From raymond@cps.msu.edu Sat Mar 20 13:24:09 1993
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From: raymond@cps.msu.edu
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To: cubelovers@ai.mit.edu
Subject: Seeking magic dodecahedron
Hello cube lovers,
I haven't seen any activity on this mailing list in a long time!
Cubing is not dead, is it?
Anyway, I'm trying to find a good quality magic dodecahedron. Does
anyone know where I can get one?
Thanks,
Carl
raymond@cps.msu.edu
From news@nntpserver.caltech.edu Sun Mar 21 14:11:53 1993
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Path: joelong
From: joelong@cco.caltech.edu (Joseph Louis Long)
Newsgroups: mlist.cubelovers
Subject: Re: Seeking magic dodecahedron
Date: 21 Mar 1993 19:10:17 GMT
Organization: California Institute of Technology, Pasadena
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raymond@cps.msu.edu writes:
>Hello cube lovers,
> I haven't seen any activity on this mailing list in a long time!
I'll say... when I saw this post it reminded me that I've been watching
this group for about two months hoping to see some mention of Square1,
but have been left disapointed.. (wimper wimper whine.) :)
So let me ask... Does anyone have a solution to Square1? Is there
a simple ``operator'' based method, like there is for the cube?
If it is simple enough to explain in text, could someone please
post it? Has there been a ``solutions book'' published?
obviously in the dark on recent cubic developments,
joe
From pbeck@pica.army.mil Mon Mar 22 07:59:36 1993
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Date: Mon, 22 Mar 93 7:58:16 EST
From: Peter Beck (BATDD)
To: cubelovers@life.ai.mit.edu
Cc: pbeck@pica.army.mil
Subject: cubing
MessageId: <9303220758.aa08917@COR4.PICA.ARMY.MIL>
NO cubing isn't dead.
CFF will hold its annual meeting the
day after IPP, which is in amsterdam this year.
their newsletter is still published.
There is a renewed interest as indicated by
sales of magic solids. In particular
ISHI PRESS, san jose is comerrcially
selling:
5xs
super novas  hungarian made regular dodecahedrons
I believe that there are 2 books on square 1 in
the editing stage. I don't know when/if they will
go to print.
there was discussion and a solution to square 1
posted to this list, in addition the test issue
of a puzzling magazine from ISHI press featured
square 1.
check the literature.
pete beck
THE FUTURE IS PUZZLING,
BUT CUBING IS FOREVER !!
From cosell@world.std.com Mon Mar 22 15:05:44 1993
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From: cosell@world.std.com (Bernie Cosell)
InReplyTo: <1oieipINNs0k@gap.caltech.edu>
(from joelong@cco.caltech.edu (Joseph Louis Long))
(at 21 Mar 1993 19:10:17 GMT)
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Path: world.std.com!cosell
Organization: Fantasy Farm Fibers
To: joelong@cco.caltech.edu (Joseph Louis Long)
Subject: Re: Seeking magic dodecahedron
Cc: cubelovers@life.ai.mit.edu
ContentLength: 774
In <1oieipINNs0k@gap.caltech.edu> on Mar 21, Joseph Louis Long wrote:
} So let me ask... Does anyone have a solution to Square1? Is there
} a simple ``operator'' based method, like there is for the cube?
} If it is simple enough to explain in text, could someone please
} post it? Has there been a ``solutions book'' published?
Dunno about the former, but the answer to the latter is 'yes'. In the April
GAMES magazine there is an ad:
BAFFLED BY SQUARE 1
Now you can solve the world's most challenging
cube puzzle. Clear, easy to unerstand book shows
you how. Send $5 to
Turn to Square 1
PO Box 1451
Westford, MA 01886
/Bernie\

Bernie Cosell cosell@world.std.com
Fantasy Farm Fibers, Pearisburg, VA (703) 9212358
From pbeck@pica.army.mil Thu Mar 25 08:40:16 1993
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Date: Thu, 25 Mar 93 8:27:15 EST
From: Peter Beck (BATDD)
To: cubelovers@life.ai.mit.edu
Subject: square1
MessageId: <9303250827.aa21198@COR4.PICA.ARMY.MIL>
PRODUCT ANNOUNCEMENT:
SQUARE 1 SOLUTION BOOK & PROPOSED NEWSLETTER
RICHARD SNYDER
POB 1451
WESTFORD, MA 01886
6172460700 VOICE
6172461167 FAX
has written a book. He says it is do back
from printing on april 10. If you are
interested CONTACT him directly.
THe world is probably waiting for the
difinitive book review.
From myrberger@e.kth.se Fri Apr 16 13:08:19 1993
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From: myrberger@e.kth.se
To: CubeLovers@ai.mit.edu
MessageId: <0096B212.D5231500.31997@e.kth.se>
Subject: 3x3x3 puzzles and other lists?
Hi,
I have recently found this mailing list and have just finished reading
through the earlier postings.
Perhaps this don't really belong to this list, but I have a list of
different puzzles that in their final state is a 3x3x3 cube. Among them are,
of course, Rubik's cube. The Soma cube and related puzzles are also there.
If you'd like a copy of the list, please MAIL me.
I also wonder if you know about other mailing lists or such which deals with
puzzles (preferrable mechanical). (I know about USENET/NEWS group rec.puzzles.)
Thanks
Johan
MAIL: myrberger@e.kth.se
From cosell@world.std.com Wed May 26 22:37:04 1993
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From: cosell@world.std.com (Bernie Cosell)
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To: cubelovers@life.ai.mit.edu
Subject: Ishi Intternational Puzzles. NEW! (fwd)
ContentLength: 2127
On May 20, Anton Dovydaitis wrote:
[ text of forwarded message follows ]
Ishi Press International is now directly accessable via email:
our online InterNet address is 'ishius@ishius.com' and for
European customers it's 'ishi@cix.compulink.co.uk'.
Ishi Press International sells a wide variety of puzzles from
simple glass, wood and metal puzzles, to collector's items.
Our line consists of several hundred puzzles, including:
Toyo Puzzle City Glass, Hikimi Puzzland Wood and Cast Iron puzzles by Nob
Arjeu wood puzzles from France
Wood, string and wire disentanglement puzzles by JeanClaude Constantin
Magic Bottle puzzles
Handcrafted English Puzzles, including handblown glass Klein bottles,
Single and Double Hourglass Paradoxes, and Wooden Trench Puzzles
Wooden puzzles by Bill Cutler and Stuart Coffin
Puzzle books by Slocum and Nob
Yamanaka Kumiki Works wooden burr puzzles
Traditional inlaid wood Trick Puzzle Boxes by Okiyama and Ninomiya
Yamanaka Kumiki Works wooden burr puzzles
Kamei puzzle boxes, including the Top Box, Die Box, Book, Cup and
Saucer and the Fan
Bolt puzzles by Strijbos
Puzzling People Puzzles, wooden 3D jigsaw puzzles from England,
including the Flummox
Wire Puzzle Sculptures by Rick Irby, including a 3' Dragon,
The Hong Kong Horror and puzzle earrings
and more. Our puzzle line is always increasing: this summer we will be
carrying wood puzzles from Pentangle.
To receive our puzzle catalog with color photographs, please email your
real mailing address to 'ishius@ishius.com'.
To receive regular email on our latest offerings, email us at
'ishius@ishius.com'.
Please be sure to put the word PUZZLE in the subject header, as many of
our customers are interested in GO, not puzzles.
Please write me if you have any questions.
================================================
Anton Dovydaitis Ishi Press International
ishius@ishius.com 76 Bonaventura Drive
Tel: 800/8592086 San Jose, CA 95134
FAX: 408/9449110
[ end of forwarded message ]
From dik@cwi.nl Sun Jun 13 19:53:20 1993
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Date: Mon, 14 Jun 93 01:53:16 +0200
From: Dik.Winter@cwi.nl
MessageId: <9306132353.AA19557.dik@boring.cwi.nl>
To: cubelovers@life.ai.mit.edu
Subject: Contents of CFF31
Last Friday I received issue #31 of Cubism For Fun. A short summary of
the contents:
1. Short articles by Jan de Geus and Frans de Vreugd about Cube Day 1992.
2. An article by Herbert Kociemba about a classification of pretty
patterns on the cube.
3. Reflections by Tom Verhoeff about puzzles and computers.
4. Announcement by Koos Verhoef and Tom Verhoeff of a contest *.
5. A short article by Jaques Haubrich about Rubik's Tangle and how to
position 24 parts in a cube like way (four on a side).
6. An article by Jan Verbakel about the creation of castles with solid
pentominoes.
7. A short article by Trevor Wood on the pecking of octacubes.
8. A short article by Jaques Haubrich about a difficult packing problem.
9. An article by David Singmaster about a gathering in Atlanta in honor
of Martin Gardner. (Nearly the whole puzzling world appears to have
been there.)
10. A new contest by Anton Hanegraaf.
11. Announcement of the 13th Dutch cube day on August 22 in Amsterdam.
This day is next to the 13th International Puzzle Party.
* This is an interesting puzzle indeed. Consider the densest sphere
packing in 3D. This is the packing where you start with a lattice
of spheres based on a triangular lattice, and put on top of it another,
similar, lattice such that each sphere of the new layer fits in a hole
in the lower layer. Add more layers. Pick from that all possible
configurations of 4 connected spheres. There are 25 such
configurations. The puzzle is to create from these 25 pieces a
pyramid with a side of 8 spheres (which contains 120 spheres), with
a hole at the center that consists of a pyramid with a side of 4
spheres (remember those sums of triangular numbers!). It is not
known whether there is a solution. The authors tell how they have
a TRS80 now running 5 years on this problem, using backtracking
techniques. Until now the first 6 pieces did not move. The could
fit 24 pieces already 521,010 times. The puzzle was first announced
at the previous Cube Day.
CFF is a newsletter published by the Nederlandse Kubus Club NKC (Dutch
Cubists Club). It appears a bit irregular, but a few times a year.
Yearly membership fee is now NLG 25. (Dutch Guilders) which amounts to
approximately $ 15.. Information:
Anton Hanegraaf
Heemskerkstraat 9
6662 AL Elst
The Netherlands
(sorry, there is no email address).

dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland
home: bovenover 215, 1025 jn amsterdam, nederland; email: dik@cwi.nl
From @cunyvm.cuny.edu:Matt_Drobel@Novell.COM Tue Jun 29 11:02:15 1993
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Date: Mon, 28 Jun 93 13:55:57 MDT
TotalTo: cubelovers%ai.mit.edu@cunyvm.cuny.edu
signoff cubelovers
From @cunyvm.cuny.edu:rbm8p@darwin.clas.virginia.edu Tue Jun 29 23:15:45 1993
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From hoey@aic.nrl.navy.mil Tue Jul 6 17:29:01 1993
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From: hoey@aic.nrl.navy.mil
MessageId: <9307062128.AA04699@sun13.aic.nrl.navy.mil>
To: CubeLovers@life.ai.mit.edu
Subject: Rubik's Shrimp
I hear on Usenet that there's a show on BBC1 called Wildlife 100,
where people saw a Mantis Shrimp playing with a Rubik's Cube. Reports
are inconclusive as to whether it was able to actually turn faces, or
whether it just waved it around, or even just took it apart. Now if
they could get it to turn faces, presumably they could film it and
play it back in reverse....
Dan Hoey
Hoey@AIC.NRL.Navy.Mil
From CPELLEY@delphi.com Tue Jul 20 21:30:11 1993
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From: CPELLEY@delphi.com
Subject: New idea for a puzzle
To: cubelovers@life.ai.mit.edu
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I have a great concept for a new variant on Rubik's Cube. Where can I
contact the people who manufacture these puzzles today? I understand
JeanClaude Constantin and Uwe Meffert are still around.
The idea is for a dodecahedral puzzle that is sliced up differently than
a Skewb or Megaminx.
From ronnie@cisco.com Wed Jul 28 19:37:15 1993
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To: CubeLovers@life.ai.mit.edu
Subject: Hint wanted for 4x4x4
Date: Wed, 28 Jul 1993 16:37:08 0700
From: "Ronnie B. Kon"
I've been beating my head against the order 4 Rubik's cube for long
enough, and I want a hint. (Not a solutionI have a solution book if
I wanted to use it).
My problem is I cannot flip a pair of adjacent edges (this is
equivalent to not being able to exchange a pair of knightsmove
separated edges). All my other transformations have no side effects,
so I can solve the edges first. But I can't see how to just affect
two of them.
I tend to solve using commutators, but I don't see a way here. The
move I use on the top moves the marked pieces clockwise (this pattern
. . 0 .
. . . .
. . . 0
. 0 . .
rotates and reflects, of course). There is no way to combine these
into a pair exchange (after doing the move, you still have two pieces
out of placenothing changed from the original).
I tried to find a move that would exchange three pieces, the third
being the correctly placed piece next to one of the incorrectly placed
pieces (ie., treat a right edge cubie as if it should be a left edge
cubie) but this can easily be shown as impossible:
Define the parity of a piece as being left if it is a left
edge cubie when the red facelet is up, right if it is a right
edge cubie when the red facelet is up. The parity is
undefined if there is no red facelet. There are only three
moves available that affect an edge cubienone of them alter
the parity. QED
So, what am I missing? As I said before, I really just want a hint
here.
Ronnie
From diamond@jit081.enet.dec.com Wed Jul 28 20:13:19 1993
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From: 29Jul1993 0914
To: cubelovers@life.ai.mit.edu
ApparentlyTo: cubelovers@life.ai.mit.edu
Subject: Re: Hint wanted for 4x4x4
Ronnie B. Kon asked for a hint but not a solution. So here is a hint.
If I understood correctly your descriptions of two transforms which you
asserted to be equivalent, then in fact they are not equivalent.
(Of course, if I didn't understand your descriptions correctly, then
this isn't a hint.)
 Norman Diamond diamond@jit081.enet.dec.com
[Digital did not write this.]
From dik@cwi.nl Wed Jul 28 20:17:38 1993
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Date: Thu, 29 Jul 93 02:17:25 +0200
From: Dik.Winter@cwi.nl
MessageId: <9307290017.AA08740.dik@boring.cwi.nl>
To: CubeLovers@life.ai.mit.edu, ronnie@cisco.com
Subject: Re: Hint wanted for 4x4x4
> I've been beating my head against the order 4 Rubik's cube for long
> enough, and I want a hint. (Not a solutionI have a solution book if
> I wanted to use it).
Yes, it is not really simple.
> ...
> I tend to solve using commutators, but I don't see a way here. The
> move I use on the top moves the marked pieces clockwise (this pattern
> . . 0 .
> . . . .
> . . . 0
> . 0 . .
> rotates and reflects, of course). There is no way to combine these
> into a pair exchange (after doing the move, you still have two pieces
> out of placenothing changed from the original).
Still you are halfway there if you are willing to forgo the pattern of
the centers (which can always be done later). Turn the front face (at
the bottom in the frawing), the right face and the back face one quarter
turn before your turn, and back after. Observe that that constitutes a
cycle of three edge cubies in a single middle slice. Combine with a
quarter turn of that middle slice.
> So, what am I missing? As I said before, I really just want a hint
> here.
I hope that is enough of a hint and not enough of a giveaway.
(I thought there was a shorter sequence, but I disremember at the moment.)
My solution for the 4x4x4 always was: first corners, next edges and finally
centers. Because there are many identical pieces for the centers those are
reasonably simple. It would be much more difficult if each center had its
own place.
dik

dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland
home: bovenover 215, 1025 jn amsterdam, nederland; email: dik@cwi.nl
From hoey@aic.nrl.navy.mil Thu Jul 29 08:36:32 1993
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From: hoey@aic.nrl.navy.mil (Dan Hoey)
MessageId: <9307291236.AA08504@Sun0.AIC.NRL.Navy.Mil>
To: CubeLovers@life.ai.mit.edu
Cc: Dik.Winter@cwi.nl, ronnie@cisco.com (Ronnie B. Kon)
Subject: Re: Hint wanted for 4x4x4
Newsgroups: ml.cubelovers
InReplyTo: <9307290017.AA08740.dik@boring.cwi.nl>
Organization: Navy Center for Applied Research in AI
Cc:
ronnie@cisco.com (Ronnie B. Kon) asks for hints for exchanging a pair
of edges:
> > I tend to solve using commutators, but I don't see a way here....
The key is that commutators are always odd permutations. So do the
move that is an odd permutation of the edges, then use commutators.
Dik.Winter@cwi.nl (dik t. winter) shows a neat way of moving most of
the cubies affected by the odd permutation into the top slice, where
they can be cycled using Ronnie's commutator, which cycles the
TB(R), TR(F), and TF(L) cubies:
> > . . 0 .
> > . . . .
> > . . . 0
> > . 0 . .
(I'm naming them by their edge and their near side.) I suspect Ronnie
is using something like (F Ti F') T (F Ti' F) T (F Ti F') T^2 (F Ti' F).
(For this I'm using "i" to mark inside slabs).
But you can cycle the FL(T), FR(T), RB(T) cubies directly, using a
different commutator. With more effort, there is a commutator that
doesn't mess up face centers. We are getting to the part where it's
hard to distinguish between the hintable and the obvious, but if
people send me email about not being able to figure out what
commutators I'm talking about I'll answer, and post them if such
nobility is common.
>My solution for the 4x4x4 always was: first corners, next edges and finally
>centers. Because there are many identical pieces for the centers those are
>reasonably simple. It would be much more difficult if each center had its
>own place.
As I mentioned years ago, I've made places for mine by cutting corners
of to clusters of face centers and their neighboring edges on each
face.
+++++
    
    
++( )++
    
    
+( )+++
    
    
+++++
    
    
+++++
It's not that hard to fix the face centers, just timeconsuming. It's
a good thing we do the edges first, though, or it would be hard to
figure where the cuts go.
Dan Hoey
Hoey@AIC.NRL.Navy.Mil
( So much discussion on this quiescent list will probably flush out
someone who wants to unsubscribe. Remember to send your note to
cubeloversrequest@ai.ai.mit.edu
to avoid annoyance.)
From ncramer@bbn.com Thu Jul 29 09:40:49 1993
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Date: Thu, 29 Jul 93 8:59:44 EDT
From: Nichael Cramer
To: "Ronnie B. Kon"
Cc: CubeLovers@life.ai.mit.edu
Subject: Re: Hint wanted for 4x4x4
Hi Ronnie
Let me try to restate the problem slightly to make sure we are talking
about the same problem.
Basically, you can solve the entire cube, except that the pieces "1" and
"2" in the diagram are flipped/exchanged:
XXXX
XXXX
XXXX
X12X
Assuming this is the problem, the hint is as follows:
The problem here is that some of the face pieces are not really in their
right places. In short, one of the center slices is 1/4 turn out of phase.
The simplest way to proceed (at least for me) is to move to the following
state:
XX2X (i.e. rotate one of the central slices 1/4 turn)
XXOX
XXOX
X1OX
^

Now, you can solve for pieces "1" and "2" and using these pieces as a
landmark proceed from there.
Hint #2: You can help avoid this problem by solving the face pieces last.
Extra Credit: Actually, the state as shown in the first diagram above is
pretty interesting in that the analogous position on a 3X3X3 cube (i.e. a
single flipped edge cube) is of course impossible.
From this state it is relatively easy to get to another, very interesting
state: namely the 4X4X4 appears to be completely solved except that two
opposite corners are exchanged. (Again, this is obviously impossible on
the 3X3X3.)
Left as an exercise for the reader. ;)
Nichael
ncramer@bbn.com Dr Pepper: It's not just for breakfast any more.
From hoey@aic.nrl.navy.mil Thu Jul 29 10:11:15 1993
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Date: Thu, 29 Jul 93 10:11:13 EDT
From: hoey@aic.nrl.navy.mil (Dan Hoey)
MessageId: <9307291411.AA09096@Sun0.AIC.NRL.Navy.Mil>
To: CubeLovers@life.ai.mit.edu
Subject: Oops... Re: Hint wanted for 4x4x4
Cc: Dik.Winter@cwi.nl, ronnie@cisco.com (Ronnie B. Kon)
I wrote:
? The key is that commutators are always odd permutations. So do the
? move that is an odd permutation of the edges, then use commutators.
But I *Meant*:
! The key observation is that commutators are always even
! permutations. So you to perform an odd permutation on edge, you
! should do the move that is an odd permutation of the edges, then use
! commutators.
Dan
From tomgm@physics.purdue.edu Thu Jul 29 11:02:30 1993
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From: Tom G. Miller
Subject: Re: Hint wanted for 4x4x4
To: ronnie@cisco.com (Ronnie B. Kon)
Date: Thu, 29 Jul 93 10:05:22 EST
Cc: CubeLovers@life.ai.mit.edu
InReplyTo: <199307282337.AA06770@lager.cisco.com>; from "Ronnie B. Kon" at Jul 28, 93 4:37 pm
XMailer: ELM [version 2.3 PL11]
Ronnie,
It's been so long since I've messed around with my 4x4x4 that
I can't answer your question directly, however when I
see descriptions for the method in which people solve the
4x4x4 it is usually different from the way I first solved
it:
What I did was to pair up the middle two edgies, and the
four central face cubes. There are few enough restraints
that this is not too hard to do for someone who has never
touched a 4x4x4 cube. One then has a cube like with faces
similar to the following:
r b b g
y o o b
y o o b
r y y w
One can then "pretend" it is a 3x3x3 cube and then solve it.
Unfortunately you will occasionally end up in an orbit of the
"pseudo3x3x3" that is impossible to solve. Oh well... scramble
it and try it again.
Using this technique I was able to solve a scrambled 4x4x4 cube within an
hour or so of when I set my hands on one. Needless to say, this is NOT a
good technique for solving a 4x4x4 cube if one is interested only in the
4x4x4. In fact I suspect it is a pretty awful algorithm, especially since
you frequently end up in an unsolveable orbit using your standard 3x3x3
techniques. But it is a useful trick for maximizing the hard work one used in
learning the 3x3x3.
As most people who have a 4x4x4 realize, if you never make any twists of a
solved 4x4x4 cube except along the center, you of course have a 2x2x2. And
as I described, if you only make moves 1deep, it is equivalent to a 3x3x3,
and of course it is also a 4x4x4.
Tom Miller
tomgm@physics.purdue.edu
From ronnie@cisco.com Fri Jul 30 00:46:55 1993
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To: CubeLovers@life.ai.mit.edu
Subject: Re: Hint wanted for 4x4x4
Date: Thu, 29 Jul 1993 21:46:48 0700
From: "Ronnie B. Kon"
Thanks to all who responded. I haven't yet got what I consider a
solution for my problem (shift a slice and resolve is my current
method which is slow and ugly) but at least I understand my problem
slightly better.
A few questions:
1. What is the definition of parity by which commutators are even,
but slice turns are odd? I haven't been able to come up with a
cubewide parity. (I know no group theory).
2. How many orbits does the order 4 cube have? I can only think of
three (twirling a corner cubie). Then again, I haven't painted the
facelets yet, so there could be orbits I haven't begun to see
involving them.
3. Would an order 6 cube have any challenge beyond the order 4? I
think the answer is noif you are able to solve the 3cube and the
4cube you can solve any cube.
From dik@cwi.nl Mon Aug 2 20:52:23 1993
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Date: Tue, 3 Aug 93 02:52:19 +0200
From: Dik.Winter@cwi.nl
MessageId: <9308030052.AA23253.dik@boring.cwi.nl>
To: cubelovers@life.ai.mit.edu
Subject: Fitting puzzle solved
Some time ago I posted an article (c.q. mailed a message) describing
the contents of Cubism For Fun, the newsletter published by the Dutch
Cubists Club (NKC). In that article (message) I gave a more elaborate
description about a problem involving fitting pieces.
Briefly:
The base problem is as follows. Build a tetrahedron consisting of balls,
8 balls on an edge. When you look at the lattice induced by this
tetrahedron after some thinking you will find there are 25 ways to pick
4 connected balls. Now take those 25 ways and make "pieces" from it.
Again, go back to the tetrahedron and inside it create a hollow
tetrahedron with 4 balls on an edge. The remainder requires 100 balls
to fill. Try to do that with the 25 "pieces" you just created.
This has been a fairly longstanding problem but it is now (partly)
solved.
I just had word that Jan de Ruiter from Purmerend (the Netherlands)
found a number of solutions. Details will likely be presented in
a forthcoming issue of CFF.
An amusing sidenote. Between the 25 pieces there are two that can
be created interlocked. It is not clear whether it is possible to
separate those two pieces by hand when interlocked, so it is not
clear whether a solution that has those two pieces interlocked
really is a solution. The first solutions Jan de Ruiter found
*had* those two pieces interlocked. But after some time he found
a solution with those two pieces far away from each other, so there
is really a true solution.
Remaining questions: How many solutions are there? How many do not
have those two pieces interlocked? Is it possible to separate those
two pieces when interlocked? (The last puzzle resembles one of those
chinese metal separation puzzles.)
(Information about CFF can be obtained from Anton Hanegraaf,
Heemskerkstraat 9, 6662 AL Elst, The Netherlands. Email is
now also possible: gm@phys.uva.nl.)
dik

dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland
home: bovenover 215, 1025 jn amsterdam, nederland; email: dik@cwi.nl
From dik@cwi.nl Mon Aug 2 21:10:32 1993
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Date: Tue, 3 Aug 93 03:10:22 +0200
From: Dik.Winter@cwi.nl
MessageId: <9308030110.AA23300.dik@boring.cwi.nl>
To: cubelovers@life.ai.mit.edu
Subject: Diameter of cube group?
I have now running (for about 60 days already) a program that implements
Kociemba's algorithm to solve the cube. It tries to solve random
configurations and stops when a solution of 20 turns or less is found.
The random configurations are created by doing 100 random turns.
Until now, with 9000 configurations tried, all proved to be solvable
in 20 turns or less.
This strongly suggests that the diameter of the cube group is at most
21, or perhaps 22; but not more.
The figure of 9000 configurations in 60 days indicates that solution
of one configuration takes slightly less than 10 minutes. This is
contrary to what I thought was possible. Whenever I tried configurations
they were mostly solved within 2 or 3 minutes. This suggests that the
random configurations are more difficult to solve than what I and many
others brought up as possible difficult patterns.
But I still need to do some analysis on the ouput (now 3 Mb of data).
Continuing and waiting for a config that requires 21 turns, dik

dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland
home: bovenover 215, 1025 jn amsterdam, nederland; email: dik@cwi.nl
From W.Taylor@math.canterbury.ac.nz Mon Aug 2 22:20:16 1993
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Date: Tue, 3 Aug 93 14:19:52 NZS
From: W.Taylor@math.canterbury.ac.nz (Bill Taylor)
Subject: re: Diameter of cube group?
To: CubeLovers@life.ai.mit.edu
MessageId: <9308030219.AA23489@math.canterbury.ac.nz>
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Fascinating news from Dik Winter about the solving of 9000 configurations.
Can someone please remind us exactly what Kociemba's algorithm is; or at
least a breif outline of how it works. I know I've heard the name before,
but can't remember anything about it.
Thanks, Bill Taylor. wft@math.canterbury.ac.nz
From dik@cwi.nl Tue Aug 3 18:44:13 1993
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From: Dik.Winter@cwi.nl
MessageId: <9308032244.AA26535.dik@boring.cwi.nl>
To: cubelovers@life.ai.mit.edu
Subject: Kociemba's algorithm
I have received a few requests for information about the algorithm and
for the program. I have put the program available for ftp. In the
set of files there is also a Description I edited from a number of
messages I mailed a long time ago to this mailing list. They give
a reasonable description of the algorithm.
Ftp to ftp.cwi.nl. File is: /pub/dik/cube.tar.Z. Do not forget
to set binary mode.
dik

dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland
home: bovenover 215, 1025 jn amsterdam, nederland; email: dik@cwi.nl
From acw@bronze.lcs.mit.edu Thu Aug 5 17:03:17 1993
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Date: Thu, 5 Aug 93 17:02:45 EDT
From: acw@bronze.lcs.mit.edu (Allan C. Wechsler)
MessageId: <9308052102.AA22841@bronze.lcs.mit.edu>
To: Dik.Winter@cwi.nl
Cc: cubelovers@life.ai.mit.edu
InReplyTo: Dik.Winter@cwi.nl's message of Tue, 3 Aug 93 03:10:22 +0200 <9308030110.AA23300.dik@boring.cwi.nl>
Subject: Diameter of cube group?
I wonder about the validity of your Monte Carlo analysis. It seems
to be based on an intuition about how fast the number of configurations
falls off with the distance from SOLVED. I share the intuition, but
I'm not sure I can rigorize it, and that makes me cautious.
What prevents a group from having a "pointy tail", that is, a "corridor"
of elements at increasing distances from the identity? In fact, does
the number of elements as a function of distance have to be unimodal?
Could this function have a "waist"? Intuitively, this sounds
impossible, but I am wondering what constraints on such functions are known.
From ronnie@cisco.com Thu Aug 5 19:55:48 1993
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To: acw@bronze.lcs.mit.edu (Allan C. Wechsler)
Cc: Dik.Winter@cwi.nl, cubelovers@life.ai.mit.edu
Subject: Re: Diameter of cube group?
InReplyTo: Your message of "Thu, 05 Aug 1993 17:02:45 EDT."
<9308052102.AA22841@bronze.lcs.mit.edu>
Date: Thu, 05 Aug 1993 16:55:36 0700
From: "Ronnie B. Kon"
Disclaimer: this sounds more authoritative than is intendedI really
don't know what I'm talking about.
It couldn't be very pointy. From the most distant configuration,
there are 6 positions immediately before it. There are 6^2 two steps
away, 6^3 three steps, etc. (well, 6^2  1 and 6^3  ?) actually.
This is necessarily so, as if any of the configurations reachable with
two twists (for example) are closer in than (max  2) steps then you
could reach the maximum configuration by getting there and then doing
the two steps.
This gives me the feeling that Monte Carlo is fairly valid. (How's
that for rigor?)
Ronnie
> I wonder about the validity of your Monte Carlo analysis. It seems
> to be based on an intuition about how fast the number of configurations
> falls off with the distance from SOLVED. I share the intuition, but
> I'm not sure I can rigorize it, and that makes me cautious.
>
> What prevents a group from having a "pointy tail", that is, a "corridor"
> of elements at increasing distances from the identity? In fact, does
> the number of elements as a function of distance have to be unimodal?
> Could this function have a "waist"? Intuitively, this sounds
> impossible, but I am wondering what constraints on such functions are known.
>
From dik@cwi.nl Thu Aug 5 19:55:54 1993
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Date: Fri, 6 Aug 93 01:55:50 +0200
From: Dik.Winter@cwi.nl
MessageId: <9308052355.AA05297.dik@boring.cwi.nl>
To: acw@bronze.lcs.mit.edu
Subject: Re: Diameter of cube group?
Cc: cubelovers@life.ai.mit.edu
> I wonder about the validity of your Monte Carlo analysis. It seems
> to be based on an intuition about how fast the number of configurations
> falls off with the distance from SOLVED. I share the intuition, but
> I'm not sure I can rigorize it, and that makes me cautious.
I am not sure (that is obvious). But when looking at other (similar)
puzzles I did I think it is a save guess.
> What prevents a group from having a "pointy tail", that is, a "corridor"
> of elements at increasing distances from the identity?
The groups I did calculate in full do *not* have a pointy tail. This holds
for 2x2x2, 3x3x3 corners only, magic domino. I think it would be a big
surprise if there is a pointy tail. Obviously we can say a priory that
there is not a single configuration opposite from start, so the tail is
not very pointy, if it is at all. For instance for the magic domino the
tail of the list of number of configuration a certain distance from start is:
14: 508704668
15: 232904952
16: 14508468
17: 129376
18: 112
With the maximum at 14 turns. (Here I took the table where only a single
solution is allowed; i.e. no full rotations of the domino.) 1 in 2 (approx.)
configurations requires 14 turns or more. 1 in 100 requires 16 turns or
more. Of course the number of configurations of the cube is quite a bit more.
Still after doing about 9000 configurations not a single one is found that
requires more than 20 turns. If we assume a picture similar to the domino
(which in my opinion is a save guess), there might be configurations that
retuire 21 or perhaps 22 turns, but more is extremely unlikely.
However, there is a remaining question; is the random choice of configuration
unbiased? I think it is. To create a random configuration I do 100 random
turns chosen from 18 possible turns (quarter turns, half turns and reverse
turns). The random number generator is (as far as I know) quite good
(Berkeley Unix's random).
dik
From dik@cwi.nl Thu Aug 5 20:01:34 1993
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From: Dik.Winter@cwi.nl
MessageId: <9308060001.AA05312.dik@boring.cwi.nl>
To: ronnie@cisco.com
Subject: Re: Diameter of cube group?
Cc: cubelovers@life.ai.mit.edu
The last remark first:
> This gives me the feeling that Monte Carlo is fairly valid. (How's
> that for rigor?)
Not very ;).
> It couldn't be very pointy. From the most distant configuration,
> there are 6 positions immediately before it. There are 6^2 two steps
> away, 6^3 three steps, etc. (well, 6^2  1 and 6^3  ?) actually.
This still can create a pointy tail; just as pointy as the front. My
experience is that the tail is much more blunt than the front. That there
are already more than a single configuration at maximum distance makes that
reasonable.
From CPELLEY@delphi.com Fri Aug 6 02:47:53 1993
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Date: Fri, 06 Aug 1993 00:53:23 0400 (EDT)
From: CPELLEY@delphi.com
Subject: Square1 Puzzle Party
To: cubelovers@life.ai.mit.edu
MessageId: <01H1EMUZ0T1E91XFD4@delphi.com>
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Richard Snyder's book on Square1 is now being published. He sent me the
following press release about a forthcoming Square1 Puzzle Party:
Square1 is the most challenging puzzle since Rubik's Cube! When you turn
it, it forms many unfamiliar shapes, and it seems impossible to get it back
to the cube shape! And if you do somehow manage to turn it into a cube,
it is scrambled and needs to be solved Rubikstyle.
It's really two puzzles in one! Harder than Rubik's, it's so hard that
only 5 people in the whole world have ever been able to come up with a
complete solution to it!
Richard Snyder of Dracut is the only person in the USA who has written a
book which shows how to solve Square1!
His book is clear and easy to follow, leading you step by step from any
scrambled state to the completely solved cube! Then he gives formulas
for over 100 colored patterns which you can make on Square1's
symmetrical shapes, and teaches you how to make your own symmetrical
patterns! There's no other book quite like it in the world!
Richard will be presenting his new book, Turn to Square1, to Boston in a
great puzzle party, which will be held at 1PM on Sat., Aug. 7, 1993, at
The Games People Play, 1105 Massachusetts Ave., Cambridge, MA (near
Harvard Square)
Richard will demonstrate solving Square1, making Square1 patterns, and he
will also demonstrate solving Rubik's Cube, the Skewb, and other cube puzzles.
He will be autographing copies of his new book, and presenting many other
fine puzzles and books that are carried by The Games People Play.
The Press and the Media will be there, and you are invited to come.
Bring your Square1, your Rubik's Cube, and your other Rubik's puzzles that
you haven't been able to solve! Richard will solve them, and show you his
solution to Rubik's Cube, the world's best, fastest, and most concise Rubik's
Cube solution!
But most of all, be prepared to be astounded as Richard shows you how you
too can Turn to Square1!
From ccw@eql12.caltech.edu Fri Aug 6 22:37:43 1993
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From: ccw@eql12.caltech.edu (Chris Worrell)
MessageId: <930806124838.23c011ac@EQL12.Caltech.Edu>
Subject: Re: Square1 Puzzle Party
InReplyTo: Your message <01H1EMUZ0T1E91XFD4@delphi.com> dated 6Aug1993
To: CPELLEY@delphi.com
Cc: ccw@eql12.caltech.edu, cubelovers@life.ai.mit.edu
Sorry. I can't let this one pass by without comment.
CPELLEY@delphi.com says
> Richard Snyder's book on Square1 is now being published. He sent me the
> following press release about a forthcoming Square1 Puzzle Party:
> It's really two puzzles in one! Harder than Rubik's, it's so hard that
> only 5 people in the whole world have ever been able to come up with a
> complete solution to it!
Unless Snyder or his agent is talking about a God's Algorithm for
Square1, this statement is ridiculous.
I doubt that this number includes myself, as I have only told a few family
members and friends that I have solved this. (I expect many of you can say
the same thing.)
Harder than Rubik's? This is a matter of opinion and definition.
Do they mean conceptually harder, harder to derive a solution method,
harder to prove a solution method, or harder to achieve an individual
solution attempt? Or does harder just mean more time? More time
to derive a solution method, more time to prove a solution method,
or more time to achieve an individual solution attempt?
I don't really doubt the last. Except for the Pyraminx (and the 2Cube),
all of the puzzles of this type take me longer to solve than the Cube.
I think that the Rubik's cube still holds the record as the puzzle that
took me longest to derive a solution method. (Of course all of the others
borrowed substantially from the cube.)
>Bring your Square1, your Rubik's Cube, and your other Rubik's puzzles that
>you haven't been able to solve!
Sorry, I don't have any. Except the 10x10 Rubik's Tangle.
Chris Worrell
ccw@eql.caltech.edu
From dn1l+@andrew.cmu.edu Fri Aug 6 23:27:13 1993
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MessageId:
Date: Fri, 6 Aug 1993 23:22:28 0400 (EDT)
From: "Dale I. Newfield"
To: cubelovers@life.ai.mit.edu
Subject: Tangle (Was: Re: Square1 Puzzle Party)
InReplyTo: <930806124838.23c011ac@EQL12.Caltech.Edu>
Excerpts from mail: 6Aug93 Re: Square1 Puzzle Party by Chris
Worrell@eql12.calt
>>Bring your Square1, your Rubik's Cube, and your other Rubik's puzzles that
>>you haven't been able to solve!
>Sorry, I don't have any. Except the 10x10 Rubik's Tangle.
I only have one quarter of that puzzle...(section 4).
I worked on it for a considerable amount of time, and concluded that the only
solution method was trial and error.
So I wrote a program to do it for me.
I know all 16 solutions (2 unique)*(2 identical exchanged pieces)*(4
orientations).
Has anyone come up with a method, besides trial and error, that solves
this thing? (or the 10x10?)
(hmmmI wonder how much the other 3 would cost?)
Dale Newfield
dn1l@{cs,andrew}.cmu.edu
From @uccvma.ucop.edu:MJTOL@UCCMVSA.BITNET Sat Aug 7 04:25:18 1993
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Date: Thu, 05 Aug 93 16:23:29 PST
From: "Michael Thwaites"
To: cubelovers@life.ai.mit.edu
Subject: cube tail?
> What prevents a group from having a "pointy tail", that is,
> a "corridor" of elements at increasing distances from the
> identity? In fact, does the number of elements as a
> function of distance have to be unimodal? Could this
> function have a "waist"? Intuitively, this sounds
> impossible, but I am wondering what constraints on such
> functions are known.
>
It seems to me it can't be too pointy. Working backwards, the
number of arrangements working from the end has to explode
(probably in symetry) with the number of arrangements form the
start.
From weber@src.dec.com Sat Aug 7 17:33:08 1993
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To: "Dale I. Newfield"
Cc: cubelovers@life.ai.mit.edu
Subject: Tangle (Was: Re: Square1 Puzzle Party)
InReplyTo: Message of Fri, 6 Aug 1993 23:22:28 0400 (EDT)
from "Dale I. Newfield"
Date: Sat, 07 Aug 93 14:32:53 0700
From: weber@src.dec.com
XMts: smtp
>>>Bring your Square1, your Rubik's Cube, and your other Rubik's puzzles that
>>>you haven't been able to solve!
>>Sorry, I don't have any. Except the 10x10 Rubik's Tangle.
>
>I only have one quarter of that puzzle...(section 4).
>
>I worked on it for a considerable amount of time, and concluded that the only
>solution method was trial and error.
I was thinking about the Rubik's Tangle, and what was puzzling me was
WHY there should be only one solution (apart from the obvious symmetries).
After all, all pieces are identical except for coloring, and a set consists
of all 24 possible coloring, and 1 duplicate, and this doesn't sound like
an artificial construction. Is there any mathematical reason for the
uniqueness of the solution? What possible "Tanglelike" puzzles have
unique solutions?
Sam
From dn1l+@andrew.cmu.edu Sun Aug 8 00:21:54 1993
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MessageId:
Date: Sat, 7 Aug 1993 19:36:39 0400 (EDT)
From: "Dale I. Newfield"
To: cubelovers@life.ai.mit.edu
Subject: Re: Tangle (Was: Re: Square1 Puzzle Party)
Cc:
InReplyTo: <9308072132.AA01481@chaucer>
Excerpts from mail: 7Aug93 Tangle (Was: Re: Square1 P.. by
weber@src.dec.com
> I was thinking about the Rubik's Tangle, and what was puzzling me was
> WHY there should be only one solution (apart from the obvious symmetries).
> After all, all pieces are identical except for coloring, and a set consists
> of all 24 possible coloring, and 1 duplicate, and this doesn't sound like
> an artificial construction. Is there any mathematical reason for the
> uniqueness of the solution? What possible "Tanglelike" puzzles have
> unique solutions?
The section I had (4) had 2 distinct solutions (apart from the exchange
of the 2 identical pieces, and the 4 orientations).
In fact, the box that the puzzle came in said it should have 2 solutions.
Dale
From @mail.uunet.ca:mark.longridge@dosgate.canrem.COM Mon Aug 9 12:02:01 1993
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To: CUBELOVERS@ai.mit.edu
Subject: SQUARE'S GROUP ANALYSIS
From: Mark Longridge
MessageId: <60.250317.104.0C180A7E@canrem.com>
Date: Sun, 8 Aug 1993 15:40:00 0400
Organization: CRS Online (Toronto, Ontario)
After reading Dik's post I figured I'd add my 2 cents worth:
Mark's Notes on the Squares Group

On studying the squares group I have found 16 antipodal cases
requiring
the maximum 15 moves. Two of these cases cycle all 8 corners and leave
the edges in place. A third case "2 DOT/Inverted T's" is pleasingly
symmetric. Also I have noted that cycling only the 4 edges in the
U or D layer requires 1 move less that cycling only the 4 corners in U
or D when using only moves in the square's group, 12 moves for edges
and 13 moves for corners.
If we define "symmetry level" as the number of distinct patterns
generated by rotating the cube through it's 24 different orientations in
space then most known antipodes are symmetry level 6. Thus the lower the
number the higher the level of symmetry. The least symmetric positions
have level 24, and this is very common. The most symmetric positions
have level 1, the two positions START and 6 X order 2.
I have also found positions with levels 3, 8 and 12.
Given the fact that 8 antipodal cases have symmetry level 6
and 8 cases have symmetry level 12 we can now account for ALL
8 * 6 + 8 * 12 = 144 of the 144 cases!
Cases with symmetry level 6:
p66 Double 4 corner sw L2 B2 R2 F2 L2 F2 T2 R2 (T2 D2 F2 T2) F2 L2 D2
(15)
p67 Antipode 2 R2 B2 D2 F2 D2 F2 T2 L2 (T2 D2 F2 T2) L2 T2 B2
(15)
p80 2 DOT, Invert T's R2 B2 D2 R2 B2 L2 B2 L2 (T2 D2 F2 T2) F2 L2 T2
(15)
p99 2 DOT, 4 ARM R2 B2 D2 L2 B2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2
(15)
p100 2 Cross, 4 ARCH 1 R2 B2 T2 R2 F2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2
(15)
p130 2 Cross, 4 ARCH 2 L2 B2 D2 B2 L2 D2 F2 L2 (T2 D2 F2 L2) F2 L2 T2
(15)
p135 2 X, 4 T L2 B2 D2 F2 T2 F2 T2 L2 (T2 D2 F2 T2) L2 D2 F2
(15)
p137 2 X, 4 ARM L2 F2 T2 B2 T2 F2 T2 L2 (T2 D2 F2 T2) L2 D2 F2
(15)
Cases with symmetry level 12:
p108 2 DOT, 2 T, 2 ARM L2 F2 T2 R2 B2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2
(15)
p128 2 H, 2 T, 2 CRN L2 B2 R2 F2 L2 F2 T2 R2 (T2 D2 F2 T2) F2 L2 T2
(15)
p129 2 H, 2 T, 2 ARCH R2 F2 L2 F2 L2 F2 T2 R2 (T2 D2 F2 T2) F2 L2 T2
(15)
p131 2 H, 2 ARM, 2 ARCH L2 F2 R2 D2 B2 L2 D2 L2 (T2 D2 F2 T2) F2 L2 F2
(15)
p132 2 Cross,2 ARCH,2CRN L2 F2 D2 R2 F2 L2 B2 L2 (T2 D2 F2 T2) F2 L2 D2
(15)
p133 2 Cross, 2 T, 2 ARM L2 F2 D2 F2 D2 F2 T2 L2 (T2 D2 F2 T2) L2 T2 B2
(15)
p134 2 CRN, 2 X, 2 ARCH L2 F2 D2 B2 T2 F2 T2 L2 (T2 D2 F2 T2) L2 T2 B2
(15)
p136 2 H, 2 ARM, 2 CRN R2 F2 L2 T2 B2 L2 T2 L2 (T2 D2 F2 T2) F2 L2 B2
(15)
5 of the 16 known antipodes are within 4 and 2 face turns (or 2 and 1
slice turns) of each other:
p66 + L2 R2 T2 D2 = p80 (allowing for whole cube rotations)
p66 + F2 B2 = p100
p80 + T2 D2 = p99
P66 + T2 D2 = P128
Using full group moves these antipodes can be reduced to:
P66a alternate method F2 R2 U2 F2 R2 U3 D3 B2 L2 F2 B2 U1 D1
(13)
p67a alternate method F2 R2 F2 U3 D3 L2 B2 D2 L2 B2 U1 D1 B2
(13)
p80a alternate method U1 F2 R2 L2 U2 D2 F2 U2 D3
(9)
p99a alternate method U1 R2 F2 B2 U2 D2 R2 D1
(8)
P100a alternate method F2 U2 D2 F2 R2 L2 D1 F2 R2 L2 B2 U1
(12)
p108a alternate method R2 F2 B2 L2 D1 R2 U2 R2 L2 U2 R2 D1
(12)
p130a alternate method F2 R2 F2 B2 U1 D1 F2 R2 D2 F2 L2 U3 D3
(13)
p133a alternate method R2 U1 F2 R2 L2 U2 D2 F2 U2 D3 R2
(11)
Both p80a and p99a are surprisingly compact, p99a being a full 7
turns
less than it's square's group equivalent. Note that in p99a a square's
group sequence is sandwiched between 2 turns on opposite faces. It is
the final turn D1 which brings it back into a sq group state! In general
U1 (sq group sequence) D1 does not lead to a sq group sequence.
Another interesting discovery was comparing the full group sequences:
L1 R1 D2 L3 R3 (antislice, 5 moves)
L1 R3 D2 L3 R1 (slice , 5 moves)
F1 B1 D2 F1 B1 (clockwise, 5 moves)
... to their square's group equivalents:
R2 F2 T2 L2 B2 L2 T2 R2 F2 (9 moves)
R2 B2 L2 D2 R2 F2 L2 T2 F2 (9 moves)
R2 B2 T2 F2 L2 F2 T2 R2 F2 (9 moves)
Also it was found possible to permute 3 edges only using:
L2 T2 R2 B2 R2 T2 L2 F2 (8 moves)
or L3 R1 F2 L1 R3 D2 (6 moves)
In general any sequence L1 R1 (any squares group moves) L3 R3
will always result in a squares group position, for example:
L1 R1 (D2 F2 B2) R3 L3
F1 B1 (T2 B2 F2 L2) F3 B3
p66a (F2 R2 U2 F2 R2) U3 D3 (B2 L2 F2 B2) U1 D1 (13 moves)
The longest irreducible square's group sequence discovered so far,
which is an embedded part of longest Phase 2 sequence (p94):
(Thus it can't be reduced by using full group moves using current
techniques)
R2 B2 U2 B2 L2 D2 L2 F2 (8)
Later on I discovered this irreducible sequence by chance:
T2 B2 T2 B2 D2 F2 R2 T2 L2 F2 (10)
Edges only (with corners in place) can be 14 moves at most, e.g.
D2 L2 F2 D2 F2 L2 T2 L2 T2 D2 F2 R2 F2 R2 (14)
This answers the question David Singmaster posed in
"Notes on Rubik's Magic Cube" on Thistlethwaite's last stage.
That is: "Are there any positions in the square's group with
corners fixed of length 14 or can they be done in less moves?"
A few observations...
 It is not possible to swap just 1 pair of edges and corners
 It is only possible to have 4, 6 or 8 corners out of place
 Known antipodal cases can be solved in <=13 moves using full
group
 In reaching an antipode one may start with any of the 6 turns
(since antipodes are global maxima, any turn will get you one
move closer)
 If the corners are fixed, the position is NOT an antipode
 Longest order appears to be 12
 All known (probably all!) antipodes have symmetry level 6 or 12
 Although only conjectural, it is now believed that one turn of
a face MUST lead to a new state which is either 1 move closer
or 1 move farther from START
Question: Are there any irreducible square's group sequences that
are longer then 10 moves? Are these truly irreducible
or only irreducible under Dik Winter's Kociemba
inspired program?
Oh well, the full group beckons. I still want to try and come up
with my own algorithm though.
> Mark <
From @mail.uunet.ca:mark.longridge@dosgate.canrem.COM Mon Aug 9 12:39:08 1993
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To: CUBELOVERS@ai.mit.edu
Subject: CUBES (OF COURSE!)
From: Mark Longridge
MessageId: <60.250315.104.0C180A7C@canrem.com>
Date: Sun, 8 Aug 1993 15:34:00 0400
Organization: CRS Online (Toronto, Ontario)
Well, I finally know what all the square's group antipodes look like.
Next message I'll post a detailed summary on these patterns.
I wrote a colour printer driver for my cube program today, tested
it and it turned out pretty slick. I'm using a Star NX2420 rainbow
printer and I'm (out of necessity) using
FACE Star NX2420 Code Real Cube
   
TOP/DOWN BLACK/CYAN 0/2 WHITE/BLUE
LEFT/RIGHT VIOLET/ORANGE 3/5 RED/ORANGE
FRONT/BACK YELLOW/GREEN 4/6 YELLOW/GREEN
...which works pretty good except violet is more like blue. I could
have also used magenta (sort of a pink) for red but it does not
contrast well with orange.
Should make the DOTC (Domain of the Cube) Newsletter look a lot
better if I can ever finish the damn thing!
More notes to follow....
> Mark Longridge, still cubing after all of these years <
From @mail.uunet.ca:mark.longridge@dosgate.canrem.COM Tue Aug 10 11:07:52 1993
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To: cubelovers@life.ai.mit.edu
Subject: Cubes (of course)
From: Mark Longridge
MessageId: <60.251050.104.0C180C04@canrem.com>
Date: Tue, 10 Aug 1993 01:22:00 0400
Organization: CRS Online (Toronto, Ontario)
Well, I finally know what all the square's group antipodes look like.
Next message I'll post a detailed summary on these patterns.
I wrote a colour printer driver for my cube program today, tested
it and it turned out pretty slick. I'm using a Star NX2420 rainbow
printer and I'm (out of necessity) using
FACE Star NX2420 Code Real Cube
   
TOP/DOWN BLACK/CYAN 0/2 WHITE/BLUE
LEFT/RIGHT VIOLET/ORANGE 3/5 RED/ORANGE
FRONT/BACK YELLOW/GREEN 4/6 YELLOW/GREEN
...which works pretty good except violet is more like blue. I could
have also used magenta (sort of a pink) for red but it does not
(contrast well with orange.
Should make the DOTC (Domain of the Cube) Newsletter look a lot
better if I can ever finish the damn thing!
More notes to follow....
> Mark Longridge, still cubing after all of these years <
From @mail.uunet.ca:mark.longridge@dosgate.canrem.com Tue Aug 10 11:11:29 1993
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To: cubelovers@life.ai.mit.edu
Subject: Square's group
From: Mark Longridge
MessageId: <60.251051.104.0C180C07@canrem.com>
Date: Tue, 10 Aug 1993 01:29:00 0400
Organization: CRS Online (Toronto, Ontario)
After reading Dik's post I figured I'd add my 2 cents worth:
Mark's Notes on the Squares Group

On studying the squares group I have found 16 antipodal cases
requiring
the maximum 15 moves. Two of these cases cycle all 8 corners and leave
the edges in place. A third case "2 DOT/Inverted T's" is pleasingly
symmetric. Also I have noted that cycling only the 4 edges in the
U or D layer requires 1 move less that cycling only the 4 corners in U
or D when using only moves in the square's group, 12 moves for edges
and 13 moves for corners.
If we define "symmetry level" as the number of distinct patterns
generated by rotating the cube through it's 24 different orientations in
space then most known antipodes are symmetry level 6. Thus the lower the
number the higher the level of symmetry. The least symmetric positions
have level 24, and this is very common. The most symmetric positions
have level 1, the two positions START and 6 X order 2.
I have also found positions with levels 3, 8 and 12.
Given the fact that 8 antipodal cases have symmetry level 6
and 8 cases have symmetry level 12 we can now account for ALL
8 * 6 + 8 * 12 = 144 of the 144 cases!
Cases with symmetry level 6:
p66 Double 4 corner sw L2 B2 R2 F2 L2 F2 T2 R2 (T2 D2 F2 T2) F2 L2 D2
p67 Antipode 2 R2 B2 D2 F2 D2 F2 T2 L2 (T2 D2 F2 T2) L2 T2 B2
p80 2 DOT, Invert T's R2 B2 D2 R2 B2 L2 B2 L2 (T2 D2 F2 T2) F2 L2 T2
p99 2 DOT, 4 ARM R2 B2 D2 L2 B2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2
p100 2 Cross, 4 ARCH 1 R2 B2 T2 R2 F2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2
p130 2 Cross, 4 ARCH 2 L2 B2 D2 B2 L2 D2 F2 L2 (T2 D2 F2 L2) F2 L2 T2
p135 2 X, 4 T L2 B2 D2 F2 T2 F2 T2 L2 (T2 D2 F2 T2) L2 D2 F2
p137 2 X, 4 ARM L2 F2 T2 B2 T2 F2 T2 L2 (T2 D2 F2 T2) L2 D2 F2
Cases with symmetry level 12:
p108 2 DOT, 2 T, 2 ARM L2 F2 T2 R2 B2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2
p128 2 H, 2 T, 2 CRN L2 B2 R2 F2 L2 F2 T2 R2 (T2 D2 F2 T2) F2 L2 T2
p129 2 H, 2 T, 2 ARCH R2 F2 L2 F2 L2 F2 T2 R2 (T2 D2 F2 T2) F2 L2 T2
p131 2 H, 2 ARM, 2 ARCH L2 F2 R2 D2 B2 L2 D2 L2 (T2 D2 F2 T2) F2 L2 F2
p132 2 Cross,2 ARCH,2CRN L2 F2 D2 R2 F2 L2 B2 L2 (T2 D2 F2 T2) F2 L2 D2
p133 2 Cross, 2 T, 2 ARM L2 F2 D2 F2 D2 F2 T2 L2 (T2 D2 F2 T2) L2 T2 B2
p134 2 CRN, 2 X, 2 ARCH L2 F2 D2 B2 T2 F2 T2 L2 (T2 D2 F2 T2) L2 T2 B2
p136 2 H, 2 ARM, 2 CRN R2 F2 L2 T2 B2 L2 T2 L2 (T2 D2 F2 T2) F2 L2 B2
5 of the 16 known antipodes are within 4 and 2 face turns (or 2 and 1
slice turns) of each other:
p66 + L2 R2 T2 D2 = p80 (allowing for whole cube rotations)
p66 + F2 B2 = p100
p80 + T2 D2 = p99
P66 + T2 D2 = P128
Using full group moves these antipodes can be reduced to:
P66a alternate method F2 R2 U2 F2 R2 U3 D3 B2 L2 F2 B2 U1 D1
p67a alternate method F2 R2 F2 U3 D3 L2 B2 D2 L2 B2 U1 D1 B2
p80a alternate method U1 F2 R2 L2 U2 D2 F2 U2 D3
p99a alternate method U1 R2 F2 B2 U2 D2 R2 D1
P100a alternate method F2 U2 D2 F2 R2 L2 D1 F2 R2 L2 B2 U1
p108a alternate method R2 F2 B2 L2 D1 R2 U2 R2 L2 U2 R2 D1
p130a alternate method F2 R2 F2 B2 U1 D1 F2 R2 D2 F2 L2 U3 D3
p133a alternate method R2 U1 F2 R2 L2 U2 D2 F2 U2 D3 R2
Both p80a and p99a are surprisingly compact, p99a being a full 7
turns
less than it's square's group equivalent. Note that in p99a a square's
group sequence is sandwiched between 2 turns on opposite faces. It is
the final turn D1 which brings it back into a sq group state! In general
U1 (sq group sequence) D1 does not lead to a sq group sequence.
Another interesting discovery was comparing the full group sequences:
L1 R1 D2 L3 R3 (antislice, 5 moves)
L1 R3 D2 L3 R1 (slice , 5 moves)
F1 B1 D2 F1 B1 (clockwise, 5 moves)
... to their square's group equivalents:
R2 F2 T2 L2 B2 L2 T2 R2 F2 (9 moves)
R2 B2 L2 D2 R2 F2 L2 T2 F2 (9 moves)
R2 B2 T2 F2 L2 F2 T2 R2 F2 (9 moves)
Also it was found possible to permute 3 edges only using:
L2 T2 R2 B2 R2 T2 L2 F2 (8 moves)
or L3 R1 F2 L1 R3 D2 (6 moves)
In general any sequence L1 R1 (any squares group moves) L3 R3
will always result in a squares group position, for example:
L1 R1 (D2 F2 B2) R3 L3
F1 B1 (T2 B2 F2 L2) F3 B3
p66a (F2 R2 U2 F2 R2) U3 D3 (B2 L2 F2 B2) U1 D1 (13 moves)
The longest irreducible square's group sequence discovered so far,
which is an embedded part of longest Phase 2 sequence (p94):
(Thus it can't be reduced by using full group moves using current
techniques)
R2 B2 U2 B2 L2 D2 L2 F2 (8)
Later on I discovered this irreducible sequence by chance:
T2 B2 T2 B2 D2 F2 R2 T2 L2 F2 (10)
Edges only (with corners in place) can be 14 moves at most, e.g.
D2 L2 F2 D2 F2 L2 T2 L2 T2 D2 F2 R2 F2 R2 (14)
This answers the question David Singmaster posed in
"Notes on Rubik's Magic Cube" on Thistlethwaite's last stage.
That is: "Are there any positions in the square's group with
corners fixed of length 14 or can they be done in less moves?"
A few observations...
 It is not possible to swap just 1 pair of edges and corners
 It is only possible to have 4, 6 or 8 corners out of place
 Known antipodal cases can be solved in <=13 moves using full
group
 In reaching an antipode one may start with any of the 6 turns
(since antipodes are global maxima, any turn will get you one
move closer)
 If the corners are fixed, the position is NOT an antipode
 Longest order appears to be 12
 All known (probably all!) antipodes have symmetry level 6 or 12
 Although only conjectural, it is now believed that one turn of
a face MUST lead to a new state which is either 1 move closer
or 1 move farther from START
Question: Are there any irreducible square's group sequences that
are longer then 10 moves? Are these truly irreducible
or only irreducible under Dik Winter's Kociemba
inspired program?
The full group beckons....
> Mark <
From hoey@aic.nrl.navy.mil Fri Aug 13 19:19:41 1993
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From: hoey@aic.nrl.navy.mil
MessageId: <9308132226.AA28300@sun13.aic.nrl.navy.mil>
To: Mark Longridge , cubelovers@life.ai.mit.edu
Subject: Re: Squares group
InReplyTo: <60.251051.104.0C180C07@canrem.com>
Mark Longridge has some interesting things
to say about the antipodes of the group generated by halfturns:
> If we define "symmetry level" as the number of distinct patterns
> generated by rotating the cube through it's 24 different
> orientations in space then most known antipodes are symmetry level
> 6. Thus the lower the number the higher the level of symmetry. The
> least symmetric positions have level 24, and this is very common.
This approach is somewhat unfortunate in two ways. First, it would be
better to use the full 48element symmetry group M of the cube,
because some patterns are not recognized as transformed images of each
other if you only use the 24element group C of rotations. For
instance, the positions reached by processes F2R2T2 and F2T2R2 cannot
be related with C, so you would see four classes of positions at
distance three rather than three. But the antipodes you give are all
mirrorsymmetric, so there is no new coalescence there.
Relating processes that are conjugates by a reflection is usually
somewhat tricky, since the moves of the process must be changed in
direction (replacing clockwise by counterclockwise) but in the squares
group this is a nonproblem.
The second deficiency of your approach is that you lose information by
specifying only the index of the symmetry subgroup (the ``number of
distinct patterns generated ...''). It makes sense to find out
exactly which subgroup of M is the symmetry group of your positions.
I've done that, below. Each of these symmetry groups comes in three
conjugates, so I've transformed some of the processes (marked x) so
they all use the same particular symmetry group(s). The group
elements are given as cycles of the cube faces, so (TD)(FRBL) means to
reflect T<>D and rotate F>R>B>L>F.
> Cases with symmetry level 6:
These are cases where the symmetry group has order 8.
> p66x Double 4 corner sw L2 D2 R2 T2 L2 T2 F2 R2 (F2 B2 T2 F2) T2 L2 B2
> p80 2 DOT, Invert T's R2 B2 D2 R2 B2 L2 B2 L2 (T2 D2 F2 T2) F2 L2 T2
> p99 2 DOT, 4 ARM R2 B2 D2 L2 B2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2
> p100 2 Cross, 4 ARCH 1 R2 B2 T2 R2 F2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2
p66x, p80, p99, and p100 have symmetry group P=<(TD)(FRBL),(FB),(LR)>.
> p67x Antipode 2 R2 D2 B2 T2 B2 T2 F2 L2 (F2 B2 T2 F2) L2 F2 D2
> p130 2 Cross, 4 ARCH 2 L2 B2 D2 B2 L2 D2 F2 L2 (T2 D2 F2 L2) F2 L2 T2
p67x and p130 have symmetry group Q=<(TD),(FRBL)>.
> p135x 2 X, 4 T D2 B2 L2 F2 R2 F2 R2 D2 (R2 L2 F2 R2) D2 L2 F2
> p137 2 X, 4 ARM L2 F2 T2 B2 T2 F2 T2 L2 (T2 D2 F2 T2) L2 D2 F2
p135x and p137 have symmetry group S=<(TD),(FB)(LR),(FR)(BL)>.
> Cases with symmetry level 12:
These have 4element symmetry groups.
> p108 2 DOT, 2 T, 2 ARM L2 F2 T2 R2 B2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2
> p128x 2 H, 2 T, 2 CRN L2 D2 R2 T2 L2 T2 F2 R2 (F2 B2 T2 F2) T2 L2 F2
> p129x 2 H, 2 T, 2 ARCH R2 T2 L2 T2 L2 T2 F2 R2 (F2 B2 T2 F2) T2 L2 F2
> p131x 2 H, 2 ARM, 2 ARCH L2 T2 R2 B2 D2 L2 B2 L2 (F2 B2 T2 F2) T2 L2 T2
> p132 2 Cross,2 ARCH,2CRN L2 F2 D2 R2 F2 L2 B2 L2 (T2 D2 F2 T2) F2 L2 D2
> p136x 2 H, 2 ARM, 2 CRN R2 T2 L2 F2 D2 L2 F2 L2 (F2 D2 T2 F2) T2 L2 D2
p108, p128x, p129x, p131x, p132, and p136x have symmetry group
HP=<(FB),(LR)>.
> p133x 2 Cross, 2 T, 2 ARM L2 T2 B2 T2 B2 T2 F2 L2 (F2 B2 T2 F2) L2 F2 D2
> p134x 2 CRN, 2 X, 2 ARCH L2 T2 B2 D2 F2 T2 F2 L2 (F2 B2 T2 F2) L2 F2 D2
p133x and p134x have symmetry group HS=<(TD),(FB)(LR)>.
In case you have trouble forming the closure of these groups:
P = {I, (FB)(LR), (TD)(FRBL), (TD)(FLBR),
(FB), (LR),
(TD)(FR)(BL), (TD)(FL)(BR)}
Q = {I, (FB)(LR), (TD), (TD)(FB)(LR),
(TD)(FRBL), (TD)(FLBR),
(FLBR), (FRBL)}
S = {I, (FB)(LR), (TD), (TD)(FB)(LR),
(TD)(FR)(BL), (TD)(FL)(BR),
(FR)(BL), (FL)(BR)}
HP = {I, (FB)(LR), (FB), (LR)}
HS = {I, (FB)(LR), (TD), (TD)(FB)(LR)}.
I should note that the subgroup names M, C, P, Q, S, HP, and HS are
part of a general classification of subgroups of M that I worked out
some time ago. I have a chart of them I can send; just ask by email.
> A few observations...
>  It is not possible to swap just 1 pair of edges and corners
Certainly, all the generators are even permutations on the edges and
on the corners.
>  It is only possible to have 4, 6 or 8 corners out of place
That is a nice, concise way of putting it. To elaborate, if you
permute one of the corner orbits in a 3cycle, the other will also be
permuted in a 3cycle; otherwise, any pair of cycle structures of the
same parity is possible.
>  In reaching an antipode one may start with any of the 6 turns
> (since antipodes are global maxima, any turn will get you one move
> closer)
Careful! This also relies on the fact you call a conjecture, below.
Otherwise you could have two neighboring global maxima, and their
inverses would be antipodes that do not have this property. For
instance, consider the corner group as generated by the 24 pairs of
neighboring squares (F2R2, etc). This is a 48element group with
diameter 2, trivial enough to be analyzed by hand. Antipodes
(L2B2)(D2R2) and (D2R2)(T2F2) are neighbors, because
(L2B2)(D2R2)(F2T2)=(D2R2)(T2F2). So there is no length2 process
equivalent to (F2T2)(R2D2) that starts with T2F2.
>  If the corners are fixed, the position is NOT an antipode
>  All known (probably all!) antipodes have symmetry level 6 or 12
I presume these comments are left over from before you found them all.
>  Longest order appears to be 12
Appears? The orbits are all of size 4 (two orbits of corners, three
orbits of edges), so 12=LCM(2,3,4) is an easy upper bound. Finding
one is easy given the processes Singmaster lists.
>  Although only conjectural, it is now believed that one turn of a
> face MUST lead to a new state which is either 1 move closer or 1
> move farther from START
Conjectural? It's immediate from the fact that each generator is an
odd permutation of the corner orbit {FTR,FDL,BTL,BDR}.
> Question: Are there any irreducible square's group sequences that
> are longer then 10 moves? Are these truly irreducible or only
> irreducible under Dik Winter's Kociemba inspired program?
Well, that could be searched for; a matter of checking 600K positions
for each of the 15K or so pattern representatives. I hope I can find
the time to hack it up.
Dan Hoey
Hoey@AIC.NRL.Navy.Mil
From hoey@aic.nrl.navy.mil Mon Aug 16 20:20:25 1993
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From: hoey@aic.nrl.navy.mil
MessageId: <9308162205.AA12648@sun13.aic.nrl.navy.mil>
To: Mark Longridge , cubelovers@life.ai.mit.edu
Subject: Squares group, correction
I should proofread these things better. I got the processes for p130,
p135x, p137, and p136x wrong in my last message. Here is the
corrected list of squaresgroup antipodes and their symmetry groups.
SG Pos Name Process
P p66x Double 4 corner sw L2 D2 R2 T2 L2 T2 F2 R2 (F2 B2 T2 F2) T2 L2 B2
P p80 2 DOT, Invert T's R2 B2 D2 R2 B2 L2 B2 L2 (T2 D2 F2 T2) F2 L2 T2
P p99 2 DOT, 4 ARM R2 B2 D2 L2 B2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2
P p100 2 Cross, 4 ARCH 1 R2 B2 T2 R2 F2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2
Q p67x Antipode 2 R2 D2 B2 T2 B2 T2 F2 L2 (F2 B2 T2 F2) L2 F2 D2
Q p130x 2 Cross, 4 ARCH 2 T2 B2 R2 B2 T2 R2 F2 T2 (L2 R2 F2 T2) F2 T2 L2
S p135x 2 X, 4 T L2 D2 B2 T2 F2 T2 F2 L2 (F2 B2 T2 F2) L2 B2 T2
S p137x 2 X, 4 ARM L2 T2 F2 D2 F2 T2 F2 L2 (F2 B2 T2 F2) L2 B2 T2
HP p108 2 DOT, 2 T, 2 ARM L2 F2 T2 R2 B2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2
HP p128x 2 H, 2 T, 2 CRN L2 D2 R2 T2 L2 T2 F2 R2 (F2 B2 T2 F2) T2 L2 F2
HP p129x 2 H, 2 T, 2 ARCH R2 T2 L2 T2 L2 T2 F2 R2 (F2 B2 T2 F2) T2 L2 F2
HP p131x 2 H, 2 ARM, 2 ARCH L2 T2 R2 B2 D2 L2 B2 L2 (F2 B2 T2 F2) T2 L2 T2
HP p132 2 Cross,2 ARCH,2CRN L2 F2 D2 R2 F2 L2 B2 L2 (T2 D2 F2 T2) F2 L2 D2
HP p136x 2 H, 2 ARM, 2 CRN R2 T2 L2 F2 D2 L2 F2 L2 (F2 B2 T2 F2) T2 L2 D2
HS p133x 2 Cross, 2 T, 2 ARM L2 T2 B2 T2 B2 T2 F2 L2 (F2 B2 T2 F2) L2 F2 D2
HS p134x 2 CRN, 2 X, 2 ARCH L2 T2 B2 D2 F2 T2 F2 L2 (F2 B2 T2 F2) L2 F2 D2
Sorry if anyone was led astray.
Dan
From acw@bronze.lcs.mit.edu Thu Aug 19 15:06:28 1993
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Date: Thu, 19 Aug 93 15:04:18 EDT
From: acw@bronze.lcs.mit.edu (Allan C. Wechsler)
MessageId: <9308191904.AA27266@bronze.lcs.mit.edu>
To: ronnie@cisco.com
Cc: Dik.Winter@cwi.nl, cubelovers@life.ai.mit.edu
InReplyTo: "Ronnie B. Kon"'s message of Thu, 05 Aug 1993 16:55:36 0700 <199308052355.AA23583@lager.cisco.com>
Subject: Diameter of cube group?
Date: Thu, 05 Aug 1993 16:55:36 0700
From: "Ronnie B. Kon"
Disclaimer: this sounds more authoritative than is intendedI really
don't know what I'm talking about.
Don't worry. Mathematical reasoning stands on its own merits.
It couldn't be very pointy. From the most distant configuration,
there are 6 positions immediately before it. There are 6^2 two steps
away, 6^3 three steps, etc. (well, 6^2  1 and 6^3  ?) actually.
Very good. This is a necessary insight, regardless of the exact
numerical details. (For example, you mean 12, not 6.) But the
possible flaw is that there might be more than one maximally distant
state; if their sets of neighbors overlap viciously enough, this
effect could make the tail pointier. You can make valence12 graphs
(not of groups, just arbitrary graphs) that have fairly bumpy
distancevspopulation functions. Any argument that rigorously
constrains N(d) must somehow appeal to the fact that the cube graph is
a Cayley graph, that is, the graph of a group.
[...]
This gives me the feeling that Monte Carlo is fairly valid. (How's
that for rigor?)
It's a start. But we have to use groupness somehow.
From @mitvma.mit.edu:DWR2560@TAMZEUS.BITNET Fri Aug 20 09:47:26 1993
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Date: Fri, 20 Aug 93 05:56 CST
From:
Subject: pointy tails
To: cubelovers@life.ai.mit.edu
XOriginalTo: cubelovers@life.ai.mit.edu, DWR2560
Allan C. Wechsler writes:
> It couldn't be very pointy. From the most distant configuration,
> there are 6 positions immediately before it. There are 6^2 two steps
> away, 6^3 three steps, etc. (well, 6^2  1 and 6^3  ?) actually.
>
>Very good. This is a necessary insight, regardless of the exact
>numerical details. (For example, you mean 12, not 6.) But the
>possible flaw is that there might be more than one maximally distant
>state; if their sets of neighbors overlap viciously enough, this
>effect could make the tail pointier. You can make valence12 graphs
[deletia]
All this misses the point (so to speak) which is that 12^N is _exceedingly_
pointy for our purposes. If one samples only 1000 positions out of ~10E19,
then one could very well miss a 12^N tail of length 14 moves!
The estimate of 22 as an upper limit relies on the intuition that
the distribution is MUCH blunter than this.
Dave Ring
dwr2560@zeus.tamu.edu
From reid@math.berkeley.edu Mon Aug 23 04:10:58 1993
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From: reid@math.berkeley.edu (michael reid)
MessageId: <9308230810.AA19047@math.berkeley.edu>
To: cubelovers@life.ai.mit.edu
Subject: Re: Diameter of cube group?
> Continuing and waiting for a config that requires 21 turns, dik
here's a pattern to try:
first do 6 checkerboards of order 2 (F2 B2 R2 L2 U2 D2) and then do
superfliptwist. in other words, the group product of these two elements.
From dik@cwi.nl Tue Aug 24 20:43:14 1993
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Date: Wed, 25 Aug 93 02:42:58 +0200
From: Dik.Winter@cwi.nl
MessageId: <9308250042.AA11725.dik@boring.cwi.nl>
To: cubelovers@life.ai.mit.edu, reid@math.berkeley.edu
Subject: Re: Diameter of cube group?
> > Continuing and waiting for a config that requires 21 turns, dik
> here's a pattern to try:
> first do 6 checkerboards of order 2 (F2 B2 R2 L2 U2 D2) and then do
> superfliptwist. in other words, the group product of these two elements.
As they commute I did it the other way around. But I am highly
suspicious that you tried it yourself. 10 minutes and only down
to 22 turns. But continuing, possibly for weeks/months.
On another machine I am trying to prove that 20 is minimal for
superfliptwist. 90 hours gone, still nothing. Most of the time
is not spend with phase 1 set to 16 turns. Phase 1 to 13 got it
doen to 20. Nothing new with phase 1 to 14 or 15. 16 turns in
phase 1 allows at most 3 turns in phase 2. The latter can be time
consuming. I do not know in how many cases actually something is
done in phase 2. When I get to 17 turns in phase 1, I suspect in
most cases in phase 2 it is immediately clear that it can not be
solved. But I am patient.
dik
From dik@cwi.nl Wed Aug 25 16:00:57 1993
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Date: Wed, 25 Aug 93 22:00:22 +0200
From: Dik.Winter@cwi.nl
MessageId: <9308252000.AA14852.dik@boring.cwi.nl>
To: cubelovers@life.ai.mit.edu, reid@math.berkeley.edu
Subject: Re: Diameter of cube group?
> here's a pattern to try:
> first do 6 checkerboards of order 2 (F2 B2 R2 L2 U2 D2) and then do
> superfliptwist. in other words, the group product of these two elements.
Was certainly one of the hardest to do. After 17 hours the best was 22
turns, but then results came in, after 18 hours 21 turns, and finally
after 19 hours 20 turns:
F1 R1 L2 U3 R2 L3 U3 D2 R2 F1 D1 B1 D1 F2 U3 R3 D3 F2 D2 L2
This on an SGI R4K Indigo.
dik

dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland
home: bovenover 215, 1025 jn amsterdam, nederland; email: dik@cwi.nl
From dik@cwi.nl Wed Aug 25 16:10:23 1993
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Date: Wed, 25 Aug 93 22:10:06 +0200
From: Dik.Winter@cwi.nl
MessageId: <9308252010.AA14880.dik@boring.cwi.nl>
To: cubelovers@life.ai.mit.edu
Subject: CFF 32 table of contents
Last Sunday (on Cube Day) I was handed issue #32 of Cubism For Fun.
A summary the contents:
1. Short articles about the solution of the puzzle by Koos en Ton Verhoeff.
(I described the puzzle before and announced the solution by Jan de
Ruiter a few weeks ago.)
2. Articles about "Bob's Binary Boxes" by Hans Dockhorn and Bob Kootstra. *
3. Description by Harold Cataquet of "Alice"; a wooden packing puzzle.
4. Description by Wim Zwaan of a packing puzzle he entered in the "Hikimi
Wooden Puzzle Competition".
5. Article by Jan Verbakel about "WirrelWarrel" puzzles (it has a
different name in the US that escapes me).
6. Article by Tom Hilligers about "Kaos", a puzzle with balls in pipes.
The orientation of the pipes with respect to each other can change.
7. Article by Ronald Fletterman about pretty "sculptures" with Square 1.
8. A contest announcement by Bernard Wiezorke figuring the sliding puzzle
Vorsicht! (I do not know whether it is available in the US.)
9. An article by Ralph Gasser about Orbik, a puzzle introduced by
Edward Hordern.
10. Results of a number of contests.
* An interesting design. These are wooden boxes with in it binary
switches. On top a ball can be put in, on the bottom there are a
number of exits. When a ball reaches a switch it passes the
switch in the given direction and puts the switch in opposite
direction. The design is such that successive balls come out in
successive exits (in circular numerical order). Bob Kootstra built
a few of those switches, but now is asking for an optimal design of
a box with 7 exits.
CFF is a newsletter published by the Nederlandse Kubus Club NKC (Dutch
Cubists Club). It appears a bit irregular, but a few times a year.
Yearly membership fee is now NLG 25. (Dutch Guilders) which amounts to
approximately $ 15.. Institutional membership is also possible.
Information is available from the editor:
Gerald Maurice
Groen van Prinstererstraat 72
1051 ED Amsterdam
The Netherlands
Phone: +31206822943
Email: gm@phys.uva.nl
From alan@parsley.lcs.mit.edu Wed Aug 25 21:13:44 1993
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From: Alan Bawden
Sender: Alan@lcs.mit.edu
To: CubeLovers@ai.mit.edu
Subject: Tools lost in the mists of time...
Despite being your moderator for the past 13 years, it has been quite a
while since I actually played with a cube (of any order). Until last
weekend, that is, when I picked up a 5x5x5 cube that someone had loaned me
quite some time ago. Imagine my surprise to discover that I had in fact
forgotten one of the tools I needed to solve even a 3x3x3! In particular
my tool for inverting two edge cubies in place in a 3x3x3 was completely
gone. It didn't take me long to develop a replacement, but I'm certain
that it is nothing like what I was using years ago. I'm amazed  there
was a time when I thought I could never forget any of those tools.
And by the way, it was a lot of fun to relearn the 3x3x3 and then go on to
solve the 5x5x5  if anyone else out there has, like me, been neglecting
their cube hacking for some years, go pick up your cube again, you may be
pleasantly surprised.
 Alan (aka CubeLoversRequest)

Alan Bawden Alan@LCS.MIT.EDU
MIT Room NE43538 (617) 2537328
545 Technology Square
Cambridge, MA 02139 06BF9EB8FC4CFC24DC75BDAE3BB25C4B
From hoey@aic.nrl.navy.mil Thu Aug 26 10:31:05 1993
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Date: Thu, 26 Aug 93 10:30:57 EDT
From: hoey@aic.nrl.navy.mil (Dan Hoey)
MessageId: <9308261430.AA21721@Sun0.AIC.NRL.Navy.Mil>
To: Alan@lcs.mit.edu, cubelovers@ai.mit.edu
Subject: Tartan reborn (Re: Tools lost in the mists of time...)
InReplyTo: <25Aug1993.204955.Alan@LCS.MIT.EDU>
Organization: Navy Center for Applied Research in AI
Alan Bawden mentioned the joy of rediscovering his lost cubesolving
techniques. This happened to me about three years ago for an unusual
reason. I've become active in science fiction fandom, and fans
determine where the World Science Fiction Convention (Worldcon) is
held each year by running miniature political campaigns. A friend of
mine was bidding for Glasgow, and she asked if I had any `plaid
things'. I told her I had a plaid Rubik's cube, and a political
strategy was born. The plaid cube is of course the Tartan, which Jim
Saxe and I discovered and described in this group on 16 February 1981
(see archives). I blanked some old cubes, and figured out how to use
spray paint to efficiently create Tartan cubes. I produced a half
dozen or so, and they make good conversation pieces at conventions.
Unfortunately, I seem to be the only conventiongoing science fiction
fan who can *solve* a Tartan (with the possible exception of Phil
Servita who as I recall figured out an
effective method but wearied in its execution). So I would see a
scrambled Tartan at a convention party, and fix it, and put it down,
and five minutes later it would be scrambled again. I quickly found
out how rusty I was, and through the enforced practice I've gotten
about as good as I was a decade ago. But some of the Glasgow
promoters took Tartan cubes over to the UK, and those cubes just never
get solved. I sent them instructions for solving it, but I don't know
if any of them have figured out the instructions.
Well, eventually they told me they really wanted something mere
mortals could deal with, and I painted some pieces of wood plaid that
they could use for doorstops. I was surprised, though, to find that
to make a plaid pattern going around a corner, if you only have four
colors of paint, it seems the *only* thing you can do is use a
coloring locally identical to the Tartan.
As for the cubes in the UK, I expect to get there in 1995. For it
seems the clever ploy worked, and the fans voted to have the 1995
Worldcon in Glasgow. I'm sure they owe it all to the Tartan. Sure.
Dan Hoey
Hoey@AIC.NRL.Navy.Mil
From reid@math.berkeley.edu Fri Aug 27 22:23:04 1993
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Date: Fri, 27 Aug 93 19:22:49 PDT
From: reid@math.berkeley.edu (michael reid)
MessageId: <9308280222.AA22114@math.berkeley.edu>
To: Dik.Winter@cwi.nl, cubelovers@life.ai.mit.edu
Subject: Re: Diameter of cube group?
dik winter says
> > first do 6 checkerboards of order 2 (F2 B2 R2 L2 U2 D2) and then do
> > superfliptwist. in other words, the group product of these two elements.
>
> As they commute I did it the other way around. But I am highly
> suspicious that you tried it yourself. 10 minutes and only down
> to 22 turns. But continuing, possibly for weeks/months.
ok, you caught me; i'd already tried this one myself. :)
but apparently i wasn't as patient as you. i just remember that it ran
for a long time without doing better than 22 face turns.
the point to be made here is the following: the length of time the
program takes for a given position depends significantly on how far it
must search in stage 1. this seems to make any claim about how long the
program takes to crunch an average position meaningless. my experience
is that it varies greatly depending upon the position. i think it would
be more informative to stratify this information. i.e., how long it
takes to search 12 moves in stage 1, and how short a solution is produced.
and then the same info for 13 turns, then 14, etc.
what i've been amazed by (and continue to be) is that searching only 13
or so moves in stage 1 is sufficient to produce very short solutions for
many positions.
something i'd thought about trying, but never got around to is trying
random positions created by twist sequences such as:
F1 R1 B1 L1 F1 R1 B1 L1 F1 R1 B1 L1 F1 R1 B1 L1 F1 R1 B1 L1
or some random string of about 20 quarter turns of the faces F,L,B,R.
a string of length 12 or 13 will be solved quickly (by the inverse of
the original string). however, for length 17 or so, the program won't
find the inverse of the original string until it is searching 17 moves
deep in stage 1. this suggests that perhaps these positions may be
harder for the program to handle. but are they harder than random
positions? i don't know.
mike
From dik@cwi.nl Sat Aug 28 20:17:32 1993
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From: Dik.Winter@cwi.nl
MessageId: <9308290017.AA01278.dik@boring.cwi.nl>
To: cubelovers@life.ai.mit.edu, reid@math.berkeley.edu
Subject: Re: Diameter of cube group?
> ok, you caught me; i'd already tried this one myself. :)
> but apparently i wasn't as patient as you. i just remember that it ran
> for a long time without doing better than 22 face turns.
So did it here. 22 in a few minutes, 20 in a lot of hours.
> the point to be made here is the following: the length of time the
> program takes for a given position depends significantly on how far it
> must search in stage 1.
This is right, and it appears (though I have not yet thoroughly verified)
that configurations that take a long time in stage 1 are a large distance
from start.
> this seems to make any claim about how long the
> program takes to crunch an average position meaningless.
Depends on how you interpret that claim. If the claim is that it turns
up with a sequence that is 20 turns or shorter you are right. The claim
might even be incorrect! The actual claim is that it takes a fairly short
time to give a fairly short sequence (where fairly short is deliberately
left unquantified). And this is true. For my set of >10000 random
positions the program came up with a sequence of 27 turns or less in
a short time indeed. (Actually the first solution found was 26 turns or
less for all but three configurations.) Bringing that down to 20 took in
a number of cases extremely long (in the order of one day). But that is
still far less than when we had done a normal single phase backtracking
process I think.
> i think it would
> be more informative to stratify this information. i.e., how long it
> takes to search 12 moves in stage 1, and how short a solution is produced.
> and then the same info for 13 turns, then 14, etc.
Some quantification is not so very difficult I think. Without treepruning
the time would be proportional to 18^n + 10^m for a nturn phase1 and a
mturn phase2 solution. The treepruning performed is (I think) proportional
to the number of turns in each phase; it will chop branches that are to
large according to predefined tables. Also there are some shortcuts that
make 18 not really 18 and 10 not really 10, but the reasoning remains the
same.
> what i've been amazed by (and continue to be) is that searching only 13
> or so moves in stage 1 is sufficient to produce very short solutions for
> many positions.
I do not think this is so very amazing. Each configuration can be brought in
12 turns or less to a configuration for phase 2. The proven diameter of the
group of phase 2 is 25, my estimate is 1921. So, based on my estimate a
worst case would be 12 turns required in phase 1 and 21 in phase 2 giving
33 turns in an estimated time of 18^12 + 10^21, this is less than 18^17,
and hence is found faster than if we had gone to 17 turns in phase 1.
Actually both 12 and 21 are rare; most cases do phase 1 in 10 turns or less
and phase 2 in 15 turns or less.
> something i'd thought about trying, but never got around to is trying
> random positions created by twist sequences such as:
> F1 R1 B1 L1 F1 R1 B1 L1 F1 R1 B1 L1 F1 R1 B1 L1 F1 R1 B1 L1
> or some random string of about 20 quarter turns of the faces F,L,B,R.
> a string of length 12 or 13 will be solved quickly (by the inverse of
> the original string). however, for length 17 or so, the program won't
> find the inverse of the original string until it is searching 17 moves
> deep in stage 1. this suggests that perhaps these positions may be
> harder for the program to handle. but are they harder than random
> positions? i don't know.
I do not know, but I think not. Yes, asking the program to find the
reverse of the string takes a long time. Asking the program to find
an inverse of the sequence takes much less time (although the inverse
found may both be shorter or longer than the original). I just tried,
and after initialization it found a 10+14 turn solution in 20 seconds,
going down to 11+10 after less than a minute. Getting this down to 20
might of course take considerable time (if the original sequence is
minimal etc.).
But I have not the time right now to check. I am busy trying to prove
that 20 is minimal for superfliptwist. I have already found that there
is no 19 turn solution with 16 turns in phase 1. That took about 48
hours (distributed over 6 machines). Now I am doing the same for 17
turns in phase 1; which wil obviously take much longer. (And yes, I
took the precaution of allowing as the first turn only F, F2, R, R2,
U, U2 in phase 1. When I go to 19 turns in phase 1, I can skip 18,
I need only F, F2, R and R2, I think.)
dik

dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland
home: bovenover 215, 1025 jn amsterdam, nederland; email: dik@cwi.nl
From reid@math.berkeley.edu Sun Aug 29 04:26:41 1993
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Date: Sun, 29 Aug 93 01:26:31 PDT
From: reid@math.berkeley.edu (michael reid)
MessageId: <9308290826.AA05585@math.berkeley.edu>
To: Dik.Winter@cwi.nl, cubelovers@life.ai.mit.edu
Subject: Re: Diameter of cube group?
> But I have not the time right now to check. I am busy trying to prove
> that 20 is minimal for superfliptwist. I have already found that there
> is no 19 turn solution with 16 turns in phase 1. That took about 48
> hours (distributed over 6 machines). Now I am doing the same for 17
> turns in phase 1; which wil obviously take much longer. (And yes, I
> took the precaution of allowing as the first turn only F, F2, R, R2,
> U, U2 in phase 1. When I go to 19 turns in phase 1, I can skip 18,
> I need only F, F2, R and R2, I think.)
in fact, you can eliminate the possibility of starting with F2, R2 or U2,
since these each commute with superfliptwist, and may be done in stage 2.
in other words, if F2 sequence = superfliptwist, then also
sequence F2 = superfliptwist.
also, you need not consider 19 turns in stage 1. by symmetry, you may
suppose the last face turned is U, which is done in stage 2.
if you use the fact that U and D commute, L and R commute and F and B
commute, then the number of sequences of length n in stage 1 grows
exponentially, with ratio approximately 13.35. if the runtime is
proportional to the number of sequences tested in stage 1, (which
may or may not be the case) that would mean testing 18 turns deep
would take approximately 178.18 times as long. (eliminating the
possibility of starting with F2, R2 or U2 would cut that in half.)
here's something you may have already considered. if your prune tables
in stage 1 consider only pairs (flip, twist), (flip, location) and
(twist, location), some search paths may be pruned 8 turns early.
(each of these pairs had positions 9 twists from start.) at the
expense of a lot more memory, you can do some pruning 11 turns early,
by storing tables for triples (flip, twist, location). you'd probably
have to store these tables in very compressed form, and divide out by
symmetries of the cube that preserve the UD axis. it may turn out
that the overhead of processing this compressed information does not
adequately compensate for the improved foresight, but it's worth
considering.
it would be excellent if you could show that 20 face turns is minimal
for superfliptwist! even finding a shorter solution would be great!
mike
From dik@cwi.nl Mon Aug 30 21:21:35 1993
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From: Dik.Winter@cwi.nl
MessageId: <9308310121.AA08781.dik@boring.cwi.nl>
To: cubelovers@life.ai.mit.edu, reid@math.berkeley.edu
Subject: Re: Diameter of cube group?
> in fact, you can eliminate the possibility of starting with F2, R2 or U2,
> since these each commute with superfliptwist, and may be done in stage 2.
> in other words, if F2 sequence = superfliptwist, then also
> sequence F2 = superfliptwist.
Right. I had not considered this in the program (it is still fairly
general), but it only does mean early termination.
> also, you need not consider 19 turns in stage 1. by symmetry, you may
> suppose the last face turned is U, which is done in stage 2.
But I have now had different thoughts. Currently phase 1 checks in 3
dimensional space. When a solution is found the program calculates the
current position for phase two after which it finds a solution in a
different 3 dimensional space. (I just though how I might speed up the
calculations to get to the starting position for the second phase,
but will not yet elaborate on that; I will first try it out.) But this
does not help finding whether there are solutions of 19 turns or less.
What I am now considering is to have a phase 1 program only, where phase
1 is done in an additional dimension: the permutation of the corner cubes.
So to prove the nonexistence of a solution of 19 turns or less, this
program would seek for a phase 1 solution in 4 dimensional space of at
most 19 turns and next check whether this also solves the edge cubes.
This would eliminate quite a few dead alleys where the current phase 1
finds a solution and has still things to do.
> if you use the fact that U and D commute, L and R commute and F and B
> commute, then the number of sequences of length n in stage 1 grows
> exponentially, with ratio approximately 13.35. if the runtime is
> proportional to the number of sequences tested in stage 1, (which
> may or may not be the case) that would mean testing 18 turns deep
> would take approximately 178.18 times as long. (eliminating the
> possibility of starting with F2, R2 or U2 would cut that in half.)
If I use a single phase algorithm, I can eliminate much more! What I
see for runtime is not entirely proportional. When looking at the
number of configurations done in phase 2, this goes up by factors that
start in the neighbourhood of 30 and diminish to (probably) ultimately
the factor you mention. This indicates that tree pruning is much more
effective with fewer turns in phase 1.
> here's something you may have already considered. if your prune tables
> in stage 1 consider only pairs (flip, twist), (flip, location) and
> (twist, location), some search paths may be pruned 8 turns early.
> (each of these pairs had positions 9 twists from start.) at the
> expense of a lot more memory, you can do some pruning 11 turns early,
> by storing tables for triples (flip, twist, location). you'd probably
> have to store these tables in very compressed form, and divide out by
> symmetries of the cube that preserve the UD axis. it may turn out
> that the overhead of processing this compressed information does not
> adequately compensate for the improved foresight, but it's worth
> considering.
I think the overhead is much to large computationwise and memorywise.
The size of the table would be uncompressed 2217093120 integers in the
range from 0 to 12. Factoring out symmetries would reduce it by a factor
of about 32 (slightly less). [4 for rotational symmetry, 4 for mirroring
both UD and FB, 2 for inversion.] Using 3.5 bits per configuration
this means > 30 MByte. The machines I am using currently are not able
to handle that amount of information. But it is feasable. If we skip
inversion (which is most difficult to do) we are at > 60 MByte. The
problem remains to adequately number the remaining positions from 1 to
max. Some configurations are inert with respect to the rotations and/or
mirroring. On the other hand, we need the table in core (do not try
to do this through disk access!). Some insightful thoughts are needed
here.
> it would be excellent if you could show that 20 face turns is minimal
> for superfliptwist! even finding a shorter solution would be great!
I agree to that! I have a number of machines, still going strong.
From ROSEJM58@snyoneva.cc.oneonta.edu Tue Sep 21 14:33:15 1993
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From raymond@cps.msu.edu Fri Oct 1 12:37:21 1993
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From: raymond@cps.msu.edu
MessageId: <9310011637.AA28543@pacific>
To: cubelovers@ai.mit.edu
Subject: Seeking 5x5x5 cube
ContentLength: 115
Hello cube lovers,
I am looking for a 5x5x5 cube. Does anybody know where I can
get one?
Thanks,
Carl Raymond
From queiroz@eepost.uta.edu Fri Oct 1 16:25:59 1993
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Date: Fri, 1 Oct 1993 15:15:49 0500 (CDT)
From: Ricardoh Queiroz
Subject: Re: Seeking 5x5x5 cube
To: raymond@cps.msu.edu
Cc: cubelovers@ai.mit.edu
InReplyTo: <9310011637.AA28543@pacific>
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On Fri, 1 Oct 1993 raymond@cps.msu.edu wrote:
> Hello cube lovers,
>
> I am looking for a 5x5x5 cube. Does anybody know where I can
> get one?
>
> Thanks,
> Carl Raymond
Hi,
I also have interest in a regular 3x3x3 and I can't find it.
If anyone has any idea, please let us know.
Thanks,
Ricardo
queiroz@eepost.uta.edu
From raymond@cps.msu.edu Fri Oct 1 18:06:21 1993
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From: raymond@cps.msu.edu
MessageId: <9310012206.AA03566@pacific>
To: queiroz@eepost.uta.edu, raymond@cps.msu.edu
Subject: Re: Seeking 5x5x5 cube
Cc: cubelovers@ai.mit.edu
ContentLength: 456
During the Christmas toy season last year, a local supermarket/department
store chain (Meijer in Michigan) had 3x3x3 cubes. I don't recall the
manufaturer, but they used the "Rubik's Cube" brand name. They also
had a picture on the center cubies on each face that had to be
correctly rotated for a "proper" solution. I can't recall the price,
but it was reasonable. Maybe they will be easier to find as Christmas
gets closer.
Good luck,
Carl Raymond
From ncramer@bbn.com Mon Oct 4 08:51:15 1993
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Date: Mon, 4 Oct 93 8:40:53 EDT
From: Nichael Cramer
To: Ricardoh Queiroz
Cc: raymond@cps.msu.edu, cubelovers@ai.mit.edu
Subject: Re: Seeking 5x5x5 cube
>Date: Fri, 1 Oct 1993 15:15:49 0500 (CDT)
>From: Ricardoh Queiroz
>Subject: Re: Seeking 5x5x5 cube
>
>Hi,
>I also have interest in a regular 3x3x3 and I can't find it.
>If anyone has any idea, please let us know.
>Thanks,
>Ricardo
>queiroz@eepost.uta.edu
Games People Play in Harvard Square had a number of 3X3X3 that go by the
name "Fourth Dimension" or something like that (they various have logos and
a picture that looks like a profile of Rubik on four of the center faces).
It was something on the order of $10.
It also seemed more cheaply made than my other, older cubes. It felt,
well, "lighter" in my hand and is rather more difficult to turn than I'm
used to. It's not that they're unusable, and it's just not that they're
just stiff, rather that they seem to be slightly out of alignment. Perhaps
if you are someone who knows how to fine tune these things...
N
From ronnie@cisco.com Mon Oct 4 11:28:32 1993
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To: cubelovers@ai.mit.edu
Subject: Re: Seeking 5x5x5 cube
InReplyTo: Your message of "Mon, 04 Oct 1993 08:40:53 EDT."
<9310041251.AA19338@life.ai.mit.edu>
Date: Mon, 04 Oct 1993 08:28:28 0700
From: "Ronnie B. Kon"
Try mailing to Peter Beck (pbeck@pica.army.mil).
Ronnie
From @mizzou1.missouri.edu:HOWSER@LUA6.LU.EDU Wed Oct 6 01:40:53 1993
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Date: 04 OCT 93 17:32
From:
To:
Subject: Stiff and/or misaligned cubes
Comments: Automatic Return Receipt Requested
MessageId:
Back in the 'good old days' when cubing was very popular, I had a cube that was
very prone to hanging up when you turned it in certain directions. I solved the
problem by disassembling the cube and working on the cublets individually to
remove any excess plastic and to smooth any rough spots by scraping with a razor
blade and/or sanding with model car sandpaper. I raced many people with that
cube and still have it after all these years. I found that the time I spent
working on the bad cublets has lead to that cube wearing much more evenly than
the ones I have that I never got around to working on. I also find that it gets
more consistant in its movements as time goes by. As it was one of the first
cubes on the market (before the BIG craze, actually) it is rather heavy but not
as precisely made as the later cubes.

Gerry Howser
INTERNET: howser@lua6.lu.edu
Postmaster@lua6.lul.edu
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From dml@hpfrcu03.france.hp.com Thu Oct 7 16:48:50 1993
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From: Patrick DEMICHEL
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Subject: help
To: CUBELOVERS@life.ai.mit.edu
Date: Thu, 7 Oct 93 17:59:18 MET
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Mailer: Elm [revision: 72.14]
help
From diamond@jit081.enet.dec.com Thu Oct 7 21:54:26 1993
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From: 08Oct1993 1054
To: "dml@hpfrcu03.france.hp.com"@jrdmax.enet.dec.com
Cc: cubelovers@life.ai.mit.edu
ApparentlyTo: cubelovers@life.ai.mit.edu
Subject: RE: help
dml@hpfrcu03.france.hp.com (Patrick DEMICHEL) writes:
>help
First you turn one side, then you turn another side. Keep it up, and
soon you'll be done.
 Norman Diamond diamond@jit081.enet.dec.com
[Digital did not write this.]
From pbeck@pica.army.mil Fri Oct 8 08:05:19 1993
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Date: Fri, 8 Oct 93 7:52:20 EDT
From: Peter Beck (BATDD)
To: cubelovers@life.ai.mit.edu
Cc: pbeck@pica.army.mil
Subject: golden solids
MessageId: <9310080752.aa13373@COR6.PICA.ARMY.MIL>
i am trying to find a copy of the following:
THE AESTHETICS OF THE SACRED, A HARMONIC GEOMETRY OF CONSCIOUSNESS &
PHILOSOPHY OF SACRED ARCHITECTURE
by ROBERT C MEURANT
THE OPOUTERE PRESS
BOULDER & AUCKLAND
ISBN 0908809026
if you know of a bookseller who might carry this
please let me know.
if you have the address of opoutere press in NZ
that would also be helpful.
thanks
From @mail.uunet.ca:mark.longridge@canrem.com Thu Oct 28 19:41:34 1993
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To: cubelovers@life.ai.mit.edu
Subject: Cube Patterns
From: mark.longridge@canrem.com (Mark Longridge)
MessageId: <60.305920.104.0C188656@canrem.com>
Date: Thu, 28 Oct 1993 19:20:00 0400
Organization: CRS Online (Toronto, Ontario)
Comments on Rubik's Cube Patterns

First some positions of theoretical interest:
(F R B L)^5 = F1 L3 D2 F3 B2 R1 L3 F2 B3 R2 B1 U2 D2 R3 D2 L2
B2 L2 F2 (19 moves)
So in the ht metric this is compressible.
I've been thinking about new approaches to finding new patterns.
To improve on the "oldfashioned" method of simply taking a cube
and twisting it I wrote a module to test for legality of position
and another module for arrangement entry. Thus I can doodle around with
a cube pattern much more efficiently.
This approach led to the discovery of the ML Checkerboard, which is
to date the most involved of the pretty patterns:
ML's Checkerboard = B1 U2 R1 L1 D2 B3 L2 F2 R1 F3 U3 D3 F3 B3 R2 U1
R2 D3 L2 (19 moves)
Also by combining the 8 twist and the first discovered square's
group antipode, a new corner's only pattern:
Antwist = R1 F2 B2 D2 R1 L3 B2 R1 B2 U1 F2 U2 F2 D2 F2 R2
L2 D3 (18 moves)
Also I have reevaluated what is a complex cube position. Cube positions
have different degrees (or types) of difficulty.
A. A position is difficult if it is visually hard to recognize, e.g.
no pattern is apparent, the cube is well mixed and random.
However the pattern superfliptwist, although being 20 moves
deep, IS easy to recognize.
B. A position is easy with respect to computer analysis if it is
cyclically decomposable. That is to say it by looking at a
position a program finds it is generated by (F R B L)^5,
so this position is EASY.
C. A position is easy with respect to the human hand if the sequence
required to solve the position can be executed rapidly. To a
degree such positions are similar to positions in point (B)
in that only a subset of all cube operators are required, and
the sequence does not require turning all 6 sides and so the
sequence is easier to memorize as well.
As a result of thinking along these lines I am going to write a
module to do cyclic decomposition.
> Mark <
Email: mark.longridge@canrem.com
....more patterns to follow...
From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Sat Dec 4 09:49:17 1993
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Date: Sat, 4 Dec 1993 09:15:56 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: First Post
This is my first post to CubeLovers, so I will introduce myself
briefly. I have been cubing since about 1979 or 1980 or so when
the cubes first appeared on the market. I have been cubing with
a computer since about 1985, and have been active on the Internet
since about 1985 (purely a coincidence of dates). I have looked
for years for a cubing list, and never found one until now.
I always looked for "Rubik" (or sometimes "Rubic"). For some silly
reason, it never occurred to me to look for "cube".
I have long since read Hofstadter's two Scientific American articles,
as well as the reprints in METAMAGICAL THEMAS. The reprints, by
the way, are excellent because of the additional information in the
appendices. I also have a copy of Singmaster and Frey's
HANDBOOK OF CUBIC MATH. I have tried unsuccessfully for years to
get copies of Singmaster's earlier work  the circulars, for
example. However, I suspect that the HANDBOOK includes most if not
all of the earlier material. Also, (and you won't believe this) I have
just read all thirteen years of the archives of CubeLovers.
My primary interest has been in calculating God's Algorithm. I am
interested in brute force breadth first tree searches. In other
words, my work is akin to the solutions of the 2x2x2 and the corners
of the 3x3x3 posted by Dan Hoey and others. It is not akin to
Thistlethwaite's methods. I am interested to see, however, that
major recent progress appears to have been made on Thistlewaite's
method.
I have calculated God's Algorithm for the 2x2x2 cube and the corners
of the 3x3x3. My results agree with those that have been posted here,
with the exception that my search is 48 times smaller (24*2), due
to the exploitation of a rotation and reflection group of the cube.
I have also calculated God's Algorithm for the edges of the 3x3x3.
This is a much larger problem, and took about a year running
continuously on two machines. The resulting output file is about
4.2 gigabytes of data, and is stored on 31 reels of magnetic tape.
This result includes the "48 times smaller" factor, else it would
have been 204 gigabytes of data stored on 1464 reels of magnetic
tape.
I understand that this list has been very quiet of late. But
assuming some modicum of interest, I will post more details
of my results in subsequent messages.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 2935192
Associate Director, WVNET (304) 2935540 fax
837 Chestnut Ridge Road BRYAN@WVNVM
Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU
If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?
From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Sat Dec 4 21:07:09 1993
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Date: Sat, 4 Dec 1993 21:04:23 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: God's Algorithm for the 2x2x2 Pocket Cube
I want to post my God's Algorithm results for the 2x2x2 cube.
These results generally speaking replicate other results that have
been posted here as far back as ten to twelve years ago.
In order to make my numbers make sense, I need to explain how I count
the states of the 2x2x2 cube. As has been posted here several times
previously, the number is (7!)(3^6)=3,674,160. Actually, I prefer
the formulation (7!)(3^7)/3=3,674,160 because the latter formulation
clearly reflects that all the cubies can be rotated but that
rotational orientation of the last one is determined by the
rotational orientation of the others. But in any case, this
calculation is based on the following. Let any one cube be fixed
in location and rotational orientation. Then, there are
7! ways to arrange the other seven cubes, and (3^7)/3 ways to
rotate them.
But there is another way to look at it. Fix none of the cubes.
Rather, choose one to be the upper,left,front one, pick a second
one to be the upper,right,front one, etc., so that there are 8!
ways to arrange the eight cubes and (3^8)/3 ways to rotate them.
We have 8!(3^8)/3=88,179,840, which is exactly twentyfour
times larger than 3,674,160.
The reason is that the 3,674,160 figure implicitly assumes
that cubes that differ only in orientation of the overall
cube are equivalent, and there are twentyfour ways to orient
the cube in space (i.e., the order of the rotation group of
the cube is 24).
Conversely, the 88,179,840 figure implicitly assumes that
cubes that differ only in orientation of the overall cube
are distinct. They can be made equivalent by applying the
rotation group of the cube to form equivalence classes, and
there will be exactly 3,674,160 equivalence classes. Hence,
the two ways of counting are isomorphic. However, I do
prefer to characterize the "things" that the 3,674,160 figure
counts as equivalence classes, and I call 3,674,160 the
number of nodes using 24fold symmetry.
Finally, I apply a second order24 rotation group (I will explain
how you can have a two order24 rotation groups on the same cube
in a followup post) and an order2 reflection group. Hence, the
number of nodes to represent the entire search tree for the 2x2x2
cube should be 88,179,840/(24*24*2)=76,545, where the 76,545 figure
represents the number of equivalence classes and each equivalence
class includes 24*24*2=1152 elements. As it turns out, a few of
the equivalence classes contain fewer than 1152 elements, so that
the total number of nodes in the search tree is slightly larger
than 76,545, namely 77,802.
The tables of results below include figures both for 24fold symmetry
and for 1152fold symmetry. My search tree was for 1152fold symmetry
only. I then sort of "backed in" to the results for 24fold symmetry
by calculating the size of each equivalence class. Calculating a
search tree with 77,802 nodes representing equivalence classes, then
calculating the size of each equivalence class, was much faster than
calculating a search tree with 88,179,840 nodes or one with
3,674,160 nodes.
The little exercise with calculating the size of each equivalence
class was very gratifying in at least two respects. First, it let
me explain the disconcerting difference between 76,545 and 77,802.
Second, it let me confirm that my results were the same as everyone
else who had gone before.
Results Using Both qturns and hturns
Distance Number of Number of
from Nodes Nodes
Start using using
24fold 1152fold
symmetry symmetry
0 1 1
1 9 2
2 54 5
3 321 19
4 1847 68
5 9992 271
6 50136 1148
7 227536 4915
8 870072 18364
9 1887748 39707
10 623800 13225
11 2644 77
  
Total 3674160 77802
Results Using Only qturns
Distance Number of Number of
from Nodes Nodes
Start using using
24fold 1152fold
symmetry symmetry
0 1 1
1 6 1
2 27 3
3 120 6
4 534 17
5 2256 59
6 8969 217
7 33058 738
8 114149 2465
9 360508 7646
10 930588 19641
11 1350852 28475
12 782536 16547
13 90280 1976
14 276 10
  
Total 3674160 77802
Results Using Only hturns
Distance Number of Number of
from Nodes Nodes
Start using using
24fold 1152fold
symmetry symmetry
0 1 1
1 3 1
2 6 1
3 11 2
4 3 2
  
Total 24 7
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 2935192
Associate Director, WVNET (304) 2935540 fax
837 Chestnut Ridge Road BRYAN@WVNVM
Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU
If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?
From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Sat Dec 4 23:18:20 1993
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XAcknowledgeTo:
Date: Sat, 4 Dec 1993 23:15:30 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: God's Algorithm for the Corners of the 3x3x3
Here are my God's Algorithm results for the corners of the
3x3x3 cube. I explained in the last post what I mean by
1152fold symmetry and 24fold symmetry. The 1152fold
symmetry is what I actually calculated. In this particular
case, I did not do the 24fold symmetry calculations myself
based on the size of the equivalence classes as I did
with the 2x2x2 cube. Rather, I went back and found the
figures in the CubeLover archives (Dik Winter's post).
Results Using Both qturns and hturns
Distance Number of Number of
from Nodes using Nodes using
Start 1152fold 24fold
symmetry symmetry
0 1 1
1 2 18
2 9 243
3 71 2,874
4 637 28,000
5 4,449 205,416
6 24,629 1,168,516
7 113,049 5,402,628
8 433,611 20,776,176
9 947,208 45,391,616
10 316,823 15,139,616
11 1,481 64,736
Results Using Only qturns
Distance Number of Number of
from Nodes using Nodes using
Start 1152fold 24fold
symmetry symmetry
0 1 1
1 1 12
2 5 114
3 24 924
4 149 6,539
5 850 39,528
6 4,257 199,926
7 16,937 806,136
8 57,800 2,761,740
9 180,639 8,656,152
10 466,052 22,334,112
11 676,790 32,420,448
12 392,558 18,780,864
13 45,744 2,166,720
14 163 6,624
Results Using Only hturns
Distance Number of
from Nodes using
Start 1152fold
symmetry
0 1
1 1
2 2
3 4
4 3
It turns out that the maximum distance from Start is the same
for the corners of the 3x3x3 cube as it is for the 2x2x2 cube.
I found this rather surprising, although the archives of
CubeLovers do provide a reasonable explanation. I am just
going to have to go back and read it five or ten times until
I fully understand it. In any case, I was curious about the
following question. Suppose you are N moves from Start on the
corners of the 3x3x3. How many moves from Start would you be
on the 2x2x2 if the 2x2x2 was in the same configuration as the
corners of the 3x3x3 where you currently were. As it turns out,
I stored the results for the 2x2x2 in the same data base as
I stored the results for the corners of the 3x3x3, so the
question was easy to answer. Here are the results.
Corresponding Distances from Start
Using Both qturns and hturns
2x2x2 Corner of 3x3x3 Number
Distance from Distance from of Nodes
Start Start
0
0 1
2 2
4 2
1
1 2
2 2
3 4
4 3
2
2 5
3 12
4 18
5 3
3
3 55
4 106
5 41
4
4 508
5 457
6 38
5
5 3,948
6 1,237
7 2
6
6 23,354
7 1,992
8 20
7
7 111,055
8 3,242
9 20
8
8 430,349
9 5,460
10 62
9
9 941,728
10 3,770
11 20
10
10 312,991
11 45
11
11 1,416
Corresponding Distances from Start
Using Only qturns
2x2x2 Corner 3x3x3 Number
Distance from Distance from of Nodes
Start Start
0
0 1
2 1
4 2
6 1
1
1 1
3 2
5 2
2
2 4
4 10
6 6
3
3 22
5 46
7 4
4
4 137
6 145
5
5 802
7 356
6
6 4,105
8 474
7
7 16,577
9 83
8
8 57,326
10 24
12 24
9
9 180,556
11 148
10
10 466,028
12 192
11
11 676,642
13 144
12
12 392,342
13
13 45,600
14
14 163
Corresponding Distances from Start
Using Only hturns
2x2x2 Corner of 3x3x3 Number
Distance from Distance from of Nodes
Start Start
0
0 1
2 1
1
1 1
3 1
2
2 1
3
3 3
4
4 3
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 2935192
Associate Director, WVNET (304) 2935540 fax
837 Chestnut Ridge Road BRYAN@WVNVM
Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU
If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?
From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Sun Dec 5 00:03:57 1993
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XAcknowledgeTo:
Date: Sun, 5 Dec 1993 00:01:08 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: God's Algorithm for the Edges of the 3x3x3
Here are my God's Algorithm results for the edges of the
3x3x3 cube. I explained in the last post what I mean by
1152fold symmetry. All results below are for 1152fold
symmetry. I am working on the 24fold case, but I am not
quite done. The 24fold case is just a matter of determining
the sizes of the equivalence classes in the 1152fold case.
One item of terminology needs to be explained. Several
people, including myself, have posted results for the
2x2x2 cube and for the corners of the 3x3x3 cube. If you
take the term "corners of the 3x3x3 cube" absolutely
literally, it is completely isomorphic to the 2x2x2 cube.
When people have posted results for the "corners of the
3x3x3 cube", they all (including myself) really mean
"corners plus centers of the 3x3x3". See below:
  
 x  x   x   x   x   x 
          
  
 x  x        x  
          
  
2x2x2  x   x   x   x 
       
 
Corners of 3x3x3 Corners + Centers
Thus, when I say I have solved the "edges of the 3x3x3", I need
to clarify what I mean. I have solved the "edges without the
centers". I suppose my next project will be "edges with the
centers". Unfortunately, "edges with the centers" is a twentyfour
times larger problem than is "edges without the centers". "Edges
without the centers" took about a year running 24 hours a day,
7 days a week, on two machines. I am going to have to rethink
"edges with the centers" before I start. I don't want it to take
24 years.
 
  x     x  
       
 
 x   x   x  x  x 
       
 
  x     x  
       
 
Edges without Centers Edges with Centers
Results using qturns only
Distance Number of
from Start Nodes using
1152fold Symmetry
0 1
1 1
2 5
3 25
4 215
5 1,860
6 16,481
7 144,334
8 1,242,992
9 10,324,847
10 76,993,295
11 371,975,385
12 382,690,120
13 8,235,392
14 54
15 1
Results using qturns and hturns
Distance Number of
from Start Nodes using
1152fold Symmetry
0 1
1 2
2 9
3 75
4 919
5 11,344
6 139,325
7 1,664,347
8 18,524,022
9 167,864,679
10 582,489,607
11 80,930,364
12 314
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 2935192
Associate Director, WVNET (304) 2935540 fax
837 Chestnut Ridge Road BRYAN@WVNVM
Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU
If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?
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From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: Addendum to God's Algorithm for the 2x2x2 Cube
I had intended to include the following table in my first
post concerning God's Algorithm for the 2x2x2 cube, but
I forgot. It addresses the question of how large are the
equivalence classes in the search tree, where the equivalence
classes are generated by the two rotational symmetry groups
and the one reflectional symmetry group. Most of the
equivalence classes have 24*24*2=1152 elements, but some have
fewer.
Size of Number Total Number
Equivalence of of
Class Nodes Permutations
Represented
24 1 24
48 1 48
72 3 216
96 1 96
144 14 2,016
192 15 2,880
288 135 38,880
384 32 12,288
576 2,208 1,271,808
1,152 75,392 86,851,584
  
Total 77,802 88,179,840
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 2935192
Associate Director, WVNET (304) 2935540 fax
837 Chestnut Ridge Road BRYAN@WVNVM
Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU
If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?
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From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: Equivalence Classes for God's Algorithm for Edges of 3x3x3
Number of Size of Total
Equivalence Equivalence States
Classes Class
4 24 96
2 48 96
12 72 864
16 96 1,536
110 144 15,840
70 192 13,440
1,544 288 444,672
1,252 384 480,768
128,858 576 74,222,208
851,493,140 1152 980,920,097,280
851,625,008 980,995,276,800
Note that 980,995,276,800=12!(2^12)/2, so the proper total was
obtained.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 2935192
Associate Director, WVNET (304) 2935540 fax
837 Chestnut Ridge Road BRYAN@WVNVM
Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU
If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?
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From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: God's Algorithm, 24fold Symmetry, Edges of 3x3x3
I just finished tabulating the results with 24fold symmetry for
the edges of the 3x3x3 cube. I have added them to the table
I posted earlier today which had 1152fold symmetry.
A couple of reminders. In the case of 1152fold symmetry, most but not
all of the equivalence classes have exactly 1152 elements. In the case
of 24fold symmetry, every equivalence class has exactly 24 elements.
Thus, you can almost, but not quite, get from the 1152fold column
to the 24fold column by multiplying by 48. Also, my program actually
generated the 1152fold column. However, it did not generate the
24fold column. That would have taken far too long. Rather, I generated
the 24fold column from the 1152fold column by determining the sizes
of all the equivalence classes. Finally, note that the total figure
for the 24fold symmetry column can be calculated as
40,874,803,200 = [12!(2^12)/2] / 24, so the total is correct.
Results using qturns only
Distance Number of Number of
from Start Nodes using Nodes using
1152fold Symmetry 24fold Symmetry
0 1 1
1 1 12
2 5 114
3 25 1,068
4 215 9,759
5 1,860 88,144
6 16,481 786,500
7 144,334 6,916,192
8 1,242,992 59,623,239
9 10,324,847 495,496,593
10 76,993,295 3,695,351,994
11 371,975,385 17,853,871,137
12 382,690,120 18,367,613,703
13 8,235,392 395,043,663
14 54 1,080
15 1 1
Total 851,625,008 40,874,803,200
Results using qturns and hturns
Distance Number of Number of
from Start Nodes using Nodes using
1152fold Symmetry 24fold Symmetry
0 1 1
1 2 18
2 9 243
3 75 3,240
4 919 42,359
5 11,344 538,034
6 139,325 6,666,501
7 1,664,347 79,820,832
8 18,524,022 888,915,100
9 167,864,679 8,056,929,021
10 582,489,607 27,958,086,888
11 80,930,364 3,883,792,136
12 314 8,827
Total 851,625,008 40,874,803,200
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 2935192
Associate Director, WVNET (304) 2935540 fax
837 Chestnut Ridge Road BRYAN@WVNVM
Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU
If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?
From hoey@aic.nrl.navy.mil Mon Dec 6 10:20:06 1993
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From: hoey@aic.nrl.navy.mil
MessageId: <9312061519.AA01483@sun13.aic.nrl.navy.mil>
To: "Jerry Bryan"
Cc: "Cube Lovers List"
Subject: Re: God's Algorithm, 24fold Symmetry, Edges of 3x3x3
These results look very interesting, though I haven't had time to
examine them closely, nor even (in a few cases) quite understand them.
I especially like to see the categorization by symmetry class.
I was somewhat startled to see the unique antipode of the 3x3x3 edges
in the quarterturn metric. Do you know what pattern that is?
Dan
From punjanza@dunx1.ocs.drexel.edu Mon Dec 6 11:37:16 1993
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From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Mon Dec 6 14:03:43 1993
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Date: Mon, 6 Dec 1993 11:06:48 EST
From: "Jerry Bryan"
To: , "Cube Lovers List"
Subject: Re: God's Algorithm, 24fold Symmetry, Edges of 3x3x3
InReplyTo: Message of 12/06/93 at 10:19:00 from hoey@aic.nrl.navy.mil
On 12/06/93 at 10:19:00 hoey@aic.nrl.navy.mil said:
>These results look very interesting, though I haven't had time to
>examine them closely, nor even (in a few cases) quite understand them.
>I especially like to see the categorization by symmetry class.
>I was somewhat startled to see the unique antipode of the 3x3x3 edges
>in the quarterturn metric. Do you know what pattern that is?
I was extremely surprised as well. With all my previous work, there
was no unique antipode. I don't know what it is yet, but I can find
out without a whole lot of trouble. It is somewhere on the 31st
tape, so I just need to spin that tape, looking for a record at level
15, and print it out. I will try to get to that sometime in the
next few days.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 2935192
Associate Director, WVNET (304) 2935540 fax
837 Chestnut Ridge Road BRYAN@WVNVM
Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU
If you don't have time to do it right today, what makes you think you are
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From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Mon Dec 6 18:34:42 1993
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From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: Unique Antipodal of the 3x3x3 Edges
In answer to the question by Dan Hoey, I printed out the unique
antipodal of the 3x3x3 edges  the one configuration that is
15 moves from Start using only qturns on the edges of the 3x3x3.
It is really quite extraordinary and wonderful. I already knew
that there were only four equivalence classes with 24 elements.
Well, two of them are Start itself and its antipodal. Without
further ado:
*6* *6*
6*6 3*4
*6* *1*
*2* *5*
2*2 3*4
*2* *2*
*3**1**4* *1**1**1*
3*31*14*4 5*23*42*5
*3**1**4* *6**6**6*
*5* *2*
5*5 3*4
*5* *5*
Start Antipodal
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 2935192
Associate Director, WVNET (304) 2935540 fax
837 Chestnut Ridge Road BRYAN@WVNVM
Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU
If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?
From @mail.uunet.ca:mark.longridge@canrem.com Mon Dec 6 19:20:24 1993
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Subject: Unique antipode of edges only
From: mark.longridge@canrem.com (Mark Longridge)
MessageId: <60.581.5834.0C18D396@canrem.com>
InReplyTo: <9312061519.AA01483@sun13.aic.nrl.navy.mil>
Date: Mon, 6 Dec 1993 10:45:00 0500
Organization: CRS Online (Toronto, Ontario)
> I was somewhat startled to see the unique antipode of the 3x3x3 edges
> in the quarterturn metric. Do you know what pattern that is?
>
> Dan
It's got to be all edges flipped in place.
I would like to see the process generating the position!
I don't understand it all either :<
But at least we got some new cube mail.
> Mark <
From alan@parsley.lcs.mit.edu Mon Dec 6 20:16:32 1993
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From: Alan Bawden
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InReplyTo: Jerry Bryan's message of Mon, 6 Dec 1993 18:32:15 EST <9312062334.AA01891@life.ai.mit.edu>
Subject: Unique Antipodal of the 3x3x3 Edges
Date: Mon, 6 Dec 1993 18:32:15 EST
From: Jerry Bryan
... It is really quite extraordinary and wonderful. I already knew
that there were only four equivalence classes with 24 elements. Well,
two of them are Start itself and its antipodal. Without further ado:...
This is very interesting indeed!
So the next natural question would seem to be: What are the other two?