From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Mon Dec 6 21:32:55 1993
ReturnPath: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU>
Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA10000; Mon, 6 Dec 93 21:32:55 EST
MessageId: <9312070232.AA10000@life.ai.mit.edu>
Received: from MITVMA.MIT.EDU by mitvma.mit.edu (IBM VM SMTP V2R2)
with BSMTP id 3272; Mon, 06 Dec 93 21:32:56 EST
Received: from WVNVM.WVNET.EDU (NJE origin MAILER@WVNVM) by MITVMA.MIT.EDU
(LMail V1.1d/1.7f) with BSMTP id 0517; Mon, 6 Dec 1993 21:32:56 0500
Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU
(LMail V1.1d/1.7f) with BSMTP id 6339; Mon, 6 Dec 1993 21:30:26 0500
XAcknowledgeTo:
Date: Mon, 6 Dec 1993 21:30:25 EST
From: "Jerry Bryan"
To:
Subject: Re: Unique antipode of edges only
InReplyTo: Message of 12/06/93 at 10:45:00 from mark.longridge@canrem.com
On 12/06/93 at 10:45:00 mark.longridge@canrem.com said:
>> I was somewhat startled to see the unique antipode of the 3x3x3 edges
>> in the quarterturn metric. Do you know what pattern that is?
>>
>> Dan
>It's got to be all edges flipped in place.
I had to stare at my picture for a couple of minutes to be sure, but
yes it is. How did you know?
>I would like to see the process generating the position!
This is doable, but it is a little harder said than done. My "data base"
is just a simple flat file with the states and the level associated with
each state. In the case of the 2x2x2, the file is about 625K, and
I have programs to search it readily. If you use the file in
"Solver mode", my "Solver program" just generates all successors of the
current node, finds each successor in the data base (it is a simple binary
search, the file is sorted), chooses one at level N1 (there is always
at least one), and makes that the new current node. It stops when
N=0.
I have a "Solver program" for the "corners plus centers of the
3x3x3" as well, but again the data base is small. It is actually
the original 625K file for the 2x2x2 case, plus three additional
625K files. This simple file structure was chosen to keep the file
small. There are no pointers, trees, or processes stored in the
data base.
The "edges of the 3x3x3 without centers" is a little tougher. Early
in the project, I generated a data base for the first few levels
(six or seven, I think), and I have a "Solver program" that will
work up to that level. However, the full "edges of the 3x3x3 without
centers" is a 4.2 gigabyte file on tape, so it is hard to process.
Also, the size of the equivalence classes is not in the data base,
only the level. I have to calculate the size of each equivalence
class, and it is an expensive calculation.
I made a pass at the
file and calculated the number of equivalence classes (took
125 hours on a mainframe), but I only saved a summary. I did not
save the number of equivalence classes for each state. I found
the antipodal by looking for level 15, since I knew there was
only one occurrence, and since the level was in the data base.
I did save the summaries by tape, so I should only have to look
on two tapes to find the other two equivalence classes which
have 24 elements.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 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 mouse@collatz.mcrcim.mcgill.edu Tue Dec 7 07:38:26 1993
ReturnPath:
Received: from Collatz.McRCIM.McGill.EDU by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA25101; Tue, 7 Dec 93 07:38:26 EST
Received: from localhost (root@localhost) by 16886 on Collatz.McRCIM.McGill.EDU (8.6.4 Mouse 1.0) id HAA16886 for cubelovers@ai.mit.edu; Tue, 7 Dec 1993 07:38:09 0500
Date: Tue, 7 Dec 1993 07:38:09 0500
From: der Mouse
MessageId: <199312071238.HAA16886@Collatz.McRCIM.McGill.EDU>
To: cubelovers@ai.mit.edu
Subject: Re: Unique Antipodal of the 3x3x3 Edges
> In answer to the question by Dan Hoey, I printed out the unique
> antipodal of the 3x3x3 edges [...].
> It is really quite extraordinary and wonderful. [...]. Without
> further ado:
Someone else remarks that it's "got to be all edges flipped in place",
and Jerry Bryan remarks that it is.
> *6* *6*
> 6*6 3*4
> *6* *1*
> *2* *5*
> 2*2 3*4
> *2* *2*
> *3**1**4* *1**1**1*
> 3*31*14*4 5*23*42*5
> *3**1**4* *6**6**6*
> *5* *2*
> 5*5 3*4
> *5* *5*
I disagree. Look at the 12 edge. If it's "flipped in place", then
since it appears to be fixed, the cube must flip around it. But then
the four 3 faces would be where the 4 faces actually are. No, it's
more complicated than just alledgesflipped.
"[Q]uite extraordinary and wonderful" it is.
der Mouse
mouse@collatz.mcrcim.mcgill.edu
From ccw@eql12.caltech.edu Tue Dec 7 08:25:59 1993
ReturnPath:
Received: from EQL12.Caltech.Edu by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA26306; Tue, 7 Dec 93 08:25:59 EST
Date: Mon, 6 Dec 93 19:13:20 PST
From: ccw@eql12.caltech.edu (Chris Worrell)
MessageId: <931206185340.20400b26@EQL12.Caltech.Edu>
Subject: Re: Unique antipode of edges only
InReplyTo: Your message <9312070232.AA10000@life.ai.mit.edu> dated 6Dec1993
To: BRYAN%WVNVM.WVNET.EDU%WVNVM.WVNET.EDU@mitvma.mit.edu,
cubelovers@ai.mit.edu
On 12/06/93 at 10:45:00 mark.longridge@canrem.com said:
>> I was somewhat startled to see the unique antipode of the 3x3x3 edges
>> in the quarterturn metric. Do you know what pattern that is?
>>
>> Dan
>It's got to be all edges flipped in place.
Unfortunately, this is wishfull thinking.
This antipode is 15 qtw from Home, an odd distance.
All edges flipped is an even distance from Home in the qtw metric.
Looking at Jerry Bryan's pictures, I see 5 two edge swaps.
>
> *6* *6*
> 6*6 3*4
> *6* *1*
> *2* *5*
> 2*2 3*4
> *2* *2*
> *3**1**4* *1**1**1*
> 3*31*14*4 5*23*42*5
> *3**1**4* *6**6**6*
> *5* *2*
> 5*5 3*4
> *5* *5*
>
> Start Antipodal
>
If we assume face 1 is F, I get
(FU) (BD) (FD,BU) (FL,LU) (FR, RU) (LD,BL) (RD,BR)
Is the 1152 number the result of factoring out the 24 spatial rotations
and 2 reflections of the centers?
Are there any estimates of how many distinct sequences actually generate
this Antipodal Class?
Ideally, it would be interesting to have a total list of these sequences.
From formail.TCPBRIDGE.FS3.FAA1.ERICM%smte@formail.formation.com Tue Dec 7 09:06:46 1993
ReturnPath:
Received: from uu3.psi.com by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA28712; Tue, 7 Dec 93 09:06:46 EST
Received: from formail.formation.com by uu3.psi.com (5.65b/4.0.071791PSI/PSINet) via SMTP;
id AA20412 for CUBELOVERS@AI.MIT.EDU; Tue, 7 Dec 93 09:06:37 0500
Received: by formail.formation.com (4.1/SMI4.1)
id AA19052; Tue, 7 Dec 93 09:01:04 EST
MessageId: <9312071401.AA19052@formail.formation.com>
Received: from smte id: 2D048EC7.CAB
(WordPerfect SMTP Gateway V3.1a 04/27/92)
Received: from formail (WP Connection)
Received: from TCPBRIDGE (WP Connection)
Received: from FS3 (WP Connection)
Received: from FAA1 (WP Connection)
From: (Moyer, Eric )
To:
Subject: cube availability
Date: Tue Dec 7 09:10:15 1993
Greetings.
I have been away from cubing since the early 80's, which was
before I went to school and before I did much computer work.
After reading the recent archives I rushed out to find a square1
and fell in love all over again, only this time, I'm armed. I
was somewhat amazed to find, however, that not a single other
cube puzzle was available at ToysRUs or at any of the stores I
tried first. I went back and reread Hofstadter's articles after
Jerry Bryan's recommendation and came across the address for Uwe
M'effert Novelties, Princewell (Far East), Ltd., P.O. Box 31008,
Causeway Bay, Hong Kong. Does anyone know if this company still
exists? Additionally, does anyone know of any mail order company
where cubes and cubevariants can be purchased? Thanks.
From andyl@harlequin.com Tue Dec 7 10:33:00 1993
ReturnPath:
Received: from hilly.harlequin.com by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA02571; Tue, 7 Dec 93 10:33:00 EST
Received: from epcot.harlequin.com by hilly.harlequin.com; Tue, 7 Dec 1993 10:35:13 0500
Received: from phaedrus.harlequin.com (phaedrus) by epcot.harlequin.com; Tue, 7 Dec 1993 10:37:38 0500
From: Andy Latto
Date: Tue, 7 Dec 1993 10:37:37 0500
MessageId: <6474.199312071537@phaedrus.harlequin.com>
To: Alan@lcs.mit.edu
Cc: BRYAN@wvnvm.bitnet, CubeLovers@ai.mit.edu
InReplyTo: Alan Bawden's message of Mon, 6 Dec 93 20:16:26 0500 <6Dec1993.195513.Alan@LCS.MIT.EDU>
Subject: Unique Antipodal of the 3x3x3 Edges
Date: Mon, 6 Dec 93 20:16:26 0500
From: Alan Bawden
Sender: Alan@lcs.mit.edu
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?
Switch each edge with its antipode, with or without flipping all twelve edges.
From tom@scumby.clipper.ingr.com Tue Dec 7 11:07:36 1993
ReturnPath:
Received: from scumby.clipper.ingr.com by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA04696; Tue, 7 Dec 93 11:07:36 EST
Received: by scumby.clipper.ingr.com (5.65c/1.921207)
id AA18849; Tue, 7 Dec 1993 08:11:28 0800
From: tom@scumby.clipper.ingr.com (Tom Granvold)
MessageId: <199312071611.AA18849@scumby.clipper.ingr.com>
Subject: Re: cube availability
To: cubelovers@ai.mit.edu,
formail.TCPBRIDGE.FS3.FAA1.ERICM%smte@formail.formation.com (Moyer Eric)
Date: Tue, 7 Dec 93 8:11:26 PST
InReplyTo: <9312071401.AA19052@formail.formation.com>; from "Moyer, Eric" at Dec 7, 93 9:10 am
XMailer: ELM [version 07.00.00.00 (2.3 PL11)]
>After reading the recent archives I rushed out to find a square1
>and fell in love all over again,
Congratulation.
>only this time, I'm armed.
Watch out, he is dangerous. :)
>I was somewhat amazed to find, however, that not a single other
>cube puzzle was available at ToysRUs or at any of the stores I
>tried first.
Instead of toy stores, I'd try game stores. I don't know if you'll
be able to find a Square1 or not. It has been a couple of years since
they come out.
>came across the address for Uwe
>M'effert Novelties, Princewell (Far East), Ltd., P.O. Box 31008,
>Causeway Bay, Hong Kong. Does anyone know if this company still
>exists?
I believe that this company has not been around for several years.
I did buy a couple of their "cubes" about ten years ago. But at some
point they seemed to have disapeared. Too bad they had some unique
variations.
>Additionally, does anyone know of any mail order company
>where cubes and cubevariants can be purchased? Thanks.
Yes. It seems that just recently Ishi Press has made some of the
cubevariants available. I saw a 5x5x5 cube in a game store recently
from Ishi Press. Game stores are a good place to look since Ishi has
long been providing products for the games; Go and Shogi. Note that
there is even an email address!
Ishi Press International
76 Bona Ventura
San Jose Ca 95134
(408)9449900
fax  (408)9449110
email  ishius@ishius.com
Have fun,
Tom Granvold
tom@clipper.ingr.com
From senya@rainbow.ldgo.columbia.edu Tue Dec 7 11:15:30 1993
ReturnPath:
Received: from lamont.ldgo.columbia.edu (ldgo.columbia.edu) by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA04949; Tue, 7 Dec 93 11:15:30 EST
Received: from rainbow.ldgo.columbia.edu by lamont.ldgo.columbia.edu (4.1/SMI3.2)
id AA20725; Tue, 7 Dec 93 11:15:20 EST
Received: by rainbow.ldgo.columbia.edu (920110.SGI/890607.SGI)
(for @lamont.ldgo.columbia.edu:cubelovers@ai.mit.edu) id AA29505; Tue, 7 Dec 93 11:14:55 0500
Date: Tue, 7 Dec 93 11:14:55 0500
From: senya@rainbow.ldgo.columbia.edu (Semyon Basin)
MessageId: <9312071614.AA29505@rainbow.ldgo.columbia.edu>
To: cubelovers@ai.mit.edu
Subject: Needed: Basic elements of solving Rubic Cube:
Could you gentelmen suggest me the place where I can find the basic
(elementary) combinations to solve the cube?
Like the sequence to swap two "internal" side's boxes while not
changing the positions of all other "internal" boxes but probably only
rotating them?
Or could you tell me about the method to rotate some of edge elements
without swapping them?
___________________________________________________________________________
____ ______ __ __ ______
Semyon Basin, / ___\ /\ ___\ /\ \ /\ \ / ____ \
LamontDoherty Earth Observatory, /\ \/_/ \/\ \/_/ \/\ \_\/\ \ /\ \___\ \
Route 9W, \/\ \ \/\ _\ \/\ ____ \\/\__ __ \
Palisades, NY 10964 \/\ \____\/\ \/___\/\ \/_/\ \\/_/ /_/\ \
\/\_____\\/\_____\\/\_\ \/\_\ /\_\ \/\_\
Internet:senya@rainbow.ldgo.columbia.edu \/_/_/_/ \/_/_/_/ \/_/ \/_/ \/_/ \/_/
________________________________________________________________________________
From dseal@armltd.co.uk Tue Dec 7 12:02:56 1993
ReturnPath:
Received: from eros.britain.eu.net by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA08106; Tue, 7 Dec 93 12:02:56 EST
Received: from armltd.co.uk by eros.britain.eu.net with UUCP
id ; Tue, 7 Dec 1993 16:42:43 +0000
Received: by armltd.co.uk (5.51/Am23) id AA04366; Tue, 7 Dec 93 16:13:04 GMT
Date: Tue, 07 Dec 93 16:13:51 GMT
From: dseal@armltd.co.uk (David Seal)
To: (Cube) cubelovers@ai.mit.edu
Subject: Re: Unique antipode of edges only
MessageId: <2D04ABBF@dseal>
> Someone else remarks that it's "got to be all edges flipped in place",
> and Jerry Bryan remarks that it is.
>
> > *6* *6*
> > 6*6 3*4
> > *6* *1*
> > *2* *5*
> > 2*2 3*4
> > *2* *2*
> > *3**1**4* *1**1**1*
> > 3*31*14*4 5*23*42*5
> > *3**1**4* *6**6**6*
> > *5* *2*
> > 5*5 3*4
> > *5* *5*
>
> I disagree. Look at the 12 edge. If it's "flipped in place", then
> since it appears to be fixed, the cube must flip around it. But then
> the four 3 faces would be where the 4 faces actually are. No, it's
> more complicated than just alledgesflipped.
>
> "[Q]uite extraordinary and wonderful" it is.
It is in fact the position arrived at by flipping all edges in place, *then*
reflecting the entire configuration. I believe this also tells us what the
other two equivalence classes with just 24 elements are: they are the
results of doing each of these two operations separately.
David Seal
dseal@armltd.co.uk
From andyl@harlequin.com Tue Dec 7 12:24:30 1993
ReturnPath:
Received: from hilly.harlequin.com by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA08855; Tue, 7 Dec 93 12:24:30 EST
Received: from epcot.harlequin.com by hilly.harlequin.com; Tue, 7 Dec 1993 12:27:09 0500
Received: from phaedrus.harlequin.com (phaedrus) by epcot.harlequin.com; Tue, 7 Dec 1993 12:29:34 0500
From: Andy Latto
Date: Tue, 7 Dec 1993 12:29:32 0500
MessageId: <12292.199312071729@phaedrus.harlequin.com>
To: ccw@eql12.caltech.edu
Cc: BRYAN%WVNVM.WVNET.EDU%WVNVM.WVNET.EDU@mitvma.mit.edu,
cubelovers@ai.mit.edu
InReplyTo: Chris Worrell's message of Mon, 6 Dec 93 19:13:20 PST <931206185340.20400b26@EQL12.Caltech.Edu>
Subject: Unique antipode of edges only
Unfortunately, this is wishfull thinking.
This antipode is 15 qtw from Home, an odd distance.
All edges flipped is an even distance from Home in the qtw metric.
Looking at Jerry Bryan's pictures, I see 5 two edge swaps.
>
> *6* *6*
> 6*6 3*4
> *6* *1*
> *2* *5*
> 2*2 3*4
> *2* *2*
> *3**1**4* *1**1**1*
> 3*31*14*4 5*23*42*5
> *3**1**4* *6**6**6*
> *5* *2*
> 5*5 3*4
> *5* *5*
>
> Start Antipodal
>
The antipodal position is an interesting one. If you take the antipodal
position, and flip all the edges, you get:
*5*
5*5
*5*
*1*
1*1
*1*
*3**2**4*
3*32*24*4
*3**2**4*
*6*
6*6
*6*
Antipodal with edges flipped.
This looks like a rotation of the solved state at first glance, since
all the faces on a given side of the cube are the same color. But
look again! This is not the solved state of the original cube, but
of the mirror image cube. If you added in the centers or the corners,
there would be no way to add them to make this a solved state.
Andy Latto
andyl@harlequin.com
From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Tue Dec 7 13:42:56 1993
ReturnPath: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU>
Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA13489; Tue, 7 Dec 93 13:42:56 EST
MessageId: <9312071842.AA13489@life.ai.mit.edu>
Received: from MITVMA.MIT.EDU by mitvma.mit.edu (IBM VM SMTP V2R2)
with BSMTP id 2007; Tue, 07 Dec 93 13:42:57 EST
Received: from WVNVM.WVNET.EDU (NJE origin MAILER@WVNVM) by MITVMA.MIT.EDU
(LMail V1.1d/1.7f) with BSMTP id 5141; Tue, 7 Dec 1993 13:42:57 0500
Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU
(LMail V1.1d/1.7f) with BSMTP id 9473; Tue, 7 Dec 1993 13:40:23 0500
XAcknowledgeTo:
Date: Tue, 7 Dec 1993 13:40:20 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: Re: Unique antipode of edges only
InReplyTo: Message of 12/07/93 at 12:29:32 from andyl@harlequin.com
On 12/07/93 at 12:29:32 Andy Latto said:
>The antipodal position is an interesting one. If you take the antipodal
>position, and flip all the edges, you get:
> *5*
> 5*5
> *5*
> *1*
> 1*1
> *1*
> *3**2**4*
> 3*32*24*4
> *3**2**4*
> *6*
> 6*6
> *6*
>Antipodal with edges flipped.
>This looks like a rotation of the solved state at first glance, since
>all the faces on a given side of the cube are the same color. But
>look again! This is not the solved state of the original cube, but
>of the mirror image cube. If you added in the centers or the corners,
>there would be no way to add them to make this a solved state.
Indeed. I spoke too quickly when I said the antipodal was simply
Start with the edges flipped. I stared at it, flipped the edges in
my mind, and it "looked" solved, so I assumed it was Start.
I am not yet for sure what they look like, but of the other two states
with order24 equivalence classes, one is at level 9 and the other
is at level 12. Since the only one at an even level is at level 12,
I am assuming that will be the one which is Start with the edges all
flipped. The one at level 9 will probably be the mirror image of Start.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 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 Tue Dec 7 20:13:23 1993
ReturnPath:
Received: from Sun0.AIC.NRL.Navy.Mil by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA05026; Tue, 7 Dec 93 20:13:23 EST
Received: by Sun0.AIC.NRL.Navy.Mil (4.1/SMI4.0)
id AA29049; Tue, 7 Dec 93 20:13:08 EST
Date: Tue, 7 Dec 93 20:13:08 EST
From: hoey@aic.nrl.navy.mil (Dan Hoey)
MessageId: <9312080113.AA29049@Sun0.AIC.NRL.Navy.Mil>
To: "Cube Lovers List"
Subject: Re: Unique antipode of edges only
"Jerry Bryan" writes:
> I spoke too quickly when I said the antipodal was simply
> Start with the edges flipped. I stared at it, flipped the edges in
> my mind, and it "looked" solved, so I assumed it was Start.
It's interesting to note that this is AllEdgesFlipped composed with
a mirror reflection of Start. Begging the question: *which* mirror
reflection? The answer is, it doesn't matter: since these are the
edges of a cube without centers, all reflections are the same
position. As long as we get to choose which reflection, the canonical
one would be the central reflection. When composed with AllEdges
Flipped, it makes the following antipode. (I think using BFTDLR
notation instead of 123456 makes these diagrams a lot easier to read).
+ T + + F +
T T R L
+ T + + B +
+ L + + F + + R + + D + + D + + D +
L L F F R R => F B R L B F
+ L + + F + + R + + T + + T + + T +
+ D + + B +
D D R L
+ D + + F +
+ B + + T +
B B R L
+ B + + D +
> I am not yet for sure what they look like, but of the other two states
> with order24 equivalence classes, one is at level 9 and the other
> is at level 12. Since the only one at an even level is at level 12,
> I am assuming that will be the one which is Start with the edges all
> flipped. The one at level 9 will probably be the mirror image of Start.
If an order24 equivalence class means what I think it does, I'm
pretty sure those two states have to be MirrorStart and AllEdges
Flipped, being the only sufficiently symmetric positions. But as to
their depth, the parity argument (which Chris Worrell also cited) is
not valid here. Remember that the cube has no face centers, so the
position is not changed by rotating the assemblage of edges in space
(i.e., with respect to the absent face centers). But a quarterturn
of the cube in space is an odd permutation of the edges. So permuta
tion parity is not an intrinsic property of edge positions. We can
show that there is no sort of parity here by explicitly constructing
an odd cycle. Just use a process that would permute the edges of a
cube with faces as (FR,FT,FL,FD)(BR,BT,BL,BD)(RT,TL,LD,DR). This has
to be an odd process, but it's an identity on the faceless cube.
My (very cheap) guess for where we will find the other two Msymmetric
positions is opposite to Jerry Bryan's. On a cube with faces, the
central reflection of the edges with respect to the faces is Pons
Asinorum, which has the easy 12qt tight lower bound we've seen before
(or if not, you can of course get it from me with email). I'm
guessing that this bound happens to be tight on the cube without
faces, as well. But I have no proof of this guess, and I'm very
grateful we won't have to settle for guesses for very long.
Dan Hoey
Hoey@AIC.NRL.Navy.Mil
From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Wed Dec 8 10:04:52 1993
ReturnPath: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU>
Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA02610; Wed, 8 Dec 93 10:04:52 EST
MessageId: <9312081504.AA02610@life.ai.mit.edu>
Received: from MITVMA.MIT.EDU by mitvma.mit.edu (IBM VM SMTP V2R2)
with BSMTP id 1780; Wed, 08 Dec 93 10:04:53 EST
Received: from WVNVM.WVNET.EDU (NJE origin MAILER@WVNVM) by MITVMA.MIT.EDU
(LMail V1.1d/1.7f) with BSMTP id 0159; Wed, 8 Dec 1993 10:04:54 0500
Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU
(LMail V1.1d/1.7f) with BSMTP id 4368; Wed, 8 Dec 1993 10:02:17 0500
XAcknowledgeTo:
Date: Wed, 8 Dec 1993 10:02:15 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: Re: Unique antipode of edges only
InReplyTo: Message of 12/07/93 at 20:13:08 from hoey@aic.nrl.navy.mil
On 12/07/93 at 20:13:08 hoey@aic.nrl.navy.mil said:
>My (very cheap) guess for where we will find the other two Msymmetric
>positions is opposite to Jerry Bryan's. On a cube with faces, the
>central reflection of the edges with respect to the faces is Pons
>Asinorum, which has the easy 12qt tight lower bound we've seen before
>(or if not, you can of course get it from me with email). I'm
>guessing that this bound happens to be tight on the cube without
>faces, as well. But I have no proof of this guess, and I'm very
>grateful we won't have to settle for guesses for very long.
>Dan Hoey
>Hoey@AIC.NRL.Navy.Mil
Dan Hoey is correct. MirrorImageofStart is at level 12.
EdgesFlipped is at level 9. MirrorImageofStartandEdgesFlipped
is at level 15. And, of course, Start is at Level 0. This exhausts
the list of configurations with order24 symmetry.
I am still thinking about the easiest way to extract sequences of
operators from my data base. I gather from Dan's comments that a
12qt operator is known for MirrorImageofStart. Are operators
known for the other two cases? This is going to be sufficiently
timeconsuming that I don't want to try to find operators that
are already known.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
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 senya@rainbow.ldgo.columbia.edu Wed Dec 8 12:21:51 1993
ReturnPath:
Received: from lamont.ldgo.columbia.edu (ldgo.columbia.edu) by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA12094; Wed, 8 Dec 93 12:21:51 EST
Received: from rainbow.ldgo.columbia.edu by lamont.ldgo.columbia.edu (4.1/SMI3.2)
id AA03963; Wed, 8 Dec 93 12:21:28 EST
Received: by rainbow.ldgo.columbia.edu (920110.SGI/890607.SGI)
(for @lamont.ldgo.columbia.edu:cubelovers@ai.mit.edu) id AA20676; Wed, 8 Dec 93 12:20:38 0500
Date: Wed, 8 Dec 93 12:20:38 0500
From: senya@rainbow.ldgo.columbia.edu (Semyon Basin)
MessageId: <9312081720.AA20676@rainbow.ldgo.columbia.edu>
To: cubelovers@ai.mit.edu
Subscribe me please
From @mail.uunet.ca:mark.longridge@canrem.com Wed Dec 8 14:21:29 1993
ReturnPath: <@mail.uunet.ca:mark.longridge@canrem.com>
Received: from mail.uunet.ca (uunet.ca) by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA19082; Wed, 8 Dec 93 14:21:29 EST
Received: from portnoy.canrem.com ([198.133.42.251]) by mail.uunet.ca with SMTP id <54067(8)>; Wed, 8 Dec 1993 13:52:15 0500
Received: from canrem.com by portnoy.canrem.com (4.1/SMI4.1)
id AA09857; Wed, 8 Dec 93 13:50:53 EST
Received: by canrem.com (PCBUUCP 1.1f)
id 18D8B1; Wed, 8 Dec 93 13:41:19 0400
To: cubelovers@life.ai.mit.edu
ReplyTo: CRSO.Cube@canrem.com
Sender: CRSO.Cube@canrem.com
Subject: Antipodal Edge Position
From: mark.longridge@canrem.com (Mark Longridge)
MessageId: <60.600.5834.0C18D8B1@canrem.com>
Date: Wed, 8 Dec 1993 12:33:00 0500
Organization: CRS Online (Toronto, Ontario)
>>It's got to be all edges flipped in place.
Oops. Well I figured if all edges flipped was one of the hardest
know cube states that in the case of edgesonly it would be the
antipode. I'm now sure (I think) that it is really:
all edges flipped + 4 X
(with the 4 X on sides F, R, B, L which should match Dan's diagram)
Hmmmm, I don't know if this is a standard form of representation,
but this picture looks like a folded out cube:
+ T + + F +
T T R L
+ T + + B +
+ L + + F + + R + + D + + D + + D +
L L F F R R => F B R L B F
+ L + + F + + R + + T + + T + + T +
+ D + + B +
D D R L
+ D + + F +
+ B + + T +
B B > R L
+ B +  + D +

+ D +
In my program I would have L R on the screen for the bottom face.
+ T +
The idea is you are always looking at a cube face headon (just to
clarify the difference in diagrams).
More quotes for Jerry Bryan:
>The "edges of the 3x3x3 without centers" is a little tougher. Early
>in the project, I generated a data base for the first few levels
>(six or seven, I think), and I have a "Solver program" that will
>work up to that level. However, the full "edges of the 3x3x3 without
>centers" is a 4.2 gigabyte file on tape, so it is hard to process.
>Also, the size of the equivalence classes is not in the data base,
>only the level. I have to calculate the size of each equivalence
>class, and it is an expensive calculation.
>
>I made a pass at the
>file and calculated the number of equivalence classes (took
>125 hours on a mainframe), but I only saved a summary. I did not
>save the number of equivalence classes for each state. I found
>the antipodal by looking for level 15, since I knew there was
>only one occurrence, and since the level was in the data base.
>
>I am not yet for sure what they look like, but of the other two states
>with order24 equivalence classes, one is at level 9 and the other
>is at level 12. Since the only one at an even level is at level 12,
>I am assuming that will be the one which is Start with the edges all
>flipped. The one at level 9 will probably be the mirror image of
Start.
I'd still like to see the process for alledgesflipped (not
caring about the centres or corners). So "level" is the number of moves
required to solve the position? That means edges flipped in place
can be done in 12 qtw.
From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Wed Dec 8 14:42:11 1993
ReturnPath: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU>
Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA19673; Wed, 8 Dec 93 14:42:11 EST
MessageId: <9312081942.AA19673@life.ai.mit.edu>
Received: from MITVMA.MIT.EDU by mitvma.mit.edu (IBM VM SMTP V2R2)
with BSMTP id 5941; Wed, 08 Dec 93 14:11:24 EST
Received: from WVNVM.WVNET.EDU (NJE origin MAILER@WVNVM) by MITVMA.MIT.EDU
(LMail V1.1d/1.7f) with BSMTP id 1649; Wed, 8 Dec 1993 14:11:24 0500
Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU
(LMail V1.1d/1.7f) with BSMTP id 9868; Wed, 8 Dec 1993 14:08:47 0500
XAcknowledgeTo:
Date: Wed, 8 Dec 1993 14:08:15 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: 1152fold Symmetry and 24fold Symmetry
I guess it's time to try to explain what I mean by 1152fold symmetry
and 24fold symmetry.
Let me start with two or three very simple ideas. First, consider
two equally colored and oriented cubes at Start. To one of them,
apply F, and to the other one apply R. The obvious solution to the
first one is then F' and the obvious solution to the second on is
then R'. But take both cubes and toss them through the
air to someone else, so that the spatial orientation is lost.
Almost anyone would solve either cube by finding the one face that
was twisted clockwise and simply twisting it counterclockwise.
No distinction between F and R would be made, and it would be
"obvious" to any reasonable person that the cubes were equivalent.
As a slightly more formal application of this idea, consider again
Start to which R has been applied. We could rotate the whole
cube in space using Singmaster's scriptU operation. That is, grasp
the Up (top) of the cube and turn the whole cube in space clockwise.
Now, the cube looks like F has been applied rather than R, and the
solution looks like F' rather than R'. If we applied F', the cube
would be solved, but the colors would be oriented wrong. We could
restore the colors by scriptU'. Thus, (scriptU F' script U') is
exactly the same thing as R' (we are just using conjugates in a
very simple way).
Continuing in this vein, take any two equally colored and oriented
cubes at Start. To one of them, apply some long sequence of
permutations P. To the second one, apply (scriptU P scriptU').
At this point, the two cubes are definitely not "equal" in some
sense  you could clearly tell them apart. Yet, they are
clearly "equivalent" in some sense, because if P' is a solution to
the first cube, then (scriptU P' scriptU') is a solution to the
second one. Furthermore, it should be obvious that it is not really
necessary to use the (scriptU scriptU') conjugate on the second
cube. Rather we can think of some rotation as having been performed
on P to give Q, and then of Q as having been performed on Start, so
that the same rotation that was applied to P could be applied to P'
to give Q', and Q' is equivalent to (scriptU P' scriptU').
If I can wax sophomorically philosophical for a minute, I tend to
think of there being two kinds of permutations in mathematics.
The first is the "permutations and combinations" kind of thing you
run into in probability and statistics. The second is the permutation
operator kind of thing you run into in abstract algebra or group
theory. With this kind of thinking, the cube itself represents the
first kind of permutation, where the cube is an object being operated
on, and the twists of the cube are the second kind of permutation,
where the twists are permutation operators and are doing the operating.
Well, at some deep level, the two kinds of permutations are very much
the same thing, so it is very natural to think of operating on
(rotating, in this particular case) a permutation P, where P itself
is an operator.
I need one more simple idea. Again, think of a cube in Start, and
think of Singmaster's scriptU operator. We can (informally) write
scriptU = (Front > Left > Back > Right > Front). But suppose
the cube is colored as Font=Red, Left=White, Back=Orange, Right=Blue).
We could just as well write scriptU = (Red > White > Orange
> Blue > Red). It looks as if for any fixed coloring, we can
freely interchange Singmaster's notation with a notation based on
colors. But we can't really. For example, colored as I described it
above, scriptF is equivalent to scriptRed. Either is the same as
grasping the front of the cube and rotating the whole cube clockwise.
But first perform scriptU. Now, scriptF is the same as
scriptBlue). The Front/Back/Up/Down/Left/Right notation is fixed in
space, but the color notation is not.
Now, we try to put all this together. Once again, consider two
equally colored and equally oriented cubes in space, and apply F
to the first one and (R scriptU) to the second one. Both
cubes can now be described as "Start with the front twisted clockwise
by 90 degrees), but the colors are not the same. They are clearly
equivalent, but under my internal computer model for the cube, they
are not equal. Furthermore, no amount of additional application of
Singmaster's whole cube "script" operators will make them equal.
The only thing that will make them equal will be to rotate the colors.
The exact same thing applies to reflections. Consider two equally
colored and oriented cubes in Start, and apply F to one and F' to
the other. The cubes are equivalent but not equal. Hold up the
cube to which F' has been applied to a mirror. The mirrorimage
now has F applied instead of F', but the colors are wrong (they
have been reflected). To make the cubes equal, it is necessary to
reflect the colors of the mirrorimage.
Hence, my program generates equivalence classes by applying
a cube rotation, a color rotation, a cube reflection, and a color
reflection. There are 24 cube rotations and 24 color rotations
(one of each is the identity), and any cube rotation can occur with
any color rotation. There are 2 cube reflections and 2 color
reflections (one each is the identity), but the cube reflection
identity must occur with the color reflection identity and vice
versa. Thus, there are (in general) 24x24x2 elements in each
equivalence class. I only store one element of each equivalence
class in my data base. Some of the equivalence classes have fewer
than 24x24x2 elements, namely those for which the cube configuration
inherently has a high degree of symmetry.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 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 Wed Dec 8 15:31:43 1993
ReturnPath: <@mail.uunet.ca:mark.longridge@canrem.com>
Received: from mail.uunet.ca (uunet.ca) by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA22765; Wed, 8 Dec 93 15:31:43 EST
Received: from portnoy.canrem.com ([198.133.42.251]) by mail.uunet.ca with SMTP id <54139(4)>; Wed, 8 Dec 1993 15:01:42 0500
Received: from canrem.com by portnoy.canrem.com (4.1/SMI4.1)
id AA16889; Wed, 8 Dec 93 15:00:21 EST
Received: by canrem.com (PCBUUCP 1.1f)
id 18D8D1; Wed, 8 Dec 93 14:56:34 0400
To: cubelovers@life.ai.mit.edu
ReplyTo: CRSO.Cube@canrem.com
Sender: CRSO.Cube@canrem.com
Subject: More corrections
From: mark.longridge@canrem.com (Mark Longridge)
MessageId: <60.601.5834.0C18D8D1@canrem.com>
Date: Wed, 8 Dec 1993 13:52:00 0500
Organization: CRS Online (Toronto, Ontario)
Mark gaffs again:
> I'm now sure (I think) that it is really:
>
> all edges flipped + 4 X
> (with the 4 X on sides F, R, B, L which should match Dan's diagram)
* sign * No, I see I entered the position into my program wrong.
A central reflection of the edges with respect to the faces is
simply 6 X or checkerboard order 2, solvable in 12 qtw or 6 htw.
So the edgesonly antipode is: alledgesflipped + 6 X.
Jerry Byran quote:
>Dan Hoey is correct. MirrorImageofStart is at level 12.
>EdgesFlipped is at level 9. MirrorImageofStartandEdgesFlipped
>is at level 15. And, of course, Start is at Level 0. This exhausts
>the list of configurations with order24 symmetry.
Ok, only 9 qtw.... it's got to play havoc with corners. I got it now.
* Hmmm, what are all the possible orders of symmetry? *
Also I note my "Symmetry Level" is the opposite of Jerry's OrderN
symmetry:
> If we define "symmetry level" as the number of distinct patterns
>generated by rotating the cube through it's 24 different orientations
in
>space then most known antipodes are symmetry level 6. Thus the lower
the
>number the higher the level of symmetry. The least symmetric positions
>have level 24, and this is very common. The most symmetric positions
>have level 1, the two positions START and 6 X order 2.
Of course alledgesflipped I never included, as at the time I was
looking at the square's group.

As a small postfix to my cyclic decomposition article, I found the
following patterns. I'm fond of pattern 16 myself. I am looking for
CDtype processes for 6 X order 3 and 6 X order 6. I find when I
am physical cubing (as opposed to computer cubing or old fashioned
mental cubing!) it really helps having a CDtype process memorywise.
Memorizing the computer generated processes is like memorizing
prime numbers.
p161 Mark's Pattern 16 (F1 R1 L1 B1) ^3 + F2 B2 D2 F2 B2 T2 (18)
p162 2 X, 4 H full (F1 T2 B1) ^4 (12)
p163 4 ARM Full (F2 T1 B2) ^4 + T1 D3 (14)
p164 4 Y's Rotated (F1 T2 D2) ^6 + F1 (19)
p165 2 Swap, 4 H full (F1 L2 T2 R2 B1) ^2 + L2 R2 T2 D2 (14)
p166 2 H adj swap (F1 L2 T2 R2 B1) ^2 + L2 T2 R2 D2 L2 T2 (16)
No doubt these are compressible and hence not as efficient, but if
you consider ease of execution....
> Mark <
From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Thu Dec 9 03:38:45 1993
ReturnPath: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU>
Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA21581; Thu, 9 Dec 93 03:38:45 EST
MessageId: <9312090838.AA21581@life.ai.mit.edu>
Received: from MITVMA.MIT.EDU by mitvma.mit.edu (IBM VM SMTP V2R2)
with BSMTP id 2761; Thu, 09 Dec 93 03:38:43 EST
Received: from WVNVM.WVNET.EDU (NJE origin MAILER@WVNVM) by MITVMA.MIT.EDU
(LMail V1.1d/1.7f) with BSMTP id 1298; Thu, 9 Dec 1993 03:38:42 0500
Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU
(LMail V1.1d/1.7f) with BSMTP id 7106; Wed, 8 Dec 1993 22:41:29 0500
XAcknowledgeTo:
Date: Wed, 8 Dec 1993 22:41:28 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: Pretty Pattern at Level 13
Here is a pretty pattern at level 13 using qturns. The size
of the equivalence class is 48.
+ B +
R L
+ F +
+ D +
L R
+ U +
+ B + + F + + B +
U D R L D U
+ F + + B + + F +
+ U +
L R
+ D +
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 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 Thu Dec 9 04:19:46 1993
ReturnPath: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU>
Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA22353; Thu, 9 Dec 93 04:19:46 EST
MessageId: <9312090919.AA22353@life.ai.mit.edu>
Received: from MITVMA.MIT.EDU by mitvma.mit.edu (IBM VM SMTP V2R2)
with BSMTP id 2764; Thu, 09 Dec 93 03:38:48 EST
Received: from WVNVM.WVNET.EDU (NJE origin MAILER@WVNVM) by MITVMA.MIT.EDU
(LMail V1.1d/1.7f) with BSMTP id 1305; Thu, 9 Dec 1993 03:38:48 0500
Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU
(LMail V1.1d/1.7f) with BSMTP id 7383; Wed, 8 Dec 1993 23:16:51 0500
XAcknowledgeTo:
Date: Wed, 8 Dec 1993 23:16:50 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: Pretty Pattern (again)
Pretty pattern, 8 qturns from Start, size of equivalence
class is 48.
+ B +
F F
+ B +
+ U +
D D
+ U +
+ L + + F + + R +
R R B B L L
+ L + + F + + R +
+ D +
U U
+ D +
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 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 Thu Dec 9 04:56:35 1993
ReturnPath: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU>
Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA22759; Thu, 9 Dec 93 04:56:35 EST
MessageId: <9312090956.AA22759@life.ai.mit.edu>
Received: from MITVMA.MIT.EDU by mitvma.mit.edu (IBM VM SMTP V2R2)
with BSMTP id 2770; Thu, 09 Dec 93 03:39:11 EST
Received: from WVNVM.WVNET.EDU (NJE origin MAILER@WVNVM) by MITVMA.MIT.EDU
(LMail V1.1d/1.7f) with BSMTP id 1322; Thu, 9 Dec 1993 03:39:11 0500
Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU
(LMail V1.1d/1.7f) with BSMTP id 7496; Wed, 8 Dec 1993 23:39:35 0500
XAcknowledgeTo:
Date: Wed, 8 Dec 1993 23:39:35 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: That's all the 48Fold Symmetries
The two patterns I just posted are the only two for which the size
of the equivalence class is 48. The next size up is 72, and there
are twelve patterns whose equivalence class size is 72. Things
become less interesting as the equivalence class size increases because
they are less symmetrical overall. Also, there are more patterns.
Finally, these things take a good deal of time to chase down.
Therefore, I am going to stop chasing down "pretty patterns"
for now unless somebody is just dying to see the patterns whose
equivalence class size is 72.
Finally, nobody really answered an earlier question, but is the
following true: 1) MirrorImageofStart is 12 qturns from
Start, and a sequence is known (Dan Hoey sent me a wellknown
sequence), 2) AllEdgesFlipped is 9 qturns from Start, and a
sequence is known, and 3) MirrorImageofStartandAllEdgesFlipped
(the antipodal of Start) is 15 qturns from Start, and a sequence
is *not* known? If this is true, then I will start working on
the 15 move sequence.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 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 mouse@collatz.mcrcim.mcgill.edu Fri Dec 10 09:16:34 1993
ReturnPath:
Received: from Collatz.McRCIM.McGill.EDU by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA25485; Fri, 10 Dec 93 09:16:34 EST
Received: from localhost (root@localhost) by 4504 on Collatz.McRCIM.McGill.EDU (8.6.4 Mouse 1.0) id JAA04504; Fri, 10 Dec 1993 09:16:29 0500
Date: Fri, 10 Dec 1993 09:16:29 0500
From: der Mouse
MessageId: <199312101416.JAA04504@Collatz.McRCIM.McGill.EDU>
To: senya@rainbow.ldgo.columbia.edu
Subject: Re: Needed: Basic elements of solving Rubic Cube:
Cc: cubelovers@ai.mit.edu
> Subject: Needed: Basic elements of solving Rubic Cube:
(Does anyone happen to know whether Ern Rubik knows what's happened to
his name? I don't mean just here, but how common a term it's become.
And I apologize for using instead of what I think is the proper long
accent, but Latin1 doesn't have the proper one....)
> Could you gentelmen suggest me the place where I can find the basic
> (elementary) combinations to solve the cube?
There are no *the* basic combinations. I would guess there are as many
different sets of combinations as there are cube solvers.
> Like the sequence to swap two "internal" side's boxes while not
> changing the positions of all other "internal" boxes but probably
> only rotating them?
> Or could you tell me about the method to rotate some of edge elements
> without swapping them?
Well, for what it's worth, when I solve a cube, I do it as follows.
(Slice turns: if I write FB, I mean turn the FB slice in the direction
one would turn F for an F turn, similarly for other slice turns: turn
the slice as if it were carried along by the turn named by the first
letter. Thus LR and RL are inverses. Is there some standard
representation for slice turns?)
 Solve one layer adhoc. (This refers to a layer of cubies, not just
one face of the cube.) I often don't worry about edge flips at this
stage. Some simple operators I use:
To get corners in place: F D' F, or F' D F, depending on corner
orientation.
To get edges in place: If the cubie is on the D face,
FB/BF/RL/LR, D/D', inverse of the slice turn. If it's
on the middle layer, F/B/R/L, UD/DU, inverse of face
turn.
 Turn the cube so the solved face is L. Solve what then becomes the
RL slice layer with a combination of R2 U2 R2 U2 R2 U2, to move
cubies around within the slice layer, and either of two sequences to
move cubies between the R layer and the slice layer:
R2 D R' B2 R2 B2 R2 B2 R' D'
FB D R' B2 R2 B2 R2 B2 R' D' BF
(The first one is a sequence that normally ends with R2, but since
the R layer is scratch at this point I often don't bother.) These
are, of course, interspersed with R, R2, and R' turns to get edges in
the correct places for them.
At this point you will have two layers solved, except possibly for some
flipped edges.
Next, corners of the "scratch" layer. Check them for placement,
ignoring orientation. They can be:
1) All in place. This is the easy case. :)
2) Two swapped. Make one quarterturn to reach case 3. (They
can't be diagonal, they must be adjacent  or some joker has
taken your cube apart.)
3) One in place, other three permuted.
4) Two pairs, each swapped. If the swaps are diagonal, turn the
layer a halfturn to reach case 1.
In case 3 or 4, the corners can be put in place by holding the cube
with the unsolved layer as U and repeating
L F L' F' L F L' F' L F L' F' twice, turning U so as to place
appropriate pairs of cubies in the UFL and UBL corners.
To orient the corners correctly, hold the cube with the unsolved layer
as F and use R B2 R' U' B2 U and its inverse U' B2 U R B2 R' with a
turn of the F face in between; this will allow you to twist the corners
into correct orientation.
All that remains at this point is positioning the edges on the last
layer, and possibly some edge flips. To position the edges, I normally
use (with U as the unsolved layer)
R2 D R' B2 R2 B2 R2 B2 R' D' R2
FB D R' B2 R2 B2 R2 B2 R' D' BF
R2 U2 R2 U2 R2 U2
with appropriate turns of U in between, swapping the FR and BR edges
repeatedly as auxiliaries while swapping pairs of edges on U to get
them in place. (The difference between the first two sequences is that
the first one swaps UB and UR, the second UB and RU.)
Edge flips are all that's left at this point. Judicious choice of
which of the two sequences above can often drastically reduce the work
to be done here, but there's often some left anyway. The basic
sequence I use for this is RL U RL U RL U RL U, which flips four edge
cubies inplace: UB, UL, DB, and DF. (A similar sequence U RL U RL U
RL U RL is similar but flips UR instead of UL; this can be thought of
as U X U', where X is the firstgiven sequence.) My use of this
sequence is usually adhoc; sequences such as X F X F' will let you
flip arbitrary pairs of edges.
Presto! You have a solved cube. :) In practice, I often take
shortcuts; for example, if X represents the R B2 R' U' B2 U sequence,
then X B2 X B2 X B2 will twist three corners on B....
der Mouse
mouse@collatz.mcrcim.mcgill.edu
From hoey@aic.nrl.navy.mil Mon Dec 13 22:31:38 1993
ReturnPath:
Received: from Sun0.AIC.NRL.Navy.Mil by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA02377; Mon, 13 Dec 93 22:31:38 EST
Received: by Sun0.AIC.NRL.Navy.Mil (4.1/SMI4.0)
id AA25974; Mon, 13 Dec 93 22:31:31 EST
Date: Mon, 13 Dec 93 22:31:31 EST
From: hoey@aic.nrl.navy.mil (Dan Hoey)
MessageId: <9312140331.AA25974@Sun0.AIC.NRL.Navy.Mil>
To: CubeLovers@ai.mit.edu
Subject: Symmetry
I suppose it's time for a few observations on symmetry. After all,
tomorrow is the thirteenth anniversary of "Symmetry and Local Maxima."
As Jerry Bryan notes, we can perform the "R" turn by rotating the cube
to put the R face in front, performing "F", and undoing the rotation.
But we can also perform "R'" by reflecting the cube in a lefttoright
mirror, performing "L", and undoing the reflection. Thus conjugation
can be extended to use the 48element group of rotations and
reflections, which we call M.
In the absence of face centers, there is another kind of reduction
that takes account of the 24 possible positions of the resulting
collection of edges in space. So two positions X and Y are considered
equivalent if
X = m' Y m c
where m is a rotation or reflection in M, and c is a rotation.
My understanding of Jerry Bryan's method is that he combines "m c"
into a single rotation or reflection, and factors out the reflection
on both sides. It seems to me that what he calls a a "color rotation"
is premultiplication, while a "cube rotation" is postmultiplication.
[I am somewhat uncertain of this, because it doesn't explain how there
can be a 1252element symmetry group when face centers are present, so
perhaps I'm missing something crucial.]
But I think we are at least conceptually better off dealing with M
when dealing with conjugation, because it takes account of the
essential similarity between clockwise and anticlockwise turns. Alan
Bawden mentioned back in 1980 that certain positions with sufficient
symmetry were local maxima (in terms of distance from start), on the
grounds that any clockwise or anticlockwise turn gives us essentially
the same position. Jim Saxe and I formalized the notion in a paper
entitled "Symmetry and Local Maxima" that we posted on 14 December
1980. [You can find it in /pub/cubelovers/cubemail1.Z on
FTP.AI.MIT.Edu].
We had some hope that some of these local maxima might turn out to be
global maxima. My hopes for that have been somewhat low in recent
years. That is perhaps my best excuse for not noticing immediately
that the single global maximum for the edge group turns out to be one
of these symmetric local maxima. In fact, all four of the positions
with 24element equivalence classes appear in the list of Msymmetric
positions.
The paper on Symmetry and Local Maxima also catalogues the positions
that have 48element equivalence classes and 72element equivalence
classes. The The former are the Hsymmetric positions, "SixH" and
"SixH with all edges flipped". The latter are the twelve Tsymmetric
positions. For Tsymmetry, the set of flipped edges may be any of
{none, girdleedges, offgirdleedges, or all}; the set of edges
exchanged with their antipodes may be any of the four as well. But if
we choose "none" or "all" for all both choices we get one of the four
Msymmetric positions with 24element equivalence classes, so only
twelve of the sixteen possibilities have 72element equivalence
classes.
With regard to the edge cube, I should mention that no one has
mentioned a 9 QT process for the allflip nor a 15 QT process for the
ponsasinorumallflip. Of course, the latter would be somewhat more
interesting, being the longest optimal sequence.
Dan Hoey
Hoey@AIC.NRL.Navy.Mil
From avm@bgerug51.bitnet Tue Dec 14 02:42:21 1993
ReturnPath:
Received: from mserv.rug.ac.be ([157.193.40.37]) by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA09383; Tue, 14 Dec 93 02:42:21 EST
Received: by mserv.rug.ac.be id AA28828
(5.65c/IDA1.4.4 for cubelovers@ai.mit.edu); Tue, 14 Dec 1993 08:42:05 +0100
Received: from BGERUG51.BITNET by BGERUG51.BITNET (PMDF #12055) id
<01H6GP6445K0000WTQ@BGERUG51.BITNET>; Tue, 14 Dec 1993 08:36 N
Date: Tue, 14 Dec 1993 08:36 N
From: AnneMie Vandermeeren  RUG
Subject: Unsunscribe rob@bgerug51.bitnet
To: cubelovers@ai.mit.edu
MessageId: <01H6GP6445K0000WTQ@BGERUG51.BITNET>
XEnvelopeTo: cubelovers@ai.mit.edu
XVmsTo: IN%"cubelovers@ai.mit.edu"
Hi,
Please remove rob@bgerug51.bitnet from your mailing list cubeLovers
Thanks,
AnneMie Vandermeeren
Postmaster for BGERUG51
From anandrao@hk.super.net Tue Dec 14 03:18:49 1993
ReturnPath:
Received: from hk.super.net by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA10381; Tue, 14 Dec 93 03:18:49 EST
Received: by hk.super.net id AA15191
(5.65c/IDA1.4.4 for Cube Lovers ); Tue, 14 Dec 1993 16:18:36 +0800
Date: Tue, 14 Dec 1993 16:11:47 +0800 (HKT)
From: Mr Anand Rao
Subject: Tangle
To: Cube Lovers
MessageId:
MimeVersion: 1.0
ContentType: TEXT/PLAIN; charset=USASCII
I managed to pick up all four Tangle puzzles in an obscure shop in
Jakarta, Indonesia. The puzzles are similar, except that the extra(25th)
piece is different in each. The solutions for each puzzle are very
different and I could not see any pattern. I solved all 4 using
'intelligent brute force', i.e. made the search as efficient as I could.
But the 10*10 puzzle seems intractable. The 5*5 could be solved using a
486DX266 PC in about 20 minutes. The 10*10 will take several months using
my algorithm.
Does anyone have a more intelligent, or a more brute method? Once this
puzzle has been put in the public domain, we MUST find a solution. So, any
ideas are welcome!
Thanks
From dn1l+@andrew.cmu.edu Tue Dec 14 11:29:57 1993
ReturnPath:
Received: from po4.andrew.cmu.edu by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA24359; Tue, 14 Dec 93 11:29:57 EST
Received: from localhost (postman@localhost) by po4.andrew.cmu.edu (8.6.4/8.6.4) id LAA04307 for CubeLovers@ai.mit.edu; Tue, 14 Dec 1993 11:29:35 0500
Received: via switchmail; Tue, 14 Dec 1993 11:29:21 0500 (EST)
Received: from loiosh.andrew.cmu.edu via qmail
ID ;
Tue, 14 Dec 1993 11:28:51 0500 (EST)
Received: from loiosh.andrew.cmu.edu via qmail
ID ;
Tue, 14 Dec 1993 11:28:42 0500 (EST)
Received: from mms.4.60.Nov..4.1993.10.47.44.sun4c.411.EzMail.2.0.CUILIB.3.45.SNAP.NOT.LINKED.loiosh.andrew.cmu.edu.sun4c.411
via MS.5.6.loiosh.andrew.cmu.edu.sun4c_411;
Tue, 14 Dec 1993 11:28:41 0500 (EST)
MessageId:
Date: Tue, 14 Dec 1993 11:28:41 0500 (EST)
From: "Dale I. Newfield"
To: Cube Lovers
Subject: Re: Tangle
Cc:
InReplyTo:
Could you explain what your algorithm was?
I have one of the puzzles, number 4, I believe, and spent a large amount
of time trying to find a solution that was not trial and error. I could
not.
The algorithm that I used to have the computer solve it for me was to
fill the 5x5 in the following manner, recursively, returning when no
possible pieces fit.
1 2 4 7 11
/ / / /
/ / / /
3 5 8 12 16
/ / / /
/ / / /
6 9 13 17 20
/ / / /
/ / / /
10 14 18 21 23
/ / / /
/ / / /
15 19 22 24 25
(wrapping at the edges to keep incrementing properly)
I did that because given any pieces diagonal from one another, there are
at most two pieces that can fill the gap (line up with both correctly).
(When the four colors are different, there are two tiles
When there is a single repeated color, there is one tile
When there are 2 pairs of colors there is no tile
And in all these cases, if the tile(s) was already used, or didn't
exist, that is the bottom of that branch of the search tree)
Is this better or worse than the algorithm you used?
Has anyone found a nonbruteforce solution scheme?
Dale
From hoch@chem.wisc.edu Tue Dec 14 12:13:15 1993
ReturnPath:
Received: from ernie.chem.wisc.edu by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA26560; Tue, 14 Dec 93 12:13:15 EST
Received: by ernie.chem.wisc.edu;
id AA19022; AIX 3.2/UCB 5.64/42; Tue, 14 Dec 1993 11:13:16 0600
Date: Tue, 14 Dec 1993 11:13:16 0600
From: Douglas E. Hoch
MessageId: <9312141713.AA19022@ernie.chem.wisc.edu>
To: cubelovers@ai.mit.edu
Subject: List removal
Please remove hoch@pigggy.chem.wisc.edu from your cubelovers list.
Thanks.
From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Tue Dec 14 21:23:57 1993
ReturnPath: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU>
Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA24330; Tue, 14 Dec 93 21:23:57 EST
MessageId: <9312150223.AA24330@life.ai.mit.edu>
Received: from MITVMA.MIT.EDU by mitvma.mit.edu (IBM VM SMTP V2R2)
with BSMTP id 1767; Tue, 14 Dec 93 20:53:27 EST
Received: from WVNVM.WVNET.EDU (NJE origin MAILER@WVNVM) by MITVMA.MIT.EDU
(LMail V1.1d/1.7f) with BSMTP id 0071; Tue, 14 Dec 1993 20:53:27 0500
Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU
(LMail V1.1d/1.7f) with BSMTP id 7374; Tue, 14 Dec 1993 20:50:53 0500
XAcknowledgeTo:
Date: Tue, 14 Dec 1993 20:50:51 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: Re: Symmetry
InReplyTo: Message of 12/13/93 at 22:31:31 from hoey@aic.nrl.navy.mil
On 12/13/93 at 22:31:31 hoey@aic.nrl.navy.mil said:
>In the absence of face centers, there is another kind of reduction
>that takes account of the 24 possible positions of the resulting
>collection of edges in space. So two positions X and Y are considered
>equivalent if
> X = m' Y m c
>where m is a rotation or reflection in M, and c is a rotation.
>My understanding of Jerry Bryan's method is that he combines "m c"
>into a single rotation or reflection, and factors out the reflection
>on both sides. It seems to me that what he calls a "color rotation"
>is premultiplication, while a "cube rotation" is postmultiplication.
>[I am somewhat uncertain of this, because it doesn't explain how there
>can be a 1252element symmetry group when face centers are present, so
^^^^
should be 1152
>perhaps I'm missing something crucial.]
I just reread "Symmetry and Local Maxima". Let's see if I can make
some sense of this. I believe "premultiplication" and
"postmultiplication" are correct. In my computer model, the
corner facelets are simply numbered from 1 to 24, and any configuration
of the corners is an order24 row vector. The rotation and reflection
operators are also order24 row vectors, again with each cell simply
containing a number from 1 to 24.
In almost anybody's programming language you would copy an order24
row vector with something like
For i = 1 to 24 B(i) = A(i)
Well, if P is a rotation operator, you could perform a rotation
two ways. I guess one is premultiplication and one is
postmultiplication.
1) For i = 1 to 24 B(i) = A(P(i))
2) For i = 1 to 24 B(i) = P(A(i))
(As an aside, this illustrates the question I raised in my previous
post about "which is the operator and which is the thing being
operated on?" Is P operating on A, or is A operating on P?)
In fact, if what I am doing is properly called pre and post
multiplication, then I am doing both as a part of a single,
composite operator. I.e.,
For i = 1 to 24 B(i) = P(A(P(i)))
More completely, there are 24 rotations, P1 through P24, so the
actual loop looks something like
For j = 1 to 24 for k = 1 to 24
for i = 1 to 24 Bj,k(i) = Pj(A(Pk(i)))
Finally, if Q is a reflection (actually, if Q1 is the identity and
Q2 is the reflection), then we have
For j = 1 to 24 for k = 1 to 24 for m = 1 to 2
for i = 1 to 24 Bj,k,m(i) = Qm(Pj(A(Qm(Pk(i)))))
I believe this loop calculates Dan Hoey's M. In my data base, I
store the minimum of Bj,k,m over j = 1 to 24, k = 1 to 24, and
m = 1 to 2. I tend to call the minimum of Bj,k,m a canonical
form. I am not sure if that is the best terminology. The
minimal element is not any simpler than any other. It is just
that I need a function to choose an element from a set, and
picking the minimal element seems very natural. Any other
element would do as well, provided I could always be sure of
picking the same element.
Also, my criterion for equivalence is slightly
different (but isomorphic, I think) than the one described by
Dan Hoey. Suppose A and B are two cubes.
Rather than mapping A to B or B to A in M, I map both A and B
to their respective canonical forms. A and B are equivalent if
their respective canonical forms are equal.
I hasten to add that the actual loop in the program is a bit
more complex than the one shown above. The one above would
be far too slow. The actual loop is several hundred times
faster.
Now, as to the centers. I still sometimes have a certain doubt
about the centers. They are fixed, so how can you reduce the
problem (i.e., increase the size of the equivalence classes)
by both rotating the cube and rotating the colors (by both pre
and postmultiplication)?
In my computer model for the centers, I simply number center facelets
from 1 to 6, and the centers are stored as an order6 row vector.
The centers are disjoint from the corners (as well as from the edges),
so there is no problem in numbering one set of objects from 1 to 24 and
another from 1 to 6.
I define a set of 24 rotation operators P* on the centers, corresponding
to the 24 rotation operators P on the corners, and a set of 2 reflection
operators Q* on the centers, corresponding to the 2 reflection operators
Q on the corners. Then, if C is an order6 row vector representing
the centers, I calculate Dj,k,m = Q*m(P*j(C(Q*m(P*k)))) anytime
I calculate Bj,k,m = Qm(Pj(A(Qm(Pk)))).
(Read the asterisks above as superscripts. I am not intending
the multiplication operation which the asterisk denotes in
many programming languages.)
Hence, I rotate and reflect the centers right along with the corners.
But there are only 24 distinct states for the centers, and each can
occur with any canonical form for the corners. Hence, the "corners
plus centers" data base is exactly 24 times larger than
the "2x2x2" data base.
My model for the cube seems to start out 24 times larger than
everybody else's. However, by storing only the canonical form
for each equivalence class, and since most of the equivalence
classes have 1152 elements, my data base seems to end up about
48 times smaller than everybody else's. This fact seems to
remain true, even when the "fixed centers" are added in.
I am not sure if this answers Dan's question about my model
with centers added. Effectively, I am using a "fixed corners"
representation of the cube, and rotating the centers. Each
equivalence class for the corners under M has (up to) 1152
elements, and each equivalence class for the centers under
M has only 24 elements. But it doesn't seem to matter.
(Up to) 48 different configurations of the corners within
M share each configuration of the centers.
Since I am in this deep, let me finish explaining certain details
of my model. I don't really store all 24 elements of each
row vector. I really just store 8. That is, I store the
facelets for the front and back face. The other 16 facelets
can be reconstructed from the first 8. In effect, storing
a number from 1 to 24 stores both the location of each cubie
and its twist. Finally, I really, really only store 7 elements.
In the canonical form, the first element is always 1, so there
is no reason to store it. Thus, a data base record for the
2x2x2 looks like
CCCCCCC,L
where the CCCCCCC are the seven elements representing the canonical
form, and L is the corresponding level.
When you add the centers, I started out with notion that the
order6 row vector for center only has 24 possible states. Thus,
it can be encoded as a number from 1 to 24. This lead to the
following
CCCCCCC,L,R
where CCCCCCC and L are as before, and R is an index encoding
the orientation of the centers. But this can be improved upon
even further. With my model for the corners plus centers, each
distinct value of CCCCCCC will occur exactly 24 times, and each
distinct value of CCCCCCC is already represented in my data
base for the 2x2x2. Hence, I can have the exact same number of
(longer) records, and encode the corners of the 3x3x3 as
CCCCCCC,L,L1 L2 L3 .... L23 L24
where CCCCCCC is as before, L is the level of CCCCCCC in the
2x2x2, and L1 through L24 are the levels of CCCCCCC in the corners
of the 3x3x3 when the index of the position of the centers with
respect to the corners is 1 through 24, respectively. Hence, my
data base for the corners of the 3x3x3 has the same number of
records as the data base for the 2x2x2, and is physically only
four times larger.
>With regard to the edge cube, I should mention that no one has
>mentioned a 9 QT process for the allflip nor a 15 QT process for the
>ponsasinorumallflip. Of course, the latter would be somewhat more
>interesting, being the longest optimal sequence.
I will work on these two cases, but it will take some time. My model
is very good at storing a great many states of the cube very
compactly, but it does not encode processes at all. I will have
to extract the processes by hand. This is quite easy in my
data bases for the 2x2x2 and corners of 3x3x3. But it is quite
hard for the edges because the data base is 4.2 gigabytes.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 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 Don.Woods@eng.sun.com Wed Dec 15 06:04:15 1993
ReturnPath:
Received: from Sun.COM by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA07720; Wed, 15 Dec 93 06:04:15 EST
Received: from Eng.Sun.COM (zigzag.Eng.Sun.COM) by Sun.COM (4.1/SMI4.1)
id AA22455; Tue, 14 Dec 93 14:48:16 PST
Received: from colossal.Eng.Sun.COM by Eng.Sun.COM (4.1/SMI4.1)
id AA21191; Tue, 14 Dec 93 14:47:01 PST
Received: by colossal.Eng.Sun.COM (5.0/SMISVR4)
id AA22891; Tue, 14 Dec 93 14:48:17 PST
Date: Tue, 14 Dec 93 14:48:17 PST
From: Don.Woods@eng.sun.com (Don Woods)
MessageId: <9312142248.AA22891@colossal.Eng.Sun.COM>
To: CubeLovers@ai.mit.edu
Subject: Re: Tangle
XSunCharset: USASCII
ContentLength: 2501
Anand Rao writes:
> The puzzles are similar, except that the extra(25th)
> piece is different in each. The solutions for each puzzle are very
> different and I could not see any pattern.
Look again. The puzzles are identical except for a remapping of the colors.
For example, if you take Tangle #1 and paint all the Blue ropes Yellow, all
the Red ropes Blue, all the Green ropes Red, and all the (originally) Yellow
ropes Green, you'll have Tangle #2. So you can solve Tangle #1 by imagining
the ropes recolored as above, constructing your solution for #2, and then
restoring the original colors.
Note: The particular recoloring given above is based on colors given in a
message sent by CCW@eql.caltech.edu (Chris Worrell) to cubelovers on April
27, 1992. I own only #1 myself and so cannot confirm or deny the accuracy
of the colors. But the basic idea applies, given that each puzzle (a) has
the same pattern of ropes on all pieces and (b) has each permutation of
colors exactly once except for one permutation which appears twice.
Solving the 10x10 is another kettle of fish, and I haven't tried it. I do
have a program that solves the 5x5 in about 45 seconds on a SparcStation II,
but I haven't looked into how much longer it would take on the 10x10.
"Dale I. Newfield" writes:
> Could you explain what your algorithm was?
> Has anyone found a nonbruteforce solution scheme?
My solution was bruteforce. I posted to cubelovers (again, in April '92)
asking if anyone had found a more logical approach to the puzzle, but got no
affirmative responses.
Dale's method is a little inefficient in the order in which it tries tiles.
Mine used the sequence: Dale's used:
1 3 5 7 9 1 2 4 7 11
2 4 6 8 10 3 5 8 12 16
11 12 13 14 15 6 9 13 17 20
16 17 18 19 20 10 14 18 21 23
21 22 23 24 25 15 19 22 24 25
The first three tiles in our two methods are equally constrained, but the
next seven in Dale's methods are constrained along 1211221 edges,
while mine are constrained along 2121212 edges. So I suspect my
search tree gets trimmed a bit more quickly. Another way in which the
search can be made more efficient is in finding the pieces to try in each
position. For each pair of colors that can appear along an edge, my program
precomputes a table listing all tiles that can match that pair of colors,
including how to rotate the tiles.
 Don.
From @mail.uunet.ca:mark.longridge@canrem.com Wed Dec 15 11:07:13 1993
ReturnPath: <@mail.uunet.ca:mark.longridge@canrem.com>
Received: from mail.uunet.ca (uunet.ca) by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA17433; Wed, 15 Dec 93 11:07:13 EST
Received: from portnoy.canrem.com ([198.133.42.251]) by mail.uunet.ca with SMTP id <55767(2)>; Wed, 15 Dec 1993 10:52:33 0500
Received: from canrem.com by portnoy.canrem.com (4.1/SMI4.1)
id AA18531; Wed, 15 Dec 93 10:50:45 EST
Received: by canrem.com (PCBUUCP 1.1f)
id 18E5EA; Wed, 15 Dec 93 10:46:24 0400
To: cubelovers@life.ai.mit.edu
ReplyTo: CRSO.Cube@canrem.com
Sender: CRSO.Cube@canrem.com
Subject: Lib of Congress
From: mark.longridge@canrem.com (Mark Longridge)
MessageId: <60.618.5834.0C18E5EA@canrem.com>
Date: Wed, 15 Dec 1993 09:38:00 0500
Organization: CRS Online (Toronto, Ontario)
I Pulled this list from the Library of Congress. The hungarian
book I've never seen before (#2), and #3 and #4 are new to me.
"Zen of Cubing" may be interesting, has anyone read this one?
ITEMS 14 OF 11 SET 15: BRIEF DISPLAY FILE: LOCI
(DESCENDING ORDER)
1. 868699: Buvos kocka. English. Rubik's cubic compendium / Oxford
; New York : Oxford University Press, 1987. xi, 225 p. : ill.
(some col.) ; 23 cm.
LC CALL NUMBER: QA491 .B8813 1987
2. 85109601: Mezei, Andras. Magyar kocka, avagy, Meg mindig ilyen
gazdagok vagyunk? / Budapest : Magveto, c1984. 473 p. : ill.
; 21 cm.
LC CALL NUMBER: QA491 .M49 1984
3. 8272610: Feder, Happy Jack. Zen of cubing : in search of the
seventh side / 1st ed. South Bend, Ind. : And Books, c1982.
100 p. : ill. ; 21 cm.
LC CALL NUMBER: PN6231.R78 F42 1982
4. 823755: O'Grady, Miles. You can kick the cube] : the cube hater's
handbook / New York, N.Y. : Penguin Books, 1982. p. cm.
NOT IN LC COLLECTION
5. 821264: Bandelow, Christoph. Inside Rubik's cube and beyond /
Boston : Birkhauser, c1982. 120, [5] p., [6] leaves of plates :
ill. (some col.) ; 23 cm.
LC CALL NUMBER: QA491 .B2613 1982
6. 8185850: Schlafly, Roger. The complete cube book / Chicago :
Regnery Gateway, c1982. vi, 51 p. : ill. ; 21 cm.
LC CALL NUMBER: QA491 .S34 1982
7. 8181556: Taylor, Don. Mastering Rubik's cube : the solution to the
20th century's most amazing puzzle / 1st American ed. New York :
Holt, Rinehart and Winston, 1981, c1980. 31 p. : ill. ; 22 cm.
LC CALL NUMBER: QA491 .T38 1981
8. 8121650: Varasano, Jeffrey, 1966 Conquer the cube in 45 seconds /
New York : Bell Pub. Co. : Distributed by Crown Publishers, c1981.
48 p. : ill. ; 21 cm.
NOT IN LC COLLECTION
9. 8116682: Varasano, Jeffrey, 1966 Jeff conquers the cube in 45
seconds : and you can too] / New York : Stein and Day, 1981. p.cm.
NOT IN LC COLLECTION
10. 8112525: Frey, Alexander H. Handbook of cubik math / Hillside,
N.J. : Enslow Publishers, c1982. viii, 193 p. : ill. ; 23 cm.
LC CALL NUMBER: QA491 .F73 1982
11. 8027751: Singmaster, David. Notes on Rubik's magic cube /
Hillside, N.J. : Enslow Publishers, 1981. vi, 73 p. : ill.; 24 cm.
LC CALL NUMBER: QA491 .S58 1981
More to follow
> Mark <
From @mail.uunet.ca:mark.longridge@canrem.com Wed Dec 15 12:23:33 1993
ReturnPath: <@mail.uunet.ca:mark.longridge@canrem.com>
Received: from mail.uunet.ca (uunet.ca) by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA22894; Wed, 15 Dec 93 12:23:33 EST
Received: from portnoy.canrem.com ([198.133.42.251]) by mail.uunet.ca with SMTP id <55811(9)>; Wed, 15 Dec 1993 10:52:43 0500
Received: from canrem.com by portnoy.canrem.com (4.1/SMI4.1)
id AA18542; Wed, 15 Dec 93 10:50:49 EST
Received: by canrem.com (PCBUUCP 1.1f)
id 18E5EB; Wed, 15 Dec 93 10:46:25 0400
To: cubelovers@life.ai.mit.edu
ReplyTo: CRSO.Cube@canrem.com
Sender: CRSO.Cube@canrem.com
Subject: 6 X order 3
From: mark.longridge@canrem.com (Mark Longridge)
MessageId: <60.619.5834.0C18E5EB@canrem.com>
Date: Wed, 15 Dec 1993 09:42:00 0500
Organization: CRS Online (Toronto, Ontario)
A few messages back I mentioned a cyclicly decomposable process for
the pattern 6 X order 3. Success! Those familar with Christoph
Bandelow's "Inside Rubik's Cube and Beyond" will recognize the
notation, but for those who don't:
Mr is the middle slice adjacent to face R
Mu is the middle slice adjacent to face U (or T)
Mf is the middle slice adjacent to face F
Thus Mr1 rotates the middle slice in the same direction as r1,
etc. ...fairly intuitive.
The 28 slice moves are rather lengthy, but one can follow the
progression to 6 X order 3 easily. Before the discovery of
process p1b, memorization and execution of this pattern was
difficult. By memorization I don't mean retention for days or
weeks or even months as I wanted a CDtype process with which I
could always reconstruct it in my head.
Perhaps this could be improved upon, nevertheless now
the checkerboard order 3 is easy to execute and easy to remember!
(rotates edges 120 degrees around the FTR corner and BDL corner)
p1b alternate method 2 (Mr2 D3 Mr2 U1) ^3 TOP becomes LEFT (28s)
(Mr2 D3 Mr2 U1) ^3 LEFT becomes TOP
Mr3 Mt1 Mr1 Mt3
From @mail.uunet.ca:mark.longridge@canrem.com Wed Dec 15 13:04:34 1993
ReturnPath: <@mail.uunet.ca:mark.longridge@canrem.com>
Received: from mail.uunet.ca (uunet.ca) by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA25770; Wed, 15 Dec 93 13:04:34 EST
Received: from portnoy.canrem.com ([198.133.42.251]) by mail.uunet.ca with SMTP id <55713(9)>; Wed, 15 Dec 1993 11:01:33 0500
Received: from canrem.com by portnoy.canrem.com (4.1/SMI4.1)
id AA20478; Wed, 15 Dec 93 11:00:22 EST
Received: by canrem.com (PCBUUCP 1.1f)
id 18E5F0; Wed, 15 Dec 93 10:58:09 0400
To: cubelovers@life.ai.mit.edu
ReplyTo: CRSO.Cube@canrem.com
Sender: CRSO.Cube@canrem.com
Subject: Part 2
From: mark.longridge@canrem.com (Mark Longridge)
MessageId: <60.620.5834.0C18E5F0@canrem.com>
Date: Wed, 15 Dec 1993 09:58:00 0500
Organization: CRS Online (Toronto, Ontario)
Hmmmm, a little inconsistent (sp?) with the notation there.
I usually only use U when quoting the processes of others.
I'm going to try tackling the 1152fold symmetry idea
later tonight.
When looking for CDtype processes I find it helps to think
in terms of distinct states you pass through in approaching
the goal state. Sort of like factoring a composite pattern
into simpler ones you add together.
Rotating the cube in space definitely helps too.
> Mark
From dn1l+@andrew.cmu.edu Wed Dec 15 13:17:11 1993
ReturnPath:
Received: from po4.andrew.cmu.edu by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA26346; Wed, 15 Dec 93 13:17:11 EST
Received: from localhost (postman@localhost) by po4.andrew.cmu.edu (8.6.4/8.6.4) id NAA04816; Wed, 15 Dec 1993 13:16:56 0500
Received: via switchmail; Wed, 15 Dec 1993 13:16:53 0500 (EST)
Received: from loiosh.andrew.cmu.edu via qmail
ID ;
Wed, 15 Dec 1993 13:15:58 0500 (EST)
Received: from loiosh.andrew.cmu.edu via qmail
ID ;
Wed, 15 Dec 1993 13:15:46 0500 (EST)
Received: from mms.4.60.Nov..4.1993.10.47.44.sun4c.411.EzMail.2.0.CUILIB.3.45.SNAP.NOT.LINKED.loiosh.andrew.cmu.edu.sun4c.411
via MS.5.6.loiosh.andrew.cmu.edu.sun4c_411;
Wed, 15 Dec 1993 13:15:45 0500 (EST)
MessageId:
Date: Wed, 15 Dec 1993 13:15:45 0500 (EST)
From: "Dale I. Newfield"
To: cubelovers@ai.mit.edu
Subject: Re: Description of Tangle, Part 2
Cc: don.woods@eng.sun.com, acw@riverside.scrc.symbolics.com
InReplyTo: <920425084746.2bc000e4@EQL.Caltech.Edu>
Just to make sure everyone knows what we are talking about, here is a
message from the archives:
Excerpts from mail: 25Apr92 Description of Tangle, Part 2 by Chris
Worrell@eql.caltec
> Annotating Don.Woods diagram (which is in the correct orientation)
> 2 3
> 
>  @ # 
>  @ # 
> 1 $$ @ # %%%% 4
>  $ @ %#% 
>  $ @ %% # 
>  $ %@ # 
>  $ %% @@# 
>  %%% #@@ 
> 4 %%%% $ # @@@ 2
>  $ # 
>  $ # 
> 
> 1 3
>
> The duplicate piece in each tangle is:
> 1 2 3 4
> Tangle 1 Blue Red Yellow Green
> Tangle 2 Yellow Blue Green Red
> Tangle 3 Green Yellow Blue Red
> Tangle 4 Red Green Yellow Blue
>
> All 4 Tangles are the same puzzle, just colored differently.
> Each has all 24 color permutations, plus a duplicate.
I had kind of hoped that the connectivity on the different puzzles was
different, instead of just the colors.
(Actually, the sequence I sent before was slightly wronghere is the
one I actually used. Using Don's format)
>Don used the sequence: Dale used:
>
> 1 3 5 7 9 1 2 6 10 15
> 2 4 6 8 10 3 4 7 11 16
> 11 12 13 14 15 5 8 12 17 20
> 16 17 18 19 20 9 13 18 21 23
> 21 22 23 24 25 14 19 22 24 25
But yes, Don's fillpattern still gets more constraints in earlierhere
is the number of constraints at each step
Don's: 0 1 1 2 1 2 1 2 1 2 1 2 2 2 2 1 2 2 2 2 1 2 2 2 2
Mine: 0 1 1 2 1 1 2 2 1 1 2 2 2 1 1 2 2 2 2 2 2 2 2 2 2
As you can see, I had my 1's clustered more toward the beginning, which
is nonoptimal.
Assuming that there is only a change in color(and not in connectivity),
as was posted by Chris in april of 92, I would think modifying code to
attempt the 10x10 would be fairly simple...(seeing as my code went poof
sometime last year, when a disk crashed(not that it was
complicated))...wanna try?
(Thanks for the pointers to the Apr 92 discussion)
I agree with the concensus expressed in the archives that this puzzle is
inherently "not that great" because no nonbruteforce method has been
found/seems to exist.
Dale
From anandrao@hk.super.net Wed Dec 15 20:14:26 1993
ReturnPath:
Received: from hk.super.net by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA17907; Wed, 15 Dec 93 20:14:26 EST
Received: by hk.super.net id AA22615
(5.65c/IDA1.4.4 for CubeLovers@ai.mit.edu); Thu, 16 Dec 1993 09:14:05 +0800
Date: Thu, 16 Dec 1993 09:09:21 +0800 (HKT)
From: Mr Anand Rao
Subject: Re: Tangle
To: Don Woods
Cc: CubeLovers@ai.mit.edu
InReplyTo: <9312142248.AA22891@colossal.Eng.Sun.COM>
MessageId:
MimeVersion: 1.0
ContentType: TEXT/PLAIN; charset=USASCII
My method is essentially the same as yours  I have several intermediate
tables which cut down the processing required within the innermost loop. I
tried the same for 10*10 but it was taking eons. I tried making tables of
2*2 arrangements and solving for 5*5 of these(thereby solving the original
10*10, but the number of 2*2 arrangements makes the problem intractable on
my measly little computer. I even trie putting together all possible 5*5
solutions and assembling them in the 10*10 pattern, but the number of 5*5
solutions with the 100 tiles is in millions!
Do you have any further insight?
On Tue, 14 Dec 1993, Don Woods wrote:
> Anand Rao writes:
> > The puzzles are similar, except that the extra(25th)
> > piece is different in each. The solutions for each puzzle are very
> > different and I could not see any pattern.
>
> Look again. The puzzles are identical except for a remapping of the colors.
> For example, if you take Tangle #1 and paint all the Blue ropes Yellow, all
> the Red ropes Blue, all the Green ropes Red, and all the (originally) Yellow
> ropes Green, you'll have Tangle #2. So you can solve Tangle #1 by imagining
> the ropes recolored as above, constructing your solution for #2, and then
> restoring the original colors.
>
> Note: The particular recoloring given above is based on colors given in a
> message sent by CCW@eql.caltech.edu (Chris Worrell) to cubelovers on April
> 27, 1992. I own only #1 myself and so cannot confirm or deny the accuracy
> of the colors. But the basic idea applies, given that each puzzle (a) has
> the same pattern of ropes on all pieces and (b) has each permutation of
> colors exactly once except for one permutation which appears twice.
>
> Solving the 10x10 is another kettle of fish, and I haven't tried it. I do
> have a program that solves the 5x5 in about 45 seconds on a SparcStation II,
> but I haven't looked into how much longer it would take on the 10x10.
>
> "Dale I. Newfield" writes:
> > Could you explain what your algorithm was?
> > Has anyone found a nonbruteforce solution scheme?
>
> My solution was bruteforce. I posted to cubelovers (again, in April '92)
> asking if anyone had found a more logical approach to the puzzle, but got no
> affirmative responses.
>
> Dale's method is a little inefficient in the order in which it tries tiles.
> Mine used the sequence: Dale's used:
>
> 1 3 5 7 9 1 2 4 7 11
>
> 2 4 6 8 10 3 5 8 12 16
>
> 11 12 13 14 15 6 9 13 17 20
>
> 16 17 18 19 20 10 14 18 21 23
>
> 21 22 23 24 25 15 19 22 24 25
>
> The first three tiles in our two methods are equally constrained, but the
> next seven in Dale's methods are constrained along 1211221 edges,
> while mine are constrained along 2121212 edges. So I suspect my
> search tree gets trimmed a bit more quickly. Another way in which the
> search can be made more efficient is in finding the pieces to try in each
> position. For each pair of colors that can appear along an edge, my program
> precomputes a table listing all tiles that can match that pair of colors,
> including how to rotate the tiles.
>
>  Don.
>
>
From anandrao@hk.super.net Wed Dec 15 20:27:19 1993
ReturnPath:
Received: from hk.super.net by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA18806; Wed, 15 Dec 93 20:27:19 EST
Received: by hk.super.net id AA23068
(5.65c/IDA1.4.4 for cubelovers@ai.mit.edu); Thu, 16 Dec 1993 09:26:47 +0800
Date: Thu, 16 Dec 1993 09:17:27 +0800 (HKT)
From: Mr Anand Rao
Subject: Re: Description of Tangle, Part 2
To: "Dale I. Newfield"
Cc: cubelovers@ai.mit.edu, don.woods@eng.sun.com,
acw@riverside.scrc.symbolics.com
InReplyTo:
MessageId:
MimeVersion: 1.0
ContentType: TEXT/PLAIN; charset=USASCII
Just because no nonbruteforce method has been found, does not make this
puzzle any less intersting. As we have been told that there is a
solution, it is exciting to search for one, even by brute force methods.
The real challenge is to find a bruteforce method with sufficient insight
to solve the problem within a reasonable timeframe. All the algorithms so
far are exponential. We may never find a linear algorithm for this
problem. The idea is to find one algorithm that can be used in actual
practice. We can then bury this puzzle into the archives, for the next
generation to pick up!
> > (Thanks for the pointers to the Apr 92
discussion)
> I agree with the concensus expressed in the archives that this puzzle is
> inherently "not that great" because no nonbruteforce method has been
> found/seems to exist.
>
> Dale
>
Is this the reason why Rubik has gone into hiding? I haven't seen any
puzzle from him after this set of 4 released in 1990/1991. I tried to
contact the Hong Kong office of Matchbox which gets Rubik's puzzles in
China, but they have closed shop. Matchbox UK said that they have
discontinued this line. If anyone has found another source for Rubik's
puzzles, or discovered anyone else who has taken the responsibility of
giving us sleepless nights, please let me know!
From Don.Woods@eng.sun.com Wed Dec 15 20:39:18 1993
ReturnPath:
Received: from Sun.COM by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA19011; Wed, 15 Dec 93 20:39:18 EST
Received: from Eng.Sun.COM (zigzag.Eng.Sun.COM) by Sun.COM (4.1/SMI4.1)
id AA12769; Wed, 15 Dec 93 17:39:17 PST
Received: from colossal.Eng.Sun.COM by Eng.Sun.COM (4.1/SMI4.1)
id AA06793; Wed, 15 Dec 93 17:38:04 PST
Received: by colossal.Eng.Sun.COM (5.0/SMISVR4)
id AA26306; Wed, 15 Dec 93 17:39:20 PST
Date: Wed, 15 Dec 93 17:39:20 PST
From: Don.Woods@eng.sun.com (Don Woods)
MessageId: <9312160139.AA26306@colossal.Eng.Sun.COM>
To: cubelovers@ai.mit.edu
Subject: Re: Description of Tangle, Part 2
XSunCharset: USASCII
ContentLength: 897
> Is this the reason why Rubik has gone into hiding? I haven't seen any
> puzzle from him after this set of 4 released in 1990/1991.
Hm, didn't "Square1" come out later than the Tangles?
Regarding solving the Tangle, I forgot one other minor optimisation: When
my program is picking a corner piece other than the first, it requires that
the piece "number" be less than or equal to that of the first corner. I.e.,
it refuses to search for solutions that are rotations of other solutions.
I've modified my program to try the 10x10, but indeed, it's taking a long
time. (Current estimate is it will take over a year to finish.) I suspect
that fact that pieces aren't "used up" as fast  i.e., since there's at
least four of any given piece, there will usually be at least one of whatever
you're looking for for quite a ways down the search tree  makes this
approach intractible.
 Don.
From dik@cwi.nl Wed Dec 15 21:03:56 1993
ReturnPath:
Received: from charon.cwi.nl by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA20400; Wed, 15 Dec 93 21:03:56 EST
Received: from boring.cwi.nl by charon.cwi.nl with SMTP
id AA18235 (5.65b/3.12/CWIAmsterdam); Thu, 16 Dec 1993 03:03:19 +0100
Received: by boring.cwi.nl
id AA10975 (4.1/2.10/CWIAmsterdam); Thu, 16 Dec 93 03:03:19 +0100
Date: Thu, 16 Dec 93 03:03:19 +0100
From: Dik.Winter@cwi.nl
MessageId: <9312160203.AA10975.dik@boring.cwi.nl>
To: anandrao@hk.super.net
Subject: Re: Description of Tangle, Part 2
Cc: cubelovers@ai.mit.edu
My memory may be extremely faulty of course, but was there not more than
one single solution for the 5x5? (Not unprecedented, I have one puzzle
that promises a single solution but there are hundreds.) And, is there
a solution for the 10x10? I seem to remember that there was (or I had)
a convincing argument that such a thing did not exist. I should go through
the lesser used parts of my memory one of these days.
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland
home: bovenover 215, 1025 jn amsterdam, nederland; email: dik@cwi.nl
From dik@cwi.nl Wed Dec 15 21:40:59 1993
ReturnPath:
Received: from charon.cwi.nl by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA21903; Wed, 15 Dec 93 21:40:59 EST
Received: from boring.cwi.nl by charon.cwi.nl with SMTP
id AA18555 (5.65b/3.12/CWIAmsterdam); Thu, 16 Dec 1993 03:40:43 +0100
Received: by boring.cwi.nl
id AA11474 (4.1/2.10/CWIAmsterdam); Thu, 16 Dec 93 03:40:42 +0100
Date: Thu, 16 Dec 93 03:40:42 +0100
From: Dik.Winter@cwi.nl
MessageId: <9312160240.AA11474.dik@boring.cwi.nl>
To: Don.Woods@eng.sun.com
Subject: Re: Description of Tangle, Part 2
Cc: cubelovers@ai.mit.edu
> > Is this the reason why Rubik has gone into hiding? I haven't seen any
> > puzzle from him after this set of 4 released in 1990/1991.
>
> Hm, didn't "Square1" come out later than the Tangles?
Square1 is not by Rubik. But he came this year with two new puzzles (at
least, they are in his name). Rubik's Maze and Rubik's Hat.
In the first there are 6 connected cubes with a black/yellow pattern on
them. The cubes can turn around each other fairly freely. The purpose
is to get a 1x2x3 where there is a single black continuous line along
the cubes. Not very difficult, interesting.
Rubik's Hat is in the form of a hat with six rings on it. You can look
trough it (and through the rings by implication). By turning rings you
see more or less rabbits. The purpose is to see a rabbit in every position.
I think the puzzle is based on light polarization, with different
polarizations coming through the segments of the rings.
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland
home: bovenover 215, 1025 jn amsterdam, nederland; email: dik@cwi.nl
From anandrao@hk.super.net Thu Dec 16 01:12:12 1993
ReturnPath:
Received: from hk.super.net by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA29768; Thu, 16 Dec 93 01:12:12 EST
Received: by hk.super.net id AA08737
(5.65c/IDA1.4.4 for cubelovers@ai.mit.edu); Thu, 16 Dec 1993 14:12:01 +0800
Date: Thu, 16 Dec 1993 14:08:39 +0800 (HKT)
From: Mr Anand Rao
Subject: Re: Description of Tangle, Part 2
To: Dik.Winter@cwi.nl
Cc: cubelovers@ai.mit.edu
InReplyTo: <9312160203.AA10975.dik@boring.cwi.nl>
MessageId:
MimeVersion: 1.0
ContentType: TEXT/PLAIN; charset=USASCII
On Thu, 16 Dec 1993 Dik.Winter@cwi.nl wrote:
> My memory may be extremely faulty of course, but was there not more than
> one single solution for the 5x5? (Not unprecedented, I have one puzzle
> that promises a single solution but there are hundreds.) And, is there
> a solution for the 10x10? I seem to remember that there was (or I had)
> a convincing argument that such a thing did not exist. I should go through
> the lesser used parts of my memory one of these days.
For each Tangle, there are 2 solutions and no more. I have searched the
tree thoroughly and verified this. Counting 4 rotations and that 2 pieces
are identical, the total search gives 16 'solutions'.
The colourful little pamphlet that comes with the puzzle says that there
IS a solution to the 10*10 puzzle.
From anandrao@hk.super.net Thu Dec 16 01:19:38 1993
ReturnPath:
Received: from hk.super.net by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA29860; Thu, 16 Dec 93 01:19:38 EST
Received: by hk.super.net id AA09218
(5.65c/IDA1.4.4 for cubelovers@ai.mit.edu); Thu, 16 Dec 1993 14:19:24 +0800
Date: Thu, 16 Dec 1993 14:12:56 +0800 (HKT)
From: Mr Anand Rao
Subject: Re: Description of Tangle, Part 2
To: Don Woods
Cc: cubelovers@ai.mit.edu
InReplyTo: <9312160139.AA26306@colossal.Eng.Sun.COM>
MessageId:
MimeVersion: 1.0
ContentType: TEXT/PLAIN; charset=USASCII
>
> I've modified my program to try the 10x10, but indeed, it's taking a long
> time. (Current estimate is it will take over a year to finish.) I suspect
> that fact that pieces aren't "used up" as fast  i.e., since there's at
> least four of any given piece, there will usually be at least one of whatever
> you're looking for for quite a ways down the search tree  makes this
> approach intractible.
>
True. The 5*5 puzzle search truncates much faster because you run out of
pieces that could fit into a specific slot. The same does not apply to the
10*10 one.
Has anyone tried to solve the 10*10 for just 1 colour. That leaves you
with only 4 tile types with 24,25 or 26 of each type. The solution may
give some indication of the resultant pattern of the selected colour. If
there aren't too many solutions, maybe we can build the 4 colour solution
from this my permuting and rotating thr tiles. Any idea on how this will work?
From andyl@harlequin.com Thu Dec 16 10:45:09 1993
ReturnPath:
Received: from hilly.harlequin.com by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA01251; Thu, 16 Dec 93 10:45:09 EST
Received: from epcot.harlequin.com by hilly.harlequin.com; Thu, 16 Dec 1993 10:36:13 0500
Received: from phaedrus.harlequin.com (phaedrus) by epcot.harlequin.com; Thu, 16 Dec 1993 10:38:46 0500
From: Andy Latto
Date: Thu, 16 Dec 1993 10:38:45 0500
MessageId: <21332.199312161538@phaedrus.harlequin.com>
To: Don.Woods@eng.sun.com
Cc: cubelovers@ai.mit.edu
InReplyTo: Don Woods's message of Wed, 15 Dec 93 17:39:20 PST <9312160139.AA26306@colossal.Eng.Sun.COM>
Subject: Description of Tangle, Part 2
Date: Wed, 15 Dec 93 17:39:20 PST
From: Don.Woods@eng.sun.com (Don Woods)
XSunCharset: USASCII
ContentLength: 897
> Is this the reason why Rubik has gone into hiding? I haven't seen any
> puzzle from him after this set of 4 released in 1990/1991.
Hm, didn't "Square1" come out later than the Tangles?
Did Rubik have anything to do with Square1?
In any case, it's a great puzzle, and I recommend it to anyone on the
list who hasn't tried it. While there's a group structure lurking here
as usual, this is the only puzzle I've seen where the set of
attainable positions is not a subgroup. This means lots of the usual
ways of thinking about puzzles like this (e.g. conjugation) don't
always work, which makes it quite challenging.
Andy Latto
andyl@harlequin.com
From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Thu Dec 16 17:11:29 1993
ReturnPath: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU>
Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA21529; Thu, 16 Dec 93 17:11:29 EST
MessageId: <9312162211.AA21529@life.ai.mit.edu>
Received: from MITVMA.MIT.EDU by mitvma.mit.edu (IBM VM SMTP V2R2)
with BSMTP id 5892; Thu, 16 Dec 93 15:39:39 EST
Received: from WVNVM.WVNET.EDU (NJE origin MAILER@WVNVM) by MITVMA.MIT.EDU
(LMail V1.1d/1.7f) with BSMTP id 5420; Thu, 16 Dec 1993 15:39:38 0500
Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU
(LMail V1.1d/1.7f) with BSMTP id 7331; Thu, 16 Dec 1993 15:37:04 0500
XAcknowledgeTo:
Date: Thu, 16 Dec 1993 15:36:58 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: Duality of Operators and Operatees
I have mentioned several times my discomfort about "an operator"
as opposed to "the thing being operated on" when it comes to
groups. I am never quite sure just which of the two it is
that people are talking about, even (or especially) when I am listening
to myself talk. There is clearly an essential duality between the
two, but I am not sure I have quite a strong enough group theory
background to fully understand it. I am very comfortable when the
operators form a group, but I am not very comfortable when the things
being operated on form a group.
I am presently rereading (hopefully VERY SLOWLY
AND VERY CAREFULLY) Dan Hoey and Jim Saxe's seminal paper from
December 1980 entitled Symmetry and Local Maxima. Here is a quote
from their paper.
We will sometimes (particularly towards the end of this message)
take the liberty of identifying a transformation with the position
reached by applying that transformation to SOLVED.
Well, I am beginning to think that the source of my discomfiture
is simply that everybody does the same thing all the time, and that
nobody ever makes the identification explicit. However, I think that
maybe the duality is there, whether the identification is explicit,
implicit, or not made at all. Let me see if I can make clear what I
mean with some noncubing programming examples.
When I first started computer cubing, I was struck by the fact that
(at least with my model of the cube), the computer code to implement
a permutation operation looked exactly like the computer code to
translate between various character codes. For example, I have had
frequent occasion to translate between ASCII and EBCDIC (in both
directions). The code to translate between the ASCII string X and
the EBCDIC string Y is something like
for i = 1 to n Y(i) = T(X(i))
where T is the translate table. To make this clear by an example,
the ASCII code for the letter A is decimal 33 and the EBCDIC code
for the letter A is decimal 193. Hence, the 33rd position of T
contains decimal 193, and the 193rd position of T' contains 33.
Beyond this simple little loop above, many (if not most) programming
languages have a function (often called TRANSLATE or TRANSFORM) which
does exactly the same thing. There are also hardware architectures
which implement the TRANSLATE in hardware. For example, you might
have something like
Y = TRANSLATE(X,T)
where X is the string to be translated and T is the translate table.
X and T are clearly not interchangeable as input to the TRANSLATE
function. However, (and repeating myself) I think there is an
essential duality between X and T. For example, consider what
would happen if you reversed the role of X and T as follows. Let X
be the hexadecimal string 010201020301020403. Then,
Y = TRANSLATE(X,' ABC') would yield the string ' A AB ACB'. Such
a role reversal for the "permutation operator" and "permutation operatee"
can be a very powerful programming technique. For example, I have
used it to redistribute data, creating wellformatted print lines or
wellformatted display screens (text mode) with one fell swoop (with
only a single invocation of the TRANSLATE function).
I am going to continue reading, but perhaps I could pose a question to
Dan Hoey anyway: is reversing the role of X and T in the TRANSLATE
function above essentially the same thing as switching between
premultiplication and postmultiplication?
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
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 anandrao@hk.super.net Thu Dec 16 20:14:55 1993
ReturnPath:
Received: from hk.super.net by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA00360; Thu, 16 Dec 93 20:14:55 EST
Received: by hk.super.net id AA28316
(5.65c/IDA1.4.4 for cubelovers@ai.mit.edu); Fri, 17 Dec 1993 09:14:38 +0800
Date: Fri, 17 Dec 1993 09:13:20 +0800 (HKT)
From: Mr Anand Rao
Subject: Re: Description of Tangle, Part 2
To: Dik.Winter@cwi.nl
Cc: Don.Woods@eng.sun.com, cubelovers@ai.mit.edu
InReplyTo: <9312160240.AA11474.dik@boring.cwi.nl>
MessageId:
MimeVersion: 1.0
ContentType: TEXT/PLAIN; charset=USASCII
On Thu, 16 Dec 1993 Dik.Winter@cwi.nl wrote:
> Square1 is not by Rubik. But he came this year with two new puzzles (at
> least, they are in his name). Rubik's Maze and Rubik's Hat.
>
> In the first there are 6 connected cubes with a black/yellow pattern on
> them. The cubes can turn around each other fairly freely. The purpose
> is to get a 1x2x3 where there is a single black continuous line along
> the cubes. Not very difficult, interesting.
>
> Rubik's Hat is in the form of a hat with six rings on it. You can look
> trough it (and through the rings by implication). By turning rings you
> see more or less rabbits. The purpose is to see a rabbit in every position.
> I think the puzzle is based on light polarization, with different
> polarizations coming through the segments of the rings.
> dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland
> home: bovenover 215, 1025 jn amsterdam, nederland; email: dik@cwi.nl
Where can we get these puzzles from? Do you know of anyone who can take
credit card orders and mail?
From dik@cwi.nl Thu Dec 16 20:22:01 1993
ReturnPath:
Received: from charon.cwi.nl by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA00492; Thu, 16 Dec 93 20:22:01 EST
Received: from boring.cwi.nl by charon.cwi.nl with SMTP
id AA13191 (5.65b/3.12/CWIAmsterdam); Fri, 17 Dec 1993 02:21:55 +0100
Received: by boring.cwi.nl
id AA15294 (4.1/2.10/CWIAmsterdam); Fri, 17 Dec 93 02:21:54 +0100
Date: Fri, 17 Dec 93 02:21:54 +0100
From: Dik.Winter@cwi.nl
MessageId: <9312170121.AA15294.dik@boring.cwi.nl>
To: anandrao@hk.super.net
Subject: Re: Description of Tangle, Part 2
Cc: Don.Woods@eng.sun.com, cubelovers@ai.mit.edu
I would not know sources for Rubik's Maze and Rubik's Hat. They are on
sale in the local shops here. I have looked, the distributer is no
longer Matchbox but Parker, so that would imply availability in the
US I think.
dik

dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland
home: bovenover 215, 1025 jn amsterdam, nederland; email: dik@cwi.nl
From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Fri Dec 17 00:56:31 1993
ReturnPath: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU>
Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA11504; Fri, 17 Dec 93 00:56:31 EST
MessageId: <9312170556.AA11504@life.ai.mit.edu>
Received: from MITVMA.MIT.EDU by mitvma.mit.edu (IBM VM SMTP V2R2)
with BSMTP id 0779; Fri, 17 Dec 93 00:56:36 EST
Received: from WVNVM.WVNET.EDU (NJE origin MAILER@WVNVM) by MITVMA.MIT.EDU
(LMail V1.1d/1.7f) with BSMTP id 9437; Fri, 17 Dec 1993 00:56:36 0500
Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU
(LMail V1.1d/1.7f) with BSMTP id 3544; Fri, 17 Dec 1993 00:54:02 0500
XAcknowledgeTo:
Date: Fri, 17 Dec 1993 00:54:00 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: Some Additional Distances in the Edge Group
It is now known that using the qturn metric, Start has a
unique antipode in the edge group, namely MirrorImage
ofEdgesFlipped. The antipode is 15 qturns from Start.
Also, I have a complete data base of equivalence classes
in the edge group documenting the distance from Start for
each configuration of the edges.
It seems to me that given these two facts, some additional
distances can be determined. For example, it is possible
to determine the distance from any configuration to
MirrorImageofEdgesFlipped. Let Z be a sequence of operators
that converts Start to MirrorImageofEdgesFlipped, and let
A be any configuration of the edges. Then apply Z' to A, look
up the result in the data base of distances from Start, and
that will be the distance from A to MirrorImageofEdgesFlipped.
The reason is quite simple. Let P be a sequence which takes
Z'(A) to Start. Then, Z'PZ takes A to MirrorImageofEdgesFlipped.
This is a very nice use of conjugates.
Another consequence of this result is the following: suppose you
began with MirrorImageofEdgesFlipped and performed a
breadthfirst exhaustive search. Start would be antipodal, and
the number of nodes at each level of the tree would be identical
to the existing tree which begins at Start.
In addition, all of the above applies to MirrorImageofStart
and EdgesFlipped with respect to each other. They are
mutually antipodal, and are 15 qturns apart. A tree built with
either at the root would have exactly the same number of nodes
at each level as the existing tree with Start at the root.
Finally, the distance of any configuration from MirrorImageofStart
or EdgesFlipped can be determined. Let Y be a sequence of operators
which converts Start to MirrorImageofStart, and let X be a sequence
of operators that converts Start to EdgesFlipped. Let A be any
cube. Then, the distance of A from MirrorImageofStart
is the same as the distance of Y'(A) from Start, and
the distance of A from EdgesFlipped is the same
as the distance of X'(A) from Start.
I have the sensation in describing this that the Edge group is
square, with Start and MirrorImageofEdgesFlipped 180 degrees
apart, and MirrorImageofStart and EdgesFlipped at the other
two corners of the square.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
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 Fri Dec 17 11:24:53 1993
ReturnPath: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU>
Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA27504; Fri, 17 Dec 93 11:24:53 EST
MessageId: <9312171624.AA27504@life.ai.mit.edu>
Received: from MITVMA.MIT.EDU by mitvma.mit.edu (IBM VM SMTP V2R2)
with BSMTP id 5191; Fri, 17 Dec 93 11:24:42 EST
Received: from WVNVM.WVNET.EDU (NJE origin MAILER@WVNVM) by MITVMA.MIT.EDU
(LMail V1.1d/1.7f) with BSMTP id 0812; Fri, 17 Dec 1993 11:24:42 0500
Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU
(LMail V1.1d/1.7f) with BSMTP id 8836; Fri, 17 Dec 1993 11:22:06 0500
XAcknowledgeTo:
Date: Fri, 17 Dec 1993 11:21:51 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: Size of the Cube Group
In 1984, Dan Hoey posed a question as follows:
>This discussion of symmetry recalls a question I have meant to propose
>to CubeLovers for some time: How many positions are there in Rubik's
>Cube? We know from Ideal that the number is somewhat over three
>billion. Most cube lovers will tell you a number of about 43
>quintillion. But I really don't see why we should count twelve
>distinct positions at one quartertwist from solvedall twelve are
>essentially the same position. So the question, suitably rephrased, is
>of the number of positions that are distinct up to conjugacy in M, the
>48element symmetry group of the cube. I think this is an interesting
>question, but I don't see any particularly easy way of answering it.
>My best guess is that it involves a casebycase analysis of the 98
>subgroups of M, or at least the 33 conjugacy classes of those
>subgroups. In ``Symmetry and Local Maxima'', Jim Saxe and I examined
>five of the classes, which we called M, C, AM, H, and T.
>
>Even finding the numbers for the pocket cube is a little tricky. If we
>limit ourselves to symmetry in S, I believe the pocket cube has 2
>positions with a sixelement symmetry group, 160 positions with a
>threeelement symmetry group, 3882 positions with a twoelement
>symmetry group, and 3670116 positions with a oneelement symmetry
>group, for 613062 positions distinct up to Sconjugacy. But the
>numbers for Mconjugacy are still elusive; I am not even sure how to
>deal with factoring out wholecube moves in the analysis. I hope to
>find time to write a program for it.
>
>I expanded my pocket cube program to deal with the corner group of
>Rubik's cube. This group is 24 times as large as the group of the
>pocket cube, having 3^7 * 8! = 88179840 elements. The number of
>elements P(N) and local maxima L(N) at each (quartertwist) distance N
>from solved are given below.
>
> N P(N) L(N)
> 0 1 0
> 1 12 0
> 2 114 0
> 3 924 0
> 4 6539 0
> 5 39528 0
> 6 199926 114
> 7 806136 600
> 8 2761740 17916
> 9 8656152 10200
> 10 22334112 35040
> 11 32420448 818112
> 12 18780864 9654240
> 13 2166720 2127264
> 14 6624 6624
>
>The alert reader will notice that rows 10 through 14 contain values
>exactly 24 times as large as those for the pocket cube. This is not
>surprising, given that the groups are identical except for the position
>of the entire assembly in space, and each generator of the corner cube
>is identical to the inverse of the corresponding generator for the
>opposite face except for the wholecube position. Thus when solving a
>cornercube position at 10 qtw or more from solved, it can be solved as
>a pocket cube, making the choice between opposite faces in such a way
>that the wholecube position comes out right with no extra moves.
>
I wish to propose an answer to Dan's question. I will propose an
approximation then (hopefully) the exact answer.
The approximation is simply 4.3*(10^19) / 1152, or about
3.7*(10^16). 1152=24*24*2, and is based on my version of Dan's
M symmetry group. I remain convinced that my version of M is
isomorphic to Dan's, but the subject deserves some more thought
and discussion.
But we can do better. We already know (under my version of M) how
many equivalence classes there are for the corner group (namely,
77,802). But each of the equivalence classes for the corners can
be rotated 24 ways with respect to the centers, so we have
77,802*24. We also already know (under my version of M) how many
equivalence classes there are for the edge group (namely
851,625,008). But each of the equivalence classes for the edges
can be rotated 24 ways with respect to the centers, so we have
851,625,008*24. Hence, we have
(77,802*24) * (851,625,008*24) = 38,164,682,230,511,620
This figure is gratifyingly close to 3.7*(10^16), and I believe it
is the correct answer to Dan's question. It is slightly larger
than the approximation because some of the equivalence classes
have fewer than 1152 elements, and consequently there are a few
more equivalence classes than the approximation suggests.
However, the alert reader should have noticed a problem. Why did I
not divide by 2 to take into account the fact that odd edge
permutations can only occur with odd corner permutations and vice
versa? Actually, I did, but the division by 2 cancelled. The reason
it canceled is slightly tricky. Also, remember that we are talking
about equivalence classes, not specific cube configurations. Any
equivalence class has both even and odd members, depending on how
the members are rotated. Hence, any corner equivalence class can be
matched up with any edge equivalence class, assuming the rotations
are compatible. But you still have to worry about "dividing by 2",
as follows.
Let G be the number of states of the whole cube without M, namely
the 4.3*(10^19) figure, and similarly let C be the number of states of
the corners without M and let E be the number of states of the edges
without M. Then, we have the trivial relation G = C * E / 2.
Here, the division by 2 does properly reflect the odd/even parity
of the corners vs. the edges.
Let Gm = G / (24*24*2), Cm = C / (24*24*2), and Em = E / (24*24*2).
Hence, G = Gm * (24*24*2), C = Cm * (24*24*2), and E = Em * (24*24*2).
What I have available (approximately) is Cm and Em, and what I want
is Gm. Hence,
Gm = G / (24*24*2)
Gm = (C * E / 2) / (24*24*2)
Gm = ((Cm * (24*24*2)) * (Em * (24*24*2)) / 2) / (24*24*2)
Gm = (Cm*24) * (Em*24)
Therefore, I replace Cm by the real figure for the number of corner
equivalence classes, replace Em by the real figure for the number
of equivalence classes, and Gm becomes the real figure for the total
states of the cube. The "division by 2" is in the formula, but it is
invisible because of all the cancellations.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 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 Fri Dec 17 14:25:29 1993
ReturnPath: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU>
Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA08679; Fri, 17 Dec 93 14:25:29 EST
MessageId: <9312171925.AA08679@life.ai.mit.edu>
Received: from MITVMA.MIT.EDU by mitvma.mit.edu (IBM VM SMTP V2R2)
with BSMTP id 7700; Fri, 17 Dec 93 14:25:17 EST
Received: from WVNVM.WVNET.EDU (NJE origin MAILER@WVNVM) by MITVMA.MIT.EDU
(LMail V1.1d/1.7f) with BSMTP id 8890; Fri, 17 Dec 1993 14:25:14 0500
Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU
(LMail V1.1d/1.7f) with BSMTP id 2064; Fri, 17 Dec 1993 14:22:36 0500
XAcknowledgeTo:
Date: Fri, 17 Dec 1993 14:22:34 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: Re: Size of the Cube Group
InReplyTo: Message of 12/17/93 at 11:21:51 from ,
BRYAN%WVNVM.BITNET@mitvma.mit.edu
On 12/17/93 at 11:21:51 Jerry Bryan said:
>However, the alert reader should have noticed a problem. Why did I
>not divide by 2 to take into account the fact that odd edge
>permutations can only occur with odd corner permutations and vice
>versa? Actually, I did, but the division by 2 cancelled. The reason
>it canceled is slightly tricky. Also, remember that we are talking
>about equivalence classes, not specific cube configurations. Any
>equivalence class has both even and odd members, depending on how
^^^^^^^^^^^^^^^^^^^^^^^^^
>the members are rotated. Hence, any corner equivalence class can be
>matched up with any edge equivalence class, assuming the rotations
>are compatible. But you still have to worry about "dividing by 2",
>as follows.
It is pretty bad when you have to followup with
corrections to your own posts. I hurried to complete the previous
post before lunch, and just didn't think clearly enough  till I
had time to think *during* lunch. Let's try this again.
A qturn of the whole cube (a 90 degree rotation of the whole cube)
is odd. However, if you think of a qturn rotation of the whole cube
as disjoint between edges and corners, a qturn rotation of the
corners is even, and a qturn rotation of the edges is odd. Hence,
for any equivalence class of the corners under M, either the whole
equivalence class is even, or the whole equivalence class is odd.
For any equivalence class of the edges under M, half of the equivalence
class is even and half is odd. Thus, any equivalence class of the
corners can occur with any equivalence class of the edges, but with only
half the members of the edge equivalence class  namely those with
the same parity.
I believe my calculations were correct, but a piece of the justification
was not. I hope I am not still missing something. You do have to
"divide by 2", and my calculations do indeed "divide by 2" as previously
described, but the parity of edges vs. the parity of corners was
incorrect in the previous post.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 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 18 17:08:38 1993
ReturnPath: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU>
Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA00538; Sat, 18 Dec 93 17:08:38 EST
MessageId: <9312182208.AA00538@life.ai.mit.edu>
Received: from MITVMA.MIT.EDU by mitvma.mit.edu (IBM VM SMTP V2R2)
with BSMTP id 5020; Sat, 18 Dec 93 14:04:37 EST
Received: from WVNVM.WVNET.EDU (NJE origin MAILER@WVNVM) by MITVMA.MIT.EDU
(LMail V1.1d/1.7f) with BSMTP id 8380; Sat, 18 Dec 1993 14:04:37 0500
Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU
(LMail V1.1d/1.7f) with BSMTP id 0957; Sat, 18 Dec 1993 14:02:02 0500
XAcknowledgeTo:
Date: Sat, 18 Dec 1993 14:02:01 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: Second Addendum  Size of Cube Group under M
I feel like I am pestering the list to death with corrections.
I still believe that the figure that I proposed for the size of
the cube group under M is correct. The first post included
a "correct" but I think unsatisfactory explanation. The second post
improved upon one point that was unsatisfactory in the first post.
Now, let's see if I can get it completely correct.
The size of the corner group under (my version of) M is known.
The size of the edge group under (my version of) M is known as
well. Let C be the size of the corner group, and E be the size
of the edge group. Remember, the elements of the groups are
equivalence classes induced by (my version of) M. Here is an incorrect
formula for G, the size of the entire cube group under (my
version of) M.
G = (C*24) * (E*24) / 2
The division by 2 is introduced to account for parity between the
corner group and the edge group. But the value for G produced by
this formula is only half as big as it should be. The problem is
that M induces equivalence classes based on both rotations and
reflections, not just base on rotations. Hence, we are led to the
following (still incorrect) formula:
G = (C*24*2) * (E*24*2) / 2
As before, the division by 2 takes care of parity between the corner
group and the edge group. In addition, the multiplication by 2 takes
care of reflecting each group. But the value for G produced by this
formula is twice as big as it should be. The problem is that while
any corner rotation can occur with any edge rotation (subject to
parity), you must either reflect both groups, or else reflect neither
group. Thus, we have the following (correct) formula:
G = ((C*24) * (E*24) / 2) * 2
The division by 2 takes care of parity between the groups, and the
multiplication by 2 takes care of reflection of the two groups
as a unit. If we wish, we can cancel the multiplication and the
division to yield
G = (C*24) * (E*24)
This is the same formula I originally posted, and I did say in the
original post that the division by 2 cancelled out. However, I
think that this post provides a better explanation of the
cancellation than did the original post.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 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 xirion!jandr@relay.nl.net Mon Dec 20 06:35:54 1993
ReturnPath:
Received: from sun4nl.NL.net by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA00411; Mon, 20 Dec 93 06:35:54 EST
Received: from xirion by sun4nl.NL.net via EUnet
id AA05737 (5.65b/CWI3.3); Mon, 20 Dec 1993 10:39:28 +0100
Received: by xirion.xirion.nl id AA03876 (5.61/UK2.1);
Mon, 20 Dec 93 10:38:37 +0100
From: Jan de Ruiter
Date: Mon, 20 Dec 93 10:38:37 +0100
MessageId: <3876.9312200938@xirion.xirion.nl>
XOrganization: Xirion Unix Software & Consultancy bv
Burgemeester Verderlaan 15 X
3454 PE De Meern
The Netherlands
XPhone: +31 3406 61990
XFax: +31 3406 61981
To: cubelovers@ai.mit.edu
To: cubelovers@ai.mit.edu
Subject: Re: Search order of Tangle
I saw the discussion of Dale and Don about the search order
(fillpattern) for rubiks tangle come by, and wondered why they both
missed an even better search order (the best?):
Don: Dale: Jan: Equivalent to:
1 3 5 7 9 1 2 6 10 15 1 2 5 10 17 17 16 15 14 13
2 4 6 8 10 3 4 7 11 16 3 4 6 11 18 18 5 4 3 12
11 12 13 14 15 5 8 12 17 20 7 8 9 12 19 19 6 1 2 11
16 17 18 19 20 9 13 18 21 23 13 14 15 16 20 20 7 8 9 10
21 22 23 24 25 14 19 22 24 25 21 22 23 24 25 21 22 23 24 25
The number of constraints is illustrative:
don: 0 1 1 2 1 2 1 2 1 2 1 2 2 2 2 1 2 2 2 2 1 2 2 2 2
dale: 0 1 1 2 1 1 2 2 1 1 2 2 2 1 1 2 2 2 2 2 2 2 2 2 2
jan: 0 1 1 2 1 2 1 2 2 1 2 2 1 2 2 2 1 2 2 2 1 2 2 2 2
I disliked the irregularity in both don and dales search orders, and
in search for a more regular order, I found this one, which is better.
It is readily extendible to the 10 by 10 tangle.
 Jan D. de Ruiter
From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Mon Dec 20 06:43:01 1993
ReturnPath: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU>
Received: from by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AB00195; Mon, 20 Dec 93 06:43:01 EST
MessageId: <9312201143.AB00195@life.ai.mit.edu>
Received: from MITVMA.MIT.EDU by mitvma.mit.edu (IBM VM SMTP V2R2)
with BSMTP id 0849; Mon, 20 Dec 93 00:46:03 EST
Received: from WVNVM.WVNET.EDU (NJE origin MAILER@WVNVM) by MITVMA.MIT.EDU
(LMail V1.1d/1.7f) with BSMTP id 3167; Mon, 20 Dec 1993 00:46:03 0500
Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU
(LMail V1.1d/1.7f) with BSMTP id 3756; Mon, 20 Dec 1993 00:43:28 0500
XAcknowledgeTo:
Date: Mon, 20 Dec 1993 00:43:27 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: Process for Antipodal of the Edge Group
This is the first process I have found for the antipodal of Start
in the edge group (edges without corners and without centers).
There are certainly many more, but I have not yet cataloged them
all.
FR'LFL'B'R'FL'FRBL'BL'
Note that this process (as with any process for the antipodal) is
its own inverse. Hence, you can use it once to get from Start to
the antipodal, and again to get from the antipodal to Start. Also,
the "natural" inverse (namely, LB'LB'R'F'LF'RBLF'L'RF') is also
a process which will go in either direction, Start to the antipodal,
or antipodal to Start.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 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 xirion!jandr@relay.nl.net Mon Dec 20 07:31:08 1993
ReturnPath:
Received: from sun4nl.NL.net by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA04598; Mon, 20 Dec 93 07:31:08 EST
Received: from xirion by sun4nl.NL.net via EUnet
id AA15711 (5.65b/CWI3.3); Mon, 20 Dec 1993 13:31:05 +0100
Received: by xirion.xirion.nl id AA04506 (5.61/UK2.1);
Mon, 20 Dec 93 13:30:27 +0100
From: Jan de Ruiter
Date: Mon, 20 Dec 93 13:30:27 +0100
MessageId: <4506.9312201230@xirion.xirion.nl>
XOrganization: Xirion Unix Software & Consultancy bv
Burgemeester Verderlaan 15 X
3454 PE De Meern
The Netherlands
XPhone: +31 3406 61990
XFax: +31 3406 61981
To: cubelovers@life.ai.mit.edu
To: cubelovers@life.ai.mit.edu
Subject: Re: Search order of Tangle
I saw the discussion of Dale and Don about the search order
(fillpattern) for rubiks tangle come by, and wondered why they both
missed an even better search order (the best?):
Don: Dale: Jan: Equivalent to:
1 3 5 7 9 1 2 6 10 15 1 2 5 10 17 17 16 15 14 13
2 4 6 8 10 3 4 7 11 16 3 4 6 11 18 18 5 4 3 12
11 12 13 14 15 5 8 12 17 20 7 8 9 12 19 19 6 1 2 11
16 17 18 19 20 9 13 18 21 23 13 14 15 16 20 20 7 8 9 10
21 22 23 24 25 14 19 22 24 25 21 22 23 24 25 21 22 23 24 25
The number of constraints is illustrative:
don: 0 1 1 2 1 2 1 2 1 2 1 2 2 2 2 1 2 2 2 2 1 2 2 2 2
dale: 0 1 1 2 1 1 2 2 1 1 2 2 2 1 1 2 2 2 2 2 2 2 2 2 2
jan: 0 1 1 2 1 2 1 2 2 1 2 2 1 2 2 2 1 2 2 2 1 2 2 2 2
I disliked the irregularity in both don and dales search orders, and
in search for a more regular order, I found this one, which is better.
It is readily extendible to the 10 by 10 tangle.
 Jan D. de Ruiter
From @cannon.ecf.toronto.edu:malone@ecf.toronto.edu Mon Dec 20 14:17:33 1993
ReturnPath: <@cannon.ecf.toronto.edu:malone@ecf.toronto.edu>
Received: from cannon.ecf.toronto.edu by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA24464; Mon, 20 Dec 93 14:17:33 EST
Received: by cannon.ecf.toronto.edu id <7382>; Mon, 20 Dec 1993 14:17:19 0500
From: MALONE MATTHEW JAMES
To: cubelovers@ai.mit.edu
Subject: Please remove ...
MessageId: <93Dec20.141719edt.7382@cannon.ecf.toronto.edu>
Date: Mon, 20 Dec 1993 14:17:10 0500
Please remove malone@ecf.toronto.edu from the cubelovers list.
Thanks
Matt
From Don.Woods@eng.sun.com Mon Dec 20 19:21:46 1993
ReturnPath:
Received: from Sun.COM by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA10335; Mon, 20 Dec 93 19:21:46 EST
Received: from Eng.Sun.COM (zigzag.Eng.Sun.COM) by Sun.COM (4.1/SMI4.1)
id AA04625; Mon, 20 Dec 93 16:21:45 PST
Received: from colossal.Eng.Sun.COM by Eng.Sun.COM (4.1/SMI4.1)
id AA17679; Mon, 20 Dec 93 16:20:26 PST
Received: by colossal.Eng.Sun.COM (5.0/SMISVR4)
id AA10356; Mon, 20 Dec 93 16:21:50 PST
Date: Mon, 20 Dec 93 16:21:50 PST
From: Don.Woods@eng.sun.com (Don Woods)
MessageId: <9312210021.AA10356@colossal.Eng.Sun.COM>
To: cubelovers@ai.mit.edu
Subject: Re: Search order of Tangle
Cc: jandr@xirion.nl
XSunCharset: USASCII
ContentLength: 1571
> I saw the discussion of Dale and Don about the search order
> (fillpattern) for rubiks tangle come by, and wondered why they both
> missed an even better search order (the best?):
>
> Don: Dale: Jan: Equivalent to:
> 1 3 5 7 9 1 2 6 10 15 1 2 5 10 17 17 16 15 14 13
> 2 4 6 8 10 3 4 7 11 16 3 4 6 11 18 18 5 4 3 12
> 11 12 13 14 15 5 8 12 17 20 7 8 9 12 19 19 6 1 2 11
> 16 17 18 19 20 9 13 18 21 23 13 14 15 16 20 20 7 8 9 10
> 21 22 23 24 25 14 19 22 24 25 21 22 23 24 25 21 22 23 24 25
I missed it on the 5x5 because my program was fast enough that I didn't
look further. When I modified my program to try the 10x10 last week, I
did come up with the ordering Jan suggests. It shaved about 1/3 the
running time off my 5x5 search, but it actually doesn't seem to make
that big a difference in the 10x10.
It turns out the 10x10 search isn't quite as bad as I thought, because
the tree does get trimmed rather early. When a piece is constrained on
two edges, there are on average only 2/3 choices for that piece. I've
got my program chugging along, and so far it has eliminated 4 of the 96
choices for piece (w/ orientation) for the upper left corner. There are
4896 choices for the first 4 points in the search order, and it's going
through one choice per 25 minutes on average, so it'll finish in a mere
3 months, if I have the patience for it. (I may try to dig up some
otherwise idle workstations to leave running over the holiday break.)
 Don.
From anandrao@hk.super.net Mon Dec 20 20:16:04 1993
ReturnPath:
Received: from hk.super.net by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA13155; Mon, 20 Dec 93 20:16:04 EST
Received: by hk.super.net id AA23740
(5.65c/IDA1.4.4 for cubelovers@ai.mit.edu); Tue, 21 Dec 1993 09:15:43 +0800
Date: Tue, 21 Dec 1993 09:13:38 +0800 (HKT)
From: Mr Anand Rao
Subject: Re: your mail
To: Jan de Ruiter
Cc: cubelovers@ai.mit.edu
InReplyTo: <3876.9312200938@xirion.xirion.nl>
MessageId:
MimeVersion: 1.0
ContentType: TEXT/PLAIN; charset=USASCII
Your concept is theoretically extendable to the 10*10 tangle, but even
with this optimisation the puzzle would take a long time to solve. How
long do you take for the 5*5 Tangle on your computer?
On Mon, 20 Dec 1993, Jan de Ruiter wrote:
> To: cubelovers@ai.mit.edu
> Subject: Re: Search order of Tangle
>
> I saw the discussion of Dale and Don about the search order
> (fillpattern) for rubiks tangle come by, and wondered why they both
> missed an even better search order (the best?):
>
> Don: Dale: Jan: Equivalent to:
> 1 3 5 7 9 1 2 6 10 15 1 2 5 10 17 17 16 15 14 13
> 2 4 6 8 10 3 4 7 11 16 3 4 6 11 18 18 5 4 3 12
> 11 12 13 14 15 5 8 12 17 20 7 8 9 12 19 19 6 1 2 11
> 16 17 18 19 20 9 13 18 21 23 13 14 15 16 20 20 7 8 9 10
> 21 22 23 24 25 14 19 22 24 25 21 22 23 24 25 21 22 23 24 25
>
> The number of constraints is illustrative:
> don: 0 1 1 2 1 2 1 2 1 2 1 2 2 2 2 1 2 2 2 2 1 2 2 2 2
> dale: 0 1 1 2 1 1 2 2 1 1 2 2 2 1 1 2 2 2 2 2 2 2 2 2 2
> jan: 0 1 1 2 1 2 1 2 2 1 2 2 1 2 2 2 1 2 2 2 1 2 2 2 2
>
> I disliked the irregularity in both don and dales search orders, and
> in search for a more regular order, I found this one, which is better.
> It is readily extendible to the 10 by 10 tangle.
>
>  Jan D. de Ruiter
From pbeck@pica.army.mil Tue Dec 21 00:24:43 1993
ReturnPath:
Received: from COR6.PICA.ARMY.MIL by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA22931; Tue, 21 Dec 93 00:24:43 EST
Date: Mon, 20 Dec 93 8:33:55 EST
From: Peter Beck (BATDD)
To: CubeLovers@ai.mit.edu
Cc: pbeck@pica.army.mil
Subject: test
MessageId: <9312200833.aa09624@COR6.PICA.ARMY.MIL>
i am having trouble posting.
please excuse this message
From xirion!jandr@relay.nl.net Tue Dec 21 02:45:28 1993
ReturnPath:
Received: from sun4nl.NL.net by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA24967; Tue, 21 Dec 93 02:45:28 EST
Received: from xirion by sun4nl.NL.net via EUnet
id AA28145 (5.65b/CWI3.3); Tue, 21 Dec 1993 08:45:07 +0100
Received: by xirion.xirion.nl id AA00997 (5.61/UK2.1);
Tue, 21 Dec 93 08:43:59 +0100
From: Jan de Ruiter
Date: Tue, 21 Dec 93 08:43:59 +0100
MessageId: <997.9312210743@xirion.xirion.nl>
XOrganization: Xirion Unix Software & Consultancy bv
Burgemeester Verderlaan 15 X
3454 PE De Meern
The Netherlands
XPhone: +31 3406 61990
XFax: +31 3406 61981
To: anandrao@hk.super.net, cubelovers@ai.mit.edu
Subject: Re: Rubiks tangle
To: anandrao@hk.super.net
Cc: cubelovers@ai.mit.edu
>Your concept is theoretically extendable to the 10*10 tangle, but even
>with this optimisation the puzzle would take a long time to solve. How
>long do you take for the 5*5 Tangle on your computer?
I am sorry to say I haven't implemented the search yet.
The 5x5 is solved, so that isn't that interesting anymore; the 10x10
has such a huge search space, that it will need a very efficient
algorithm and/or clever representation. I just haven't decided on the
representation yet. I did decide on the search order though.
 Jan D. de Ruiter
From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Tue Dec 21 09:27:36 1993
ReturnPath: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU>
Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA01929; Tue, 21 Dec 93 09:27:36 EST
MessageId: <9312211427.AA01929@life.ai.mit.edu>
Received: from MITVMA.MIT.EDU by mitvma.mit.edu (IBM VM SMTP V2R2)
with BSMTP id 2781; Tue, 21 Dec 93 08:56:49 EST
Received: from WVNVM.WVNET.EDU (NJE origin MAILER@WVNVM) by MITVMA.MIT.EDU
(LMail V1.1d/1.7f) with BSMTP id 6664; Tue, 21 Dec 1993 08:56:48 0500
Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU
(LMail V1.1d/1.7f) with BSMTP id 0289; Tue, 21 Dec 1993 08:54:13 0500
XAcknowledgeTo:
Date: Tue, 21 Dec 1993 08:54:12 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: 9 Qturn Sequence for AllEdgesFlipped in the Edge Group
RUB DRB LDB (the spaces are for readability only).
Remember that this is for the "edges without corners and without centers"
case. Hence, the edges are all flipped and are all properly configured
with respect to each other, but they are not flipped "in place" with
respect to a fixed coordinate system of centers. They are rotated with
respect a fixed coordinate system of centers.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 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 pbeck@pica.army.mil Tue Dec 21 18:32:39 1993
ReturnPath:
Received: from COR6.PICA.ARMY.MIL by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA14683; Tue, 21 Dec 93 18:32:39 EST
Date: Tue, 21 Dec 93 7:49:23 EST
From: Peter Beck (BATDD)
To: CubeLovers@ai.mit.edu
Cc: pbeck@pica.army.mil
Subject: puzzle party
MessageId: <9312210749.aa08453@COR6.PICA.ARMY.MIL>
ROBERT HOLBROOK
11837 LINDEN CHAPEL ROAD
CLARKSVILLE, MD 21029
4105316135
IS Planning a puzzle party
for FEB 19,20 1994
at his home.
IF YOU ARE INTERESTED PLEASE CONTACT BOB
directly.
Clarksville is 1/2 way between DC and Baltimore.
VENUE is low keyed with trading, buying, selling and
TALKING.
THE FUTURE IS PUZZLING,
BUT CUBING IS FOREVER !!!
From pbeck@pica.army.mil Tue Dec 21 19:05:28 1993
ReturnPath:
Received: from COR6.PICA.ARMY.MIL by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA15488; Tue, 21 Dec 93 19:05:28 EST
Date: Tue, 21 Dec 93 7:51:39 EST
From: Peter Beck (BATDD)
To: cubelovers@ai.mit.edu
Cc: pbeck@pica.army.mil
Subject: puzzle party
MessageId: <9312210751.aa09222@COR6.PICA.ARMY.MIL>
ROBERT HOLBROOK
11837 LINDEN CHAPEL ROAD
CLARKSVILLE, MD 21029
4105316135
IS Planning a puzzle party
for FEB 19,20 1994
at his home.
IF YOU ARE INTERESTED PLEASE CONTACT BOB
directly.
Clarksville is 1/2 way between DC and Baltimore.
VENUE is low keyed with trading, buying, selling and
TALKING.
THE FUTURE IS PUZZLING,
BUT CUBING IS FOREVER !!!
From hoey@aic.nrl.navy.mil Wed Dec 22 13:58:45 1993
Received: from Sun0.AIC.NRL.Navy.Mil by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA11979; Wed, 22 Dec 93 13:58:45 EST
Received: from sun1.aic.nrl.navy.mil by Sun0.AIC.NRL.Navy.Mil (4.1/SMI4.0)
id AA20803; Wed, 22 Dec 93 13:58:43 EST
ReturnPath:
Received: by sun1.aic.nrl.navy.mil; Wed, 22 Dec 93 13:58:42 EST
Date: Wed, 22 Dec 93 13:58:42 EST
From: hoey@aic.nrl.navy.mil
MessageId: <9312221858.AA08479@sun1.aic.nrl.navy.mil>
To: CubeLovers@ai.mit.edu
Subject: The 4^3 and 3^4 Rubik puzzles
Organization: Naval Research Laboratory, Washington, DC
[ CubeLovers,
There has recently been a discussion on Usenet group rec.puzzles
about some cube topics. There were a few pieces of new information,
such as that you can now get Ishi's 5^3 cubes in a lot of places (I
got mine in Learningsmith's) for about $35. Here's a message I sent
that's relevant to some CubeLovers topics.
By the way, I'm still working through Jerry Bryan's articles on his
bruteforce program and his approach to symmetry. I hope to get a
reply out soon.
]
eric@gsb002.cs.ualberta.ca (Holleman Eric) wrote:
> By the way, I found the Revenge somewhat easier than the Cube, and I
> don't think that it was because of my familiarity with the earlier
> puzzle.
x87bennett@gw.wmich.edu (Joe) wrote:
> From my experience, if you can solve a Rubik's Revenge, you can
> solve the Cube very easily. Once you get each of the middle 2 cubes
> on each edge to match, and all 4 center cubes on each face to match,
> it works exactly like a Rubik's cube.
and alan@saturn.cs.swin.oz.au (Alan Christiansen) wrote:
> I have both. I solved both. The 4x4x4 is a superset of the 3x3x3.
> ie by fixing all the face centres and then pairing all edges you are
> left with a 3x3x3 cube, except that when you have solved this 3x3x3
> there may be a single pair of edges flipped. This is impossible
> on a real 3x3x3. Fixing this requires a middle layer to be rotated
> 1/4 revolution and then all the bits put back.
> I cant see how it can be [any] easier than a 3x3x3.
I, too, found the 3^3 easier than the 4^3. But I can imagine ways in
which a solver could find the 4^3 easier. Let us first consider a 4^3
with the faces fixed, the edges together, and the correct simulated
edge flip parity. I would solve this as if it were a 3^3, and a lot
of people do. But another solver might find it easier to take
advantage of the extra moves that are not possible on a 3^3. To take
a concrete example, it could be that the solver has a hard time with
flipping edges by pairs, as is needed to solve the 3^3. On the 4^3
you can flip one edge at a time. So the solver would find the 4^3
position easier than the corresponding position on a 3^3. If the
solver finds this so much easier that it overcomes the difficulty of
putting the faces and edges togetheror in fact puts the faces and
edges together in the course of solving the corners and the edge
positionsthen the 4^3 could be easier. It depends on the solution
procedure.
alan@saturn.cs.swin.oz.au (Alan Christiansen) continues:
> ANyway the real reason I am writing this is that I have written
> a cube simulator.
> It can simulate 3x3x3 4x4x4 5x5x5 .... cubes.
> I am working on 4x4x4x4 cube simulation.
This is interesting, as there is more than one way to model the
fourdimensional cube problem. Consider the 3^4 cube. It has eight
hyperfaces, each in the shape of a cube. One model of this puzzle is
that you could turn any face of any hyperface as if it were a face of
a 3^3 Rubik's cube. In a second model, you cannot move part of a
hyperface, but can turn each hyperface as if it were a solid cube in
space. A third model allows either kind of move.
These models are different from each other. The second model permits
the face centers of the hyperfaces to move around, whereas in the
first model only edges and corners move. In the first model, odd
permutations of corners are possible, which is not true in the second
model. Of course, the third model is the closure of the first two.
According to Hofstatder's column reprinted in _Metamagical_Themas_,
there is an unpublished 1982 manuscript by H J Kamack and T R Keane
entitled ``The Rubik Tesseract''. They calculated the size of the
group of the 3^4 puzzle, but I don't know which model was used.
Alan Christiansen indicates he has gone directly to the 4^4 puzzle. I
don't know which model he plans, or if the models become more similar
with the extra possibilities inherent in the larger cube. I don't
even know whether he plans to figure out how big the groups are or
whether they are identical.
Dan Hoey
Hoey@AIC.NRL.Navy.Mil
From xirion!jandr@relay.nl.net Thu Dec 23 02:48:30 1993
ReturnPath:
Received: from sun4nl.NL.net by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA08367; Thu, 23 Dec 93 02:48:30 EST
Received: from xirion by sun4nl.NL.net via EUnet
id AA04747 (5.65b/CWI3.3); Thu, 23 Dec 1993 08:48:27 +0100
Received: by xirion.xirion.nl id AA04326 (5.61/UK2.1);
Thu, 23 Dec 93 08:47:44 +0100
From: Jan de Ruiter
Date: Thu, 23 Dec 93 08:47:44 +0100
MessageId: <4326.9312230747@xirion.xirion.nl>
XOrganization: Xirion Unix Software & Consultancy bv
Burgemeester Verderlaan 15 X
3454 PE De Meern
The Netherlands
XPhone: +31 3406 61990
XFax: +31 3406 61981
To: cubelovers@ai.mit.edu
Subject: Re: Rubiks tangle
To: anandrao@hk.super.net
Cc: cubelovers@ai.mit.edu
>Your concept is theoretically extendable to the 10*10 tangle, but even
>with this optimisation the puzzle would take a long time to solve. How
>long do you take for the 5*5 Tangle on your computer?
Your question prompted me to actually write the program, and to
squeeze as much efficiency from the program as I could.
You wrote on december the 14th, your program took about 20 minutes
on a 486DX266,
Don Woods writes on the same date, that his program takes 45 seconds
on a SparcStation II,
And now I am proud to present my timing: trrrrr (drum roll)
7 seconds on a Compacq Deskpro 386/33. (and still only brute force!)
Now I am ready to try the 10x10.
Some thoughts in the mean time:
If the algorithm treats the duplicate pieces just as ordinary pieces,
i.e. as different, this will cause the program to find 4 solutions
for the 5x5 where only 2 exist (by exchanging the duplicate pieces).
This factor of 2 may not be dramatical, but if the same algorithm
tries the 10x10, then for every 1 solution that exists, the program
will find (5!)^4 x (4!)^20 identical versions (combinations of
duplicate exchanges).
My program views duplicate pieces as one, which may be placed several
times. So for some position X a piece with duplicates will only be
tried once.
Don Woods writes:
>Regarding solving the Tangle, I forgot one other minor optimisation:
>When my program is picking a corner piece other than the first, it
>requires that the piece "number" be less than or equal to that of the
>first corner. I.e., it refuses to search for solutions that are
>rotations of other solutions.
My program prevents finding rotations of solutions, by excluding the
rotations of just one piece. The list of possibilities to try on any
position includes this one piece just once, and every other piece four
times. You can choose any piece for this, except the duplicated one.
Regrettably this approach works only for the 5x5: the 10x10 will
probably have to use Don Woods method.
 Jan D. de Ruiter
From anandrao@hk.super.net Thu Dec 23 04:29:34 1993
ReturnPath:
Received: from hk.super.net by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA10668; Thu, 23 Dec 93 04:29:34 EST
Received: by hk.super.net id AA16480
(5.65c/IDA1.4.4 for cubelovers@ai.mit.edu); Thu, 23 Dec 1993 17:29:17 +0800
Date: Thu, 23 Dec 1993 17:26:41 +0800 (HKT)
From: Mr Anand Rao
Subject: Re: your mail
To: Jan de Ruiter
Cc: cubelovers@ai.mit.edu
InReplyTo: <4326.9312230747@xirion.xirion.nl>
MessageId:
MimeVersion: 1.0
ContentType: TEXT/PLAIN; charset=USASCII
On Thu, 23 Dec 1993, Jan de Ruiter wrote:
> And now I am proud to present my timing: trrrrr (drum roll)
> 7 seconds on a Compacq Deskpro 386/33. (and still only brute force!)
Were you travelling at the speed of .999c?
> Now I am ready to try the 10x10.
>
With this algorithm, you should have the solution before the year is out!
Best luck!
From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Sat Dec 25 22:48:40 1993
ReturnPath: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU>
Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA20316; Sat, 25 Dec 93 22:48:40 EST
MessageId: <9312260348.AA20316@life.ai.mit.edu>
Received: from MITVMA.MIT.EDU by mitvma.mit.edu (IBM VM SMTP V2R2)
with BSMTP id 2435; Sat, 25 Dec 93 22:48:38 EST
Received: from WVNVM.WVNET.EDU (NJE origin MAILER@WVNVM) by MITVMA.MIT.EDU
(LMail V1.1d/1.7f) with BSMTP id 7378; Sat, 25 Dec 1993 22:48:38 0500
Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU
(LMail V1.1d/1.7f) with BSMTP id 3493; Sat, 25 Dec 1993 22:46:06 0500
XAcknowledgeTo:
Date: Sat, 25 Dec 1993 22:46:06 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: Withdrawal of Proposal
I wish to withdraw, for the time being, my proposed answer to Dan
Hoey's question about how large is the cube group when symmetries
are taken into account. Notwithstanding two rounds of "correction",
I believe my proposal is fundamentally incorrect, and it will take
some time to come up with something better.
I believe that my proposed approximation is incorrect by a factor
of 24. That is, my proposed approximation would be correct for
corners plus edges (without centers), but would need to be
multiplied by 24 in order to be correct for corners plus edges
(with centers). My proposed approximation was
4.3 * 10^19 / (24*24*2). I now believe it should be
4.3 * 10^19 / (24 * 2), with the former figure correct only if
centers are omitted.
Secondly, I believe that my proposed procedure to calculate an
exact value from the known sizes of corner and edge groups is
incorrect. My procedure would be correct if all equivalence classes
had exactly 1152 elements. But they don't. It is not presently
clear to me whether the size of the equivalence classes when
corners and edges are combined can be calculated from the known
sizes of equivalence classes for corners and edges separately,
or whether a computer search will have to be performed for the
case where corners and edges are combined.
I will get back to this in a week or two. In the meantime, my
apologies if I have wasted your time, and I look forward to
any words of wisdom that any of you all might have.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 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 27 02:33:51 1993
ReturnPath: <@mail.uunet.ca:mark.longridge@canrem.com>
Received: from mail.uunet.ca (uunet.ca) by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA27687; Mon, 27 Dec 93 02:33:51 EST
Received: from portnoy.canrem.com ([198.133.42.251]) by mail.uunet.ca with SMTP id <53769(1)>; Mon, 27 Dec 1993 02:01:14 0500
Received: from canrem.com by portnoy.canrem.com (4.1/SMI4.1)
id AA18888; Mon, 27 Dec 93 02:00:05 EST
Received: by canrem.com (PCBUUCP 1.1f)
id 18F5BD; Mon, 27 Dec 93 01:58:20 0400
To: cubelovers@life.ai.mit.edu
ReplyTo: CRSO.Cube@canrem.com
Sender: CRSO.Cube@canrem.com
Subject: Local Maxima Revisited
From: mark.longridge@canrem.com (Mark Longridge)
MessageId: <60.656.5834.0C18F5BD@canrem.com>
Date: Mon, 27 Dec 1993 00:58:00 0500
Organization: CRS Online (Toronto, Ontario)
Some thoughts on Local Maxima

I have verified that the position I call "6 H order 2 type 2"
is a local maximum.
p175 6 H order 2 type 2 T2 B2 L2 T2 D2 L2 F2 T2 (8)
A rare example of a pattern with symmetry level 2, perhaps even
the only one of this type, and also the most symmetric of the
6 H patterns.
Nothing new here, as this was noted formerly by David Singmaster
in one of the Cubic Circulars and by Jim Saxe & Dan Hoey in the
archives.
Somewhat more interesting is the conclusion that the pattern
4 H order 2, or H's on the F,R,B,L faces (oriented like the
letter H) is also a local maximum, at least in the square's
group.
p160 4 H order 2 Type 2 B2 D2 (L2 R2 F2) ^2 T2 F2 (10)
From the archives:
> We include a description of 71 local maxima, which we believe
> to be all of the local maxima that can be proven using known
> techniques other than exhaustive search.
Oh well, I used an exhaustive search. p160 is 10 moves long in the htw
metric, and each of the moves ( T2, D2, F2, B2, L2, R2 ) all bring
one to a position requiring nine 180 degree twists, thusly....
4 H + T2 = L2, R2, F2 B2, T2, L2 R2, B2, F2 (9)
4 H + D2 = L2, R2, F2 B2, D2, L2 R2, B2, F2 (9)
4 H + F2 = B2, D2, L2 R2, F2, L2 R2, F2, T2 (9)
4 H + B2 = F2, D2, L2 R2, F2, L2 R2, F2, T2 (9)
4 H + L2 = R2, D2, L2 F2, B2, L2 F2, B2, T2 (9)
4 H + R2 = L2, D2, L2 F2, B2, L2, F2, B2, T2 (9)

I did discover an interesting property of the "Cube in a cube" pattern
I didn't notice before.
p7a Cube in a cube U2 F2 R2 U3 L2 D1 (B1 R3) ^3 B1 D3 L2 U3 (15)
Let's say you are entertaining some cube guests at a cube party and
the topic is (cube) patterns. Your guests are impressed with the
efficiency of the wellmemorized process. You would like to go on
to the next pattern but you don't quite remember how the inverse
goes. No problem! Rotate the whole cube so TOP becomes BACK then
BACK becomes DOWN, and finally FRONT becomes RIGHT. Simply repeat
the process p7a and your reputation as a cube expert is saved. ;>
This same idea works for the 6 X order 3 pattern as well.
And now for an unsymmetric local maximum!!
(Just kidding)
> Mark <
From @mail.uunet.ca:mark.longridge@canrem.com Mon Dec 27 12:31:47 1993
ReturnPath: <@mail.uunet.ca:mark.longridge@canrem.com>
Received: from mail.uunet.ca (uunet.ca) by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA10387; Mon, 27 Dec 93 12:31:47 EST
Received: from portnoy.canrem.com ([198.133.42.251]) by mail.uunet.ca with SMTP id <57298(4)>; Mon, 27 Dec 1993 12:31:39 0500
Received: from canrem.com by portnoy.canrem.com (4.1/SMI4.1)
id AA26935; Mon, 27 Dec 93 12:30:32 EST
Received: by canrem.com (PCBUUCP 1.1f)
id 18F603; Mon, 27 Dec 93 12:23:56 0400
To: cubelovers@life.ai.mit.edu
ReplyTo: CRSO.Cube@canrem.com
Sender: CRSO.Cube@canrem.com
Subject: Cube Rotations
From: mark.longridge@canrem.com (Mark Longridge)
MessageId: <60.659.5834.0C18F603@canrem.com>
InReplyTo: <19166.199312271606@phaedrus.harlequin.com>
Date: Mon, 27 Dec 1993 11:12:00 0500
Organization: CRS Online (Toronto, Ontario)
> goes. No problem! Rotate the whole cube so TOP becomes BACK then
> BACK becomes DOWN, and finally FRONT becomes RIGHT. Simply repeat
> the process p7a and your reputation as a cube expert is saved. ;>
>
> The faces FRONT and BACK are opposite each other. After your
> rotation, they become RIGHT and DOWN, which are not opposite each
> other. This would certainly establish a reputation for you, but if
> you did it with my cube, it might not be the sort of reputation you
> wanted to have :)
> Andy Latto
> andyl@harlequin.com
Perhaps my description of the rotations was unclear...
Rotate the entire cube so that TOP > DOWN
FRONT > LEFT
Ok, before I meant rotate the cube in space in 3 steps so
that the TOP face becomes BACK, then the face that is the
BACK at this point becomes DOWN, and the face that is the
FRONT at this point becomes the RIGHT.
The reason I used this type of description is because there
are multiple ways for the TOP to become the DOWN face....
TOP becomes BACK becomes DOWN and
TOP becomes RIGHT becomes DOWN and
TOP becomes LEFT becomes DOWN etc...
Perhaps it is better to use the form
old FACE A > new FACE A
old FACE B > new FACE B
Where the faces A & B are adjacent.
Mark
Email: mark.longridge@canrem.com
....wait a second, I don't think faces A & B have to be
adjacent for the rotation to be unambiguous. Any 2 faces
should do!
From hoey@aic.nrl.navy.mil Mon Dec 27 17:52:28 1993
Received: from Sun0.AIC.NRL.Navy.Mil by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA23079; Mon, 27 Dec 93 17:52:28 EST
Received: from sun30.aic.nrl.navy.mil by Sun0.AIC.NRL.Navy.Mil (4.1/SMI4.0)
id AA08057; Mon, 27 Dec 93 17:52:17 EST
ReturnPath:
Received: by sun30.aic.nrl.navy.mil; Mon, 27 Dec 93 17:52:16 EST
Date: Mon, 27 Dec 93 17:52:16 EST
From: hoey@aic.nrl.navy.mil
MessageId: <9312272252.AA22049@sun30.aic.nrl.navy.mil>
To: CubeLovers@life.ai.mit.edu
Cc: Jerry Bryan
Subject: Group theory basics (Re: Symmetry)
Jerry Bryan asked a bunch of
questions a couple of weeks ago, and I'll try to get to them all.
The first bunch has to do with some fairly basic stuff, that I thought
had been pretty well understood since the beginning of the mailing
list, but maybe we need a refresher, or an explicit statement.
In the message of Tue, 14 Dec 1993 20:50:51 EST, Jerry describes his
representation of cube positions and transformations.
> In my computer model, the corner facelets are simply numbered from
> 1 to 24, and any configuration of the corners is an order24 row
> vector. The rotation and reflection operators are also order24 row
> vectors, again with each cell simply containing a number from 1 to
> 24.
That is the most usual way of doing it, but it's important to specify
what you represent by those vectors. When I do it, I number the
corner facelet locations from 1 to 24, and these locations retain
their numbers through manipulations of the cube. I use a vector A to
specify a position in which the facelet whose home location is i has
been moved to location A(i), for each i. I use a vector P to specify
the transformation that moves the facelet in location i to location
A(i), for each i. I'll assume you're doing the same, though you
could, for instance, be representing the inverse of the operators, or
the locations from which the facelets originate. Note that a position
is represented by the same vector that represents the transformation
that takes SOLVED to that position.
> Well, if P is a rotation operator, you could perform a rotation
> two ways. I guess one is premultiplication and one is
> postmultiplication.
> 1) For i = 1 to 24 B(i) = A(P(i))
I would write this as B = P A, and say that A is premultiplied by P,
or equivalently that P is postmultiplied by A. In a general group, we
could have B = P A where the multiplication is not considered to be
the composition of permutations. But it turns out we can restrict our
attention to permutation groups without loss of generality. For
instance, when we are dealing with the supergroup, we can consider the
orientation of a face center to be a permutation of the corners of the
face center.
> 2) For i = 1 to 24 B(i) = P(A(i))
Here B = A P, A is postmultiplied by P, and P is premultiplied by A.
(Note that the operator or position name appears in the reverse order
from the prefix format. Algebraists sometimes avoid this by writing
(i)B = ((i)A)P. I kid you not.)
> (As an aside, this illustrates the question I raised in my previous
> post about "which is the operator and which is the thing being
> operated on?" Is P operating on A, or is A operating on P?)
Well, the answer is ``both''. I agree it's easy to get confused,
which is why proofs are a good idea.
> Finally, if Q is a reflection (actually, if Q1 is the identity and
> Q2 is the reflection), then we have
> For j = 1 to 24 for k = 1 to 24 for m = 1 to 2
> for i = 1 to 24 Bj,k,m(i) = Qm(Pj(A(Qm(Pk(i)))))
> I believe this loop calculates Dan Hoey's M.
On the the theory that proofs are a good idea, let's see what this
loop calculates. I'm going to put brackets around the subscripts.
Then I'll substitute "R" for "Q", because I use Q for the set of
quarterturns of faces. Furthermore, I'll use "C" instead of "P",
because the P[j] are just the elements of C, the group of cube
rotations. So you are computing
B[j,k,m] = C[k] R[m] A C[j] R[m] (1)
for j in {1,...,24}, k in {1,...,24}, and m in {1,2}. Now every
member of M (the group of cube rotations and reflections) has a unique
representation as M[n] = C[k] R[m]. Let us define Cind() and Rind()
as the functions for which M[n]=C[Cind(M[n])] R[Rind(M[n])]. So we
can write (1) as
B[j,k,m] = M[n] A M'[n] (M[n] C[j] R[Rind(M[n])])
Note that (M[n] C[j] R[Rind(M(n))] must be an element of C. So B is a
set of elements of the form M[n] A M'[n] C[o]. To see that we have
all such elements, first observe that (M[n]' C[o] R[Rind(M[n])]') is
an element of C, say C[j]. So equation (1) includes:
C[Cind(M[n])] R[Rind(M[n])] A C[j] R[Rind(M[n])]
= M[n] A (M[n]' C[o] R[Rind(M[n])]') R[Rind(M[n])]
= M[n] A M'[n] C[o].
Thus the set of all B[j,k,m] is the set of all M[n] A M'[n] C[o]. Or
in English, that's the set of all Mconjugates of A, operated on by
all wholecube rotations.
> In my data base, I store the minimum of Bj,k,m over j = 1 to 24,
> k = 1 to 24, and m = 1 to 2. I tend to call the minimum of Bj,k,m a
> canonical form. I am not sure if that is the best terminology. The
> minimal element is not any simpler than any other. It is just that
> I need a function to choose an element from a set, and picking the
> minimal element seems very natural. Any other element would do as
> well, provided I could always be sure of picking the same element.
It's pretty common terminology. You might be slightly better off
calling it a ``representative element,'' as that connotes that the
element is ordinary except in that it represents the equivalence class
(like representatives in the U.S. Congress).
> Also, my criterion for equivalence is slightly
> different (but isomorphic, I think) than the one described by
> Dan Hoey. Suppose A and B are two cubes.
> Rather than mapping A to B or B to A in M, I map both A and B
> to their respective canonical forms. A and B are equivalent if
> their respective canonical forms are equal.
This is straightforward once we show that Mconjugacy is an
equivalence relation, and B[j,k,m] is an equivalence class.
If A ~ Representative[A] = Representative[B] ~ B, then by transitivity
A ~ B. Conversely, if A ~ B, then Class[A] = Class[B], and therefore
Representative[A] = Representative[B]. This shows that the criteria
are equivalent.
> Now, as to the centers. I still sometimes have a certain doubt
> about the centers. They are fixed, so how can you reduce the
> problem (i.e., increase the size of the equivalence classes)
> by both rotating the cube and rotating the colors (by both pre
> and postmultiplication)?
What you have done is to increase the size of the whole cube problem
by a factor of 24, by dealing with all rotations of the cube, and the
equivalence classes expand by the same factor, from 48 to 1152. This
has allowed you to calculate something like Mconjugacy classes for
cube problems that lack face centers. But the size of the equivalence
classes doesn't shrink the problem for cubes that have face centers.
You could have just calculated Mconjugates and got the same answer.
> I am not sure if this answers Dan's question about my model
> with centers added.
It's clear now. I hadn't realized you were rotating the cube in space
when the face centers were present. I expected that to be a wasted
effort. But I am impressed by the way it allows you to shrink the
database by storing positions together that differ only by wholecube
moves of the face centers. I think it should be possible to shrink
the database without the effort, though.
In your message of Thu, 16 Dec 1993 15:36:58 EST, on the ``Duality of
Operators and Operatees'':
> I have mentioned several times my discomfort about "an operator" as
> opposed to "the thing being operated on" when it comes to groups. I
> am never quite sure just which of the two it is that people are
> talking about, even (or especially) when I am listening to myself
> talk.
It is hard to keep it straight. Sometimes we all get it wrong. The
best way to avoid errors, as far as possible, is to avoid such
language and talk about group multiplication. But then we have to
explain what is going on with the cube, so we get caught into talking
about operators again. It's a discomfort that must be endured.
> The code to translate between the ASCII string X and
> the EBCDIC string Y is something like
> for i = 1 to n Y(i) = T(X(i))
> where T is the translate table.
Yes, or Y = X T as above.
> I am going to continue reading, but perhaps I could pose a question to
> Dan Hoey anyway: is reversing the role of X and T in the TRANSLATE
> function above essentially the same thing as switching between
> premultiplication and postmultiplication?
Yes.
Dan Hoey
Hoey@AIC.NRL.Navy.Mil
From dik@cwi.nl Mon Dec 27 18:43:04 1993
ReturnPath:
Received: from charon.cwi.nl by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA24660; Mon, 27 Dec 93 18:43:04 EST
Received: from boring.cwi.nl by charon.cwi.nl with SMTP
id AA15198 (5.65b/3.12/CWIAmsterdam); Tue, 28 Dec 1993 00:43:03 +0100
Received: by boring.cwi.nl
id AA25571 (4.1/2.10/CWIAmsterdam); Tue, 28 Dec 93 00:43:02 +0100
Date: Tue, 28 Dec 93 00:43:02 +0100
From: Dik.Winter@cwi.nl
MessageId: <9312272343.AA25571.dik@boring.cwi.nl>
To: CubeLovers@life.ai.mit.edu
Subject: Re: Group theory basics (Re: Symmetry)
One additional remark:
> > Well, if P is a rotation operator, you could perform a rotation
> > two ways. I guess one is premultiplication and one is
> > postmultiplication.
> > 1) For i = 1 to 24 B(i) = A(P(i))
> I would write this as B = P A, and say that A is premultiplied by P,
> or equivalently that P is postmultiplied by A.
There is quite a bit of confusion about this. When permutation
groups are considered; even textbooks do not agree. When A and P
are permutations you can find both that P A means: apply P first, A next,
but also: apply A first, P next. (The first meaning comes from the pure
group theorists, the second meaning more from the algebra inclined.)
Sorry to confuse the issue, but when I read such texts I have always to
think hard to get at the intended meaning. I think the functional
notation is much clearer and leads to less confusion.
Of course, doing notations for cube rotations the group theorists
notation is applied, but when doing abstract operations...
From hoey@aic.nrl.navy.mil Tue Dec 28 14:17:02 1993
Received: from Sun0.AIC.NRL.Navy.Mil by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA24468; Tue, 28 Dec 93 14:17:02 EST
Received: from sun30.aic.nrl.navy.mil by Sun0.AIC.NRL.Navy.Mil (4.1/SMI4.0)
id AA19858; Tue, 28 Dec 93 14:16:50 EST
ReturnPath:
Received: by sun30.aic.nrl.navy.mil; Tue, 28 Dec 93 14:16:49 EST
Date: Tue, 28 Dec 93 14:16:49 EST
From: hoey@aic.nrl.navy.mil
MessageId: <9312281916.AA25640@sun30.aic.nrl.navy.mil>
To: CubeLovers@life.ai.mit.edu
Cc: Jerry Bryan
Subject: Re: Some Additional Distances in the Edge Group
In his message of Fri, 17 Dec 1993 00:54:00 EST, Jerry Bryan
makes some observations on the
distances between the following positions in the edge group:
I = Solved,
P = Pons Asinorum (or Mirror),
E = All edges flipped, and
PE = P E = Pons Asinorum with all edges flipped.
[I _will_ continue to use permutation multiplication as we have done
so in this group since its inception. I realize that this agrees with
some textbooks and is backwards from others, but it would be far more
confusing to write these functionally all the time.] Jerry's
bruteforce search has shown that d(I,PE)=15, and he notes that
conjugation by E shows us that d(P,E)=15 as well. He concludes:
> I have the sensation in describing this that the Edge group is
> square, with Start and MirrorImageofEdgesFlipped 180 degrees
> apart, and MirrorImageofStart and EdgesFlipped at the other
> two corners of the square.
Well, it's not quite a square, since d(I,P)=12 and d(I,E)=9, according
to Jerry's message of Wed, 8 Dec 1993 10:02:15 EST. Conjugation will
similarly show that d(E,PE)=12 and d(P,PE)=9. So we are dealing with
a rectangle. The sides of the rectangle are 9 and 12, and the
diagonal is 15: a most fortuitous set of numbers, in that we can
actually embed such a rectangle in the Euclidean plane!
We can map the positions of the edge group to 4tuples of distances.
For any position X, let
f(X)=(d(I,X), d(E,X), d(P,X), d(PE,X)).
If f(X)=(a,b,c,d), then conjugation shows us that f(X E)=(b,a,d,c),
f(X P)=(c,d,a,b), and F(X PE)=(d,c,b,a). So the set of quadruples has
the symmetries of the rectangle.
We know f(I)=(0,9,12,15). What is more, the earlier results on
symmetry show us that I is at a local maximum distance from E, P, and
PE. So, letting I1 be the unique (up to Mconjugacy) position
adjacent to I, we have F(I1)=(1,8,11,14). (This destroys Euclidean
embeddability.) An analogous result holds for the unique neighbor of
each corner of the rectangle.
We also have Jerry's results of Wed, 8 Dec 1993 22:41:28 EST and
23:16:50 EST that H (the 6H pattern) and HE=H E are at distances 8
and 13 from start, respectively. Since H is an Mconjugate of P H,
this gives us f(H)=(8,13,8,13). [Note: there are two distinct
Mconjugates of H, call them H and Hbar. This distinction is
important when we compose permutations: H H = I, but H Hbar = P. So
we have to be careful when conflating Mconjugates.] We can by
symmetry find f(H1)=(7,12,7,12) for H's unique neighbor H1.
What quadruples are possible? If f(X)=(a,b,c,d), and X is not one of
the eight corners and neighbors, we have
max(2,9b,12c,15d) <= a <= min(14,9+b,12+c)
with constraints on b, c, and d from symmetry. A quick hack tells me
there are 7836 such quadruples. I wonder how many of them are
realized? If it's fairly few, I would like to see a diagram of
quadruples, with lines between those quadruples that represent
adjacent positions (adjacent quadruples differ by at most one in each
coordinate). Maybe with the number of positions for each quadruple,
too. I have an idea that such a diagram might tell us something about
the problem, or at least look pretty.
Dan Hoey
Hoey@AIC.NRL.Navy.Mil
From hoey@aic.nrl.navy.mil Tue Dec 28 18:42:22 1993
Received: from Sun0.AIC.NRL.Navy.Mil by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA04472; Tue, 28 Dec 93 18:42:22 EST
Received: from sun30.aic.nrl.navy.mil by Sun0.AIC.NRL.Navy.Mil (4.1/SMI4.0)
id AA06105; Tue, 28 Dec 93 18:40:52 EST
ReturnPath:
Received: by sun30.aic.nrl.navy.mil; Tue, 28 Dec 93 18:40:52 EST
Date: Tue, 28 Dec 93 18:40:52 EST
From: hoey@aic.nrl.navy.mil
MessageId: <9312282340.AA25691@sun30.aic.nrl.navy.mil>
To: CubeLovers@ai.mit.edu
Subject: Re: Cube Rotations
Cc: CRSO.Cube@canrem.com
mark.longridge@canrem.com (Mark Longridge) writes:
> Perhaps my description of the rotations was unclear...
Yes.
> ...Perhaps it is better to use the form
> old FACE A > new FACE A
> old FACE B > new FACE B
> Where the faces A & B are adjacent.
That will serve to uniquely identify a rotation, but it's somewhat
verbose. Worse, it does not suffice to uniquely identify a symmetry
from the group of rotations and reflections, M. I find it's far more
informative to identify a rotation or reflection as a permutation of
the faces, in cycle format. There are only ten kinds:
Even rotations: I=Identity (1),
(FRT)(BLD)=120degree rotation (8),
(FB)(RL)=180degree orthogonal rotation (3).
Odd rotations: (FRBL)=90degree rotation (6),
(FB)(TR)(DL)=180degree diagonal rotation (6).
Even reflections: (FR)(BL)=diagonal reflection (6),
(FRBL)(TD)=90degree glide reflection (6),
Odd reflections: (FB)=orthogonal reflection (3),
(FRTBLD)=60degree glide reflection (8),
(FB)(RL)(TD)=central reflection (1).
In case it isn't clear, the cycle notation for (e.g.) a 120degree
rotation (FTL)(BDR) means that the F, T, L, B, D, and R faces move to
the T, L, F, D, R, and B, locations, respectively. The only thing I'm
afraid of with this notation is that someone will think I'm describing
a magiccube process rather than a wholecube move.
So when you say Top>Down, Front>Left, I would say (TD)(FL)(BR) for
the 180degree diagonal rotation, to distinguish it from (TD)(FLBR)
the 90degree glide reflection.
> ....wait a second, I don't think faces A & B have to be
> adjacent for the rotation to be unambiguous. Any 2 faces
> should do!
No, you're back to your original bogosity. Knowing the destinations
of two opposite faces doesn't give you any more information than
knowing the destination of one (unless you go breaking the axles).
Dan Hoey
Hoey@AIC.NRL.Navy.Mil
From hoey@aic.nrl.navy.mil Wed Dec 29 17:43:47 1993
ReturnPath:
Received: from Sun0.AIC.NRL.Navy.Mil by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA12501; Wed, 29 Dec 93 17:43:47 EST
Received: by Sun0.AIC.NRL.Navy.Mil (4.1/SMI4.0)
id AA05575; Wed, 29 Dec 93 17:43:28 EST
Date: Wed, 29 Dec 93 17:43:28 EST
From: hoey@aic.nrl.navy.mil (Dan Hoey)
MessageId: <9312292243.AA05575@Sun0.AIC.NRL.Navy.Mil>
To: CubeLovers@life.ai.mit.edu
Cc: Jerry Bryan
Subject: Correction Re: Some Additional Distances in the Edge Group
A couple of days ago, I said that proofs are a good idea. I'll say it
again today with a redder face.
Yesterday I discussed the edge group positions
I = Solved,
P = Pons Asinorum (or Mirror),
E = All edges flipped, and
PE = P E = Pons Asinorum with all edges flipped
and the function from the edge group to 4tuples of distances
f(X)=(d(I,X), d(E,X), d(P,X), d(PE,X)).
I wrote:
?? If f(X)=(a,b,c,d), then conjugation shows us that ??
?? f(X E)=(b,a,d,c), f(X P)=(c,d,a,b), and F(X PE)=(d,c,b,a). ??
?? So the set of quadruples has the symmetries of the rectangle. ??
The first sentence is incorrect, though the argument as a whole is
reparable.
First, I'll do what I should have done yesterday, and define the
distance function d(X,Y). We want the minimum length process Z such
that X Z = Y. But premultiplying both sides by X', we have Z = X' Y.
So I define d(X,Y)=Length(X' Y). From the properties of the length
function (Length(I)=0, Length(X)=Length(X'), and
Length(X Y)<=Length(X) + Length(Y)) we can conclude that d(X,Y) is a
metric.
Suppose f(X)=(a,b,c,d). I claim f(E X)=(b,a,d,c), f(P X)=(c,d,a,b),
and F(PE X)=(d,c,b,a).
Proof: To show f(E X)=(b,a,d,c), first observe that I=I', E=E', and
P E = E P.
d(I,E X) = Length(I' E X) = Length(E' X) = d(E,X),
so d(E,E X) = d(I, E E X) = d(I,X);
d(P,E X) = Length(P' E X) = Length((PE)' X) = d(PE,X)
so d(PE,E X)=d(P,E E X)=d(P,X).
To show that f(P X)=(c,d,a,b), exchange P and E in the above argument.
To show that f(PE X)=(d,c,b,a), use both occurrences of the argument.
QED.
So the idea of yesterday's message is correct, but I had X E, X P, and
X PE instead of E X, P X, and PE X, respectively. I would show you a
counterexample to yesterday's formulation, but it turns out there is
none. I claim that f(X,E)=f(E,X), f(X,P)=f(P,X), and f(X,PE)=f(PE,X).
Proof: Recall that E commutes with every element of the Rubik cube
group, so f(X E)=f(E X). It turns out that ``up to Mconjugacy'', P
commutes with every element of the edge group as well. For P performs
a mirrorreflection of the edges, and so can be regarded as an element
of M acting on the edge group. So P' X P = Xbar is an Mconjugate of
X, and X P = P Xbar. Since Length(X) agrees on Mconjugates, so does
d(X,Y), and so f(X), so f(X P)=f(P Xbar) = f(P X). Finally,
f(X PE) = f(X P E) = f(E X P) = f(P E X) = f(PE X). QED.
So it turns out it that the statement about f was true. But I am no
less embarrassed for asserting it, for I had no reason to think it
would be true. It's only rescued by the surprising commutativity of
the Pons Asinorum.
Finally, I would like to note something that I nearly included in
yesterday's message, but yanked when I decided it was false:
f(X')=f(X). Now I'll prove it:
Proof: For W among {I,E,P,PE}, we have X W = W Xbar, for Xbar an
Mconjugate of X. So
d(X,W)=Length(X'W)=Length(W'Xbar')=Length(W'X')=d(W,X').
QED.
Dan Hoey
Hoey@AIC.NRL.Navy.Mil
From Don.Woods@eng.sun.com Sun Jan 2 20:10:16 1994
ReturnPath:
Received: from Sun.COM by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA12259; Sun, 2 Jan 94 20:10:16 EST
Received: from Eng.Sun.COM (zigzag.Eng.Sun.COM) by Sun.COM (4.1/SMI4.1)
id AA11819; Sun, 2 Jan 94 17:10:12 PST
Received: from colossal.Eng.Sun.COM by Eng.Sun.COM (4.1/SMI4.1)
id AA03228; Sun, 2 Jan 94 17:08:39 PST
Received: by colossal.Eng.Sun.COM (5.0/SMISVR4)
id AA13229; Sun, 2 Jan 94 17:10:24 PST
Date: Sun, 2 Jan 94 17:10:24 PST
From: Don.Woods@eng.sun.com (Don Woods)
MessageId: <9401030110.AA13229@colossal.Eng.Sun.COM>
To: cubelovers@ai.mit.edu
Subject: 10x10 Tangle
ContentLength: 1000
Hm. Well, I split up the 10x10 Tangle exhaustive search and ran it on
several machines over Christmas break, getting the 90 days of compute
time done in about 10.
And turned up no solutions.
There could of course be a bug in my program, but the same code with
minor changes finds the same solutions as others have found for the 5x5.
I also tried adding some extra tiles for the 10x10, and it began finding
solutions okay. I did doublecheck that the 100 tiles matched the info
posted to CubeLovers re which tiles are duplicated in the four 5x5s;
I have no way of checking whether that info was correct.
Has anyone out there ever heard definitely that someone has found a
solution to the 10x10? Is it possible that the makers of Tangle (Matchbox,
using Rubik's name under license) merely claimed that such a solution
exists, without actually verifying it? (Seems pretty sleazy if so,
but then, having Tangles 24 be merely color permutations of #1 is
pretty weak in the first place.)
 Don.
From dik@cwi.nl Sun Jan 2 21:52:23 1994
ReturnPath:
Received: from charon.cwi.nl by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA16116; Sun, 2 Jan 94 21:52:23 EST
Received: from boring.cwi.nl by charon.cwi.nl with SMTP
id AA27901 (5.65b/3.12/CWIAmsterdam); Mon, 3 Jan 1994 03:52:21 +0100
Received: by boring.cwi.nl
id AA07139 (4.1/2.10/CWIAmsterdam); Mon, 3 Jan 94 03:52:20 +0100
Date: Mon, 3 Jan 94 03:52:20 +0100
From: Dik.Winter@cwi.nl
MessageId: <9401030252.AA07139.dik@boring.cwi.nl>
To: Don.Woods@eng.sun.com, cubelovers@ai.mit.edu
Subject: Re: 10x10 Tangle
> Has anyone out there ever heard definitely that someone has found a
> solution to the 10x10?
As I wrote before, I have embedded in my memory that there is an easy
argument that the 10x10 is *not* solvable. I do not know whether I
found it myself (and ever did mail it to other people) or whether I
found it somewhere on the net; it is a long time ago. When I find the
time I will do a check. (I know very sure that I have had a program
running at that time but that I abandoned the search because it would
be fruitless.)
> Is it possible that the makers of Tangle (Matchbox,
> using Rubik's name under license) merely claimed that such a solution
> exists, without actually verifying it?
Yes, very probable. You should never trust the number of solutions the
manufacturers give. Sometimes it is much more, in this case it is less.
An actual example is a puzzle that consists of of nine rings (eh, this
is from memory, I do not have access to the puzzle at this time). Five
rings contain digits; three rings contain operators; one ring contains
equal signs. All in four positions around the rings. The idea is to
create correct sums (like 5 + 1  4 + 1 = 3) on all four positions of the
rings. The claim was that there was only a single solution. Actually
there are many. If there is interest I can hunt down the rings and
describe them in more detail. (An interesting detail is that my father
was the first to find the puzzle; he had correct solutions like:
1 + 3 : 2 + 1 = 3. He was a physicist. The accomanying leaflet did not
give details about operator priorities. Hence it actually makes two
puzzles; one with regards to priorities, the other just going left to
right.)
> (Seems pretty sleazy if so,
> but then, having Tangles 24 be merely color permutations of #1 is
> pretty weak in the first place.)
Indeed, the mass manufacturers are sleazy.
Cheers.
I will mail when I find back the argument disallowing 10x10.

dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924098
home: bovenover 215, 1025 jn amsterdam, nederland; email: dik@cwi.nl
From xirion!jandr@relay.nl.net Mon Jan 3 02:29:28 1994
ReturnPath:
Received: from sun4nl.NL.net by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA21848; Mon, 3 Jan 94 02:29:28 EST
Received: from xirion by sun4nl.NL.net via EUnet
id AA18380 (5.65b/CWI3.3); Mon, 3 Jan 1994 08:29:22 +0100
Received: by xirion.xirion.nl id AA22110 (5.61/UK2.1);
Mon, 3 Jan 94 08:29:37 +0100
From: Jan de Ruiter
Date: Mon, 3 Jan 94 08:29:37 +0100
MessageId: <22110.9401030729@xirion.xirion.nl>
XOrganization: Xirion Unix Software & Consultancy bv
Burgemeester Verderlaan 15 X
3454 PE De Meern
The Netherlands
XPhone: +31 3406 61990
XFax: +31 3406 61981
To: cubelovers@ai.mit.edu
To: Don.Woods@eng.sun.com, cubelovers@ai.mit.edu
Subject: RE: 10x10 Tangle
>Hm. Well, I split up the 10x10 Tangle exhaustive search and ran it on
>several machines over Christmas break, getting the 90 days of compute
>time done in about 10.
>
>And turned up no solutions.
My program is a bit faster, but as I have less machines at my disposal
and I started a bit later, my programs are still running.
Up until now they did not produce a solution either. I am starting
to get worried.
>There could of course be a bug in my program, but the same code with
>minor changes finds the same solutions as others have found for the 5x5.
The same goes for me.
> I did doublecheck that the 100 tiles matched the info
>posted to CubeLovers re which tiles are duplicated in the four 5x5s;
>I have no way of checking whether that info was correct.
I have the puzzles myself, and checked the info in the message from
Dale I Newfield (15 Dec 1993), which quotes the archives.
I can assure you: those are indeed the duplicate pieces.
>
>Has anyone out there ever heard definitely that someone has found a
>solution to the 10x10? Is it possible that the makers of Tangle (Matchbox,
>using Rubik's name under license) merely claimed that such a solution
>exists, without actually verifying it? (Seems pretty sleazy if so,
>but then, having Tangles 24 be merely color permutations of #1 is
>pretty weak in the first place.)
>
I thought about that too, but considered that the choice for precisely
those four duplicate pieces could be dictated by the desire to have
a solution for the 10x10.
>I also tried adding some extra tiles for the 10x10, and it began finding
>solutions okay.
Question: did you add pieces at random, or did you add more duplicate
pieces?
In the latter case you may have found the duplications that should have
been made to get a solvable 10x10.
That in turn would show there could not exist an argument disallowing
10x10 (as claimed by Dik Winter), unless that argument is based on the
particular colouring of the four duplicated pieces...
 Jan
From Don.Woods@eng.sun.com Mon Jan 3 05:45:40 1994
ReturnPath:
Received: from Sun.COM by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA23952; Mon, 3 Jan 94 05:45:40 EST
Received: from Eng.Sun.COM (zigzag.Eng.Sun.COM) by Sun.COM (4.1/SMI4.1)
id AA10025; Mon, 3 Jan 94 02:45:24 PST
Received: from colossal.Eng.Sun.COM by Eng.Sun.COM (4.1/SMI4.1)
id AA09005; Mon, 3 Jan 94 02:43:43 PST
Received: by colossal.Eng.Sun.COM (5.0/SMISVR4)
id AA14295; Mon, 3 Jan 94 02:45:35 PST
Date: Mon, 3 Jan 94 02:45:35 PST
From: Don.Woods@eng.sun.com (Don Woods)
MessageId: <9401031045.AA14295@colossal.Eng.Sun.COM>
To: cubelovers@ai.mit.edu, jandr@xirion.nl
Subject: Re: 10x10 Tangle
ContentLength: 1209
> My program is a bit faster, but as I have less machines at my disposal
> and I started a bit later, my programs are still running.
Incidentally, I would be interested in seeing your program. (And am
willing to send you mine.) I'm always willing to learn something about
how to make combinatorial searches more efficient.
> >I also tried adding some extra tiles for the 10x10, and it began finding
> >solutions okay.
>
> Question: did you add pieces at random, or did you add more duplicate
> pieces?
I just gave it 5 of each piece, instead of 4 of most pieces and 5 of some.
It churned out positions pretty quick that way! But since this involved
giving it more than 100 tiles to draw from, it says nothing about Dik Winter's
claimed impossibility proof.
It's a shame, really. I'll bet that it would be possible to come up with
four Tangles that (a) really are different instead of being simple color
permutations of each other, (b) each have a unique solution (not counting
rotations) instead of two, and (c) can be combined to form a 10x10 that has
a unique solution. Well, strike the "unique" from (c) and I'd make the bet;
but with the "unique" I certainly wouldn't bet against it!
 Don.
From xirion!jandr@relay.nl.net Mon Jan 3 08:17:10 1994
ReturnPath:
Received: from sun4nl.NL.net by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA26254; Mon, 3 Jan 94 08:17:10 EST
Received: from xirion by sun4nl.NL.net via EUnet
id AA06106 (5.65b/CWI3.3); Mon, 3 Jan 1994 14:17:08 +0100
Received: by xirion.xirion.nl id AA22788 (5.61/UK2.1);
Mon, 3 Jan 94 14:16:52 +0100
From: Jan de Ruiter
Date: Mon, 3 Jan 94 14:16:52 +0100
MessageId: <22788.9401031316@xirion.xirion.nl>
XOrganization: Xirion Unix Software & Consultancy bv
Burgemeester Verderlaan 15 X
3454 PE De Meern
The Netherlands
XPhone: +31 3406 61990
XFax: +31 3406 61981
To: cubelovers@ai.mit.edu
To: Don.Woods@Eng.Sun.COM
Subject: Re: 10x10 Tangle
>Incidentally, I would be interested in seeing your program. (And am
>willing to send you mine.) I'm always willing to learn something about
>how to make combinatorial searches more efficient.
Will be sent separately (and yes, I would like to see yours too!)
>It's a shame, really. I'll bet that it would be possible to come up with
>four Tangles that (a) really are different instead of being simple color
>permutations of each other, (b) each have a unique solution (not counting
>rotations) instead of two, and (c) can be combined to form a 10x10 that has
>a unique solution. Well, strike the "unique" from (c) and I'd make the bet;
>but with the "unique" I certainly wouldn't bet against it!
When you limit yourself to 4 ropes with 4 colours, you always get 24 pieces,
and when you want to build a puzzle of 25 pieces, you will have to duplicate
one, which causes (a).
Using 5 colours instead, creates a set of 120 pieces, from which you could
probably pick 25 pieces (without duplication) which would satisfy both (a)
and (b), and probably 4 such sets could be found to satisfy (c) as well,
but such a puzzle would be less attractive, because the choice of 25 from
120 is somewhat arbitrary, and a puzzler would probably be more inclined
to use all 120 pieces... Of course it is all a matter of taste.
 Jan
From Don.Woods@eng.sun.com Mon Jan 3 17:11:06 1994
ReturnPath:
Received: from Sun.COM by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA21618; Mon, 3 Jan 94 17:11:06 EST
Received: from Eng.Sun.COM (zigzag.Eng.Sun.COM) by Sun.COM (4.1/SMI4.1)
id AA09153; Mon, 3 Jan 94 14:10:57 PST
Received: from colossal.Eng.Sun.COM by Eng.Sun.COM (4.1/SMI4.1)
id AA19324; Mon, 3 Jan 94 14:09:21 PST
Received: by colossal.Eng.Sun.COM (5.0/SMISVR4)
id AA15405; Mon, 3 Jan 94 14:11:09 PST
Date: Mon, 3 Jan 94 14:11:09 PST
From: Don.Woods@eng.sun.com (Don Woods)
MessageId: <9401032211.AA15405@colossal.Eng.Sun.COM>
To: cubelovers@ai.mit.edu, jandr@xirion.nl
Subject: Re: 10x10 Tangle
XSunCharset: USASCII
ContentLength: 978
> >It's a shame, really. I'll bet that it would be possible to come up with
> >four Tangles that (a) really are different instead of being simple color
> >permutations of each other, ...
>
> When you limit yourself to 4 ropes with 4 colours, you always get 24 pieces,
> and when you want to build a puzzle of 25 pieces, you will have to duplicate
> one, which causes (a).
Not so. There's nothing that says all permutations must be present.
Back in '92 when I first wrote the program to solve Tangle #1, I fiddled
with it a bit and found that removing a particular tile and adding a
duplicate of a second particular tile caused the solution to become
unique. It didn't take long to find such a combination, so I'm confident
there are many many more that have unique solutions.
Hm, using just the set of 24 distinct tiles, I wonder if it's possible to
tile the faces of a 2x2x2 cube such that colors match at the edges of the
cube as well as within the faces?...
 Don.
From Don.Woods@eng.sun.com Mon Jan 3 19:50:59 1994
ReturnPath:
Received: from Sun.COM by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA28614; Mon, 3 Jan 94 19:50:59 EST
Received: from Eng.Sun.COM (zigzag.Eng.Sun.COM) by Sun.COM (4.1/SMI4.1)
id AA24247; Mon, 3 Jan 94 16:50:55 PST
Received: from colossal.Eng.Sun.COM by Eng.Sun.COM (4.1/SMI4.1)
id AA22695; Mon, 3 Jan 94 16:49:20 PST
Received: by colossal.Eng.Sun.COM (5.0/SMISVR4)
id AA16535; Mon, 3 Jan 94 16:51:07 PST
Date: Mon, 3 Jan 94 16:51:07 PST
From: Don.Woods@eng.sun.com (Don Woods)
MessageId: <9401040051.AA16535@colossal.Eng.Sun.COM>
To: cubelovers@ai.mit.edu
Subject: tangled cube
XSunCharset: USASCII
ContentLength: 2449
Well, a pleasant surprise! It _is_ possible to take a set of 24 distinct
tiles from any Rubik's Tangle, and use them to tile the surface of a 2x2x2
cube such that all touching ropes match. And the solution is unique!
I'll include the solution below, after some blank lines to avoid spoiling it
for anyone who wants to try solving the puzzle without seeing the answer...
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
First, a hint. If you look at any face of the cube, and look at the two
pairs of colors at any edge of that face, the two pairs will be the same.
That is, if one tile touches a cube edge with colors redblue, the adjacent
tile on that face touching the same edge will also touch the edge with red
blue. I see no obvious reason why the solution should have this property,
but it does.
Solution below.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Solution: Note that there is also a correlation between the color pairs that
occur within a face, and the color pairs that occur at the edges of that face.
Also, the orientation of every tile is the same relative to the adjacent of
the cube that the tile touches. This makes it relatively easy to reconstruct
the solution manually.
BR BR
G..G G..Y
R..Y Y..B
YB GR
YB GR
G..R R..Y
R..B B..B
YG YG
GR GR YG YG BY BY RB RB
Y..Y Y..B B..B B..R R..R R..G G..G G..Y
R..B B..G G..R R..Y Y..G G..B B..Y Y..R
BG YR RY BG GB RY YR GB
BG YR RY BG GB RY YR GB
Y..R R..B B..G G..R R..Y Y..G G..B B..Y
R..G G..G G..Y Y..Y Y..B B..B B..R R..R
BY BY RB RB GR GR YG YG
RB RB
Y..Y Y..G
B..G G..R
GR YB
GR YB
Y..B B..G
B..R R..R
GY GY
From jbharris@tenet.edu Mon Jan 3 20:04:38 1994
ReturnPath:
Received: from abernathy.tenet.edu (KayAbernathy.tenet.edu) by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA28829; Mon, 3 Jan 94 20:04:38 EST
Received: by abernathy.tenet.edu id AA16334
(5.65c/IDA1.4.4 for CUBELOVERS@AI.AI.MIT.EDU); Mon, 3 Jan 1994 19:02:39 0600
Date: Mon, 3 Jan 1994 19:02:13 0600 (CST)
From: Judi Harris
Subject: Volunteers Requested
To: CUBELOVERS@life.ai.mit.edu
MessageId:
MimeVersion: 1.0
ContentType: TEXT/PLAIN; charset=USASCII
WOULD YOU BE WILLING TO SHARE WHAT YOU KNOW WITH
PRECOLLEGE STUDENTS AND TEACHERS BY ELECTRONIC MAIL?
Recent estimates indicate that there are now more than
300,000 classroom teachers from primary, middle, and
secondary schools who hold accounts on the Internet.
This makes a very special kind of learning available to
them: one which directly involves subject matter experts
communicating with students and teachers about their
specialties, via electronic mail.
With support from the Texas Center for Educational
Technology, we (at the University of Texas at Austin)
have piloted and are now expanding an Internetbased
service (the "Electronic Emissary") that brings together
precollege students, their teachers, and subject matter
experts (SMEs) electronically, helping them to create
telecomputing exchanges centered around the students'
learning in the SMEs' disciplines. For example,
* A class studying South America could learn about
recent global environmental research results from a
scientist who studies rainforest deforestation in
Brazil.
* A class studying geometry might "talk" electronically
with Euclid, who is actually a mathematics professor.
* A class studying the future of education might
converse with an emerging technologies specialist from
California's Silicon Valley.
* A class studying American History might
electronically interview Harry Truman, who is really a
curator with the National Archives.
* A class exploring the rapidlychanging governmental
structures that are emerging in what was once the Soviet
Union might correspond with a group of graduate
political science students at a university in the CIS.
* Or, a class reading _Huckleberry Finn_ might
correspond with an AfricanAmerican studies scholar
about the repercussions resulting from the enacting of
the Emancipation Proclamation.
In successive phases of the project, increasing numbers
of SMEs or SME groups are needed to correspond regularly
(approximately 4 times per week) with primary, middle
school, or secondary students and their teachers (1 SME
or expert group per class, study group, or "special
student"). Each electronic exchange will begin with
approximately 2 weeks of project planning via electronic
mail between the SMEs and the teachers. Communications
with students will begin on mutually convenient dates,
and will continue for previouslyarranged periods of
time, usually between 2 and 10 weeks.
Subject matter expert volunteers are sought in all
disciplines, but there is immediate need for SMEs with
expertise in:
~ gravity and satellite motion
~ heat transfer
~ Hitler's rise to power during World War II
~ the Indian Wars (1870's & 1880's)
~ 20th century fragmentation due to weapons of war,
especially the atom bomb
~ Maya Angelou (and other women in literature)
~ Native American literature, specifically Cherokee
~ George Orwell's _Animal Farm_ & Russian revolutions
~ personal finance
~ geometry
==> If you would like to find out more about
==> participating in this project, please send
==> electronic mail to Judi Harris, jbharris@tenet.edu.
==> Please include your name, institution, and areas of
==> expertise.
==> PLEASE RESPOND ASAP; teacherSME pairs in the
==> specific areas requested above will be formed on
==> 1/12/93.
From xirion!jandr@relay.nl.net Tue Jan 4 02:30:20 1994
ReturnPath:
Received: from sun4nl.NL.net by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA10206; Tue, 4 Jan 94 02:30:20 EST
Received: from xirion by sun4nl.NL.net via EUnet
id AA24141 (5.65b/CWI3.3); Tue, 4 Jan 1994 08:30:19 +0100
Received: by xirion.xirion.nl id AA23801 (5.61/UK2.1);
Tue, 4 Jan 94 08:30:56 +0100
From: Jan de Ruiter
Date: Tue, 4 Jan 94 08:30:56 +0100
MessageId: <23801.9401040730@xirion.xirion.nl>
XOrganization: Xirion Unix Software & Consultancy bv
Burgemeester Verderlaan 15 X
3454 PE De Meern
The Netherlands
XPhone: +31 3406 61990
XFax: +31 3406 61981
To: cubelovers@ai.mit.edu
To: Don.Woods@eng.sun.com, cubelovers@ai.mit.edu
Subject: RE tangle cube
>Well, a pleasant surprise! It _is_ possible to take a set of 24 distinct
>tiles from any Rubik's Tangle, and use them to tile the surface of a 2x2x2
>cube such that all touching ropes match. And the solution is unique!
>
>I'll include the solution below, after some blank lines to avoid spoiling it
>for anyone who wants to try solving the puzzle without seeing the answer...
You may have noticed it yourself, but the solution you promised was missing
from your message. But I take your word for it that you found it, because
(sorry to spoil your scoop) a solution for the tanglecube as you described
was published before in CFF (Cubism For Fun) the periodical of the NKC
(Nederlandse Kubus Club = Dutch Cubist Club). Contrary to what the name
suggests members are not solely interested in cubes.
Membership to that club is open to anyone interested in puzzles like these
and highly recommended! The periodical CFF is published in English, and
appears three or four times a year.
Further information can be obtained via gm@phys.uva.nl
 Jan.
From mouse@collatz.mcrcim.mcgill.edu Tue Jan 4 07:25:33 1994
ReturnPath:
Received: from Collatz.McRCIM.McGill.EDU ([132.206.78.1]) by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA14281; Tue, 4 Jan 94 07:25:33 EST
Received: from localhost (root@localhost) by 11684 on Collatz.McRCIM.McGill.EDU (8.6.4 Mouse 1.0) id HAA11684; Tue, 4 Jan 1994 07:20:43 0500
Date: Tue, 4 Jan 1994 07:20:43 0500
From: der Mouse
MessageId: <199401041220.HAA11684@Collatz.McRCIM.McGill.EDU>
To: jandr@xirion.nl
Cc: cubelovers@ai.mit.edu
>> Well, a pleasant surprise! It _is_ possible to [tile a 2x2x2 cube
>> with distinct Rubik's Tangle pieces, uniquely]
>> I'll include the solution below, after some blank lines to avoid
>> spoiling it for anyone who wants to try solving the puzzle without
>> seeing the answer...
> You may have noticed it yourself, but the solution you promised was
> missing from your message.
Not from the copy I got  though the lines in question weren't blank;
they each contained a dot. Perhaps someone's mailer isn't doing the
hiddendot algorithm correctly?
der Mouse
mouse@collatz.mcrcim.mcgill.edu
From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Tue Jan 4 11:12:37 1994
ReturnPath: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU>
Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA23149; Tue, 4 Jan 94 11:12:37 EST
MessageId: <9401041612.AA23149@life.ai.mit.edu>
Received: from MITVMA.MIT.EDU by mitvma.mit.edu (IBM VM SMTP V2R2)
with BSMTP id 1951; Tue, 04 Jan 94 11:12:39 EST
Received: from WVNVM.WVNET.EDU (NJE origin MAILER@WVNVM) by MITVMA.MIT.EDU
(LMail V1.1d/1.7f) with BSMTP id 0432; Tue, 4 Jan 1994 11:12:39 0500
Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU
(LMail V1.1d/1.7f) with BSMTP id 0016; Tue, 4 Jan 1994 11:10:03 0500
XAcknowledgeTo:
Date: Tue, 4 Jan 1994 11:10:02 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: Which is the Real Start?
The net is so wonderful about answering questions, here are a few
more:
1. Take a standard 3x3x3 Rubik's cube, and remove the corner and
center labels to make an Edges Cube. (I am assuming that the
underlying plastic is black. If the underlying plastic is
white and one of the colors on the labels is also white, the
Edges Cube is not so pretty). Scramble the cube. Give it to
a cubemeister to solve. How will the cubemeister know if the
cube is solved? In other words, how will the cubemeister
distinguish Start from Pons Asinorum?
One answer is that the cubemeister cannot. Unless the cubemeister
saw the cube before it was scrambled, or unless the cubemeister
was told which reflection of the colors was Start, there would
be no way to tell. Another answer is that either one is Start 
that there are two Starts. However, if you like this answer,
and if you identify the identity with Start, you are in the
disquieting situation of having a group with two distinct
identities (grin!).
It is obvious that this problem does not arise if the labels are
left on the centers. Almost as obvious is the fact that the
problem does not arise if the labels are left on the corners, even
if the labels are removed from the centers. The corner group
cannot be turned inside out by a reflection as can be the edge
group.
2. As silly as my second answer is, it leads to a second question.
Just what is the 2x2x2 cube? Or more correctly, how do you
know when it is solved? With any size of cube, if you restrict
yourself to quarterturns, by definition you cannot rotate the cube
in space as a single operation. Yet, a simple quarterturn sequence
such as RL' does rotate the 2x2x2 cube because it is faceless. Is
Start of the 2x2x2 operated on by RL' solved? If so, you can argue
that the 2x2x2 has 24 Starts. Most people would not. They would
argue that there is only one Start, and that 2x2x2 cubes that differed
only by a rotation are equivalent.
3. Combining #1 and #2, I *think* that most people would argue that
Start and Pons Asinorum on the Edge Cube are not equivalent,
but that simple rotations of the 2x2x2 are equivalent. If I am
correct about "most people", why? Is a rotation symmetry
intrinsically a stronger or weaker symmetry than a reflection
symmetry?
4. When I was first posting my results about the Edge Group, and
particularly when it first began to sink in what the four equivalence
classes with only 24 elements really were, I had a moment of
panic. Since Start and Pons Asinorum differ only by a simple
reflection, why had not my version of Mconjugation declared them
to be equivalent? (I speak of "my version of Mconjugation", but
the question is no different if you look at Dan Hoey's original
Mconjugation). I think I know the answer, but I will leave the
problem as an exercise for the student. Furthermore, I think the
answer to #4 is really the same as the answer to #3.
5. What is a reflection, really? Here is an exercise to illustrate
the question. Take two identically colored and oriented 3x3x3
cubes. On one, perform F and on the other perform F'. Examine
the two cubes, plus their images in a mirror. Why are there
four distinct cubes rather than only two? At one level of
abstraction, the answer is simple. Of the four, one is not
reflected, one is prereflected, one is postreflected, and one
is both pre and postreflected. Is this a sufficient answer,
or is there something deeper?
At this point, I can't help but note Martin Gardner's famous
mirror question in Scientific American many years ago: why
does a mirror reverse left and right but not up and down?
6. I found Dan Hoey's postings about the four special states of the
Edge Group to be delightful. Some of the results were based on
a computer search of the group, for example the fact that
f(I)=(0,9,12,15) could only reasonably be determined from a
computer search. However, the thought occurred to me that most
of Dan's results were independent of the computer search, and
I was curious precisely which results would stand without the
search? For example, if we identified the group as being
rectangular, would we be led to saying which of the four special
states were diagonally opposed without the computer search?
Without the search, I might be tempted to say that Start and
Pons Asinorum were diagonally opposed.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 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 mouse@collatz.mcrcim.mcgill.edu Tue Jan 4 13:48:23 1994
ReturnPath:
Received: from Collatz.McRCIM.McGill.EDU by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA01572; Tue, 4 Jan 94 13:48:23 EST
Received: from localhost (root@localhost) by 12863 on Collatz.McRCIM.McGill.EDU (8.6.4 Mouse 1.0) id NAA12863 for cubelovers@ai.mit.edu; Tue, 4 Jan 1994 13:48:03 0500
Date: Tue, 4 Jan 1994 13:48:03 0500
From: der Mouse
MessageId: <199401041848.NAA12863@Collatz.McRCIM.McGill.EDU>
To: cubelovers@ai.mit.edu
Subject: Re: Which is the Real Start?
> The net is so wonderful about answering questions, here are a few
> more:
I am hardly more than a dilettante of the Cube, but I can perhaps offer
a few suggestions, since this seems to me to be largely psychology and
3D geometry rather than Cube group theory.
> 1. Take a standard 3x3x3 Rubik's cube, and remove the corner and
> center labels to make an Edges Cube. [...] Scramble the cube.
> Give it to a cubemeister to solve. How will the cubemeister know
> if the cube is solved? In other words, how will the cubemeister
> distinguish Start from Pons Asinorum?
> One answer is that the cubemeister cannot. [...] Another answer
> is that either one is Start  that there are two Starts.
Obviously, it is correct to state that without some a priori knowledge
of the cube's coloring, the cubemeister cannot tell. As for whether
you call them both Start:
> However, if you like this answer, and if you identify the identity
> with Start, you are in the disquieting situation of having a group
> with two distinct identities (grin!).
Not at all. All you have to do is consider the group elements to be
equivalence classes under not only wholecube rotation but also
reflection.
If you take a(n ordinary) cube and rotate the whole thing a
quarterturn, the result is not essentially different from the
original  most programs and virtually all humans would consider them
"the same". Taking the stand that Start and P.A. are the same on the
Edge Cube means only that on the Edge Cube you consider a single group
element to consist of not only a position and all those reachable by
wholecube rotations, but also those reachable by reflections as well.
The grouptheoretic identity is then neither Start nor Pons Asinorum,
but rather the equivalence class containing both those (and 46 other
elements).
> 2. [...] Just what is the 2x2x2 cube? Or more correctly, how do you
> know when it is solved?
When you have achieved any of the 24 elements of the class that we lump
together as Start.
> With any size of cube, if you restrict yourself to quarterturns,
> by definition you cannot rotate the cube in space as a single
> operation.
I'd argue the 1x1x1 breaks this statement :)
What's more, it's not clear what "quarterturn" includes: it usually
doesn't include slice turns on the 3Cube, but on the 4Cube and
higher, they must of necessity be included.
> Yet, a simple quarterturn sequence such as RL' does rotate the
> 2x2x2 cube because it is faceless. Is Start of the 2x2x2 operated
> on by RL' solved?
Yes, I would say so. I would hope most people would.
> If so, you can argue that the 2x2x2 has 24 Starts. Most people
> would not. They would argue that there is only one Start, and
> that 2x2x2 cubes that differed only by a rotation are equivalent.
Right. I would say that RL' produces a cube that is precisely as
solved as that produced by RR' is  that on the 2x2x2, R and L are in
some sense the same thing. My position would be that there is only one
Start on the 2Cube, and it is an equivalence class with 24 members.
> 3. Combining #1 and #2, I *think* that most people would argue that
> Start and Pons Asinorum on the Edge Cube are not equivalent, but
> that simple rotations of the 2x2x2 are equivalent. If I am
> correct about "most people", why?
I would say that Start and Pons Asinorum on the Edge Cube can be looked
at as mathematically equivalent (though they need not be, if you
choose) but are not intuitively equivalent. Physical objects generally
cannot be turned into reflected versions of themselves; they normally
*can* be turned into rotated versions of themselves. Thus, rotations
"feel" equivalent, but reflections don't.
> 4. [...] Since Start and Pons Asinorum differ only by a simple
> reflection, why had not my version of Mconjugation declared them
> to be equivalent?
I'm too lazy to answer this; I no longer have the messages describing
exactly what your Mconjugation operation is online. Presumably, you
implemented some intuitivelyreasonable operation, and it produced
identical results for rotations but not reflections.
> 5. What is a reflection, really?
Ouch. Mathematically, this is easy enough: given a center of reflection
P in Cartesian 3space, one computes the reflection of a point p as
P+(Pp). All the things one thinks of as reflections can be
represented as this operation compounded with rotation and/or
translation.
> Here is an exercise to illustrate the question. Take two
> identically colored and oriented 3x3x3 cubes. On one, perform F
> and on the other perform F'. Examine the two cubes, plus their
> images in a mirror. Why are there four distinct cubes rather than
> only two?
There are certainly four cubes  or at least four cube images. For
there to be only two distinct cubes, one would have to identify some of
them with one another. However, the only operations (on the cube as a
whole) that will allow identifying two of them are (1) reflection and
(2) recoloring. If your mathematical treatment considers reflections
or recolorings to be equivalent, then mathematically, there are only
two distinct cubes. Neither of these operations "feels" trivial,
though, so the four cubes all "feel" distinct.
> At this point, I can't help but note Martin Gardner's famous
> mirror question in Scientific American many years ago: why does a
> mirror reverse left and right but not up and down?
(rot13 for those who would rather think about this for a while.)
Nf abgrq va jungrire vg jnf V ernq gung dhrfgvba va, vg qbrfa'g  vg
erirefrf sebaggbonpx (jurer "sebag" naq "onpx" ner qrsvarq va grezf
bs gur fhesnpr qbvat gur ersyrpgvat). Jul guvf *nccrnef* gb nzbhag gb
erirefvat yrsg naq evtug vf n zber vagrerfgvat dhrfgvba, naq vg nzbhagf
gb nfxvat jung xvaqf bs fcngvny bcrengvbaf jr cresbez jvgubhg abgvpvat
(pbafvqrevat gurz abbcf). Va gur pnfr bs n ersyrpgvba bs n crefba, gur
bcrengvba jr'er cresbezvat jvgubhg abgvpvat vf gung bs znccvat
crefbavzntr bagb frysvzntr ol ebgngvba, fb nf gb (1) znc urnq bagb
urnq naq srrg bagb srrg naq (2) znc obqlsebag gb obqlsebag naq
obqlonpx gb obqlonpx. Ersyrpgvba, pbzcbhaqrq jvgu guvf ebgngvba,
*qbrf* erirefr yrsggbevtug.
> 6. [...]
I'm not qualified to comment.
der Mouse
mouse@collatz.mcrcim.mcgill.edu
From hoey@aic.nrl.navy.mil Tue Jan 4 19:05:27 1994
ReturnPath:
Received: from Sun0.AIC.NRL.Navy.Mil by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA17502; Tue, 4 Jan 94 19:05:27 EST
Received: by Sun0.AIC.NRL.Navy.Mil (4.1/SMI4.0)
id AA07265; Tue, 4 Jan 94 19:05:25 EST
Date: Tue, 4 Jan 94 19:05:25 EST
From: hoey@aic.nrl.navy.mil (Dan Hoey)
MessageId: <9401050005.AA07265@Sun0.AIC.NRL.Navy.Mil>
To: "Jerry Bryan" ,
Subject: Combining conjugacy classes
Last month Jerry Bryan posted a
sequence of articles about counting the number of Mconjugacy classes
of Rubik's cube positions. Having calculated the number of conjugacy
classes of the corner and edge groups separately, his idea was to
combine these to calculate the number of conjugacy classes of the
entire group. Eventually, he withdrew the calculation, after
realizing that he had not found enough information to determine the
answer. This article is about how we might calculate the answer from
separate searches of the edge and corner groups.
My first idea is to formalize the concept of symmetry in the conjugacy
classes that Jerry used in his searches. Recall that the conjugacy
class of a position X is defined to be the set of all positions m'Xmc,
where m is an element of M, the 48element symmetry group of the cube,
and c is an element of C, the 24element subgroup of M consisting of
rigid rotations of the cube in space. The reason some positions have
more symmetry than others is that for some positions, there are
nontrivial elements m and c such that m'Xmc=X. The way in which this
arises can be formalized as a kind of symmetry group of X.
For an edgegroup position X, let CSymm(X) be the set of all f in M
such that X'f'Xf is an element of C. First, I'll claim that CSymm(X)
is a subgroup of M [see proof 1, below]. Second, I note that CSymm(X)
is the set of all m in M such that there exists c in C with m'Xmc=X
[proof: m'Xmc=X iff X'm'Xm=c']. Third, I'll claim that if m'Xmc=Y,
then CSymm(X) and CSymm(Y) are conjugate subgroups of M [proof 2]. So
when Jerry says that a position X has orderN symmetry, he is saying
that CSymm(X) has 1152/N elements. But the identity of CSymm(X) has
more information than just its size, and I believe this information is
crucial if we are to combine symmetry groups. It looks to me as if it
would be sufficient to record the conjugacy class of CSymm(X), and
there are only 33 possibilities.
Now the usual symmetry group of X, Symm(X), is defined to be the group
consisting of all f in M such that X'f'Xf=I [or, equivalently, Xf=fX.
Symm(X) is the largest group such that X is Symm(X)symmetric, in the
sense of the Symmetry and Local Maxima article]. The first step in
combining the corner and edge sets is to calculate the symmetry groups
of the rotations of a position X, AllSymms(X)={Symm(Xc) : c in C}.
This corresponds to computing the symmetry groups of the edgesand
centers group from the symmetry groups of the edges group. I suspect
there is a way of computing this from CSymm(X), but I do not know it.
I am not even sure that AllSymms(X) is determined by CSymm(X). One
useful experiment would be to calculate CSymm(X) and AllSymms(Xc) for
all elements of the corner group and see what the correspondence is.
Barring an ability to calculate AllSymms(X) from CSymm(X), we could
calculate AllSymms(X) directly. This involves a great number of
calculations, though: 24 symmetry group calculations for each element
of the edge group. My first thought was to try to split the problem
up further, to deal with the group of permutations separately from the
group of orientations. But I abandoned this when I realized there is
a problem that shows up when we try this with the corner group. The
permutation of the corners that takes each cubie to its antipode is
clearly Msymmetric, and no matter how we decide to measure
orientation, there is a way to perform this permutation leaving the
cubies in their `home' orientation. But there is no way to compose
the two together in an Msymmetric way. I suspect the same problem
arises in the edge group.
But there may be some help from the edge search available in calcu
lating AllSymms(X). For take a position Y in the edgesandcenters
group; Y is also a rotated position in the edges group, so Y=m'Xmc for
some X in Jerry Bryan's list. So for f in Symm(Y), Y'f'Yf=I is an
element of C, so f is in CSymm(Y). This says that Symm(Y) is a
subgroup of CSymm(Y), which is a conjugate of CSymm(X). So if Symm(Y)
is nontrivial, then CSymm(X) will also be nontrivial. So to find the
symmetry groups of the edgesandcenters group we need only look at
those positions that have nontrivial groups in Jerry's list (i.e. less
than order1152 symmetry), as all the others will have Symm(Y)=I. So,
Jerry, do you have the data on how many positions of the edge group
have less than order1152 symmetry, and which positions those are?
So, on to finding the symmetry groups of the Rubik's group positions.
We need to calculate Symm(X) for every element X of the edgesandcen
ters group and Symm(Y) for every element Y of the cornersandcenters
group, while keeping track of the permutation parity of X and Y. (The
permutation parity will be constant over each Symm(X), Symm(Y)). The
symmetry groups in the Rubik's group will be the intersections of
symmetry groups of edge and corner positions of the same parity. We
need not keep track of the particular positions here, only the numbers
for each parity and each (conjugacy class of) symmetry group. I have
a program that could produce a table easily enough. Recently, I took
a look in Paul B. Yale's _Geometry_and_Symmetry_ and it looks like
this is the sort of problem we could use the PolyaBurnside theorem
on. Unfortunately, I don't understand it yet, and it looks like the
number of cases here might be too large to conveniently carry out by
hand. So it would help to go after this problem computationally.
The rest of this article has the proofs for the claims I mentioned in
the second paragraph.
================================================================
Proof 1: Suppose f, g are elements of CSymm(X); it suffices to show
that f'g is an element of CSymm(X).
X'(f'g)'X(f'g)=X(g'f)X(f'g)
=X'g'(Xgg'ff'X')fXf'g
=(X'g'Xg) g'f (f'X'fX) f'g,
=(X'g'Xg) (f'g)' (X'f'Xf)' (f'g).
Since we assumed f, g in CSymm(X), X'g'Xg and X'f'Xf must be in C.
(f'g)' and (f'g) are elements of M that are either both in C or
neither. So the product is in C, so f'g is in CSymm(X). Therefore
CSymm(X) is a group, QED.
Proof 2: Suppose Y=m'Xmc. First let f be an element of CSymm(X), so
that X'f'Xf is in C. I will first show that m'fm is an element of
CSymm(Y).
Y'(m'fm)'Y(m'fm)=(m'Xmc)'(m'fm)'(m'Xmc)(m'fm)
=(c'm'X'm)(m'f'm)(m'Xmc)(m'fm)
=c'm'(X'f'X)(mcm'fm)
=c'm'(X'f'Xf)(f'mcm'fm)
All of which are elements of M, with an even number in C. Therefore
the expression is in C, so m'fm is in CSymm(Y).
Now let g be an element of CSymm(Y), so that Y'g'Yg is in C. Let
f=mgm', so f is an element of M such that m'fm is in CSymm(Y). I will
show that f is an element of CSymm(X):
X'f'Xf=(mc)(mc)'X'(mm')f'(mm')Xf(f'mcm'fm)(f'mcm'fm)'
=(mc)(m'Xmc)'(m'fm)'(m'Xmc)(m'fm)(f'mcm'fm)'
=(mc)Y'(m'fm)'Y(m'fm)(f'mcm'fm)'
=(mc)Y'g'Yg(f'mcm'fm)',
which is in C, so f is in CSymm(X).
I've shown that for every element f of CSymm(X), m'fm is an element of
CSymm(Y), and that every element of CSymm(Y) is m'fm for some f in
CSymm(X). Therefore CSymm(Y)=m' CSymm(X) m, QED.
================================================================
Dan Hoey
Hoey@AIC.NRL.Navy.Mil
From hoey@aic.nrl.navy.mil Tue Jan 4 21:36:19 1994
ReturnPath:
Received: from Sun0.AIC.NRL.Navy.Mil by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA23585; Tue, 4 Jan 94 21:36:19 EST
Received: by Sun0.AIC.NRL.Navy.Mil (4.1/SMI4.0)
id AA07661; Tue, 4 Jan 94 21:36:18 EST
Date: Tue, 4 Jan 94 21:36:18 EST
From: hoey@aic.nrl.navy.mil (Dan Hoey)
MessageId: <9401050236.AA07661@Sun0.AIC.NRL.Navy.Mil>
To: "Jerry Bryan" , CubeLovers@ai.mit.edu
Subject: Re: Which is the Real Start?
Jerry Bryan has some more questions.
> Take a standard 3x3x3 Rubik's cube, and remove the corner and
> center labels to make an Edges Cube.... Scramble..., how will
> the cubemeister distinguish Start from Pons Asinorum?
> ... if you identify the identity with Start, you are in the
> disquieting situation of having a group with two distinct identities
> (grin!).
The problem is that we would not be dealing with a _group_ then, but a
collection of cosets of M. Just as in the edge `group', we deal with
either 1) a lesssymmetric group in which one of the edges never
moves, or 2) a larger group in which we distinguish positions that
differ by rigid motions of the cube, or
3) a nongroup in which we consider cosetsequivalence
classes of group #2, where group elements that differ by rigid motions
are equivalent. You have got a lot of mileage out of working with
group #2 to save duplication among symmetries, then reducing to
nongroup #3. But what you lose is the group structure of the object
you are studying. Instead, you have to work in the large group and
then deduce information about the cosets. All in all, though, I'm
very glad of it, for the lost symmetries of group #1 were sorely
missed.
For most of the other questions, mouse@collatz.mcrcim.mcgill.edu
provides satisfactory answers. However, strictly speaking we should
not call an equivalence class to be a group element (unless it is a
coset of a normal subgroup, and neither C nor M is normal in the large
group). I'll admit I've also abused the term when considering
distances in the ``edge group'', as if all 24 rotations of a position
were the same element of some group. But when we start dealing with
the distinction between fixed and movable cubes I think we need to
start being more careful.
[ mouse also mentions that quarterturn ``usually doesn't include
slice turns on the 3Cube, but on the 4Cube and higher, they must
of necessity be included.'' I'll take that as an argument for
eccentric slabism: a QT rotates any 1xNxN slab except a central slab
of an oddedged cube. As opposed to cutism, where a QT consists of
a rotation of part of the cube with respect to the other. ]
Other questions:
> ...since Start and Pons Asinorum differ only by a simple
> reflection, why had not my version of Mconjugation declared them
> to be equivalent?
Your versino treats positions X,Y for which m'Xmc=Y (m in M, c in C)
as equivalent. If you instead determine when m'Xmn=Y (m,n in M) you
would find them equivalent. This is equivalent to changing the loop
in your version of Mconjugacy.
> For j = 1 to 24 for k = 1 to 24 for m = 1 to 2
> for i = 1 to 24 Bj,k,m(i) = Qm(Pj(A(Qm(Pk(i)))))
so that the two occurrences of Qm need not be the same.
> (I speak of "my version of Mconjugation", but the question is no
> different if you look at Dan Hoey's original Mconjugation).
No, I didn't use Mconjugation except for a cube with a fixed
orientation in space [or equivalently, with face centers]. So in the
original concept of Mconjugation that Jim Saxe and I put together,
Start and Pons Asinorum don't just differ by a reflection.
> I found Dan Hoey's postings about the four special states of the
> Edge Group to be delightful.... However, [without the results on
> distances] if we identified the group as being rectangular, would we
> be led to saying which of the four special states were diagonally
> opposed without the computer search? Without the search, I might be
> tempted to say that Start and Pons Asinorum were diagonally opposed.
Well, really the `group' is in the shape of a sphenoid, a word I
learned yesterday for a tetrahedron whose three pairs of opposite
edges are equal. [Or equivalently, a tetrahedron whose edges are face
diagonals of a rectangular prism.] But it might be more accurate to
consider it as a large ball of string with a bunch of symmetries.
Calling it a rectangle or sphenoid may lead us to ignore the structure
that is not representable in Euclidean space.
Dan Hoey
Hoey@AIC.NRL.Navy.Mil
From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Thu Jan 6 04:11:20 1994
ReturnPath: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU>
Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA29330; Thu, 6 Jan 94 04:11:20 EST
MessageId: <9401060911.AA29330@life.ai.mit.edu>
Received: from MITVMA.MIT.EDU by mitvma.mit.edu (IBM VM SMTP V2R2)
with BSMTP id 1228; Thu, 06 Jan 94 04:11:19 EST
Received: from WVNVM.WVNET.EDU (NJE origin MAILER@WVNVM) by MITVMA.MIT.EDU
(LMail V1.1d/1.7f) with BSMTP id 0545; Thu, 6 Jan 1994 04:08:07 0500
Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU
(LMail V1.1d/1.7f) with BSMTP id 9949; Wed, 5 Jan 1994 23:34:59 0500
XAcknowledgeTo:
Date: Wed, 5 Jan 1994 23:34:58 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: Square Brackets
This posting has very little to do with cubes of any sort, but I
think you may find it of interest anyway. If not, you can just
delete it.
In his analysis of my operator which combines Mconjugation with
a whole cube rotation, Dan Hoey inserted square brackets to make
his exposition more readable. And therein lies a story.
As it turns out, I had composed most of my original note at work,
and I had used square brackets. Actually, I had used square
brackets to delimit the indexes for the rotation and reflection
operations, and I had used parenthesis to delimit the indexes for
the individual cells in the row vectors. I wanted to make a
distinction between the two kinds of indexing. Dan avoids the
necessity for a distinction by simply not detailing the
indexing of the individual cells.
Anyway, I completed the note at home. Much to my dismay, all of my
square brackets had disappeared! I pretty much understand the
problem. My Email system is an IBM mainframe which uses EBCDIC as
its basic code. EBCDIC does not deal very well with square brackets.
There are at least two "standards" for encoding square brackets in
EBCDIC.
There are any number of ways to access an IBM mainframe, but the
native terminal support is 3270 terminals using EBCDIC. Both at home
and at work, I use a PC running TN3270 to access our mainframe.
TN3270 is a 3270 version of TELNET. However, the TN3270 I use at work
is considerably different from the TN3270 I use at home. One TN3270
implements one of the square bracket standards, and the other
TN3270 implements the other.
For similar reasons, mail gateways often have difficulties with
square brackets. They may have to translate EBCDIC to ASCII or
ASCII to EBCDIC, and it is difficult to know how best to set up
the translate tables. My experience is that some gateways get it
"right" and others get it "wrong". I therefore had a great fear
that if I posted my note with square brackets, that the square brackets
might appear as gibberish to at least some of you. Thus, I sort
of temporized and faked the subscripts with upper and lower
case letters (e.g., Bk to mean Bsubk), omitting square brackets
entirely.
It is probably no accident that old programming languages such
as FORTRAN and COBOL use parentheses for indexing. These languages
originated in the 50's. At that time, the dominant character
code was BCD, which did not include square brackets. EBCDIC is just
extended BCD, and the original EBCDIC did not include square brackets,
either. Square brackets are a latter day addition to EBCDIC, and
the implementation of square brackets in EBCDIC is inconsistent.
ASCII has always included square brackets. "Modern" languages
(say, starting in the 70's) such as Pascal and C (and their
descendents) grew up in the ASCII world, and tend to use
square brackets for indexing and parentheses for function
arguments. FORTRAN compilers to this day have difficulty figuring
out with things like Y=X(I) or Y=F(X)  which are functions and
which are subscripted arrays. Also, Pascal and C tend not to
coexist very well in the EBCDIC world because of these kinds of
character set difficulties.
There are several other characters with similar difficulties. For
example, if G is the cube group, you might want to refer to the
size of the cube group as G. But the delimiting vertical bars
can be very different between EBCDIC and ASCII. Finally, FORTRAN
used ** for exponentiation. More modern languages tend to use
some sort of uparrow or carat. But such characters don't
translate well between EBCDIC and ASCII. For example, if I
write G = 4.3 * 10^19, it is highly problematic whether the
character between the 10 and the 19 which I am using to express
exponentiation will make any sense on your particular system.
For whatever it is worth, here are my home and work versions of
square brackets:
home left square bracket [[[[[[[ x'AD'
home right square bracket ]]]]]]] x'BD'
work left square bracket x'BA'
work right square bracket x'BB'
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 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 Thu Jan 6 13:52:03 1994
ReturnPath: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU>
Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA24385; Thu, 6 Jan 94 13:52:03 EST
MessageId: <9401061852.AA24385@life.ai.mit.edu>
Received: from MITVMA.MIT.EDU by mitvma.mit.edu (IBM VM SMTP V2R2)
with BSMTP id 6454; Thu, 06 Jan 94 13:21:25 EST
Received: from WVNVM.WVNET.EDU (NJE origin MAILER@WVNVM) by MITVMA.MIT.EDU
(LMail V1.1d/1.7f) with BSMTP id 5918; Thu, 6 Jan 1994 13:21:24 0500
Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU
(LMail V1.1d/1.7f) with BSMTP id 6741; Thu, 6 Jan 1994 13:18:51 0500
XAcknowledgeTo:
Date: Thu, 6 Jan 1994 13:18:50 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: An Alternate Analysis of B
Here is a definition. If X is any cube, then the function B
is defined as B(X) = min (C[i] R[j] X R[j] C[k]) for i in {1 to 24},
j in {1 to 2}, and k in {1 to 24}, and where C is the set of rotations
of the cube and R is the set of reflections of the cube. Then, two
cubes X and Y are said to be Bequivalent if B(X)=B(Y). Note that
the B function calculates the "canonical form" or the "representative
element" of an equivalence class under Mconjugation and rotation of
the cube in space.
Dan Hoey already gave a thorough analysis of this situation. I would
like to provide an alternative analysis which I hope to be equivalent
to Dan's. My analysis will be fairly informal as compared to Dan's.
Here are a couple of preliminaries. First, multiplication of
permutations is generally not commutative. For example, if X is any
cube, then it is generally not the case that m'Xm=mXm', where m
is in M, the set of all cube rotations composed with reflections.
However, we can calculate all Mconjugates of X as B[n]=m[n]' X m[n]
for n in {1 to 48}, or we can calculate all Mconjugates of X as
B[n]=m[n] X m[n]'. Either way, B will be the same set of cubes. It
will be in a different order, but it will be the same set. The reason
is that the set of all m[n] is the same as the set of all m[n]',
namely just M, but m[n] is in a different order than m[n]'. This means
that as long as we are calculating all Mconjugates as opposed to
a specific Mconjugate, we can sort of "violate" the normal rules
about multiplication commutivity.
Second, if X is any cube, consider the set of all rotations of
X, namely B[i] = X c[i] for i in {1 to 24}, and where c[i] is in C,
the set of all cube rotations. Having generated the set of all
rotations of X, we can rotate as many times as we wish, for example
B[j] = X c[i] c[j] for i in {1 to 24} and j in {1 to 24}, or even
B[k] = X c[i] c[j] c[k] for i in {1 to 24}, j in {1 to 24}, and
k in {1 to 24}. No matter how many times we multiply, B will be the
same set, it will just be in a different order. Conversely, if we
have any number of adjacent rotations in the multiplication, we can
eliminate all but one rotation, and B will be the same set, and again
will just be in a different order.
With the preliminaries out of the way, we note that the set of all
Mconjugates of X is generated as B[i]=m[i]' X m[i] for i in
{1 to 48}. But we can also generate the same set in a different order
as B[i]=m[i] X m[i]' for i in {1 to 48}. We can decompose M and
calculate all Mconjugates as B[i,j]=c[i] r[j] X r[j]' c[i]' for i in
{1 to 24} and j in {1 to 2}.
But r[1]'=r[1] (r[1] is the identity) and r[2]'=r[2] (the reflection
is its own complement). So we have B[i,j]=c[i] r[j] X r[j] c[i]'
for i in {1 to 24} and j in {1 to 2}. The set of all c[i] is the
same as the set of all c[i]' (just in a different order), so we
define i' as the index for which c[i']=c[i]'. Hence the calculation
of an Mconjugate can be written as B[i,j]=c[i] r[j] X r[j] c[i']
for i in {1 to 24} and j in {1 to 2}.
Finally, we wish to multiply the Mconjugate by the set of all rotations,
so we have B[i,j,k]=c[i] r[j] X r[j] c[i'] c[k] for i in {1 to 24},
j in {1 to 2}, and k in {1 to 24}. But as we noted in our second
preliminary note, we can collapse multiple rotations into one, and we
have B[i,j,k]=c[i] r[j] X r[j] c[k], and B will be the set of all
Mconjugates of X multiplied by all rotations. I guess I am overusing
the letter "B" a bit, because the "B" function is simply the minimum
of the "B" matrix. But in any case, we have shown that the "B" loop
in my program is simply calculating the set of all Mconjugates multiplied
by all rotations. This is the exact result already proven by Dan Hoey,
but I found the above derivation a little easier to follow.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 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 pbeck@pica.army.mil Thu Jan 6 14:22:12 1994
ReturnPath:
Received: from COR6.PICA.ARMY.MIL by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA26466; Thu, 6 Jan 94 14:22:12 EST
Date: Thu, 6 Jan 94 14:06:54 EST
From: Peter Beck (BATDD)
To: CubeLovers@ai.mit.edu
Subject: Mickey's Challenge
MessageId: <9401061406.aa23113@COR6.PICA.ARMY.MIL>
NEW PUZZLE "MICKEY'S CHALLENGE" is at your Disney
store now, price $10.
This is a legal MACHBALL, ie, a spherical
SKEWB. It comes with a solution book.
Christoph Bandelow (a longer time cuber)
wrote the solution.
I haven't bought one or it played with it
yet.
GOOD PUZZLING
pete beck
pbeck@pica.army.mil
From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Thu Jan 6 14:31:34 1994
ReturnPath: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU>
Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA26764; Thu, 6 Jan 94 14:31:34 EST
MessageId: <9401061931.AA26764@life.ai.mit.edu>
Received: from MITVMA.MIT.EDU by mitvma.mit.edu (IBM VM SMTP V2R2)
with BSMTP id 6601; Thu, 06 Jan 94 13:28:44 EST
Received: from WVNVM.WVNET.EDU (NJE origin MAILER@WVNVM) by MITVMA.MIT.EDU
(LMail V1.1d/1.7f) with BSMTP id 6254; Thu, 6 Jan 1994 13:28:44 0500
Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU
(LMail V1.1d/1.7f) with BSMTP id 6837; Thu, 6 Jan 1994 13:26:10 0500
XAcknowledgeTo:
Date: Thu, 6 Jan 1994 13:26:09 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: Character Test, Please Ignore
Forward slashes ///////////////
Back slashes \\\\\\\\\\\\\\\
Left Braces {{{{{{{{{{{{{{{
Right Braces }}}}}}}}}}}}}}}
Carat ^^^^^^^^^^^^^^^
Tildes ~~~~~~~~~~~~~~~
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
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 Fri Jan 7 10:35:45 1994
ReturnPath: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU>
Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA09588; Fri, 7 Jan 94 10:35:45 EST
MessageId: <9401071535.AA09588@life.ai.mit.edu>
Received: from MITVMA.MIT.EDU by mitvma.mit.edu (IBM VM SMTP V2R2)
with BSMTP id 5292; Fri, 07 Jan 94 10:35:42 EST
Received: from WVNVM.WVNET.EDU (NJE origin MAILER@WVNVM) by MITVMA.MIT.EDU
(LMail V1.1d/1.7f) with BSMTP id 3068; Fri, 7 Jan 1994 10:35:42 0500
Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU
(LMail V1.1d/1.7f) with BSMTP id 7903; Fri, 7 Jan 1994 10:33:09 0500
XAcknowledgeTo:
Date: Fri, 7 Jan 1994 10:33:04 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: Some Proposed Terminology
I wish to propose some terminology and definitions to make certain
concepts a bit more precise. For example, when we are talking about
"corners only", it is not always clear whether we are talking about
"corners with centers without edges" or "corners without centers
without edges". In this note, I have tried to be consistent with
previous usage on the list, but I welcome any historical corrections
that might be deemed necessary.
Let G be the standard cube group for the 3x3x3 cube, and let G be
the size G. Hence, we have G = (8!(3^8)/3 * 12!(2^12)/2) / 2,
which is the famous 4.3 * 10^19.
Let GC be the corners with centers without edges group for the 3x3x3
cube, and let GC be the size of GC. Hence, we have
GC = 8!(3^8)/3. (I welcome a suggestion other than "GC" as the
name for this group. I did not find one in the archives.) This group
could be modeled by removing the edge labels from a standard
3x3x3 cube.
Let GE be the edges with centers without corners group for the 3x3x3
cube, and let GE be the size of GE. Hence, we have
GE = 12!(2^12)/2. (As before, I welcome a suggestion other than
"GE" for the name for this group.) This group could be modeled by
removing the corner labels from a standard 3x3x3 cube.
Note that G = GC * GE / 2.
Let G\C be the corners with edges without centers group. I intend
for the notation to indicate G reduced by C, where C is the rotation
group for the cube. It should be the case that G\C = G / 24,
but I want to return to this question a little later. This group
could be modeled by removing the center labels from a standard
3x3x3 cube.
Let GC\C be the corners without edges without centers group. This is
the 2x2x2 cube. We should have GC\C = GC / 24, but again I want
to return to this question a little later. In addition to being the
2x2x2 cube, this group could be modeled by removing the center and
edge labels from a standard 3x3x3 cube.
Let GE\C be the edges without corners without centers group. We
should have GE\C = GE / 24, but again I want to return to this
question a little later. This group could be modeled by removing
the center and corner labels from a standard 3x3x3 cube.
Let G\M be the set of Mconjugate classes for G. In this case,
G\M is approximately 48 times smaller than G. I believe that
when Dan Hoey asked in 1984 the question "how big is G, really?",
that he was really asking how big is G\M, and that he was asking
for the approximation to be resolved to an exact number.
Let GC\M be the set of Mconjugate classes for GC. In this case,
GC\M is approximately 48 times smaller than GC.
Let GE\M be the set of Mconjugate classes for GE. In this case,
GE\M is approximately 48 times smaller than GE.
Recall that B is the function which calculates the canonical form
for a cube under the composed operations of Mconjugation plus
rotation. My programs calculate equivalence classes under B.
Let G\B be the set of Bclasses for G. Let GC\B be the set of Bclasses
for GC. Let GE\B be the set of Bclasses for GE. So far, my programs
have built complete search trees for GC\B and GE\B.
Let Gx denote any of G, GC, and GE. Then, we have Gx\B=(Gx\C)\M=(Gx\M)\C.
In English, we can decompose B into a multiplication by C and M (in
either order).
Also, note that Gx\C=(Gx\C)\C=((Gx\C)\C)\C=.... Similarly,
(G\M)\C=((G\M)\C)\C=.... In English, having reduced once by C, we can
reduce again by C as many times as we wish, but we simply get the same
set back again each time.
This notation can help us address the question of whether B actually
accomplishes a "times 48" or a "times 1152" reduction in the size of
the cube. If we are dealing with Gx, then Gx\B is a "times
1152" reduction. However, information is lost. For example, consider
GC and GC\B. GC is "corners plus centers", and Breduction of GC
removes the centers and calculates Mconjugates of the corners.
But you really don't have the same problem any more because the centers
are gone. If on the other hand we are dealing with Gx\C, then (Gx\C)\B
is a "times 48" reduction. All we have really done is calculate
Mconjugates. The reduction by the C that is composed into B is duplicate
effort which accomplishes nothing.
I have come to realize that my program for the 2x2x2 actually models
GC (corners with centers without edges) rather than GC\C (corners
without centers without edges). My program does not explicitly encode
the centers. However, it encodes all eight corner cubies, and when
it makes qturns, any of the eight cubies can move. Hence, rotational
information is encoded, even if the centers themselves are not explicitly
encoded. If I wanted to model GC\C, I would have had to either model
only seven of the cubies, or else modeled all eight but moved only seven
of them. Since what I really wanted was (GC\C)\M, and since what I had
was GC, I had to invent this funny B thing, where GC\B=(GC\C)\M. If I
had been clever enough to model GC\C in the first place, I never would
have had to invent B. Similar comments apply to my model for the edges.
To convince yourself that eight corner cubies model GC and seven corner
cubies model GC\C, just think about calculating GC and GC\C.
For GC, there are eight ways to pick the first cubie, seven ways to
pick the second cubie, and so forth yielding the familiar
GC=8!*(3^8)/3. For GC\C, we let one of the cubies be fixed,
then there are seven ways to pick the second cubie, and so forth yielding
GC\C=7!*(3^7)/3, and GC = GC\C * 24. Hence, the "corners of the
3x3x3" problem is 24 times larger than the "2x2x2" problem.
I will discuss the "times 24" reduction that is accomplished by
reducing by C in a followup note.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 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 Jan 8 10:21:12 1994
ReturnPath: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU>
Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA26746; Sat, 8 Jan 94 10:21:12 EST
MessageId: <9401081521.AA26746@life.ai.mit.edu>
Received: from MITVMA.MIT.EDU by mitvma.mit.edu (IBM VM SMTP V2R2)
with BSMTP id 4208; Sat, 08 Jan 94 08:48:56 EST
Received: from WVNVM.WVNET.EDU (NJE origin MAILER@WVNVM) by MITVMA.MIT.EDU
(LMail V1.1d/1.7f) with BSMTP id 0001; Sat, 8 Jan 1994 08:48:56 0500
Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU
(LMail V1.1d/1.7f) with BSMTP id 7240; Sat, 8 Jan 1994 08:46:21 0500
XAcknowledgeTo:
Date: Sat, 8 Jan 1994 08:46:20 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: Some Terminology Concerning B
I have started to use "B" to indicate various aspects of the
conjugacy class generated by m'Xmc. The choice of B is sort
of an accident. I used "B" in the program fragment which
I posted to the list, and Dan Hoey analyzed the program fragment.
I have called it "B" in my mind ever since. However, I have used
B in several inconsistent ways. This is a proposal to
rectify that inconsistency.
Let X be any cube. Then the set of Bconjugacy classes of X is
the set of all m'Xmc for all m in M and all c in C. We denote
this set as BClass(X). B is the function B(X)=min(BClass(X)).
Note that we could have defined BClass(X) equivalently as the
set of all mXm'c, or as the set of all cm'Xm, or as the set of
all cmXm'. It is in general not the case that
m'Xmc = mXm'c = cm'Xm = cmXm' for any fixed value of m and c.
(Quite the contrary!). However, when we say "the set of all...",
the four ways of generating BClass(X) become equivalent. This is
the justification for the assertion in a previous note that
Gx\B = (Gx\M)\C = (Gx\C)\M.
Two cubes X and Y are Bequivalent if BClass(X) = BClass(Y).
Equivalently, two cubes X and Y are Bequivalent if B(X) = B(Y).
X is the length of X (the distance of X from Start). We have
B(X) = X for centerless cubes, but it is generally not the
case that B(X) = X for cubes with centers. In fact, let
X and Y be cubes with centers such that B(X)=B(Y). It is not
necessarily the case that X = Y. For example, consider
the set GC of cubes with corners with centers without edges.
We have B(RL')=B(I), but RL'=2 and I=0.
BClass(X) is the number of elements in BClass(X). If
BClass(X) = N, then X is said to have orderN symmetry. (I
sincerely regret ever using this terminology. As has been noted
on the list, it seems "backwards" somehow. But given that this
usage exists, the value 1152/N is generally more useful than the
value N.)
We note the following:
1. B(X) is a cube.
2. BClass(X) is a set of cubes.
3. B(B(X)) = B(X)
4. BClass(B(X)) = BClass(X).
5. Both X and B(X) are in BClass(X).
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 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 Jan 8 11:25:43 1994
ReturnPath: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU>
Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA28874; Sat, 8 Jan 94 11:25:43 EST
MessageId: <9401081625.AA28874@life.ai.mit.edu>
Received: from MITVMA.MIT.EDU by mitvma.mit.edu (IBM VM SMTP V2R2)
with BSMTP id 4466; Sat, 08 Jan 94 10:55:08 EST
Received: from WVNVM.WVNET.EDU (NJE origin MAILER@WVNVM) by MITVMA.MIT.EDU
(LMail V1.1d/1.7f) with BSMTP id 0645; Sat, 8 Jan 1994 10:55:08 0500
Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU
(LMail V1.1d/1.7f) with BSMTP id 7645; Sat, 8 Jan 1994 10:52:33 0500
XAcknowledgeTo:
Date: Sat, 8 Jan 1994 10:52:22 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: Calculating G\M
Armed with Dan Hoey's note of 4 January "Combining conjugacy classes",
I wish to propose once again a procedure for calculating G\M, the
number of Mconjugate classes of G, which I think in some sense is the
"real" size of G. My proposal draws *very* heavily on Dan's note.
My first (incorrect) proposal was based on the following idea. By
computer search, we already have a database of GC\B and GE\B, the
corners and edges of G, respectively, reduced by Bconjugacy. Hence,
we know GC\B and GE\B. For each X in GC\B and each Y in GE\B,
we calculate BClass(X) and BClass(Y). Then, we can combine
BClass(X) and BClass(Y) in all legal ways (minding parity considerations).
Call the combinations BClass(X) * BClass(Y). For any fixed X and Y,
BClass(X) * BCLass(Y) is a set of cubes in G. Hence we can calculate
(BClass(X) * BClass(Y))\M and (BClass(X) * BClass(Y))\M. My idea
then was just to sum (BClass(X) * BClass(Y))\M over all values
of X and Y to calculate G\M. And we know how many X's and Y's there are!
But, alas, (BClass(X) * BClass(Y))\M is not the same across all X's and
Y's because, well because of symmetry. All X's and Y's are not equally
symmetrical. I was assuming that (BClass(X) * BClass(Y))\M was
constant, and of course it is not.
My next (incorrect) proposal was never posted to the list. It was
a slight improvement on the first idea. We have a data base of all X's
in GC\B and of all Y's in GE\B. For each X in GC\B and for each
Y in GE\B, we know BClass(X) and BClass(Y). (Actually, we don't.
I have to calculate it. I have done so, and I have posted
summaries of those calculations. However, I did not store
the order of the equivalence classes in the data base. I kick myself
for not doing so, but this is a minor problem, so let us continue).
There are only 10 distinct values for BClass(X) and for BClass(Y),
namely 24, 48, 72, 96, 144, 192, 288, 384, 576, and 1152. (By the
way, I have never figured out why it is *exactly* the same set of
values for both the corners and for the edges. It is easy to see
why it is approximately the same set of values, but the structure
of the corners is enough different from the structure of the edges
that I see no obvious reason the set of values should be exactly
the same in both cases.)
Let GC[m] be the set of all X for which BClass(X) = m and let
GE[n] be the set of all Y for which BClass(Y) = n. Hence,
GC\B is partitioned into GC[m]\B for m=24,48..., and GE\B is
partitioned into GE[n]\B for n=24,48,... Now, we form the sets
BClass(X)[m] * BClass(Y)[n] for all X in GC[m]\B and for all
Y in GE[n]\B, and for all legal values of m and n. There will
be 100 such sets.
For any fixed X, Y, m and n, BClass(X)[m] * BCLass(Y)[n] is a
set of cubes in G. Hence we can calculate
(BClass(X)[m] * BClass(Y)[n])\M and (BClass(X)[m] * BClass(Y)[n])\M.
My idea then was just to sum (BClass(X)[m] * BClass(Y)[n])\M
over all X in GC[m]\B and over all Y in GE[n]\B to calculate
(BClass(X)[m] * BClass(Y)[n])\M. We know how many X's there are
in GC[m]\B and we know how many Y's there are in GE[n]\B, so the
calculation seemed possible. I then intended to sum again over all
m and over all n, and I would be done.
But, alas, in order to perform the sum over all X and all Y, I
needed a theorem which I couldn't prove and which I now believe
is not true anyway. I needed to be able to prove that for a fixed
m and n, that (BClass(X)[m] * BClass(Y)[n] had the same value
for all X in GC[m]\B and all GE[n]\B. For a while I thought it was
true, but right now I can't think of any reason why it should be.
But perhaps Dan Hoey comes to the rescue with his CSymm function.
I still need a theorem which I cannot (yet) prove, but I believe it
is true. If it can be proven, my basic overall scheme can be
rescued.
In my second proposal, I used the values of all possible BClass
sizes as indexes  24, 48, 72... It would perhaps be more
convenient to make these sizes a set {24, 48, 72, ...},
and to think of the indexes m and n taking on the values from
1 to 10, where the values from 1 to 10 index the set
{24, 48, 72, ....}. With this understanding, all the above results
are valid, and the indexing is more convenient.
We can now say that GC\B is partitioned into GC[1]\B, GC[2]\B, ...
through GC\B[10] and similarly for GE\B. Unfortunately, using
the Bequivalence class sizes to partition GC\B and GE\B did not
permit us perform the calculations we wanted to perform. However,
suppose we partition GC\B and GE\B a different way, namely using
CSymm. Suppose, for each X in GC\B and for each Y in GE\B,
we calculate CSymm(X) and CSymm(Y). (We would have to do this
by computer). CSymm(X) and CSymm(Y) are sets, but there are only
a (relatively) small number of such sets. Let each distinct value
CSymm(X) and CSymm(Y) be mapped to an index. We can call such a
mapping function CSind, and we can calculate CSind(CSymm(X)) and
CSind(CSymm(Y)). Actually, there is no reason not to define
CSind in such a way that the domain is the set of X's and Y's, so
that we can calculate CSind(X) and CSind(Y).
Now, we use m=CSind(X) and n=CSind(Y) to form a partition of
GC\B and GE\B. All our results from before are valid. The only
issue is, can we now perform the sum? In order to perform the sum,
we need the following to be true:
For a fixed m and n, BClass(X)[m] * BClass(Y)[n] is constant
for all X in GC[m]\B and all Y in GE\[n]\B, where GC[m] is
the set of all X in GC such that CSind(X)=m and GE[n] is the set
of all Y in GE such that CSind(Y)=n.
It really seems true to me, and I shall strive to prove it.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 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 Jan 8 18:47:56 1994
ReturnPath: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU>
Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA13806; Sat, 8 Jan 94 18:47:56 EST
MessageId: <9401082347.AA13806@life.ai.mit.edu>
Received: from MITVMA.MIT.EDU by mitvma.mit.edu (IBM VM SMTP V2R2)
with BSMTP id 5459; Sat, 08 Jan 94 15:13:25 EST
Received: from WVNVM.WVNET.EDU (NJE origin MAILER@WVNVM) by MITVMA.MIT.EDU
(LMail V1.1d/1.7f) with BSMTP id 3153; Sat, 8 Jan 1994 15:13:25 0500
Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU
(LMail V1.1d/1.7f) with BSMTP id 8475; Sat, 8 Jan 1994 15:10:53 0500
XAcknowledgeTo:
Date: Sat, 8 Jan 1994 15:10:52 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: Calculating G\M (Three Typos)
I have found three typos in my article from this morning. Two are
trivial, and I would not have bothered to report them. One of them
is fundamental, and I feel obliged to report a correction.
>for all X in GC[m]\B and all GE[n]\B. For a while I thought it was
and for all Y in GE[n]\B
>We can now say that GC\B is partitioned into GC[1]\B, GC[2]\B, ...
>through GC\B[10] and similarly for GE\B. Unfortunately, using
GC[10]\B
> For a fixed m and n, BClass(X)[m] * BClass(Y)[n] is constant
(BClass(X)[m] * BClass(Y)[n])\M
> for all X in GC[m]\B and all Y in GE\[n]\B, where GC[m] is
> the set of all X in GC such that CSind(X)=m and GE[n] is the set
> of all Y in GE such that CSind(Y)=n.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 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 CPELLEY@delphi.com Mon Jan 10 17:31:46 1994
ReturnPath:
Received: from bos1a.delphi.com (delphi.com) by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA11105; Mon, 10 Jan 94 17:31:46 EST
Received: from delphi.com by delphi.com (PMDF V4.211 #4520) id
<01H7INM9NDRK91X4TE@delphi.com>; Mon, 10 Jan 1994 12:49:21 EDT
Date: Mon, 10 Jan 1994 12:49:21 0400 (EDT)
From: CPELLEY@delphi.com
Subject: Mickey's Challenge
To: CubeLovers@ai.mit.edu
MessageId: <01H7INM9NNEQ91X4TE@delphi.com>
XVmsTo: INTERNET"CubeLovers@ai.mit.edu"
MimeVersion: 1.0
ContentType: TEXT/PLAIN; CHARSET=USASCII
ContentTransferEncoding: 7BIT
I visited the local Disney Store and picked up a Mickey's Challenge puzzle for
$10. It's really cute, and the book that it comes with is excellent.
Included are color photos of Christophe Bandelow, Uwe Meffert, and the puzzle
disassembled into all its parts. Plus it gives a solution for the puzzle and
has a short bio on Uwe Meffert. It also shows color photos of the Megaminx,
Pyraminx (not the Tomy version, but a black one), and the 5x5x5 which they
refer to as "Professor's Cube."
Some general notes on Mickey's Challenge. It is a spherical Skewb, and it
actually turns much more smoothly than my cubical Skewb. It has the same
delightful "clicking" mechanism that the Skewb and original Pyraminx had.
It is a bit easier than the Skewb, since there are a few blank pieces that
can be exchanged without noticing the difference. In fact, the book's solution
actually leaves Mickey intact while solving Donald. After you're bored with
solving it, the concept of making patterns takes on strange dimensions, as you
can make Mickey and Donald exchange body parts and look like Disney on acid!
All in all, it is an excellent little puzzle and I am very glad to see the
Skewb widely available to puzzle enthusiasts everywhere.
One final note: the booklet gives no credit whatsoever to Tony Durham, who
was credited with the Skewb's invention in Hofstadter's Sci Am articles years
ago. They instead credit Meffert, since the Skewb's mechanism is based on
the Pyraminx.
From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Mon Jan 10 23:08:35 1994
ReturnPath: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU>
Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA26716; Mon, 10 Jan 94 23:08:35 EST
MessageId: <9401110408.AA26716@life.ai.mit.edu>
Received: from MITVMA.MIT.EDU by mitvma.mit.edu (IBM VM SMTP V2R2)
with BSMTP id 5041; Mon, 10 Jan 94 23:08:38 EST
Received: from WVNVM.WVNET.EDU (NJE origin MAILER@WVNVM) by MITVMA.MIT.EDU
(LMail V1.1d/1.7f) with BSMTP id 4686; Mon, 10 Jan 1994 23:08:38 0500
Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU
(LMail V1.1d/1.7f) with BSMTP id 9272; Mon, 10 Jan 1994 23:06:01 0500
XAcknowledgeTo:
Date: Mon, 10 Jan 1994 23:06:00 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: G\M  Some Trivial Partial Results
It occurs to me that a small part of my incorrect attempt during
December to calculate G\M can be salvaged. In particular,
for those cases where Bconjugate classes are of order 1152,
the calculations are trivial. About 99.9923% of the edge conjugate
classes and about 96.924% of the corner conjugate classes are of
order 1152, so we can calculate the correct number of Mconjugates
of G for a very large percentage of the cases.
Consider some fixed X in GC\B and some fixed Y in GE\B where
BClass(X)=1152 and BClass(Y)=1152. Form
BClass(X) * BClass(Y). Now, BClass(X) * BClass(Y) =
BClass(X) * BClass(Y) / 2 = 1152 * 1152 / 2.
(The division by 2 takes care of parity). Finally, form
(BClass(X) * BClass(Y))\M, and we have (BClass(X) * BClass(Y))\M =
BClass(X) * BClass(Y) / 48 = (1152 * 1152) / 2 / 48 = 13,824.
We know the number of BClasses of GC of order 1152 from computer
search (namely 75,392), and we know the number of BCLasses of GE of
order 1152 from computer search (namely 851,493,140). Hence, for
the special case of both BClasses being of order 1152, we have the
total number of elements of G\M being 851,493,140 * 75,392 * 13,824 =
887,442,335,689,605,120.
We can derive similar results if only one of BCLass(X) and BClass(Y)
are of order 1152. For example, there are 4 elements of GE\B
for which BClass(Y)=24. Choose such a Y, and choose X in GC\B
such that BClass(X)=1152. Form BClass(X) * BClass(Y). It will
be the case that BClass(X) * BClass(Y) = 1152 * 24 / 2 = 13,824.
Hence, (BClass(X) * BClass(Y))\M = 13,824/48 = 288. There are 75,392
values of X for which BClass(X)=1152, 4 values of Y for which
BCLass(Y)=24, and hence there are 75,392 * 4 * 288 = 86,851,584
elements of G\M of the form BClass(X) * BClass(Y) for which
BCLass(X) = 1152 and BClass(Y) = 24.
There are nineteen cases in all for which at least one of BClass(X)
and BCLass(Y) are of order 1152, and this note calculates only
two of the nineteen. Completing the other
seventeen would be trivial but tedious. However, a total solution
to the problem will require coming up with some way to deal with
the cases where neither BClass(X)=1152 nor BClass(Y)=1152.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 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 tonyd@earwax.pd.uwa.edu.au Tue Jan 11 02:06:32 1994
ReturnPath:
Received: from earwax.pd.uwa.edu.au by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA01183; Tue, 11 Jan 94 02:06:32 EST
Received: from [130.95.156.19] (chaos [130.95.156.19]) by earwax.pd.uwa.edu.au (8.1C/8.1) with SMTP id PAA20165; Tue, 11 Jan 1994 15:09:05 +0800
MessageId: <199401110709.PAA20165@earwax.pd.uwa.edu.au>
MimeVersion: 1.0
ContentType: text/plain; charset="usascii"
Date: Tue, 11 Jan 1994 15:08:16 +0800
To: CubeLovers@ai.mit.edu
From: tonyd@earwax.pd.uwa.edu.au
Subject: Rubik chaos?
On sci.nonlinear...
In article <1994Jan5.120409@oxygen.aps1.anl.gov> Thomas D. Orth,
orth@oxygen.aps1.anl.gov writes:
>A friend of mine has written a few papers on the subject of
>the Rubik's Cube Group, and the elements of Chaos within
>it, or Pseudochaos as she calls it.
The papers are being submitted to the journal CHAOS.
cheers, Tony
From @mail.uunet.ca:mark.longridge@canrem.com Tue Jan 11 16:07:24 1994
ReturnPath: <@mail.uunet.ca:mark.longridge@canrem.com>
Received: from mail.uunet.ca (uunet.ca) by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA01871; Tue, 11 Jan 94 16:07:24 EST
Received: from portnoy.canrem.com ([198.133.42.251]) by mail.uunet.ca with SMTP id <57202(4)>; Tue, 11 Jan 1994 14:27:37 0500
Received: from canrem.com by portnoy.canrem.com (4.1/SMI4.1)
id AA09812; Tue, 11 Jan 94 13:52:09 EST
Received: by canrem.com (PCBUUCP 1.1f)
id 190BA3; Tue, 11 Jan 94 13:41:58 0400
To: cubelovers@life.ai.mit.edu
ReplyTo: CRSO.Cube@canrem.com
Sender: CRSO.Cube@canrem.com
Subject: Andras Mezei's Book
From: mark.longridge@canrem.com (Mark Longridge)
MessageId: <60.692.5834.0C190BA3@canrem.com>
Date: Tue, 11 Jan 1994 12:18:00 0500
Organization: CRS Online (Toronto, Ontario)
A while back I reported on the list of cube books available at the
Library of Congress. At the time, I did not realize the
significance of #2:
2. 85109601: Mezei, Andras. Magyar kocka, avagy, Meg mindig ilyen
gazdagok vagyunk? / Budapest : Magveto, c1984. 473 p. : ill.
; 21 cm.
Digging through some old magazines I reread the cube article in
the March 1986 issue of "Discover" magazine. In this issue
John Tierney talks to Rubik himself. The article itself is
excellent, showing pictures of Rubik's first wooden prototype,
and having discussions on the Golden Age of the Cube when
only Rubik had access to his invention.
I learned that Andras Mezei (a Budapest writer) wrote a book and a
play called "The Hungarian Cube". This chronicles the debacle
that occured when the Hungarians attempted to expand their
operations to meet the huge demand, rather than farm the work
to other factories. Andras writes: "Everyone made money on the
cube except the Hungarians".
Does anyone on CubeLovers have this book? Judging by the size
of the book, and the fact that it's illustrated, I think it
would be a worthy addition to any cubist's library, even more
so if there exists an english translation. If not, I feel
encouraged enough to get an EnglishHungarian dictionary and
read it anyway!
> Mark
From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Wed Jan 12 23:46:28 1994
ReturnPath: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU>
Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA20019; Wed, 12 Jan 94 23:46:28 EST
MessageId: <9401130446.AA20019@life.ai.mit.edu>
Received: from MITVMA.MIT.EDU by mitvma.mit.edu (IBM VM SMTP V2R2)
with BSMTP id 1427; Wed, 12 Jan 94 23:46:31 EST
Received: from WVNVM.WVNET.EDU (NJE origin MAILER@WVNVM) by MITVMA.MIT.EDU
(LMail V1.1d/1.7f) with BSMTP id 7322; Wed, 12 Jan 1994 23:46:31 0500
Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU
(LMail V1.1d/1.7f) with BSMTP id 4696; Wed, 12 Jan 1994 23:43:55 0500
XAcknowledgeTo:
Date: Wed, 12 Jan 1994 23:43:54 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: G\M  the Nineteen Cases We Know
A  BClass(X) for X in GC\B (order of Bconjugate class for corners)
B  BClass(Y) for Y in GE\B (order of Bconjugate class for edges)
C  GC[A]\B (number of Bconjugate classes of size A for corners)
D  GE[B]\B (number of Bconjugate classes of size B for edges)
E  BClass(X) * BClass(Y) = BClass(X) * BClass(Y) / 2
F  BClass(X) * BClass(Y) * GC[A]\B * GE[B]\B / 48
(number of Mconjugates of GC[A]\B * GE[B]\B)
A B C D E F
1152 24 75,392 4 13,824 86,851,584
1152 48 75,392 2 27,648 86,851,584
1152 72 75,392 12 41,472 781,664,256
1152 96 75,392 16 55,296 1,389,625,344
1152 144 75,392 110 82,944 14,330,511,360
1152 192 75,392 70 110,592 12,159,221,760
1152 288 75,392 1,544 165,888 402,296,537,088
1152 384 75,392 1,252 221,184 434,952,732,672
1152 576 75,392 128,858 331,776 67,149,128,466,432
1152 1152 75,392 851,493,140 663,552 887,442,335,689,605,120
24 1152 1 851,493,140 13,824 245,230,024,320
48 1152 1 851,493,140 27,648 490,460,048,640
72 1152 3 851,493,140 41,472 2,207,070,218,880
96 1152 1 851,493,140 55,296 980,920,097,280
144 1152 14 851,493,140 82,944 20,599,322,042,880
192 1152 15 851,493,140 110,592 29,427,602,918,400
288 1152 135 851,493,140 165,888 397,272,639,398,400
384 1152 32 851,493,140 221,184 125,557,772,451,840
576 1152 2,208 851,493,140 331,776 12,995,229,448,765,440
Total 901,082,361,368,033,280
Note that we have covered over 99.99 percent of edge positions (which
are combined with all corner positions), and over 96.9 percent of
the remaining corner positions (which are combined with all edge
positions). Hence, we have covered about 99.99969 percent of all
positions. However, that last fraction of a percent is going to
be devilishly difficult.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 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 Thu Jan 13 05:03:10 1994
ReturnPath: <@mail.uunet.ca:mark.longridge@canrem.com>
Received: from mail.uunet.ca (uunet.ca) by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA26601; Thu, 13 Jan 94 05:03:10 EST
Received: from portnoy.canrem.com ([198.133.42.251]) by mail.uunet.ca with SMTP id <61630(3)>; Thu, 13 Jan 1994 04:51:08 0500
Received: from canrem.com by portnoy.canrem.com (4.1/SMI4.1)
id AA23090; Wed, 12 Jan 94 18:30:28 EST
Received: by canrem.com (PCBUUCP 1.1f)
id 190EF0; Wed, 12 Jan 94 18:20:47 0400
To: cubelovers@life.ai.mit.edu
ReplyTo: CRSO.Cube@canrem.com
Sender: CRSO.Cube@canrem.com
Subject: 4x4x4 Cube moves
From: mark.longridge@canrem.com (Mark Longridge)
MessageId: <60.694.5834.0C190EF0@canrem.com>
Date: Wed, 12 Jan 1994 17:07:00 0500
Organization: CRS Online (Toronto, Ontario)
Some comments on flipping a single pair of edges on the 4x4x4 cube:
Singmaster notation on the 4x4x4 (same notation as Jeffery Adams)

L left face l inner left slice
r inner right slice R right face
F front face f inner front slice
b inner back slice B back face
U up face u inner up slice
d inner down slice D down face
So L1 would be turn the left face 90 degrees clockwise and l1 would
be turn the inner left slice 90 degrees clockwise.
I'll use the suffix "2" to be for 180 degree turns and the suffix
"3" to be for 270 degree turns clockwise or 90 degree turns
counterclockwise.
This is the shortest sequence I found for flipping 2 adjacent
edges on the 4x4x4 cube (LD pair):
(r3 D3) ^3 + (r1 D1) ^4 + Rr3 D3 R1 D1 r3 D3 R3 D1 R1 D3
Note the use of Rr to represent both the turns R face & r inner
slice. Counting slice turns the sequence is 25 turns, or
24 "hyper moves". This sequence moves some centre pieces around.
However, on checking David Singmaster's Cubic Circular, in issues
5 & 6, Autumn & Winter 1982 there is a shorter process on
page 15, (UB pair):
r2 D2 l3 D1 R3 U1 R3 U3 l3 U1 R1 U3 l1 R1 D1 r2
This process, although more difficult to memorize, is only 16 slice
moves. It also disturbs centre pieces, although in a simpler way.
I always solve the centre pieces last on the 4x4x4.
Hope this helps!
> Mark
Email: mark.longridge@canrem.com
From Mikko.Haapanen@otol.fi Thu Jan 13 09:19:35 1994
ReturnPath:
Received: from lassie.eunet.fi by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA02187; Thu, 13 Jan 94 09:19:35 EST
Received: from sulo.otol.fi by lassie.eunet.fi with SMTP id AA29207
(5.67a/IDA1.5 for ); Thu, 13 Jan 1994 16:19:16 +0200
Received: from rhea.otol.fi by sulo.otol.fi with SMTP (PP)
id <015420@sulo.otol.fi>; Thu, 13 Jan 1994 16:19:11 +0200
Received: from otol.fi by rhea.otol.fi id <267620@rhea.otol.fi>;
Thu, 13 Jan 1994 16:19:00 +0200
Date: Thu, 13 Jan 1994 16:17:01 +0200 (EET)
From: "M. Haapanen"
Sender: "M. Haapanen"
ReplyTo: "M. Haapanen"
Subject: Re: 4x4x4 Cube moves
To: cubelovers@life.ai.mit.edu
InReplyTo: <60.694.5834.0C190EF0@canrem.com>
MessageId:
MimeVersion: 1.0
ContentType: TEXT/PLAIN; CHARSET=USASCII
> ...
> This is the shortest sequence I found for flipping 2 adjacent
> edges on the 4x4x4 cube (LD pair):
>
> (r3 D3) ^3 + (r1 D1) ^4 + Rr3 D3 R1 D1 r3 D3 R3 D1 R1 D3
>
> Note the use of Rr to represent both the turns R face & r inner
> slice. Counting slice turns the sequence is 25 turns, or
> 24 "hyper moves". This sequence moves some centre pieces around.
>
> However, on checking David Singmaster's Cubic Circular, in issues
> 5 & 6, Autumn & Winter 1982 there is a shorter process on
> page 15, (UB pair):
>
> r2 D2 l3 D1 R3 U1 R3 U3 l3 U1 R1 U3 l1 R1 D1 r2
>
> This process, although more difficult to memorize, is only 16 slice
> moves. It also disturbs centre pieces, although in a simpler way.
> I always solve the centre pieces last on the 4x4x4.
Thank you. But what is the shortest way to flip 2 adj. edges without
messing the center pieces? I can't find shorter than 49 turns.
==================================
= Mikko Haapanen == hazard57@sulo.otol.fi == (981) 530 7768 =
========== Haapanatie 2C411 90150 OULU ==========
From ishius@ishius.com Thu Jan 13 12:38:38 1994
ReturnPath:
Received: from holonet.net (giskard.holonet.net) by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA13888; Thu, 13 Jan 94 12:38:38 EST
Received: from DialupEudora (ishius@localhost) by holonet.net (Anton Dovydaitis) with SMTP
id JAA25594; Thu, 13 Jan 1994 09:37:53 0800
Date: Thu, 13 Jan 1994 09:37:53 0800
MessageId: <199401131737.JAA25594@holonet.net>
To: cubelovers@life.ai.mit.edu
From: ishius@ishius.com (Ishi Press International)
Subject: Skewb, 5x5x5 cubes available
I've been watching all the discussion here, and I thought some people on this
list might appreciate knowing that 5x5x5 Rubik's Cubes and the Skewb are
available from Ishi Press International. We also have hundreds of other
mechanical puzzles.
If you would like to be on our puzzle email list, write us. If you would
like a color catalog of our puzzles, send us your snailmail address.
I apologize for intruding with a commercial message, but it did seem to me
that at least a few people on this list would like to get their hands on
some of these puzzles, and I don't know of any other distributor for these
two items.
Anton Dovydaitis
Customer Support
========================================================================
Ishi Press International 800/8592086 voice, 408/9449110 FAX
76 Bonaventura Drive ishius@ishius.com The Americas
San Jose, CA 95134 ishi@cix.compulink.co.uk Europe
From CPELLEY@delphi.com Thu Jan 13 23:25:54 1994
ReturnPath:
Received: from bos2a.delphi.com (delphi.com) by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA15306; Thu, 13 Jan 94 23:25:54 EST
Received: from delphi.com by delphi.com (PMDF V4.211 #4520) id
<01H7NGQCPVOG99DX0P@delphi.com>; Thu, 13 Jan 1994 23:18:47 EDT
Date: Thu, 13 Jan 1994 23:18:47 0400 (EDT)
From: CPELLEY@delphi.com
Subject: Mickey's Challenge
To: cubelovers@life.ai.mit.edu
MessageId: <01H7NGQCR7WI99DX0P@delphi.com>
XVmsTo: INTERNET"cubelovers@life.ai.mit.edu"
MimeVersion: 1.0
ContentType: TEXT/PLAIN; CHARSET=USASCII
ContentTransferEncoding: 7BIT
I visited the local Disney Store and picked up a Mickey's Challenge puzzle for
$10. It's really cute, and the book that it comes with is excellent.
Included are color photos of Christophe Bandelow, Uwe Meffert, and the puzzle
disassembled into all its parts. Plus it gives a solution for the puzzle and
has a short bio on Uwe Meffert. It also shows color photos of the Megaminx,
Pyraminx (not the Tomy version, but a black one), and the 5x5x5 which they
refer to as "Professor's Cube."
Some general notes on Mickey's Challenge. It is a spherical Skewb, and it
actually turns much more smoothly than my cubical Skewb. It has the same
delightful "clicking" mechanism that the Skewb and original Pyraminx had.
It is a bit easier than the Skewb, since there are a few blank pieces that
can be exchanged without noticing the difference. In fact, the book's solution
actually leaves Mickey intact while solving Donald. After you're bored with
solving it, the concept of making patterns takes on strange dimensions, as you
can make Mickey and Donald exchange body parts and look like Disney on acid!
All in all, it is an excellent little puzzle and I am very glad to see the
Skewb widely available to puzzle enthusiasts everywhere.
One final note: the booklet gives no credit whatsoever to Tony Durham, who
was credited with the Skewb's invention in Hofstadter's Sci Am articles years
ago. They instead credit Meffert, since the Skewb's mechanism is based on
the Pyraminx.
From ishius@ishius.com Fri Jan 14 14:17:29 1994
ReturnPath:
Received: from holonet.net (giskard.holonet.net) by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA25577; Fri, 14 Jan 94 14:17:29 EST
Received: from DialupEudora (ishius@localhost) by holonet.net (Anton Dovydaitis) with SMTP
id LAA17654; Fri, 14 Jan 1994 11:13:42 0800
Date: Fri, 14 Jan 1994 11:13:42 0800
MessageId: <199401141913.LAA17654@holonet.net>
To: CubeLovers@ai.mit.edu
From: ishius@ishius.com (Ishi Press International)
Sender: ishius@ishius.com (Unverified)
Subject: 4x4x4 and 5x5x5 cubes.
I've been getting a lot of requests for 4x4x4 cubes, and we're looking into
getting them. However, I have a couple questions.
1) Why are 4x4x4 cubes so interesting? Do the additional symmetries make for
interesting questions, are they more fun, or easier to solve?
2) It appears to me that if you know how to solve the 3x3x3 Rubik's cube,
then you can easily solve the 5x5x5 rubiks (i.e., the solution is
derivative). For example, you can treat the inner 3x3 faces of the 5x5x5
as a single 3x3x3 cube. Alternately, you can treat the edges/faces along
with the the middle three slices combined into a single slice as its own
3x3x3 cube, and this would not really disturb the "inner face" 3x3x3 cube.
Is this really so, or am I missing something? Is the 5x5x5 cube simply
the group product of two 3x3x3 cubes and one or two subgroups of a 3x3x3,
or is it more complex than that? How does this relate to the 4x4x4?
I do have a Bachelor's degree in mathematics and am familiar with abstract
algebra. I appreciate any light you can shed on these questions. I would
like to be able to converse intelligently about the cubes; that is why I
subscribed to this list.
Anton Dovydaitis
========================================================================
Ishi Press International 800/8592086 voice, 408/9449110 FAX
76 Bonaventura Drive ishius@ishius.com The Americas
San Jose, CA 95134 ishi@cix.compulink.co.uk Europe
From mouse@collatz.mcrcim.mcgill.edu Fri Jan 14 15:45:34 1994
ReturnPath:
Received: from Collatz.McRCIM.McGill.EDU by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA01604; Fri, 14 Jan 94 15:45:34 EST
Received: from localhost (root@localhost) by 960 on Collatz.McRCIM.McGill.EDU (8.6.4 Mouse 1.0) id PAA00960 for cubelovers@ai.mit.edu; Fri, 14 Jan 1994 15:44:48 0500
Date: Fri, 14 Jan 1994 15:44:48 0500
From: der Mouse
MessageId: <199401142044.PAA00960@Collatz.McRCIM.McGill.EDU>
To: cubelovers@ai.mit.edu
Subject: Re: 4x4x4 and 5x5x5 cubes.
> I've been getting a lot of requests for 4x4x4 cubes, and we're
> looking into getting them.
I'd (probably) buy one  I don't know what's become of the one I had.
Depends on price, of course, but I found the 5Cube price acceptable.
> 1) Why are 4x4x4 cubes so interesting? Do the additional symmetries
> make for interesting questions, are they more fun, or easier to
> solve?
I doubt it. If one ignores the center slice in each dimension on a
5x5x5, one has a 4x4x4. I think it's the completist instinct any
collector has. :)
Now what I'd *really* like is something topologically equivalent to a
2x2x2x2 Cube. Obviously a 2x2x2x2 Cube can't really be built, but it
should be possible to build something topologically equivalent. (A
3x3x3x3 would be nice too, but perhaps too difficult.) The hard part
is designing an emulation that has some aesthetic appeal; it's easy
enough to write a program that lets you permute appropriate overlapping
4cycles of objects without any intuitivelyobvious structure.
> 2) It appears to me that if you know how to solve the 3x3x3 Rubik's
> cube, then you can easily solve the 5x5x5 rubiks (i.e., the
> solution is derivative).
No, not really. If you can do the 3Cube *and the 4Cube*, then the
5Cube has no new challenges to offer (nor, I believe, does any size).
But the 4Cube and 5Cube do have challenges the 3Cube doesn't, namely
edge cubies and facecenter cubies. All the 3Cube experience in the
world won't help you if you get your 5Cube solved except for two edge
cubies which are swapped. (Or rather, general Cubetypepuzzle
experience will help  for example, how to use conjugates  but
3Cubespecific experience won't.)
> For example, you can treat the inner 3x3 faces of the 5x5x5 as a
> single 3x3x3 cube. Alternately, you can treat the edges/faces
> along with the the middle three slices combined into a single
> slice as its own 3x3x3 cube, and this would not really disturb the
> "inner face" 3x3x3 cube. Is this really so, or am I missing
> something?
You're missing something, but not much. :) As you say, there are two
ways of emulating a 3Cube on the 5Cube, namely to paste slices 212
along each dimension and to paste them 131. (I hope that's not too
abbreviated to be comprehensible  I mean, along each dimension, paste
the 5Cube slices together into three groups, taking two, then one,
then two slices, or one, three, and one.) However, it is entirely
possible to scramble the 5Cube in ways that cannot be solved by
treating the 5Cube as virtual 3Cubes, except in the trivial sense
that any 5Cube turn can be viewed as one or more turns of
appropriatelychosen virtual 3Cubes.
For example, I can swap two edge cubies (and also permute center cubies
in invisible ways); alternatively, I can permute the face cubies so
that the 212 virtual 3Cube has two identical corner vcubies.
der Mouse
mouse@collatz.mcrcim.mcgill.edu
From ncramer@bbn.com Fri Jan 14 17:05:26 1994
ReturnPath:
Received: from LABSN.BBN.COM by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA05939; Fri, 14 Jan 94 17:05:26 EST
MessageId: <9401142205.AA05939@life.ai.mit.edu>
Date: Fri, 14 Jan 94 7:47:37 EST
From: Nichael Cramer
To: Ishi Press International
Cc: cubelovers@life.ai.mit.edu
Subject: Re: Skewb, 5x5x5 cubes available
>Date: Thu, 13 Jan 1994 09:37:53 0800
>MessageId: <199401131737.JAA25594@holonet.net>
>From: Ishi Press International
>Subject: Skewb, 5x5x5 cubes available
>
>I've been watching all the discussion here, and I thought some people on this
>list might appreciate knowing that 5x5x5 Rubik's Cubes and the Skewb are
>available from Ishi Press International. We also have hundreds of other
>mechanical puzzles.
>
>If you would like to be on our puzzle email list, write us. If you would
>like a color catalog of our puzzles, send us your snailmail address.
Anton
Yes to both the above, please.
email : ncramer@bbn.com
landmail: Nichael Cramer
123 B Spring St
Cambridge MA 02141
Thanks much
Nichael
From @mail.uunet.ca:mark.longridge@canrem.com Fri Jan 14 23:47:23 1994
ReturnPath: <@mail.uunet.ca:mark.longridge@canrem.com>
Received: from mail.uunet.ca (uunet.ca) by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA23113; Fri, 14 Jan 94 23:47:23 EST
Received: from portnoy.canrem.com ([198.133.42.251]) by mail.uunet.ca with SMTP id <58749(9)>; Fri, 14 Jan 1994 23:44:39 0500
Received: from canrem.com by portnoy.canrem.com (4.1/SMI4.1)
id AA19059; Fri, 14 Jan 94 23:21:00 EST
Received: by canrem.com (PCBUUCP 1.1f)
id 191357; Fri, 14 Jan 94 23:16:58 0400
To: cubelovers@life.ai.mit.edu
ReplyTo: CRSO.Cube@canrem.com
Sender: CRSO.Cube@canrem.com
Subject: Higher Order Cubes, correction
From: mark.longridge@canrem.com (Mark Longridge)
MessageId: <60.704.5834.0C191357@canrem.com>
Date: Fri, 14 Jan 1994 22:10:00 0500
Organization: CRS Online (Toronto, Ontario)
> (fm2 + u2) ^2 + (fm2 + lm2) ^2 (corrects the centres)
This should be:
(fm2 + u2) ^2 + (fm2 + l2) ^2
> Mark
From @mail.uunet.ca:mark.longridge@CANREM.COM Sat Jan 15 00:52:48 1994
ReturnPath: <@mail.uunet.ca:mark.longridge@CANREM.COM>
Received: from mail.uunet.ca (uunet.ca) by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA25240; Sat, 15 Jan 94 00:52:48 EST
Received: from portnoy.canrem.com ([198.133.42.251]) by mail.uunet.ca with SMTP id <58628(8)>; Sat, 15 Jan 1994 00:47:19 0500
Received: from canrem.com by portnoy.canrem.com (4.1/SMI4.1)
id AA19041; Fri, 14 Jan 94 23:20:55 EST
Received: by canrem.com (PCBUUCP 1.1f)
id 191356; Fri, 14 Jan 94 23:16:57 0400
To: cubelovers@life.ai.mit.edu
ReplyTo: CRSO.Cube@canrem.com
Sender: CRSO.Cube@canrem.com
Subject: Higher Order Cubes
From: mark.longridge@canrem.com (Mark Longridge)
MessageId: <60.703.5834.0C191356@canrem.com>
Date: Fri, 14 Jan 1994 21:59:00 0500
Organization: CRS Online (Toronto, Ontario)
Anton Dovydaitis writes:
>I've been getting a lot of requests for 4x4x4 cubes, and we're
>looking into getting them. However, I have a couple questions.
>
>1) Why are 4x4x4 cubes so interesting? Do the additional symmetries
>make for interesting questions, are they more fun, or easier to
>solve?
A well lubed 4x4x4 cube is still relatively easy to physically
manipulate. As der Mouse suggests, it is arguably the largest
interesting cube from a solver's point of view. Once one starts
actually twisting with a 5x5x5 cube, the physical problems
become more severe, e.g. the stickers come off easier,
turning the slice you want to is more of a challenge, etc.
In the virtual realm of computer cubing the patterns you can
create are more elaborate, although I find in practice that
finding pretty patterns on the 5x5x5 can become wearisome
due to fact there are 9 centre pieces per side!
>2) It appears to me that if you know how to solve the 3x3x3 Rubik's
> cube, then you can easily solve the 5x5x5 rubiks (i.e., the
> solution is derivative). For example, you can treat the inner 3x3
> faces of the 5x5x5 as a single 3x3x3 cube.
Using the 4x4x4 cube we can produce a single exchange of centres
and an exchange of edge pairs, and we can invert a single edge pair.
Thus we can construct all the impossible 3x3x3 patterns except those
involving a twist of a single corner! That is why I think the 4x4x4
cube is a good cube to have. The individual centre cubies can
naturally wander all over the cube, and on the 3x3x3 cube they are
fixed.
In the case of the 5x5x5 cube, lots of the 3x3x3 knowledge does
help. When dealing with the 5x5x5's middlemost slice (let's call
one such slice "fm" for middlemost front slice) some of the
3x3x3 move sequences will move the appropriate edges, but now
these sequences will also move centre pieces, specifically the
ones which have no counterpart on the 3^3 and 4^3.
To solve cubes 4x4x4 and greater requires new sequences to
efficiently move centre cubies at will, and in the case of
the 5^3 there really is no standard language to interchange
move sequences and label individual cubies.
I find having a letter as a mnemonic helps, so I'll suggest the
following as an extension of Singmaster's 4x4x4 notation
for the 5x5x5 cube:
L left face l inner left slice lm left middlemost slice
R right face r inner right slice rm right middlemost slice
F front face f inner front slice fm front middlemost slice
B back face b inner back slice bm back middlemost slice
U up face u inner up slice um up middlemost slice
D down face d inner down slice dm down middlemost slice
Again, we follow the alphabetic component by a number to
signify the rotation (1 = 90, 2 = 180, 3 = 270 or 90)
This is overkill, and we can dispense with rm, bm and dm.
Thus we could flip 2 middlemost edges at FD and BD with:
(fm1 D1) ^3 + fm1 D2 + (fm1 D1) ^3 + fm1 (disturbs some centres)
followed by:
(fm2 + u2) ^2 + (fm2 + lm2) ^2 (corrects the centres)
I believe this is correct, and I will doublecheck on my physical
5x5x5 at home. Definitely 5^3 cubing is a sport for the
specialist ;>
> Mark
Email: mark.longridge@canrem.com
From ncramer@bbn.com Sat Jan 15 09:22:26 1994
ReturnPath:
Received: from LABSN.BBN.COM by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA03599; Sat, 15 Jan 94 09:22:26 EST
MessageId: <9401151422.AA03599@life.ai.mit.edu>
Date: Sat, 15 Jan 94 9:12:37 EST
From: Nichael Cramer
To: CRSO.Cube@canrem.com
Cc: cubelovers@life.ai.mit.edu
Subject: Re: Higher Order Cubes
>Subject: Higher Order Cubes
>From: Mark Longridge
>Date: Fri, 14 Jan 1994 21:59:00 0500
>
>A well lubed 4x4x4 cube is still relatively easy to physically
>manipulate. As der Mouse suggests, it is arguably the largest
>interesting cube from a solver's point of view. Once one starts
>actually twisting with a 5x5x5 cube, the physical problems
>become more severe, e.g. the stickers come off easier,
>turning the slice you want to is more of a challenge, etc.
This is interesting, because it's almost exactly the opposite of my
experience.
The problem seems to be the difference between the internal mechanisms of
the odd and even ordered cubes. The 3X and 5X have that "fixed" center
piece attached to the core whereas the center face cubelets of the 4X are
held together "under tension". My experience has been that this adjustment
is critical, but often out of whack. As a consequence, of the four 4X's
I've owned, only one was really useable; two were so stiff they were very
difficult to turn (even with lubrication) and one was so loose that it
never lasted more than about 20 minutes before dissolving into a pile of
cubelets (it currently lives in a sack in my office drawer). These were
all real "brandnamed" cubes, not cheap twizo knockoffs.
On the other hand all of the 5X's I've owned have been _very_ easy to turn
without any special customization. Except for the tendency (as Mark
mentions) for the stickers to come off of one of them, they're consistently
more comfortable to the hand than any of the 3X's I've owned.
Nichael
ncramer@bbn.com  Captain and left quarter guard, BBN Calvinball Team
From mouse@collatz.mcrcim.mcgill.edu Sat Jan 15 12:33:29 1994
ReturnPath:
Received: from Collatz.McRCIM.McGill.EDU by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA10873; Sat, 15 Jan 94 12:33:29 EST
Received: from localhost (root@localhost) by 2409 on Collatz.McRCIM.McGill.EDU (8.6.4 Mouse 1.0) id MAA02409 for cubelovers@ai.mit.edu; Sat, 15 Jan 1994 12:33:21 0500
Date: Sat, 15 Jan 1994 12:33:21 0500
From: der Mouse
MessageId: <199401151733.MAA02409@Collatz.McRCIM.McGill.EDU>
To: cubelovers@ai.mit.edu
Subject: Re: Higher Order Cubes
>> A well lubed 4x4x4 cube is still relatively easy to physically
>> manipulate. As der Mouse suggests, it is arguably the largest
>> interesting cube from a solver's point of view.
Probably true; it's the largest cube that actually offers new
challenges. However, bigger cubes are better in that they offer more
variety for making pretty patterns. :)
>> Once one starts actually twisting with a 5x5x5 cube, the physical
>> problems become more severe, e.g. the stickers come off easier,
>> turning the slice you want to is more of a challenge, etc.
> This is interesting, because it's almost exactly the opposite of my
> experience.
> The problem seems to be the difference between the internal
> mechanisms of the odd and even ordered cubes.
This brings up an interesting point. Perhaps it would be possible to
build a 4Cube that was internally a 5Cube but for which the middle
slice was not actually visible on the surface? Or a 2Cube that's
internally a 3Cube? I wonder if it might make for smootherturning
cubes.
> [O]f the four 4X's I've owned, only one was really useable; two were
> so stiff they were very difficult to turn (even with lubrication) and
> one was so loose that it never lasted more than about 20 minutes
> before dissolving into a pile of cubelets [...].
I have owned only one 4Cube, and it's been long enough since I knew
where it was that I don't recall how easy it was to turn. I now have
two 3Cubes and a 5Cube. One of the 3Cubes is a joy to turn; it's
lubed enough that it turns readily and easily, even when the turn has a
good deal of skew to correct, but it's not so loose that it turns when
I don't want it to. (The other 3Cube is (a) missing one center cubie
face and (b) much more difficult to turn.) The 5Cube (one of the
recent Ishi Press cubes, btw) is mechanically quite good, though the
orange stickers did tend to come off (no other color did, and contact
cement worked just fine for putting them back on). Not as good as my
good 3Cube, though.
I've wondered whether it would be possible to build higherorder cubes.
The corners of the 5Cube still catch by a respectable amount as the
face turns, but by little enough that it makes me wonder if the 6Cube
or 7Cube is actually feasible. (Oh, for a really good
forcereflecting dataglove...then such a thing could be done virtually
with no problem at all!)
der Mouse
mouse@collatz.mcrcim.mcgill.edu
From @mail.uunet.ca:mark.longridge@canrem.com Sat Jan 15 17:03:53 1994
ReturnPath: <@mail.uunet.ca:mark.longridge@canrem.com>
Received: from mail.uunet.ca (uunet.ca) by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA20492; Sat, 15 Jan 94 17:03:53 EST
Received: from portnoy.canrem.com ([198.133.42.251]) by mail.uunet.ca with SMTP id <53851(9)>; Sat, 15 Jan 1994 17:02:08 0500
Received: from canrem.com by portnoy.canrem.com (4.1/SMI4.1)
id AA29781; Sat, 15 Jan 94 17:00:39 EST
Received: by canrem.com (PCBUUCP 1.1f)
id 1914B9; Sat, 15 Jan 94 16:53:33 0400
To: cubelovers@life.ai.mit.edu
ReplyTo: CRSO.Cube@canrem.com
Sender: CRSO.Cube@canrem.com
Subject: 4x4x4 & 5x5x5 processes
From: mark.longridge@canrem.com (Mark Longridge)
MessageId: <60.710.5834.0C1914B9@canrem.com>
Date: Sat, 15 Jan 1994 15:44:00 0500
Organization: CRS Online (Toronto, Ontario)
Here are a couple of processes for larger cubes, plus the
requested edge pair flip without disturbing centres (p2),
as well as a minor correction for the 5x5x5 process:
4x4x4 processes (measured in slice moves)

p1 Flip LD edge pair (r3 D3) ^3 + (r1 D1) ^4 + Rr3 D3 R1 D1 r3 D3 R3
(disturbs centres) D1 R1 D3 (25)
p2 Flip UB edge pair r2 D2 l3 D1 R3 U1 R3 U3 l3 U1 R1 U3 l1 R1 D1
(retain centre positions ) + U2 r1 (u2 r2 l2) ^2 + r3 U2 r2 (26)
5x5x5 processes (measured in slice moves)

Flip 2 middlemost edges at FD and BD with:
p1 (fm1 D1) ^3 + fm1 D2 + (fm1 D1) ^3 + fm1 (disturbs some centres)
followed by:
D1 + (fm2 u2) ^2 + (fm2 l2) ^2 + D3 (corrects the centres) (25)
> Mark
Email: mark.longridge@canrem.com
From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Mon Jan 17 09:09:39 1994
ReturnPath: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU>
Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA27830; Mon, 17 Jan 94 09:09:39 EST
MessageId: <9401171409.AA27830@life.ai.mit.edu>
Received: from MITVMA.MIT.EDU by mitvma.mit.edu (IBM VM SMTP V2R2)
with BSMTP id 0725; Mon, 17 Jan 94 09:09:39 EST
Received: from WVNVM.WVNET.EDU (NJE origin MAILER@WVNVM) by MITVMA.MIT.EDU
(LMail V1.1d/1.7f) with BSMTP id 6845; Mon, 17 Jan 1994 09:09:37 0500
Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU
(LMail V1.1d/1.7f) with BSMTP id 5415; Mon, 17 Jan 1994 09:06:59 0500
XAcknowledgeTo:
Date: Mon, 17 Jan 1994 09:06:59 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: Number of MConjugate Classes for GC\M
On 4 December 1993, I posted results from a breadthfirst exhaustive
search of GC\M, the corners of the 3x3x3, reduced by Mconjugation.
The posting included a summary of how many conjugate classes there
were at each level of the search tree (i.e., distance from Start).
It occurred to me that I had not also posted a summary of Mconjugates
for the corners of the 3x3x3 by the size of the conjugate classes. I
searched my records, only to discover that I had never calculated
such sizes. If I had, I probably would have been forced to analyze
properly the distinction between Mconjugation and Bconjugation,
because Bconjugation makes no sense for the corners of the 3x3x3.
Bconjugation *can* be performed for the corners of the 3x3x3, but you
end up with the 2x2x2 instead because Bconjugation effectively
removes the centers.
Anyway, I have now calculated Mconjugate class sizes for GC\M via
computer search, and here are the results.
MClass Number
Size of
Classes
1 1
2 1
3 3
4 1
6 34
8 33
12 301
16 104
24 9064
48 1832428
Total 1841970
Notice that with Mconjugation, the maximum class is size is 48,
rather than 1152 as it is with Bconjugation. Hence, my posting
of 4 December 1993 incorrectly identified the results as being
for "1152 fold symmetry". The results are correct, but they
should be labeled as being for "48 fold symmetry", i.e., for
Mconjugation rather than for Bconjugation.
In calculating Mconjugate class sizes for GC\M, I did not "start
from scratch". Rather, I used the existing results for Bconjugate
classes as a base. In the case of Bconjugate classes of order
1152, no calculations are required. Each such Bclass can simply
be partitioned into 24 Mclasses of order 48. Hence, I had to
perform calculations for less than 4% of the Bclasses. Here is
a summary matrix, showing for each Bclass size the number of
each Mclass size which are derived.
MClass Size
1 2 3 4 6 8 12 16 24 48 Total
24 1 0 1 0 2 1 0 0 0 0 5
BClass 48 0 1 0 0 1 2 2 0 0 0 6
Size 72 0 0 2 0 11 0 2 0 5 0 20
96 0 0 0 1 0 1 3 0 2 0 7
144 0 0 0 0 20 0 42 0 30 14 106
192 0 0 0 0 0 29 0 8 73 16 126
288 0 0 0 0 0 0 252 0 682 406 1340
384 0 0 0 0 0 0 0 96 0 224 320
576 0 0 0 0 0 0 0 0 8272 22360 30632
1152 0 0 0 0 0 0 0 0 0 1809408 1809408
Total 1 1 3 1 34 33 301 104 9064 1832428 1841970
The first row of the matrix exemplifies the process of calculating
MClass sizes from BClass sizes. In the case of corners, there
is only one Bclass of order 24, namely Start. The 24 elements of
the Bclass are the 24 elements of the form Ic, where c is in C,
the 24 rotations of the cube. Under Bconjugation, these 24
elements are equivalent (i.e., in a centerless cube such as the
2x2x2, the 24 rotations of I are indistinguishable). But in a cube
with centers, such as the corners of the 3x3x3, the 24 elements
are not equivalent.
For example, the Mclass of order 1 is {I}. One of the Mclasses
of order 6 is {FB', UD', RL', LR', BF', DU'}. The Mclass of order 3
is {FFB'B', RRL'L', UUD'D'}. That's as many as I can do in my head,
but I think the pattern is clear.
Mclasses are a partition of the Bclasses.
In the case of Bclasses of order 1152, the partition is regular 
i.e., you get exactly 24 Mclasses of order 48. However, all
partitions are not regular. In the partition of the Bclass of I
which we just discussed, there is 1 Mclass of order 1, 1 Mclass
of order 3, 2 Mclasses of order 6, and 1 Mclass of order 8, for
a total of 24 Mclasses. Many other partitions are not regular,
as well.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
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 dseal@armltd.co.uk Mon Jan 17 14:14:08 1994
ReturnPath:
Received: from eros.britain.eu.net by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA11921; Mon, 17 Jan 94 14:14:08 EST
Received: from armltd.co.uk by eros.britain.eu.net with UUCP
id ; Mon, 17 Jan 1994 18:21:12 +0000
Received: by armltd.co.uk (5.51/Am23) id AA07766; Mon, 17 Jan 94 18:16:09 GMT
Date: Mon, 17 Jan 94 18:15:33 GMT
From: dseal@armltd.co.uk (David Seal)
To: (Cube) cubelovers@ai.mit.edu
Subject: Re: Higher Order Cubes
MessageId: <2D3AD5C5@dseal>
InReplyTo: <199401151733.MAA02409@Collatz.McRCIM.McGill.EDU>
> This brings up an interesting point. Perhaps it would be possible to
> build a 4Cube that was internally a 5Cube but for which the middle
> slice was not actually visible on the surface? Or a 2Cube that's
> internally a 3Cube? I wonder if it might make for smootherturning
> cubes.
Yes, I think you could build such a 4Cube. Likewise, you could build a
2Cube as a 3Cube with invisible middle slices. But I don't believe you'd
want one: it could get completely jammed much too easily.
The reason: If you take a 3Cube and rotate its left and right slices 45
degrees each, you cannot rotate any of its other faces. This isn't
surprising, since you don't expect to be able to perform one rotation
halfway through another. If its middle slices were hidden, however, it would
*appear* to be a 2Cube which is not halfway through a rotation, and the
fact that you couldn't move any faces but the left and right ones would be
surprising  and undesirable.
Unfortunately, I believe such a situation could probably arise quite easily.
If you were to take the 2Cube concerned and rotate its right face 90
degrees relative to its left face, you're going to be OK if the hidden
middle layer rotates 0 or 90 degrees relative to the left face, but not OK
if it rotates any other amount. I suspect most mechanisms would be more
liable to rotate it an intermediate amount!
There may be a way out, though. If you can anchor the place where the three
axes meet to one of the corner cubelets in some way, the problem is solved:
if the "anchor cubelet" is in the left face, then the hidden layer will
rotate 0 decrees; if it is in the right face, then 90 degrees.
David Seal
dseal@armltd.co.uk
From mouse@collatz.mcrcim.mcgill.edu Mon Jan 17 16:23:05 1994
ReturnPath:
Received: from Collatz.McRCIM.McGill.EDU by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA17923; Mon, 17 Jan 94 16:23:05 EST
Received: from localhost (root@localhost) by 5806 on Collatz.McRCIM.McGill.EDU (8.6.4 Mouse 1.0) id QAA05806 for cubelovers@ai.mit.edu; Mon, 17 Jan 1994 16:22:50 0500
Date: Mon, 17 Jan 1994 16:22:50 0500
From: der Mouse
MessageId: <199401172122.QAA05806@Collatz.McRCIM.McGill.EDU>
To: cubelovers@ai.mit.edu
Subject: Re: Higher Order Cubes
>> Perhaps it would be possible to build a 4Cube that was internally a
>> 5Cube but for which the middle slice was not actually visible on
>> the surface? Or a 2Cube that's internally a 3Cube?
> Yes, I think you could build such a 4Cube. Likewise, you could build
> a 2Cube as a 3Cube with invisible middle slices. But I don't
> believe you'd want one: it could get completely jammed much too
> easily.
> The reason: If you take a 3Cube and rotate its left and right slices
> 45 degrees each, you cannot rotate any of its other faces.
Duh, yeah; that never occurred to me.
> There may be a way out, though. If you can anchor the place where the
> three axes meet to one of the corner cubelets in some way, the
> problem is solved: [...].
Yes. I think this may be possible, too...consider a normal 3Cube, and
restrict yourself to R, U, and F turns. Then ignore the center and
edge cubies  the ones that get invisibilized. You're left with a
2Cube. Three edge cubies never move with respect to the center cubies
or the corner cubie they surround; glue those together. Presto!
The same treatment is not possible for making a 4Cube out of a 5Cube,
but an alternative occurs to me, that I *think* will work for higher
cubes: key three of the (invisible) center cubies to the center
sixpronged piece, so that they can't turn. Then half the face turns
will cause the invisible center slice to turn with them; nonface
slices (which don't exist on the 2/3Cube) work normally.
I notice with this construction for (say) a 4Cube, the puzzle core
turns whenever certain face slices do. With the 4Cube I owned (and
presumably still own, if I could find it), the puzzle core turns
whenever certain nexttocenter slices do. I suspect the latter would
make for a smootherturning puzzle. Perhaps someone will someday build
a 5Cubeturned4Cube and this can be determined.
In the (IMO unlikely) event I originated any of the above ideas, I
hereby place it/them in the public domain. Go wild, Ishi Press. :)
der Mouse
mouse@collatz.mcrcim.mcgill.edu
From Mikko.Haapanen@otol.fi Wed Jan 19 15:09:31 1994
ReturnPath:
Received: from lassie.eunet.fi by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA10391; Wed, 19 Jan 94 15:09:31 EST
Received: from sulo.otol.fi by lassie.eunet.fi with SMTP id AA03827
(5.67a/IDA1.5 for ); Wed, 19 Jan 1994 22:07:44 +0200
Received: from rhea.otol.fi by sulo.otol.fi with SMTP (PP)
id <278530@sulo.otol.fi>; Wed, 19 Jan 1994 22:07:41 +0200
Received: from otol.fi by rhea.otol.fi id <259380@rhea.otol.fi>;
Wed, 19 Jan 1994 22:07:27 +0200
Date: Wed, 19 Jan 1994 21:53:24 +0200 (EET)
From: "M. Haapanen"
Subject: Re: 4x4x4 & 5x5x5 processes
To: cubelovers@ai.mit.edu
InReplyTo: <60.710.5834.0C1914B9@canrem.com>
MessageId:
MimeVersion: 1.0
ContentType: TEXT/PLAIN; charset=USASCII
Sender: Mikko.Haapanen@otol.fi
Hello cube lovers!
> requested edge pair flip without disturbing centres (p2),
> as well as a minor correction for the 5x5x5 process:
> ...
> (retain centre positions ) + U2 r1 (u2 r2 l2) ^2 + r3 U2 r2 (26)
^^^^
>
> 5x5x5 processes (measured in slice moves)
>
> Flip 2 middlemost edges at FD and BD with:
> p1 (fm1 D1) ^3 + fm1 D2 + (fm1 D1) ^3 + fm1 (disturbs some centres)
> followed by:
> D1 + (fm2 u2) ^2 + (fm2 l2) ^2 + D3 (corrects the centres) (25)
> > Mark ^^^^
> Email: mark.longridge@canrem.com
The following might be trivial, but i write it here anyway. This was
invented about 10 years ago:
5x5x5

(R1 um2 R2 um1 R1) + U2 + (R3 um3 R2 um2 R3) + U2 > 12 (18) turns :)
============== Finland ================
= Mikko Haapanen == hazard57@rhea.otol.fi == (981) 530 7768 =
========== Haapanatie 2C411 90150 OULU ==========
From hoey@aic.nrl.navy.mil Fri Jan 21 18:32:30 1994
ReturnPath:
Received: from Sun0.AIC.NRL.Navy.Mil by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA17275; Fri, 21 Jan 94 18:32:30 EST
Received: by Sun0.AIC.NRL.Navy.Mil (4.1/SMI4.0)
id AA13137; Fri, 21 Jan 94 18:32:15 EST
Date: Fri, 21 Jan 94 18:32:15 EST
From: hoey@aic.nrl.navy.mil (Dan Hoey)
MessageId: <9401212332.AA13137@Sun0.AIC.NRL.Navy.Mil>
To: CubeLovers@ai.mit.edu
Cc: Jerry Bryan
Subject: Re: Some Proposed Terminology
I welcome Jerry Bryan's efforts to
improve the terminology of the groups associated with Rubik's cube.
But there is some additional clarification I think is necessary.
> Let G be the standard cube group for the 3x3x3 cube....
> Let GC be the corners with centers without edges group....
> Let GE be the edges with centers without corners group....
That much will do, mod quibbles about what name is best.
> Let G\C be the corners with edges without centers group. I intend
> for the notation to indicate G reduced by C, where C is the rotation
> group for the cube....
> Let GC\C be the corners without edges without centers group....
> Let GE\C be the edges without corners without centers group....
First, these are not, strictly speaking, groups. Well, you can make
them groups, by defining what the group operation is. But I don't
know any way of doing that without losing the symmetrical nature of
the problem.
Second, I would suggest that G/C, GC/C, and GE/C are more standard
names for these objects. The elements are nominally 24element sets,
each of which is an equivalence class when two positions are
considered equivalent when they differ by their position with respect
to the corners. The classes are called the cosets of C in G, GC, and
GE, respectively.
> Let G\M be the set of Mconjugate classes for G.....
> Let GC\M be the set of Mconjugate classes for GC....
> Let GE\M be the set of Mconjugate classes for GE....
The partition of a group into conjugacy classes is not at all the same
as the partition into cosets. So I would prefer to use different
symbology, like "\" for conjugacy and "/" for cosets, but....
> Recall that B is the function which calculates the canonical form
> for a cube under the composed operations of Mconjugation plus
> rotation. My programs calculate equivalence classes under B.
> Let G\B be the set of Bclasses for G [ and likewise for GE, GC ].
Well, if you are using "\" for a generic partition into equivalence
classes, then we should really do something like G\Conj(M) for
partitions into conjugacy classes. At least then you can say
G/C=G\Cosets(C).
> Then, we have Gx\B=(Gx\C)\M=(Gx\M)\C. In English, we can decompose
> B into a multiplication by C and M (in either order).
No, that's _multiplication_ by C and _conjugation_ by M. A good
example of why it's important not to use confusing symbols. M and C
are not at all treated the same, except inasmuch as they are used to
induce partitions into equivalence classes. Say instead that
Gx\B = (Gx/C)\Conj(M) = (Gx\Conj(M))/C.
> If I wanted to model GC\C, I would have had to either model only
> seven of the cubies, or else modeled all eight but moved only seven
> of them. Since what I really wanted was (GC\C)\M, and since what I
> had was GC, I had to invent this funny B thing, where GC\B=(GC\C)\M.
> If I had been clever enough to model GC\C in the first place, I
> never would have had to invent B. Similar comments apply to my
> model for the edges.
Well, the part about moving only seven (corner) cubies is the approach
that's been taken before on this list to deal with cubes that don't
have face centers. It has the advantage that the object being treated
is a group. But the problem is that the group is no longer cubically
symmetrical (in some vague sense). This led me, at least, to lose
track of the structure that would allow analysis of Mconjugacy. So I
have to admire your tackling GE as a whole, instead of trying to stick
to GE/C. At first blush, it looks like GE/C is 24 times smaller. But
since GE/C\Conj(M) is almost 48 times smaller still, it's important to
work in GE at least enough to be able to use the conjugation. Which
is beside the point that I'm actually very interested in the structure
of Gx/Conj(M) itself. And that is what I was really getting at in
1984 when I asked about how many positions there really are.
Dan Hoey
Hoey@AIC.NRL.Navy.Mil
From walace@ntiaa.embrapa.ansp.br Fri Jan 21 21:35:11 1994
ReturnPath:
Received: from fpsp.fapesp.br by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA24242; Fri, 21 Jan 94 21:35:11 EST
Received: from ntiaa.embrapa.ansp.br by fpsp.fapesp.br with PMDF#10108; Sun, 16
Jan 1994 23:06 BDB (0200 C)
Received: by ntiaa.embrapa.ansp.br  Sat, 15 Jan 1994 12:44:15 0300
Date: Sat, 15 Jan 1994 12:44:15 0300
From: walace@ntiaa.embrapa.ansp.br (Walace Sartori Bonfim)
Subject: ICSI94
To: cubelovers@life.ai.mit.edu
MessageId: <4CD9C4A5C000C949@fpsp.fapesp.br>
XEnvelopeTo: cubelovers@ai.ai.mit.edu
Dear reader,
Due to the wide spectrum of people that might be interested in the
subjects to be discussed during the III International Conference on Systems
Integration, we decided to post this call for papers in your mailing list.
We encourage you to participate in this event as a paper author.
The paper arrival deadline is March 3, 1994.
Please forward this message to whoever you think it might be of interest and we
appreciate your effort to post it.
Thanks,
Prof. Fuad Gattaz Sobrinho
Conference Chairman

Call for Papers
The Third International Conference for Systems Integration
Sao Paulo City  Brazil
July 30th  August 6th, 1994

The Integration of Society for the Social, Economical, Scientific and
Technological Development. This conference focuses on the integration
of technologies, processes and systems, and the development of mechanisms
and tools enabling solutions to complex multidisciplinary problems dealing
with agriculture, housing, telecommunications, financing and business,
public services, education and software. The conference will provide an
international and interdisciplinary forum in which researchers, educators,
managers, practitioners and politicians, involved within the production
process, can share novel research and development, education, production,
trading, management and political experiences. Papers should deal with
recent effort in theory, design, implementation, methodology, technics,
tools and experiences of integration. Topics to be addressed include, but
are not limited to:
Technical and Scientific Aspects:
 Integration, Modeling, Characterization and Automation of Process
and Systems
 Reengineering and Simplification of Processes
 Computational Environments and Software Factories for Engineerind,
Design, Manufacturing and System Development
 Rol of Human Engineering in Integration
 Experiences within National or Continental Software Projects
 The Implication of Systems Integration for Manpower Skills
 Quality Control and Certification in Organizational and Process
Integration.
Social, Political and Economical Aspects:
 Experiences in Modeling, Development, Evolution and Integration
of Enterprises
 Experiences in Management and Identification of ValueAdd Chains
within Agriculture, Housing, Telecommunications, Financing and
Business, Public Services, Education and Software
 Public Policies and City Management
 Management of Multidimensional Integration.
Infrastructure Aspects:
 Qualified Information Resources
 Education and Training
 Science and Technology
 Enterprise Development.
Information and Instructions for Authors: All papers must be in English
or Portuguese, typed in double spaced format, and may not exceed 6,000
words. Each submission should provide a cover page containing author(s),
affiliation(s), complete address(es), identification of principal author,
and telephone number. Also include SIX copies of complete text with a
title and abstract. Notice of acceptance will be mailed to the principal
author(s) by March 15, 1994. If accepted, the author(s) will prepare the
final manuscript, in English, in time for inclusion in the conference
proceedings and will present the paper at the conference; otherwise, the
author(s) will incur a page charge. Authors of accepted papers must sign
a copyright release form. The proceedings will be published by the IEEE
Computer Society Press.
Send SIX copies of your paper(s) to:
Prof. Peter A. Ng
IIISis  USA Office  New Jersey Institute of Technology
University Heights
Newark, NJ 07102
USA
For Further Information, Contact:
Prof. Peter A. Ng Prof. Fuad Gattaz Sobrinho
Fone:(1) (201) 5963387 OR Phone:(55)(192) 414504
Fax: (1)(201) 5965777 Fax: (55)(192) 413098
Email: ng_p@vienna.njit.edu Email: iiisis@ccvax.unicamp.br

>>>>>>>>>> Paper Arrival Deadline: March 3rd, 1994 <<<<<<<<<<<<<<<<

CONFERENCE COMMITTEE
Conference Chair Fuad Gattaz Sobrinho
IIISis
Program Chair Peter A. Ng
NJIT
Finance & Business CoChair Alcir A. Calliari
Banco do Brasil
Agriculture CoChair Ney B. Araujo
ABAG
European CoChair Herbert Weber
University of Dortmund
Pac!fic CoChair Fumihiko Kamijo
IPA
Middle East CoChair Asuman Dogac
METU
South America CoChair Julio C. S. P. Leite
PUC/RJ
North America CoChair Bruce Berra
Syracuse University
Tutorials CoChairs Oscar Ivan Palma Pacheco
EMBRAPA
Murat M. Tanik
SMU
Organization CoChairs Rita de Cassia A. Marchiore
IIISis
Carole Poth
NJIT
Steering Committee Chair Peter A. Ng
NJIT
Honorary Advisors Raymond T. Yeh
C. V. Ramamoorthy
Laurence C. Seifert
Honorary Conference Chair Irma Rossetto Passoni
Sc&Tech, Info. and Comm. Comission of
Brazilian Congress.
Sponsored by IIISis  International Institute for Systems Integration,
BB  Banco do Brasil, TELEBRAS, FINEP, CNPq, FBB, with colaboration of
NJIT, SUCESU, EMBRAPA, ABAG, ACM e IEEE Computer Society.
Instituto Internacional de Integracao de Sistemas  IIISis  Brazil.
From hoey@aic.nrl.navy.mil Mon Jan 24 19:15:23 1994
Received: from Sun0.AIC.NRL.Navy.Mil by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA07945; Mon, 24 Jan 94 19:15:23 EST
Received: from sun13.aic.nrl.navy.mil by Sun0.AIC.NRL.Navy.Mil (4.1/SMI4.0)
id AA07954; Mon, 24 Jan 94 19:15:16 EST
ReturnPath:
Received: by sun13.aic.nrl.navy.mil; Mon, 24 Jan 94 19:15:15 EST
Date: Mon, 24 Jan 94 19:15:15 EST
From: hoey@aic.nrl.navy.mil
MessageId: <9401250015.AA05746@sun13.aic.nrl.navy.mil>
To: CubeLovers@ai.mit.edu
Cc: "Jerry Bryan"
Subject: Concerning B, CSymm, and Symm
In his message of Sat, 8 Jan 1994 08:46:20 EST, Jerry Bryan
considers his use of the term "B"
``to indicate various aspects of the conjugacy class generated by
m'Xmc.''
I don't think that's properly called a conjugacy class, but a
different sort of equivalence class. A conjugacy class is a special
kind of equivalence class (just as a coset is a special kind of
equivalence class) but this B is a little bit of both, so I don't
think it is correct to call it either.
> Let X be any cube. Then the set of Bconjugacy classes of X is
> the set of all m'Xmc for all m in M and all c in C. We denote
> this set as BClass(X). B is the function B(X)=min(BClass(X)).
That's a little unfortunateI'd prefer to use B(X) for the
equivalence class, and min(B(X))or repr(B(X))for the canonical
representative. That's because the representative is not the
important thing here, it's just a convenient way to represent (!) the
class in a computer.
> Note that we could have defined BClass(X) equivalently as the set of
> all mXm'c, or as the set of all cm'Xm, or as the set of all
> cmXm'.... This is the justification for the assertion in a previous
> note that Gx\B = (Gx\M)\C = (Gx\C)\M.
Not quite. The justification for (Gx\Conj(M))/C = (Gx/C)\Conj(M)
is that instead of m'Xmc we could choose m'Xcm, a possibility you
didn't list.
In his message of "Sat, 8 Jan 1994 10:52:22 EST", Jerry continues with
discussion on combining conjugacy classes. We've exchanged some
private email on the subject material, but in case anyone on the list
is following this stuff....
> There are only 10 distinct values for BClass(X) and for
> BClass(Y), namely 24, 48, 72, 96, 144, 192, 288, 384, 576, and
> 1152. (By the way, I have never figured out why it is *exactly* the
> same set of values for both the corners and for the edges. It is
> easy to see why it is approximately the same set of values....
I'm not sure what kind of approximation you mean, but certainly those
ten values are all that are possible:
Proof: For if m1,m2 are in the same coset of M/CSymm(X), then
(m1 m2') is in CSymm(X), so X' (m1 m2')' X (m1 m2') = c0 in C so
m1' X m1 = m2' X c0 m2. It's then clear that
{ m1' X m1 c : c in C } = { m2' X m2 c : c in C } (*)
are equal 24element sets. The same manipulation in reverse shows
that if (*) holds for some m1,m2 in M, then m1 and m2 are in the
same coset of M/CSymm(X). So BClass(X)=24 M/CSymm(X).
M/CSymm(X) must be a divisor of M=48, QED.
It wouldn't have been all that surprising to see one of the possible
sizes of CSymm(X) fail to appear as a symmetry group of the corners
or edges, but it's not surprising that they all do, either.
> [For the original approach] I needed to be able to prove that for a
> fixed m and n, that (BClass(X)[m] * BClass(Y)[n] had the same
> value for all X in GC[m]\B and all GE[n]\B.
That is to say, that the sizes CSymm(X) and CSymm(Y) might
determine {Symm(X*Y)} for X in GC\B, Y in GE\B, and so (X*Y) in
G\Conj(M). It doesn't, but the situation is even worse. Jerry goes
on to suppose that perhaps CSymm(X) and CSymm(Y) themselves might
determine {Symm(X*Y)}, and even that isn't true. I've discovered
this by a computer search of GC\B. (A search of GE\B is in progress,
but for the current result we can take Y=I in GE\B). I have found
that AllSymms(X) is not determined, even up to subgroup sizes, by
CSymm(X).
According to the search, the following are the only positions of GC\B
for which CSymm(X)=16.
X1 X2 X3
++ ++ ++
F F B F F B
B B F B F B
++++ ++++ ++++
R RD DL L L LD TR R L LD TR R
R RT TL L L LT DR R L LD TR R
++++ ++++ ++++
B B F B B F
F F B F B F
++ ++ ++
T T T D T D
D D D T T D
++ ++ ++
Coincidentally, I have been (privately) calling the CSymm(Xi)
subgroups the "X subgroups" of M, an X subgroup being the subgroup
that maps an orthogonal axis of the cube (in the above examples, the
LR axis) to itself. X1 is a notable position, in that each corner
has been swapped with its opposite corner. Symm(X1) is an X subgroup
as well, and there is another X subgroup in AllSymms(X1). There is,
however, no 16element subgroup in AllSymms(X2) or AllSymms(X3). (We
have seen X2 before: it is the corners of the Laughter (or 4/)
position). In fact, my program says that
AllSymms(X1) contains two occurrences of 16element X subgroups,
two occurrences of the 8element HX subgroup,
two occurrences of 8element R subgroups,
two occurrences of 8element S subgroups,
eight occurrences of 4element HS subgroups, and
eight occurrences of the 2element HV subgroup.
AllSymms(X2) and AllSymms(X3) each contain
two occurrences of 8element CX subgroups,
two occurrences of 8element AX subgroups,
two occurrences of 8element P subgroups,
two occurrences of 8element Q subgroups,
eight occurrences of 2element ES subgroups, and
eight occurrences of 2element HW subgroups.
The names of these groups are part of a taxonomy of the subgroups of M
I've developed, which I won't go into just now. But the point that I
find surprising here is that AllSymms(X1) and AllSymms(X2) are
completely disjoint. While that can't happen all the time (smaller
CSymm() groups have many occurrences of the oneelement "I" subgroup)
I think the tendency to disjointness is too pronounced to be simple
anticoincidence.
Dan Hoey
Hoey@AIC.NRL.Navy.Mil
From cuf@aol.com Thu Feb 10 23:09:33 1994
ReturnPath:
Received: from mailgate.prod.aol.net by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA10256; Thu, 10 Feb 94 23:09:33 EST
Received: by mailgate.prod.aol.net
(1.37.109.4/16.2) id AA23795; Thu, 10 Feb 94 23:14:51 0500
From: cuf@aol.com
XMailer: America Online Mailer
Sender: "cuf"
MessageId: <9402102217.tn53025@aol.com>
To: cubelovers@life.ai.mit.edu
Date: Thu, 10 Feb 94 22:17:05 EST
Subject: Computer & Health
The Computer User Family (CUF) is concerned about the health problem
associated with computers. Video Display Terminals, emit UV and ELF
radiation and may cause cancer, immune system irregularities, miscarriages
and eye fatigue.
Computer noise from fans, disk and CD drives is also becoming a source of
anxiety, stress and general discomfort . We usually don't realize how loud
our computers are: 50dB and more.
These problems should be dealt with and addons should be provided for
present computers to avoid putting us at risks. Some safe screens and quiet
power supplies are coming out but they are marginal and prices are
prohibitive.
Meanwhile the general guidelines for the users are:
1. Position yourself approximately 22 inches to 28 inches (arm's length) from
the screen and four feet from the sides and rear of other terminals.
2. Eliminate sources of glare and lower light levels in the room. Don't sit
facing a bright window. If necessary, use screen hoods, glare shields over
the screen or wear antiUV/antiglare glasses.
3. Put a noise absorbing mat under your computer. Pull your computer away
from the wall or any hard surface that reflects noise and vibration back to
you.
4. Rest occasionally during periods of intense concentration. Closing your
eyes helps.
5. Turn off the VDT when not in use.
From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Sun Feb 13 16:59:40 1994
ReturnPath: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU>
Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI4.10) for /com/archive/cubelovers id AA17813; Sun, 13 Feb 94 16:59:40 EST
MessageId: <9402132159.AA17813@life.ai.mit.edu>
Received: from MITVMA.MIT.EDU by mitvma.mit.edu (IBM VM SMTP V2R2)
with BSMTP id 1093; Sun, 13 Feb 94 16:59:35 EST
Received: from WVNVM.WVNET.EDU (NJE origin MAILER@WVNVM) by MITVMA.MIT.EDU
(LMail V1.1d/1.7f) with BSMTP id 4178; Sun, 13 Feb 1994 16:59:35 0500
Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU
(LMail V1.1d/1.7f) with BSMTP id 1509; Sun, 13 Feb 1994 16:59:25 0500
XAcknowledgeTo:
Date: Sun, 13 Feb 1994 16:59:22 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: Re: Some Proposed Terminology
InReplyTo: Message of 01/21/94 at 18:32:15 from hoey@AIC.NRL.Navy.Mil
On 01/21/94 at 18:32:15 hoey@AIC.NRL.Navy.Mil said:
>I welcome Jerry Bryan's