -60<y<60, z>0, 200<r<240
    AND all points such that 30<x<z, -30<y<30, z>0, 100<r<200

INTERNAL CORNER PIECE consists of:

        all points such that Dx>sqrt(x^2+60^2), Dy>sqrt(y^2+60^2),
                             Dz>sqrt(z^2+60^2), x>0, y>0, z>0,
                             200<r<240
    AND all points such that Dx>sqrt(x^2+30^2), Dy>sqrt(y^2+30^2),
                             Dz>sqrt(z^2+30^2), x>0, y>0, z>0,
                             100<r<200
    AND all points such that x>120, y>120, z>120, 240<r<280

SPACER PIECE 1 consists of:

        all points such that x>120, y>120, 0<z<120, 240<r<280

SPACER PIECE 2 consists of:

        all points such that 70<x<100, 175<y<200, 175<z<325, 320<r<360

INTERNAL FRAME consists of

        all points such that -60<x<60, -60<y<60, 200<r<240
    AND all points such that -60<x<60, -60<z<60, 200<r<240
    AND all points such that -60<y<60, -60<z<60, 200<r<240
    AND all points such that Dx<30, r<200
    AND all points such that Dy<30, r<200
    AND all points such that Dz<30, r<200
    AND all points such that r<100

--Tenie Remmel (tenie1@juno.com)

From cube-lovers-errors@mc.lcs.mit.edu  Mon Apr 13 12:07:55 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sat Apr 11 21:10:52 1998
Message-Id: <01BD658D.8DD543C0@jburkhardt.ne.mediaone.net>
From: John Burkhardt <jburkhardt@mediaone.net>
To: "Cube-Lovers@ai.mit.edu" <Cube-Lovers@ai.mit.edu>
Subject: RE: A workable 6x6x6 cube design (probably) - correction
Date: Sat, 11 Apr 1998 21:05:25 -0400

So who gets to try and make one?  I understood that the dies for the
5x5x5 cube are too expensive to build now due to "lack of interest".  On
the other hand, we should try to build one because we can.  If we can
that is :)

-JRB

From cube-lovers-errors@mc.lcs.mit.edu  Wed Apr 15 15:02:21 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Apr 15 13:07:56 1998
Date: Wed, 15 Apr 1998 18:07:57 +0100
From: David Singmaster <david.singmaster@sbu.ac.uk>
To: cube-lovers@ai.mit.edu
Message-Id: <009C4C21.E208C3B3.8@ice.sbu.ac.uk>
Subject: Hamiltonian circuits on the cube

	The discussion of isoglyphs, etc., has reminded me of a problem which I
worked on in the early 1980s but never resolved.  I took an all white cube and
traced a Hamitonian circuit through all the 54 facelets.  If you jumble this
up, it is essentially impossible to restore.  Indeed there are probably many
solutions to the problem.  This led me to ask some questions about such
Hamiltonian circuits through the 54 facelets.
	A.  How many are there?
	B.  Are there any such circuits where the pattern is the same on each
face?  I thought I could prove that such did not exist, but I think I assumed
that the circuit entered and left each face once, but this need not be the
case.
	I was able to find a circuit with two types of face pattern and the two
types were mirror images.  If you index the facelets on a face by  11, 12, ...,
33,  then the path on the face is:  11, 12, 22, 21, 31, 32, 33, 23, 13.
	If the circuit enters and leaves each face just once, then the sequence
of faces visited forms a Hamiltonian circuit on the faces of the cube, which is
better viewed as the vertices of an octahedron.  It is easy to see that there
are just two such circuits on the octahedron (up to isomorphism).  One of these
circuits has two kinds of vertex behavior and hence is not suitable.
	Does this question interest anyone?  The reason for the second question
was that if just one type of face pattern could be used, then it would be easy
to print up stickers for sale - one would just do the same pattern six times!

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499;
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Wed Apr 15 16:23:24 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Apr 15 15:35:06 1998
Date: Wed, 15 Apr 1998 15:38:00 -0400 (Eastern Daylight Time)
From: Dale Newfield <din5w@cs.virginia.edu>
Reply-To: DNewfield@cs.virginia.edu
To: cube-lovers@ai.mit.edu
Subject: Re: Hamiltonian circuits on the cube
In-Reply-To: <009C4C21.E208C3B3.8@ice.sbu.ac.uk>
Message-Id: <Pine.WNT.3.96.980415153024.-386749A-100000@biff.cs.virginia.edu>

On Wed, 15 Apr 1998, David Singmaster wrote:
> 	The discussion of isoglyphs, etc., has reminded me of a problem which I
> worked on in the early 1980s but never resolved.  I took an all white cube and
> traced a Hamitonian circuit through all the 54 facelets.  If you jumble this
> up, it is essentially impossible to restore.  Indeed there are probably many
> solutions to the problem.  This led me to ask some questions about such
> Hamiltonian circuits through the 54 facelets.

This is quite reminiscent of "Oddmaze,"
(http://www.edoc.com/zarf/custom-cubes.html) which is a creation by Andrew
Plotkin realized using Kristin Looney's "Custom Cube Technology"
(http://www.wunderland.com/WTS/Kristin/Technology.html).

On its surface is a labyrinth with no branches or dead ends.  Each
facelet has exactly two paths through it.  In the "start" position, at
least, the path obeys the Celtic knotwork property (over/under
alternations).  It is really quite interesting, and well described on the
above mentioned page.

(This doesn't help answer your questions, but might put you in contact
with another that has given them some thought.)

-Dale Newfield
 Dale@Newfield.org

From cube-lovers-errors@mc.lcs.mit.edu  Wed Apr 15 17:12:21 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Apr 15 16:25:48 1998
Date: Wed, 15 Apr 1998 16:29:26 -0400 (EDT)
From: Nicholas Bodley <nbodley@tiac.net>
To: John Burkhardt <jburkhardt@mediaone.net>
Cc: "Cube-Lovers@ai.mit.edu" <Cube-Lovers@ai.mit.edu>
Subject: RE: A workable 6x6x6 cube design (probably)
In-Reply-To: <01BD658D.8DD543C0@jburkhardt.ne.mediaone.net>
Message-Id: <Pine.BSF.3.96.980415160127.28862D-100000@shell2.tiac.net>

 Am I missing something? The geometrical description seemed plausible
and fine, but unless I'm far off base, it seems that some quite-clever
mechanical design is essential. Fairly sure that Douglas Hofstadter
noted in passing (I think in Go"del (G"odel ? :), Escher, Bach...) that
a physical prototype of the 6^3 has been built.

 I have pulled apart and studied all "sizes" from the 2^3 to the 5^3,
and the innards of each are rather different; the 5 is based on the 3,
but the 4 (Rubik's Revenge) has a ball inside, as probably most List
readers know. The innards of the 2 are quite distinctive, again; (also,
borderline impossible to assemble/disassemble!). It's remarkable how a
simple increment of one, so to speak, has such a profound effect on the
basic internal design.

 My awareness of most abstruse corners of math. is quite comparable with
that of, let's say, a turtle. However, I do know modest bits about
formal kinematics, four-bar linkages, and some underlying principles of
the linkage variety of mechanical analog computers, for instance, so my
ignorance is somewhat better that that of a rock. I also know the
innards of mechanical calculators rather well.

 However, with such non-qualifications, I suspect that there is no
theory of such mechanisms as we find inside our cubes and related
puzzles. Mathematicians seem to be able to handle braids (Emil Artin?)
rather well, and knots seem to be doing well, but I really doubt that
there's any significant theory that can be used to develop a design such
as the innards of a 5^3.

 Ordinary geometry, I feel fairly confident, is of relatively little
help. One can at least define the geometry of the requisite constraints
and "freedoms" of motion, but to create the requisite shapes, seems to
me, requires a special and clever kind of mind.

 Honestly, I'd welcome having big holes figuratively shot through my
contentions! I'm sure I'd learn something.

 For limited (and probably very costly) prototype runs, the technology
that goes by various names such as 3-D printing, rapid prototyping, and
(ugh!) stereolithography should do well to create the shapes. (Seems to
me it's a fairly formidable challenge to a CAD program to create some
of the weird shapes, but I plead ignorance!

 (The "stereo" part of that long word is fine, but it's really
stretching a point to think of it as writing on stone.)

My best to all,

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*  When will the non-word "alot" first be listed
|*  Amateur musician  *|*  in a dictionary? Maybe 2030?
--------------------------------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Wed Apr 15 18:36:13 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Apr 15 16:32:14 1998
Date: Wed, 15 Apr 1998 16:35:57 -0400 (EDT)
From: Nicholas Bodley <nbodley@tiac.net>
To: John Burkhardt <jburkhardt@mediaone.net>
Cc: "Cube-Lovers@ai.mit.edu" <Cube-Lovers@ai.mit.edu>
Subject: RE: A workable 6x6x6 cube design (probably) - another comment
In-Reply-To: <01BD658D.8DD543C0@jburkhardt.ne.mediaone.net>
Message-Id: <Pine.BSF.3.96.980415163057.28862E-100000@shell2.tiac.net>

On Sat, 11 Apr 1998, John Burkhardt wrote:

}So who gets to try and make one?  I understood that the dies for the
}5x5x5 cube are too expensive to build now due to "lack of interest".  On

Does anyone know if the dies still exist? I wouldn't be a bit surprised
if the whole set weighs several tons, even if they are single-cavity
types. Tooling for injection molding is fiercely expensive! (Tooling for
a decent ("serious") plastic soprano recorder runs probably a third to a
half $US million, for instance. (Mostly bigger parts, a few very
critical tolerances, and far fewer parts, also.))

Best,

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*  I might need to switch to shore.net, but will
|*  Amateur musician  *|*  do my best to minimize the nuisance if so.
--------------------------------------------------------------------------



From cube-lovers-errors@mc.lcs.mit.edu  Mon Apr 20 15:57:40 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Mon Apr 20 11:51:47 1998
To: cube-lovers@ai.mit.edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Re: Hamiltonian circuits on the cube
Date: 20 Apr 1998 15:55:44 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <6hfr60$lfq@gap.cco.caltech.edu>
References: <cube-lovers.009C4C21.E208C3B3.8@ice.sbu.ac.uk>

David Singmaster <david.singmaster@sbu.ac.uk> writes:
>	The discussion of isoglyphs, etc., has reminded me of a problem which I
>worked on in the early 1980s but never resolved.  I took an all white cube and
>traced a Hamitonian circuit through all the 54 facelets.  If you jumble this
>up, it is essentially impossible to restore.  Indeed there are probably many
>solutions to the problem.  This led me to ask some questions about such
>Hamiltonian circuits through the 54 facelets.
>	A.  How many are there?

>	B.  Are there any such circuits where the pattern is the same on each
>face?  I thought I could prove that such did not exist, but I think I assumed
>that the circuit entered and left each face once, but this need not be the
>case.

The answer to B is "Yes"!!

I was pretty surprised to come up with this within ten minutes of reading
the question:

+--+--+--+
|42|43|44|
+--+--+--+
|47|46|45|
+--+--+--+
|54| 3| 4|
+--+--+--+--+--+--+--+--+--+--+--+--+
| 1| 2| 5| 6| 7| 8|26|27|40|41|48|53|
+--+--+--+--+--+--+--+--+--+--+--+--+
|14|13|12|11|10| 9|25|28|39|38|49|52|
+--+--+--+--+--+--+--+--+--+--+--+--+
|15|16|17|18|21|22|24|29|36|37|50|51|
+--+--+--+--+--+--+--+--+--+--+--+--+
|33|32|19|
+--+--+--+
|34|31|20|
+--+--+--+
|35|30|23|
+--+--+--+

X=====X=====X=====X
H     H     H     H
---------------+  H
H     H     H  |  H
X=====X=====X==|==X
H     H     H  |  H
---------------+  H
H     H     H     H
X=====X=====X=====X
H     H     H     H
---+  H  +-----+  H
H  |  H  |  H  |  H
X==|==X==|==X==|==X=====X=====X=====X=====X=====X=====X==|==X==|==X==|==X
H  |  H  |  H  |  H     H     H     H     H     H     H  |  H  |  H  |  H
H  +-----+  H  +-----------------+  H  +-----+  H  +-----+  H  |  H  |  H
H     H     H     H     H     H  |  H  |  H  |  H  |  H     H  |  H  |  H
X=====X=====X=====X=====X=====X==|==X==|==X==|==X==|==X=====X==|==X==|==X
H     H     H     H     H     H  |  H  |  H  |  H  |  H     H  |  H  |  H
H  +-----------------------------+  H  |  H  |  H  +-----+  H  |  H  |  H
H  |  H     H     H     H     H     H  |  H  |  H     H  |  H  |  H  |  H
X==|==X=====X=====X=====X=====X=====X==|==X==|==X=====X==|==X==|==X==|==X
H  |  H     H     H     H     H     H  |  H  |  H     H  |  H  |  H  |  H
H  +-----------------+  H  +-----+  H  |  H  |  H  +-----+  H  +-----+  H
H     H     H     H  |  H  |  H  |  H  |  H  |  H  |  H     H     H     H
X=====X=====X=====X==|==X==|==X==|==X==|==X==|==X==|==X=====X=====X=====X
H     H     H     H
H  +-----+  H  +---
H  |  H  |  H  |  H
X==|==X==|==X==|==X
H  |  H  |  H  |  H
H  |  H  |  H  +---
H  |  H  |  H     H
X==|==X==|==X=====X
H  |  H  |  H     H
H  |  H  |  H  +---
H  |  H  |  H  |  H
X==|==X==|==X==|==X

--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
Smoking cigarettes are bad for you, so smoking cigarettes is bad for you.

From cube-lovers-errors@mc.lcs.mit.edu  Wed Apr 22 11:53:02 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Apr 22 11:43:05 1998
Date: Wed, 22 Apr 98 11:42:49 EDT
Message-Id: <9804221542.AA10123@sun28.aic.nrl.navy.mil>
From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: whuang@ugcs.caltech.edu
Cc: cube-lovers@ai.mit.edu
In-Reply-To: <6hfr60$lfq@gap.cco.caltech.edu>
Subject: Re: Hamiltonian circuits on the cube

whuang@ugcs.caltech.edu (Wei-Hwa Huang) writes:

> I was pretty surprised to come up with this within ten minutes of reading
> the question:

Wow, I'm impressed.  I thought I'd have to write a program to find
them, and here's a nice symmetric solution.  The symmetry is more
visible in a different unfolding:

+-@-+-@-+-@-+---+---+---+
| @@@@@   @@|@@@@@@@@@@ |
+   +   +   +   +   + @ +
| @@@@@@@@@@|@@@@@@@@@@ |
+ @ +   +   +   +   +   +
| @@@@@@@@@@|@@   @@@@@ |
+---+---+---+-@-+-@-+-@-+---+---+---+
            | @@@@@   @@|@@@@@@@@@@ |
            +   +   +   +   +   + @ +
            | @@@@@@@@@@|@@@@@@@@@@ |
            + @ +   +   +   +   +   +
            | @@@@@@@@@@|@@   @@@@@ |
            +---+---+---+-@-+-@-+-@-+---+---+---+
                        | @@@@@   @@|@@@@@@@@@@ |
                        +   +   +   +   +   + @ +
                        | @@@@@@@@@@|@@@@@@@@@@ |
                        + @ +   +   +   +   +   +
                        | @@@@@@@@@@|@@   @@@@@ |
                        +---+---+---+-@-+-@-+-@-+

It shouldn't be that hard to solve a cube with these markings--there
are only two different kinds of corner cubies, three kinds of edge
cubies, and the face centers need only be oriented mod 180 degrees.
Working from one of the symmetric corners, it's not hard to see that
this is the only continuous solution.

I've noticed a minor modification to your pattern that also admits an
isoglyphic Hamiltonian path:

+-@-+-@-+-@-+-@-+---+---+
|@@   @@@@@ | @@@@@@@@@ |
+   +   +   +   +   + @ +
| @@@@@@@@@@|@@@@@@@@@@ |
+ @ +   +   +   +   +   +
| @@@@@@@@@ | @@@@@   @@|
+---+---+-@-+-@-+-@-+-@-+-@-+---+---+
            |@@   @@@@@ | @@@@@@@@@ |
            +   +   +   +   +   + @ +
            | @@@@@@@@@@|@@@@@@@@@@ |
            + @ +   +   +   +   +   +
            | @@@@@@@@@ | @@@@@   @@|
            +---+---+-@-+-@-+-@-+-@-+-@-+---+---+
                        |@@   @@@@@ | @@@@@@@@@ |
                        +   +   +   +   +   + @ +
                        | @@@@@@@@@@|@@@@@@@@@@ |
                        + @ +   +   +   +   +   +
                        | @@@@@@@@@ | @@@@@   @@|
                        +---+---+-@-+-@-+-@-+-@-+

Anyone who's working on an exhaustive search to see if there are any
others, send me e-mail before I hack again!

Dan
Hoey@AIC.NRL.Navy.Mil

From cube-lovers-errors@mc.lcs.mit.edu  Wed Apr 22 12:36:05 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Apr 22 12:19:02 1998
Message-Id: <353E1961.6231@sgi.com>
Date: Wed, 22 Apr 1998 09:22:57 -0700
From: Derek Bosch <bosch@sgi.com>
To: Dan Hoey <Hoey@aic.nrl.navy.mil>
Cc: cube-lovers@ai.mit.edu
Subject: Re: Hamiltonian circuits on the cube - kind of
References: <9804221542.AA10123@sun28.aic.nrl.navy.mil>

On a similar note, has anyone stickers with:

  |
 /
-   -
   /
  |

or

  |
  |
-----
  |
  |

(or any of those rotations?)  Kind of a cross between a
rubik's Tangle and a rubik's cube?  Especially if each of
the lines has a different color?

D
-- 
Derek Bosch        "A little nonsense now and then
(650) 933-2115      is relished by the wisest men"... W.Wonka
bosch@sgi.com

From cube-lovers-errors@mc.lcs.mit.edu  Wed Apr 22 14:41:52 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Apr 22 14:20:19 1998
Date: Wed, 22 Apr 1998 14:24:21 -0400 (Eastern Daylight Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: Hamiltonian circuits on the cube
In-Reply-To: <9804221542.AA10123@sun28.aic.nrl.navy.mil>
To: Dan Hoey <Hoey@aic.nrl.navy.mil>
Cc: whuang@ugcs.caltech.edu, cube-lovers@ai.mit.edu
Message-Id: <Pine.WNT.3.96.980422141817.-200953J-100000@GN209A.PSTCC.CC.TN.US>

On Wed, 22 Apr 1998, Dan Hoey wrote:

> whuang@ugcs.caltech.edu (Wei-Hwa Huang) writes:
>
> > I was pretty surprised to come up with this within ten minutes of reading
> > the question:
>
> Wow, I'm impressed.  I thought I'd have to write a program to find
> them, and here's a nice symmetric solution.  The symmetry is more
> visible in a different unfolding:
>

Not to minimize the difficulty of the problem or the beauty of the
solution (quite the contrary), but the solution seems almost trivial
when viewed in the light of Dan's particular unfolding of the surface of
the cube.  The same comment is true of Dan's isoglyphic solution.

It makes me wonder of you actually saw Dan's unfolding in your mind's
eye, as it were, as you worked out your solution.  Or another way to put
it, did you work out your solution in 2-D or in 3-D?  It also makes me
wonder if there is any other unfolding that would lead as naturally to a
Hamiltonian circuit.  I tend to think not, but I could well be wrong.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990

From cube-lovers-errors@mc.lcs.mit.edu  Thu Apr 23 11:51:20 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Thu Apr 23 11:42:59 1998
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Message-Id: <199804231547.IAA09346@gluttony.ugcs.caltech.edu>
Subject: Re: Hamiltonian circuits on the cube
To: jbryan@pstcc.cc.tn.us (Jerry Bryan)
Date: Wed, 22 Apr 1998 16:58:41 -0700 (PDT)
Cc: cube-lovers@ai.mit.edu
In-Reply-To: <9804231425.AA10935@sun28.aic.nrl.navy.mil> from "Dan Hoey" at Apr 23, 98 10:25:30 am
Reply-To: whuang@ugcs.caltech.edu

Jerry Bryan typed something like this in a previous message:
> It makes me wonder of you actually saw Dan's unfolding in your mind's
> eye, as it were, as you worked out your solution.  Or another way to put
> it, did you work out your solution in 2-D or in 3-D?  It also makes me
> wonder if there is any other unfolding that would lead as naturally to a
> Hamiltonian circuit.  I tend to think not, but I could well be wrong.
>
Actually, I didn't visualize any unfolding at all, so I guess I
did it in 3-D.  Here's approximately the line of reasoning that
led to my solution.

As Dr. Singmaster notes, there is only one way to draw a Hamiltonian on
a 1x1x1 cube where all the faces are identical, and that is with a right
angle on each face.  Naturally one's first impulse is to find a path that
enters each 3x3 face in one place and exits in another -- and these two
ends must be on edges 90-degree apart.  One quickly sees that the two exits
must be on edge cubies, since if any were on corner cubies there would
be a parity problem between "inner corners" and "outer corners."  But if
they were edge cubies, then no Hamiltonian path exists (as the inner corner
must join to the ends already).

However, another extension is the "three parallel paths" pattern:  put this
on each face:

 A  B  C
 |  |  |
 |  |  +-D
 |  +----E
 +-------F
This leads to three paths on the cube, where the center one is the
traditional 1x1x1 Hamiltonian.  If this can be rearranged to a solution,
we must try to reconnect the ends so that there is some "interaction"
between the three paths.  C must connect to D, but we can connect A to B
instead -- and this leads to a solution, which surprised me when I
visualized it on a 3-d cube.  (I most definitely find visualizing in
3-D easier than visualizing the links in an unfolded cube.)

-- 
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
Smoking cigarettes are bad for you, so smoking cigarettes is bad for you.

From cube-lovers-errors@mc.lcs.mit.edu  Thu Apr 23 20:24:58 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Thu Apr 23 20:22:45 1998
Date: Thu, 23 Apr 98 20:21:11 EDT
Message-Id: <9804240021.AA11374@sun28.aic.nrl.navy.mil>
From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: cube-lovers@ai.mit.edu
Cc: whuang@ugcs.caltech.edu
Subject: Re: Hamiltonian circuits on the cube

I wrote: 
    "...send me e-mail before I hack again!"

Too late.  The only chiral Hamiltonian isopaths are the two we've
already seen, and:
                        +---+---+-@-+---+-@-+---+
                        | @@@@@   @@|@@@@@@   @@|
                        + @ + @ +   +   +   + @ +
                        | @   @@@@@@|@@@@@@   @ |
                        + @ +   +   +   + @ + @ +
                        |@@   @@@@@@|@@   @@@@@ |
            +---+---+-@-+---+-@-+---+-@-+---+---+
            | @@@@@   @@|@@@@@@   @@|
            + @ + @ +   +   +   + @ +
            | @   @@@@@@|@@@@@@   @ |
            + @ +   +   +   + @ + @ +
            |@@   @@@@@@|@@   @@@@@ |
+---+---+-@-+---+-@-+---+-@-+---+---+
| @@@@@   @@|@@@@@@   @@|
+ @ + @ +   +   +   + @ +
| @   @@@@@@|@@@@@@   @ |
+ @ +   +   +   + @ + @ +
|@@   @@@@@@|@@   @@@@@ |
+---+-@-+---+-@-+---+---+

I actually generated all the continuous chiral isopaths, and the
following is the other extreme--the only one with nine disjoint paths.
Yet one of the paths goes through one third of the facelets.

                        +-@-+-@-+---+-@-+-@-+-@-+
                        |@@   @@@@@@|@@   @   @ |
                        +   +   +   +   + @ + @ +
                        |@@   @@@@@@|@@@@@@   @@|
                        + @ + @ +   +   +   +   +
                        | @   @   @@|@@@@@@   @@|
            +-@-+-@-+---+-@-+-@-+-@-+---+-@-+-@-+
            |@@   @@@@@@|@@   @   @ |
            +   +   +   +   + @ + @ +
            |@@   @@@@@@|@@@@@@   @@|
            + @ + @ +   +   +   +   +
            | @   @   @@|@@@@@@   @@|
+-@-+-@-+---+-@-+-@-+-@-+---+-@-+-@-+
|@@   @@@@@@|@@   @   @ |
+   +   +   +   + @ + @ +
|@@   @@@@@@|@@@@@@   @@|
+ @ + @ +   +   +   +   +
| @   @   @@|@@@@@@   @@|
+-@-+-@-+-@-+---+-@-+-@-+

Dan
Hoey@AIC.NRL.Navy.Mil

From cube-lovers-errors@mc.lcs.mit.edu  Fri Apr 24 09:41:36 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Fri Apr 24 09:38:22 1998
Date: Fri, 24 Apr 98 09:38:06 EDT
Message-Id: <9804241338.AA11821@sun28.aic.nrl.navy.mil>
From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: cube-lovers@ai.mit.edu
Cc: whuang@ugcs.caltech.edu
Subject: Re: Hamiltonian circuits on the cube

I wrote:
> I actually generated all the continuous chiral isopaths, and the
> following is the other extreme--the only one with nine disjoint paths.

Which was bogus.  I actually generated only the continuous chiral
isopaths in which no circuit lies entirely on one face.  That's fine
for the Hamiltonian circuit problem, but for the maximum number of
disjoint circuits we probably want the 14-circuit pattern

                        +-@-+-@-+-@-+---+---+---+
                        |@@   @@@@@ | @@@@@   @@|
                        +   +   +   + @ + @ + @ +
                        |@@   @@@@@ | @@@@@   @@|
                        + @ + @ + @ +   +   +   +
                        |@@   @@@@@ | @@@@@   @@|
            +-@-+-@-+-@-+---+---+---+-@-+-@-+-@-+
            |@@   @@@@@ | @@@@@   @@|
            +   +   +   + @ + @ + @ +
            |@@   @@@@@ | @@@@@   @@|
            + @ + @ + @ +   +   +   +
            |@@   @@@@@ | @@@@@   @@|
+-@-+-@-+-@-+---+---+---+-@-+-@-+-@-+
|@@   @@@@@ | @@@@@   @@|
+   +   +   + @ + @ + @ +
|@@   @@@@@ | @@@@@   @@|
+ @ + @ + @ +   +   +   +
|@@   @@@@@ | @@@@@   @@|
+---+---+---+-@-+-@-+-@-+

which should be familiar to Tartan fans.

Dan
Hoey@AIC.NRL.Navy.Mil

From cube-lovers-errors@mc.lcs.mit.edu  Sat Apr 25 20:15:48 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Fri Apr 24 14:24:49 1998
Date: Fri, 24 Apr 1998 14:21:43 -0400 (Eastern Daylight Time)
From: Dale Newfield <din5w@cs.virginia.edu>
Reply-To: DNewfield@cs.virginia.edu
To: cube-lovers@ai.mit.edu
Subject: 4x4x4 pieces, and in quantity
Message-Id: <Pine.WNT.3.96.980424135218.-329659L-100000@biff.cs.virginia.edu>

[ Moderators note: Dale Newfield passes on this notice.
  Contact Mike Green <mgreen@speakeasy.org> for details. ]

Date: Fri, 24 Apr 1998 01:17:15 -0700
From: Mike Green <mgreen@speakeasy.org>
To: Dale Newfield <DNewfield@cs.virginia.edu>
Cc: Dale Newfield <din5w@cs.virginia.edu>, Dale Newfield <Dale@Newfield.org>
Subject: "Rubik's Revenge" - 4x4x4

Dale,

Thank you for your inquiry.  We do have a limited number of "Rubik's
Revenge" parts for those of you who have a broken cube: 

         ITC-030a 4x4x4 Center Cubie - Ideal Toy Co. $ 2.50 each 
         ITC-030b 4x4x4 Ball Center - Ideal Toy Co.  $10.00 each 
         ITC-030c 4x4x4 Corner Cubie - Ideal Toy Co. $ 2.00 each 
         ITC-030d 4x4x4 Edge Cubie - Ideal Toy Co.   $ 2.00 each 
         ITC-030e 4x4x4 Sticker - Ideal Toy Co.      $  .50 each

You want 1 corner and 2 centers?  You will reuse your stickers?  How will
you pay?  Postage will probably be $2.00. 

Recently the price of a "Rubik's Revenge" has hit as high as $200.00 each
on the "Web". Can you believe that! The last five we sold, fortunately for
our customers, went for $65.00 each. How would you like to see it back in
the market for less than $30.00? Possibly even less than $25.00. Would you
buy more than one?  For us to bring it back we have to place a minimum
order of between 10,000 to 30,000 pieces and pay for new tooling - all up
front. Tell your friends and have them tell their friends, and their
friend's friends to get on our wish list. Have your local puzzle retailer
contact us as well. By using the power of the "Internet", e-mail, and word
of mouth I'm sure we can get the numbers up there and make this happen in
less than a year. I'm ready and willing are you? 

In the meantime, we also carry as standard stock the Rubik's 2x2x2 for
$5.99, Rubik's 3x3x3 for $10.99, 3x3x3 Magic Cube for $6.99, 5x5x5 for
$38.99, Square 1 for $14.99, and Skewb for $32. 

We also pull in on a fairly regular basis Megaminx, Impossiball, Pyraminx,
Mickey's Challenge, Masterballs, and various other sequential movement
puzzles when we can. Prices and quantities vary, but we're always on the
hunt.

We'd very much like to bring the 4x4x4 back to market. You can help
greatly by spreading the word. 

Thank you.

Sincerely,

Mike D. Green 
President

From cube-lovers-errors@mc.lcs.mit.edu  Sat Apr 25 21:20:14 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sat Apr 25 20:50:39 1998
Date: Sat, 25 Apr 98 20:50:27 EDT
Message-Id: <25Apr1998.202137.Hoey@AIC.NRL.Navy.Mil>
From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: bosch@sgi.com
Cc: cube-lovers@ai.mit.edu
In-Reply-To: <353E1961.6231@sgi.com> (message from Derek Bosch on Wed, 22 Apr
	1998 09:22:57 -0700)
Subject: Re: Hamiltonian circuits on the cube - kind of

Derek Bosch <bosch@sgi.com> asks for a cross between a Rubik's tangle
and a Rubik's cube.  Here's a Hamiltonian chiral isotangle.

                                    .__._____._____.__.__._____._____.__.
                                    |  \  :  /  :  \  |  \  :  |  :  \  |
                                    +-. `-+-' .-+-. `-+-. `-+--+--+-. `-+
                                    |..\..:../..:..\..|..\..:..|..:..\..|
                                    |  |  :  /  :  /  |  /  :  /  :  |  |
                                    +--+--+-' .-+-' .-+-' .-+-' .-+--+--+
                                    |..|..:../..:../..|../..:../..:..|..|
                                    |  \  :  |  :  \  |  \  :  /  :  \  |
                                    +-. `-+--+--+-. `-+-. `-+-' .-+-. `-+
                  .__._____._____.__|__\__:__|__:__\__|__\__:__/__:__\__|
                  |  \  :  /  :  \  |  \  :  |  :  \  |
                  +-. `-+-' .-+-. `-+-. `-+--+--+-. `-+
                  |..\..:../..:..\..|..\..:..|..:..\..|
                  |  |  :  /  :  /  |  /  :  /  :  |  |
                  +--+--+-' .-+-' .-+-' .-+-' .-+--+--+
                  |..|..:../..:../..|../..:../..:..|..|
                  |  \  :  |  :  \  |  \  :  /  :  \  |
                  +-. `-+--+--+-. `-+-. `-+-' .-+-. `-+
.__._____._____.__|__\__:__|__:__\__|__\__:__/__:__\__|
|  \  :  /  :  \  |  \  :  |  :  \  |
+-. `-+-' .-+-. `-+-. `-+--+--+-. `-+
|..\..:../..:..\..|..\..:..|..:..\..|
|  |  :  /  :  /  |  /  :  /  :  |  |
+--+--+-' .-+-' .-+-' .-+-' .-+--+--+
|..|..:../..:../..|../..:../..:..|..|
|  \  :  |  :  \  |  \  :  /  :  \  |
+-. `-+--+--+-. `-+-. `-+-' .-+-. `-+
|__\__:__|__:__\__|__\__:__/__:__\__|

There's only one path, so it's all one color.

Dan
Hoey@AIC.NRL.Navy.Mil

From cube-lovers-errors@mc.lcs.mit.edu  Wed Apr 29 10:54:31 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Apr 29 00:53:36 1998
Message-Id: <3546B17A.3419@idirect.com>
Date: Wed, 29 Apr 1998 00:50:02 -0400
From: Mark Longridge <cubeman@idirect.com>
To: cube-lovers@ai.mit.edu
Cc: cubeman@idirect.com
Subject: Various Cube Thoughts

Ok, I'm back into cubing again... a few interesting, if somewhat
disjoint observations:

Summary of the 3 different types of optimal superflip sequences:

1)  Superflip with minimal q turns & symmetric moves
    Process has central reflection symmetry
    R3 U2 B1 L3 F1 U3 B1 D1 F1 U1 D3 L1 D2 F3 R1 B3 D1 F3 U3 B3 D3 U1
    (24q, 22f)

2)  Superflip with minimal q turns & asymmetric moves
    U1 R2 F3 R1 D3 L1 B3 R1 U3 R1 U3 D1 F3 U1 F3 U3 D3 B1 L3 F3 B3 D3 L3
    (24q, 23f)

3)  Superflip with minimal f turns & asymmetric moves
    U1 R2 F1 B1 R1 B2 R1 U2 L1 B2 R1 U3 D3 R2 F1 D2 B2 U2 R3 L1
    (28q, 20f)

------------------------------------------------------------------

No matter which cube you start searching from, e.g. pons asinorum,
12 flip, or any random cube, the dispersion of cubes is the same:

1, 12, 114, 1068, 10011... etc

So much for trying to search backwards from the 12-flip to number
the positions from (perhaps) antipode to start!

------------------------------------------------------------------

 I have got Mike Reid's optimal solver to work under the dos shell
in windows 95. I finally managed to compile it using WATCOM 11.0
thusly:

     wcl386 /k10000000 search.c

 I had to give it a 10 megabyte stack for it to work!

 It found the sequence ( F R B L )^5 to require 20 q turns, so there
is nothing better. Next I tried ( F R B L )^6 to see if that would
be 24 q but a 20 q solution was found. Mike Reid confirmed the
result on another computer running Linux.

-------------------------------------------------------------------

Lastly, some non-mathematical ideas on how to do optimal searches
of rubik's cube patterns. Using my own human solving algorithm
I solve the 4 down edge cubes last. One of the patterns I
get was solved optimally by Mike's program thusly:

D' R' D' F B' D' L' D L D F' B D R

If we assign a value of 1 to each face and add them we get:

D = 6  U = 0
F = 2  B = 2
L = 2  R = 2

Note that most of the action occurs with the D face, which I find
suggestive. After all, nothing is moved except the 4 bottom edge cubes.
Also all the other faces have an even number of turns!
My idea is perhaps with some pre-processing of a goal state it is
possible to prune the number of moves down to such a degree that
the number of moves actually tried is quite small. Also note that
this particular goal state has only 2 pairs of cubes swapped, and
all the other cubes are in place.

Now I may be using too much hindsight, but to me it is counter-
intuitive that it is possible to have 3 separate turns of the
D face! So, sequences with 3 uses of the D face should not be
considered. My theory is that ultimately with enough pre-processing
only the optimal sequences will be even considered!

-> Mark <-

From cube-lovers-errors@mc.lcs.mit.edu  Thu Apr 30 10:09:06 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Thu Apr 30 09:59:19 1998
Date: Thu, 30 Apr 1998 09:57:20 -0400 (Eastern Daylight Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: Various Cube Thoughts
In-Reply-To: <3546B17A.3419@idirect.com>
To: Mark Longridge <cubeman@idirect.com>
Cc: cube-lovers@ai.mit.edu
Message-Id: <Pine.WNT.3.96.980430083712.-225563B-100000@GN209A.PSTCC.CC.TN.US>

On Wed, 29 Apr 1998, Mark Longridge wrote:

> ------------------------------------------------------------------
> 
> No matter which cube you start searching from, e.g. pons asinorum,
> 12 flip, or any random cube, the dispersion of cubes is the same:
> 
> 1, 12, 114, 1068, 10011... etc
>
> So much for trying to search backwards from the 12-flip to number
> the positions from (perhaps) antipode to start!
>
> ------------------------------------------------------------------

This has been discussed before on Cube-Lovers.  There are several ways
to look at why it is true.  I think at the most basic level that it
depends on the inverse property of groups.  Let A be any non-empty
subset (not necessarily a subgroup) of G, and let x be any element of G.
Then xA contains the same number of elements as A.  Hence, if A is (for
example) the set of all positions which are n moves from Start, then xA
is the set of all positions which are n moves from x, and xA is the same
size as A (remember that the distance from Start to a is the same as the
distance from x to xa for any a in A).

Notice that if A is a subgroup of G rather than just being a subset,
then xA is a coset. The fact that cosets are either equal or disjoint,
combined with the fact that A is the same size as xA, constitute the
basis for the proof that the size of a subgroup must divide evenly the
size of the group.

The inverse property is involved in showing that A and xA are the same
size as follows. Suppose we have A={a,b,c} which contains three
elements.  Then we have xA={xa,xb,xc} which also appears to contain
three elements.  The only way that xA would not have three elements
would be if some of the apparently distinct elements were really the
same, for example if xa and xc were really two different names for the
same element.  But if xa=xc, then we have x'(xa)=x'(xc) so that
(x'x)a=(x'x)c so that ia=ic so that a=c.  We know by definition that a
and c are distinct.  Hence, xa and xc must be distinct.

Just to give one more illustration of the importance of the inverse
property in showing that A and xA are the same size, here is a false
counterexample.  Consider the multiplicative group of the real numbers
or of the rational numbers.  Suppose A={ 2/3, 3/4, 7} and x=4.  Then,
xA={ 8/3, 3, 28}.  So far, so good because both A and xA have three
elements.  But suppose x=0.  Then xA={0, 0, 0}={0} which has only one
element. Here we have A with three elements and xA with only one
element.  So what is wrong.  The problem is that any multiplicative
group of what we might call "normal"  numbers (e.g., real or rational or
complex) must omit zero because 0 does not have a multiplicative
inverse.  That is, there is no solution to the equation 0*x=1. So when I
let x=0, I was cheating by multiplying by a number which is not in the
multiplicative group and which does not have a multiplicative inverse.

The reason I know that this has been discussed before was that I was
involved in the discussion.  At one point I incorrectly asserted that
what you are calling "the dispersion of the cubes" did depend on which
position was at the root of the search.  Cube-Lovers was quick to
correct me, of course.  However egregious was my error, it was still an
honest error.  The reason for the honest error is that I accomplish
nearly all my searches by counting patterns (M-conjugacy classes) rather
than by counting positions.  And when you count by patterns, "the
dispersion of the cubes" does depend upon which pattern is at the root
of the search.  So my mistake was to make a statement about positions
which should have been applied only to patterns.

Your note reminded me of a question I have thought about off and on ever
since that previous discussion.  Suppose you are searching by patterns.
Under what circumstances can you start the search with two different
patterns and still have the "dispersion of the cubes" be the same?  I
suspect that there is a very simple answer, but I am having trouble
ascertaining what it is.  I suspect that the only possibility is if the
two positions differ by superflip, that is if one of them is x then the
other one must be xf=fx, where f is the superflip.  But I am simply not
sure if there are any more possibilities.  Note that having the two
different patterns be M-conjugate is not an answer to the question
because if two patterns are M-conjugate then they are really just one
pattern.

As a last comment, readers of Cube-Lovers should be familiar with the
sequence 1, 12, 114... for positions in quarter turn searches.  A search
for patterns in quarter turns begins 1, 1, 5...  The first 1 is Start.
The second 1 (1q from Start) is Q, the set of twelve quarter turns.  The
5 (2q from Start) represents the following five patterns: 1) any face
twisted twice in the same direction, 2) any two opposite faces twisted
once each in the same direction (an antislice), 3) any two opposite
faces twisted once each in the opposite direction (a slice), 4) any two
adjacent faces twisted once each in the same direction (e.g., UF or
U'F'), and 5) any two adjacent faces twisted once each in the opposite
direction (e.g., UF' or U'F).  Beyond 2q from Start, it becomes too
complicated to calculate the patterns in my head.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990

From cube-lovers-errors@mc.lcs.mit.edu  Thu Apr 30 14:16:03 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Thu Apr 30 13:01:01 1998
Message-Id: <v02140b02b16e6b37786e@[205.230.130.11]>
Date: Thu, 30 Apr 1998 12:59:19 -0500
To: cube-lovers@ai.mit.edu
From: kristin@wunderland.com (Kristin Looney)
Subject: Garden Variety Rubik's Cube

Cube Lovers -

a new cube pic on the image wall...  for your viewing pleasure...

http://wunderland.com/EBooks/ImageWall/Pages/GardenVarietyCube.html

Peace -

-K.
kristin@wunderland.com
http://www.wunderland.com/wts/kristin

To all the fishies in the deep blue sea, Joy.

From cube-lovers-errors@mc.lcs.mit.edu  Fri May  1 10:45:38 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Fri May  1 01:54:06 1998
From: Andrew John Walker <ajw01@uow.edu.au>
Message-Id: <199805010552.PAA00579@wumpus.its.uow.edu.au>
Subject: Square like groups
To: cube-lovers@ai.mit.edu
Date: Fri, 1 May 1998 15:52:34 +1000 (EST)

	Does anyone have any information on patterns where each
face only contains opposite colours, but are not in the square
group?  L' R U2 L R' may be an example. If square moves
are applied to such patterns to form new groups, how many such groups
exist?
	Andrew Walker

From cube-lovers-errors@mc.lcs.mit.edu  Fri May  1 19:58:06 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Fri May  1 19:57:13 1998
Date: Fri, 1 May 98 19:56:56 EDT
Message-Id: <9805012356.AA16835@sun28.aic.nrl.navy.mil>
From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: ajw01@uow.edu.au
Cc: cube-lovers@ai.mit.edu
In-Reply-To: <199805010552.PAA00579@wumpus.its.uow.edu.au> (message from
	Andrew John Walker on Fri, 1 May 1998 15:52:34 +1000 (EST))
Subject: Re: Square like groups

Andrew Walker <ajw01@uow.edu.au> asks:

>         Does anyone have any information on patterns where each
> face only contains opposite colours, but are not in the square
> group?

We may call this the "pseudosquare" group P.  It consists of
orientation-preserving permutations that operate separately on the
three equatorial quadruples of edge cubies and the two tetrahedra of
corner cubies, and for which the total permutation parity is even.  So
Size(P) = 4!^5 / 2 = 3981312.

> L' R U2 L R' may be an example.

No, that's in the square group, says GAP.  Also, Mark Longridge
noticed (8 Aug 1993) that the square group is mapped to itself under
conjugation by an antislice (though I don't recall a proof--is there
an easy one?).  Your position is (L R)' R2 T2 R2 (L R), so this result
would apply.  Does anyone have a square process for it?

> If square moves are applied to such patterns to form new groups, how
> many such groups exist?

Consider the subgroup of P consisting of positions in which the
parity of the corner permutation is even.  (The edge permutation will
then also be even, and the parity of the permutations of the two edge
tetrahedrons will be equal).  Call it AP, for "alternating P".
Size(AP) = Size(P)/2 = 1990656.

The square group S is a subgroup of index 3 in AP, so
Size(S)=Size(AP)/3=663552.  I don't have a very criterion for choosing
elements of AP to be in S, except that it has to do with a correlation
between the permutations of the two tetrahedrons of corners, provided
those permutations are of the same parity (as they must be for the
position to be in AP).

According to GAP, these are the only three possibilities.  To be
explicit, let us label the cube's corners

  1 D   B 3
  C 2   4 A

Then we can partition S4 into six cosets:

    C1 = { (),      (3,4)(1,2), (1,4)(2,3), (2,4)(1,3) }
    C3 = { (1,2,3), (1,4,2),    (1,3,4),    (2,4,3)    }
    C2 = { (1,3,2), (1,4,3),    (2,3,4),    (1,2,4)    }
    C4 = { (1,2),   (1,4,2,3),  (1,3,2,4),  (3,4)      }
    C5 = { (2,3),   (1,4),      (1,3,4,2),  (1,2,4,3)  }
    C6 = { (1,3),   (2,4),      (1,4,3,2),  (1,2,3,4)  }

and similarly D1,D2,...,D4 for S4 acting on {A,B,C,D}.  Now let c be
an arbitrary permutation in P that fixes {A,B,C,D} elementwise, and
let Coset(c) be the coset to which c's operation on {1,2,3,4} belongs.
Let d be an arbitrary permutation in P that fixes {1,2,3,4}
elementwise, and let Coset(d) be the coset to which d's operation on
{A,B,C,D} belongs.  Then the group generated by <S, c*d> depends only
on Coset(c) and Coset(d):

     Coset(d)
             D1 D2 D3   D4 D5 D6
Coset(c)
        C1    S AP AP    P  P  P
        C2   AP  S AP    P  P  P
        C3   AP AP  S    P  P  P

        C4    P  P  P    S AP AP
        C5    P  P  P   AP  S AP
        C6    P  P  P   AP AP  S

There may be some wisdom to be gained in seeing that C1 is normal in
S4, so S4/C1 is isomorphic to S3.  We can represent the Ci and Di by
their action on {1,2,3,A,B,C}.  The above table shows whether the
group <Coset(c)*Coset(d), C1*D1, C2*D2, C3*D3, C4*D4, C5*D5, C6*D6>,
has order 6, 18, or 24.

I'd love to hear a more explanatory description of this phenomenon,
especially if it explains the absence of a subgroup of index 3 in P.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From cube-lovers-errors@mc.lcs.mit.edu  Sat May  2 17:23:32 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Fri May  1 22:41:24 1998
Message-Id: <354A8671.730D@idirect.com>
Date: Fri, 01 May 1998 22:35:29 -0400
From: Mark Longridge <cubeman@idirect.com>
Reply-To: cubeman@idirect.com
To: Dan Hoey <Hoey@aic.nrl.navy.mil>
Cc: cube-lovers@ai.mit.edu
Subject: Re: Square like groups
References: <9805012356.AA16835@sun28.aic.nrl.navy.mil>

Dan Hoey wrote:
>
> Andrew Walker <ajw01@uow.edu.au> asks:
>
> >         Does anyone have any information on patterns where each
> > face only contains opposite colours, but are not in the square
> > group?
>
> We may call this the "pseudosquare" group P.  It consists of
> orientation-preserving permutations that operate separately on the
> three equatorial quadruples of edge cubies and the two tetrahedra of
> corner cubies, and for which the total permutation parity is even.  So
> Size(P) = 4!^5 / 2 = 3981312.
>
> > L' R U2 L R' may be an example.

 R2 F2 R2 U2 R2 F2 R2 U2 F2

>
> No, that's in the square group, says GAP.  Also, Mark Longridge
> noticed (8 Aug 1993) that the square group is mapped to itself under
> conjugation by an antislice (though I don't recall a proof--is there
> an easy one?).  Your position is (L R)' R2 T2 R2 (L R), so this result
> would apply.  Does anyone have a square process for it?

I almost forgot about all that info back in 1993!

But I hardly think a proof is necessary. After the moves (L' R) all the
following moves are in the square's group. Then we are just doing
the inverse of (L` R) at the end. Not very rigourous, but...

I'll search for a counter-example.

-> Mark <-

From cube-lovers-errors@mc.lcs.mit.edu  Sat May  2 18:35:38 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sat May  2 18:31:11 1998
Date: Sat, 2 May 98 18:30:59 EDT
Message-Id: <9805022230.AA17631@sun28.aic.nrl.navy.mil>
From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: cube-lovers@ai.mit.edu
Subject: Re: Square like groups

With respect to the square group, I wrote:

> I'd love to hear a more explanatory description of this phenomenon,
> especially if it explains the absence of a subgroup of index 3 in P.

I should really have waited until I got back home to Singmaster's
book, which has a marvelous explanation of why the squares group has
index 6 in the pseudosquare group.

First, the edges are permuted in in all possible ways consistent with
    1. remaining in their "equators" of four edges,
    2. not being flipped, and
    3. having a permutation parity equal to that of the corners.
so we need only consider the 2x2x2 cube, and then we fix the BLD
corner in place.  Corners don't get twisted, so we consider only the
permutation.

We express the generators as permutations of the seven movable
corners, expressed as follows:

            2-------A
           /       / \
          /   T   /   \      F^2 = B^2 = (1,4)(B,C),
         /       /     \     R^2 = L^2 = (1,3)(A,C),
        B-------1   R   3    T^2 = D^2 = (1,2)(A,B).
         \       \     /
          \   F   \   /
           \       \ /
            4-------C

The neat part is to notice that the permutation on {A,B,C} is
determined by the permutation on {1,2,3,4}.  We do this by
representing these generators as symmetries on a tetrahedron, labelled
as follows.

            1-----------C-----------2
             \`-.               .-'/
              \  `A.         .B'  /
               \    `-.   .-'    /
                \      `4'      /
                 \      :      /
                  B     :     A
                   \    :    /
                    \   C   /
                     \  :  /
                      \ : /
                       \:/
                        3

Notice that the symmetry that permutes the tetrahedron's vertex labels
as (1,4) also permutes the edge labels as (B,C), corresponding to F^2
in the cube's action.  Similarly (1,3) implies (A,C) and (1,2) implies
(A,B).

With respect to Mark Longridge's having noticed that the square group
is mapped to itself under conjugation by an antislice (L R), the proof
turns out to be pretty easy.  First, we notice that we may consider
conjugation by a slice (L R') since that differs by a square (R^2)
from the antislice.  Now we work in the group that includes whole-cube
orientations, and perform the slice in the mechanically easy way, as a
4-cycle of face centers and an equatorial 4-cycle of edges.  Note that
all the edges of the equator are flipped (with respect to the
orientation that is preserved by the psueudo-square and square groups)
by the slice.

So if S is a square-group process that rotates the edges in an equator
E, the process

     Slice' S Slice S'

has the following actions:

1. Identity on the corners and the two equators other than E,
   because they are not moved by the slice,
2. Identity on the face centers, because they are not moved by S,
3. Flips each edge of E twice (once in Slice' and once in Slice), so
   restores the orientation, and
4. Is an even permutation of the edges in E (odd in Slice, odd in
   Slice', and equal in S and S').

The even permutation (4) of the edges in E is a slice group process,
as Mark noted, as for instance the 3-cycle (R^2 F^2 R^2 T^2)^2 F^2.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From cube-lovers-errors@mc.lcs.mit.edu  Mon May  4 10:31:18 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sun May  3 17:19:27 1998
Message-Id: <199805032117.RAA07495@life.ai.mit.edu>
Date: Sun, 3 May 1998 17:18:53 -0400
From: michael reid <reid@math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: Re:  Square like groups

andrew walker asks

> 	Does anyone have any information on patterns where each
> face only contains opposite colours, but are not in the square
> group?  L' R U2 L R' may be an example.

the set of such patterns is what i called the "target subgroup" for my
optimal solver.  it is the intersection of the three subgroups
<U, D, F2, R2, B2, L2>,  <F, B, U2, R2, D2, L2>  and  <R, L, F2, U2, B2, D2>
(or the intersection of any two of them).

the position he mentions is in the square group (mark longridge gives
a minimal maneuver for it).  dan hoey remarks that the square group has
index 6 in this "pseudo-square" group.  christoph bandelow's book
"inside rubik's cube and beyond" gives a nice criterion for a pseudo-
square pattern to be in the square group.

bandelow's criterion (slightly paraphrased) is

     the four  U  corners must be coplanar, the four  F  corners
     must be coplanar, and the four  R  corners must be coplanar.
     (equivalently, all twelve sets of four coplanar corners
     remain coplanar.)

in fact, this forces the parity of the corner permutation to be
even (and thus the same for the edge permutation).

this reminds me of an interesting idea i had for a puzzle: a 3x4x5 box,
whose faces and slices are restricted to 180 degree turns.  this sort
of thing could also be done with any dimensions.

mike

From cube-lovers-errors@mc.lcs.mit.edu  Mon May  4 11:24:21 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Mon May  4 10:36:17 1998
From: "Noel Dillabough" <noel@mud.ca>
To: <Cube-Lovers@ai.mit.edu>
Subject: Revenge and the 5x5x5
Date: Mon, 4 May 1998 10:35:52 -0400
Message-Id: <000001bd7769$f0ced480$02c0c0c0@nat>

Since we all know that Rubik's Revenge (4x4x4) puzzles are nearly impossible
to find (all of mine have long ago broken) and the 5x5x5 cubes fall apart so
easily that they are basically unusable.

Well, as a solution to this, I took a Virtual Cube simulation and added
sizing buttons (the cube program supports 2x2x2 to 5x5x5 sized cubes), a
keyboard interface, and allowed it to receive sequences in standard UDFBLR
notation.  I also added locking of the center pieces to make using a paired
up Revenge easier.

The cube is located at http://www.mud.ca/cube/cube.html.  Any thoughts,
comments, suggestions about the program should be sent to:
mailto://noel@mud.ca.

Enjoy,
-Noel Dillabough

From cube-lovers-errors@mc.lcs.mit.edu  Mon May  4 14:18:15 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Mon May  4 12:28:19 1998
Message-Id: <19980504162440.4037.qmail@hotmail.com>
From: "Philip Knudsen" <philipknudsen@hotmail.com>
To: Cube-Lovers@ai.mit.edu
Subject: Re: Revenge and the 5x5x5
Date: Mon, 04 May 1998 09:24:39 PDT

Noel writes:

> Since we all know that Rubik's Revenge (4x4x4) puzzles are
> nearly impossible to find (all of mine have long ago broken)
> and the 5x5x5 cubes fall apart so easily that they are basically
> unusable.

You are right about the 4x4x4 availability. I have, however,
never had any problems with my 5x5x5 cube. Actually the 5x5x5
mechanism is quite ingenious. I never heard of any broken one.
The only problem i can think of is the orange sticker tendency
fall off.

Philip K

From cube-lovers-errors@mc.lcs.mit.edu  Mon May  4 15:28:57 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Mon May  4 12:40:08 1998
Message-Id: <199805041642.MAA16954@nineCo.com>
To: Cube-Lovers@ai.mit.edu
Subject: 4x4x4 (Rubik's Revenge) puzzles for sale
Reply-To: yanowitz@gamesville.com
Date: Mon, 04 May 1998 12:42:51 -0400
From: Jason Yanowitz <yanowitz@Taz.nineCo.com>

Hi,

I have 6 Rubik's Revenge puzzles (in the original packaging) that I'm
considering selling.  If people are interested in purchasing one, send
me an offer (yanowitz@gamesville.com).

I apologize for the commercial nature of this post, but I've seen a
few other commercial posts.

thanks,
-- Jason

[ Moderator's note: Announcements of on-topic stuff for sale is generally
  okay, up until it starts clogging the list.  I usually snip any
  detailed descriptions of the auction process, catalogues of other
  products, corporate history, etc.--you can get that from Jason
  (though he thoughtfully omitted excess in his message). ]

From cube-lovers-errors@mc.lcs.mit.edu  Mon May  4 15:58:52 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Mon May  4 15:11:31 1998
Date: Mon, 4 May 1998 15:09:33 -0400 (EDT)
From: Nichael Cramer <nichael@sover.net>
To: Cube-Lovers@ai.mit.edu
Subject: Re: Revenge and the 5x5x5
In-Reply-To: <19980504162440.4037.qmail@hotmail.com>
Message-Id: <Pine.BSI.3.95.980504150725.4632C-100000@granite.sover.net>

Philip Knudsen wrote:
> You are right about the 4x4x4 availability. I have, however,
> never had any problems with my 5x5x5 cube.

BTW, for interested (and near-by) folks, Games People Play in Harvard Sq
had several 5Xs on the shelf when I dropped through the store last Thurs.

Nichael

From cube-lovers-errors@mc.lcs.mit.edu  Wed May  6 09:18:29 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Tue May  5 13:57:15 1998
Message-Id: <19980505174802.8836.qmail@hotmail.com>
From: "Philip Knudsen" <philipknudsen@hotmail.com>
To: Cube-Lovers@ai.mit.edu
Subject: Re: Revenge and the 5x5x5
Date: Tue, 05 May 1998 10:48:00 PDT

I suggest people with 5x5x5 that tend to fall apart try and fasten the
small screw underneath the center caps. This might help, at least it did
on mine. Mine didn't fall apart though, it just got loose, and sometimes
the pieces between the corners and the centres would sort of make a
wrong twist. After i tightened the screws that problem disappeared.

Philip K

From cube-lovers-errors@mc.lcs.mit.edu  Tue May 12 15:55:03 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Tue May 12 14:24:43 1998
Message-Id: <35576405.5EC25A05@frontiernet.net>
Date: Mon, 11 May 1998 16:48:05 -0400
From: John Bailey <jmb184@frontiernet.net>
To: Cube-Lovers <Cube-Lovers@ai.mit.edu>
Subject: Solving a 4 Dimensional Rubik's type Cube

Announcing a web page at
http://www.frontiernet.net/~jmb184/solution.html which gives the
explicit steps to solve a challenge configuration for a 2x2x2x2 (that's
four dimensions) Rubik type cube.

The challenge configuration is available at
http://www.frontiernet.net/~jmb184/Nteract4.html.
These pages do NOT require a Java enabled browser however, they do
require Netscape 4.0 or Microsoft Explorer 4.0.

This note is to solicit your judgments regarding the difficulty of the 4
Dimensional Rubik with no edge cubes (2x2x2x2.)  I believe it is
relatively easy, provided only that the simulation provides for the
cyclic permutation move, (NE-->SW, SW-->NW, NW--->NE)

Background:
Posted on rec.puzzles dl April 21, 1998:

A four dimensional articulated cube is on the web at
http://www.frontiernet.net/~jmb184/4cube.html

The result of marrying a Rubik's cube with a tesseract, this cube is
2x2x2x2. It has 16 corners and 24 faces.  It does not have edge cubes
and the corners have no orientation requirement.  Only 4 colors are
used. The solution space is thus roughly equivalent to that of a 3x3x3
Rubik if not smaller.  It is rendered in Javascript and will run on
Netscape 3.0 and 4.0

This posting caused about 25 hits to the page, but got no follow-up
dialog on the rec.puzzles news dl.

Note that in this first version, the corners are only identified by
color, not by correct position.

I wrote the page without having a clue as to how to solve it.  In the
process of just testing code I discovered that it is remarkably
unchallenging, once you get a sense of which corners the various
buttons rotate. (Flipping a glove from left-handed to right-handed can
be done in 4-space, but is impossible in 3-space.)  I may not be an
unprejudiced solver, but I would rate the challenge only slightly
harder than a 15 square slider puzzle. To increase the level of
difficulty, a second version of the puzzle was developed.  In this
version, the solution requires that the corners are returned to their
correct location.  They still do not requires 4-space orientation.
This version was announced in the following posting.


Posted on the rec.puzzles dl May 2, 1998:
A Four dimensional Rubik's Cube with solution.
At http://www.frontiernet.net/~jmb184/Nteract4.html

Re-designed to allow importing of 3D Rubik methods, this version uses
(a slightly extended version of)  standard Rubik cube naming of moves
and positions, has a shortcut button for one of the common permutation
moves and a scramble button to provide a challenge position.
I rate the challenge as equivalent to solving two faces of a 3D Rubik
cube.  I am looking forward to your comments, opinions,  and
suggestions.  I am especially interested in positions which cannot be
solved  or cannot be solved without extensive permutation moves other
than the one included.

This page has received about 50 hits.  But again, there was no
responding dialog on rec.puzzles news dl.

The difficulty of the second version is higher, but I rated the
challenge as equivalent to solving two layers of a 3x3x3 cube. The only
obstacle, an ordinary solver might face,  is finding the longish
sequence required to permutate 3 of 4 corners.  That's why I provided
the shortcut button (which applies the actions: L'URU'R'LRUR'U' with one
click.)

Discussion:
My concern is that people assume the puzzle is really hard and not worth
the effort.  It  may be seen as somewhat like the sequences from one
time pads which would be cryptographers who post and ask if anyone can
decrypt them.  To make it clear that a solution is not that difficult, I
have now made a page which gives an explicit solution, with
illustrations of each step and even some animation at
http://www.frontiernet.net/~jmb184/solution.html

There are obviously shorter sequences to obtain a solution, however this
one has the value of providing clear checkpoints along the way, such
that a solver can determine if they have missed a twist.

I want and would welcome your judgment about how easy or hard the puzzle
is.

John Bailey
jmb184@frontiernet.net
http://www.frontiernet.net/~jmb184
May 11, 1998

From cube-lovers-errors@mc.lcs.mit.edu  Tue May 12 17:33:46 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Tue May 12 14:24:43 1998
Message-Id: <35576405.5EC25A05@frontiernet.net>
Date: Mon, 11 May 1998 16:48:05 -0400
From: John Bailey <jmb184@frontiernet.net>
To: Cube-Lovers <Cube-Lovers@ai.mit.edu>
Subject: Solving a 4 Dimensional Rubik's type Cube

Announcing a web page at
http://www.frontiernet.net/~jmb184/solution.html which gives the
explicit steps to solve a challenge configuration for a 2x2x2x2 (that's
four dimensions) Rubik type cube.

The challenge configuration is available at
http://www.frontiernet.net/~jmb184/Nteract4.html.
These pages do NOT require a Java enabled browser however, they do
require Netscape 4.0 or Microsoft Explorer 4.0.

This note is to solicit your judgments regarding the difficulty of the 4
Dimensional Rubik with no edge cubes (2x2x2x2.)  I believe it is
relatively easy, provided only that the simulation provides for the
cyclic permutation move, (NE-->SW, SW-->NW, NW--->NE)

Background:
Posted on rec.puzzles dl April 21, 1998:

A four dimensional articulated cube is on the web at
http://www.frontiernet.net/~jmb184/4cube.html

The result of marrying a Rubik's cube with a tesseract, this cube is
2x2x2x2. It has 16 corners and 24 faces.  It does not have edge cubes
and the corners have no orientation requirement.  Only 4 colors are
used. The solution space is thus roughly equivalent to that of a 3x3x3
Rubik if not smaller.  It is rendered in Javascript and will run on
Netscape 3.0 and 4.0

This posting caused about 25 hits to the page, but got no follow-up
dialog on the rec.puzzles news dl.

Note that in this first version, the corners are only identified by
color, not by correct position.

I wrote the page without having a clue as to how to solve it.  In the
process of just testing code I discovered that it is remarkably
unchallenging, once you get a sense of which corners the various
buttons rotate. (Flipping a glove from left-handed to right-handed can
be done in 4-space, but is impossible in 3-space.)  I may not be an
unprejudiced solver, but I would rate the challenge only slightly
harder than a 15 square slider puzzle. To increase the level of
difficulty, a second version of the puzzle was developed.  In this
version, the solution requires that the corners are returned to their
correct location.  They still do not requires 4-space orientation.
This version was announced in the following posting.


Posted on the rec.puzzles dl May 2, 1998:
A Four dimensional Rubik's Cube with solution.
At http://www.frontiernet.net/~jmb184/Nteract4.html

Re-designed to allow importing of 3D Rubik methods, this version uses
(a slightly extended version of)  standard Rubik cube naming of moves
and positions, has a shortcut button for one of the common permutation
moves and a scramble button to provide a challenge position.
I rate the challenge as equivalent to solving two faces of a 3D Rubik
cube.  I am looking forward to your comments, opinions,  and
suggestions.  I am especially interested in positions which cannot be
solved  or cannot be solved without extensive permutation moves other
than the one included.

This page has received about 50 hits.  But again, there was no
responding dialog on rec.puzzles news dl.

The difficulty of the second version is higher, but I rated the
challenge as equivalent to solving two layers of a 3x3x3 cube. The only
obstacle, an ordinary solver might face,  is finding the longish
sequence required to permutate 3 of 4 corners.  That's why I provided
the shortcut button (which applies the actions: L'URU'R'LRUR'U' with one
click.)

Discussion:
My concern is that people assume the puzzle is really hard and not worth
the effort.  It  may be seen as somewhat like the sequences from one
time pads which would be cryptographers who post and ask if anyone can
decrypt them.  To make it clear that a solution is not that difficult, I
have now made a page which gives an explicit solution, with
illustrations of each step and even some animation at
http://www.frontiernet.net/~jmb184/solution.html

There are obviously shorter sequences to obtain a solution, however this
one has the value of providing clear checkpoints along the way, such
that a solver can determine if they have missed a twist.

I want and would welcome your judgment about how easy or hard the puzzle
is.

John Bailey
jmb184@frontiernet.net
http://www.frontiernet.net/~jmb184
May 11, 1998

From cube-lovers-errors@mc.lcs.mit.edu  Thu May 14 10:52:20 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Thu May 14 09:18:42 1998
Date: Thu, 14 May 1998 14:03:21 +0100
From: David Singmaster <david.singmaster@sbu.ac.uk>
To: Cube-Lovers@AI.MIT.Edu
Cc: zingmast@ice.sbu.ac.uk
Message-Id: <009C62C9.843B05E9.31@ice.sbu.ac.uk>
Subject: New radio programme

TO:  FRIENDS AND COLLEAGUES

        I am participating in a new weekly program called 'Puzzle Panel' on
BBC Radio 4, beginning on Thursday, 4 June at 1:30.  We recorded a pilot in
January and the commissioning producers were delighted with it.  There will be
a group of three to five panelists and we will discuss both mathematical and
verbal puzzles.  Some will be sent in by listeners and some will be set to the
listeners by the panellists.  At the pilot, the panel was myself, Chris
Maslanka (of the Guardian) as chair, William Hartston (of the Independent,
etc.) and Ann Bradford (compiler of a Crossword dictionary), but the membership
may vary.  I'll let you know of any changes of time/date, etc.

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499;
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Thu May 21 13:24:10 1998
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Message-Id: <35646320.2295@ping.be>
Date: Thu, 21 May 1998 18:23:45 +0100
From: Geoffroy Van Lerberghe <Geoffroy.VanLerberghe@ping.be>
To: Cube-Lovers <Cube-Lovers@ai.mit.edu>
Subject: Cristoph's Jewel internal mechanism

The Christoph's Magic Jewel is a disguised Rubik's cube (cf. Metamagical
Themas by Douglas R. Hofstadter p.339 : Stan Isaacs's coloring scheme)
but what about the internal mechanism? Is it simply a Rubik's cube with
only edge and centre cubes or is the mechanism different from the
classic cube.
I haven't managed to disassemble the Magic Jewel yet.


Geoffroy.VanLerberghe@ping.be

From cube-lovers-errors@mc.lcs.mit.edu  Thu May 21 17:26:51 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Thu May 21 14:06:58 1998
Message-Id: <19980521180258.25636.qmail@hotmail.com>
From: "Philip Knudsen" <philipknudsen@hotmail.com>
To: Cube-Lovers@ai.mit.edu
Subject: Re: Cristoph's Jewel internal mechanism
Date: Thu, 21 May 1998 11:02:57 PDT

The Jewel is basically an octahedron, but the vertex pieces are absent.
This does not make the puzzle easier. Apart from the jewel i also have a
taiwanese and a polish made octahedreon (with vertex pieces). A third
version exists, made by Uwe Meffert, but quite rare. The turning quality
of the jewel is very close to that of the polish made octahedron, so i
believe that is where the jewel originates (Correct me if i'm wrong,
Christoph!)
The disassembled polish octahedron has a mechanism very close to that of
the Pyraminx puzzle, also by Uwe Meffert. It is not a cube mechanism.

>The Christoph's Magic Jewel is a disguised Rubik's cube (cf.
Metamagical
>Themas by Douglas R. Hofstadter p.339 : Stan Isaacs's coloring scheme)
>but what about the internal mechanism? Is it simply a Rubik's cube with
>only edge and centre cubes or is the mechanism different from the
>classic cube.
>I haven't managed to disassemble the Magic Jewel yet.
>
>Geoffroy.VanLerberghe@ping.be

____________________________________
Philip K
recording and performing artist
Vendersgade 15, 3th
DK - 1363 Copenhagen K
Phone:  +45 33932787
Mobile: +45 21706731
E-mail: philipknudsen@hotmail.com
E-mail: skouknudsen@get2net.dk
E-mail: skouknudsen@email.dk
E-mail: 4521706731@sms.tdm.dk (leave subject blank!)

From cube-lovers-errors@mc.lcs.mit.edu  Fri May 22 12:26:20 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Fri May 22 06:24:16 1998
Message-Id: <19980522101332.6763.qmail@hotmail.com>
From: "Philip Knudsen" <philipknudsen@hotmail.com>
To: cube-lovers@ai.mit.edu
Subject: spare piece for domino variant
Date: Fri, 22 May 1998 03:13:31 PDT

I just received puzzle from a fellow collector:
It is like a Magic Domino, but only about 47 mm along the long
edges. The pieces are red and white. The 9 red pieces have a
drawing of Superman and the 9 white pieces a drawing of Superwoman!
Unfortunately the puzzle was broken on arrival.
Does anyone on the list have a similar broken puzzle,
and maybe could spare a piece (edge)?

____________________________________
Philip K
recording and performing artist
Vendersgade 15, 3th
DK - 1363 Copenhagen K
Denmark
Phone:  +45 33932787
Mobile: +45 21706731
E-mail: skouknudsen@get2net.dk
E-mail: philipknudsen@hotmail.com
E-mail: skouknudsen@email.dk      (soon to expire)
E-mail: 4521706731@sms.tdm.dk     (leave subject blank!)

From cube-lovers-errors@mc.lcs.mit.edu  Fri May 22 19:06:21 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Fri May 22 19:02:24 1998
Date: Fri, 22 May 1998 18:59:24 -0400 (EDT)
From: Nicholas Bodley <nbodley@tiac.net>
To: Cube Mailing List <Cube-lovers@ai.mit.edu>
Subject: Magic Jack
Message-Id: <Pine.BSF.3.96.980522182519.9037C-100000@shell2.tiac.net>


 Sorry if my memory's faulty, but I don't recall any recent mention of
the Magic Jack. This is a 3-cubed, 3-D array of 26 small cubes
constrained by an outer cage to slide past their neighbors. At first
glance, it looks like a Rubik's Cube, but immediately one realizes it's
quite different. It's about the same size. Disassembly looks impossible
unless the outer "cage" is cut. 

 As you'd expect, it's a 3^3 array, but with one position empty. It's a
3-D analog of the 15 Puzzle. The individual cubes are not connected in
any sense to their neighbors. While the moves in a 15 Puzzle are in one
plane and easily defined by amateur mathematicians, in the Magic Jack,
there are many more possible ways of moving a given cube to another
position.

 Also, not surprisingly, cube moves are strictly translational.

 The fun begins when one attempts to create patterns. Each cube has
specific surface markings. The simplest configuration creates an
exterior in which all cubes have a random, fine-grained, glittery
diffraction-grating-like surface. More complicated, and difficult, are
the colored patterns, which when solved, create (iirc) a continuous path
around the whole puzzle. There are three, I'm fairly sure; one creates a
message. Solving is made more difficult by the fact that most cube
faces are obscured by their neighbors.

 As to its intrinsic mathematical difficulty, I'm not close to being
well informed/educated enough to judge. The practical problem of hidden
faces does add to the practical difficulty, and the number of "degrees"
of freedom for a given cube (from 3 to 6, depending on position)
certainly increases the available choices.

 I saw this puzzle at Games People Play in Cambridge; it's a German
import. Quality of construction was good, although there was no
detenting, and it could be easier to move the cubes. It might actually
be easier to constrain potential interferers, and let gravity do the
work. The difficulty was essentially caused by other cubes' getting out
of position, not poor quality. Price in the store is $25.

 Not sure whether they're interested in mail orders, but it might be
worth a try. While I have no connections with G.P.P., perhaps it
wouldn't be out of order to give some info.:

 The Games People Play
 1100 Massachusetts Ave. (Abbreviation = Mass. is OK!)
 Cambridge, Mass. 02138
 (617) 492-0711

 Afaik, they had possibly as many as a dozen in stock.

 G.P.P. also periodically imports 5^3s from Germany, perhaps not from
Dr. Bandelow. They have a nice collection of movable-piece puzzles.

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*  Are you designing an icon for a GUI?
|*  Amateur musician  *|*  China has been doing it for millennia.
--------------------------------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Mon May 25 15:42:05 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Fri May 22 19:38:36 1998
Date: Fri, 22 May 1998 19:35:00 -0400
Message-Id: <22May1998.192434.Alan@LCS.MIT.EDU>
From: Alan Bawden <Alan@lcs.mit.edu>
Sender: Alan@lcs.mit.edu
To: nbodley@tiac.net
Cc: Cube-lovers@ai.mit.edu
In-Reply-To: <Pine.BSF.3.96.980522182519.9037C-100000@shell2.tiac.net>
	(message from Nicholas Bodley on Fri, 22 May 1998 18:59:24 -0400
	(EDT))
Subject: Re: Magic Jack

   Date: Fri, 22 May 1998 18:59:24 -0400 (EDT)
   From: Nicholas Bodley <nbodley@tiac.net>
   ...
    Not sure whether they're interested in mail orders, but it might be
   worth a try. While I have no connections with G.P.P., perhaps it
   wouldn't be out of order to give some info.:

    The Games People Play
    1100 Massachusetts Ave. (Abbreviation = Mass. is OK!)
    Cambridge, Mass. 02138
    (617) 492-0711

    Afaik, they had possibly as many as a dozen in stock.

Check your local puzzle outlet first -- Magic Jack may be pretty widely
available.  When I was in the hospital last summer, my father brought one
of these with him when he came to vist me from Philadelphia.  I don't
recall the name of the store there where he purchased it.

I still haven't solved it.  The first step would clearly be to just catalog
the 26 different cubies, but I haven't even done that...

				- Alan

From cube-lovers-errors@mc.lcs.mit.edu  Mon May 25 16:14:10 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Fri May 22 20:01:43 1998
Date: Fri, 22 May 1998 19:58:40 -0400 (EDT)
From: Nicholas Bodley <nbodley@tiac.net>
To: Alan Bawden <Alan@lcs.mit.edu>
Cc: Cube-lovers@ai.mit.edu
Subject: Magic Jack website (!)
In-Reply-To: <22May1998.192434.Alan@LCS.MIT.EDU>
Message-Id: <Pine.BSF.3.96.980522195247.9037F-100000@shell2.tiac.net>

 Sorry, all; the 'Net still has its surprises. Guess what: The Magic
Jack has its own Web site:

www.magicjack.com

They list the retailers who carry it; there are very roughly a dozen or
so. The site looks worth a visit.

Gosh, Alan, I guess we all should welcome you back, if my recollection's
clear! May you continue to be well!

My best to all,

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*  Are you designing an icon for a GUI?
|*  Amateur musician  *|*  China has been doing it for millennia.
--------------------------------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Mon May 25 16:48:52 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sat May 23 03:33:16 1998
From: canttype@earthlink.net
Message-Id: <l03130300b18c2787a716@[38.29.35.221]>
In-Reply-To: <Pine.BSF.3.96.980522182519.9037C-100000@shell2.tiac.net>
Date: Sat, 23 May 1998 00:34:41 -0700
To: <Cube-lovers@ai.mit.edu>
Subject: Re: Magic Jack vs. Vadasz Cube

Nicholas Bodley <nbodley@tiac.net> wrote

> Sorry if my memory's faulty, but I don't recall any recent mention of
>the Magic Jack....

check out
http://members.aol.com/islandcom/
for information about the Vadasz Cube which is a variation of the Magic
Jack described above. I have a "3x3x3 Classic Cube Solid" and have been
able to solve it.
The Vadasz Cube allows you to easily disassemble it, if desired. Also, each
of the 26 cubies can be disassembled and reconfigured allowing you to
create variations of the puzzle. The cubies are made out of plastic tiles
so that you can re-arrange the construction and colors of each of the 26
cubies if you desire.

Five different puzzles are available:
2x2x2
3x3x3
4x4x4
and
multi 3x3x3
multi 4x4x4

From cube-lovers-errors@mc.lcs.mit.edu  Wed May 27 07:01:51 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed May 27 03:39:50 1998
Message-Id: <19980527073534.13389.qmail@hotmail.com>
From: "Philip Knudsen" <philipknudsen@hotmail.com>
To: cube-lovers@ai.mit.edu
Subject: Re: Magic jack
Date: Wed, 27 May 1998 00:35:33 PDT

Two comments on the Magic Jack:

Apart from Magic Jack and Vadasz Cube, there also exists a german
produced puzzle called "IQUBE". Like the Magic Jack, this is a 3x3x3
sliding puzzle with 26 smaller cubes. Cubes have colours red, green and
yellow, and it is possible to arrange them so the entire surface is
either red or green. Yellow is possible with red or green centres. IQUBE
comes with a leaflet that suggests a total 12 different solution
possibilities. The puzzle is suitable for the blind, since the three
different colours also feel differently. I bought mine from Spielkiste
 (http://www.twfg.de/puzzle/default.htm).

A 2x2x2 sliding puzzle is mentioned in "Rubik's Cubic Compendium", in
the part that is written by David Singmaster (quote):
"The only such (three dimensional moving-piece puzzle) puzzle that i
know of is a sliding cube puzzle of Piet Hein which is so rare that both
Rubik and I recently re-invented it before learning that it had been
done by Hein."
There is also an illustration which shows a 2x2x2 sliding cube puzzle
similar to the small Vadasz Cube.

Philip K

From cube-lovers-errors@mc.lcs.mit.edu  Wed May 27 12:52:33 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed May 27 11:37:46 1998
Message-Id: <01BD8995.A6F13540.Johan.Myrberger@ebc.ericsson.se>
From: Johan Myrberger <Johan.Myrberger@ebc.ericsson.se>
Reply-To: <Johan.Myrberger@ebc.ericsson.se>
To: <cube-lovers@ai.mit.edu>
Subject: RE: Magic jack
Date: Wed, 27 May 1998 17:34:05 +0200
Organization: Ericsson Business Networks AB

On 27 May 1998 09:36, Philip Knudsen wrote:

> Apart from Magic Jack and Vadasz Cube, there also exists a german
> produced puzzle called "IQUBE". Like the Magic Jack, this is a 3x3x3
> sliding puzzle with 26 smaller cubes. Cubes have colours red, green and
> yellow, and it is possible to arrange them so the entire surface is
> either red or green. Yellow is possible with red or green centres. IQUBE
> comes with a leaflet that suggests a total 12 different solution
> possibilities....

Some years ago (around 1989?) I made a computer search on this kind of
puzzle.

The idea was "is there a way of colouring the 27 cubies (and then remove
one) so that a 3x3x3 cube can be arranged (with sliding block moves) to
show all external sides of either of three colours".

Since a 3x3x3 cube shows 9x6=54 cubie sides at one time, and 27 cubies have
in all 27x6=54x3 cubie sides all "cubie sides" would be used in one
configuration each.

So - I hunted for the answers to:
1) Is such a colouring possible?
2) Which cubie would be nicest to remove?

My search showed that 1) was indeed possible, and that there is one
distinct way for the colouring (not counting reflections etc) and 2) It is
possible to choose a cubie to remove so that the space will be positioned
in one of the space diagonals for each of the three solutions.

If anyone is interested I can dig out the specific colouring.

Regards
/Johan Myrberger
mailto:Johan.Myrberger@ebc.ericsson.se
http://home.bip.net/johan.myrberger

From cube-lovers-errors@mc.lcs.mit.edu  Thu May 28 12:18:52 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed May 27 16:46:38 1998
From: David A Bagley <bagleyd@gwyn.tux.org>
Message-Id: <199805272042.QAA15526@gwyn.tux.org>
Subject: Updated 3x3x3x3 (Rubik's Tesseract)
To: cube-lovers@ai.mit.edu
Date: Wed, 27 May 1998 16:42:53 -0400 (EDT)
Cc: charlied@erols.com

Hi All

A new version of Charlie Dickman's Rubik Tesseract program and its
accompanying documentation is now available from http://www.tux.org/~bagleyd/
(under the heading of "Neat 4D stuff I wish I wrote" :) ).
This latest version contains a general solution to unscramble an
arbitrarily scrambled Rubik Tesseract as well as some improved bells and
whistles.  The solution is given in the docs and is also implemented in
the Macintosh program.

All mail about the Tesseract docs and program should be addressed to
Charlie Dickman <charlied@erols.com>.

-- 
Cheers,
 /X\  David A. Bagley
(( X  bagleyd@bigfoot.com  http://www.tux.org/~bagleyd/
 \X/  xlockmore and more   ftp://ftp.tux.org/pub/tux/bagleyd

From cube-lovers-errors@mc.lcs.mit.edu  Thu May 28 13:26:48 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Thu May 28 07:41:10 1998
Message-Id: <01BD8A0B.2B00FBC0@jburkhardt.ne.mediaone.net>
From: John Burkhardt <jburkhardt@mediaone.net>
To: Cube Mailing List <Cube-lovers@ai.mit.edu>
Subject: RE: Magic Jack
Date: Thu, 28 May 1998 07:35:19 -0400

I bought one of these from Games People Play and it is unsolvable on one
of the colors.  At least, getting the piping to wander all the way
around on the red colors was not possible.  It looks like one of the
stickers is oriented incorrectly.  I told the folks at FunTech and they
told me to send it back to them and they would look.  Also, when I
solved the message version I couldn't get the rest of it to line up.
Some of the pipes on the other edges didn't line up and it wasn't
possible to make the rest silver.  This was kind of disappointing.


From cube-lovers-errors@mc.lcs.mit.edu  Fri Jun  5 19:41:11 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Mon Jun  1 19:17:58 1998
Message-Id: <357329D5.FD5A7E89@t-online.de>
Date: Tue, 02 Jun 1998 00:23:17 +0200
Organization: BERA Softwaretechnik GmbH
To: Cube Mailing List <Cube-Lovers@ai.mit.edu>
Subject: New member
From: Rainer.adS.BERA_GmbH@t-online.de (Rainer aus dem Spring)

Dear cube lovers,

a few days ago I have subscribed to the list. I have downloaded all
the archives and also found some software (great stuff).

Some of you may know my name from Mr. Bandelow's book - I'm the guy
who won a prize with the shortest maneuver fro the super-flip-twist.

I always thought the cube is dead. Now I am really surprised to see
this completely wrong :) The level of discussion is really amazing.

I have are some questions concerning terminology:

What does M-conjugacy mean (this doesn't seem to be a standard term
from group theory).
[ Moderator's note: See "Symmetry and Local Maxima", 14 December 1980.
  This also introduces the group M and some of its subgroups, which
  are helpful in a lot of the cube-lovers discussion.  Jerry Bryan
  also tried an explanation of M-conjugacy on 3 October 1996.  ]

Some mails in the archives mention numbers like p102 for patterns ?!?!?
[ M: I believe Mark Longwood uses numbers of that form to catalogue
  patterns. ]

Does anybody know the current upper bound for God's algorithm (in q/f
metric).
[ M: 29 face turns, 42 quarter turns.  The best known lower bounds are
  20 face turns, or 24 quarter turns (both achieved by superflip).
  This was true on 13 February 1996, and I don't think there has been
  an advance since then. ]

Is there any serious research on the 4*4*4 or 5*5*5 cube. Computer
search is probably beyond available/affordable hardware :(

Are there any maneuver search programs that can handle slice metric ?
I think that slice metric makes sense since q and f metric have no
natural extension for "higher" cubes.
[ M: See 1 June 1983 for "Eccentric Slabism", a genereralization of
  the q metric that could be adapted to a f metric. ]

Does anybody have some nice patterns on the 4^3 or 5^3 cube ?
[ M: I reported some 4^3 patterns on 15 June 1982.  Have there been
  others?  Any 5^3 patterns? ]

I think I have some in my old (and thick) cube folder (paper,  not on
my PC :))
[ M: Could you type some in? ]

Some of these questions probably have been discussed already. Sorry, I
haven't read ALL old mails.
[ M: Note that I left quite a few of these questions unanswered--other
  replies are welcome, either pointers to archive messages I forgot or
  new answers.  But this highlights a major failing of the archives:
  We don't have a FAQ, or even an index to the major articles.  Is
  anyone interested in working on something like this?  I have very
  little time for it just now. ]

adS

-- 
--------------------------------------------------------------------------
Rainer aus dem Spring       email  Rainer.adS.BERA_GmbH@t-online.de
(home)
Schimmelbuschstr. 10        email  TEEADS@TEE.toshiba.de (business only)
40699 Erkrath               tel. +49 (0)02104 35157  (private)
Germany                     tel. +49 (0)02104 936150 (business)
---------------------------------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Fri Jun  5 21:42:40 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Fri Jun  5 20:53:28 1998
Date: Fri, 5 Jun 1998 20:36:06 -0400
Message-Id: <0020A7D5.001706@scudder.com>
From: jdavenport@scudder.com (Jacob Davenport)
Subject: Re: New member
To: Cube Mailing List <Cube-Lovers@ai.mit.edu>

>Does anybody have some nice patterns on the 4^3 or 5^3 cube ?

I made a chess board out of four 5^3 cubes, which you can check out along
with our other cube sculptures on www.wunderland.com/WTS/Jake/CubeArt.  If
anyone has any good pattern ideas for four 5^3 cubes, I'd like to hear
them, particularly before I peel off the stickers on one of them and give
it a color pattern similar to Colorspace created by Andy Plotkin (which you
can see at www.wunderland.com/WTS/Kristin/CustomCubes.html).

-Jacob

From cube-lovers-errors@mc.lcs.mit.edu  Fri Jun  5 22:31:09 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Fri Jun  5 21:44:38 1998
Message-Id: <v03007800b19e40ee2e37@[10.10.10.2]>
In-Reply-To: <357329D5.FD5A7E89@t-online.de>
Date: Fri, 5 Jun 1998 21:41:23 -0400
To: Rainer.adS.BERA_GmbH@t-online.de (Rainer aus dem Spring),
        Cube Mailing List <Cube-Lovers@ai.mit.edu>
From: Kristin <kristin@wunderland.com>
Subject: Re: New member

Rainer aus dem Spring wrote:

>Some of you may know my name from Mr. Bandelow's book - I'm the guy
>who won a prize with the shortest maneuver fro the super-flip-twist.
>
>I always thought the cube is dead. Now I am really surprised to see
>this completely wrong :) The level of discussion is really amazing.

<smile>Your words perfectly describe a day a couple of years ago
when I first found this list.</smile>


Peace -

-Kristin
kristin@wunderland.com
wunderland.com/home/rubik.html

From cube-lovers-errors@mc.lcs.mit.edu  Mon Jun  8 15:52:11 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sat Jun  6 09:39:26 1998
Message-Id: <3.0.5.16.19980606155910.0f372bfe@ryle.get2net.dk>
Date: Sat, 06 Jun 1998 15:59:10
To: cube-lovers@ai.mit.edu
From: Philip Knudsen <skouknudsen@get2net.dk>
Subject: 4*4*4 patterns [Re: New member]

> Does anybody have some nice patterns on the 4^3 or 5^3 cube ?
> [ M: I reported some 4^3 patterns on 15 June 1982.  Have there been
> others?  Any 5^3 patterns? ]

Maybe you find the following pattern for the 4*4*4 interesting (I'm not
sure I am using the proper notation, but by Capital letters I mean side
moves, small letters are slice moves):

R r U2 u2 R2 r2 U3 u3
F3 L D2 L3 D3 F2 U2 F2 D L D2 L3 F
L3 F U2 F3 U3 L2 D2 L2 U F U2 F3 L
D F2 B2 D2 F2 B2 D

The last three lines alone make a similar pattern on a 3*3*3 cube. I have a
shorter sequense for it somewhere that I can't remember by head. One can
also make a similar pattern on the 5*5*5, I'll try and dig it out...

Philip K

From cube-lovers-errors@mc.lcs.mit.edu  Fri Jun 19 11:15:07 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Fri Jun 19 03:22:16 1998
Message-Id: <3.0.5.16.19980619094150.0c8f65de@ryle.get2net.dk>
Date: Fri, 19 Jun 1998 09:41:50
To: cube-lovers@ai.mit.edu
From: Philip Knudsen <skouknudsen@get2net.dk>
Subject: New Puzzle: "Dogic"

I just received a new Puzzle called "Dogic - Test Your Logic".
It's in the shape of an Icosahedron, and moves in the following manner: 5
triangles can rotate around their common vertex somewhat like the Impossiball.
Each triangle is again subdivided into 4 smaller triangles which move
separately, i.e. one can also rotate 5 smaller triangles around the same
vertex. Thus there are 60 "vertex" triangles and 20 "middle" triangles, the
latter are in fact equivalent to the Impossiball. The "vertex" triangles
are unicolored, the "middle" triangles have three colours. The whole Puzzle
has twelve colours, one for each vertex.

  I count the number of distinguishable positions:

  20! 3^19 60!
  ------------  =  2,199110779324 x 10^82
   2 5!^12 60

I'm not sure these calculations are correct, but if they are, this Puzzle
is at the very top of Mark Longridge's "Great Cosmic Ranking List", even
above the good old 5x5x5 Cube!
The Puzzle is very well "Made In Hungary". A true must for anyone who likes
cube-type Puzzles. Available from Spielkiste/Germany, check out:
www.twfg.de/puzzle/default.htm


Philip Knudsen
Recording and Performing Artist
Vendersgade 15, 3th
DK - 1363 Copenhagen K
Denmark
Phone:  +45 33932787
Mobile: +45 21706731
E-mail: skouknudsen@get2net.dk
E-mail: philipknudsen@hotmail.com

From cube-lovers-errors@mc.lcs.mit.edu  Tue Jun 23 10:18:07 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Mon Jun 22 09:08:14 1998
Message-Id: <199806221304.AA08381@world.std.com>
To: "Cube-Lovers@ai.mit.edu" <Cube-Lovers@ai.mit.edu>
Subject: Puzzles newly available in US
Date: Mon, 22 Jun 98 09:02:41 -0500
From: "Michael C. Masonjones" <mcmj@blazetech.com>

I apologize that I don't have more information, since I am away from the
stuff that I bought and I can't look at the packaging.

I was in Toy Works in W. Springfield, MA, and was very surprised to find a
series of puzzles I believe are newly available in the US after a long
hiatus.  The puzzles are Pyraminx, Skewb, and Meffert's Ball (with the four
colored rings arranged on a spherical Skewb device.  I think only the Skewb
was called by its real name.  All were in the same basic packaging, looked
pretty authentic, and I think they all had Meffert patent/copyright info on
them (at least the ball did).  And they were all marked down from $10 to $6.
Not bad at all.

There must be more out there, but my nearest KayBee didn't have anything.
I'm not sure if Toy Works is a big chain or not.  It seems to be from the
stuff they carry and the size of the store.

Happy hunting.

Mike Masonjones

From cube-lovers-errors@mc.lcs.mit.edu  Fri Jul 10 12:57:34 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Fri Jul 10 08:48:35 1998
Date: Fri, 10 Jul 1998 08:48:05 -0400 (Eastern Daylight Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Ten Face Moves from Start
To: cube-lovers@ai.mit.edu
Message-Id: <Pine.WNT.3.96.980710083950.-3929923B-100000@GN209A.PSTCC.CC.TN.US>

Face Moves          Patterns       Positions   Branching   Positions/
   from                                           Factor     Patterns
   Start

      0                     1               1                 1.0
      1                     2              18     18          9.0
      2                     9             243     13.5       27.0
      3                    75            3240     13.333     43.2
      4                   934           43239     13.345     46.294
      5                 12077          574908     13.296     47.604
      6                159131         7618438     13.252     47.875
      7               2101575       100803036     13.231     47.965
      8              27762103      1332343288     13.217     47.991
      9             366611212     17596479795     13.207     47.998
     10            4838564147    232248063316     13.199     47.999


This run took about three weeks on a Pentium 300.  The next level from
Start is going to be difficult.  With the current algorithm and
hardware, it would take about thirty to forty weeks.  In addition, the
memory requirements will go up considerably.  Currently, I store only
the positions up to five moves from Start in memory.  To calculate the
next level, I will have to store the positions up to six moves from
Start.

I still suggest (see "How Big is Big?" in the archives) that the problem
can be calculated all the way to the bitter end, eventually.  The Cube
problem simply is not as big as, for example, Chess or Go.

As a possible strategy, if we could add one level per decade, we could
probably calculate the problem all the way to the end within about 100
years. Moore's Law (the power of computers doubles about every eighteen
months) suggests that such a schedule might be possible.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990

From cube-lovers-errors@mc.lcs.mit.edu  Fri Jul 10 16:23:48 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Fri Jul 10 13:31:51 1998
Date: Fri, 10 Jul 1998 13:31:22 -0400
Message-Id: <199807101731.NAA02872@corwin.ece.cmu.edu>
From: "Jonathan R. Ferro" <jferro@corwin.ece.cmu.edu>
Organization: Electrical and Computer Engineering, CMU
To: cube-lovers@ai.mit.edu
In-Reply-To: 
	<Pine.WNT.3.96.980710083950.-3929923B-100000@GN209A.PSTCC.CC.TN.US>
	(message from Jerry Bryan on Fri, 10 Jul 1998 08:48:05 -0400 (Eastern
	Daylight Time))
Subject: Re: Ten Face Moves from Start

"Jerry" == Jerry Bryan <jbryan@pstcc.cc.tn.us> writes:
Jerry> As a possible strategy, if we could add one level per decade, we
Jerry> could probably calculate the problem all the way to the end
Jerry> within about 100 years. Moore's Law (the power of computers
Jerry> doubles about every eighteen months) suggests that such a
Jerry> schedule might be possible.

This method is called Zarf's Linearization: For any exponential-time
problem, just wait the linear amount of time for the current generation
of computation to make it possible to solve your instance in one hour,
then solve your instance in one hour.

From cube-lovers-errors@mc.lcs.mit.edu  Tue Jul 14 17:15:12 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Tue Jul 14 13:46:15 1998
Message-Id: <19980714163444.6844.rocketmail@send1a.yahoomail.com>
Date: Tue, 14 Jul 1998 09:34:44 -0700 (PDT)
From: Eddy Liao <liao_1@yahoo.com>
Subject: Cubes for sale
To: Cube-Lovers@ai.mit.edu

Dear Madam/Sir,

I have the following items for sale:

Rubic's cube(6-color) - $5.50
Rubic's cube(poker) - $5.50
Magic snake - $5.50
Rubic's cube keychain (1.5 inch) - $4.50
Rubic's cube keychain (3/4 inch) - $3.50
Pyramid key chain (1.5 inch) - $4.50
Magic snake keychain - $4.50

List your orders plus $1 shipping charge of entire order
(plus $5 if you prefer COD(Cash on Delivery))


Please send check or money order to:

Eddy Liao
694 Yorkhaven Rd.
Cincinnati, OH 45246

If you have any questions, please Email me at: liao_1@yahoo.com 

From cube-lovers-errors@mc.lcs.mit.edu  Thu Jul 16 12:05:34 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Tue Jul 14 13:46:15 1998
Date: Tue, 14 Jul 1998 09:34:44 -0700 (PDT)
From: Dan Hoey <Cube-Lovers-Requst@AI.MIT.Edu>
Subject: Spammer says: Cubes for sale
To: Cube-Lovers@AI.MIT.Edu

I greatly regret allowing the advertisement from Eddy Liao entitled
"cubes for sale".  Having just received his ad spammed in my personal
mailbox, I must conclude he is an abuser of the network.  So if you
buy anything from him, you're supporting network abuse, and for all I
know he may steal your money as blithely as he steals the network's
resources.

Dan Hoey, Moderator
Cube-Lovers-Request@AI.MIT.Edu

From cube-lovers-errors@mc.lcs.mit.edu  Fri Jul 24 13:30:22 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Thu Jul 23 18:21:31 1998
Date: Thu, 23 Jul 1998 18:21:02 -0400
From: michael reid <reid@math.brown.edu>
Message-Id: <199807232221.SAA07643@hilbert.math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: patterns on 5x5x5 cube

a while ago, rainer asked for patterns on the 5x5x5 cube.  here
are some i know (the hardest part seems to be finding the scraps
of paper on which the maneuvers are written).

standard notation uses  R  and  r  for 90 degree clockwise
twists of the outer layer and second layer, respectively.  i've
found it convenient to have notation for  Rr , so i use  _R_
(that is, capital R underlined).  i guess this notation is more
convenient for handwritten maneuvers, but not so convenient for
e-mail.  i'll use  _R_  to denote  R  underlined and

        _( F L U B ... )_

to mean that the whole thing inside the parentheses is underlined.

the first pattern is a "double" snake; it meanders onto each face twice.

_R'_  b'  _U_  F2  _U'_  b  _U_  F2  _(U' R2 F')_  u2  _(F U L')_
u2  _(L U' R' L)_  f  _D'_  B2  _D_  f'  _D'_  B2  _(D L2 B)_  d2
_(B' D' R)_  d2  _(R' D L)_

if i understand the terminology correctly, this is a continuous
non-chiral isoglyph with the pattern

     ...*.
     **.**
     .*...
     .*...
     .*...

i still remember that when i found this pattern some 10-12 years
ago, i saw the URF faces together.  then i turned the cube around,
and was surprised by how it continued on the other three faces.
(i shouldn't have been surprised, but you know how that goes ... )

i came across this pattern accidently.  then i went snake hunting,
and found several others:

snake 2
_(R F2 R2 U2)_  r2  _(U2 R2 F2 R' D' F2 B2 D R F2 R2 U2)_  r2
_(U2 R2 F2 R' D')_  r2  _(F2 B2)_  L2  _(R2 U' D F2 B2 U)_

those two have the property that the two snake segments on each face
have the same color.  if this condition is relaxed, we also have

snake 3
_(R L' F U2 R F2 R2 U2)_  r2
_(D2 L2 F B' D' F B' U' D F R L D' B2 L' F B' D')_
f2  _(U2 D2)_  f2  _(U' D2)_

this one can be modified slightly; change the U and D faces

        .*.*.        .*.*.
        .*.*.        .***.
        .*.*.        .....
from    .*.*.   to   .***.
        .*.*.        .*.*.

if only one is changed, then we get two separate snakes.

there's also

snake 4
_(D F2 B2)_  l2  _(F2 B2 R')_  R2  _(F2 R2 U2)_  r2
_(U2 R2 F2 R' D' F2 B2 D R F2 R2 U2)_  r2
_(U' D' F' U2 D2 B U' D L2 B2 L' U2 D F2 B2)_


another interesting pattern is

U R' U' F'  _U'_  R'  _U_  f  _U'_  R  _(U F')_  F2 U R U'
_B_  l'  _D2_  l  _D_  f'  _D2_  f  _(D' B')_
D' L D B  _D_  L  _D'_  b'  _D_  L'  _(D' B)_  B2 D' L' D
_F'_  r  _U2_  r'  _U'_  b  _U2_  b'  _(U F)_

which gives a continuous non-chiral isoglyph with the pattern

     .*...
     .*...
     .*...
     *****
     ...*.

the same maneuver produces an analogous pattern on the 4x4x4 cube,
but there's probably an easier maneuver.  another isoglyph (also
continuous and non-chiral) with the same pattern is

R  f'  U2  f  U  l'  U2  l  U' R'  _D'_
L  b2  L'  _B'_  U  b2  U'  _(B D)_
L'  b  D2  b'  D'  r  D2  r'  D L  _U_
R'  f2  R  _F_  D'  f2  D  _(F' U')_

modifying this pattern is how i found the first double snake.

mike

From cube-lovers-errors@mc.lcs.mit.edu  Sun Jul 26 14:10:13 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sat Jul 25 17:52:37 1998
Message-Id: <35BA52EF.35D831D6@t-online.de>
Date: Sat, 25 Jul 1998 23:49:35 +0200
Reply-To: Rainer.adS.BERA_GmbH@t-online.de
Organization: BERA Softwaretechnik GmbH
To: michael reid <reid@math.brown.edu>
Cc: cube-lovers@ai.mit.edu
Subject: Re: patterns on 5x5x5 cube
References: <199807232221.SAA07643@hilbert.math.brown.edu>
From: Rainer.adS.BERA_GmbH@t-online.de (Rainer aus dem Spring)

Thanks for the patterns,

the problem is - my 5*5*5 cube is scrambled and I have to figure out
how to unscramble it. I haven't touched it for 15 years.

I also have several hundred pages of hand-written stuff.

I have several 4*4*4 patterns. I wonder if enough people on the mailing
list have a (La)TeX system so that I can post the patterns in LaTeX
format.

Concerning this I need feedback !!

By the way, my last 4*4*4 cube starts to fall apart. Does anybody know
if it is still available ?


michael reid wrote:
> 
> a while ago, rainer asked for patterns on the 5x5x5 cube.  here
> are some i know (the hardest part seems to be finding the scraps
> of paper on which the maneuvers are written).
> 
snip
....

Rainer adS 

PS

Does anybody else have patterns for the 4 or 5 cube ?

If so, pls. send them to me. I will create a document in ps and/or pdf
format about
patterns.

[ Moderator's note: My sense of this is that short notes in latex can
  be made readable enough as text that it might be workable on this
  list.  Postscript and PDF are not acceptable in this medium, though
  they can be placed in the archive at
  ftp://ftp.ai.mit.edu/pub/cube-lovers/contrib/ --Dan]

From cube-lovers-errors@mc.lcs.mit.edu  Mon Jul 27 17:33:49 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sun Jul 26 15:51:06 1998
Message-Id: <35BB827C.10B7C9A7@t-online.de>
Date: Sun, 26 Jul 1998 21:24:44 +0200
Reply-To: Rainer.adS.BERA_GmbH@t-online.de
Organization: BERA Softwaretechnik GmbH
To: Cube Mailing List <Cube-Lovers@ai.mit.edu>
Subject: 4*4*4 patterns
From: Rainer.adS.BERA_GmbH@t-online.de (Rainer aus dem Spring)

Dear cube lovers,

as promised the other day here comes my collection of 4*4*4 patterns.
My favorites are the single twisted rings. I still find it surprising
to see that there is no second ring on the "other" side.

The maneuvers use all sorts of slice moves which are probably not
accepted as moves by most cubeologists. I am too lazy to rewrite them.

Does anybody have any idea which format I should post for people
without a TeX system?
Plain ASCII is not really what one needs to display cube maneuvers.

I can offer Mathematica notebook, postscript, WINWORD (arrrrgh)and
(perhaps?) pdfd.

Does anybody know of a 4*4*4 emulator or even solver?  Anything like
a (sub)optimal solver is probably beyond the current PC powers.

Rainer

PS
I am NOT a LaTeX expert :) hints are welcome !

[Moderator's note: The notation is fairly straightforward, but may
 be new to cube-lovers.  Subscripts (encoded with underscores) show
 which 1x4x4 slabs relative to the given face are turned; omitted
 subscripts apparently mean the outer face, as "_1" would mean if it
 were used.  Exponents have their usual meanings as repetition.  The
 only advantage to running this through TeX seems to be that you get
 true superscripts and subscripts and somewhat nicer fonts.

 I've added a few commands that seem required by LaTeX.  I've also
 replaced a number of narrow spaces (coded as backslash-comma) with
 ordinary spaces, so that this is more readable as text and so that
 some of the worst long-line problems are reduced.  Perhaps some of
 these processes would be more readable if printed on multiple lines,
 or punctuated somehow.  I wonder if there could be some
 simplification with the [X,Y] = X Y X^{-1} Y^{-1} commutator notation
 or the X^Y = Y^{-1} X Y conjugate notation, or if this would make the
 processes too hard to follow. --Dan]

\documentstyle{article}
\begin{document}
\section{Patterns}
\subsection{Dot Patterns}
\subsubsection*{2 Dots (u,d)}
$(R^2_2 F_{23}^2)^2$

Since this pattern exists it is obviously possible to create all dot
patterns, i.e.\ all $6! = 720$ elements of the dot permutation group.

\subsubsection*{2 Dots (f,r)}
$D_2 R_{23}^2 D_2^{-1} L^2 B^2 U_2^{-1} R_{23}^2 U_2
 R_{23}^2 B^2 L_{123}^2$

\subsubsection*{3 Dots (f,b,r)}
$L_{123}^2 B^2 U_2^2 R_{23}^2 U_2 R_{23}^2 U_2 B^2 L^2 D_2 R_{23}^2
 D_2^{-1}$

\subsubsection*{3 Dots (f,u,r)}
$F_2^{-1} U^2 F_2 D_{23} F_2^{-1} U^2 F_2 D_{23}^{-1}
 B_2 U^2 B_2^{-1} D_{23} B_2 U^2 B_2^{-1} D_{23}^{-1}$

\subsubsection*{4 Dots (f,r)(l,b)}
$D_2 R_{23}^2 D_2^{-1} U_2^{-1} R_{23}^2 U_2$

\subsubsection*{4 Dots (f,b)(r,l)}
$R_{23}^2 D_{23} R_{23}^2 D_{23}^{-1}$

\subsubsection*{4 Dots (f,u)(r,l)}
$R_2^{-1} U^2 R_2^2 B^2 R_2 R_{23}^2 D_{23} R_{23}^2 D_{23}^{-1}
 R_2^{-1} B^2 R_2^2 U^2 R_2$

\subsubsection*{6 Dots (f,b)(r,l)(u,d)}
$D_{23} F_{23}^2 D_{23}^{-1} R_2^2 F_{23}^2 R_2^2$

\subsection{Brick Patterns}
\subsubsection*{Exchanged 1*1*1 Cubes}
$B^{-1} U^{-1} B L^2 F^{-1} D R_2^2 B_2^2 R_{12}^2 B_2^2 R^2 B_2^2
 F^{-1} D^{-1} F^2 L^2$

\subsubsection*{Exchanged 1*1*2 Bricks}
$R^2 U^2 R_{123}^{-1} D_{12}^{-1} R_{123} U^2 R_{123}^{-1} D_{12}
 R_{123} U^2 F_{12} U^2 F_{12} U^2 F_{12}^{-1} U^2 R_2^2 F_{12}^2
 R_2^2 F_{12}^2 R_{12}^2$

\subsubsection*{Exchanged 1*1*3 Bricks}
$B D^{-1} U^2 B^{-1} R^{-1} B U^2 F^{-1} L F^{-1} L^{-1} F^2 D B^{-1}$

Of course, this is a 3*3*3 maneuver.

\subsubsection*{Exchanged 1*2*2 Bricks}
$R_{12} B R_{12}^{-1} F_{12}^2 R_{12} B^{-1} R_{12} D R_{12}^2
 F_{12}^2$

\subsubsection*{Exchanged 1*2*3 Bricks}
$D^{-1} B_{12}^{-1} L^2 U^2 F_{12}^{-1} R_{123}^{-1} D_{12}^{-1}
 R_{123} U^2 R_{123}^{-1} D_{12} R_{123} U^2 F_{12} U^2 F_{12} L^2
 B_{12} D$

\subsubsection*{Exchanged 1*3*3 Bricks}
$F_2^2 R_{23}^2 F_2^2 B^{-1} U^{-1} B L^2 F^{-1} D L_2^2 B_2^2
 L_{12}^2 B_2^2 L^2 B_2^2 F^{-1} D^{-1} F^2 L_{123}^2$

\subsubsection*{Exchanged 2*2*2 Cubes}
$B_{12}^{-1} U_{12}^{-1} B_{12} L_{12}^2 F_{12}^{-1} D_{12}
 F_{12}^{-1} D_{12}^{-1} F_{12}^2 L_{12}^2$

Of course, this is a 2*2 maneuver.

\subsubsection*{Exchanged 2*2*3 Bricks}
$U_2 L_{12}^2 U_2^{-1} D_2^{-1} L_{12}^2 D_2 R_{12} B R_{12}^{-1}
 F_{12}^2 R_{12} B^{-1} R_{12} D R_{12}^2 F_{12}^2$

\subsubsection*{Exchanged 2*3*3 Bricks}
???

\subsubsection*{Exchanged 3*3*3 Cubes}
$F^2 L^2 D F^{-1} B_{12}^2 R^2 B_2^2 R_{12}^2 B_2^2 R_2^2 U^{-1} R B^2
 R^{-1} D_{23} R F_{23}^2 R^{-1} D R $

\subsubsection*{4 Chess Boards}
$U^2 D_2^2 R^2 L_2^2 F^2 B_2^2 R^2 L_2^2$

\subsubsection*{6 Bars}
$F^2 R^2 F_{23}^2 L^2 F_2^2 D_{23}^2 F_{12}^2 D_{23}^2$

\subsubsection*{2 Twisted Rings}
$L_{12}^2 F_{12}^{-1} L_{12}^{-1} F_{12} L_{12}^{-1} U_{12}
 B_{12}^{-1} U_{12} B_{12} U_{12}^2 (B^{-1} U^2 B R^{-1} U^2 R)^2$

Certainly not the shortest maneuver.

\subsubsection*{1 Small Twisted Ring}
$F^{-1} L_2^2 F R^2 B_2 U^{-1} B_2^{-1} D_{12}^{-1} B_2 U B_2^{-1}
 D_{12} R^2 F^{-1} L_2^2 F$

\subsubsection*{1 Large Twisted Ring}
$F_{12}^{-1} R_{12} D_2^2 R_{12}^{-1} U_{12}^{-1} R_{12} D_2^2 U L_2
 D_2^{-1} L_2^{-1} U^{-1} L_2 D_2 L_2^{-1} R_{12}^{-1} U_{12} F_{12}$

\subsubsection*{4 Diagonals}
$U(R^2 F R^2 D_{23}^2)^2 U_{12}^{-1} F^2 R_{12}^2 D_{23} R_2^2
 D_{23}^{-1} R^2 F^2 U_2$

\subsubsection*{2 Small Twisted Peaks}
$B_2^2 D_2^2 L_{12}^2 U F^2 L^2 D^{-1} L^{-1} D L^{-1} F U^{-1} F
 L_{12}^2 D_2^2 B_2^2$

\subsubsection*{2 Large Twisted Peaks}
$D_{12}^2 R_{12}^{-1} D_{12}^{-1} R_{12} D_{12}^{-1} F_{12}
 L_{12}^{-1} F_{12} L_{12} F^2 U_2^2 R_2^2 F^{-1} R D_{23}^2 R^{-1}
 U^{-1} R D_{23}^2 R^{-1} U F R_2^2 U_2^2 F_2^2$

\subsection{Snakes}
\subsubsection*{Snake 1}
$L_{12}^{-1} U_2^2 L_{12} F_{12}^2 L_2 F_{12}^2 R_2^{-1} D_{12}^2 R_2
 U_{12}^2 R_2^{-1} D_{12}^2 R_2 U_{12}^2$

\subsubsection*{Snake 2}
$F^2 B^2 D^2 L_2 D^2 L_2^{-1} D^2 R_2 D^2 R_2^{-1} B^2 L_2 F^2 R_{12}
 U_2^2 R_{12}^{-1}$

\subsubsection*{Snake 3}
$D_{12} R_{12}^2 F_{12} R_{12}^{-1} B_2 R_{12} F_{12}^{-1} R_{12}^2
 D_{12}^{-1} R_{12}^{-1} B_{12}^2 D_{12}^{-1} B_{12} L_2^{-1}
 B_{12}^{-1} D_{12} B_{12}^2 R_{12}$

\end{document}

From cube-lovers-errors@mc.lcs.mit.edu  Mon Jul 27 19:57:45 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Mon Jul 27 09:23:29 1998
Date: Mon, 27 Jul 1998 09:23:10 -0400 (EDT)
From: Nicholas Bodley <nbodley@tiac.net>
To: Rainer aus dem Spring <Rainer.adS.BERA_GmbH@t-online.de>
Cc: michael reid <reid@math.brown.edu>, cube-lovers@ai.mit.edu
Subject: Restoring a 5^3 to solved state (Was: Re: patterns on 5x5x5 cube)
In-Reply-To: <35BA52EF.35D831D6@t-online.de>
Message-Id: <Pine.SUN.3.95.980727084518.1715H-100000@sunspot.tiac.net>

[Also an endorsement for Acrobat (*.PDF) at the end of this msg.]

On Sat, 25 Jul 1998, Rainer aus dem Spring wrote:

{Snips}

}Thanks for the patterns,
}
}the problem is - my 5*5*5 cube is scrambled and I have to figure out
}how to unscramble it. I haven't touched it for 15 years.

Although it's using your mind in a different fashion, you could
disassemble it, sort the pieces (takes a while!) and reassemble it in
the solved state. With any reasonable degree of care, you won't harm a
5^3, I'm just about sure.* Have a clean work surface. I have done it
maybe 5 or 6 times on mine (from Meffert, ca. 1987). If you have a cat,
don't even think of letting it in the same apartment or house while it's
apart! :) The insides are really quite amazing to see. The internal
"foot" that retains a corner cubie is an amazing shape.

 *A 4^3 requires much more care. The center cubies are fragile! There
was a message a while back from someone who's selling parts for 4^3s.

 To start the disassembly, align all layers (obviously), so it's a cube.
Then rotate one face, leaving the other four layers aligned. Rotate it
either less or more than 45 degrees, so that a left or right edge cubie
of the rotated face is aligned with the edge of the other four layers.
Plainly, it doesn't matter which you choose, because of physical
symmetry internally.

With the rotated face on top, pry up the left (or right) edge cubie,
away from the edge you aligned it with. Use your thumb, thumbnail facing
down. Once it disengages, the rest won't fall apart uncontrollably; a
few pieces will fall out, but most will need to be actually removed, one
by one. Study the structure as you pull it apart. Amazement is one
reason, and the other is to get a better idea of how to reassemble it.

Sort the pieces (it might take longer than you think!).

Your color references for rebuilding will obviously be the center and
corner cubies. Build one face completely, place that face down onto your
work surface, and build progressively up from there. The last cubie will
be in the same position you removed to start.

My 3-D sense happens to be extremely good (apparently hereditary), so I
had very little trouble figuring out what goes where and how. It might
be harder for some others.

=====
[ Moderator's note: Nicholas Bodley's and Rainer aus dem Spring's
  discussions of the merits of Acrobat and other graphics languages
  are not on topic for Cube-Lovers.  Send them e-mail for the
  discussion--the addresses are in the headers of this message.  We
  may eventually get some in the archives, which will be announced. ]

My regards to all,

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*  The personal computer industry will have become
|*  Amateur musician  *|*  mature when crashes become unacceptable.
--------------------------------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Mon Jul 27 21:55:55 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Mon Jul 27 14:45:54 1998
Date: Mon, 27 Jul 1998 14:45:33 -0400 (Eastern Daylight Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: Ten Face Moves from Start
In-Reply-To: <Pine.WNT.3.96.980710083950.-3929923B-100000@GN209A.PSTCC.CC.TN.US>
To: cube-lovers@ai.mit.edu
Message-Id: <SIMEON.9807271433.J@GN209A.PSTCC.CC.TN.US>

On Fri, 10 Jul 1998 08:48:05 -0400 (Eastern Daylight Time)
Jerry Bryan <jbryan@pstcc6.pstcc.cc.tn.us> wrote:

> This run took about three weeks on a Pentium 300.

Here is a bit of a follow-up.  I didn't say so explicitly,
but only the results 10f from Start were new.  The search
had been calculated through 9f from Start previously.

In some ways, there was nothing new in the program to
calculate 10f from Start vs. 9f from Start, because the
memory requirements are the same either way (all the
positions up through 5f from Start have to be stored either
way). Basically, the only difference was to let the program
run longer. A faster machine helped a great deal. Also, I
did add a simple checkpointing capability which helped a
great deal.

I received some private E-mails suggesting using the net as
a massively parallel computer to calculate the problem
further from Start, similar to what has already been done
on the net to break certain ciphers.  The checkpointing I
added to the program is a step in the direction of parallel
processing.

As has been described in the Cube-Lovers archives, the
program uses an algorithm that produces permutations in
lexicographic order.  Such an algorithm inherently
decomposes easily into parallel processing.  So by analogy
to processing a phone book or a dictionary, it should be
possible for one machine to process the A's, another
machine to process the B's, a third machine to process the
C's, etc., and then to add the results together.

(Actually, you would use finer decomposition than that.
One machine would process the AB's, another would process
the AC's, etc., or perhaps you would use even a finer
decomposition.  Note that there are no AA's because these
are permutations we are talking about -- no letter repeats
within a word.)

What is really needed are some scripts to drive and control
such a process.  I really don't have time right now  --
maybe in the future.  Also, to go past 10f from Start, the
machines working on the project would have to have a good
bit of memory, maybe something in the 100MB range would
have to be dedicated to the program to calculate 11f or 12f
from Start.

The existing program is in C.  It was suggested in a
private E-mail that writing the program in Java might make
it easier to run "on the net".  Maybe, but I am dubious at
this point if Java is ready to handle the size of problem
we are talking about.

> As a possible strategy, if we could add one level per decade, we could
> probably calculate the problem all the way to the end within about 100
> years. Moore's Law (the power of computers doubles about every eighteen
> months) suggests that such a schedule might be possible.

With respect to the E-mail about waiting for faster
machines to deal with exponential problems, my real point
was not that waiting on technology is a wonderful way to
attack the Cube problem. Rather, I was suggesting that the
Cube problem is small enough, even at about 10^19, that it
can ultimately be defeated by technology (i.e., by Moore's
law). Chess is about 10^75 and Go is about 10^120.  Moore's
law is therefore pretty well bound to fail before Chess
or Go can be solved.  (Deep Blue played very good chess
against Kasparov, but not perfect chess.)

There are two strong local maxima 9f from Start, and
they have already been posted to Cube-Lovers.  Six more
strong local maxima showed up at 10f from Start.
Regrettably, my "simple checkpointing" did not include
printing out the permutations for the strong local maxima
-- I just counted them.  I have improved the checkpointing,
and am rerunning a part of the program to print out the six
strong local maxima.  So far, the only one of the six which
has been printed turns out to be a 4-H pattern.

D  B2 L2 B2 D  U' R2 F2 R2 U'
F2 R2 F2 D' U  R2 F2 R2 U  D'
L' R' D' U' B2 F2 D' U' L' R'
L' R' D' U' B2 F2 D' U' R' L'
B' F' D' U' L2 R2 D' U' B' F'
B' F' D' U' L2 R2 D' U' F' B'
D' F2 R2 F2 D' U  R2 F2 R2 U
B2 L2 B2 D  U' R2 F2 R2 U' D
L  R  D' U' B2 F2 D' U' L  R
L  R  D' U' B2 F2 D' U' R  L
B  F  D' U' L2 R2 D' U' B  F
B  F  D' U' L2 R2 D' U' F  B
B2 F2 L2 R2 U2 B2 F2 L2 R2 U2
B2 F2 L2 R2 D2 B2 F2 L2 R2 D2
L2 D2 B2 F2 R2 B2 F2 R2 U2 R2
R2 D2 B2 F2 R2 B2 F2 R2 U2 L2
B2 D2 F2 L2 R2 F2 L2 R2 U2 F2
F2 D2 F2 L2 R2 F2 L2 R2 U2 B2

----------------------
Jerry Bryan
jbryan@pstcc.cc.tn.us

From cube-lovers-errors@mc.lcs.mit.edu  Tue Jul 28 10:38:07 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Mon Jul 27 19:41:11 1998
Message-Id: <35BD0149.9F7384B5@t-online.de>
Date: Tue, 28 Jul 1998 00:38:01 +0200
Reply-To: Rainer.adS.BERA_GmbH@t-online.de
Organization: BERA Softwaretechnik GmbH
To: cube-lovers@ai.mit.edu
Subject: Re: Restoring a 5^3 to solved state (Was: Re: patterns on 5x5x5 cube)
References: <Pine.SUN.3.95.980727084518.1715H-100000@sunspot.tiac.net>
From: Rainer.adS.BERA_GmbH@t-online.de (Rainer aus dem Spring)

Nicholas Bodley wrote:
> Although it's using your mind in a different fashion, you could
> disassemble it, sort the pieces (takes a while!) and reassemble it in
> the solved state. With any reasonable degree of care, you won't harm a
> 5^3, I'm just about sure.* Have a clean work surface. I have done it
> maybe 5 or 6 times on mine (from Meffert, ca. 1987). If you have a cat,
> don't even think of letting it in the same apartment or house while it's
> apart! :) The insides are really quite amazing to see. The internal
> "foot" that retains a corner cubie is an amazing shape.

I have found an old booklet by Endl (terrible) that contains a Mickey
Mouse solution for the 5x5x5 cube.

Thank God - I have TWO cats :)


>  *A 4^3 requires much more care. The center cubies are fragile! There
> was a message a while back from someone who's selling parts for 4^3s.

Yeah, mine is very flabby.

A real cube meister will never disassemble his cube :)

Rainer adS

From cube-lovers-errors@mc.lcs.mit.edu  Wed Jul 29 11:39:38 1998
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Message-Id: <35BDCC35.97F9BD3@nadn.navy.mil>
Date: Tue, 28 Jul 1998 09:03:49 -0400
From: David Joyner <wdj@nadn.navy.mil>
Reply-To: wdj@nadn.navy.mil
Organization: Math Dept, USNA
To: Cube Mailing List <Cube-Lovers@ai.mit.edu>
Cc: Rainer.adS.BERA_GmbH@t-online.de
Subject: Re: 4*4*4 patterns
References: <35BB827C.10B7C9A7@t-online.de>

Rainer aus dem Spring wrote:

> Dear cube lovers,
>
> as promised the other day here comes my collection of 4*4*4 patterns.
> My favorites are the single twisted rings. I still find it surprising
> to see that there is no second ring on the "other" side.
>
> The maneuvers use all sorts of slice moves which are probably not
> accepted as moves by most cubeologists. I am too lazy to rewrite them.
>
> Does anybody have any idea which format I should post for people
> without a TeX system?

I have sent Rainier an html conversion of his file. With his permission
and approval I'll post on my web page
http://www.nadn.navy.mil/MathDept/wdj/rubik.html

> ... Does anybody know of a 4*4*4 emulator or even solver?  Anything
> like a (sub)optimal solver is probably beyond the current PC powers.

Yes. MAPLEV5 (Mathematica's main competitor) released a 4x4 Rubik's
cube emulator (as well as a masterball emulator and a 3x3 Rubik's cube
emulator) in their "share package" included with the software. The
share package is actually free but MAPLEV5 is not! Incidently, the
emulators work on some older versions of MAPLE as well. The pictured
linked to on the bottom of the above-mentioned web page were obtained
using this emulator. - David Joyner

> Rainer
>
> PS
> I am NOT a LaTeX expert :) hints are welcome !
>
> [Moderator's note: ... I wonder if there could be some
>  simplification with the [X,Y] = X Y X^{-1} Y^{-1} commutator notation
>  or the X^Y = Y^{-1} X Y conjugate notation, or if this would make the
>  processes too hard to follow. --Dan]

It would be theoretically interesting, IMHO, to have the expressions
rewritten using commutators but more confusing in practice to follow.

>

(Latex file deleted)

--
David Joyner, Assoc Prof of Math
US Naval Academy, Annapolis, MD 21402
(410)293-6738
wdj@nadn.navy.mil
http://web.usna.navy.mil/~wdj/homepage.html
++++++++++++++++++++++++++++++++++++++++++++
"A Mathematician is a machine for turning
coffee into theorems." Alfred Renyi

From cube-lovers-errors@mc.lcs.mit.edu  Wed Jul 29 13:55:23 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Tue Jul 28 12:10:01 1998
Message-Id: <002001bdba41$af584480$99118bc0@tellus.switchview.com>
From: "Michael Swart" <Michael.Swart@switchview.com>
To: <cube-lovers@ai.mit.edu>
Subject: Re: Restoring a 5^3 to solved state
Date: Tue, 28 Jul 1998 12:06:31 -0400

>A real cube meister will never disassemble his cube :)


That's true but it is better than the alternative:  the dreaded _sticker
peeling_! I took a course in university called "Intro to Public Speaking".
In it we had to give a persuasive speech and mine was Called "Why you
shouldn't peel stickers of a Cube". Here are some of the reasons I gave.

1. The stickers weren't designed to be peeled. So until the people at 3M
come out with a post-it note version of the cube, then peeling will only
wear the cube out faster

2. Cheating defeats the purpose of the puzzle. It reduces it to a simple
jigsaw puzzle. But if you find this simpler puzzle challenging - an unlikely
scenario for cube-lovers - then by all means peel away.

3. Douglas R. Hofstadter once noted that there were two mysteries to the
cube: 1. How does one solve the cube and 2. How does the cube stay together.
If frustration gets the better of you and you must cheat then disassembling
the cube is the preferred way because even though it does nothing to shed
light on the first mystery it does give insight to the second mystery.
Besides disassembling the cube and reassembling it don't cause (much) damage
if you're careful.

4. Chances are greater that you'll leave the cube unsolvable. Kids in my
grade school in the 80's used to peel stickers because they were so close to
completing two sides that they resorted to peeling one or two stickers to
get the job done. This behaviour inevitably left the cube unsolvable. If you
disassemble a cube and then assemble it randomly, there is a 1 in 12 chance
that you'll be able to solve the cube. But if you take all the stickers off
a cube and restick them randomly, then you have a better chance at winning
the lotto 6/49 twice in a row on your next two tickets, than your chances
and getting a solvable cube. (I'll post my math if anyone asks)

Michael Swart
Switchview Inc.
Michael.Swart@switchview.com

From cube-lovers-errors@mc.lcs.mit.edu  Thu Jul 30 11:17:20 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Tue Jul 28 14:39:58 1998
Message-Id: <35BE13D0.874D8E6D@t-online.de>
Date: Tue, 28 Jul 1998 20:09:20 +0200
Reply-To: Rainer.adS.BERA_GmbH@t-online.de
Organization: BERA Softwaretechnik GmbH
To: Cube Mailing List <Cube-Lovers@ai.mit.edu>
Subject: 4x4x4 patterns
From: Rainer.adS.BERA_GmbH@t-online.de (Rainer aus dem Spring)

Dear cube-lovers,

Of course, HTML is the format I should have used. I'll try to get a
latex->html converter for NT.

Anybody my post the patterns in any format.

The R_123 means turn 3 slices - this is the same as turning L and then
turn the whole cube R.

This notation was inspired by Bandelow's usage of slice moves.  The
advantage is, it makes sense for any cube. The disadvantage is, most
cubologists don't accept these "moves".

I have checked all patterns on my (physical) 4x4x4 cube.

I don't think using conjugation and commutatotrs is very user-friendly.
The maneuvers are not optimized and anybody will be able to figure out
how they were constructed using conjugation and commutators, though.

As far as I remember the patterns were the last thing I did in those
golden cube days :)

Does anybody know of a cube simulator for Mathematica ?

Rainer adS
-- 
--------------------------------------------------------------------------
Rainer aus dem Spring       email  Rainer.adS.BERA_GmbH@t-online.de (home)
Schimmelbuschstr. 10        email  TEEADS@TEE.toshiba.de (business only)
40699 Erkrath               tel. +49 (0)02104 35157  (private)
Germany                     tel. +49 (0)02104 936150 (business)
---------------------------------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Thu Jul 30 12:19:17 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Tue Jul 28 21:13:28 1998
Date: Tue, 28 Jul 1998 11:32:48 -0400
Message-Id: <00269731.001706@scudder.com>
From: Jacob_Davenport@scudder.com (Jacob Davenport)
Subject: Solving the 5^3
To: cube-lovers@ai.mit.edu,
        Rainer.adS.BERA_GmbH@t-online.de (Rainer aus dem Spring)

I have a fairly straight forward solution to the 5x5x5, and if there is 
some interest in having me post it, I will.

-Jacob

From cube-lovers-errors@mc.lcs.mit.edu  Thu Jul 30 15:25:22 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Jul 29 15:20:36 1998
Date: Wed, 29 Jul 1998 15:17:00 -0400 (EDT)
From: David Barr <Davebarr@concentric.net>
X-Sender: Davebarr@viking.cris.com
Reply-To: davidbarr@iname.com
To: cube-lovers@ai.mit.edu
Subject: Meffert's Challenge
Message-Id: <Pine.GSU.4.02.9807291459520.29412-100000@viking.cris.com>

I recently bought a "Meffert's Challenge" puzzle, and I see there hasn't
been discussion of this puzzle on this list.  Maybe it is new.  It is a
round Skewb, like Mickey's Challenge, but with different markings.  When
solved, it has four colored rings on it.  The triangular pieces each have
about a quarter ring on them.  Four of the square pieces have two separate
quarter ring markings, and the other two square pieces are blank (actually,
they say "Meffert's Challenge").  I think it is fun because in addition to
solving the puzzle, you can try to make different "snake" patterns on it.
It took me a while to figure out how to make a snake that uses all but one
of the segments.

I threw away the packaging, and I don't remember who makes it.  I think it
was Pressman.  I bought it at a toy store in the Supermall in Auburn, WA,
and they also had a normal Skewb made by the same company.

David

From cube-lovers-errors@mc.lcs.mit.edu  Thu Jul 30 18:19:26 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Thu Jul 30 13:37:06 1998
Message-Id: <35C0AEFD.3D67E038@nadn.navy.mil>
Date: Thu, 30 Jul 1998 13:35:57 -0400
From: David Joyner <wdj@nadn.navy.mil>
Reply-To: wdj@nadn.navy.mil
Organization: Math Dept, USNA
To: Rainer.adS.BERA_GmbH@t-online.de
Cc: Cube Mailing List <Cube-Lovers@ai.mit.edu>
Subject: Re: 4x4x4 patterns
References: <35BE13D0.874D8E6D@t-online.de>

Rainer aus dem Spring wrote:

> Dear cube-lovers,
>
> Of course, HTML is the format I should have used. I'll try to get a
> latex->html converter for NT.

> Anybody my post the patterns in any format.

 Rainer's patterns (with some pictures) are on the web page
http://web.usna.navy.mil/~wdj/4x4patterns_b.htm
- David Joyner

--
David Joyner, Assoc Prof of Math
US Naval Academy, Annapolis, MD 21402
(410)293-6738
wdj@nadn.navy.mil
http://web.usna.navy.mil/~wdj/homepage.html
++++++++++++++++++++++++++++++++++++++++++++
"A Mathematician is a machine for turning
coffee into theorems." Alfred Renyi

From cube-lovers-errors@mc.lcs.mit.edu  Fri Jul 31 10:28:00 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Thu Jul 30 14:17:23 1998
Message-Id: <199807301816.OAA14197@life.ai.mit.edu>
Date:     Thu, 30 Jul 98 14:16:38 EDT
From: Nichael Cramer <ncramer@bbn.com>
To: cube-lovers@ai.mit.edu
Cc: nichael@sover.net
Subject:  "Hints" for Solving the 5X

Since you 1] posted to this list and 2] have a 5X and have solved it
the past, I'm going to make the assumption that really want is not
necessarily a cook-book for solving the 5X, but enough hints to get
you started in the right direction.

If I'm wrong about this, you can stop here.  ;-)  If not, below is
high-level description of a scheme for solving the 5X that assumes
1] that you're 3X3X3-literate and 2] leaves unspecified the details of
the three other simple operations that you'll need.

(Now, this is far from elegant; and certainly not anything like a
maximal solution.  But at least it will 1] get your cube solved and 2]
at least you thinking about these things again.)




---
Step 1:

First, ignore everything except the corner cubies, the center (face)
cubies and the center-edge cubes.

Now, paying attention to those cubies only, pretend that you're
dealing with a 3X and "solve" it.


---
Step 2: Solve the non-center edge-cubies [NCEC].

First devise an operator that allows you to cyclically-swap three of
the NCECs (i.e. without messing with any of the previously solved
cubies).  With a little clever permutation, this single operation will
allow you to complete this step (but see Step 2A below).

(Note that since a NCEC _has_ to be in its correct orientation when it
is in its corrected location --i.e. a correctly placed NCEC can't be
simply flipped as can a center-edge cubie in a 3X-- you can complete
this step using this single operation.)


Step 2A: There is one wrinkle at this point.

It is possible to be in an "orbit" in which you can apparently "solve"
all of the NCECs except for two.  If you reach this point, leave the
two unsolved NCEC alone for the moment.  They will be easier to solve
after completing the next step.


---
Step 3: Solve the remaining non-center face cubies [NCFC].

Similar to the above, devise two operators: one that allows you to
cyclically-swap three "corner-like" NCFCs and one that allows you to
cyclically-swap three "edge-like" NCFCs (i.e. without in either case
disturbing the previously-solved cubies).

Again with a little clever permutation you should be able to complete
this step with this single operation.

(Note that since, for a given color, all four "center-like" NCFCs are
pretty much interchangable --as are all four "edge-like" NCFCs.  This
means that that any "bogus" symetries are invisible.  What this means
is that, by saving this step 'til last, you don't risk getting all the
way to the and finding out you're in some non-standard "orbit" that
you have to back out of.)


---
Step 2A Continuted:

Assuming that you've got this far, you should now be in a state where
the entire cube is solved except for --at most-- two NCECs.

In short, this state of affairs means that your cube is a wrong
"orbit"; i.e. there is a "hidden" symmetry among that cubies that
allows your cube to appear to be more nearly solved than it is.

The quickest way to get your cube in the "correct orbit" is as
follows:

Choose one of the "internal" slices that contains one of non-solved
NCECs (by "internal", I mean a slice that is neither a face slice nor
one that contains a center cubie).

Now rotate that internal slice by a quarter turn (i.e. by 90 degrees)
in either direction.

Now what you want to do is solve remaining cubies from its current
situation.

The tricky part here is keeping everything straight.  There are a
couple of things that you can do help this.

1] If possible, you can first manipulation the NCECs in such a way
that the two unsolved NCEC share the same slice and are on the same
face.  Then it will be possible --when performing the quarter-turn as
described above-- to bring one of the unsolved NCEC into its correct
location.

Once that is done, you will now have exactly three unsolved
NCECs.  Since these must necessarily be a cyclic permuation, you
should be able to solve these without further ado.

Now, all that remains is solving the remaining newly scrambled
face-cubies.


2] If you are unable to position the two unsolved NCEC as described
above, proceed as follows:

>From the current state, (i.e. after performing the quarter-turn
on the internal slice) re-solve the newly shuffled face-cubies
*while*being*sure*not*to*disturb*any*other*cubies!

Once you have all the faces solved, you should now have four NCEC in
the wrong locations, along the single internal slice.  Using the
operation that you developed in Step 2 above, this should be easy to
solve.


---
Step 4:  Place your newly-solved cube in a prominent place on your
desk and assume a smug demeanor when asked about it.



Hope this helps
Nichael


--
Nichael Cramer
work: ncramer@bbn.com
home: nichael@sover.net
http://www.sover.net/~nichael/

From cube-lovers-errors@mc.lcs.mit.edu  Fri Jul 31 11:16:22 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Thu Jul 30 17:34:13 1998
Message-Id: <35C0C750.9D29907D@t-online.de>
Date: Thu, 30 Jul 1998 21:19:44 +0200
Reply-To: Rainer.adS.BERA_GmbH@t-online.de
Organization: BERA Softwaretechnik GmbH
To: Cube Mailing List <Cube-Lovers@ai.mit.edu>
Subject: Corrections
References: <35BE13D0.874D8E6D@t-online.de>
From: Rainer.adS.BERA_GmbH@t-online.de (Rainer aus dem Spring)

Sorry:

\subsubsection*{2 Twisted Rings}
$L_{12}^2 F_{12}^{-1} L_{12}^{-1} F_{12} L_{12}^{-1} U_{12} B_{12}^{-1}
U_{12}
B_{12} U_{12}^2  (B^{-1} U^2 B R^{-1} D^2 R)^2$

not ... U^2 R)^2$

Rainer adS

U was "unten", i.e., German for down.

From cube-lovers-errors@mc.lcs.mit.edu  Fri Jul 31 12:04:34 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Thu Jul 30 17:52:36 1998
Message-Id: <35C0D64D.B84386E@t-online.de>
Date: Thu, 30 Jul 1998 22:23:41 +0200
Reply-To: Rainer.adS.BERA_GmbH@t-online.de
Organization: BERA Softwaretechnik GmbH
To: Cube Mailing List <Cube-Lovers@ai.mit.edu>
Subject: The hunt is up
From: Rainer.adS.BERA_GmbH@t-online.de (Rainer aus dem Spring)

Dear cube lovers,

I am glad to see that my patterns started such a long thread.
The cube is alive :)

Mike Reid sent some improvements. I am sure he doesn't mind me to send
them to the mailing list.

What about other improvements ? Anybody mad enough to search other
patterns ?

Mike's improvements:


\subsubsection*{Exchanged 2x3x3 Bricks}
$D_{12}^2 L^2 B^2  D_{12}^{-1} R^2  D_{12} R^2  U_{12}^{-1} R^2  U_{12}
R^2 B^2
U_{12} L^2 D_{12} L^2  D_{12}$ (Michael Reid)

improves my ????? :)

$U_{12}^2  R^2 B^2  D_{12}  L^2  D_{12}^{-1}  L^2  U_{12}  L^2
U_{12}^{-1} L^2
B^2 U_{12}^{-1} R^2 U_{12} R^2 U_{12}$ (Michael Reid)

improves the 1x1x2 Brick pattern

$U R^{-1} U^{-1}  F_2  D_{12}^{-1} F^2 D_{12} F_2^{-1}  D_{12}^{-1}  F^2
D_{12} U R U^{-1}$ (Michael Reid)

improves "1 Small Twisted Ring"

Rainer

-- 
--------------------------------------------------------------------------
Rainer aus dem Spring       email  Rainer.adS.BERA_GmbH@t-online.de
(home)
Schimmelbuschstr. 10        email  TEEADS@TEE.toshiba.de (business only)
40699 Erkrath               tel. +49 (0)02104 35157  (private)
Germany                     tel. +49 (0)02104 936150 (business)
---------------------------------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Fri Jul 31 13:39:50 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Thu Jul 30 18:29:08 1998
From: Douglas Zander <dzander@solaria.sol.net>
Message-Id: <199807302228.RAA17538@solaria.sol.net>
Subject: Meffert's Challenge (fwd)
To: cube-lovers@ai.mit.edu (cube)
Date: Thu, 30 Jul 98 17:28:21 CDT

David Barr wrote:

> I recently bought a "Meffert's Challenge" puzzle, and I see
 <snip>
> I threw away the packaging, and I don't remember who makes it.  I
> think it was Pressman.  I bought it at a toy store in the Supermall
> in Auburn, WA, and they also had a normal Skewb made by the same
> company.

> David

 yes, I just happened to buy both of these puzzles myself today.  They
 are made by Pressman.  It is interesting to note that the copyright
 says, "(copyright) 1997 Uwe Meffert patent #5,358,247" for *both*
 puzzles.  (same mechanism)
 They also are selling the Pyraminx in the same type packaging.  Is
 the Pyraminx also a Skewb mechanism?  (I didn't buy one since I
 already had one) Now I wish they would bring back some other puzzles!
 :-)

-- 
 Douglas Zander                | 
 dzander@solaria.sol.net       |
 Shorewood, Wisconsin, USA     |

From cube-lovers-errors@mc.lcs.mit.edu  Fri Jul 31 16:44:32 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Fri Jul 31 16:24:59 1998
Date: Fri, 31 Jul 1998 16:24:49 -0400
From: michael reid <reid@math.brown.edu>
Message-Id: <199807312024.QAA10616@euclid.math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: new optimal solver

lately i've been working on a new optimal solver.  this is similar
to the previous program, but uses different subgroups.  let  H
be the subgroup in which

        the four edges  FR, FL, BR  and  BL  are all in place,
        and are correctly oriented
and
        the four  U  corners are on the  U  face (and thus the
        four  D  corners are on the  D  face, and they are oriented
        so that the  U  [respectively D]  facelet is on the  U
        [respectively D]  face.

then the cosets  H \ G  are described by triples  (e, cl, ct)  where

e  describes the location and orientation of the four edges  FR, FL,
    BR  and  BL,
cl  describes the location of the four  U  corners, and
ct  describes the orientation of the eight corners.

there are

24 * 22 * 20 * 18 = 190080  different  e  coordinates,
/ 8 \
\ 4 /  =   70  different  cl  coordinates, and
3^7    = 2187  different  ct  coordinates.

all combinations are possible, so there are  190080 * 70 * 2187 =
29099347200  cosets.  the subgroup  H  has 16-fold symmetry; it is
invariant under any symmetry of the cube that preserves the  U-D
axis.  therefore the coset space  H \ G  also has this symmetry.
up to symmetry, there are  12094  e  coordinates.  thus, we can
reduce the coset space to  12094 * 70 * 2187 = 1851470460
configurations.  store each configuration in half a byte of memory
(storing its distance from start).  the whole thing can be stored
in a tiny array of 925735230 bytes, approximately 883 megabytes.

the number of cosets (actual numbers, not reduced by symmetry)
at each distance is

distance   quarter turns       face turns
 0                     1                1
 1                     8               12
 2                    76              162
 3                   696             2044
 4                  6418            25442
 5                 57912           316290
 6                514318          3899553
 7               4496206         46650252
 8              38304572        517476714
 9             308312232       4480840746
10            2142297548      16776040760
11            9789496784       7259620140
12           14800845359         14475084
13            2014724044
14                291026

i have this running on one processor of a sun ultra enterprise 450,
configured with 1024Mb of RAM.  startup time is significant: it takes
about 85 minutes for quarter turns, 125 minutes for face turns, to
exhaustively search the coset space.

some rough estimates are that it is 6.7 times faster than my previous
optimal solver for quarter turns, 3.4 times faster for face turns.
this is not nearly as good as i'd hoped.  there seems to be some
performance issue with this machine.  it appears to be significantly
slower when accessing large amounts of memory at random, despite the
fact that it is all real memory, so no swapping is occurring.  the
performance drop off starts at about 256Mb.  my program runs slower
by a factor of 3 or maybe even 4 because of this.  my sysadmin has
reproduced the same behavior on a small test program, so the problem
is unlikely to be caused by my code.  i'm told that it is probably
some gross inefficiency in the cache paging system of the operating
system (solaris).  the os seems to have plenty of options, so perhaps
one of them will fix this problem and speed up my program by a factor
of 3 or maybe 4.  it seems ridiculous to me that things work this way,
but apparently they do.

nevertheless, the program is already fast enough for the tasks at hand.

mike

From cube-lovers-errors@mc.lcs.mit.edu  Sat Aug  1 23:44:12 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Fri Jul 31 22:28:10 1998
Date: Fri, 31 Jul 1998 22:28:02 -0400
From: michael reid <reid@math.brown.edu>
Message-Id: <199808010228.WAA11081@euclid.math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: all 24q maneuvers for superflip

with my new optimal solver, i've calculated all 24q maneuvers for
superflip.  there are three transformations we can apply to a
maneuver for superflip, none of which change its length.

     we may conjugate by any cube symmetry.
     we may cyclically permute the maneuver, i.e. replace

          sequence_1 sequence_2  by  sequence_2 sequence_1

     we may invert the maneuver.

in a previous message (august 7, 1997), i showed that, using these
three transformations, any maneuver for superflip can be transformed
into one that begins with one of the ten sequences

     U  R2        U  D' R      U  D  R      U  D  R'     U  R  F
     U  R  F'     U  R' F      U  R' F'     U' R  F'     U' R' F'

my program took 101 hours to exhaustively search these ten cases.
there are four inequivalent maneuvers; two were previously known:

R' U2 B  L' F  U' B  D  F  U  D' L  D2 F' R  B' D  F' U' B' U  D'    (24q*)
U  R2 F' R  D' L  B' R  U' R  U' D  F' U  F' U' D' B  L' F' B' D' L' (24q*)

the two new ones are:

U  D' R  F  U' D' L  D' F  R  U' R  U' D' F  U' F  L  B' U  F' B' L  B' (24q*)
U  D' R  F' D  L' B  L' U' R' D' B' U' D  L' F  D' R  B' R  U  L  D  B  (24q*)

this last one can be written as

     (U  D' R  F' D  L' B  L' U' R' D' B'   R_rl)^2

where  R_rl  denotes reflection through the R-L plane.

we can also count the total number of 24q maneuvers for superflip.
note that  U2 = U U  also is  U' U' , so can be cyclically shifted
in an extra way.  similarly,  U  D' = D' U , so this also accounts
for an extra cyclic shift.  and the same is true for  U' D'.
the total number of maneuvers therefore is

28 * 24 * 2 + 28 * 48 * 2 + 28 * 48 * 2 + 26 * 24 * 2 = 7968

where the first factor is the number of cyclic shifts, the second
factor is the number of cube symmetries we can apply, and the third
factor is 2, for inversion.  the first and last maneuvers only get
a factor of 24 for the number of cube symmetries, because a cyclic
shift by 12q gives the same maneuver in a different orientation.

the total number of 24q sequences is 274575811926317204506464.
the total number of even positions is 21626001637244928000.
so even positions have an average of 12696.56 different 24q maneuvers.

mike

From cube-lovers-errors@mc.lcs.mit.edu  Sun Aug  2 17:46:24 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sun Aug  2 08:47:54 1998
Date: Sun, 2 Aug 1998 08:47:44 -0400
From: michael reid <reid@math.brown.edu>
Message-Id: <199808021247.IAA08734@cauchy.math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: superflip  composed with  four spot

with my new optimal solver, i can show that the position

     superflip  composed with  four spot

is exactly 26 quarter turns from start.  this gives a new lower bound
for the diameter of the cube group.  the previous lower bound, 24q, was
from the position superflip, and was first established by jerry bryan.

let  F2 B2 U  D' R2 L2 U  D'  be our choice of orientation of
four spot.  although four spot is not central, the position

     F2 B2 U  D' R2 L2 U  D'   C_U2

moves only face center cubies:  (F, B) (R, L).  (here  C_U2  denotes
whole cube rotation by 180 degrees about the U-D axis.)  since quarter
turns do not move face center cubies, we see that the sequence above
commutes with any sequence of quarter turns.  the same is also true
for
     superflip . four spot . C_U2

in terms of singmaster's fixed face model, this means that we can
cyclically shift a maneuver for superflip composed with four spot,
but the part that is cyclically shifted gets conjugated by the cube
rotation  C_U2.  for example:

(B  U2 L) (U' D  L2 F2 R2 B  U2 R' L' D  R2 D  F2 U  R2 D  B)

creates this position.  if we cyclically shift the first three twists
to the end, we get another maneuver for this position:

(U' D  L2 F2 R2 B  U2 R' L' D  R2 D  F2 U  R2 D  B) (F  U2 R)

this observation about cyclic shifting enables us to prove

proposition 1.  superflip  composed with  four spot  is a local
                maximum in the quarter turn metric.

proof.  we need to show that any quarter turn takes us closer to
        start.  the 12 different twists split up into two different
        types under the symmetry of this position:  {U, U', D, D'}
        and {R, R', F, F', L, L', B, B'}.  we claim that any maneuver
        for superflip composed with four spot must contain twists of
        both types.  a maneuver consisting only of twists in
        {U, U', D, D'} clearly cannot produce this position.  also,
        a maneuver consisting only of twists in
        {R, R', F, F', L, L', B, B'} cannot flip any edges.  thus
        both twist types must occur.  now consider a minimal
        maneuver for superflip composed with four spot.  we may
        cyclically shift (and apply symmetry) so that the last twist
        is  U'.  thus, applying  U  cancels this last twist and
        brings us closer to start.  similarly, we can cyclically shift
        to get a minimal maneuver ending with  R', so applying  R
        also brings us closer to start.  since any twist is equivalent
        to  U  or  R , we have proved local maximality.     qed

the significance of this proposition is that this is the first case
beyond the hoey-saxe local maxima in which we can prove local
maximality without computer searching.  (please correct me if i'm
wrong about this.)

dan hoey noted (a long time ago) that the position  four spot  is a
local maximum.  however, i don't see that this can be proved without
computer search.  the sticking point is that four spot can be achieved
using only {R, R', F, F', L, L', B, B'}.  however, no minimal maneuver
consists only of these twists, a fact determined by computer search.

similar to the transformations for superflip, we have three
transformations to apply to maneuvers for superflip composed with
four spot.

     we may conjugate by any of the 16 cube symmetries that fix
     the U-D axis.
     we may cyclically shift the maneuver, as described above.
     we may invert the maneuver.

proposition 2.  by using the three transformations above, any maneuver
                for superflip composed with four spot can be transformed
                into one that begins with one of the six sequences

                R  U          R' U  D        R' U  F'
                R' U  R'      R' U  B'       R' U  L'

proof.  as shown in prop. 1, any sequence for superflip composed with
        four spot contains both types of twists.  thus, the two types
        occur as consecutive twists.  by cyclic shifting, and applying
        symmetry, we may suppose that the first two quarter turns are
        either  R U  or  R' U.  (this would already be enough reduction
        for my program).  we can cut down the case  R' U  further.
        there are eleven possibilities for the third quarter turn;
        only  U'  is not allowed.  the case  R' U U = R' U2  is
        equivalent under symmetry to  R U2, which is part of the case
        beginning with  R U.  the case  R' U D'  is equivalent under
        symmetry to  R D' U = R U D', again part of the case beginning
        with  R U.  the case  R' U B  inverts to  B' U' R, and this is
        equivalent to  R U B', which is part of the case beginning with
        R U.  similarly, the cases beginning with  R' U R ,  R' U F
        and  R' U L  invert to  R U R' , R U F'  and  R U L',
        respectively.  this leaves only the sequences listed above.  qed

my program exhaustively searched the positions

     superflip. four spot . R  U    through 22q  and

     superflip. four spot . R' U  D   \
     superflip. four spot . R' U  F'   \
     superflip. four spot . R' U  R'    >  all through 21q
     superflip. four spot . R' U  B'   /
     superflip. four spot . R' U  L'  /

and found no maneuvers.  thus  superflip  composed with  four spot
requires more than 24 quarter turns.  the total search time was about
153 hours.  to see that  superflip  composed with  four spot  can be
achieved in 26 quarter turns, use

U2 D2 L  F2 U' D  R2 B  U' D' R  L  F2 R  U  D' R' L  U  F' B'  (26q*, 21f)

it might be reasonable to ask for all 26q maneuvers.  this is probably
out of reach for now.  however, i suspect that there will be so many
different 26q maneuvers that it would not be of much use to see a long
list of maneuvers.  (i have a bunch already.)

superflip  composed with  four spot  also requires 20f.

proposition 3.  any maneuver for superflip  composed with  four spot
                of length <= 20f can be transformed to one that begins
                with one of the sequences   U2 R ,   R2 F  or   R2 U .

the proof is very similar to the reductions for superflip in the face
turn metric.

using this, a complete search for 20f maneuvers is straightforward.
there are two inequivalent 20f maneuvers for superflip composed with
four spot:

F  U2 R  L  D  F2 U  R2 D  F2 D  F' B' U2 L  F2 R2 B2 U' D   (20f*, 28q)
F  U2 R  L  D  F2 U  R2 D  F2 D  F' B' U2 L  U' D  R2 B2 L2  (20f*, 28q)

this also shows that no maneuver is simultaneously minimal in both
metrics.

mike

From cube-lovers-errors@mc.lcs.mit.edu  Mon Aug  3 13:19:38 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Mon Aug  3 02:10:51 1998
Date: Mon, 3 Aug 1998 00:24:35 -0400
Message-Id: <00273C38.001706@scudder.com>
From: Jacob_Davenport@scudder.com (Jacob Davenport)
Subject: Web address for solving the 5^3
To: cube-lovers@ai.mit.edu

I have created a page with a description of how I solve the 5x5x5.
Please check out www.wunderland.com/wts/jake.  Although I did spend a
fair amount of time on this page, I certainly consider it a first
draft, and I would appreciate any comments about it, either those
involving clarity of the explanation, or even better moves that would
perform the same functions.  Be warned that it is a long page,
although I'm sure you expected that.

One person wrote to me and said that he had all but two cubies solved.
I suspect his difficulty was "parity" which I cover in my Sixth Step.
I'm sure there are many good solutions to the 5x5x5, just as there are
for the 3x3x3, so if you have a half-solved cube you may need to scrap
your work if you want to use my solution.

Good luck.

-Jacob

From cube-lovers-errors@mc.lcs.mit.edu  Thu Aug  6 11:20:35 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Aug  5 13:41:35 1998
Date: Wed, 05 Aug 1998 13:41:10 -0400 (Eastern Daylight Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: superflip  composed with  four spot
In-Reply-To: <199808021247.IAA08734@cauchy.math.brown.edu>
To: michael reid <reid@math.brown.edu>
Cc: cube-lovers@ai.mit.edu
Message-Id: <SIMEON.9808051310.B@muahost.pstcc.cc.tn.us>

On Sun, 02 Aug 1998 08:47:44 -0400 michael reid
<reid@math.brown.edu> wrote:

> with my new optimal solver, i can show that the position
>
>      superflip  composed with  four spot
>
> is exactly 26 quarter turns from start.  this gives a new lower bound
> for the diameter of the cube group.  the previous lower bound, 24q, was
> from the position superflip, and was first established by jerry bryan.

Nobody has said so yet on the list, but I think this is
exciting news for Cube-Lovers, both the fact that a new
lower bound has been discovered for the diameter of the
cube group, and the fact that a new (and very long) local
maximum has been found by means other than computer search.

It seems to me that Mike's proof might provide an outline
for a method for looking for other local maxima.  I have
not at this point been able to use his method to find other
local maxima, but here is how it might work.


> proposition 1.  superflip  composed with  four spot  is a local
>                 maximum in the quarter turn metric.
>
> proof.  we need to show that any quarter turn takes us closer to
>         start.  the 12 different twists split up into two different
>         types under the symmetry of this position:  {U, U', D, D'}
>         and {R, R', F, F', L, L', B, B'}.  we claim that any maneuver
>         for superflip composed with four spot must contain twists of
>         both types.  a maneuver consisting only of twists in
>         {U, U', D, D'} clearly cannot produce this position.  also,
>         a maneuver consisting only of twists in
>         {R, R', F, F', L, L', B, B'} cannot flip any edges.  thus
>         both twist types must occur.

More generally, for any position x, calculate Symm(x)=K,
where K is as usual the subgroup of all k in M such that
x=k'xk, and where M is the group of 48 symmetries of the
cube.  Conjugation by K and grouping the elements of Q into
conjugate equivalence classes induces a partition on Q, the
set of twelve quarter turns.  In Mike's case, the partition
is {Q1,Q2} where Q1={U, U', D, D'} and Q2={R, R', F, F', L,
L', B, B'}.

The process I am going to describe is much simpler if we
confine ourselves to 2-way partitions of Q, such as Mike's
case.  I think the process I am descibing can be
generalized to more than 2-way partitions of Q, but some of
the steps get more complicated.  So for now we confine
ourselves to subgroups K of M which induce at most a 2-way
partition of Q.

Roughly speaking, this means that we need to find positions
that are fairly symmetric.  I have been meaning for a long
time to calculate a table of partitions of Q for each of the
possible 98 subgroups of M. Perhaps Mike's new result will
provide sufficient motivation to perform the calculations.

The next hurdle is that we must find positions x such that
x is not in <Q1> or <Q2>, so that a maneuver for x must
contain quarter turns from both Q1 and Q2.  Mike's position
certainly satisfies this criterion.   Notice that if we get
this far, we can say that a maneuver for x must contain at
least one element from Q1 and at least one element from Q2,
but the elements from Q1 and Q2, respectively, need not
necessarily appear at the end of the maneuver.  Also, by
the definition of Q1 and Q2, *any* maneuver from Q1 and Q2
can appear in a maneuver for x by K-conjugation.

So far, so good.  I would go about this type of a search by
determining which subgroups K of M induce a 2-way partition
of Q, and then by thinking about what a K-symmetric
position must look like.  But here's the rub -- the part I
cannot figure out *in general".  In order to get the
elements of Q1 and Q2 to the end of the maneuver for x, we
need positions which may be cyclically shifted, either in
the normal sense or in Mike's new sense where the part of
the maneuver that is shifted is conjugated by K.

There is a good bit of discussion in the archives about
cyclical shiftiness.  I'm going to go back and re-read that
discussion to see if it helps with this problem.  But any
position x whose symmetry group induces a 2-way partition
{Q1,Q2} on Q, where x is not in <Q1> or <Q2>, and where x
is cyclically shiftable (possibly with K-conjugation of
the shifted part) is a local maximum in the quarter turn
metric.

----------------------
Jerry Bryan
jbryan@pstcc.cc.tn.us



From cube-lovers-errors@mc.lcs.mit.edu  Thu Aug  6 12:19:33 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Aug  5 21:22:00 1998
Message-Id: <35C8FF13.A6997C43@frontiernet.net>
Date: Wed, 05 Aug 1998 20:55:47 -0400
From: John Bailey <jmb184@frontiernet.net>
To: Submissions Cube-Lovers <Cube-Lovers@ai.mit.edu>
Subject: Four dimensional cube solution and two dimensional cube simulator

Earlier this year I announced:
http://www.frontiernet.net/~jmb184/Nteract4.html
a four dimensional Rubik Cube (2x2x2x2)  While that post referenced a
sketch of a solution, it seemed that a clearer, more explicit solution
was needed to show that the tesseract was indeed a tractable cubing
problem.
An explicit solution of the four dimensional analog of the Rubik cube
is posted at:
http://www.frontiernet.net/~jmb184/solution.html
This page includes extensive graphics which are intended to make the
solution clear and visible.

Also, during the process of developing a detailed explaination,  I
realized that by using similar display techniques, a 2D analog of the
cube provided an interesting model of cube solutions.  This 2
dimensional 3X3 cube simulator  is at:
http://www.frontiernet.net/~jmb184/3x3cube.html

All of these are written in Javascript, which means they do not require
extended interaction with the server to manipulate.  They are read in
directly and then can be kept for running off-line.

John Bailey

From cube-lovers-errors@mc.lcs.mit.edu  Wed Aug 12 10:43:50 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Thu Aug  6 23:33:59 1998
Date: Thu, 6 Aug 1998 23:33:52 -0400
From: michael reid <reid@math.brown.edu>
Message-Id: <199808070333.XAA03183@chern.math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: superflip composed with six spot

another position to consider is  superflip  composed with  six spot,
since its maneuvers also have a corresponding cyclic shifting property.
until recently, i didn't know any 24q maneuvers for this position, so
i had planned to do an exhaustive search through 24q.  however, by
looking at the new 24q maneuvers for superflip, i was able to modify
one to get a 24q maneuver for superflip composed with six spot:

D' R  L' F  L' F  B  U' B  L' F' U  F' U  D  R' U  R' F' D  L' U  D  F' (24q*)

as a result, i only did a partial 24q search, namely for maneuvers that
contain a half turn.  up to cyclic shifting and symmetry, the only such
maneuvers are

R' U  D  R' U  F' D  R' B  U' L' U' F' D  F' B' D' R' F  D  F  D' R2 (24q*)

and a suitable reorientation of the inverse maneuver.  (the inverse
position is the same pattern, but in a different orientation.)

superflip composed with six spot also requires 20f:

U2 F  B' R  F  L2 F2 D  B2 D2 R2 B' L2 F' D2 R2 D' B  R  B2  (20f*)

no maneuver is simultaneously minimal in both metrics; this is a
consequence of the partial 24q search.

mike

From cube-lovers-errors@mc.lcs.mit.edu  Mon Aug 17 14:50:07 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sun Aug 16 11:40:37 1998
Message-Id: <c=US%a=_%p=INEA_Corporation%l=INEADEV-980816153228Z-2353@proxy.ineacorp.com>
From: Noel Dillabough <Noel.Dillabough@ineacorp.com>
To: "'cube-lovers@ai.mit.edu'" <cube-lovers@ai.mit.edu>
Subject: New puzzle simulator
Date: Sun, 16 Aug 1998 11:32:28 -0400

I wrote up a puzzle simulator called Puzzler, containing the cubes
(2x2x2, 3x3x3, 4x4x4, 5x5x5), a pyramid, a sphere, a skewb, and a
dodecahedron.  While not the same as a physical puzzle, its still pretty
fun to use.

Its just a beta version, just compiled yesterday, so there are bound to
be problems, and features that should/could be implemented.  Anyone
interested should download the program at
http://www.mud.ca/puzzler/puzzler.html, and let me know of any problems,
enhancements etc.

One of my friends asked me to implement the square 1 in the same
program.  In order to do so I would have to change a lot of the backend
code so I won't do so unless there is enough interest.

-Noel

P.S. I forgot, the program is for win95, win98 and winNT.

From cube-lovers-errors@mc.lcs.mit.edu  Tue Aug 18 17:36:34 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Tue Aug 18 16:28:43 1998
Date: Tue, 18 Aug 1998 16:28:28 -0400
From: michael reid <reid@math.brown.edu>
Message-Id: <199808182028.QAA24353@euclid.math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: minimal maneuvers for X symmetric positions

X  is the subgroup of the cube symmetry group which preserves
the U-D axis.  there are 128 positions which have X symmetry:

     the UR edge can go in any of the 8 positions  UR, RU, DR,
     RD, UL, LU, DL, LD; this determines the location of the
     edges UB, UL, UF, DR, DB, DL, DF.

     the FR edge can go in any of the 4 positions FR, RF, BL, LB;
     this determines the location of the edges FL, BR, BL.

     the UFR corner can go in any of the 4 positions UFR, UBL,
     DRF, DLB; this determines the location of all the corners.

any combination of these is possible, which gives 128 positions.
4 of the positions have more symmetry, namely M symmetry.
(these positions are  start, superflip, pons asinorum, and
pons asinorum composed with superflip.)

minimal maneuvers for the other positions are:

1.  F2 R  L' D2 F2 D2 R' L  F2 D2  (16q*, 10f*)

2.  U  F' B' R2 U' D' F' B  U  D  R2 F  B  D'  (16q*, 14f)
    U  F2 U2 F2 R  L  F2 U2 F2 U2 R' L' U   (13f*, 20q)

3.  U  F  B  R2 U  D  F  B' U' D' R2 F' B' D'  (16q*, 14f)
    U  F  B  U2 R2 U2 R2 F' B' R2 U2 R2 U   (13f*, 20q)

4.  F2 B2 U  D' R2 L2 U' D   (12q*, 8f*)

5.  F2 R2 F2 B2 R2 B2  (12q*, 6f*)

6.  U  F' B' R' L' F' B' R  L  F  B  R  L  U'  (14q*, 14f)
    U  F2 U2 F2 R  L  B2 D2 B2 U2 R' L' U   (13f*, 20q)

7.  U  F  B  R  L  F  B  R' L' F' B' R' L' U'  (14q*, 14f)
    U  F  B  U2 R2 D2 R2 F' B' L2 U2 L2 U   (13f*, 20q)

8.  F  R' U  B2 L' F  U  D' L' B  R2 U' F  L' U' D   (18q*, 16f*)

9.  F  U' F  R' D  F' D  F' R  L  B' U  B' U  L' B  D' B  U  D   (20q*, 20f)
    F2 R  F  B' D  B2 D' F2 B2 U  F2 U' F  B' R' B2  (16f*, 22q)

10.  F  R  F  D' F' B  R  F' U' B' R  L' F  U' D' F' B' R2 U   (20q*, 19f)
     U  F' B' R  F2 U  D' L2 F' U' D' F2 R' U' D  R  L' D2  (18f*, 22q)

11.  F  R  D  R' F' U  B' L  U' D  L' F  D' B  R  U' R' B'  (18q*, 18f*)

12.  F  R' B  R' L  U' R  L' U  B  R  L' D' B' L  F'  (16q*, 16f)
     F  R2 F2 U' D  R' U2 D2 L' F2 L2 U' D  F   (14f*, 20q)

13.  F  R' L' U' F  B' R' L  F2 U2 F  B  D' R  L  U2 B'  (20q*, 17f)
     F  U2 F2 B2 R' L  F2 U  F2 B2 U  F2 R  L' D2 F   (16f*, 24q)

14.  U  F  B' U' R  F' R' B  R' U2 R' F  R' B' R  U' F' B  U   (20q*, 19f)
     U  F2 R' L' F  D2 R' L  B2 D' F' B' R' U  D' R' L  U2  (18f*, 22q)

15.  F  B  R  F2 U' D  R2 B' U' D' L' U  D' R' L  U  R2  (20q*, 17f*)

16.  F  B  U  D  R2 L2 U  D  F  B   (12q*, 10f*)

17.  U2 D2  (4q*, 2f*)

18.  U  F  B  R2 U  D  F' B  U' D' R2 F' B' D'  (16q*, 14f)
     U  F  B  D2 R2 D2 R2 F' B' R2 D2 R2 U   (13f*, 20q)

19.  U  F' B' R2 U' D' F  B' U  D  R2 F  B  D'  (16q*, 14f)
     U  F2 D2 F2 R  L  F2 D2 F2 D2 R' L' U   (13f*, 20q)

20.  F2 R2 F2 B2 R2 B2 U2 D2  (16q*, 8f*)

21.  F2 B2 U  D' R2 L2 U  D'  (12q*, 8f*)

22.  U  F  B  R' L' F  B  R' L' F  B  R' L' U'  (14q*, 14f)
     U  F  B  D2 R2 U2 R2 F' B' L2 D2 L2 U   (13f*, 20q)

23.  U  F' B' R  L  F' B' R  L  F' B' R  L  U'  (14q*, 14f)
     U  F2 D2 F2 R  L  B2 U2 B2 D2 R' L' U   (13f*, 20q)

24.  F  R' L' U' L2 B2 R  L  F' B  R' L  D' R  L  U2 F'  (20q*, 17f)
     F  U2 F2 B2 R  L' B2 U' R2 L2 U  F2 R  L' D2 B'  (16f*, 24q)

25.  F  R2 D  F' U' R  U2 D2 L' U  B  D' L2 B'  (18q*, 14f*)

26.  F  U  R' U  L' U' L  D  R  L  U  R  D' R' D  L' D  B   (18q*, 18f)
     F  B  R  F2 U' D  R2 B' U' D' R' U' D  R  L' D  L2  (17f*, 20q)

27.  F  R' F' B  L' F  U' F  B  U2 F  B  U' R' B  R  L' F  R'  (20q*, 19f)
     U  F' B' R  F2 U  D' L2 F' U' D' F2 R' U' D  R  L' U2  (18f*, 22q)

28.  F  U' R  B  R  F  B' U' F  U' D  L' D  R  L' B2 R' U  F'  (20q*, 19f)
     F2 R  F  B' D  B2 U' R2 L2 D  B2 U' F' B  L' F2  (16f*, 22q)

29.  F  R2 U' R  U  F' U  D' L  U' F' U  F2 L' U  D'  (18q*, 16f*)

30.  F  R' F' B  L' F  U  F  B  U2 F  B  U  L' F  R' L  B  L'  (20q*, 19f)
     F  B' R  F  B  D  L2 F  B' U2 L' U  D  R2 F2 B' R  L'  (18f*, 22q)

31.  F  B' U  R  D  B' U  D' R  D' F' D' R  B2 U' D' R' L' U   (20q*, 19f)
     U  F2 R' L' F  D2 R' L  B2 D' F' B' R' U  D' R' L  D2  (18f*, 22q)

32.  U  D  F2 B2 U  D' F2 B2 D2  (14q*, 9f)
     F2 R2 F2 B2 R2 F2 R2 L2  (8f*, 16q)

33.  F2 B2 U  D' F2 B2 U' D   (12q*, 8f*)

34.  F  B  R  L  F  B  R' L' F' B' R' L' U' F2 B2 R2 L2 D'  (22q*, 18f)
     F  B  R  F2 B2 U2 D2 L' F' B' U2 R' L' F2 R' L' U2  (17f*, 24q)

35.  F  R  F  L  D  F' L' F  B' R' L  F' B  R  F  U  L' B' L  F   (20q*, 20f)
     F2 U  F  B  D2 R2 B2 D2 L2 F' B' U2 R2 D  F2  (15f*, 24q)

36.  U2 F2 B2 R2 L2 D2  (12q*, 6f*)

37.  F2 B2 R2 L2  (8q*, 4f*)

38.  F  U' B' D' F2 U' F' U' D2 F  U2 F' D' F  U' B' L2 D'  (22q*, 18f)
     F2 R2 L2 U  R  L  F2 U  D  L2 U  D  B2 R  L  U  B2  (17f*, 24q)

39.  F2 U  F  B  D2 F  B  R2 L2 D' B2 U' D' R2 U' D'  (22q*, 16f*)

40.  U  F  R  U  F' B2 R  F  D' R' F' L' F2 R  D  R  L2 D2  (22q*, 18f)
     U2 F  U2 R' L  F2 U' B2 R' L  D2 B' U2 D2 B  D2  (16f*, 24q)

41.  F  U2 R' L  F2 U' B2 R' L  D2 B' U2 D2 B   (20q*, 14f*)

42.  F  B  R  F2 U  D' L2 F' U  D  L  U' D  R  L' U' L2  (20q*, 17f*)

43.  F  B  R  F  B' R' L  B' U' D' R' U2 B2 U2 R  L  F  B' U   (22q*, 19f*)

44.  F  U  D  F  B' U2 R' L  B' R' L  F  B' R' U' D' F'  (18q*, 17f)
     F  R2 U  D  R  U' D  F2 R  L' D  R  L' U2 R2 F'  (16f*, 20q)

45.  F  R  L  U  B2 R2 U  D' R' L' U  D' R  U' D' L2 F'  (20q*, 17f)
     F  R2 U  D' B2 R2 L  F2 B2 L' B2 U' D  R2 F   (15f*, 22q)

46.  F  B  R  F  B' R' L  B' R2 U' D' R2 L  U2 F  B' R2 U   (22q*, 18f*)

47.  F  R' F  R2 L' B' D  R' U' B  L  F' R2 B2 L2 D' F2  (22q*, 17f*)

48.  F2 B2 U  D' F2 B2 U  D'  (12q*, 8f*)

49.  U  D  F2 B2 U  D' F2 B2 U2  (14q*, 9f*)

50.  F  B  R  F2 B2 U2 R  L  B2 R  L  U2 L  F  B  R2  (22q*, 16f)
     F  B  R  F2 R2 B2 U2 L2 F2 R2 U2 L' F  B  L2  (15f*, 24q)

51.  F  B  R' L' F  B  U  F2 B2 R2 L2 D  R  L  F' B' R  L   (22q*, 18f)
     F  B  R  F2 B2 U2 D2 L' F' B' U2 R' L' F2 R' L' D2  (17f*, 24q)

52.  U2 F2 B2 R2 L2 U2  (12q*, 6f*)

53.  F  B  R  F2 R' L' D2 F2 B2 R  L  B2 L  F  B  L2  (22q*, 16f)
     F2 U  F2 R  U2 F2 B2 R2 L2 U2 L  B2 D  B2  (14f*, 24q)

54.  F  U' F  L' F' D  F' R' L  U' D  F  R' L  D  R' D  L' F' L  U' L
                                                                  (22q*, 22f)
     F  B  R  F2 B2 U2 D2 L' U2 R2 U2 F  B  U2 R2 U2 R2  (17f*, 28q)

55.  F  U2 R' L  B2 U  F2 B2 U' F2 R' L  U2 B'  (20q*, 14f*)

56.  F  U  D  R  U' D  R' L  U  F' B  R2 F' B  R' L  U2 B'  (20q*, 18f)
     F  R  F2 R  U  L  D  B2 U' R' D  F2 B2 D2 L2 B'  (16f*, 22q)

57.  F  R  L  U  R' D' F' B' U  L  F' B' R  F' B' D' F  R  U  R   (20q*, 20f)
     F  U' L  U  L  F2 D  B2 U' D  B  D' R' B' L2 U' B  L2  (18f*, 22q)

58.  F  B  R  F  B' R' L  B' U' D' L  F  B' D2 F2 L2 F2 U'  (22q*, 18f*)

59.  F  R2 L2 F' U2 R' L  B2 D  F2 R  L' U2 B'  (20q*, 14f*)

60.  F  U  D  F  B' U2 R' L  B' R' L  F  B' R' U' D' F' U2 D2  (22q*, 19f)
     U2 F  R2 L2 F' U2 R' L  B2 D  F2 R  L' U2 B' D2  (16f*, 24q)

61.  F  B  R  F  B' R' L  B' U' D' L  F  B' D2 B2 L2 B2 U'  (22q*, 18f*)

62.  F  B  R  F  B' R' L  B' R2 U' D' L  D2 F  B' L2 U'  (20q*, 17f*)

63.  F  B  U  D  R  L  F  B  U  D  R  L   (12q*, 12f*)

64.  F  B  U  D  R  L  F' B' U' D' R' L' U2 D2  (16q*, 14f*)

65.  U  F2 R  L' F  B' U2 B2 R  L  D2 R2 D'  (18q*, 13f*)

66.  U  F2 R' L  F' B  D2 B2 R' L' U2 R2 D'  (18q*, 13f*)

67.  F  B  U2 F  B  R  L' F  B  D2 R' L' F2 U' D'  (18q*, 15f)
     F  B' R  L  F' B  R2 B2 L2 U' D' L2 F2 R2  (14f*, 20q)

68.  F  B  U  D  R  L  F  B  U  D' R2 L2 D2 R' L'  (18q*, 15f)
     F  B' R  L  F' B  R2 F2 R2 U  D  R2 B2 R2  (14f*, 20q)

69.  U  F2 R  L  F  B  R  L  U  D  F2 U'  (14q*, 12f*)

70.  U  F2 R' L' F' B' R' L' U' D' F2 U'  (14q*, 12f*)

71.  F  R' F' B  R' B' L' F  B  U  R' L  D  B  U  L  D  L  D' R'  (20q*, 20f)
     F  U  R  F' B  U2 D  F' U' D  R  F  U' D  L' U' F' R2 L2  (19f*, 22q)

72.  F  R  B  D  F  U  B  R' L  F' D' F  B' R' B  R2 D   (18q*, 17f*)

73.  U  F' U  D' F' B  D  F  B  R' U2 F' B  L2 D' R' L' U   (20q*, 18f)
     F  B' R  U2 R  D2 R' U  D' F  R2 D2 F  R' L' U2 D'  (17f*, 22q)

74.  F  R2 U' D  B2 L  U' D' F  B  R2 B  U  D' F  B' U  R' L'  (22q*, 19f*)

75.  F  U  L  U  F' R' L' U  B' L' F' U' D  B' U' D' R2 U' D'  (20q*, 19f*)

76.  U  F  R  L' F' B  L' U' D' F  D2 R  L' F2 D  R  L  U   (20q*, 18f)
     U  F  B  R' U2 F2 R' U  D' B  L2 F' R2 B' R' L  D2  (17f*, 22q)

77.  F2 R  L' U2 B  R' L' B2 D' R  L' U' D  R  F  B  D   (20q*, 17f*)

78.  F  B  U  D  R  L  F' B' U' D' R' L'  (12q*, 12f*)

79.  F  B  U  D  R  L  F  B  U  D  R  L  U2 D2  (16q*, 14f*)

80.  U  F2 U  D' F' B' R' L' F' B' D2 B2 U'  (16q*, 13f*)

81.  U  F2 U  D' F  B  R  L  F  B  U2 B2 U'  (16q*, 13f*)

82.  F  U' B' L' B' U  F  U  D' L  D  R' B' R' D' L  D2  (18q*, 17f)
     F  B' R  L  F' B  R2 F2 R2 U' D' L2 F2 L2  (14f*, 20q)

83.  F  B  U  D  R  L  F  B  U2 F2 B2 U  D' R' L'  (18q*, 15f)
     F  B' R  L  F' B  R2 B2 L2 U  D  R2 B2 L2  (14f*, 20q)

84.  U  F  B  R' L' U  D  F' B' U  D  F' B' U'  (14q*, 14f)
     U  F2 R' L' F' B' R' L' U' D' F2 U  D2  (13f*, 16q)

85.  U  F  B  R  L  U' D' F  B  U' D' F' B' U'  (14q*, 14f)
     U  F2 R  L  F  B  R  L  U  D  F2 U  D2  (13f*, 16q)

86.  F  R2 U' D  B2 L  F' B' U' R  L' U' D  R' B2 R  L  U' D'  (22q*, 19f*)

87.  F  R2 U' D  B2 L  U' D' F  B  R2 F  U' D  F' B  D  R' L'  (22q*, 19f*)

88.  U  F' R  U  L  U' R  F' U' R  U  R  L' B' R' L' F' D   (18q*, 18f)
     F  B' R  U2 R  D2 R' U  D' F  R2 D2 F  R' L' D   (16f*, 20q)

89.  F  R' F' R  U  D' F' D' F' R' U' L' B' U  L' F' D' L   (18q*, 18f)
     U  F  B  R' U2 F2 R' U  D' F  U2 F' D2 F' R' L  U2  (17f*, 22q)

90.  F  U  L  U  F' R' L' U  B' L' F' U' D  B' U' D' R2 U  D   (20q*, 19f)
     F  U  R  D' R2 U2 F  U  D  B  D2 L2 U' L  D  B  R  L   (18f*, 22q)

91.  U2 D2 L  F2 U' D  R2 B  U' D' R  L  F2 R  U  D' R' L  U  F' B' (26q*, 21f)
     F  U2 R  L  D  F2 U  R2 D  F2 D  F' B' U2 L  F2 R2 B2 U' D   (20f*, 28q)

92.  F  B' R  F2 L  B2 R' U' D  F  L2 U2 F  R' L' U   (20q*, 16f*)

93.  F  B' U2 L  F  B  L2 U  F  B' U' D  B' R' L' U  R2  (20q*, 17f*)

94.  F  B  U  D  R' L' U2 F' B' U' D' R  L  U2  (16q*, 14f*)

95.  F  B  U  D  R' L' U2 F  B  U  D  R' L' D2  (16q*, 14f*)

96.  F  B  R  L  F  B  R2 U' F2 B2 R2 L2 D' R2 U' D'  (22q*, 16f*)

97.  F  R' B  L  U  B  R  U  D  L  F  D  R  F  L' B  R2 L2  (20q*, 18f)
     F2 R  L  F  B  R  L  U  D  F2 U' F2 B2 R2 L2 D'  (16f*, 22q)

98.  F  B  U  D  R' L' F' B' U  D  R  L  U2 D2  (16q*, 14f*)

99.  F  B  U  D  R' L' F  B  U' D' R' L'  (12q*, 12f*)

100.  F  R  L  B' R' L' B  U' D' F' B' U' D' F' R' L' F  R  L  B'  (20q*, 20f)
      F  B  R  L  F  B  U  D' F2 U' F2 B2 R2 L2 D' F2 U2  (17f*, 24q)

101.  F  B  R  F' B' R' L  F  B  R  L  U  D  F  B  R  F  B  L2  (20q*, 19f)
      F2 U  F  B' U2 R  L  F  B' L2 F2 U' D' B2 D' F2  (16f*, 22q)

102.  F  U  D' F  U  D' R2 U2 L' F2 U  D' L2 F  R  L  U  B'  (22q*, 18f*)

103.  F  R  B  L  D  F' B  L' U' D  F' B2 U  D' B' D' R' B' U'  (20q*, 19f*)

104.  F  R' U' D' R  D' F' D' R  U' D' L  U' B' U' L  U' D' L' B   (20q*, 20f)
      F  B' R  D2 L  F2 B2 R' F2 R' U' D  B  U2 F2 R' L  D'  (18f*, 24q)

105.  F  B' U2 R  F  B  R2 U' D2 F  B' U' D  B' R' L' U  L2  (22q*, 18f*)

106.  F  R  U' D2 F' B  R' B' R  L' D  B  R  L' U' L' F'  (18q*, 17f*)

107.  F  U2 R' L  F2 U  R  L  F' B' D2 F' R' L  F  B' R' U  D   (22q*, 19f*)

108.  F  R  L  U' R  U2 B  R' F  L  B' L' B  L  D' F  B' U' F  D' F'
                                                                   (22q*, 21f)
      F  B' R  U  D' B2 R2 B' R  L' D  F2 D  F2 B2 U' L2 D'  (18f*, 24q)

109.  F  B' U2 L  F' B' D' F' B  U' D  B  U2 R  L  D  L2  (20q*, 17f*)

110.  F  B  U  D  R' L' U2 F  B  U  D  R' L' U2  (16q*, 14f*)

111.  F  B  U  D  R' L' U2 F' B' U' D' R  L  D2  (16q*, 14f*)

112.  F2 U  R  L' D2 F  B  R  L' B2 R2 U' D' L2 D' B2  (22q*, 16f*)

113.  F  B  R  L  F  B  R2 U' F2 B2 R2 L2 D' R2 U  D   (22q*, 16f*)

114.  F  B  U  D  R' L' F  B  U' D' R' L' U2 D2  (16q*, 14f*)

115.  F  B  U  D  R' L' F' B' U  D  R  L   (12q*, 12f*)

116.  F  B  R  F  B  R  L' F' B' R' L' U  D  F' B' L  F  B  R2  (20q*, 19f*)
      F2 U  F  B' U2 R  L  F  B' L2 F2 U  D  F2 D' B2  (16f*, 22q*)

117.  F  B  R  L  F  B  U  D' F2 U' F2 B2 R2 L2 D' F2 D2  (24q*, 17f*)

118.  F  B  R  F  B' R' L  F  L2 F  B  U' D' L  D2 F  B' L2 D   (22q*, 19f*)

119.  F  B  R2 F  B  U' F  B' D' L' U' B  R' L' D' B  L  B  D   (20q*, 19f*)

120.  F  B' U2 R  F  B  R2 U  F' B  U' D  F' R' L' U  R2  (20q*, 17f*)

121.  F  R  L' U' R2 D  B' L' B' D  B' U' F  D  B' D  R  U  R' U' D
                                                                   (22q*, 21f)
      F  B' R  U  D' F  L2 F  R2 L2 B' D2 F' R  L' U  L2 U2  (18f*, 24q)

122.  F  R  L' B  R  L' U2 R2 D' F2 R' L  D2 F  U  D  R  F'  (22q*, 18f*)

123.  F  B' U2 R  F  B  R2 D  F  B' U  D' F2 B  R' L' U  R2  (22q*, 18f*)

124.  F  R' D' R  F  R' L2 U  B  U' R' U  R' L' U' D' F' B' D  F2  (22q*, 20f)
      F  B  R' F2 U2 R' U' D  B  R2 B' U2 F' U2 D2 R  L' D   (18f*, 24q)


as usual, when there are two maneuvers are given, this means that no
maneuver is simultaneously minimal in both metrics.  superflip composed
with four spot is #91. many of these are locally maximal.

local maxima in the quarter turn metric:

#1, 4, 8, 9, 10, 15, 20, 21, 24, 25, 27, 29, 30, 31, 33, 34, 38,
39, 40, 41, 42, 43, 46, 47, 48, 50, 51, 53, 55, 58, 60, 61, 62,
64, 67, 68, 71, 72, 73, 75, 76, 77, 79, 82, 83, 86, 90, 91, 92,
93, 94, 95, 96, 98, 102, 103, 105, 107, 108, 110, 111, 112, 113,
114, 117, 118, 119, 121, 122, 123, 124.

(strong) local maxima in the face turn metric:

#8, 10, 11, 16, 29, 30, 31, 34, 38, 39, 43, 46, 51, 54, 58, 61, 64,
71, 72, 75, 77, 79, 86, 91, 93, 94, 95, 98, 100, 103, 104, 107, 110,
111, 114, 117, 118, 119, 121.

mike

From cube-lovers-errors@mc.lcs.mit.edu  Wed Aug 19 22:57:36 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Aug 19 17:10:15 1998
Date: Wed, 19 Aug 1998 17:09:54 -0400 (Eastern Daylight Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: minimal maneuvers for X symmetric positions
In-Reply-To: <199808182028.QAA24353@euclid.math.brown.edu>
To: michael reid <reid@math.brown.edu>
Cc: cube-lovers@ai.mit.edu
Message-Id: <SIMEON.9808191754.D@GN209A.PSTCC.CC.TN.US>


On Tue, 18 Aug 1998 16:28:28 -0400 michael reid
<reid@math.brown.edu> wrote:

> X  is the subgroup of the cube symmetry group which preserves
> the U-D axis.  there are 128 positions which have X symmetry:
>
>      the UR edge can go in any of the 8 positions  UR, RU, DR,
>      RD, UL, LU, DL, LD; this determines the location of the
>      edges UB, UL, UF, DR, DB, DL, DF.
>
>      the FR edge can go in any of the 4 positions FR, RF, BL, LB;
>      this determines the location of the edges FL, BR, BL.
>
>      the UFR corner can go in any of the 4 positions UFR, UBL,
>      DRF, DLB; this determines the location of all the corners.
>
> any combination of these is possible, which gives 128 positions.
> 4 of the positions have more symmetry, namely M symmetry.
> (these positions are  start, superflip, pons asinorum, and
> pons asinorum composed with superflip.)
>
> minimal maneuvers for the other positions are:
>
> 1.  F2 R  L' D2 F2 D2 R' L  F2 D2  (16q*, 10f*)
>
> 2.  U  F' B' R2 U' D' F' B  U  D  R2 F  B  D'  (16q*, 14f)
>     U  F2 U2 F2 R  L  F2 U2 F2 U2 R' L' U   (13f*, 20q)
>

I don't think Mike has said so explicitly, but he appears to
have adopted a very useful convention from Herbert Kociemba's
Cube Explorer 1.5.  To wit, Cube Explorer 1.5 flags the length
of a maneuver with an asterisk when the length has been shown to
be minimal. Cube Explorer 1.5 operates only in face turns, so it
omits the q or f designation of units.

But for example, Cube Explorer 1.5 might show the length of a
cube upon which it is operating as (13) meaning 13f, then later
in the search show the length as (12), and still later show the
length as (12)* to show that 12 face moves have been proven to
be minimal. The only difference between Mike's style and Cube
Explorer's style is that Cube Explorer puts the asterisk outside
the parentheses.

I loaded Mike's E-mail into Cube Explorer to take a quick look
at the X symmetric positions.  Many of them are familiar to
readers of this list, and all of them are quite pretty.
(Loading Mike's E-mail into Cube Explorer "just worked".  I
didn't have to edit it at all to remove extraneous text.  Cube
Explorer's maneuver reader seems to have a remarkable ability to
extract maneuvers in BFUDLR notation which are imbedded in other
extraneous text.)  If you have Cube Explorer 1.5 (and you
should!), I would encourage you similarly to load Mike's X
symmetric patterns into it and take a look.

The patterns look as expected for patterns which preserve the
U-D axis. The U and D faces are the same pattern.  The F, R, B,
and L faces are the same pattern and may be described as
being in the same orientation with respect to rotations of the
square.

For positions #1 through #62, the U and D faces may be described
as being symmetric with respect to the symmetries of the square.
They range from being solid, to having one dot, to being a +, to
being an X, etc.  All are glyphs.

Positions #63 through #124 are essentially the first 62
positions composed with superflip.  I had never noticed it, and
I don't *think* it has been described on the list, but for every
symmetry group, half of the corresponding positions can be
described as "basic" positions and the other half can be
described as the basic positions composed with superflip.  That
is, if Symm(x)=K, then Symm(xf)=Symm(fx)=K, where x is any
position and f is the superflip.

----------------------
Jerry Bryan
jbryan@pstcc.cc.tn.us

From cube-lovers-errors@mc.lcs.mit.edu  Wed Aug 19 23:44:32 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Aug 19 23:34:51 1998
Date: Wed, 19 Aug 98 23:34:34 EDT
Message-Id: <19Aug1998.231259.Hoey@AIC.NRL.Navy.Mil>
From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: jbryan@pstcc.cc.tn.us
Cc: reid@math.brown.edu, cube-lovers@ai.mit.edu
In-Reply-To: <SIMEON.9808191754.D@GN209A.PSTCC.CC.TN.US> (message from Jerry
	Bryan on Wed, 19 Aug 1998 17:09:54 -0400 (Eastern Daylight Time))
Subject: Re: minimal maneuvers for X symmetric positions

Jerry Bryan <jbryan@pstcc.cc.tn.us> writes:

> Positions #63 through #124 are essentially the first 62 
> positions composed with superflip.  I had never noticed it, and 
> I don't *think* it has been described on the list, but for every 
> symmetry group, half of the corresponding positions can be 
> described as "basic" positions and the other half can be 
> described as the basic positions composed with superflip.  That 
> is, if Symm(x)=K, then Symm(xf)=Symm(fx)=K, where x is any 
> position and f is the superflip.

This is easy to see if we consider that  Symm(x)  is the set of all
m  in  M  that commute with  x, because

         m' x m = x     if and only if     x m = m x.

At some times since 1981 I've wondered whether symmetry discussions
are better done with commutativity rather than conjugacy.

So if  c  is any element of the center of G* -- i.e.,  c  commutes
with all elements of  M  and  G  -- then  Symm(x)=Symm(x c).  As is
well known to cube-lovers, the center of the usual cube group consists
of the identity and the superflip.  In the supergroup, we may also
compose these with Big Ben (all face centers rotated 90 degrees) and
Noon (Big Ben squared).

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From cube-lovers-errors@mc.lcs.mit.edu  Thu Aug 20 14:36:11 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Thu Aug 20 11:37:31 1998
Date: Thu, 20 Aug 1998 11:37:14 -0400 (Eastern Daylight Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: minimal maneuvers for X symmetric positions
In-Reply-To: <199808182028.QAA24353@euclid.math.brown.edu>
To: michael reid <reid@math.brown.edu>
Cc: cube-lovers@ai.mit.edu
Message-Id: <SIMEON.9808201114.B@GN209A.PSTCC.CC.TN.US>

On Tue, 18 Aug 1998 16:28:28 -0400 michael reid
<reid@math.brown.edu> wrote:

> local maxima in the quarter turn metric:
>
> #1, 4, 8, 9, 10, 15, 20, 21, 24, 25, 27, 29, 30, 31, 33, 34, 38,
> 39, 40, 41, 42, 43, 46, 47, 48, 50, 51, 53, 55, 58, 60, 61, 62,
> 64, 67, 68, 71, 72, 73, 75, 76, 77, 79, 82, 83, 86, 90, 91, 92,
> 93, 94, 95, 96, 98, 102, 103, 105, 107, 108, 110, 111, 112, 113,
> 114, 117, 118, 119, 121, 122, 123, 124.
>
> (strong) local maxima in the face turn metric:
>
> #8, 10, 11, 16, 29, 30, 31, 34, 38, 39, 43, 46, 51, 54, 58, 61, 64,
> 71, 72, 75, 77, 79, 86, 91, 93, 94, 95, 98, 100, 103, 104, 107, 110,
> 111, 114, 117, 118, 119, 121.


I am curious how the local maxima were determined.  4-spot
composed with superflip was based on sort of an "extended
symmetry" argument, but what about all the others?

If I had to guess, I would suspect that you found all minimal
maneuvers for each position and observed that there was a
maneuver terminating with each quarter (respectively, face)
turn for each position.  Or equivalently, perhaps you found all
minimal maneuvers unique to symmetry for each position and
observed that conjugation of the maneuvers would yield a
maneuver terminating with each required kind of turn.  Was it
something like this?

(All you would really need for the conjugation argument, since
you already know that the maneuvers in question preserve the U-D
axis, would be to find at least one minimal maneuver ending with
any of {U, U', D, D'} and to find another minimal maneuver ending
with any of {R, R', F, F', L, L', B, B'}.)

It is interesting that you found strong local maxima in the face
turn metric, rather than just "plain" local maxima.  In my
experience, finding strong local maxima with a computer search
is easier than finding "plain" local maxima.  Finding "plain"
local maxima includes finding weak local maxima (where at least
one face turn does not change the distance of the position from
Start).  If my guess about how you are identifying local maxima
is correct, then your method would not identify weak local
maxima.

Finally, I have mused previously to Cube-Lovers that strong
local maxima in the face turn metric may be extremely rare.  I
think I might be wrong.  My God's algorithm searches in the face
turn metric have already turned up more strong local maxima than
I expected, and your search of the X-symmetric positions turned
up more strong local maxima than I would have expected.

----------------------
Jerry Bryan
jbryan@pstcc.cc.tn.us

From cube-lovers-errors@mc.lcs.mit.edu  Thu Aug 20 18:08:04 1998
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Date: Thu, 20 Aug 98 18:06:30 EDT
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From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: jbryan@pstcc.cc.tn.us
Cc: reid@math.brown.edu, cube-lovers@ai.mit.edu
Subject: Group centers (oops)

I wrote:

> So if  c  is any element of the center of G* -- i.e.,  c  commutes
> with all elements of  M  and  G  -- then  Symm(x)=Symm(x c).  As is
> well known to cube-lovers, the center of the usual cube group consists
> of the identity and the superflip.  In the supergroup, we may also
> compose these with Big Ben (all face centers rotated 90 degrees) and
> Noon (Big Ben squared).

In short, I should not have included Big Ben in that paragraph, only Noon.

The long explanation is that both of these are in the center of the
usual supergroup, as is any position that differs from Solved only by
face center orientation.  But for the  Symm(x)=Symm(x c)  argument to
work,  c  must be in the center of the group generated by the union of
the supergroup with  M.  This is equivalent to saying that  c  must be
in the center of the supergroup and be M-symmetric.  Big Ben is only
C-symmetric.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From cube-lovers-errors@mc.lcs.mit.edu  Fri Aug 21 23:33:56 1998
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Date: Fri, 21 Aug 1998 23:07:24 -0400
From: michael reid <reid@math.brown.edu>
Message-Id: <199808220307.XAA10899@cauchy.math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: minimal maneuvers for E symmetric positions

E  is the subgroup of cube symmetries consisting of rotations
(no reflections) that preserve the tetrad of corners  UFR,
UBL, DFL  and  DBR.  of course it preserves the other tetrad as
well.  there are 72 positions that have E symmetry:

     each corner must remain in place, but can be twisted.
     corners in the same tetrad must be twisted in the same
     direction; therefore, by conservation of twist, adjacent
     corners are twisted in opposite directions.

     the UR edge can go in any location in any orientation.
     this determines the location and orientation of all edges.

this gives 3 * 24 = 72 positions.  if the UR edge remains in the
F-B slice, then the position has more symmetry, namely H symmetry
(at least).  this accounts for 24 of the 72 positions; 20 of which
are H symmetric, and 4 of which are M symmetric.

E is a normal subgroup of M; in fact, it's the commutator subgroup.
therefore, any M conjugate of an E symmetric position is also E
symmetric.  the 48 remaining positions form 12 equivalence classes
under M conjugacy, of 4 positions each.  minimal maneuvers for these
are

1.  F' B' U  R' U' D  R  L' U  R' D' L  U' D  R  L' D' L  F' B'  (20q*, 20f)
    F  R  L  F  U' D' F  R2 L2 U2 D2 B  R' L' B  U  D  B   (18f*, 22q)

2.  F' R2 U  D  R' L' U' F  B  R2 F' U' D' F' B  U' D' B'  (20q*, 18f*)

3.  F' B' R' L' F  B  U  D  R' L' U' D'  (12q*, 12f*)

4.  F  U2 R  F  B  R' U2 B  R  L  B  U' D' R' L  U' L2  (20q*, 17f*)

5.  F  R' F  L  U  D' L' U  R  U' D  F' D' B  U  D' B' R' F  R'  (20q*, 20f)
    F2 U  D2 L2 F  U  D  F' L2 U  F  B  U  R' L' F' B  R   (18f*, 22q)

6.  F  B  R' L' F  B  U' D' R  L  U' D'  (12q*, 12f*)

7.  F  R  U2 F  L  B  U  D' F  L  D  F' B  D  R  B  D   (18q*, 17f*)

8.  F  U2 F' B  D  B' L  B  L  D' B2 R' D  F  D  F' R  D2 F'  (22q*, 19f)
    F' R2 D2 F  U' D  F2 R  B2 L  F2 R  D2 F  B' U' F' B'  (18f*, 24q)

9.  F' R  F  U  R2 U  F' L' D  F  U2 R' F' B2 R' D' R' B' U'  (22q*, 19f*)

10.  F  R2 B' L' U' L' F' B  D  B  R  B  R  F  U' R  U2 D2 F'  (22q*, 19f*)

11.  F  B  R2 U' F  L  F  U' F  B' R  F' U' B' R  L' B' R  L   (20q*, 19f)
     F  R2 U  D' F2 L' U  D  L2 B  U  D' F  B' U' D2 F  B   (18f*, 22q)

12.  F  R' D  B  U  D  F' L  B  D  R' F  R' L  B' U' F' B' L  F' D  L
                                                                   (22q*, 22f)
     F  B  U  F  L2 D2 F2 B  R  F  B  R  F' L2 D2 R2 U2 B  U   (19f*, 26q)

as usual, i give a maneuver that is simultaneously minimal in both
metrics, unless one does not exist.  some of these are local maxima.

local maxima in the quarter turn metric: #1, 4, 7, 8, 9, 10, 11, 12.

(strong) local maxima in the face turn metric: #10, 11, 12.

mike

From cube-lovers-errors@mc.lcs.mit.edu  Sun Aug 23 01:09:13 1998
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Date: Sat, 22 Aug 1998 19:27:50 -0400
From: michael reid <reid@math.brown.edu>
Message-Id: <199808222327.TAA13228@cauchy.math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: Re: minimal maneuvers for X symmetric positions

jerry asks

> I am curious how the local maxima were determined.  4-spot
> composed with superflip was based on sort of an "extended
> symmetry" argument, but what about all the others?
>
> If I had to guess, I would suspect that you found all minimal
> maneuvers for each position and observed that there was a
> maneuver terminating with each quarter (respectively, face)
> turn for each position.  Or equivalently, perhaps you found all
> minimal maneuvers unique to symmetry for each position and
> observed that conjugation of the maneuvers would yield a
> maneuver terminating with each required kind of turn.  Was it
> something like this?

yes, this is essentially what i did.  i added automatic symmetry
reduction to my program (this was a challenge to program, but it
makes things so much more convenient).  so now the program finds
all minimal maneuvers up to symmetry, from which local maxima can
be spotted easily.

i did not find all minimal maneuvers for #91 (superflip composed
with four spot) nor for #117 in the quarter turn metric, because
these are too far from start (26q, 24q respectively).  so for these
positions, which are locally maximal, it suffices to find minimal
maneuvers ending with each quarter turn.  as you see, symmetry is
helpful here.  also, all the X symmetric positions have order 2,
so any maneuver can be inverted.  this is also helpful.

> It is interesting that you found strong local maxima in the face
> turn metric, rather than just "plain" local maxima.  In my
> experience, finding strong local maxima with a computer search
> is easier than finding "plain" local maxima.  Finding "plain"
> local maxima includes finding weak local maxima (where at least
> one face turn does not change the distance of the position from
> Start).  If my guess about how you are identifying local maxima
> is correct, then your method would not identify weak local
> maxima.

yes, this is exactly correct.  i will leave it to someone who's more
interested in "weak" local maxima to determine those.

mike

From cube-lovers-errors@mc.lcs.mit.edu  Mon Aug 24 14:00:58 1998
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From: Noel Dillabough <Noel.Dillabough@ineacorp.com>
To: "Cube-Lovers (E-mail)" <cube-lovers@ai.mit.edu>
Subject: New version of Puzzler
Date: Sun, 23 Aug 1998 14:30:24 -0400

The latest version of Puzzler can be found at
http://www.mud.ca/puzzler/puzzler.html.  In addition to the bug fixes
that were put in, I have added the most requested features, the ability
to take back moves, and the ability to enter move macros for cube
puzzles in standard cubist (UDFBLR) notation.

For those who haven't used the program before, Puzzler is a collection
of sequential movement puzzles including the cubes (all sizes from 2x2x2
to 5x5x5), the pyraminx, the impossiball, the skewb, and the megaminx
puzzle.

Enjoy,

-Noel

From cube-lovers-errors@mc.lcs.mit.edu  Mon Aug 24 14:37:46 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sat Aug 22 20:18:25 1998
Message-Id: <35DF587E.6C0D@ameritech.net>
Date: Sat, 22 Aug 1998 18:47:10 -0500
From: Hana Bizek <hbizek@ameritech.net>
Reply-To: hbizek@ameritech.net
To: cube-lovers@ai.mit.edu
Subject: designs from Rubik's cubes

Hi, fellow cube-lovers
I am a recent member of the cube-lovers mailing list. I would like to
help me answer this question: has any of you constructed, or does anyone
of you know someone who has constructed, a composite, pleasant,
geometrical design on a set of Rubik's cubes? The design is a three
dimensional {but not necessarily cubical} structure that exhibits some
symmetry on all its faces. Such designs are quite different from the
picture-like structures built by Jacob Davenport. I saw his pictures
when I consucted an ongoing web search to answer the above question. So
far I was not successful, and so I seek your help.
I am the author of these designs. I know they can be done because I have
done them. I do not have a web page of my own yet, but a friend of mine
kindly offered to put three of these designs on his web page. They may
be seen at http://www.ssie.binghamton.edu/~jirif/cube.html. This should
open my friend's speed cubing page. My designs are there under the
heading "Hana Bizek's cube art."  Yes, Jirka Fridrich is a speed cubist,
which is an art in itself. You will find other interesting things there,
including a signature of Erno Rubik.
A photo of a design has one flaw; you can only see three faces of the
design. What does the rest of the design look like? Answer: sometimes
opposite faces of the design are exactly identical, both in color and
geometrical pattern. One of my designs in Jirka's page, the so-called
ctyrsprezi design, is such a design. It has four colors only on its six
faces. Why this should be so is a cornerstone of the design theory. The
reason is explained in my book,"Mathematics of the Rubik's cube design,"
published last year. amazon.com has it online.
Well, O better end this message, or it will itself deberate into a book.
Any help you can offer in my search for a "cube sculptor" will be
gratefully appreciated. And of course I stand by these designs and
will answer any questions. My name is Hana M. Bizek and my email address
is hbizek@ameritech.net. Thank you very much.
I will be looking forward to hearing from you.  Best regards,
Hana

From cube-lovers-errors@mc.lcs.mit.edu  Mon Aug 24 15:32:46 1998
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Date: Sun, 23 Aug 1998 23:19:35 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: "Basic" vs. Superflipped Positions
To: Cube-Lovers <cube-lovers@ai.mit.edu>
Message-Id: <Pine.PMDF.3.95.980823231032.2685A-100000@PSTCC6.PSTCC.CC.TN.US>

I recently commented on the fact that half of Mike Reid's X-symmetric
positions were "basic" and the other half were superflipped versions of
the first half.  I further commented that this was true for all positions
associated with any symmetry group -- half are "basic", and the other half
are superflipped versions of the first half.  Well, I am not quite sure
that this is true in general.  Or more correctly, I am not sure you can
always tell the "basic" version apart from the superflipped version.

Consider any two positions x and xf, where f is the superflip.  We would
say that x is the "basic" position and xf is the superflipped position.
But if we define y=xf, then y is the "basic" position and yf is the
superflipped position, and it is also true that x=yf.

So which is the "basic" position, x or y? It appears that there is no way
to tell.  Yet when you look at X-symmetric positions, it is trivial for
the eye to see which ones are "basic", and which ones are superflipped.
So what is going on here?

I briefly (*very* briefly) hoped to find a unique subgroup H of G with
index 2 which did not contain superflip.  Then, it would have been natural
to call H the "basic" positions and Hf the superflipped positions.  But it
is well known to Cube-Lovers that the only subgroup of G with index 2 is
the subgroup consisting of those positions where are an even number of
quarter turns from Start.  And this subgroup does contain the superflip.
Therefore, there seems to me to be little possibility of a general way to
distinguish between "basic" positions and superflipped positions.

Upon further reflection, it seems to me that there is a natural way to
tell "basic" positions apart from superflipped positions for some symmetry
groups but not for others.  I have not examined all 98 symmetry groups (33
symmetry classes) of the cube in this regard, but I have looked at a few
of them, and can give a few examples.

Before looking at examples, we need to look at a subtle but important
point.  We may think of a position x as consisting of corners and edges
separately, so that x=x[c]*x[e].  Similarly, we may look at the symmetry
of the corners and the edges separately, as in Symm(x[c]) and Symm(x[e]).
The equation that relates the symmetries is
Symm(x)=Symm(x[c]*x[e])=Symm(x[c]) intersect Symm(x[e]).  But because
Superflip affects only the edges, we need consider only Symm(x[e]) when we
compare "basic" positions to superflipped positions.

Example 1.  Suppose Symm(x[e])=M.  Then it seems natural to view the
position as "basic" if all four edge facelets on each face are the same
color, and to view the position as superflipped otherwise. The eye sees
this distintion very clearly.

Example 2.  Suppose Symm(x[e])=X1.  X1 is the symmetry group in Dan Hoey's
taxonomy which preserves the U-D axis.  X2 and X3 are conjugate subgroups
preserving the F-B and R-L axes, respectively.  X is the symmetry class
consisting of X1, X2, and X3.  All of Mike Reid's X-symmetric positions
are in particular X1-symmetric.  For X1, it seems natural to view the
position as "basic" if all four edge facelets on the U and D faces are the
same color, and to view the position as superflipped otherwise.  For X2,
the same rule would apply to the F and B faces.  FOr X3, the same rule
would apply to the R and L faces.  The eye sees this distinction very
clearly.

Example 3.  Suppose Symm(x[e])={i,v}, where v is the central inversion.
For such a position, any particular edge cubie could be placed anywhere,
but each edge cubie would have to be placed diametrically opposite its
diametrically opposed edge cubie.  For example, if cubie uf were placed in
the rd cubicle, then cubie db would have to be placed in the lu cubicle,
etc.  Also, for Symm(x[e]) to be {i,v} the edges could not have any
additional symmetry.  In this case, I don't think there is any natural way
to distinguish between a "basic" position and a superflipped position.

Example 4. Suppose Symm(x[e])=I={i}.  In other words, the edges have no
symmetry.  In this case, I don't think there is any natural way to
distinguish between a "basic" position and a superflipped position.

Example 5. Suppose Symm(x[e])={i,c_u2}, where c_u2 is a 180 degree whole
cube rotation around the U-D axis.  In this case, the position would be
"basic" if opposite edge facelets on the U face were the same color and if
opposite edge facelets on the D face were the same color, and would be
superflipped otherwise.  The eye sees this distinction very clearly.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990

From cube-lovers-errors@mc.lcs.mit.edu  Mon Aug 24 17:55:26 1998
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Date: Mon, 24 Aug 1998 10:05:52 -0400 (Eastern Daylight Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: minimal maneuvers for E symmetric positions
In-Reply-To: <199808220307.XAA10899@cauchy.math.brown.edu>
To: michael reid <reid@math.brown.edu>
Cc: cube-lovers@ai.mit.edu
Message-Id: <SIMEON.9808241052.A@GN209A.PSTCC.CC.TN.US>

On Fri, 21 Aug 1998 23:07:24 -0400 michael reid
<reid@math.brown.edu> wrote:

> E  is the subgroup of cube symmetries consisting of rotations
> (no reflections) that preserve the tetrad of corners  UFR,
> UBL, DFL  and  DBR.  of course it preserves the other tetrad as
> well.  there are 72 positions that have E symmetry:
>
>      each corner must remain in place, but can be twisted.
>      corners in the same tetrad must be twisted in the same
>      direction; therefore, by conservation of twist, adjacent
>      corners are twisted in opposite directions.
>
>      the UR edge can go in any location in any orientation.
>      this determines the location and orientation of all edges.
>

There are generally several different (equivalent) ways to
characterize a subgroup of the cube symmetries.  For example, of
the 48 symmetries, 24 of them are even and 24 of them are odd,
and 24 of them are rotations and 24 of them are reflections.
The E symmetries may be characterized as the intersection of the
even symmetries with the rotational symmetries, and hence
consist of the 12 even rotations.

The 12 even rotations consist of the identity, the three 180
degree rotations around the face axes (c_u2 around the U-D axis,
c_f2 around the F-B axis, and c_r2 around the R-L axis), and the
eight 120 degree rotations around the four major diagonal axes
(c_urf and c_ufr; c_ufl and c_ulf; c_ulb and c_ubl; and c_ubr
and c_urb).  It is the eight major axis rotations which give E
its tetradic nature.

In addition to the characterizations of the E positions which
Mike gave (the corners must stay home, perhaps twisted, etc.),
we can describe the E positions informally by the appearance of
the faces.  Each face must have the same pattern as its opposite
face, and each pattern must have the 180 rotational symmetry of
the square.

The hardest part (to me, at least) in thinking about what a
position x with Symm(x)=E must look like is to subtract out or
ignore those positions which are E-symmetric but which have more
symmetry.  Indeed, many of the Symm(x)=E positions look very
much like slightly broken versions of positions with stronger
symmetry.  For example, #3 and #6 look like slightly broken
6-spots.  #7, #10, and #12 look like slightly broken 6-H's.
#1, #2, and #4 look like slightly broken Pons Asinorums.  Etc.

This visual effect is the strongest if your cube adopts the
"opposite faces differ by yellow" convention, so that white is
opposite yellow, green is opposite blue, and red is opposite
orange.  Your eye will then tend to identify white with yellow,
green with blue, and red with orange.  With these
identifications having taken place, most (if not all) of the
Symm(x)=E positions look exactly like positions with more
symmetry.

----------------------
Jerry Bryan
jbryan@pstcc.cc.tn.us

From cube-lovers-errors@mc.lcs.mit.edu  Wed Aug 26 12:59:48 1998
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Date: Wed, 26 Aug 1998 13:20:41 +0100
From: David Singmaster <david.singmaster@sbu.ac.uk>
Organization: Computing & Maths South Bank Univ
To: cube-lovers@ai.mit.edu
Message-Id: <009CB47C.C9309B92.16@ice.sbu.ac.uk>
Subject: depicting a cube

	With reference to Hana Bizek's reference to how one can show all six
faces of a cube, I found the following the most satisfactory.  View the F, U
and R faces along the diagonal.  Now imagine the back faces 'exploded' out,
i.e. moved outward along the axes.  When they are moved far enough, they can be
seen.  The effect is that the cube seems to be suspended in front of a corner
and the three back seem to have been projected onto the walls and floor.  I'll
try to make a drawing.

	   /|           |\
	  / |    / \    | \
	 /  |   /   \   |  \
	|  /   /     \   \  |
	| /   |\     /|   \ |
	|/    | \   / |    \|
              |  \ /  |
               \  |  /
                \ | /
                 \|/

                 / \
                /   \
               /     \
               \     /
                \   /
                 \ /

This is a bit crude, but it may be better when printed?  if one puts in more
horizontal space, it might look better.

                   /|               |\
                 /  |      / \      |  \
               /    |    /     \    |    \
              |    /   /         \   \    |
              |  /    |\         /|    \  |
              |/      |  \     /  |      \|
                      |    \ /    |
                       \    |    /
                         \  |  /
                           \|/

                           / \
                         /     \
                       /         \
                       \         /
                         \     /
                           \ /

Well, that's a bit better, but one can't get it perfect on an orthogonal grid.



DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499; 
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Thu Aug 27 21:08:23 1998
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Date: Thu, 27 Aug 1998 17:56:00 -0500
From: Hana Bizek <hbizek@ameritech.net>
Reply-To: hbizek@ameritech.net
To: cube-lovers@ai.mit.edu
Subject: Re: depicting a cube
References: <009CB47C.C9309B92.16@ice.sbu.ac.uk>

David Singmaster wrote:

>         With reference to Hana Bizek's reference to how one can show
> all six faces of a cube, I found the following the most
> satisfactory.  View the F, U and R faces along the diagonal.  Now
> imagine the back faces 'exploded' out, i.e. moved outward along the
> axes.  When they are moved far enough, they can be seen.  The effect
> is that the cube seems to be suspended in front of a corner and the
> three back seem to have been projected onto the walls and floor.


A mirror can be placed on those walls and floor so that the design's B,
L and D faces can be reflected off those surfaces. The design would need
to stand on a glass-topped table, so that the D face can be reflected
off the  mirrorred floor. The whole setup could be photographed.

Unfortunately, I do not have resources to implement this. I don't even
own a glass-topped table!

Here is a challenge for the programmers out there. Can you write an
applet that will slowly rotate my design in order for a viewer to see F,
B, R, L and U faces, then tilt it upward to expose the D face? Do these
moves any way you want, just make sure a viewer can see it all.Thank tou
very much.

You can find three of my designs at
http://www.ssie.binghamton.edu/~jirif/cube.html. Two designs there are
cubical. Opposite faces are identical, both in color and geometrical
pattern. To wit: e. g. F face is exactly identical to B face, etc. This
property holds for a majority of these designs, but there are
exceptions.

Hana Bizek {hbizek@ameritech.net}

From cube-lovers-errors@mc.lcs.mit.edu  Thu Aug 27 22:49:16 1998
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Message-Id: <006a01bdd226$d0495920$551a2bcb@mercury>
Reply-To: "Bill Webster" <Haddock@bluep.com>
From: "Bill Webster" <Haddock@bluep.com>
To: <cube-lovers@ai.mit.edu>
Subject: Re: Depicting a cube
Date: Fri, 28 Aug 1998 11:54:36 +1000

Hana Bizek wrote:

>Unfortunately, I do not have resources to implement this. I don't even
>own a glass-topped table!

>Here is a challenge for the programmers out there. Can you write an
>applet that will slowly rotate my design in order for a viewer to see F,
>B, R, L and U faces, then tilt it upward to expose the D face? Do these
>moves any way you want, just make sure a viewer can see it all.Thank you
>very much.


If static, generated images are acceptable, (i.e. if the pattern is more
important to your sculpture than its physical realisation in plastic),
you could achieve this reasonably easily with a ray-tracer - build the
cube model and situate it in a scene with three plane mirrors, or
perhaps models of some other reflecting objects for enhanced aesthetics
- perhaps even a glass topped table! A free ray tracer is available at
www.povray.org I have some source and sample images for cube models if
you are interested.

Cheers, Bill Webster (haddock@bluep.com)

From cube-lovers-errors@mc.lcs.mit.edu  Fri Aug 28 07:25:33 1998
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Message-Id: <3.0.32.19980828092911.009738f0@mail.spc.nl>
Date: Fri, 28 Aug 1998 09:29:12 +0200
To: cube-lovers@ai.mit.edu
From: Christ van Willegen <c.v.willegen@spcgroup.nl>
Subject: Re: Depicting a cube

Bill Webster (haddock@bluep.com) wrote:

>If static, generated images are acceptable, (i.e. if the pattern is more
>important to your sculpture than its physical realisation in plastic),
>you could achieve this reasonably easily with a ray-tracer....

Hi,

my first mailing on this list. I've seen some stuff, and thought it
too hard (at the moment). _This_ discussion, however, I can handle!


Another way to do this is in VRML. It's quite easy to build up a model
of a cube in 3D, including colors.

For a sample of what can be done with VRML in combination with a computer
program to generate the stuff, go to:

http://www.iaehv.nl/users/richtofe/

Follow the link about 'Triplets'. These are 3D models, inspired by Douglas
Hofstadter. I have included some examples on that page, as well.

It wouldn't be hard to do the same for a cube model. Writing a text file
with LRUDTB and ' in it to describe the model, then and generating VRML is
not too hard.

This has the advantage over PovRAY that you can really rotate the model in
space, and look at it in all directions.

The mirror idea is nice, but it will mess up the design (Left-Right swap).

More thoughts/ideas?

Christ van Willegen

From cube-lovers-errors@mc.lcs.mit.edu  Mon Aug 31 17:11:38 1998
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Date: Sat, 29 Aug 1998 20:25:01 -0500
From: Hana Bizek <hbizek@ameritech.net>
Reply-To: hbizek@ameritech.net
To: cube-lovers@ai.mit.edu
Subject: Re: Depicting a cube
References: <006a01bdd226$d0495920$551a2bcb@mercury>

Bill Webster wrote:

>
> If static, generated images are acceptable, (i.e. if the pattern is more
> important to your sculpture than its physical realisation in plastic),


Oh no! Please remember that those are Rubik's cubes. Their "physical
realization" is usually that they are stacked together to form larger
cubes. If *that* was all to the design problem, I wouldn't have the
nerve to make a posting to the cube-lovers msiling list. Some of its
members are first-class mathematicians.

I feel that an explanation of what I call the design problem is in
order. Ther goal of this problem is to create, by conventional cube
manipulation, a composite pleasant geometrical design on a set of
Eubik's cubes. The basic algorithm consists of three simple steps:
1} construct patterns on individual cubes
2} make sure that the colors match properly from cube to cube {color
control}
3} stack the cubes together.
You start with a set of solved cubes. If you have scrambled cubes, you
need to solve them. That is just one excellent reason why you *must*
solve the Rubik's cube comopletely. Being able to solve only one side is
woefully inadequate.

Don't  forget color control. Without it you don't have a design. This
unavoidable aspect of the design problem further complicates the design
algorithm. It is a little bit like chess. You try to consider two or
three moves ahead of your opponent to achieve a winning strategy..,. or
create a viable design from a set of Rubik's cubes.

The last step is easy. It is sort of like a three-dimensional jigsaw
puzzle. The patterned cubes you constructed are part of this jigsaw.

Here, in a nutshell, is a  description of the design problem. Please,
get our your Rubik's cubes and start twiddling.

Good luck,  Hana

From cube-lovers-errors@mc.lcs.mit.edu  Tue Sep  1 10:26:49 1998
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Date: Tue, 01 Sep 1998 09:59:50 -0400 (Eastern Daylight Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Strong Local Maxima 9f and 10f from Start
To: Cube Lovers <cube-lovers@ai.mit.edu>
Message-Id: <SIMEON.9809010950.C@GN209A.PSTCC.CC.TN.US>

#1. D2 L2 B2 F2 U2 B2 F2 U2 R2 U2  (10f*)
#2. D2 F2 L2 D' U  L2 F2 D' U'     (9f*)
#3. U2 B2 L2 D  U' R2 B2 D' U'     (9f*)
#4. B2 D' U' L2 B2 L2 D' U' B' F'  (10f*)
#5. L2 U2 F2 L2 D2 F2 U2 R2 B' F'  (10f*)
#6. D' U' B2 R2 D2 L2 D  U  B' F'  (10f*)
#7. D  U  L2 D2 R2 F2 D' U' B' F'  (10f*)
#8. B2 F2 D  U' B' F  L  R' D  U'  (10f*)

This completes the list of strong local maxima 9f and 10f from
Start in the face turn metric.  I posted #1, #2, and #3
previously, but the rest are new.  9f is the shortest strong
local maximum.

I continue to think that all eight of these positions share a
special kind of symmetry that is related to the fact that they
are strong local maxima, but I can't quite get my arms around a
good description for this symmetry.  Generally speaking, they
look more symmetric if you look at corner cubies and edge cubies
separately than if you look at them in combination.  Also, they
look more symmetric if you look only at the colors of the
facelets (looking at two dimensional 3x3 faces) rather than
if you look at the location of entire cubies.

They do all share the following in common.  Looking just at the
colors of the facelets, all pairs of opposed 3x3 faces have the
same pattern for all eight positions. Hence, there are (up to)
three different face patterns for each position.  Also, if the
cube is colored according the "opposite faces differ by yellow"
convention, then the pairs of opposed face patterns for all
eight positions are the "yellow complements" of each other.
Finally, all the face patterns (and some of them are fairly
complicated, having as many as four colors) are symmetric with
respect a reflection across either a vertical or horizontal axis
of the 3x3 square making up the face.

Even though none of these strong local maxima are q-transitive
in the classic Saxe-Hoey sense, the "face symmetry" they all
share seems too unusual to me to be just a coincidence.

I think #8 is an especially interesting position.  All six faces
have the same face pattern, sort of a three colored checkerboard
(if that is not a contradiction in terms).  The position is
basically Pons Asinorum with the edge and corner cubies rotated
as a unit along a major diagonal axis relative to the fixed face
centers.

----------------------
Jerry Bryan
jbryan@pstcc.cc.tn.us

From cube-lovers-errors@mc.lcs.mit.edu  Wed Sep  9 13:14:07 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Tue Sep  1 12:41:00 1998
Date: Tue, 1 Sep 1998 12:36:17 -0400 (Eastern Daylight Time)
From: Dale Newfield <din5w@cs.virginia.edu>
Reply-To: DNewfield@cs.virginia.edu
To: cube-lovers@ai.mit.edu
Subject: Re: Depicting a cube
In-Reply-To: <35E8A9ED.72BA@ameritech.net>
Message-Id: <Pine.WNT.3.96.980901123211.-301813C-100000@biff.cs.virginia.edu>

On Sat, 29 Aug 1998, Hana Bizek wrote:
> Here, in a nutshell, is a  description of the design problem. Please,
> get our your Rubik's cubes and start twiddling.

My supply of spare cubes seriously dwindled as I constructed a series of
"Bandaged Cubes" (XXX         XXX XXX).  Do you have a suggestion
                  XXX         XXX XXX
                  XXXXX , and XXXXXXX
                    XXX         XXX
                    XXX       XXXXXXX
                              XXX XXX
                              XXX XXX
regarding the acquisition of a large enough supply to create interesting
art without going broke?

-Dale Newfield
 Dale@Newfield.org

From cube-lovers-errors@mc.lcs.mit.edu  Wed Sep  9 13:47:36 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sun Sep  6 23:27:56 1998
Message-Id: <35F350B1.626F@ameritech.net>
Date: Sun, 06 Sep 1998 22:19:13 -0500
From: Hana Bizek <hbizek@ameritech.net>
Reply-To: hbizek@ameritech.net
To: cube-lovers@ai.mit.edu
Subject: Rubik's cube kingdom

Hello, cube-lovers,

the following miniessay talks about that realm of human knowledge, where
the rubik's cube reigns supreme. The gates of this kingdom are open to
anybody, but only those who love the cube, venture beyond its gates. The
rest of humanity are either unaware of it, or ignore its existence.
Those folks are truly missing a lot. After finishing {if indeed you
choose to finish} this epistle, you may want to do several things:

  a} emit a few chuckles thinking what an insane idea this is

  b} pause to think about the contents, debating whether all this is
     worth your precious time

  c} email your ideas, commenrts, etc, to me at hbizek@ameritech.net

Thank you. 

                WELCOME TO THE RUBIK'S CUBE KINGDOM.

This is not a kingdom of people, it is a kingdom of ideas. Its king is
the Rubik's cube. We pay homage to our king by trying to learn more
about him and see if this knowledge could be extended to other areas of
human pursuit. I am now going to tell you what I think those areas are
and point to a couple of web sites where results may be found. This is
by no means a finite list. As new ideas occur to all of us, they should
be added to the kingdom.  The Rubik's cube kingdom is there for anyone
to benefit from, just as any other field. Please feel welcome to sample
and browse.

Initially, of course, one should master the Rubik's cube solution. The
ability to solve one side should be the absolute minimum. It is far
better and morre satisfying to be able to solve the cube completely,
that is, get all six sides a solid color. I have seen numerous solution
algorithms on the web. Someone might consider collecting those varied
algorithms in a handy volume for cube lovers and others to use. I have
seen some solutions on the web, in which you are presented with a
solved cube in a little square field. You are instructed to press 's'
to scramble the cube and 'r' to restore {i. e. solve} it. That is not
solving the cube! You have to understand the steps of the solution
algorithm.

The areas where the cube has any impact are art, mathematics and
science. Let me look at art first. Quite recently there was a small item
in TIME magazine stating  that the cube has entered Hollywood and is
the subject of some movies. I heard that there are also songs about the
cube. By the above definition those human expressions too belong in the
Rubik's cube kingdom. However, I am going to zero on two aspects:
pictures and sculptures.

At http://www.wunderland.com/WTS/Jake/CubeArt one may see creations by
the people at Wunderland company {the spelling is correct} that depict
mostly 2-dimensional picture-like creations from a set of many Rubik's
cubes. They just show the pictures, thwy do not  describe their method
in any book, as far as I can determine {of course, I can be wrong}. But
from my  observation it seems to me they need to be worried about
continuity from one side of the cube to the one side of the next cube,
which is not too complicated. 

The sculptures are 3-dimensional structures and require some symmetry
on all the  faces of the cube, simultaneously. In this case, the
complete solution of the cube is a must. The required algorithm is
described in the book, "Mathematics of the Rubik's cube design," written
by me and published last year. As far as I know, I am the author of
these designs, described in my previous postings to the cube lovers. I
have stated a web site where three of these designs may be seen. I
repeat it here for completeness:
http://www.ssie.binghamton.edu/~jirif/hana1.html

Next, I am going to talk about science. I have to warn you: those
ideas are, as far as I know, unknown and undeveloped, as is, indeed,
the design Problem itself. First on the agenda is fractals and fractal
design prototypes. Some of the designs in my book, such as the Menger
sponge, are such fractals. One can think of the Rubik's cube as a
three dimensionl version of a Cantor set, which is a {one dimensional}
line.  Actually these fractals are neither three nor one dimensional;
they have fractional dimensions. But the 0th iteration are. One can
formulate rules for geometrical fractal iteration. Remember that
iteration preserves fractal dimension.  By the same token, for integer
dimension, it really doesn't matter if you subdivide by m or 10000n;
the dimension is always the same. Between the integer dimension there
are fractals of fractional dimension. These fractals can be reached by
breaking up of integer dimension or by some other
manipulation. Fractal design made from cubes suggest one such
manipulation, as witnessed by box fractal. But this is supposed to be
a miniessay, nor a book.

Another idea consists of computerizing the design algorithm, in some
comprehensive way so that the Rubik's cubes are used as 3-dimensional
cellular automata. It would be deliciously complicated game, perhaps
employing some of the patterns in the design theory.

Finally there is the question of what happens if internal combined faces
of a design that touch are colored the same? I will let you figure that
one out. It is not too hard. Personally, I had much more difficulty to
properly formulate the question than to provide the answer.

Finally, there is math. In this respect the ideas were formulated
partially by mathematicians in the 1980s during the heyday of the cube,
partially by cube lovers today. I would like to include my book in this
category.  At least its title indicates that there is some math, if that
has to be the only reason. And, as every cube lover knows, all possible
elements of a Rubik's cube form a mathematical group. A visitor to the
Rubik's cube kingdom will surely encounter some joys of group theory on
his travels.

Any objections, criticism, etc are welcome. Free speach prevails in the
Rubik's cube kingdom. Suppress free speach and not much is left.  

Hana

[ Moderator's note: I am somewhat concerned at the low
  information-to-woowoo ratio of Hana's "miniessay", but I think
  there are enough real ideas there that I've passed it on to the
  list.  I must, however, note that while "free speach" may prevail on
  the Internet, the contents of the cube-lovers mailing list is
  subject to editing for topicality, format, sensibility, and content.
  Which is to say that if the silliness level gets too high, you may
  have to find somewhere else to make your "kingdom".  I encourage
  guidance from the readership on where to draw the line; send your
  opinions to cube-lovers-request@ai.mit.edu. --Dan ]

From cube-lovers-errors@mc.lcs.mit.edu  Wed Sep  9 17:02:21 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Tue Sep  8 21:02:47 1998
Message-Id: <INEADEV-980909005322Z-6150@proxy.ineacorp.com>
From: Noel Dillabough <Noel.Dillabough@ineacorp.com>
To: "Cube-Lovers (E-mail)" <cube-lovers@ai.mit.edu>
Subject: Dogic
Date: Tue, 8 Sep 1998 20:53:22 -0400

A while back we heard about the puzzle "Dogic", an icosahedron puzzle.
After playing around with it for a while, I decided to model it in
Puzzler (http://www.mud.ca/puzzler/puzzler.html), since I dare not mix
it up until I have some moves to work with :)

That brings me to the question, has anyone made up a notation for moves
with the Dogic puzzle?  Perhaps similar to the Megaminx (R++, R+- R--
etc) moves.  I have yet to seriously sit down and try to solve it, but
eventually I will find some time and a few rudimentary moves would be
very helpful.

Also, if anyone wants a physical puzzle to play with, Hendrik Haak
(mailto:HendrikHaak@t-online.de) still has some available (that's where
I got mine)

-Noel

P.S. To those using the puzzler version, moves can be made along any of
the 12 axis in both minor (just the tip pieces) or major (the entire
slice) by dragging a cubie from one place to its eventual destination (I
didn't bother putting an entry for it in the helpfile).  Also, I had a
few requests for more detailed information on the puzzles and solving
them.  I have very nice solutions for the Megaminx, Pyraminx and Cubes,
but nothing written down for the Skewb, Masterball, or Dogic puzzles
(the skewb and masterball are quite easy so perhaps a solution is
unnecessary).  Any notes or information on these puzzles would be
appreciated.

From cube-lovers-errors@mc.lcs.mit.edu  Wed Sep  9 17:32:25 1998
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Date: Wed, 9 Sep 1998 14:59:27 -0400
Message-Id: <002BFC91.001706@scudder.com>
From: Jacob_Davenport@scudder.com (Jacob Davenport)
Subject: Re: Rubik's cube kingdom
To: cube-lovers@ai.mit.edu

While it is true that most of the cube art on our web pages 
(http://www.wunderland.com/WTS/Jake/CubeArt) is two dimensional and 
therefore pretty easy to make, I have made a few designs that were bloody 
difficult.  I'm rather proud of writing "WTS" on both sides of a hundred 
some cubes.  I'm particularly happy with the chessboard made of four 5x5x5 
cubes with a symmetrical design on the sides.  Some of my failed 
experiments were still tough to make, even if they didn't look very good.  
If anyone has any good 3d design suggestions, I'd like to hear them.

Hana, here is my favorite pattern for a single 3x3x3 cube.  There is no 
easy set of twists from solved to this pattern.  I had fun doing this 
pattern on a 5x5x5 cube, and you should be able to create an analogous 
pattern with all of your cubes:

   -------
  |\ * * *\ 
  | \ o o *\
  |x \ x o *\
  | * \------     
  |x o|* x o|
  | * |     |
  |x *|x x o|
   \x |     |
    \x|o o o|
     \|------
              
The same pattern should be on the other three faces with the other three 
colors.  My ASCII art isn't the greatest but I hope this is clear enough.

-Jacob

From cube-lovers-errors@mc.lcs.mit.edu  Sun Sep 13 16:29:34 1998
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Date: Sat, 12 Sep 1998 13:01:52 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Weak Local Maxima, 6f from Start
To: Cube-Lovers <cube-lovers@ai.mit.edu>
Reply-To: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Message-Id: <Pine.PMDF.3.95.980912124038.38277A-100000@PSTCC6.PSTCC.CC.TN.US>

I finally have had enough time to add support to my God's Algorithm
program to calculate weak local maxima.  The shortest strong local maximum
in the face turn metric is 9f, but the shortest weak local maximum has not
previously been verified.  It has long been known that Pons Asinorum is a
weak local maximum at 6f from Start.  I have been curious to know if Pons
is the shortest, and if there are any other short weak local maxima.

It turns out that 6f is indeed the shortest.  There are two such positions
unique to symmetry which are 6f from Start, the Pons and one other.  The
other one is quite pretty:

L2 R2 D2 U2 B' F  (6f*)


The eighteen neighbors are as follows.

L2 R2 D2 U2 F2 B'  (6f*)
L2 R2 D2 U2 F      (5f*)
B' F  L2 R2 D2 U'  (6f*)
B' F  L2 R2 D' U2  (6f*)
B' F  D2 U2 L2 R'  (6f*)
B' F  D2 U2 L' R2  (6f*)
L2 R2 D2 U2 B'     (5f*)
L2 R2 D2 U2 B2 F   (6f*)
B' F  L2 R2 D2 U   (6f*)
B' F  L2 R2 D  U2  (6f*)
B' F  D2 U2 L2 R   (6f*)
B' F  D2 U2 L  R2  (6f*)
L2 R2 D2 U2 B' F'  (6f*)
L2 R2 D2 U2 B  F   (6f*)
B' F  L2 R2 D2     (5f*)
B' F  L2 R2 U2     (5f*)
B' F  D2 U2 R2     (5f*)
B' F  D2 U2 L2     (5f*)


 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990

From cube-lovers-errors@mc.lcs.mit.edu  Mon Sep 14 11:21:10 1998
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Date: Sun, 13 Sep 1998 20:37:04 -0700
From: "Jin 'Time Traveler' Kim" <chrono@ibm.net>
Reply-To: chrono@ibm.net
To: cube-lovers@ai.mit.edu
Subject: Rubik's Cube-type puzzles FAQ
References: <9705131050.AA12905@mentda.me.ic.ac.uk>

Back in February or so I became fairly active in the bulletin board at
www.rubiks.com.  I found the same questions being repeated over and over
so with the help of some people I took it upon myself to write a FAQ for
people.  Due to the type of questions answered I refrained from talking
about it on the cube lovers list, but someone told me to give it a try
anyway.  So here it is:

http://www.slamsite.com/chrono

or more specifically,

http://www.slamsite.com/chrono/other/rcfaq006.txt

It's just a very big text file.  Kept it simple to give it that "old
school" flavor.

Corrections, additions, comments, & criticisms are welcome.  After all,
what's a FAQ if it provides the wrong answers.



-- 
Jin "Time Traveler" Kim
chrono@ibm.net
http://www.slamsite.com/chrono
'95 PGT - SCPOC



From cube-lovers-errors@mc.lcs.mit.edu  Thu Sep 17 17:18:28 1998
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Date: Mon, 14 Sep 1998 14:14:22 -0400 (Eastern Daylight Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: Weak Local Maxima, 6f from Start
In-Reply-To: <14Sep1998.111621.Hoey@AIC.NRL.Navy.Mil>
To: Cube Lovers <cube-lovers@ai.mit.edu>
Message-Id: <SIMEON.9809141422.E@GN209A.PSTCC.CC.TN.US>

> It turns out that 6f is indeed the shortest.  There are two such positions
> unique to symmetry which are 6f from Start, the Pons and one other.  The
> other one is quite pretty:
>
> L2 R2 D2 U2 B' F  (6f*)
>

I didn't notice it originally, but this position is in the slice
subgroup, and is only one slice move from Pons.  Half turns such
as L2 can be written equally well as either LL or as L'L',
so we can write (L2 R2) as (L'R)(L'R) and (D2 U2) as (D'U)(D'U).
Thus, the weak local maximum 6f from Start can be written as five
slices, one slice short of Pons.

L'R  L'R  D'U  D'U  B'F

All we would have to do to get the Pons would be to add one more
B'F slice.

----------------------
Jerry Bryan
jbryan@pstcc.cc.tn.us

From cube-lovers-errors@mc.lcs.mit.edu  Thu Sep 17 20:27:13 1998
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From: "Chris and Kori Pelley" <ck1@home.com>
To: "Cube Lovers" <cube-lovers@ai.mit.edu>
Subject: DOGIC solution
Date: Mon, 14 Sep 1998 18:01:56 -0400
Message-Id: <000a01bde02b$49b5aae0$da460318@CC623255-A.srst1.fl.home.com>

Using Noel Dillabough's PUZZLER program for MS Windows, I was able to verify
the basic moves needed to solve the new DOGIC puzzle.

SPOILER WARNING!  If you wish to solve the puzzle yourself, read no further.


What seems fairly obvious is that the DOGIC is essentially a superset of
Impossi-Ball, which is basically the corners of MegaMinx.  On DOGIC,
however, these corners have been flattened and could more properly be called
"centers."  Using classic 3x3x3 techniques, you can position and orient
these pieces using the following moves:

Center 3-cycle:

(R' U L U') (R U L' U')

Center orientation (pair):

(R' D R) (F D F') U'
(F D' F') (R' D' R) U

Note that these faces refer to the large pentagons and must be "translated"
to fit the dodecahedral nature of DOGIC.  R, U, and L form a horseshoe, and
F intersects all three.  The D face is not really D, in fact it touches the
U face at one point.

The remaining triangular pieces turn out to be fairly trivial, and any two
can be swapped with the simple sequence:

R u R' u'

In this case, R is a large pentagon and u is any intersecting smaller
pentagon.

A general strategy would be to manually place the top "centers" followed by
their adjacent centers (if you have solved ImpossiBall this should not be
difficult).  Then apply the first two moves above to complete the remaining
centers.  Finally, place all the smaller triangles with the third move.

Despite having more permutations than most magic puzzles, DOGIC seems to be
fairly easy to solve.

Chris Pelley
ck1@home.com
http://www.chrisandkori.com

From cube-lovers-errors@mc.lcs.mit.edu  Fri Sep 18 14:29:06 1998
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Date: Tue, 15 Sep 1998 12:21:58 -0400 (EDT)
From: der Mouse  <mouse@rodents.montreal.qc.ca>
Message-Id: <199809151621.MAA19286@Twig.Rodents.Montreal.QC.CA>
To: cube-lovers@ai.mit.edu
Subject: Two-face and three-face subgroups

I've been playing with the two-face subgroup [%] of the 3-Cube and got
to wondering - how much work has been done on the two-face and
three-face subgroups?  Certainly the two-face subgroup "feels" like a
much smaller object than even the 2-Cube (though perhaps more tedious
for human solution), perhaps about the size of the Pyraminx.

[%] Okay, strictly speaking there are two different two-face subgroups,
    but one of them is not even the least bit interesting.

And what about the three-face subgroups?  Certainly the three- and
four-face subgroups are smaller than the whole Cube group, though ISTR
that the five-face (sub)group is actually the whole thing.  But how
much smaller, and how difficult of human solution?  I'd expect one of
the three-face groups (the one involving two opposite faces - call it
the L-F-R one) to be more tedious but no more difficult than the
two-face group, whereas the other one (involving one face from each
pair of opposite faces - U-F-R, say) should have more interest.

In particular, the two-face subgroup is smaller than the set of all
position that leave unchanged the cubies that the two-face subgroup
never touches.  (To put it another way, I'm saying that the subgroup
generated by {R,F} is smaller than the set of positions of the full
group that leaves unmoved the 11 cubies that don't touch either of
those two faces - 7 if you don't count face cubies.)  I can see a
factor of 128 smaller, since it's not possible to flip edge cubies in
the two-face group, but I haven't thought about the corners, so it may
be even smaller than that.

What about the three-face subgroups?  The L-F-R subgroup is also
smaller, if for no other reason than an inability to flip edge cubies,
like the two-face group.  But is the U-F-R subgroup the same as the
subset of the full group that leaves untouched the 7 (4 if you don't
count face centers) cubies in the DBL corner?

What about human solvability?  I've taught myself to solve the two-face
group, and with the tools I developed (largely powers, reorientations,
and inverses of F' R' F R) I feel confident I can handle the L-F-R
three-face group or even the L-F-R-B four-face group.  Can anyone
comment on how humanly difficult the U-F-R group, or for that matter
the U-F-R-L four-face group, is?

					der Mouse

			       mouse@rodents.montreal.qc.ca
		     7D C8 61 52 5D E7 2D 39  4E F1 31 3E E8 B3 27 4B

From cube-lovers-errors@mc.lcs.mit.edu  Tue Sep 22 16:10:10 1998
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Date: Fri, 18 Sep 1998 15:52:33 -0400 (EDT)
From: Nicholas Bodley <nbodley@tiac.net>
To: Hana Bizek <hbizek@ameritech.net>
Cc: cube-lovers@ai.mit.edu
Subject: Re: Rubik's cube kingdom
In-Reply-To: <35F350B1.626F@ameritech.net>
Message-Id: <Pine.SUN.3.95.980918154427.13428O-100000@sunspot.tiac.net>

My apologies for a delayed reply.

Hana's essay was rather philosophical, and contained some uncommon
points of view; it was appropriate, in my opinion.

One aspect of the Cube (and related puzzles) that seemed to be ignored
is the remarkable ingenuity of their internal mechanisms. I maintain
that the mechanism of the original (i.e., 3^3) Rubik's Cube is one of
the most ingenious ever invented.

I recall being very fatigued, riding the West Side IRT subway in NYC
about 2 AM, perhaps, and catching sight of someone manipulating what
must have been one of the very first Cubes, probably from Hungary*. I
was fairly sure I wasn't hallucinating, but was very troubled that what
I'd seen simply appeared impossible. I've been a somewhat-casual student
of mechanisms all my life.

*This was probably several weeks, or more, before the Scientific
American article, and the later explosion of its popularity.

My regards to all,

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*  The personal computer industry will have become
|*  Amateur musician  *|*  mature when crashes become unacceptable.
--------------------------------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Tue Sep 22 18:09:16 1998
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Date: Thu, 17 Sep 1998 00:03:47 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: More on Calculating Weak Local Maxima
To: Cube-Lovers <cube-lovers@ai.mit.edu>
Message-Id: <Pine.PMDF.3.95.980916234641.7009A-100000@PSTCC6.PSTCC.CC.TN.US>

In the process of adding the code to my God's Algorithm program to
calculate weak local maxima in the face turn metric, I realized that the
algorithm I posted previously to do so was incomplete in one subtle but
very important respect.  This message will provide the missing piece to
the algorithm.

I have posted much of this before, but my program is in general jumping
ahead by more than one move at a time.  For example, suppose we can store
all the positions up to five moves from Start.  Then, we can determine all
the positions which are eight move s from Start by calculating all the
products xy where x is a position of length five and y is a position of
length three.

Obviously, just because the length of x is five and the length of y is
three does not mean that the length of xy is eight.  In fact, the length
of xy could be anywhere from two through eight.  To determine the true
length of xy, we compare xy to the stored positions of length two, three,
four, and five.  In addition, we compare xy to the calculated positions of
length six and seven, which are calculated in the same manner as is xy.
If xy fails to match all such shorter positions, then its length is indeed
eight.

Next we focus on the quarter turn metric.  For some fixed q in Q, the set
of twelve quarter turns, what is the length of xyq if the length of xy is
eight with the length of x equal to five and the length of y equal to
three?  First of all, it must be either seven or nine.  Second of all, the
length of yq must be either two or four.  If the length of yq is two, then
we know that the length of xyq must be seven.  But if the length of yq is
four, then we are still not sure.  The reason is that there might be some
u not equal to x of length five and some v not equal to y of length three
such that xy=uv, but where the length of vq is two.  If so, then the
length of xyq is the same as the length of uvq which is seven.

The basic idea is that if z=xy where the length of x is five and the
length of y is three, then there may be many, many x and y pairs of length
five and three respectively whose product yields z.  The length of zq is
nine only if for every such y the length of yq is four.  Even if all but
one yq is of length four, it only takes one yq of length two to spoil the
pudding, as it were.

The mechanism which I have posted previously to capture this concept is
the Ends-with function E(z).  E(z) is defined to the be set of all moves
with which a minimal maneuver for z can end.  So in the case at hand,
since the length of z is eight, the length of zq is nine only if E(z) does
not contain q'.  E(z) can be calculated in the case at hand as the union
of E(y) taken over all the y values of length three which can be composed
with an x of length five to create z.  Therefore, to say that E(z) does
not contain q' is the same thing as saying that none of the E(y) contain
q'.

So far, so good and there is nothing new here which I haven't posted
before.  But let's consider the exact same issue in the face turn metric.
If the length of x is five and the length of y is three, then the length
of xy can be in the range of two through eight as before.  And as before,
if we compare xy with all positions of length two through seven without
finding a match, then the length of xy is indeed eight.

But this time we need to consider xyf, where f is some fixed face turn in
the set Q+H of twelve quarter turns and six half turns.  What is the
length of xyf?  For starters, it is either seven or eight or nine.  Also,
the length of yf is two or three or four.

If the length of yf is two, then the length of xyf is guaranteed to be
seven.

If the length of yf is three, then the length of xyf is guaranteed to be
no more than eight.  But the length nevertheless might be seven, because
as in the quarter turn case, there may be some u of length five and some v
of length three such that uv=xy, but such that the length of vf is only
two.  If so, the length of xyf is the same as the length of uvf which is
guaranteed to be seven.

The definition of Ends-with is the same in the face turn case as in the
quarter turn case, namely E(z) is the set of all face turns with which a
minimal maneuver for z can end.  If z=xy then E(z) can be calculated as
the union of E(y) over all the y value s of length three which can be
combined with an x value of length five to form z.  To say that the length
of zf is at least eight is the same thing is saying that E(z) does not
contain f' which is the same thing as saying that none of the E(y) contain
f'.

Next, let's suppose that indeed E(z) does not contain f'.  We are still
left with the issue of whether the length of z is eight or nine, having
eliminated seven as a possibility.  The test is still the length of all
the yf, with a length of two having been eliminated as a possibility.  If
all of the yf are of length 4, then xyf is of length nine.  But if even so
many as one of the yf are of length three, then xyf is of length 8.

The mechanism I have posted before to capture this concept is the
Ends-with2 function.  E2(z) is a little tricky to describe.  Informally,
we might say that E2(z) is the set of all f in Q+H with which z can end
without changing it's length.  It is probably better to say that E2(z) is
the set of all f in Q+H such that the length of zf' is the same as the
length of z.  The technique which I have posted before (and which I must
now correct) to calculate E2(z) is to form the union of E2(y) over all y
values of length three which can be combined with an x value of length
five to form z.

If the length of zf is eight or nine, then this mechanism is fine.  But if
the length of zf turns out to be seven, there is a problem.  That is,
there may be one y where the length of yf is two and where E(y) contains
f, and there may be another y where the length of yf is is three and where
E2(y) contains f.  In such a case, both E(z) and E2(z) would contain f.
Hence, we must always calculate E(z) prior to calculating E2(z), and we
must omit from E2(z) any f values which are already contained in E(z).

With this correction, everything works.  A local maximum is a position z
for which |E(z)|+|E2(z)|=18, a strong local maximum is a local maximum z
for which |E(z)|=18 and |E2(z)|=0, and a weak local maximum is a local
maximum z for which |E(z)| < 18 and |E 2(z)| > 0.  All my examples have
been specific to y values of length 3 for clarity of exposition, but the
calculation of E(z) and E2(z) is totally general, and is the union of E(y)
and E2(y), respectively, over all y values which can be used to form a z
of the form z=xy, and with any values which are in E(z) omitted from
E2(z).

Finally, my programs also calculate a Starts-with and a Starts-with2
function, which are defined analogously.  The same correction must be made
to the Starts-with2 function as was made for the Ends-with2 function.
Equivalently, we can define S(z)=E'(z') and S2(z)=E2'(z'), where E' and
E2' are the set of all inverses of the elements of E and E2, respectively.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990

From cube-lovers-errors@mc.lcs.mit.edu  Tue Sep 22 19:05:00 1998
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Date: Sat, 19 Sep 1998 09:13:58 -0400 (Eastern Daylight Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: Two-face and three-face subgroups
In-Reply-To: <199809151621.MAA19286@Twig.Rodents.Montreal.QC.CA>
To: der Mouse <mouse@rodents.montreal.qc.ca>
Cc: Cube Lovers <cube-lovers@ai.mit.edu>
Message-Id: <SIMEON.9809190958.A@GN209A.PSTCC.CC.TN.US>

On Tue, 15 Sep 1998 12:21:58 -0400 (EDT) der Mouse
<mouse@rodents.montreal.qc.ca> wrote:

> I've been playing with the two-face subgroup [%] of the 3-Cube and got
> to wondering - how much work has been done on the two-face and
> three-face subgroups?  Certainly the two-face subgroup "feels" like a
> much smaller object than even the 2-Cube (though perhaps more tedious
> for human solution), perhaps about the size of the Pyraminx.
>

The <U,R> subgroup has been explored fairly thoroughly.  For
example, look in the archives 8/31/1994 for a summary of the
first complete God's Algorithm search of this particular
subgroup.  There are a number of articles in the archives
thereafter.  <U,R> has been searched in both the quarter turn
metric and the face turn metric, and local maxima have been
investigated as a part of the search.

<U,R> has a very small branching factor and a corresponding
large diameter of 25 in the quarter turn metric, at least I
think it's a large diameter for such a small group. Until Mike
Reid recently showed that the diameter of G in the quarter turn
metric was at least 26, the diameter of <U,R> was the largest
known for the 3x3x3 cube or any of its subgroups.

Frey and Singmaster's book discusses both two face and three
face subgroups, among other things giving their sizes.

To my knowledge, no God's Algorithm searches have been performed
for the three face subgroups.

We have |<U,R>|=73483200, so <U,R> is slightly smaller than the
corners group at 88179840.  The 2x2x2 is 24 times smaller than
the corners group, at 3674160.  However, I am of the school
of thought that tends not to equate the size of the group (or
search space, for problems that are not actually groups) with
the difficulty of the problem.

----------------------
Jerry Bryan
jbryan@pstcc.cc.tn.us

From cube-lovers-errors@mc.lcs.mit.edu  Tue Sep 22 19:50:07 1998
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Date: Mon, 21 Sep 1998 16:34:29 -0400 (Eastern Daylight Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Local Maxima which Fix the Corners, 12q from Start
To: Cube Lovers <cube-lovers@ai.mit.edu>
Message-Id: <SIMEON.9809211629.A@GN209A.PSTCC.CC.TN.US>

I am making a run to calculate God's Algorithm out to 12 moves
from Start in the quarter turn metric.  It has been running
several weeks, and will probably run several more.

I have made some changes to my program to make it easier to
extract the positions for local maxima, and to checkpoint the
local maxima data.  As a part of the checkpointing, I can
actually see the local maxima as they are generated without
having to wait for the program to end.

It is becoming apparent that there are a *lot* of local maxima
12q from Start.  It is already known that there are only four
(unique to symmetry) which are 10q from Start (the shortest ones
in the quarter turn metric), and that there are none 11q from
Start. So I am a little surprised that I am seeing so many.

I have looked at quite a few of them, and most of them are not
all that interesting.  But the ones which fix the corners are
all quite pretty.  Because the positions are being produced in
lexicographic order, and because I am sorting by corners first,
edges second, the positions which fix the corners are the first
ones to appear.  There are eight of them as follows.

1.  F2 L2 F2 B2 R2 B2
2.  F  B' U2 D2 F' B  R2 L2
3.  F  B  R2 F' B' U  D  L2 U' D'
4.  D' F  B' R  F  R' F' B  U  F'  U' D
5.  F  B  R2 L2 F  B  U2 D2
6.  R  L' F2 B2 R  L' F2 B2
7.  F2 B2 U2 D2 R2 L2
8.  R  L' U  D' F  B' R2 L2 U  D'

#1 is a 2-H pattern (only four edge cubies are moved).

#2 is a 4-H.

#3 moves four edge cubies, leaving eight of the nine
facelets the same color on four faces, and a solid color on the
other two faces.

#4 moves three edge cubies, leaving eight of the nine facelets
the same color on all six faces.

#5 has 2 H's, 2 checkerboards, and 2 solid faces  -- with the
respective H's, checkerboards, and solid faces opposing each
other.

#6 has 4 H's and 2 checkerboards, with the 2 checkerboards
opposing each other.

#7 is the Pons Asinorum, and is included only for completeness
because we already knew that the Pons was a local maximum of
length 12q.

#8 has all six faces being sort of a "three colored
checkerboard".

Some of these positions may have appeared on Cube-Lovers in some
other context, but the only one I recognize for sure is the Pons.
In some ways, #4 is the most interesting to me, because it a
simple 3-cycle on the edges, and who would have thought that
such a position would turn out to be a local maximum?  #1 and #3
both consist of two 2-cycles on the edges, and are about as
striking to me as is #4.

----------------------
Jerry Bryan
jbryan@pstcc.cc.tn.us

From cube-lovers-errors@mc.lcs.mit.edu  Wed Sep 23 12:37:03 1998
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Date: Tue, 22 Sep 1998 13:47:17 -0700 (PDT)
From: "Jorge E. Jaramillo" <kingeorge@rocketmail.com>
Subject: Moves to this pattern
To: Cube-Lovers@ai.mit.edu

Hi I just joined the list!

I was wondering if someone could help me with a
pattern that has been bugging me for a while.
I have been able to solve the cube to this pattern so
i know it is a valid one.
I used one of those on-line solvers entered the
pattern and then reversed the sequence the solver
gave me and it indeed gets the pattern i want but
somehow it seemed too many moves for me for such a
simple pattern.
the pattern I am talking about is: the 4 cubelets
that make the vertex formed by FDR are exchanged with
the vertex from BDL

Can any one give me the set of moves to get to this
pattern from a solved cube? Does this pattern have a
name?

[Here is a] set of moves (they work but I am
sure there is a shorter way):

D2 F B- L2 F- B D- F B- L F- B D2 F B- L2 F- B D F B-
L- F- B F B- L- F- B D2 F B- L- F- B D2 R- D- R D- R-
D2 R D2 L- D- L B D- B- L- D L2 D L- D- F- D- F D B-
D- B D R D R- D F- B D B- F R D- R- T B- T- B- D- B-

This long set of moves reminds me of something else:
There are many Rubik cube annimations that you move
the faces with either the keypad or the mouse. Does
anyone know of one that follows sets of orders you
write?

It would be neat to try this long patterns in one of
those simulators.

Thanks
===
Jorge E Jaramillo

From cube-lovers-errors@mc.lcs.mit.edu  Wed Sep 23 19:54:38 1998
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Date: Tue, 22 Sep 1998 14:32:21 -0700 (PDT)
From: "Jorge E. Jaramillo" <kingeorge@rocketmail.com>
Subject: Is it only mine?
To: cube <Cube-Lovers@ai.mit.edu>

In order to solve the cube faster I developed a
method that would turn a lot the middle faces. I
don't know how you call them in this list I am
talking about the face between Top and Bottom, the
face between Right and Left and even the face between
Front and Back.
Well after twisting it a few times my cube came
undone, fell apart and I thought "Damn made in Taiwan
cubes" (although the ones I can buy here do not say
where are they made). I was going through my second
cube in a short while (I just regained interest in
the cube a short while ago after say 15 years) and it
fell apart again. I took off the plastic color of the
center cubelet that came apart and found a screw and
a spring that keeps the screw tight. I re screwed it
but did not have any glue at hand so I kept on
playing without the color of the center cubelet I was
doing one of these center face moves and saw how the
screw was turning counterclockwise in other words the
way that makes the cube fall apart. Needless to say I
had to re design my method.
This long story is to ask if all the cubes are built
this way or only the ordinary ones I can buy here?
Thanks.


===
Jorge E Jaramillo

From cube-lovers-errors@mc.lcs.mit.edu  Fri Sep 25 18:05:43 1998
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Date: Wed, 23 Sep 1998 23:15:22 +0200
Reply-To: Rainer.adS.BERA_GmbH@t-online.de
Organization: BERA Softwaretechnik GmbH
To: "Jorge E. Jaramillo" <kingeorge@rocketmail.com>
Cc: Cube-Lovers@ai.mit.edu
Subject: Re: Moves to this pattern
References: <19980922204717.20206.rocketmail@web1.rocketmail.com>
From: Rainer.adS.BERA_GmbH@t-online.de (Rainer aus dem Spring)

Jorge E. Jaramillo wrote:

> the pattern I am talking about is: the 4 cubelets
> that make the vertex formed by FDR are exchanged with
> the vertex from BDL.

R2 U  R2 U2 B2 D  L2 U2 L2 B2 D' B2 U2

Rainer adS

PS
If you have a WINTEL system you should download Herbert Kociemba's cube
program.

[ Moderator's note: Jerry Bryan provides another 13f process,
R2 U' B2 R2 U2 R2 U' B2 U2 R2 U' B2 U2, which hasn't been proven optimal.
Steve LoBasso has a longer one. ]

From cube-lovers-errors@mc.lcs.mit.edu  Sat Sep 26 00:05:04 1998
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Date: Fri, 25 Sep 1998 08:36:22 -0400 (Eastern Daylight Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Summary of Local Maxima, Face Turn Metric
To: Cube Lovers <cube-lovers@ai.mit.edu>
Message-Id: <SIMEON.9809250822.A@GN209A.PSTCC.CC.TN.US>

I have posted maneuvers for a number of specific strong and weak 
local maximal positions (the strong local maxima at 9f and 10f, 
and the weak local maxima at 6f) , but I haven't really posted a 
summary of the numbers. Here are the numbers I have so far.  In 
order to complete the table through 10f from Start for weak 
local maxima, I would have to repeat a rather long run.  As 
might be expected, it appears that the number of weak local 
maxima will greatly exceed the number of strong local maxima.

As is the usual case, patterns are M-conjugacy classes (symmetry 
classes), and represent the number of positions which are unique 
up to symmetry.



Distance   Strong    Strong     Weak      Weak
  from     Lclmax    Lclmax     Lclmax    Lclmax
  Start   Patterns   Positions  Patterns  Positions

   0f          0        0          0          0
   1f          0        0          0          0
   2f          0        0          0          0
   3f          0        0          0          0
   4f          0        0          0          0
   5f          0        0          0          0
   6f          0        0          2          7
   7f          0        0          1          6
   8f          0        0         37        739
   9f          2       32        327      13014
  10f          6      107          

----------------------
Jerry Bryan
jbryan@pstcc.cc.tn.us

From cube-lovers-errors@mc.lcs.mit.edu  Tue Oct  6 15:05:42 1998
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Date: Sun, 27 Sep 1998 20:35:33 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Corners Only, Ignoring Twist
To: Cube-Lovers <cube-lovers@ai.mit.edu>
Reply-To: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Message-Id: <Pine.PMDF.3.95.980927193742.73560A-100000@PSTCC6.PSTCC.CC.TN.US>

I have been playing around with the idea of trying to calculate God's
Algorithm all the way to the bitter end for the group which results from
ignoring all twists of the corners and flips of the edges.  It's a pretty
big group.  The order is |G|/(3^7)/(2^11), which is about 9.7*10^12, call
it about 10^13 to make it a round number.  (Another way to calculate it is
8!12!/2.)

This is probably right at the bare edge, maybe even slightly past the bare
edge, of the size of problem I can handle right now, which makes it a
worthy endeavor.  Also, it would provide a lower limit on the diameter of
G (although the limit might not be any better than the ones we already
have), which again makes it a worthy endeavor.

Such a result might be suitable as the estimator function required by IDA*
searches.  The distance from Start in the no-twist, no-flip group would
certainly be a lower bound for every position where twist and flip *are*
considered.  My only hesitation about suggesting this group as a suitable
IDA* estimator function is that there are obvious pathological cases such
as the superflip where this function would be a very poor estimator.

In developing a no-twist, no-flip version of the program, I decided to try
it out on the corners only case.  Here are the results.


Distance
 from           Patterns     Positions
Start


  0q                   1               1
  1q                   1              12
  2q                   5             114
  3q                  24             876
  4q                 119            4931
  5q                 301           12972
  6q                 364           15066
  7q                 166            6300
  8q                   3              48



Distance
 from           Patterns       Positions
Start

 0f                    1               1
 1f                    2              18
 2f                    9             243
 3f                   68            2646
 4f                  302           12516
 5f                  418           17624
 6f                  170            7080
 7f                   14             192

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990

From cube-lovers-errors@mc.lcs.mit.edu  Tue Oct  6 16:52:26 1998
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Date: Fri, 2 Oct 1998 23:18:37 +0200 (METDST)
From: Martin Moller Pedersen <tusk@daimi.aau.dk>
Message-Id: <199810022118.XAA19053@stargazer.daimi.aau.dk>
To: cube-lovers@ai.mit.edu
Subject: cubes at spielmessen in Essen

There will soon be a big gathering for games in Germany - Essen a so-called 
Spielmessen.

I am attending this gathering for the first time in three years so I am
looking for companies who will came to the spielmessen and who sells cubes.

The places is big so it would be nice for me to have same names to look for.

and hopefully I will have a real 4x4x4 and 5x5x5 cube to play with in a few weeks :-)

/Martin

From cube-lovers-errors@mc.lcs.mit.edu  Tue Oct  6 18:38:32 1998
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Date: Sun, 4 Oct 1998 11:12:27 -0600
To: cube-lovers@ai.mit.edu
From: Steve LoBasso <slobasso@dtint.com>
Subject: Cube Solver for Macintosh

I just wrote a Macintosh port to Dik T. Winter's cube solving code.
I put it on my web page at the link below. I haven't had a chance to make
an info page for it so the link below is just the application.

It will run on both 68k and PPC Native.
Be warned the 68k version runs fairly slow and the initialization phase
takes quite a while.

Cube Solver

By Steve LoBasso
  slobasso@dtint.com

Written using algorithm code by
Dik T. Winter based on algorithm described by Herbert Kociemba.

http://members.tripod.com/~slobasso/downloads/Cube_Solver.hqx

--
Steve LoBasso                          mailto:slobasso@dtint.com
Digital Technology International    or mailto:slobasso@hotmail.com
500 West 1200 South, Orem, UT, 84058   http://members.tripod.com/~slobasso
(801)226-6142  ext.265                 FAX (801)221-9254

From cube-lovers-errors@mc.lcs.mit.edu  Tue Oct  6 20:00:34 1998
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From: "Chris and Kori Pelley" <ck1@home.com>
To: <cube-lovers@ai.mit.edu>
Subject: That's Incredible!
Date: Sun, 4 Oct 1998 22:37:28 -0400
Message-Id: <002701bdf009$180cb5e0$da460318@CC623255-A.srst1.fl.home.com>

I recently obtained (courtesy of Peter Beck) the Rubik's Cube-a-Thon video
from the TV show "THAT'S INCREDIBLE" and digitized it in RealVideo format.
The file is rather large (18.1 megabytes) but it's worth a download if
you're into cubic nostalgia.  Eleven and a half minutes long, it features
Minh Thai, Jeff Varasano, Kris Wunderlich, and others that may be on this
list.

Here's the URL:

http://www.chrisandkori.com/incredible.htm

It requires the RealPlayer 5.0 or later to view it.  Note that you must
download the file, then view it.  I do not have a streaming video server.
If anyone would like to host the file on a streaming server, please contact
me.

Chris Pelley
ck1@home.com

From cube-lovers-errors@mc.lcs.mit.edu  Thu Oct  8 19:04:05 1998
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Date: Thu, 8 Oct 1998 15:11:54 +0100
From: David Singmaster       <david.singmaster@sbu.ac.uk>
To: cube-lovers@ai.mit.edu
Message-Id: <009CD656.72A81016.35@ice.sbu.ac.uk>
Subject: Davenport's pattern

	The pattern given by Jacob Davenport is what I called a cube in a cube
in a cube.  I discovered this in 1979 or 1980 and was very pleased with it.
Indeed, I used the cube in a cube as the logo of the late and much lamented
David Singmaster Ltd. in 1980-1982 (approx. dates since I'm not where my
records are).  The pattern is in my Notes.  There
are various ways to generate the pattern, but the one that I can remember uses
what Roger Penrose called the Y-commutator, which has the form  FR'F'R.  The
reason this is the Y-commutator is that it affect the three edges adjacent to a
corner and the corner and its three adjacent corners.  I.e. the affected pieces
form a Y, while the pieces affected by the ordinary commutator  FRF'R'  form a
Z.  Combining three Y-commutators as follows:  FR'F'R  RU'R'U  UF'U'F  gives a
process that twists the corner and the three adjacent edges as a unit and
twists an adjacent corner the opposite direction.  NOTE - I'm doing this from
memory and I have a suspicion that the middle group may need to be inverted??
	By moving the odd corner to the right place adjacent to the opposite
corner and applying the inverse of the above, one gets the same sort of pattern
at the opposite corner and the odd corner has been restored.  Now one 3-cycles
the centers, as is easily done by a commutator of slice moves, and one has the
cube in a cube.  Now one can twist the two opposite corners to get the cube in
a cube in a cube, though I find this not as visually dramatic as the cube in a
cube.
	Someone - Mike Reid ? - sent me a minimal method for one of these
patterns, but it's not very memorable.

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499;
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Thu Oct  8 19:47:11 1998
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Date: Thu, 8 Oct 1998 16:45:31 +0100
From: David Singmaster       <david.singmaster@sbu.ac.uk>
To: cube-lovers@ai.mit.edu
Message-Id: <009CD663.86DEF2D6.16@ice.sbu.ac.uk>
Subject: Nicholas Bodley's message of 22 Sep 1998.

	Nicholas Bodley's message reminds me of when I wrote about the Cube in
1978 or early 1979, I think in the Observer, which seems to have been the first
article outside Hungary.  I mentioned that the mechanical problem seemed even
harder than the mathematical problem and this led to about six submissions of
mechanisms from readers.  All but one of these were clearly impossible, but the
last was Rubik's mechanism with slight differences - e.g. he had the undersides
of the centers rounded.  The submitter of this was a UK patent agent with
obvious mechanical aptitude.
	However, one of my students, who had bought a cube from me, told me
that a friend rang her up and asked if she had seen the hoax article about a
cube that moved in all directions.  The friend had just proven that such an
object was impossible.  My student had to disabuse her.
	When the cubes first came out of Hungary, we didn't know what the
mechanism was and they were too precious to fiddle with.  Roger Penrose said he
had one face center piece come undone and he carefully wrapped thread around
the exposed part of the screw and worked the screw into place and pulled on the
thread to screw the screw  back into the central piece.  Sometime in late 1978,
a friend had trouble with his cube and took a screwdriver to it and discovered
the cover plates and the screw heads inside!
	Enough for now.

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499;
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Fri Oct  9 18:40:38 1998
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Message-Id: <361E8D3F.6597@ameritech.net>
Date: Fri, 09 Oct 1998 17:25:03 -0500
From: Hana Bizek <hbizek@ameritech.net>
Reply-To: hbizek@ameritech.net
To: cube-lovers@ai.mit.edu
Subject: comments on "davenport's pattern"

David Singmaster claims to have discovered this pattern in 1979 or
1980, so it should be credited to him. In my own book I present a
number of patterns, but I would never dare to claim authorship to any
of them. Singmaster's comments prompted me to look at my own books. In
CUBE GAMES (Taylor and Rylands} this pattern appears on the top of page
37.I have a strong suspicion this pattern could be a combination of the
two cyclicity-three patterns on page 36 therein. One may use this
pattern as a corner in a 3-color design.

Design-construction is a step beyond pattern-construction. My question
has not yet been satisfactorily answered. Has anyone seen construction
of 3-dimensional "sculpture-like" designs? People referred me to
Davenport's creations, but my own designs are quite different. 

Hana

From cube-lovers-errors@mc.lcs.mit.edu  Fri Oct 30 13:56:08 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Thu Oct 15 11:02:13 1998
Message-Id: <19981015145704.8314.rocketmail@attach1.rocketmail.com>
Date: Thu, 15 Oct 1998 07:57:04 -0700 (PDT)
From: "Jorge E. Jaramillo" <kingeorge@rocketmail.com>
Subject: Moves to this other pattern
To: cube <Cube-Lovers@ai.mit.edu>

Please if this is not one of the purposes of this
list someone let me know I don't meant to be rude.

Could someone please tell me the moves to get from a
solved cube to the following pattern:
The top and bottom faces keep their colors.
The 4 columns in the middle of every side face stay
with their color.
The left column on the front face moves to the right
and the right column moves to the left.
The left column on the back face moves to the right
and the right column moves to the left.

Thanks

===
Jorge E Jaramillo

[ Moderator's note: We have a lot of requests for processes for various
  and have got a lot of optimal processes.  Maybe the hard part is
  figuring out how to look them up.  This is one of the "4-" patterns,
  and probably appears among the quasi-continuous partial isoglyphs. --Dan ]

From cube-lovers-errors@mc.lcs.mit.edu  Fri Oct 30 14:17:00 1998
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Date: Mon, 26 Oct 1998 23:58:27 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: 12q From Start
To: Cube-Lovers <cube-lovers@ai.mit.edu>
Message-Id: <Pine.PMDF.3.95.981026235420.29075A-100000@PSTCC6.PSTCC.CC.TN.US>


 |x|     Patterns     Lcl    Positions    Lcl    Branching
                      Max                 Max      Factor


  0q             1     0              1     0
  1q             1     0             12     0    12
  2q             5     0            114     0     9.5
  3q            25     0           1068     0     9.368
  4q           219     0          10011     0     9.374
  5q          1978     0          93840     0     9.374
  6q         18395     0         878880     0     9.366
  7q        171529     0        8221632     0     9.355
  8q       1601725     0       76843595     0     9.347
  9q      14956266     0      717789576     0     9.341
 10q     139629194     4     6701836858    42     9.337
 11q    1303138445     0    62549615248     0     9.333
 12q   12157779067   103   583570100997  2913     9.330

The last time a new level was calculated for the quarter turn metric was 4
February 1995.

The cumulative number of positions now identified is 653625391832, or
about 6.5*10^11.  This is well past the "geometric halfway point"  of
sqrt(|G|), which is about 6.5*10^9.  However, it is known that the
diameter of G is at least 26q, strongly indicating that there is a bit of
a tail to the distribution of positions by length.

Of the 103 local maxima of length 12q, 70 of them also have their inverse
as local maxima.  For the other 33, the inverse is not a local maximum.
For one of them, the inverse has 11 moves which go closer to Start.  For
seven of them, the inverse has 10 moves which go closer to Start.  For
eleven of them, the inverse has 8 moves which go closer to Start.  For six
of them, the inverse has 6 moves which go closer to Start.  For two of
them, the inverse has 4 moves which go closer to Start.  And for six of
them, the inverse has only 2 moves which go closer to Start.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us

From cube-lovers-errors@mc.lcs.mit.edu  Mon Nov  2 09:40:50 1998
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Message-Id: <363B1799.3534@hrz1.hrz.tu-darmstadt.de>
Date: Sat, 31 Oct 1998 14:58:49 +0100
From: Herbert Kociemba <kociemba@hrz1.hrz.tu-darmstadt.de>
Reply-To: kociemba@hrz1.hrz.tu-darmstadt.de
To: cube-lovers@ai.mit.edu
Subject: Unauthorized Use of RUBIK'S CUBE and CUBE Design Marks?

About 2 weeks ago I received the following message and it seems to me
that it might be interesting for you too:

> Subject:
>         Use of RUBIK'S mark
>   Date:
>         Sun, 18 Oct 1998 21:05:31 EDT
>   From:
>         CK4IPLAW@aol.com
>     To:
>         kociemba@hrz1.hrz.tu-darmstadt.de
> 
> 
> CLEARY, KOMEN & LEWIS, LLP
> 600 Pennsylvania, Avenue, S.E.
> Suite 200
> Washington, D.C. 20003-4316
> Telephone: 202 675-4700
> Telecopier: 202 675-4716
> E-Mail: ck4iplaw@aol.com
> 
> 
> October 18, 1998
> 
> Via Electronic Mail
> 
> Herbert Kociemba
> kociemba@hrz1.hrz.th-darmstadt.de
> 
> Re:     Unauthorized Use of RUBIK'S CUBE and CUBE Design Marks
> 
> Dear Mr. Kociemba:
> 
> This firm is intellectual property counsel to Seven Towns Limited ("Seven
> Towns"), the manufacturer and worldwide distributor of the RUBIK'S CUBE three-
> dimensional puzzle and provider of an electronic version of the puzzle via its
> official web site, which is located at http://rubiks.com.
> 
> The RUBIK'S CUBE mark is famous throughout the world.  The distinctive
> overall appearance of the RUBIK'S CUBE puzzle also is a famous trademark owned
> by Seven Towns.  These marks are registered or are the subject of pending
> trademark applications in most of the major countries of the world.
> 
> It has come to our attention that your web site features a program under the
> name of Rubik's Cube Explorer.   I must advise that your unauthorized use of
> the RUBIK'S CUBE mark owned by Seven Towns constitutes trademark infringement.
> Specifically, the use of this mark in designating the origin of your program
> confuses the public into believing mistakenly that it derives from, is
> associated with, or is endorsed or sponsored by the owner of this commercial
> symbol (i.e., Seven Towns).  Moreover, apart from causing consumer confusion,
> your use of the well-known mark dilutes its distinctive value in violation of
> the federal and state anti-dilution laws.
> 
> Seven Towns appreciates your interest in the RUBIK'S CUBE puzzle, and it
> certainly does not wish to inhibit legitimate discussion of the puzzle on the
> Internet or in any other medium.  However, it also must be vigilant in
> maintaining the value and integrity of its intellectual property.  It cannot
> afford to lose control over its commercial reputation, or damage to its
> substantial goodwill, by permitting another party to use its trademarks or
> trade dress in a manner that causes source confusion or otherwise dilutes
> their selling power.  Thus, Seven Towns requests that you remove from your web
> site the electronic version of the RUBIK'S CUBE manipulative puzzle, and that
> you discontinue any further use of the term RUBIK'S CUBE or any similar
> designation in the prominent, source-indicating manner of a trademark.
> 
> I hope that you are understanding of our client's position, and I thank you
> in advance on behalf of Seven Towns for your prompt attention to this matter.
>  
>                                                         Sincerely yours,
> 
>                                                         //sjm//
> 
>                                                         Scott J. Major



Indeed the headline of my homepage at

http://home.t-online.de/home/kociemba/cube.htm

was "Rubik's Cube Explorer 1.5". So I removed the word "Rubik's" and
added some note at the bottom of the page (...blah blah is not derived
from, is not associated with blah blah...).

I definitely will not remove the program from my homepage. This seems
ridiculous. I know, that other cube fans received similar mail because
they have some similar statements on their homepages now.

What do you think about that?


Herbert Kociemba

From cube-lovers-errors@mc.lcs.mit.edu  Mon Nov  2 14:41:00 1998
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Precedence: bulk
Date: Sat, 31 Oct 1998 23:06:34 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: All the Local Maxima at 12q
To: Cube-Lovers <cube-lovers@ai.mit.edu>
Message-Id: <Pine.PMDF.3.95.981031222133.81424A-100000@PSTCC6.PSTCC.CC.TN.US>

I had originally decided not to post all the 12q local maxima because
there are so many and because not all of them look all that interesting. 
But I have been looking at them a bit more, and I think it's worth the
effort.  Some of them will be very familiar and some of them not.  I would
highlight the following ones. 

#9  is a partial isoglyph  --  four short T's (or four short U's).
#57 is the well-known four spot.
#68 is one of the more striking of several pseudo two spots.
#80 is worthy of some study.  It's the only one where the maximality
    of the inverse is 11  --  almost but not quite a local maximum.
#97 along with #98 and #99 are very striking pseudo six spots.  #97
    and #99 are almost the same pattern, and it took me a while to
    see how they differ.

In the table below, the right most column gives the maximality of the
inverse of the pattern, where the maximality is the number of moves which
go closer to Start.  The maximality of the inverse gives an indication of
how close the inverse comes to being a local maximum, with 12 indicating a
local maximum.  (The maximality of the pattern itself is not given, since
it is 12 by definition.)  The table gives the 103 local maxima of length
12q which are unique up to symmetry.

   1.  F  F  L  L  F  F  B  B  R  R  B  B    12
   2.  F  B' U  U  D  D  F' B  R  R  L  L    12
   3.  F  B  R  R  F' B' U  D  L  L  U' D'   12
   4.  U' R' L  B  L  B' R  L' D  L' U  D'   12
   5.  F  B  R  R  L  L  F  B  U  U  D  D    12
   6.  R  L' F  F  B  B  R  L' F  F  B  B    12
   7.  F  F  B  B  U  U  D  D  R  R  L  L    12
   8.  R  L' U' D  F' B  R  R  L  L  U' D    12
   9.  F  B  U  U  D  D  F' B  R  R  L  L    12
  10.  U  B' D  F  D' B' D  F' D' B  B  U'   12
  11.  L  B  B  R  D  D  R' L  B  B  L  L    12
  12.  U  R' U  L' U' R' U  L  U' R  R  U'   12
  13.  F  F  U' D  R  R  U  D  D  B  B  D'   12
  14.  F  B  D  D  F  B' L  L  U  U  F  F    12
  15.  U  D' F  U  D  R' L' U  D  F  U  D'   12
  16.  U  U  R  L  F' B' U  D' R  R  L  L    10
  17.  R  R  U  D  F' B  U' D  F  B' U' D     8
  18.  U  U  D  D  F  B' R  R  L  L  B  B    12
  19.  R' L  F' B  U  U  F  F  B  B  U' D'   12
  20.  U  D' F  F  U' D  R  L  F  B' U' D    12
  21.  F  B  R' L  D  D  F  B' R' L  U' D    12
  22.  L  L  U  D' B  B  R  L  U' D  B  B    12
  23.  F  L  L  F' D  D  F  B' L  L  B  B    12
  24.  F' R  R  F' U  U  F' B  L  L  B  B    12
  25.  F  B  U  U  L  L  B  B  L  L  U  U     8
  26.  U' D' R  L' F  F  B  B  L  L  U' D     4
  27.  U  D  R  L' F  F  B  B  L  L  U  D'    4
  28.  F  B' U  U  B  B  L  L  B  B  U  U    12
  29.  F' B  U  D  L  L  B  B  R  R  U  D    12
  30.  F' B  U  D  R  R  F  F  L  L  U  D    12
  31.  F  B' U  U  F  F  R  R  F  F  U  U    12
  32.  F  B' U  U  L  L  B  B  L  L  U  U    12
  33.  F  B' U  U  R  R  F  F  R  R  U  U    12
  34.  L  L  U  D  L  L  F' B  R  L  U  U    10
  35.  F' R  R  F  U  U  F  B' R  R  B  B    12
  36.  L  L  F' B  D  D  R  R  F  F  B  B    12
  37.  L  L  F' B  D  D  F  F  B  B  L  L    12
  38.  D  D  F' B  U  D' F  F  B  B  U  D'   10
  39.  F  F  R' L  F  B' U' D  F  F  L  L    12
  40.  U  R' L  F  U  D' R' L  F  F  B  B     8
  41.  U  D  R  L  F  B' U  D' R' L  F  F    12
  42.  U  D  F  F  U' D  R  L  F  B' U' D    12
  43.  U  D  B  B  U' D  R' L' F  B' U' D    12
  44.  F  F  R  L' F  B' U  D' F  F  L  L    12
  45.  U  D  F' B  R  L  U' D  B  B  U' D    12
  46.  U  D  F' B  R' L' U' D  F  F  U' D    12
  47.  F  B' D  R' L' U  U  R  L  D  B  B     8
  48.  B' U' D  R' L  B' U' D  R  R  L  L     6
  49.  B' U  D' R  L' F  U  D' R  R  L  L     6
  50.  U  D  F' B' U' D  L  L  F  B' U' D'    8
  51.  U' F' U' D  F  B' U' D  R  R  L  L     2
  52.  U' F  F  B  R' L  U' D  F  F  B  B     8
  53.  D  R  R  L  F' B  U' D  R  R  L  L     2
  54.  R  U' F' B  R  L' U  D' F  F  U  D'    2
  55.  L  F  F  B  R  L' F  B' R  R  L  L     2
  56.  L  F  F  B  U  D' R' L  F  F  B  B     8
  57.  F  B' U  U  D  D  F  B' R  R  L  L    12
  58.  F  B' R  L' U' D  F' B' R  L' U' D'   12
  59.  F  B  R  L  U  D' F  B  R' L' U  D'   12
  60.  F  B' U  U  D  D  F' B  U  U  D  D    12
  61.  R  L' U' D  F' B  L  L  U' D  F  F    12
  62.  R  L  B  B  U  D' R  R  F  B  U  D'   12
  63.  F  B' U  D' R  L' F  F  B  B  U  D'   12
  64.  F  B  R  R  L  L  U  D' R' L' U  D'   12
  65.  F  B' U  D' R  L' B  B  U  D' L  L    12
  66.  F  F  U' D  R  R  U  D  F  F  U  U    12
  67.  F  L  L  B  U  U  F' B  L  L  F  F    12
  68.  B  R  L  F  F  R' L' B  U  U  L  L    12
  69.  U' F  B' L' U  D' R  L' F  F  B  B     2
  70.  F  B  R' L' U  D' F' B' R' L' U' D'   12
  71.  F  B  R  L  U  D' F  B  R' L' U' D'   12
  72.  U' R  L' U' D  F  B' R  R  L  L  U'   12
  73.  F' B  R' L  U' D  B  R' L  U' D  D    12
  74.  U' R' L  F  B' U' D  F  B  B  U' D     6
  75.  U  R' L  F' B  U  D' B  F  F  U  D'    6
  76.  F  B  U  D  R' L  F  F  R' L  U' D    12
  77.  R' L' F' B  R  R  L  L  U  D  B  B    12
  78.  R  L  F  B' R  R  L  L  U' D' F  F    12
  79.  F  F  R  R  F' B' U' D' L  L  U  D     8
  80.  U  U  R' L  F' B' U' D  R  R  L  L    11
  81.  U  U  D  R  L' B  U  D' R  R  L  L     6
  82.  U  D  F' B  U' D  F  B' U' D  R  R    12
  83.  F  B' U  D  R  L' B  B  R  L' U' D    12
  84.  U  D' F  F  U  D' R' L' F  B' U' D    12
  85.  B  R  L' D' F  B' U  D' R  R  L  L     2
  86.  F' R' L  U' F  B' R' L  U  D' B  B    12
  87.  F' R' L  U' F' B  R' L  U' D  B  B    12
  88.  R' L  F  B' U  D  R  R  L  L  D  D    12
  89.  F  B  R  L' F' B  R  L' U  D' F  F    12
  90.  U' D' R  L' F  F  U  D' R  R  L  L    12
  91.  F  F  R' L  F' B' U' D' L  L  B  B    10
  92.  F  U  D  B  B  U' D' F  R  L  U  D'   10
  93.  R  L  U  D' F  B' R' L' F  B' U' D     8
  94.  F  B' U' D  L  L  F  B' U' D' F  F    10
  95.  F  B' U  U  D  D  F  B' R' L' U  D'   10
  96.  R' L  L  F' B  U  R  L' F  F  B  B     6
  97.  F  B' U' D  R  L  F' B  U  U  B  B    12
  98.  F  B  U' D  R' L  F' B  D  D  B  B    12
  99.  F  B' U' D  R  L  F  B' D  D  F  F    12
 100.  U' D  F  B' U  D' F  F  B  B  U  D'   12
 101.  U  R  L' F' U  D' R  L' F  F  B  B     8
 102.  U  D' R  L' F' B  U  D' F  F  B  B    12
 103.  F' B' L  L  U' D  F  B  U' D  R  R     8

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us
Pellissippi State                            (423) 539-7198
10915 Hardin Valley Road                     (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990

From cube-lovers-errors@mc.lcs.mit.edu  Mon Nov  2 18:00:09 1998
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Message-Id: <3.0.32.19981102164235.0094e100@mail.spc.nl>
Date: Mon, 02 Nov 1998 16:42:36 +0100
To: cube-lovers@ai.mit.edu
From: Christ van Willegen <c.v.willegen@spcgroup.nl>
Subject: Re: Unauthorized Use of RUBIK'S CUBE and CUBE Design Marks?

At 14:58 31-10-1998 +0100, you wrote:
>About 2 weeks ago I received the following message and it seems to me
>that it might be interesting for you too:

[Message deleted for brevity]
>
>Indeed the headline of my homepage at
>
>http://home.t-online.de/home/kociemba/cube.htm
>
>was "Rubik's Cube Explorer 1.5". So I removed the word "Rubik's" and
>added some note at the bottom of the page (...blah blah is not derived
>from, is not associated with blah blah...).
>
>I definitely will not remove the program from my homepage. This seems
>ridiculous. I know, that other cube fans received similar mail because
>they have some similar statements on their homepages now.
>
>What do you think about that?
>

I had the same 'problem' a couple of months ago.

A handheld computer users group I was active in, once published a
program that could be described as 'well, it looks a bit like Tetris
((R), if those lawyers are reading this, as well :-), but it's a long
way off the mark'.

The program was published in our magazine, and was also placed on a
web-page.

A couple of months ago, I received a letter from a Belgian lawyer firm,
addressed to the user's group. This user's group, by the way, died about
5 years (!) ago.

They told us to 'cease and decist' (a couple of things), including publishing
this article on 'our web-page'. Well, since this was someone else's web-
page, there was nothing we could do about it.

We told them (in friendly terms) that the user's group no longer existed,
that we were not affiliated to the user having the article on his web-page,
and that we nover sold the program.

We haven't heard from them (not even a letter stating that they received our
reply!) since.

It's kinda sad: The program was for an HP28S. I would have _loved_ to see
those lawyers type the program in (a couple of kilobytes, via the keyboard!),
only to find out that it was 'almost, but not quite, entirely unlike Tetris'
(Douglas, if you're reading this, quotes are ok, no?).

Just tell them you will take down the name, and they will probably be
off your back.

Christ van Willegen

From cube-lovers-errors@mc.lcs.mit.edu  Tue Nov  3 06:37:22 1998
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Date: Mon, 2 Nov 1998 15:32:29 -0500
From: michael reid <reid@math.brown.edu>
Message-Id: <199811022032.PAA24281@euclid.math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: Re: Unauthorized Use of RUBIK'S CUBE and CUBE Design Marks?
Cc: kociemba@hrz1.hrz.tu-darmstadt.de

> > Re:     Unauthorized Use of RUBIK'S CUBE and CUBE Design Marks
> >
> > Dear Mr. Kociemba:

[ ... ]

> What do you think about that?

i think it's not good business strategy to attack the people who
are the biggest fans of their product!  to claim ownership of
"the distinctive overall appearance of the RUBIK'S CUBE puzzle"
is ludicrous!

the changes you've described to your web page seem quite
reasonable and appropriate (given the circumstances), without
compromising too much.  i hope you have no further trouble with
seven towns.  but if you do, please let me know about it.

to make this situation even more ridiculous, i just checked out
their "official" website, which features a java cube
(http://rubiks.com/VRCUBE.html).  their applet is stolen
from karl ho"rnell!  (http://www.tdb.uu.se/~karl/java/rubik.html)

mike

From cube-lovers-errors@mc.lcs.mit.edu  Mon Nov  9 17:39:21 1998
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Date: Tue, 3 Nov 1998 19:29:38 -0800 (PST)
From: Han Wen <hansker@yahoo.com>
Subject: Query for Corners-First Method Rubik Solution
To: cube-lovers@ai.mit.edu


Hi,

Does anyone know of any websites that describe the Corners-first
method of the solving the rubik's cube?  I know of many layer-first
methods such as Jiri Fridrich's (for which I have spent many hours
learning), but I really haven't seen a comprehensive explanation of
the corners-first method.  I'm really curious to understand how anyone
can solve the cube under 30secs by solving the corners first.

-Han-

From cube-lovers-errors@mc.lcs.mit.edu  Mon Nov  9 18:35:15 1998
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Date: Wed, 04 Nov 1998 16:41:26 -0500 (Eastern Standard Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Local Maxima Whose Inverses are not  Local Maxima
To: Cube Lovers <cube-lovers@ai.mit.edu>
Message-Id: <SIMEON.9811041626.D@GN209A.PSTCC.CC.TN.US>

On 30 June 1997 I reported that if you could find a local
maximum whose inverse was not a local maximum, then you could
also find a longer local maximum.  For example, suppose x is a
local maximum in the quarter turn metric and x' is not.  Then,
there exists q in Q such that |x'q| = |x'| + 1 = |x| + 1.  But
we know that q'x is a local maximum and we also know that
|q'x| = |x| + 1 because |q'x| is the same as |x'q|.

Because we now have at 12q a good number of local maxima  whose
inverses are not local maxima as specimens, I have begun to
wonder if the same process might be able to be repeated several
times to yield progressively longer local maxima.  For example,
if x is a local maximum and (q1)x is a local maximum, might also
(q2)(q1)x be a local maximum and also (q3)(q2)(q1)x etc.  It
seems to me that good candidates to investigate in this regard
might be those local maxima at 12q whose inverses have a very
small maximality.  For example, if x is a local maximum where
the maximality of x' is 2 (and there are several such cases),
then we know that there are 10 local maxima of the form qx.

I am not sure if I have time to investigate this question
further, but I certainly would love to hear from anyone who has
the time and the computing resources to do so.

----------------------
Jerry Bryan
jbryan@pstcc.cc.tn.us



From cube-lovers-errors@mc.lcs.mit.edu  Tue Nov 10 06:10:31 1998
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Sender: bosch@sgi.com
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Date: Mon, 09 Nov 1998 15:37:18 -0800
From: Derek Bosch <bosch@sgi.com>
To: Han Wen <hansker@yahoo.com>
Cc: cube-lovers@ai.mit.edu
Subject: Re: Query for Corners-First Method Rubik Solution
References: <19981104032938.14284.rocketmail@send102.yahoomail.com>

Here's the method I use to solve Rubik's Cube in 30 seconds or less.

My notation is as follows:

  F  = turn front face clockwise 90 degrees.
  F' = turn front face counter-clockwise 90 degrees.
  F" = turn front face 180 degrees.

  U = turn top face clockwise 90 degrees.
  R = turn right face   "      "    "
  L = turn left face    "      "    "
  B = turn back face    "      "    "
  D = turn bottom face  "      "    "

  ^  = move middle slice 90 degrees up.
  v  = move middle slice 90 degrees down.

OK.  Now the order I do things is corners, edges on two OPPOSITE sides (Right
and Left), followed by the middle slice edges.

(1) Corners:
  I solve my corners a bit wierdly, but I find it is really fast.  I position
any four corners of the same color on a side.  I don't care what colors are
on the adjoining faces right now, as I fix them later.

(1a)
  Once I have the 4 corners of the same color, I turn the cube so that those
colors are on the down face.  Now there are a few combinations that can occur
on the top face:

  All corners on top face same color:   Goto (1b)

  Three corners need to rotate clockwise (position like below o=no rotate)
                                         (+ = needs clockwise rotation)
					 (- = needs counter-clkwise rot.)
       + +     Move: R'U"RUR'UR    and goto (1b)
       + o

  Three corners need counter-clockwise rotate:

       - o     Move: RU"R'U'RU'R'  and goto (1b)
       - -
  One corner needs clockwise rotate, One needs counter-clockwise rotate:

  3 cases:
       + -     Move: RU"RU"RUR"    and goto (1b)
       o o

       - +     Move: RUR'U'F'U'F   and goto (1b) 
       o o

       o -     Move: R'URUBU'B'    and goto (1b)
       + o

   Two corners need clockwise rotate, Two need counter clockwise:

   2 cases:
       + -     Move R"U"RU"R"      and goto (1b)
       - +

       - +     Move RUR"F'R"UR'    and goto (1b)
       - +

(1b)
   Now, you should have two opposite sides, with the corners of those
   two sides the proper color.

   We have to correct the 4 remaining sides to get corners in the right
   place, before we can move onto edges.  To do this, count the number
   of sides that have the upper pair of corners the same color.  Also
   counter the number of sides that have the lower pair of corners the
   same color.

   All four sides (upper and lower) corner pairs match.  Goto (2)

   No sides' corner pairs match.   Do Move R"F"R".  Goto (2)
    
   One Bottom corner pair matches.
      Move that corner pair to the Down-Left position.
      Move R"UR"U'R"UR"U'R.  Goto (2)
   One Top corner pair matches.  Turn Cube over, and do previous moves.

   One Top and one Bottom pair matches.
      Move both corner pairs to the front face.
      Move R"UR"U"F"UF".  Goto (2)

   All Bottom pairs match.
      Move R"UR"U"F"UF"U"L"UL".  Goto (2)
   All Top pairs match.  Turn Cube over, and do previous move.

   All Bottom pairs match.  One Top pair match.
      Move Top Pair to Left-Upper position.
      Move R"UR"U'R"F"U'F"UF".  Goto(2)
   All Top pairs match.  One Bottom pair match.  Turn Cube over, and do prev.

(2)  Solving two Opposite Sides.
   Now, all the corners should be solved.  You should move the center of
   each cube to its respective corners, to get an X on each side (at least
   on two opposite sides).  From now on orient the cube so that the two 
   opposite sides are right and left.

(2a)
   Solve three edges on the left face with the following moves.   

   U'^U -  moves the edge piece in the Front-Down position to the Up-Left
             position.
   UvU' -  moves the edge piece in the Back-Down position to the Up-Left
             position.

   This is easier done with a cube in your hand, and try and see how this
works.  This will mess up the centers and edges in the middle slice, as well
as the Up-Right edge.  Don't worry about this.  As long as you keep this 
orientation, and rotate the left face to get ready for a new edge to be moved
you can solve three out of four of the edges on the left face.

(2b)
  Solve four edges on the right face:
  First, rotate the left face, so that the unsolved edge is in the Up-Left
     position.

  Then, using the following moves, solve all four edges (similarly to step 2a).

   U^U' -  moves the edge piece in the Front-Down position to the Up-Right
             position.
   U'vU -  moves the edge piece in the Back-Down position to the Up-Right
             position.

(2c)
  Solve remaining edge on left face:
  2 cases (other than already solved):
    edge in place, needs to be flipped:  Use U'vUvUvU'

    otherwise, move edge to Down-Front position, using v or ^.
       if the front color (of the DF edge) is the same as the left face color,
					 Use U'vU"vU'
       else Use vU^U"^U

(3)
  Solve middle slice edges.
  First use ^ or v to position middle slice centers in proper faces.

(3a) Position edges:
  3 cases (other than all in place):
     only three edges out of place:
        position cube such that DF needs to go to UB and UB needs to go to UF.
        Use ^U"vU".
     all four edges need to move:
        if UF needs to go to DB, Use ^F"B"vF"B".
     otherwise, position so that UF needs to go to UB, Use U'^^U'^^.

(3b) Flip required edges:
   3 cases (other than no flips needed).
     all four need flipping, use FR'F'^U^U^U^UFRF'

     two edges need flipping, both on same face.  Turn cube so that these
       edges are the UF and UB edges.  use ^U^U^U"vUvUvU"
     otherwise, turn cube so UB and DF need flipping, use F"^U^U^U"vUvUvU"F"

That should do it.  I apologize for the roughness of this solution.  I think
diagrams would help it a lot.  If you have any criticism or ideas that could
help this solution become more readable, let me know.

Note, this solution is very close to Jeff Verasano (sp) and
Minh Thai's methods...

D
-- 
Derek Bosch        "A little nonsense now and then
(650) 933-2115      is relished by the wisest men"... W.Wonka
bosch@sgi.com

From cube-lovers-errors@mc.lcs.mit.edu  Tue Nov 10 07:33:39 1998
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Date: Mon, 9 Nov 1998 19:15:58 -0500 (EST)
From: Alchemist Matt <monroem@email.unc.edu>
Reply-To: Alchemist Matt <monroem@email.unc.edu>
To: Han Wen <hansker@yahoo.com>
Cc: cube-lovers@ai.mit.edu
Subject: Re: Query for Corners-First Method Rubik Solution
In-Reply-To: <19981104032938.14284.rocketmail@send102.yahoomail.com>
Message-Id: <Pine.GSO.3.95L.981109190803.17496B-100000@sunny.isis.unc.edu>


        My page at http://www.unc.edu/~monroem/rubik.html describes a
method that is sort of "corners first".  Although, in my first step I say
to solve the first layer before going on, one could effectively simply
place only the four corners in the top layer, then move on to the four
corners in the bottom layer (specified in steps 2 and 3), then begin
filling in the gaps on the top and bottom layers (steps 4 and 5), and
lastly finish the middle layer.

        In fact, a chemistry professor at my current school, Holden Thorp,
competed in one of the Rubik's cube playoff contests that was aired on the
TV show That's Incredible.  Someone posted the video of it about a month
ago (and mentioned it in this discussion list), and he saw it here at my
school after I downloaded it.  He then looked at my page and mentioned
that the winner of the contest actually used the solution shown on my page
(probably modified slightly).  I can only solve a well-scrambled cube in 2
to 3 minutes using the solution, but I'm sure someone quite adept, nimble,
and fast could push it to under one minute.  (Please note this isn't "my"
solution; I simply learned it from a book many years ago.  Further, I have
never been in a cube solving competition).

                Matt

-----------------------------------------------------------------------
Matthew Monroe				Monroem@UNC.Edu
Analytical Chemistry      		http://www.unc.edu/~monroem/
UNC - Chapel Hill, NC                   This tagline is umop apisdn

From cube-lovers-errors@mc.lcs.mit.edu  Thu Nov 12 14:18:47 1998
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Message-Id: <19981111170341.14375.rocketmail@send105.yahoomail.com>
Date: Wed, 11 Nov 1998 09:03:41 -0800 (PST)
From: Han Wen <hansker@yahoo.com>
Subject: RE: Query for Corners-First Method Rubik Solution
To: Noel Dillabough <Noel.Dillabough@ineacorp.com>
Cc: cube-lovers@ai.mit.edu

Hi,

Thanks for the link to your Puzzler program.

You're not going to believe this, but you can still purchase the
Professor's Cube (5x5x5) and the Megaminx! Since it's difficult... no,
impossible to find anyone that sell these puzzles, I think it's worth
mentioning. You can get them from Meffert's site:
http://ue.net/mefferts-puzzles/

Your Puzzler program is a tremendously useful tool to develop moves.
I've got 11/12 sides of the Megaminx solved.  But for the last side, I
need to figure out corner/edge twisting/permuting moves.  You're
Puzzler program's great for that.  I'm surprised how many of my
Rubik's cube moves can be applied with minor modifications to the
Megaminx.

-Han-

---Noel Dillabough <Noel.Dillabough@INEACORP.COM> wrote:

> I actually solve all the cubes this way (or at least centers ->
> corners -> edges for larger cubes) I just find it more logical and
> easier to memorize than other methods.

> You can check out my solution at
> http://www.mud.ca/puzzler/puzzler.html.  Its in the puzzler help
> file under "solving the cube".  I will be adding other solutions
> soon that are clearer, let me know if you would like them I could
> mail them to you.

> -Noel.

From cube-lovers-errors@mc.lcs.mit.edu  Thu Nov 12 15:00:46 1998
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Date: Thu, 12 Nov 1998 15:01:40 +0000
From: David Singmaster <david.singmaster@sbu.ac.uk>
To: cube-lovers@ai.mit.edu
Message-Id: <009CF1D5.D1348312.50@ice.sbu.ac.uk>
Subject: Use of the name Rubik's Cube

        The lawyers are being obsessively zealous as the name is
certainly well on its way to becoming a common noun.  It was included
in the Oxford English Dictionary in the mid-1980s.  Other examples are
Kleenex and Aspirin, which were both originally tradenames and their
owners fought to retain them but eventually lost.  Xerox is fighting a
rear-guard action on its name.  If you don't want to get involved in
legal hassle, I suggest that you use the name  Magic Cube  which was
the original name and is such a common term that they can't claim it
is a trademark.

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499;
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

[ Moderator's note: I am still dropping messages that consist mainly of
  generic comments on intellectual property issues.  There a great variety
  of individualistic and contentious debate on these topics that you may
  follow in dedicated fora such as the Usenet group misc.int-property.  I
  am not yet persnickety enough to elide the third and fourth sentences
  from the above, but they are on the edge.

  I will also note that the term "Magic Cube" is also used to refer to a
  cubical array of natural numbers whose orthogonal and diagonal rows sum
  to the same number, as a generalization of "Magic Square", so it is
  advisable to include context such as "The geometrical puzzle originally
  known as the Hungarian Magic Cube." ]

From cube-lovers-errors@mc.lcs.mit.edu  Wed Nov 18 12:52:47 1998
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Date: Thu, 12 Nov 1998 17:05:37 -0500 (EST)
From: Nicholas Bodley <nbodley@tiac.net>
To: Han Wen <hansker@yahoo.com>
Cc: Noel Dillabough <Noel.Dillabough@ineacorp.com>, cube-lovers@ai.mit.edu
Subject: Solutions (Was: RE: Query for Corners-First Method Rubik Solution)
In-Reply-To: <19981111170341.14375.rocketmail@send105.yahoomail.com>
Message-Id: <Pine.SUN.3.95.981112170248.21046A-100000@sunspot.tiac.net>

On Wed, 11 Nov 1998, Han Wen wrote:

{snips}

}mentioning. You can get them from Meffert's site:
}http://ue.net/mefferts-puzzles/

It might be of interest to mention that Meffert has solutions to many
puzzles at his Web site. The Contributors section gives generous credit
to a number of experts; they wrote the solutions.

Best,

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*  The personal computer industry will have become
|*  Amateur musician  *|*  mature when crashes become unacceptable.
--------------------------------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Wed Nov 18 13:27:34 1998
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Message-Id: <3.0.32.19981113091853.00948790@mail.spc.nl>
Date: Fri, 13 Nov 1998 09:18:55 +0100
To: cube-lovers@ai.mit.edu
From: Christ van Willegen <c.v.willegen@spcgroup.nl>
Subject: MegaMinx/5^3 (Was: RE: Query for Corners-First Method Rubik Solution)

At 09:03 11-11-1998 -0800, you wrote:
>Hi,
>
>Thanks for the link to your Puzzler program.
>
>You're not going to believe this, but you can still purchase the
>Professor's Cube (5x5x5) and the Megaminx! Since it's difficult... no,
>impossible to find anyone that sell these puzzles, I think it's worth
>mentioning. You can get them from Meffert's site:
>http://ue.net/mefferts-puzzles/

Also, a store in The Netherlands sells these! Last time I was there
(last saturday), they had;
- a couple of 3^3's
- a couple of 5^3's
- skewb

I can't recall if they had a Megaminx at that time.

The store is based in Eindhoven. If anybody wants some, I can buy
them and send them out.

5^3 costs F. 50 (about $25). 3^3 costs F. 10 (about $5).

Before anybody gets this wrong:
- I do not work for them, I'm a happy customer
- I don't get paid to do this
- I make no money out of this

You can call them at: +31-40-2461376

Business hours are 0900 to 1800. The Netherlands is at CET
(differs +6 hours with NY, +9 with CA)

Christ van Willegen

From cube-lovers-errors@mc.lcs.mit.edu  Wed Nov 18 17:23:08 1998
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Message-Id: <19981114070437.9789.rocketmail@attach1.rocketmail.com>
Date: Fri, 13 Nov 1998 23:04:37 -0800 (PST)
From: "Jorge E. Jaramillo" <kingeorge@rocketmail.com>
Subject: RE: Moves to this other pattern
To: David Singmaster <david.singmaster@sbu.ac.uk>, <Cube-Lovers@ai.mit.edu>

Maybe (although I don't think so since some people
already answered what I was asking) I made a mistake
when describing the position I wanted to accomplish.
What I wanted was:
      L B R
      L B R
      L B R
F F F T T T F F F D D D
L L L T T T R R R D D D
B B B T T T B B B D D D
      L F R
      L F R
      L F R

And until now the best solution is:
F B L F2 T D- L2 B- T- D- F2 R- T- D L- R D

===
Jorge E Jaramillo

From cube-lovers-errors@mc.lcs.mit.edu  Wed Nov 18 18:05:42 1998
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Message-Id: <365115AC.22369936@erco.com>
Date: Tue, 17 Nov 1998 07:20:28 +0100
From: "michael ehrt" <m.ehrt@erco.com>
Reply-To: m.ehrt@erco.org
To: Cube Lovers Mail <Cube-Lovers@ai.mit.edu>
Subject: Getting 2x2x2 cubes

If anyone is interested in getting 2x2x2 cubes, during my holiday in the
UK two weeks ago I found a shop in Sheffield which has them in stock.
It's called "The Puzzle Shop" and in situated in Meadowhall shopping
centre. The cubes are GBP 5 each, and they have a few other things like
keyring 3x3x3s etc.

Michael

From cube-lovers-errors@mc.lcs.mit.edu  Thu Nov 19 13:41:00 1998
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Message-Id: <19981117105414.18936.rocketmail@attach1.rocketmail.com>
Date: Tue, 17 Nov 1998 02:54:14 -0800 (PST)
From: "Jorge E. Jaramillo" <kingeorge@rocketmail.com>
Subject: The Cylinder
To: cube <Cube-Lovers@ai.mit.edu>

I was checking the Rubik official website and I was
surprised not to find one product that I seem to find
here (I live in Colombia South America) fairly
easily. I am talking about the cylinder.
When I first saw it I bought it and thought it was
going to be some amazing and tricky to solve puzzle,
it ended up being a 3x3 cube with the corners cut, so
corner cubelets only have 2 colors and there are two
types of borders, 8 borders with the usual two colors
and 4 with only one.

Does it mean that this cube was "invented" by some
manufacturer other than Mr Rubik and that is not so
common?
===
Jorge E Jaramillo

[Moderator's note: I own such a puzzle, but I would call its shape
 an octagonal prism, rather than a cylinder.  On mine, the solved
 position is not an octagonal prism because one beveled face is
 rotated 90 degrees, forming a decahedron whose faces are six
 rectangles and four irregular hexagons. I don't remember whether
 it was originally manufactured this way or whether I altered the
 color tabs. ]

From cube-lovers-errors@mc.lcs.mit.edu  Thu Nov 19 17:36:55 1998
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Date: Thu, 19 Nov 1998 15:00:06 -0500 (Eastern Standard Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re : The Cylinder
In-Reply-To: <19981117105414.18936.rocketmail@attach1.rocketmail.com>
To: "Jorge E. Jaramillo" <kingeorge@rocketmail.com>
Cc: cube <Cube-Lovers@ai.mit.edu>
Message-Id: <SIMEON.9811191506.A@GN209A.PSTCC.CC.TN.US>

Jorge E. Jaramillo's message of 17 Nov 1998 included:

>[Moderator's note: I own such a puzzle, but I would call its shape
>  an octagonal prism, rather than a cylinder.  On mine, the solved
>  position is not an octagonal prism because one beveled face is
>  rotated 90 degrees, forming a decahedron whose faces are six
>  rectangles and four irregular hexagons. I don't remember whether
>  it was originally manufactured this way or whether I altered the
>  color tabs. ]

I also own such a puzzle, although I have never seen one in 
a store.  I got mine at a garage sale for $0.25.

I haven't played with it in a long time.  But my best
recollection is that it can be solved basically the same
way as a 3x3x3 cube, except that *I think* (don't remember
for sure) that the color scheme permits invisible swaps of
identically colored pieces which can make the puzzle seem
"impossible" to solve unless you realize that the
identically colored pieces must be swapped.  It is also my
best recollection that such a puzzle is mentioned briefly
and is pictured in one of Douglas Hofstadters's cube
articles in Scientific American back in the early 80's.  So
I don't think it is any kind of new invention.

----------------------------------------
Jerry Bryan
jbryan@pstcc.cc.tn.us
Pellissippi State Technical Community College

From cube-lovers-errors@mc.lcs.mit.edu  Fri Nov 20 11:11:16 1998
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Date: Mon, 16 Nov 1998 23:01:39 -0800 (PST)
From: Han Wen <hansker@yahoo.com>
Subject: Fwd: Re: HOW TO SOLVE THE PROBLEM OF THE PROFESSOR CUBE 
To: cube-lovers@ai.mit.edu

I thought this would be of value to Rubik fans out there with
Professor Cubes (5x5x5)...

note: forwarded msgs attached.
_________________________________________________________

---Uwe Meffert <uwe@ue.net> wrote: [reprinted here with his permission]

> Dear Mr. Wen

> Thank you for your interest in my puzzles, I am sorry to hear that
> you are having problems with the Prof. Cube.

> You received one of the last ones from the last production batch and
> the next production is not for another month.

> What unfortunately happened is that when gluing the center small
> caps excess glue fixed the screw to the plastic centre piece that it
> should turn in.  So when you turn these sections it will tighten /
> loosen that one screw.

> If you are skilful enough you can try and carefully remove the
> centre label and then pry open and remove the centre cap of the blue
> and orange side.  Then try to remove the excess glue from around the
> screw with a sharp object and try turning the screw with a
> screwdriver firmly holding the plastic piece so you can break the
> glue bond.  Once the screw can freely turn inside the plastic part,
> re-tighten it to the same tension as it was originally, so as to
> allow smooth turning without any pieces falling out during play.
> Then carefully using only very little glue fix the centre cap back
> into place and re-attach the color label.

> Good Luck and Happy Puzzeling.

> Please let me know the outcome of this recommended procedure.


> With warm regards
> Uwe Meffert
________________________________________________________________

Date: Mon, 16 Nov 1998 22:57:20 -0800 (PST)
From: Han Wen <hansker@yahoo.com>
Subject: Re: HOW TO SOLVE THE PROBLEM OF THE PROFESSOR CUBE 
To: Uwe Meffert <uwe@ue.net>
Cc: Jing Meffert <jing@ue.net>

Hi,

My cube is all fixed.

Thank you for your prompt reply.  You were right, the glue used to fix
the caps also fixed the spindle screw! Actually, your instructions
gave me the perfect excuse to take your cube apart.  I was dying to
find out how the heck all these pieces are held together.  

Anyways, I popped of the caps carefully using a razor blade, scraped
off all the excess glue, greased the screw head and before screwing it
back together, I took all the pieces for one face out just to see and
understand the engineering holding all the pieces together. Wow, what
an amazing bit of engineering.  It's like a cube spindle inside
another cube spindle!  Amazing. 

Actually, the center caps don't really need glue.  They fit nice and
snug, and it also leaves me the option to adjust the screw again in
case it becomes loose.

Now that I understand the mechanism, I've decided to only rotate faces
clockwise to minimize the possibility that a counter-clockwise
rotation will actually loosen one of the spindle screws.

I still haven't messed the faces up though. I'm so close to finishing
the Megaminx.  I just have two edge pieces to swap on the last face! 
The other 11 sides were fairly straightforward to solve. I also got
the corners of the last face fairly quickly by using Sune's move to
twist corner pieces (a standard Rubik's cube move).  However, getting
those edge pieces was a different story.  I had to develop quite a few
moves to rotate and twist the edge pieces around.  I'm close... so
close..! :)

I hope you keep inventing and making new puzzles.  I eagerly click on
your new releases on your web page quite regularly, hoping to find a
worthy successor to the Megaminx or the Professor's Cube.  Maybe a
7x7x7?!! Or a Buckyball?  One can only imagine...

-Han-

From cube-lovers-errors@mc.lcs.mit.edu  Fri Nov 20 14:44:20 1998
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Message-Id: <3654ED01.E6D38EE8@hurstlinks.com>
Date: Thu, 19 Nov 1998 23:16:01 -0500
From: "Guy N. Hurst" <gnhurst@hurstlinks.com>
Organization: HurstLinks Sites On the Internet <http://www.hurstlinks.com>
To: "Jorge E. Jaramillo" <kingeorge@rocketmail.com>
Cc: cube <Cube-Lovers@ai.mit.edu>
Subject: Re: The Cylinder
References: <19981117105414.18936.rocketmail@attach1.rocketmail.com>

I have seen one or both of these puzzles, and they were very different
from each other.  The cylinder, or prism, was actually the first cube
I learned to solve, when my cousin from Luxembourg visited back in
1981. I have pleasant memories of it, because it was very well made
and pleasing to view. It is harder to solve than the cube since the
four of the edges are "cut", so it is impossible to match edges to
centers - leaving the possibility of having to backtrack later and
figure out which "corner" (and matching edge) is in the wrong place!
But I had it down and could quickly readjust (usually had to swap
corners diagonally in the top two layers if I found a single flipped
edge left in the bottom layer when almost done solving it, if I
remember).
I liked it so much, I requested and obtained 4 more after my cousin
returned to Europe! I would take them to school, one at a time, until
(unfortunately) they all eventually disappeared. At least two were
stolen out of my (locked) locker on different occasions.  Someone else
liked them, too.
Anyway, I never found that puzzle in the US, and could only get it
from my cousin in Europe. (Who I think may have gotten them from
England).
But the other puzzle, as described by the moderator, was available in
the US back then, I think in the following year or so after my cousin
visited, since one of my friends had one. But it wasn't as nice
looking or well made.  So I didn't care for it. It was more like the
cube with its corners cut, forming rectangles and triangles in a
spherical symmetry, as opposed to the prism from Europe which has four
of its 3-piece-edges cut, forming only rectangles and having a
cylindrical symmetry.

Guy N. Hurst

From cube-lovers-errors@mc.lcs.mit.edu  Fri Nov 20 15:16:19 1998
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Date: Fri, 20 Nov 1998 08:46:30 +0100 (CET)
From: Bas de Bakker <basde.bakker@pica.nl>
To: Cube-Lovers@ai.mit.edu
In-Reply-To: <SIMEON.9811191506.A@GN209A.PSTCC.CC.TN.US> (message from Jerry
	Bryan on Thu, 19 Nov 1998 15:00:06 -0500 (Eastern Standard Time))
Subject: Re: The Cylinder
References:  <SIMEON.9811191506.A@GN209A.PSTCC.CC.TN.US>

>>>>> "Jerry" == Jerry Bryan <jbryan@pstcc.cc.tn.us> writes:

[About the octagonal "cube"]

    Jerry> I haven't played with it in a long time.  But my best
    Jerry> recollection is that it can be solved basically the same
    Jerry> way as a 3x3x3 cube, except that *I think* (don't remember
    Jerry> for sure) that the color scheme permits invisible swaps of
    Jerry> identically colored pieces which can make the puzzle seem
    Jerry> "impossible" to solve unless you realize that the
    Jerry> identically colored pieces must be swapped.

Your recollection is not exact.  There are no identically colored
pieces to swap, but you can swap complete columns consisting of two
"corners" (what would have been corners on the cube) and one "edge"
without noticing.

In fact, if you create an even permutation of those columns, there is
no problem.  But if you create an odd permutation, it will become
impossible to solve the upper layer.

Presuming you solve cubes in layers, the easiest way out is to not
start at one of the octagonal layers (which seems the most natural
way), but to start with a "side" layer.  If you do it this way, it
will always be possible to solve the last layer.

I hope I'm making myself at least somewhat clear,
Bas.

From cube-lovers-errors@mc.lcs.mit.edu  Fri Nov 20 17:41:06 1998
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Message-Id: <3.0.1.32.19981120105340.009ac040@icex5.cc.ic.ac.uk>
Date: Fri, 20 Nov 1998 10:53:40 +0000
To: jbryan@pstcc.cc.tn.us
From: "Andrew R. Southern" <a.southern@ic.ac.uk>
Subject: Space Shuttle.
Cc: Cube-Lovers@ai.mit.edu, kingeorge@rocketmail.com

I got a similar puzzle in the same way. Mine was called the "Space
Shuttle". The lack of the name Rubik probably meant it wasn't from Rubiks.
But since they only have a patent in Hungary, and everywhere else they are
protected by copyrights on the name and the external appearence, this is
probably legally legitimate. The colour scheme allowed the puzzle to be
solved in more ways than the cube and so I reckon its easier.

Each of the chamfered sides was coloured in a colour which did not relate
to the rest of the puzzle, and so these were (within the boundaries of a
2-swap) possible to position ("correctly") in a few different positions.

The mid-edges of the chamfered edges were rotationally symmetrical every
180 degreees and so, since they were all one colour, it was possible to
have one (or three) of them rotated, and hence one of the other mid edges
rotated and it would look majorly FUBAR'd. Once you'd realised what
happened, the puzzle was easier than the cube though.


-Andy Southern.

From cube-lovers-errors@mc.lcs.mit.edu  Mon Nov 23 13:45:28 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Fri Nov 20 22:15:27 1998
Date: Fri, 20 Nov 1998 22:13:38 -0500 (EST)
From: Nicholas Bodley <nbodley@tiac.net>
To: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Cc: "Jorge E. Jaramillo" <kingeorge@rocketmail.com>,
        cube <Cube-Lovers@ai.mit.edu>
Subject: Re: Re : The Cylinder
In-Reply-To: <SIMEON.9811191506.A@GN209A.PSTCC.CC.TN.US>

Uwe Meffert had a printed color catalogue around 1985 that showed some
very interesting moving-piece "group theory" puzzles (is that a proper
term?). Although I have a copy safely stashed somewhere, I don't know
where.

I'm just about sure that one was a cylinder, possibly in three layers
like layer cake; it also, iirc, had maybe three more "cutting planes"
that were spherical sectors bounded by the cylinder. Rotating the pieces
would exchange top and bottom.


|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*  The personal computer industry will have become
|*  Amateur musician  *|*  mature when crashes become unacceptable.
--------------------------------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Mon Nov 23 16:35:08 1998
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From: roger.broadie@iclweb.com (Roger Broadie)
To: "cube" <Cube-Lovers@ai.mit.edu>
Cc: "Jorge E. Jaramillo" <kingeorge@rocketmail.com>
Subject: Re: The Cylinder
Date: Sat, 21 Nov 1998 14:29:22 -0000

I was given a cylinder here in England in 1981.  I no longer have the
packaging, but I suspect it was Taiwanese, unless the Hungarians made
this variant.  It was my first cube puzzle, and its shape was so
unappealing when disturbed that I put it on one side and got a genuine
cube to learn on - well, almost genuine: it came from a street trader
in Regent Street.

The apparently impossible state is a monoflip of a top or bottom edge
piece.  There will be a matching flip of a middle-layer edge piece,
but that will be invisible, since the piece has only one face.

I wondered if it would be possible to get the puzzle into the solved
shape and then restore the positions of the pieces without losing the
shape, that is, only allowing turns from the group <U, D, F2, L2, B2,
R2, S, A>, where S and A are slice and anti-slice moves of the middle
layers (I needed them).  In fact it is not.  There may always be a
hidden flip in the middle layer and you can't correct that without
moving the piece out of the middle layer, which needs a turn like F,
and that destroys the shape.  But if you cheat a little and make sure
the flips are got right before the shape is finally restored, then it
can be done.

Andy  Southern has already made the point about the flips.  He also
pointed out that the configuration is not unique because columns
corresponding to the vertical edges on a normal cube can be swapped.
As a rider to that point, the pretty pattern stripes on the normal
cube is not distinguishable on the octagonal prism, because it's
striped already.

It should be possible to work out whether our moderator's puzzle came
in the form he now has by counting stickers.  The octagonal prism has
2 sets of 9 (the top and bottom) and 8 of 3 (the side columns).  If
I've grasped his configuration correctly, if it came in that form
originally it should have 4 sets of 6 and 6 of 3.

Roger Broadie

[ Yes, my decahedron's stickers are incompatible with an octagonal
  prism solution.  I just can't remember whether I replaced some of
  the stickers to make this new shape. --Dan ]

From cube-lovers-errors@mc.lcs.mit.edu  Mon Nov 23 18:13:16 1998
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Date: Sun, 22 Nov 1998 23:19:31 -0800 (PST)
From: Han Wen <hansker@yahoo.com>
Subject: Method for Solving the Megaminx
To: Cube-Lovers@ai.mit.edu

Hi,

Well, it took me a 2-3 days, but I finally solved the Megaminx. Whew.
I know, big deal. It's been done.. many many times... over a decade
ago. But, I thought some of the moves I found may of interest to some
of the Megaminx aficionados out there.  So, here it is, I apologize
for its length:

Solving the Megaminx faces 1-11 are fairly straightforward.
Ironically, the larger number of faces makes it easier to solve than
the Rubik's cube, because they provide a lot more "free lanes" to move
pieces around.  There's actually just one move you need to remember to
solve these faces.  It's the same move when solving the middle layer
of the Rubik's cube, when you want to move edge pieces from the bottom
layer to their respective position in the middle layer.  Namely,

D'R'DRF'RFR'

Solving the last face, however, is another matter. The general
strategy I followed is the same as some of the standard methods for
solving the bottom layer of the Rubik's cube.  Namely, I first solve
the 5 corners, then I solve the 5 edge pieces.  To solve the corners,
I simply used Sune's move applied with slight modification to the
Megaminx.  For the remaining edge pieces, I had to develop moves that
only moved the edge pieces around, while leaving the corners
unchanged.  Noel Dillabough's Puzzler program was an invaluable tool
for helping me experiment with various edge moves.  Anyways, the
following are my notes describing some of the more useful moves I've
found.  I'm pretty sure they're not the most efficient method for
solving the Megaminx, but they're the best I could come up with.
___________________________________________
Notation for Solving the Last Face corner pieces:

F=Front Face, D=Lower Face, L=Left Face, R=Right Face
The F and D faces are adjacent

The last layer containing the corners you need to flip/permute should
be positioned at the D-face
____________________________________________
Move for Solving the Last Face corner pieces:

Name: Sune's Double-Swap
Description: Sune's Rubik's Cube move applied to the Megaminx
Number of pairs of corners swapped: 2
Number of corners twisted counterclockwise: 3
Move: R'D'RD'R'D'3R
____________________________________________
Strategy for Solving the Last Face corner pieces:

- Position the corners
- Twist the corners in place by applying Sune's Double-Swap
   move twice

============================================
Notation for Solving the Last Face edge pieces:

F=Front Face, U=Upper Face, L=Left Face, R=Right Face
The F and U faces are adjacent

X= L'R U2 LR' F2
X'=L'R U'2 LR' F'2
X2= X X = (L'R U2 LR' F2) (L'R U2 LR' F2)
Xa= L'R U2 LR' F'2
Y= LR' F2 L'R U2

The last layer containing the edges you need to flip/permute should be
positioned as the F-face or the U-face depending on the move described
below:
_____________________________________________
Moves for Solving the Last Face edge pieces:

Name: F Tricycle 1
Description: Permutes 3 adjacent edges clockwise on the lower left of
the F-face
No. Edges permuted: 3
No. Edges flipped: 2
Move: (Xa3 X'2)^2

Name: F Tricycle 2
Description: Permutes 3 adjacent edges clockwise on the upper half of
the F-face
No. Edges permuted: 3
No. Edges flipped: 2
Move: (Xa3 X2)^2

Name: U Tricycle 1
Description: Permutes 3 edges clockwise on the U-face
No. Edges permuted: 3
No. Edges flipped: 2
Move: F' X2 Y'2 F

Name: U Tricycle 2
Description: Permutes 3 edges counterclockwise on the U-face
No. Edges permuted: 3
No. Edges flipped: 2
Move: F X'2 Y2 F'

Name: Cross-country Tricycle
Description: Permutes 3 edges across the U and F faces
No. Edges permuted: 3
No. Edges flipped: 1
Move: (X2 X'2)^4

Name: U Bi-Flip 1
Description: Flips two opposite edges on the U-face
No. Edges permuted: 0
No. Edges flipped: 2
Move: (Xa3 X'2)^3

Name: U Bi-Flip 2
Description: Flips two adjacent edges on the U-face
No. Edges permuted: 0
No. Edges flipped: 2
Move: (X2 X2 X'2)^5

Name: Cross-country Bi-Flip
Description: Flips two edges, one on the U-face, one on the F-face
No. Edges permuted: 0
No. Edges flipped: 2
Move: (Xa3 X2)^3

Name: "W"-Cycle
Description: Permutes all edges on the F-face in a "W" pattern
No. Edges permuted: 5
No. Edges flipped: 2
Move: (X2 X2 X'2)^2

Name: "Figure 8"-Cycle
Description: Permutes all edges on the F-face in a "Figure 8" pattern
No. Edges permuted: 5
No. Edges flipped: 4
Move: (X2 X2 X'2)^4
____________________________________________
Strategy for Solving the Last Face edge pieces:

- You should only need to use F Tricycle and the Bi-Flip moves to
  completely solve the edges.  The F Tricycle move usually needs
  to be applied twice.

If anything is vague/unclear please feel free to request clarification.

-Han Wen-

From cube-lovers-errors@mc.lcs.mit.edu  Tue Nov 24 16:09:26 1998
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To: cube-lovers@ai.mit.edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Re: The Cylinder
Date: 24 Nov 1998 18:53:56 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <73evc4$cq0@gap.cco.caltech.edu>
References: <cube-lovers.001e01be155b$9a456940$6bc4b0c2@home.icl.web>

roger.broadie@iclweb.com (Roger Broadie) writes:
>I was given a cylinder here in England in 1981.  I no longer have the
>packaging, but I suspect it was Taiwanese, unless the Hungarians made
>this variant.  It was my first cube puzzle, and its shape was so
>unappealing when disturbed that I put it on one side and got a genuine
>cube to learn on - well, almost genuine: it came from a street trader
>in Regent Street.

There is a Taiwanese manufacture of the octagonal prism.  I have
part of one in my collection.  (Got it when I was 10, and many
cubies have disappeared since then.)

I also have one of the "truncated cubes" mentioned earlier in this
thread.

I find the discussion on these two quite strange, since I always
thought of these as cubes with weird cubies -- no more special than,
say, that spherical "cube" they had a few years back.

--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
StethoPHONE, not stethoSCOPE.  What do doctors SEE in those things anyway?

From cube-lovers-errors@mc.lcs.mit.edu  Tue Dec  1 14:27:08 1998
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To: Cube-Lovers@ai.mit.edu
From: "Andrew R. Southern" <a.southern@ic.ac.uk>
Subject: Uwe Meffert's Re-issueing of Prof. Cube
Message-Id: <E0zkc71-00072b-00.1998-11-30-22-48-11@mail2.svr.pol.co.uk>
Date: Mon, 30 Nov 1998 22:48:11 +0000

Dear Cube Lovers,

I have written a website for Uwe Meffert (with input from both W. David
Joyner and David Byrden) that can be found at:

http://www.ue.net/mefferts-puzzles/

and was speaking with him earlier today.

Uwe is going to make another batch of Professor Cubes (5x5x5) in the next
week or so, and is taking orders through his site.

This is a subject that is often raised on the newsgroup, and I hope people
don't think of this as taking too much of a liberty.

The website contains a credit card order page, information about the puzzles
(including a solution to all of his popular puzzles) and multiple links to
other pages. Whilst I am not involved with the day to day running of the
website, if people would like their pages added to the links, please forward
the URL to this address WITH A SHORT SUMMARY that will appear with the link.

Puzzle available include some of the more recent ones (Orbix, Pyramorphix,
Megaminx, Prof Cube).

I am told that orders *may* still be on time for delivery before Chirstmas.

I hope this has been of use to you guys,

Andy Southern.
a.southern@ic.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Tue Dec  1 15:45:22 1998
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Date: Mon, 30 Nov 1998 22:09:18 -0800 (PST)
From: Han Wen <hansker@yahoo.com>
Subject: Method for Solving the Professor's Cube (5x5x5)
To: Cube Lovers <Cube-Lovers@ai.mit.edu>
Cc: Charles Lin <clin@ibm.net>, Keith Miller <Keith_A_Miller@amat.com>

Hi,

Okay, another so what, big deal.  I finally solved the Professor's
Cube. For those who may not be familiar, the Professor's Cube is a
5x5x5 Rubik's cube. Whew that was hard.  It took me a good 4 days to
figure out all the moves.  Gees, it made the Megaminx seem like
child's play in comparison.  Once again, Noel Dillabough's Puzzler
program was an invaluable tool to visualize and experiment with
various moves.  Thanks Noel!

For those brave souls who would like to conquer this beast, the
following solution may provide some enlightenment.  It's a layers
solution, in contrast to the corners-first solution that I have seen
posted on various web sites. Good luck to you.  The Professor's Cube
is a truly challenging puzzle.
______________________________________________________
Method for Solving the Professor's Cube (5x5x5)

I will use Noel Dillabough's system for referring to various slices or
layers, as described in his Puzzler's F1 help.
________________________________________
Notation:

U - The upper slice
u - One slice away from the upper slice
e - The equator slice
d - One slice away from the lower slice
D - The lower slice

L - The leftmost slice
l - One slice away from the leftmost slice
m - The middle slice
r - One slice away from the rightmost slice
R - The rightmost slice

F - The facing slice
f - Once slice away from the facing slice
M - The facing middle slice
b - One slice away from the back face
B - The back slice

I will use the words "slice" and "layer" synonymously. A "face" is one
of the six outer slices; namely, U, D, L, R, F or B.  Rotations of the
middle slices e, m or M will be in the same direction as the U, R and
F faces, respectively.

Let y denote one of the slices.
y - represents a clockwise 1/4 turn of the y-slice
y' - represents a counterclockwise 1/4 turn of the y-slice
y2 - represents a clockwise 1/2 turn of the y-slice
(For example, Rrm represents clockwise 1/4 turns using the RIGHT-hand
of the R, r and m slices.  Ll represents clockwise 1/4 turns using the
LEFT-hand of the L and l slices.)

Finally, let's consider the pieces or cubelets on any given face.
There are four types of cubelets: corners, edges, centrals and a
center.  For a given face, there are 4 corners, 12 edges, 8 centrals
and 1 center.  With these four types and the intersection of any two
slices using Dillabough's notation, we can specify the location of any
cubelet.  For example, consider the F-face:

LU-corner: the corner cubelet on the upper left-hand corner of the
F-face
re-central: the central cubelet adjacent and to the right of the
center cubelet of the F-face
_______________________________________
First Layer (U slice):

Solving the first layer is fairly straightforward.  Basically the same
as solving the Rubik's cube.  The central pieces are the only thing
really different.
_______________________________________
Second Layer (u slice):

1. First, solve for the mid-central pieces (F-face mu, B-face mu,
L-face Mu, R-face Mu).  Get one of the mid-central piece on the same
color face, and then rotate it into position by using the "free lane"
from the opposite face.  For example, let's say we want have a
mid-central piece at the re position of the F-face.  Use the D-"free
lane" of the B face to position the mid-central piece without
affecting your newly completed U slice, by moving: B2 U2 F' U2 B2.

2. Now, solve for the left and right central pieces (F-face lu, ru,
L-face bu, fu, etc).  Here's where we'll use a genuinely new move.
Position one of the left/right central pieces on the D-face so that it
and the position you want to move the cubelet into lie in the save
vertical slice.  For example, let's say we want to move the left
central cubelet into the F-face lu position.  Position the left
central cubelet at the D-face lb position and perform the following
u-layer DF Swing move:

>From the D-face lb position: l d' l' d' l d2 l'

See how that works?  The corresponding move at the D-face rb position
is:

>From the D-face rb position: r' d r d r' d2 r

This same concept is used to move the left/right central pieces into
position for both the Second (u-slice) and Fourth (d-slice) layers.

"Hey, what if my left/right central piece is on the F face? How do I
move the piece to the D face so that I can apply this move?"  Good
question.  Position the piece on the F-face ld or rd position and
apply the corresponding move described above.  That should move the
cubelet to the D face where you can then apply the move again to move
it into the correct left/right central position.

3. Finally, solve for the left and right edges (F-face and B-face Lu,
Ru).  Use the classic Rubik's cube move to rotate an D-edge piece into
one of the middle layer edge positions.  Namely, if the cubelet is at
the F-face rD or lD position and the destination position is F-face Ru
or Lu then perform the following:

F-Edge Swing Moves:
Destination position F-face Ru: D' R' D R F' R F R'
Destination position F-face Lu: D L D' L' F L' F' L
_______________________________________
Third Layer (e slice):

1. Solve for the left/right central pieces (F-face le, re, L-face be,
fe, etc).  You'll notice that the DF Swing moves will not work here.
Darn.  Instead, we'll use the F-Edge Swing move adapted for the l and
r slices.  Position the cubelet at the F-face md position then perform
the following:

F-Central Swing Moves:
Destination position F-face Re: d'r'dD rR f'F' r fF r'R'
Destination position F-face Le: d l d'D' l'L' fF l' f'F' lL

"Hey,  what if my left/right central piece is on the D face? How do I
move the piece to the F face so that I can apply this move?"  Same
problem.  Position the cubelet at the D-face rM position then apply
the Re F-Central Swing move.

2. Solve for the left and right edges (F-face and B-face Le, Re).
Again, a slight variation of the F-Edge Swing move will do.  Position
the edge piece on the F-face mD position and perform the following:

e-Layer F-edge Swing Moves:
Destination position F-face Re: D' R' D rR F' R F r'R'
Destination position F-face Le: D L D' L'l' F L' F' Ll
______________________________________
Fourth Layer (d slice):

1. First, solve for the mid-central pieces (F-face md, B-face md,
L-face Md, R-face Md). This is one of the most difficult steps.  The
mid-central pieces will be on either the d-slice or on the D-face. To
move them into there correct positions, you'll need to use a few
modified Rubik's cube moves:

Place the D-face as the U-face when applying these moves:

The following sets of cubelets are affected by these moves:
cL = (central L-face Lu, edge U-face LM and central U-face lM)
cR = (central R-face Ru, edge U-face RM and central U-face rM)
cF = (central F-face mu, edge U-face mF and central U-face mf)

Mid-central Tricycle:
move: T2(U) = F2 f2 Uu Ll r'R' F2 f2 L'l' rR Uu F2 f2
action: Permutes the three sets of cubelets (cL, cR, cF) clockwise:

Mid-central Bi-Flip Tricycle:
move:  S2(B) = L'l' rR bB Ll r'R' U2u2 L'l' rR Bb Ll r'R'
action: Permutes the three sets of cubelets (cL, cR, cF) clockwise and
flips the cR and cF sets.  Let's clarify "flipping".  Let's say for
the cR set you have the colors: blue, (blue, yellow), yellow
corresponding to the three cubelets. After flipping the cR set you'll
have the colors: yellow, (yellow, blue), blue.

Use these two moves to position all the mid-central pieces for the
Fourth Layer.  Now, if you're lucky, and Murphy's Law says that you
will be, you may end up in a configuration where you'll have three of
the mid-central pieces positioned properly, but the fourth mid-central
position will be on the D-face.  Okay, now we're going to start having
fun.  Position the central cubelet at the D-face lM position (i.e. on
the left-hand side).  Place the D-face as the U-face and then apply
the following sequence of moves:

S2(B) T2(U') U2 T2(U) S2(B') U' S2(B')

Yes, all that trouble just to move one mid-central cubelet from the
U-face to the F-face.

2. Whew, congratulate yourself if you've made it this far.  Now, solve
for the left/right central cubelets, (F-face ld, rd, L-face bd, fd,
etc).

Position the left central cubelet at the D-face lf or rf position and
perform the following d-layer DF Swing move:

>From the D-face lf position: l d l' d l d2 l'
>From the D-face rf position: r' d' r d' r' d2 r

3. Solve for the left and right edges (F-face and B-face Ld, Rd).
Again, a slight variation of the F-Edge Swing move will do.  Position
the edge piece on the F-face lD or rD position and perform the
following:

d-Layer F-edge Swing Moves:
Destination position F-face Rd: D' R' D mrR F' R F m'r'R'
Destination position F-face Ld: D L D' L'l'm F L' F' Llm'
______________________________________
Fifth Layer (D slice):

1. Solve for the corner cubelets using standard Rubik's cube moves.
First, position the corners in their correct locations using the usual
corner swappers:

Adjacent corners swap: R' D' R F D F' R' D R D2
Diagonal corners swap: R' D' R F D2 F' R' D R D

And then rotate or twist the corners in position using Sune's move:

Sune's 3-corner twister: : R' D' R D' R' D2 R D2

2. Solve for the mid-edges (mF, RM, mB, LM) using a slight
modification to the Tricycle moves.

Place the D-face as the U-face when applying these moves:

Mid-edge Tricycle:
move: F2 U Ll r'R' F2 L'l' rR U F2
action: Permutes the three edges (LM, RM, mF) clockwise:

Mid-edge Bi-Flip Tricycle:
move: L'l' rR B Ll r'R' U2 L'l' rR B Ll r'R'
action: Permutes the three edges (LM, RM, mF) clockwise and flips the
RM and mF.

3. Solve for the left/right edges (lF,rF, Rf, Rb, lB, rB, Lf, Lb).
Now, we're going to have some serious fun. The hardest part of this
step is not getting lost while performing the long sequence of moves.
Also while spinning all these slices, another difficulty is preventing
the cube from exploding and keeping the central pieces from twisting
around.

Again, place the D-face as the U-face with applying these collection
of moves:

LR-edge Tricycle:
move: F2 U Lm'R' F2 L'mR U F2
action: Permutes the three pairs of edges ((Lf,Lb), (Rf,Rb), (lF,rF))
clockwise:

LR-edge Bi-Flip Tricycle:
move: L'mR B Lm'R' U2 L'mR B Lm'R'
action: Permutes the three pairs of edges ((Lf,Lb), (Rf,Rb), (lF,rF))
clockwise and flips the Rf, Rb, lF and rF.

To get those last remaining cubelets in place, a few more exotic moves
are necessary:

Definitions:
T(x) = F2 U x F2 x' U F2
x1 = L r' R'
x2 = L l R'
x3 = L m' R'
(T(x) is a generalized form of the Mid-edge Tricycle)

X1 = T(x1) T(x1) T(x1)
X2 = T(x2) T(x2) T(x2)
X3 = T(x3)

Name: Double pair F swap
Description: Swap two pairs of edges: (lF - Lb)  and (rF - Rb)
Move: X2 X1

Name: Double pair F cross swap
Description: Swap two pairs of edges: (lF -Lf ) and (rF -Rf )
Move: X1 X2

Name: Double pair R swap
Description: Swap two pairs of edges: (Rb - Lf) and (Rf - lF)
Move: X2 X1 X3

Name: Double pair R cross swap
Decription: Swap two pairs of edges:  (Rb - rF) and (Rf - Lb)
Move: X1 X2 X3

Name: Double pair L swap
Description: Swap two pairs of edges: (Lb - Rf) and (Lf - rF)
Move: X3 X2 X1

Name: Double pair L cross swap
Description: Swap two pairs of edges: (Lb - lF) and (Lf - Rb)
Move: X3 X1 X2

Name: LRL-edge Bi-Flip Tricycle
Description: Permutes (lF, Lf, Rf) edges clockwise and flip lF and Lf
edges
Move: X3 X1

Name: LLR-edge Bi-Flip Tricycle
Description: Permutes (lF, Lb, Rb) edges clockwise and flip Lb and Rb
edges
Move: X1 X3

Name: RRL-edge Bi-Flip Tricycle
Description: Permutes (rF, Lf, Rf) edges clockwise and flip Lf and Rf
edges
Move: X2 X3

Name: RLR-edge Bi-Flip Tricycle
Description: Permutes (rF, Lb, Rb) edges clockwise and flip rF and Lb
edges
Move: X3 X2

With these collection of moves, you should be able to finish off the
Professor's Cube!   *Sigh*

-Han-

P.S. Thanks "Professor" Meffert.  For those folks like myself who have
wrestled and completed your 5x5x5 cube, we can only ask and plead,
"What's Next?!!"  :)

From cube-lovers-errors@mc.lcs.mit.edu  Tue Dec  1 19:18:47 1998
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Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Date: Mon, 30 Nov 1998 23:13:51 -0500
From: michael reid <reid@math.brown.edu>
Message-Id: <199812010413.XAA11878@euclid.math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: new types of cyclic shifters

a few months ago, i introduced the position  superflip composed with
four spot  and showed that it had a new type of cyclic shifting
property.  i've now found some new ways to generalize cyclic shifting,
and this in turn suggests some new positions to consider.

first, some brief review.

the position  superflip  is central, so it commutes with all turns.
therefore, if  x y  produces superflip, so does  y x .  we can shift
one turn at a time, removing the first turn of the sequence and
shifting it to the end.  in other words, if  m  produces superflip,
then  x m = m x  for all turns  x .  clearly, any position  m  with
the property that  x m = m x  for all turns  x ,  is central, so the
only such positions are  superflip  and  start.

i showed in an earlier message, that if  x  is any turn and  m  is
superflip composed with four spot, then  x m = m y , where  y  is
x  conjugated by the cube rotation  C_U2 .  more generally, we can
ask for positions  m  such that for any turn  x , there is another
turn  y  satisfying  x m = m y .  for such a position, we can
cyclically shift any maneuver, one turn at a time, by replacing
the turn  x^(-1)  at the beginning with the corresponding  y^(-1)
at the end.  some other positions with this property are:  four
spot, six spot, six spot composed with superflip.

a new way to generalize this is to consider positions  m  such that
for any turn  x ,  we have  x m = n x , where  n  is the same
pattern as  m , but perhaps in a different orientation.  for such
a position, we can cyclically shift any maneuver, by shifting the
first turn to the end, and then conjugating by the appropriate cube
symmetry.  for example, consider the position in which the UFR corner
is twisted clockwise, and the other seven corners are twisted
counterclockwise.  (i'll call this "1-7-twist" for now, but this
pattern needs a better name.)  this position is created by

     U  F2 B' U  B  U  D2 R2 U2 B  U' D2 B  U  F2 B  L2 B2

now, cyclically shift the  U  at the beginning to the end to get

     F2 B' U  B  U  D2 R2 U2 B  U' D2 B  U  F2 B  L2 B2 U

which produces a different orientation of the same position; this
time, the ULF corner is twisted clockwise.  now conjugate this
maneuver by  C_U  to get

     R2 L' U  L  U  D2 B2 U2 L  U' D2 L  U  R2 L  F2 L2 U

which produces the original position, in its original orientation.
actually, there are 3 cube symmetries by which one could conjugate,
since the position has 3-fold symmetry.  another position with
this type of cyclic shifting property is  1-7-twist composed with
superflip.

we can combine both types of generalizations, and ask for positions
m  that have the property that for any turn  x , we have  x m = n y ,
where  y  is another turn, and  n  is the same pattern as  m , but
perhaps in a different orientation.  for such positions, we can
cyclically shift any maneuver by replacing  x^(-1)  at the beginning
of the maneuver by the corresponding  y^(-1)  at the end, and then
conjugating the whole maneuver by the appropriate cube symmetry.
two such positions are:  1-7-twist composed with four spot, and
1-7-twist composed with four spot composed with superflip.

here's all the examples of cyclic shifters that i know, along with
minimal maneuvers:

1. central positions

start  (0q*, 0f*)

superflip
 R' U2 B  L' F  U' B  D  F  U  D' L  D2 F' R  B' D  F' U' B' U  D' (24q*, 22f)
 F' B' D2 L' B2 L2 F2 U' D  B' D2 R  L  D' F2 U' L2 D' F2 D'  (20f*, 28q)

2.
four spot   F2 B2 U  D' R2 L2 U  D'  (12q*, 8f*)
six spot    F  B' U  D' R  L' F  B'  (8q*, 8f*)

four spot composed with superflip
 U2 D2 L  F2 U' D  R2 B  U' D' R  L  F2 R  U  D' R' L  U  F' B'  (26q*, 21f)
 F  U2 R  L  D  F2 U  R2 D  F2 D  F' B' U2 L  F2 R2 B2 U' D   (20f*, 28q)

six spot composed with superflip
 R' U  D  R' U  F' D  R' B  U' L' U' F' D  F' B' D' R' F  D  F  D' R2
                                                               (24q*, 23f)
 U2 F  B' R  F  L2 F2 D  B2 D2 R2 B' L2 F' D2 R2 D' B  R  B2  (20f*, 30q)

3.
1-7-twist
 F  R' U' L' F' U' B' L' U' R2 F  L' D' R' F' D' B' R' D' L2  (22q*, 20f)
 F  R2 L' F  L  F  B2 U2 F2 L  F' B2 L  F  R2 L  D2 L2  (18f*, 26q)

1-7-twist composed with superflip
 F  R  F' R  U  B  L  D  B  D' L' D  L  F  B' R  B  D  F  R'  (20q*, 20f)
 F  R  L' B  L  D  F' L2 B  D2 B' L' F2 B' D  B  U  L  B   (19f*, 22q)

4.
1-7-twist composed with four spot
 F  B2 L' U' B' L  U' D  L2 U  R' U' R  F' U' L' F  D  B' U'  (22q*, 20f*)

1-7-twist composed with four spot composed with superflip
 F  U' R' L  F' U  R  L' U2 D' B  R  L  F' R  D' R' F2 L  U'  (22q*, 20f*)

as usual, i give a maneuver which is minimal in both metrics whenever
this is possible.  i don't claim that i've found all positions in
these categories, but these are all that i know.  if you find any others,
they'd be good candidates for positions far from start.

mike

From cube-lovers-errors@mc.lcs.mit.edu  Wed Dec  2 14:51:54 1998
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Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Date: Mon, 30 Nov 1998 23:39:25 -0500
From: michael reid <reid@math.brown.edu>
Message-Id: <199812010439.XAA11918@euclid.math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: asymmetric local maxima

based on the previous analysis, i can now give 2 asymmetric positions
that are local maxima, namely

     1-7-twist composed with four spot, and
     1-7-twist composed with four spot composed with superflip.

all previous examples of local maxima had some symmetry (although
jerry bryan recently gave a bunch of new local maxima at distance 12q;
perhaps these contain some other examples.)

to show that a position is locally maximal, i must give a minimal
maneuver that ends with each possible quarter turn.  bear in mind
that  F2 = F F = F' F'.

1-7-twist composed with four spot:
 F  L' U' B' L  U' D  L2 U  R' U' R  F' U' L' F  D  B' U' F2        (22q*)
 U' R' D  L  F' U' L' B  U' B' U  F2 U' D  F  R' U' F' L  R2        (22q*)
 D' L  B' D' L' F  B2 D' F' U  B  R' D' B' L  D' L' D  R2 U         (22q*)
 F  B2 L' U' B' L  U' D  L2 U  R' U' R  F' U' L' F  D  B' U'        (22q*)
 R' U  L' F  B  D' F  L' D  B' L  F' L' F  D' B  D' B  L' U  R' B   (22q*)
 U' F  L' U' B' L  D  R' U' R  L2 F' U' L' F  U' D  F2 U  B'        (22q*)
 U' R' D  L  F' U' L' B  U' B' U  F2 U' D  F  R' U' F' R2 L         (22q*)
 U' F' L  U' L' U  R2 U' D  R  B' U' R' F  B2 U' F' D  B  L'        (22q*)
 F' U' F  B2 L' U' B' L  U' D  L2 U  R' U' R  F' U' L' F  D         (22q*)
 L  B' D' L' F  B2 D' F' U  B  R' D' B' L  D' L' D  R2 U  D'        (22q*)

1-7-twist composed with four spot composed with superflip:
 R' D' R  B' R  L  F  U2 D' R  L' U  B' R' L  U' B  U' R  F2     (22q*)
 B  U' R  U' F  B' R' U  F' B  U2 D' L  F  B  R' B  D' B' R2     (22q*)
 B  D  R  B  R  L' U  F' B' L' F  L  D' B2 U  R' B  L  B  D' U   (22q*)
 F  U' R' L  F' U  R  L' U2 D' B  R  L  F' R  D' R' F2 L  U'     (22q*)
 U' D2 L  F  B  R' F  U' F' R2 B  D' L  D' F  B' L' D  F' B      (22q*)
 U' F  L' F  B  R  U' D2 F  B' D  L' F' B  D' L  D' F  R2 B'     (22q*)
 U' F  L2 B' D' B  L' F  B  R  U2 D' F' B  U  L' F  B' U' L      (22q*)
 B' D  R  L' U' D2 F  R  L  B' R  U' R' B2 L  D' F  D' R  L'     (22q*)
 R  L' D  F' R' L  D' F  D' R  B2 L' U' L  B' R  L  F  U' D2     (22q*)

these positions are also strong local maxima in the face turn metric.

1-7-twist composed with four spot:
 U  D' B2 D  F' D' F  R' D' B' R  U  L' D' R2 L  F' D' R' F   (20f*)
 F  L' U' B' L  U' D  L2 U  R' U' R  F' U' L' F  D  B' U' F2  (20f*)
 U' R' B  D  F' U' F  B2 L' U' B' L  U' D  L2 U  R' U' R  F'  (20f*)
 F' U' L' B  U' B' U  F2 U' D  F  R' U' F' R2 L  U' L' D  R   (20f*)
 D  R' D' R  B' D' L' B  U  F' D' F  B2 R' D' B' R  U  D' R2  (20f*)
 F  B2 D' F' U  B  R' D' B' L  D' L' D  R2 U  D' R  F' D' R'  (20f*)
 D' L  B' D' L' F  B2 D' F' U  B  R' D' B' L  D' L' D  R2 U   (20f*)
 D' B  D  F2 B  R2 B2 D  F2 B' D  B  D' R2 U2 B2 U2 D2 F' U2  (20f*)
 F  B2 L' U' B' L  U' D  L2 U  R' U' R  F' U' L' F  D  B' U'  (20f*)
 L2 U  R2 B  D2 B  L2 U' F' B2 D  F  D' R2 L2 D2 B2 L2 D' B   (20f*)
 B  L2 U  F' L2 U' F' B  U  F  D' L2 U  F2 D  F2 D' R2 U2 B2  (20f*)
 U' F  L' U' B' L  D  R' U' R  L2 F' U' L' F  U' D  F2 U  B'  (20f*)
 U' R' D  L  F' U' L' B  U' B' U  F2 U' D  F  R' U' F' R2 L   (20f*)
 D2 F2 U' L2 U  L2 D  B2 U' L  D  R  L' D' B2 L' D  B2 R  L2  (20f*)
 U' F' L  U' L' U  R2 U' D  R  B' U' R' F  B2 U' F' D  B  L'  (20f*)
 F' U' F  B2 L' U' B' L  U' D  L2 U  R' U' R  F' U' L' F  D   (20f*)
 F2 B  R2 D  F' R2 D' F' B  D  F  U' R2 D  F2 U  F2 U' L2 D2  (20f*)
 L  B' D' L' F  B2 D' F' U  B  R' D' B' L  D' L' D  R2 U  D'  (20f*)

1-7-twist composed with four spot composed with superflip:
 R  L  F' L  U' L' F2 R  D' B  D' R' L  B' D  R  L' U' D2 F   (20f*)
 R' D' R  B' R  L  F  U2 D' R  L' U  B' R' L  U' B  U' R  F2  (20f*)
 R  L  F  U' D2 R' L  D  B' R  L' D' B  D' L  F2 R' U' R  F'  (20f*)
 R  F' L  U' L' F2 R  D' B  D' R' L  B' D  R  L' U' D2 F  R   (20f*)
 B  U' R  U' F  B' R' U  F' B  U2 D' L  F  B  R' B  D' B' R2  (20f*)
 F' B  D' L  D' F  R2 B' U' B  R' F  B  L  U' D2 F' B  D  R'  (20f*)
 F  D' B' R2 D' F  R  U  B' L  D2 B  R2 U2 B2 U' R  L2 B  U   (20f*)
 R2 U' F  B2 R  U  R  D' L' B2 D' R  B  U  L' F  D2 L  B2 U2  (20f*)
 F  U' R' L  F' U  R  L' U2 D' B  R  L  F' R  D' R' F2 L  U'  (20f*)
 U' D2 L  F  B  R' F  U' F' R2 B  D' L  D' F  B' L' D  F' B   (20f*)
 B  L  U  L  D' R' F2 D' L  F  U  R' B  D2 R  F2 U2 R2 U' B2  (20f*)
 U' F  L' F  B  R  U' D2 F  B' D  L' F' B  D' L  D' F  R2 B'  (20f*)
 U' F  L2 B' D' B  L' F  B  R  U2 D' F' B  U  L' F  B' U' L   (20f*)
 D' F2 B  R  D  R  U' L' F2 U' R  F  D  L' B  U2 L  F2 D2 L2  (20f*)
 B' D  R  L' U' D2 F  R  L  B' R  U' R' B2 L  D' F  D' R  L'  (20f*)
 R  L' U' D2 F  R  L  B' R  U' R' B2 L  D' F  D' R  L' F' D   (20f*)
 R  L' D  F' R' L  D' F  D' R  B2 L' U' L  B' R  L  F  U' D2  (20f*)
 F  D' R  B2 L' U' L  B' R  L  F  U' D2 R' L  D  B' R  L' D'  (20f*)

mike

From cube-lovers-errors@mc.lcs.mit.edu  Wed Dec  2 20:31:57 1998
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From: Jacob_Davenport@scudder.com (Jacob Davenport)
Subject: Re: Method for Solving the Professor's Cube (5x5x5)
To: Cube Lovers <Cube-Lovers@ai.mit.edu>

It is difficult.  I also took a long time to solve it, but I used a very
different solution.  Check out www.wunderland.com/WTS/Jake/5x5x5.html for
my solution.  Perhaps you can combine the best moves of both solutions to
find a way of solving this interesting puzzle that you find most pleasing.
If you do use my solution and have any comments about how I can make it
better, either in my writing or my moves, please let me know.

-Jacob

From cube-lovers-errors@mc.lcs.mit.edu  Thu Dec  3 15:12:43 1998
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To: cube-lovers@ai.mit.edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Re: Method for Solving the Professor's Cube (5x5x5)
Date: 2 Dec 1998 18:21:22 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <7440f2$q5v@gap.cco.caltech.edu>
References: <cube-lovers.19981201060918.6975.rocketmail@send104.yahoomail.com>

Han Wen <hansker@yahoo.com> writes:
>For those brave souls who would like to conquer this beast, the
>following solution may provide some enlightenment.  It's a layers
>solution, in contrast to the corners-first solution that I have seen
>posted on various web sites. Good luck to you.  The Professor's Cube
>is a truly challenging puzzle.
>______________________________________________________
>Method for Solving the Professor's Cube (5x5x5)

[snip]

Well, since we're sharing solutions, here's my solution to the 5x5x5:

First, a preliminary exercise that
should be mastered before the solution is attempted:

Let's ignore all the corners and all the cubies adjacent or
diagonally adjacent to the corners.  (In other words, ignore the
"supercorners," where "super" is a prefix meaning "two layers deep.")
Ignore the centers, too.  Paint all of them black, if you want.  :-)

Now, all we have left are the 12 "superedges."  Each superedge is
composed of a normal edge piece and two attached edge centers.  Or,
in other words, each of the 24 edge centers are attached to
an edge piece face.  In a normal messed-up cube, these edge
centers will not match their edge piece faces.  Our goal in this
exercise will be to match all the edge centers with their edge piece
faces.

Note that superface turns never destroy an edge center pairing.

Now, consider the following sequence of moves:
1.  Rotate any face (NOT superface) 180 degrees.
2.  Turn any center slice (as much as you want).
3.  Rotate the same face in step 1 180 degrees.
    (i.e., perform the inverse move of step 1.)

Now, if you chose the center slice to be parallel to the face,
obviously this sequence doesn't do anything.  Ditto for when you
turned the center slice some multiple of 360 degrees.

In all other cases, this will essentially perform two swaps of
edge centers.  Step 1 swaps two pairs of edge centers all around the
face, but one of those swaps gets undone by Step 3.  Step 2
moves the other pair out of the way and puts another pair in its
place to be swapped again.  So, if you choose wisely, you can
increase the number of correctly matched edge center pairs by this
move.

Many of these moves, interspersed with superface turns, will
allow you to match all the edge center pairs.  Practice this on
your cube.

Thus ends the preliminary exercise.  Note that all the moves in
the exercise do not disturb the individual supercorners (well, one
move does for a bit, but then it undoes the damage) but does change
their orientation with respect to each other.

Now, the solution!
Step 1.  Ignore the superedges and the centers.  You now have what
  is equivalent to a 4x4x4.  Solve it.
Step 2.  Match the edge centers with the edges as detailed in the
  preliminary exercise.
Step 3.  You now have a cube with correct supercorners (as done in
  step 1) and correct superedges (as done is step 2).  This means
  that your cube is equivalent to a 3x3x3, using only superface
  turns.  Solve it.
Step 4.  Tada!  Your 5x5x5 is now solved.

--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
O*e T*o: "Thre* *our fi*e s*x; se*en *ight *ine, *en!"

From cube-lovers-errors@mc.lcs.mit.edu  Thu Dec  3 16:32:04 1998
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Date: Wed, 2 Dec 1998 23:15:57 -0700
To: Cube Lovers <Cube-Lovers@ai.mit.edu>
From: Steve LoBasso <slobasso@dtint.com>
Subject: Re: Method for Solving the Professor's Cube (5x5x5)

Although I use a different method, centrals first, edge combinations, edge
parity corrections, finish using 3x3x3 solution.

I was playing with a layered solution last week also, amazing coincidence.
Much of my solution is the same as Han's.

Most of the differences are in the 4th and 5th layers.
To move the 4th layer mid centrals into place:

central D-face bm to central F-face dm: F l D l' D' F'

If there are no central pieces in the bottom central area, simply move a
bottom central up causing another central to go down.

To move 4th layer edges into place:

edge L-face Db to edge F-Rd: R' D' r D R D' r'
edge R-face Db to edge F-Ld: L D l' D' L' D l

My 5th layer edge moves are a bit different but I haven't had time to write
them down with this terminology.

--
Steve LoBasso                          mailto:slobasso@dtint.com
Digital Technology International    or mailto:slobasso@hotmail.com
500 West 1200 South, Orem, UT, 84058   http://members.tripod.com/~slobasso
(801)226-6142  ext.265                 FAX (801)221-9254

From cube-lovers-errors@mc.lcs.mit.edu  Thu Dec  3 20:49:15 1998
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Date: Thu, 3 Dec 1998 17:47:33 -0500
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From: Jacob_Davenport@scudder.com (Jacob Davenport)
Subject:  (5x5x5) edge parity corrections
To: Cube Lovers <Cube-Lovers@ai.mit.edu>

I don't like the edge parity correction move that I use in my solution, and
I'm hoping that someone can give me a better one.

The parity problem is found in 5x5x5 cubes (and 4x4x4 cubes, I understand)
when two of the edges right next to the corners (which I call "wings") are
switched.  Some fairly simple moves can get all three edges in line with
each other, but half the time two wings need to be switched.  By the time I
figure this out when doing a 5x5x5 cube, I've solved most of it, and my
parity fixing move messes up many of the edges I've been working on.

How do other people fix this problem?

-Jacob

From cube-lovers-errors@mc.lcs.mit.edu  Fri Dec  4 11:30:15 1998
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Date: Thu, 3 Dec 1998 20:32:41 -0800 (PST)
From: Han Wen <hansker@yahoo.com>
Subject: Re: Method for Solving the Professor's Cube (5x5x5)
To: Steve LoBasso <slobasso@dtint.com>
Cc: Cube-Lovers@ai.mit.edu

Hi,

> To move the 4th layer mid centrals into place:
>
> central D-face bm to central F-face dm: F l D l' D' F'
>
This is a stunningly elegant move.  You've reduce the difficulty in
solving the 4th layer by an order of magnitude.  I tried this move
out, actually it swaps two centrals:

F-face md <-> D-face lM (mid central swap)
F-face rd <-> D-face lf (right central swap)

Beautiful move.

There is one particular move that I haven't figured out yet.  It pops
up occasionally when I solve the edges of the D-layer.  Sometimes I
end up with every cubie in place except for two right centrals on
adjacent faces. For example: F-face rD and L-face Lf.  The two pieces
only need to be swapped.  No flipping is needed.  Does anyone know how
to perform this move?  I've been beating the Puzzler program for a
while, but I have been unsuccessful so far.
______________________________________________
Han Wen
Applied Materials
3050 Bowers Ave, MS 1145
Santa Clara, CA  95054
e-mail: Han_Wen@amat.com / hansker@yahoo.com

From cube-lovers-errors@mc.lcs.mit.edu  Fri Dec  4 13:17:06 1998
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Date: Thu, 3 Dec 1998 23:58:19 -0500 (EST)
From: der Mouse  <mouse@rodents.montreal.qc.ca>
Message-Id: <199812040458.XAA26116@Twig.Rodents.Montreal.QC.CA>
To: Cube-Lovers@ai.mit.edu
Subject: Re: (5x5x5) edge parity corrections

> The parity problem is found in 5x5x5 cubes (and 4x4x4 cubes, I
> understand) when two of the edges right next to the corners (which I
> call "wings") are switched.

Yes, it does occur equally on the 4-Cube.  Though I have never seen
one, I feel certain that similar parity problems will occur on all
higher-order Cubes as well, though above order 5 there will be multiple
distinct types of "wings", each of which will have its own comparable
potential problem.

Note that the problem goes away entirely if cube faces are marked such
that symmetrically placed face cubies are not visually
indistinguishable, because the parity problem in question always occurs
in conjunction with a similar parity problem on face cubies, but the
latter is invisible on most cubes.

> [...] half the time two wings need to be switched.

> How do other people fix this problem?

Most briefly, how I do it is to make a single quarter-turn of a slice
containing one of the wing pieces involved, then fix up the damage by
moving wings back into place using commutators rather than slice moves.

					der Mouse

			       mouse@rodents.montreal.qc.ca
		     7D C8 61 52 5D E7 2D 39  4E F1 31 3E E8 B3 27 4B

From cube-lovers-errors@mc.lcs.mit.edu  Fri Dec  4 14:26:23 1998
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To: cube-lovers@ai.mit.edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Method for Solving the 4x4x4
Date: 4 Dec 1998 17:26:00 GMT
Organization: California Institute of Technology, Pasadena
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References: <cube-lovers.7440f2$q5v@gap.cco.caltech.edu>

I have been told that my solution for the 5x5x5 includes
knowing how to solve the 4x4x4, which is of course not
trivial.  With the post asking about the parity problem, I
thought I might as well post my solution to the 4x4x4.

Yes, the biggest barrier is the parity problem where two
adjacent edge cubies are flipped.  My earliest attempt at a 4x4x4
solution was the following:
 1.  Match all the centers.
 2.  Match all the edges.
 3.  You now have a 3x3x3.  Solve.
Unfortunately, with the parity problem you can often
end up with an unsolvable 3x3x3 by the time you get to step 3.
Any simple moves that fix the parity problem tend to mess up the
rest of the cube quite badly -- I wrestled with this problem
a long time until I realized one thing:

Most solutions of the 3x3x3 treat the centers as static, using them
as "anchors" for the entire cube.  But this is entirely unnecessary!
If you solve the 3x3x3 while IGNORING the centers, you will eventually
get a solved cube where the centers are either in the "6 dots" or
"4 dots" situation well known to cubists -- and these have rather
simple solutions, essentially consisting of a slice turn conjugated
with another slice turn.

So, my most favorite 4x4x4 solution is now:
 1.  Match all the edges.
 2.  Solve the parity problem, if necessary (postpone until after
      step 3 if desired).
 3.  Ignore the centers and treat the cube as a 3x3x3.  Solve.
 4.  Solve the centers.

Okay.  Now to qualify the solution.  Part 1 is simple and can be
 done anyway you wish (the move rF2r'F2 will be rather useful in
 the later stages).  Part 3 is simple, with the caveat that you
 may be treating the "centers" in the wrong manner!

Part 2 stems from the fact that the cube apparently has an "even"
permutation (a 2-cycle involving two edge pieces), an apparent
paradox since 2-cycles should not exist (e.g., on the 3x3x3 it is
impossible to swap exactly two edges).  The reason this is only
an "apparent" paradox, however, is because of the misassumption that
the centers of the 4x4x4 are static, which they certainly are not!
In fact, just rotate one slice incident on your 2-cycle, and you
have magically turned the 2-cycle into a 5-cycle, which is perfectly
solvable!

Personally, I solve the 5-cycle by two or more 3-cycles, which generally
take on the form:
  FR'F' r FRF' r'
This move performs a cycle on the three edges fUR, FUr, and FDr,
without disturbing the corners, but doing rather annoying things to
the centers.  (This move is an extension of the perhaps-not-so-well-known
sequence for the 3x3x3: FR'F'LR'DRD'L'R that rotates 3 edges.)
  For FR'F' you may substitute any sequence of moves that brings
your desired edge piece (in this case fUR) to the FUr position,
as long as it does not disturb any other edges on the r slice (specifically,
the edges FDr, BUr, and BDr).  You will also have to substitute
the inverse of your sequence for FRF'.  As an example, the move
  F'L2F r F'L2F r'
cycles BdL, FUr, and FDr.  You may also use r2 instead of r and r', which
means that BDr is affected instead of FDr.

And finally, step 4: the centers.  This is solved by a
generalization of the "6-dots" rule.
This move creates "6 dots":
  u'r'ur
This permutes bUr, Fur, and buR, as well as their "opposites"
fDl, Bdl, and fdL, while affecting no other cubies.  This is two 3-cycles
on 6 faces, which is rather unwieldy, so I conjugate (is that the
right word?) it with a simple face turn to get
  u'r'ur F r'u'ru F'
which permutes Fur, Fdr, and buR in a simple 3-cycle.  Both Fur and Fdr
are on the same face, which makes this move rather easy to deal with.
Especially, if one of those is the right color already, it can be
involved in the 3-cycle without "increasing error."

I think I may have extended my personal jargon a bit more into this
post -- if you wish to understand anything in this post or, conversely,
would like to teach me more "Standard" jargon, please e-mail me.

--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
O*e T*o: "Thre* *our fi*e s*x; se*en *ight *ine, *en!"

From cube-lovers-errors@mc.lcs.mit.edu  Fri Dec  4 15:28:52 1998
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Message-Id: <c=US%a=_%p=INEA_Corporation%l=INEADEV-981204174407Z-7528@proxy.ineacorp.com>
From: Noel Dillabough <Noel.Dillabough@ineacorp.com>
To: "'Cube Lovers'" <Cube-Lovers@ai.mit.edu>
Cc: "'Jacob_Davenport@scudder.com'" <Jacob_Davenport@scudder.com>,
        "'noel@mud.ca'" <noel@mud.ca>
Subject: RE: (5x5x5) edge parity corrections
Date: Fri, 4 Dec 1998 12:44:07 -0500

The parity problem can be solved on a 4x4x4 or 5x5x5 by using the
following move (can be pasted into puzzler's move macro):

r2D2l1D2l1D1l3r3d2l1r1D3l3r3d2B2r1B2l3B2l1B2r2

For the 4x4x4, this is all that is needed, but for the 5x5x5, two
crosses (centre edges) are swapped.  So you'll need to use the following
to solve the crosses:

First, get the crosses across from each other with:

F2l3F2e1l2e3l2F2l1F2

Now swap the opposite crosses with:

R2e1l2e3l2R2e1l2e3l2

Parity problem solved...

If anyone has a better solution to this rather long one, let me know,
I'm sure some moves could be shaved off.

From cube-lovers-errors@mc.lcs.mit.edu  Fri Dec  4 16:11:19 1998
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Message-Id: <199812041806.NAA21024@pike.sover.net>
Date: Fri, 04 Dec 1998 13:07:36 -0500
To: Jacob Davenport <Jacob_Davenport@scudder.com>
From: Nichael Lynn Cramer <nichael@sover.net>
Subject: Re: (5x5x5) edge parity corrections
Cc: Cube Lovers <Cube-Lovers@ai.mit.edu>
In-Reply-To: <00096300.C22092@scudder.com>

Jacob Davenport wrote:
>I don't like the edge parity correction move that I use in my solution, and
>I'm hoping that someone can give me a better one.
>
>The parity problem is found in 5x5x5 cubes (and 4x4x4 cubes, I understand)
>when two of the edges right next to the corners (which I call "wings") are
>switched.  Some fairly simple moves can get all three edges in line with
>each other, but half the time two wings need to be switched.  By the time I
>figure this out when doing a 5x5x5 cube, I've solved most of it, and my
>parity fixing move messes up many of the edges I've been working on.
>
>How do other people fix this problem?
>
>-Jacob

Hi Jacob

In both cases (4X and 5X) I solve this problem in the following way:

1] I solve the rest of the cube, leaving me with the two "switched wings"
(in your terminology).

2] I then arrange things so both  "wings" are on the same
"off-center-slice".  (Also it will always be the case that both of these
winds are now on the same face.)

This will be easy to do using the 3-wing swapping operators.

3] At this point I now rotate the "off-center-slice" containing the
"switched wings" by a quarter turn.

As a result of this move it will be the case that that the
"off-center-slice" now has one of the previously "switched wings" in its
"correct cubicle".  The other three "wings" will be now be in a cyclic
permutation.

4] Since --from your note above-- I assume you understand how to cycle
three "wings", all you have to do now is put the "wings" in the right place
and replace the damage to the off-center central faces that were messed up
during that initial quarter-turn above.  (And since they are in "paired"
clusters, this should be pretty straightforward.)

(In short, the quarter-turn of the non-central slice puts the cube back in
the proper "orbit" for finishing up.)


Now clearly this is far from maximal.  And it's certainly not terribly
fast.  But I find it a very simple, and an easy (and easy-to-remember [and
easy-to-explain]) way to clean up this potentially messy situation.

Hope this helps
Nichael
--
Nichael Cramer
nichael@sover.net                        deep autumn--
http://www.sover.net/~nichael/                   my neighbor what does she do

From cube-lovers-errors@mc.lcs.mit.edu  Fri Dec  4 16:49:43 1998
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Date: Fri, 4 Dec 1998 11:26:10 -0700
To: Jacob_Davenport@scudder.com (Jacob Davenport)
From: Steve LoBasso <slobasso@dtint.com>
Subject: Re: (5x5x5) edge parity corrections
Cc: Cube-Lovers@ai.mit.edu

This should solve the edge parity problem by swapping the
edge F-Ru and edge F-Rd pieces.

R2 d L2 d L2 d' R2 u' F2 u2 F2 u' F2 L2 F l' F' L2 F l F

This move swaps only these two pieces and some centrals, but only within
their face. A variant of this move should be scalable to solve parity
issues in any NxNxN cube.

The only way I can think of to not have the parity problem, or at least not
require such a long series, is to solve centrals last. Another other idea
would be to spot the parity problem much earlier by counting edge flips.
Not very easy for a person to do, but I have seen it done in software for
normal 3x3x3 cubes. If it were were known very early in either the centers
first or layered solution, it would be trivial to fix.

>I don't like the edge parity correction move that I use in my solution, and
>I'm hoping that someone can give me a better one.
>
>The parity problem is found in 5x5x5 cubes (and 4x4x4 cubes, I understand)
>when two of the edges right next to the corners (which I call "wings") are
>switched.  Some fairly simple moves can get all three edges in line with
>each other, but half the time two wings need to be switched.  By the time I
>figure this out when doing a 5x5x5 cube, I've solved most of it, and my
>parity fixing move messes up many of the edges I've been working on.
>
>How do other people fix this problem?
>
>-Jacob



--
Steve LoBasso
Digital Technology International    mailto:slobasso@dtint.com
500 West 1200 South              or mailto:slobasso@hotmail.com
Orem, UT    84058                   http://members.tripod.com/~slobasso
(801)226-6142 ext.265               FAX (801)221-9254

From cube-lovers-errors@mc.lcs.mit.edu  Fri Dec  4 17:13:44 1998
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Message-Id: <36683B51.A50833E4@switchview.com>
Date: Fri, 04 Dec 1998 14:43:13 -0500
From: Michael Swart <Michael.Swart@switchview.com>
Organization: Switchview
To: Cube-Lovers@ai.mit.edu
Subject: Re: (5x5x5) edge parity corrections
References: <199812040458.XAA26116@Twig.Rodents.Montreal.QC.CA>

I got this from the archives, it may be relevant to repost it. It's a
way of solving the parity problem:
r2 U2 r l' U2 r' U2 r U2 r l U2 l U2 r U2 l r2 U2
I'm confident you can't do too much better than this.
Mike

From cube-lovers-errors@mc.lcs.mit.edu  Tue Dec  8 11:22:45 1998
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Message-Id: <199812042200.RAA02240@pike.sover.net>
Date: Fri, 04 Dec 1998 16:40:51 -0500
To: der Mouse <mouse@rodents.montreal.qc.ca>
From: Nichael Lynn Cramer <nichael@sover.net>
Subject: Re: (5x5x5) edge parity corrections
Cc: Cube-Lovers@ai.mit.edu
In-Reply-To: <199812040458.XAA26116@Twig.Rodents.Montreal.QC.CA>

der Mouse wrote:
>> The parity problem is found in 5x5x5 cubes (and 4x4x4 cubes, I
>> understand) when two of the edges right next to the corners (which I
>> call "wings") are switched.
>
>Yes, it does occur equally on the 4-Cube.  [...]

The appearance is particularly striking on the 4X cube.  Especially in the
situation where the two out-of-place "wings" are side-by-side.

It looks very similar to a solved 3X cube with a single edge-cubie flipped.
This is an interesting state to leave your cube in, when it is just lying
around your office, for visitors to find.

N

From cube-lovers-errors@mc.lcs.mit.edu  Tue Dec  8 12:38:33 1998
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Message-Id: <199812051241.VAA08521@soda3.bekkoame.ne.jp>
Date: Sat, 5 Dec 1998 21:47:39 +0900
To: Cube-Lovers@ai.mit.edu
From: Ishihama Yoshiaki <ishmnn@cap.bekkoame.ne.jp>
Subject: 4DRubik Cube

I have simulated 4DRubikCube for Macintosh.
It is madeup of 2x2x2x2 hypercubes.
It is on my HomePage.


//----------------------------------------//
Ishihama Yoshiaki
Tokyo Chofu

E-mail: ishmnn@cap.bekkoame.or.jp (Until 1999/3/31)
            ishmnn@cap.bekkoame.ne.jp ( This is correct address)
HomePage : http://www.asahi-net.or.jp/~hq8y-ishm/
//--------------------------------------//

From cube-lovers-errors@mc.lcs.mit.edu  Tue Dec  8 13:31:28 1998
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Message-Id: <366C1ED9.C11@hrz1.hrz.tu-darmstadt.de>
Date: Mon, 07 Dec 1998 19:30:49 +0100
From: Herbert Kociemba <kociemba@hrz1.hrz.tu-darmstadt.de>
Reply-To: kociemba@hrz1.hrz.tu-darmstadt.de
To: cube-lovers@ai.mit.edu
Subject: Optimal Cube Solver

New Optimal Cube Solver

I wrote an optimal Cube Solver and experimented with coordinates
different of those I use in my Cube Explorer program or of those in Mike
Reid's Optimal Cube Solver. Its pruning tables are not very large (about
25MB), so the performance is relatively low (at least in comparison with
Mike's program), but I think it is worth to give you some information
about it.

Some general considerations on the use of "coordinates" in cube solving
algorithms first. Instead of representing a state of the cube by the
positions of corners or edges, the use of coordinates not only increases
the speed of computing a face-turn but also serves as an index for the
pruning tables.

If we have an arbitrary subgroup H of the Cube Group G, we map the right
cosets Ha to natural numbers from  0 to ord(G)/ord(H)-1). A face-turn T
(which also is an element from G) now induces a map on these numbers,
which can be implemented as a simple lookup-table. For this to work we
have to ensure that if x=h1*a and y=h2*a are in the same coset Ha, then
x*T and y*T are in the same coset Hb. But this is true because
(x*T)*(y*T)^-1 = (h1*a*T)*(h2*a*T)^-1 = h1*h2^-1 is in H.

If  we take for example H1={all g from G with corner orientations 0,
corner permutations and edges arbitrary} the resulting coordinate
(0<=x<2187) represents the orientation of the corners.

It also should be possible to reduce the size of the coordinates by the
48 symmetries of the cube (or at least by a subgroup of the symmetry
group M). This is done by defining equivalence classes on the cosets.
Two cosets Ha and Hb are called equivalent, if there is an m from M with
Hb = m*Ha*m^-1. But to make this definition work we have to ensure, that
the elements of a coset Ha are really all mapped to the same coset Hb by
the conjugation with m. This only is true, if

(1) mHm^-1=H

The subgroup H1 from above  for example does have this property only for
symmetries which do not change the UD-axis in the way the orientations
of the corners are usually defined. So the corner orientation coordinate
can only be reduced by 16 symmetries. Is it possible to define the
corner orientations in another way, so that (1) holds for all 48
symmetries? I do not believe it, but I do not know how to prove this.
For the analogous case of the edge orientations there is a possibility
to define the orientations in a way (different to the way usually used)
which allows reduction by all 48 symmetries: every quarter turn changes
the orientation of any involved edge.

In my program I use 3 coordinates. The first (let's call it the
X2-coordinate) is defined by the subgroup, where the edges are arbitrary
and the corners are generated by <U2,D2,R2,L2,F2,B2>. There are 918540
different cosets. Because (1) holds for all m, they can be reduced by
all 48 symmetries and we get 19926 equivalence classes.

The second coordinate is the edge orientation defined by the subgroup
{all g from G with edge orientations 0, edge permutations and corners
arbitrary}. There are  2048 cosets. I do not reduce them by symmetries
because the number is relative small.

The third coordinate describes the edge permutation. Because there are
12! coordinate values, even reduction by 48 symmetries still gives too
many coordinate values. So for use in a turntable we define two edge
permutations a and b equivalent, if a=m1*b*m2, were m1 and m2 are in M.
In this way we get 208816 equivalence classes c. If now m1*c*m2 is a
(not necessarily unique) representation of an edge permutation applying
a faceturn T is done like that:

(m1*c*m2)*T = m1*c*[m2*T*m2^-1]*m2 = m1*[c*T']*m2=
[m1*m1']*c'*[m2'*m2]=m1''*c'*m2''

The operations in square brackets are done by table lookups:
[m2*T*m2^-1]:=T',  [c*T']:=m1'*c'*m2', [m1*m1']:=m1''  and [m2'*m2]:=
m2''.

A cube, which has all three coordinates zero, is in a subgroup with 96
elements, were the edges are in place and the corner orientations are
correct. To find such states, I use two pruning-tables. The first
combines the X2-coordinate and the edge-orientation coordinate which
takes 19926*2048/2=20404224Bytes of memory (we only need 4 bit per
entry). The maximal table entry is 12, with an average of about 9.5. The
second is a pruning table for the edge-permutation. It takes
208816*48/2=5011584Bytes, the maximal table entry is 10 (so it takes not
more then 10 faceturns to position all edges ignoring the orientations).

The program produces about 1 million nodes per second on a P350 and a
depth 15 search is done in about 4 minutes (depending on the situation).
So a complete depth 18 search will need a few days which of course is
not very satisfying. A possible improvement could be to use the subgroup
<U2D2,R2L2,F2B2> instead of <U2,D2,R2,L2,F2,B2> for the first
coordinate. The subgroup has only 4 elements, so the coset-space has 24
times the size. The pruning table will need about 480MB instead of 20MB
which is above that what is possible for me in the moment. But a
complete depth 18 search should be done in about 1/24 of the time which
will be a few hours then.

Herbert

From cube-lovers-errors@mc.lcs.mit.edu  Tue Dec  8 14:34:56 1998
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From: jmb184@frontiernet.net (John Bailey)
To: ishmnn@cap.bekkoame.or.jp (Ishihama Yoshiaki)
Cc: Submissions Cube-Lovers <Cube-Lovers@ai.mit.edu>
Subject: Re: 4DRubikCube
Date: Sat, 05 Dec 1998 13:10:10 GMT
Message-Id: <36692f00.213266943@mail.frontiernet.net>
References: <ishmnn-0512981857030001@p849359.mitk.ap.so-net.ne.jp>

On Sat, 05 Dec 1998 18:57:03 +0900, in rec.puzzles you wrote:

>I have created 4Dimension Rubik Cube for Macintosh.
>URL: http://www.asahi-net.or.jp/~hq8y-ishm/

I went there the instant I read your post.  Unfortunately, I am
running a Pentium based machine.  Could you put a gif image of your
cube on the page? Maybe even a screen copy bitmap of the user
interface. We IBM-PC types can stare and drool.
If you haven't checked out my 2x2x2x2 cube, it's at
http://www.ggw.org/donorware/4D_Rubik

John
http://www.frontiernet.net/~jmb184

From cube-lovers-errors@mc.lcs.mit.edu  Tue Dec  8 16:03:32 1998
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Message-Id: <199812042200.RAA02263@pike.sover.net>
Date: Fri, 04 Dec 1998 17:00:42 -0500
To: cube-lovers@ai.mit.edu
From: Nichael Lynn Cramer <nichael@sover.net>
Subject: Re: Method for Solving the Professor's Cube (5x5x5)
In-Reply-To: <7440f2$q5v@gap.cco.caltech.edu>
References: <cube-lovers.19981201060918.6975.rocketmail@send104.yahoomail.com>


>>Method for Solving the Professor's Cube (5x5x5)
>
>[snip]

This is not a formal solution, but --say when I want to kill some time-- I
often find it entertaining to solve the 5X cube in "ascending spirals".

By which I mean:  Start with the center face on a particular color (I
always start with blue).  Next solve the non-center face cubies, one by
one, in order moving clockwise around the "loop".  When that loop is done,
then solve one of the blue-faced corners and then solve the remaining
blue-sided edge cubies (in order).  Then move up, solving each
parallel-to-the-blue-face internal slice in order; and so on.

Needless to say, this is hardly an optimal solution (in either time or
number of moves).  But think of it as a way to "practice scales"  (Or as I
say, just a good way to kill some time.  ;-)

There are obvious variations on this.  For example, solve the individual
faces in "ascending spirals" like the above, but instead of starting on a
center face cubie, start on a corner cubie and work your way diagonally, in
slices, across the cube toward the opposite corner.

Or, for the truly masochistic, solve the cube --again a cubie at a time--
in a checkboard pattern (i.e. the result of putting the 5X cube through the
Pons Asinorum transformation) doing first the half of the cubies in the
first "phase" and then the cubies in the other.
--
Nichael Cramer
work: ncramer@bbn.com
home: nichael@sover.net
http://www.sover.net/~nichael/

From cube-lovers-errors@mc.lcs.mit.edu  Tue Dec  8 19:08:13 1998
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Date: 	Tue, 8 Dec 1998 18:37:02 -0500 (EST)
From: Alchemist Matt <monroem@email.unc.edu>
Reply-To: Alchemist Matt <monroem@email.unc.edu>
To: Herbert Kociemba <kociemba@hrz1.hrz.tu-darmstadt.de>
Cc: cube-lovers@ai.mit.edu
Subject: Re: Optimal Cube Solver
In-Reply-To: <366C1ED9.C11@hrz1.hrz.tu-darmstadt.de>
Message-Id: <Pine.GSO.3.95L.981208183158.13948B-100000@sunny.isis.unc.edu>

This question is directed to both Herbert and Mike Reid in case he's
reading this list:  With all this discussion of the "Professor Cube"
lately, how hard would it be to extend either Optimal cube solving program
to handle 4x4x4 and 5x5x5 cubes in addition to the traditional 3x3x3?
Considering reasonable table files (50 - 100 mb), how much longer would
the computation time be extended by.  If either of you would find the time
to implement this modification, I would be very interested in trying out
the program.

		Matt

-----------------------------------------------------------------------
Matthew Monroe				Monroem@UNC.Edu
Analytical Chemistry      		http://www.unc.edu/~monroem/
UNC - Chapel Hill, NC                   This tagline is umop apisdn
-----------------------------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Wed Dec  9 12:45:11 1998
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Date: Tue, 8 Dec 1998 23:24:09 -0500
From: michael reid <reid@math.brown.edu>
Message-Id: <199812090424.XAA00740@euclid.math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: Re: Optimal Cube Solver

matt monroe asks

> This question is directed to both Herbert and Mike Reid in case he's
> reading this list:  With all this discussion of the "Professor Cube"
> lately, how hard would it be to extend either Optimal cube solving program
> to handle 4x4x4 and 5x5x5 cubes in addition to the traditional 3x3x3?
> Considering reasonable table files (50 - 100 mb), how much longer would
> the computation time be extended by.  If either of you would find the time
> to implement this modification, I would be very interested in trying out
> the program.

i think it's reasonable to say that an optimal solver for the 4x4x4
(or 5x5x5) is currently far out of reach.  one could write a program
that theoretically finds optimal solutions after running for enough
time.  but it would be feasible only for positions a few turns from
start; other positions would take years, centuries, millenia, ...

on the other hand, a sub-optimal solver is certainly possible.
just teach the computer your favorite method.  this would be more
"sub" than it is "optimal", so next we'd ask to make it as good as
possible.  the real question is: are computer methods superior to
human methods for the larger cubes?  so far, probably not, but not
much work has been done on sub-optimal solvers for larger cubes.

mike

From cube-lovers-errors@mc.lcs.mit.edu  Wed Dec  9 15:20:18 1998
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Date: Tue, 8 Dec 1998 23:36:23 -0500
From: michael reid <reid@math.brown.edu>
Message-Id: <199812090436.XAA00765@euclid.math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: meffert's web site and puzzles

i'm glad to hear that uwe effert is still making puzzles.
i hope this means he'll make that master pyraminx (which
was once planned) in which edges can also turn!

and while i'm dreaming ... how about a higher order pyramid,
preferably also of the edge-turning as well as peak-turning
variety?  david byrden's web site has some puzzles i'd like
to see made: the icosahedron, the octahedral "oddity", ...

any chance of making any of these?

mike

From cube-lovers-errors@mc.lcs.mit.edu  Wed Dec  9 16:00:55 1998
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Message-Id: <199812090645.BAA17397@terminus.idirect.com>
From: "Mark Longridge" <cubeman@idirect.com>
To: <cube-lovers@ai.mit.edu>
Subject: My rubik's cube webpage
Date: Wed, 9 Dec 1998 01:49:33 -0500

Hello cube-lovers,

My site has moved to: http://www.snipercade/com/cubeman/index.html

the old site: http://web.idirect.com/~cubeman will be up for a few days
still.

web masters please update your links!

Thanks,

-> Mark

From cube-lovers-errors@mc.lcs.mit.edu  Wed Dec  9 17:00:48 1998
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To: cube-lovers@ai.mit.edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Creative ways of solving the cube
Date: 9 Dec 1998 15:48:18 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <74m642$lb5@gap.cco.caltech.edu>
References: <cube-lovers.199812042200.RAA02263@pike.sover.net>

Nichael Lynn Cramer <nichael@sover.net> writes:
>This is not a formal solution, but --say when I want to kill some time-- I
>often find it entertaining to solve the 5X cube in "ascending spirals".

Although I don't play with my 5x5x5 much, I do play with the 3x3x3 a lot
and have entertained myself by solving it in many different ways.

The canonical methods:
  1.  First level, second level, third level
  2.  Centers, corners, edges

After much more understanding, however, I now try different techniques
for entertainment.  In order of approximate difficulty:
  0.  Solve to a particular state (pons asinorum, super-flip)
  1.  Corners, edges, centers
  2.  Edges, corners, centers (rather disorienting)
  3.  First level, third level, center slice
  4.  One face at a time, with no regard to correct cubie placement
        as long as the color is correct (this is fun)
  5.  Solve to a particular subgroup (half-turn group, anti-slice group)
        then stay in that subgroup

--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
"I'd like to have the same quest again, sir."

From cube-lovers-errors@mc.lcs.mit.edu  Mon Dec 14 12:44:15 1998
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Message-Id: <199812111354.IAA18055@terminus.idirect.com>
From: "Mark Longridge" <cubeman@idirect.com>
To: <cube-lovers@ai.mit.edu>
Subject: New URL Correction
Date: Fri, 11 Dec 1998 08:59:00 -0500

Hello folks...

Sorry, but the URL I posted for my new web page is wrong.

The correct URL is: http://www.snipercade.com/cubeman

The old site http://web.idirect.com/~cubeman will be up for a few days yet.

The virtual URL http://welcome.to/cubeman should always point to the
current site! :-)

More interesting stuff to follow

-> Mark

From cube-lovers-errors@mc.lcs.mit.edu  Mon Dec 14 13:45:36 1998
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Message-Id: <199812111354.IAA18055@terminus.idirect.com>
From: "Mark Longridge" <cubeman@idirect.com>
To: <cube-lovers@ai.mit.edu>
Subject: New URL Correction
Date: Fri, 11 Dec 1998 08:59:00 -0500

Hello folks...

Sorry, but the URL I posted for my new web page is wrong.

The correct URL is: http://www.snipercade.com/cubeman

The old site http://web.idirect.com/~cubeman will be up for a few days yet.

The virtual URL http://welcome.to/cubeman should always point to the
current site! :-)

More interesting stuff to follow

-> Mark

From cube-lovers-errors@mc.lcs.mit.edu  Mon Dec 14 15:00:05 1998
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Date: Thu, 10 Dec 1998 23:04:10 -0500
From: michael reid <reid@math.brown.edu>
Message-Id: <199812110404.XAA03343@cauchy.math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: fixing edge parity on 4x4x4

several people have posted maneuvers for "fixing" the edge parity
on rubik's revenge.  i haven't seen any maneuvers as short as
mine (although there might be some disagreement about "length").
recall that i am using the notation  _R_  (R underscored) to mean
turn the outer two layers together.

to switch the two UF edges:

_R2_  B2 L U2  l  U2  r'  U2  r  U2 F2  r  F2  _L'_  B2  _R2_

side effects: rotates the set of 4 U centers by 180 degrees.
also makes a 4-cycle of internal (0 faces visible) cubies.

if you're not concerned about moving centers, use

_(R2 B)_  u  _(B' D2 B)_  u'  _B_  l  _(B2 D2 R2)_

here,  _( ... )_  means the whole thing inside the parentheses
is underlined.

the maneuver that i normally use, since it's appropriate for
my solving method, is

U2  (r  U2)^5

which makes a 4-cycle of edges in the  r-slice  (and also
rotates the set of 4 U centers by 180 degrees).  this one
is short and easy to remember!

these maneuvers all work well on the 5x5x5 also.

mike

From cube-lovers-errors@mc.lcs.mit.edu  Tue Dec 15 08:40:06 1998
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Sender: bosch@sgi.com
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Date: Mon, 14 Dec 1998 11:14:04 -0800
From: Derek Bosch <bosch@sgi.com>
To: Cube-Lovers@ai.mit.edu
Subject: re-assembling a 2x2x2?

Well, I accidentally managed to pop apart my Rubik's Mini-Cube,
aka the 2x2x2...  Are there any easy instructions on getting it
back together?  I'd rather not force it...

D
-- 
Derek Bosch        "A little nonsense now and then
(650) 933-2115      is relished by the wisest men"... W.Wonka
bosch@sgi.com

From cube-lovers-errors@mc.lcs.mit.edu  Tue Dec 15 10:30:58 1998
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Date: Tue, 15 Dec 1998 17:30:24 +0900 (JST)
Message-Id: <199812150830.RAA03209@soda2.bekkoame.ne.jp>
To: cube-lovers@ai.mit.edu
From: Ishihama Yoshiaki <ishmnn@cap.bekkoame.ne.jp>
Subject: 4D Rubik Cube(2x2x2x2) Java

I have converted my "4DRubikCube" (for Macintosh) to java applet.
I have uploaded it to my java page.
I have not yet added direct drag mode, only rotate cubes by buttons.
Please check this applet.



//----------------------------------------//

Ishihama Yoshiaki ( Tokyo Japan)
E-mail: ishmnn@cap.bekkoame.ne.jp
HomePage : http://www.asahi-net.or.jp/~hq8y-ishm/

//----------------------------------------//

From cube-lovers-errors@mc.lcs.mit.edu  Tue Dec 15 12:13:33 1998
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Date: Tue, 15 Dec 1998 11:35:13 -0500 (EST)
From: Nicholas Bodley <nbodley@tiac.net>
To: Derek Bosch <bosch@sgi.com>
Cc: Cube-Lovers@ai.mit.edu
Subject: Re: re-assembling a 2x2x2?
In-Reply-To: <3675637C.6231@sgi.com>
Message-Id: <Pine.BSF.3.96.981215112038.5993D-100000@shell2.tiac.net>


 I've pulled mine apart a *few* times. Imho, it's probably impossible to
reassemble without some forcing. If it were made of cheap plastic, I
very much doubt that it could be assembled. Study the structure, so
you won't try to assemble it wrong; you probably wouldn't make such a
mistake, though. Hope you didn't lose any pieces! (Be *sure* to match
colors properly before assembling; of course, you know that, too.)

 My hopeful guess is that you'll succeed, but be rather amazed by the
force it takes, and also that the plastic can take such stress. I have
seen 2^3s on sale fairly recently, btw, so there has been a stock of
them, possibly a new production run.

 I've been a mechanical tech. at times for several decades, so I'm
reasonably sure of what I say. I've pulled apart many movable-part
puzzles, and the 2^3 is surely the most intractable of all I've dealt
with.  Alexanders' Star and the 4^3 have some parts that are easy to
break.

 I'd love to know how it's done at the factory. I hope it's not some
subtle ultrasonic welding. (Maybe someone could ask Dr. Christoph
Bandelow, Dr. Uwe Meffert, or (Dr.?*) David Singmaster.)

*Sorry if I forget!

Best regards, and good luck!

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*  The personal computer industry will have become
|*  Amateur musician  *|*  mature when crashes become unacceptable.
--------------------------------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Wed Dec 16 13:02:37 1998
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Date: Tue, 15 Dec 1998 18:41:19 -0500
To: Cube-Lovers@ai.mit.edu
From: Charlie Dickman <charlied@erols.com>
Subject: Re: re-assembling a 2x2x2?

>Well, I accidentally managed to pop apart my Rubik's Mini-Cube,
>aka the 2x2x2...  Are there any easy instructions on getting it
>back together?  I'd rather not force it...

You should use a small Phillips screwdriver to remove one of the triangular
flanges that forms the tracks that the "cubies" ride in. Then, put the
"cubie" pieces in place and then g_e_n_t_l_y spread the space between the 4
"cubie" surfaces that hide the stump that holds the triangular flange and
put the screw back in.

Be careful not to break the shaft between the cubie face and its anchor.


Charlie Dickman
charlied@erols.com

From cube-lovers-errors@mc.lcs.mit.edu  Thu Dec 17 12:58:39 1998
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Date: Wed, 16 Dec 1998 23:46:46 -0500 (EST)
From: Nicholas Bodley <nbodley@tiac.net>
To: Charlie Dickman <charlied@erols.com>
Cc: Cube-Lovers@ai.mit.edu
Subject: Newer mechanism? (Was: Re: re-assembling a 2x2x2?)
In-Reply-To: <v04020a00b29ca31d7c8f@[207.172.132.155]>
Message-Id: <Pine.BSF.3.96.981216232654.16519H-100000@shell2.tiac.net>


 Charlie, I'm just about dead sure my 2^3s (from about 15 (?) years go)
have no screws. I studied your description, and it seems that the
mechanism has been redesigned! I described the mechanism of mine in
considerable (if not painful!) detail, maybe a year and a half ago; it's
probably in the archives. The keyword "jack" should help to locate the
post.

 Perhaps a continuing market combined with the difficulty of assembling
the original design created a need for a new one.

 Would really *love* to know whether there is a newer and different
mechanism. As a somewhat casual student of these mechanisms, I've come
to realize that for all "sizes", more than one mechanism is possible.

 I have great admiration for the designers who create these marvelous
mechanisms. I love the 5^3 as much for its innards (which I regard as
thoroughly astonishing) as for its essential function.

 I also admire the mathematicians, programmers, and practical users of
group theory on this List; I have only a faint awareness of what they're
talking about, but their amazing posts keep my mind properly stretched.
I feel a bit like a dog listening to his human family discussing, say, a
trip to Australia (the dog didn't go). However, that's perfectly OK with
me! My mind is quite good, and to some degree it's circumstance that I'm
not "with it".

My regards to all,

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*  The personal computer industry will have become
|*  Amateur musician  *|*  mature when crashes become unacceptable.
--------------------------------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Thu Dec 17 14:01:55 1998
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Date: Thu, 17 Dec 1998 09:32:42 -0500 (Eastern Standard Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re : Optimal Cube Solver
In-Reply-To: <366C1ED9.C11@hrz1.hrz.tu-darmstadt.de>
To: kociemba@hrz1.hrz.tu-darmstadt.de
Cc: Cube Lovers <cube-lovers@ai.mit.edu>
Message-Id: <SIMEON.9812170942.C@GN209A.PSTCC.CC.TN.US>

On Mon, 07 Dec 1998 19:30:49 +0100 Herbert Kociemba
<kociemba@hrz1.hrz.tu-darmstadt.de> wrote:

>
> The third coordinate describes the edge permutation. Because there are
> 12! coordinate values, even reduction by 48 symmetries still gives too
> many coordinate values. So for use in a turntable we define two edge
> permutations a and b equivalent, if a=m1*b*m2, were m1 and m2 are in M.
> In this way we get 208816 equivalence classes c. If now m1*c*m2 is a
> (not necessarily unique) representation of an edge permutation applying
> a faceturn T is done like that:
>
> (m1*c*m2)*T = m1*c*[m2*T*m2^-1]*m2 = m1*[c*T']*m2=
> [m1*m1']*c'*[m2'*m2]=m1''*c'*m2''
>

This is remindful of a technique I used many years ago to
reduce the size of the search space for the 2x2x2 problem,
and the issue would apply to any cube such as the 4x4x4
with an even number of cubies per edge.  That is, in the
(2n)x(2n)x(2n) problem you can treat as equivalent any
positions of the form (m1)*x*(m2) for m1 and m2 in M,
provided only that both of m1 and m2 are rotations or that
both of m1 and m2 are reflections.

Another (and in some ways better) way to model a
(2n)x(2n)x(2n) problem and to reduce the size of the search
space is to fix one of the corners and to use the symmetry
group which preserves the major diagonal axis which includes
the corner which is fixed, but that is a different issue.

Dan Hoey showed that (m1)*x*(m2) is equivalent to m'xmc for
suitable choices of m and c, for m in M and for c in C (the
set of 24 rotations).  Requiring that m1 and m2 both be
rotations or both be reflections is necessary because you
really can't turn the corners inside out on a physical
cube.

Herbert does not impose the same restriction on both of m1
and m2 being rotations or reflections because his third
coordinate applies only to the edges, and the edges can
indeed be turned inside out on a physical cube, namely with
the Pons Asinorum maneuver. So for this case, (m1)*x*(m2)
is equivalent to m'xmc if m1 and m2 are both rotations or
both reflections, and is equivalent to m'xmcv if they are
not, where v is the central inversion of the edges
(essentially, the Pons Asinorum applied to the edges).

I used to talk about 1152-fold symmetry for the 2x2x2
(1152=24*48).  Herbert's approach for the third coordinate
yields a 2304-fold reduction in the search space
(2304=48*48).  However, the reductions in the search space
in the two cases are not really dealing with quite the same
issue. In the case of 1152-fold symmetry for the 2x2x2,
there are (up to) 1152 equivalent positions which are the
same distance from Start. If I am understanding Herbert's
technique correctly, when positions are equivalent in the
third coordinate, there are (up to) 2304 positions of
the edges for which the distance from Start has the same
lower bound.  (Maybe I should say "the same non-trivial
lower bound", because (for example) zero would be a lower
bound for all positions.)

----------------------------------------
Jerry Bryan
jbryan@pstcc.cc.tn.us
Pellissippi State Technical Community College

From cube-lovers-errors@mc.lcs.mit.edu  Thu Dec 17 15:55:31 1998
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Date: Thu, 17 Dec 1998 10:02:24 -0500 (Eastern Standard Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re : Re: Optimal Cube Solver
In-Reply-To: <Pine.GSO.3.95L.981208183158.13948B-100000@sunny.isis.unc.edu>
To: Alchemist Matt <monroem@email.unc.edu>
Cc: Cube Lovers <cube-lovers@ai.mit.edu>
Message-Id: <SIMEON.9812171024.D@GN209A.PSTCC.CC.TN.US>

On Tue, 08 Dec 1998 18:37:02 -0500 (EST) Alchemist Matt
<monroem@email.unc.edu> wrote:

> This question is directed to both Herbert and Mike Reid in case he's
> reading this list:  With all this discussion of the "Professor Cube"
> lately, how hard would it be to extend either Optimal cube solving program
> to handle 4x4x4 and 5x5x5 cubes in addition to the traditional 3x3x3?
> Considering reasonable table files (50 - 100 mb), how much longer would
> the computation time be extended by.  If either of you would find the time
> to implement this modification, I would be very interested in trying out
> the program.

Mike Reid has already answered this question in the
negative with respect to optimal solvers, based on the
huge size of the search spaces that would be involved. For
several years, I have wondered about the same thing with
respect to a God's Algorithm search of a Start rooted tree
(how many positions are one move from Start, how many are
two moves from Start, etc.).

You could obviously get a few moves from Start, but I don't
think you would get very far.  For example, with my
existing program, I think maybe I could get five or six
moves from Start with the 4x4x4 or the 5x5x5.

However, I have been reluctant to deal with either the
4x4x4 or the 5x5x5 for several reasons.  One is that the
programming is not quite as easy as it might seem, or at
least not for my program the way it is written. In
principle, all I would have to do is replace the existing
tables for the permutations which generate the 3x3x3 with
the corresponding tables for the 4x4x4 and the 5x5x5 and
everything should just work.  However, my program contains
optimizations previously described on Cube-Lovers which are
very dependent on the edge and corner structure of the
3x3x3.  For the larger problems, I would have to add a bit
(not a lot, but a bit) of new code to deal with new kinds
of pieces.

Secondly, in the case of the 4x4x4 I would have to deal
with might be called rotational equivalences, for example
that RrL'l' (capital letters denote moving the outer layers
and lower case letters denote moving the inner layers)
would normally treated as being equivalent to the Start
state.  Both ways I know how to do it would require some
reprogramming, especially in light of the same existing
optimizations I talked about before with respect to the
3x3x3.  Namely, I could treat rotations as being
equivalent, so that x is equivalent to all positions of the
form xc for c in C.  Or I could fix one of the corners.

Thirdly, I would have to deal with what might be called
invisible equivalences, where pieces can be moved without
the movement being visible on a physical cube.  In the case
of the 4x4x4 (for example), the four "face center" facelets
on each face can move with respect to each other (subject
to parity constraints) without the movement being visible. 
I would think that you would want to treat such differences
as being equivalent.

Actually, I think that an optimal solver for the 4x4x4 or
the 5x5x5 would need to deal with some of the same issues,
in addition to the huge size of the search spaces that was
pointed out by Mike Reid in his original response to this
question.

----------------------------------------
Jerry Bryan
jbryan@pstcc.cc.tn.us
Pellissippi State Technical Community College

From cube-lovers-errors@mc.lcs.mit.edu  Thu Dec 17 19:18:53 1998
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Date: Thu, 17 Dec 1998 21:39:14 +0100
From: Herbert Kociemba <kociemba@hrz1.hrz.tu-darmstadt.de>
Reply-To: kociemba@hrz1.hrz.tu-darmstadt.de
To: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Cc: Cube Lovers <cube-lovers@ai.mit.edu>
Subject: Re: Optimal Cube Solver
References: <SIMEON.9812170942.C@GN209A.PSTCC.CC.TN.US>

Jerry Bryan wrote:
>
> On Mon, 07 Dec 1998 19:30:49 +0100 Herbert Kociemba
> <kociemba@hrz1.hrz.tu-darmstadt.de> wrote:
>
> >
> > The third coordinate describes the edge permutation. Because there are
> > 12! coordinate values, even reduction by 48 symmetries still gives too
> > many coordinate values. So for use in a turntable we define two edge
> > permutations a and b equivalent, if a=m1*b*m2, were m1 and m2 are in M.
> > In this way we get 208816 equivalence classes c. If now m1*c*m2 is a
> > (not necessarily unique) representation of an edge permutation applying
> > a faceturn T is done like that:
> >
> > (m1*c*m2)*T = m1*c*[m2*T*m2^-1]*m2 = m1*[c*T']*m2=
> > [m1*m1']*c'*[m2'*m2]=m1''*c'*m2''
> >

> I used to talk about 1152-fold symmetry for the 2x2x2
> (1152=24*48).  Herbert's approach for the third coordinate
> yields a 2304-fold reduction in the search space
> (2304=48*48).  However, the reductions in the search space
> in the two cases are not really dealing with quite the same
> issue. In the case of 1152-fold symmetry for the 2x2x2,
> there are (up to) 1152 equivalent positions which are the
> same distance from Start. If I am understanding Herbert's
> technique correctly, when positions are equivalent in the
> third coordinate, there are (up to) 2304 positions of
> the edges for which the distance from Start has the same
> lower bound.  (Maybe I should say "the same non-trivial
> lower bound", because (for example) zero would be a lower
> bound for all positions.)

I do not use the equivalence in the third coordinate as an index in a
pruning table. On the contrary, I have to "expand" the coordinate again
by a factor of 48 to get equivalence classes, which have the same
distance from start and from which I built the pruning table. But due to
the large size (12!) of edge permutations, it seems a good way (and I
see no other way) to keep track of the edge-permutation-coordinate with
only a few table-lookups.

I now have enough RAM (128MB) to implement a pruning table for all
possible coordinates of the first phase of my Two-Phase-Algorithm, which
brings the cube into the subgroup H=<U,D,R2,L2,F2,B2>. This is what Mike
Reid already did about one year ago and which seems powerful enough even
to be used as an Optimal Solver (omitting phase 2, in which the edge-
and cornerpermutations are restored). Due to this power I think of
implementing a "static" phase 2 only with a table which stores the edge-
and corner permutations of all positions up to maybe 5 face-turns in H
from start.

Using the approach for the edge permutation from above,the computation
of the starting position of phase 2 should be very fast. In the
implementation currently used, I have to switch back from the
coordinate-representation of the cube in phase 1 to a more "physical"
representation using edges and corners, apply the maneuver generated in
phase 1 and then compute the starting coordinates of phase 2. In the new
approach only coordinates could be uses.

Herbert

From cube-lovers-errors@mc.lcs.mit.edu  Fri Dec 18 11:33:50 1998
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Message-Id: <002c01be2a01$f6ba7020$7ac4b0c2@home>
From: roger.broadie@iclweb.com (Roger Broadie)
To: <Cube-Lovers@ai.mit.edu>
Cc: "Nicholas Bodley" <nbodley@tiac.net>,
        "Charlie Dickman" <charlied@erols.com>
Subject: Re: Newer mechanism? (Was: Re: re-assembling a 2x2x2?)
Date: Thu, 17 Dec 1998 21:11:22 -0000

Nicholas Bodley <nbodley@tiac.net> wrote (17 December 1998)


>
> Would really *love* to know whether there is a newer and different
>mechanism. As a somewhat casual student of these mechanisms, I've
>come to realize that for all "sizes", more than one mechanism is
>possible.
>


According to the reports on the patent case brought against Ideal for
infringement  of the Nichols patent (Moleculon Research Corp v. CBS,
Inc) there were two Ideal 2x2x2 cubes, both sold as the Rubik's Pocket
Cube, but one from Taiwan and one from Japan.  The Japanese version
used an internal sphere, which could well be the version with the
Philips screw referred to by Charlie Dickman, since it sounds like the
inside of a 4x4x4.  The Taiwanese version is less clearly described -
the internal faces are said to form a tongue and groove mechanism -
but probably also had an internal spider like the conventional 3x3x3 -
is this Nicholas Bodley's version?

Incidentally, by the time the case had been up and down to the Appeals
court a couple of times, the final decision, in 1989, was that just
these two forms infringed the patent.  The 3x3x3 and 4x4x4 were held
not to infringe.

Roger Broadie

From cube-lovers-errors@mc.lcs.mit.edu  Fri Dec 18 14:53:20 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Thu Dec 17 22:07:25 1998
Date: Thu, 17 Dec 1998 22:06:01 -0500
From: michael reid <reid@math.brown.edu>
Message-Id: <199812180306.WAA21451@cauchy.math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: Re: Optimal Cube Solver

herbert writes

> I do not use the equivalence in the third coordinate as an index in a
> pruning table. On the contrary, I have to "expand" the coordinate again
> by a factor of 48 to get equivalence classes, which have the same
> distance from start and from which I built the pruning table. But due to
> the large size (12!) of edge permutations, it seems a good way (and I
> see no other way) to keep track of the edge-permutation-coordinate with
> only a few table-lookups.

> I now have enough RAM (128MB) to implement a pruning table for all
> possible coordinates of the first phase of my Two-Phase-Algorithm, which
> brings the cube into the subgroup H=<U,D,R2,L2,F2,B2>. This is what Mike
> Reid already did about one year ago and which seems powerful enough even
> to be used as an Optimal Solver (omitting phase 2, in which the edge-
> and cornerpermutations are restored). Due to this power I think of
> implementing a "static" phase 2 only with a table which stores the edge-
> and corner permutations of all positions up to maybe 5 face-turns in H
> from start.

> Using the approach for the edge permutation from above,the computation
> of the starting position of phase 2 should be very fast. In the
> implementation currently used, I have to switch back from the
> coordinate-representation of the cube in phase 1 to a more "physical"
> representation using edges and corners, apply the maneuver generated in
> phase 1 and then compute the starting coordinates of phase 2. In the new
> approach only coordinates could be uses.


herbert, you might be interested in what my sub-optimal program
(the one based on your two-stage algorithm) does about edge
permutations.  i have this extra coordinate i call "sliceedge",
(really this is just another coset space) which considers the
locations of four distinguishable edges.  there are 12*11*10*9 = 11880
possibilities for this coordinate.  when the cube is entered, it
calculates the corresponding coordinate for edges in the U-D slice,
also for edges in the F-B slice, and also for the R-L slice.
then i have lookup tables to tell me how this coordinate transforms
under turns.  this lookup table is 18 * 11880 shorts = 427680 bytes.

when stage 2 is reached, i have a lookup table that maps this
coordinate into permutations of the four U-D slice edges.  actually,
only 24 of the entries are valid, but only these occur, since we've
reached stage 2.  this lookup table is 11880 chars.
there's also a lookup table to transform the "sliceedge" coordinate
into another coordinate, which gives the locations of four
distinguishable edges among the eight U and D edges.  this coordinate
has 8*7*6*5 = 1680 possibilities, and the lookup table is 11880 shorts.

the big lookup table is the one that takes two of these last coordinates
and transforms it into a permutation of the eight U and D edges.
this table has 1680 * 1680 shorts = about 5.5 megabytes.  most of
the entries are garbage, only 40320 = 8! actually occur, since we've
reached stage 2.

so for about 6 megabytes of space, all the edge permutations are done
with lookup tables.  i haven't actually calculated how much of a speed
up this is, but it's probably good.

mike

From cube-lovers-errors@mc.lcs.mit.edu  Fri Dec 18 15:30:11 1998
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From: "CRAWFORD, SCOTT" <CRAWFORDS@evangel.edu>
To: Cube-Lovers@ai.mit.edu
Subject: Snake
Date: Thu, 17 Dec 1998 19:50:21 -0600

This may be a little off topic, but I've recently fell in love with the
snake, making many shapes I'd never even thought of.  Are there any websites
or archives of different snake patterns?

Thanks
Scotte

From cube-lovers-errors@mc.lcs.mit.edu  Mon Dec 21 14:00:52 1998
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Message-Id: <367BDD1D.DE6@hrz1.hrz.tu-darmstadt.de>
Date: Sat, 19 Dec 1998 18:06:37 +0100
From: Herbert Kociemba <kociemba@hrz1.hrz.tu-darmstadt.de>
Reply-To: kociemba@hrz1.hrz.tu-darmstadt.de
To: cube-lovers@ai.mit.edu
Cc: michael reid <reid@math.brown.edu>
Subject: Re: Optimal Cube Solver
References: <199812180306.WAA21451@cauchy.math.brown.edu>

michael reid wrote:

> herbert, you might be interested in what my sub-optimal program
> (the one based on your two-stage algorithm) does about edge
> permutations.  i have this extra coordinate i call "sliceedge",
> (really this is just another coset space) which considers the
> locations of four distinguishable edges.  there are 12*11*10*9 = 11880
> possibilities for this coordinate.  when the cube is entered, it
> calculates the corresponding coordinate for edges in the U-D slice,
> also for edges in the F-B slice, and also for the R-L slice.
> then i have lookup tables to tell me how this coordinate transforms
> under turns.  this lookup table is 18 * 11880 shorts = 427680 bytes.
>
> when stage 2 is reached, i have a lookup table that maps this
> coordinate into permutations of the four U-D slice edges.  actually,
> only 24 of the entries are valid, but only these occur, since we've
> reached stage 2.  this lookup table is 11880 chars.

I already made some experience with the "sliceedge"-coordinate before. I
built it in the way:
24*position of the 4 edges + permutation of the 4 edges,
where the position range is from 0 to 494 and permutation ranges from 0
to 23.
In this way when reaching stage 2, the "sliceedge"-coordinate
automatically is in the range from 0 to 23 and you need no lookup table
at all.

> there's also a lookup table to transform the "sliceedge" coordinate
> into another coordinate, which gives the locations of four
> distinguishable edges among the eight U and D edges.  this coordinate
> has 8*7*6*5 = 1680 possibilities, and the lookup table is 11880 shorts.
>
> the big lookup table is the one that takes two of these last coordinates
> and transforms it into a permutation of the eight U and D edges.
> this table has 1680 * 1680 shorts = about 5.5 megabytes.  most of
> the entries are garbage, only 40320 = 8! actually occur, since we've
> reached stage 2.

This seems an interesting approach. Using the
edge-permutation-coordinate in the way I described it before, I need
about 20MB for the lookup-table which tells the coordinate-tranformation
under turns, which is quite a lot. Maybe I also try your method.

Herbert

From cube-lovers-errors@mc.lcs.mit.edu  Mon Dec 21 14:52:35 1998
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Date: Sun, 20 Dec 1998 16:41:43 -0500 (EST)
From: Nicholas Bodley <nbodley@tiac.net>
To: Roger Broadie <roger.broadie@iclweb.com>
Cc: Cube Mailing List <Cube-Lovers@ai.mit.edu>,
        Charlie Dickman <charlied@erols.com>, Mark Glusker <glusk@sgi.com>
Subject: Re: Newer mechanism? (Was: Re: re-assembling a 2x2x2?)
In-Reply-To: <002c01be2a01$f6ba7020$7ac4b0c2@home>
Message-Id: <Pine.BSF.3.96.981220161825.27653I-100000@shell2.tiac.net>


On Thu, 17 Dec 1998, Roger Broadie wrote:

(Interesting to read about the lawsuit...)

}Nicholas Bodley <nbodley@tiac.net> wrote (17 December 1998)

}inside of a 4x4x4.  The Taiwanese version is less clearly described -
}the internal faces are said to form a tongue and groove mechanism -
}but probably also had an internal spider like the conventional 3x3x3 -
}is this Nicholas Bodley's version?

In the mechanism I know for a 2X2X2 Cube, at its center is a piece like
a jack, that is, one of the pieces in the traditional game, but without
the knobs at the ends. You could also think of it as three rods
intersecting at a common point, and mutually orthogonal; it's as if you
had plus and minus x, y, and z axes defined by the directions of the
rods. These create the axes of revolution for one half relative to the
other.

The cubies are hollow, and their mating faces have curved cutaways. To
keep the cubies from moving too far from each other, 12 "clips" extend
from the center outward. If you think of a deeply-grooved pulley, cut
pie-style into quarters, you have a general idea. The curved-cutout
edges of the cubies fit between two curved, parallel sides of the
"clips".

Finally, the "clips" are kept engaged with the cubies either directly by
square cross-section extensions of the center "jack", or by hollow
square rods that pivot on (smaller) cylindrical extensions. Just as the
ball in a 4^3 is locked to one half, the "jack" is, also.

The big problem with this mechanism is that unless the parts can deform
sufficiently without breaking (the actual case; they can do so), it's
impossible to assemble or to disassemble as molded. If it were made of
metal, it couldn't be assembled without design changes.

Illustrations are really needed here; this mechanism is a challenge to
describe understandably!

I'd think that the internal ball uses a variant of the tongue and groove
scheme, if it's like the 4^3.

|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*  The personal computer indusztry will have become
|*  Amateur musician  *|*  mature when crashes become unacceptable.
--------------------------------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Tue Dec 22 12:31:20 1998
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Date: Sun, 20 Dec 1998 21:18:09 -0500 (EST)
From: Nicholas Bodley <nbodley@tiac.net>
To: "CRAWFORD, SCOTT" <CRAWFORDS@evangel.edu>
Cc: Cube-Lovers@ai.mit.edu
Subject: Re: Snake
In-Reply-To: <5B9619E72C59D211B22100A0C99CC4632EEE9E@www.evangel.edu>
Message-Id: <Pine.BSF.3.96.981220211300.2706L-100000@shell2.tiac.net>


 The Snake is delightful; you can do some interesting investigations by
starting with a straight config., and twisting each consecutive joint
according to a pattern. Just as long as you don't get physical
interferencies, you see some modestly-interesting shapes.

 It's also worth a bit of casual effort to create a "ball". All of these
things are, however, trivially easy to many of the subscribers to this
list. The snake is for mental relaxation!


|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*  The personal computer industry will have become
|*  Amateur musician  *|*  mature when crashes become unacceptable.
--------------------------------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Tue Dec 22 15:28:37 1998
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Date: Tue, 22 Dec 1998 16:40:58 +0900 (JST)
Message-Id: <199812220740.QAA00692@soda3.bekkoame.ne.jp>
To: cube-lovers@ai.mit.edu
From: Ishihama Yoshiaki <ishmnn@cap.bekkoame.ne.jp>
Subject: 5DRubikCube

I have made simulation of 5D RubikCube(2x2x2x2x2).
This is consisted of 2x2x2x2x2=32  5DCubes.
This is for Macintosh only.
I will not convert this program to java because it is too troublesome.
It is on my HomePage.

//----------------------------------------//

Ishihama Yoshiaki (Tokyo Japan)
E-mail: ishmnn@cap.bekkoame.ne.jp
HomePage : http://www.asahi-net.or.jp/~hq8y-ishm/

//----------------------------------------//

[Moderator's note: This is the third announcement of this website this
 month.  For all further developments, check the website. ]

From cube-lovers-errors@mc.lcs.mit.edu  Tue Dec 22 18:44:01 1998
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Date: Tue, 22 Dec 1998 18:37:57 -0500
From: michael reid <reid@math.brown.edu>
Message-Id: <199812222337.SAA26516@adams.math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: Re: Optimal Cube Solver

i'm glad that herbert brought up this issue of edge transformations.
because now that i think about this again, i realize that my tables
can be reduced dramatically.  i described my tables:

> there's also a lookup table to transform the "sliceedge" coordinate
> into another coordinate, which gives the locations of four
> distinguishable edges among the eight U and D edges.  this coordinate
> has 8*7*6*5 = 1680 possibilities, and the lookup table is 11880 shorts.
>
> the big lookup table is the one that takes two of these last coordinates
> and transforms it into a permutation of the eight U and D edges.
> this table has 1680 * 1680 shorts = about 5.5 megabytes.  most of
> the entries are garbage, only 40320 = 8! actually occur, since we've
> reached stage 2.

since this big table is sparse, we don't need most of it.  what i
should do is have another table (11880 char's) to transform "sliceedges"
into permutations of four edges.  the location of the four R-L slice
edges determines the location of the four F-B slice edges, so we only
need to know how they're permuted.  thus the big table can be replaced
by one with 1680 * 24 shorts that gives the permutation of the eight
U and D edges.  it no longer would have error-checking (i.e. making
sure we don't get an invalid entry in the big table), but that could
be installed with another simple table lookup, if desired.

with this new mechanism, only about 543K of tables would be needed,
the largest being a lookup table which tells how "sliceedges"
transform under face turns.  this is much better than the 6 megabytes
of tables i'm currently using.  i don't know why i didn't think of
this earlier!

mike

From cube-lovers-errors@mc.lcs.mit.edu  Wed Dec 23 17:40:01 1998
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Date: Wed, 23 Dec 1998 23:19:05 +0100
From: Herbert Kociemba <kociemba@hrz1.hrz.tu-darmstadt.de>
Reply-To: kociemba@hrz1.hrz.tu-darmstadt.de
To: cube-lovers@ai.mit.edu
Cc: michael reid <reid@math.brown.edu>
Subject: Re: Optimal Cube Solver
References: <199812222337.SAA26516@adams.math.brown.edu>

michael reid wrote:

> since this big table is sparse, we don't need most of it.  what i
> should do is have another table (11880 char's) to transform "sliceedges"
> into permutations of four edges.  the location of the four R-L slice
> edges determines the location of the four F-B slice edges, so we only
> need to know how they're permuted.  thus the big table can be replaced
> by one with 1680 * 24 shorts that gives the permutation of the eight
> U and D edges.  it no longer would have error-checking (i.e. making
> sure we don't get an invalid entry in the big table), but that could
> be installed with another simple table lookup, if desired.
>
> with this new mechanism, only about 543K of tables would be needed,
> the largest being a lookup table which tells how "sliceedges"
> transform under face turns.  this is much better than the 6 megabytes
> of tables i'm currently using.  i don't know why i didn't think of
> this earlier!
>
> mike

Is it necessary to use the table for the map from the "sliceedges" to
the 1680 "4 out of 8"-coordinate at all? I think you constructed this
"helper"-coordinate, because a 11880*11880 table-size was too big and
1680*1680 was reasonable. But 11880*24 also is small (just twice as much
as the lookup table which tells how "sliceedges" transform under face
turns).

In the way I construct the sliceedge-coordinate x, the x mod 24 gives
the permutation part, x/24 the location part. So I could compute the
edge-coordinate  at the start of phase 2 with M[x][y mod 24], where x
and y are the RL- and FB-sliceedge coordinates and M is a table with
11880*24 shorts. So I need only one tablelookup to get the coordinate.

Herbert

From cube-lovers-errors@mc.lcs.mit.edu  Thu Dec 24 11:35:18 1998
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Date: Wed, 23 Dec 1998 19:07:26 -0500
From: michael reid <reid@math.brown.edu>
Message-Id: <199812240007.TAA28326@adams.math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: Re: Optimal Cube Solver

herbert writes

> Is it necessary to use the table for the map from the "sliceedges" to
> the 1680 "4 out of 8"-coordinate at all?

no, i guess the "4 out of 8" coordinate is not really needed.
good point.  the tradeoff would be one extra table lookup versus
having tables that are 455K larger.  i don't know if there's a clear
choice between these two options, but either is much better than the
6 megabytes i'm using now!

mike

From cube-lovers-errors@mc.lcs.mit.edu  Mon Dec 28 12:21:48 1998
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sun Dec 27 13:14:59 1998
Message-Id: <003301be31c4$60a798e0$335755ca@Uwe.ue.net>
From: uwe@ue.net (Uwe Meffert)
To: "Cube-Lovers" <Cube-Lovers@ai.mit.edu>
Subject: ADDITIONAL FEATURES FOR OUR PUZZLE SITE
Date: Mon, 28 Dec 1998 01:47:54 +0800


Requesting Help:

For 1999 I will be adding a lot of additional interactive puzzles &
games to the "Meffert's World of Puzzles" site in several puzzle
categories enabling a person to challenge a friend or send an
interactive  puzzle greeting.

Each Puzzle Challenge will be made available for 30 days, with actual
time taken to solve it being relayed back to the challenger.  After the
Puzzle Challenge is received, the challenge can be accepted or rejected
or a counter challenge made.

Puzzle Games will also be available online.  I am hoping to have a very
large range of categories from very simple games & puzzles such as
electronic tik tak toe (from single layer to triple layer) with some
additional new features, the traditional 8 and 15 piece sliding puzzles
with talking help function, the Orbix & Orbix Junior (12 & 6 lights in 3
colors) Electronic Reversy etc. etc. etc. to more complex puzzles &
games.

Whilst I have developed some of these already I will need a lot of help
from puzzlers worldwide to make this the best FREE interactive really
Great Puzzle and Games site for 1999.

Please spread the word to anyone you know that can contribute and other
puzzle site Web Editors that may let me use some of their existing
puzzles.

I hope to have very unique graphics and concepts to really popularize
puzzles again, this time as a Free service over the Internet.   To
appeal to the majority of the people the puzzles must not be too hard,
yet still be challenging.

Also, I am presently looking for a new Web Editor for our site, our
present Webmaster Andrew Southern is unfortunately fully tied up until
July with his studies,  please pass the word along to anyone you think
my be suitable.

Many Thanks


Warm regards and a Very Happy New Year to All.
Uwe

Uwe Meffert
P.O. Box 24455, Aberdeen, Hong Kong.
Tel. 852-2518-3080, Fax. 852-2518-3282
Email:- uwe@ue.net
Sites: www.bloodpressure.org,  www.cmd-diagnostics.com
           www.ue.edu www.ue.net www.ue.net/mefferts-puzzles

From cube-lovers-errors@mc.lcs.mit.edu  Mon Dec 28 19:00:54 1998
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Message-Id: <3685267B.D1459EC7@geocities.com>
Date: Sat, 26 Dec 1998 10:10:03 -0800
From: Jono <bagelboyj@geocities.com>
Reply-To: BagelBoyJ@geocities.com
Organization: Fine Finger Design
To: cube-lovers@ai.mit.edu
Subject: Other Cubes

Hi, cube lovers. I have a few questions.
Does anyone know if Erno Rubik is still alive?
About 4 years I vaguely remember seeing a large star-shaped
rubiks puzzle. Does anyone know where I can find one?
I am also looking for a 4x4x4 and a 5x5x5 cube. Where can I
find one?

Thanks to all.


-Jono

[ Moderator's note: Cube-lovers-request gets a lot of requests for
  information on finding 4^3 and 5^3 puzzles.  I'm pretty sure there
  is no source of 4^3 puzzles, except for the occasional auction.
  Last I heard Uwe Meffert sells 5^3 puzzles at
  http://www.ue.net/mefferts-puzzles/ --Dan ]

From cube-lovers-errors@mc.lcs.mit.edu  Tue Dec 29 15:00:41 1998
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Date: Tue, 29 Dec 1998 02:22:08 -0500 (EST)
From: Nicholas Bodley <nbodley@tiac.net>
Reply-To: Nicholas Bodley <nbodley@tiac.net>
To: Jono <bagelboyj@geocities.com>
Cc: cube-lovers@ai.mit.edu
Subject: Re: Other Cubes
In-Reply-To: <3685267B.D1459EC7@geocities.com>
Message-Id: <Pine.BSF.3.96.981229015425.28800A-100000@shell2.tiac.net>


On Sat, 26 Dec 1998, Jono wrote:

}Hi, cube lovers. I have a few questions.
}Does anyone know if Erno Rubik is still alive?

Quite likely. There's a Website that might help you find out:

http://www.rubiks.com

Try a Web search (I like the meta-search engines; Metasearch and
Metafind are a couple).

}About 4 years I vaguely remember seeing a large star-shaped
}rubiks puzzle.

It might well have bees Alexander's Star. As to finding one, sorry to
say, I can't help. Rather sure it wasn't a Rubik design, though. It was
harder to manipulate mechanically than one might like. Not sure, but I
think I've seen it it a store. You might try a Web puzzle dealer.
 (...Puzzletts.com ?)

} Does anyone know where I can find one? I am also looking for a 4x4x4
}and a 5x5x5 cube. Where can I find one?

For the 5^3, in addition to Meffert, as Dan said, Dr. Christoph
Bandelow, in Germany, was selling them, as well, I'm almost certain. The
store [The Games People Play] on Mass. Ave. in Cambridge, Mass., might
have them in stock. Sorry, but I've lost track of Dr. Bandelow's
address.

My general impression is that there's enough interest in the 5^3 to have
revived production, although it could have been just one big run. Fwiw,
(honestly, not much!), my experience regarding the 4^3 agrees with Dan's
comments.


}[ Moderator's note: Cube-lovers-request gets a lot of requests for
}  information on finding 4^3 and 5^3 puzzles.  I'm pretty sure there
}  is no source of 4^3 puzzles, except for the occasional auction.
}  Last I heard Uwe Meffert sells 5^3 puzzles at
}  http://www.ue.net/mefferts-puzzles/ --Dan ]
}


|*  Nicholas Bodley   *|*  Electronic Technician {*} Autodidact & Polymath
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*  The personal computer industry will have become
|*  Amateur musician  *|*  mature when crashes become unacceptable.
--------------------------------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Tue Dec 29 20:15:59 1998
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Message-Id: <199812292316.SAA06321@life.ai.mit.edu>
From: "Christoph Bandelow" <bandecbv@mailhost.rz.ruhr-uni-bochum.de>
To: cube-lovers@ai.mit.edu
Date: Wed, 30 Dec 1998 00:14:50 +0000
Subject: Availability of 5x5x5 cubes and other Rubik type puzzles

On Tue, 29 Dec 1998, Nicholas Bodley wrote about the availability of
5x5x5 cubes  and other puzzles:

> Sorry, but I've lost track of Dr. Bandelow's address.

May I help without being too much vituperated for making an unseemly
advertisement? Christoph Bandelow's email address is

Christoph.Bandelow@ruhr-uni-bochum.de

He does not only sell excellent 5x5x5 Magic Cubes (the good old ones
made in Hong Kong, not in mainland China), but also Magic
Dodecahedra, Skewbs, Pyraminxes, Impossiballs, various Puzzle Balls,
Mach Balls and other Rubik type puzzles and books about those
puzzles. It is a pleasure to deal with him for he is quick,
absolutely reliable, fair and competent. Free mail order catalog.

Christoph
--
Christoph Bandelow
mailto: Christoph.Bandelow@ruhr-uni-bochum.de
http://www.ruhr-uni-bochum.de/mathematik3/bandelow.htm

[ Objective testimonials written in the third person are always
  appreciated. --Dan ]

From cube-lovers-errors@mc.lcs.mit.edu  Tue Dec 29 21:21:03 1998
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Errors-To: cube-lovers-errors@mc.lcs.mit.edu
From: Douglas Zander <dzander@solaria.sol.net>
Message-Id: <199812282147.PAA06517@solaria.sol.net>
Subject: ADDITIONAL FEATURES FOR OUR PUZZLE SITE (fwd)
To: cube-lovers@ai.mit.edu (cube)
Date: Mon, 28 Dec 98 15:47:11 CST

 I would like to voice my concern over this site.  The thing is that not
 everyone has the latest version of Netscape or IE installed.  Many
 libraries (from which I myself access the WWW) do not allow Java to be
 installed on their machines nor do they allow sound.  The highest browser
 versions I have access to in any library here in Milwaukee is 3.0
 I had gone to this site and I could not play the puzzle that was supposed
 to be there; nothing showed up on the screen. 

 I just wished to point this out to the creator of the site; in fact, to
 all creators of puzzle sites.  There are, I believe, a significant number
 of people who use public browsers without Java or sound.  Please keep this
 in mind when designing your sites.  Thank you.

--
 Douglas Zander                |
 dzander@solaria.sol.net       |
 Shorewood, Wisconsin, USA     |

[Moderator's note: Cube-lovers is not really the place to debate the
 philosophy of web design.  I must agree that for maximum usability,
 the text on a page should be displayed without requiring images, and
 that the static images should be displayed with requiring Java.
 Still, setting up a puzzle simulator without Java is more problematic
 than we can reasonably ask of a free server, and having a simulation
 that requires Java is certainly better than none. --Dan ]

From cube-lovers-errors@mc.lcs.mit.edu  Wed Dec 30 10:43:10 1998
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Date: Tue, 29 Dec 1998 23:48:14 -0500 (EST)
From: der Mouse  <mouse@rodents.montreal.qc.ca>
Message-Id: <199812300448.XAA02148@Twig.Rodents.Montreal.QC.CA>
To: cube-lovers@ai.mit.edu
Subject: Re: Availability of 5x5x5 cubes and other Rubik type puzzles

>> Sorry, but I've lost track of Dr. Bandelow's address.

> Christoph Bandelow's email address is
> Christoph.Bandelow@ruhr-uni-bochum.de
> He does not only sell [...] but also [...].  It is a pleasure to deal
> with him for he is quick, absolutely reliable, fair and competent.

> Christoph Bandelow
> mailto: Christoph.Bandelow@ruhr-uni-bochum.de
> http://www.ruhr-uni-bochum.de/mathematik3/bandelow.htm

Heehee!  Had me going there until I saw the signature. :-)

I have had only one experience dealing with Mr. Bandelow.  In that one,
I did indeed find him quick, reliable, fair, and competent.  (I bought
a 5x5x5 and the Cube-family puzzle whose name I forget based on the
dodecahedron, the one that turns based on slices made parallel to the
faces, passing through edge centres.  Interestingly, my experience with
the dodecahedral puzzle taught me enough that I can now solve a
two-face scramble on the 3x3x3 without leaving the two-face subgroup,
which I previously couldn't.)

					der Mouse

			       mouse@rodents.montreal.qc.ca
		     7D C8 61 52 5D E7 2D 39  4E F1 31 3E E8 B3 27 4B

From cube-lovers-errors@mc.lcs.mit.edu  Tue Jan  5 17:17:11 1999
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Mail-from: From cube-lovers-request@life.ai.mit.edu Wed Dec 30 20:17:39 1998
Message-Id: <v04011700b2b074d73b09@[207.172.215.201]>
In-Reply-To: <3685267B.D1459EC7@geocities.com>
Date: Wed, 30 Dec 1998 20:26:43 -0400
To: cube-lovers@ai.mit.edu
From: Charlie Dickman <charlied@erols.com>
Subject: Re: Other Cubes

>[...]I am also looking for a 4x4x4 and a 5x5x5 cube. Where can I
>find one?

Jono,

I recently saw a 4x4x4 for sale at auction at www.ebay.com. Check it out.

Charlie Dickman
charlied@erols.com

From cube-lovers-errors@mc.lcs.mit.edu  Tue Jan  5 17:47:44 1999
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From: "Chris and Kori Pelley" <ck1@home.com>
To: <cube-lovers@ai.mit.edu>
Subject: Rubik's old-timer
Date: Fri, 1 Jan 1999 22:44:07 -0500
Message-Id: <001b01be3602$2685f3e0$da460318@CC623255-A.srst1.fl.home.com>

Puducky@aol.com asked me to forward the following to this list:

I have a 96 year old friend that can do the rubiks cube in less than five
minutes, and near the end he doesn't even look at it! I thought this pretty
amazing for a man of his age. Do you know anyone I can contact about him?
Was wondering if there is a record for his age? Thank you.

Puducky@aol.com

From cube-lovers-errors@mc.lcs.mit.edu  Tue Jan  5 19:10:03 1999
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Message-Id: <19990102200254.8149.rocketmail@send104.yahoomail.com>
Date: Sat, 2 Jan 1999 12:02:54 -0800 (PST)
From: Han Wen <hansker@yahoo.com>
Subject: Original Rubik's Cube Query
To: Cube Lovers <Cube-Lovers@ai.mit.edu>

Hi,

This may be inappropriate for this list, but I thought a few cube
lovers may be interested in this.

I was talking to my college buddy the other day, and he was telling me
about a curious habit of his mom's of buying x-mas gifts and not
giving them away.  Instead, the gifts were stowed away in one of their
closets like museum pieces.  One of these "pieces" is a Rubik's cube.

Hmm... Now I became interested. I asked him about it.. he thought it
wasn't worth much, the packaging was pristine, looked new.  I asked
him to look carefully at the packaging.  The manufacturer was Ideal,
and it has a "1980" and a "Made in Hungary" stamped on it.  (i.e. an
Original Rubik's Cube, untouched in it's original packaging for almost
two decades). I told him it was probably worth more than he thinks.

Just curious.  Does anyone know how much this little gem is worth?

==
_________________________________________________________
Han Wen
Applied Materials
3050 Bowers Ave, MS 1145
Santa Clara, CA  95054
e-mail: Han_Wen@amat.com / hansker@yahoo.com

[Moderator's note: For the past few years there have been rumors
 flying around the net, and perhaps the news services, about thousands
 of dollars being paid for vintage unopened cubes, either the Ideal
 "Rubik's cubes", or the earlier "Hungarian Magic Cube".  Some of
 these rumors may even be true.  However, I feel I should warn people
 against careless speculation, since the Internet is prone to fraud of
 various forms. ]

From cube-lovers-errors@mc.lcs.mit.edu  Tue Jan  5 19:59:00 1999
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Message-Id: <3692A126.3FA0@hrz1.hrz.tu-darmstadt.de>
Date: Wed, 06 Jan 1999 00:32:54 +0100
From: Herbert Kociemba <kociemba@hrz1.hrz.tu-darmstadt.de>
Reply-To: kociemba@hrz1.hrz.tu-darmstadt.de
To: cube-lovers@ai.mit.edu
Cc: michael reid <reid@math.brown.edu>
Subject: Your Optimal Solver

Mike,

I now have a Linux-partition on my PC and I compiled your optimal cube 
solver on it. It really runs fast, about 30%-40% faster than my own
optimal solver which uses the same coordinates. I then compiled your
source code with the Microsoft Visual C++ compiler with similar results.
(By the way, if there other users of the Wintel platform who are
interested in Mikes program I could send the program code to the Cube
Lovers Archives, its only 50KB).

The main reason for the different performance is the fact, that during
the tree search I only hold one cube in memory and I do not use an array
for the cube-coordinates.
But then I had another idea, which was not implemented in my program and
which does not seem to be implemented in yours and which significantly
increased the performance of my program (about 20%) with a few lines of
code (but I think it only works in face-turn-metric):

You use the lines similar to

if (p_node[1].remain_depth<DIST) continue;

which means tree pruning and we apply the next move in this depth.

An analysis shows, that the case (p_node[1].remain_depth<DIST-1) also
happens quite often (while p_node[1].remain_depth<DIST-n can never occur
for n>1 except at the very beginning of the search). In this case, if we
for example had applied the move R, we need not to check R2 and R' any
more but we can continue with another axis.
In the case of the quarter-turn-metric, if we had applied R, we still
had to check R' because the distance of the two resulting states from
start can differ by 2.

Herbert

From cube-lovers-errors@mc.lcs.mit.edu  Thu Jan  7 13:53:48 1999
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Date: Wed, 6 Jan 1999 23:02:32 -0500
From: michael reid <reid@math.brown.edu>
Message-Id: <199901070402.XAA18140@cauchy.math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: Re: Your Optimal Solver

herbert writes

> You use the lines similar to
>
> if (p_node[1].remain_depth<DIST) continue;
>
> which means tree pruning and we apply the next move in this depth.
>
> An analysis shows, that the case (p_node[1].remain_depth<DIST-1) also
> happens quite often (while p_node[1].remain_depth<DIST-n can never occur
> for n>1 except at the very beginning of the search). In this case, if we
> for example had applied the move R, we need not to check R2 and R' any
> more but we can continue with another axis.

interesting idea.  when i get a chance i'll see if i can also get a
performance boost using this idea.

for quarter turns, there is something similar i can do, but this is
only because of the method i used for quarter turns.  namely, i
don't ever do  R R , instead, i do  R2  and count it as two moves.
if applying the move  R  results in a branch of the tree that gets
pruned, then we do not have to try  R2.

however, if i used a different method for quarter turns, where i only
make one move at a time, then the  R2  branch would be a sub-branch
of the  R  branch.  thus it would be pruned automatically.
this suggests that it might be better to use this latter method for
searching the tree.  (the only reason i didn't do this is that i
wanted to use one function for both quarter turns and face turns.)

another idea, suggested to me by rich korf, is to use the line

     if ((node.remain_depth < ELEVEN) && (node.remain_depth < DIST))
        continue;  /*  prune this branch  */

where  ELEVEN  is just the constant 11, and  DIST  is the macro to
look into the big table for the distance of the current coset.
if the first part of the expression is  TRUE , then we evaluate
the second part.  in this case we did a tiny bit of extra work
to evaluate the first part.  but if the first part is  FALSE ,
then we save some work by not looking into the table.  we lose
a little bit of pruning (there are some cosets at distance 12)
but this is very small.  rich explained (if i understood correctly)
that every look into the big table is expensive, because it will
pull a small piece of the table into cache.  but this piece
is unlikely to be used again soon, so it probably displaced some
more useful stuff from cache.  the DIST macro is also a complicated
expression, so it is also expensive in that way.  when i tried this,
i didn't measure any significant performance boost (< 1%).  but the
cache benefit would be more noticeable for longer searches, so
perhaps my test was just too short.  it also depends upon your DIST
macro (or corresponding code); i think rich had more processing to
do besides looking into the table.  and it may also depend on the
size of your secondary cache.

i do have this in my huge optimal solver, so it must have given
some improvement there, but i don't remember how much.  i had to
do lots of tweaking for performance issues on this program.

herbert, if you have a program that uses the exact same coordinates
as mine, you will find it amusing to try the positions

*  position created by  R2 F' R2 F2 D2 F' R2 F2 R2 D2 F' D2 F'
*  inverse of the above position.

and noticing the huge difference.  because of this, i thought
about maybe solving either the input position or its inverse,
depending upon which should be faster.  my experiments showed
that it wasn't easy to predict which would be easier by looking
only at the distances of the 3 initial cosets.  but perhaps
doing a mini-search on each, and looking at how many nodes
they spawn would give a better guess.

of course, we can't expect to get the kind of performance boost
suggested by the extreme example above, but we might get something.

then i had a more ambitious idea:  maybe we could prune the search
tree by having the program realize "the inverse of this position is
too far from start".  the conclusion i eventually arrived at was that
it wouldn't be possible to keep track of the coset of the inverses
by using transformation tables.  so this idea probably won't work.

mike

From cube-lovers-errors@mc.lcs.mit.edu  Wed Jan 13 18:08:03 1999
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Date: Tue, 12 Jan 1999 23:18:11 +0000
From: David Singmaster <david.singmaster@sbu.ac.uk>
To: cube-lovers@ai.mit.edu
Message-Id: <009D220A.48E392F4.2@ice.sbu.ac.uk>
Subject: Clinton and Rubik?

Date: Sun, 10 Jan 1999 09:28:19 -1000
From: drogers@math.hawaii.edu (Douglas Rogers)
Message-ID: <199901101928.JAA01661@knuth.hawaii.edu>
To: wpr3@tutor.open.ac.uk
Subject: From Monica Lewinsky to the Roubik Cube


Bill,

   Here is a quotation that might be appropriate for The
Mathematical Gazette.

   Senator Trent Lott (R-Missouri), Senate Majority Leader,
speaking on 6th January, 1999, to reporters about arrangements
for the impeachment of the President, declared, ``All sides
of this Rubik's cube have been talked about. We hope to have
this all resolved tomorrow''. [As quoted in The New York Times
for 7th January, p. A1.]

   The natural inference here is that Senators have a version of
the Rubik cube on which the US President and friends are depicted,
the object of the exercise being to twist things so as to get
the US President into or out of compromising positions - it was
thoughtful of the Senate Majority Leader to spare us the details,
and to leave this to our imagination.

   DGR.

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499;
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Thu Jan 14 13:12:53 1999
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To: cube-lovers@ai.mit.edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Re: Clinton and Rubik?
Date: 14 Jan 1999 14:43:30 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <77kvqi$o6k@gap.cco.caltech.edu>
References: <cube-lovers.009D220A.48E392F4.2@ice.sbu.ac.uk>

David Singmaster <david.singmaster@sbu.ac.uk> writes:
>   Senator Trent Lott (R-Missouri), Senate Majority Leader,
>speaking on 6th January, 1999, to reporters about arrangements
>for the impeachment of the President, declared, ``All sides
>of this Rubik's cube have been talked about. We hope to have
>this all resolved tomorrow''. [As quoted in The New York Times
>for 7th January, p. A1.]

Of course, claiming that they can REsolve it implies that
it was already SOLVED at some point in the past ... :-)

--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
"... I guess that explains why you're automatically dogmatic!"

From cube-lovers-errors@mc.lcs.mit.edu  Mon Jan 25 14:16:47 1999
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sun Jan 24 00:08:55 1999
Date: Sun, 24 Jan 1999 00:07:11 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: Corners Only, Ignoring Twist
In-Reply-To: <Pine.PMDF.3.95.980927193742.73560A-100000@PSTCC6.PSTCC.CC.TN.US>
To: Cube-Lovers <cube-lovers@ai.mit.edu>
Reply-To: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Message-Id: <Pine.PMDF.3.95.990123235414.196055A-100000@PSTCC6.PSTCC.CC.TN.US>

On Sun, 27 Sep 1998, Jerry Bryan wrote:

> In developing a no-twist, no-flip version of the program, I decided to try
> it out on the corners only case.  Here are the results.

Here is one more tidbit on this subject.  The program which performed the
God's Algorithm search for the no-twist corners only case produced a
summary by symmetry class.  I was surprised to note that there were two
positions for which Symm(x)=M.

Normally, there is only one position for the corners where Symm(x)=M,
namely the position which fixes all the corners (i.e., Start).  However,
if twists are ignored, then the central inversion of the corners has the
property that Symm(x)=M.  This is analogous to the Pons Asinorum position
where the edges are centrally inverted.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                jbryan@pstcc.cc.tn.us

From cube-lovers-errors@mc.lcs.mit.edu  Mon Feb  1 15:08:06 1999
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Mail-from: From cube-lovers-request@life.ai.mit.edu Sun Jan 24 13:57:32 1999
Message-Id: <19990124185842.5788.rocketmail@send103.yahoomail.com>
Date: Sun, 24 Jan 1999 10:58:42 -0800 (PST)
From: Han Wen <hansker@yahoo.com>
Subject: Query on Octagon Cube Edge Parity Problem
To: Cube Lovers <Cube-Lovers@ai.mit.edu>

Hi,

I ran into an unusual scenario with the Octagon cube recently where
only ONE edge piece was flipped and all the other pieces were
positioned and oriented properly. This is bizarre of course, because
with a Rubik's cube, this is an impossible scenario; there must be a
minimum of TWO edge pieces flipped.

Does anyone understand the redundancy that allows this strange edge
parity problem?  And I guess, how to solve it.  I lamely mixed the
cube up again, resolved until the problem "went away".

For those who may not be familiar, the Octagon cube is a variant of
the Rubik's Cube. The cube is organized by color into 8 columns of
three cublets: corner, edge, corner. If you look at the top face you
see that the half of the corner cublets have been cut away so the the
face forms an octagon instead of a square. This octagon shape is
extended down through the middle and bottom layer, so that the puzzle
looks like an octagon "tube". There are a total of 10 colors, two for
the top and bottom faces, and 8 for the eight columns of
(top-middle-bottom) cublets.

==
_________________________________________________________
Han Wen
Applied Materials
3050 Bowers Ave, MS 1145
Santa Clara, CA  95054
e-mail: Han_Wen@amat.com / hansker@yahoo.com

From cube-lovers-errors@mc.lcs.mit.edu  Tue Feb  2 17:09:46 1999
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Mail-from: From cube-lovers-request@life.ai.mit.edu Tue Jan 26 09:45:56 1999
Date: Tue, 26 Jan 1999 14:45:54 +0000
From: "Collins, Lindon" <lindon.collins@lmco.com>
Subject: Middle layer last?
To: "'Cube Lovers'" <cube-lovers@ai.mit.edu>
Message-Id: <B1F50FC23BE7D1119DA700805FBB672C870B@swnd01>

Sorry if I'm retreading old ground but I'm new to this.

1.
	I remember reading somewhere an article that advocated solving
the middle layer of the cube last.  I have got some fairly short moves
to place cubes from the middle layer onto the bottom layer, but when it
comes to solving the middle layer, I seem to be trusting to luck that I
have reached a favourable position.  I cannot see how I am going to
reduce my average number of moves to solve the cube using this method.
I think there are two possibilities:-

	1.  There are some cool moves for solving the middle layer last
that I have missed.
	2.  I should forget about solving the middle layer last, and
stick to my usual method
	(ie. top,middle,bottom)

2.
	A more general question is: What is the shortest (practical)
method for solving the cube that anyone knows of?  (keyword: "practical"
- don't say 22 moves)

Thanks,
Lindon Collins
Swindon, UK

From cube-lovers-errors@mc.lcs.mit.edu  Tue Feb  2 17:56:32 1999
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Date: Mon, 1 Feb 1999 17:16:54 -0500
From: michael reid <reid@math.brown.edu>
Message-Id: <199902012216.RAA07304@chern.math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: Re: Query on Octagon Cube Edge Parity Problem
Cc: hansker@yahoo.com

> I ran into an unusual scenario with the Octagon cube recently where
> only ONE edge piece was flipped and all the other pieces were
> positioned and oriented properly. This is bizarre of course, because
> with a Rubik's cube, this is an impossible scenario; there must be a
> minimum of TWO edge pieces flipped.

the "octagon" puzzle has some "edges" with only one visible face.
namely, these are the  U-D layer edges on a cube, which were shaved
when the cube was modified in shape.  these edges have no visible
orientation, so flip one of these along with the edge that's
definitely in the wrong orientation.

in a similar way, it's possible to get positions that appear to have
the wrong permutation parity.  there are four vertical columns of
two corners and an edge each.  these do not have any fixed "home"
location, so that any permutation of these also constitutes a "solved"
state.  (well, at least most people would consider it to be solved.)
but swapping two of these columns creates an odd permutation parity.
thus you can swap two columns, and also swap a pair of edges or
corners, which gives the impression of incorrect parity.

for a simple example, do  R2 F2 R2  from the solved position.
the edges  UF  and  DF  have been swapped, and it looks like nothing
else has happened.  in fact, the  FL  column has been swapped with
the  BR  column as well.

mike

[Moderator's note: I hadn't noticed that this had such an obvious
 answer.  Thanks also to Jon Ferro, Steve LoBasso, der Mouse,
 Guy N. Hurst, Michael Ehrt, and Christ van Willegen for also
 providing the answer.  I've selected Mike Reid's, since he points out
 the other notable ambiguity of the Octagon.

 What wasn't noted is that the Spratt wrench can be used to flip the
 noted edge along with three of the ambiguous edges. --Dan]

From cube-lovers-errors@mc.lcs.mit.edu  Tue Feb  2 19:27:15 1999
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Message-Id: <36B77B8B.2A223AA5@sgi.com>
Date: Tue, 02 Feb 1999 14:26:19 -0800
From: Derek Bosch <bosch@sgi.com>
To: "Collins, Lindon" <lindon.collins@lmco.com>
Cc: "'Cube Lovers'" <cube-lovers@ai.mit.edu>
Subject: Re: Middle layer last?
References: <B1F50FC23BE7D1119DA700805FBB672C870B@swnd01>

well, if you want "simple", the only two moves you really need
are:

 the edge 3-cycle...  U2^U2v  (UB->DF->UF)
 and the 2-edge flip...  ^U^U^U2vUvUvU2  (flips UF and UB)

assuming you are holding the cube so that the left and right faces are solved
(where ^ is moving the middle slice up, and v is moving the middle slice down)

note, once you get used to those moves, you can create variations that move
and flip, like:

U'^U'^U2vU'vU

these are also REALLY ergonomic - very easy to do rapidly...
-- 
Derek Bosch        "A little nonsense now and then
(650) 933-2115      is relished by the wisest men"... W.Wonka
bosch@sgi.com

From cube-lovers-errors@mc.lcs.mit.edu  Tue Feb  2 21:16:59 1999
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Date: Tue, 2 Feb 1999 20:27:26 -0500
Message-Id: <00131725.C22092@scudder.com>
From: Jacob_Davenport@scudder.com (Jacob Davenport)
Subject: Re: Middle layer last?
To: "'Cube Lovers'" <cube-lovers@ai.mit.edu>

1.  I learned the solution of top, middle, and then bottom many years ago.
I forgot most of it, so when I asked Kristin Looney to remind me how to do
the cube, she showed me her solution, which is top corners, bottom cFrom cube-lovers-errors@mc.lcs.mit.edu  Sat Feb  6 01:46:02 1999
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Date: Tue, 2 Feb 1999 20:27:26 -0500
Message-Id: <00131725.C22092@scudder.com>
From: Jacob_Davenport@scudder.com (Jacob Davenport)
Subject: Re: Middle layer last?
To: "'Cube Lovers'" <cube-lovers@ai.mit.edu>

1.  I learned the solution of top, middle, and then bottom many years ago.
I forgot most of it, so when I asked Kristin Looney to remind me how to do
the cube, she showed me her solution, which is top corners, bottom corners,
top and bottom edges at the same time, then middle edges.  It is a much
easier solution to do and to teach.  A variant of it can be found at
http://www.unc.edu/~monroem/rubik.html, although in this variant he is more
rigid about how to solve the top and bottom edges.  So, I'd forget your
original method of top/middle/bottom, because it does not revel in the
corner/edge dichotomy of the cube.  I only use top/middle/bottom when
solving cubes with complicated patterns, such as those at
http://www.wunderland.com/WTS/Kristin/CustomCubes.html.

2.  The shortest, pratical, method for solving the cube, in my opinion, is
the one discussed above.  Kristin, who does not have incredibly fast hands,
did very well in speed cube competitions when pitted against people who had
hot hands but the top/middle/bottom solution.  Also, I can teach someone
this solution in an hour.

-Jacob

From cube-lovers-errors@mc.lcs.mit.edu  Tue Feb  9 15:30:14 1999
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Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Date: Sun, 7 Feb 1999 21:18:58 +0000
From: David Singmaster <david.singmaster@sbu.ac.uk>
To: reid@math.brown.edu
Cc: CUBE-LOVERS@ai.mit.edu
Message-Id: <009D3667.F01254DB.8@ice.sbu.ac.uk>
Subject: Re: Query on Octagon Cube Edge Parity Problem

	Similar parity problems can be produced by recolouring a cube.  I once
sold a cube to someone who came back a few minutes later with two centers
exchanged.  I accused him of taking it apart, but then I fiddled with it and
got it back right, which amazed me even more.  Then I discovered that two
opposite faces of the cube were colored red!  Some time later, I had an example
with two adjacent sides having the same color.
	In 1980, Tamas Varga showed me some cubes with just two colors and I
then made up numerous such color variants.  E.g. using just two colors, have
the three faces of one color meet at an corner, or not meet at a corner (these
are the only two ways of coloring the faces with two colors, three faces of
each color).  Also fun is a three color cube - two opposite sides red, two
opposite sides white and two opposite sides blue.  Then every corner is  red,
white, blue  - except half of them are red, blue, white.  All these are
difficult to solve for people who have only done ordinary cubes.
	In the mid 1980s, Edward Hordern showed me a cube which I recall he
said Nob Yoshigahara had invented, but my example was made by Marcel Gillen.
This appear to be a  4^3,  but when turned, it moves eccentrically.
Examination shows that it is a  3^3 with three layers of pieces glued to three
adjacent faces.  Edward's original example had no colors, so it took some time
to solve as one didn't know where pieces went.  Further, the eccentric movement
causes parts to protrude, making it hard to hold and to move.  All in all, a
most enjoyable variant.

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499;
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Tue Feb  9 20:27:47 1999
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Message-Id: <19990208030401.14755.rocketmail@send1e.yahoomail.com>
Date: Sun, 7 Feb 1999 19:04:01 -0800 (PST)
From: Han Wen <hansker@yahoo.com>
Subject: Moves for Solving the Pyramorphix puzzle
To: Cube Lovers <Cube-Lovers@ai.mit.edu>


Hi

For those cube masters out there getting bored, you may want to play
with Meffert's latest puzzle, the Pyramorphix.  It only takes about
1-2 hours to solve, but it provides some Square-1-like entertainment.
For those who may not be familiar, the Pyramorphix looks like the
Pyraminx's little brother with 4 pieces instead of 9 pieces for each
face.  However, unlike its big brother, the Pyramorphix turns a lot
differently (90 deg rotations), morphing into different shapes as you
twist and turn.  As you know, with the Pyraminx, you're always
rotating little pyramids about one of the 4 tips. However, with the
Pyramorphix, when starting out with the pyramid shape, you rotate an
edge and its two adjacent corners (i.e. the whole edge of a given
face).  The rotation axes lie along the intersection of the 4 center
triangle pieces.

There appears to be 6 general shapes you can "morph" the Pyramorphix
into I'll affectionately call: the Pyramid, the Butterfly, the Crown,
the Rocket, the Airplane and the truncated Star of David.  (There are
actually two truncated Star of Davids, mirror images of one another).
You'll know what I mean if you actually have the little creature in
your hands. The hardest part of this puzzle is figuring out how to
morph between all these different shapes.

I solve the Pyramorphix by first solving the 4 corner pieces and then
orienting the 4 center pieces.

First, a few notation definitions. Hold the Pyramorphix so that you
have one of the faces facing you.  I'll call that face Front (F).  The
face on the bottom will be called Down (D), and the faces left and
right of the F face will be called Left (L) and Right (R) respectively.

An edge will refer to a center piece and its two adjacent corner
pieces. I'll specify which edge by indicating the two faces the edge
intersects (e.g. R-D edge is the edge formed by the interesection of
the Right and Down faces).
Now, unfortunately, I can't just refer to rotations of edges, because
as you'll see, sometimes you need to rotate strange shapes that look
nothing like an edge.  Instead, it's better to refer to slices.
Namely the plane about which one of the edges rotates on.  So, if you
look at the F face, you'll see three slices: the R-D slice (60 deg),
the L-D slice (120 deg) and the horizontal F-slice (0 deg).  When the
Pyramorphix is in its pyramid shape, you can rotate the two edges on
the R-D and L-D slices, but you cannot rotate the corner piece sitting
on the F-slice.  Don't worry, when we morph the pyramid into the
Rocket, you'll see that you'll be able to rotate the tip of the rocket
about the F-slice.

Now, that I've thoroughly confused you, here are my notations for
actual moves:
R - 90 deg clockwise rotation about the R-D slice
L - 90 deg clockwise rotation about the L-D slice
F - 90 deg clockwise rotation about the F-slice
R2 - 180 degree rotation about the R-D slice
R' - 90 degree counterclockwise rotation about the R-D slice

Finally, hold the Pyramorphix in your hands, so the the F-face is
facing towards you with the tip of the triangle pointing up.  If you
now rotate the puzzle so the the D-face is facing towards you, you
should see an upside-down triangle with the tip pointing down.  I will
refer to the three corner pieces that you see as:

DL - left corner piece
DR - right corner piece
DM - middle corner piece
_______________________________________________________
To solve the four corner pieces, first get them in their proper
positions by performing 180 rotations of the edges. Now, you need to
orient the corners by making appropriate clockwise or counterclockwise
twists.  Here are some moves to do this:

Name: Single corner twister
Move: (R L' R' L) ^2
Shapes: Butterfly - Star of David - Airplane - Airplane - Star of
David - Star of David - Butterfly - Pyramid
Action: Clockwise (CW) twist of DM corner

Name: Left-side double corner twister
Move: (R L R' L' ) ^2
Shapes: Butterfly - Star of David - Crown - Star of David - Crown -
Star of David - Butterfly - Pyramid
Action: CCW twist of DL corner, CW twist of the DM corner

Name: Right-side double corner twister
Move: (R' L' R L) ^2
Shapes: Butterfly - Star of David - Crown - Star of David - Crown -
Star of David - Butterfly - Pyramid
Action: CCW twist of DR corner, CW twist of the DM corner

Name: Triple corner twister
Move: (R' L R L') ^2
Shapes: Butterfly - Star of David - Star of David - Airplane -
Airplane - Star of David - Butterfly - Pyramid
Action: CW twist of DL corner, CW twist of DR corner, CCW twist of DM
corner
_______________________________________________________
Now, to orient the center pieces in place, here are some moves to do
this:

Name: Double edge swapper
Move: R' L2 F2 L2 R
Shapes: Butterfly - Rocket - Rocket - Butterfly - Pyramid
Action: Swap F <--> L and R<-->D center pieces

Name: Tricycle swap
Move: (L R2 F' R2 L') (R' L2 F L2 R)
Shapes: (Butterfly - Rocket - Rocket - Butterfly - Pyramid ) ^2
Action: Permute F --> R --> L --> F center pieces

Name: Edge swapper
Move: (R' L2 F L2 R) (L R2 F R2 L') (R' L2 F L2 R)
Shapes:  (Butterfly - Rocket - Rocket - Butterfly - Pyramid ) ^3
Action: Swap F <--> D center pieces

With these short collection of moves, you should be able to readily
solve the Pyramorphix...
_______________________________________________________
Epilogue:

So, you've solved the Pyramorphix, bored silly and you want to know
what's next?!  Check out Meffert's Puzzle Ball.  It's actually a bit
more challenging.  Happy puzzling.... :)



==
_________________________________________________________
Han Wen
Applied Materials
3050 Bowers Ave, MS 1145
Santa Clara, CA  95054
e-mail: Han_Wen@amat.com / hansker@yahoo.com

From cube-lovers-errors@mc.lcs.mit.edu  Tue Feb  9 22:42:43 1999
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Message-Id: <714F77ADF9C1D111B8B60000F863155102DD6CBA@tbjexc2.tbj.dec.com>
From: Norman Diamond <Norman.Diamond@dec-j.co.jp>
To: <cube-lovers@ai.mit.edu>
Subject: Re: Moves for Solving the Pyramorphix puzzle
Date: Wed, 10 Feb 1999 11:34:47 +0900

Hey!  Everyone on this list _already_ knows moves to solve
Pyramorphix, although it often requires careful staring at the
thing in order to recognize the exact configuration each time.

You see, everyone on this list knows how to solve 3x3x3
Rubik's cube.  And everyone knows how to solve 2x2x2
Rubik's cube because it's a subset of the 3x3x3, without
edges or centers.  And everyone knows how to solve
Pyramorphix because it's a subset of the 2x2x2, where
some of the corners don't need orienting.

-- Norman.Diamond@dec-j.co.jp
[Not speaking for Compaq]

From cube-lovers-errors@mc.lcs.mit.edu  Thu Feb 11 21:17:42 1999
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Message-Id: <3.0.3.32.19990210083221.0092a700@cscan02.caddscan.com>
Date: Wed, 10 Feb 1999 08:32:21 -0500
To: Cube-Lovers@ai.mit.edu
Reply-To: <bmain@caddscan.com>
From: Bryan Main <bmain@caddscan.com>
Subject: Puzzle Stores in London area

I'll be going to the london/cambrige area in a few days and was wondering
if anyone knew where there are any puzzle stores that have cube puzzles.
I'm not really looking for 3x3x3 cubes since I have enough, mainly I'm
looking for the other ones like pyrimix, megamix etc.  thanx in advance.

bryan

[Please respond directly to Bryan; if there are particularly
worthwhile and non-repettive suggestions he can send them to
cube-lovers --Moderator ]

From cube-lovers-errors@mc.lcs.mit.edu  Fri Feb 19 12:26:12 1999
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Date: Thu, 18 Feb 1999 23:45:18 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Edges only, Ignoring Flips, Face Turn Metric
To: Cube-Lovers <cube-lovers@ai.mit.edu>
Message-Id: <Pine.PMDF.3.95.990218232951.21673A-100000@PSTCC6.PSTCC.CC.TN.US>

I have completed a God's Algorithm run in the face turn metric for the
group consisting of edges only ignoring flips.  The size of the group is
therefore 12!  The results are as follows:

  Distance    Patterns   Positions    Branching
    From                               Factor
   Start

     0f            1              1
     1f            2             18     18.000
     2f            9            243     13.500
     3f           75           3240     13.333
     4f          920          42535     13.128
     5f        11406         542234     12.748
     6f       136423        6529891     12.043
     7f      1386164       66478628     10.181
     8f      6481303      310957078      4.678
     9f      1969536       94443600      0.304
    10f          129           4132      0.000

  Total      9985968      479001600

I should mention that Herbert Kociemba was the first to calculate the
"patterns" column.  He did it as a part of an investigation into
developing an IDA* optimal solver, with the patterns column being part of
a patterns data base used by the IDA* algorithm.

I used my usual program where (for example) I calculate the set of all
positions which are 10f from Start by forming the products of all
positions which are 5f from Start in lexicographic order, and throwing
away the duplicates and the ones that are shorter than 10f.

This technique is reasonably efficient when the branching factor is fairly
constant, as it is at this distance from Start for larger problems such as
the whole cube.  However, it is very inefficient for this particular
problem.  I have to calculate about (542234^2)/48 products just to get the
129 products I keep at 10f from Start.  (The "divide by 48" takes symmetry
into account.)  In a "one level at a time" search by contrast, the tail of
the distribution usually goes very quickly because there are so few
positions in the tail.

I have come to believe that any corners only (with or without twist) or
edges only (with or without flip) group, or the group which keeps both
corners and edges but without twists and flips, will be a fairly poor
pattern data base for IDA*.  The problem is that any such search space
will have a diameter which is too small, and more importantly will have an
average distance from Start which is too small.

Here is why I think the diameter and average distance from Start will be
too small for these groups.  Consider the quarter turn metric.  We know
immediately that the maximum branching factor is 12 because there are 12
quarter turns.  We know almost as immediately that the maximum branching
factor beyond one move from Start is 11 because there is always at least
one quarter turn that goes closer to Start.  Finally, readers of
Cube-lovers know that the maximum branching factor is asymptotic to about
9.3 because of Dan Hoey's syllable analysis.  Syllable analysis takes into
account moves which commute because they are on opposite faces such as
RL=LR.

(Similar analysis for the face turn metric yields an asymptotic maximum
branching factor of about 13.3)

I have come to think of syllable analysis not just as an upper limit for
the branching factor but as a predictor for the branching factor. Indeed,
the actual branching factor differs from the branching factor "predicted"
by syllable analysis only because of duplicate positions which arise from
processes which are not accounted for by syllable analysis.

Such duplicate positions must exist by the finiteness of the problem, else
a God's algorithm search would be infinite.  But such duplicate positions
are non-trivial and are generally not very close to Start. With the full
cube, they are quite rare as close to Start as has been searched so far
(10f and 12q, respectively).

What happens with a typical search is that the branching factor stays
relatively constant until within a couple of levels of the mode of the
frequency distribution of the distances from Start.  The branching factor
then declines rapidly due to duplicate positions, and there is a short
tail in the frequency distribution just past the mode.

The key point is that syllable analysis is identical for all groups
involving corners only, edges only, corners and edges, and/or with or
without twists and flips.  Hence, the basic branching factor is the same
for all such groups.  Therefore, the mode of the distribution is reached
much sooner when the group is smaller and the average distance from Start
is much smaller.

What would be desirable for a pattern data base for IDA* would be a
subgroup of G whose branching factor was smaller so that the mode, the
diameter, and the average distance from Start would be larger.  That is,
what would be desirable would be a subgroup which was not constrained by
standard syllable analysis.  Failing that, it seems to me that the only
way to improve the pattern data base is to make it larger.  In this light,
I interpret what Mike Reid (and more recently) Herbert Kociemba have done
with their IDA* programs is to find clever ways to make their pattern data
bases as large as possible, but they do it in such a way (using symmetry
and other equivalence classes) that their large data bases are small
enough to store in memory'

As I think has been mentioned before, a group such as <U,R> has a small
branching factor but is not suitable as a pattern data base for IDA*
because the distance from Start for a particular position in <U,R> may be
larger than the distance from Start for the same position in G.

Jerry Bryan

From cube-lovers-errors@mc.lcs.mit.edu  Fri Feb 19 17:17:38 1999
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From: Douglas Zander <dzander@solaria.sol.net>
Message-Id: <199902192002.OAA04117@solaria.sol.net>
Subject: 2by alternative mechanism
To: cube-lovers@ai.mit.edu
Date: Fri, 19 Feb 99 14:02:03 CST

Hello,  I remember someone asked if there were two different mechanisms 
to the 2x2x2 cube but I can't remember if it was answered.  Last night
while doing a patent search I came across a mechanism totally different
than what my own cube looks like.  It is U.S. Patent # 5826871  Oct '98
Basically, it is a standard Rubik's Cube with the central layers hidden
underneath and one quadrant is fixed in place.  I am wondering if there 
exists a web site with all the diagrams of all different puzzles from
their patent pages.  As you may know, patents and patent diagrams are 
*not* covered by copyright issues, you may photocopy them for personal
viewing and even display them on a web page.  I think such a web page
would be nice to have; a web page of all the diagrams of the working 
mechanisms of all the puzzles discussed on this list.  (The thing is 
though that soon the U.S. Patent office will be adding the diagrams to
their web pages. www.uspto.gov)   Anyone wish to start such a page?
I have access to a patent depository.  :-)

-- 
 Douglas Zander                | 
 dzander@solaria.sol.net       |
 Shorewood, Wisconsin, USA     |

From cube-lovers-errors@mc.lcs.mit.edu  Mon Feb 22 13:23:12 1999
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Date: Mon, 22 Feb 1999 17:23:57 +0000
From: David Singmaster <david.singmaster@sbu.ac.uk>
To: cube-lovers@ai.mit.edu
Message-Id: <009D4210.9784799A.12@ice.sbu.ac.uk>
Subject: Fwd: Request for spectacular cube-solving  -  Can anyone help?

From:	mytv-film@t-online.de
To:	MX%"david.singmaster@sbu.ac.uk"
Subj:	rubik's cube
Date: Wed, 17 Feb 1999 17:49:19 +0100
To: david.singmaster@sbu.ac.uk
Subject: Rubik's cube

Hello Mr. Singmaster,

my name is Gvksen and I'm working for the new German TV show "Guinness -Show
of Records" on behalf of the German Broadcast Station ARD.

In one of our next shows we would like to present a person who is able to solve
a classical Rubik's Cube in a spectacular way. For example by studying the cube
first for a couple of minutes and then solving it without looking at it. On an
internet homepage with different links to Rubik's cube fans we've read that there
is a young man in U.K. called John White who is able to do that: Studying the
cube for 10 minutes and than solving it behind his back in 136 seconds. Now,
we're looking for this young man, who is or was a mathematics student at the
University of Warwick.

Our investigation up to now brought us to you, so you are considered to be one
of the leading "Rubik's cube experts" in the world. In addition you were a judge
during the "Rubik's Cube World Championship". So you may know a lot of persons
and beyond that a lot of different variations of solving a Rubik's cube in an
impressive way.

We would be very happy if you could give us a hint or some advice concerning our
idea of presenting this category of record in our show.

One characteristic feature of our show is the fact that we present a record as a
competition. That means: We would invite the record holder and a challenger who
believes himself capable of being able to solve a cube in a similar way. In the
best case this challenger should be a person who speaks German. Do you know
anyone who could be suitable?

I hope that these are not too much questions and I would like to thank you even
now for your help and kind co-operation.

Our e-mail address: mytv-film@t-online.de

kind Regards

Gvksen
MyTV Film and Tv Production

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499;
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Tue Feb 23 11:10:32 1999
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Message-Id: <19990222194015.5998.rocketmail@send105.yahoomail.com>
Date: Mon, 22 Feb 1999 11:40:15 -0800 (PST)
From: pete beck <just_puzzles@yahoo.com>
Subject: Cubes in the news (was: Request for spectacular cube-solving...)
To: David Singmaster <david.singmaster@sbu.ac.uk>, cube-lovers@ai.mit.edu

It must be the age of the CUBE.

Pete Beck
----------------
from Yahoo! News Technology Headlines Monday February 22 1:19 PM ET

Corrected: Rubik Cube Whiz Offers Millennium Bug Solution

LONDON (Reuters) - A man who solved the riddle of Rubik's cube has
invented a test kit to detect where the millennium computer bug will
strike.

At the age of 12, Patrick Bossert shot to fame when he worked out his
own solution to the mystifying cube and wrote a bestseller about it
that sold 1.5 million copies.

Now 30, he and a team of software experts at London-based WSP Business
Technology have developed Delta-T Probe, a program that can work out
whether microchips embedded in electronic equipment are likely to fail
on January 1, 2000.

Delta-T works by electronically detecting equipment to identify chips
that process date and time, making it likely to malfunction when 1999
becomes 2000.

``Only a small percentage of systems will fail to recognize the next
millennium, but finding out which ones might go wrong is a huge and
costly process,'' said Bossert, technical director at WSP Business
Technology, a unit of consulting engineering group WSP Group Plc.

Bossert estimates hundreds of millions of chips are buried deep inside
equipment in Britain. The chips control devices such as security
systems, fire alarms, production lines, medical equipment and
telecommunications.

Bossert expects one in 500 embedded systems will take equipment back
in time to Jan. 1, 1900, causing equipment to fail.

British supermarket chain Sainsbury's Plc is among major companies
that have tested Delta-T. Sainsbury's said a trial run at one of its
stores in Devon, southwestern England, had been a success.



Hi, this is Pete Beck's personal e-mail site, a.k.a. Just Puzzles is a
hobby mail order seller of "Mechanical puzzles" specializing in
Rubik's Cube type puzzles. HOME is at
<www.freeyellow.com/members4/justpuzzles/> - post POB 267, Wharton NJ
07885, answering machine is 973-625-4191. Current as of 10 Nov 1998.

From cube-lovers-errors@mc.lcs.mit.edu  Wed Feb 24 12:52:23 1999
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Date: Mon, 22 Feb 1999 12:13:16 -0800 (PST)
From: Tim Smith <tzs@tzs.net>
To: cube-lovers@ai.mit.edu
Subject: Re: Fwd: Request for spectacular cube-solving  -  Can anyone help?
In-Reply-To: <009D4210.9784799A.12@ice.sbu.ac.uk>
Message-Id: <Pine.LNX.4.04.9902221210120.6439-100000@52-a-usw.rb1.blv.nwnexus.net>

> In one of our next shows we would like to present a person who is able to
> solve a classical Rubik's Cube in a spectacular way.

A spectacular way I'd like to see someone do the cube is while juggling.
Instead of that old routine where the juggler juggles some fruit and
eats the fruit while juggling it, juggle three (or more!) cubes, and
solve them at the same time.

--Tim Smith

From cube-lovers-errors@mc.lcs.mit.edu  Wed Feb 24 13:57:12 1999
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Message-Id: <19990223234105.19048.rocketmail@web4.rocketmail.com>
Date: Tue, 23 Feb 1999 15:41:05 -0800 (PST)
From: "Jorge E. Jaramillo" <kingeorge@rocketmail.com>
Subject: Number of moves
To: cube <Cube-Lovers@ai.mit.edu>

A lot has been said about world records solving the
cube when it comes to time but I don't recall seeing
anywhere how many moves the person who holds the
record did.
Does anyone know?


===
Jorge E Jaramillo

From cube-lovers-errors@mc.lcs.mit.edu  Wed Feb 24 15:56:12 1999
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Message-Id: <36D3B818.E20EFCD6@ibm.net>
Date: Wed, 24 Feb 1999 00:28:08 -0800
From: "Jin 'Time Traveler' Kim" <chrono@ibm.net>
Reply-To: chrono@ibm.net
To: cube-lovers@ai.mit.edu
Subject: Acquiring rare puzzles
References: <199902192002.OAA04117@solaria.sol.net>

This may be common knowledge to many of you on this list, but I thought
I'd mention these and give everyone a fair shot (as well as maybe
angering some who used it as a resource to locate some harder to find
pieces).

1) If you are looking for some rare pieces like the Revenge, Megaminx,
Alexander's Star, these items often come up for auction at

http://www.ebay.com

In the three weeks or so that I've been actively participating in
auctions, I've seen several of each of the following: Rubik's Revenge,
Alexander's Star, Missing Link, Megaminx, and Whippit.  I've also seen
Skewb, 5x5x5, Rubik's Magic, Pyraminx, Mickey's Challenge, and any
number of others come up for bidding.  Haven't seen a Rubik's Domino
(2x3x3), though.  For a general idea of what's there, try doing a search
for the key word "rubik" at the site.  That's just for starters.  I even
saw a Cuboctahedron (yes, just a shaved down 3x3x3 cube) go for over 60
dollars.  Incredible.  Using eBay I have personally added two more
Revenges to my collection.



2) There is also someone on the net who seems to be contacting people
directly and trying to sell a stock of puzzles he ran across.  His
email is jo_schumacher@[see moderator's note].  Try emailing him for
availability and pricing.  For example, he was last selling mint
Revenges for $109 and Alexander's Stars for $70.  I haven't bought
anything from him yet so consider it an "at your own risk" venture.
  
-- 
Jin "Time Traveler" Kim
chrono@ibm.net
http://www.slamsite.com/chrono
'95 PGT - SCPOC

[ Moderator's note: At least one person on the list has complained
  about getting spam from schumacher, and it sure looked like spam to
  me.  I certainly can't countenance sending e-mail to offer goods for
  sale to list members who haven't indicated a willingness to receive
  such an offer.  When it comes to spammers, "at your own risk"
  includes "at the risk of helping to make this medium unusable.

  I usually steer the numerous requests from people seeking Rubik's
  Revenge to ebay (and to yard sales, and to simulations).  But note
  that the FTC says that internet auctions are the growth zone for
  Internet fraud.  This can be somewhat mitigated by sending your
  money through an escrow agent--see ebay for information on that. ]

From cube-lovers-errors@mc.lcs.mit.edu  Wed Feb 24 17:03:02 1999
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Message-Id: <A417D02E1D96D111AF910080D88001E7D58B62@GONZO.cle.ab.com>
From: "Schmidt, Greg" <Greg.Schmidt@gonzo.cle.ab.com>
To: "'Tim Smith'" <tzs@tzs.net>, cube-lovers@ai.mit.edu
Subject: RE: Fwd: Request for spectacular cube-solving  -  Can anyone help ?
Date: Wed, 24 Feb 1999 13:53:02 -0500

While we're at it, I might as well add that I think the juggler should
be blindfolded!

-- Greg

[ And holding her breath underwater! And counting to one hundred
  backwards with her toes!

  Greg also notes that you can't trust ebay's auction ending dates.
  Caveat browsor.   -- Moderator ]

From cube-lovers-errors@mc.lcs.mit.edu  Thu Feb 25 11:25:26 1999
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Message-Id: <4.1.19990224195654.0092e1c0@mail.vt.edu>
Date: Wed, 24 Feb 1999 20:38:44 -0500
To: cube-lovers@ai.mit.edu
From: Kevin Young <keyoung3@vt.edu>
Subject: Oddzon version of the cube
In-Reply-To: <36D3B818.E20EFCD6@ibm.net>
References: <199902192002.OAA04117@solaria.sol.net>

I am curious if anyone has had the same problems with the current official
Rubik's cube that is in production.  I bought a Rubik's cube last year in a
toy store.  This was the newest version made by Oddzon.  I noticed that
there was a clear sticker overtop of each of the colored stickers.  In less
than a month this clear sticker peeled up.  I just don't remember the
original cubes by Ideal wearing down that fast.  In fact I still have a
couple cubes from Ideal with stickers still intact.

I contacted Oddzon by email and have received several responses from them,
forwarding my email to the appropriate department.  Currently my concern is
with under review by the Quality Control and Marketing department.  They
have claimed that this concern about the stickers is not common.  Maybe I
just got a bad cube, I don't know.  But, in order for them to make a better
product, I'm sure it's going to take a strong voice from the people
actually buying the products.  I just encourage everyone if they have
experienced the same thing with the stickers with the Oddzon product, to
contact them.

Most of the time, I still play with my cube I got from Ideal from 1982.  It
was called "Rubik's Cube Deluxe".  It was the official Rubik's cube that had
colored tiles instead of stickers.  That cube is holding up like new.  I
also made a motion to Oddzon to put the "Rubik's Cube Deluxe" back into
production.  That motion is now with the Senior Marketing Manager at
Oddzon.  It's his voice to decides which products go into production.  I
also encourage anyone that wants to see that version back into production,
to contact Oddzon.

Oddzon has been incredibly helpful and incredibly nice.  They have informed
me that two new Rubik's products will be coming out this year from Oddzon.
They should be out by the middle of the summer.  They werent at liberty to
say what products they were.  I can think of some that I'd like to see back
into production.  I'll have to wait and see.

On another note.  I do not endorse ebay, however, I have had a bit of
success with ebay.  I won two different auctions on ebay.  They were both
for Cubes by Ideal, never opened or played with, cellophane still intake.
I was lucky enough to get them for under 20 bucks apiece.  But, use
caution.  I wouldn't recommend ever auctioning on an item from a seller
with negative or low feedback.  Definantly be careful.  I've been really
lucky with the products I've won.

Cheers,
Kevin

From cube-lovers-errors@mc.lcs.mit.edu  Thu Feb 25 17:28:33 1999
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Date: Thu, 25 Feb 1999 08:58:31 +0100
To: cube-lovers@ai.mit.edu
From: Christ van Willegen <c.v.willegen@spcgroup.nl>
Subject: RE: Fwd: Request for spectacular cube-solving  -  Can anyone  help ?

At 13:53 24-2-1999 -0500, you wrote:
>While we're at it, I might as well add that I think the juggler should
>be blindfolded!

Hey! _I'm_ already teaching my blind friend how to solve a cube!
We marked the colors of the cube with braille letters spelling
1 - 6 dots. We're having a terribly hard time to teach him to solve
it. It's fun, though...


Would a team of 3-4 blind people competing to solve the cube be
considered 'spectacular'? I think juggling is too hard. I'll ask
my gf, though (she knows how to juggle a bit, _and_ how to solve
the cube).

Christ van Willegen

From cube-lovers-errors@mc.lcs.mit.edu  Mon Mar  8 21:24:30 1999
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Date: Thu, 25 Feb 1999 09:02:15 +0100
To: cube-lovers@ai.mit.edu
From: Christ van Willegen <c.v.willegen@spcgroup.nl>
Subject: Megaminx solving times?

Hi,

I've been practising the Megaminx, and I can now solve it
without resorting to formulas written down on paper. I can
do it in about 10 minutes. How does this compare to other
people's times?

And, what method do you use? The method I developed relies
heavily upon the standard cube moves, and I solve the Mega-
minx going down from one flat top in rings. I needed to
adapt 1 (one!) standard cube formula to get it to work on
the Megaminx.

Christ van Willegen

From cube-lovers-errors@mc.lcs.mit.edu  Mon Mar  8 22:00:14 1999
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Date: Thu, 25 Feb 1999 10:56:09 -0700 (MST)
From: Paul Hart <hart@iserver.com>
To: cube-lovers@ai.mit.edu
Subject: Re: Oddzon version of the cube
In-Reply-To: <4.1.19990224195654.0092e1c0@mail.vt.edu>
Message-Id: <Pine.BSF.3.96.990225103436.94998B-100000@anchovy.orem.iserver.com>

On Wed, 24 Feb 1999, Kevin Young wrote:

> I noticed that there was a clear sticker overtop of each of the colored
> stickers.  In less than a month this clear sticker peeled up.  I just
> don't remember the original cubes by Ideal wearing down that fast.  In
> fact I still have a couple cubes from Ideal with stickers still intact.

[...]

> They have claimed that this concern about the stickers is not common.
> Maybe I just got a bad cube, I don't know.

No, it's not just you.  I had this exact problem with an Oddzon cube.
Eventually all Oddzon cubes will develop this problem after any extended
use, I believe.  The "stickers" appear to be actually nothing more than
some sort of colored paper glued to the cube surface, with a thin
transparent plastic film covering the paper.

Like Kevin mentions, this plastic film begins to peel up at the corners
and edges of the stickers after extended usage of the cube, perhaps due to
fingerprint oils.  Eventually the film will fall off (or it gets to the
point where it must be deliberately removed), and the underlying colored
paper is completely ruined not long after.  I noticed this very same
phenomenon with one of the "C4" cubes, those cubes with Rubik's profile in
the center of the one of the sides.

It seems that these newly manufactured cubes are not up the level of
quality of cubes from "back in the day".  All of the original Ideal cubes
that I have seen have stickers that appear to be made of straight colored
plastic.  Even the clone knock-offs from that era used these stickers.
These solid plastic stickers seem to hold up much better than the
paper-based ones used on the Oddzon products.  After my disappointing
results with my Oddzon cube, I pledged to never again buy one of their
products until they change or improve their sticker design.

Paul Hart

--
Paul Robert Hart        ><8>  ><8>  ><8>        Verio Web Hosting, Inc.
hart@iserver.com        ><8>  ><8>  ><8>        http://www.iserver.com/

From cube-lovers-errors@mc.lcs.mit.edu  Tue Mar  9 12:30:38 1999
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In-Reply-To: <4.1.19990224195654.0092e1c0@mail.vt.edu>
References: <36D3B818.E20EFCD6@ibm.net> <199902192002.OAA04117@solaria.sol.net>
Date: Thu, 25 Feb 1999 10:06:33 -0800
To: Kevin Young <keyoung3@vt.edu>
From: Patrick Weidhaas <weidhaas1@llnl.gov>
Subject: Re: Oddzon version of the cube
Cc: RUBIK cube group <cube-lovers@ai.mit.edu>

Kevin Young wrote:

>...  I bought a Rubik's cube last year in a
>toy store.  This was the newest version made by Oddzon.  I noticed that
>there was a clear sticker overtop of each of the colored stickers.  In less
>than a month this clear sticker peeled up....

Kevin,

I do not have an answer for you, but your email made me wonder why stickers
are being used at all? As far as I know, nobody has produced a cube (or
variation) where the plastic "cubies" are colored appropriately without
relying on stickers. Is that process so much more expensive, or do the
toy-makers want to give their customers a chance to cheat by switching the
stickers in case they can't get the puzzle solved?

Patrick

------------------------------------------------------------------
Patrick P. Weidhaas                     e-mail: weidhaas1@llnl.gov
Parallel I/O Project                    voice: 925-422-7704
Lawrence Livermore National Laboratory  fax:   925-422-6287
P.O. Box 808,  L-560
Livermore, CA 94551-0808


From cube-lovers-errors@mc.lcs.mit.edu  Tue Mar  9 13:06:35 1999
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Date: Fri, 26 Feb 1999 07:10:24 +1100
From: Wayne Johnson <sausage@zeta.org.au>
Reply-To: sausage@zeta.org.au
To: Kevin Young <keyoung3@vt.edu>
Cc: cube-lovers@ai.mit.edu
Subject: Re: Oddzon version of the cube
References: <199902192002.OAA04117@solaria.sol.net> <4.1.19990224195654.0092e1c0@mail.vt.edu>

Kevin Young wrote:

> I am curious if anyone has had the same problems with the current official
> Rubik's cube that is in production...

However, it must be said that they are not suddenly aware of this. This
has been a problem for some time. Frankly, I have a Rubik's original
from 5 years ago with the same propblem. Oddzon don't seem to built the
items or make them in colours that Rubik himself even approves of (take
the new colours of the magic for instance - terrible).

I am currently on the lookout for an asian fake cube because they have
always been (dare I say it) better quality in the sticker dept. My Fake
is 15 years old with all stickers nicely attached.

Come on Oddzon... a cube that wears out in a month?.... I think this is
very clever and deliberate marketing.

Regards,
Wayne

From cube-lovers-errors@mc.lcs.mit.edu  Tue Mar  9 13:31:19 1999
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Date: Thu, 25 Feb 1999 18:12:51 -0500
To: cube-lovers@ai.mit.edu
From: Kevin Young <keyoung3@vt.edu>
Subject: Re: Request for spectacular cube-solving  -  Can anyone help ?
In-Reply-To: <3.0.32.19990225085830.00963b40@mail.spc.nl>

I am currently trying to learn how to do it without looking.  Have a long
ways to go.  I use to be able to do it, watching the cube of course,  in
less than a minute.  After 17 years and not being a school age boy anymore,
and forgetting some of my tricks, I can still do it everytime in approx. 90
seconds.  Long ways away from being a world champion.  But, currently I'm
working on the no looking thing.  Right now, I can put two pieces at a time
in their appropriate position, behind my back, however, after those pieces
are set, I have to look.  Good luck to all that are trying to learn how to
do it blindfolded.

Regards,

Kevin Young

From cube-lovers-errors@mc.lcs.mit.edu  Tue Mar  9 14:01:38 1999
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From: Norman Diamond <Norman.Diamond@dec-j.co.jp>
To: cube-lovers@ai.mit.edu
Subject: RE: Fwd: Request for spectacular cube-solving  -  Can anyone  help ?
Date: Fri, 26 Feb 1999 09:24:41 +0900

Christ van Willegen [c.v.willegen@spcgroup.nl] wrote:

>We marked the colors of the cube with braille letters spelling
>1 - 6 dots.

I have a magic domino which was actually manufactured that way.
Bought it from Christoph Bandelow about 13 years ago.

Dr. Bandelow, you're on this list, right?  Were there ordinary Rubik's
cubes with the same feature?

(Question to self:  How can the words "ordinary" and "Rubik's"
be placed next to each other in a sentence?)

-- Norman.Diamond@dec-j.co.jp
[Not speaking for Compaq]

From cube-lovers-errors@mc.lcs.mit.edu  Wed Mar 10 14:54:38 1999
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Message-Id: <199902261047.CAA19201@f15.hotmail.com>
From: "Philip Knudsen" <philipknudsen@hotmail.com>
To: cube-lovers@ai.mit.edu
Subject: RE: Request for spectacular cube-solving
Date: Fri, 26 Feb 1999 02:47:40 PST

What about solving the cube while performing a full-length rap?
I even know someone who can do that: MYSELF ;-)
Actually I am a singer, but I _do_ rap from time to time _and_ I also 
happen to speak german. But I can't joggle :-(

____________________________________
Philip K
Vendersgade 15, 3th
DK - 1363 Copenhagen K
Denmark
Phone:  +45 33932787
Mobile: +45 21706731
E-mail: philipk@bassandtrouble.com
E-mail: philipknudsen@hotmail.com

[Moderator's note: Philip also mentions his good experiences with ebay.
 The cube-lovers list isn't running a poll on them, so no one else needs to
 send reports of their performance.  Pay your money, take your chances, seek
 advice from appropriate sources. ]

From cube-lovers-errors@mc.lcs.mit.edu  Wed Mar 10 15:49:18 1999
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Date: Fri, 26 Feb 1999 09:06:55 -0500 (Eastern Standard Time)
From: Dale Newfield <din5w@cs.virginia.edu>
Reply-To: DNewfield@cs.virginia.edu
To: cube-lovers@ai.mit.edu
Subject: RE: Fwd: Request for spectacular cube-solving  -  Can anyone  help ?
In-Reply-To: <3.0.32.19990225085830.00963b40@mail.spc.nl>
Message-Id: <Pine.WNT.4.03.9902260905060.-424545@biff.cs.virginia.edu>

On Thu, 25 Feb 1999, Christ van Willegen wrote:
> Hey! _I'm_ already teaching my blind friend how to solve a cube!
> We marked the colors of the cube with braille letters spelling
> 1 - 6 dots. We're having a terribly hard time to teach him to solve
> it. It's fun, though...

How difficult is it to read braille when the characters are in arbitrary
orientations?  Have you thought of using some other coding technique that
might prove more easy to distinguish in any orientation?  (Any idea what
that might be?)

-Dale

[Moderator's note: I'm inordinately proud of my own invention, which I
thought I mentioned years ago but can't find in the archives: Wire
symbols glued to a destickered cube, polished to a high gloss.  The
symbols are blank opposite dot, square opposite circle, and plus
opposite X.  The supergroup is marked by a cutout at a corner of each
face center and the adjacent cubies.  I can solve it behind my back,
but when I lent it to a blind computer scientist, he gave up. --Dan]

From cube-lovers-errors@mc.lcs.mit.edu  Wed Mar 10 19:40:41 1999
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To: cube-lovers@ai.mit.edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Re: Fwd: Request for spectacular cube-solving - Can anyone help ?
Date: 26 Feb 1999 20:52:13 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <7b71ht$3pp@gap.cco.caltech.edu>
References: <cube-lovers.3.0.32.19990225085830.00963b40@mail.spc.nl>

Christ van Willegen <c.v.willegen@spcgroup.nl> writes:
>Would a team of 3-4 blind people competing to solve the cube be
>considered 'spectacular'? I think juggling is too hard. I'll ask
>my gf, though (she knows how to juggle a bit, _and_ how to solve
>the cube).

How about team solving?  n people, n cubes, everyone makes one move
and passes the cube to the left.  Repeat.  :-)

--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
"Pop", "Soda", or "Coke"?  http://www.ugcs.caltech.edu/~almccon/pop_soda/

From cube-lovers-errors@mc.lcs.mit.edu  Wed Mar 10 20:15:40 1999
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Date: Sat, 27 Feb 1999 05:00:50 +0100 (MET)
From: Martin Moller Pedersen <tusk@daimi.au.dk>
Message-Id: <199902270400.FAA482087@bonestell.daimi.au.dk>
To: cube-lovers@ai.mit.edu
Subject: help on 5x5x5 wings

I am trying to solve my new cube the 5x5x5 cube.

I have managed to solve all of it except the wings.

The wings are the y's in the following diagram:


ZyZyZ
yZZZy
ZZZZZ
yZZZy
ZyZyZ

I have many moves for the 3x3x3 but I can't figure out how to apply these moves
to the wings.

Thanks for all help.

/Martin

From cube-lovers-errors@mc.lcs.mit.edu  Wed Mar 10 20:47:02 1999
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Date: Thu, 4 Mar 1999 23:09:00 -0500
From: michael reid <reid@math.brown.edu>
Message-Id: <199903050409.XAA24597@cauchy.math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: Re: Edges only, Ignoring Flips, Face Turn Metric

jerry writes

> I have completed a God's Algorithm run in the face turn metric for the
> group consisting of edges only ignoring flips.  The size of the group is
> therefore 12!  The results are as follows:
[ ... ]

very interesting.  i hope that you'll also do the quarter turn metric.

> I have come to believe that any corners only (with or without twist) or
> edges only (with or without flip) group, or the group which keeps both
> corners and edges but without twists and flips, will be a fairly poor
> pattern data base for IDA*.  The problem is that any such search space
> will have a diameter which is too small, and more importantly will have an
> average distance from Start which is too small.

another shortcoming of this coset space for ida* is that transformations
aren't easy to compute.  for the cosets spaces i've used, they always
split up as a product of smaller coset spaces.  then i use transformation
tables for everything.

ida* spend a lot of time moving from a position to its neighbors.
instead of keeping the cube position, i just keep track of which
coset i'm in.  then i need to find out what coset i'll be in if i
apply the turn  F  (for example).  i always do this by using
transformation tables.  to simplify things, suppose that my coset
space had 1000000 cosets.  i could use a table with 18 * 1000000
entries that tells me which coset i go to by applying a given turn.
if my coset space is a product of two spaces, each with 1000 cosets,
then i only need a tranformation table with 18 * 1000 entries for
the first coordinate and one of the same size for the second coordinate.

this is really addressing implementation issues of ida*, not so much
the effectiveness of it.

mike

From cube-lovers-errors@mc.lcs.mit.edu  Wed Mar 10 21:22:07 1999
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Message-Id: <36E4A993.5AB5@zeta.org.au>
Date: Tue, 09 Mar 1999 15:54:43 +1100
From: Wayne Johnson <sausage@zeta.org.au>
Reply-To: sausage@zeta.org.au
To: cube-lovers@ai.mit.edu
Subject: Speed cube times

Hello All,

There doesn't seem to be any records these days kept of people's current
solving times for the cube. Perhaps they should be shared here?

My current time for solving the 3x3x3 Rubik's cube is 47 seconds using
the Petrus method and involved:

Building the 2x2x2,
finishing the bottom and mid layer, 
1 edge alignment, 
1 corner swap, 
2 sunes, 
and 1 clockwise edge rotation.

How are other people's speeds fairing?

Wayne Johnson
http://www.zeta.org.au/~sausage/rubikscube.html

[Moderator's note: Since this may result in a large number of very
 small messages, please send your answers to Wayne, who I hope will
 summarize the results for the list. ]

From cube-lovers-errors@mc.lcs.mit.edu  Wed Mar 10 22:03:13 1999
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To: cube-lovers@ai.mit.edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Re: Megaminx solving times?
Date: 9 Mar 1999 15:46:55 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <7c3fpf$mte@gap.cco.caltech.edu>
References: <cube-lovers.3.0.32.19990225090214.009654a0@mail.spc.nl>

Christ van Willegen <c.v.willegen@spcgroup.nl> writes:
>I've been practising the Megaminx, and I can now solve it
>without resorting to formulas written down on paper. I can
>do it in about 10 minutes. How does this compare to other
>people's times?

I've never solved mine for speed, because I'm afraid of more
stickers falling off (already 3 are missing and I have to "deduce"
what they are).  10 minutes sounds reasonable -- I'm not sure I've
ever resorted to formulas written on paper.   (For one thing, I'm
not sure I know of any notation!)

>And, what method do you use? The method I developed relies
>heavily upon the standard cube moves, and I solve the Mega-
>minx going down from one flat top in rings. I needed to
>adapt 1 (one!) standard cube formula to get it to work on
>the Megaminx.

This brings up, actually, a rather embarrasing point for me as a puzzle
solver.  At first I had no idea how to generalize the standard cube moves
I used to the Megaminx.  So, eventually I figured out a new method,
which was:
  1.  Solve a large chunk of it by normal moves, perhaps leaving
    only three faces unsolved;
  2.  Solve the edges of the remaining faces (if you can solve an
    Alexander's Star, you can do this);
  3.  Solve the corners.

I found this quite effective.

About a year later, when my Megaminx was in storage and I was
playing with the Cube, I suddenly realized that my method for the
Megaminx would work perfectly well for the Cube!  (O, for that
matter, anything with a similar structure of "corners" with three
faces and "edges" with two faces.) I chastized myself
heavily for not realizing this "obvious generalization", and
with it was able to work out more moves for the Cube.

--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
"Pop", "Soda", or "Coke"?  http://www.ugcs.caltech.edu/~almccon/pop_soda/

From cube-lovers-errors@mc.lcs.mit.edu  Wed Mar 10 22:50:04 1999
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Date: Tue, 9 Mar 1999 19:08:55 +0000
From: David Singmaster <david.singmaster@sbu.ac.uk>
To: weidhaas1@llnl.gov
Cc: cube-lovers@ai.mit.edu
Message-Id: <009D4DE8.BDBAD646.12@ice.sbu.ac.uk>
Subject: Re: Oddzon version of the cube

	Re: coloured cubies.
	I only ever saw one version of the cube where the black plastic ccubies
had painted faces.  The colours were rather paler than on the ordinary cubes
and one colour was violet.  I can't find it immediately, but I should find it
if I stop an look.

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499;
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Wed Mar 10 23:23:35 1999
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Date: Tue, 9 Mar 1999 19:24:29 +0000
From: David Singmaster <david.singmaster@sbu.ac.uk>
To: sausage@zeta.org.au
Cc: cube-lovers@ai.mit.edu
Message-Id: <009D4DEA.EA4E5AAE.11@ice.sbu.ac.uk>
Subject: Re: Oddzon version of the cube

	It's depressing that manufacturers can't provide a decent cube.  When
the C4 cube was introduced about 10(?) years, I found the mechanism very poor
and poeple told me that their examples broke within an hour of buying it.  As
you say, the Asian pirates had become very good in 1982 or so and I believe
Ideal was actually buying production from some of the same companies.
	Regarding cubes with printed colours, I have located mine.  It has
violet replacing orange, but is otherwise the usual colours and arrangement.
The colours are pretty good, but because the plastic surface is not perfectly
smooth, it gives the effect of a matte finish, rather than a glossy finish,
which is why I remember the colours as a bit paler.  My records indicate this
was bought in a regular Rubik Cube packaging, but I don't recall when and I've
never seen the technique used again.  I suppose one has to place squares of
coloured material against the cube and then fuse the colour into the surface of
the plastic and this seems likely to be more expensive than the use of
stickers.  The elimination of orange may be due to the fact that many orange
colours are based on cadmium which is toxic and not permitted on objects for
children.

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499;
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Thu Mar 11 18:02:02 1999
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Message-Id: <714F77ADF9C1D111B8B60000F863155102DD6D8E@tbjexc2.tbj.dec.com>
From: Norman Diamond <Norman.Diamond@dec-j.co.jp>
To: cube-lovers@ai.mit.edu
Subject: Re: Oddzon version of the cube
Date: Wed, 10 Mar 1999 09:36:17 +0900

Patrick Weidhaas [weidhaas1@llnl.gov] wrote:

>As far as I know, nobody has produced a cube (or variation) where the
>plastic "cubies" are colored appropriately without relying on stickers.

I think I have mentioned on this list before that I bought one in India
in 1996.

>Is that process so much more expensive,

I think it is not.  I am not an expert on manufacturing so can't really
say if it's more expensive to make a multicolored plastic piece than
it is to make (or buy) adhesive tapes and punch stickers out of them
for attachment to unicolored pieces.  But I do think, when the version
with multicolored plastic pieces could be retailed for 35 rupees, the
cost of manufacture must be less than 35 rupees, the difference between
this method of manufacture and the more common method must be less than
35 rupees, and I'd lay odds on a distributor not even noticing the
difference in costs if they went that way.

(35 rupees was about 120 yen then.  35 rupees would be about 100
yen now, though the product's price in rupees might have risen.)

>or do the toy-makers want to give their customers a chance to cheat
>by switching the stickers in case they can't get the puzzle solved?

Interesting.
Is this the reason why the stickers come off by themselves :-)
Maybe they don't know that some people used to disassemble cubes and
rearrange the cubies :-)  Of course early Rubik's Revenges used to
disassemble themselves that way.

-- Norman.Diamond@dec-j.co.jp
[Not speaking for Compaq]

From cube-lovers-errors@mc.lcs.mit.edu  Thu Mar 11 18:27:53 1999
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Date: Tue, 9 Mar 1999 22:40:45 -0500 (EST)
From: Nicholas Bodley <nbodley@tiac.net>
To: Patrick Weidhaas <weidhaas1@llnl.gov>
Cc: Kevin Young <keyoung3@vt.edu>, RUBIK cube group <cube-lovers@ai.mit.edu>
Subject: Re: Oddzon version of the cube
In-Reply-To: <v04020a07b2fb4082cb5b@[134.9.13.42]>


On Thu, 25 Feb 1999, Patrick Weidhaas wrote:

{snips}

}Kevin,
}
}I do not have an answer for you, but your email made me wonder why stickers
}are being used at all? As far as I know, nobody has produced a cube (or
}variation) where the plastic "cubies" are colored appropriately without
}relying on stickers.

Ideal once made a deluxe Cube that had individual colored plastic tiles
attached to the cubies. It also had a mechanism (same general principle,
just different details) that would self-align as you began a maneuver
with a slight misalignment, instead of jamming. In other words, it was a
good bit more tolerant of misalignment. It would be delightful if they'd
reissue it!

}Is that process so much more expensive, or do the
}toy-makers want to give their customers a chance to cheat by switching the
}stickers in case they can't get the puzzle solved?

It was maybe almost twice the price of the standard Cube, iirc.
Swapping stickers is silly! Learn how to disassemble (but reassemble as
a solved Cube; iirc, there are 11 wrong ways to do it, essentially).

Most of the movable-piece puzzles can be disassembled, but not all.

}Patrick


|*  Nicholas Bodley   *|*  Autodidact & Polymath * Electronic Tech. (ret.)
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*  Frequent crashes are unacceptable in a mature
|*  Amateur musician  *|*  computer industry.
--------------------------------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Thu Mar 11 19:29:29 1999
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Message-Id: <3.0.32.19990310090119.00960d80@mail.spc.nl>
Date: Wed, 10 Mar 1999 09:01:20 +0100
To: cube-lovers@ai.mit.edu
From: Christ van Willegen <c.v.willegen@spcgroup.nl>
Subject: Stickers Re: Oddzon version of the cube

>I do not have an answer for you, but your email made me wonder why stickers
>are being used at all? As far as I know, nobody has produced a cube (or
>variation) where the plastic "cubies" are colored appropriately without
>relying on stickers. Is that process so much more expensive, or do the
>toy-makers want to give their customers a chance to cheat by switching the
>stickers in case they can't get the puzzle solved?

Switching stickers is a no-no. You can so this a couple of times, after
that they'll just fall off...

I've seen cubes that have real plastic colors! They are 3by's in black,
with colored square bricks on the faces that form the colors. They
are about $5 (I think). I haven't bought one to check quality (yet?).
Perhaps if the mechanism is alright, these might be better suited for
the cube-addict. But I'm afraid that the mechanism won't be able to stand
lots of use.

Perhaps I'll just go ahead and try the experiment. After all, it's only
$5...

Christ van Willegen

[ Moderator's note: I take it you mean they are available in the
  Netherlands?  Anywhere else? ]

From cube-lovers-errors@mc.lcs.mit.edu  Thu Mar 11 20:18:19 1999
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From: "Philip Knudsen" <philipknudsen@hotmail.com>
To: cube-lovers@ai.mit.edu
Subject: Re: Oddzon version of the cube
Date: Wed, 10 Mar 1999 00:51:10 PST

Patrick Weidhaas wrote:

>As far as I know, nobody has produced a cube
>(or variation)where the plastic "cubies" are
>colored appropriately without relying on stickers....

Cubes like that are probably more expensive to manufacture. However, a
"Deluxe" version, using plastic tiles, was produced by Ideal in the
early 80's. Pretty hard to get now, i'm afraid.
I also have one that seems newer, this also has plastic tiles. It came
on a card, that had the name "Old Brand Magic Cube" on it. The back of
the card has a (poor) solution printed, and there is also a picture of
an Octagon, supposedly by same manufacturer. Apart from the plastic
tiles, there is nothing deluxe about this last one - the turning is o.k.
but not VERY smooth.

____________________________________
Philip K
Vendersgade 15, 3th
DK - 1363 Copenhagen K
Denmark
Phone:  +45 33932787
Mobile: +45 21706731
E-mail: philipk@bassandtrouble.com
E-mail: philipknudsen@hotmail.com

From cube-lovers-errors@mc.lcs.mit.edu  Thu Mar 11 21:43:05 1999
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From: "Philip Knudsen" <philipknudsen@hotmail.com>
To: cube-lovers@ai.mit.edu
Subject: Re: Oddzon version of the cube
Date: Wed, 10 Mar 1999 01:02:17 PST

Paul Hart wrote:

>After my disappointing results with my Oddzon cube,
>I pledged to never again buy one of their products
>until they change or improve their sticker design.

You're right about their Cube, but the other Rubik products by OddzOn
are fine, or at least acceptable. For instance, I think the Eclipse is
MUCH better in design than the Magic Strategy Game by Matchbox, 10 years
earlier. Same game, but much more attractive and somehow also better
gameplay.
Another example is Rubik's Bricks, which is Rubik's version of the Soma
Cube. There is a short mention of it in "Cubic Compendium", but OddzOn
were the first to market it - and their version is excellent!

BTW has anyone seen/tried any of the other new stuff by OddzOn, like
"Rubik's Infinity" or Rubik's Double Tangram" ???

__________________________________
Philip K
Vendersgade 15, 3th
DK - 1363 Copenhagen K
Denmark
Phone:  +45 33932787
Mobile: +45 21706731
E-mail: philipk@bassandtrouble.com
E-mail: philipknudsen@hotmail.com

From cube-lovers-errors@mc.lcs.mit.edu  Thu Mar 11 22:12:57 1999
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Date: Wed, 10 Mar 1999 10:43:27 -0500 (EST)
From: der Mouse  <mouse@rodents.montreal.qc.ca>
Message-Id: <199903101543.KAA28632@Twig.Rodents.Montreal.QC.CA>
To: cube-lovers@ai.mit.edu
Subject: Re: Oddzon version of the cube

> [...] made me wonder why stickers are being used at all?  As far as I
> know, nobody has produced a cube (or variation) where the plastic
> "cubies" are colored appropriately without relying on stickers.

A while ago, I took the smoothest-acting Cube I have, peeled off all
the stickers, took the thing apart, and painted all the facicles.
Et voila! no more sticker problems!

Now, I just need to do that with one of the 5-Cubes I have, the one
that's suffering from the Dread Orange Sticker Disease; it's already
lost one orange sticker completely, and about four more are so loose
that only a piece of masking tape is keeping them with the Cube.
(Assuming I can figure out how to get it apart non-destructively.)

I agree, it would be much more pleasant if the plastic itself were
coloured.  But that would require at least six different plastics,
instead of one, which is probably why it's not done commercially.
Low volume already makes the things expensive....

On the other hand, I wonder how much more it really would cost to do
coloured plastics.  Anyone with enough experience in the industry to
say?

					der Mouse

			       mouse@rodents.montreal.qc.ca
		     7D C8 61 52 5D E7 2D 39  4E F1 31 3E E8 B3 27 4B

From cube-lovers-errors@mc.lcs.mit.edu  Thu Mar 11 23:48:09 1999
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Date: Wed, 10 Mar 1999 22:04:21 +0000
From: David Singmaster Computing <david.singmaster@sbu.ac.uk>
To: Norman.Diamond@dec-j.co.jp
Cc: CUBE-LOVERS@ai.mit.edu
Message-Id: <009D4ECA.6A22DEF3.10@ice.sbu.ac.uk>
Subject: RE: Fwd: Request for spectacular cube-solving  -  Can anyone  help ?

	Yes there were ordinary cubes made with markings fo the blind.  Rainier
Seitz, product manager fro Arxon, the German distributor, made the first
examples by heating a needle and making dice-like marking of one to six spots.
Several versions were made commercially or by specialist firms.  I have
examples with zero to five spots in this style, alos with brass studs of five
sizes and then Ideal (perhaps only Arxon) produced a version with moulded
plastic facelets having symbols on them: +, -, hollow circle, square, triangle
and solid circle.  Meffert made pyramids with four different textures of
surface  I asked a blind friend if they were easyily distinguished and he said
yes.

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499;
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Fri Mar 12 00:15:29 1999
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Date: Wed, 10 Mar 1999 22:19:21 -0500 (Eastern Standard Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re : Re: Edges only, Ignoring Flips, Face Turn Metric
In-Reply-To: <199903050409.XAA24597@cauchy.math.brown.edu>
To: michael reid <reid@math.brown.edu>
Cc: Cube Lovers <cube-lovers@ai.mit.edu>
Message-Id: <SIMEON.9903102221.A@GN209A.PSTCC.CC.TN.US>

On Thu, 04 Mar 1999 23:09:00 -0500 michael reid
<reid@math.brown.edu> wrote:

> jerry writes
>
> > I have completed a God's Algorithm run in the face turn metric for the
> > group consisting of edges only ignoring flips.  The size of the group is
> > therefore 12!  The results are as follows:
> [ ... ]
>
> very interesting.  i hope that you'll also do the quarter turn metric.
>

I have completed a run out to 11q (took a long time).
Regrettably, the diameter proved to be greater than 11q.  I
now have a run in progress out to 12q.  It's going *very*
slowly.  The problem I described where my method is
inefficient calculating the tail of the distribution is
even worse for the quarter turn metric than for the face
turn metric for this particular problem.  Also, to
calculate to 11q or 12q I have to store all the positions
out to 6q, which I can do.  I don't think I can store out
to 7q on the computer I have.  Even if I could, a run to
13q or 14q would be too slow, I think.

I know from parity considerations that the diameter is
greater than 12q, so in some ways my run to 12q is a fool's
errand.  That is, there are less than 12!/2 odd positions
through 11q, so there must be at least a few at 13q.  My
only hope is that all the even positions will show up by
the time I get to 12q.  If so, I would know that the rest
of the odd ones must be at 13q.  Otherwise, I am doomed for
now.

I have an idea of how to approach the inefficiency in the
tail.  Since I am calculating ends-with for each position I
calculate, I know for sure for each position I calculate
which quarter-turns go further from Start and which go
closer.  The idea is that once I get to the tail of the
distribution, I once again begin storing calculated
positions in memory (those which are at the maximal
distance which I am able to calculate).  From there, I
continue further from Start in a more traditional fashion,
leaping one level at a time rather than many levels at a
time.  This works because I have knowledge of which quarter
turns go closer to Start, and hence I don't have to worry
about comparing against those positions closer to
Start which I am not able to store.  If I had time to put
this plan into action, the run time for the tail of the
distribution should be only a few minutes or a few seconds.

----------------------------------------
Jerry Bryan
jbryan@pstcc.cc.tn.us

From cube-lovers-errors@mc.lcs.mit.edu  Fri Mar 12 14:16:07 1999
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Date: Wed, 10 Mar 1999 23:47:56 -0500
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From: Jacob_Davenport@scudder.com (Jacob Davenport)
Subject: Re: Request for spectacular cube-solving
To: cube-lovers@ai.mit.edu

Wei-Hwa Huang suggested:
>How about team solving?  n people, n cubes, everyone makes one move
>and passes the cube to the left.  Repeat.  :-)

Well, I'm left handed and Kristin Looney is right handed, and her solution 
is the one I use as well.  So we have solved it together, each contributing 
one hand.  Of course it helps a little that both of us can solve it one 
handed, but hey....

-Jacob Davenport

From cube-lovers-errors@mc.lcs.mit.edu  Fri Mar 12 14:57:16 1999
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Message-Id: <36E7D039.265049BA@marlboro.edu>
Date: Thu, 11 Mar 1999 09:16:25 -0500
From: Jim Mahoney <mahoney@marlboro.edu>
Organization: Marlboro College
To: Martin Moller Pedersen <tusk@daimi.au.dk>
Cc: cube-lovers@ai.mit.edu
Subject: Re: help on 5x5x5 wings
References: <199902270400.FAA482087@bonestell.daimi.au.dk>

Martin Moller Pedersen wrote:
>
> I am trying to solve my new cube the 5x5x5 cube.
> I have managed to solve all of it except the wings.
> The wings are the y's in the following diagram:
>
> ZyZyZ
> yZZZy
> ZZZZZ
> yZZZy
> ZyZyZ
>
> I have many moves for the 3x3x3 but I can't figure out how to apply these moves
> to the wings.

Hi Martin.

Here's an excerpt from a longer disucussion of the NxNxN cube
which I posted to cube-lovers some time ago.  Other folks have
done similar work, and published consistent results.

In what follows a "cubie" is one of the small, colored cubes
that make up the NxNxN, a "slice" is an NxN plane of the cube
(even if the inside cubies don't exist), and an "orbit" is
a set of cubies which can be moved into each other's places,
like the corners or edges.  The method below can be made
to work for any kind of orbit, including the "wings" you ask about.

Good luck,

 Jim Mahoney

-- excerpt from http://www.marlboro.edu/~mahoney/cube/NxN.txt --

=====================================================================
(VI) How to Cycle Three Cubies  =====================================
=====================================================================

The basic idea is to find a move sequence that will (1) take a chosen
cubie off from its "hot seat" on a chosen slice *without* (here's
the trick) disturbing any other cubie on that slice.  The rest
of the cube can be completely scrambled by this operation.  Then (2)
rotate the chosen slice, (3) undo step (1), putting the original
cubie back into its original slice and undo whatever changes were
made to the other cubies, and (4) undo step 2.

The sequence always of the form
   A R A' R'
where "A" is step 1, "R" is a rotation of a single slice, and
the ' mark means, as usual, the inverse operation.

Here's a detailed example, using the Corner orbit of a 3x3x3 cube,
with the top layer as the "chosen slice" and the cubie marked "1" in
the unfolded sketch of a cube below as the focus of attention. In
eight moves the cubies in locations 1, 2, and 3 will trade places.

The starting position:

             U
    a  -  1  - 2 - d -        (a,1,2,d,e,3,g,h) are a Corner orbit.
    |  L  |  F | R | B
    e  -  3  - g - h -        (U, D, L, R, F, B) are the possible
             D                clockwise rotations.


 (1) Get "b" off the chosen slice, without disturbing any other
     cubie on that slice.  Replace it with the cubie that you
     want to put in its place.

                 e  -  a  - 2 - d -
 ->   L  ->      |     |    |   |
                 3  -  1  - g - h -


                 e  -  a  - 2 - d -
 ->   D  ->      |     |    |   |
                 h  -  3  - 1 - g -


                 a  -  3  - 2 - d -
 ->   L'  ->     |     |    |   |           After L D L'
                 e  -  h  - 1 - g -

      The top layer was (a,b,c,d); now it is (a,f,c,d).
      "b" has been taken off the top slice, and "f" is in its place.

 (2) Rotate the chosen slice to place a new cubie in the hot seat.

                 3  -  2  - d - a -
 ->   U  ->      |     |    |   |           After (L D L') U
                 e  -  h  - 1 - g -


 (3) Undo step 1, which pops the chosen cubie "b" back to its
      original slice, *and* (here's the key part), restore (nearly) all
      other cubies to their original locations, since none of the
      disturbed ones were on the slice that rotated in step (2).

                    3  -  1  - d - a -
 -> L D' L' ->      |     |    |   |        After (L D L') U (L D' L')
                    e  -  2  - g - h -


 (4) Undo step 2, restoring the chosen slice back to its original
position.

                    a  -  3  - 1 - d -
   ->  U'  ->       |     |    |   |        After (L D L') U (L D' L') U'
                    e  -  2  - g - h -

So the move sequence to cycle corners (1,2,3) is simply
(L D L') U (L D' L') U'   (reading left to right).

With a few extra moves before this sequence (which should be undone
afterwards) to arrange the cubies which should be moved into the
places which are actually modified by this operation (or a similar
one), this trick and its variations can be used to put back all 8
corners into their proper places.

And with a bit of exploration, this same idea can be used to cycle
three cubies of any type, in any orbit, on any layer, without
disturbing anything else.  For the Edge-Singles on the 3x3x3, for
example, to bring an edge off the top slice without disturbing
anything else on top, step (1) can be S D S', where "S" vertical is a
rotation of a center slice.

From cube-lovers-errors@mc.lcs.mit.edu  Fri Mar 12 15:58:21 1999
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Message-Id: <36E7D3A7.1796@ameritech.net>
Date: Thu, 11 Mar 1999 08:31:03 -0600
From: Hana Bizek <hbizek@ameritech.net>
Reply-To: hbizek@ameritech.net
To: cube-lovers@ai.mit.edu
Subject: color

Hello, cube-lovers,

May I address an issue of cube colors, brought here by Dr. Singmaster?

Color problem is crucial to those of us who engage in creating
multi-cube designs, particularly if those designs are 3-dimensional. In
such designs it is possible to suppress colors and create designs of,
e.g., three colors only on its faces. No, I have not gone insane, I know
what I am talking about.

It is most unpleasant for a designer to have stickers falling off his or
her cubes! The cubes should be well-made and their colors distinct. I
find the orange and red colors to be nearly identical in hue. The red is
light and the orange is dark, which is bad. I try to solve the problemn
by suppressing orange in my 3-, 4- and 5-color designs, but it is
irritating. Why  can't the cube makers replace red-orange colors by
pink-dark red for better contrast?

Another issue is parity pairs. In solution algoirithms you  don't kave
to know them, but they are crucial in 3-dimensional-design algorithms.
They make the above-mentioned color suppression possible. Now why can't
the manufacturers sell such pairs? If they don't know what parity pairs
are, I will tell them.

Hana Bizek
physicist, and
3-d Rubik's cube designer

[Moderator's note: By parity pairs, I rather suspect he means mirror-image
pairs.]

From cube-lovers-errors@mc.lcs.mit.edu  Fri Mar 12 16:53:45 1999
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Date: Wed, 10 Mar 1999 23:53:19 -0500
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From: Jacob_Davenport@scudder.com (Jacob Davenport)
Subject: Re: help on 5x5x5 wings
To: cube-lovers@ai.mit.edu, Martin Moller Pedersen <tusk@daimi.au.dk>

I have a decent solution to the 5x5x5 on my web page at
www.wunderland.com/WTS/Jake.  I call those cubies "wings"
also, and I foolishly assumed that I invented the term.
Either that, or you have looked at my solution and are
still having trouble.  Please let me know if this page
is helpful or not.  If not, I would be happy to explain
further how I solve the wings.

Indeed, 3x3x3 moves will not help you with the wings.

-Jacob

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Date: Thu, 11 Mar 1999 08:49:39 +0100
To: DNewfield@cs.virginia.edu, cube-lovers@ai.mit.edu
From: Christ van Willegen <c.v.willegen@spcgroup.nl>
Subject: RE: Fwd: Request for spectacular cube-solving  -  Can anyone help ?

At 09:06 26-2-1999 -0500, Dale Newfield wrote:
>On Thu, 25 Feb 1999, Christ van Willegen wrote:
>> Hey! _I'm_ already teaching my blind friend how to solve a cube!
>> We marked the colors of the cube with braille letters spelling
>> 1 - 6 dots. We're having a terribly hard time to teach him to solve
>> it. It's fun, though...
>
>How difficult is it to read braille when the characters are in arbitrary
>orientations?  Have you thought of using some other coding technique that
>might prove more easy to distinguish in any orientation?  (Any idea what
>that might be?)

It's not the characters that count, just the numbers of dots. And it
seems to be reasonably easy to read the number of dots in any
position.

Besides, it was the easiest and quickest thing we could think of.

The problem with (most) blind people is lack of 3D concepts (I'd
almost put a Duh! here). I know I solve the first layer of the
cube with insight in the movement of cubelets. I've had lots of
trouble describing what I do to my friend! Diagrams are impossible
to draw, so you'd have to describe in words _exactly_ where cubelets
are to be placed w.r.t. the rest of the cube before a formula can
be applied.

Too bad we didn't have any time to practise, yet. But we will!

>
>-Dale
>
>[Moderator's note: I'm inordinately proud of my own invention, which I
>thought I mentioned years ago but can't find in the archives: Wire
>symbols glued to a destickered cube, polished to a high gloss.  The
>symbols are blank opposite dot, square opposite circle, and plus
>opposite X.  The supergroup is marked by a cutout at a corner of each
>face center and the adjacent cubies.  I can solve it behind my back,
>but when I lent it to a blind computer scientist, he gave up. --Dan]

That's also a nice idea. I might try that if the braille dots thing
doesn't work. Do you have pictures of this thing on-line?

[Moderator's reply: No pics, but the above description should get you
 pretty close.  The only symbols you might want help with are
    Dot: Made with a short tight spiral of wire,
    Plus and X: Double-outlined so that they can be made with a
         continuous strand of wire, end to end.
 After gluing them down, I put an extra cover of cement to protect the
 wire and blunt the sharp ends.  Silicone protectant ("Armor All" TM in
 the US) gives it a good feel.  I used steel wire about .2mm thick,
 like a paperclip, but perhaps I should stay vague so you will invent
 your own variation. ]

Dr. Bandelow! Please adjust your machines to make cubes like this :-)

[But building it is half the fun! --Dan]

Christ van Willegen

From cube-lovers-errors@mc.lcs.mit.edu  Fri Mar 12 18:03:02 1999
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In-Reply-To: <009D4DE8.BDBAD646.12@ice.sbu.ac.uk>
Date: Thu, 11 Mar 1999 10:14:14 -0800
To: David Singmaster <david.singmaster@sbu.ac.uk>
From: Patrick Weidhaas <weidhaas1@llnl.gov>
Subject: Re: Oddzon version of the cube
Cc: cube-lovers@ai.mit.edu

Thanks for your info re coloured cubies. I also got some info from Kevin
Young, see below.
Patrick

Date: Thu, 25 Feb 1999 15:58:16 -0500
To: Patrick Weidhaas <weidhaas1@llnl.gov>
From: Kevin Young <keyoung3@vt.edu>
Subject: Re: Oddzon version of the cube

Patrick,

As a long time cubist, I can assure you that at one time Ideal did make a
cube called "Rubik's Cube Deluxe".  They used tiles instead of stickers.
They were colored appropriately.  In fact, that is still the cube
that I use almost all of the time.  The tiles are colored with blue on
one side opposite of white, red opposite of orange, and yellow opposite of
green.  On the center red tile, was written in Gold Lettering "Rubik's
Cube Deluxe".  This placement of the Rubik's Cube logo differs from the
Ideal sticker version, where that version has the logo on a sticker on top
of the white center cube.  The gold lettering has faded over the past 17
years, but the cube is like new other than that.  If I remember correctly
this was a limited release by Ideal.  If you can find another one out there
that someone is willing to sell to you, I recommend getting it.

Sincerely,

Kevin Young

From cube-lovers-errors@mc.lcs.mit.edu  Fri Mar 12 18:39:11 1999
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From: WaVeReBeL@webtv.net
Date: Thu, 11 Mar 1999 10:34:35 -0800 (PST)
To: Cube-Lovers@ai.mit.edu
Subject: Re:  Speed cube times
Message-Id: <21178-36E80CBB-632@mailtod-121.bryant.webtv.net>

My records:

LAYER BY LAYER - 33 sec.

I can get it under a minute almost every time.  My average is about 55
sec.

I use a very eclectic approach in order to get as few moves as possible.
I'll start off w/ Jiri's or Lars Petrus' method for the top, depending
on which might get fewer moves.  For the bottom, I use a whole bunch of
different methods that I have learned on the web to finish it w/ as few
series as possible.  Usually it takes 3 series of moves to finish off
the last layer.  Often only 2 series, at the most 4.  I average about 65
turns.

My hands aren't super fast, so I try to make up for it by looking ahead
and limiting the number of moves.  If I take the time to practice the
hand movements and fingering, I think I can increase my time
dramatically.  My goal is to get it down to average under 30 seconds
(yeah right). =)


CORNERS FIRST - 47 sec.

My average is about 1 min.  I basically use Matthew Monroe's method w/ a
few tricks of my own.  I just wish there is more info on this method on
the web.  The fingering is much easier and faster to do than the layer
method, but it takes me longer because I have to use so many more turns.
I'm guessing there is lots of info I don't know about in the books that
have been written on this method, but I don't know how to get my hands
on 'em.  Could somebody point me to a web page or a book store that
could help?  I'm looking for more series of moves on this method.


ONE HANDED! - 1 min 12 sec.

I have been so addicted to cubing recently that I would spend hours at a
time.  So much so that my arms and hands would get tierd, and I would
sometimes get pains in my wrists & forearms.  (But as an addict I still
keep on going)  One day my right hand gave up, so I thought...what the
hell, why not try w/ my left by itself!  At first it took me about 5
min, then I got it down to 3 min.  After a little bit more practice, I
now average about 1 min, 45 sec!  Again, I'm not super fast, I just look
ahead and use as few moves as possible.  On the web I saw that the
record for one hand is 53 seconds.  I'm hoping to beat that someday.
Does anybody else on this list specialize one handed?  Got any good one
handed records?


MY FEET!!! - 7 min 19 sec.

I didn't think I could do it, but after my success w/ one hand, I had to
try it.  I also remember seeing somebody do it on that's incredible.

At first it took me about 20-30 minutes...I didn't bother timing myself,
because I didn't think I could do it.  The hardest part is doing long
sequences of moves in the last layer.  I've messed up a lot at that
point.  Sometime later I tried it again, and did it in 10 min, then
finally my record.  I would practice this a lot more, but my legs get
tierd lifting and manipulating the cube after just a couple of tries.
With some practice, I could probably get it down to 5 minutes.

-----

This brings me to a couple of ideas for a spectacular feat. (No pun
intended)  How about solving two cubes at once...one in eace
hand...solving w/ feet and hands at the same time?

-----

I have to say I'm pretty proud of these accomplishments.  I'm new to
this.  4 months ago I couldn't even solve a cube.  I just barely learned
how in November.  Now I'm addicted.  I wore out the stickers on two
brand new cubes in my first 3 weeks. (A problem which is currently being
addressed on this mailing list)  It has gotten so bad that It's
interfered w/ my school work.  (As I'm writing this, I should be
studying for my Trig exam)

-Alex Montilla-

From cube-lovers-errors@mc.lcs.mit.edu  Mon Mar 15 13:18:24 1999
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From: WaVeReBeL@webtv.net
Reply-To: WaVeReBeL@webtv.net
Date: Thu, 11 Mar 1999 13:47:33 -0800 (PST)
To: Cube-Lovers@ai.mit.edu
Subject: Local Cubists
Message-Id: <10765-36E839F5-696@mailtod-122.bryant.webtv.net>

I feel like I was born in the wrong era.  I'm 20 years old and in
college right now.  If I had been born 10-15 years earlier, I might be
training to be in competitions.  I was a little too young when it was in
its prime in the 80's.  I got my hands on a cube this past November &
now am hooked for life.  I went from 30 mins to 3 mins in the 1st month,
& in 2 more months got it down to under a minute.  I bring a cube
everywhere I go.  Tons of people come up to to talk to me about it.  99%
of the time they would say:  "You know what I used to do?" and then
either: 1) peel off the stickers or 2) take it apart.  I used to laugh
along at this amusing anecdote, for I myself admit to both.  But, after
the 100th person...Its just annoying.

There is no one I can share my hobby with.  Of all these people, in the
4 months I've been into this, I have only met one person who actually
knew how, but it was years ago, and has since lost interest.  I got one
of my friends into it, but he isn't at a level where he can compete w/
me yet.  Plus, he's too busy to put some serious time into it.  With the
exception of my one friend, I feel all alone when it comes to cubes.

Right now I am looking for any cube enthusiasts, beginner through
advanced, in my area who want to get together to compete, buy/sell/trade
books & cube related stuff, share secrets/techniques that can only be
taught in person (there are limits as to what text & 2D images on a web
page & in books can convey about manipulating a 3D cube), etc.

All of my knowledge on cubes is from people's web pages, but practically
all of them are made to teach the beginner.  My current goal is a 30
second average.  At a 55 sec average w/ a 33 sec record, I'm not too far
from my goal.  I achieved this using moves I've gotten from the web.  I
would like to talk to people personally, and see some of the books no
longer in print.  I read somewhere that there are over a hundred
written.  Please contact me if you have books I can look at, or know
where I can get a hold of them.

I would like also to join any local clubs if any still exist or, If
there are enough people interested, start a new one.  A lot of people
who have seen me playing with mine in public have asked me if the
Rubik's Cube has "come back" like a new craze.  Although there isn't, I
would sure like to see one.  And how about a new championship?  That
would be cool.  Not a world wide thing, but it wouldn't be hard to set
up a small local one with a small prize like a Megaminx.  Any thoughts
on this?  Has this been attempted lately?

I live in Carson, California.  So, if anyone in the LA/South Bay area is
interested, please contact me.

-Alex Montilla-
waverebel@webtv.net
ipiiika@aol.com

From cube-lovers-errors@mc.lcs.mit.edu  Mon Mar 15 14:13:32 1999
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Reply-To: <noel@mud.ca>
From: "Noel Dillabough" <noel@mud.ca>
To: "'Christ van Willegen'" <c.v.willegen@spcgroup.nl>
Cc: <cube-lovers@ai.mit.edu>
Subject: RE: Stickers Re: Oddzon version of the cube
Date: Thu, 11 Mar 1999 20:47:24 -0500
Message-Id: <000801be6c2a$49a18160$030a0a0a@noel>
In-Reply-To: <3.0.32.19990310090119.00960d80@mail.spc.nl>

I too found a "tiled" cube that was about $5 USD with plastic tiles rather
than stickers.  At first this seemed ideal since there can be no wear and
tear on the stickers.  However quite soon afterwards the tiles start to come
off.  Perhaps if they were glued better...

> with colored square bricks on the faces that form the colors. They
> are about $5 (I think). I haven't bought one to check quality (yet?).
> Perhaps if the mechanism is alright, these might be better suited for
> the cube-addict. But I'm afraid that the mechanism won't be
> able to stand
> lots of use.
>

I would be willing to pay more for puzzles of better quality.  On a similar
note, old versions of the Rubik's Revenge puzzle are fetching prices over
$75 USD.  The puzzles are basically unusable, as they are so stiff and
brittle from age that they fall apart.  The best buy is to buy a used
Rubik's Revenge, one that was actually used during the 80s and was worn in.
These puzzles are not so stiff and are actually usable.

Anyway back to my point, I was wondering how many people on this list would
be interested in getting a manufacturer to do a good quality run of 4x4x4
cubes?  A place in the U.S. called "Puzzlets" was supposed to have a sign up
list to create a production run of these cubes, but I have heard that they
are no longer in business.  Perhaps there is enough interest in the cube
lovers' list?

[Moderator's note: Has anyone else found aged, unused 4^3s more brittle than
 the originals? Even the early ones were usually stiff; I needed to take
 them apart and apply wax or other lubricant.  And still they broke much
 too easily, due to the tiny necks on the face centers.  Perhaps the only
 advantage aged, used cubes have is that the stiff ones whose owners didn't
 lubricate them are long broken. --Dan]

From cube-lovers-errors@mc.lcs.mit.edu  Mon Mar 15 15:58:49 1999
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Date: Fri, 12 Mar 1999 01:01:52 -0500
From: michael reid <reid@math.brown.edu>
Message-Id: <199903120601.BAA15248@euclid.math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: Re : Re: Edges only, Ignoring Flips, Face Turn Metric

i guess i'm not sure what you're doing, jerry.  but i don't think
it should be *that* difficult.  the number of configurations is
12! = about 480 million.  if you divide out by symmetry, you get
about 10 million configurations.  this should be small enough to
store in memory and do a complete breadth-first search of the space.

mike

From cube-lovers-errors@mc.lcs.mit.edu  Mon Mar 15 16:49:33 1999
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Message-Id: <36E8B44F.3A68D055@erco.com>
Date: Fri, 12 Mar 1999 07:29:35 +0100
From: "michael ehrt" <m.ehrt@erco.com>
Reply-To: m.ehrt@erco.org
To: cube-lovers@ai.mit.edu
Cc: der Mouse <mouse@rodents.montreal.qc.ca>
Subject: Re: Oddzon version of the cube
References: <199903101543.KAA28632@Twig.Rodents.Montreal.QC.CA>

> Now, I just need to do that with one of the 5-Cubes I have, the one
> that's suffering from the Dread Orange Sticker Disease; it's already
> lost one orange sticker completely, and about four more are so loose
> that only a piece of masking tape is keeping them with the Cube.
> (Assuming I can figure out how to get it apart non-destructively.)

That's not a big problem. Underneath each center piece there's a screw
(just like in some (or all?) 3x3s, although we don't need to open it to
take it apart). Just open one of them and the whole thing comes apart.
And if you've never taken it apart, enjoy the beautiful mechanism.
Putting it together again is a pretty nice puzzle itself :-)

Michael

From cube-lovers-errors@mc.lcs.mit.edu  Mon Mar 15 17:46:20 1999
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Message-Id: <006c01be6c55$1fdab740$ca685dcb@uwe>
From: "UMroaming" <uwe@ue.net>
To: "Cube-Lovers" <cube-lovers@ai.mit.edu>
Subject: MULTI COLORED PLASTIC SURFACES
Date: Fri, 12 Mar 1999 14:27:24 +0800

der Mouse <mouse@rodents.montreal.qc.ca> wrote:

>> [...] made me wonder why stickers are being used at all?  As far as I
>> know, nobody has produced a cube (or variation) where the plastic
>> "cubies" are colored appropriately without relying on stickers.
>
>A while ago, I took the smoothest-acting Cube I have, peeled off all
>the stickers, took the thing apart, and painted all the facicles.
>Et voila! no more sticker problems!


It is not to difficult to make cubes or other puzzles without stickers and to
spray paint the surfaces, the price is about the same.  I once made a
Pyraminx test run using this method.  As David Singmaster mentioned the
plastic surface under the sticker has some flow lines which are unavoidable
and the sticker serves in part to hide these imperfections.

>>The one which I bought in India did not have spray-painted
>>surfaces, it was made out of multicolored plastic.  So I think that
>>manufacturers can afford it.

The problem with molding the cubes out of colored plastic is the corner
pieces.  as they have to be molded in 3 different colored plastic, such
tooling and molding procedure is extremely expensive and I can not imagine
that India has the technology to produce such an item.  What would make more
sense is to make the stickers out of small clip on plastic tiles such as is
used in my Impossiball this would not increase production cost, but totally
new tooling would need to be made costing around US$50,000.00.  I made a
survey for my Pyraminx many years ago.  Using either colored plastic tiles
or the none slip  fluorescent stickers and the stickers won.

>Now, I just need to do that with one of the 5-Cubes I have, the one
>that's suffering from the Dread Orange Sticker Disease; it's already
>lost one orange sticker completely, and about four more are so loose
>that only a piece of masking tape is keeping them with the Cube.
>(Assuming I can figure out how to get it apart non-destructively.)
>
>I agree, it would be much more pleasant if the plastic itself were
>colored.  But that would require at least six different plastics,
>instead of one, which is probably why it's not done commercially.
>Low volume already makes the things expensive....

No problem to use different colored plastics only the corner pieces pose a
problem.

>On the other hand, I wonder how much more it really would cost to do
>colored plastics.  Anyone with enough experience in the industry to
>say?

As mentioned above there would be no cost increase apart from new tooling
having to be made for the corner pieces and the flow marks on some of the
surfaces will be noticeable.

The best solution would be spray painting, as is being done with my Orbix
and the 4 colored rings on my 3D-Puzzle Balls.

I hope that this clears up your discussion on why Manufacturers use
stickers.


Regards
Uwe

HAPPY PUZZLING
Uwe Meffert
P.O. Box 24455, Aberdeen, Hong Kong.
Tel. 852-2518-3080, Fax. 852-2518-3282
Email:- uwe@ue.net
www.ue.edu www.ue.net www.mefferts-puzzles.com

From cube-lovers-errors@mc.lcs.mit.edu  Mon Mar 15 18:31:44 1999
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Message-Id: <3.0.32.19990312090954.009582d0@mail.spc.nl>
Date: Fri, 12 Mar 1999 09:09:55 +0100
To: cube-lovers@ai.mit.edu
From: Christ van Willegen <c.v.willegen@spcgroup.nl>
Subject: Re: Plastic colors Re: Stickers Re: Oddzon version of the cube
Reply-To: Christ van Willegen <c.v.willegen@spcgroup.nl>

At 09:01 10-3-1999 +0100, I wrote:
>I've seen cubes that have real plastic colors! They are 3by's in black,
>with colored square bricks on the faces that form the colors. They
>are about $5 (I think). I haven't bought one to check quality (yet?).
>Perhaps if the mechanism is alright, these might be better suited for
>the cube-addict. But I'm afraid that the mechanism won't be able to stand
>lots of use.
>
>Perhaps I'll just go ahead and try the experiment. After all, it's only
>$5...

I was wrong... They are not $5, but $2.50 :-)  They're using some quite
agressive glue to put the plastic colors on the faces. The color sceme
is weird (White <-> Red, Green <-> Yellow, Blue <-> Orange, White, Green
and Yellow are clockwise). When I bought it, two cubelets were glued
together! Also, there are traces of the glue everywhere.

The machanism is not too well, but it doesn't fall apart easily. The
spring loaded mechanism is quite strong! When I took it apart, the springs
pulled the centre pieces together quite a bit.

I think that after some time, it may turn quite well. It will never be
a professional cube, however...

On a side note, my gf claims she saw these cubes at Intertoys or Toys-R-Us
for about $1.50...

>
>[ Moderator's note: I take it you mean they are available in the
>  Netherlands?  Anywhere else? ]

As far as I could tell, these are manufactured somewhere in Europe. I could
buy lots and send them to someone willing to re-ship them locally (as in:
In the States). I know the moderator usually doesn't approve of scemes like
this, but if people _really_ want one, it's the only way... Shipping is
_way_ over $1.50 apiece.

If people really want one, let me know, and also if you'd be willing to
re-ship from the USA (to someone else there).

Christ van Willegen

From cube-lovers-errors@mc.lcs.mit.edu  Mon Mar 15 19:01:32 1999
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Date: Fri, 12 Mar 1999 11:10:18 +0000
From: David Singmaster <david.singmaster@sbu.ac.uk>
To: whuang@ugcs.caltech.edu
Cc: cube-lovers@ai.mit.edu
Message-Id: <009D5001.60154AD6.25@ice.sbu.ac.uk>
Subject: Re: Fwd: Request for spectacular cube-solving - Can anyone help ?

	Very hard to have a person do just one move and pass it on.  Perhaps
allow five seconds?
	A bit of spectacular solving would be to have someone make five or six
moves and let the solver work out how to unscramble it in the same number of
moves.  Kate Fried in Budapest could do four moves, perhaps five, regularly.

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499;
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Mon Mar 15 19:37:47 1999
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Mail-from: From cube-lovers-request@life.ai.mit.edu Fri Mar 12 07:39:01 1999
Date: Fri, 12 Mar 1999 12:34:43 +0000
From: David Singmaster <david.singmaster@sbu.ac.uk>
To: c.v.willegen@spcgroup.nl
Cc: cube-lovers@ai.mit.edu
Message-Id: <009D500D.2B3620CA.27@ice.sbu.ac.uk>
Subject: RE: Stickers Re: Oddzon version of the cube

	Cubes with tiles instead of stickers came out very early on.  I recall
one called Gyro Cube, which I think was Korean, about 1981?.  Then Ideal took
on the idea as the Deluxe Cube.  I've got an example made in China (real PRC,
not Taiwan, and for the Chinese market).  Recently, I've bought an example at a
newsagent's in London for somewhere in the #1 to #2 range and it worked
moderately well (the previous two which I bought recently were virtually
immovable!).  A large version of this (about 90mm or 3 1/2 in on an edge) was
on sale from a street vendor in December, but I don't know if he has any left
and I haven't seen it elsewhere.

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499;
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Tue Mar 16 13:06:51 1999
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From: WaVeReBeL@webtv.net
Reply-To: WaVeReBeL@webtv.net
Date: Fri, 12 Mar 1999 10:31:40 -0800 (PST)
To: Cube-Lovers@ai.mit.edu
Subject: Re: Oddzon version of the cube
Message-Id: <1095-36E95D8C-2128@mailtod-121.bryant.webtv.net>
In-Reply-To: "Philip Knudsen" <philipknudsen@hotmail.com>'s message of Wed, 10 Mar 1999 00:51:10 PST

I too have seen many of  these cubes w/ plastic tiles.  I have many
different brands including the "Old Brand Magic Cube".  I find them at
swap meets.  Most sell for $2, but I recently found a vendor who sells
'em for ONLY $1!  I immediately bought 5 on the spot.  The next time I
go down there, I think I'll buy out the shop's stock of cubes.  Either
that, or ask how to get them myself.

The quality is pretty bad.  The turning is pretty sticky, but w/ some
lube and a little wear and tear, it's alright.  Plus, they break apart
pretty easily.  When they're really worn out, cubies start popping out
all the time.  If you use WD-40, it'll eat away at the plastic resulting
in really smooth turning.  It'll be great for about 3-4 weeks of daily
cubing, but the WD-40 will take its toll, and the cube will start
falling apart.  But for a few bucks every 3-4 weeks is worth it to me.

The good thing about these is that you can make your own "deluxe" cube.
First, with a little bit of effort, and a good razor blade, you can pry
off the plastic tiles.  Next, strip off all the stickers (If they
haven't already fallen off by themselves) and wipe off the sticky residue
on your smoothest, sturdiest cube.  Then, super glue the tiles in any
color arrangement you want!  (the tiles come in the standard colors)
The tiles may have a slightly rough surface after you pry them off, so
you might have to strip off any scarring/hardened glue w/ a razor to
make sure it's nice and flat when glued down.

This process isn't easy.  It took me hours.  But in the end, it's all
worth it.  I've only done one so far, but it works so well, I don't need
to do another one yet.  I've been cubing daily, hours at a time for
about two months with the same cube.

-Alex Montilla-

P.S.
I live in Carson, CA.  E-mail me if you want to know where I get em.

From cube-lovers-errors@mc.lcs.mit.edu  Tue Mar 16 13:34:31 1999
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In-Reply-To: <001B5B91.C22092@scudder.com>
Date: Fri, 12 Mar 1999 15:17:53 -0400
To: cube-lovers@ai.mit.edu
From: Kristin Looney <kristin@wunderland.com>
Subject: Re: Request for spectacular cube-solving
Cc: alison@wunderland.com

Jake wrote:

>Well, I'm left handed and Kristin Looney is right handed, and her solution
>is the one I use as well.  So we have solved it together, each contributing
>one hand.  Of course it helps a little that both of us can solve it one
>handed, but hey....

wow!  that explains it!   I had forgotten that you are left handed!
I started teaching Alison to solve the cube last night, and we did
a few trials at the solving-it-together-with-on-hand-each thing,
and had a really hard time at it.

-Kristin Looney
kristin@wunderland.com
http://wunderland.com/Home/Rubik.html

To all the fishies in the deap blue sea, Joy.

From cube-lovers-errors@mc.lcs.mit.edu  Tue Mar 16 14:07:10 1999
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Message-Id: <00e001be6cdf$a0cabaa0$75c4b0c2@home>
From: roger.broadie@iclweb.com (Roger Broadie)
To: <cube-lovers@ai.mit.edu>
Cc: "Martin Moller Pedersen" <tusk@daimi.au.dk>
Subject: Re: help on 5x5x5 wings
Date: Fri, 12 Mar 1999 23:25:23 -0000

Martin Moller Pedersen <tusk@daimi.au.dk> wrote

>I am trying to solve my new cube the 5x5x5 cube.

>I have managed to solve all of it except the wings.

>The wings are the y's in the following diagram:


>ZyZyZ
>yZZZy
>ZZZZZ
>yZZZy
>ZyZyZ

Here's a set of explicit processes:  a lower-case letter means a turn
of the layer containing the wing piece next to the outer layer denoted
by the corresponding capital letter, and in the same sense.  The
brackets show the movement of the pieces in the upper layer.

l F' L F l' F' L' F          (Bl, lF, Lf)

F2 r2 D R2 D' r2 D R2 D' F2  (Br, Fl, Rf)

r' U b U' F2 U b' U' F2 r    (Bl, lF, Fr)

R2 U2 l D' l' U2 l D l' R2   (Br, Fl, bR)

b L2 D l D' L2 D l' D' b'    (Lb, Fl, fL)

l2 U2 r' l U2 l' U2 l U2 r l U2 r' U2 l U2 r l2 U2  (Fl, rF)

The final sequence swaps a pair of pieces in the front face.  There's
been a lot of discussion of this move in Cube-lovers over the years.
The process I've quoted changes pieces in the central nine on the back
face, but nothing will show if all the cube except the top layer is
solved.  The other processes change no other pieces.

Roger Broadie

From cube-lovers-errors@mc.lcs.mit.edu  Tue Mar 16 14:43:02 1999
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Message-Id: <36EAE6E9.6B1F0608@okanagan.net>
Date: Sat, 13 Mar 1999 14:30:04 -0800
From: Karen Loewen <loewens@okanagan.net>
Reply-To: Karen Loewen <loewens@okanagan.net>
To: cube-lovers@ai.mit.edu
Subject: Speed Cubing

I have I question for any one willing to answer. I was wondering if the
people who can get the rubik's cube under 45 seconds did you actually
figure it all out by yourself. Or did you find out how through books,
email, websites etc? My best time is 90 seconds but I can't seem to beat
it the way I do it. I don't want any one telling me ways to do it faster
because I want to find out for myself. But I am wondering are there
certain ways to achieve faster times. Please just answer yes or no. Also
I have just ordered the 5x5x5 and I was wondering how much harder is it
than the 4x4x4. Thanks.

[Moderator's note: Send responses to Karen; I hope she will send
 cube-lovers a summary of the results of her survey. ]

From cube-lovers-errors@mc.lcs.mit.edu  Tue Mar 16 15:39:51 1999
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Message-Id: <36EB0B28.1169@ameritech.net>
Date: Sat, 13 Mar 1999 19:04:40 -0600
From: Hana Bizek <hbizek@ameritech.net>
Reply-To: hbizek@ameritech.net
To: cube-lovers@ai.mit.edu
Subject: parity pairs
References: <36E7D3A7.1796@ameritech.net>

> [Moderator's note: By parity pairs, I rather suspect he means mirror-image
> pairs.]

Let me tell you what I mean by parity pairs, why very few have probably
heard about this concept and why they are crucial in 3-dimensional (3-d)
cube art.

Suppose one has two cubes of identical color scheme such that the color
on both cubes' up, down, front and back faces are exactly the same. If
the color of the left face of one cube is identical to the color of the
right face of the other cube, such a pair of cubes is said to form a
parity pair. The color scheme is still identical, but the ORIENTATION of
the faces is reversed for one of the members of the pair. One cannot
obtain parity pairs by conventional cube manipulations, but must obtain
them either from the manufacturer, or switch the faces themselves
manually. I would prefer to buy such pairs from the toymaker, for it
pains me to tamper illegally with  those stickers. I have devised a
simple algorithm to do it as painlessly as possible, but it still is a
pain. But will a manufacturer sell me parity pairs?

The reason so few people know about parity pairs is that such pairs are
moot in solution algorithms. You do not need to concern yourself at all
with parity pairs, you just have one cube and painlessly solve it. Ditto
for 2-dimensional (2-d) designs (unless you treat them as lxmx1)
designs. However, they are essential in 3-d cube art. They are
responsible for reflection-equivalent designs, designs of fewer than six
colors and ultimately fractal design prototypes. They also determine
special symmetries in a 3-d design. They are the cornerstone of 3-d
design theory. Without their presence all of the 3-d designs I have
constructed would not be possible.

Why all this  self-serving fuss about parity pairs and 3-d designs? The
point is this: given four parity pairs, one can construct a 2x2x2 larger
clean design, that has three colors only on its six faces. The internal
faces that touch are colored the same. Those colors are hidden inside
the design or suppressed. Such an array of cubes, when used as corners,
produce, e.g., reflection-equivalence in a design. Go to your cube
collection, extract four parity pairs and see for yourselves.

So I think you got the idea, Mr. Moderator. Just one slight correction;
I am a "she," not a "he." You will find this almost incredible, but
women too, love the cube.

Hana Bizek
(female) physicist and
3-d Rubik's cube designer

[Moderator's note: On the contrary, there are several women on
 cube-lovers, and Dame Kathleen Ollerenshaw is well-known as one of
 the earliest writers about Rubik's cube and one of the first victims
 of Cubist's Thumb.  I just didn't know that "Hana" was a woman's
 name, and I had forgotten that this information was presumed by a
 mention of you in the archives.  I apologize for the oversight.

 As for nomenclature, the reason no one knows about "parity pairs" is
 that the term is ambiguous--"parity" could refer to representatives
 of any even division of a set into two parts.  If you wish to enable
 people to know what you mean without going through your somewhat
 confusing description, then you should use the term "enantiomorphic
 pairs", "chiral pairs", or "mirror-image pairs".  I believe these are
 the standard terms used by chemists, physicists, and everyone else,
 respectively.

 There is an interesting question, though, which your hobby may give
 you a particular ability to answer.  According to _Rubik's Cubic
 Compendium_, the most common color scheme has red opposite orange,
 blue opposite green, and white opposite yellow.  This permits two
 mirror-image color schemes, distinguished by whether red, white, and
 blue go clockwise or anticlockwise around a corner.  The question is
 whether there is a tendency for one of these schemes to predominate,
 and if so, which and by how much?  For instance, one enantiomorph
 predominates extremely strongly in the manufacture of dice, though I
 don't know why. --Dan ]

From cube-lovers-errors@mc.lcs.mit.edu  Tue Mar 16 16:11:07 1999
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Message-Id: <19990314063522.18949.rocketmail@send103.yahoomail.com>
Date: Sat, 13 Mar 1999 22:35:22 -0800 (PST)
From: Han Wen <hansker@yahoo.com>
Subject: Temporary Fix for OddzOn Sticker Peel
To: Cube Lovers <Cube-Lovers@ai.mit.edu>

Hi,

For those folks out there, like myself, frustrated with OddzOn sticker
peel problem, there is an effective method to prolonging the life of
these pitiful stickers. I got this technique from one of the posts on
the Rubik's website.

Currently, the plastic laminate that makes up the surface of the
stickers for OddzOn Rubik's Cubes starts to peel at the edges after
only a few weeks of intensive playing.

Well, get yourself some long-lasting acrylic nail polish and paint
over all the stickers. (Make sure you do this when the cube is brand
new.) I know, it's a pain in the ass, but it's worth it. The first
coat lasts for about 2-3 weeks before the edges start peeling again.
What I do then is take a razor blade and cut off the peeled away
sections and then nail polish over the stickers again.  You'll
probably get another 2-3 weeks of intensive play again before the cut
laminate starts peeling again.  However, after I repeated this process
once more (razor blade cut/nail polish over), the stickers have now
lasted over several months without additional peel. Hope this helps...

==
_________________________________________________________
Han Wen
Applied Materials
3050 Bowers Ave, MS 1145
Santa Clara, CA  95054
e-mail: Han_Wen@amat.com / hansker@yahoo.com

From cube-lovers-errors@mc.lcs.mit.edu  Tue Mar 16 16:52:10 1999
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Date: Mon, 15 Mar 1999 13:27:05 -0700 (MST)
From: Paul Hart <hart@iserver.com>
To: Noel Dillabough <noel@mud.ca>
Cc: cube-lovers@ai.mit.edu
Subject: RE: Stickers Re: Oddzon version of the cube
In-Reply-To: <000801be6c2a$49a18160$030a0a0a@noel>
Message-Id: <Pine.BSF.3.96.990315131452.7422L-100000@anchovy.orem.iserver.com>

On Thu, 11 Mar 1999, Noel Dillabough wrote:

> On a similar note, old versions of the Rubik's Revenge puzzle are
> fetching prices over $75 USD.  The puzzles are basically unusable, as
> they are so stiff and brittle from age that they fall apart.

Really?  I haven't noticed this in my own collection as far as I can see.

> Anyway back to my point, I was wondering how many people on this list would
> be interested in getting a manufacturer to do a good quality run of 4x4x4
> cubes?

I think this would be an excellent idea.

> A place in the U.S. called "Puzzlets" was supposed to have a sign up
> list to create a production run of these cubes, but I have heard that
> they are no longer in business.

Is Puzzletts out of business?  Their web site is still up at least.  Check
it out at:

    http://www.puzzletts.com/

[Moderator's note: If you get a response from puzzletts, please contact
 cube-lovers-request@ai.mit.edu.  I've had several people say they don't
 answer.]

> Has anyone else found aged, unused 4^3s more brittle than the originals?
> Even the early ones were usually stiff; I needed to take them apart and
> apply wax or other lubricant.  And still they broke much too easily, due
> to the tiny necks on the face centers.

A year or two ago I had the very good fortune of stumbling across a number
of 4x4x4 cubes that were brand new in their unopened original boxes from
1982.  Of the two cubes that I personally opened and used extensively, I
did not notice any unusual stiffness.  The 4x4x4 does suffer from the
known weak neck in the center pieces that is prone to snapping, but aside
from that the cubes were in excellent condition and have held up very
well.  I'm not sure if lubricating the cube will remedy the tendency for
the neck to break, but perhaps it would since it seems to happen when the
cube jams slightly.  The first of the two cubes to suffer a broken center
piece became my spare parts cube that I use to keep the other in good
running condition.

Paul Hart

--
Paul Robert Hart        ><8>  ><8>  ><8>        Verio Web Hosting, Inc.
hart@iserver.com        ><8>  ><8>  ><8>        http://www.iserver.com/

From cube-lovers-errors@mc.lcs.mit.edu  Tue Mar 16 17:24:14 1999
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Message-Id: <01BE6F31.56407CA0.Jean.LEBLANC@wanadoo.fr>
From: Jean Leblanc <Jean.LEBLANC@wanadoo.fr>
To: "Cube-Lovers (Adresse de messagerie)" <cube-lovers@ai.mit.edu>
Subject: Re:speed cube times
Date: Mon, 15 Mar 1999 22:14:43 +0100

Bonjour les fous du cube

Honestly, I think I am an unrecognized champion. I CAN solve the cube 
within 5 minutes or more, especially when I throw it through the window 
into the garden (my dog loves cubes, too).
If somebody can do worse, please tell me !
I didn't make up a method to solve 3*3*3, nor 4*4*4.
My cube (my only 3*3*3) is a poor clone of the 80's ; it creaks and get 
jammed but it still works !
I'm a poor lonesome cubist...
My wife and my children are not interested in cubes; shall I sacrifice my 
family to a plastic coloured God ?
"Il faut savoir raison garder !"
After all, I should be very interested in a new fabrication of 4*4*4, 
because mine is broken.

Jean Leblanc
Muret
 France.

From cube-lovers-errors@mc.lcs.mit.edu  Tue Mar 16 17:55:07 1999
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Message-Id: <714F77ADF9C1D111B8B60000F863155102DD6DAD@tbjexc2.tbj.dec.com>
From: Norman Diamond <Norman.Diamond@dec-j.co.jp>
To: cube-lovers@ai.mit.edu
Subject: Old 4^3s
Date: Tue, 16 Mar 1999 09:45:38 +0900

Noel Dillabough [noel@mud.ca] didn't write, but his message contained:

>[Moderator's note: Has anyone else found aged, unused 4^3s more brittle than
> the originals? Even the early ones were usually stiff; I needed to take
> them apart and apply wax or other lubricant.  And still they broke much
> too easily, due to the tiny necks on the face centers.  Perhaps the only
> advantage aged, used cubes have is that the stiff ones whose owners didn't
> lubricate them are long broken. --Dan]

Nob Yoshigahara told me that he had designed a correction for the original
design of the 4^3 cubes so that they would not fall apart.

In my experience, early 4^3s easily fell apart, and then when a cubie hit the
floor it easily broke.  In my experience, later 4^3s don't easily fall apart.
I would guess that the manufacturer accepted Nob-sensei's advice.

-- Norman.Diamond@dec-j.co.jp
[Speaking for Norman Diamond not for Compaq]

From cube-lovers-errors@mc.lcs.mit.edu  Tue Mar 16 18:24:27 1999
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From: "Jorge E. Jaramillo" <jejarami@usa.net>
To: "cube" <Cube-Lovers@ai.mit.edu>
Subject: Easy to find tile cubes
Date: Mon, 15 Mar 1999 21:38:27 -0500

I live in Colombia South America and here it is very easy to find cubes with
tiles instead of colored stickers. They are also very cheap (less than U$ 2).
They are (I guess) those Asian cubes that are not that durable. They have
this mechanism of one screw under the center tile. If you twist them many
times they unscrew and come apart but since they are so cheap when I break
one I just buy another one

Jorge E. Jaramillo
kingeorge@hotmail.com
Cut the chain and chase the dream
Savatage 1984

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Date: Tue, 16 Mar 1999 08:59:41 +0000 (GMT)
From: the terminal sloth <alex@slothdom.demon.co.uk>
To: Cube-Lovers <cube-lovers@ai.mit.edu>
Subject: Re: MULTI COLORED PLASTIC SURFACES
In-Reply-To: <006c01be6c55$1fdab740$ca685dcb@uwe>
Message-Id: <Pine.LNX.4.05.9903160855180.3329-100000@slothdom.demon.co.uk>

On Fri, 12 Mar 1999, Uwe Meffert wrote:
> ...
> It is not to difficult to make cubes or other puzzles without stickers and to
> spray paint the surfaces, the price is about the same.  I once made a
> Pyraminx test run using this method.  As David Singmaster mentioned the
> plastic surface under the sticker has some flow lines which are unavoidable
> and the sticker serves in part to hide these imperfections.

Why can't you take a modelling knife or a file and remove these lines? And
use a bit more paint where necessary (easier with acrylic than with spray
paints).
Obviously this isn't practical for large-scale manufacture.

<snip>
> I hope that this clears up your discussion on why Manufacturers use
> stickers.


Alex
-- 
Alexander Lewis Jones - the terminal sloth
sometimes I sits and thinks, sometimes I just sits

From cube-lovers-errors@mc.lcs.mit.edu  Tue Mar 16 19:32:55 1999
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From: "Klodshans" <klodshans@get2net.dk>
To: cube-lovers@ai.mit.edu
Date: Tue, 16 Mar 1999 10:36:42 +0000
Subject: Re: OddzOn version of the cube
Reply-To: klodshans@get2net.dk

Wayne Johnson <sausage@zeta.org.au> wrote:

> I've got the new magic.
> Works exactly the same except that the colours are pretty ugly.
> Matchbox did a nice one that was rainbow on black.
> The new ones are yellow on red.

I agree. I have all the different Magics and the OddzOn version is
not an improvement. The weird thing is that their website
www.rubiks.com
shows a picture of the Magic thats looks like one of the Matchbox
ones (BTW on the same site I read the other day that OddzOn is
planning to make their own deluxe cube).

Anyway, in the message that Wayne responded to, I was not
talking about Magic, but the "Magic Strategy Game" by Matchbox
which is a completely different thing. As I wrote, this has been re-
launched by OddzOn under the new name "Eclipse", and in this
case the OddzOn guys have actually improved the original design,
in my opinion.


______________________________________
Philip K
E-mail: philipk@bassandtrouble.com
E-mail: klodshans@get2net.dk
web:    http://hjem.get2net.dk/philip-k

From cube-lovers-errors@mc.lcs.mit.edu  Tue Mar 16 20:02:41 1999
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Date: Tue, 16 Mar 1999 09:41:52 +0000 (GMT)
From: "M. P. Baker" <mpb2@mcs.le.ac.uk>
Reply-To: "M. P. Baker" <mpb2@mcs.le.ac.uk>
Subject: Re: Stickers vs. tiles
To: cube-lovers@ai.mit.edu

David Singmaster <david.singmaster@sbu.ac.uk> wrote:

> Recently, I've bought an example [of a tiled cube] at a
> newsagent's in London for somewhere in the #1 to #2 range and it worked
> moderately well (the previous two which I bought recently were virtually
> immovable!).  A large version of this (about 90mm or 3 1/2 in on an edge) was
> on sale from a street vendor in December, but I don't know if he has any left
> and I haven't seen it elsewhere.

These cubes appear to be available from street vendors all over the UK. I've
recently bought some in Leicester and Plymouth. They also sell them in a
"gadget" shop at the end of my street, the sort of place that also sells blow-up
aliens etc. The large ones are really nice for displaying patterns on, and came
with a solution entertainingly translated from some Chinese type script :-)
--------------------------------------------
Matthew Baker
Dept. of Mathematics and Computer Science
University of Leicester

mpb2@mcs.le.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Tue Mar 16 20:35:46 1999
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Date: Tue, 16 Mar 1999 05:14:55 -0500 (EST)
From: Nicholas Bodley <nbodley@tiac.net>
To: Cube Mailing List <Cube-lovers@ai.mit.edu>
Subject: Taking apart the 5^3
Message-Id: <Pine.SUN.3.96.990316050216.6308M-100000@sunspot.tiac.net>


As nearly as I can remember, you can begin dismantling one of these by
rotating the top slice by maybe 30 degrees or so, then prying upward on
one of the "wing" cubies (between the center and the corner cubies). Use
your thumb, nail side down, and lift.

Experiment with different amounts of rotation; you'll find a position
where the "wing" cubie's "foot" will push aside many others. Once it's
disengaged, life becomes easier.

This method, if you choose the proper cubie to pry, and align it
properly with the "loosest" neighbor below it, is harmless.

Just possible that I'm suggesting the wrong cubie to pry, but iirc, the
center cubies are more directly held than the "wings".

Believe me, the insides of a 5^3 are utterly amazing. The scheme used
for the 3^3 can't hold a 5^3 together unaided; the mechanism is an
extension of that in the 5^3, but has an additional set of retaining
surfaces, generally spherical in their geometry.  The shape of a "foot"
on a corner cubie is something to behold; it could be a bit of a
challenge to define it in a CAD program.

As I've said before, don't even think of allowing your cat to watch
the process!

Sorting the pieces for reassembly is part of the fun.

|*  Nicholas Bodley   *|*  Autodidact & Polymath * Electronic Tech. (ret.)
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*
|*  Amateur musician  *|*
--------------------------------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Wed Mar 17 13:35:41 1999
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Message-Id: <199903170235.VAA11730@garnet.sover.net>
Date: Tue, 16 Mar 1999 21:37:49 -0500
To: Nicholas Bodley <nbodley@tiac.net>
From: Nichael Lynn Cramer <nichael@sover.net>
Subject: Re: Taking apart the 5^3
Cc: Cube Mailing List <Cube-lovers@ai.mit.edu>
In-Reply-To: <Pine.SUN.3.96.990316050216.6308M-100000@sunspot.tiac.net>

Nicholas Bodley wrote:
>As nearly as I can remember, you can begin dismantling one of these by
>rotating the top slice by maybe 30 degrees or so, then prying upward on
>one of the "wing" cubies (between the center and the corner cubies). Use
>your thumb, nail side down, and lift.

Oh course, if you want to wimp out, the center face cubie is held on by a
screw (at least his is true on my 5Xs).

Just take off the sticker (this is probably easier on the orange side...)
and the rest becomes pretty easy.
---
Nichael Cramer                 loose shoes,
nichael@sover.net                 a tight schedule,
http://www.sover.net/~nichael/       and a warm place to write Lisp...

From cube-lovers-errors@mc.lcs.mit.edu  Wed Mar 17 14:07:04 1999
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Message-Id: <015c01be7033$e646df80$8ccfaf8b@uwe>
From: "UMroaming" <uwe@ue.net>
To: "Christ van Willegen" <c.v.willegen@spcgroup.nl>
Cc: "Cube-Lovers" <cube-lovers@ai.mit.edu>, "Jing Meffert" <jing@ue.net>
Subject: Re: MULTI COLORED PLASTIC SURFACES
Date: Wed, 17 Mar 1999 13:05:23 +0800

Dear Christ

Thanks for your email, I will answer / comment in text

> From: Christ van Willegen <c.v.willegen@spcgroup.nl>
> Date: Wednesday, March 17, 1999 12:29 AM
> Subject: Re: MULTI COLORED PLASTIC SURFACES

> Uwe Meffert

> I saw one of your newer cubes in Eindhoven, The Netherlands
> last week. They are black, and have plastic tiles as the colors.
> By the way, 'newer' as in: I've never seen them before.
> Prices vary from $1.50 to $2.50 (about 3 to 5 DM, your name
> makes me assume you're German, but I may be wrong, of course).
> I saw them at 'Jolie/Promida' in Eindhoven, my gf saw them at
> Intertoys (she's not sure...) in Veldhoven.

YES I AM GERMAN LIVING IN HONG KONG

> I've modified one of those cubes (with glue and electricity wire)
> to bear symbols that can easily be discerned by hand (as you may
> be aware from previous discussions on this list, one of my friends
> is blind, and I wanted to learn him how to solve the cube).

> I assume (from glue residues found on the cube (even glueing together
> two cubelets!)) that the tiles are glued on. How 'hard'/'expensive'
> would it be to make these tiles not square, but pre-formed? It
> would make the cubes nicer, and even more useful for blind people
> (and seeing people who would like to learn to solve the cube
> behind their backs). The forms would be:

THE CUBES THAT YOU REFER TO ARE NOT MINE BUT A CHEAP COPY OUT OF CHINA.

Different textured labels including tile labels should be easily purchasable
from a good stationary store and you can easily up a few samples by hand.

About 12 years ago I made 50k pieces of my Pyraminx with 4 different texture
material for the Blind which I donated to several Blind Intitutes around the
world and I understand that some of the players where able to solve the
pyraminx by themselves without any help.

> - Filled square
> - Open square
> - Filled circle
> - Open circle
> - Plus sign
> - Star (6-points)

> Do you like the idea? Is it marketable? Is it produceable? It's

Produceable yes marketable NO.

> probably a bit harder to produce. I don't know if the tiles are
> glued on by hand, or if they are glued on by machines. It would
> make a big difference...

All the self adhesive labels are glued on by hand one side at a
time using a sort of special scotch tape with its adhesive
properties being lower then the adhesive on the labels.

> The cubes I have (2 of them) suggest hand production. A few tiles
> are glued on a bit skew, and the colors are not in the same order.
> I'd assume a machine would glue them on in perfect alignment and
> always in the same order.

Thats because the cubes are from a small copy company that does
not care about quality

> By the way: I can solve the ImpossiBall! I'm using about the same
> sequences of rotations as I do on the cube.

CONGRATULATIONS

With warm regards
Uwe


HAPPY PUZZLING
Uwe Meffert
P.O. Box 24455, Aberdeen, Hong Kong.
Tel. 852-2518-3080, Fax. 852-2518-3282
Email:- uwe@ue.net
www.ue.edu www.ue.net www.mefferts-puzzles.com

From cube-lovers-errors@mc.lcs.mit.edu  Wed Mar 17 14:36:56 1999
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From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
To: Cube Lovers <cube-lovers@ai.mit.edu>
Subject: Re : Re: Edges only, Ignoring Flips, Face Turn Metric
In-Reply-To: <199903120601.BAA15248@euclid.math.brown.edu>
Message-Id: <SIMEON.9903170012.A@pavilion.pstcc.cc.tn.us>
Date: Wed, 17 Mar 1999 00:34:12 -0500 (Eastern Standard Time)

On Fri, 12 Mar 1999 01:01:52 -0500 michael reid <reid@math.brown.edu>
wrote:

> i guess i'm not sure what you're doing, jerry.  but i don't think
> it should be *that* difficult.  the number of configurations is
> 12! = about 480 million.  if you divide out by symmetry, you get
> about 10 million configurations.  this should be small enough to
> store in memory and do a complete breadth-first search of the space.
>

The way you describe the search is how Herbert Kociemba did it, but
it is not how my program does it.  I think his program only took an
hour or two. I am applying my program to a problem to which it is not
well suited because I do not have time to write one more like
Herbert's.

I tend to think that the most fundamental design decision in a
program which does a Start rooted breadth first search for a cube
space is to decide whether the search space can fit in memory.  If it
can, and if there is an easy way to index the search space, then the
permutations themselves do not have to be stored.  All that has to be
stored is the distance from Start for each permutations.  These
distances are usually stored one per byte, or sometimes one per
half-byte.  There is even some discussion the Cube-lovers archives
about how the storage can be reduced to two bits per permutation.

If the search space cannot fit in memory, then it seems to me to be
the case that some representation of the permutations themselves must
be stored in addition to the distance from Start for each
permutation.  My program is designed to search as much as possible of
the 4.3*10^19 search space for the entire cube group, so it stores
permutations.  To make it into a program to search edges only without
flips, I simply fixed the corners and the flips, plus I made the
lexicographic ordering consider edges before corners.  But it still
stores the permutations.  It's sort of a quick and dirty solution
which runs very slowly for the problem at hand.

When a search space consists of the elements of a cube group, it is
easy to index the search space.  But when a cube group is reduced by
symmetry the result is generally not a group and the resultant search
space is (in my experience) not very easy to index.  The thing about
Herbert's program that I have trouble comprehending is that he is able
to reduce the search space by symmetry and still have the indexing be
well behaved.  He has posted a clear exposition of his method, so the
problem is in my understanding rather than in his explanation.

The reason reduction by symmetry results in  poorly behaved indexing
for the search space is because not all positions are equally
symmetric. There is much discussion of this phenomenon in the
archives under the general heading of "the real size of cube space".
Herbert seems to have overcome this problem for the edges problem.
But if I understand correctly, he does not believe the same solution
can be applied to the corners.

If Q[n] is the set of permutations which are n moves from Start, then
my program is calculating the product Q[6]Q[6] (all products of the
form st for s and t in Q[6]) as a way to determine Q[12].  For the
whole cube, most such products are in fact 12q from Start and most
such products are distinct.  There is very little wasted time or
energy.  But for edges only without flips, Q[12] is in the tail of
the distribution so most such products are either duplicate or are
less than 12q from Start.  Nearly all the products are a waste of
time.

My program does reduce by symmetry to certain extent.  If R[n] is the
set of representatives (patterns) which are n moves from Start, then
I only store R[n].  (R[n] is about 48 times smaller than Q[n].) Q[n]
is inferred via pointers to R[n], and is represented as Q[n]=R[n]^M,
where M is the set of 48 rotations and reflections of the cube.
Secondly, I only produce elements of R[2n] rather than elements of
Q[2n], which in theory speeds up the program by about 48 times but
which in practice only seems to speed it up by about 20 times.  But
for the edges without flips search, this kind of a speedup is utterly
dwarfed by all those wasted products from Q[6]Q[6].  My program
always runs into this problem when it gets into the tail of a
distribution.

----------------------------------------
Jerry Bryan
jbryan@pstcc.cc.tn.us
Pellissippi State Technical Community College

From cube-lovers-errors@mc.lcs.mit.edu  Wed Mar 17 15:30:12 1999
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From: "Klodshans" <klodshans@get2net.dk>
To: cube-lovers@ai.mit.edu
Date: Wed, 17 Mar 1999 17:09:42 +0000
Subject: Re: OddzOn version of the cube
Reply-To: klodshans@get2net.dk
In-Reply-To: <4.1.19990316202534.0092a160@mail.vt.edu>

Kevin Young <keyoung3@vt.edu> wrote:

> Where at on www.rubiks.com does it say that Oddzon
> is planning on making a deluxe version?

At www.rubiks.com, go into the "news" section. The 
announcement for an OddzOn deluxe Cube was added on March 
11th.


______________________________________
Philip K
E-mail: philipk@bassandtrouble.com
E-mail: klodshans@get2net.dk
web:    http://hjem.get2net.dk/philip-k

[Moderator's note: http://www.rubiks.com/deluxe.html says in one place
 that they are planning to have "super high quality durable stickers"
 and in another that "there won't be stickers" so it's anyone's guess
 what they have in mind.  They're also considering "holographic center
 labels" vs. "the classic look" and say they will be conducting a
 poll.  Rumors that the Tartan design is being promoted by one of its
 inventors are unverified, and those were gifts, not bribes.  --Dan]

From cube-lovers-errors@mc.lcs.mit.edu  Wed Mar 17 16:04:42 1999
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Message-Id: <36EFE0A5.3B50AF48@switchview.com>
Date: Wed, 17 Mar 1999 12:04:37 -0500
From: Michael Swart <Michael.Swart@switchview.com>
Organization: Switchview
To: Roger Broadie <roger.broadie@iclweb.com>, cube-lovers@ai.mit.edu
Cc: Martin Moller Pedersen <tusk@daimi.au.dk>
Subject: Re: help on 5x5x5 wings
References: <00e001be6cdf$a0cabaa0$75c4b0c2@home>

Regarding 5x5x5 wings
Roger Broadie said:
> l F' L F l' F' L' F          (Bl, lF, Lf)
> F2 r2 D R2 D' r2 D R2 D' F2  (Br, Fl, Rf)
> r' U b U' F2 U b' U' F2 r    (Bl, lF, Fr)
> R2 U2 l D' l' U2 l D l' R2   (Br, Fl, bR)
> b L2 D l D' L2 D l' D' b'    (Lb, Fl, fL)
> l2 U2 r' l U2 l' U2 l U2 r l U2 r' U2 l U2 r l2 U2  (Fl, rF)

Wow, that's a big help.
I discovered the following long-winded maneuvers a while ago. Changing
'wings' on D.

RT = (r' D' r D' r' D2 r) (right thingy)
LT = (l D l' D l D2 l') (left thingy)

1. RT D2 RT D2 LT D2 LT D2     (Fl, Bl)(fL, Fr)  
2. RT D LT D' RT LT D' RT D LT (fL, bL)(Fl, Br)

(Side note, RT and LT are used by me to get the last squares of the
fourth layer)

Michael Swart
Michael.Swart@switchview.com

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Message-Id: <36EEEF3F.2645@zeta.org.au>
Date: Wed, 17 Mar 1999 10:54:39 +1100
From: Wayne Johnson <sausage@zeta.org.au>
Reply-To: sausage@zeta.org.au
To: cube-lovers@ai.mit.edu
Subject: Speed cubing results - March 99

Hello all,

This is the list so far. There are others to go on, but I need CURRENT
best times, and CURRENT averages:

Name		Method		Best Time  Average Time

Lindon Collins	Layer by Layer	38 sec	   47.5 sec
Jiri Fridrich	Fridrich	14 sec	   20 sec  
Ryan Heise	Fridrich	34 sec	   43 sec  
Ryan Heise	Petrus		40 sec	   56 sec  
Wayne Johnson	Petrus		47 sec	   65 sec  
Wayne Johnson	Layer by Layer	58 sec	   75 sec  
Karen Loewen	Karen Loewen	90 sec	   ***	  
Clive McCaig	Layer by Layer	38 sec	   60-75 se
Alex Montilla	Layer by Layer	33 sec	   55 sec  
Alex Montilla	Corners first	47 sec	   60 sec  
Alex Montilla	One hand only	73 sec	   01:45	  
Alex Montilla	Feet only	07:19	   N/A	  
Han Wen		Fridrich	25 sec	   40 sec  

Wayne

From cube-lovers-errors@mc.lcs.mit.edu  Wed Mar 17 17:02:28 1999
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From: Douglas Zander <dzander@solaria.sol.net>
Message-Id: <199903120225.UAA13599@solaria.sol.net>
Subject: cube with colors attached
To: cube-lovers@ai.mit.edu (cube)
Date: Thu, 11 Mar 99 20:25:33 CST

I received as a present many years ago the Game of Rubik's Cube.  (I forget
the exact name)  but just in case anyone wonders about it, it is a hard
mechanism to turn.  The cubies each had plastic colored blocks fitted into
the cube surface with holes for pegs.  What you were supposed to do was
place a peg into a hole then play some game with an opponent.  Does anyone
else have this cube game?  How rare is this?

Also, I wish to start a new topic about the cube.  Has anyone ever thought
of making a large cube out of wood so that there is a lot of wood (or clay)
on the outside of the cube and then carve something like a human head out
of the material? (a bust)  Then the object is to scramble and reconstruct
the head.  (Has this been talked about before?)  I think a white bust of
an ancient Greek (like white marble statues) would be cool.

--
 Douglas Zander                |
 dzander@solaria.sol.net       |
 Shorewood, Wisconsin, USA     |

From cube-lovers-errors@mc.lcs.mit.edu  Wed Mar 17 17:40:24 1999
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Date: Tue, 9 Mar 1999 22:46:57 -0500 (EST)
From: Nicholas Bodley <nbodley@tiac.net>
Reply-To: Nicholas Bodley <nbodley@tiac.net>
To: Norman Diamond <Norman.Diamond@dec-j.co.jp>
Cc: cube-lovers@ai.mit.edu
Subject: Magic Domino: WTB; also, What's its mechanism like?
In-Reply-To: <714F77ADF9C1D111B8B60000F863155102DD6D3A@tbjexc2.tbj.dec.com>


 Highly unlikely that anyone has one that they'd like to part with, but
I'd like to buy a Magic Domino. [...]

 I'd love to know what the mechanism of a Magic Domino is like, inside.

|*  Nicholas Bodley   *|*  Autodidact & Polymath * Electronic Tech. (ret.)
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*  Frequent crashes are unacceptable in a mature
|*  Amateur musician  *|*  computer industry.

[Moderator's note: Anyone who has one for sale, please contact nbodley by
 e-mail. --Dan]

From cube-lovers-errors@mc.lcs.mit.edu  Wed Mar 17 18:39:41 1999
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From: TSWatts@aol.com
Message-Id: <2d4d92d3.36eef932@aol.com>
Date: Tue, 16 Mar 1999 19:37:06 EST
To: cube-lovers@ai.mit.edu
Subject: Puzzlets

On 3/16/99 hart@iserver.com (Paul Hart) asked if Puzzlets is still in
business.  I can verify that they are in fact still in business in the
Supermall in Auburn, Washington (near Seattle).  They recently moved from a
location in downtown Seattle which may be the source of the confusion.

I can't say anything one way or the other about their tendency to not answer
Emails, but I have spoken to the owner recently about various cube-like
puzzles and he was very knowledgeable and willing to talk to me at length
(unfortunatley I don't remember his name).  I'm surprised he isn't a part of
this Email group!

I can also tell you that he DOES still maintain a list of people who would be
interested in paying to get somebody to do another run of 4x4x4 cubes (aka
"Rubik's Revenge").  Apparently no manufacturer will do it unless they can do
a run of 30,000 of them!  Also, I know he has a list of people who would be
willing to pay a premium for a used version of the 4x4x4, if anyone's
interested.  I traded my 4x4x4 with him for some other merchandise since I
find the 5x5x5, which I only just learned existed about two months ago, to be
basically the same degree of difficulty as the 4x4x4, just bigger.  They do
have some 5x5x5's in stock.

-Tom Watts
Puyallup, WA, USA

From cube-lovers-errors@mc.lcs.mit.edu  Wed Mar 17 19:17:00 1999
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From: WaVeReBeL@webtv.net
Date: Wed, 17 Mar 1999 15:41:21 -0800 (PST)
To: Cube-Lovers@ai.mit.edu
Subject: Re: Speed cubing results - March 99
Message-Id: <18576-36F03DA1-15@mailtod-122.bryant.webtv.net>
In-Reply-To: Wayne Johnson <sausage@zeta.org.au>'s message of Wed, 17 Mar
	1999 10:54:39 +1100

As far as timing is concerened, do you include a preview before the
timer is started, or is it done "cold" (No looking at all before the
timer starts)?  If a preview is allowed, how long do you get?  Is there
a standard for this that everyone is going by?  For longer times this
shouldn't matter too much, but for record keeping and fast times such as
Jiri's 14 seconds, I can see how a few seconds in the begining can
really make a difference.  Is there a FAQ about this?

Also, are you people timing yourselves or do you have someone to do it?
I do it myself.  It doesn't affect my time too much, but I can shave off
a second or two if someone else does it.

-Alex-

From cube-lovers-errors@mc.lcs.mit.edu  Thu Mar 18 12:34:38 1999
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From: Derrick Schneider <derrics@ibm.net>
To: "cube-lovers@ai.mit.edu" <cube-lovers@ai.mit.edu>
Subject: RE: Puzzlets
Date: Wed, 17 Mar 1999 16:27:10 -0800

I also can't comment on their non-responding trend (though a recent
move might be reason enough), but the owner's name is Mike Green, I
believe, and he's the one you e-mail when going to www.puzzletts.com.

Derrick

From cube-lovers-errors@mc.lcs.mit.edu  Thu Mar 18 13:22:39 1999
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Date: Thu, 18 Mar 1999 00:53:22 -0500 (EST)
From: der Mouse  <mouse@rodents.montreal.qc.ca>
Message-Id: <199903180553.AAA24726@Twig.Rodents.Montreal.QC.CA>
To: Cube-lovers@ai.mit.edu
Subject: Re: Taking apart the 5^3

> As nearly as I can remember, you can begin dismantling one of these
> by rotating the top slice by maybe 30 degrees or so, then prying
> upward on one of the "wing" cubies (between the center and the corner
> cubies).  Use your thumb, nail side down, and lift.

Well, I fiddled with it and finally managed to get one of my 5-Cubes
apart (I used the one with the loose stickers).

I found it more effective to turn a "thick slice" (ie, the outer two
slices turned together) about 45 degrees, then pry with my thumb
between the corner and wing of the turned slice.  (This is perhaps
ambiguous.  Start with a solved 5-Cube, turn the U face 45 degrees
clockwise, so the URF cubie and the RF wing cubie next to it are just
above the middle of the F face.  Then stick your thumb between those
two cubies, nail towards the URF corner cubie, and lever the wing cubie
down.)  It's harder to get the last wing cubie back in than it is to
take the first wing cubie out, but by reversing the move I described
above I find it not too difficult.

Now I just need to find paints that will stick well to the plastic
these things are made of.  (The paints I used for the 3-Cube I painted
don't stick as well as I'd like.)

> Believe me, the insides of a 5^3 are utterly amazing.  [...]  The
> shape of a "foot" on a corner cubie is something to behold;

True.  Quite impressive to look at.  Indeed, once you've taken out the
off-center face cubies and the wing cubies, you're left with something
that looks like a ricketey skeletal 3-Cube (and indeed can, if you're
careful, be manipulated as such).

Amusingly, I realized that as long as you get that "ricketey 3-Cube"
put together in its solvable orbit, it's impossible to put the rest of
the 5-Cube together unsolvably!  (Unless you've marked the face cubies
so they're distinguishable, of course.)

					der Mouse

			       mouse@rodents.montreal.qc.ca
		     7D C8 61 52 5D E7 2D 39  4E F1 31 3E E8 B3 27 4B

From cube-lovers-errors@mc.lcs.mit.edu  Thu Mar 18 13:40:33 1999
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Message-Id: <36F09AFF.752D8854@ibm.net>
Date: Wed, 17 Mar 1999 22:19:44 -0800
From: "Jin 'Time Traveler' Kim" <chrono@ibm.net>
Reply-To: chrono@ibm.net
To: skouknudsen@get2net.dk
Cc: cube-lovers@ai.mit.edu
Subject: Some more Rubik's Revenge notes
References: <199902251022.LAA13413@bednorz.get2net.dk>

Check that, the Korea cube is definitely of poorer quality than either
Macau or Hong Kong yet at the same time has its own merits.  The reason
that the cube felt more like it was going to break was because the
middle "wing" cubelets and corner cubelets are "hollow."  The material
is also softer, but somehow it feels like the corner pieces are sturdier
than the earlier ones made in Macau and Hong Kong.  I am assuming it was
made later because seldom are puzzles made of higher quality in later
runs, especially when the tooling is probably more expensive and when
the print run is so short.  But all 3 versions wear the same "(c) I.T.C.
1982" (Ideal Toy Company?) with the different "Made in..." markings.

Anyway, the puzzle was easily disassembled without the need to remove a
screw (There IS no screw unlike the others).  Just like popping open a
standard cube.  Looser tolerances went into making this piece so this
was not a difficult task.  Like I said, the corner cubelets actually
feel sturdier than the others but the overall the plastic used is much
softer and tolerances aren't tight.  Of course, this makes for an easier
turning puzzle but tends to "stick" a lot because of the hollow pieces.
If these "hollows" were somehow smoothed in, this would be a pleasure to
work on instead of the stiffer (and more break prone) Macau and Hong
Kong cubes.  As it is, the center windows tend to bind easily against
each other and I think ultimately it will break just as easily (if not
more easily) than the Macau or Hong Kong deals.  Oh yeah, the Korea
Revenge also feels slightly lighter than the Macau or Hong Kong cubes
for obvious reasons.

Hmmm... I think I'll cc the cube list with this.  Might be interesting
to some.  I wonder how I could squeeze some of this into the Cube FAQ...

--
Jin "Time Traveler" Kim
chrono@ibm.net
http://www.slamsite.com/chrono
'95 PGT - SCPOC

From cube-lovers-errors@mc.lcs.mit.edu  Thu Mar 18 14:09:55 1999
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From: "Johan Myrberger (EBC)" <Johan.Myrberger@ebc.ericsson.se>
Reply-To: johan.myrberger@bigfoot.com
To: "'cube-lovers@ai.mit.edu'" <cube-lovers@ai.mit.edu>
Subject: RE: cube with colors attached
Date: Thu, 18 Mar 1999 09:24:44 +0100

> ... Has anyone ever thought of making a large cube out of wood so
> that there is a lot of wood (or clay) on the outside of the cube and
> then carve something ...

Something like this has been manufactured. I believe it was by Disney corp.
They made cubes which loked like the head of Donald Duck and Mickey Mouse if
my memory serves me.

regards
/Johan Myrberger
mailto:Johan.Myrberger@bigfoot.com

From cube-lovers-errors@mc.lcs.mit.edu  Thu Mar 18 14:41:45 1999
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From: C.McCaig@queens-belfast.ac.uk
Date: Thu, 18 Mar 1999 09:56:55 GMT
To: Cube-Lovers@ai.mit.edu
Message-Id: <009D54AE.1E20C937.3@a1.qub.ac.uk>
Subject: Re: Speed cubing results - March 99


> As far as timing is concerened, do you include a preview before the
> timer is started, or is it done "cold" (No looking at all before the
> timer starts)?  If a preview is allowed, how long do you get?  Is there
> a standard for this that everyone is going by?

i usually do it cold.  i use the layer-by-layer method, so a preview
isnt really all that advantageous.  i've tried jiri's method, but i
couldnt get used to it, and only once managed to break 60 secs with it.

> Also, are you people timing yourselves or do you have someone to do it?
> I do it myself.  It doesn't affect my time too much, but I can shave off
> a second or two if someone else does it.

i just use my watch, with the alarm set to go off, and the display reading
the seconds (so the alarm goes off at :00) and then look when i've
completed the cube.

> -Alex-

Clive

--
Clive McCaig
Dept. Applied Mathematics
Queens University Belfast
Northern Ireland

From cube-lovers-errors@mc.lcs.mit.edu  Thu Mar 18 19:04:31 1999
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Date: Thu, 18 Mar 1999 07:35:57 -0500 (EST)
From: Jiri Fridrich <fridrich@binghamton.edu>
To: WaVeReBeL@webtv.net
Cc: Cube-Lovers@ai.mit.edu
Subject: Re: Speed cubing results - March 99
In-Reply-To: <18576-36F03DA1-15@mailtod-122.bryant.webtv.net>
Message-Id: <Pine.SOL.L3.93.990318072843.17846C-100000@bingsun1>


I recommend that we accept the same rules as during the 1st (and the last)
world championship. We had a chance to pick up the cube and look at it for
15 seconds. It was then returned to the table and the actual solving
followed. Most competitors actually needed only 5-10 sec. to figure out
the first couple of moves.

Timing? I think most of us when we practice do the timing ourselves.

I have one more point regarding the average. We should standardize
this as well. For example, one can solve the cube 12 times, remove the
worst and the best time and average the remaining 10. Or, do you want to
list all the times during a practice and average them together?

Jiri

*********************************************
 Jiri FRIDRICH, Research Scientist
 Center for Intelligent Systems
 SUNY Binghamton
 Binghamton, NY 13902-6000
 Ph/Fax: (607) 777-2577
 E-mail: fridrich@binghamton.edu
 http://ssie.binghamton.edu/~jirif/jiri.html
*********************************************

From cube-lovers-errors@mc.lcs.mit.edu  Thu Mar 18 19:37:40 1999
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Message-Id: <36F18BF0.5950@hrz1.hrz.tu-darmstadt.de>
Date: Fri, 19 Mar 1999 00:27:44 +0100
From: Herbert Kociemba <kociemba@hrz1.hrz.tu-darmstadt.de>
Reply-To: kociemba@hrz1.hrz.tu-darmstadt.de
To: Cube Lovers <cube-lovers@ai.mit.edu>
Subject: Re: Re : Re: Edges only, Ignoring Flips, Face Turn Metric
References: <SIMEON.9903170012.A@pavilion.pstcc.cc.tn.us>

Jerry Bryan wrote:
>
> On Fri, 12 Mar 1999 01:01:52 -0500 michael reid <reid@math.brown.edu>
> wrote:
>
> > i guess i'm not sure what you're doing, jerry.  but i don't think
> > it should be *that* difficult.  the number of configurations is
> > 12! = about 480 million.  if you divide out by symmetry, you get
> > about 10 million configurations.  this should be small enough to
> > store in memory and do a complete breadth-first search of the space.
> >

> When a search space consists of the elements of a cube group, it is
> easy to index the search space.  But when a cube group is reduced by
> symmetry the result is generally not a group and the resultant search
> space is (in my experience) not very easy to index.  The thing about
> Herbert's program that I have trouble comprehending is that he is able
> to reduce the search space by symmetry and still have the indexing be
> well behaved.  He has posted a clear exposition of his method, so the
> problem is in my understanding rather than in his explanation.

I think you are right to say that the indexing of a cube group reduced
by symmetries does not behave very well. For this reason I must build a
table which maps the index to a representative of the corresponding
equivalence class. I have no method to directly compute the index.

About 10 million entries would be possible but quite a lot, so I
defined two edge permutations x and y as "equivalent" if x = MyN with
two symmetries M and N. So I reduced by another factor of about 48 and
got 208816 classes. If x is a representative of such a class with index i,
Mx with an arbitrary symmetry M is a representative of a "real" symmetry
class. The "well behaved" index of the latter is computed by 48*i +
Index(M), where index(M) enumerates the symmetries from 0 to 47.
The problem with that which I did not realize first is, that Mx and M'x
could be equivalent, which led to wrong results when computing the God's
Algorithm for positions more than 3 face turns from start (I compared my
results with Jerry's, who made a quick run for positions up to 6 face
turns). With some exta computation this problem could be fixed.

Herbert Kociemba

From cube-lovers-errors@mc.lcs.mit.edu  Fri Mar 19 10:32:05 1999
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From: WaVeReBeL@webtv.net
Date: Fri, 19 Mar 1999 02:07:36 -0800 (PST)
To: Cube-Lovers@ai.mit.edu
Cc: IBILUIE@aol.com, IPIIIKA@aol.com
Subject: Re: Speed cubing results - March 99
Message-Id: <9085-36F221E8-14@mailtod-122.bryant.webtv.net>
In-Reply-To: Jiri Fridrich <fridrich@binghamton.edu>'s message of Thu, 18
	Mar 1999 07:35:57 -0500 (EST)

I think that if we are to be keeping track records & averages, we should
ALL stick to one standard.  Using tournament rules sounds like a good
idea.  This makes comparing times more accurate.

Almost everybody responded w/ a different preview time (anywhere from no
preview to 15 seconds).  People like me who started cold had a
disadvantage to those who had a preview.  A 15 second preview sounds
good to me.  This gives enough time to familiarize oneself w/ the cube,
look for pieces, and plan out the first few moves.  I've been timing
myself cold which means much time is wasted at the beginning.  Having a
preview helps a lot.

When it comes to averages, I guess there is no standard.  I agree w/
disregarding the high & low extremes though.  They can distort the
average (arithmetic mean).  This should give a more accurate
representation.  Also, the more entries calculated into the average the
better.

I hope I'm not going overboard.  It's not like we're in a tournament.
If all that is necessary is an informal rough estimate, then you can
disregard this entire message. =)

-Alex Montilla-

From cube-lovers-errors@mc.lcs.mit.edu  Fri Mar 19 11:06:11 1999
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Message-Id: <199903191105.MAA28298@bednorz.get2net.dk>
From: "Klodshans" <klodshans@get2net.dk>
To: cube-lovers@ai.mit.edu
Date: Fri, 19 Mar 1999 12:04:43 +0000
Subject: RE: cube with colors attached
Reply-To: klodshans@get2net.dk
In-Reply-To: <0F8184664EA9D21192F70008C75D16925CAEFF@esealnt145>

Johan Myrberger <Johan.Myrberger@bigfoot.com> wrote:

> > ... Has anyone ever thought of making a large cube out of wood so
> > that there is a lot of wood (or clay) on the outside of the cube and
> > then carve something ...
> 
> Something like this has been manufactured. I believe it was by Disney corp.
> They made cubes which looked like the head of Donald Duck and Mickey Mouse if
> my memory serves me.

These "cubes" were made by Disney in Spain. They work with a 
2x2x2 mechanism. The mechanism seems to be different to the 
one used in the Rubik's Mini Cube - you can see a screw inside 
the "cube" when turning slightly on two axis simultaniously. A bit 
similar to the Pyramorphix - maybe these were also manufactured 
by Meffert ?

I have seen them for sale at Puzzle-shop
www.puzzle-shop.de
and from Pete Beck/Just Puzzles
www.freeyellow.com/members4/justpuzzles/

They are quite fun to operate - one can turn Mickey's ears so they 
point backwards instead of upwards, or turn his eyes so they look 
like i don't know what. Pretty perverse ;-)

Philip

______________________________________
Philip K
Vendersgade 15, 3th
DK - 1363 Copenhagen K
Denmark
Phone:  +4533932787
Mobile: +4521706731
E-mail: philipk@bassandtrouble.com
E-mail: klodshans@get2net.dk
web:    http://hjem.get2net.dk/philip-k

From cube-lovers-errors@mc.lcs.mit.edu  Fri Mar 19 12:02:52 1999
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Date: Fri, 19 Mar 1999 11:14:58 +0000
From: David Singmaster Computing <david.singmaster@sbu.ac.uk>
To: c.v.willegen@spcgroup.nl
Cc: cube-lovers@ai.mit.edu
Message-Id: <009D5582.3042380C.305@ice.sbu.ac.uk>
Subject: RE: Fwd: Request for spectacular cube-solving  -  Can anyone help ?

	It is true that some blind people have limited 3-D perception, but a
colleague once told me he came into a graduate student room and heard the only
blind student in his class explaining subdivisions in n-dimensions to the other
students!

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499;
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Fri Mar 19 12:47:00 1999
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Date: Fri, 19 Mar 1999 12:16:04 +0000
From: David Singmaster <david.singmaster@sbu.ac.uk>
To: hbizek@ameritech.net
Cc: cube-lovers@ai.mit.edu
Message-Id: <009D558A.B96AC759.280@ice.sbu.ac.uk>
Subject: RE: parity pairs

	Conway noted the two mirror-image orientations of the standard colour
pattern (W/Y, B/G, R/O).  One of the corners has  BOY  at a corner and he called
this a  BOY,  versus the mirror-image  YOB.  I think he read the colours clockwise?
Certainly most of the production was  BOY  and one had to hunt a bit for  YOBs.
Some cubists were particularly keen to have one orientation rather than the
other.

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499;
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Fri Mar 19 13:23:41 1999
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Message-Id: <001001be7220$d2f8e740$0237a8c0@uwe>
From: uwe@ue.net (Uwe Meffert)
To: "der Mouse" <mouse@rodents.montreal.qc.ca>
Cc: "Cube-Lovers" <cube-lovers@ai.mit.edu>
Subject: Re: Taking apart the 5^3
Date: Fri, 19 Mar 1999 23:53:58 +0800


>>From: der Mouse <>
>>To: Cube-lovers@ai.mit.edu <Cube-lovers@ai.mit.edu>
>>Date: Friday, March 19, 1999 11:03 AM

>>>I found it more effective to turn a "thick slice" (ie, the outer two
>>>slices turned together) about 45 degrees, then pry with my thumb
>>>between the corner and wing of the turned slice....

That procedure is not recommended as it voids the implied warranty and has
the danger of permanently stripping the thread inside the center of the
cube.

If you must take the cube apart do so by prying off one of the center small
squares and then loosening one of the screws, which later after re-assembly
should be re-tightened.

Regards
Uwe


HAPPY PUZZLING
Uwe Meffert
P.O. Box 24455, Aberdeen, Hong Kong.
Tel. 852-2518-3080, Fax. 852-2518-3282
Email:- uwe@ue.net
www.ue.edu www.ue.net www.mefferts-puzzles.com

From cube-lovers-errors@mc.lcs.mit.edu  Fri Mar 19 15:00:22 1999
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Message-Id: <4.1.19990319114446.0095a7d0@mail.vt.edu>
Date: Fri, 19 Mar 1999 11:48:07 -0500
To: Cube-Lovers@ai.mit.edu
From: Kevin Young <keyoung3@vt.edu>
Subject: Mustering Interest in the Rubik's Cube

Hi-

I've been a cubist since elementary school in 1980.  My interest increases
and decreases in waves, however, it never dies.  I'm now back in school at
Virginia Tech as a computer science major.  Does anyone have any
suggestions on how to muster serious  interest with some of my peers at the
University?

Thank you,

Kevin Young

From cube-lovers-errors@mc.lcs.mit.edu  Fri Mar 19 19:31:58 1999
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Date: Fri, 19 Mar 1999 16:21:33 -0500 (Eastern Standard Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re : RE: parity pairs
In-Reply-To: <009D558A.B96AC759.280@ice.sbu.ac.uk>
To: Cube Lovers <cube-lovers@ai.mit.edu>
Message-Id: <SIMEON.9903191633.A@GN209A.PSTCC.CC.TN.US>

On Fri, 19 Mar 1999 12:16:04 +0000 David Singmaster
<david.singmaster@sbu.ac.uk> wrote:

> 	Conway noted the two mirror-image orientations of the standard colour
> pattern (W/Y, B/G, R/O).

W/Y, B/G, R/O is the "differ by yellow" standard, which I prefer as
"the" standard.  However, there are also references in Cube-Lovers
archives to W/B, R/O, and Y/G as a standard or as the tournament
standard.  I have no idea who gets to be the standards body to select
"the" standard.

But as one example of why I like the W/Y, B/G, R/O standard, many of
the local maxima at 12q from Start are only "somewhat symmetric", but
the eye's sense of symmetry in looking at them can be much stronger.
The reason is that the eye (or my eye, at least) can easily identify
W/Y as the "same" color, B/G as the "same" color, and R/O as the
"same" color.  And when such identifications are made, the symmetry
of many of the 12q local maxima is much stronger than it would be
otherwise.  I really haven't looked at them with any other color
scheme, but I can't imagine that the apparent symmetry would look as
strong otherwise.

Also, in all the various discussions about stickers, falling off and
otherwise, there have been comments about cubes where it is hard to
tell the colors apart, depending on the exact colors which are used,
how faded the colors are with age, etc.  I guess my experience has
been pretty positive in that my stickers have not fallen off and with
one notable exception, the colors seem easy to distinguish.  The
exception is that with my 2x2x2 Pocket Cube, it is very difficult to
distinguish the orange from the red stickers unless I have very, very
good lighting conditions.  This particular cube has always been this
way.  I can think of no reason that a 2x2x2 should be this way as
compared to a 3x3x3 or a 4x4x4, but it does seem to be the case.

----------------------------------------
Jerry Bryan

From cube-lovers-errors@mc.lcs.mit.edu  Mon Mar 22 13:38:42 1999
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Date: Sat, 20 Mar 1999 06:50:39 -0500 (EST)
From: Nicholas Bodley <nbodley@tiac.net>
To: Uwe Meffert <uwe@ue.net>
Cc: der Mouse <mouse@rodents.montreal.qc.ca>,
        Cube-Lovers <cube-lovers@ai.mit.edu>
Subject: Re: Taking apart the 5^3
In-Reply-To: <001001be7220$d2f8e740$0237a8c0@uwe>
Message-Id: <Pine.BSF.3.96.990320064432.15379L-100000@shell2.tiac.net>


On Fri, 19 Mar 1999, Uwe Meffert wrote:

}>I found it more effective to turn a "thick slice" (ie, the outer two
}>slices turned together) about 45 degrees, then pry with my thumb
}>between the corner and wing of the turned slice....

}That procedure is not recommended as it voids the implied warranty and has
}the danger of permanently stripping the thread inside the center of the
}cube.
}
}If you must take the cube apart do so by prying off one of the center small
}squares and then loosening one of the screws, which later after re-assembly
}should be re-tightened.

Since I was the first to suggest this method, I'll retract the advice.
Indeed, it would be really unfortunate to strip the threads in the
plastic that hold a screw in place.

Mr. Meffert is, and has been, a formidable inventor and manufacturer of
cube-like puzzles for quite some time, for those who don't recognize his
name. I'd follow his advice; definitely! Do note that he said "loosen",
not "remove" the screw. I posted recently about special care in
reinserting a removed screw.

My regards to all...
NB

|*  Nicholas Bodley   *|*
|*   Waltham, Mass.   *|*
|*  nbodley@tiac.net  *|*
|*  Amateur musician  *|*



From cube-lovers-errors@mc.lcs.mit.edu  Thu Mar 25 14:32:49 1999
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Date: Sun, 21 Mar 1999 11:59:19 -0500 (EST)
From: der Mouse  <mouse@rodents.montreal.qc.ca>
Message-Id: <199903211659.LAA04577@Twig.Rodents.Montreal.QC.CA>
To: cube-lovers@ai.mit.edu
Subject: Re: Taking apart the 5^3

>>>> I found it more effective to turn a "thick slice" (ie, the outer
>>>> two slices turned together) about 45 degrees, then pry with my
>>>> thumb between the corner and wing of the turned slice....

> That procedure is not recommended as it voids the implied warranty

(a) doesn't this depend on the manufacturer?  (b) doesn't taking it
apart at all do that anyway?

> and has the danger of permanently stripping the thread inside the
> center of the cube.

I can't see how.  Based on how I saw my cube move as I did this, I
don't believe it's putting significant stress on any of the face-center
cubies.  Remember, what I said to do was to push the wing cubie away
from the corner cubie, not vice versa; the only face cubie anywhere
near the operation is the one you're prying towards, and the wing cubie
is pivoting around that, not prying past it.

> If you must take the cube apart do so by prying off one of the center
> small squares and then loosening one of the screws,

After disassembling my cube, I remembered this advice, and tried to get
one of the face centers off.  Even with the cube disassembled, I
desisted for fear of breaking the plastic rather than the glue join.

Perhaps there are multiple production runs in existence and some of
them come apart more easily this way than others?

					der Mouse

			       mouse@rodents.montreal.qc.ca
		     7D C8 61 52 5D E7 2D 39  4E F1 31 3E E8 B3 27 4B

From cube-lovers-errors@mc.lcs.mit.edu  Thu Mar 25 14:42:29 1999
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From: "Dallas Foster" <fostindi@ecicnet.org>
Reply-To: "Dallas Foster" <fostindi@ecicnet.org>
To: <cube-lovers-request@ai.mit.edu>
Subject: Instructions needed
Date: Thu, 25 Mar 1999 13:01:23 -0500
Message-Id: <01be76e9$7e7e0c40$03c3d4d1@dallasfo>

I have a Rubik's Magic Strategy Game that consists of the playing board,
16 playpieces and instructions.  Unfortunately we have lost the
instructions and our son, who's game it is no longer lives at home and
can't remember how it was played.

Would you have or know where I might obtain a copy of these
instructions. The game is of no use to us with out them.  The game was
distributed by Matchbox toys in 1987.

Any help you can lend would be greatly appreciated.

Thank You,

Mrs. D.D. Foster
2734 W. Maple Street
Anderson, IN 46013 USA

fostindi@ecicnet.com

From cube-lovers-errors@mc.lcs.mit.edu  Thu Mar 25 19:29:58 1999
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Date: Thu, 25 Mar 1999 21:59:53 +0000
From: David Singmaster <david.singmaster@sbu.ac.uk>
To: jbryan@pstcc.cc.tn.us
Cc: cube-lovers@ai.mit.edu
Message-Id: <009D5A93.46B93D17.11@ice.sbu.ac.uk>
Subject: RE: Re : RE: parity pairs

	If one really wants trouble with cube colors being indistinguishable
try solving under a sodium vapor street lamp.  I found this gave two colors:
grey and greyer!

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499;
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Fri Mar 26 15:39:42 1999
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Date: Thu, 25 Mar 1999 23:58:38 -0500 (Eastern Standard Time)
From: Dale Newfield <din5w@cs.virginia.edu>
Reply-To: DNewfield@cs.virginia.edu
To: cube-lovers@ai.mit.edu
Subject: RE: Re : RE: parity pairs
In-Reply-To: <009D5A93.46B93D17.11@ice.sbu.ac.uk>
Message-Id: <Pine.WNT.4.10.9903252355560.-498359@biff.cs.virginia.edu>

On Thu, 25 Mar 1999, David Singmaster wrote:
> 	If one really wants trouble with cube colors being indistinguishable
> try solving under a sodium vapor street lamp.  I found this gave two colors:
> grey and greyer!

Or corner Andrew Plotkin and get him to let you try out "The Liquid
Crystal Cube" (http://www.eblong.com/~zarf/custom-cubes.html) -- Apparently
quite a challenge.

-Dale

[Moderator's note: I was going to say "Mood cube", but of course zarf has
 already heard that one. Tastefully, he resists the nomenclature. --Dan ]

From cube-lovers-errors@mc.lcs.mit.edu  Fri Mar 26 18:20:51 1999
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Message-Id: <19990326231259.15935.qmail@hotmail.com>
From: "Don Harper" <donharper@hotmail.com>
To: Cube-Lovers@ai.mit.edu
Subject: Subject: RE: Re : RE: parity pairs
Date: Fri, 26 Mar 1999 15:12:58 PST

I always had problems solving the cube in bars.  I would always have a
"friend" that would want to drag me into a bar and make money off of my
cube solving skills, but the lighting was always low and I couldn't tell
the colors apart.

Along those lines, I have a brass cube.  Easy to solve, or nearly
impossible?  It is heavy, even though I believe it is only "plated".

I have a picture of it on my web site at:
http://www.geocities.com/Colosseum/Sideline/2953/

I also have a "cubo 15" cube you all may find interesting.

Thanks!
Don Harper

From cube-lovers-errors@mc.lcs.mit.edu  Fri Mar 26 18:37:05 1999
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Date: Fri, 26 Mar 1999 03:46:13 -0500 (EST)
From: Nicholas Bodley <nbodley@tiac.net>
To: David Singmaster <david.singmaster@sbu.ac.uk>
Cc: jbryan@pstcc.cc.tn.us, cube-lovers@ai.mit.edu
Subject: Colors under Na vapor lamps (Was: RE: Re : RE: parity pairs)
In-Reply-To: <009D5A93.46B93D17.11@ice.sbu.ac.uk>
Message-Id: <Pine.BSF.3.96.990326034101.27875Z-100000@shell2.tiac.net>

On Thu, 25 Mar 1999, David Singmaster wrote:

}	If one really wants trouble with cube colors being indistinguishable
}try solving under a sodium vapor street lamp.  I found this gave two colors:
}grey and greyer!

Here in the northeastern US, high-pressure sodium (Na) vapor lamps are
common; they have a much broader spectrum that the essentially
monochromatic low-pressure lamps. The latter are rather strange! They're
also more efficient, I'm fairly sure. I haven't looked at a cube under
the h.p. lamps, though.

|*  Nicholas Bodley   *|*  Autodidact & Polymath * Electronic Tech. (ret.)
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*
|*  Amateur musician  *|*

From cube-lovers-errors@mc.lcs.mit.edu  Mon Mar 29 12:46:44 1999
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To: cube-lovers@ai.mit.edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Re: Instructions needed
Date: 26 Mar 1999 23:51:28 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <7dh6i0$ekh@gap.cco.caltech.edu>
References: <cube-lovers.01be76e9$7e7e0c40$03c3d4d1@dallasfo.SOMEWHERE>

"Dallas Foster" <fostindi@ecicnet.org> writes:
>I have a Rubik's Magic Strategy Game that consists of the playing board,
>16 playpieces and instructions.  Unfortunately we have lost the
>instructions and our son, who's game it is no longer lives at home and
>can't remember how it was played.

>Would you have or know where I might obtain a copy of these
>instructions. The game is of no use to us with out them.  The game was
>distributed by Matchbox toys in 1987.

The simplest thing you can do is to go to your local store and buy a 
copy of "Rubik's Eclipse".  The rules are identical.

I may have a copy of the original game lying around if this is unsucessful.

--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
This new video game can challenge feet (8,7)

From cube-lovers-errors@mc.lcs.mit.edu  Mon Mar 29 13:47:07 1999
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From: WaVeReBeL@webtv.net
Date: Fri, 26 Mar 1999 23:57:49 -0800 (PST)
To: Cube-Lovers@ai.mit.edu
Subject: Wrist pains
Message-Id: <10861-36FC8F7D-5994@mailtod-121.bryant.webtv.net>

Hi all,

I've been having some marathon cubing sessions lately, and sometimes my
wrists, fingers, or forearms would hurt.  It's really hard to stop
because I'm so addicted, but I have to because I'm scared I might get
carpal tunnel syndrome or some sort of repetitive stress problem.  I
love computers, and since I'm studying computer science, and because
I'll have many many hours on a keyboard ahead of me, this scares me even
more.  Does anyone else have have this problem?  Any advice?  Thanks.

-Alex Montilla-

[ Moderator's note: Dame Kathleen Ollerenshaw is mentioned in Singmaster's
  notes as being one of the first to develop cubist's thumb; Roger
  Frye also got a wrist sprain as he mentioned on this list in 1981.
  Repetitive stress injury is nothing to play with, and continuing
  only makes recovery take longer, so stop now! Switch to a simulator.
  Simulators I've seen are somewhat less convenient to use than a real
  cube (though I haven't seen the recent ones), but if that's the case
  I hope this motivates you to change that sorry state. -- Dan ]

From cube-lovers-errors@mc.lcs.mit.edu  Mon Mar 29 14:18:29 1999
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Message-Id: <36FEEE8F.CE5E8EBE@binghamton.edu>
Date: Sun, 28 Mar 1999 22:07:59 -0500
From: Mirek Goljan <bg22976@binghamton.edu>
Organization: SUNY Binghamton
To: sausage@zeta.org.au
Cc: cube-lovers@ai.mit.edu
Subject: Speed cubing

Hello, Wayne and other speed cubists,
    if you see about 15 year backwards you would realize that a few
speed cubists have their average time below 20 sec and quite a lot of others
below 25 sec.
But you are interested in CURRENT Best Time and Average Time, I see.
After a few weeks of practicing my times are:
18 sec, 24 sec, (not bad after some years of 'abstinence' :-))
My method is Jiri's.
Beside that, I practice in solving the cube in minimum moves (face moves)
for which I use slightly different method for first two layers:
2x2 subcube first, 2x2x3 next, 2x3x3 - two layers and for the last layer
I may or may not use some more moves than Jiri's method uses.
My average number of moves is about 48 within 2 min restriction.

    Mirek
********************************
Miroslav Goljan
Watson School of Engineering and
    Applied Science, Dept. of EE
 State University of New York
 PO BOX 00238
 Binghamton, NY 13902-6000
e-mail: bg22976@binghamton.edu
********************************

From cube-lovers-errors@mc.lcs.mit.edu  Mon Mar 29 15:19:18 1999
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To: cube-lovers@ai.mit.edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Re: Taking apart the 5^3
Date: 26 Mar 1999 23:43:35 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <7dh637$ea1@gap.cco.caltech.edu>
References: <cube-lovers.199903211659.LAA04577@Twig.Rodents.Montreal.QC.CA>

der Mouse  <mouse@rodents.montreal.qc.ca> writes:
>Perhaps there are multiple production runs in existence and some of
>them come apart more easily this way than others?

This is definitely true.  Mine fell off after a few days -- I eventually
glued them back on.

--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
This new video game can challenge feet (8,7)

From cube-lovers-errors@mc.lcs.mit.edu  Mon Mar 29 18:40:48 1999
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From: wheeler@cipr.rpi.edu (Frederick W. Wheeler)
Message-Id: <14080.1711.289304.134028@cipr.no_spam.rpi.edu>
Date: Mon, 29 Mar 1999 18:03:11 -0500 (EST)
To: Cube-Lovers@ai.mit.edu
Subject: Inventing your own techniques


I've been reading Cube-Lovers e-mail for a few months now and am
enjoying it very much.  It is one of the great benefits of the
Internet that people with a common interest, but spread so far apart,
can so easily communicate like this.

For me, the most fun, and the ultimate challenge, in cubing comes from
figuring out how to solve the puzzle in the first place.  I avoid
published and posted techniques.  I'd really like to hear from people
on this list on how you go about inventing new moves and techniques or
how you feel about learning to solve a puzzle on your own.

I vaguely remember how I learned to solve the 3x3x3 back in the early
80's.  I was in 8th grade at the time; now I'm in 25th grade.
According to my family I quite thoroughly infatuated by the puzzle at
the time.  Solving one side was faily easy and then I was able to get
2 and 3 sides, but with a disorganized and perhaps even random method.

I heard from a friend (who had a solution book) that the key to
solving the top and bottom was a set of special moves that allowed you
to manipulate the bottom corner pieces without affecting the top side.
I set out to find these moves on my own and did.  I would carefully
record the position and orientation of each corner piece, then move a
top corner out of position and then back into position in a different
way and check how the bottom side changed.  This led to a few
sequences which I could repeatedly apply to solve the bottom corners.

The rest was fairly easy, except for the situation in which two edges
were flipped.  I had to have someone show me a move to get past this
point.  I couldn't figure it out.  Otherwise, I had a 50% chance of
solving on any given attempt.

Now I have a 4x4x4 and a 5x5x5 cube as well.  I've been able to solve
these primarily using extensions of the techniques I learned for the
3x3x3 and a few new extras, but only to a point.  I'm now stuck if one
pair of "wing" pieces are switched.  If two pairs are switched, I can
solve it, but not if only one are switched.  Again, I solve it 50% of
the times I set out.

Of course, there was at least on posted solution for this very problem
to this list a couple of weeks ago.  I saved it to a folder just in
case I decide to resort to it, but in the mean time want to figure
this out on my own.

Regards,
Fred Wheeler

--
Fred Wheeler
wheeler@cipr.rpi.edu
www.cipr.rpi.edu/wheeler

From cube-lovers-errors@mc.lcs.mit.edu  Tue Mar 30 14:02:52 1999
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Reply-To: <noel@mud.ca>
From: "Noel Dillabough" <noel@mud.ca>
To: "Cube Lovers (E-mail)" <Cube-Lovers@ai.mit.edu>
Cc: <WaVeReBeL@webtv.net>
Subject: RE: Wrist pains
Date: Mon, 29 Mar 1999 23:42:38 -0500
Message-Id: <000001be7a68$23427d20$040a0a0a@laptop>
In-Reply-To: <10861-36FC8F7D-5994@mailtod-121.bryant.webtv.net>

<cut>
[ Moderator's note: Dame Kathleen Ollerenshaw is mentioned in Singmaster's
  notes as being one of the first to develop cubist's thumb; Roger
  Frye also got a wrist sprain as he mentioned on this list in 1981.
  Repetitive stress injury is nothing to play with, and continuing
  only makes recovery take longer, so stop now! Switch to a simulator.
  Simulators I've seen are somewhat less convenient to use than a real
  cube (though I haven't seen the recent ones), but if that's the case
  I hope this motivates you to change that sorry state. -- Dan ]


As a programmer, I am no stranger to repetitive stress related aches and
pains.  I even had cubist's thumb way back :)  My simulator, puzzler, has
mainly a mouse interface and this given enough time can cause you to be
rather sore as well.  I was wondering if anyone had an idea for a "hands on
keyboard" approach that would allow you to naturally move the puzzles'
slices.

There is a macro interface for the cubes to enter moves in UDFBLR notation,
but I am thinking of something that you could, with practice, manipulate a
cube in realtime using a keyboard.

Any ideas would be appreciated and I'll try my best to implement the best
one.  If simulators "felt" like a cube when you used them they would be more
fun (and move without jamming etc)

-Noel

From cube-lovers-errors@mc.lcs.mit.edu  Tue Mar 30 15:11:03 1999
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Date: Tue, 30 Mar 1999 12:53:02 +0100
From: David Singmaster <david.singmaster@sbu.ac.uk>
To: WaVeReBeL@webtv.net
Cc: cube-lovers@ai.mit.edu
Message-Id: <009D5E34.B59FB7EA.32@ice.sbu.ac.uk>
Subject: RE: Wrist pains

	Beryl Fletcher, who organized my first cube at the International
Congress of Mathematicians in Helsinki in 1978, also developed Cubist's Thumb.
The early cubes were stiff and one held them with a corner pressing on the
tendon of the left thumbb, in the fleshy part of the thumb.  With olde people,
the tendon sheath, or rather the inner lubricant, has become a bit aged and the
constant pressure leads to a chronic inflammation.  This is readily treated by
a small operation which cuts open the tendon sheath.  Both Dame Kathleen
Ollerneshaw and Beryl Fletcher had this.
	When I was working on my Notes, I did a lot of checking of move
sequences and typing and got definite pains in the wrist.  Since then, word
processing has also occasionally produced RSI problems.  Fortunately, rest lets
it go away, but I have had physiotherapy several times.  There were several
periods when I had to stop typing for several weeks!  The most unusual
diagnosis was stiff neck due to tension while typing.  Physiotherapy helped
that.  All the warnings/instructions about RSI are worth heeding.  Take breaks.
Stretch regularly.  Make sure your work station is comfortable.  Etc.  RSI is
real and I've known several people semi-permanently disabled by it.

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499;
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Tue Mar 30 16:55:26 1999
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To: cube-lovers@ai.mit.edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Re: Inventing your own techniques
Date: 30 Mar 1999 14:55:07 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <7dqokb$43e@gap.cco.caltech.edu>
References: <cube-lovers.14080.1711.289304.134028@cipr.no_spam.rpi.edu>

wheeler@cipr.rpi.edu (Frederick W. Wheeler) writes:
>For me, the most fun, and the ultimate challenge, in cubing comes from
>figuring out how to solve the puzzle in the first place.  I avoid
>published and posted techniques.  I'd really like to hear from people
>on this list on how you go about inventing new moves and techniques or
>how you feel about learning to solve a puzzle on your own.

After I understood conjugation well enough, I have never invented a
move that I can in all honesty call "new" -- although they may appear
"new" to others.  The only new part is just applying it to different
types of moves and seeing what the result is.

--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
She ran by screaming "No, I run by moving my feet rapidly, you idiot!"

From cube-lovers-errors@mc.lcs.mit.edu  Tue Mar 30 18:05:07 1999
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Date: Tue, 30 Mar 1999 16:25:51 -0500 (EST)
From: Nicholas Bodley <nbodley@tiac.net>
To: Noel Dillabough <noel@mud.ca>
Cc: "Cube Lovers (E-mail)" <Cube-Lovers@ai.mit.edu>, WaVeReBeL@webtv.net
Subject: RE: Wrist pains
In-Reply-To: <000001be7a68$23427d20$040a0a0a@laptop>
Message-Id: <Pine.BSF.3.96.990330162309.7585E-100000@shell2.tiac.net>


On Mon, 29 Mar 1999, Noel Dillabough wrote:

}Any ideas would be appreciated and I'll try my best to implement the best
}one.  If simulators "felt" like a cube when you used them they would be more
}fun (and move without jamming etc)

Can you imagine force-feedback joystick technology? Hooray for the
reset button! I very recently tried a demo at a computer store;
pleasant surprise.

|*  Nicholas Bodley   *|*  Autodidact & Polymath * Electronic Tech. (ret.)
|*   Waltham, Mass.   *|*
|*  nbodley@tiac.net  *|*
|*  Amateur musician  *|*
--------------------------------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Tue Mar 30 19:30:45 1999
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Message-Id: <370168C9.9ECB5992@whitewolf.com.au>
Date: Wed, 31 Mar 1999 10:14:01 +1000
From: Ryan Heise <ryan@whitewolf.com.au>
To: noel@mud.ca
Cc: "Cube Lovers (E-mail)" <Cube-Lovers@ai.mit.edu>, WaVeReBeL@webtv.net
Subject: Keyboard cube [was: Re: Wrist pains]
References: <000001be7a68$23427d20$040a0a0a@laptop>

Noel Dillabough wrote:

> As a programmer, I am no stranger to repetitive stress related aches and
> pains.  I even had cubist's thumb way back :)  My simulator, puzzler, has
> mainly a mouse interface and this given enough time can cause you to be
> rather sore as well.  I was wondering if anyone had an idea for a "hands on
> keyboard" approach that would allow you to naturally move the puzzles'
> slices.

I had some ideas on this once. The first keyboard layout below is easier
to learn but the second would be faster once you had mastered it.

[QWERTY keyboard]

1)

U = r
U'= u
D'= f
D = j

R = 7
R'= m
L'= 4
L = v

F = h
F'= g
B'= y
B = t

2)

Any mappings using this set of keys:

  er  ui
asdf  jkl;

The simulator input key should be configurable so you can try out these
variations once its written. If you want help implementing it, feel free
to ask me - just in case I ever have some free time on my hands!

--
Ryan Heise

http://www.progsoc.uts.edu.au/~rheise/

From cube-lovers-errors@mc.lcs.mit.edu  Tue Mar 30 20:15:57 1999
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From: Norman Richards <orb@cs.utexas.edu>
To: "Cube Lovers (E-mail)" <Cube-Lovers@ai.mit.edu>
Subject: Keyboard cube [was: Re: Wrist pains]
In-Reply-To: <000001be7a68$23427d20$040a0a0a@laptop>
Date: Tue, 30 Mar 1999 17:47:33 -0600


> [...] I was wondering if anyone had an idea for a "hands on
> keyboard" approach that would allow you to naturally move the puzzles'
> slices.
>
> There is a macro interface for the cubes to enter moves in UDFBLR notation,
> but I am thinking of something that you could, with practice, manipulate a
> cube in realtime using a keyboard.


  Oddly enough, I have been thinking about a method to manipulate a
cube by use of the numeric keypad.  It seems most moves can be
completed rather naturally, but I do not know if it works in practice.

  Anyways, take a keypad like this:

7 8 9
4 5 6
1 2 3

  This corresponds to a face of a cube quite nicely.  Suppose you
wanted to rotate the right face clockwise.  One could enter 36 or 69,
for example, which you could conceptually think of as the direction
your hand would move to rotate the right face clockwise.  Counter
clockwise would be the other direction.  (96 or 63 or even 93) The
same technique could be applied to the left face or top face or bottom
face.  The middle slices could be rotated just as easil: 52 would rotate
the middle vertical slice down.

  The question is how to effect rotating the front and rear faces.
For me, 19 and 91 seem natural for F and F' because they basically
mimic the twisting motion.  The rear is more troublesome, but perhaps
for symetry 73 and 37 might be used?

  Anyways, that could take care the turns, Cube rotations could be as
simple as a shift followed by a direction.  shift-5-2 might rotate the
cube along the X axis such that U is now F and F is now D, etc...

  I do not know if keypad entry is any more or less prone to these
types of entry, but I think that the general mechanism might work.

___________________________________________________________________________
orb@cs.utexas.edu                                           soli deo gloria

From cube-lovers-errors@mc.lcs.mit.edu  Wed Mar 31 12:38:08 1999
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Message-Id: <4.1.19990330184411.00932b80@mail.vt.edu>
Date: Tue, 30 Mar 1999 18:52:17 -0500
To: Cube-Lovers@ai.mit.edu
From: Kevin Young <keyoung3@vt.edu>
Subject: Future Rubiks Products Your Vote Counts!
In-Reply-To: <Pine.BSF.3.96.990330162309.7585E-100000@shell2.tiac.net>
References: <000001be7a68$23427d20$040a0a0a@laptop>

Hello-

We talked in the past about Oddzon (Current distributor of Rubik's Products
for Seven Towns - the holder of the Rubik's brand) and the version of the
Rubik's Cube on the market.  If you are tired of those stickers that fall
apart and would like a well built cube...one that has a sprung indexing
mechanism which would mean that a layer would gently "click" into place
when aligned along with tiles instead of stickers, then it's time for you
to place your vote and influence them.  They are currently looking into
making a "Deluxe" Rubik's Cube, but, need to see if the market will support
such a product.  Anyway, you can place your vote at the following link:

http://www.rubiks.com/poll.html?q=6

Be sure to browse their web site while you are there.  Check out their news
link to find out more on the production of the "Deluxe" Rubik's Cube.
Click on the following link to go directly to their web site:

http://www.rubiks.com

Happy Cubing!

Kevin Young

From cube-lovers-errors@mc.lcs.mit.edu  Wed Mar 31 13:17:28 1999
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Date: Wed, 31 Mar 1999 08:46:33 -0500 (Eastern Standard Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re : Keyboard cube [was: Re: Wrist pains]
In-Reply-To: <199903302344.SAA11205@life.ai.mit.edu>
To: Cube Lovers <cube-lovers@ai.mit.edu>
Message-Id: <SIMEON.9903310833.A@GN209A.PSTCC.CC.TN.US>

On Tue, 30 Mar 1999 17:47:33 -0600 Norman Richards <orb@cs.utexas.edu> wrote:
>
> > [...] I was wondering if anyone had an idea for a "hands on
> > keyboard" approach that would allow you to naturally move the puzzles'
> > slices.
> >

I fought with this issue back in about 1985 when I was originally working on
the 2x2x2 and the corners of the 3x3x3.  I created a data base for each, with a
program which would allow you to manipulate the cube on the screen, and the
program would always show you your exact distance from Start.  Actually, the
screen would show you two renderings of the cube  --  one was the cube you were
manipulating and the other was the representative of the same cube under
reduction by symmetry.  It was the representative that was looked up in the
data base.

I tried all kinds of input mappings, none of which were very satisfying.  One
obvious thing to try is U for Up, D for Down, etc., but on a QWERTY keyboard
this is not very easy to deal with.

Keyboards in 1985 only had a numeric pad, not the additional arrow and page
up/down keys of modern keyboards.  Within the numeric pad, I ended up using up
arrow for U, down arrow for D, right arrow for R, and left arrow for L.  That
left the question of F and B.  It *sounds* kind of dumb, but using the
numeric pad + key for B and the numeric pad 0/INS key for F worked about as
well as anything.  With these mappings, I could manipulate the cube without
looking at the keyboard and with very little movement of my hand.

I used the shift keys for things like U' and U2.  So left shift plus up arrow
was U' and right shift plus up arrow was U2, etc.  I read the shift keys
directly from the keyboard hardware (this was early DOS, and you could do such
things).  I wouldn't necessarily recommend accessing the hardware so directly
these days.  However, I found that I almost never used the shift keys.  Rather,
I would tap the up arrow key quickly three times for U', and I would tap the up
arrow key quickly two times for U2.

I did not implement an interface for whole cube moves.  Rather, to rotate the
whole cube I would do something like RL' or R'L or UD' etc.  Well, this
works for the 2x2x2, but not for the corners of the 3x3x3.

It's tricky to make the interface simple and intuitive, and also to make it
functionally rich at the same time.

----------------------------------------
Jerry Bryan
jbryan@pstcc.cc.tn.us
Pellissippi State Technical Community College

From cube-lovers-errors@mc.lcs.mit.edu  Wed Mar 31 14:47:54 1999
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Date: Wed, 31 Mar 1999 09:48:07 -0600
To: cube-lovers@ai.mit.edu
From: Tom Magliery <mag@ncsa.uiuc.edu>
Subject: Re: Keyboard cube [was: Re: Wrist pains]
In-Reply-To: <199903302344.SAA11205@life.ai.mit.edu>
References: <000001be7a68$23427d20$040a0a0a@laptop>

i already sent this to noel directly, but since others are sending
suggestions here i thought i'd toss in my thoughts about a keyboard cube
interface...

mag

At 11:42 PM 3/29/99 -0500, Noel Dillabough unabashedly said:
>There is a macro interface for the cubes to enter moves in UDFBLR notation,
>but I am thinking of something that you could, with practice, manipulate a
>cube in realtime using a keyboard.

this is a wonderful user interface design challenge.  i've thought about it
before, because i have never been comfortable with any of the mouse-based
interfaces for moving 3d objects around.  i've never used one that seemed
fully intuitive to me.

i'd love to see an interface like this:

* 5 faces of the cube visible at once (all except B), perhaps something
like this (except square, and with the individual cubies showing of course):

      +------------------+
      |\                /|
      | \      U       / |
      |  \            /  |
      |   +----------+   |
      |   |          |   |
      | L |    F     | R |
      |   |          |   |
      |   |          |   |
      |   +----------+   |
      |  /            \  |
      | /      D       \ |
      |/                \|
      +------------------+

* unshifted keys for turning each of the visible faces either CW or CCW, a
total of 10 keys in all, and repeated on the left- and right-hand side of
the keyboard.  the following diagram shows the cube operations thus
associated with various keys (though the diagram is laid out mostly like
the keyboard, i've added space between the hands for clarity):

     w:L'  e:U'  r:U   t:R         y:L'  u:U'  i:U   o:R

       s:F'        f:F                     j:F'        l:F

         x:L   c:D'  v:D   b:R'          n:L   m:D'  ,:D   .:R'

* shifted keys for turning the cube itself.  again, available on both
hands.  3 axes of rotation, two directions each ==> only 6 keys needed on
each hand.  but why not have even more duplication?  for example, suppose
(in real life) you want to roll the cube forward away from you.  you might
do it by either grabbing it from the R side or the L side.  hence the same
10 keys as above, when shifted, work to turn the entire cube in the same
direction as they turn the faces when unshifted.

--
///X  Tom Magliery, Research Programmer           217-333-3198  .---o
\\\   NCSA, 605 E. Springfield      O-       mag@ncsa.uiuc.edu  `-O-.
///   Champaign, IL 61820       http://sdg.ncsa.uiuc.edu/~mag/  o---'

From cube-lovers-errors@mc.lcs.mit.edu  Wed Mar 31 21:24:53 1999
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Date: Tue, 30 Mar 1999 21:57:43 -0500 (EST)
From: Nichael Cramer <ncramer@bbn.com>
To: Norman Richards <orb@cs.utexas.edu>
Cc: "Cube Lovers (E-mail)" <Cube-Lovers@ai.mit.edu>
Subject: Re: Keyboard cube [was: Re: Wrist pains]
In-Reply-To: <199903302344.SAA11205@life.ai.mit.edu>

Norman Richards wrote:
>   Oddly enough, I have been thinking about a method to manipulate a
> cube by use of the numeric keypad.  It seems most moves can be
> completed rather naturally, but I do not know if it works in practice.
>
>   Anyways, take a keypad like this:
>
> 7 8 9
> 4 5 6
> 1 2 3
>
>   This corresponds to a face of a cube quite nicely.  Suppose you
> wanted to rotate the right face clockwise.  One could enter 36 or 69,
> for example, which you could conceptually think of as the direction
> your hand would move to rotate the right face clockwise.  Counter
> clockwise would be the other direction.  (96 or 63 or even 93) The
> same technique could be applied to the left face or top face or bottom
> face.  The middle slices could be rotated just as easil: 52 would rotate
> the middle vertical slice down.

Or a extension of your scheme might be to combine arrow keys with the
keypad, in a two-handed approach.

I.e. 6^ or 6v rather than 36 or 69; no order dependence this way.
(Likewise, I find the arrow keys more mnemomic.)  The model is that you're
"grabbing" the specified cubie and rotating it in the specified direction.

In its most natural form, this assumes that the numeric key-pad and the
arrow keys are separate (as they are on my keyboard), although one could
certainly use some other set of "directional" keys aside from the standard
arrow keys.

And as I say, this also assumes --most naturally, although not
necessarily-- a two-handed approach.

>   The question is how to effect rotating the front and rear faces.
> For me, 19 and 91 seem natural for F and F' because they basically
> mimic the twisting motion.  The rear is more troublesome, but perhaps
> for symetry 73 and 37 might be used?

For completeness, too, there is also the slice between the front and back
face.

Rather than muddy the paradigm, perhaps other (non-assigned) keys should
be used.

The handy "0" key might make a natural candidate for the front face.

On the other hand since the back face and middle slice are rather
"pathological" cases in this paradigm, might it perhaps make sense to use
keys "outside" the model?  For example "R" and "M".  (Or, less
mnemonically --and depending on the set-up of your keyboard-- "/" or "*"
or ".", which on my keyboard are right beside the numeric keypad keys.)

(For these last cases, perhaps on the <- and -> keys should by
operational.  The up and down arrows being less meaningful here.)

>   Anyways, that could take care the turns, Cube rotations could be as
> simple as a shift followed by a direction. [...]

This seems like a good scheme.

N

From cube-lovers-errors@mc.lcs.mit.edu  Thu Apr  1 14:42:07 1999
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Date: Thu, 1 Apr 1999 14:07:30 -0500 (EST)
From: Nicholas Bodley <nbodley@tiac.net>
To: Nichael Cramer <ncramer@bbn.com>
Cc: Norman Richards <orb@cs.utexas.edu>,
        "Cube Lovers (E-mail)" <Cube-Lovers@ai.mit.edu>
Subject: Spaceball (tm) input to a Cube simulator
In-Reply-To: <199904010224.VAA00984@mc.lcs.mit.edu>
Message-Id: <Pine.BSF.3.96.990401140302.3465Q-100000@shell2.tiac.net>


If anyone has a Spaceball, it should make quite a nice input device.
It senses both torque about all 3 orthog. axes, and linear
displacement forces ditto; a total of 6 channels. Displacement could
select a layer, which could be highlighted (anyone for alpha-channel
translucency?), and torque would rotate the selected layer.


|*  Nicholas Bodley   *|*
|*   Waltham, Mass.   *|*
|*  nbodley@tiac.net  *|*
|*  Amateur musician  *|*

From cube-lovers-errors@mc.lcs.mit.edu  Thu Apr  1 17:01:45 1999
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From: Andrew John Walker <ajw01@uow.edu.au>
Message-Id: <199904010348.NAA17158@wumpus.its.uow.edu.au>
Subject: Mike Reid's Solver
To: cube-lovers@ai.mit.edu
Date: Thu, 1 Apr 1999 13:48:23 +1000 (EST)

If anyone has compiled a PC version of Mike Reid's program
set to find optimal solutions in the face turn metric please
contact me (or better still the list).

Andrew Walker


From cube-lovers-errors@mc.lcs.mit.edu  Fri Apr  2 03:11:29 1999
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Date: Thu, 1 Apr 1999 15:56:33 -0500 (EST)
From: Dale Newfield <din5w@cs.Virginia.edu>
Reply-To: DNewfield@cs.Virginia.edu
To: "Cube Lovers (E-mail)" <Cube-Lovers@ai.mit.edu>
Subject: Re: Spaceball (tm) input to a Cube simulator
In-Reply-To: <Pine.BSF.3.96.990401140302.3465Q-100000@shell2.tiac.net>

On Thu, 1 Apr 1999, Nicholas Bodley wrote:
> If anyone has a Spaceball, it should make quite a nice input device.
> It senses both torque about all 3 orthog. axes, and linear
> displacement forces ditto; a total of 6 channels. Displacement could
> select a layer, which could be highlighted (anyone for alpha-channel
> translucency?), and torque would rotate the selected layer.

Except that it is typically quite difficult to separately control
translation and rotation with these devices.

-Dale

From cube-lovers-errors@mc.lcs.mit.edu  Sun Apr  4 18:06:09 1999
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To: cube-lovers@ai.mit.edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Keyboard cube [WAS Re: Wrist pains]
Date: 3 Apr 1999 00:22:20 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <7e3mvs$q14@gap.cco.caltech.edu>
References: <cube-lovers.000001be7a68$23427d20$040a0a0a@laptop.SOMEWHERE>

"Noel Dillabough" <noel@mud.ca> writes:
>Any ideas would be appreciated and I'll try my best to implement the best
>one.  If simulators "felt" like a cube when you used them they would be more
>fun (and move without jamming etc)

Immediately after reading this mail, my concept of a "keyboard interface"
almost popped up immediately.  It is similar to the 1-9 keypad interface
mentioned already.

[Assume that the person is using a standard American QWERTY keyboard;
 remap if necessary.]

First, represent the front face by

q w e
a s d
z x c

A "move" is represented by any two ordered keystrokes.  If the two keystrokes
are in the same row, it represents a horizontal layer moving, else
if they're in the same row, it represents a vertical layer moving.

Example:
 "ed" = R'
 "cd" = R
 "sd" = slice move
 "we" = U'
 "qe" = U'
 "qw" = U'

This method does not have moves for rotating the three front to back layers
nor 180-degree moves.

Extension A:  Make a gap of a key represent a 180-degree move.
 So, "qe" is now U2 instead of just U'.

Extension B:  Represent front-to-back layers by the unused key combinations.
 This will necessarily be idiomatic (so I don't like it).  For example,
 you could have "knight moves" represent F and B turns, and combinations
 with the "s" be slice moves.

Extension C:  Add another grid for the right side of the cube:
 r t y
 f g h
 v b n
 This solves the problem of the front-to-back layers nicely, and adds
 some redundancy.  Most cube programs should display more than one face
 to the user anyway!

Extension D:  Use the arrow keys to rotate the entire cube.  This can
 be done by the right hand while the left hand is just turning faces, which
 is similar to how a lot of people solve the cube anyway.  This also
 alleviates the problem of the front-to-back layers.

I would implement all extensions, but leave A and B as "optional" for the
user to turn off as desired.

--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
She ran by screaming "No, I run by moving my feet rapidly, you idiot!"

From cube-lovers-errors@mc.lcs.mit.edu  Mon Apr  5 14:06:24 1999
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Date: Sat, 03 Apr 1999 21:19:54 -0800
From: karen loewen <loewens@okanagan.net>
Reply-To: loewens@okanagan.net
To: cube-Lovers@ai.mit.edu
Subject: square-1 and megeminx

Is there a world record for the the 4x4x4 and the 5x5x5 rubiks cubes?
If so what is it?

I was wondering what in your opinion is the hardest rubiks like cube
available right now.

On a web page several people from rated the greatest to least in
difficulty:

1)square-1
2)megaminx
3)5x5x5
4)4x4x4
5)?
6)?

Anyway it went something like that. Would you agree? I learnt how to
do square-1 all by myself.

Would you consider Square-1 one of the hardest cubes to do? If so would
this be considered good of me to learn? Although I only tried about 4
times my record when it was in a cube is 2 minutes. Do you know of a
world record for Square-1 or any good websites? Do you know what an
average would be for the best time? (perhaps 90 seconds for those
really into it?)

I was wondering how this compares to others times and acomplishments. At
http://byrden.com/puzzles/ I almost figured out the megaminx. I got
everything except the last row. Is the Megaminx one of the hardest
Rubiks like puzzles available? After I get bored of my 5x5x5 that I
orderd thinking about buying the megaminx. Do you know of a world record
on the megaminx? What would be a average time (perhaps 3-4 minutes?)

If you are not sure of answers thats fine. Then do you know of any good
websites I can go to?

Thank you very much.

From cube-lovers-errors@mc.lcs.mit.edu  Mon Apr  5 14:57:39 1999
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Message-Id: <3707EF11.3F51A90@t-online.de>
Date: Mon, 05 Apr 1999 01:00:33 +0200
Reply-To: Rainer.adS.BERA_GmbH@t-online.de
Organization: BERA Softwaretechnik GmbH
To: Andrew John Walker <ajw01@uow.edu.au>
Cc: cube-lovers@ai.mit.edu
Subject: Re: Mike Reid's Solver
References: <199904010348.NAA17158@wumpus.its.uow.edu.au>
From: Rainer.adS.BERA_GmbH@t-online.de (Rainer aus dem Spring)

I have-the version that requires 80MB, NOT the version that requires 800MB :)

Rainer adS

Andrew John Walker wrote:
>
> If anyone has compiled a PC version of Mike Reid's program
> set to find optimal solutions in the face turn metric please
> contact me (or better still the list).
>
> Andrew Walker

-- 
--------------------------------------------------------------------------
Rainer aus dem Spring       email  Rainer.adS.BERA_GmbH@t-online.de (home)
Schimmelbuschstr. 10        email  TEEADS@TEE.toshiba.de (business only)
40699 Erkrath               tel. +49 (0)02104 35157  (private)
Germany                     tel. +49 (0)02104 936150 (business)
---------------------------------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Mon Apr  5 21:30:54 1999
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Message-Id: <3707EE0F.73ED160F@ibm.net>
Date: Sun, 04 Apr 1999 15:56:15 -0700
From: "Jin 'Time Traveler' Kim" <chrono@ibm.net>
Reply-To: chrono@ibm.net
To: cube-lovers@ai.mit.edu
Subject: Re: Spaceball (tm) input to a Cube simulator
References: <Pine.BSF.3.96.990401140302.3465Q-100000@shell2.tiac.net>

If anybody wants a SpaceOrb 360 (for experimentation please, I generally
don't just give away stuff out of the kindness of my heart, although
that's not unknown either) just drop me an email with a mailing
address.  I still have the box around here somewhere, just don't
remember where.  If I dig around long enough I know I can find the
instructions too.  I don't have it in front of me, but I'm sure it's
just a serial device like a mouse.



Nicholas Bodley wrote:
>
> If anyone has a Spaceball, it should make quite a nice input device.
> It senses both torque about all 3 orthog. axes, and linear
> displacement forces ditto; a total of 6 channels. Displacement could
> select a layer, which could be highlighted (anyone for alpha-channel
> translucency?), and torque would rotate the selected layer.

--
Jin "Time Traveler" Kim
chrono@ibm.net
http://www.chrono.org
'95 PGT - SCPOC

From cube-lovers-errors@mc.lcs.mit.edu  Mon Apr  5 22:59:11 1999
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Reply-To: <jburkhardt@mediaone.net>
From: "John Burkhardt" <jburkhardt@mediaone.net>
To: "Cube-Lovers (E-mail)" <cube-lovers@ai.mit.edu>
Subject: Different versions of the Ideal cube
Date: Mon, 5 Apr 1999 08:23:22 -0400
Message-Id: <000e01be7f5f$19623600$3cca8018@octopod.ne.mediaone.net>

Recently I've acquired a couple of older cubes from ebay.  One was sold as
"the original rubik's cube" by Ideal.  When I opened it up I noticed that
the quality was not the same as my original cube that I bought in 1980 or
so.  The stickers were not as bright for one thing.  On my "original" cube
the orange stickers are almost fluorescent, whereas on this version the
color is dull, and its not a faded version of the original color.  Its
clearly a different shade of orange.  The yellow and green colors are not as
bright.

And then there is the center white sticker.  On some older Ideal cubes there
is a logo on it, on this one I just got the logo was a separate decal.  On
my 1980's cube there is no logo at all.

Next we come to differences in the actual pieces.  On some Ideal cubes the
cubies are solid, well, probably hollow in the center, but there is no
openining.  But on others there is an opening.

Does anyone know anything about the history of the Ideal cube?  Were there
different runs of production?  Did they start "cutting corners" to lower
cost?  Is there any way to tell from the packaging, which cube is the
version that has the nice bright colors?

Sorry if this has been discussed in the past, I'm relatively new to this
list...

[ Moderator: The archives have quite a bit about various types of
  cubes.  Are you sure the open-corner cubes were from Ideal?  I
  thought those were the pirated knockoffs. ]

-JRB

From cube-lovers-errors@mc.lcs.mit.edu  Tue Apr  6 14:31:56 1999
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From: wheeler@cipr.rpi.edu (Frederick W. Wheeler)
Message-Id: <14088.45800.718995.311244@cipr.no_spam.rpi.edu>
Date: Mon, 5 Apr 1999 08:56:08 -0400 (EDT)
To: Cube-Lovers@ai.mit.edu
Subject: Re: Inventing your own techniques
In-Reply-To: <14080.1711.289304.134028@cipr.no_spam.rpi.edu>
References: <14080.1711.289304.134028@cipr.no_spam.rpi.edu>

Dear Cube-Lovers list:

I received several very interesting replies to my e-mail last week
regarding inventing techniques to solving cube puzzles.  Here are some
excerpts of note that were e-mailed to me but not the list.

First, part of what I wrote ...

Fred Wheeler writes:
> For me, the most fun, and the ultimate challenge, in cubing comes
> from figuring out how to solve the puzzle in the first place.  I'd
> really like to hear from people on this list on how you go about
> inventing new moves and techniques or how you feel about learning to
> solve a puzzle on your own.

Wei-Hwa Huang sent me this teaser about conjugation.

whuang@ugcs.caltech.edu (Wei-Hwa Huang) wrote:
> After I understood conjugation well enough, I have never invented a
> move that I can in all honesty call "new" -- although they may
> appear "new" to others.  The only new part is just applying it to
> different types of moves and seeing what the result is.

Later, at my request, Wei-Hwa Huang was kind enough to elaborate on
conjugation.

whuang@ugcs.caltech.edu (Wei-Hwa Huang) wrote:
> I keep on meaning to write a more detailed explanation but can never
> seem to find the time.
>
> Essentially, by conjugation I mean taking two routines (call 'em A
> and B), consider their reverses (a and b), and juxtapose them (do
> the move ABab).  When A and B have a small intersection the results
> of the conjugation is a simple permutation.  And pretty much more
> cube puzzles can be solved if you have the simplest permutations.
>
> Eg, to rotate two corner pieces, let
>  A = R'DRFDF'  (rotate one corner in the top face without affecting
>     the rest of the top face)
>  B = U (rotate the top face)
>
> As A and B have a small intersection (one corner cubie), the move
>  ABA'B' is quite useful.
>
> Note that A is itself a move arrived at by conjugation.

Tom Magliery also has a system for discovering solution techniques.

Tom Magliery <mag@ncsa.uiuc.edu> wrote:
> rather than telling you my actual operation for fixing the "switched wings"
> problem on the 4x and 5x cubes, i'll tell you how i discovered it:  one of
> the things i experiment with is repeated applications of short(ish)
> sequences of moves.  for example, i'll just take a particular 2 or 3-move
> sequence, and apply it over and over again until the cube arrives back at a
> state very similar to (but hopefully slightly different than) where it was
> when i started.  i was doing this (starting from a solved cube) one day
> when i discovered with much jubilation that i had arrived at a state with
> one switched wing pair.  (there was also another slight jumble, but i
> already knew an independent move to fix that by itself.)

I also received a good suggestion that discovering new sequences may
be easier on a computer simulator.  This, way the cube can be reset to
the solved state quickly before each new attempt to make it easier to
see what the trial sequence actually changes.  Unfortunately, I lost
this particular e-mail and forget who sent it, so I cannot attribute
it.  My apologies to the author.

Regards,
Fred Wheeler

--
Fred Wheeler
wheeler@cipr.rpi.edu
www.cipr.rpi.edu/wheeler

From cube-lovers-errors@mc.lcs.mit.edu  Wed Apr  7 16:04:40 1999
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From: roger.broadie@iclweb.com (Roger Broadie)
To: "Cube Lovers (E-mail)" <Cube-Lovers@ai.mit.edu>
Subject: Re: Keyboard cube [was: Re: Wrist pains]
Date: Mon, 5 Apr 1999 21:57:13 +0100

For anyone who is interested in trying out a simulator that can be
operated in an instant-response keyboard mode, I have christened my
web-pages by uploading a copy of my cube simulator.  In this mode, the
function keys do face, centre-slice and whole-cube turns for the
3x3x3, and alphabetic keys can also be used for the same purpose.

L, U, F and R are assigned to F5 to F8, so it's reasonably easy to
work up a rhythm.  Squares and inverses are done with Shift and
Control, though as Jerry Bryan pointed out repeated jabbing is at
least as easy.  These keys also work for 4x4x4 and 5x5x5 cubes, with
the alphabetic keys b f u d l r used for the off-centre slices.

I mainly wrote the simulator to show me the effects of turns in
Singmaster notation and common variants such as are found in
cube-lovers.  It's in Qbasic, which is the limit of my knowledge, and
is a lot less sophisticated in appearance than the simulators
available on the web - no mouse operations and no nice visible turning
of the layers while you watch for instance, and distinctly rudimentary
interactions with external files - so I am not offering it to the
world in general, but it does have some features I haven't seen
elsewhere that may appeal to Cube-lovers readers.

Among them are that it can be set to show the effect of the turns on
the centre pieces: it shows their orientation as little clock-hands
that point to 12 o'clock in their home positions.  The orientation of
each piece's pointer then allows its identity to be deduced.  I don't
really feel I've understood the effects of a sequence unless I can see
what it does to the centres.

Also (this is a party trick, really) it can be set to work as a cube
with any number of pieces per edge from two to 15, using an extension
of normal Singmaster notation - no keyboard mode for these cubes.

The program can be downloaded from

    home.iclweb.com/icl1/roger.broadie

I won't necessarily keep it there all that long.

Roger Broadie

From cube-lovers-errors@mc.lcs.mit.edu  Wed Apr  7 16:33:55 1999
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From: "John Burkhardt" <jburkhardt@mediaone.net>
To: "'Cube-Lovers (E-mail)'" <cube-lovers@ai.mit.edu>
Subject: RE: Different versions of the Ideal cube
Date: Tue, 6 Apr 1999 06:29:31 -0400
Message-Id: <001001be8018$5c165350$3cca8018@octopod.ne.mediaone.net>

> [ Moderator: The archives have quite a bit about various types of
>   cubes.  Are you sure the open-corner cubes were from Ideal?  I
>   thought those were the pirated knockoffs. ]
>

Yes, I'm sure the open corner cubes I have are Ideal.  Unless the pirates
put them in Ideal boxes.  I own three Ideal cubes.  I also have a siamese
cube, which looks like it started life as regular Ideal cubes and those
corners are closed.  And the new OddzOn cubes are solid too.

-JRB

From cube-lovers-errors@mc.lcs.mit.edu  Wed Apr  7 17:25:37 1999
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Date: Sun, 04 Apr 1999 18:12:32 -0500
To: cube-lovers@ai.mit.edu
From: Tom Magliery <mag@ncsa.uiuc.edu>
Subject: Fwd: Keyboard cube [WAS Re: Wrist pains]

>From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
>First, represent the front face by
>
>q w e
>a s d
>z x c
>
>A "move" is represented by any two ordered keystrokes.  If the two keystrokes
>are in the same row, it represents a horizontal layer moving, else
>if they're in the same row, it represents a vertical layer moving.
                        ^^^
                        you meant to say "column" here

>
>Example:
> "ed" = R'
> "cd" = R
> "sd" = slice move
> "we" = U'
> "qe" = U'
> "qw" = U'

this is sort of a grab-and-move interface (in my mental picture, one
keypress "grabs" a face, and the next "moves" it).  i think i might like
this paradigm better than my own suggestion of before (unshifted single
keystrokes for face moves, shifted single keystrokes for cube rotations).
one nice advantage is that it makes it as easy on the keyboard interface as
it is in real life to do RL' and R'L, for which i suspect most people's
mental model is really a single turn of the middle slice.

i think i'd prefer some different key assignments than wei-hwa suggests,
though.  for example, i think i'd like to "grab" with the middle key in the
line, and then "move" with the outer one.  for example:
    "de" = R
    "dc" = R'
(according to wei-hwa's definition of "move", our two suggestions are
compatible.  mine only adds redundancy.)

>This method does not have moves for rotating the three front to back layers
>nor 180-degree moves.

another suggestion for 180-degree moves:  allow the user to tap the second
key in the sequence 2 (or 3) times for 180 (or 270 aka 90') degree moves.
thus, e.g., "wee" = U2.  this is intuitive not only to me:  someone else
posted that in their keyboard interface of years ago, they didn't use the
specially-assigned 180 keys, but in practice just tapped the 90-keys twice
anyway.

an additional suggested extension:  let me do things with both hands!
(this is my favorite aspect of my own keyboard interface, posted a few days
ago.)

perhaps a similar nonad(?) of keys could be used for R-face-relative
operations analogous to the F-face-relative ones above.  this is like
wei-hwa's extension C, except i'd rather use my right hand in its "normal"
typing position.

how about a third set, in the middle of the keyboard somewhere, for doing
D-relative things?

mag

-- 
///X  Tom Magliery, Research Programmer           217-333-3198  .---o
\\\   NCSA, 605 E. Springfield      O-       mag@ncsa.uiuc.edu  `-O-.
///   Champaign, IL 61820       http://sdg.ncsa.uiuc.edu/~mag/  o---'

From cube-lovers-errors@mc.lcs.mit.edu  Wed Apr  7 18:00:36 1999
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Date: Mon, 5 Apr 1999 00:49:14 -0400 (EDT)
From: der Mouse  <mouse@rodents.montreal.qc.ca>
Message-Id: <199904050449.AAA08526@Twig.Rodents.Montreal.QC.CA>
To: cube-lovers@ai.mit.edu
Subject: Re: Keyboard cube [WAS Re: Wrist pains]

> [Assume that the person is using a standard American QWERTY keyboard;
>  remap if necessary.]

Gotcha.  And your idea seems sound.  But...

> Extension D:  Use the arrow keys to rotate the entire cube.

Arrow keys are sufficiently nonstandard I would very much prefer
something else - say, a shifted slice move to rotate the whole cube.
(Under some reasonable circumstances it can be hard to even tell what
you'll see for "the arrow keys", if they even exist.)

In passing,

> Extension C:  Add another grid for the right side of the cube:
>  r t y
>  f g h
>  v b n

I'd probably prefer
   u i o
   j k l
   m , .

I might even argue in favor of w e r / s d f / x c v for the other set,
simply because they're the home keys for the three principal fingers of
the left hand.

Other than that, it sounds like an eminently reasonable approach.

					der Mouse

			       mouse@rodents.montreal.qc.ca
		     7D C8 61 52 5D E7 2D 39  4E F1 31 3E E8 B3 27 4B

From cube-lovers-errors@mc.lcs.mit.edu  Wed Apr  7 18:42:26 1999
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Date: Tue, 6 Apr 1999 16:58:47 -0700 (PDT)
From: Han Wen <hansker@yahoo.com>
Subject: Re: Keyboard cube [Layout for Speed Cubing]
To: noel@mud.ca
Cc: Cube Lovers <Cube-Lovers@ai.mit.edu>

Hi,

This is a very interesting problem.  I've been scratching my head for a
while, and I've cooked up with a keyboard combination that could work
for virtual speed cubing.  I tried to follow a few guidelines: speed
cubing friendly; ergonomic and intuitive.

There are of course, two types of rotations that are required to solve
the cube: body rotations (rotate the whole cube) and slice rotations
(rotate individual faces).  For each type of rotation there are three
additional sub-types.  For body rotations, there are three axes x, y
and z.  For slice rotations, there are 1/4-turn clockwise, 1/4-turn
counterclockwise and 1/2-turn.

Let's define the three body axes as follows: the z-axis will intersect
the U and D faces.  The x-axis will intersect the F and B faces. The
y-axis will intersect the R and L faces.

 Okay, now ergonomics.  When I solve the cube, I notice that my left
hand does most of the body rotations and my right hand does most of the
slice rotations.  I tried to keep this intuition in assigning keyboard
keys by making the left hand perform body rotations and the right hand
perform slice rotations (well, sort of...).  The most intuitive form
that I could think of for your hands to be in is the fetal or "clawed"
position.  This can be satisfied by placing the left-hand fingers on
the A, W, E and F keys and right-hand fingers on the J, I, O and ; keys
(assuming a QWERTY keyboard layout).  To be speed cubing friendly,
these 8 fingers should never have to leave these keyboard key
assignments.
__________________________________________________________________________________
For body rotations, the keyboard key assignments could be:
F - clockwise body rotation about the y-axis (i.e. if we were rotating
    a face it would be R)
E - clockwise body rotation about the x-axis (i.e. if we were rotating
    a face it would be F)
W - clockwise body rotation about the z-axis (i.e. if we were rotating
    a face it would be U)

To perform a counter-clockwise body rotation, hit the SPACEBAR key with
the right thumb in combination with the key assigment (e.g. F and
SPACEBAR)
__________________________________________________________________________________
For the slice rotations, the keyboard 2-key assignments could be:
F and J - 1/4-turn clockwise of the L-face
F and I - 1/4-turn clockwise of the R-face
F and O - 1/4-turn clockwise of R-face and 1/4-turn counterclockwise of
the L-face

E and J - 1/4-turn clockwise of the F-face
E and I - 1/4-turn clockwise of the B-face
E and O - 1/4-turn clockwise of the F-face and 1/4-turn
counterclockwise of the B-face

W and J - 1/4-turn clockwise of the U-face
W and I - 1/4-turn clockwise of the D-face
W and O - 1/4-turn clockwise of the U-face and 1/4-turn
counterclockwise of the D-face

To perform 1/4-turn counterclockwise slice rotations, hit the SPACEBAR
key with the right thumb in combination with the 2-key assignment (e.g.
F and J and SPACEBAR)

To perform 1/2-turn slice rotations, hit the N key with the right thumb
in combination with the 2-key assignment
(e.g. F and J and N)
__________________________________________________________________________________
I've ordered the keys in order of their importance.  As you can see the
J key is responsible for "LEFT-LIKE" slice rotations
I key is responsible for "RIGHT-LIKE" slice rotations
O key is responsible for middle slice rotations

You might ask, why make O the middle slice and not I, since it's the
middle key?  Well, when you actually solve the cube, you never move the
middle slice.  You only move the left and right slices to effectively
move the middle slice.  Furthermore, this operation is performed much
more seldomly than normal R and L slice rotations.

Anyways, you can see with this keyboard mapping you only need a total
of  8 keys to perform all of possible rotations you need to solve the
cube.

If we really want to make this speed cubing friendly we could also do
the following:
-body rotations should be performed in real-time; the speed of rotation
should be an adjustable constant that the user can tweak to his or her
preference
-slice rotations should be instantaneous.  There's really no point in
watching a slice rotate.  Just wasting time...

> "Noel Dillabough" <noel@mud.ca> writes:
> >Any ideas would be appreciated and I'll try my best
> to implement the best
> >one.  If simulators "felt" like a cube when you
> used them they would be more
> >fun (and move without jamming etc)
>

===
_________________________________________________________
Dr. Han Wen
Applied Materials
3100 Bowers Ave, MS 1158
Santa Clara, CA  95054
e-mail: Han_Wen@amat.com / hansker@yahoo.com

From cube-lovers-errors@mc.lcs.mit.edu  Wed Apr  7 21:38:40 1999
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From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
To: Cube Lovers <cube-lovers@ai.mit.edu>
Subject: Re : Re: Inventing your own techniques
In-Reply-To: <14088.45800.718995.311244@cipr.no_spam.rpi.edu>
Message-Id: <SIMEON.9904062037.C@pavilion.pstcc.cc.tn.us>
Date: Tue, 6 Apr 1999 20:49:37 -0400 (Eastern Daylight Time)

On Mon, 05 Apr 1999 08:56:08 -0400 (EDT) "Frederick W. Wheeler"
<wheeler@cipr.rpi.edu> wrote:


> Wei-Hwa Huang sent me this teaser about conjugation.
>
> whuang@ugcs.caltech.edu (Wei-Hwa Huang) wrote:
> > After I understood conjugation well enough, I have never invented a
> > move that I can in all honesty call "new" -- although they may
> > appear "new" to others.  The only new part is just applying it to
> > different types of moves and seeing what the result is.
>
> Later, at my request, Wei-Hwa Huang was kind enough to elaborate on
> conjugation.
>
> whuang@ugcs.caltech.edu (Wei-Hwa Huang) wrote:
> > I keep on meaning to write a more detailed explanation but can never
> > seem to find the time.
> >
> > Essentially, by conjugation I mean taking two routines (call 'em A
> > and B), consider their reverses (a and b), and juxtapose them (do
> > the move ABab).  When A and B have a small intersection the results
> > of the conjugation is a simple permutation.  And pretty much more
> > cube puzzles can be solved if you have the simplest permutations.
> >
> > Eg, to rotate two corner pieces, let
> >  A = R'DRFDF'  (rotate one corner in the top face without affecting
> >     the rest of the top face)
> >  B = U (rotate the top face)
> >
> > As A and B have a small intersection (one corner cubie), the move
> >  ABA'B' is quite useful.
> >
> > Note that A is itself a move arrived at by conjugation.

There are two separate ideas here.

A process of the form XYX'Y' is called a commutator rather than a conjugate.
As you say, a commutator which moves very few cubies can be a very useful
process.  In fact, the number of cubies moved by XYX'Y' can be used as a sort
of informal measure of how close X and Y come to commuting. In the extreme case
where X and Y do commute, we have XYX'Y'=YXX'Y'=YY'=I so that no cubies are
moved. And conversely, two processes X and Y which "nearly" commute and/or
which intersect in very few cubies are good candidates for forming a useful
commutator.

A process of the form Y'XY is called a conjugate, and in particular is called
the conjugate of X by Y.  Note that YXY' is also a conjugate, and in particular
is called the conjugate of X by Y'.  This can be a little confusing because a
few books (incorrectly in my opinion) call YXY' the conjugate of X by Y.

Of Y and Y', which is the "real" process and which is the inverse is totally
arbitrary.  For example, if Z=Y', then Z'=Y.  So we could write a conjugate as
YXZ (the conjugate of X by Z) and another conjugate as ZXY (the conjugate of X
by Y) if we know that Y and Z are respectively the inverses of each other.

It is sometimes said that the conjugate Y'XY results in X shifted by Y, which
is the real utility of using conjugates to solve a cube.  Use a process you
know, but shift it to apply to a slightly different set of cubies.

Your process A=R'DRFDF' consists of the conjugates R'DR (the conjugate of D by
R) and FDF' (the conjugate of D by F').  It is often the case that useful
processes can be formed from both commutators and conjugates.

I am perversely proud that my own personal solution technique for solving the
cube consists of only two processes  --  one for the corners and one for the
edges  --  plus conjugates of those two processes.  I think it is indicative
of the power of conjugates that a cube can be solved with so few processes
provided only that they are combined with conjugation.  I am "perversely proud"
because my "two processes" technique probably yields one of the slowest
solution times of anybody on Cube-Lovers. I am always embarrassed to read about
those people who can do it in under 30 seconds.  I have taught myself some of
the faster techniques, but I always find that after a few months the only
technique my hands can remember is the old, slow one which I invented myself
many years ago.

----------------------------------------
Jerry Bryan
jbryan@pstcc.cc.tn.us

From cube-lovers-errors@mc.lcs.mit.edu  Thu Apr  8 11:48:13 1999
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From: WaVeReBeL@webtv.net
Date: Wed, 7 Apr 1999 09:02:51 -0700 (PDT)
To: Cube-Lovers@ai.mit.edu
Subject: Re: Keyboard cube [WAS Re: Wrist pains]
Message-Id: <28684-370B81AB-994@mailtod-123.bryant.webtv.net>
In-Reply-To: whuang@ugcs.caltech.edu (Wei-Hwa Huang)'s message of 3 Apr
	1999 00:22:20 GMT

How about giving the player a choice between using a default
configuration, or customizing his/her own buttons, like in certain home
video games such as Street Fighter?

Another cool feature would be two simultaneous cubes, so players can
battle head to head, also like Street Fighter.  (I really like Street
Fighter)

-Alex Montilla-

From cube-lovers-errors@mc.lcs.mit.edu  Thu Apr  8 12:10:06 1999
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Reply-To: <noel@mud.ca>
From: "Noel Dillabough" <noel@mud.ca>
To: <cube-lovers@ai.mit.edu>
Subject: SpaceOrb 360 and Simulators
Date: Wed, 7 Apr 1999 13:16:14 -0400
Message-Id: <000001be811a$572c33c0$020a0a0a@pprodual>

I added support for the SpaceOrb 360 by SpaceTec to Puzzler, grab this
version at:

http://www.mud.ca/puzzler/Puzzler.EXE

You must have a SpaceOrb 360 to try it out, but it is very very nice to use
compared to the mouse.  Twist the orb to rotate the puzzle, and press the
buttons and twist to do moves.  After a little practice you'll be able to
solve a cube like it was in your hands!

Only the cubes (all sizes) have support for the SpaceOrb, I may add support
for the other puzzles at a later date.

-Noel Dillabough

From cube-lovers-errors@mc.lcs.mit.edu  Thu Apr  8 13:14:00 1999
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From: Andrew John Walker <ajw01@uow.edu.au>
Message-Id: <199904080113.LAA02581@wumpus.its.uow.edu.au>
Subject: Solvers
To: cube-lovers@ai.mit.edu
Date: Thu, 8 Apr 1999 11:13:10 +1000 (EST)

Regarding Mike Reid's program, I compiled it fine with DJGPP, an MSDOS compiler,
but it didn't run. Any ideas?
I'll probably get a windows compiler soon.
Anyway, two other points I was thinking of recently.

Firstly, do any of the kociemba algorithm search programs use the fact
that you can perform a depth n search by 3 depth n-1 searches using the
3 orthogonal orientations? (if my logic is correct!)
This is because if you are using the <U,D,L2,R2,B2,F2> group for the final
phase, the last move of any depth n sequence must end in a square move,
in which case the n-1 will easily find it, or else a quarter turn in which
case the three orientations are required to make it found in the second
phase. Unfortuneately I doubt the n-1 searches could be replaced by n-2
searches.

Also when a cube is being scrambled adjacent cubes tend to stay
together for a while. Has this been of any use in search methods?
(eg. to help prune the search tree). Obvously a sequence like
F2 B2 U2 D2 L2 R2 separates all adjoining pairs, but there is still a
high degree of order with next to adjacent cubes, so maybe they could be
used as well.

	Andrew Walker

From cube-lovers-errors@mc.lcs.mit.edu  Thu Apr  8 20:23:30 1999
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Date: Thu, 8 Apr 99 19:45:22 EDT
Message-Id: <9904082345.AA16179@aic.nrl.navy.mil>
From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: Jerry Bryan <jbryan@pstcc.cc.tn.us>, Cube Lovers <cube-lovers@ai.mit.edu>
Subject: Conjugation done right [Re: Inventing your own techniques]

First, let's make sure everyone remembers that we're using  X' as an
abbreviation for  X^(-1)  for inverses of permutations.  People really
should read the archives, at <ftp://ftp.ai.mit.edu/pub/cube-lovers>
so they know this sort of thing, but that's getting to be a lot to
ask.  Still, remember that address, because it's a good place to go
for things you forgot about the list (in fact, it would be nice if the
README mentioned that cube-lovers-request@ai.mit.edu is the e-mail
address for administrative requests to the list management, just in
case someone loses their greeting message.)

Jerry Bryan wrote:

> A process of the form Y'XY is called a conjugate, and in particular is called
> the conjugate of X by Y.  Note that YXY' is also a conjugate, and in particular
> is called the conjugate of X by Y'.  This can be a little confusing because a
> few books (incorrectly in my opinion) call YXY' the conjugate of X by Y.

I tried to explain this a while ago, but it's such a subtle,
counterintuitive point that I had better try again.  One form of
conjugate is correct, and the other form is incorrect, but just which
is correct depends on how you write function composition.  The point is
that there are two schools of function composition, "leftward" and
"rightward", and the choice of your function composition determines
how you define conjugates.  It's surprising that a notational
convention can have this sort of effect, but we'll see it does.

First, I'll describe the two schools of composition.  It will be
convenient to consider a set  X  and two permutations  f  and  g  on
X.  Let  h:X->X  be the unique permutation that satisfies  h(x) = f(g(x))
for all  x  in  X.  We could let  f, g,  and  h  be any functions, not
just permutations, but we will need for them to be permutations later,
when we use the group structure.

How do we write  h  in terms of  f  and  g?  The rightward school says
h = g f.  This is the way we have been writing things on cube-lovers
all along: we write  g f  for applying a permutation  g  to something
and then applying  f  to the result.  But remember that we write
h(x) = f(g(x)),  which is to say that  (g f)(x) = f(g(x)).  The fact
that the order of  f  and  g  depends on the parenthesization is often
considered ugly, so some seriously rightward people write the function
name after the arguments: That is to say,  they write  (x)f  instead
of  f(x),  (x)g  instead of  g(x),  and  (x)h = ((x)g)f = (x)(g f).
This makes function composition a kind of associative law.  If you're
seeing this for the first time, I'm sure you consider it a bizarre and
useless and gratuitously confusing complication, but I assure you that
rightward functions are in wide use in some branches of the
mathematical community, chiefly in abstract algebra.  But cube-lovers
was started by computer programmers, not algebraists, and programmers
have  f(x)  very tightly wired into their minds and parsers.  So
cube-lovers uses rightward composition with leftward functions, and we
say  (g f)(x) = f(g(x)).  As for swapping the order of  f  and  g,  we
just get over it, but there are some people out there who will call us
disfunctional.

The leftward composition school takes a different approach: they say
h(x)=f(g(x))  means  h = f g.  When you follow a cube process written
by these people, you have to perform it from the right to the left.
This is also a little hard to get used to, but at least we have an
"associative" rule,  (f g)(x) = f(g(x)),  with  f  and  g  in the same
order, without having to write our functions after the arguments.  For
this reason, most mathematicians other than algebraists find leftward
composition to be more natural.  You probably learned leftward
composition in calculus or whenever.  But on cube-lovers no-one wanted
to write all their cube processes from right to left, so we've pretty
much forgotten about leftward composition on the list.  Remember,
though, leftward composition is pretty standard for a lot of
mathematics, and it works better for the way we write functions, so
you can't really call it wrong.  And there are people who say that if
we are going to write our functions to the left we also ought to
compose them to the left.

So far so good.  The rightward and leftward schools write the
composition of functions in opposite orders, but either way the
permutations still form a group under composition.  As long as you
don't mix them, it shouldn't change anything else, should it?  But it
really does change the definition of conjugation.  (Remember
conjugation?  This is a message about conjugation.)

Suppose we have a group  G,  not necessarily a permutation group.
Conjugation is one way of mapping  G  to a permutation group, where
the set being permuted is the set of group elements of G.  For an
element  s,  I'll define the right conjugate of  s,  R_s,  as the
permutation for which  R_s(g) = s' g s  for all  g  in  G.  Similarly,
the left conjugate of  s,  L_s  is defined by  L_s(g) = s g s'  for
all  g.  It's important to notice that in either case, conjugation by
a product is the composition of conjugations.  For letting  s  and  t
be two specific elements of  G,  we can carry out manipulations that
hold for all elements  g  of G.  I'll calculate with the left
conjugate in the left column and the right conjugate in the right
column:

    L_s(g) = s g s';                R_s(g) = s' g s;

    L_t(g) = t g t';                R_t(g) = t' g t;

    L_st(g) = (st) g (st)'          R_st(g) = (st)' g (st)
            = s t g t' s'                   = t' s' g s t
            = L_s(t g t')                   = R_t(s' g s)
            = L_s(L_t(g))                   = R_t(R_s(g))    (*)


These calculations were carried out using the group operation of  G,
independently of how we write function composition.  But let's look at
how we write the composition in our two notations.

In the rightward composition that cube-lovers has been using all
along, (*) shows that  L_st = L_t L_s  and  R_st = R_s R_t.  So the
mapping  s |-> L_s  is an _antihomomorphism_--it reverses the order of
multiplication--but  s |-> R_s  is a homomorphism.  Homomorphisms are
a lot nicer than antihomomorphisms, so we should use right conjugation
all the time, right?

But consider the people who use leftward composition, (f g)(x)=f(g(x)).
So the function composition in (*) is now written  L_st = L_s L_t  and
R_st = R_t R_s.  So with leftward composition, _left_ conjugation is
the homomorphism, and _right_ conjugation is the antihomomorphism.

It is so very convenient for conjugation to be a homomorphism that
people who use rightward composition always use right conjugation, and
people who use leftward composition always use left conjugation (unless
they think it doesn't matter and guess wrong).  We're rightward
composers on cube-lovers, so conjugation by  s  is  g |-> s' t s,  but
remember that most math texts (other than algebra) will use the
leftward composition, and so they will correctly use left conjugation,
g |-> s g s'.

I learned this from Jim Saxe, when I tried using left conjugation in
the Symmetry and Local Maxima message.  Jim told me that unless I
wanted to start using leftward composition I had better use right
conjugation, but I was pretty sure it really didn't matter.  Jim just
splained and splained until he got across how much it really does
matter, and why the only right answer is different in different books.
Now I've done it for you, and I hope it helps.

And they said that consistency was the hobgoblin of little minds....

Dan
Hoey@AIC.NRL.Navy.Mil

From cube-lovers-errors@mc.lcs.mit.edu  Fri Apr  9 13:06:46 1999
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From: "Jorge E. Jaramillo" <jejarami@usa.net>
To: "cube" <Cube-Lovers@ai.mit.edu>, <Roger.Broadie@iclweb.com>
Subject: Cube program by Roger Broadie
Date: Thu, 8 Apr 1999 22:24:23 -0500

I'd like to say that the cube program from Mr. Broadie is quite nice,
actually is something I have been searching for for a while and something I
requested to several Web based cube programers and never found, the ability
to have a sequence interfase is something very useful mostly when you are
looking for patterns and you find these lists on the web that don't show the
actual result but just describe it with some sort of complicated notation,
trying to do it with a real cube is a real pain since you have to do all the
moves and if you make a mistake you have to start all over again by solving
the cube and then making the moves that would take you to the final pattern.
Well done Mr. Broadie and keep up the good work, please let the list know
when you have the Windows version  ready. ;-)
=======

Jorge E. Jaramillo
jejarami@usa.net
Cut the chain and chase the dream
Savatage 1984

From cube-lovers-errors@mc.lcs.mit.edu  Fri Apr  9 13:52:13 1999
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From: WaVeReBeL@webtv.net
Date: Sat, 27 Mar 1999 00:03:32 -0800 (PST)
To: Cube-Lovers@ai.mit.edu
Subject: Lubricant
Message-Id: <10861-36FC90D4-6028@mailtod-121.bryant.webtv.net>

Many people have suggested different lubricants on their web pages.
Many have referred to silicon.  Where can I get this?  What other ones
are good?  Where can I find them?

-Alex Montilla-

[Moderator's note: I skipped sending this message two weeks ago
 because discussing cube lubrication with someone complaining of RSI
 symptoms seemed like a bad idea.  Then I got a large pile of
 interface-related messages.  But now that the backlog is mostly sent
 out, we might want to return to considering what lubricants to use.

 The lubes I remember from the archives are vaseline (greasy, erodes
 plastic slowly), soap (temporary), candle wax (drips flakes), and
 molybdenum disulfide (doesn't help, destroys cube quickly).  The only
 silicone product I've used is Armor-All, which is a protectant that
 provides a very little lubrication.  Does anyone know what silicone
 Alex is discussing? Any other good ideas on lubricants? -- Dan ]

From cube-lovers-errors@mc.lcs.mit.edu  Fri Apr  9 17:58:59 1999
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Message-Id: <370E49B3.E058A9D8@u.washington.edu>
Date: Fri, 09 Apr 1999 18:40:51 +0000
From: David Barr <davidb@u.washington.edu>
To: cube-lovers@ai.mit.edu
Subject: Re: Lubricant
References: <10861-36FC90D4-6028@mailtod-121.bryant.webtv.net>

>  [ ... Does anyone know what silicone
>  Alex is discussing? Any other good ideas on lubricants? -- Dan ]

I think most auto parts stores sell a silicone spray lubricant.  I don't
know what the brand names are.  I have used this on my cubes and it
works well.  After you first apply it, the cube will seem stiff, but if
you keep turning it for a while it starts working.

David

From cube-lovers-errors@mc.lcs.mit.edu  Fri Apr  9 19:43:01 1999
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From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
To: "Jorge E. Jaramillo" <jejarami@usa.net>
Cc: cube <Cube-Lovers@ai.mit.edu>, Roger.Broadie@iclweb.com
Subject: Cube Explorer 1.5 by Herbert Kociemba
In-Reply-To: <000701be8238$8c026820$92121fc8@lucentmd>
Message-Id: <SIMEON.9904091829.A@pavilion.pstcc.cc.tn.us>
Date: Fri, 9 Apr 1999 18:04:29 -0400 (Eastern Daylight Time)

On Thu, 08 Apr 1999 22:24:23 -0500 "Jorge E. Jaramillo" <jejarami@usa.net>
wrote:

> Well done Mr. Broadie and keep up the good work, please let the list know
> when you have the Windows version  ready. ;-)

Another alternative for Windows is Herbert Kociemba's Cube Explorer 1.5, which
is easily found with Web based search engines, and which I think can also be
found on the download section of the Cube-Lovers site.

Cube Explorer 1.5 does not allow you to type in a maneuver directly, which I
think is what you are after. But it will do something just about as good, and
in some ways even better. It will read in a standard ASCII text file which
contains maneuvers in standard Singmaster notation (the BFUDLR notation), and
it will then show you the end product.  The file can contain any (reasonable)
number of maneuvers, and the program will show you the end product for each
maneuver all in one fell swoop (or swell foop, if you prefer).

One thing Cube Explorer 1.5 does not support that you may be looking for is to
manipulate the cube and look at each intermediate position.  But that's usually
not what I am looking for.  I am usually looking for the end product of a
maneuver.  (Of course, you could put each intermediate position into your ASCII
text file.)

You can create the ASCII text file for Cube Explorer 1.5 by typing maneuvers
into the file with the text editor of your choice.  But most typically I just
cut and paste maneuvers out of an E-mail into the text file, and then have Cube
Explorer 1.5 read the file.  It's much easier and less error prone than typing
the maneuvers myself.

Here is a case where Cube Explorer 1.5 may be "even better".  Take a
Cube-Lovers message with lots of maneuvers, cut and paste the whole thing into
an ASCII text file, and read the file into Cube Explorer 1.5.  Instantly, you
see the positions for all the maneuvers.  You don't even have to worry about
deleting the extraneous non-maneuver text from the E-mail.  For example, take
Mike Reid's lists of minimal maneuvers for highly symmetric positions, or take
my lists of local maxima and put them into Cube Explorer 1.5 in this fashion.
The results are quite pretty.  There are many other messages in the Cube-Lovers
archives with lots of pretty patterns, but these two come to mind quickly.

(By the way, converting BFUDLR strings to a graphical representation of a
position is an extremely useful feature of Cube Explorer 1.5, but it has
several other nice features. For example, you can give it a graphical
representation of a position and get back very quickly --  in a matter of
seconds -- a very good suboptimal maneuver for that position.)

----------------------------------------
Jerry Bryan
jbryan@pstcc.cc.tn.us

From cube-lovers-errors@mc.lcs.mit.edu  Mon Apr 12 12:16:29 1999
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Date: Fri, 9 Apr 1999 19:06:08 -0400 (EDT)
From: der Mouse  <mouse@rodents.montreal.qc.ca>
Message-Id: <199904092306.TAA25503@Twig.Rodents.Montreal.QC.CA>
To: Cube-Lovers@ai.mit.edu
Subject: Re: Lubricant

> Many people have suggested different lubricants on their web pages.
> Many have referred to silicon.

Silicon in its (relatively) pure form is somewhat difficult to get hold
of and probably not a very good lubricant.  You probably mean silicone.

> [Moderator's note: [...] Any other good ideas on lubricants? -- Dan ]

I don't know how well it works - I haven't perceived a need for
lubricants for my Cubes - but one I wouldn't write off without trying
is graphite, available easily and cheaply as pencil lead, the softer
the pencil the better.  The major downside I would expect is black dust
rubbing off on fingers and to a lesser extent everything you set the
Cube down on.  (But it's a kind of black dust that washes off easily.)

					der Mouse

			       mouse@rodents.montreal.qc.ca
		     7D C8 61 52 5D E7 2D 39  4E F1 31 3E E8 B3 27 4B

[Moderator's note: John Bailey also advises graphite, applied carefully and
 sparingly.]

From cube-lovers-errors@mc.lcs.mit.edu  Mon Apr 12 14:10:19 1999
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Date: Sat, 10 Apr 1999 00:38:26 -0400 (EDT)
From: Nicholas Bodley <nbodley@tiac.net>
To: WaVeReBeL@webtv.net
Cc: Cube-Lovers@ai.mit.edu
Subject: Re: Lubricant
In-Reply-To: <10861-36FC90D4-6028@mailtod-121.bryant.webtv.net>
Message-Id: <Pine.BSF.3.96.990410002626.16640C-100000@shell2.tiac.net>


On Sat, 27 Mar 1999 WaVeReBeL@webtv.net wrote:

}Many people have suggested different lubricants on their web pages.
}Many have referred to silicon.  Where can I get this?  What other ones
}are good?  Where can I find them?

White powder, used like graphite, sometimes for locks? Possibly
Tri-Flow (tm), which (iirc) is a Teflon (tm) particle suspension.

I experimented with quite a variety of lubricants, found a good one,
and then one of the periodic disruptions of my life came along and I
failed to note which it was! :(

"Teflon" is a Du Pont trade name for PTFE, polytetrafluoroethylene.

Btw, in case you do a Web search, watch "silicon" vs. "silicone".
The chemist who coined the term "silicone" was wildly optimistic
about people's ability to keep the two straight. Silicon, the
chemical element, is gray, opaque to visible light, brittle like
glass, and a lousy lubricant. You don't normally see it, nor can you
normally buy it.

Silicones are chemical compounds that include the chemical element,
silicon.

|*  Nicholas Bodley   *|*  Autodidact & Polymath * Electronic Tech. (ret.)
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*  Silicon oil and grease do not exist.
|*  Amateur musician  *|*  IC chips made of silicone do not exist.
--------------------------------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Mon Apr 12 17:15:33 1999
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Date: Sat, 10 Apr 1999 02:39:35 -0400 (EDT)
From: der Mouse  <mouse@rodents.montreal.qc.ca>
Message-Id: <199904100639.CAA26973@Twig.Rodents.Montreal.QC.CA>
To: Cube-Lovers@ai.mit.edu
Subject: Cube-manipulation programs

Well, with the current discussion going on about Cube manipulation
programs, I'll toss out my minor contribution the genre.

It's a C program that takes simple text lines describing maneuvers and
prints out a text representation of the resulting cube, together with
its cycle structure.  (I specifically chose to make it text-only
because I'm comparatively often on text-only links.)  It also allows
you to define names for operations and then use them as primitives.

A sample transcript, which defines the Spratt wrench as an operator and
then uses it to flip the four F-face edges:

> .set WRENCH (SLICER F)4
`WRENCH' defined
> WRENCH F2 WRENCH F2
Cube:
              u u u
              u u u
              u f u
        l l l f u f r r r b b b
        l l f l f r f r r b b b
        l l l f d f r r r b b b
              d f d
              d d d
              d d d
Cycles: (uf)+ (lf)+ (fr)+ (fd)+ [2]
Already centered
>

Another example:

> F SLICER F'
Cube:
              u f u
              u f u
              u u u
        l l l f f f r r r b l b
        l l f r d d b r r b u b
        l l l f f f r r r b u b
              d d d
              d b d
              d b d
Cycles: (u,b,d,f) (ub,bd,rf,fl) [4]
Centred: (ul,fl,fu,df,bd,rd,rb,ru,rf,dl,bl) (ulb,flu,dlf,bld) (ubr,fur,dfr,bdr) [44]
>

The numbers in [ ] are the smallest power to which that operator must
be raised to get the identity.

While I don't expect it ever attain the popularity of slick graphics
programs, there may be a few people interested in it; if anyone is, you
have only to drop me a line asking for a copy....

					der Mouse

			       mouse@rodents.montreal.qc.ca
		     7D C8 61 52 5D E7 2D 39  4E F1 31 3E E8 B3 27 4B

From cube-lovers-errors@mc.lcs.mit.edu  Mon Apr 12 21:07:29 1999
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Reply-To: <noel@mud.ca>
From: "Noel Dillabough" <noel@mud.ca>
To: "Cube Lovers List (E-mail)" <cube-lovers@ai.mit.edu>
Subject: Cube moves macros in simulators
Date: Sat, 10 Apr 1999 10:17:49 -0400
Message-Id: <000a01be8362$9646ab40$040a0a0a@laptop>
In-Reply-To: <SIMEON.9904091829.A@pavilion.pstcc.cc.tn.us>

It seems my help file is poorly written, I've gotten some surprise replies
from people who I have told...

On Friday, April 09, 1999 6:04 PM Jerry Bryan [mailto:jbryan@pstcc.cc.tn.us]
wrote:

>Cube Explorer 1.5 does not allow you to type in a maneuver directly, which I
>think is what you are after.

Puzzler does support cube move macros, in standard UDFBLR notation (use
udfblr in lowercase for inner slices and meM for central slices in the 5x5x5
cube).  To use this feature hit enter and enter the move...F2 will move the
F slice 2 times, F3 and F' are equivalent and will move counterclockwise 1
time, F1 and F move clockwise 1 time.  You can enter any number of moves in
sequence in a string then by hitting enter again the move will be executed.

One more note: these macros assume the cube is oriented so that you can see
three faces (exactly like the initial orientation of the cube).  So F refers
the the bottom left face, R is the bottom right face and U is the upper
face.

  /\
 /  \
|\U /|
| \/ |
\F|R/
 \|/

Last note: I assumed that UDFBLR was the notation used by most people.
However if there is another notation that is used or if some other notation
is required for experimentation, let me know and I'll put it in.

-Noel

From cube-lovers-errors@mc.lcs.mit.edu  Tue Apr 13 12:28:58 1999
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Message-Id: <37134515.4AD5@zeta.org.au>
Date: Tue, 13 Apr 1999 23:22:29 +1000
From: Wayne Johnson <sausage@zeta.org.au>
Reply-To: sausage@zeta.org.au
To: Cube-Lovers@ai.mit.edu
Subject: Building a twisty puzzle

I have been in the process of building plans for a twisty puzzle that is
no longer available.

I have been re-doing the design for a few weeks now and I want to start
molding or crafting the pieces soon. This is where I am coming undone.

I believe I will have some success crafting the three different pieces
required using clay, and filing back the pieces once dry.

>From here, I believe making plaster cast molds of the clay pieces will
allow me to start making all the identical pieces.

Now, the problem... what can I use to fill the mold? Can I use plastics
or a type of resin?

I would greatly appreciate any ideas or input. I'll be posting the
design on my site as soon as they are nearing completion. Maybe I can
get comments on whether the whole thing will work at all.

Many thanks,
Wayne
www.zeta.org.au/~sausage

[ Moderator's note: Should this turn out to be a puzzle covered by a
  patent, I'll have to drop discussion of the subject. It is my under-
  standing that construction of patented objects is an infringement
  even if the object is made for personal use, and even if the patent
  holder has ceased production.  But I have no objections to the
  general topic of materials and construction techniques suitable for
  puzzles, though if any subscribers find this drifting too far from
  the topic, please let cube-lovers-request@ai.mit.edu know.  --Dan ]

From cube-lovers-errors@mc.lcs.mit.edu  Tue Apr 13 13:25:11 1999
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Date: Tue, 13 Apr 1999 16:19:14 +0100
From: David Singmaster Computing <david.singmaster@sbu.ac.uk>
To: noel@mud.ca
Cc: cube-lovers@ai.mit.edu
Message-Id: <009D6951.D5AB9641.72@ice.sbu.ac.uk>
Subject: RE: Wrist pains

	When I went to a meeting on RSI problems, the speaker mentioned 'mouse
finger' which is about as debilitating as the white finger suffered by people
who use pneumatic drills (or jack-hammers).  When the speaker went to one
small computer design firm, she found two people suffering from it, one having
essentially lost the use of his right hand and having had to change to using
his left hand!

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499;
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Wed Apr 14 22:38:03 1999
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Date: Tue, 13 Apr 1999 16:46:33 +0100
From: David Singmaster Computing <david.singmaster@sbu.ac.uk>
To: mag@ncsa.uiuc.edu
Cc: cube-lovers@ai.mit.edu
Message-Id: <009D6955.A6FEB505.56@ice.sbu.ac.uk>
Subject: Re: Keyboard cube [was: Re: Wrist pains]

	Re: Tom Magliery's representation of the cube.
	A form of this was used by Kathleen Ollerenshaw in the early 1980s, but
she also had all but the center facelet of the back face folded out to form an
additional square ring around the figure so one knew what everything was.

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499;
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

[Moderator's note:  I couldn't help myself:

 .................................................
 :               :               :               :
 :               :               :               :
 :     +---------+---------------+---------+     :
 :     |`.        \             /        .'|     :
 :     |  `+.......\.........../.......+'  |     :
 :     |   :`.      \         /      .':   |     :
 :     |   :  `+.....\......./.....+'  :   |     :
 :.....+   :   :`.    \     /    .':   :   +.....:
 :     |`..:   :  `+---+---+---+'  :   :..'|     :
 :     |   :`..:   |   :   :   |   :..':   |     :
 :     |   :   :`..+...:...:...+..':   :   |     :
 :     |   :   :   |   :   :   |   :   :   |     :
 :     |   :   : ..+...:...:...+.. :   :   |     :
 :     |   : ..:'  |   :   :   |  `:.. :   |     :
 :     | ..:'  :   +---+---+---+   :  `:.. |     :
 :.....+'  :   : .'   /     \   `. :   :  `+.....:
 :     |   :   :'..../.......\....`:   :   |     :
 :     |   : .'     /         \     `. :   |     :
 :     |   :'....../...........\......`.   |     :
 :     | .'       /             \       `. |     :
 :     +'--------+---------------+--------`+     :
 :               :               :               :
 :               :               :               :
 :...............:...............:...............:

I think something like this appears in Winning Ways, too. --Dan ]

From cube-lovers-errors@mc.lcs.mit.edu  Thu Apr 15 16:59:24 1999
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Date: Wed, 14 Apr 1999 15:44:11 +0100
From: David Singmaster Computing <david.singmaster@sbu.ac.uk>
To: jburkhardt@mediaone.net
Cc: cube-lovers@ai.mit.edu
Message-Id: <009D6A16.1B1880B4.26@ice.sbu.ac.uk>
Subject: RE: Different versions of the Ideal cube

	Ideal certainly did use oriental suppliers when the Great Cube Craze
was on and I recall several qualities of cube appeared here.  I can't remember
for sure if they ever sold the open cornered versions.  I definitely don't
recall examples with markedly different colors.  When Rubik's Fourth Dimension
was launched here, I saw that they were using distinctly poor quality cubes,
presumably from China or thereabouts.  One person at the launch said hers broke
within an hour.

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499;
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Thu Apr 15 19:31:56 1999
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From: Douglas Zander <dzander@solaria.sol.net>
Message-Id: <199904142338.SAA14966@solaria.sol.net>
Subject: Building a twisty puzzle (fwd)
To: cube-lovers@ai.mit.edu (cube)
Date: Wed, 14 Apr 99 18:38:20 CDT

 In the US there exists a magazine called, "Inventor's Digest"  which
 I have a subscribtion to and which is currently running a series of
 articles about building prototypes from molds.  It would probably be
 just what you are looking for.  It explains the whole process.  If you
 want, I can get subscription information for you or information on how
 to order individual articles.  You may wish to look in your local
 library to see if they carry the magazine (au = australia?) but I'm
 not sure your country would carry it.  If you want I will send you
 excerpts just ask.  (is this something the whole list would want to find
 out about?)

--
 Douglas Zander                | 
 dzander@solaria.sol.net       |
 Shorewood, Wisconsin, USA     |

[ Moderator's note: I doubt we have that many molder/fabricator people
  on the list (but I could be wrong).  It's also somewhat questionable
  how large typed-in excerpts can be before we run into copyright
  problems.  So I'd suggest people who are looking for excerpts should
  contact Douglas Zander at the above address.  Subscription/ordering
  information is more generally useful and less problematic, so please
  send cube-lovers that info. --Dan ]

From cube-lovers-errors@mc.lcs.mit.edu  Thu Apr 15 20:25:17 1999
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Message-Id: <199904151407.KAA00624@livia.East.Sun.COM>
Date: Thu, 15 Apr 1999 10:07:53 -0400 (EDT)
From: Guy Steele - Sun Microsystems Labs <gls@livia.East.Sun.COM>
Reply-To: Guy Steele - Sun Microsystems Labs <gls@livia.East.Sun.COM>
Subject: Winning Ways
To: cube-lovers@ai.mit.edu
Cc: Guy.Steele@east.sun.com


   Date: Tue, 13 Apr 1999 16:46:33 +0100
   From: David Singmaster Computing <david.singmaster@sbu.ac.uk>
   ...
   [Moderator's note:  I couldn't help myself:
     [ Picture ]
   I think something like this appears in Winning Ways, too. --Dan ]

I just checked my copy; it does not contain this sort of
diagram for the cube, at least on the pages cited in the index
for "Rubik's Cube", but it does have some very nice isometric
3-D "cutaway/see-through" diagrams of the cube.

--Guy

From cube-lovers-errors@mc.lcs.mit.edu  Thu Apr 15 21:06:50 1999
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Message-Id: <37165923.7ACC5D48@u.washington.edu>
Date: Thu, 15 Apr 1999 21:24:51 +0000
From: David Barr <davidb@u.washington.edu>
To: cube-lovers@ai.mit.edu
Subject: Re: Keyboard cube [was: Re: Wrist pains]
References: <009D6955.A6FEB505.56@ice.sbu.ac.uk>

David Singmaster Computing wrote:
>
>         Re: Tom Magliery's representation of the cube.
>         A form of this was used by Kathleen Ollerenshaw in the early 1980s, but
> she also had all but the center facelet of the back face folded out to form an
> additional square ring around the figure so one knew what everything was.

I've been playing around with something like this myself.  It doesn't
work yet,
but you can look at

  http://home1.gte.net/davebarr/Cube/

I got inspired by all of the recent discussion of input methods and
started thinking about visual representations.

David

From cube-lovers-errors@mc.lcs.mit.edu  Fri Apr 16 13:08:36 1999
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From: Douglas Zander <dzander@solaria.sol.net>
Message-Id: <199904160046.TAA20588@solaria.sol.net>
Subject: Inventor's Digest info
To: cube-lovers@ai.mit.edu (cube)
Date: Thu, 15 Apr 99 19:46:42 CDT

Forwarded message:
> [ Moderator's note: ...  Subscription/ordering
>   information is more generally useful and less problematic, so please
>   send cube-lovers that info. --Dan ]

 Inventors' Digest
 Subscription Dept.
 PO BOX 70
 Guffey CO 80820

 FAX: (617) 723-6988
 Email: InventorsD@aol.com
 WWW: www.inventorsdigest.com

 Volume XV, No. 1, January/February 1999
 Part 1/3: "Casting Plastic Prototypes at Room Temperature"
 Author: Jack Lander

 Volume XV, No. 2, March/April 1999
 Part 2/3: "Making a Rubber Mold For Casting Plastic"

 Volume XV, No.3, May/June 1999
 Part 3/3: "Casting Plastic in Rubber Molds"

--
 Douglas Zander                |
 dzander@solaria.sol.net       |
 Shorewood, Wisconsin, USA     |

From cube-lovers-errors@mc.lcs.mit.edu  Mon Apr 19 12:14:43 1999
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Message-Id: <3718413B.251EE6E0@okanagan.net>
Date: Sat, 17 Apr 1999 00:07:29 -0800
From: karen loewen <loewens@okanagan.net>
Reply-To: loewens@okanagan.net
To: Cube-Lovers@ai.mit.edu
Subject: Rubiks Cube, Megaminx and Square-1 

Is there a world record for the the 4x4x4 and the 5x5x5 rubiks cubes?
If so what is it?

I was wondering what in your opinion is the hardest rubiks like cube
available right now.

On a web page several people from rated the greatest to least in
difficulty:

1)square-1
2)megaminx
3)5x5x5
4)4x4x4
5)?
6)?

Anyway it went something like that.

Would you consider Square-1 one of the hardest cubes to do? If so would
this be considered good of me to learn all by myself? My best time is
83 sec. As if now I am not sure what my average is.  Do you know of
a world record for Square-1 or any good websites? Do you know what an
average would be for the best time?

I was wondering how this compares to others times and accomplishments. At
http://byrden.com/puzzles/ I almost figured out the megaminx. I got
everything except the last row. Is the Megaminx one of the hardest
Rubiks like puzzles available? After I get bored of my 5x5x5 that I
ordered thinking about buying the megaminx. Do you know of a world record
on the megaminx? What would be a average time (perhaps 3-4 minutes?)

Currently I am more interested in the Rubiks cube. Lately I have been
trying to improve my time. I have a personal record of 50 seconds, and
an average time of about 80 seconds. However, I learned how to do the cube
all by myself. I didn't get help from others, websites, books...
I was wondering If you learned by yourself or if you got help? Then what
is your time?

I don't believe that I will ever get much faster. Learning by yourself
is a really hard task. I was wondering if there is any one out there who
agrees that getting help from resources is cheating?

In my opinion they make the people who haved learned by themselves look
bad. This is just my opinion, you don't have to agree.

If you are not sure of answers that's fine. Then do you know of any good
websites I can go to?

Thank you very much.

From cube-lovers-errors@mc.lcs.mit.edu  Mon Apr 19 17:34:05 1999
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Date: Mon, 19 Apr 1999 15:13:27 +0100
From: David Singmaster <david.singmaster@sbu.ac.uk>
To: jbryan@pstcc.cc.tn.us
Cc: cube-lovers@ai.mit.edu
Message-Id: <009D6DFF.A3A1ABCB.24@ice.sbu.ac.uk>
Subject: RE: Re : Re: Inventing your own techniques

	The technique of solving the cube by use of commutators and conjugates
is what most people worked out in the early days.  However, I can testify that
it took us some time to realise that one could use second level commutation.
That is, FRF'R' = [F,R] only affects seven pieces, but in fact it only affects
one piece in the L face, so taking the commutator with the L face produces a
3-cycle of corners.  Likewise  [F,R]^2  only twists one corner in the L face
and combining it with turns of L allows you to twist three corners the same way
or two corners opposite ways.

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499;
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From david.singmaster@sbu.ac.uk  Mon Apr 19 18:35:09 1999
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Date: Mon, 19 Apr 1999 15:13:27 +0100
From: David Singmaster <david.singmaster@sbu.ac.uk>
To: jbryan@pstcc.cc.tn.us
Cc: cube-lovers@ai.mit.edu
Message-Id: <009D6DFF.A3A1ABCB.24@ice.sbu.ac.uk>
Subject: RE: Re : Re: Inventing your own techniques
X-MDaemon-Deliver-To: cube-lovers-outbound@mc.lcs.mit.edu

	The technique of solving the cube by use of commutators and conjugates
is what most people worked out in the early days.  However, I can testify that
it took us some time to realise that one could use second level commutation.
That is, FRF'R' = [F,R] only affects seven pieces, but in fact it only affects
one piece in the L face, so taking the commutator with the L face produces a
3-cycle of corners.  Likewise  [F,R]^2  only twists one corner in the L face
and combining it with turns of L allows you to twist three corners the same way
or two corners opposite ways.

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499;
email:  zingmast  or  David.Singmaster  @sbu.ac.uk



From cube-lovers-errors@mc.lcs.mit.edu  Tue Apr 20 14:22:17 1999
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Date: Mon, 19 Apr 1999 15:48:36 +0100
From: David Singmaster <david.singmaster@sbu.ac.uk>
To: WaVeReBeL@webtv.net
Cc: cube-lovers@ai.mit.edu
Message-Id: <009D6E04.8CAE2B95.34@ice.sbu.ac.uk>
Subject: RE: Lubricant

	I used to use silicone grease of the type used for O-ring seals in
diving equipment, etc.  I haven't seen any damage caused by this.  WD-40 has
also been suggested and I have used it, but I think it may damage some
plastics.  Note that the central spindle is often made of a different plastic
than the other pieces and I think it is more subject to attack by some of the
lubricants.

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499;
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Thu Apr 22 11:18:36 1999
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From: WaVeReBeL@webtv.net
Date: Mon, 19 Apr 1999 16:30:06 -0700 (PDT)
To: Cube-Lovers@ai.mit.edu
Cc: loewens@okanagan.net
Subject: Re: Rubiks Cube, Megaminx and Square-1
Message-Id: <8843-371BBC7E-10617@postoffice-123.bryant.webtv.net>
In-Reply-To: karen loewen <loewens@okanagan.net>'s message of Sat, 17 Apr
	1999 00:07:29 -0800

Hi,

I think it's great that you learned to do the cube all by yourself.  I
give much respect to anyone who solved any of these twisty puzzles.
Very few people have the ability and the time to figure them out.  I'm
curious if you are willing to share your methods?

I personally learned from a book.  Afterwards, I kind of wished I
hadn't.  Since then, I promised myself not to look at any more solutions
for any new puzzles I encounter.  So far, all I've managed to solve  on
my own is the pyraminx. (very easy compared to the cube)

I can see how you might feel a little disgruntled about others achieving
quick solving times with aid from outside resources.  I am an example of
that.  In about 4 months I achieved a 36 second average after scouring
the net for every bit of information I could find.  You shouldn't think
it makes you look bad.  If anything, I feel bad knowing that people like
you must put a lot of hard work into it, while I had a much easier time.
It still takes a lot of hard physical and mental work on my part to
achieve a fast average, but a lot of the mental work was already done
for me.

As far as cheating is concerned, in my and many others' defense, I must
respond, as the majority of people in this group have referred to
outside information at one point or another.  Consider speed cubing as a
sport just like any other in the Olympics.  Everyone is going to use
every resource available to win.  No one in the Olympics is going to try
and win (or succeed for that matter) all by themselves w/out any help or
coaching at all.  The use of outside help brings those who use it to a
higher level playing field.  These people are in an equal playing field
among themselves having access to the same resources.  If people choose
to learn on their own, it is only fair to put them in a different group
rather than lumping everyone in the same group.  But just because people
have accessed other resources, doesn't mean they will all of a sudden
reach less than a minute averages, it still takes a lot of hard work to
get there.  I know your average is still better than a lot of people who
got help.

On the other hand, getting help on learning how to solve the cube might
be considered cheating just as it would be when one gets hints to a
riddle or looks in the back of the book for all the answers.

I don't think there is anything wrong with help at all, unless your goal
is to do it all on your own without any help.  You said that you don't
believe that you will ever get much faster.  If your skill level
plateaus, I don't see anything wrong with using any resources available
to exceed your limits.  That's what help is for!  You can still be proud
of your accomplishments, but everyone has their limits.  There are very
few people who have achieved what you have.


-Alex Montilla-

From cube-lovers-errors@mc.lcs.mit.edu  Thu Apr 22 12:22:10 1999
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Message-Id: <3.0.32.19990420085812.009b2880@mail.spc.nl>
Date: Tue, 20 Apr 1999 08:58:14 +0200
To: cube-lovers@ai.mit.edu
From: Christ van Willegen <c.v.willegen@spcgroup.nl>
Subject: RE: Re : Re: Inventing your own techniques

At 15:13 19-4-1999 +0100, you wrote:
>	The technique of solving the cube by use of commutators and conjugates
>is what most people worked out in the early days.  However, I can testify
that
>it took us some time to realise that one could use second level commutation.
>That is, FRF'R' = [F,R] only affects seven pieces, but in fact it only
affects
>one piece in the L face, so taking the commutator with the L face produces a
>3-cycle of corners.  Likewise  [F,R]^2  only twists one corner in the L face
>and combining it with turns of L allows you to twist three corners the
same way
>or two corners opposite ways.

When I learned how to solve the cube (some 20 years back, I think...),
I was taught a layer-by-layer solution, consisting of 'modular' moves.
Most of the modularity occurs on the 3rd layer, namely:
- In flipping the edges, I use (RE)^4 to flip the UR edge. Then, I move
  the top layer so that there's another edge to be flipped at the UR
  position and I repeat the same procedure.
- In rotating the edges, I was taught the above-mentioned {R,F]^2 to
  rotate an edge by 120 degrees. Later, I 'discovered' that [F,R]^2
  does the same thing backward.

Most other layer-by-layer methods I've seen have formulas to flip 2
edges in the top layer, but you'll need two for the different confi-
gurations. Also, these methods mostly have two, three or more formulas
to rotate the edges in the toplayer, depending on configuration.

My method may not be the fastest around (it takes me about 2 minutes
to solve The Cube), but I don't need many formulas for it:

1st layer: on my own (make a cross, fill in the corners)
2nd layer: 2 formulas (standard layer 3 to layer 2 swing, two directions)
3rd layer: 4 (exchange edges, flip one edge, exchange corners, rotate
           one corner)

Christ van Willegen

From cube-lovers-errors@mc.lcs.mit.edu  Thu Apr 22 18:11:24 1999
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Message-Id: <B11CE48E1A39D211BB540080C8646221013D92@is.intralan.demon.co.uk>
From: Paddy Duncan <PaddyD@intralan.co.uk>
To: cube-lovers@ai.mit.edu
Subject: Re: Lubricant
Date: Tue, 20 Apr 1999 21:27:31 +0100

WD-40 will make it stick REALLY badly! Please don't do it.  Not
because it will dissolve the plastic (although it might), but because
it's too thin. Also I have seen in the past a problem with the kind of
solvents in this kind of product where it seems to 'realise' any small
cracks in plastic, causing it to just fall to bits the next day.
Silicone grease sounds good to me.

         Paddy Duncan
      Systems Engineer
       Intralan (UK) Ltd

[ Moderator's note: I'll remind you that last month Alex Montilla
  suggested (for low-quality tiled cubes), "If you use WD-40, it'll
  eat away at the plastic resulting in really smooth turning.  It'll
  be great for about 3-4 weeks of daily cubing, but the WD-40 will
  take its toll, and the cube will start falling apart.  But for a few
  bucks every 3-4 weeks is worth it to me."  So WD-40 is bad for the
  cube, but possibly good for the turning, at least temporarily.
  That's the way the cubie crumbles.  --Dan ]

From cube-lovers-errors@mc.lcs.mit.edu  Fri Apr 23 11:39:54 1999
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Message-Id: <371FCBE9.6DE8069@ibm.net>
Date: Thu, 22 Apr 1999 18:24:57 -0700
From: "Jin 'Time Traveler' Kim" <chrono@ibm.net>
Reply-To: chrono@ibm.net
To: cube-lovers@ai.mit.edu
Subject: Re: Rubiks Cube, Megaminx and Square-1
References: <3718413B.251EE6E0@okanagan.net>

The only record for a 4x4x4 I saw was 117 or 170+ seconds, I can't
remember which.  BTW, that was my web page you visited. :)

The reason I think that the Square-1 is rated so high is because it
"mutates" in form and shape as you twist it.  All other puzzles remain
mechanically symmetrical to themselves (even the pyramorphix, while
changing shape, is still the same basic architecture underneath).  The
Square-1's possible moves change as you mix it, causing an extra level
of confusion.  Does this make it tougher to solve?

Learning to solve puzzles by yourself is great.  Unlike some others on
the cube list, I've always relied on intuition and mental picturing to
make all of my moves.  This has allowed me to solve puzzles since I was
about six years old.  The only puzzles I have been unable to solve this
way are the Rubik's Cube series (2^3 through 5^3), specifically the
corner moves.  Everything else I was able to do independently at one
Time or another including the center pieces on the 4x4x4 and 5x5x5.

A quick anecdote.  For the longest Time I didn't bother scrambling my
Geomaster Masterball for fear of messing it up.  That barrier was
overcome when a friend of mine inadvertently scrambled it for me.  14
hours of hair pulling later I came up with a "logical" solution for it
(that actually helped me solve the Dogic later).  Of course, shortly
afterwards they sold Masterballs with solutions far simpler than mine,
but I still prefer using my method, because it's mine.

-- 
Jin "Time Traveler" Kim
chrono@ibm.net
http://www.chrono.org
'95 PGT - SCPOC

From cube-lovers-errors@mc.lcs.mit.edu  Mon Apr 26 19:06:53 1999
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From: "Marius Loots" <MLOOTS@medic.up.ac.za>
Organization:  University of Pretoria
To: Cube-Lovers@ai.mit.edu
Date:          Fri, 23 Apr 1999 09:36:26 CAT-2
Subject:       Re: Lubricant
Reply-To: mloots@medic.up.ac.za
Message-Id: <A50C51237C@medic.up.ac.za>

> Many people have suggested different lubricants on their web pages.

>  The lubes I remember from the archives are vaseline (greasy, erodes
>  plastic slowly), soap (temporary), candle wax (drips flakes), and
>  molybdenum disulfide (doesn't help, destroys cube quickly).  The only

>From these lubricants the only one I have ever used was Vaseline.  At
some stage I did try car grease but this was too much of a mess on the
hands.  For a real slippery slide, we sometimes dipped the Vaseline Cube
in water.  This would give you a few minutes of realy high speed cubing.
This shouldn't have any effect on the plastic, I am not sure about the
mechanisms although I still have my very first cube and it is still in
working order (albeit a bit loose because of age).


Groetnis
Marius

mloots@medic.up.ac.za    +27-12-319-2144
Add some Chaos to your Life and put the World in Order
http://www.geocities.com/Athens/6398/

From cube-lovers-errors@mc.lcs.mit.edu  Mon Apr 26 19:39:09 1999
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In-Reply-To: <B1A5BAB06743C03A8725675B007E3CD7.007E3D498725675B@dtint.com>
Date: Fri, 23 Apr 1999 09:12:13 -0600
To: cube-lovers@ai.mit.edu
From: Steve LoBasso <slobasso@dtint.com>
Subject: Re: Lubricant

White lithium grease is pretty good, but I don't know about its long term
effects on plastic. If anyone has ever changed a distributor / points on an
old car, there is a small tube of grease that is intended to be used on a
plastic metal contact point. It is actually absorbed by the plastic and
allows it to remain quite frictionless. If you want some good advice on
this stuff look for an auto parts store with people that know more than
just what they read from the packages or an older mechanic.

   -Steve


--
Steve LoBasso
Digital Technology International    mailto:slobasso@dtint.com
500 West 1200 South              or mailto:slobasso@hotmail.com
Orem, UT    84058                   http://members.tripod.com/~slobasso
(801)226-6142 ext.265               FAX (801)221-9254

[ Moderator's note: Be sure to read the tube or find a MSDS for
  warnings about prolonged contact with skin.  If you can absorb Li+
  from this, it's serious bad news. --Dan ]

From cube-lovers-errors@mc.lcs.mit.edu  Mon Apr 26 21:15:50 1999
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To: cube-lovers@ai.mit.edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Reinventing (and some edge-flipping techniques)
Date: 26 Apr 1999 18:03:02 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <7g29om$4ua@gap.cco.caltech.edu>

I skimmed through Singmaster's "Notes on Rubik's Magic Cube" this weekend.
It was rather disappointing to realize that the techniques I had invented
five years ago had all been independently invented twenty years ago!

In any case, I thought I might share my thoughts about monoflips with the
list.

--------------

[I shall use "E" to denote moving the center slice (between the U and D
faces) to the left (so that the F center becomes the L center).  E' I
will use to denote its inverse.
In other words, E is the same as U'D followed by rotating the entire
cube clockwise from the point of view of the U face.
Similarly, I will use M to denote moving the L-R slice "up".
Traditionally, a slice is counted as two turns, although it is usually
easier to perform than two orthogonal turns.]

As you may know, a "monoflip" is a sequence of moves that flips one edge
on a given face (usually the U face) while not affecting the rest of the
face.  For instance, this is a monoflip:

  F E F M FF M' (10 turns)

This flips the FU edge cubie.  If you do this sequence of moves, it is
particularly easy to look at the FUL, FU, and FUR cubies and see how they
move around to perform this monoflip.

What makes a monoflip so useful is the fact that it can me commutated with
moves of the face (U) to flip two edges.  E.g.,

  FEFMFFM' U MFFM'F'E'F' U'

will flip the FU and RU edges.  (The first FEFMFFM' flips FU while messing
up the rest of the cube, then the U moves the RU to where the FU was,
then MFFM'F'E'F' flips the "new" FU as well as undoing what was done to
the rest of the cube.  U' then restores what was done.)

Commutating with other moves will, of course, allow you to flip other edges.
Using UU instead of U will flip the FU and BU edges, for instance.
If you add conjugation, you can get pretty much any two edge pieces to
flip -- e.g.,

  R' FEFMFFM' U MFFM'F'E'F' U' R

will flip the FU and RB edges.

--------

The drawback of FEFMFFM', however, is that it is not that easy to remember,
since you also have to remember its inverse MFFM'F'E'F'.  (Although the
embedded conjugate MFFM' is its own inverse, which is nice.)

Therefore, the monoflip I see more often in other people's repertoire is

  F E' F E' F E' F E' (12 moves)

which has many advantages:
 1.  It is easier to remember;
 2.  It is its own inverse;
 3.  Its results are "elegant".  (flips of FU,FL,BR,BL)

Of course, this move is special enough that you don't NEED to use it as
a monoflip if you rotate and reflect this move.  To flip two adjacent
edges, simply do

 FE'FE'FE'FE' R'ER'ER'ER'E

and you'll flip the FU and RU edges.  (The second part is just a
reflection along the BL - FR plane.)  Other variants of this move
can pretty much let you flip any combinations of edge cubies you
want.

--------------

I used to use the previous "monoflip" a lot.
The only drawback with the previous monoflip, however, is that it is
too slow, especially those slice moves.  Therefore, when I need
to do edge flipping these days, I use this monoflip:

 R F' U R' F (5 moves)

At first this doesn't look like a monoflip, since each face seems to be
pretty messed up.  But this is not your normal monoflip -- this is a
SLICE-BASED monoflip, where the E slice is intact except for the FR edge.

In fact, I'd conjecture that this is the shortest monoflip there is.  To
flip the FR edge, RUF is pretty much the quickest you can to it in.  But
this messes up two diagonally opposite edges on the slice, so at
least two more moves are needed to restore this -- hence, the F' and R'.

You can think of this as RF' conjugated with U -- I'm not sure it
helps much though.

The inverse, F' R U' F R', is just a reflection of the original move and
so is pretty easy to remember.

In any case, this makes edge flipping nice and quick:

 RF'UR'F E' F'RU'FR' E  (flips FR and FL in 12 moves)
 RF'UR'F EE F'RU'FR' EE  (flips FR and BL in 16 moves)

But how about the other edge pair combinations?  Conjugates, of course.
I use these moves:

 1. Adjacent edges, same face:

   F'R'  RF'UR'F E' F'RU'FR' E RF  which reduces a bit to:
   FFUR'F E' F'RU'FR' E RF  (flips FU and RU in 16 moves)

 2. Adjacent edges, same slice:
   (done above, 12 moves)

 3. Opposite edges:
   (done above, 16 moves)

 4. Skew edges:

   R'  RF'UR'F E' F'RU'FR' E R  which reduces a bit to:
   F'UR'F E' F'RU'FR' E R  (flips FL and RU in 14 moves)

These are the current edge flips in my repertoire, and they achieve a
balance between being easy to remember and being fast to do.

----------

I'd appreciate if others could share the moves they use for 2-edge flips, as
well as know of any results known of God's algorithm for 2-edge flips.

--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
She ran by screaming "No, I run by moving my feet rapidly, you idiot!"

From cube-lovers-errors@mc.lcs.mit.edu  Tue Apr 27 11:11:51 1999
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From: "Chris and Kori Pelley" <ck1@home.com>
To: <Cube-Lovers@ai.mit.edu>
Subject: RE: Lubricant
Date: Mon, 26 Apr 1999 22:38:57 -0400
Message-Id: <LLBBJDCFCKLJHGBBOAGOGEFCCEAA.ck1@home.com>
In-Reply-To: <A50C51237C@medic.up.ac.za>

> -----Original Message-----
> From: Marius Loots [mailto:MLOOTS@medic.up.ac.za]
> ...For a real slippery slide, we sometimes dipped the Vaseline Cube
> in water.  This would give you a few minutes of realy high speed cubing.

I concur with Marius.  I used to practice solving underwater in my swimming
pool, back when I averaged 33 seconds!  I would use a cube with just a small
amount of vaseline.  After taking the cube out and letting it dry a couple
hours, it would turn VERY WELL.  Also, the deleterious effect on the plastic
seems to be negligible, as I still have those same cubes and they still
operate smoothly some 15+ years later.

Chris Pelley
ck1@home.com
http://www.chrisandkori.com/cubes.htm

From cube-lovers-errors@mc.lcs.mit.edu  Tue Apr 27 12:19:57 1999
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Message-Id: <3725DDD8.327E@hrz1.hrz.tu-darmstadt.de>
Date: Tue, 27 Apr 1999 17:55:04 +0200
From: Herbert Kociemba <kociemba@hrz1.hrz.tu-darmstadt.de>
Reply-To: kociemba@hrz1.hrz.tu-darmstadt.de
To: cube-lovers@ai.mit.edu
Subject: Re: Reinventing (and some edge-flipping techniques)
References: <7g29om$4ua@gap.cco.caltech.edu>

Wei-Hwa Huang wrote:

>
> I'd appreciate if others could share the moves they use for 2-edge flips,
> as well as know of any results known of God's algorithm for 2-edge flips.
>

I did a quick run with my optimal solver program. Here are all maneuvers
with shortest length (face-turn metric) for the four possible two-flips.
Note that many of them are basically identical due to the symmetry of
the two-flip patterns itself:

uf,ur flips:

F U2 F2 D' U' L' U L D F2 U' F' U'  (13f*)
F D U R2 U2 R' U R U R2 D' F' U'  (13f*)
F' U' F2 D R U R' D' U' F2 U2 F U'  (13f*)
F' D' L2 U L U L' U2 L2 D U F U'  (13f*)
R D B2 U' B' U' B U2 B2 D' U' R' U  (13f*)
R U R2 D' F' U' F D U R2 U2 R' U  (13f*)
R' U2 R2 D U B U' B' D' R2 U R U  (13f*)
R' D' U' F2 U2 F U' F' U' F2 D R U  (13f*)
U' R U2 R2 D' U' F' U F D R2 U' R'  (13f*)
U' R D U B2 U2 B' U B U B2 D' R'  (13f*)
B F U F' U' B' R2 F R F R' F2 R2  (13f*)
F R' F' R' F2 L D R D' L' R' F2 R2  (13f*)
L F R' F' L' U2 R U R U' R2 U2 R  (13f*)
U' R' U' R2 D B U B' D' U' R2 U2 R  (13f*)
L F2 R2 F' R F R F2 L' U' R' U R  (13f*)
U' R' D' F2 U F U F' U2 F2 D U R  (13f*)
F2 R2 B' F' D' F D B R2 F' R' F' R  (13f*)
B F U2 F2 U' F U F U2 B' R' F' R  (13f*)
F R' F' L' U2 R U R U' R2 U2 L R  (13f*)
F2 R2 F' R F R F2 L' U' R' U L R  (13f*)
B' R2 F2 R F' R' F' R2 B U F U' F'  (13f*)
U F D R2 U' R' U' R U2 R2 D' U' F'  (13f*)
B' R' F R B U2 F' U' F' U F2 U2 F'  (13f*)
U F U F2 D' L' U' L D U F2 U2 F'  (13f*)
R2 F2 L R D R' D' L' F2 R F R F'  (13f*)
L' R' U2 R2 U R' U' R' U2 L F R F'  (13f*)
R2 F2 R F' R' F' R2 B U F U' B' F'  (13f*)
R' F R B U2 F' U' F' U F2 U2 B' F'  (13f*)
L' R' U' R U L F2 R' F' R' F R2 F2  (13f*)
R' F R F R2 B' D' F' D B F R2 F2  (13f*)
U F' U2 F2 D U R U' R' D' F2 U F  (13f*)
U F' D' U' L2 U2 L U' L' U' L2 D F  (13f*)
R' U2 R2 U R' U' R' U2 L F R F' L'  (13f*)
R' U' R U L F2 R' F' R' F R2 F2 L'  (13f*)
F U F' U' B' R2 F R F R' F2 R2 B  (13f*)
F U2 F2 U' F U F U2 B' R' F' R B  (13f*)


uf,ub flips:

L F' U L' B' F U R' F U' R B F' U'  (14f*)
L' R B U' L B' U L R' F' U L' F U'  (14f*)
R B' U R' B F' U L' B U' L B' F U'  (14f*)
L R' F U' R F' U L' R B' U R' B U'  (14f*)
R2 F D U' R B2 U2 B2 R' D' U F' R2 U2  (14f*)
R2 F L R' D L2 B2 L2 D' L' R F' R2 U2  (14f*)
R2 F' D' U L' F2 U2 F2 L D U' F R2 U2  (14f*)
R2 F' L R' D' L2 B2 L2 D L' R F R2 U2  (14f*)
R2 B D U' L B2 U2 B2 L' D' U B' R2 U2  (14f*)
R2 B L' R D L2 F2 L2 D' L R' B' R2 U2  (14f*)
R2 B' D' U R' F2 U2 F2 R D U' B R2 U2  (14f*)
R2 B' L' R D' L2 F2 L2 D L R' B R2 U2  (14f*)
F U2 L R F2 D R2 D' F2 L' R' U2 F' U2  (14f*)
F R2 D U B2 R B2 R' B2 D' U' R2 F' U2  (14f*)
F' U2 L' R' F2 D' L2 D F2 L R U2 F U2  (14f*)
F' L2 D' U' B2 L' B2 L B2 D U L2 F U2  (14f*)
L2 F D U' R F2 U2 F2 R' D' U F' L2 U2  (14f*)
L2 F L R' D R2 B2 R2 D' L' R F' L2 U2  (14f*)
L2 F' D' U L' B2 U2 B2 L D U' F L2 U2  (14f*)
L2 F' L R' D' R2 B2 R2 D L' R F L2 U2  (14f*)
L2 B D U' L F2 U2 F2 L' D' U B' L2 U2  (14f*)
L2 B L' R D R2 F2 R2 D' L R' B' L2 U2  (14f*)
L2 B' D' U R' B2 U2 B2 R D U' B L2 U2  (14f*)
L2 B' L' R D' R2 F2 R2 D L R' B L2 U2  (14f*)
B U2 L R B2 D L2 D' B2 L' R' U2 B' U2  (14f*)
B L2 D U F2 L F2 L' F2 D' U' L2 B' U2  (14f*)
B' U2 L' R' B2 D' R2 D B2 L R U2 B U2  (14f*)
B' R2 D' U' F2 R' F2 R F2 D U R2 B U2  (14f*)
L' R B' U R' B U' L R' F U' R F' U  (14f*)
L' B U' L B' F U' R B' U R' B F' U  (14f*)
R' F U' R B F' U' L F' U L' B' F U  (14f*)
L R' F' U L' F U' L' R B U' L B' U  (14f*)
L F U' R F' U L' R B' U R' B U' R'  (14f*)
F U' R B F' U' L F' U L' B' F U R'  (14f*)
L F' U L' F U' L' R B U' L B' U R'  (14f*)
F U' R F' U L' R B' U R' B U' L R'  (14f*)
F' U L' F U' L' R B U' L B' U L R'  (14f*)
U F' L U' F L' R U' B L' U B' L R'  (14f*)
U' F R' U F' L' R U B' R U' B L R'  (14f*)
F' L' U B' U B L F R U' B U' B' R'  (14f*)
L U F' L U' F L' R U' B L' U B' R'  (14f*)
F' U' F U' R B L F U F' U L' B' R'  (14f*)
U B F' L' U B' L U' B' F R U' B R'  (14f*)
L U' F R' U F' L' R U B' R U' B R'  (14f*)
F D U' R B2 U2 B2 R' D' U F' R2 U2 R2  (14f*)
F L R' D L2 B2 L2 D' L' R F' R2 U2 R2  (14f*)
F' D' U L' F2 U2 F2 L D U' F R2 U2 R2  (14f*)
F' L R' D' L2 B2 L2 D L' R F R2 U2 R2  (14f*)
B D U' L B2 U2 B2 L' D' U B' R2 U2 R2  (14f*)
B L' R D L2 F2 L2 D' L R' B' R2 U2 R2  (14f*)
B' D' U R' F2 U2 F2 R D U' B R2 U2 R2  (14f*)
B' L' R D' L2 F2 L2 D L R' B R2 U2 R2  (14f*)
U2 R2 F D U' R B2 U2 B2 R' D' U F' R2  (14f*)
U2 R2 F L R' D L2 B2 L2 D' L' R F' R2  (14f*)
U2 R2 F' D' U L' F2 U2 F2 L D U' F R2  (14f*)
U2 R2 F' L R' D' L2 B2 L2 D L' R F R2  (14f*)
U2 R2 B D U' L B2 U2 B2 L' D' U B' R2  (14f*)
U2 R2 B L' R D L2 F2 L2 D' L R' B' R2  (14f*)
U2 R2 B' D' U R' F2 U2 F2 R D U' B R2  (14f*)
U2 R2 B' L' R D' L2 F2 L2 D L R' B R2  (14f*)
L' B U' L B' U L R' F' U L' F U' R  (14f*)
B' U R' B F' U L' B U' L B' F U' R  (14f*)
L' B' U R' B U' L R' F U' R F' U R  (14f*)
L' U B' R U' B L R' U' F R' U F' R  (14f*)
U' B F' L U' F L' U B' F R' U F' R  (14f*)
L' U' B L' U B' L R' U F' L U' F R  (14f*)
B L U' F U' F' L' B' R' U F' U F R  (14f*)
B U B' U R' F' L' B' U' B U' L F R  (14f*)
B U' L B' U L R' F' U L' F U' L' R  (14f*)
B' U R' B U' L R' F U' R F' U L' R  (14f*)
U B' R U' B L R' U' F R' U F' L' R  (14f*)
U' B L' U B' L R' U F' L U' F L' R  (14f*)
L' B' R' U F' U F R B L U' F U' F'  (14f*)
U2 F U2 L R F2 D R2 D' F2 L' R' U2 F'  (14f*)
L' R U B' R U' B L R' U' F R' U F'  (14f*)
R U' B F' L U' F L' U B' F R' U F'  (14f*)
F' D' L' U' L D U F2 U2 F' U F U F'  (14f*)
L2 U2 L2 F D U' R F2 U2 F2 R' D' U F'  (14f*)
R2 U2 R2 F D U' R B2 U2 B2 R' D' U F'  (14f*)
L' B' U' B U' L F R B U B' U R' F'  (14f*)
U2 F R2 D U B2 R B2 R' B2 D' U' R2 F'  (14f*)
L2 U2 L2 F L R' D R2 B2 R2 D' L' R F'  (14f*)
R2 U2 R2 F L R' D L2 B2 L2 D' L' R F'  (14f*)
F' U F U F' U2 F2 D U R U' R' D' F'  (14f*)
F D R U R' D' U' F2 U2 F U' F' U' F  (14f*)
R2 U2 R2 F' D' U L' F2 U2 F2 L D U' F  (14f*)
L2 U2 L2 F' D' U L' B2 U2 B2 L D U' F  (14f*)
L' R U' B L' U B' L R' U F' L U' F  (14f*)
L' U B' F R' U F' R U' B F' L U' F  (14f*)
U2 F' U2 L' R' F2 D' L2 D F2 L R U2 F  (14f*)
R B L U' F U' F' L' B' R' U F' U F  (14f*)
L2 U2 L2 F' L R' D' R2 B2 R2 D L' R F  (14f*)
R2 U2 R2 F' L R' D' L2 B2 L2 D L' R F  (14f*)
F U' F' U' F U2 F2 D' U' L' U L D F  (14f*)
U2 F' L2 D' U' B2 L' B2 L B2 D U L2 F  (14f*)
R B U B' U R' F' L' B' U' B U' L F  (14f*)
R B U' L B' U L R' F' U L' F U' L'  (14f*)
R B' U R' B U' L R' F U' R F' U L'  (14f*)
B U' L B' F U' R B' U R' B F' U L'  (14f*)
B' R' U F' U F R B L U' F U' F' L'  (14f*)
R U B' R U' B L R' U' F R' U F' L'  (14f*)
B' U' B U' L F R B U B' U R' F' L'  (14f*)
R U' B L' U B' L R' U F' L U' F L'  (14f*)
U B' F R' U F' R U' B F' L U' F L'  (14f*)
F D U' R F2 U2 F2 R' D' U F' L2 U2 L2  (14f*)
F L R' D R2 B2 R2 D' L' R F' L2 U2 L2  (14f*)
F' D' U L' B2 U2 B2 L D U' F L2 U2 L2  (14f*)
F' L R' D' R2 B2 R2 D L' R F L2 U2 L2  (14f*)
B D U' L F2 U2 F2 L' D' U B' L2 U2 L2  (14f*)
B L' R D R2 F2 R2 D' L R' B' L2 U2 L2  (14f*)
B' D' U R' B2 U2 B2 R D U' B L2 U2 L2  (14f*)
B' L' R D' R2 F2 R2 D L R' B L2 U2 L2  (14f*)
U2 L2 F D U' R F2 U2 F2 R' D' U F' L2  (14f*)
U2 L2 F L R' D R2 B2 R2 D' L' R F' L2  (14f*)
U2 L2 F' D' U L' B2 U2 B2 L D U' F L2  (14f*)
U2 L2 F' L R' D' R2 B2 R2 D L' R F L2  (14f*)
U2 L2 B D U' L F2 U2 F2 L' D' U B' L2  (14f*)
U2 L2 B L' R D R2 F2 R2 D' L R' B' L2  (14f*)
U2 L2 B' D' U R' B2 U2 B2 R D U' B L2  (14f*)
U2 L2 B' L' R D' R2 F2 R2 D L R' B L2  (14f*)
F' U L' B' F U R' F U' R B F' U' L  (14f*)
R' F U' R F' U L' R B' U R' B U' L  (14f*)
R' F' U L' F U' L' R B U' L B' U L  (14f*)
U' B' F R U' B R' U B F' L' U B' L  (14f*)
R' U F' L U' F L' R U' B L' U B' L  (14f*)
R' U' F R' U F' L' R U B' R U' B L  (14f*)
F R U' B U' B' R' F' L' U B' U B L  (14f*)
F U F' U L' B' R' F' U' F U' R B L  (14f*)
R' F' L' U B' U B L F R U' B U' B'  (14f*)
U2 B U2 L R B2 D L2 D' B2 L' R' U2 B'  (14f*)
L2 U2 L2 B D U' L F2 U2 F2 L' D' U B'  (14f*)
R2 U2 R2 B D U' L B2 U2 B2 L' D' U B'  (14f*)
L U' B' F R U' B R' U B F' L' U B'  (14f*)
L R' U F' L U' F L' R U' B L' U B'  (14f*)
B' D' R' U' R D U B2 U2 B' U B U B'  (14f*)
L2 U2 L2 B L' R D R2 F2 R2 D' L R' B'  (14f*)
R2 U2 R2 B L' R D L2 F2 L2 D' L R' B'  (14f*)
B' U B U B' U2 B2 D U L U' L' D' B'  (14f*)
R' F' U' F U' R B L F U F' U L' B'  (14f*)
U2 B L2 D U F2 L F2 L' F2 D' U' L2 B'  (14f*)
R' U B F' L' U B' L U' B' F R U' B  (14f*)
L R' U' F R' U F' L' R U B' R U' B  (14f*)
R2 U2 R2 B' D' U R' F2 U2 F2 R D U' B  (14f*)
L2 U2 L2 B' D' U R' B2 U2 B2 R D U' B  (14f*)
B D L U L' D' U' B2 U2 B U' B' U' B  (14f*)
U2 B' U2 L' R' B2 D' R2 D B2 L R U2 B  (14f*)
L F R U' B U' B' R' F' L' U B' U B  (14f*)
L2 U2 L2 B' L' R D' R2 F2 R2 D L R' B  (14f*)
R2 U2 R2 B' L' R D' L2 F2 L2 D L R' B  (14f*)
U2 B' R2 D' U' F2 R' F2 R F2 D U R2 B  (14f*)
L F U F' U L' B' R' F' U' F U' R B  (14f*)
B U' B' U' B U2 B2 D' U' R' U R D B  (14f*)

uf,rb flips:

R' U' R U2 R2 D' U' F' U F D R2 U'  (13f*)
B2 D L U L' D' U' B2 U2 B U' B' U'  (13f*)
R2 D' U' F' U F D R2 U' R' U' R U2  (13f*)
R2 U' R U R U2 L' B' R' B L R U2  (13f*)
L F2 R' F' R' F R2 F2 L' R' U' R U  (13f*)
R U2 L' B' R' B L R U2 R2 U' R U  (13f*)
U' R' U R2 U2 L' R' B' R B L U2 R'  (13f*)
F2 L D R D' L' R' F2 R2 F R' F' R'  (13f*)
U2 L' R' B' R B L U2 R' U' R' U R2  (13f*)
U2 R' U R U R2 D' F' U' F D U R2  (13f*)
D B2 U' B' U' B U2 B2 D' U' R' U R  (13f*)
U R2 D' F' U' F D U R2 U2 R' U R  (13f*)
L' U' R' U L R F2 R2 F' R F R F2  (13f*)
R F R F' R2 F2 L R D R' D' L' F2  (13f*)
R' U' R D U B2 U2 B' U B U B2 D'  (13f*)
B2 U' B' U' B U2 B2 D' U' R' U R D  (13f*)
U' R' U L R F2 R2 F' R F R F2 L'  (13f*)
F2 R' F' R' F R2 F2 L' R' U' R U L  (13f*)
D' R' U' R D U B2 U2 B' U B U B2  (13f*)
U B U B' U2 B2 D U L U' L' D' B2  (13f*)

uf,db flips:

L2 R2 D B' L D' B L2 R2 F' U L' F U'  (14f*)
U' B' U' B' U B2 U2 B' F' L' B L F U'  (14f*)
U' F R B R' B' F' U2 B2 U B' U' B' U'  (14f*)
U B U B U' B2 U2 B F R B' R' F' U  (14f*)
L2 R2 D' B R' D B' L2 R2 F U' R F' U  (14f*)
U F' L' B' L B F U2 B2 U' B U B U  (14f*)
L2 D B' L D' B L2 R2 F' U L' F U' R2  (14f*)
L2 D' B R' D B' L2 R2 F U' R F' U R2  (14f*)
L2 B D' R B' D L2 R2 U' F R' U F' R2  (14f*)
L2 B' D L' B D' L2 R2 U F' L U' F R2  (14f*)
L2 U F' L U' F L2 R2 B' D L' B D' R2  (14f*)
L2 U' F R' U F' L2 R2 B D' R B' D R2  (14f*)
D B' L D' B L2 R2 F' U L' F U' L2 R2  (14f*)
D' B R' D B' L2 R2 F U' R F' U L2 R2  (14f*)
B D' R B' D L2 R2 U' F R' U F' L2 R2  (14f*)
B' D L' B D' L2 R2 U F' L U' F L2 R2  (14f*)
U F' L U' F L2 R2 B' D L' B D' L2 R2  (14f*)
U' F R' U F' L2 R2 B D' R B' D L2 R2  (14f*)
F U' R F' U L2 R2 D' B R' D B' L2 R2  (14f*)
F' U L' F U' L2 R2 D B' L D' B L2 R2  (14f*)
L2 F U' R F' U L2 R2 D' B R' D B' R2  (14f*)
L2 F' U L' F U' L2 R2 D B' L D' B R2  (14f*)
L2 R2 B D' R B' D L2 R2 U' F R' U F'  (14f*)
F' D' F' D' F D2 F2 D' U' R' D R U F'  (14f*)
F' U L D L' D' U' F2 D2 F D' F' D' F'  (14f*)
F D F D F' D2 F2 D U L D' L' U' F  (14f*)
L2 R2 B' D L' B D' L2 R2 U F' L U' F  (14f*)
F U' R' D' R D U F2 D2 F' D F D F  (14f*)
D' B R F R' B' F' D2 F2 D F' D' F' D'  (14f*)
L2 R2 U F' L U' F L2 R2 B' D L' B D'  (14f*)
D' F' D' F' D F2 D2 B' F' L' F L B D'  (14f*)
D B' L' F' L B F D2 F2 D' F D F D  (14f*)
D F D F D' F2 D2 B F R F' R' B' D  (14f*)
L2 R2 U' F R' U F' L2 R2 B D' R B' D  (14f*)
R2 D B' L D' B L2 R2 F' U L' F U' L2  (14f*)
R2 D' B R' D B' L2 R2 F U' R F' U L2  (14f*)
R2 B D' R B' D L2 R2 U' F R' U F' L2  (14f*)
R2 B' D L' B D' L2 R2 U F' L U' F L2  (14f*)
R2 U F' L U' F L2 R2 B' D L' B D' L2  (14f*)
R2 U' F R' U F' L2 R2 B D' R B' D L2  (14f*)
R2 F U' R F' U L2 R2 D' B R' D B' L2  (14f*)
R2 F' U L' F U' L2 R2 D B' L D' B L2  (14f*)
B' D L U L' D' U' B2 U2 B U' B' U' B'  (14f*)
L2 R2 F U' R F' U L2 R2 D' B R' D B'  (14f*)
B' U' B' U' B U2 B2 D' U' R' U R D B'  (14f*)
B D' R' U' R D U B2 U2 B' U B U B  (14f*)
B U B U B' U2 B2 D U L U' L' D' B  (14f*)
L2 R2 F' U L' F U' L2 R2 D B' L D' B  (14f*)

A few remarks to my optimal solver program. Its input is the
output-textfile of my Cube Explorer program and it optimizes the
maneuvers of this textfile.

It is basically the phase1 of the Cube Explorer algorithm, but with Mike
Reid's idea of reducing the number of cube-states by symmetries it is
possible to put the whole phase1 in a pruning table of about 69MB (Mike
already did this more than a year ago in his own optimal solver
program). My optimal solver needs to generate about 20% less nodes than
Mike's when searching the tree and in this way does the search a bit
faster.

I do some more cosmetical work on the source code and then give it to
the public. The hardware requirements are about the same as for Mike's
solver, but it runs on the windows platform.

Herbert

From cube-lovers-errors@mc.lcs.mit.edu  Tue Apr 27 18:00:36 1999
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Date: Tue, 27 Apr 1999 15:07:29 -0400 (Eastern Daylight Time)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Conjugation done right [Re: Inventing your own techniques]
To: Cube Lovers <cube-lovers@ai.mit.edu>
Message-Id: <SIMEON.9904271529.A@GN209A.PSTCC.CC.TN.US>

On Thu, 08 Apr 1999 19:45:22 -0400 (EDT) Dan Hoey <Hoey@AIC.NRL.Navy.Mil> wrote:

> Jerry Bryan wrote:
>
> > A process of the form Y'XY is called a conjugate, and in particular is called
> > the conjugate of X by Y.  Note that YXY' is also a conjugate, and in particular
> > is called the conjugate of X by Y'.  This can be a little confusing because a
> > few books (incorrectly in my opinion) call YXY' the conjugate of X by Y.

  ... and Dan wrote:

>
> I tried to explain this a while ago, but I guess it didn't quite work.
> One form of conjugate is right, and the other form is wrong, but just
> which is right depends on how you write function composition.

My apologies for leaving out a couple of things in my note about conjugation
and commutators.  But if I hadn't left them out, I doubt that we would have
received Dan's very nice little message about rightward and leftward
composition.

I should have mentioned the rightward vs. leftward composition issue. As Dan
points out, it is certainly the case that Y'XY can be correct in one book while
YXY' can be correct in another.  However, what I *meant* to say was that I had
seen books in which (unless I was missing something obvious) the author's
definition of conjugation did not correspond to the author's composition
direction.

All I was really trying to say was that irrespective of the author's chosen
direction, YXY' and Y'XY are both conjugates.  It's just that in one case, you
have the conjugate of X by Y and in the other you have the conjugate of X by
Y', and which is which depends on the right vs. left system the author is using.

I cited the reason "X shifted by Y" as a reason for preferring the conjugate of
X by Y to be Y'XY in the Cube-Lovers system.  I should also have mentioned the
homomorphism vs. antihomomorhism issue.  There are two reasons for preferring
Y'XY to YXY' for the conjugate of X by Y in the Cube-Lovers system, and
regrettably I only mentioned one of them.  Homomorphism is the other.

Dan covered the homomorphism issue extremely well, so I would like to make some
additional comments about the "X shifted by Y" interpretation of conjugation.

Let us suppose that we have a maneuver

    A = L2 F2 L2 U  L' R B R2 D2 R2 B' L R' U'

which flips the uf cubie and the df cubie while leaving the rest of the cube
unchanged (the rest of the cubies are said to be fixed by A). The uf and df
cubies are the edge cubies which are in the middle of the top row and the
middle of the bottom row of the front face, respectively. (The given maneuver
for A is minimal in the face turn metric, but the exact maneuver doesn't matter
for our purposes.)

Suppose that instead we want to flip the ul cubie and the df cubie.  A maneuver
which will do so is U'AU.  The U' move brings the ul cubie into the uf cubie's
place while leaving the df cubie where it was.  The A maneuver flips the
*contents* of the uf cubicle which is now the ul cubie and flips the df cubie
as usual.  The U moves returns the (now flipped) ul cubie to the ul cubicle.

We might write the actions of the A maneuver as follows:

     A:  uf ->  fu  (flips the uf cubie)
         df ->  fd  (flips the df cubie)

We might write the actions of the U  and U' moves on the edge cubies as follows:

     U:  uf ->  ul          U':  uf -> ur
         ul ->  ub               ur -> ub
         ub ->  ur               ub -> ul
         ur ->  uf               ul -> uf

Hence, if we just consider the AU part of U'AU, we have that the uf cubie goes
to fu which in turn goes to lu.  (If U performs uf -> ul, then it equivalently
performs fu -> lu).  So the uf cubie is carried to the ul cubicle and flipped
to be lu.  This is the general idea of what we want (to flip the cubie which is
in the ul cubicle), but it is the wrong cubie in the ul cubicle.  So preceding
AU by U' "cancels" the movement of the cubie (and also the movement of all the
other cubies) and retains only the flip of the cube.  The net result is that the
cubie which is flipped is shifted from being the uf cubie to being the ul cubie,
as desired.  A is shifted by U, which is what we wanted.

It should be clear that UAU' flips the ur and the df cubies.  UAU' flips the ur
rather than the uf cubie, which we can describe as saying that A has been
shifted by U'.

Essentially any two edge cubies can be flipped by variations on this basic
theme.  Our next example will involve whole cube moves.  We denote the whole
cube move of grasping the right face and turning the whole cube clockwise and
counterclockwise by c_R and c_R', respectively.  So c_R' would bring the bu and
fu cubies into the uf and df cubicles (respectively), A would flip them, and
c_R would return them to their original locations.  The net result is that the
maneuver (c_R' A c_R) flips the bu and the fu cubies.

It is standard on Cube-Lovers to denote the 24 rotations of the cube by C, and
we might write a C-conjugate as c'Ac where c is some fixed but arbitrary
element of C.  c_R and c_R' are just two particular elements of C.  Working
with a real cube, you probably wouldn't even think about C-conjugation in this
particular context  --  you would just do it.  That is, if your hands knew how
to perform the A maneuver to flip the uf and df cubies, and if you needed to
flip two edge cubies which were on opposite sides of the same face, you would
just rotate the whole cube in space to bring the two cubies which needed to be
flipped into the uf and df locations and then you would perform the A maneuver
--  simpler to do than to describe.

It is more common on Cube-Lovers to talk about M-conjugation than to talk about
C-conjugation, where M is the group of 24 rotations and 24 reflections of the
cube.  C is a subgroup of M.  So c_R and c_R' just as well elements of M as
they are of C, and our (c_R' A c_R) maneuver is a good example of
M-conjugation.  M-conjugation lets us deal with reflections in addition to
rotations, which in effect means it let's us treat clockwise and
counterclockwise moves as equivalent when appropriate for symmetry purposes.

But when we are dealing with whole cube rotations of a real cube, we are just
dealing with C-conjugation.  In the case of our A maneuver, C-conjugation means
that c'Ac lets us flip any two cubies anywhere on the cube which are opposite
edge cubies on the same face of the cube.

Finally, whole cube rotations are a convenient way to apply the maneuver A to
any face of a real cube.  But mathematically, we really do not have to perform
whole cube rotations.  We can use C-conjugation (and more generally,
M-conjugation) to apply the "same" maneuver to a different face.  Consider
again (c_R' A c_R).  If we write out A, we get

     c_R' (L2 F2 L2 U  L' R B R2 D2 R2 B' L R' U') c_R

But the maneuver (c_R c_R') is equal to the identity, so we can insert it between
each face move thusly.

    c_R' L2 (c_R c_R') F2 (c_R c_R') L2 (c_R c_R') U (c_R c_R')  etc.

Now, we can re-associate thusly so that we have the c_R-conjugate of each face
turn.

    (c_R' L2 c_R) (c_R' F2 c_R) (c_R' L2 c_R) (c_R' U c_R) etc.

Finally, if we actually perform the calculations, we discover that conjugation
by c_R leaves L, L', L2, R, R', and R2 alone; it takes F, F', and F2 to U, U',
and U2, respectively; it takes U, U', and U2 to B, B', and B2, respectively;,
it takes B, B', and B2 to D, D', and D2, respectively; and it takes D, D', and
D2 to F, F', and F2, respectively.  Hence, conjugation by r_C gives us a
maneuver to flip the bu and fu cubies thusly.

            A = L2 F2 L2 U  L' R B R2 D2 R2 B' L R' U' (flip uf and df)
   c_R' A C_R = L2 U2 L2 B  L' R D R2 F2 R2 D' L R' B' (flip bu and fu)

----------------------------------------
Jerry Bryan
jbryan@pstcc.cc.tn.us

From cube-lovers-errors@mc.lcs.mit.edu  Tue Apr 27 18:45:51 1999
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Date: Tue, 27 Apr 1999 13:23:11 -0700 (PDT)
From: Tim Browne <ue451@victoria.tc.ca>
To: cube-lovers@ai.mit.edu
Subject: Square-1 problem
Message-Id: <Pine.GSO.3.95.iB1.0.990427130942.3580A-100000@vtn1>

Hey there. I've got a problem with my Square-1 puzzle. A few days ago, I
got a whole bunch of sugar grains stuck inside which are screwing up the
mechanism big time. At first it seemed as though the it was getting better
as the sugar was ground down and flushed out during each rotation, but now
it's getting affected by the dampness and the sugar's turning into a
pretty decent adhesive. The only real solution seems to be to take the
puzzle apart, clean it, dry it, and reassemble it. Someone on rec.puzzles
suggested soaking it in water, but given that the stickers are laminated
paper instead of plastic and the possibility of rusting the mechanism, I'd
rather not do this. Besides, I'd really like to see the internal workings
of this Machiavellian torture machine. :-) If anyone on this list knows
how to disassemble/reassemble the puzzle, I'd really appreciate it if you
could let me know how to do it. :-) Thanks in advance. L8r.

                     Cubic Puzzles - The SIMPLEST Solutions
                  http://www.victoria.tc.ca/~ue451/solves.html

From cube-lovers-errors@mc.lcs.mit.edu  Tue Apr 27 20:31:40 1999
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Date: Thu, 22 Apr 1999 21:48:58 -0400 (EDT)
From: Nicholas Bodley <nbodley@tiac.net>
Reply-To: Nicholas Bodley <nbodley@tiac.net>
To: Paddy Duncan <PaddyD@intralan.co.uk>
Cc: cube-lovers@ai.mit.edu
Subject: Re: WD-40 as a lubricant
In-Reply-To: <B11CE48E1A39D211BB540080C8646221013D92@is.intralan.demon.co.uk>
Message-Id: <Pine.BSF.3.96.990422213052.27206G-100000@shell2.tiac.net>

 Although I'm not really sure of my info, I believe WD-40 was
originally a Water Displacer, hence "WD". Fairly sure it was meant
to displace water from mechanisms made of metal. It's possible that
its lubricating properties were secondary. As a semi-amateur
mechanical technician, I gather that it's not intended to be used as
a long-term lubricant. Used as intended, it seems to be a very good
product.

 As to compatibility with plastics, I once was cleaning the inside
of a computer printer that had a clear plastic rack which engaged a
gear on the print head drive motor (which, amazingly enough, was on
the print head itself; no kidding). I was using rubbing alcohol, as
I remember. Whether it was the alcohol itself, or an adiitive to
dilute it, I don't know, but the rack crystallized, cracked, and
fell apart in only a few minutes. Astonishing to watch.

 Alcohol is usually a safe solvent for electronic work, but
apparently not for some plastic mechanical parts! Of course, I had
knowingly ignored the warning not to use any solvents for cleaning.

 Lesson from this is that there's some risk in applying liquids of
unknown compatibility to some plastics. HDPE won't be bothered by
any liquid you're likely to apply. Silicones (not spelled "silicons")
should be safe, very likely, and Teflon powder would be. (Just don't
get that powder near a cigarette or flame; decomposition products
are very poisonous.)

|*  Nicholas Bodley   *|*  Autodidact & Polymath * Electronic Tech. (ret.)
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*
|*  Amateur musician  *|*

From cube-lovers-errors@mc.lcs.mit.edu  Tue Apr 27 21:05:47 1999
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From: "Chris and Kori Pelley" <ck1@home.com>
To: "Tim Browne" <ue451@victoria.tc.ca>, <cube-lovers@ai.mit.edu>
Subject: RE: Square-1 problem
Date: Tue, 27 Apr 1999 20:50:46 -0400
Message-Id: <LLBBJDCFCKLJHGBBOAGOMEFFCEAA.ck1@home.com>

> -----Original Message-----
> From: Tim Browne [mailto:ue451@victoria.tc.ca]
> of this Machiavellian torture machine. :-) If anyone on this list knows
> how to disassemble/reassemble the puzzle, I'd really appreciate it if you

I know how to disassemble it!  First you get it into the original shape that
it comes in, and turn 180 degrees.  Then you twist it about 22 degrees and
pry out one of the dart (edge) pieces.  This can best be seen in a diagram,
and I have a rough sketch at this address:

http://members.home.net/chrisandkori/sq-disassem.jpg

Once one piece is out, they all come out easily, similar to Rubik's Cube.
Putting it back together is similar-- do the same thing in reverse.  The
mechanism is really neat!

Chris Pelley
ck1@home.com

From cube-lovers-errors@mc.lcs.mit.edu  Fri Apr 30 20:25:46 1999
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Mail-from: From cube-lovers-request@life.ai.mit.edu Fri Apr 30 07:35:29 1999
Date: Fri, 30 Apr 1999 12:33:43 +0100
From: David Singmaster <david.singmaster@sbu.ac.uk>
To: whuang@ugcs.caltech.edu
Cc: cube-lovers@ai.mit.edu
Message-Id: <009D768E.2615E470.8@ice.sbu.ac.uk>
Subject: RE: Reinventing (and some edge-flipping techniques)

	I've just spent three days being a Rubik's Cube demonstrator at a trade
fair, using the method given the my Step by Step solution in the middle of the
fifth edition of my Notes.  The problems that Wei-Hwa Huang has described are
ones that I found in trying to develop an easy method.  Basically there is no
'simple' way to flip edges, where 'simple' means easily understood and
remembered.  There are simple ways to move edges, move corners and twist
corners.  Consequently I decided to get the edge orientations correct at the
beginning of work on the last face, so that I wouldn't have to worry about what
happened to the rest of the face.  In case you haven't got my notes at hand, I
used  BLUL'U'B'  which is a simple conjugate of the commutator  [L,U].  This
exchanges the four  U  corners as two pairs of exchanges and cycles three  U
edge, but effectively flips two of them on the way.  Using the inverse process
BULU'L'B'  does the same thing, but one of them effectively flips two adjacent
edges and the other flips two opposite edges.  Two applications will flip all
four edges.  Then we know simple processes which do 3-cycles or pairs of
2-cycles of edges, preserving orientation, or of corners and we have simple
processes for twisting corners.  So we could carry out these three steps in any
order giving six possible algorithms and I'm pretty sure I received examples of
all of these.  Indeed, of one considers doing the last face as having a fourth
stage of orienting edges, there are 24 possible algorithms and at one point I
was classifying algorithms into these 24 cases.  I don't think I kept up with
this long enough to have all 24 cases, but I expect all of them exist!
	Some personal comments and recollections, much of which is in my Notes.
	The ideas of monotwist and monoflip, though blindingly obvious, took
well over a year to emerge!  Despite the fact that lots of quite bright
mathematicians were working on the cube (e.g. Conway, Penrose, Rubik), I
remember first hearing about the idea in Jan 1980 (?) from Peter McMullen who
said they were using the idea at Cambridge.  At first it seemed unreasonable as
we generally were looking for moves that only affected the  U  face and the
mono-moves are almost all elsewhere.  However once I realised that the idea
gives a way of building simple moves, I realised that the commutator  [F,R]
was a mono-move in the  L  face and its square was a mono-twist in the  L
face.  The Cambridge group had been using shorter, but less simple, moves.
With these mono-moves, it was now pretty easy to build the algorithm that is my
Step by Step solution and I think I did it within a few weeks as I recall the
5th ed. of my Notes was produced by March.  (Remebering dates from 20 years
ago is always a bit dodgy - check what's in the Notes, which I don't have a
copy of here.)

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499;
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Mon May  3 13:47:11 1999
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Message-Id: <372A817B.5825@zeta.org.au>
Date: Sat, 01 May 1999 14:22:19 +1000
From: Wayne Johnson <sausage@zeta.org.au>
Reply-To: sausage@zeta.org.au
To: cube-lovers@ai.mit.edu
Subject: Rubiks Magic Strategy Rules

Does anyone know how to play the Rubiks Magic Strategy game? I just
bought it second hand and while there's some rough instructions on the
back, the manual is missing and we don't know how to start off.

I'd appreciate any help.

Thanks,
Wayne Johnson,
www.zeta.org.au/~sausage

From cube-lovers-errors@mc.lcs.mit.edu  Mon May  3 15:29:40 1999
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Date: Sat, 1 May 1999 13:37:32 -0400 (EDT)
From: der Mouse  <mouse@rodents.montreal.qc.ca>
Message-Id: <199905011737.NAA19339@Twig.Rodents.Montreal.QC.CA>
To: cube-lovers@ai.mit.edu
Subject: Re: Reinventing (and some edge-flipping techniques)

> Basically there is no 'simple' way to flip edges, where 'simple'
> means easily understood and remembered.

On what point does the Spratt wrench fail?  That's always been my
favored edge-flipper.  (On the 3-Cube, that is; when solving a
dodecehedral puzzle I bought from Mr. Bandelow, I was forced to develop
other edge-flippers, and the one I ended up with maps easily onto the
Cube.  In 3-Cube terms, it's based on R F' R' F, which induces a
3-cycle (fr,fd,dr) on edges.  By applying this, rotating the cube 120 degrees
about its rfd-lbu long diagonal, and applying the inverse, you can get
a two-edge flipper at the price of disturbing four corners.  The
dodecahedron I solve by doing edges first with this procedure, then
using (the dodecehedral analog of) (R F' R' F) 3, which leaves edges
alone and produces two corner swaps, to fix up the corners.  But when
solving the 3-Cube, I still find the Spratt wrench more convenient.)

What I'd like is a puzzle like the dodecahedron, but with an additional
turning mode: 72 degrees on a cut through the center.  The puzzle as it stands
has face-cut lines that make it clear such turns are conceivable,
though designing a mechanism for them would be interesting.  Perhaps
when we get force-reflecting datagloves.... :-)

					der Mouse

			       mouse@rodents.montreal.qc.ca
		     7D C8 61 52 5D E7 2D 39  4E F1 31 3E E8 B3 27 4B

From cube-lovers-errors@mc.lcs.mit.edu  Mon May  3 17:26:27 1999
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Message-Id: <199905031954.VAA24148@bednorz.get2net.dk>
From: "Klodshans" <klodshans@get2net.dk>
To: sausage@zeta.org.au
Date: Mon, 3 May 1999 20:52:12 +0000
Subject: Re: Rubiks Magic Strategy Rules
Reply-To: klodshans@get2net.dk
Cc: cube-lovers@ai.mit.edu
In-Reply-To: <372A817B.5825@zeta.org.au>

Wayne wrote:
> Does anyone know how to play the Rubiks Magic Strategy game? I just
> bought it second hand and while there's some rough instructions on the
> back, the manual is missing and we don't know how to start off.

The game is similar/equivalent to Rubik's Eclipse. The rules are as
follows:

-------------------------------------------------------------------------

Set Up:
At the beginning of the game, the board is empty. One player
takes the 8 circle pieces, and the other player takes the 8 square
pieces.

Objective:
To win the game you must get 3 of your game pieces of the same
color (black or silver) in a "locked" row, either vertically, horizontally
or diagonally. A row of 3 is "locked" when none of the game pieces
in it can be flipped onto an adjacent square, because all the
adjacent squares are occupied.

Rules of play:
Players alternate in taking turns. The Circle player starts the game
by placing a circle piece, either side up, somewhere on the board.
Subsequently, starting with the Square players's first move, each
player's turn consist of two actions: Flip and Place,  _in that order._

Flip:
On each turn, the player must flip _any_ one of their opponent's
pieces already on the board. Flipping a piece means moving it to
any adjacent vacant square (horizontally or vertically, but not
diagonally) while _reversing_ the sides. The grey becomes black
and vice versa.

Place:
Each player must, on each turn, then place one of their pieces,
either side up, on any vacant square on the board, _even if this
results in a winning line for the opponent._

Note: A player _must_ flip an opponent's piece, even if this may
result in a winning line for the opponent.

If it is not possible to flip any of their opponent's pieces, the player
still goes on to placing a piece of their own.

Play goes on in this manner, with each player first flipping any of
their opponent's pieces (not necessarily the last one placed),
before placing a piece of their own. This means once placed on the
board,, players will never touch their own pieces again, but they
can change position and color of their opponent's pieces.

A player may leave an opponent's row of 3 without loosing the
game (provided it is not "locked").

Winning the Game:
The first player to archieve a "locked" vertical, horizontal, or
diagonal line of 3 of their _own_ game pieces, all with the same
color (black or silver), is the winner. A "locked" line is one in which
none of the 3 game pieces can be flipped because all the adjacent
squares are occupied.

If the game continues until the last piece is placed on the board,
and both players end up with a winning line, or neither player has
won, the game is a draw.

-------------------------------------------------------------------------
______________________________________
Philip K
E-mail: philipk@bassandtrouble.com
E-mail: klodshans@get2net.dk
web:    http://hjem.get2net.dk/philip-k

From cube-lovers-errors@mc.lcs.mit.edu  Tue May  4 15:15:59 1999
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Message-Id: <009301be95b5$fc951340$70c4b0c2@home>
From: roger.broadie@iclweb.com (Roger Broadie)
To: <cube-lovers@ai.mit.edu>
Cc: "David Singmaster" <david.singmaster@sbu.ac.uk>
Subject: Re: Reinventing (and some edge-flipping techniques)
Date: Mon, 3 May 1999 23:40:28 +0100

David Singmaster wrote (1 May 1999)

>Consequently I decided to get the edge orientations correct at the
>beginning of work on the last face, so that I wouldn't have to worry
>about what happened to the rest of the face. In case you haven't got
>my notes at hand, I used BLUL'U'B' which is a simple conjugate of the
>commutator [L,U].

When I read David Singmaster's algorithm in his Notes on Rubik's Magic
Cube I thought his idea of curing the orientations of the edge-piece
before positioning them was a great one, and adopted it in my own
algorithm.  One may be able to get away with only 6 turns to cope with
the orientation, which is quite a gain on the number of turns needed
if one takes the obvious route and orients once the pieces are in
position.

I then went on to apply the same principle to the corner pieces, by
orienting them next, which can be done in 7 or 14 face turns by
selecting the first of the following (for three twists) or the first
followed by the second suitably applied (for two or four twists):

F U2 F' U' F U' F'
L' U2 L U L' U L

These processes, which I mentioned in my last post, move the corner
pieces as well as twist them, and also move the edge pieces but
preserve their orientation.  So, if applied after the edge pieces have
been oriented but not positioned, they get the top face of the cube
the right colour, and it can then be solved by orientation-preserving
moves of the edge and corner pieces.  I may be rather bad at
perceiving patterns, but I find the simplification of having to look
only at the faces of the top-layer pieces that lie in the side faces
of the cube in order to work out what needs to go where enough of an
advantage to make this order of the stages worthwhile even if it did
not take fewer moves in total.

Roger Broadie

From cube-lovers-errors@mc.lcs.mit.edu  Tue May  4 16:04:16 1999
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From: roger.broadie@iclweb.com (Roger Broadie)
To: <Cube-Lovers@ai.mit.edu>
Cc: "Frederick W. Wheeler" <wheeler@cipr.rpi.edu>
Subject: Re: Inventing your own techniques
Date: Mon, 3 May 1999 23:40:04 +0100

I'd like to return to Fred Wheeler's interesting question (30 March
1999), for its own sake and partly as a prelude to my next posting.

The processes I originally used to solve the cube - before I had been
introduced to commutators by David Singmaster's little blue book -
were based on a principle that formed itself in my head as I explored
the cube.  I called it "Out and back by a different route".  If pieces
were moved away from their original position by one process and then
restored by a different one, the result would be a process that would
only move pieces other than those moved out and back.  Then it struck
me that the different routes could be based on the obvious processes
used in a bottom-up algorithm to move a bottom corner piece into
position from the top corner vertically above it, depending on the
orientation of the piece. Thus, taking the DLF piece as the one
subject to the out-and-back movement, the out trip would take the
piece to FLU using front face turns, and the return route would bring
it back using left face turns:

F U' F'. U' L' U L  (1)

This process, I discovered, moved an edge piece out of the top layer
into the middle, UB > FL, but had no other effect on the middle layer.
So, at once, it formed the basis for solving the middle layer.

Besides inverses and reflections there is one other essentially
different process of this type involving the DLF corner.  It takes the
piece to the far corner RUB on the out trip:

F U2 F' .  U2 L' U' L  (2)

And it too moves just one edge piece out of the top into the middle
layer, from a different source position but to the same target
position as (1), UF > FL.

We can now repeat this approach, using these two processes and their
inverses and reflections to take a piece out of the top layer and then
return it.  That generates a set of upper-layer processes that led me
to my first solution.  It was not very efficient, but one of these
process is very attractive.  It results from following (2) above with
the reflection of (1) in the diagonal plane FL-BR:

F U2 F' U2 L' U' L  . L' U  L U F U' F'    =  F U2 F' U' F U' F' (3)

It is short, because of the cancellation, very easy to do, because all
the turning can be done with one hand, and has a very useful cycle
pattern, with an untwisted 3-cycle of edge pieces and a twisting pair
of swaps of corner pieces.

That leads on to another basic technique: looking at the cycle pattern
of a process and seeing what can be done to suppress or simplify some
of the cycles.

If we take (3) above, by combining it with its reflection suitably
applied, we can suppress either the movement of the edge pieces to
give a double twist of the corners (4), or the movement of the corner
pieces to give an untwisted edge 3-cycle (5):

F U2 F' U' F U' F' B' U2 B U B' U B UF(L-, R+) (4)
F U2 F' U' F U' F' L' U2 L U L' U L U(F, L, R) (5)

There are shorter alternatives, of course, but (4) remains my
favourite for the purpose, done as

  F U2 F' U' F U' F' [twist whole cube parallel to U'] L' U2 L U L' U
L

Of course, taking powers of a process is one way selectively to
eliminate constituent cycles, and (1) above, if done four times, is a
triple corner twist, because the edge pieces move through a 4-cycle
which is eliminated, one corner piece undergo a twist with no change
of position, which is preserved, and the other two undergo a twisted
2-cycle that leaves them in position but twisted.  But powers tend to
be rather lengthy, as this example illustrates.

Then one can look for patterns in processes and apply them elsewhere.
(3) above can be thought of as taking DLF up to the top and then round
and home again, with the + and - turns of the F face cancelling out.
One other process I discovered on the cube using this principle is the
following, where M is the turn of the centre slice parallel to R:

M2 U M2 U2 M U M2  (UF,UB) (UR, UL) (6)

And the structure of (3) transfers directly to the tetrahedron as

F U F' U F U F' 3-cycle round the vertical axis

and the dodecahedron as

F U2 F' U' F U' F' and  F U2 F' U F U2 F' edge 3-cycles + corner
double-swaps

In these I am taking the puzzle to be sitting on a table, with U being
the vertex (for the tetrahedron) or face (for the dodecahedron) at the
top and F being a vertex or face adjacent to the top pointing towards
you.

For a discussion of designing 3-cycles using commutators and
conjugates, see the message I posted in November 1997, which looked at
processes of the type

[P, TQT']

where P and Q are turns of layers that are parallel to one another,
and T is a turn of a layer transverse to P and Q.   These processes
yield a result that can be expressed as second-level commutation, as
mentioned by David Singmaster.

Roger Broadie

From cube-lovers-errors@mc.lcs.mit.edu  Tue May  4 16:30:21 1999
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From: WaVeReBeL@webtv.net
Date: Mon, 3 May 1999 22:17:36 -0700 (PDT)
To: Cube-Lovers@ai.mit.edu
Subject: Attention Star Wars Fans!
Message-Id: <2346-372E82F0-11792@postoffice-123.bryant.webtv.net>

I just got back from Target & saw a new Rubik's product.  It is a 2x2x2
puzzle of Darth Maul's head! (he's from Star Wars: Episode 1)  It comes
in the usual cardboard & plastic box except it is colored in Black &
red.  I think it is $14.99.

I haven't had a chance to buy one and try out the mechanism though,
because I spent all my money hoarding the action figures....

-Alex-

May the force be with you...

From cube-lovers-errors@mc.lcs.mit.edu  Wed May  5 14:09:02 1999
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Date: Tue, 4 May 1999 21:56:22 +1000 (EST)
Message-Id: <199905041156.VAA09100@pcug.org.au>
Content-Type: text/plain; charset="us-ascii"
To: Cube-Lovers@ai.mit.edu
From: pfoster@pcug.org.au (Peter Foster)
Subject: Re: Speed cubing results - March 99

>Almost everybody responded w/ a different preview time (anywhere from no
>preview to 15 seconds).  People like me who started cold had a
>disadvantage to those who had a preview.  A 15 second preview sounds
>good to me.  This gives enough time to familiarize oneself w/ the cube,
>look for pieces, and plan out the first few moves.  I've been timing
>myself cold which means much time is wasted at the beginning.  Having a
>preview helps a lot.

I would prefer that there is no preview time. Solving the cube isn't just 
turning the faces, it's also deciding which faces to turn. Surely, time spent 
in examining the cube is therefore part of the solving time.

>When it comes to averages, I guess there is no standard.  I agree w/
>disregarding the high & low extremes though.  They can distort the
>average (arithmetic mean).  This should give a more accurate
>representation.  Also, the more entries calculated into the average the
>better.

I used to do a run of 11, then take the median (middle) value. 

Cheers
_______________________________________________________________
Peter Foster        
pfoster@pcug.org.au

[Moderator's note: While would I avoid previews in principle, I
 understand that the official contests were carried out with a free
 preview period.  It seems unavoidable that we qualify all our records
 with the conditions under which they are measured--with or without
 preview; best score, average, or median; by sight or by touch; in air
 or underwater; solo or mixed doubles; etc. --Dan ]

From cube-lovers-errors@mc.lcs.mit.edu  Wed May  5 14:48:26 1999
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To: cube-lovers@ai.mit.edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Algorithm for the Antislice Group
Date: 4 May 1999 14:11:57 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <7gmv7d$ncj@gap.cco.caltech.edu>

In a small fit of playing around last week, I finally worked out a
method of solving the Antislice Group.

Of course, God's Algorithm is known, but since it's pretty hard for
me to memorize it, here's mine.

For those of you that don't know, the antislice group is all
configurations that the cube can be in, provided that the only
move you make is FB, with rotations and reflections.

So, with fixed centers, the only allowed moves are FB, RL, and UD.
I'll abbreviate these to F, R, and U.   I'll even use the "clockwise
quarter turn" metric, which means that F' = FFF is three moves, not one.

Okay.  The algorithm has four steps:
 A.  Get the cube to be a subset of the half-face turn group.
      (up to 4 moves)
 B.  Match the corners.
      (up to 6 moves)
 C.  "HOt HOt HOt" (double-slice group).
      (up to 12 moves)
 D.  Solve the cube.
      (up to 6 moves)

This is a maximum of 28 moves.  In practice, one can often avoid future
extra moves by modifying the algorithm slightly.

A.  Get the cube to be a subset of the half-face turn group.

  This means that the UD faces only contain the UD colors, the FB faces
  only contain the FB colors, and the RL faces only contain the RL colors.
  Here's how to do it:

 A0.  Pick any set of colors on opposite face centers.  Let's call it the
  color "grud".  (For instance, if blue is opposite white on the solved cube,
  "grud" can mean "blue or white".)  We'll try to solve "grud" first.
 A1.  Find a face with grud at the corners and reorient the cube
  so that it is the U face.
 A2.  If the U center is grud but none of the U edges are grud, perform
  R, repeat step A1, and go to step A5.
 A3.  If the U center is not grud but all of the U edges are grud, then
  either the F center is grud or the R center is grud.  If the F center is
  grud, perform F; if the R center is grud, perform R.  Then repeat step
  A1 and go to step A5.
 A4.  If the U center is grud and two of the U edges are grud (so there
  is a grud "H" shape on the U side), then either F has two grud edges or
  R has two grud edges.  If the former, perform F; if the latter, perform R.
  Then repeat step A1 and go to step A6.
 A5.  If the U center is not grud, two of the U edges are grud, and
  every face has some grud on it, perform U and go to step A7.
 A6.  If the U center is not grud and none of the U edges are grud,
  either the F center is grud or the R center is grud.  If the F center is
  grud, perform F; if the R center is grud, perform R.  Then repeat step
  A1 and go to step A7.
 A7.  If the U center is not grud, two of the U edges are grud, and
  either F or R does not have grud on it, then perform F or R (whichever
  one did not have grud on it), repeat step A1, and go to step A8.
 A8.  The U face should now be all grud.  If the other four faces are
  "unsolved" (have colors of adjacent faces), perform U and it should
  be solved.

B.  Match the corners.

 In other words, does each face have four corners of the same color?  If so,
 go on to step C.  Otherwise, at least one face will have four corners of
 the same color.  Orient the cube so that it is the U face, and
 then perform FRFRFR.  This should match the corners.

C.  "HOt HOt HOt" (double-slice group)

 This step is to get rid of any face that looks like an "H" (all one color
 except for two opposite edges), an "O" (all one color except for the
 center), or a "t" (all one color except for the corners).  This may
 have to be done up to three times, and after they are removed
 the cube should be in the double-slice group.

 Orient the cube so that the top face is "HOt".  Then, perform FRRF.
 Repeat until no faces are "HOt".

D.  Solve the cube.

 The cube should now be in a very easy to solve state, if not already solved
 already.  If the cube is not already solved, there must be some
 corner which does not match an edge next to it.  Orient the cube so
 that said corner is the UFR corner and the edge is the RU edge.  Then,
 perform FF.  Repeat until the cube is solved.


--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
..make lemonade and hope it also gives you a source of protein and vitamin A.

From cube-lovers-errors@mc.lcs.mit.edu  Fri May  7 11:26:58 1999
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Date: Fri, 7 May 1999 12:27:25 +0100
From: David Singmaster <david.singmaster@sbu.ac.uk>
To: mouse@rodents.montreal.qc.ca
Cc: cube-lovers@ai.mit.edu
Message-Id: <009D7C0D.6D45DA64.17@ice.sbu.ac.uk>
Subject: Re: Reinventing (and some edge-flipping techniques)

	Der Mouse's message reminds me of a point which I had meant to include.
In my algorithm, most of the moving involves just two adjacent faces, which I
find easier to remember and to carry out.  The Spratt wrench (which I don't
really know, so I'll have to try it, BTW, who was/is Spratt?) doesn't have
quite this simple structure.

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499;
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Fri May  7 17:25:58 1999
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From: "Chris and Kori Pelley" <ck1@home.com>
To: "Cube Lovers" <cube-lovers@ai.mit.edu>
Subject: Darth Maul 2x2x2 Cube
Date: Fri, 7 May 1999 08:05:46 -0400
Message-Id: <LLBBJDCFCKLJHGBBOAGOAEHKCEAA.ck1@home.com>

I've put a couple photos of the new Darth Maul Cube on my home page, in case
anybody hasn't seen it yet.  The cube turns very smoothly, and is basically
just an oversized 2x2x2.  I was surprised to discover they are made by
OddzOn.

http://www.chrisandkori.com/darthmaul.htm

Chris Pelley
ck1@home.com

From cube-lovers-errors@mc.lcs.mit.edu  Fri May  7 20:53:13 1999
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From: WaVeReBeL@webtv.net
Date: Fri, 7 May 1999 13:02:42 -0700 (PDT)
To: Cube-Lovers@ai.mit.edu
Subject: Thanks all!
Message-Id: <9973-373346E2-9532@postoffice-121.bryant.webtv.net>

Hello all,

I just want to say thanks to all the great help & advice everyone has
given me.  I wish I had time to E-mail everyone personally.  I'm so glad
I found out about this mailing list & all of the great web pages.  (Yay
internet!)  You have helped me reach my goal of beating the 30 second
barrier.

It all started w/ a little book fair where I picked up "The Simple
Solution to Rubik's Cube" by James G. Nourse for $.25 last year.  It
then snowballed, and I learned everything I could about the cube, spent
tons of money on different puzzles, & put a lot of time into it.  After
the first 4 months of intense cubing, my records are:

LAYER METHOD
30" best
36" average

CORNERS FIRST
26" best
37" average

I've been on a break for a while now ever since my hands started to
hurt.  Already I've started to loose my edge, & forgot some of the
moves.  I'm slower now, but I can still solve it under 45" guaranteed.
When the summer comes, school will be over and I'll have more free time
to train again  & hopefully reach my new goal of consistently finishing
under 30".  For now I'll give my hands a well deserved rest.

I just bought that book for a quarter on a whim.  Who knew a little cube
would have such a great impact in my life.  Thanks again for all of the
support.

-Alex Montilla-

From cube-lovers-errors@mc.lcs.mit.edu  Fri May  7 21:12:30 1999
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Message-Id: <372B283E.428F@hrz1.hrz.tu-darmstadt.de>
Date: Sat, 01 May 1999 18:13:50 +0200
From: Herbert Kociemba <kociemba@hrz1.hrz.tu-darmstadt.de>
Reply-To: kociemba@hrz1.hrz.tu-darmstadt.de
To: Cube Lovers <cube-lovers@ai.mit.edu>
Subject: Optimal solver for Windows now available

I finished my optimal solver program for Windows now. It optimizes the
maneuvers written to the output file of the Cube Explorer program.

I am quite satisfied with the performance. I searches 1,100,000 nodes
per second on a P350. 90MB of RAM should be enough (though I run it on a
machine with 256MB). A depth 17 search on random cubes typically takes
less than one hour. I made a test run with 10 random cubes and optimal
solutions for all of them were found within 35 hours (6 had length 18
and 4 had lenght 17).

The exe file and the source code are available at

http://home.t-online.de/home/kociemba/cube.htm

Herbert Kociemba

[ Moderator's note: This is also available in

     ftp://ftp.ai.mit.edu/pub/cube-lovers/contrib/opt09.zip

  If you've been waiting for the cube-lovers archives, the server is
  serving again. -- Dan ]

From cube-lovers-errors@mc.lcs.mit.edu  Tue May 11 13:03:08 1999
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From: WaVeReBeL@webtv.net
Date: Fri, 7 May 1999 13:52:12 -0700 (PDT)
To: Cube-Lovers@ai.mit.edu
Subject: "Negative" Turning
Message-Id: <11000-3733527C-9578@postoffice-123.bryant.webtv.net>

When the cube locks up on me, I have to turn the layer in the opposite
of the desired direction anywhere from a little jiggle to over 90 degrees in
order to straighten out the pieces.  I'm sure all of you have to do this
too.  I'm calling it "negative" turning.  Is there another term for
this?  I do this rather unconsciously now.

All this locking up wastes time, especially with numerous unsuccessful
efforts.  I'm guessing when fractions of seconds count, time spent
fixing could be spent doing more turns.

I'm wondering, do any of you perform negative turning before starting
any turns to prevent a lock up before it starts rather than try and fix
it after it happens?  Or do you incorporate these negative movements
within combos in anticipation of frequent sticky situations?  When
practicing frequently used combos I have incorporated negative turns
into them with some success.  Does anyone do this to all moves?  It
would take a long time to achieve this w/ all of my combos, so I'm
looking for someone who has.  Would it take longer to do combos w/
negative moves "built in", or just fix the problem when in comes.  Or
maybe there is someone out there who makes turns so accurately he never
experiences this.

Thanks.

-Alex Montilla-

From cube-lovers-errors@mc.lcs.mit.edu  Tue May 11 15:16:14 1999
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Date: Tue, 11 May 1999 11:25:03 -0700
From: David Barr <davidbarr@iname.com>
To: WaVeReBeL@webtv.net, cube-lovers@ai.mit.edu
Subject: Re: "Negative" Turning
References: <11000-3733527C-9578@postoffice-123.bryant.webtv.net>

The reason that "negative turning" is necessary is because the prior
turn did not get the faces lined up exactly.  Let's look at the sequence
"R U".  You turn the right face a quarter turn, then you turn the upper
face a quarter turn.  If you make the R turn a fraction of inch too far,
then the cube will correct itself when you try to make the U turn.  But
if you make the R turn a fraction of an inch too short, the upper face
will lock if you first try to turn it clockwise.  The "negative turn"
doesn't lock up because it is counterclockwise.  When you then start to
turn the U face counterclockwise it straightens up the R face so that
the U face is free to turn in either direction.

"Negative turning" is sometimes necessary, but a better solution is to
always either overturn or underturn by a fraction of an inch depending
on which way you plan on making the next turn.  So if you are planning
on turning "R U", make sure your "R" is a fraction of an inch past a
quarter turn, but if you are planning to turn "R U'", make sure your "R"
is a fraction of an inch short of a quarter turn.

David

From cube-lovers-errors@mc.lcs.mit.edu  Tue May 11 17:17:19 1999
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Date: Thu, 6 May 1999 12:41:20 -0700
From: Gene Johannsen <gej@spamalot.mfg.sgi.com>
To: Wei-Hwa Huang <whuang@ugcs.caltech.edu>
Cc: cube-lovers@ai.mit.edu
Subject: Re: Algorithm for the Antislice Group
In-Reply-To: <7gmv7d$ncj@gap.cco.caltech.edu>
Message-Id: <Pine.SGI.4.05.9905061235400.222639-100000@spamalot.mfg.sgi.com>



On 4 May 1999, Wei-Hwa Huang wrote:

> [snip]
>
> B.  Match the corners.
> 
>  In other words, does each face have four corners of the same color?  If so,
>  go on to step C.  Otherwise, at least one face will have four corners of
>  the same color.  Orient the cube so that it is the U face, and
>  then perform FRFRFR.  This should match the corners.

	I am having problems with this step.  My cube is in a
	configuration that this maneuver does not solve:

                YYY
		YYG  <- Back face
		GGG

	    BBW ORO BBW ROR
	    BBB ROR WWW ORO  <- Bottom face
	    BWW ORO BWW ROR

		YYY
		YGG  <- Front face
		GGG


	FRFRFR does not solve the corners for this position.

gene




From cube-lovers-errors@mc.lcs.mit.edu  Wed May 12 16:54:45 1999
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To: cube-lovers@ai.mit.edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Re: Algorithm for the Antislice Group
Date: 11 May 1999 22:54:04 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <7hacec$rs6@gap.cco.caltech.edu>
References: <cube-lovers.Pine.SGI.4.05.9905061235400.222639-100000@spamalot.mfg.sgi.com>

Gene Johannsen <gej@spamalot.mfg.sgi.com> writes:

>	I am having problems with this step.  My cube is in a
>	configuration that this maneuver does not solve:

As I e-mailed Gene, I do not believe his configuration is part of the
anti-slice group -- would any members care to give a quick heuristic to
determine if a cube is in the anti-slice group?
--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
..make lemonade and hope it also gives you a source of protein and vitamin A.

From cube-lovers-errors@mc.lcs.mit.edu  Wed May 12 18:20:43 1999
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Date: Wed, 12 May 1999 00:19:21 -0400 (EDT)
From: Nicholas Bodley <nbodley@tiac.net>
To: WaVeReBeL@webtv.net
Cc: Cube-Lovers@ai.mit.edu
Subject: Re: "Negative" Turning
In-Reply-To: <11000-3733527C-9578@postoffice-123.bryant.webtv.net>
Message-Id: <Pine.BSF.3.96.990512001439.27538J-100000@shell2.tiac.net>


Sorry to say (because it's no longer made), the Deluxe Cube from
Ideal had differently-shaped details inside so that with moderate
amounts of misalignment, it would self-align as you began a move. It
also had plastic colored tiles attached.

"Negative" turning seemed to be unnecessary if you used ordinary
care, but I wasn't trying for speed.

|*  Nicholas Bodley   *|*  Autodidact & Polymath * Electronic Tech. (ret.)
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*  Keep smiling! It makes people wonder
|*  Amateur musician  *|*  what you've been up to.
--------------------------------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Tue May 18 20:21:45 1999
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Reply-To: <jburkhardt@mediaone.net>
From: "John Burkhardt" <jburkhardt@mediaone.net>
To: <cube-lovers@ai.mit.edu>
Subject: "The" Hungarian cube ??
Date: Tue, 18 May 1999 05:41:10 -0400
Message-Id: <000301bea112$904d7070$3cca8018@octopod.ne.mediaone.net>

I recently bought a rubik's cube from someone.  It was in a blue cardboard
box that has Hungarian all over it.  The closest thing to a logo says:
"Politoys" and under that it reads: Hungary Budapest.  On one side of the
box there is a long three paragraph description which I can't read and it's
signed by E. Rubik.

When I opened it up I found a cube that looks in every way identical to the
Ideal version of the rubik's cube.  Same colors.  Same "Rubik's cube" logo
in the center white square.

Does anyone know if this is, in fact, a very early version of the cube or
did someone put an Ideal cube in this box?  What was the first and original
cube?  The guy who sold it to me is also very curious.  Could it be that
Ideal took over and this is a later production that was begin sold in
Hungary?  I'm wondering then why there is no indication of "Ideal" anywhere
on it.

-John

[ See ftp://ftp.ai.mit.edu/pub/cube-lover/cube-mail-0.gz for
  discussions of early Hungarian and American cubes.  -- Dan ]

From cube-lovers-errors@mc.lcs.mit.edu  Tue May 18 20:55:53 1999
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Date: Tue, 18 May 1999 19:39:16 +0100
From: David Singmaster <david.singmaster@sbu.ac.uk>
To: cube-lovers@ai.mit.edu
Cc: chrisit@seventowns.com
Message-Id: <009D84EE.93F50834.1@ice.sbu.ac.uk>
Subject: Speed cubing

	A UK TV program is looking for a speedy cubist.  I suppose anything
under a minute would be considered speedy.  Is there anyone in the UK who has
this kind of current speed?  If so, could they email  chrisit@seventowns.com 
and send a copy to me since I've had some other possible inquiries.

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499;
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Tue May 18 21:21:14 1999
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From: "Jason Werner" <mrhip@sgi.com>
Message-Id: <9905181213.ZM1550@sgi.com>
Date: Tue, 18 May 1999 12:13:44 -0700
To: Cube-Lovers@ai.mit.edu
Subject: Star Wars mania

Picked this up at Target a couple days ago:

	http://reality.sgi.com/mrhip/dmrc.jpg

If you haven't seen it in person, it appears to be just a 2x2x2.  Haven't
opened the package; don't plan to. :)

	-Jason

--
Jason K. Werner                      Phone: 650.933.9393
Systems Administrator                Fax:   650.932.9393
SGI, CSBU Division                   Pager: 650.317.7954, mrhip_p@sgi.com
mrhip@sgi.com                        URL:   http://reality.sgi.com/mrhip

[ Moderator's note: This is the toy Chris Pelley mentioned a few days ago,
  but pictured in a box. ]

From cube-lovers-errors@mc.lcs.mit.edu  Wed May 19 13:49:11 1999
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From: Christine Trussell <ChrisiT@SEVENTOWNS.COM>
To: <cube-lovers@ai.mit.edu>
Subject: Customised Cubes
Date: Wed, 19 May 1999 15:59:20 +0100

Have you customised your cube?

If so please let me know what you have done to it to make it run
faster, smoother etc.

We will pay for this information.

Thanks and look forward to hearing from you
Regards,
Chrisi Trussell
Seven Towns Ltd
7 Lambton Place
London W11 2sH
Tel:	44 171 727 5666
Fax:	44 171 727 5666
Email:  ChrisiT@SEVENTOWNS.COM

[Moderator's note: This is more hucksterish than I usually pass
 through to cube-lovers, but in case anyone wants to know who's in the
 cube biz, this is one of them.  She also wants to buy Rubik's Revenges.
 (Who doesn't?  I've gotten a dozen such requests, either ignored or
 answered with the explanation that no one is selling them now except
 in one-off auctions.  Be sure to let cube-lovers know if that changes.)

 Anyway, this is Chrisi's ad; if you want to read more ads from her,
 her addresses are above; the ads won't run on cube-lovers. --Dan ]

From cube-lovers-errors@mc.lcs.mit.edu  Thu May 20 13:56:11 1999
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Message-Id: <37440E6E.244C@zeta.org.au>
Date: Thu, 20 May 1999 23:30:22 +1000
From: Wayne Johnson <sausage@zeta.org.au>
Reply-To: sausage@zeta.org.au
To: cube-lovers@ai.mit.edu
Subject: Huge Cube!

My cousin came back from england on his recent trip and brought me home
a cube. Most cubes are 57mm. This one is.. wait for it.. 90mm. Has
anyone else got one of these? I'd never heard or read about one before.

It's a tile cube that is of average quality. It was bought at a market
near a place called Kensington.

Wayne Johnson
www.zeta.org.au/~sausage

From cube-lovers-errors@mc.lcs.mit.edu  Thu May 20 17:04:55 1999
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Message-Id: <37441ED1.2C5B3F5E@binghamton.edu>
Date: Thu, 20 May 1999 10:40:17 -0400
From: Mirek Goljan <bg22976@binghamton.edu>
Organization: SUNY Binghamton
To: cube-lovers@ai.mit.edu
Subject: Re: "The" Hungarian cube ??
References: <199905201405.QAA28449@ms.globe.cz>

The cube is most likely the original Rubik's cube made in Hungary by Politoys
in the early 80's.  I bought many of those being in Czechoslovakia during the 80's.
Basically two types wrapped in a blue cardboard box were sold.
The first one has opposite colors w-y, g-b, r-o and it tends to fall apart after a
few months of speed cubing. A positive thing is that this type is well suited for
speed cubing, it doesn't require perfect aligning of cubicles. A negative thing is
that stickers don't last long, the glue is bad.
The second type has colors w-b, g-y, r-o, I think.  I always changed stickers to
get w-y oposite colors. There was a very slight difference between cardboard boxes.
This cube requires better aligning of layers before a next move (the construction
was a little bit improved), the glue is better, but the plastic used I rate lower.
    Mirek

>  -----Original Message-----
> From: "John Burkhardt" <jburkhardt@mediaone.net>

> > I recently bought a rubik's cube from someone.  It was in a blue cardboard
> > box that has Hungarian all over it....

********************************
Miroslav Goljan
Watson School of Engineering and
    Applied Science, Dept. of EE
 State University of New York
 PO BOX 00238
 Binghamton, NY 13902-6000
e-mail: bg22976@binghamton.edu
********************************

From cube-lovers-errors@mc.lcs.mit.edu  Thu May 20 22:44:17 1999
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From: roger.broadie@iclweb.com (Roger Broadie)
To: <cube-lovers@ai.mit.edu>
Cc: <sausage@zeta.org.au>
Subject: Re: Huge Cube!
Date: Thu, 20 May 1999 23:37:38 +0100

It is one of a series of three that are on sale in the UK in the
cheaper toy shops.  They were mentioned on Cube-lovers in March.

The big one that you have works pretty well and makes a handsome
display object.  There is also a normal-sized cube also with tiles
that is about the most horrible to turn I have ever met - the
shop-keeper even felt compelled to warn of the fact me in advance.
Then there is also a key-ring sized cube (30 cm) which turns tolerably
well and has conventional stickers, though not firmly stuck on, for
only 73 pence in my local shop.  Apparently the children like them
because they are so easy to solve by rearranging the stickers.

Roger Broadie

Wayne Johnson wrote (20 May 1999)

>My cousin came back from england on his recent trip and brought me
>home a cube. Most cubes are 57mm. This one is.. wait for
>it.. 90mm.[...]

From cube-lovers-errors@mc.lcs.mit.edu  Fri May 21 13:55:13 1999
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Date: Thu, 20 May 1999 20:41:56 -0700 (PDT)
From: Tim Browne <ue451@victoria.tc.ca>
To: cube-lovers@ai.mit.edu
Subject: Re: Huge Cube!
In-Reply-To: <015101bea311$641f58a0$60c4b0c2@home>
Message-Id: <Pine.GSO.3.95.iB1.0.990520203555.13333A-100000@vtn1>

On Thu, 20 May 1999, Roger Broadie wrote:

> The big one that you have works pretty well and makes a handsome
> display object.  There is also a normal-sized cube also with tiles
> that is about the most horrible to turn I have ever met - the
> shop-keeper even felt compelled to warn of the fact me in advance.

The most problematic ones I've come across so far are the "Wisdom" Ball
(everything has to be lined up exactly right), Alexander's Star (same),
and the Pyramorphix (same). It's a neat puzzle, but it seems to have a
mind of its own, many times forcing you to turn a half/direction you don't
want to and gets jammed constantly. When you finally do get past the jams,
it's completely unexpected and flies at super speed. On top of this, the
centres are so sharp that when it happens, it often cuts your fingers. Has
anyone managed to come up with a fix for this? My guess is all that would
need to be done to fix all of these problems would be to make the edges of
the centre pieces more rounded, but I can't be sure about that.


> Then there is also a key-ring sized cube (30 cm) [...]

Uh... I think you mean 30Mm cube, don't you? :-) L8r.

-- 
 Victoria Animart - American Prices, Canadian Currency. |       HIT Jedi
         http://www.focus-asia.com/home/animart         | Use the Force, Mike!
--------------------------------------------------------+-----------------------
            "No thanks. I'm trying to cut down." - Michael Garibaldi

From cube-lovers-errors@mc.lcs.mit.edu  Fri May 21 20:46:18 1999
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From: "David Byrden" <David@byrden.com>
To: <Cube-Lovers@ai.mit.edu>
Subject: Rubik and XML
Date: Fri, 21 May 1999 20:48:33 +0100


	I want to define an XML dialect so that all 
Rubik-related programs can exchange data...
I also want it to be useful to humans who
are saving the state of their cube.

	I am putting my ideas together at

	http://byrden.com/puzzles/

and would be glad of criticism.


				David

From cube-lovers-errors@mc.lcs.mit.edu  Sun May 23 20:46:27 1999
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Date: Sun, 23 May 99 20:37:41 EDT
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From: Dan Hoey <Hoey@aic.nrl.navy.mil>
To: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Cc: gej@spamalot.mfg.sgi.com, Cube-lovers@ai.mit.edu
Subject: Re: Algorithm for the Antislice Group

Gene Johannsen <gej@spamalot.mfg.sgi.com> wrote:

> I am having problems with this step.  My cube is in a
> configuration that this maneuver does not solve:

And whuang@ugcs.caltech.edu (Wei-Hwa Huang) replied

> As I e-mailed Gene, I do not believe his configuration is part of the
> anti-slice group -- would any members care to give a quick heuristic to
> determine if a cube is in the anti-slice group?

Gene Johannsen sent in a reply agreeing that he had probably made a
twisto in scrambling the cube. The moderator did not forward that
reply to the group, hoping instead to determine a sure answer to the
question.  I'm sorry if the delay has cast doubt on Wei-Hwa Huang's
antislice algorithm (which I have not examined in detail).

The problem is actually fairly difficult if we want a definitive
answer for any antislice-group position--Singmaster (p. 54) despairs
of presenting Morwen B. Thistlethwaite's analysis of the antislice
group, settling instead for an outline.  I must admit that I have not
yet found a complete answer that can be carried out by hand.  However,
I have a method that will recognize positions in a group four times
the size of the antislice group, and that is good enough to weed out
almost all of the near misses.  In fact, a simpler method that will
detect positions in a group 972 times the size of the antislice group
is sufficient for the case in question.

As in Singmaster we consider a corner-based representation.  That is
to say, we keep the BLD corner from moving, and perform an antislice
as a half-turn of the F, T, or R face together with a quarter-turn of
the adjacent center-slice.  [ For newcomers, I'll mention that my use
of "T" for "Top" instead of "U" for "Up" is an intentional preference;
see 22 Feb 90 and 28 Oct 94 in the archives for an explanation. ]

In this representation the face centers form a movable part of the
cube.  We could represent their position as a permutation of the six
face centers, but it is more convenient to represent them as a
permutation of the four major diagonals of the enclosing cube, as Jim
Saxe and I did for the Tartan cube (16 February 1981).  We may label
the faces and diagonal-endpoints as follows:

     Z---X
     | B |
 Z---W---Y---X---Z
 | L | T | R | D |
 Y---X---Z---W---Y
     | F |
     Y---W

The T antislice acts on the face centers as (F,L,B,R) and on the
diagonals as (W,Y,Z,X).  Similarly, the R antislice induces the face
permutation (T,B,D,F) and the diagonal permutation (W,Z,Y,X). The F
antislice induces the permutations (L,T,R,D) and (W,Y,X,Z).  In the
following, I will use note the diagonal permutations.

In addition to the face centers, there are three orbits of edge cubies
and two orbits of corner cubies (ignoring BLD).  The corner
orientation never changes, and reorientation of edges is applied to an
entire orbit at a time.  I label the edges and corners as follows:

         [Z]--Z3--Xp
          |       |
          Z2  B   Y2
          |       |
 [Z]--Z2--Wp--Y3--Y---Y2--Xp--Z3-[Z]
  |       |       |       |       |
  Z1  L   X1  T   Y1  R   W1  D   Z1
  |       |       |       |       |
  Yp--X2--X---X3--Zp--W2--W---W3--Yp
          |       |
          X2  F   W2
          |       |
          Yp--W3--W

The diagonals W-Wp, X-Xp, Y-Yp, and Z-Zp are labeled Wc, Xc, Yc, and
Zc, respectively, for the purpose of recording the face center
position.  In addition I label the orientation of the edge orbits as
P1, P2, and P3, where P1^2 = P2^2 = P3^2 = I.  (These could be
represented as permutations of 2-sets, but that seems unnecessary).

So the three antislices are:

    Fa = (W,X)(Yp,Zp) (W1,Z1,X1,Y1) (W2,X2) (W3,X3) P1 (Wc,Yc,Xc,Zc),
    Ta = (X,Y)(Wp,Zp) (X1,Y1) (W2,X2,Z2,Y2) (X3,Y3) P2 (Wc,Yc,Zc,Xc),
    Ra = (W,Y)(Xp,Zp) (W1,Y1) (W2,Y2) (W3,X3,Y3,Z3) P3 (Wc,Zc,Yc,Xc).

         corner perm   edge perm                    ori  center perm

It is immediately apparent that each antislice is an odd permutation
of each orbit of corners, of each orbit of edges, and of the center
diagonals. In addition, the number of Pi orientations is changed by
one on each antislice.  Thus we expect to see each of these seven
parities agree in any position of the antislice group.

Gene Johannsen's position (after replacing color letters with position
letters) is

        BBB
        BBF  <- Back face
        FFF

    LLR TDT LLR DTD
    LLL DTD RRR TDT  <- Down face
    LRR TDT LRR DTD

        BBB
        BFF  <- Front face
        FFF

which is represented as

    (X,Y) (Wp,Zp) (W1,Y1)(X1,Z1) (X2,Y2) (W3,Y3,Z3,X3)

in the corner-based representation.  Here the permutations on the
corner orbits and two of the edge orbits is odd.  But the
{W1,X1,Y1,Z1} orbit has an even permutation and the orientation and
center positions are the identity, of even parity.  So the position
cannot be in the antislice group.

I noticed that the difference could conceivably be caused by a single
error, say an F-slice move:

  Fs = (W1,Z1,X1,Y1) (Wc,Yc,Xc,Zc) P1,

and with a short program in GAP I was able to find the following
single-error process for Gene's position:

    Fa Ta Fs Fa Ta' Fa Ra^2.

Now I'll apologize to anyone whose head is reeling, and invite anyone
who's game to join in a little bit of slightly tougher group theory.
This will show you how far I've been able to analyze the antislice
group, and the part that remains mysterious.

Note that the parity constraints above allow:

    6 permutations of {W,X,Y},
    24 permutations of {Wp,Xp,Yp,Zp},
    24 permutations each of {W1,X1,Y1,Z1}, {W2,X2,Y2,Z2}, and {W2,X2,Y2,Z2},
    8 subsets of {P1,P2,P3}, and
    24 permutations of {Wc,Xc,Yc,Zc},

with a parity constraint on the seven components that reduces the
number by 2^6, for

    6 * 8 * 24^5 / 64 = 5971968

possible positions.  Singmaster notes that the actual size of the
antislice group is 6144 = 5971968 / 972 positions.  Clearly there are
more constraints at work than permutation parity.  Most of them are
due to life in a certain quotient group.  The group S4 of permutations
on four letters contains a normal subgroup consisting of the identity
plus the three pairs of two-cycles:

    H = { (), (W,X)(Y,Z), (W,Y)(X,Z), (W,Z)(X,Y) }.

The quotient S4/H then has six elements, and is isomorphic to S3.  We
can see this explicitly by writing down the blocks of S4/H:

   Block 0      Block 1    Block 2    Block 3    Block 4    Block 5

   ()           (W,X,Y)    (W,Y,X)    (W,X)      (W,Y)      (X,Y)
   (W,X)(Y,Z)   (W,Y,Z)    (W,Z,Y)    (Y,Z)      (X,Z)      (W,Z)
   (W,Y)(X,Z)   (W,Z,X)    (W,X,Z)    (W,Y,X,Z)  (W,X,Y,Z)  (W,X,Z,Y)
   (W,Z)(X,Y)   (X,Z,Y)    (X,Y,Z)    (W,Z,X,Y)  (W,X,Y,X)  (W,Y,Z,X)

The top line of each block shows the element of S3=S({W,X,Y})
corresponding to the block.  Note how easy it is to recognize the
blocks 3, 4, and 5:

    If you have a four-cycles, use one two-cycle from its square;
    If you have a two-cycle containing Z, use the disjoint two-cycle.

This procedure will yield the S3 representative from these blocks.

Now remember the antislices?

    Fa = (W,X)(Yp,Zp) (W1,Z1,X1,Y1) (W2,X2) (W3,X3) P1 (Wc,Yc,Xc,Zc),
    Ta = (X,Y)(Wp,Zp) (X1,Y1) (W2,X2,Z2,Y2) (X3,Y3) P2 (Wc,Yc,Zc,Xc),
    Ra = (W,Y)(Xp,Zp) (W1,Y1) (W2,Y2) (W3,X3,Y3,Z3) P3 (Wc,Zc,Yc,Xc).

Ignoring the orientation component, we see that the action of every
component of Fa is from block 3, every component of Ta from block 4,
and every component of Ra from block 5.  So for every position in
< Fa, Ta, Ra >, each permutation orbit will be in the same block, and
the orientation component must still agree in parity.  So instead
dividing by 2^6, we divide by 6^5 2 = 15552, for

    6 * 8 * 24^5 / 15552 = 24576

positions, which is only four times the size of the actual antislice
group.

The factor of four arises if we fix all but the {Wp,Xp,Yp,Zp} and
{Wc,Xc,Yc,Zc} orbits.  There are only four possibilities for the last
two orbits, rather than the sixteen that my analysis provides.  (As
Singmaster notes, when the other components are the identity, the four
possibilities consist of SOLVED and the four Zigzag/Laughter
patterns).  Unfortunately, I have not found any way to describe the
correspondence between these components in general, to make the
analysis exact.

Dan
Hoey@AIC.NRL.Navy.Mi

From cube-lovers-errors@mc.lcs.mit.edu  Mon May 24 16:21:39 1999
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Message-Id: <3748AFD3.DDEF4717@whitewolf.com.au>
Date: Mon, 24 May 1999 11:48:03 +1000
From: Ryan Heise <ryan@whitewolf.com.au>
To: David Byrden <David@byrden.com>
Cc: Cube-Lovers@ai.mit.edu
Subject: Re: Rubik and XML
References: <199905211952.PAA04633@life.ai.mit.edu>

David Byrden wrote:

>         I want to define an XML dialect so that all
> Rubik-related programs can exchange data...
> I also want it to be useful to humans who
> are saving the state of their cube.
>
>         I am putting my ideas together at
>
>         http://byrden.com/puzzles/
>
> and would be glad of criticism.

Quote: "I want RubXML files to be able to hold move sequences instead of
puzzle configurations. I will add this option after I study the
notations currently in use."

I think both ways of specifying a cube configuration are important. You
can only specify a cube configuration in move sequences if you know the
exact move sequence you used to mess it up.

Move sequences are probably more useful for storing solutions to cube
configurations. Also, Lars Petrus has a cube tutorial that shows you how
to do each step by scripting a Java applet to rotate the cube according
to stored sequences. (Sure, his "lightweight" applet would not want to
drag in a whole XML parser, but the use is there)

BTW, here is one way to specify move sequences:

<sequence name="twist3corners">
R'U'RU'R'U2R
</sequence>

If you want something flexible, how about:

<sequence name="foo">
	RURU
	<useseqence name="bar" orientation="000" direction="backwards"
reflect="x">
	U'R'U'R
</sequence>

where:

usesequence imports another sequence
orientation is a 3D coordinate (is there a better way to specify this?)
direction is which direction to execute the sequence
reflect is the axis over which the sequence should be reflected.

It's flexible but I don't ask me if it's useful :-)

-- 
Ryan Heise

http://www.progsoc.uts.edu.au/~rheise/

From cube-lovers-errors@mc.lcs.mit.edu  Tue May 25 11:17:55 1999
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From: roger.broadie@iclweb.com (Roger Broadie)
To: "Cube-Lovers" <Cube-Lovers@ai.mit.edu>
Subject: Availability of 5x5x5 and 2x2x2 in UK
Date: Mon, 24 May 1999 22:24:45 +0100

Hah!  For British readers, I can report that Toys 'R' Us have the
Meffert 5x5x5 and the Rubik's 2x2x2 - the first time I have seen
either of these sizes on sale in the UK.

Roger Broadie

From cube-lovers-errors@mc.lcs.mit.edu  Fri Jun  4 15:47:14 1999
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From: "David Byrden" <David@byrden.com>
To: <Cube-Lovers@ai.mit.edu>
Subject: Standard file format for Rubik's Cube is just about ready
Date: Fri, 4 Jun 1999 12:14:00 +0100

	Following on from my mail
of the 21st May...

	I have defined a language for writing
down the state of a standard Rubik's cube.
Many people offered suggestions, and I thank
them.

	The language now has its own website:

	http://byrden.com/rubxml/

which contains an online demo program to
read the language and draw the corresponding
cube in "unfolded" format. People who want
to record their "pretty patterns" may find this a
useful resource!

I have one question that I want to ask now. This
language needs a "default colour scheme". I
understand that one way of colouring a Rubik's
Cube was much more common than the others,
but I don't know which one. Can anyone specify it
exactly for me?


				David Byrden

From cube-lovers-errors@mc.lcs.mit.edu  Mon Jun  7 15:49:59 1999
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To: cube-lovers@ai.mit.edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Re: Standard file format for Rubik's Cube is just about ready
Date: 7 Jun 1999 13:48:49 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <7jgik1$d5d@gap.cco.caltech.edu>
References: <cube-lovers.199906041117.HAA05009@life.ai.mit.edu>

"David Byrden" <David@byrden.com> writes:
>I have one question that I want to ask now. This
>language needs a "default colour scheme". I
>understand that one way of colouring a Rubik's
>Cube was much more common than the others,
>but I don't know which one. Can anyone specify it
>exactly for me?

The most common color scheme was
 White opposite Blue,
 Green opposite Yellow,
 Red opposite Orange.
 If you hold Yellow as Front and Red as Up, then Blue is Right.

Color "purists" have complained about this, preferring
 White opposite Yellow,
 Blue opposite Green,
 Red opposite Orange.
 -- note that each pair is the same color except for some "Yellow" added.
--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
If all my friends jumped off a cliff... what reason is there for me to live?

[Moderator's note: You say white opposite blue was common, but when
 was that?  My experience and other reports lead me to believe that
 what you call the "purist" color scheme was most common in 1985 (viz
 _Rubik's_Cubic_Compendium_) and I don't recall seeing any change
 since.  It also is reportedly most common in the chirality you
 describe, which is called the BOY version because blue, orange, and
 yellow appear in that order clockwise around a corner.  The mirror
 image coloring is called YOB (a Cockney term for a yokel).  I don't
 know of any cute naming schemes for the enantiomorphs in color
 schemes in which the blue, orange, and yellow faces do not all
 meet--Dan ]

From cube-lovers-errors@mc.lcs.mit.edu  Tue Jun  8 17:22:07 1999
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Message-Id: <001201beb186$4ef27880$0401a8c0@sanyi.android.com>
Reply-To: "Michaletzky_Sandor" <michas@androsoft.com>
From: michas@androsoft.com (michas@androsoft.com)
To: <cube-lovers@ai.mit.edu>, "Wei-Hwa Huang" <whuang@ugcs.caltech.edu>
Subject: Re: Standard file format for Rubik's Cube is just about ready
Date: Tue, 8 Jun 1999 10:10:01 +0200

Hi, Friends!

The original color schema is (designed by Erno Rubik, and it is written in
Rubik's: The Magic Cube book also): blue is opposite to green, white is
opposite to yellow and red is opposite to orange. The differerence between
each opposite colour-pair is yellow: blue + yellow = green, red + yellow =
orange and white + yellow = (guess it! :-).  If blue is on the top and red
is in front of you, yellow is on the right side.


From cube-lovers-errors@mc.lcs.mit.edu  Tue Jun  8 20:03:25 1999
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From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Message-Id: <199906081306.GAA24838@necro.ugcs.caltech.edu>
Subject: Re: Standard file format for Rubik's Cube is just about ready
To: michas@androsoft.com
Date: Tue, 8 Jun 1999 06:06:28 -0700 (PDT)
Cc: cube-lovers@ai.mit.edu, whuang@ugcs.caltech.edu
In-Reply-To: <001201beb186$4ef27880$0401a8c0@sanyi.android.com> from "michas@androsoft.com" at Jun 8, 99 10:10:01 am
Reply-To: whuang@ugcs.caltech.edu

> >[Moderator's note: You say white opposite blue was common, but when
> > was that?  My experience and other reports lead me to believe that
> > what you call the "purist" color scheme was most common in 1985 (viz
> > _Rubik's_Cubic_Compendium_) and I don't recall seeing any change
> > since.  It also is reportedly most common in the chirality you
> > describe, which is called the BOY version because blue, orange, and
> > yellow appear in that order clockwise around a corner.  The mirror
> > image coloring is called YOB (a Cockney term for a yokel).  I don't
> > know of any cute naming schemes for the enantiomorphs in color
> > schemes in which the blue, orange, and yellow faces do not all
> > meet--Dan ]

I'm going by what I have observed -- which is mostly Ideal cubes.  The
Ideal Deluxe cube has white opposite blue, as well as their Revenge
and the three normal cubes I have in my collection.  I can check the
orientation of the minicubes in Rubik's Race and the "deluxe-ish"
cube supplied with Rubik's Game.

The 5x5x5 in my collection is also white opposite blue, although it
isn't Ideal (pun intended).

The OddzOn cubes are also blue opposite white, so I think it's safe
to say that the majority of cubes out there are blue opposite white.

 From http://www.rubiks.com/cubesolution_new.html :

>Note: we use the color arrangement of the original Rubik's cube (i.e.
>blue is opposite to green, red is opposite to orange, and yellow is
>opposite to white; if the blue side is on the top, then the red is
>on the left and the yellow on the right, and so on), because this
>is the one Erno Rubik prefers. The new Rubik's Cubes made by OddzOn
>since 1995 are colored differently (i.e. blue is opposite to white,
>green is opposite to yellow, and red is opposite to orange).


--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
Is there a metonym in this sentence?

From cube-lovers-errors@mc.lcs.mit.edu  Tue Jun 29 18:12:20 1999
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Date: Tue, 29 Jun 1999 16:26:03 -0400
To: Cube-Lovers@ai.mit.edu
From: pink@cartserv.rserv.uga.edu (_pink)
Subject: hello to all

	Hey! I just thought I'd drop a line to introduce myself since I've
subscribed to your mailing list. My name is Brian Davis (my friends all
call me "_pink". The underscore is silent.) I live in Athens, Georgia USA
and Manage a Computer Graphics Facility at the Univ of Georgia.
	I have been a cube fanatic since day one... I've got a pretty good
collection and several volumes of notes (as I sure most of you do) but feel
there is still more out there. I have noticed many names here on the Cube
Lovers List that have some wonderful www pages. Maybe I'll find some time
to create a www cube presence of my own that doesn't overlap too much.
(this might be hard to do as you guys have it pretty well wrapped up right
now...)
	I have "e-talked" with Tim Browne some previously (don't blame him
for me being here, he's innocent really) and when I found he was here and
the list seems to still be alive I thought I might as well subscribe.
	I'll start out by sticking my neck out a bit and see if any one
wants to hear some information I have about Square 1 moves or
inner-workings (since I've crushed a few out of curiosity and could take
picts to post on the web) or the mechanics of the Skewb since I've had to
put it back together after an employee dropped it.
	At any rate, I'm glad to be here and will try not to ramble so
incoherently in the future.

_pink

btw: anyone have a Siamese Cube that they would like to get rid of? ;) Ok,
how about just answer a few questions about so I can try to construct
one... 8)

to see a listing of cubes I'm looking for go to:
http://128.192.40.238/pink/wanted.html

From cube-lovers-errors@mc.lcs.mit.edu  Thu Jul  1 13:26:56 1999
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Date: Wed, 30 Jun 1999 13:35:04 -0400
To: Cube-Lovers@ai.mit.edu
From: pink@cartserv.rserv.uga.edu (_pink)
Subject: replies all around

Hello again,
	Thanks to all who sent me a welcoming note.

A few notes of interest:

Dale: Crazy 5 combined cubes man! I might have to try it... Did you also
attach the edge cubes or just the corners?

Chris (or Chris and Kori): I've been to your page(s) many times. Nice site!
I've been to ebay a bit but I'm not giving out my ebay nickname... ;) I
have the Pyramorphix and am not sold on the last pict of my wanted page
being one in scrambled state...

Jared: Thanks for the tip. I'll follow up and let you know what turns up.

Bob Harris: Hey! another Georgian! Glad to see I'm not the only one. I'm
sure we shall be talking a bit...

Noel: If you are talking about the octagonal prism then the answer is just
like the 3x3x3 but if you're talking about the Octahedron (double pyramid)
then unfortunately I cannot help as I have not had the chance to hold one
in my hands...

Hana: Thank you for the wonderful note. I have not had a chance to visit
your site but it will be first on my list when I get home. I am intrigued
to see what you have been doing!

Pete Beck: Did the money order I sent ever arrive? The beachball is a Skewb
of 4 colors (thus the subclass) I contacted Meffert's a few weeks back and
they informed me that the beachball is not in production but they are going
to make more available in October! I have found info on the bandaged cube
but need clearer direction as to which pairs of cubies are 'glued'
together... As for Unknown #2 on my wanted page, when tracing the
dissections I can only imagine a 2x2x2 with non cubed cubies... Who knows...

Others inquired about my Square 1 and Skewb notes. Ok. I'll do a write up
on each under two different posts as to keep things separated. Posts to
follow...

Thanks again to all for making me feel at home.

_pink

[ Moderator's note: also Wei-Hwa Huang noted that The "Unknown #1" on
  Pink's web page is just a taped cube. ]

From cube-lovers-errors@mc.lcs.mit.edu  Thu Jul  1 14:46:52 1999
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Date: Wed, 30 Jun 1999 13:58:55 -0400
To: Cube-Lovers@ai.mit.edu
From: pink@cartserv.rserv.uga.edu (_pink)
Subject: Skewb notes

Ok, while Meffert's site has the solution posted, there still seems to be
some confusion about how this thing is actually made. Well I found out
first hand when one of my employees dropped my skewb on a concrete floor. I
have plans to take picts of the disassembled skewb and post it online but
until I get around to it here's a description.

The Skewb consists of:

6 Square center pieces that have small tabs on 2 opposite undersides.
4 Pyramid shaped corners with a triangular tab on its interior side.
1 4-armed spider mechanism with 4 corner pieces (described above)
	attached to the end of each arm.

To understand the arrangement of the spider mech refer to the diagram below:

      A _____________________
       /|                   /|B
      / |                  / |
     /  |                 /  |
    /   |              A /   |
  B/____________________/    |
   |    |               |    |
   |    |               |    |
   |    |      x        |    |
   |    |               |    |
   |    |B              |    |
   |    |_______________|____|A
   |   /                |   /
   |  /                 |  /
   | /                 B| /
  A|/___________________|/

	If you connect each corner marked "A" with the "x" (with the x
being the exact center of the cube) you will have the spider mechanism. The
corners marked "B" are only attached by the various tabs on each piece.
	Skewbs are EASY to disassemble if you know how...

1.) Rotate any dissection of your skewb 60 deg so that 2 centers are
aligned edge to edge.

2.) Place a thin bladed screwdriver between the common edge of any 2 edge
pieces and gently wiggle applying increasing pressure.

3.) Catch all the pieces as it crumbles in your hands.

Putting it back together is interresting and certainly was when my employee
dropped mine but after you do it it's kinda neat.

1.) Align the colors of 2 corners on the spider and insert the center of
the corresponding color so that the tabs hold it in place.

2.) Align the corners and place 2 more center sections. Slight rotation of
one corner helps at times.

3.) Insert the proper loose corners being sure to orient each correctly.

4.) Place one edge tab of the last center into position and snap into
place. I have found it easiest to put the skewb on the table with the last
center on top and firmly press it into place.

	Meffert's puzzle ball series are the same mechanism but some are
supergroup puzzles because of the center orientations. The Meffert
Beachball is IMO a subgroup because of only having 4 colors.
	I hope this has been of value for someone and isn't just a waste of
bitspace for all.

_pink

From cube-lovers-errors@mc.lcs.mit.edu  Thu Jul  1 17:31:32 1999
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Date: Thu, 1 Jul 1999 15:37:36 -0400 (Eastern Daylight Time)
From: Dale Newfield <din5w@cs.virginia.edu>
Reply-To: DNewfield@cs.virginia.edu
To: Cube-Lovers@ai.mit.edu
Subject: Re: Skewb notes
In-Reply-To: <v04003a09b3a0062a6a87@[128.192.40.243]>
Message-Id: <Pine.WNT.4.10.9907011536110.-824133@biff. cs.virginia.edu>

On Wed, 30 Jun 1999, _pink wrote:
> 	Meffert's puzzle ball series are the same mechanism but some are
> supergroup puzzles because of the center orientations. The Meffert
> Beachball is IMO a subgroup because of only having 4 colors.

I find it interesting to think that the internal mechanism for the skewb
and the puzzle balls are the same as for the pyraminx, but that the pieces
stuffed into the gaps make it a completely different puzzle.

-Dale

From cube-lovers-errors@mc.lcs.mit.edu  Fri Jul  2 12:15:09 1999
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From: Douglas Zander <dzander@solaria.sol.net>
Message-Id: <199907020153.UAA27551@solaria.sol.net>
Subject: Q: fix a pyramorphix?
To: cube-lovers@ai.mit.edu (cube)
Date: Thu, 1 Jul 99 20:53:08 CDT

Hello, I wonder if anyone can help me.  I allowed a friend of mine to
play with my pyramorphix and it broke in his hands.  One of the inner
pieces broke apart.  It was the piece that sits on the dowel with four
little rods sticking up.  Does anyone know if any local (USA) shop fixes
this particular puzzle?  Does Meffert's Puzzle & Games fix them in Hong
Kong?  (I'd prefer a local shop)  Does anyone have a broken pyramorphix
they would sell me a piece to replace mine?  :-)   Thanks for any help.
(Is Meffert the inventor of the pyramorphix?)

--
 Douglas Zander                |
 dzander@solaria.sol.net       |
 Shorewood, Wisconsin, USA     |

From cube-lovers-errors@mc.lcs.mit.edu  Fri Jul  2 13:23:14 1999
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Reply-To: <noel@mud.ca>
From: "Noel Dillabough" <noel@mud.ca>
To: "Cube Lovers List (E-mail)" <cube-lovers@ai.mit.edu>
Subject: Solution for the size 2 octahedron
Date: Thu, 1 Jul 1999 17:48:12 -0400
Message-Id: <000001bec441$c8879680$29f5ddce@laptop>

After playing with the size 2 corner moves based octahedron, I found that
the solution for most of the edges is trivial, just a matter of putting them
into place using standard puzzle moves.  The only part of the puzzle that
needs a special move is the final step that you may be left with a pair of
faces that need to be switched:

The following represent a tip of the octahedron, and sides A and B must be
switched

     R->
     |
_____|______
     |
  A  |  B


The following move accomplishes this:

Down
Rotate Counterclockwise
Up
Rotate Twice
Down
Rotate Twice
Up
Rotate Twice
Down
Rotate Clockwise

The "Up" and "Down" moves refer to the right half of the octahedron, cut
through the middle of the tip and leaving the left half steady.  The Rotate
moves refer to rotating the tip in place.

Anyone with a good notation for this and the size 3 corners based
octahedron, or the side based size 2 or 3 octrahedra let me know.

[Moderator's note: Isn't a corners-based octahedron just a face-based cube
 with the corners flattened and the faces made pointy?  I think BFTDLR or
 BFUDLR would work for notation. --Dan]

From cube-lovers-errors@mc.lcs.mit.edu  Tue Jul  6 15:19:25 1999
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Date: Fri, 2 Jul 1999 11:57:33 -0400
To: Cube-Lovers@ai.mit.edu
From: pink@cartserv.rserv.uga.edu (_pink)
Subject: cube picts / Square 1 picts

Hello Cube Lover's

	Thanks to all who have provided information that should help me to
track down some of the items on my wanted list. I have had several requests
for pictures of cubes I have and picts of disassembled cubes. Well good
news! I purchased a digital camera yesterday and will be taking picts of
all requested items this weekend (time permitting) and I will post them
online.

A good weekend to all.

_pink

It's quick and dirty but...

http://128.192.40.238/pink/sq1asmbly.html

feel free to email with any questions.

_pink

From cube-lovers-errors@mc.lcs.mit.edu  Tue Jul  6 18:54:46 1999
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Reply-To: <noel@mud.ca>
From: "Noel Dillabough" <noel@mud.ca>
To: "Cube Lovers List (E-mail)" <cube-lovers@ai.mit.edu>
Subject: Re: Solution to a size 2 Octahedron
Date: Fri, 2 Jul 1999 19:17:23 -0400
Message-Id: <000601bec4e1$19d7a080$0ef5ddce@laptop>

[Moderator's note: Isn't a corners-based octahedron just a face-based cube
 with the corners flattened and the faces made pointy?  I think BFTDLR or
 BFUDLR would work for notation. --Dan]

Actually this notation works perfectly, here is the move in this fashion:

R' F' R F2 R' F2 R F2 R' F

From cube-lovers-errors@mc.lcs.mit.edu  Tue Jul  6 19:03:51 1999
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Message-Id: <377E55CC.3C4F@ameritech.net>
Date: Sat, 03 Jul 1999 13:26:20 -0500
From: Hana Bizek <hbizek@ameritech.net>
Reply-To: hbizek@ameritech.net
To: cube-lovers@ai.mit.edu
Subject: New web page
References: <377E509B.39A5@ameritech.net>

To my fellow cube-lovers:

Josef Jelinek and I have made a web page devoted to the cube. He
contributed cube solution algorithms, accompanied by graphics and I
contributed 12 of my 3-dimensional designs, many of them recent. That
means they were constructed after my book was published in 1997. I also
sketch another problem resulting from these designs, that of fractals.
You may find all this in our URL: http://cube.misto.cz.

Any comments, criticism, etc would be appreciated. I still did not get
my question answered, either by you or by the search engines. It is
this: Is there anyone, on or off the web, who has created something
similar to what I have done? I know about the people at Wunderland, and
have seen thir work; however, what I have done is different.

Thank you very much for any input you can provide.
Hana M. Bizek

From cube-lovers-errors@mc.lcs.mit.edu  Wed Jul  7 19:25:18 1999
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Date: Sun, 4 Jul 1999 14:45:12 -0700 (PDT)
From: Tim Browne <ue451@victoria.tc.ca>
To: Cube-lovers@ai.mit.edu
Subject: Puzzletts?
In-Reply-To: <v04003a09b3a0062a6a87@[128.192.40.243]>
Message-Id: <Pine.GSO.3.95.iB1.0.990704143754.11069A-100000@vtn1>

Has anyone been able to get through to Puzzletts at all? I've been trying
to get through to their site for the past month or so, but it's always
"down, overloaded, or unreachable" (surely they could narrow it down a
bit better than that!). Does anyone know what's going on? Has it crashed
and burned for good? L8r.

-- 

From cube-lovers-errors@mc.lcs.mit.edu  Mon Jul 12 14:25:00 1999
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Message-Id: <377A4984.5167C958@nadn.navy.mil>
Date: Wed, 30 Jun 1999 12:44:52 -0400
From: David Joyner <wdj@nadn.navy.mil>
Reply-To: wdj@nadn.navy.mil
Organization: Math Dept, USNA
To: cubelovers <cube-lovers@ai.mit.edu>
Subject: puzzle collectors web page has moved

Hello Cube Lovers:
The Directory of Mechanical Puzzle Collectors web page has moved to
http://anduin.eldar.org/~problemi/joyner/slocum.html
Please update any links. - David Joyner

--
David Joyner, Assoc Prof of Math
Math Dept, 572C Holloway Dr
US Naval Academy, Annapolis, MD 21402
office: 410-293-6738, fax: 410-293-4883
wdj@nadn.navy.mil, wdj@gwmail.usna.edu
http://web.usna.navy.mil/~wdj/homepage.html

[Moderator's note: This note was delayed because I generally give web
 page announcements lower priority than discussion.  Also, I usually
 drop announcements of changes to web pages that have been previously
 announced, since interested people can learn about the changes from
 the web page itself.  In the last 18 months or so, change notices for
 the following pages have been suppressed in this way:

     http://www.snipercade.com/cubeman
     http://www.asahi-net.or.jp/~hq8y-ishm/
     http://www.rubiks.com/
     http://www.ue.net/mefferts-puzzles/
     http://home1.gte.net/davebarr/Cube/
     http://128.192.40.238/pink/

 I'm willing to accept suggestions from the list membership about
 these policies; direct them to cube-lovers-request@ai.mit.edu. -- Dan ]

From cube-lovers-errors@mc.lcs.mit.edu  Mon Jul 12 15:38:04 1999
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Date: Mon, 12 Jul 1999 14:09:58 -0400
To: Cube-Lovers@ai.mit.edu
From: pink@cartserv.rserv.uga.edu (_pink)
Subject: Square 1 Notes

Greetings to all,

	Ok, I've been extreeeeeemely busy as of late but I did find a few
mins at lunch today to throw together some of my notes on the Square 1
puzzle.
	First my standard disclaimers:

1.) The web page design and layout for getting the Square 1 back to cube
shape have been stolen directly from Christian Eggermont. My notes were
very similar but I liked his presentation much better than mine. The only
thing I have done to his notes was to rearrange the classification of
shapes as to their 'distance' from cube shape. To see his original work
please refer to:

http://web.inter.nl.net/C.Eggermont/

2.) I expanded on Andrew Arensburger's cube preserving formulas and also
adopted his piece notation system. To see his original work please refer to:

http://www.cfar.umd.edu/~arensb/Square1/

3.) Some formula were revised and expanded upon from Richard Snyder's
Square 1 Solution Book. Nice book but sometimes a bit cryptic.

4.) I know that there has to be some errors in my notes. If you find them
please advise me so that I might make the corrections necessary. (Actually
now that I think about it, I believe that in the Edge moves section there
is a mistake where I say to turn the bottom 30deg in one direction but it
would make an impossible Right turn... if you see this just turn the bottom
the other direction... I'll try to find it and correct it asap)

5.) There are 2 or 3 pages of blank diagrams or unfinished notes that
indavertantly got included. Please disregard anything that looks unusual...
(hopefully that won't be every page you look at...) ;)

	Ok, so if you're not asleep by now you can hit

http://128.192.40.238/pink/SquareList.html

for the shapes list as to get from where you are now back to cube shape.
There is a link at the bottom of that page to 3 PDF files. One to
manipulate Corners, one for Edges and one for my initial thoughts on 2x2x2
cycling of Square 1 quadrants. I hope there is some info of use for
somebody.
	And as always... feedback is very welcome.

later all,
_pink

From cube-lovers-errors@mc.lcs.mit.edu  Thu Jul 22 13:23:38 1999
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Message-Id: <006001bed45e$d4429420$1b4b43cf@compaq>
From: "Paul Hanna" <phanna@gbonline.com>
Reply-To: phanna@gbonline.com
To: <Cube-Lovers@ai.mit.edu>
Subject: cube computer solutions using procedural languages
Date: Thu, 22 Jul 1999 11:24:45 -0500

Have any of you done any work on solving the cube with computer programs
using procedural languages such as C? I've seen books on the manual methods
of solving the cube (with the cube in your hand) such as David Singmaster's
texts but haven't seen any publications regarding computer solutions using
procedural languages.

Do you have any suggestions you can pass my way? I am a good programmer but
not a cube solution expert. I am just a novice at best when it comes to cube
algorithms and efficient cube solutions.

I am attempting to work on a project involving solving the cube
programmatically and also planning on doing analysis/comparisons of various
algorithms to solve it. Any help, suggested methods, advice, tables,
algorithms, etc. that you may be able to provide me would be very greatly
appreciated.  I can think of a number of ways to approach this task but
would also like some of your folks expertise as well.  I am having trouble
getting going in the right direction.

Also what is the theoretical least number of plane movements that are
required to solve the cube no matter what its configuration and why?

You can reply directly to me.

Thanks in advance,

Paul Hanna
Green Bay, WI
phanna@gbonline.com

[Moderator's note: There's certainly a lot in the archives
(ftp://ftp.ai.mit.edu/pub/cube-lovers/) on the topics of efficient and
optimal programmatic solutions and upper and lower bounds.
Unfortunately, it's not indexed. --Dan]

From cube-lovers-errors@mc.lcs.mit.edu  Tue Jul 27 18:50:00 1999
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From: "Chris Pelley" <ck1@home.com>
To: "Cube-Lovers" <Cube-Lovers@ai.mit.edu>
Subject: Rubik's Cube Perpetual Calendar
Date: Thu, 22 Jul 1999 21:43:52 -0400
Message-Id: <NDBBLHBJOLNCAMMDPAPACENPCAAA.ck1@home.com>

I recently saw one of the original Ideal Perpetual Calendar cubes sell on
EBAY for over $100 US dollars!  I decided it would be cheaper and more fun
to try and make my own.  Based on photos I have collected, I've tried to
reconstruct the original sticker scheme.  This turned out to be quite a fun
puzzle by itself, since there are only a few ways it can be done.  For
example, by a lucky coincidence, there just happen to be exactly the right
number of sides for the middle letters of the months (P, O, C, A, E, and U).

I did a quick scan of the cube-lovers archives to see if anyone had posted
the sticker scheme before.  I found nothing.  I was wondering if anyone who
still has their calendar cube could post either a detailed description, or
even photos from a few different angles.

My own homemade version can be seen here:

http://www.chrisandkori.com/calendarcube.htm

I just printed out the stickers I needed on paper, then cut and scotch-taped
them to a "blank" cube.  I plan to make a somewhat higher-quality version
with real stickers once I nail down the original Ideal sticker
configuration.

Thanks to anyone who can help!
Chris Pelley
ck1@home.com

From cube-lovers-errors@mc.lcs.mit.edu  Thu Aug  5 17:39:56 1999
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Date: Fri, 23 Jul 1999 14:32:38 +0200
From: "Walter van Iterson (EMN)" <Walter.van.Iterson@emn.ericsson.se>
Subject: Solving cube with robot arm
To: Cube-Lovers@ai.mit.edu
Message-Id: <3173B642ECF3D111B3FB0008C724B5609527BC@enleent100.ericsson.se>

Hi all,

Has anybody even built (or throught about building) a device 
which physically solves a cube? Something like a combination 
of a vision system to recognize the colors, a 'robot arm'
which can perform the rotations and a computer to solve the 
mathematics and to operate the robot arm.

I'm currently thinking about making such a system, so I'm 
interested if anybody else has done something similar.

Greetings,
Walter van Iterson
walter.van.iterson@hta.nl

[ Moderator's note: I'm sure there are some mentions of such efforts
  in the archives.  I seem to recall one such project at the
  University of Maryland. --Dan ]

From cube-lovers-errors@mc.lcs.mit.edu  Thu Aug  5 18:10:24 1999
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Message-Id: <3799A6A3.9A46C38D@doc.ic.ac.uk>
Date: Sat, 24 Jul 1999 12:42:27 +0100
From: Colin Waters <csw98@doc.ic.ac.uk>
To: phanna@gbonline.com
Cc: Cube-Lovers@ai.mit.edu
Subject: Re: cube computer solutions using procedural languages
References: <006001bed45e$d4429420$1b4b43cf@compaq>

Hi,

I'm currently working on my MSc dissertation on this topic. I'm
working in Java to build a Cube to which I can then apply algorithms
in Singmaster Notation. The approach I'm taking is to use an
incremental breadth first search to generate a library of
"interesting" macro moves (Richard Korf did some work on this a few
years ago).  I would then like to use a heuristic approach with the
macro moves to head towards the solution. Currently I am thinking of
using a version of the Manhattan distance but I would be grateful for
any other heuristic suggestions.

I would also be interested in a reply to Paul's question about the
least number of theoretical plane moves.

Best Regards,
Colin Waters,
Department of Computing
Imperial College, London.

[Moderator's note:  What is a "plane" move?  All cube metrics count a
 quarter-turn of a face as a single move.  Some also include a
 half-turn of a face.  Some also include a quarter-turn of a center
 slice, or perhaps also a half-turn of a center slice.  I've not seen
 anyone count an antislice move as a single move, though it seems
 fairly reasonable to me, since it's fairly easy to do with one motion
 of the wrist.  In fact, we might even include moves such as "F^2 B"
 under the rubric of "plane moves", so that there would be 45 generators.

 My recollection is that you may find fairly good bounds in the
 archives for the first two metrics--say a ratio of 3:2 or less--but
 not for the others. --Dan ]

From cube-lovers-errors@mc.lcs.mit.edu  Thu Aug  5 18:57:05 1999
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Date: Tue, 27 Jul 1999 22:41:37 -0400
Message-Id: <003B08AD.C22092@scudder.com>
From: Jacob_Davenport@scudder.com (Jacob Davenport)
Subject: Re: Rubik's Cube Perpetual Calendar
To: "Cube-Lovers" <Cube-Lovers@ai.mit.edu>

Chris,

You may want to look at www.wunderland.com/WTS/Kristin/CustomCubes.html for 
some more information about making a custom cube.  Kristin also owns a 
French calendar cube, if I remember correctly, but I imagine that if you 
have successfully made a calendar cube that you don't really need to know 
how hers works.

-Jacob

From cube-lovers-errors@mc.lcs.mit.edu  Thu Aug  5 19:42:04 1999
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Message-Id: <4.1.19990728163706.0096d9d0@gumby145.mail.iastate.edu>
Date: Wed, 28 Jul 1999 17:11:48 -0500
To: "Cube-Lovers" <Cube-Lovers@ai.mit.edu>
From: Corey Folkerts <gumby145@iastate.edu>
Subject: Re: Rubik's Cube Perpetual Calendar

	In response to Chris Pelley's message about a Rubik's Calender Cube, I
would like to say that I made one, although it may not be identical to the
one that Rubik manufactured. (I thought up the idea before I knew one had
been made commercially) I just mapped the letters and numbers out on paper
and then printed out all the characters I'd need with my printer and
affixed them to a blank cube.
I really lucked out because the scotch tape I had was exactly the same
width as the width of one cubie! I also used an exacto knife so I could affix
multiple characters at once, then cut between the cubies with the knife.
An example of how the date reads is:
 J U L
   2 8
19 9 9

The 19 (and 20 once we get there) are on one cubie.

I've attached a jpg of how I arranged all the characters. I make 
no guarantees that is works, but I'm pretty sure it does.
Sorry to those of you who can't receive files. I would have made the
drawing ASCII, but I can't make upside or sideways characters =(
In the drawing there may be some discrepancy between O the letter and zero.
The 'thinner' one is zero. As you can see, my cube is still on Jul 27. I
haven't changed it yet since yesterday. 

-Corey Folkerts

[Moderator's note: I don't send binary attachments out on this list.
 However, I was able to decode Corey's picture, and here's a diagram.
 The characters ^ > v and < denote symbols that are rotated 0, 90,
 180, or 270 degrees clockwise, respectively.

             N^ .. ..
             O< 1v 0<
             Nv Ev 0<

  R< C^ V^   D^ A^ 1<   6^ 6^ 19v  20> .. B^
  6v 3< ..   .. .. 0^   6v 2v ..    .. 0> 9^
  A< 1> T>   Pv 2> G>   Lv Uv Jv    F< 5> Sv

             C< 8< Y^
             3v .. P>
             O< 4> Mv

 I believe the corner marked 0< should be 0^.  The cube only goes
 through 2001, though the addition of 2< on the blank corner would add
 another year.

 Sorry for the delay in sending this message out. --Dan ]

From cube-lovers-errors@mc.lcs.mit.edu  Thu Aug  5 20:28:33 1999
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Message-Id: <006501beda1b$4cf43ec0$4dc4b0c2@home>
From: roger.broadie@iclweb.com (Roger Broadie)
To: "Cube-Lovers" <Cube-Lovers@ai.mit.edu>
Cc: "Chris Pelley" <ck1@home.com>
Subject: Re: Rubik's Cube Perpetual Calendar
Date: Fri, 30 Jul 1999 00:35:35 +0100

Chris Pelley wrote (23 July 1999)

>I was wondering if anyone who still has their calendar cube could
>post either a detailed description, or even photos from a few
>different angles.


This is what my Rubik's Calendar looks like (use a non-proportional
typeface).  The date is as I took it out of the box, from which I
deduce I last played with it in 1982, since I certainly haven't
touched it this decade. Curiously enough, it will be correct again
this year.

The digits in the lower right corners  of the cells indicate the
orientation in quarter turns from the vertical as shown in the
diagram.  But the orientations can be worked out anyway for all except
the centre pieces, because of the requirement to be the right way up
on the face showing the date.

Roger


FRONT
           (Up)
-------------------------
|       |       |       |
|    SUN|DAY    |       |
|       |       |       |
-------------------------
|       |       |       |
|   A   |   U   |   G   |
|       |       |       |
-------------------------
|       |       |       |
|       |       |   8   |
|       |       |       |
-------------------------

RIGHT
          (Up)
-------------------------
|       |       |       |
|   5   |   1   |  1    |
|     2 |     2 |    3  |
-------------------------
|       |       |       |
|   N   |   C   |       |
|       |     2 |       |
-------------------------
|       |       |       |
|   2   |   D   |       |
|     1 |     3 |       |
-------------------------

BACK
           (Up)
-------------------------
|       |       |       |
|   7   |   B   |       |
|     2 |     3 |       |
-------------------------
|       |       |       |
|   C   |   E   |   T   |
|     2 |       |       |
-------------------------
|       |       |       |
|  SATUR|   J   |       |
|     3 |     3 |       |
-------------------------

LEFT
           (Up)
-------------------------
|       |       |       |
|   0   |   P   |       |
|     2 |     3 |       |
-------------------------
|       |       |       |
|   S   |   O   |   L   |
|       |       |       |
-------------------------
|       |       |       |
| WEDNES|   F   |    MON|
|     3 |     3 |     2 |
-------------------------

TOP
         (Back)
-------------------------
|       |       |       |
|   6   |   Y   |   3   |
|     2 |     3 |     3 |
-------------------------
|       |       |       |
|   O   |   P   |   3   |
|       |       |     3 |
-------------------------
|       |       |       |
|  THURS|       |   9   |
|     3 |       |       |
-------------------------

BOTTOM
         (Back)
-------------------------
|       |       |       |
|   TUES|   R   |       |
|       |     3 |       |
-------------------------
|       |       |       |
|   V   |   A   |   M   |
|     2 |     2 |     2 |
-------------------------
|       |       |       |
|   4   |   2   |    FRI|
|     1 |       |     2 |
-------------------------

[Moderator's note: A picture of the cube by Corey Folkerts
 that I described in the last message  can be found at
 http://www.public.iastate.edu/~gumby145/cubes_gallery.html --Dan ]

From cube-lovers-errors@mc.lcs.mit.edu  Thu Aug  5 21:04:33 1999
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Message-Id: <37A19059.40026399@frontiernet.net>
Date: Fri, 30 Jul 1999 07:45:30 -0400
From: John Bailey <jmb184@frontiernet.net>
To: phanna@gbonline.com
Cc: Cube-Lovers@ai.mit.edu
Subject: Re: cube computer solutions using procedural languages
References: <006001bed45e$d4429420$1b4b43cf@compaq>



Paul Hanna wrote:

> Have any of you done any work on solving the cube with computer programs
> using procedural languages such as C?
>
> Do you have any suggestions you can pass my way? I am a good programmer but
> not a cube solution expert. I am just a novice at best when it comes to cube
> algorithms and efficient cube solutions.

You wouldn't believe how long ago it was that I attempted a computer
based cube solution.  At the time, I had become adept at assembly
languague for the chip in the Apple II and I worked in that.  Today I
would typically use C++.At the time, I had not found Singmasters book
( the time was roughly 25 years ago) and when I did, I realized that
what I was attempting was doomed, so I quit.  I was attempting to
apply the tree search techniques used for Chess programs to look far
ahead and by applying a heuristic scoring method, find likely paths to
solution, which would in turn be iterated.  In the process, I worked
ways to encode the orientation of each cubelet, its postion, and
transformation rules for moves etc.  As soon as I read Singmaster, I
realized how deep the iterations would have to go to reach a
convergent solution.  That's when I dropped the problem.  Some of the
recent work which the moderator referenced have moved the state of the
art further, but not yet so far as to search and find solutions in a
general sense.  (If that statement is false, it will be worth the
embarassment, if only the critic sends me a URL for source code.)

I do have some suggestions about format and linguistic conventions.
For an example,go to http://www.ggw.org/donorware/4D_Rubik/ and look
at the source code.  Yes, that's really all there is.  If that helps
or you think I should add a few comment lines, let me know.

John
http://www.frontiernet.net/~jmb184
http://www.ggw.org/donorware

From cube-lovers-errors@mc.lcs.mit.edu  Thu Aug  5 21:35:14 1999
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Reply-To: <noel@mud.ca>
From: "Noel Dillabough" <noel@mud.ca>
To: "Cube Lovers List \(E-mail\)" <cube-lovers@ai.mit.edu>
Subject: JPuzzler
Date: Sat, 31 Jul 1999 11:06:34 -0400
Message-Id: <000001bedb68$878eb170$020a0a0a@NOEL>

I have finished the last touches on JPuzzler, the Java port for Puzzler for
Windows.  Many people with other operating systems asked if I was going to
port the program and well, this is it.

Speed is an issue.  The program runs very well on a PPro200 or better, but I
believe the video card that is used makes the biggest difference because on
a PII-300 with a basic video card the program ran slower.

JPuzzler was written in 100% pure Java so it should work on any Java
compatible browser but for now it has only been tested on Netscape 4.5+ and
Internet Explorer 4+.  I don't have access to other hardware, so I am
particularly interested in hearing from players with Macintosh computers, or
any flavour of XWindows.

All 17 puzzles from Puzzler are implemented and I will probably add more in
the future; suggestions and criticism are welcome.

http://www.mud.ca/puzzler/JPuzzler/JPuzzler.html

-Noel

From cube-lovers-errors@mc.lcs.mit.edu  Fri Aug  6 11:10:06 1999
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Date: Tue, 3 Aug 1999 02:11:45 -0700 (PDT)
From: Tim Browne <ue451@victoria.tc.ca>
To: Cube-Lovers@ai.mit.edu
Subject: Notes on the Bandaged Cube... combinations and other musings.
In-Reply-To: <v04003a09b3a0062a6a87@[128.192.40.243]>

I just picked up a bandaged cube the other day, and I get the feeling that
this might be one of the few puzzles which would beat even the Square-1
for solving difficulty. For those of you who haven't seen this nightmare
of a cube, Hendrik Haak has a picture of it in the museum section of his
Puzzle Shop (he calls it a "Bicube"). After playing around with it for a
bit, I figured I'd try and work out the number of possible combinations.
The cube is constructed from a standard Rubik's Cube mechanism, so all the
standard Rubik's Cube restrictions apply here.
The cube is made up of 13 pieces. One of them is created by fusing an edge
piece between 2 centres, effectively turning it into a 2x2 piece. Because
of this piece, 2 axes of motion are effectively cut off permanently,
making a maximum of 4 axes of rotation. There are 4 pieces which are made
up by fusing an edge to a centre, 7 pieces are a corner/edge fusion,
leaving us with one standard corner piece. These fusions make the puzzle
much more difficult than it first appears, as the contortions of all of
these 1x2 pieces effectively block axes of rotation which were easily
accessible only a quarter turn ago, sometimes getting so bad as to make
the only available axis the one you just turned, in some extreme cases
even forcing you to back out following exactly the same path you used to
enter the current state. Needless to say, this makes solving the puzzle
very frustrating indeed. Anyhoo, back to the combinations...
Two of the centres are effectively locked into place for all time, leaving
us with 4 edge/centre fusions which can be rotated one of 4 ways, giving
us a factor of 4^4 combinations. Piece rotations in place are impossible.
The first potential restriction would be creating more than one area where
only the corner cube could fit, but the 2x2 piece makes this an
impossibility. The second would be potential collisions between pieces.
There are 5 pairs of adjacent sides where you have a 1:16 chance of a
conflict, and 2 ways of creating a double conflict, so we reduce this
amount by 5*256/16-2=82, leaving us with 174 possible arrangements of the
edge/centre fusions.
The edge/corner fusions and the lone corner piece will be handled
together. The corner piece can be placed into any one of 8 corner slots,
while the edge/corner fusions fit tightly into the remaining slots
surrounding it, giving us a factor of 8!=40,320 combinations. Now for the
restrictions... let's start with a simple swap first. Let's take an
example side. Like numbers in the table are bandaged together.

112
345
345

Would it be possible to swap piece 1 and piece 3? Given the restrictions
carried over from the Rubik's Cube, the only way you can swap a pair of
edges or a pair of corners is if you also swap either a second pair of the
same type, or a pair of its opposite. To see this latter case, rotate a
slice of a solved cube 90 degrees either way and then "reconstruct" it
using your favourite patterns. God's Algorithm in this case is expressly
prohibited. Since the corners and edges are fused, a swap of one expressly
implies a swap of the other, so this is OK.
How about swapping pieces 3 and 5? This one's a bit more difficult. Not
only are you swapping the pieces, but you're rotating them as well. When
you swap these 2, both edge pieces are inverted, maintaining the even
parity, so that's OK... the corner half of piece 3 is given a positive
rotation, while the corner half of piece 5 is given a negative rotation.
Modulo 3 parity is maintained, so this is also OK, meaning the whole swap
is OK. Swapping any 2 edge/corner fusions on the cube can be broken down
into a compound movement of either of these, so any of these pieces can be
swapped with any other similar piece without restriction.
The second potential restriction happens with edge/centre fusions. In
certain special cases, the centres can rotate in such a way they they box
in a 1x1x1 area, limiting it to 1 possible place. This can happen in one
of two possible ways, bringing our combinations from 174x40,320 down to
172x40,320 + 2x5,040.
The final possible restriction happens when we swap an edge/centre fusion
with an edge/corner fusion. What happpens when we swap pieces 4 and 5? So
you don't have to scroll back, here it is again...

122       122
345  -->  344  ?
345       355

There are 3 problems with this: 1) you're swapping a single pair of edge
pieces, 2) you're flipping a single edge piece, and 3) You're rotating a
single corner in place, *all* of which are definite no-no's on the Rubik's
Cube, either solo or in any combination, so you definitely can't do it
here. The corner rotation can be accounted for easily enough by rotating
the 1x1x1 piece in place to compensate, which you've probably noticed that
I conveniently left out of all calculations to this point, to save
multiplying and dividing by 3 unnecessarily. ALL of these problems can be
compensated for by either simply not doing it, or by doing the same thing
with another set, giving us 2 pairs of swapped edges and 2 edges flipped,
compensating for the corner rotation again with the 1x1x1. This
effectively slashes the potential combinations in half, bringing us down
to 86x40,320 + 1x5,040 = (86x8+1)x5,040 = 689 x 5,040 =

    5040
     689
    ----
   45360
  403200
 3024000
--------
 3472560 possible combinations.

Most puzzles of this type are difficult because of the sheer number of
combinations. This is perhaps the only one of this type which is so
difficult because of its limits. If you want to take it apart and put it
back together randomly, you've got a 1 in 6 1/1920th chance of doing so
correctly, perhaps this puzzle's one advantage over the cube. However, if
you want to tough it out and devise a pattern for it, then you'll need to
work out 172x8+2 = 1,378 patterns to get it back to its default shape,
followed by at least 6 more patterns to restore the edge/corner fusions to
their proper positions, making at least 1,384 patterns to work out for a
general solution. I've graciously decided to leave this as an exercise for
the reader. ;-) L8r.

From cube-lovers-errors@mc.lcs.mit.edu  Fri Aug  6 14:45:54 1999
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Date: Wed, 04 Aug 1999 12:04:52 -0700
From: David Barr <davidb@u.washington.edu>
To: cube-lovers@ai.mit.edu
Subject: Meffert's Assembly Cube

I saw on the http://www.mefferts-puzzles.com/ web site that he has a new
3x3x3 cube design.  Has anyone gotten their hands on one yet?

David

From cube-lovers-errors@mc.lcs.mit.edu  Fri Aug  6 16:15:54 1999
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From: WaVeReBeL@webtv.net
Date: Thu, 5 Aug 1999 16:46:13 -0700 (PDT)
To: Walter.van.Iterson@emn.ericsson.se
Cc: Cube-Lovers@ai.mit.edu
Subject: Re: Solving cube with robot arm
Message-Id: <17117-37AA2245-5060@postoffice-123.bryant.webtv.net>
In-Reply-To: "Walter van Iterson (EMN)"'s message of Fri, 23 Jul 1999 14:32:38 +0200

Hi Walter,

Have you visited the web page below?  I think this is exactly what
you're talking about.

http://www2.active.ch/~jbyland/english.html

-ALEX-

From cube-lovers-errors@mc.lcs.mit.edu  Fri Aug  6 17:22:52 1999
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From: GJB1332@aol.com
Message-Id: <5b03f8b9.24dc5396@aol.com>
Date: Fri, 6 Aug 1999 11:04:54 EDT
Subject: 5x5x5
To: Cube-Lovers@ai.mit.edu

My new 5x5x5 cube arrived a few days ago [took it 8 weeks to be delivered
'cos I was too cheap to get air-mail], and I have a couple of questions
pertaining to it:

Does anyone know a [reasonably] simple move to swap two edge pieces [the ones
on the very edge, not the center and not the corner] around?

Also, does anyone know of an equivalent to the Cube Explorer or anything like
it that can do 5x5x5 cubes?

Thank-you very much,
Gary (-;

From cube-lovers-errors@mc.lcs.mit.edu  Fri Aug  6 17:52:20 1999
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Date: Wed, 28 Jul 1999 19:59:53 -0400
From: Howard <shdrake@telegram.infi.net>
Organization: The Drake Family
To: Cube-Lovers <Cube-Lovers@ai.mit.edu>
Subject: Puzzle Shop, was Rubik's Cube Perpetual Calendar
References: <NDBBLHBJOLNCAMMDPAPACENPCAAA.ck1@home.com>

  Thanks to Chris's post, and his web page, I found a new shop
in Germany for twist-to-solve puzzles.  It is

Hendrik Haak's Puzzle Shop  http://www.puzzle-shop.de/

  Has anyone in the group ordered from this shop and what
have been your results?  I especially like the looks of the
dogic, octagon, mozaika, and octaedercube puzzles.  Has anyone
tried these, and does anyone have pictures of them in mid-twist,
on their web pages?

  Thanks for any info.

Howard Drake

From cube-lovers-errors@mc.lcs.mit.edu  Fri Aug  6 18:37:44 1999
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Message-Id: <37AB5804.6956@sgi.com>
Date: Fri, 06 Aug 1999 14:47:48 -0700
From: Derek Bosch <bosch@sgi.com>
Organization: SGI
To: Cube-Lovers@ai.mit.edu
Subject: scrambling puzzle

anyone know the "shortest" sequence of moves to "scramble"
a cube...  By scramble, my definition is:

all six colors on each face (except for 2x2x2), no 2 adjacent tiles
of the same color, and no "dominating" color on a face (ie for
a 3x3x3 cube, no more than 2 of a single color on a face, for
4x4x4, no more than 3 of a given color on a face).

D
--
Derek Bosch        "A little nonsense now and then
(650) 933-2115      is relished by the wisest men"... W.Wonka
bosch@sgi.com       
http://reality.sgi.com/bosch_engr/

From cube-lovers-errors@mc.lcs.mit.edu  Fri Aug  6 19:13:19 1999
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Date: Fri, 6 Aug 1999 18:22:01 -0400 (EDT)
From: Nicholas Bodley <nbodley@tiac.net>
To: David Barr <davidb@u.washington.edu>
Cc: cube-lovers@ai.mit.edu
Subject: Re: Meffert's Assembly Cube
In-Reply-To: <37A88ED4.B35BA92F@u.washington.edu>
Message-Id: <Pine.BSF.3.96.990806181644.6928U-100000@shell2.tiac.net>


On Wed, 4 Aug 1999, David Barr wrote:

}I saw on the http://www.mefferts-puzzles.com/ web site that he has a new
}3x3x3 cube design.  Has anyone gotten their hands on one yet?

Being a legacy-software addict, I use Lynx (2.8.2dev22) with a shell
account, so I haven't seen the images yet, but this seems to be even
more desirable than the very-nice, discontinued Ideal DeLuxe Cube.

Let's hope that it self-aligns as well as, or better than, the Ideal
I mentioned.

|*  Nicholas Bodley   *|*  Autodidact & Polymath * Electronic Tech. (ret.)
|*   Waltham, Mass.   *|*  -----------------------------------------------
|*  nbodley@tiac.net  *|*  Before 1960 or so: $100. Later: 100$
|*  Amateur musician  *|*  Before 1990 or so: 100%. Later: %100
--------------------------------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Mon Aug  9 14:41:59 1999
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Date: Fri, 6 Aug 1999 18:18:42 -0400
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From: Jacob_Davenport@scudder.com (Jacob Davenport)
Subject: Re: 5x5x5
To: Cube-Lovers@ai.mit.edu

Sure.  You can find my full solution to the 5x5x5 at
http://www.wunderland.com/WTS/Jake/5x5x5.html.  You will want
to skip ahead to section on solving the "Wings" as I've
been calling them, namely the Sixth Step--Wings to Edges.
Let us know when you get to Step Eight.

-Jacob Davenport

From cube-lovers-errors@mc.lcs.mit.edu  Mon Aug  9 15:23:00 1999
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Date: Sat, 7 Aug 1999 00:37:50 -0700 (PDT)
From: Tim Browne <ue451@victoria.tc.ca>
To: cube-lovers@ai.mit.edu
Subject: Re: Puzzle Shop, was Rubik's Cube Perpetual Calendar
In-Reply-To: <379F9979.E9CD2EBD@telegram.infi.net>
Message-Id: <Pine.GSO.3.95.iB1.0.990807001800.18553A-100000@vtn1>

On Wed, 28 Jul 1999, Howard wrote:

>   Thanks to Chris's post, and his web page, I found a new shop
> in Germany for twist-to-solve puzzles.  It is
>
> Hendrik Haak's Puzzle Shop  http://www.puzzle-shop.de/

Actually, that's been around for well over a year now... He has, however,
just recently undergone a major change on his site, including adding a
museum section with lots of really twisted puzzles. Anyone on this list
who checks it out should get a plastic keyboard cover to avoid its
shorting out due to the inevitable salivation. A few notables to check out
are the Skewb Diamond, the Extended Cube, the Mushroom, and the Rubik's
Revenge Special. If anyone can tell me how this is constructed, I'd really
like to know about it.

>   Has anyone in the group ordered from this shop and what
> have been your results?  I especially like the looks of the
> dogic, octagon, mozaika, and octaedercube puzzles.

Is the octaeder the Magic Octohedron, or the Cuboctohedron? I placed
several puzzles on order with him over a month ago, but he only shipped
them out some time last week as there were several he didn't have in
stock. Apparently, once they're shipped they take about 6 weeks to reach
North America, unless you expedite the service. I asked him about it, but
he just kept telling me "$19 for surface mail". :-/

> Has anyone
> tried these, and does anyone have pictures of them in mid-twist,
> on their web pages?

I've placed the Dogic on order... it looks especially nightmarish, but
everyone I've heard from who's tried it claimed it was really easy to
solve. Given that it's a superset of the Impossiball, I'd think it would
be particularly painful. I guess I'll find out once it arrives. The
Mozaika I haven't seen yet, but it doesn't look all that difficult... Then
again, I also thought that including a hint book with the Square-1 was an
insult to the intelligence and that anyone with as much experience as I
had with Rubik's Cube and similar puzzles should have no problem with it
before I found out just how wrong I was (it took over a year for me to
solve it once), so who's to say? For the other two... if you mean the
"Octagon barrel", it's basically a Rubik's Cube subset, The Magic
Octahedron can be solved as 2 Pyraminxen slammed back to back, and the
Cuboctohedron, being no more than a Rubik's Cube with huge chunks sliced
out of it, would be solved the same way. As for pictures showing puzzles
in mid twist, the only site I'm aware of with pictures like that would be
Chris and Kori's page. L8r.

From cube-lovers-errors@mc.lcs.mit.edu  Mon Aug  9 16:02:48 1999
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Message-Id: <000201bee0f0$f84cace0$295755ca@pca200series>
From: "MEFFERT" <meffert@ue.net>
To: <Cube-Lovers@ai.mit.edu>
Subject: Meffert's Puzzles News
Date: Sun, 8 Aug 1999 00:17:46 +0800

Dear Gary

Sorry to hear that you had to wait 8 weeks for your Prof. cube to arrive.
So here is some good news for Cube Lovers. We are offering free
Airmail Postage for all of our puzzles ordered during the Summer Holidays.
I would also like to announce our New PUZZLER CHALLENGE of the Month
competition.  Starting with the 3x3x3 cube and Orbix which can be played
both with mouse or SpaceOrb.
The fasted time in each category, puzzle plus mouse or SpaceOrb receives
a free puzzle of their choice.  There will be a different Puzzle each month.
See: http://www.mud.ca/puzzler/JPuzzler/challenge.html
You can practice on line at:
http://www.mud.ca/puzzler/JPuzzler/JPuzzler.html

For all cube lovers that do not have a SpaceOrb and would like one, we have
arranged a special deal with Spacetec who will ship the SpaceOrbs ordered
directly by UPS, at the Special Price of US$45.00 including free delivery in
the USA & Canada, add US$16.00 for all other countries, see our Puzzle Shop.

I will be announcing the start of the "Mind Sports Olympiad 2000, Puzzle
Championships soon.  There will be great prizes, puzzles must be played both
online as well as the real thing.  The finals will be held August 19 - 27
the Year 2,000 in the National Hall at Olympia in London UK, which has
9,000 square meters of floor space (approx 100,000 square feet).
Something very spectacular is being planned for the millennium year. We
intend to smash the record for the largest number of entries at any
Olympiad ever, which currently stands at 10,744 for the Olympic Games in
Atlanta 1996. This will be achieved by having a number of "biggest ever"
tournaments including the Puzzle Championships within the Mind Sports
Olympiad.

HAPPY PUZZLING TO ALL
Uwe


Uwe Meffert
MPG, "Meffert's Puzzles & Games"
2008, Remex Centre, 42 Wong Chuk Hang Road,
P.O. Box 24455, Aberdeen, Hong Kong.
Email:- Meffert@mefferts.com
www.mefferts.com

From cube-lovers-errors@mc.lcs.mit.edu  Mon Aug  9 18:30:49 1999
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To: cube-lovers@ai.mit.edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Re: Rubik's Cube Perpetual Calendar
Date: 7 Aug 1999 18:53:51 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <7ohvbv$k0t@gap.cco.caltech.edu>
References: <cube-lovers.4.1.19990728163706.0096d9d0@gumby145.mail.iastate.edu>

Corey Folkerts <gumby145@iastate.edu> writes:
>I've attached a jpg of how I arranged all the characters. I make
>no guarantees that is works, but I'm pretty sure it does.
>Sorry to those of you who can't receive files. I would have made the
>drawing ASCII, but I can't make upside or sideways characters =(

>-Corey Folkerts

>[Moderator's note: I don't send binary attachments out on this list.
> However, I was able to decode Corey's picture, and here's a diagram.
> Sorry for the delay in sending this message out. --Dan ]

For those of you who want to see the original jpg, I've put it on my
website at
http://www.ugcs.caltech.edu/~whuang/gp/cube/calender-cube.jpg

If anyone else would like to share graphics and stuff with the rest of
cube-lovers, I can provide similar services.  Offer good until I get
overwhelmed (not likely, I think).

--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
It's not sarcasm.  It's reductio ad absurdum.

From cube-lovers-errors@mc.lcs.mit.edu  Wed Aug 11 12:02:16 1999
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Date: Sun, 08 Aug 1999 12:38:42 -0500
From: Joe Johnson <jgjohnso@pressenter.com>
To: Cube-Lovers@ai.mit.edu
Subject: repairing broken cubes

I recently broke one of the face cubies of my 4X4.  I attempted to
repair it by drilling a hole for a pin and gluing it together (it broke
in the "ankle" that holds the "foot" to the top.)  I tried several
different types of glues and epoxies and nothing seemed to stick very
well.  I finally used some woodworkers glue and it is holding together
for now.  I'm being very carefull and cannot do any fast moves with it.
Does anyone know of a better way to repair these types of breaks.  I
would not have attempted the repair at all if the cubes were still
readily available, but since they are out of production I have little
choice.  This is the second 4X4 I've broken in this manner.

Joe Johnson

From cube-lovers-errors@mc.lcs.mit.edu  Wed Aug 11 13:54:40 1999
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From: gete@cheerful.com
Message-Id: <007401bee3ec$7e546640$9f1c9eca@dzine>
To: <Cube-Lovers@ai.mit.edu>
References: <37AB5804.6956@sgi.com>
Subject: Re: scrambling puzzle
Date: Wed, 11 Aug 1999 18:21:34 +0700

Derek Bosch wrote:

> anyone know the "shortest" sequence of moves to "scramble"
> a cube...  By scramble, my definition is:
>
> all six colors on each face (except for 2x2x2), no 2 adjacent tiles
> of the same color, and no "dominating" color on a face (ie for
> a 3x3x3 cube, no more than 2 of a single color on a face, for
> 4x4x4, no more than 3 of a given color on a face).


I don't, but for 3x3x3, I use software: Cube Explorer by H.Kociemba
http://home.t-online.de/home/kociemba/cube.htm

Here are some move sequences I get:

* D L2 B2 F2 R2 U' L2 . F L2 B2 R F2 D' U R2 B L2 R2 B R2  (20 moves)
* U2 F' L2 B D2 B F' U2 . R' B D R' F' L R U' F' L2 R U'  (20 moves)

Just apply them on an already-solved cube.



From cube-lovers-errors@mc.lcs.mit.edu  Thu Aug 26 17:03:07 1999
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Date: Wed, 11 Aug 1999 10:09:29 -0700
From: Darin Haines <darin_haines@adc.com>
Organization: ADC Kentrox
To: Cube-Lovers@ai.mit.edu
Cc: Joe Johnson <jgjohnso@pressenter.com>
Subject: Re: repairing broken cubes
References: <37ADC0A2.35DC8809@pressenter.com>

Joe Johnson wrote:

> I recently broke one of the face cubies of my 4X4....

Joe,

The guy to talk to is Christoph Bandelow.  His email address is

   mailto:Christoph.Bandelow@ruhr-uni-bochum.de

(I'm pretty sure he monitors the list.)  He is located in Germany and
has a bunch of other puzzles that I'm sure you will be interested in.
He is very prompt with delivering orders, and is very easy to work
with.

My Rubik's Revenge sat broken on the shelf for about 15 years.  After
contacting Christoph, my Rubik's Revenge is as good as new!  Not to
mention that I now have a few more puzzles that I didn't have before. 
Buy a few of his puzzles (5^3's etc.), and he might give you a good
deal on the 4^3 replacement parts.

Hope this helps.

-Darin

P.S.  Let us know how everything turns out.  I'm sure there are other
people in the same boat.

From cube-lovers-errors@mc.lcs.mit.edu  Thu Aug 26 18:12:39 1999
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From: "Diamond, Norman" <Norman.Diamond@jp.compaq.com>
To: Cube-Lovers@ai.mit.edu
Subject: Re: Meffert's Assembly Cube
Date: Thu, 19 Aug 1999 13:11:10 +0900

Uwe Meffert was too humble to post the following facts to the list.  However,
true cube lovers know that facts are too important.  Just as we needed to be
informed that recent remakes of old standards by Odds-On don't survive very
long in actual use, we need to know of superb manufactures such as the ones
in this thread.  Therefore I post them.

Recently my wife and I visited Mr. Meffert in Hong Kong and played with some
of his products and prototypes.  Prototypes of the assembly cubes were the
smoothest turning cubes that we have ever got our hands on.  In addition the
assembly cubes had colored tiles which look like they will not wear out during
an owner's lifetime.

Mr. Meffert added in private e-mail, with permission to repost:  the cube has a
unique patented mechanism which allows assembly disassembly of the cube
without screws or tools, which in itself is quite a challenge.

-- Norman.Diamond@jp.compaq.com
[Not speaking for Compaq]

From cube-lovers-errors@mc.lcs.mit.edu  Thu Sep  2 14:10:05 1999
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Reply-To: <noel@mud.ca>
From: "Noel Dillabough" <noel@mud.ca>
To: "Cube Lovers List \(E-mail\)" <cube-lovers@ai.mit.edu>
Cc: <arensb@cfar.umd.edu>
Subject: Square One Parity fix
Date: Fri, 20 Aug 1999 02:24:28 -0400
Message-Id: <000801beead4$b3a899d0$020a0a0a@NOEL>

I have just finished modelling the Square One puzzle in Puzzler and in
testing found a flaw in my solution.  In order to correct a parity problem
(where you have to swap only 2 pieces rather than 4) I do the following
move:

(in Arensburger notation:)

R,t3,b3,R,t1,b2,R,t2,b2,R,t6,R,b-2,t-2,R,b-2,t-1,R,b-3,t-3,R

This scrambles the cube up a bit but makes it solvable using standard moves.

There must be a better move than this!  If anyone knows a better way to swap
two pieces, let me know.

-Noel

From cube-lovers-errors@mc.lcs.mit.edu  Thu Sep  2 14:50:26 1999
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From: "Chris and Kori Pelley" <ck1@home.com>
To: "Cube-Lovers" <Cube-Lovers@ai.mit.edu>
Subject: FW: Meffert's Assembly Cube
Date: Thu, 26 Aug 1999 20:36:13 -0400
Message-Id: <NDBBLHBJOLNCAMMDPAPAIEDFCBAA.ck1@home.com>

From: Diamond, Norman [mailto:Norman.Diamond@jp.compaq.com]
> [...]  Prototypes of the assembly cubes were the
> smoothest turning cubes that we have ever got our hands on....

I agree with Norman's comments.  I recently received two of the new ASSEMBLY
CUBES and they turn very smoothly immediately after assembly.  No
lubricating is required.  The cube is neither too loose nor too tight.  The
use of the metal key to pop a center piece off is quite ingenious.  I'd say
the only cubes that rival the ASSEMBLY CUBE's quality are the 1982-vintage
Deluxe Cubes by Ideal.  They had that springy, elastic feel that was
critical for speed cubing.  Meffert's are definitely the very best
contemporary deluxe cubes around, though.

I can't wait for the transparent version to be available!

Chris Pelley

From cube-lovers-errors@mc.lcs.mit.edu  Thu Sep  2 18:21:12 1999
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Date: Sat, 28 Aug 1999 15:40:29 +0200
To: Cube-Lovers@ai.mit.edu
Subject: Rubik's 3x3x3 Cube - Pretty Patterns
From: ronald.fletterman@t-online.de (Ronald L. Fletterman)

Dear Cube Lovers,

I just joined the cube lovers' society. So, here's a bit of a cv: I
started Cubology after the first cubes were marketed in Germany. I'm a
member of the NKC, the dutch cube club. Cube Lovers who are members
too and read CFF, (standing for Cubism For Fun - the NKC's
publication) know me either from meetings at the annual Cube Days or
from the articles, I published in CFF.  Since I have a PC with an AMD
K6-III/450MHz and 128 RAM, I can run Herbert Kociemba's "Cube Explorer
and the add-on programme "Cube Optimizer", which re-activated my
interest in Pretty Patterns by looking for new species and finding
shorter manoeuvers of those that are in my almost 20 years old
collection. The results are amazing: I discover entirely new Pretties
and shorten lots of existing algorithms.
Here are a few of my recent discoveries:
R2 B2 U' L' B D' B R2 D2 L' F' D R2 U2  (14f*)
D2 U L' B L2 B' L U' B D2 B2 U2 B' F' U2  (15f*)
D F D R B L R U' B2 F' R2 B' U' B' F' U' F U' (18)
U' F U2 F2 L' D2 B L2 D R F2 L' U' R2 B' F2 D' (17)
If you are a hunter of Pretties, please let me know and my recent
findings are yours! Obviously, I'd appreciate to be informed about
cubological developments from other fellow Cube Lovers. 3x3x3, 4x4x4
and 5x5x5 cube info's are welcome indeed.

Cubology to you!
Ronald Fletterman, a Dutchman living in Warburg, Germany.

From cube-lovers-errors@mc.lcs.mit.edu  Thu Sep  2 19:21:26 1999
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Date: Fri, 27 Aug 1999 20:59:54 -0500
From: Joe Johnson <jgjohnso@pressenter.com>
To: "Cube-Lovers@AI.MIT.EDU" <Cube-Lovers@ai.mit.edu>
Subject: Re: repairing broken cubes

Thanks to all who sent suggestions.  I achieved a successful repair
using a product called 'J-B Weld' and a pin (actually a cut off box
nail.)  The 'J-B Weld' is very strong and hardens almost to the same
hardness as the

original material.  I've been using the cube now for over a week and
it is holding up very well, although it probably is going to wear out
faster now since I take it apart quite often to inspect the repair.

Joe Johnson

From cube-lovers-errors@mc.lcs.mit.edu  Fri Sep  3 13:16:50 1999
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From: roger.broadie@iclweb.com (Roger Broadie)
To: "Cube-Lovers" <Cube-Lovers@ai.mit.edu>
Subject: Re: Rubik's Cube Perpetual Calendar
Date: Tue, 17 Aug 1999 23:05:08 +0100

A calendar cube has just sold on eBay.  Quite possibly it comes from
the same source as the cube Chris Pelley mentioned.  The description
includes a picture, from which I see that it is not the same cube as
the one I described on 30 July, which was bought in the UK.  The eBay
version uses a different type-face and uses upper and lower case for
the week-day names, unlike the British version, which uses all
capitals.   It also has logos, whereas the British version has blanks
on all faces that are not used for date markings. The British version
was nonetheless a genuine Ideal cube and came in a cardboard drum
bearing the title Rubik's Calendar and the copyright notice
"[c-in-a-circle] MMLXXXI Ideal Toy Co Ltd, Wokingham, Berks".  There
are other European calendar cubes pictured on eBay which are like the
British one

There is something very strange about the cube in the eBay photo.  In
fact, I am convinced that is does not work for all dates, which drives
me to the conclusion that the stickers have been rearranged.  I'll
explain my reasons and see if others disagree.

The photo shows only one view, so we have only three faces to work on.
They look like this:

-------------------
|     |     |     |
|  2  |  M  |  5  |
|  V  |  >  |  <  |
-------------------
|     |     |     |
|  J  |  P  |  1  |
|     |  V  |  <  |
-------------------
|     |     |     |
|  Mon|  P  |  6  |
|  <  |  >  |     |
-------------------

-------------------   -------------------
|     |     |     |   |     |     |     |
|Satur|day  |     |   |  7  |  A  |  0  |
|     |     |     |   |  V  |  >  |  <  |
-------------------   -------------------
|     |     |     |   |     |     |     |
|  F  |  E  |  B  |   |     |  C  |  3  |
|     |     |     |   |     |  V  |  <  |
-------------------   -------------------
|Rbk's|     |     |   |     |     |     |
|Cube |     |  8  |   |Thurs|Ideal|  9  |
|     |     |     |   |     |     |     |
-------------------   -------------------
The orientation is shown by the Vs, which point to the local upright
for those markings that are not upright as shown - I've followed Dan
in this.

Once the markings on the various pieces of a calendar cube have been
fixed, the way the cube is assembled does not matter, since to show
the date we need solve only one face, and that can be done from any
starting position, any necessary counterbalancing twists or
permutations taking place in the other layers.  Let's assume the face
used to show the date is the front face.

Therefore we only need to worry about what markings each piece will
bear.  The straightforward approach in designing a cube of this sort,
it seems to me, is as far as possible to keep all the markings of the
same type - that is, destined for the same position on the front
face - together on the pieces carrying them.  So the week-day names
would be on one set of corner pieces, destined for the top left of the
front face, and the numbers forming the units digit of the day of the
month would be on another set, destined for the bottom right of the
front face.  In that way, clashes in which the piece would be needed
in two places at once are avoided.  If this approach is not followed,
then either there must be no clash, or markings must be duplicated.
An example of the first would be an edge piece that combined the F or
B of FEB with the 3 of the tens digit of the day of the month, a
combination that is possible because there is no FEB 30.  The picture
on the container of the British calendar cube, though not the cube
itself, illustrates the other possibility, since one edge-piece
combines J and 2.  That means that JAN 20 and similar dates cannot be
shown unless either the J or the 2 is duplicated.  As it happens,
there is one spare edge-piece face on that cube, so one duplication
could be managed, but no more, but it is impossible to see from the
picture if there is any duplication.  Probably not, because there are
some other impossibilities and inconsistencies in the pictures which
suggest they show non-functioning mock-ups.

The different markings that need to be accommodated on the edge
pieces, defined by their position on the front face, are:

Top:  DAY

Left: the eight initial letters of the month, J F M A S O N D

Right: the ten letters completing the abbreviation for the month, N B
R Y L G P T V C

Bottom: the four numbers for the tens digit of the day of the month, 0
(or blank) 1 2 3.

The edge-piece at the top will always stay there, since it is needed
to show DAY (assuming no duplicates).  So the other face of this piece
will never show on the front face and cannot carry a useful marking.
It can be blank, or carry a logo.

In total, including the face backing DAY, we now have 24 faces, and
that is exactly the number we have available if we have one face for
each of the markings above.  As it happens, the one letter that occurs
at both the start and the end of a month, N, is symmetrical in the
sans-serif typeface used, and in the British cube is made to double as
a starting and an ending letter, since it can be either way up. That
frees up one face to permit one duplication.  It is impossible to see
if the same approach is followed in the eBay cube.

As a matter of interest, the British cube mostly (but not in all
cases) puts an initial and a final letter together on the same
edge-piece.  That is possible without too much juggling to avoid the
clash of having the start and end of a month on the same edge piece,
but is not necessary - the principle of segregating the different
types of marking would lead to four edge-pieces with the initial
letters and five with final letters.

If we turn back to the eBay cube, we find that the visible edge-pieces
are as follows.

Day/P, J?, M?, 1A, F?, 3?, B/blank, blank/? logo/?, 3x??

Since Day is combined with P, we need a duplicate of one or other if
dates in SEP or APR are to be showable.  But even then we would not be
able to show APR 10 unless a duplicate of 1 or A was included.  Thus
even if one N is used for both the start of NOV and the end of JAN or
JUN, we would already have one face too many.  Yet on top of that
there are two blanks and a logo.  One blank is usable as the blank for
the tens digit in dates like JAN 1, and one among the month letters is
there is only one N.  But that still means that there is an extra
blank even if there are no duplicates.

If extra blank faces are included, then other needed markings must be
omitted and dates involving those markings could not be shown.  In
fact the only reasonable explanation I can see is that some stickers
have been removed and put back wrongly.

Roger Broadie

From cube-lovers-errors@mc.lcs.mit.edu  Fri Sep  3 19:04:22 1999
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Date: Thu, 2 Sep 1999 01:35:45 +0100
From: David Singmaster <david.singmaster@sbu.ac.uk>
To: pink@cartserv.rserv.uga.edu
CC: cube-lovers@ai.mit.edu
Message-ID: <009DCC4C.2E241665.12@ice.sbu.ac.uk>
Subject: RE: Skewb notes

	Regarding disassembly of the skewb and the various skewballs.
	The early versions disassemble quite easily and when I put them out at
exhibition, I regularly have to reassemble them.  However, I have just learned
from Yee Dian Lee at the International Puzzle Party that the more recent
skewballs are quite hard to take apart and he finds that he has to break a
piece in order to do so!  He puts them inside jars!

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499;
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Tue Sep 21 20:52:44 1999
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Date: Thu, 2 Sep 1999 12:57:37 -0700 (PDT)
From: Tim Browne <ue451@victoria.tc.ca>
To: Noel Dillabough <noel@mud.ca>
Cc: "Cube Lovers List (E-mail)" <cube-lovers@ai.mit.edu>, arensb@cfar.umd.edu
Subject: Re: Square One Parity fix
In-Reply-To: <000801beead4$b3a899d0$020a0a0a@NOEL>
Message-Id: <Pine.GSO.3.95.iB1.0.990902124509.4819A-100000@vtn1>

On Fri, 20 Aug 1999, Noel Dillabough wrote:

> I have just finished modelling the Square One puzzle in Puzzler and in
> testing found a flaw in my solution.  In order to correct a parity problem
> (where you have to swap only 2 pieces rather than 4) I do the following
> move:
>
> (in Arensburger notation:)
>
> R,t3,b3,R,t1,b2,R,t2,b2,R,t6,R,b-2,t-2,R,b-2,t-1,R,b-3,t-3,R
>
> This scrambles the cube up a bit but makes it solvable using standard moves.
>
> There must be a better move than this!  If anyone knows a better way to swap
> two pieces, let me know.

For starters, you'd be amazed what a B6 added to the beginning or end of
this pattern will do. :-) Alternatively, you could skip over that and
simply add two dual edge swaps by appending something like
t1,R,t-1,b-1,R,t6,R,t1,b1,R,t3,b3,R,t-1,b-1,R,t6,R,t1,b1,R,t-4,b-3
to the beginning or end of your pattern. L8r.

-- 
        Cubic Puzzles - The SIMPLEST Solutions          |       HIT Jedi
     http://www.victoria.tc.ca/~ue451/solves.html       | Use the Force, Mike!
--------------------------------------------------------+-----------------------
            "No thanks. I'm trying to cut down." - Michael Garibaldi

From cube-lovers-errors@mc.lcs.mit.edu  Tue Sep 21 22:01:15 1999
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Date: Sat, 04 Sep 1999 13:27:32 -0500
From: Joe Johnson <jgjohnso@pressenter.com>
To: "Cube-Lovers@AI.MIT.EDU" <Cube-Lovers@ai.mit.edu>
Subject: 5X5X5 challenge

I recently bought my first 5X5X5 Rubik's Cube (from  Meffert's) and when
I received it, the first thing I noticed was that the stickers had
little grooves in them and all of them on each face were lined up in the
same direction.  After I learned how to solve the cube for color I
decided to try to solve it so that all of the little grooves lined up.
After much hard work I finally got it and can repeat the solution.  It
is a wonderful challenge after you've grown tired of solving it the
'normal' way!  I won't give the solution unless I see that there is
interest in it here in this list

Joe Johnson.

[Moderator's note: See "Supergroup" in the archives.  If you want to
experience the full measure of complication, mark the a corner of
twelve facelets on each face in this pattern:

    +-----+-----+-----+-----+-----+
    |     |     |     |     |     |
    |     |     |     |     |     |
    +-----+-----+-----+-----+-----+
    |     |     |     |     |     |
    |     |   .'|`.   |   .'|`.   |
    +-----+-----+-----+-----+-----+
    |     |   `.|.'   |   `.|.'   |
    |     |     |     |     |     |
    +-----+-----+-----+-----+-----+
    |     |     |     |     |     |
    |     |   .'|`.   |     |     |
    +-----+-----+-----+-----+-----+
    |     |   `.|.'   |     |     |
    |     |     |     |     |     |
    +-----+-----+-----+-----+-----+

and make them line up afterwards.  You'll learn something surprising
about the center facelets. --Dan ]

From cube-lovers-errors@mc.lcs.mit.edu  Tue Sep 21 22:12:52 1999
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From: uwe@ue.net (Uwe Meffert)
Reply-To: uwe@ue.net (Uwe Meffert)
To: <Cube-Lovers@ai.mit.edu>
Subject: Color survey
Date: Thu, 16 Sep 1999 16:39:02 +0800

To all CubeLovers.

I am conducting a color survey for a revised color scheme for the new molded
tiles that I am preparing for some of our Puzzles for Xmas.

The final choice will be based on the majority of the suggestions received
from CubeLovers like you.

Megaminx: do you prefer 6 colors or 12 colors;  If 12, which colors?

Prof. cube & new improved 4x4x4 cube: any color preferences?

Looking forward to hearing from you all, your input is important as this is
most likely the last time that I will run some of these Items.

The other good news is that I have finally been persuaded to release several
new puzzles for Xmas.  More later

With warm regards to all
Uwe


Let's keep puzzling alive!

Uwe Meffert
Meffert's Puzzles & Games
2008, Remex Centre, 42 Wong Chuk Hang Rd,
P.O. Box 24455, Aberdeen, Hong Kong.
Tel. 852-2518-3080, Fax. 852-2518-3282.
Email: Uwe@Mefferts.com
Web: www.Mefferts.com

From cube-lovers-errors@mc.lcs.mit.edu  Wed Sep 22 19:31:25 1999
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Date: Wed, 15 Sep 1999 22:07:37 -0700 (PDT)
From: John Davis <jdjabberwock@yahoo.com>
Subject: bandaged cube
To: cube lovers <cube-lovers@ai.mit.edu>


I'd like to make one of these bandaged cubes out of an old 3x3x3, but
can't tell from pictures what pieces to fuse on the back of the cube.
Can someone who has one of these post a description making it clear
which pieces are fused?  Thanks, John.

From cube-lovers-errors@mc.lcs.mit.edu  Wed Sep 22 21:06:02 1999
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Date: Thu, 16 Sep 1999 23:20:21 -0700 (PDT)
From: Tim Browne <ue451@victoria.tc.ca>
To: Cube-Lovers@ai.mit.edu
Subject: Pyramorphix
In-Reply-To: <199908061510.LAA28883@mc.lcs.mit.edu>
Message-Id: <Pine.GSO.3.95.iB1.0.990916231933.11677D-100000@vtn1>

I'd like to try taking my Pyramorphix puzzle apart to see what makes it
tick. I took a brief look at it today, but the internal mechanism appears
to be somewhat delicate. of course, it could just be paranoia on my part.
How does one go about taking a Pyramorphix apart safely? L8r.

-- 
        Cubic Puzzles - The SIMPLEST Solutions          |       HIT Jedi
     http://www.victoria.tc.ca/~ue451/solves.html       | Use the Force, Mike!
--------------------------------------------------------+-----------------------
            "No thanks. I'm trying to cut down." - Michael Garibaldi

From cube-lovers-errors@mc.lcs.mit.edu  Wed Sep 22 21:17:03 1999
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Date: 	Thu, 16 Sep 1999 11:08:22 -0400 (EDT)
From: Alchemist Matt <monroem@email.unc.edu>
Reply-To: Alchemist Matt <monroem@email.unc.edu>
To: Cube-Lovers <Cube-Lovers@ai.mit.edu>
Subject: Square 1
Message-Id: <Pine.GSO.3.95L.990916110208.4489A-100000@sunny>

Hello,

	In case anyone is interested, Uwe Meffert is now selling the
Square 1.  I've been looking to buy one for a while, and this is the
perfect opportunity (www.mefferts-puzzles.com/mefferts-puzzles/index.html).  
In relation to the Square 1, I once came across a program that claimed to
help one solve the puzzle.  I could have sworn I saved it to my hard
drive, but now, when I look for it, I can't find it.  I thought it was
called Square1.Exe or Square1.Bas Does anybody know anything about this
program or have a copy lying around?

		Thanks,
		 Matt

-----------------------------------------------------------------------
Matthew Monroe				Monroem@UNC.Edu
Analytical Chemistry      		http://www.unc.edu/~monroem/
UNC - Chapel Hill, NC                   This tagline is umop apisdn
-----------------------------------------------------------------------

From cube-lovers-errors@mc.lcs.mit.edu  Thu Sep 23 19:45:54 1999
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From: roger.broadie@iclweb.com (Roger Broadie)
To: "Cube-Lovers" <Cube-Lovers@ai.mit.edu>
Cc: "Joe Johnson" <jgjohnso@pressenter.com>
Subject: Re: 5X5X5 challenge
Date: Wed, 22 Sep 1999 23:16:36 +0100

Joe Johnson wrote (4 September 1999)


>I recently bought my first 5X5X5 Rubik's Cube (from  Meffert's) and when
>I received it, the first thing I noticed was that the stickers had
>little grooves in them and all of them on each face were lined up in the
>same direction.

Yes, the grooving effect shows up quite strikingly against the light
as two distinct shades and allows some nice pretty patterns.  Since
there are only two apparent alternatives, whereas there are four
possible orientations, some possibilities look a little surprising, to
my eyes at least.  Thus a three-cycle of centre pieces shows up as a
change to just two pieces.

Roger

From cube-lovers-errors@mc.lcs.mit.edu  Thu Sep 23 21:09:43 1999
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From: "Steve Adler" <adler@xnet.com>
To: <Cube-Lovers@ai.mit.edu>
Subject: Strange way to solve cube
Date: Wed, 22 Sep 1999 01:15:45 -0500

I received a Magic catalog with the following trick:

Oral Fix-Sational

Bring out a miniature Rubik's cube. It's completely mixed up.
Place the Rubik's cube in your mouth! Yummy!
Roll the cube around in your mouth. This way, then that. Up
then down. It looks relly weird.
Now, Open your mouth and push the cube out. It is completely
solved!
Do it surronded. It resets in seconds. No shell is used.
Oral Fix-Sational from Ed Alonzo, comes with complete instructions
and the 1-1/4" gimmicked cube. $15.00

From cube-lovers-errors@mc.lcs.mit.edu  Thu Sep 23 21:49:54 1999
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Message-Id: <003b01bf057a$4fb871a0$2e5755ca@pca200series>
From: uwe@ue.net (Uwe Meffert)
To: "Cube-Lovers" <cube-lovers@ai.mit.edu>, <ue451@victoria.tc.ca>
Cc: "Jing Meffert" <jing@ue.net>
References: <001b01bf056b$22b833c0$2e5755ca@jing-notebook>
Subject: Re: Pyramorphix
Date: Thu, 23 Sep 1999 12:14:27 +0800
Organization: UE Foundation

Dear Tim

The Pyramorphix mechanism is indeed delicate and can not be taken apart
without the risk of destroying it, after assembly the last piece is glued
into place to hold the puzzle firmly together without pieces falling out
during play.

With warm regards
Uwe

Let's keep puzzling alive!

Uwe Meffert
Uwe@Mefferts.com
www.Mefferts.com

From cube-lovers-errors@mc.lcs.mit.edu  Mon Sep 27 17:23:43 1999
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Date: Wed, 22 Sep 1999 20:52:28 -0500
From: Joe Johnson <jgjohnso@pressenter.com>
To: Cube-Lovers@ai.mit.edu
Subject: Re: 5X5X5 challenge
References: <37D16494.DE65A654@pressenter.com> <m11TkUz-00076uC@fwd04.btx.dtag.de>

I snipped off the same corner of every sticker on each face so that all of the face
cubies have to be positioned correctly for a 'complete' solution.  If you have good
eyes (I recently had cataract surgery) you will not need to snip off enough to make
it obtrusive.  My second 5x5x5 just arrived in today's mail and it almost seems like
cheating to solve it in the 'normal' manner after solving for 'completeness.'

Joe Johnson

From cube-lovers-errors@mc.lcs.mit.edu  Tue Sep 28 12:32:22 1999
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Message-Id: <19990923162848.24944.rocketmail@web126.yahoomail.com>
Date: Thu, 23 Sep 1999 09:28:48 -0700 (PDT)
From: Jaap Scherphuis <jaapsch@yahoo.com>
Reply-To: jaap@org2.com
Subject: Square 1, Pyramorphix, Cheap Skewballs
To: Cube Lovers <Cube-Lovers@ai.mit.edu>


Hi all,
This is my first post to Cube-Lovers, though I have been reading the archives
for a long time now. I'm a 27 year-old mathematician.

Re: Square-1
A few years ago I figured out a solution to the Square-1. It was one of the
hardest puzzles to solve. Though I did theoretically solved it, the solution
was so long and tedious I never actually performed it. In the end I wrote a
program that searched for short sequences that go from cube shape to cube shape
that do not move the corners. In the list it produced were a few useful
sequences moving only a few edges, but all the odd permutations moved a lot of
edges. By combining one of them with some other sequences I finally got my own
parity fixing sequence that is a nice triple edge swap:

Swap FU-BU, LU-RU, FD-BD:
      /(3,3)/(1,2)/(2,-4)/(-2,4)/(-1,1)/(3,3)/(0,3)/(3,3)/(0,3)/(6,0)/(6,0)/

The notation is fairly obvious: /=half turn of right hand side, (t,b)=move
top/bottom the given number of twelfths clockwise, negative for anti-clockwise.
I find this much easier to read than any others I've seen, though it is
sometimes easy to forget the leading / if there is one.

Unfortunately I have since lost any other results I got then except for those I
have incorporated into my solution.

The square-1 solving program Matt mentioned can be found in the cube-lovers
archive in the contrib directory.


Re: my webpages
I have recently typed up a lot of my notes and put them on the web in a
text-only preliminary form. Eventually I hope to make them into proper
web-pages with pictures etc.
There are solutions there for:
Alexander's Star, Pocket Cube (2x2x2), Rubik's Cube (3x3x3), Rubik's Revenge
(4x4x4), Profesor's Cube (5x5x5), Dogic, Domino, Impossiball, Megaminx,
Octahedron, Pyraminx, Pyramorphix, Skewb, Brain ball, Rubik's Fifteen, Equator,
It, Ivory Tower (Babylon Tower), Masterball, Orb, Puck, Roundy, Square One,
Topspin, Tower (Whip-It), Rubik's Triamid, Tricky Disky, Rubik's Clock, Lights
Out, Rubik's Magic, Spinout, Crazy Tantrix.
At the moment there is not yet a links page.

You can find it here: http://www.org2.com/jaap/puzzles
I'd appreciate any feedback.


Re: Pyramorphix.
I only have the pocket Pyramorphix, and these are delicate (my first one broke
within 5 minutes). The pieces have small feet which slide through grooves in a
ball. The grooves are formed between 8 triangular pieces which are screwed onto
the ball. By pushing a small screwdriver through at a point where 4 pieces come
together you can unscrew it. It may work best if you bring the 4 flat pieces
together and use the spot between them to unscrew it.


Re: Cheap Skewballs.
This week I bought several cheap puzzleballs at the Oxford Toys'r'Us, all of
the France '98 type.  I bought the last two keychain ones (1 uk pound each),
and a couple of normal sized ones (2 uk pounds each). They still have many of
those.
I plan to paint them with diffent designs, e.g.
dodecahedron/icosahedron/octahedron, or rather the spherical projections of
these shapes.

That's all for now.
Bye,
Jaap.


=====
Jaap Scherphuis
Visit the Psion Organiser II CM, XP & LZ Homepage:
URL: http://www.org2.com      email: jaap@org2.com

From cube-lovers-errors@mc.lcs.mit.edu  Tue Sep 28 13:09:57 1999
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From: HarLikin@aol.com
Message-Id: <1bb80147.251b137f@aol.com>
Date: Thu, 23 Sep 1999 01:24:15 EDT
Subject: Skewb corner malfunctions
To: Cube-Lovers@ai.mit.edu

    After a hard day of solving the skewb for the first time it started to
become loose in my hands and eventually fall apart. As I put it back together
I noticed that the problem was that one of the four corners which are
connected to the internal mechanism would unscrew whenever it was rotated
counter clockwise, eventually causing it to move far enough away from the
mechanism as to cause the pieces to fall away. Is this a problem which has
been encountered by other regular Skewbers? If anyone knows what sort of
mechanism usually keeps the corner attached properly and/or how to fix such
an error I would be grateful.

-Terrence

From cube-lovers-errors@mc.lcs.mit.edu  Tue Sep 28 16:38:38 1999
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From: "Steve Adler" <adler@xnet.com>
To: <Cube-Lovers@ai.mit.edu>
Subject: Meffert's new puzzles
Date: Thu, 23 Sep 1999 20:48:33 -0500

I received an email from Meffert about new Puzzles :
(he mentions doing a version of a 4x4x4)

"Other puzzles that I will be making for Xmas are the Siamese Twin assembly
cube, Pyraminx Diamond, (8 sides, Skewb mechanism)  The Star of David
(Pyramorphix mechanism) and a Dodecahedron with each face being divided into
4 segments, (Skewb mechanism).

All new Puzzles will feature a molded tile finish."

WOW.....GREAT NEWS!!

From cube-lovers-errors@mc.lcs.mit.edu  Tue Oct 12 19:00:32 1999
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To: cube-lovers@ai.mit.edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Re: Square 1
Date: 24 Sep 1999 14:04:47 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <7sg0dv$t7s@gap.cco.caltech.edu>
References: <cube-lovers.Pine.GSO.3.95L.990916110208.4489A-100000@sunny.SOMEWHERE>

Alchemist Matt <monroem@email.unc.edu> writes:
>	In case anyone is interested, Uwe Meffert is now selling the
>Square 1.  I've been looking to buy one for a while, and this is the
>perfect opportunity (www.mefferts-puzzles.com/mefferts-puzzles/index.html).
>In relation to the Square 1, I once came across a program that claimed to
>help one solve the puzzle.  I could have sworn I saved it to my hard
>drive, but now, when I look for it, I can't find it.  I thought it was
>called Square1.Exe or Square1.Bas Does anybody know anything about this
>program or have a copy lying around?

A copy is in the cube-lovers archive at
ftp://ftp.ai.mit.edu/pub/cube-lovers/contrib/square1.exe.gz

The rec.puzzles archive also has some information on solving the puzzle.
--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
Piano for sale, cheap.  No strings attached.


From cube-lovers-errors@mc.lcs.mit.edu  Tue Oct 12 19:34:22 1999
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Message-Id: <016801bf0a44$caac3300$2e5755ca@pca200series>
From: uwe@ue.net (Uwe Meffert)
To: <HarLikin@aol.com>
Cc: <Cube-Lovers@ai.mit.edu>
References: <005801bf0a33$81b07be0$2e5755ca@jing-notebook>
Subject: Re: Skewb corner malfunctions
Date: Wed, 29 Sep 1999 14:35:12 +0800

> -----Original Message-----
> From: HarLikin@aol.com <>
> To: Cube-Lovers@ai.mit.edu <Cube-Lovers@ai.mit.edu>
> Date: Wednesday, September 29, 1999 10:01 AM
> Subject: Skewb corner malfunctions
>
> After a hard day of solving the skewb for the first time it started
>to become loose in my hands and eventually fall apart. As I put it
>back together I noticed that the problem was that one of the four
>corners which are connected to the internal mechanism would unscrew
>whenever it was rotated counter clockwise, eventually causing it to
>move far enough away from the mechanism as to cause the pieces to
>fall away. Is this a problem which has been encountered by other
>regular Skewbers? If anyone knows what sort of mechanism usually
>keeps the corner attached properly and/or how to fix such an error I
>would be grateful.
>
>-Terrence

Dear Terrence

I am sorry to hear that you are one of the very few persons experiencing
problem with the Skewb. I believe that you are actually the first. Even so
there have a few identical problems with the Prof.cube.

What unfortunately happened is that when gluing the corner cap excess
glue fixed the screw to the plastic corner piece that it should turn
in.  So when you turn this section it will tighten / loosen that one
screw.

If you are skilful enough you can try and carefully remove the corner
cap to expose the screw that has become loose.  Then try to remove the
excess glue from around the screw with a sharp object and try turning
the screw with a screwdriver firmly holding the plastic piece so you
can break the glue bond.  Once the screw can freely turn inside the
plastic part, re-tighten it to the same tension as it was originally,
so as to allow smooth turning without any pieces falling out during
play.  Then carefully using only very little glue fix the corner cap
back into place.

Please let me know the outcome of this recommended procedure. If you can't
fix it please let me know and I will send you a free replacement.

Sorry for any frustration and inconvenience caused.

Good Luck and Happy Puzzling.


With warm regards
Uwe

Let's keep puzzling alive!

Uwe Meffert
Meffert's Puzzles & Games
2008, Remex Centre, 42 Wong Chuk Hang Rd,
P.O. Box 24455, Aberdeen, Hong Kong.
Tel. 852-2518-3080, Fax. 852-2518-3282.
Email:- Uwe@Mefferts.com

www.Mefferts.com

From cube-lovers-errors@mc.lcs.mit.edu  Tue Oct 12 20:07:12 1999
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From: Daniel B Knights <knights@panther.middlebury.edu>
To: Cube-Lovers@ai.mit.edu
Subject: 3-Cube in 1 One-Look
Message-Id: <Pine.A41.4.02.9910041844460.182270-100000@puma.middlebury.edu>

Hi all, I'm new to this list, and new to the cube as well - only got my
first one in March.
Of course, now it doesn't leave my side.

I have seen a few Cube-Lovers emails about solving the cube in a minimum
number of looks. Here is a system that I use to solve the cube in 1 look,
with 10-25 minutes of studying time. (Please excuse my lack of knowledge
of terminology/group theory.)
_________________________________________________

When most people solve the cube they do it by decomposing the whole
problem into successively more specific subgroups. (e.g. first layer
edges, first layer corners, second layer edges, etc.) I say "successively
more specific" because the moves someone would use to position the first
few pieces are very simple and intuitive, usually changing (but not
solving) the unsolved pieces in the cube. As one approaches the solved
state, one uses much more specialized algorithms that affect only the
remaining unsolved pieces.

For "multiple-look" purposes, this is a great approach. Often the smaller
the subgroup of pieces affected by an algorithm, the larger the number of
moves in that algorithm, and since there is usually no perceived order to
the unsolved pieces, there is no benefit to preserving them with lengthy
specialized moves.

To a person visualizing an entire solution in his or her head, however,
these types of moves are very expensive in terms of memory. Instead I
begin from the start using specialized moves that affect as little of the
cube as possible. I might start off with an algorithm to permute 3 corners
(hopefully putting at least two of them in the correct place/orientation)
while leaving the other 5 corners and all 12 edges untouched.

In fact, by the time I have all of the corners solved, the edge pieces are
in exactly the same random configuration as when I started! (with the
possible exception of having interchanged exactly 2 of them.) The solution
has then been decomposed into 2 nearly independent problems.  The moves I
use are mostly single-layer permutations with
some commutators mixed in when necessary.

One can get the corners solved after applying 5 or 6 move sequences, and
then solve the edges with an additional 7 or 8 sequences. (This has
nothing to do with the number of moves used to solve the cube. In fact,
when I solve it with my eyes closed, I average 150-200 moves!)
_____________________________________________________

The Rules:
I would consider it cheating to use a pen and paper. Basically, you have
to sit down with a random cube and look at it for a while without
manipulating it. Then close your eyes and start solving. When you next
open them, it should be solved. (You don't get to "practice" the moves
before you go.)
_____________________________________________________

So, has anyone else tried this?  I'm curious to know what method someone
else uses.  I use my 15 minutes of studying time to plan out where I'm
going to need to move the pieces.  I wonder if anyone with better memory
skills can just memorize the locations of all the pieces and then work out
the entire solution with their eyes closed.

Dan Knights
Middlebury College
http://www.middlebury.edu/~knights/

From cube-lovers-errors@mc.lcs.mit.edu  Thu Oct 14 15:47:40 1999
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Date: Wed, 13 Oct 1999 09:21:25 -0700
From: David Barr <davidb@u.washington.edu>
To: Daniel B Knights <knights@panther.middlebury.edu>
Cc: Cube-Lovers@ai.mit.edu
Subject: Re: 3-Cube in 1 One-Look
References: <Pine.A41.4.02.9910041844460.182270-100000@puma.middlebury.edu>

Daniel B Knights wrote:
> So, has anyone else tried this?  I'm curious to know what method someone
> else uses.  I use my 15 minutes of studying time to plan out where I'm
> going to need to move the pieces.  I wonder if anyone with better memory
> skills can just memorize the locations of all the pieces and then work out
> the entire solution with their eyes closed.

There was discussion of blindfolded solving during March 98 on this
list.  At that time no one was able to do it.  You may be the first.

David

From cube-lovers-errors@mc.lcs.mit.edu  Thu Oct 14 16:44:45 1999
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Date: Sat, 25 Sep 1999 17:45:54 -0600
To: cube-lovers@ai.mit.edu
From: slapdash@enteract.com (Russ Perry Jr)
Subject: Rubik-like puzzles I've never seen before...

[I was referred to this list by someone who read a post I made in the
rec.puzzles newsgroup; I'm not a member of the list, but have some
questions about puzzles I've run across on eBay recently that I'm
curious of their origin.  I thought maybe you'd post my questions
on the list.  Please direct followups to my email address.]

I check out eBay regularly for Rubik's Cube type puzzles I don't have...
Recently I've come across a few I've never seen before -- does anybody
know where they came from?  Or better yet, where I might find them
cheaply, as the prices on eBay seem to be way up there...

1) The Gem/Diamond.  This is an octahedron, but each face is one solid
piece, and it appears to turn in half, turning four triangulars on the
square plan between halves.

2) ??.  This is a 14-sided puzzle, with 8 triangles and 6 squares.  Two
opposite triangles make the "centers" (in 3x3x3 terms), around which
three squares and three triangles turn.  I don't know if ALL triangles
act as centers...  It can be left "unshaped" similar to Square-1, only
not so jarringly.

3) Siamese Cube.  This was an old hack in the 80s, with two normal cubes
sharing a row of two corners and an edge, but these look professionally
made, and are made out of the new OddzOn cubes apparently.  They come
with a display case.

4) "Triple" Cube.  This is similar to the Siamese Cube, only there are
three cubes connected, and unlike the Siames Cube, each new cube rises
a level -- a row of one corner and an edge are aligned with the next
cube's edge and the other corner (ie one is bottom corner and edge,
joining the other cube's edge and top corner).  This also comes with
a display case.

The mechanism for the Gem is likely just a 2x2x2 cube mechanism, but the
pieces?  The 14-sider looks ALL new, though with 8 centers, maybe it's
some kind of octahedron shape variation?

With the Siamese and Triple cubes you could just have a few different
pieces swapping out the originals -- is someone just making the new
pieces?

Are these homemade, prototypes, original puzzles by pirate manufacturers?
Does anybody know?

//*================================================================++
||  Russ Perry Jr   2175 S Tonne Dr #105   Arlington Hts IL 60005  ||
||  847-952-9729    slapdash@enteract.com    VIDEOGAME COLLECTOR!  ||
++================================================================*//

From cube-lovers-errors@mc.lcs.mit.edu  Thu Oct 14 18:12:07 1999
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Date: Thu, 07 Oct 1999 10:21:49 -0500
From: Hana Bizek <hbizek@ameritech.net>
Reply-To: hbizek@ameritech.net
To: cube-lovers@ai.mit.edu
Subject: fad of the century

Hello, cube-lovers,
it is my pleasure to infoem you that a poll conducted by aol.com voted
the Rubik's cube as fad of the century. 10 fads were listed and our
little obsession (at least for some of us) was on the very top of the
list.

Congratulations, Rubik's cube!

Hana Bizek

PS IMHO, the cube is more than a fad!

From cube-lovers-errors@mc.lcs.mit.edu  Thu Oct 14 19:52:23 1999
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Date: Wed, 22 Sep 1999 02:22:00 -0400
From: Kevin Young <keyoung3@vt.edu>
Subject: Rubiks Magic: Create A Cube
To: Cube-Lovers@ai.mit.edu
Message-Id: <4.2.0.58.19990922022037.00a62530@mail.vt.edu>

Anyone having notes on or knows a solution to solving this version of
Rubiks magic and making it a cube, I'd appreciate info on it.

Thanks,
Kevin M. Young

From cube-lovers-errors@mc.lcs.mit.edu  Mon Oct 25 19:08:23 1999
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From: Elsa Sharp <ElsaS@Planet24.co.uk>
To: Cube-Lovers@AI.MIT.EDU
Subject: Rubik's Cube Fans
Date: Mon, 25 Oct 1999 16:57:22 +0100

Hello,

I found your email, address on the internet. I work on the Big Breakfast
Channel 4's daily entertainment show. We are inviting Erno Rubik on the show
next week and we would like to find fans of the Rubik's cube. I wonder if
you could email me the contact numbers of your members, with a view to
inviting them on the show.

My email address is: elsas@planet24.co.uk

Many thanks,
Elsa Sharp 

[Moderator's note: Interested cube lovers should contact Ms Sharp as above; I'm
 not going to release the subscription list.  And as far as e-mail addresses of
 other people you know, I suggest you pass this message to them rather than
 sending their address to some third party without their explicit consent. --Dan ]

From cube-lovers-errors@mc.lcs.mit.edu  Mon Oct 25 19:23:26 1999
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Message-Id: <3803289F.8337B0D5@zeta.org.au>
Date: Tue, 12 Oct 1999 23:25:03 +1100
From: Wayne Johnson <sausage@zeta.org.au>
To: Cube-Lovers@ai.mit.edu
Subject: Working with plastics
References: <37134515.4AD5@zeta.org.au>

If anyone has ever been interested in learning how to work with plastics to
make your own puzzle designs/prototypes, make interesting modifications to
your cubes, or you want to be able to learn how to repair your puzzles; I
have just completed the Twisty megasite.

This is located at:
http://www.zeta.org.au/~sausage/twistymegasite/

I have quite a lot of new and original resource material here, so it might
be well worth keeping an eye on it.

Thanks,
Sausage

From cube-lovers-errors@mc.lcs.mit.edu  Mon Oct 25 19:46:45 1999
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To: cube-lovers@ai.mit.edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: The Cube at the World Puzzle Championship
Date: 15 Oct 1999 15:18:04 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <7u7gjc$6d8@gap.cco.caltech.edu>

The World Puzzle Championship was held last week in Budapest, Hungary.
Erno Rubik was invited to kick off the opening ceremonies (so I got
a few pictures of him, woo hoo!).

There was only one cube-related puzzle in the competition, however.

Contestants were given a picture of an unsolved Rubik's Cube and asked
to identify the center face colors on the unseen three faces.  Perhaps
not too challenging (although I'm sure many will find problem 3 
difficult to do without pencil and paper).  You can find the puzzles
at http://www.rubiks.com/  or the specific URLs:

http://www.rubiks.com/puzzles.html?p=rubiks_cube&q=1
http://www.rubiks.com/puzzles.html?p=rubiks_cube&q=2
http://www.rubiks.com/puzzles.html?p=rubiks_cube&q=3

If you think these puzzles are too easy, I have an added question
for you -- what is the minimum number of moves needed to get a 
solved cube into the states depicted by the three problems?  You will
find problem 1 easy, problem 2 intermediate, and problem 3 difficult.

--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
"Everyone is trying to CYA their butt!" -- my supervisor

[ Moderator's note: In case some of our subscribers don't know, I will
  supersede Wei-Hwa Huang's modesty to mention that he took first place at
  the World Puzzle Championship (for the fourth time!)]

From cube-lovers-errors@mc.lcs.mit.edu  Tue Oct 26 14:00:55 1999
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Date: Sat, 16 Oct 1999 01:32:31 -0600
To: Cube-Lovers@ai.mit.edu
From: slapdash@enteract.com (Russ Perry Jr)
Subject: Re: Rubik-like puzzles I've never seen before...

First off, thanks for all the replies!

At 8:06 PM 10/14/99, Kevin Young wrote:
>As far as the siamese cubes and the triplet cubes go that you see on ebay.
>They are not professionally made.  Oddzone does not make these cubes.  And
>furthermore, they are not "rare".

That's kind of what I thought.  The OddzOn cubes are represented by
http://www.rubiks.com, right?  Or is it just on Hasbro's site somewhere?

>If you really want a quality cube, Meffert is selling assembly cubes that
>use ABS plastic instead of stickers.  Later on this year, he will have
>available pieces for assembly cubes to make siamese, triplets, and more
>cubes that can be joined together.

Yep, I found Meffert's site.  I hear he's also planning 4x4x4 & 5x5x5
(assembly?) cubes, which is pretty neat.


At 8:27 PM 10/14/99, _pink wrote:
>>1) The Gem/Diamond.

>These are created by utilizing pieces from the Pyramorphix puzzle. One can
>disassemble 2 Pyramorphixes and construct one "Gem" and one "Starburst"
>puzzles as seen on ebay...

So you get 2 "new" puzzles out of 2 Pyramorphix?  Hmm, pure profit...


At 7:30 PM 10/14/99, Noel Dillabough wrote:
>> 2) ??.  This is a 14-sided puzzle, with 8 triangles and 6
>> squares.

>This one I modelled in Puzzler and it is quite nice.  Yes it is a morphing
>puzzle, although very simply in nature.  I have been told by some of my
>users that this puzzle was a prototype and never sold, but I can't confirm
>this.  Supposedly there are only a handful of these in existance.

Someone postulated that it's a dual of the Skewb, and it strikes me that
might be true.  Opinion?

>> 4) "Triple" Cube.  This is similar to the Siamese Cube [...]

>I guess you could keep going forever, however I don't see the advantage of a
>triple cube.  I haven't modelled this one.

Well, it can be left in even more contorted shapes than the Siamese Cube...

>I am constantly on the lookout for more puzzles to add to Puzzler, let me
>know if you think of any,

I'm already trying to get a bunch of doodles ready for Noel to add... :-)


At 10:01 AM 10/15/99, Rob Hegge wrote:
>>2) ??.  This is a 14-sided puzzle, with 8 triangles and 6 squares.

>I bought either this one or the 1) GEM/Diamond or both from Christoph
>Bandelow [...] He can probably help you also with the other puzzles.
>Hope this helps

Sure!  I'll send him a note and ask.  Thanks for the pointer.


At 11:03 AM 10/15/99, Ronald Fletterman wrote:
>Hi Perry, I suggest, you contact Tony Fisher, who takes "standard" sliding
>puzzles and redesigns them in wonderful variants. His address:

Thanks for the address!  Tony was mentioned to me by a couple other sources.
Does anyone know if Tony has an email address or a URL?


At 5:19 AM 10/15/99, Klodshans wrote:
>the 14-sided thing sounds like a Cuboctahedron (3x3x3 mechanism)

Pete Beck had suggested the same thing, but that's definitely not it.  I've
seen those before, and one characteristic is that the planes cute through
the squares & triangles; on this one there are NO cuts through the faces.

>mail me the item number and I will check for sure.

It's not on any active auctions any more, but Noel put it in his Puzzler
program, and it looks right.

>Tony Fisher is also making Siamese Skewbs.

Hmm, are there pix of these anywhere?  Are the just twinned at the corners?

> At www.rubiks.com, there was a thread in the Forum a while ago about how
> to make these things.

I assume you mean Siamese Cubes, but not Siamese Skewbs.  If SS, I couldn't
find the posts...


At 11:37 PM 10/15/99, Wayne Johnson wrote:
>Check out my site, I can show you how the Siamese cube is done:
>http://www.zeta.org.au/~sausage/twistymegasite

Yep, I found that via www.rubiks.com Forum...

I'm not sure I have the talent to mold pieces, but I could mail you some
notes about 2x2x3, 3x3x4 and triangular prism mechanisms...  :-)


At 10:14 PM 10/15/99, Uwe Meffert wrote:
>Dear Russ

Hello Uwe.  I heard you hung out here; glad to make your e-acquaintance.

>These are home or professional hand made for puzzle collectors.  They will
>all be available from our puzzle shop in the near future made from high
>quality ABS with molded plastic tiles instead of labels.

Glad to hear it.  Tell me though, is there any chance we'll see some of
the puzzles from the OLD Meffert catalogs -- like the triangular,
hexagonal, pentagonal and the two circular, prisms?


At 12:57 PM 10/15/99, Tim Browne wrote:
>> 2) ??.  This is a 14-sided puzzle, with 8 triangles and 6 squares.

>This one's news to me. Do you have an eBay item number?

It occurs to me that even though there isn't an auction running currently
with that, you could search for it.  Trouble is, the title was something
lame like "blah blah Rubik blah blah EXTREMELY RARE", which will match
other stuff too...  On the other hand, it might have been auctioned by
the same ssongas (sp?) guy who's auctioning off the Siamese/Triple/Gem
puzles, so a search for him in completed auctions might find it.  I can't
get to the web right now to do it myself, but will try to look later.
Actually, it might have said "14-sided" in the title, which would be
easy to find...

>> Are these homemade, prototypes, original puzzles by pirate manufacturers?

>Well, I seriously doubt it would be "pirate manufacturers", as my company
>officially makes such cubes...

What company is that?

>at least not yet. It's kind of hard to "pirate" something that doesn't
>officially exist. :-) L8r.

Well, I guess I meant like the Siamese cubes still requiring two cubes to
make one Siamese, so the cubes to make it would be pirated, even if the SC
itself wasn't.  Same with the Pyramorphix going to the Gem & Starburst.
But point taken.  And in this case, obviously somebody with a good supply
of OddzOn cubes and ?? Pyramorphix and some time & talent out to make a
good buck or a hundred.


Again, thanks all!

//*================================================================++
||  Russ Perry Jr   2175 S Tonne Dr #105   Arlington Hts IL 60005  ||
||  847-952-9729    slapdash@enteract.com    VIDEOGAME COLLECTOR!  ||
++================================================================*//

From cube-lovers-errors@mc.lcs.mit.edu  Tue Oct 26 17:14:19 1999
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Date: Sat, 16 Oct 1999 02:46:11 -0600
To: Cube-Lovers@ai.mit.edu
From: slapdash@enteract.com (Russ Perry Jr)
Subject: 2x2x4 cube-like puzzle?

Somewhere I ran across mention of a 2x2x4 cube, made from a 4x4x4 with
some pieces glued together.  It gave a link:

>http://hjem.get2net.dk/philip-k/puzzles/puzzlist.htm

There was no pic as the referring doc said, but it says it was made by
Greg Stevens -- is Greg around here and/or are there actually pix out
there somewhere?

On the other hand, another site (http://www.puzzleshop.de ??) has pix
of what appears to actually be a 2x2x4 puzzle in its Puzzle Museum.
How does that work?

//*================================================================++
||  Russ Perry Jr   2175 S Tonne Dr #105   Arlington Hts IL 60005  ||
||  847-952-9729    slapdash@enteract.com    VIDEOGAME COLLECTOR!  ||
++================================================================*//

From cube-lovers-errors@mc.lcs.mit.edu  Tue Oct 26 20:15:00 1999
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From: Norman Richards <orb@cs.utexas.edu>
To: Cube-Lovers@ai.mit.edu
Subject: Re: 3-Cube in 1 One-Look 
In-Reply-To: Your message of "Wed, 13 Oct 1999 09:21:25 PDT."
             <3804B185.A055DA64@u.washington.edu> 
Date: Thu, 14 Oct 1999 16:47:25 -0500


> There was discussion of blindfolded solving during March 98 on this
> list.  At that time no one was able to do it.  You may be the first.

  I'd love some more elaboration of the specific sequences the
original poster used.  (or any suggestions others have to make) I
think this would be quite a trick.  I'd be happy just being able to do
a 2x2x2 cube like this or maybe one face of a 3x3x3...

___________________________________________________________________________
orb@cs.utexas.edu                                           soli deo gloria

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Message-Id: <000e01bf205f$95393700$2e5755ca@pca200series>
From: uwe@ue.net (Uwe Meffert)
To: <slapdash@enteract.com>
Cc: "MPG" <mpg@mefferts.com>, <Cube-Lovers@ai.mit.edu>
Subject: NEW PUZZLES TO BE RELEASED.
Date: Wed, 27 Oct 1999 17:42:21 +0800
Organization: UE Foundation

Hi All Cube Lovers,

Just to clarify some point asked below.  I will be releasing original
Siamese cubes, utilizing our Patented Assembly cube parts plus 3 new
connectors so that multiple cubes can be assembled / disassembled. These
will be available for shipping  around the 10th of December.  Also the 4x4x4
cube and our New Pyraminx Diamond Puzzle made from a Skewb mechanism with
new pieces so that you have 6 pyramid apexes and 8 triangles making an
Octahedron. Furthermore there will be bandaged cube with molded tiles
using the assembly cube mechanism. By using the parts of 2 Pyramorphix's I
will be offering a limited number of Star of David and Diamond Puzzles as a
collectors set, also with our new molded tile finish.  Also some New Disney
& Donald Puzzle Heads using our patented 2x2x2 mechanism will be available
soon.

Happy Puzzling
Uwe

Let's keep puzzling alive!

Uwe Meffert
Meffert's Puzzles & Games
2008, Remex Centre, 42 Wong Chuk Hang Rd,
P.O. Box 24455, Aberdeen, Hong Kong.
Tel. 852-2518-3080, Fax. 852-2518-3282.
Email:- Uwe@Mefferts.com
Web: www.Mefferts.com

From cube-lovers-errors@mc.lcs.mit.edu  Fri Nov  5 19:07:28 1999
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Date: Thu, 28 Oct 1999 10:37:46 -0700 (PDT)
Message-Id: <199910281737.KAA06812@denali.cs.ucla.edu>
From: Richard E Korf <korf@cs.ucla.edu>
To: cube-lovers@ai.mit.edu
Subject: How many Cubes have been sold?

Does anyone have a good idea how many Rubik's Cubes have been sold worldwide?
Send mail to me, and I'll post a summary of the responses. Thanks!
                             -rich


From cube-lovers-errors@mc.lcs.mit.edu  Thu Nov 11 18:02:33 1999
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Date: Thu, 28 Oct 1999 15:38:18 -0400 (EDT)
From: Daniel B Knights <knights@panther.middlebury.edu>
To: Norman Richards <orb@cs.utexas.edu>
Cc: Cube-Lovers@ai.mit.edu
Subject: 3-Cube in 1 Look
Message-Id: <Pine.A41.4.02.9910281447240.273854-100000@tiger.middlebury.edu>


>> There was discussion of blindfolded solving during March 98 on this
>> list. At that time no one was able to do it. You may be the first.

> I'd love some more elaboration of the specific sequences the
>original poster used. (or any suggestions others have to make) I
>think this would be quite a trick. I'd be happy just being able to do
>a 2x2x2 cube like this or maybe one face of a 3x3x3...


Solving one face of a 3x3x3 in one look does not require any special
system for planning.  If it can be done in 20-25 moves, then I've found
it is feasible (although cumbersome) to simply plan out all 20 moves
while keeping track of each move's effect on the relevant pieces.

If you are interested in solving the whole cube in 1-look,
(or a 2x2x2 cube) I suggest you try doing 2 looks first:
1 to solve corners (without changing edges), 1 to solve edges.
Once you can do this, it should be clear how to do it in one look.
If 2-looks for the whole cube is still too difficult at first, then
try solving just the corners with 2-looks, 1 to position them
and 1 to orient them correctly. (But don't mess up
the edges while you do it!)

You don't really have to "update" your memory as you go,
because you can basically plan out the entire solution before
you start, as follows:

1. it's only 3-5 corner permutations to get all corners in the
correct location, and you just have to keep track of each
corner's orientation (i.e., needs to be rotated clockwise or
anti-clockwise in place).
For these permutation moves, I most often use a single-layer
3-corner interchange that preserves the corner orientation
relative to that layer. The move [Ri F Ri B2 R Fi Ri B2 R2]
accomplishes this in the top layer.

2. 1-3 more sequences to orient all the corners.
I use simple moves like [(R U Ri Ui)^2 D (U R Ui Ri)^2 Di]  to
re-orient two corners in the bottom layer.

(**Now you've SOLVED a 2x2x2 Cube!**)

3. Then usually 5-7 sequences to put all edges in place,
but again keeping track of which ones are flipped.
For these permutations, I will again often use a 3-edge swap
like the following: [R2 Ui Fs R2 Bs Ui R2]

4. then a few more edge-flip sequences.  I use the (very)
inefficient two-edge flip maneuver:
[Ls Fi Ls Di Ls B2 Rs Di Rs Fi Rs U2]

Then you're done! (about 200 moves later.)
This is not an easy "trick" - I still find it quite
challenging to correctly plan out the entire corners solution
and the entire edges solution, and to then implement them
correctly from memory.

The real "trick" for me is that I don't memorize the
locations of the pieces, just the sequence of permutations
that I planned out to solve them. This way, if you can plan
out the 5 corner permutations in advance, then you only
need to remember those 5 items to solve the corners, which don't
change throughout the solution. If you instead memorize the
locations of the pieces, you have to keep memorizing new
locations throughout the solution. (which is impossible for me.)

I first plan out the entire solution with my eyes open, and memorize it.
(planning it out correctly may be the most difficult part.)
Then you only have 15 permutations to remember and execute
correctly, without any new memorization after you close your
eyes.

Well, that's enough about that. I just want to make this approach clear
because I think blindfolded cubing is well within the bounds of "normal
human memory capabilities."

Good Luck!

From cube-lovers-errors@mc.lcs.mit.edu  Thu Nov 11 20:10:51 1999
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Message-Id: <v04020a01b43e8319c5c5@[10.10.10.2]>
Date: Thu, 28 Oct 1999 19:08:57 -0400
To: Cube-Lovers@ai.mit.edu
From: Kristin Looney <kristin@wunderland.com>
Subject: Alison painted me a giant rubiks cube!

The big green composter in the yard has
been transformed!  And it looks fabulous!

ye kindred spirits who love cubes might
understand how cool it is to have a giant
two foot rotating rubiks cube decomposing
vegetable matter in my front yard.  wow!

you can view a snapshot of it at the end of
this weeks update at wunderland.com.
                     http://www.wunderland.com


       anyone out there got a bigger one?


I've got another one, that measures only about
a foot and a half:
       http://wunderland.com/Home/Rubik.html


anyway... just thought I'd share the pictures


-Kristin
http://www.wunderland.com/WTS/Kristin

To all the fishies in the deep blue sea, JOY.

From cube-lovers-errors@mc.lcs.mit.edu  Fri Nov 26 16:59:03 1999
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Message-Id: <199910311159.MAA11039@ai2.accesinternet.com>
From: "valley" <gilbert.valley@accesinternet.com>
To: <Cube-Lovers@ai.mit.edu>
Subject: Rubik's cube?
Date: Sun, 31 Oct 1999 12:58:08 +0100

Last week I found a puzzle that looks like something between my old
Rubik's cube and my pyracube.

It has 14 faces: 6 are squares and 8 triangles.

Imagine one side of the pyracube: a square in the middle, and four
little triangles in the corners.

Now when you look at the whole pyracube, the little triangles form 8
pyramids. Then virtually remove these pyramids and you have the shape
of my new puzzle.

In addition it is cut (?) like a Rubik's cube, and consequently has
the same solution.

Have you ever seen this puzzle? was it made by rubik? How is it
called?  Well, I want ot know everything about it since I've been
collecting all rubiks cube-like puzzles for a few years.

Thanks in advance,
Paul

From cube-lovers-errors@mc.lcs.mit.edu  Fri Nov 26 17:37:28 1999
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Message-Id: <381F6399.811C15C8@pressenter.com>
Date: Tue, 02 Nov 1999 16:20:09 -0600
From: Joe Johnson <jgjohnso@pressenter.com>
To: David Singmaster <david.singmaster@sbu.ac.uk>
Cc: cube-lovers@ai.mit.edu
Subject: Re: 5X5X5 challenge
References: <009E0848.8EC8F5C8.5@ice.sbu.ac.uk>

David Singmaster wrote:

>         On 22 Sep, Dan sais one would discover something interesting
> about the central pieces of the 5^3.  I think this result is
> entirely obvious once one thinks about the mechanics of these cubes.

> DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist

I'm not sure what you and Dan have discovered.

[ Moderator's Note: I should have been less coy.  I thought Joe
  Johnson might not have discovered that the supergroup contains
  positions in which a face center is rotated 180 degrees.  I still
  think this is surprising. It's so easy to rotate the face centers
  by an amount adding up to an integral number of 360-degree
  rotations (see 9 January 1981 in the archives).  Does anyone have a
  short process for rotating two face centers 90 degrees clockwise?
  --Dan ]

What I have found is that the cross cubies and wing cubies are both
involved in parity problems. i.e. if you repair the parity of the
cross cubies then there will be no parity problems with the wing
cubies, if you leave the parity problem with the cross cubies then
there will be a parity problem with the wing cubies.  One fix repairs
both.  The point cubies and edge cubies never have parity problems;
simple 3 cubie exchanges are all that are necessary to place the point
cubies (the only cubies that require orientation are the face, corner,
and edge cubies - the others are automatically oriented correctly when
they are placed correctly.)

From cube-lovers-errors@mc.lcs.mit.edu  Fri Nov 26 18:07:49 1999
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Message-Id: <3823768E.53A6755B@iname.com>
Date: Fri, 05 Nov 1999 16:30:06 -0800
From: David Barr <davidbarr@iname.com>
To: Richard E Korf <korf@cs.ucla.edu>
Cc: cube-lovers@ai.mit.edu
Subject: Re: How many Cubes have been sold?
References: <199910281737.KAA06812@denali.cs.ucla.edu>

Richard E Korf wrote:
> Does anyone have a good idea how many Rubik's Cubes have been sold
> worldwide?  Send mail to me, and I'll post a summary of the
> responses. Thanks!
>                              -rich

http://www.rubiks.com/cubemain.html claims that over 200 million have
been sold.

The Rubik's Games CD-ROM has a history of the cube that includes other
estimates.

David

From cube-lovers-errors@mc.lcs.mit.edu  Fri Nov 26 18:47:04 1999
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From: adams.gallant@pei.sympatico.ca
Message-Id: <1.5.4.32.19991106125203.00687470@pop1.pei.sympatico.ca>
Date: Sat, 06 Nov 1999 07:52:03 -0500
To: Cube-Lovers@ai.mit.edu
Subject: Rubik Books or Video?

Just curious if there exists any video tape footage of Erno Rubik
[interviews, TV specials] out there, or any easily accessible books about
his work or an extensive look into the cube phenomena? I've seen a number of
interesting books on Amazon, but nothing still in print.

Thanks,

Dave Gallant

From cube-lovers-errors@mc.lcs.mit.edu  Fri Nov 26 19:32:13 1999
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Reply-To: <noel@mud.ca>
From: "Noel Dillabough" <noel@mud.ca>
To: "Cube Lovers List \(E-mail\)" <cube-lovers@ai.mit.edu>
Subject: Sliding Puzzles
Date: Mon, 8 Nov 1999 22:37:28 -0500
Message-Id: <000401bf2a64$2ea9f650$020a0a0a@NOEL>

My latest addiction has been sliding puzzles, sequential movement puzzles
that are very Rubik-like in nature...I have found a couple of places online
that have are particularly interesting (see below).  I have solved (with the
aid of a program) 44 of the 49 puzzles, and 4 more are just a matter of time
(the states are large but not too large).

However one puzzle, the "Climb Pro 24", by Minoru Abe (link below) is beyond
my program's reach.  It has far too many combinations for my state program
to solve it.

This is where the cube lovers come in.  How do I design a program to solve
this puzzle?  Any ideas or algorithms would be appreciated, especially on
how to reduce the decision tree and know when you've repeated a state
without keeping all of the states in memory.

Regardless if I ever figure out a solution, I know lots of you will enjoy
the following,

-Noel

Classic Sliding Puzzles:

http://www.pro.or.jp/~fuji/java/puzzle/slide/V1.0/fuji.index-eng.html


Block 10 Puzzles:

http://www.pro.or.jp/~fuji/java/puzzle/slide/V2.0/block10-eng.html

Climb Puzzles:

http://www.johnrausch.com/SlidingBlockPuzzles/cg15-1.htm
http://www.johnrausch.com/SlidingBlockPuzzles/cg15-2.htm
http://www.johnrausch.com/SlidingBlockPuzzles/cg15-3.htm
http://www.johnrausch.com/SlidingBlockPuzzles/cg15-4.htm

This one, Minoru Abe's "Climb Pro 24" is the mother of all sliding puzzles:

http://www.johnrausch.com/SlidingBlockPuzzles/pro24-1.htm

From cube-lovers-errors@mc.lcs.mit.edu  Fri Nov 26 19:57:08 1999
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From: Douglas Zander <dzander@solaria.sol.net>
Message-Id: <199911120937.DAA07616@solaria.sol.net>
Subject: puzzles and stroke victims
To: cube-lovers@ai.mit.edu (cube)
Date: Fri, 12 Nov 99 3:37:48 CST

Hello Puzzle-Fans,

  I am wondering if anyone could comment on the use of twisty puzzles
(like rubik cube, etc) for those people who suffered a stroke and are
paralyzed slightly on one side and need to exercise their hands.
(The reason I'm asking is because my brother-in-law suffered a stroke
and he needs to exercise his left hand.)   I am wondering if these
types of puzzles are good therapy or a bad idea.  One exercise he does
with his wife is to close his eyes and hold an object in his left hand
and try to figure out its shape (cube, sphere, tetrahedral, etc...)
Any comments, suggestions, references?

Thanks you.
--
 Douglas Zander
 dzander@solaria.sol.net
 Shorewood, Wisconsin, USA

From cube-lovers-errors@mc.lcs.mit.edu  Wed Dec  1 14:07:45 1999
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Date: Mon, 29 Nov 1999 14:31:49 -0800 (PST)
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From: Richard E Korf <korf@cs.ucla.edu>
To: cube-lovers@ai.mit.edu
In-Reply-To: <3823768E.53A6755B@iname.com> (message from David Barr on Fri, 05
	Nov 1999 16:30:06 -0800)
Subject: Re: How many Cubes have been sold?
References: <199910281737.KAA06812@denali.cs.ucla.edu> <3823768E.53A6755B@iname.com>

I promised to summarize the responses I got to my question about how many
Rubik's Cubes were sold worldwide.  The most informative (and authoritative)
answer I got was from David Singmaster, so I've just included his response
below. 
                         -rich korf

Date: Mon, 8 Nov 1999 14:27:49 +0000
From: David Singmaster Computing & Maths South Bank Univ
      <david.singmaster@sbu.ac.uk>
To: korf@cs.ucla.edu
Subject: RE: How many Cubes have been sold?
Content-Type: text
Content-Length: 1165

	I'm doing this from memory.  If you want, I can look up what I wrote
about 20 years ago.
	Ideal told me they sold about 20 million cubes in one year - I think
this was in the US.  Polytechnika said they sold about twice (or several) times
as many cubes in Hungary as there were people, making perhaps 10 million.  Both
of these statements were while the cube craze was in midstream.  So I suspect
that somewhere between 50 and 100 million legitimiate cubes were sold.  However
the pirate cubes sold perhaps twice as many.  So I've generally given an
estimate of 200 million!  I think this may be conservative, but the number was
probably somewhere between 100 and 300 million.  This doesn't take account of
the fact that the fad started rather later in the Communist countries and they
may have produced many millions as well, though I don't think this would make a
huge difference to the estimate above.

DAVID SINGMASTER,  Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411;  fax: 0171-815 7499;
email:  zingmast  or  David.Singmaster  @sbu.ac.uk

From cube-lovers-errors@mc.lcs.mit.edu  Wed Dec  1 19:43:18 1999
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From: adams.gallant@pei.sympatico.ca
Message-Id: <1.5.4.32.19991114154546.006830dc@pop1.pei.sympatico.ca>
Date: Sun, 14 Nov 1999 10:45:46 -0500
To: Cube-Lovers@ai.mit.edu
Subject: Square-1 Question

Greetings:

After many months of trying pattern after pattern, I've developed an
algorithm for swapping two adjacent corner pieces on Square-1. I was hoping
that if I posted what I've come up with, that someone might have an easier
method. Who knows? Maybe I've just put a new spin on an old solution to this
move, or re-discovered an existing pattern. Anyhow, I'd love to get some
feedback on it. Here's what I've got:

* All moves are done with the orange side facing you, yellow on the left.
* T   = Top Slice [white in this case]
* R   = Right-Hand Slice [blue]
* B   = Bottom Slice [green]
* L   = Left-Hand Slice [yellow]
* CW  = Clock Wise
* CCW = Counter Clock Wise
* CW and CCW moves are done from the perspective of tilting the cube and
looking at the side in question.
* This series of moves will swap the two bottom corner pieces on the blue
side.

MOVES [From a perfect cube - orange on front]
*****

T [1/8 CW], R [1/2], B [1/4 CCW], T [1/2], R [1/2], B [1/4 CW], T [1/2], R
[1/2], B [1/4 CCW], T [1/2], R [1/2], B [1/4 CCW], T [1/4 CW], L [1/2], T
[1/4 CW], L [1/2], T [1/8 CW], B [1/2]

Please let me know if you have any comments or suggestions. Now that I have
my own, I won't mind borrowing a simpler one!

Dave

From cube-lovers-errors@mc.lcs.mit.edu  Wed Dec  1 20:14:28 1999
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Message-Id: <m11s5dr-0002B5C@fwd07.btx.dtag.de>
Date: Sun, 28 Nov 1999 15:49:31 +0100
To: cube-lovers@ai.mit.edu
Subject: centre slice table
From: ronald.fletterman@t-online.de (Ronald Fletterman)


cubologist proper remember how much time and effort was spent in
generating the table of all maneuvers, concerning one outer slice of
the 3x3x3 cube.(the so-called upper table). John F. Jarvis seems to be
the first to have published the table (in 1983), but his maneuver
lengths challenged the brothers dockhorn to get the max length down to
16 moves (1 move equals either a 90 degree or 180 degree face
turn). upto now, little attention has been paid to the centre slice
table per se.  digging into my files, i found a study by herbert
kociemba, including the centre slice maneuvers. to me, it belongs to
the files of every cubologist; ref. the attachement, which can be
opened by owners of kociemba's cube explorer. rgds

ronald.

[ Moderator's note: The attachment is presented here as text for the
  reading pleasure of cube-lovers.  Hopefully cube explorers can
  turn it back into something useful. --Dan]

D R F' U R' F D' U L' F U' L F' U'  (14f*)<two facing cubies flip>
D2 L B' U L' B D2 U2 F' R U' F R' U2  (14f*)<two diagonally facing
                                               cubies flip>
D B R' D F2 L' B D' U R' B L2 D' F R' U'  (16f*)<four cubies flip>
D' F2 D U' R2 U  (6f*)<cubies 3-cycle 1/8>
R2 F2 R D' U F' D2 U2 R2 B' D' U R'  (13f*)<cubies 3-cycle 2/8>
R' D U' B R2 D2 U2 F D U' R' F2  (12f*)<cubies 3-cycle 3/8>
R2 B2 R' B' F D B2 D' B F' R'  (11f*)<cubies 3-cycle 4/8>
R B F' U' B2 U B' F R' B2  (10f*)<cubies 3-cycle 5/8>
F2 R B' F D' F2 D B F' R'  (10f*)<cubies 3-cycle 6/8>
B2 R' D U' B D2 U2 R2 F D U' R'  (12f*)<cubies 3-cycle 7/8>
F U' L B F2 R' F R U' B' F2 L' F' U  (14f*)<cubies 3-cycle 8/8>
D2 L2 D2 U2 R2 U2  (6f*)<facing cubies double pair swap 1/5>
B2 U2 R' D' U B' R2 D2 L2 F' D' U R' U2  (14f*)<facing cubies double
                                               pair swap 2/5>
R U R' D2 B L2 D2 R2 F U2 R' U' R'  (13f*)<facing cubies double pair
                                               swap 3/5>
R U' L B F' D' B D' B' F L B' U R'  (14f*)<facing cubies double pair
                                               swap 4/5>
F R' L' U L2 F2 L2 F2 L2 F2 U' R L F' (14f*)<facing cubies double pair
                                               swap 5/5>
D B2 F2 D' U L2 R2 U'  (8f*)<diagonally facing cubies double pair
                                               swap 1/4>
F' D' U L' D2 F2 U2 R' D' U F' D2 R2 D2  (14f*)<diagonally facing cubies
                                               double pair swap 2/4>
R2 F' R' F U F' U F' D' F L D U2 R'  (14f*)<diagonally facing cubies
                                               double pair swap 3/4>
F' R' L' U L2 F2 L2 F2 L2 F2 U' R L F (14f*)<diagonally facing cubies
                                               double pair swap 4/4>

From cube-lovers-errors@mc.lcs.mit.edu  Tue Dec  7 16:55:07 1999
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From: HarLikin@aol.com
Message-Id: <0.a5c6bfe6.2579acd1@aol.com>
Date: Fri, 3 Dec 1999 18:31:29 EST
Subject: Megaminx 2 edge swap
To: Cube-Lovers@ai.mit.edu

For a long time i thought and heard that the 2 edge swap was impossible on a
megaminx as it is simply a platonic dodecahedron in rubiks format, and a 2
edge swap on a rubiks cube is impossible. yet i have many times solved my
megaminx down to where there are only 2 edges swapped, then randomized it,
and solved again only to find the problem nonexistant. apparently there is
some way to swap only two edges. If anyone knows this move sequence please
tell me. thank you.

-Harlequin

[Moderator's note: It is impossible to perform an odd permutation of
 the edges of the Megaminx, because 5-cycles are even.  This is not
 the same situation as with Rubik's cube, where it is possible to
 perform an odd permutation of the edges if an odd permutation of the
 corners is also performed.  Did they use the same color for more than
 one face of the Megaminx?  Is there another way to get fooled? --Dan]

From cube-lovers-errors@mc.lcs.mit.edu  Sat Dec 11 15:45:34 1999
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Date: Wed, 8 Dec 1999 10:33:39 +0100 (CET)
From: Christ van Willegen <chwi@email.spcgroup.nl>
To: Cube-Lovers@ai.mit.edu
Subject: Re: Megaminx 2 edge swap
In-Reply-To: <0.a5c6bfe6.2579acd1@aol.com>
Message-Id: <Pine.LNX.4.04.9912080842390.26996-100000@email.spcgroup.nl>

On Fri, 3 Dec 1999 HarLikin@aol.com wrote:

[About the two-edge swap on the MegaMinx]

> [Moderator's note: It is impossible to perform an odd permutation of
>  the edges of the Megaminx, because 5-cycles are even.  This is not
>  the same situation as with Rubik's cube, where it is possible to
>  perform an odd permutation of the edges if an odd permutation of the
>  corners is also performed.  Did they use the same color for more than
>  one face of the Megaminx?  Is there another way to get fooled? --Dan]

The MegaMinx in Puzzler has only 6 colors. When I contacted the author
about it, he told me that the one he got only had those 6 colors. When
I got mine (2 years back), it had 12 colors.

I'm not sure if I've ever run into that problem described above. On second
thought, I have...

Let me describe what I do to solve a MegaMinx. I hope you can follow what
I'm saying, because it's hard to describe this... I'll try to make
some pictures... Ok, I stole one from the alt.ascii-art group:

             _._
         _,-'/\ '-._
    _,.-'1  /2 \   3`-.._
 .'________/____\________'.
 :'-.  4  /      \  6  ,-':
 :   '-, /   5    \ ,-'   :
 :  7   /-.      ,-\   10,:
 `:    /   '-,.-'   \    :
  :.  /  8 ,-''-. 9  \  .:
  `: /  ,-'      '-.  \ :
 & :/,-'     11     '-.\: %
   `'-................-''


First, I solve one layer. This includes the pieces 1, 2, 3 and 5 on the
5 layers adjeacent to the top. Then, I put in the edge pieces 4 and 6.
Next, I put in the corner pieces 7 and 10. All of these can be done
using standard 3x3x3 moves :-)

Putting in 8 and 9 requires a trick. You have to rotate the layers
indicated by & and % so that the pieces 8 and 9 can be reached from the
'bottom' layer. Then, use standard 3x3x3 moves to swing the edge pieces
from the bottom layer to pieces 8 and 9. I always find this process to
take the longest (you need to put in 10 edges. Because of Murphy's law,
the edge pieces you need are probably in the 'same' positions in other
layers, so you need to take them out before you can put them in. Long
work, indeed).

Next, put in the corner piece 11.

Now, the upper half of the MegaMinx is solved.

Next, put in the edge pieces on the second-to-last layer (easy work),
and we're down to the last layer.

First, I put the edge pieces in the correct _position_. Sometimes, I need
two edges to be swapped. The formula I have excahnges three edges, so I
mess around with it until I have them all in the correct position.

Then, I align edges, position corners (using a three-corner exchange
formula), and align corners. The last two steps usually impose no
problems.

Unfortunately, my notes with Megaminx formulas is at home (we found
them back yesterday..), so I can't look up any formulas.

If people are interested in the formulas I use, or in the way I solve
it, let me know and I'll look up my notes.

Christ van Willegen

From cube-lovers-errors@mc.lcs.mit.edu  Mon Dec 13 17:24:01 1999
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Message-Id: <001701bf4163$a712fb30$d80c13ac@ste04w0001.ste04.icl.co.uk>
From: "Roger Broadie" <Roger.Broadie@iclweb.com.icl.co.uk>
To: <Cube-Lovers@ai.mit.edu>
Cc: <HarLikin@aol.com>
Subject: Re: Megaminx 2 edge swap
Date: Wed, 8 Dec 1999 10:04:43 -0000

The current Megaminx uses only six colours.  We had a discussion about
the ambiguities this creates in January 1998, to be found in the
archives in cube-mail-24.

Roger Broadie

From cube-lovers-errors@mc.lcs.mit.edu  Tue Dec 14 14:44:11 1999
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From: jmb184@frontiernet.net (John Bailey)
To: Cube-Lovers@ai.mit.edu
Cc: bcheney@teleport.com
Subject: Re: WTD: Rubik's Cube Program w/MACROS
Date: Sat, 11 Dec 1999 10:31:12 GMT
Message-Id: <385225bd.86868338@mail.frontiernet.net>
References: <3851F715.3DF7@teleport.com> <38521c82.84505154@news.frontiernet.net>
In-Reply-To: <38521c82.84505154@news.frontiernet.net>


>On Fri, 10 Dec 1999 23:02:46 -0800, bcheney <bcheney@teleport.com>
>wrote:
>
>>I'm looking for a Rubik's cube program that will allow you to
>>manipulate the cube and allow you to record MACRO sequences and
>>execute them at any time.
>>
>>I've looked at a lot of the programs on the cube web sites,
>>but none have this capability.
>>
>>If you know of a program that has this capability, could you
>>EMAIL or post a reference to it?

to which I replied:

>Good point!
>At http://www.ggw.org/donorware/3x3cube/ is a javascript program
>implementing a 3x3 cube which has two fixed macros (without bothering
>to describe them)
>and the 4D cube at:
>http://www.ggw.org/donorware/4D_Rubik/ has a very explict macro but
>again its fixed.
>
>What you suggest is there should be an ordinary 3x3x3 cube with the
>ability to record and then use a sequence which the player enters.
>
>Macros would make cube solving and searching for solutions far less
>cumbersome, without detracting from the challenge.
>
>I'll watch this thread with interest.  If no one else nibbles the
>troll, I will consider making one for the both of us.

Can anyone on Cube-lovers give a pointer to a cube manipulator with
macro capability?

John

From cube-lovers-errors@mc.lcs.mit.edu  Tue Dec 14 16:06:52 1999
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Message-Id: <000701bf442c$151f8420$ea023dd4@wschwi>
From: "Christ van Willegen" <christw@iae.nl>
To: <Cube-Lovers@ai.mit.edu>
Subject: MegaMinx notes
Date: Sat, 11 Dec 1999 22:34:41 +0100

I just found back my notes on the MegaMinx. Here's what I have.
It isn't much, because I solve it as a 'regular' cube, with
the exception of a few formulas.

The last layer needs these steps:
- Correctly flip the edge pieces.

I use the formulas F U R U' R' F' to flip the 'Front' and
'Right' upper edge pieces. F R U R' U' F' flips the 'Front'
and 'Right-back' upper edge pieces.

Repeat one of these formulas as needed.

- Correctly align the edge pieces.

I have two formulas for this as well. Formula 1 assumes that
the edge pieces 'Front' and 'Left' are correctly aligned, and
keeps these in place: R U R' U R U2 R' rotates the other three
edge pieces around.

Formula 2 assumes that the edge pieces 'Front' and 'Left-back'
are correctly aligned, and keeps these in place: R U R' U2 R U
R' rotates the other three edge pieces around.

- Correctly align the corner pieces.

I only have on of these. The two corner pieces in the 'R' layer
are kept in the same position: U2 R U2' L' U2 R' U2' L (note the
notation: U2 is not the same as U2'!) rotates the other three
corner pieces around.

- Correctly flip the corner pieces.

The formula I've used for 20 years (is this possible? When I
started cubing, I was quite small. I can't really remember
when it was) is the following: Use R F' R' F  R F' R' F to
rotate the corner piece FRU clockwise. Repeat until this piece
is correctly aligned. Then, rotate the U layer until you find
another corner piece that's not correctly flipped. Repeat the
formula until it's correctly flipped (later, I found out that
F' R F R'  F' R F R' rotates a corner piece in the ccw
direction). This also works nicely on the Megaminx.


What I like in my solution of the last layer of the Megaminx:
- The formulas are 'standard' Cube moves, adapted for another
  geometry.
- Most only use three layers. Only one (conceptually) uses 4
  layers (The one to align corner pieces). Since the three
  layers meeting at every corner of the Megaminx are identical
  in behaviour as the layers of the Cube, the formulas are easy
  to grasp (and were easy to adapt from the Cube moves).

Lots of manuals coming with cubes nowadays use the E layer for
some goals (I have to admit: My formulas for the Cube include
this one for the Edge pieces on the third layer: (R E)4, then
rotate the upper layer until another incorrectly flipped edge
piece is in the UR position). These can clearly _not_ be 'ported'
to the Megaminx.

My time for solving the Megaminx is about 10 minutes. My Cube
time is still about 3 minutes (I know, I know: Layer-by-layer
is _not_ a fast approach...)

Well, that's all I have to say about the Megaminx for now...

Happy Minxing!

Christ van Willegen

From cube-lovers-errors@mc.lcs.mit.edu  Tue Dec 21 16:15:00 1999
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Date: Tue, 14 Dec 1999 21:56:29 -0500 (EST)
From: der Mouse  <mouse@rodents.montreal.qc.ca>
Message-Id: <199912150256.VAA02345@Twig.Rodents.Montreal.QC.CA>
To: Cube-Lovers@ai.mit.edu
Subject: Re: WTD: Rubik's Cube Program w/MACROS

>>> I'm looking for a Rubik's cube program that will allow you to
>>> manipulate the cube and allow you to record MACRO sequences and
>>> execute them at any time.
>> [...]
> Can anyone on Cube-lovers give a pointer to a cube manipulator with
> macro capability?

I wrote one that might qualify, depending on what the original poster
actually wants.

You can do things like this (commentary in [[ ]] markers)

% twist
> R F2
Cube:
              u u f
              u u f
              b d d
        l l r d f f l r r u b b
        l l r d f f l r r u b b
        l l r d f f l r r u b b
              f u u
              d d b
              d d b
Cycles: (ur,br,dr,fl,fr) (uf,df) (ubr,bdr,ufl,dfr,fdl,fur) [30]
Already centered
[[ The unfolded cube view shows the result of R F2; the cycle
decomposition is also printed, along with the smallest power that it
needs to be raised to to give the identity - I can't offhand recall the
group-theory term for this number.  The "Already centered" line is
related to something I'll demonstrate later. ]]
> .set t LAST
`t' defined
[[ Define "t" as a name for this transformation. ]]
> t2
Cube:
              u u f
              u u f
              u u f
        l l r b f d r l l d b b
        l l r b f d r r r f b b
        l l l u f d r r r f b b
              f d b
              d d u
              d d u
Cycles: (ur,dr,fr,br,fl) (ubr,ufl,fdl) (urf,drb,frd) [15]
Already centered
[[ Do two `t' operations. ]]
> t6
Cube:
              u u u
              u u f
              u u u
        l l l f f f r r r b b b
        l l r d f f l r r u b b
        l l l f f f r r r b b b
              d d d
              d d b
              d d d
Cycles: (ur,br,dr,fl,fr) [5]
Already centered
[[ Do six `t' operations.  Produces a five-cycle on edges. ]]
> SLICER
Cube:
              u f u
              u f u
              u f u
        l l l f d f r r r b u b
        l l l f d f r r r b u b
        l l l f d f r r r b u b
              d b d
              d b d
              d b d
Cycles: (u,b,d,f) (ub,bd,df,fu) [4]
Centred: (ul,fl,dl,bl) (ur,fr,dr,br) (ulb,flu,dlf,bld) (ubr,fur,dfr,bdr) [4]
[[ SLICER is the R-L slice turned in the direction it would turn if you
did an R turn but turned the center slice with the face.  (SLICEL is
therefore equivalent to SLICER', the inverse of SLICER.  The "Centred"
line shows the cycle decomposition of the move resulting from taking
the given move and then concatenating it with a whole-cube move that
returns all face-center cubies to their home cubicles - in this case,
we can see that this is... ]]
> SLICER CUBEL
Cube:
              b u b
              b u b
              b u b
        l l l u f u r r r d b d
        l l l u f u r r r d b d
        l l l u f u r r r d b d
              f d f
              f d f
              f d f
Cycles: (ul,fl,dl,bl) (ur,fr,dr,br) (ulb,flu,dlf,bld) (ubr,fur,dfr,bdr) [4]
Already centered
[[ As promised.  Note the cycle decomposition matches the "Centred:"
line from the previous example. ]]
> SLICER U
Cube:
              u u u
              f f f
              u u u
        f d f r r r b u b l l l
        l l l f d f r r r b u b
        l l l f d f r r r b u b
              d b d
              d b d
              d b d
Cycles: (u,b,d,f) (ub,bd,df,lu)+ (ur,uf)+ (ulb,ubr,urf,ufl) [8]
Centred: (ul,fu,fr,dr,br,ur,fd,fl,dl,bl) (ulb,fur,lfd,ldb)+ (ubr,frd,drb)+ (ufl)+ [180]
[[ The move of which the Spratt wrench is the fourth power. ]]
> LAST 4
Cube:
              u b u
              l u u
              u u u
        l u l f f f r r r b u b
        l l l f f f r r r b b b
        l l l f d f r r r b d b
              d f d
              d d d
              d b d
Cycles: (ub)+ (ul)+ (fd)+ (bd)+ [2]
Already centered
[[ The wrench itself. ]]
> .set wrench (SLICER U) 4
`wrench' defined
[[ This could also have been ".set wrench LAST", since the last thing
we did was the fourth power of SLICER U.  Parentheses group, so that
the 4 takes the fourth power of the concatenation within them, as
opposed to "SLICER U 4", which is SLICER concatenated with U 4. ]]
> wrench U wrench U'
Cube:
              u b u
              u u u
              u f u
        l l l f u f r r r b u b
        l l l f f f r r r b b b
        l l l f f f r r r b b b
              d d d
              d d d
              d d d
Cycles: (ub)+ (uf)+ [2]
Already centered
[[ A classic double-edge-flipper.  Note how we can use "wrench" as if
it were a primitive.  To make the structure pellucid, it really should
be written "wrench U wrench' U'", but wrench is its own inverse.  For
an example where it's not... ]]
> R B2 R' U' B2 U
Cube:
              b d d
              u u u
              u u f
        r l l f f r u b b r b u
        r l l f f f r r r u b b
        d l l f f f r r u l l l
              d d d
              d d d
              b b b
Cycles: (ub,ur,br,bl,db) (ulb,rbd,bru) (urf)+ (ldb)- [15]
Already centered
> .set sct LAST
`sct' defined
> sct F sct' F'
Cube:
              u u u
              u u u
              f u f
        l l u l f r u r r b b b
        l l l f f f r r r b b b
        l l l f f f r r r b b b
              d d d
              d d d
              d d d
Cycles: (ufl)- (urf)+ [3]
Already centered
[[ This happens to be one of my personal favorites for twisting
corners, largely because it's one I developed completely on my own,
deliberately setting out to develop a way of twisting corners, and my
fingers know it well.  Note how this differs from... ]]
> sct F sct F'
Cube:
              r b b
              u u d
              l u f
        b l f u f r u b l u l d
        r l l f f f r r b u b u
        b l l f f f r r b r r d
              d d d
              d d d
              l b u
Cycles: (ub,br,db,ur,bl) (ulb,bru,rbd) (ufl)+ (urf)+ (ldb)+ [15]
Already centered
[[ ...which is a mess. ]]
>

The program has other facilities as well.  You can set a mode whereby
each input line, rather than being applied to a clean cube, is applied
to the last-printed cube; this is what you'd want if you wanted to try
actually solving a cube using it.  There are also ways to save and load
files containing user-defined transformations.  Twenty transformations
are predefined: the six face quarter-turns, the six quarter-turns of
the whole cube, the six SLICEx turns, NOOP (which has no effect), and
LAST (which refers to the operation that carries a clean cube into the
last-printed cube).  Parentheses group, numeric suffixes repeat what
they apply to, ' takes the inverse, and * `centers' - it takes what it
applies to and concatenates it with whole-cube rotations as necessary
to return the face-center cubies to their home cubicles.

The program is 1308 lines long (34106 bytes) and does require gcc to
compile (it uses nonlocal gotos, nested functions a la Pascal, and
block expressions with nested functions to get lambda functions), but
other than that it should be fairly portable - there's stuff that's
compiler-dependent but not much that's OS-dependent, I think.  I'll be
happy to mail a copy to anyone who wants; if anyone cares to put it up
to be generally fetchable, that's fine with me.

					der Mouse

			       mouse@rodents.montreal.qc.ca
		     7D C8 61 52 5D E7 2D 39  4E F1 31 3E E8 B3 27 4B

From cube-lovers-errors@mc.lcs.mit.edu  Wed Dec 22 14:34:46 1999
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Date: Wed, 15 Dec 1999 22:05:48 -0500 (EST)
From: noel <noel@titan.tcn.net>
To: John Bailey <jmb184@frontiernet.net>
Cc: Cube-Lovers@ai.mit.edu, bcheney@teleport.com
Subject: Re: WTD: Rubik's Cube Program w/MACROS
In-Reply-To: <385225bd.86868338@mail.frontiernet.net>
Message-Id: <Pine.BSI.3.95.991215220417.9151A-100000@titan.tcn.net>

There is a modified cube (2x2x2 to 5x5x5 with macro abilities at the
following url:

http://www.mud.ca/cube/cube.html

See the examples for how to enter the macros in.

-Noel

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From: ronald.fletterman@t-online.de (Ronald Fletterman)
To: cube-lovers@ai.mit.edu
Subject: solutions of 4x4x4 and 5x5x5 cubes
Date: Thu, 16 Dec 1999 17:49:15 +0100
Message-Id: <11ye5b-07iT68C@fwd01.sul.t-online.de>

dear cube lovers, i can offer methods to resolve the standard 4x4x4
and 5x5x5 cubes to anyone interested. please make this an email
message to all.

tks & b.rgds. ronald l. fletterman 

[Moderator's note: I apologize for the month-long outage that
 cube-lovers has just experienced.  I can't offer any extenuating
 circumstances, but I'll try to be more diligent in the future. -Dan ]

From cube-lovers-errors@mc.lcs.mit.edu  Fri Jan 21 21:21:35 2000
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From: "VALLEY" <gilbert.valley@accesinternet.com>
To: <cube-lovers@ai.mit.edu>
Subject: rubiks hat or rabbit
Date: Sun, 19 Dec 1999 15:49:09 +0100

Hello,

Two days ago I bought a rubiks rabbit.  It is really amazing, and is
more complicated than it looks. Here is the description found in the
archives:

"Rubik's Hat is in the form of a hat with six rings on it. You can
look through it (and through the rings by implication). By turning
rings you see more or less rabbits. The purpose is to see a rabbit in
every position.

I think the puzzle is based on light polarization, with different
polarizations coming through the segments of the rings."

I'm not entirely satisfied with the explanation. Light polarisation
doesn't have this effect on colours, and polarisation lens are too
much expensive to be used. I think this effect is due to simple light
filters (green, red, yellow or a mix of two of these colours). If the
filter is red (=retain all other wave lenghths), then you will only
see red rabbits.

But there is something I can't explain: in one segment I saw a green
rabbit. then I turn a ring and this rabbit disappears. I concluded
that the segment of the ring retains green light. Then I tried all
other colours to see what light could pass through. When i turn this
same ring on another green segment, then I was able to see the rabbit
through it! This was of course not the same rabbit, and in one case
the green rabbit was there and in the other, not, all that with the
same filter!

Could somebody explain that? Do you have any other explanation of this
puzzle?

I've tried my best to explain that in english, but feel free to ask me
for more details,

Bye,
Paul

From cube-lovers-errors@mc.lcs.mit.edu  Fri Jan 21 21:43:52 2000
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Message-Id: <001201bf4ad7$df987d40$af0333c8@98>
Reply-To: "Pedro Reissig" <preissig@movi.com.ar>
From: "Pedro Reissig" <preissig@movi.com.ar>
To: "Johan Myrberger" <Johan.Myrberger@ebc.ericsson.se>,
        <cube-lovers@ai.mit.edu>
Subject: RE: Magic jack
Date: Sun, 19 Dec 1999 11:36:27 -0300

I am a puzzle designer working in Argentina, and looking for Magic
Jack type products.  Do you know websites for the 2 other similar
products, the Vadasz and IQUBE?

thanks, Pedro Reissig

From cube-lovers-errors@mc.lcs.mit.edu  Fri Jan 21 23:22:37 2000
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Date: Fri, 31 Dec 1999 04:29:08 -0800 (PST)
From: Jaap Scherphuis <jaapsch@yahoo.com>
Reply-To: jaap@org2.com
Subject: Square-1 tables of move sequences
To: Cube-Lovers@ai.mit.edu

Dear Cube-Lovers,

I have just computed many new results for the Square-1 puzzle. I have
written a program that applies a Kociemba-like algorithm to this puzzle,
and used it to find many beautifully short sequences for nearly all
standard moves of the edges and corners.

This post is very long, and will list most of the results. First however
I should quickly describe the puzzle and highlight some of its intricacies,
since these might not be overly familiar even though it has been mentioned
on this list in the past.

--
Short description of the Square-1 puzzle:

This puzzle is a cube consisting of three layers. The top and bottom layers
are cut like a pie in 8 pieces; 4 edge pieces and 4 corner pieces, 30 and 60
degrees wide respectively. The top and bottom layers can rotate. The middle
layer is cut in only two halves along one of the lines of the other layers.
If there are no corner pieces in the way, you can twist half the cube 180
degrees so that pieces from the top and bottom layers mingle.

The puzzle is unique in that the two types of pieces intermingle. The edge
and corner pieces can freely move between the two outer layers. Of course,
the puzzle will not necessarily be a cube shape when the pieces are mixed.
The puzzle has six colours, each face has a single colour similar to the
Rubik's cube. The aim is of course to return a mixed puzzle back to its
original solved position.

The number of positions:
There are three categories of puzzle shapes.
  a. Both layers have 4 edges and 4 corners each.
  b. One layer has 3 corners, 6 edges, the other 5 corners 2 edges.
  c. One layer has 2 corners, 8 edges, the other 6 corners and no edges.

There are 1, 3, 10, 10 and 5 layer shapes with 6, 5, 4, 3 and 2 corners.
This means there are 5*1+10*3+10*10+3*10+1*5 = 170 shape combinations
for the top and bottom layers (all of them can be attained). The middle layer
has two shapes because half of it is assumed to be in a fixed position and
only the other half moves. This means that there seem to be 170*2*8!*8! =
552,738,816,000 positions if we disregard rotations of the layers. Some
layer shapes however have symmetry, and these have been counted too
many times this way.

To take account of the symmetries we can simply count the number of layer
shapes differently. Instead of the numbers 1, 3, 10, 10, 5 we use the
numbers 2, 36, 105, 112, 54, which are the number of shapes if we consider
rotations different (e.g. a square counts as 3 because it has three possible
orientations). By the same method as before we then get 19305*2*8!*8!
or 627,768,369,664,000 positions.

To exclude layer rotations, divide by 12^2 to get a total of 435,891,456,000
distinct positions.

--
Notation

I use a different notation to that found on other places on the web,
because I find this one more descriptive. Hold the puzzle so that the
yellow middle layer piece is on the left hand side with its 'Square-1'
inscription the right way up. Denote a 180 degree turn of the right hand
side of the puzzle by a / sign (a slash). Turns of the top and bottom
layers are denoted by a pair of numbers (n,m). These numbers are the
multiple of 30 degrees clockwise that the top/bottom layers are to turn
respectively. Thus (3,0) means turn only the top layer clockwise 90
degrees, and (0,-1) means turn only the bottom layer 30 degrees
anti-clockwise (i.e. one edge along).

Note that I define the LENGTH OF A SEQUENCE of moves simply as the
number of / moves in it.

By labelling the faces by the letters U, D, L, R, F, B in the standard
way for the Rubik's cube, the pieces of Square-1 can also be denoted
in the usual way; a combination of two letters for an edge piece and
three letters for a corner piece.

--
Subtleties of the puzzle

Generally the puzzle is solved by first bringing its shape back to a cube,
and then placing the pieces correctly. The reason for this is that there
are many moves that keep the top an bottom layers square, for example
(1,0)/(0,-1). Each / does of course change the middle layer shape from a
square to a kite shape, but this can be ignored because /(0,6)/(0,6)/(0,6)
affects only the middle layer. Ignoring the middle layer, a cube can be
formed in at most 7 moves.

A difficulty arises from the fact that these cube moves swaps two pairs
of corners and two pairs of edges, which is an even permutation. The pieces
can however end up in an odd permutation. To solve this, you will have to
leave the cube shape behind. One way is by doing /(3,3)/(1,2)/ which brings
together four corner pieces in each layer. Now (2,-2)/(-2,2) will swap
three pairs of corners, and by reversing the previous moves we can return
to a cube. As you can see this has taken 7 moves, and in fact there is no
shorter way of performing an odd permutation on the cube.

--
The Search Algorithm

The search algorithm in my program is very similar in design to the
Kociemba algorithm for solving the Rubik's cube, as it solves it in two
stages and uses tables to prune the search tree.

During the first stage of the search a position is found in which the
top and bottom layers are square and where the pieces lie in an even
permutation. The second stage will then solve it with moves that keep
the top and bottom layers square.

The first stage uses a single look-up table, that holds the number of
moves needed to bring the puzzle to a cube from the current shape. It is
only when the cube shape is reached that the parity of the permutation is
checked. In the future I may try to build a larger table which combines
the permutation parity with the shape.

The second stage uses in effect two look-up tables, one for the edges and
one for the corners, and the number of moves needed to solve them is given.
In reality the two tables are identical since cube-moves swap corners and
edges in the same way.

In virtually all other aspects the two phase search is performed in the
same manner as the Kociemba algorithm, so I need not explain further.
The only remaining difference is that my program continues searching for
sequences of the same length as any already found. My reason for this is
that some sequences require fewer turns of the top and bottom layers, and
are therefore better despite being of the same length as defined above.

--
Results

I suspect that God's algorithm (the shortest possible way of solving any
position) uses at most about 12 moves. Clearly this cannot be proved with
this program, but nearly all the positions I have tried can be done in 12
or fewer moves.

Most of the results that I have found using this program are on my website.
Jaap's Puzzle page:  http://www.org2.com/jaap/puzzles

The most important ones are below. Sequence E6 is especially amazing, as
it swaps three pairs of corners and nothing else in only 7 moves. Another
highlight is C4, an edge swap in 10 moves.

A. Sequences involving only edges, and where some of them change layer:
1. Swap DF-UF, DR-UR, DB-UB, DL-UL: (0,5)/(1,1)/(-4,2)/(1,1)/(2,3)
2. Swap DF-UF, DB-UB: (0,5)/(1,1)/(-1,6)
3. Swap DF-UB, DB-UF: (0,-1)/(1,1)/(-1,0)
4. Swap DR-UR, DB-UB: (0,2)/(0,3)/(1,1)/(-1,-4)/(0,-2)
5. Swap DR-UB, DB-UR: (0,2)/(0,3)/(1,-5)/(-1,5)/(0,3)/(0,-2)
6. Swap DB-UB, DR-UF: /(-3,0)/(0,5)/(6,1)/(0,3)/(-5,0)/(-1,6)
7. Swap DB-UF, DR-UB: (1,0)/(0,5)/(6,3)/(0,5)/(-5,0)/(-3,6)/(6,0)
8. Swap DR-UF, UR-UB: (1,0)/(-4,5)/(0,-3)/(1,1)/(-1,2)/(4,-5)/(-1,0)
9. Swap DR-UR, UF-UB: (1,3)/(0,3)/(0,3)/(-1,2)/(1,4)/(0,3)/(-1,0)
10. Swap DR-UB, UF-UR: (4,3)/(3,0)/(-4,5)/(1,1)/(-3,0)/(0,-3)/(2,3)
11. Cycle UF->UR->DR: (1,3)/(0,5)/(0,3)/(6,1)/(0,5)/(3,6)/(6,-3)
12. Cycle UF->UB->DR: (0,5)/(0,1)/(6,3)/(5,0)/(-5,0)/(0,3)/(-1,0)/(0,1)
13. Swap UF-DF:
/(3,3)/(5,0)/(2,0)/(-4,4)/(2,0)/(-1,3)/(0,3)/(3,3)/(2,0)/(-2,1)/(5,2)/(4,-5)/(2,6)

B. Sequences involving only edges of both layers where they do not change
layer:
1. Swap UF-UB, UR-UL, DF-DB, DR-DL: (1,0)/(-3,3)/(2,2)/(3,3)/(-2,4)/(5,0)
2. Swap UF-UL, UR-UB, DF-DL, DR-DB:
(0,2)/(-3,0)/(1,1)/(-4,2)/(1,1)/(5,-4)/(0,-2)
3. Swap UF-UB, DF-DB: (1,0)/(-1,5)/(1,-5)/(-1,6)
4. Swap UR-UB, DR-DB: (0,2)/(0,-3)/(1,1)/(-1,2)/(0,-2)
5. Swap UF-UB, DR-DB: (0,2)/(1,0)/(0,3)/(6,1)/(0,5)/(-3,0)/(5,6)/(6,-2)
6. Swap UF-UB, UL-UR, DF-DB:
/(3,3)/(1,2)/(2,-4)/(-2,4)/(2,4)/(3,3)/(3,0)/(3,3)/(3,0)

C. Sequences involving only edges of the top layer:
1. Swap UF-UB, UR-UL: /(3,-3)/(3,-3)/(6,-2)/(3,-3)/(3,-3)/(2,0)
2. Swap UF-UL, UR-UB: /(3,3)/(1,4)/(5,5)/(-3,0)/(3,3)/
3. Cycle UF->UB->UR: (1,0)/(-1,2)/(-5,1)/(0,3)/(-3,0)/(5,2)/(-5,4)/(-4,0)
4. Swap UF-UB:
/(3,3)/(3,2)/(-4,2)/(-2,4)/(-2,0)/(-4,2)/(-5,1)/(3,0)/(3,3)/(0,-3)
5. Swap UF-UR:
/(3,3)/(-3,0)/(0,4)/(-2,4)/(-4,2)/(-1,0)/(3,3)/(0,4)/(-3,0)/(0,3)/(-1,2)/(-2,1)/(-1,0)
6. Cycle UF->UR->UB->UL:
/(3,3)/(1,0)/(2,2)/(0,2)/(4,4)/(2,0)/(2,2)/(-1,0)/(-3,-3)/(0,3)


D. Sequences involving only corners, and where some of them change layer:
1. Swap UFR-DFR, UBR-DBR, UBL-DBL, UFL-DFL: (4,0)/(2,2)/(-3,3)/(-2,-2)/(-1,-3)
2. Swap UFL-DFL, UBR-DBR: (4,0)/(2,2)/(6,-2)
3. Swap UFL-DBR, UBR-DFL: (-2,0)/(2,2)/(0,-2)
4. Swap UFL-DFL, UFR-DFR: (6,5)/(-3,0)/(4,4)/(2,5)/(0,1)
5. Swap UFL-DFR, UFR-DFL: (1,0)/(3,0)/(-4,2)/(-2,4)/(0,3)/(5,6)
6. Swap UFL-DFL, UBR-DFR: /(3,0)/(6,2)/(4,0)/(-3,0)/(6,-2)/(-4,0)
7. Swap UFL-DFR, UBR-DFL: (6,0)/(3,0)/(6,2)/(4,0)/(-3,0)/(6,-2)/(2,0)
8. Swap UFR-UBR, UFL-DFR: (4,3)/(0,3)/(3,0)/(2,5)/(-5,4)/(3,0)/(5,3)
9. Swap UFL-UBR, UFR-DFR: (0,5)/(0,3)/(0,3)/(-2,1)/(2,5)/(0,3)/(0,-2)
10. Swap UFL-UFR, UBR-DFR: (-2,0)/(0,3)/(6,3)/(2,2)/(-2,1)/(-3,0)/(-4,6)
11. Cycle UFL->UFR->DFR: (1,3)/(-4,0)/(6,3)/(0,4)/(-4,0)/(3,0)/(-3,3)
12. Cycle UFL->UBR->DFR: (-5,0)/(3,0)/(5,2)/(-5,4)/(0,3)/(-1,2)/(-2,4)/(-4,6)
13. Swap UFR-DFR:
(-3,0)/(6,3)/(-1,4)/(-2,2)/(-4,4)/(-4,1)/(0,3)/(0,2)/(0,3)/(-2,4)/(-4,2)/(3,0)/(-3,4)

E. Sequences involving only edges of both layers where they do not change
layer:
1. Swap UFR-UBL, UFL-UBR, DFR-DBL, DFL-DBR:
(1,0)/(-1,5)/(3,3)/(1,1)/(-3,3)/(5,0)
2. Swap UFR-UFL, UBL-UBR, DFR-DFL, DBL-DBR:
(0,5)/(0,3)/(4,4)/(-3,3)/(2,2)/(0,-3)/(6,1)
3. Swap UFL-UBR, DFL-DBR: (0,5)/(-2,4)/(2,-4)/(0,1)
4. Swap UFL-UFR, DFL-DFR: (1,0)/(-3,0)/(2,2)/(1,-2)/(-1,0)
5. Swap UFL-UBR, DFL-DFR: (-2,0)/(0,2)/(0,3)/(-4,0)/(4,0)/(6,3)/(-2,0)/(2,6)
6. Swap UFL-UBR, UFR-UBL, DFL-DBR: /(3,3)/(-3,4)/(-2,4)/(-4,2)/(-4,3)/(3,3)/

F. Sequences involving only edges of the top layer:
1. Swap UFL-UBR, UFR-UBL: /(-3,3)/(-3,3)/(0,1)/(-3,3)/(-3,3)/(-1,6)
2. Swap UFL-UFR, UBR-UBL: /(3,3)/(-3,0)/(4,4)/(2,5)/(3,3)/
3. Cycle UFL->UBR->UFR: (-5,0)/(-4,5)/(4,1)/(-3,0)/(0,3)/(-4,2)/(-2,1)/(2,0)
4. Swap UFL-UBR:
/(3,3)/(-5,0)/(4,4)/(2,0)/(4,4)/(-2,5)/(3,3)/(0,5)/(-2,-2)/(5,0)
5. Swap UFL-UFR:
(0,3)/(1,2)/(3,2)/(-4,0)/(0,4)/(-4,3)/(5,4)/(6,3)/(2,0)/(-2,4)/(-4,2)/(6,-2)
6. Cycle UFL->UBL->UBR->UFR:
/(3,3)/(-5,0)/(4,4)/(2,0)/(4,4)/(-2,5)/(3,3)/(0,5)/(-2,-2)/(5,0)

Copyright 2000 Jaap Scherphuis.
Jaap's Puzzle Page:  http://www.org2.com/jaap/puzzles


=====
Jaap Scherphuis
Visit the Psion Organiser II CM, XP & LZ Homepage:
URL: http://www.org2.com      email: jaap@org2.com

From cube-lovers-errors@mc.lcs.mit.edu  Fri Jan 21 23:40:23 2000
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Message-Id: <3871A3E9.D780781A@spcgroup.nl>
Date: Tue, 04 Jan 2000 08:40:25 +0100
From: Christ van Willegen <c.v.willegen@spcgroup.nl>
To: Cube Lovers <Cube-Lovers@ai.mit.edu>
Subject: Meffert's web-site

Hi,

I've tried to go to Uwe Meffert's web-site, but have been unable to
open it (http://www.mefferts-puzzles.com). Also, I tried www.ue.net
(another site made by Meffert). Does anyone know why the sites are
down?

[Moderator's note: CvW followed up with a note that the site is now
 named www.mefferts.com. --Dan]

On a side note, this weekend I 'constructed' a 12-spot on a Megaminx.
It's not as simple as it looks. After that, I constructed the
pattern that looks like U2 D2 R2 L2 F2 B2 on the Cube (is this
called 'Pons Asinorum'?). Only, the colors are not always the ones
of the oposite sides of the MegaMinx.

Is it possible to make the 12-spot and the other pattern in an easy
way when I start with a 'clean' MegaMinx? I've seen formulas for
the 10-spot already.


Regards,

Christ van Willegen

From cube-lovers-errors@mc.lcs.mit.edu  Sat Jan 22 13:55:41 2000
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From: Andrew John Walker <ajw01@uow.edu.au>
Message-Id: <200001050242.NAA17628@wumpus.its.uow.edu.au>
Subject: Cubes and genetic/evolution based solving
To: cube-lovers@ai.mit.edu
Date: Wed, 5 Jan 100 13:42:44 +1100 (EST)


Would please comment on http://www.iteration-gmbh.de/Erubik.html
[ or on the German version http://www.iteration-gmbh.de/Rubik.htm ].
They have an animation which appears to be solving a 6x6x6 cube.

This is possibly related to the English page at
http://www2.informatik.uni-erlangen.de/~jacob/Evolvica/ES-GA/
ga_magic_problem.html

which shows screen shots of a solver program. I looked at ftp
site and the program doesn't appear to be available, although the
magic square program and a few others are in
ftp://ftp-bionik.fb10.tu-berlin.de/pub/ESdemo/

What then are the prospects of using these methods for cube solving?
While it may be a bit much for them to produce optimal solutions
(or maybe not!), for 4x4x4 and larger the suboptimal solutions may
still be very useful.

Andrew Walker

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Date: Mon, 10 Jan 2000 08:18:52 -0800 (PST)
From: Jaap Scherphuis <jaapsch@yahoo.com>
Reply-To: jaap@org2.com
Subject: Re: Square-1 tables of move sequences
To: Cube-Lovers@ai.mit.edu


Dear Cube-Lovers,

Here are my latest and probably final Square-One results.

I have finalised my square-1 solving program, and it can now be downloaded from
my website. I managed to speed it up so much that it can do an exhaustive depth
9 search in a few hours on a fast pc. It has thus confirmed that all the
sequences below of length <=10 are minimal (though there might be other
sequences that are easier to perform).

The only sequences that are more than 10 long are A13 and C5, and these are 11
and 12 moves respectively. I have no plans as yet to do longer searches on
these two.

Here are the move tables. I will quote them in full, even though many are the
same as in my previous post.

A. Sequences involving only edges, and where some of them change layer:
1. Swap DF-UF, DR-UR, DB-UB, DL-UL: (0,5)/(1,1)/(-4,2)/(1,1)/(2,3)
2. Swap DF-UF, DB-UB: (0,5)/(1,1)/(-1,6)
3. Swap DF-UB, DB-UF: (0,-1)/(1,1)/(-1,0)
4. Swap DR-UR, DB-UB: (0,2)/(0,3)/(1,1)/(-1,-4)/(0,-2)
5. Swap DR-UB, DB-UR: (0,2)/(0,3)/(1,-5)/(-1,5)/(0,3)/(0,-2)
6. Swap DB-UB, DR-UF: /(-3,0)/(0,5)/(6,1)/(0,3)/(-5,0)/(-1,6)
7. Swap DB-UF, DR-UB: (1,0)/(0,5)/(6,3)/(0,5)/(-5,0)/(-3,6)/(6,0)
8. Swap DR-UF, UR-UB: (1,0)/(-4,5)/(0,-3)/(1,1)/(-1,2)/(4,-5)/(-1,0)
9. Swap DR-UR, UF-UB: (1,3)/(0,3)/(0,3)/(-1,2)/(1,4)/(0,3)/(-1,0)
10. Swap DR-UB, UF-UR: (4,3)/(3,0)/(-4,5)/(1,1)/(-3,0)/(0,-3)/(2,3)
11. Cycle UF->UR->DR: (1,3)/(0,5)/(0,3)/(6,1)/(0,5)/(3,6)/(6,-3)
12. Cycle UF->UB->DR: (0,5)/(0,1)/(6,3)/(5,0)/(-5,0)/(0,3)/(-1,0)/(0,1)
13. Swap UF-DF:
(4,5)/(3,0)/(0,1)/(10,2)/(0,4)/(0,4)/(10,2)/(0,1)/(3,0)/(3,5)/(7,4)/(11,0)


B. Sequences involving only edges of both layers where they do not change
layer:
1. Swap UF-UB, UR-UL, DF-DB, DR-DL: (1,0)/(-3,3)/(2,2)/(3,3)/(-2,4)/(5,0)
2. Swap UF-UL, UR-UB, DF-DL, DR-DB:
(0,2)/(-3,0)/(1,1)/(-4,2)/(1,1)/(5,-4)/(0,-2)
3. Swap UF-UB, DF-DB: (1,0)/(-1,5)/(1,-5)/(-1,6)
4. Swap UR-UB, DR-DB: (0,2)/(0,-3)/(1,1)/(-1,2)/(0,-2)
5. Swap UF-UB, DR-DB: (0,2)/(1,0)/(0,3)/(6,1)/(0,5)/(-3,0)/(5,6)/(6,-2)
6. Swap UF-UB, UL-UR, DF-DB:
/(3,3)/(1,2)/(2,-4)/(-2,4)/(2,4)/(3,3)/(3,0)/(3,3)/(3,0)


C. Sequences involving only edges of the top layer:
1. Swap UF-UB, UR-UL: /(3,-3)/(3,-3)/(6,-2)/(3,-3)/(3,-3)/(2,0)
2. Swap UF-UL, UR-UB: /(3,3)/(1,4)/(5,5)/(-3,0)/(3,3)/
3. Cycle UF->UB->UR: (1,0)/(-1,2)/(-5,1)/(0,3)/(-3,0)/(5,2)/(-5,4)/(-4,0)
4. Swap UF-UB:
/(3,3)/(3,2)/(-4,2)/(-2,4)/(-2,0)/(-4,2)/(-5,1)/(3,0)/(3,3)/(0,-3)
5. Swap UF-UR:
(3,0)/(3,0)/(1,2)/(10,3)/(10,3)/(10,5)/(4,0)/(0,1)/(3,5)/(9,0)/(2,3)/(10,4)/(5,3)
6. Cycle UF->UR->UB->UL:
/(3,3)/(1,0)/(2,2)/(0,2)/(4,4)/(2,0)/(2,2)/(-1,0)/(-3,-3)/(0,3)


D. Sequences involving only corners, and where some of them change layer:
1. Swap UFR-DFR, UBR-DBR, UBL-DBL, UFL-DFL: (4,0)/(2,2)/(-3,3)/(-2,-2)/(-1,-3)
2. Swap UFL-DFL, UBR-DBR: (4,0)/(2,2)/(6,-2)
3. Swap UFL-DBR, UBR-DFL: (-2,0)/(2,2)/(0,-2)
4. Swap UFL-DFL, UFR-DFR: (6,5)/(-3,0)/(4,4)/(2,5)/(0,1)
5. Swap UFL-DFR, UFR-DFL: (1,0)/(3,0)/(-4,2)/(-2,4)/(0,3)/(5,6)
6. Swap UFL-DFL, UBR-DFR: /(3,0)/(6,2)/(4,0)/(-3,0)/(6,-2)/(-4,0)
7. Swap UFL-DFR, UBR-DFL: (6,0)/(3,0)/(6,2)/(4,0)/(-3,0)/(6,-2)/(2,0)
8. Swap UFR-UBR, UFL-DFR: (4,3)/(0,3)/(3,0)/(2,5)/(-5,4)/(3,0)/(5,3)
9. Swap UFL-UBR, UFR-DFR: (0,5)/(0,3)/(0,3)/(-2,1)/(2,5)/(0,3)/(0,-2)
10. Swap UFL-UFR, UBR-DFR: (-2,0)/(0,3)/(6,3)/(2,2)/(-2,1)/(-3,0)/(-4,6)
11. Cycle UFL->UFR->DFR: (1,3)/(-4,0)/(6,3)/(0,4)/(-4,0)/(3,0)/(-3,3)
12. Cycle UFL->UBR->DFR: (-5,0)/(3,0)/(5,2)/(-5,4)/(0,3)/(-1,2)/(-2,4)/(-4,6)
13. Swap UFR-DFR: /(3,0)/(1,2)/(-2,0)/(2,1)/(6,3)/(1,0)/(-2,2)/(0,1)/(3,3)/


E. Sequences involving only edges of both layers where they do not change
layer:
1. Swap UFR-UBL, UFL-UBR, DFR-DBL, DFL-DBR:
(1,0)/(-1,5)/(3,3)/(1,1)/(-3,3)/(5,0)
2. Swap UFR-UFL, UBL-UBR, DFR-DFL, DBL-DBR:
(0,5)/(0,3)/(4,4)/(-3,3)/(2,2)/(0,-3)/(6,1)
3. Swap UFL-UBR, DFL-DBR: (0,5)/(-2,4)/(2,-4)/(0,1)
4. Swap UFL-UFR, DFL-DFR: (1,0)/(-3,0)/(2,2)/(1,-2)/(-1,0)
5. Swap UFL-UBR, DFL-DFR: (-2,0)/(0,2)/(0,3)/(-4,0)/(4,0)/(6,3)/(-2,0)/(2,6)
6. Swap UFL-UBR, UFR-UBL, DFL-DBR: /(3,3)/(-3,4)/(-2,4)/(-4,2)/(-4,3)/(3,3)/


F. Sequences involving only edges of the top layer:
1. Swap UFL-UBR, UFR-UBL: /(-3,3)/(-3,3)/(0,1)/(-3,3)/(-3,3)/(-1,6)
2. Swap UFL-UFR, UBR-UBL: /(3,3)/(-3,0)/(4,4)/(2,5)/(3,3)/
3. Cycle UFL->UBR->UFR: (-5,0)/(-4,5)/(4,1)/(-3,0)/(0,3)/(-4,2)/(-2,1)/(2,0)
4. Swap UFL-UBR:
/(3,3)/(-5,0)/(4,4)/(2,0)/(4,4)/(-2,5)/(3,3)/(0,5)/(-2,-2)/(5,0)
5. Swap UFL-UFR: /(3,3)/(1,4)/(0,4)/(10,4)/(8,2)/(2,0)/(10,4)/(4,1)/(3,3)/
6. Cycle UFL->UBL->UBR->UFR:
/(3,3)/(-5,0)/(4,4)/(2,0)/(4,4)/(-2,5)/(3,3)/(0,5)/(-2,-2)/(5,0)

Jaap Scherphuis
Copyright January 2000
--------------------------
Jaap's Puzzle Site:  http://www.org2.com/jaap/puzzles
Psion Organiser II homepage:  http://www.org2.com

From cube-lovers-errors@mc.lcs.mit.edu  Sat Jan 22 15:35:46 2000
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Message-Id: <387C51CD.706ED535@ibm.net>
Date: Wed, 12 Jan 2000 02:05:01 -0800
From: "Jin 'Time Traveler' Kim" <chrono@ibm.net>
To: Cube-Lovers@ai.mit.edu
Subject: Rubik's Revenge by Oddzon
References: <000701bf442c$151f8420$ea023dd4@wschwi>

Oddzon recently decided to manufacture a limited run of Rubik's
Revenges.  My understanding is that they originally made 250 for sale
but the cubes were so popular that they made another 250.  All 500
initial cubes are gone.

Here's the URL to the Rubik's page about the status of the Revenge:  It
will return to production.

http://www.rubiks.com/revenge.html

I recently picked up a couple of these prototype Revenges to compare
them to the original Ideal runs of the early 80's.  As far as I can
tell, Ideal had two different sets of Revenges made.  The first set is
from Macau.  Its mechanism is initially quite stiff and squeaks a lot
when turned.  This can eventually loosen up to a pretty smooth turning
mechanism.  It's made of stiff plastic which leads me to believe that
this would be the one more easily broken due to the potential
brittleness of the pieces.  The other Revenge is manufactured in Korea. 
This particular cube is made from a softer plastic and turns much more
smoothly than the Macau cube when brand new.  The drawback of the Korea
cube IMHO is that the cubelets are hollow.  Although both sets of cubes
seem to be injection molded, the Korean cube has "cheaped out" and left
the cubes hollow, which in my experience causes quite a bit of binding
in teh cube unless the planes of rotation are perfectly aligned. 
Neither puzzle is what I would consider a pleasure to work with.

Enter the Oddzon Cube.  It didn't come in a fancy box at all; just in a
simple plastic baggie and taped shut.  While the turning itself is a bit
rough feeling, the puzzle doesn't seem to bind at all and very little
pressure is required.  The plastic seems to be similar to the Korean
cube's softness but the cubelets are not hollow, which means no binding
and very forgiving about not having perfect alignment.  Unfortunately it
appears that the puzzle currently suffers from Oddzon's really cheap
laminated paper stickers.  According to their web site this puzzle is
only a prototype, so maybe they will improve the mechanism as well as
stick to using purely plastic stickers in their final production cubes. 
This will be THE Revenge I play with, not just because it will be in
production, but because it is by far the best feeling.  

If anybody doubts my claim to the puzzle's smoothness, I will gladly
send them one of my Oddzon Revenges.  I am willing to part with two of
them just so you can get a feel for the mechanism.  I will choose two
people to receive the cubes (chosen at my discretion) for an
evaluation.  If you don't think the puzzle is smooth, just mail it back
to me and I'll cover the postage.  However if you find the cube as good
as I do, be so kind as to mail me the cost of the cube and postage
(about 24 bucks).  

Please don't take this as an advertisement, but rather an endorsement
from someone who has found the ultimately Revenge cube to date.  

-- 
Jin "Time Traveler" Kim
chrono@ibm.net
http://www.chrono.org
'91/'95 PGT - SCPOC

From cube-lovers-errors@mc.lcs.mit.edu  Sat Jan 22 16:07:13 2000
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From: Cycloggl@aol.com
Message-Id: <93.658734.25affe7c@aol.com>
Date: Thu, 13 Jan 2000 23:22:20 EST
Subject: Alexanders star
To: cube-lovers@ai.mit.edu

Hi   I was wondering if any one knows where I can find a solution (either
online or what book has it) to the Alexander star I can get all except for
two and it's driving me crazy
Thanks for reading
John

From cube-lovers-errors@mc.lcs.mit.edu  Sat Jan 22 16:37:11 2000
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Message-Id: <000b01bf6041$c33a93c0$31121fc8@jorgej>
From: "Jorge E. Jaramillo" <kingeorge@crosswinds.net>
To: "cube" <Cube-Lovers@ai.mit.edu>
Subject: I got a "new" old cube
Date: Sun, 16 Jan 2000 11:48:38 -0500

I haven't seen many of this posts in this list since most posts are
technical but I thought it wouldn't hurt to share a Cube experience.

Last time I was at a friend's house (a friend I don't visit very often) I
saw a cube laying somewhere and I started playing with it and telling my
friend how I liked the cube, we made a contest to see who could solve it
faster.

The cube is just like the first cube I had, one of those cubes that turn vey
nicely and that the cubies align when you twist the sides, it has stickers
instead of tiles and they are kind of worn off.  My friend says he bought
this cube with his father some 18 or 19 years ago. To me this is an original
cube since here where I live you can not (and never been able to) find those
cubes manufatured by the companies mentioned in this list.

[Moderator's note: The Wonderful Puzzler brand was mentioned in six
 cube-lovers messages from 1981.  Besides the knockoff Rubik's cube,
 they made a keychain version and an octagonal prism version.]

Yesterday I went again to this friend's house and he told me the cube would
be better in my hands and he gave it to me! It comes in its original
cardboard box that reads: Wonderful Puzzler on the top, on two sides it has
a drawing of a scrambled cube and in the other two sides a solved cube and
it reads Can you contend with more than 3,000,000,000 combination (without
the "S") to reach the solution? On the bottom it reads Made in Taiwan

Well thats all a simple story I just wanted to share

====================================================
Jorge E. Jaramillo
kingeorge@crosswinds.net

From cube-lovers-errors@mc.lcs.mit.edu  Sat Jan 22 17:55:22 2000
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Date: Sat, 22 Jan 2000 13:58:11 -0800 (PST)
From: Tim Browne <ue451@victoria.tc.ca>
To: Cube-Lovers@ai.mit.edu
Subject: Re: Rubik's Revenge by Oddzon
In-Reply-To: <387C51CD.706ED535@ibm.net>
Message-Id: <Pine.GSO.3.95.iB1.0.1000122132406.23581A-100000@vtn1>

On Wed, 12 Jan 2000, Jin 'Time Traveler' Kim wrote:

> Enter the Oddzon Cube. [...]
> This will be THE Revenge I play with, not just because it will be in
> production, but because it is by far the best feeling.  

> If anybody doubts my claim to the puzzle's smoothness, I will gladly
> send them one of my Oddzon Revenges.  I am willing to part with two of
> them just so you can get a feel for the mechanism.  I will choose two
> people to receive the cubes (chosen at my discretion) for an
> evaluation.  If you don't think the puzzle is smooth, just mail it back
> to me and I'll cover the postage.

I don't know... I found it really distubing that they couldn't even be
bothered to change the sticker to read "Rubik's Revenge" instead of
"Rubik's Cube", and I've heard from some people that the OddzOn version
fell apart as soon as they turned it once. I picked up a Meffert's
version. Not only is it $2 less than the OddzOn version and shipped free
worldwide, but it's also apparently designed for speed cubing. Instead of
using stickers of any kind, Meffert decided to go with tiles for all his
puzzles, which definitely aren't coming off any time soon. To add to the
challenge, Meffert grooved all the tiles, so if you'd like you can also
treat it as a junior version of a picture cube (junior because there's
still pairs of centres which are completely indistibguishable). L8r.


        Cubic Puzzles - The SIMPLEST Solutions
     http://www.victoria.tc.ca/~ue451/solves.html

From cube-lovers-errors@mc.lcs.mit.edu  Mon Jan 24 15:27:16 2000
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Date: Sat, 22 Jan 2000 14:00:41 -0800 (PST)
From: Tim Browne <ue451@victoria.tc.ca>
To: cube-lovers@ai.mit.edu
Subject: Re: Alexanders star
In-Reply-To: <93.658734.25affe7c@aol.com>
Message-Id: <Pine.GSO.3.95.iB1.0.1000122135831.23581B-100000@vtn1>

On Thu, 13 Jan 2000 Cycloggl@aol.com wrote:

> Hi   I was wondering if any one knows where I can find a solution (either
> online or what book has it) to the Alexander star I can get all except for
> two and it's driving me crazy
> Thanks for reading
> John

The problem is that the Alexander's Star uses only 6 colours instead of
the usual 12, so you get pairs of pieces all over the puzzle that look
identical. What you need to do is find a piece that is identical to one of
the pieces that need swapping, and then do a 3 piece rotation on them.
L8r.

-- 
        Cubic Puzzles - The SIMPLEST Solutions
     http://www.victoria.tc.ca/~ue451/solves.html

From cube-lovers-errors@mc.lcs.mit.edu  Mon Jan 24 16:09:32 2000
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Date: Sat, 22 Jan 2000 17:50:30 -0600 (CST)
From: Douglas Zander <dzander@solaria.sol.net>
Subject: Me too! :-) ( Was: I got a "new" old cube)
To: cube-lovers@ai.mit.edu
Message-Id: <Pine.3.02.10001221730.S22801-a100000@solaria.sol.net>


Since we are sharing stories, I just thought I'd mention how I found a
scrambled cube at the local Salvation Army Thrift Store.  It cost me 18
cents (0.18 USD).  I tried to solve it but I found out the plastic stickers
were removed and rearranged into an impossible position.  The neat thing
is that I believe it is an original Ideal; but I am not sure.  It really
turns smoothly and lines up easily; it is a fantastic cube for speed solving.
Is there a way I can find out what brand it is?  It doesn't say on the white
sticker, I think it wore off.  (this is a 3by cube)

 Douglas Zander
 dzander@solaria.sol.net
 Shorewood, Wisconsin, USA




From cube-lovers-errors@mc.lcs.mit.edu  Mon Jan 24 19:28:46 2000
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Date: Sun, 23 Jan 2000 02:05:50 -0500
From: "Kevin M. Young" <keyoung3@vt.edu>
Subject: RE: Rubik's Revenge by Oddzon
In-Reply-To: <387C51CD.706ED535@ibm.net>
To: Cube-Lovers@ai.mit.edu
Message-Id: <PLEJLGAOLFCFHLCEEHHBAENGCAAA.keyoung3@vt.edu>

I purchase the Oddzon Revenge during it's first initial run.  I have played
with it alot, and it's mechanism is quite nice.  It is as good as my
original Revenge.  So, I can support Jin's claim.

As far as the stickers go, we all know that the plastic laminated stickers
that Oddzon uses do not last if you plan on playing with the puzzle.  This
is still true with the "new" Revenges.  Even if you do not plan on speed
cubing with it, the oils in ones hands gets under the lamination and it ends
up folding up, making it annoying to play with.

I have found a temporary fix for this.  I found at the hardware store
(almost any commercial hardware store should do) that they sell electrical
tape in several colors.  Since electrical tape is vinyl, it will stand up to
abuse.  I bought all six (identically the same as the original) colors in
1/4 inch width size.  This is definatly a good fix, since you don't have to
"ruin" the cube (i.e. painting) the plastic cubies.

I say that this is a temporary fix, because from what I understand, starting
in Febuary, Uwe Meffert is going to offer plastic tiles for sale, so cubist
can fix up are old worn out stickers on their puzzles.  He already offers
plastic tiles for 3X3X3 cubes for one dollar per color (each color includes
9 tiles for all cubies on one side).  This is an incredible option, because
not only will we be able to restore our puzzles back to excelent condition,
we will be able to create more custom color arrangements.

Kevin

From cube-lovers-errors@mc.lcs.mit.edu  Thu Jan 27 18:52:18 2000
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Message-Id: <388AB533.798B2C6@health.on.net>
Date: Sun, 23 Jan 2000 18:30:51 +1030
From: Ghan <ghan@health.on.net>
Reply-To: ghan@health.on.net
To: Cube-Lovers <Cube-Lovers@ai.mit.edu>
Subject: Stuff...

Hi everyone!

This is my first post to Cube-Lovers! I don't really know whether most
people on this mailing list are regulars or just a bunch of strangers
who don't know each other, but I'll introduce myself anyway. I am:
Justin Ghan, recently turned 18 year old, about to start his second
year of Engineering/Science at Adelaide University, South Australia. I
wasn't really interested in all this cube stuff until the middle of
last year ... Well, when I was really young, my cousin had a pyraminx
and I solved that. But yeah, last July, I got one of my friends to
teach me how to solve the Rubik's Cube, which I have ever since
regretted, since I really would have liked to solve it myself...

Since then, I've bought a Megaminx from Meffert's and also won a
Professor Cube (thanks to the Puzzle Challenge!). I solved both of
them mostly using altered Rubik's Cube sequences, which sort of
dampened my satisfaction because I knew I didn't REALLY fully solve
them myself. I borrowed a friend's Creative Puzzle Ball (I'm not sure
what it's officially called) and managed to be able to solve it in a
trial-and-error sort of way.  I also bought a Bandaged Cube recently,
which arrived only yesterday and is annoying the hell out of me - I've
only solved it once by accident so far. And I wasn't paying attention
so I don't know how I did it. I've also got a Square-One and a Skewb
Diamond coming in the mail. I can't wait!

I was a bit intimidated when I read through the past couple of years
worth of Cube-Lovers mail because I always thought I was pretty smart
at maths (I went to the International Mathematical Olympiad twice) and
was also pretty proud of my puzzling skills but I couldn't make head
nor tail of most of the mathematical discussion! I've been trying to
learn some group theory since, so I hope one day I'll be able to join
in.

Anyway, I have quite a few unrelated questions I need help with...

First of all, the glue on my Megaminx is failing, meaning that every
time I turn a face I have to pause to readjust all the stickers which
have slid out of place. This is obviously very annoying. Any ideas on
what to do? What type of glue could I use to secure all the stickers
more permanently? By the way, I really like the grooved tiles on the
Bandaged Cube (and all the new Meffert's Puzzles, I think), although
the colours aren't as bright.

Now for the Professor Cube ... I've seen various solutions on the
internet but so far I haven't found anybody else who uses the boring
layer by layer approach. Does anybody else do it my way? It's probably
a really slow, inefficient method but it seemed the most obvious to me
when I started, since that's how I solved the Rubik's Cube and the
Megaminx. The reason I ask is that I would really like a fix for the
parity problem when I get to the last layer. When I end up with two
edge pieces swapped, I have to stuff up the last two layers
completely, which takes ages to redo.

I've been considering buying an Assembly Cube since, believe it or
not, I don't actually have a Rubik's Cube (although I can use the
Professor Cube as one). But I don't really know much about them.
What's going on with the 3-colour Assembly Cube at Meffert's? Also, if
I buy a Assembly Siamese Cube, can I use it to assemble a normal
Rubik's Cube? And what's the Alexander's Star? And I thought I saw a
"Star of David" (or something) mentioned on the Meffert's website...

Finally, I decided today to try my hand at some programming. Now, I
haven't had any programming experience except for extremely basic
BASIC, years and years ago on my Commodore 64. But I wanted to see if
I could write a program to solve the Bandaged Cube, so I tried to
learn some QBASIC. I thought it would be easier to solve a 2x2x2 cube
first, so I finally managed to write a working program to do that. It
was just a brute force search, first testing all 1 move algorithms,
then 2 move algorithms, etc.

I thought I was pretty good until I discovered that my program took 18
minutes to check all the 9 move algorithms! (I have a Pentium II 266,
although not being very computer smart I'm not sure how much
difference this makes.) I found out that more than half the
configurations required 9 moves or more. Well, this was pretty
embarrassing, considering that there are computer programs which can
solve the 3x3x3 in seconds (or so I've read, I haven't tried any of
these programs). So I'm wondering how to write a fast solver. Is my
program slow because of my bad programming skills, or is my method
just plain slow? I can't think of any other methods to find the
solution.

I hope some of you can help with my problems. I also realise that some
of my questions may have been answered in past discussions, but I
wasn't willing to read through 20 years of archives to find out! If
you can direct me to a previous relevant discussion, that would also
be appreciated.

Thanks!

Justin

From cube-lovers-errors@mc.lcs.mit.edu  Thu Jan 27 19:18:31 2000
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Date: Sun, 23 Jan 2000 16:44:37 -0500
From: Howard <shdrake@worldnet.att.net>
Reply-To: shdrake@worldnet.att.net
Organization: The Drake Family
To: "Cube-Lovers@ai.mit.edu" <Cube-Lovers@ai.mit.edu>
Cc: "Jing Meffert - Meffert's Puzzles" <jing@ue.net>
Subject: Ideas for new puzzles

Dear Cube-Lovers,

  As I twist and turn, I often think, wouldn't it be great
if this puzzle did this.  After discovering Jaap Scherphuis'
page on puzzle info, I thought I should put my ideas forward.
If these have already appeared, then please don't take offense,
I am not stealing, just reinventing the wheel.  If anyone knows
if these puzzles already exist, I would very much appreciate
hearing about them.  If these puzzles don't exist, then I
offer my ideas as a challenge to any puzzle manufacturers, and
puzzle designers, who may read this news letter.

  As I twisted my skewb, to prove to myself that it is very
similar to the Magic Pyramid, I thought, I would like to swap
corners from one group (of 4) to the other group.  Super-
imposing the cuts of a 2x2x2 cube on the skewb would do this.
Each face would consist of 8 triangles.  Allowed twists would
be the 120 degree twist of the skewb and the 90 degree twist
of the 2x cube, only.  Stopping at 45 degrees or 60 degrees would
not allow a twist in a new direction to begin.  Any number
of 90 degree cube twists could follow any number of 120 degree
of skewb twists, followed by 90 degree twists, etc.

  My second idea/wish is a Dogic, but using 20 colors, rather than
12 colors.  Each face of 3 triangles (plus insert), would be
a unique color.  The main challenge I see in making this, is
choosing 20 easily distinguishable colors.  I like the 12 Dogic
color choices very much, because, to my eyes, they are easily
distinguishable, (in contrast to one of my skewbs that has 2
shades of orange for adjoining sides).

  2 ideas that I am currently building are an 8 color cube
and 12 color cube using Mefferts assembly cubes and tiles.
The 8 color is the corner centered design, and the 12 color
is an edge centered design.  These appeared in an early
Scientific American article.

  Does anyone in Cube-Lovers have a Web Page to collect ideas
of the group for future puzzle designs?

Howard Drake

From cube-lovers-errors@mc.lcs.mit.edu  Thu Jan 27 19:43:12 2000
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From: "Chris Pelley" <ck1@home.com>
To: <Cube-Lovers@ai.mit.edu>
Subject: RE: Rubik's Revenge by Oddzon and Meffert's New Puzzles
Date: Tue, 25 Jan 2000 00:08:38 -0500
Message-Id: <NDBBLHBJOLNCAMMDPAPAKECCCDAA.ck1@home.com>
In-Reply-To: <PLEJLGAOLFCFHLCEEHHBAENGCAAA.keyoung3@vt.edu>

I'll add my two cents to this discussion, since I recently acquired some of
the new 4x4x4 units as well.  In short, both the Meffert and Oddzon reissues
are excellent.  The Oddzon initially turned the smoothest, but now my
Meffert's version is broken in and turns equally well.  Both cubes offer a
little give when the faces are not perfectly aligned.  They seem very
forgiving and you don't get the sense that you're going to break something
by playing with them.  In contrast to the old 1982 models (I have 3 or 4 of
the squeaky variety that broke easily), I'd say they're both great
improvements.

Regarding the stickers used, the Oddzon uses the blue opposite green color
scheme (yellow opposite white, orange opposite red).  The stickers on the
unit I received have not worn off like the Oddzon 3x3x3 stickers.  This
could be due to the fact that they are smaller, and less likely to be
scraped by a fingernail?  At any rate, the Meffert's 4x4x4 is clearly
superior since it uses the same six colors, but with the non-slip grooved
tiles.  His color combination places red opposite white and yellow opposite
orange, which works well.

I also got the other new Meffert cubes that feature the non-slip tiles,
including a Megaminx that has 12 unique colors (even the original had two
yellow faces on the top and bottom).  This is the best Megaminx ever!  It's
also nice having the Pyraminx, Skewb, and 5x5x5 in deluxe tile versions.
The cubic puzzles feature the same colors as his new 4x4x4 and the Pyraminx
receives red, blue, yellow, and green.

Finally, I'll mention a new puzzle called QUBLE by Geospace.  This is really
just a 3x3x3 with letters printed on the stickers.  Scrambling the letters
allows word games to be played with the cube, and there are six colors so
you can also solve it the traditional way.  Here is the URL:

http://www.kbkids.com/toys/product.html?WebID=0098549f5b33001b519f5b330041ea
9f5b330018619f5b33

If that doesn't work, just search for "rubik" at www.kbkids.com

Christopher Pelley
ck1@home.com
www.chrisandkori.com/cubes.htm

From cube-lovers-errors@mc.lcs.mit.edu  Tue Feb 15 15:14:59 2000
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Message-Id: <200001252022.PAA12063@garnet.sover.net>
Date: Tue, 25 Jan 2000 15:24:11 -0500
To: Cube-Lovers@ai.mit.edu
From: Nichael Cramer <nichael@sover.net>
Subject: A Simple Rubik's Race Page.
In-Reply-To: <PLEJLGAOLFCFHLCEEHHBAENGCAAA.keyoung3@vt.edu>
References: <387C51CD.706ED535@ibm.net>

The recent flurry of messages surround the revitalization of this list
finally prompted me to finish up (well, mostly...) something that's been
sitting around on my back burner for some time.

A few years back, I happened to pick up (for 10cents) a copy of the Rubik's
Race game at a local rummage sale.  Having never seen the game before, I
posted a description of the game to this list.

That posting --living on in the archives of this list-- seems to be just
about the only reference to the game on the 'net.  Consequently, every six
months or so I get a message from someone wanting more information about
the game or --more likely-- want to know where they can get a copy.

Anyway, I thought it might be useful to throw together a small "Rubik's
Race" webpage.  I still need to get to a scanner, so none of the promised
pictures are there (yet).  But the page contains a description of the game
and a Java Applet to allow the viewer to play a "solitaire" version of the
game.

   http://www.sover.net/~nichael/puzzles/rubrac/index.html

Enjoy
Nichael

--
Nichael Cramer
nichael@sover.net                    My Child is an Honour
http://www.sover.net/~nichael/        Student at Hogwarts

From cube-lovers-errors@mc.lcs.mit.edu  Tue Feb 15 15:48:00 2000
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From: "Klodshans" <klodshans@get2net.dk>
To: <cube-lovers@ai.mit.edu>
Subject: Sv: Magic jack
Date: Wed, 26 Jan 2000 11:07:36 +0100

Pedro Reissig wrote:

>I am a puzzle designer working in Argentina, and looking for Magic
>Jack type products.  Do you know websites for the 2 other similar
>products, the Vadasz and IQUBE?


The Vadasz Cube has a page at http://members.aol.com/islandcom/
The page says there are 5 different versions. I wrote them an email to
order some of these but there was no reply. If anyone knows where one
can order the full range of Vadasz Cubes please tell the list.

Philip Knudsen, Denmark

From cube-lovers-errors@mc.lcs.mit.edu  Tue Feb 15 16:23:34 2000
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Date: Thu, 27 Jan 2000 17:28:02 -0800 (PST)
From: Tim Browne <ue451@victoria.tc.ca>
To: Howard <shdrake@worldnet.att.net>
Cc: "Cube-Lovers@ai.mit.edu" <Cube-Lovers@ai.mit.edu>,
        "Jing Meffert - Meffert's Puzzles" <jing@ue.net>
Subject: Re: Ideas for new puzzles
In-Reply-To: <388B7645.F3DA4D27@worldnet.att.net>
Message-Id: <Pine.GSO.3.95.iB1.0.1000127172224.6201B-100000@vtn1>

On Sun, 23 Jan 2000, Howard wrote:

>   As I twisted my skewb, to prove to myself that it is very
> similar to the Magic Pyramid, I thought, I would like to swap
> corners from one group (of 4) to the other group.  Super-
> imposing the cuts of a 2x2x2 cube on the skewb would do this.
> Each face would consist of 8 triangles.  Allowed twists would
> be the 120 degree twist of the skewb and the 90 degree twist
> of the 2x cube, only.  Stopping at 45 degrees or 60 degrees would
> not allow a twist in a new direction to begin.  Any number
> of 90 degree cube twists could follow any number of 120 degree
> of skewb twists, followed by 90 degree twists, etc.

That would be quite a challenge to create. Truth be told, I'm not even
sure if such a creation would be possible. However it is possible to make
a Skewb with 8 trianges on a side, and has already been done. It's called
a Star Skewb, and is made by switching the centres of a standard Skewb
with those of a Diamond Skewb, which Meffert is currently selling.

>   My second idea/wish is a Dogic, but using 20 colors, rather than
> 12 colors.  Each face of 3 triangles (plus insert), would be
> a unique color.  The main challenge I see in making this, is
> choosing 20 easily distinguishable colors.  I like the 12 Dogic
> color choices very much, because, to my eyes, they are easily
> distinguishable, (in contrast to one of my skewbs that has 2
> shades of orange for adjoining sides).

This has also already been done. Hendrik Haak (www.puzzle-shop.de) is
currently selling the Dogic 2. If have yet to buy one myself, but it looks
like the colours should be easily distinguishable.

>   2 ideas that I am currently building are an 8 color cube
> and 12 color cube using Mefferts assembly cubes and tiles.
> The 8 color is the corner centered design, and the 12 color
> is an edge centered design.  These appeared in an early
> Scientific American article.

I don't know about the 8 colour cube, but I know for certain that the 12
colour version has already been made. I believ it's called "The Ultimate
Cube", and one was sold on eBay during the summer. L8r.

-- 
        Cubic Puzzles - The SIMPLEST Solutions
     http://www.victoria.tc.ca/~ue451/solves.html

From cube-lovers-errors@mc.lcs.mit.edu  Tue Feb 15 17:39:09 2000
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From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
To: Cube-Lovers <Cube-Lovers@ai.mit.edu>
Subject: Re : Stuff...
In-Reply-To: <388AB533.798B2C6@health.on.net>
Message-Id: <SIMEON.10001272347.B@pavilion.pstcc.cc.tn.us>
Date: Thu, 27 Jan 2000 23:47:47 -0500 (Eastern Standard Time)

On Sun, 23 Jan 2000 18:30:51 +1030 Ghan <ghan@health.on.net> wrote:
 
> I thought I was pretty good until I discovered that my program took
> 18 minutes to check all the 9 move algorithms! (I have a Pentium II
> 266, although not being very computer smart I'm not sure how much
> difference this makes.) I found out that more than half the
> configurations required 9 moves or more. Well, this was pretty
> embarrassing, considering that there are computer programs which can
> solve the 3x3x3 in seconds (or so I've read, I haven't tried any of
> these programs). So I'm wondering how to write a fast solver. Is my
> program slow because of my bad programming skills, or is my method
> just plain slow? I can't think of any other methods to find the
> solution.

First of all, you need to be careful what you define as a move.
Cubists seem to fall approximately equally into one of two camps  --
those who count a 90 degree turn of one face as a move (a quarter
turn) and those who count either a 90 degree turn (a quarter turn) or
a 180 degree turn of one face (a half turn) as one move.

Quarter turns and half turns are collectively called face turns.  You
will sometimes see a solution described as an 18q solution (18
quarter turns) or as a 16f solution (16 face turns) or something like
that. The face turn terminology distinguishes between twists of
external (faces) and internal layers (slices).  A 3x3x3 can be solved
using only face moves.  A 4x4x4 or 5x5x5 etc. require both face and
slice moves.

Second, you need to distinguish between programs which calculate
optimal solutions vs. those which calculate suboptimal solutions.  I
am not aware of any programs which can calculate optimal solutions
for the 3x3x3 in seconds.  The best programs for optimal solutions
can require an hour or two or maybe a day or two to calculate an
optimal solution, depending on the speed of the machine, the memory
size of the machine, and the difficulty of the problem at hand.  This
may seem like a long time, but really it's an amazing achievement.

It is true that there are programs which can calculate *very good*
suboptimal solutions in just a few seconds.  The best example is
Herbert Kociemba's Cube Explorer 1.5.  It is downloadable from
Herbert's Web site (you can find it with any search engine) and from
the Cube-Lovers FTP site.  If you let it run long enough it will
eventually find an optimal solution, but "long enough" may be a great
deal longer than for programs designed specifically to find optimal
solutions.

The two amazing things to me about Cube Explorer 1.5 are that it is
so extremely fast, and that its suboptimal solutions are so darn near
optimal as quickly as they are.  In fact, it sometimes finds an
optimal solution in a matter of minutes, except that it is usually
not able to prove the that the solution is optimal anywhere near as
quickly as it is able to find the solution.  I will include only an
extremely brief description of its algorithm, and I will defer that
description until later in my note.

The best programs for finding optimal solutions use an IDA* algorithm
invented by Richard Korf.  IDA* was not invented specifically for
solving the cube, but it is well adapted to solving the cube.  Cube
Explorer 1.5 uses an IDA* algorithm in part.

I will assume that you understand breadth first and depth first
searches of a search space.  Your program for the 2x2x2 was
essentially doing a breadth first search  --  all one move sequences,
all two move sequences, all three move sequences, etc.

Depth first searches generally require much less memory than breadth
first. Only the positions from your current state to the goal state
have to be stored.  To do a depth first search of up to nine moves,
you never have to store more than nine states (well, really ten, the
original one plus nine more).  However, depth first searches can kind
of run away with you and can run practically forever.

The first piece of IDA* to deal with this bad aspect of depth first
searches is called Iterative Deepening depth first (the ID of IDA*).
You do a depth first search that is first bound at one move, then
bound at two moves, then bound at three moves, etc.  The procedure
can seem sort of breadth first instead of depth search, but the
underlying search really is depth first.  You just bound the search
and gradually increase the bound.

Iterative Deepening depth first is still too slow for the cube.  Here
is where the A* bit comes in.  Suppose you were going to do an
Iterative Deepening search one move deep, then two moves, then three
moves, etc., but suppose you also knew for sure by some magic that
the solution is at least twelve moves.  Then, there is no point in
doing the one move or two move or eleven move search.  You might as
well start by bounding you Iterative Deepening search at twelve, then
going on to thirteen, fourteen, etc.

What is this "magic" by which you might know that the solution is at
least twelve moves?  Korf calls it a pattern data base.  I just call
it a table.  All it amounts to is that you create a table of
positions that are a subset of the entire cube  --  say the corners
only.  For each one of those positions, you calculate how many moves
there are in the minimal solution.  Then you take your position and
look it up in the table.  For example, for the 3x3x3 you ignore the
edges and look up the corners in the table.  If it will require
twelve moves to solve the corners, based on your table, then it will
require at least twelve moves to solve the whole cube.  So you start
your Iterative Deepening search at twelve moves because you know that
anything less is going to fail.

But then you go one step further and look in your table after every
move.  For example, you determine that your initial position is at
least twelve moves from Start, so you commence an Iterative Deepening
search which is bounded at twelve moves.  You make your first move.
If the corners are now eleven moves from Start, you let the search
continue.  But if the corners are now twelve or thirteen moves from
Start, it will be impossible to find a twelve move solution so you
cut off that branch of the search and backtrack (you have already
made one move, so twelve or thirteen more yield a total of thirteen
or fourteen which is greater than your bound of twelve).

The best IDA* programs run on a fast processor with lots of memory to
hold a very large table, and a table which is very cleverly
constructed.

Finishing my note by talking about Cube Explorer 1.5 again, it is a
two phase algorithm which finds one position intermediate between the
current state and the goal state.  It breaks the problem down into
two substeps because it takes so long to solve the problem all in one
go. But because there is an intermediate step, the solution is not
necessarily optimal.  It uses IDA* to get from the initial position
to the intermediate position.  After a suboptimal solution is found,
it looks for better solutions by making the intermediate position
further from the initial step.  If it is ever able to solve a
position all in the first substep, then that solution is proven
optimal.

>
> I hope some of you can help with my problems. I also realise that some
> of my questions may have been answered in past discussions, but I
> wasn't willing to read through 20 years of archives to find out! If
> you can direct me to a previous relevant discussion, that would also
> be appreciated.
>

I agree that the archives are becoming unmanageable simply because
there are so many of them.  I don't know what the solution is.  I
would still urge you to read as much of the archives as possible.
For this particular subject, look for Kociemba, Cube Explorer, Korf,
and IDA*.  Also, look up the discussion of mike reid's optimal
solver, and the very clever table he creates.

----------------------------------------
Jerry Bryan
jbryan@pstcc.cc.tn.us

From cube-lovers-errors@mc.lcs.mit.edu  Tue Feb 15 18:33:33 2000
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Date: Thu, 27 Jan 2000 21:38:47 -0500 (EST)
From: der Mouse  <mouse@rodents.montreal.qc.ca>
Message-Id: <200001280238.VAA02190@Twig.Rodents.Montreal.QC.CA>
To: Ghan <ghan@health.on.net>
Cc: Cube-Lovers@ai.mit.edu
Subject: Re: Stuff...

> I don't really know whether most people on this mailing list are
> regulars or just a bunch of strangers who don't know each other,

My sense of it is that - like most lists - cube-lovers consists of a
few relatively prolific posters and a whole lotta lurkers.  But only
the listowner really knows....

> but I'll introduce myself anyway.  [...]

Welcome!  Glad to see you here.

> [L]ast July, I got one of my friends to teach me how to solve the
> Rubik's Cube, which I have ever since regretted, since I really would
> have liked to solve it myself...

Heh.  Yeah, that's a danger.  I managed most of it myself, though it
took a while; the one piece someone had to show me is a double edge
flip.

> Anyway, I have quite a few unrelated questions I need help with...

> What type of glue could I use to secure all the stickers more
> permanently?

If you don't mind something really permanent, you can always take the
stickers off and paint the facicles.  I've done this on a 3-Cube and
have a 5-Cube sitting disassembled waiting for the same treatment.

> Finally, I decided today to try my hand at some programming.
> [...exhaustive search for the 2-Cube...]

> I thought I was pretty good until I discovered that my program took
> 18 minutes to check all the 9 move algorithms!

The 2-Cube *has* been completely solved; back in 1992 I wrote a program
that did so, and I'm sure dozens of others have done likewise, before
and since.  Here are position counts:

For the quarter-turn metric:
              1 at distance  0 [Wed May 13 23:18:51 1992]
              6 at distance  1 [Wed May 13 23:18:52 1992]
             27 at distance  2 [Wed May 13 23:18:53 1992]
            120 at distance  3 [Wed May 13 23:18:54 1992]
            534 at distance  4 [Wed May 13 23:18:55 1992]
           2256 at distance  5 [Wed May 13 23:18:56 1992]
           8969 at distance  6 [Wed May 13 23:18:58 1992]
          33058 at distance  7 [Wed May 13 23:19:06 1992]
         114149 at distance  8 [Wed May 13 23:19:29 1992]
         360508 at distance  9 [Wed May 13 23:20:53 1992]
         930588 at distance 10 [Wed May 13 23:24:52 1992]
        1350852 at distance 11 [Wed May 13 23:35:53 1992]
         782536 at distance 12 [Wed May 13 23:53:44 1992]
          90280 at distance 13 [Thu May 14 00:03:19 1992]
            276 at distance 14 [Thu May 14 00:04:25 1992]

For the half-turn metric:
              1 at distance  0 [Thu May 14 00:12:37 1992]
              9 at distance  1 [Thu May 14 00:12:39 1992]
             54 at distance  2 [Thu May 14 00:12:41 1992]
            321 at distance  3 [Thu May 14 00:12:42 1992]
           1847 at distance  4 [Thu May 14 00:12:44 1992]
           9992 at distance  5 [Thu May 14 00:12:56 1992]
          50136 at distance  6 [Thu May 14 00:13:08 1992]
         227536 at distance  7 [Thu May 14 00:14:26 1992]
         870072 at distance  8 [Thu May 14 00:20:43 1992]
        1887748 at distance  9 [Thu May 14 00:47:06 1992]
         623800 at distance 10 [Thu May 14 01:46:46 1992]
           2644 at distance 11 [Thu May 14 02:07:51 1992]

I don't recall why the half-turn run was so much slower; perhaps
something else (backups?) started running at midnight....

> [T]here are computer programs which can solve the 3x3x3 in seconds
> (or so I've read,

There are - but not if you also demand that the result be certain to be
an optimal solution.  Heck, a program could be written that just cranks
through a recipe such as you can find in any of dozens of books; such a
program could run in less than a second on all but the tiniest of
computers.  The interest lies in finding *good* - or otherwise
interesting - solutions.

> So I'm wondering how to write a fast solver.  Is my program slow
> because of my bad programming skills, or is my method just plain
> slow?

The latter, certainly.  The former, well, perhaps; I haven't seen the
code, so it's hard to say. :-)  Certainly your choice of language will
work against producing a high-speed program.

					der Mouse

			       mouse@rodents.montreal.qc.ca
		     7D C8 61 52 5D E7 2D 39  4E F1 31 3E E8 B3 27 4B