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
Message-Id: <9510220252.AA04563=dik@bever.cwi.nl>
To: cube-lovers@life.ai.mit.edu
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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Date: Mon, 23 Oct 1995 13:33:08 -0400
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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Reply-To: CRSO.Cube@canrem.com
Sender: CRSO.Cube@canrem.com
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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Date: Fri, 3 Nov 1995 02:39:27 -0500
Message-Id: <3Nov1995.011909.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: 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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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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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 <-
From coumes@issy.cnet.fr Mon Nov 13 04:23:07 1995
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From JBRYAN@pstcc.cc.tn.us Tue Nov 14 09:12:58 1995
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14 Nov 1995 09:13:44 -0400 (EDT)
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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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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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Date: Fri, 1 Dec 95 10:24:24 -0800
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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Date: Sun, 3 Dec 95 14:46:30 EST
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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Reply-To: CRSO.Cube@canrem.com
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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Date: Wed, 06 Dec 1995 11:14:06 +0001 (GMT)
From: "C.S.S.T. Torino" <csstto@alpcom.it>
Subject: Information request
To: cube-lovers@life.ai.mit.edu
Message-Id: <Pine.3.89.9512061120.A9263-0100000@monviso.alpcom.it>
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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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Date: Thu, 7 Dec 1995 02:44:49 -0500
Message-Id: <7Dec1995.013844.Alan@LCS.MIT.EDU>
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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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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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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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)
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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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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)
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To: Cube-Lovers@ai.mit.edu
Subject: WrisTwist
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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)-----
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%" ( @6
end
From walts@federal.unisys.com Fri Dec 15 09:03:36 1995
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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
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To: Cube-Lovers@ai.mit.edu
Subject: Million dollar cube
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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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15 Dec 1995 10:12:29 -0400 (EDT)
Resent-Date: Fri, 15 Dec 1995 10:12:29 -0400 (EDT)
Date: Fri, 15 Dec 1995 10:12:27 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: Million dollar cube
In-Reply-To: <199512151405.JAA12350@homer.MCLN.Federal.Unisys.COM>
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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)
Message-Id: <H.ekBL.jS2P_4E8vJw@dazzle.sol.net>
Organization: The DAZzleman Empire
Subject: Re: WrisTwist
To: Cube-Lovers@ai.mit.edu
Reply-To: dzander@dazzle.sol.net
X-Software: HERMES GUS 1.14.37 Rev. 16 Apr 1994
Date: Sat, 16 Dec 1995 07:13:20 CST
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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
Apparently-To: hoey@aic.nrl.navy.mil, cube-lovers@ai.mit.edu
Subject: Re: Bela Szalai (was Re: Million dollar cube)
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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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From: hoey@aic.nrl.navy.mil
To: Cube-Lovers@life.ai.mit.edu
X-To: Frank R Bernhart <frb6006@cs.rit.edu>,rusin@washington.math.niu.edu (Dave Rusin)
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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To: cube-lovers@ai.mit.edu
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)
Message-Id: <9512181739.AA13899@source.asset.com>
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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Date: Mon, 18 Dec 1995 13:47:26 -0500
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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19 Dec 1995 17:43:05 -0400 (EDT)
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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
Message-Id: <Pine.LNX.3.91.951220031715.4459B-100000@B17.Catch22.COM>
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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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20 Dec 1995 08:44:18 -0400 (EDT)
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Date: Wed, 20 Dec 1995 08:44:16 -0400 (EDT)
From: Jerry Bryan <jbryan@pstcc.cc.tn.us>
Subject: Re: Physical Cubes and Models Thereof
In-Reply-To: <199512201124.GAA04566@Collatz.McRCIM.McGill.EDU>
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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)
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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
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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
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From S005AXR@desire.wright.edu Sat Jan 6 17:20:27 1996
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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
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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
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and we wish you a Happy New Year.
Sincerely,
Pat Clawson Mark Hayton Jeff Reeder
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