From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Mon Dec 6 21:32:55 1993 Return-Path: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU> Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA10000; Mon, 6 Dec 93 21:32:55 EST Message-Id: <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 X-Acknowledge-To: Date: Mon, 6 Dec 1993 21:30:25 EST From: "Jerry Bryan" To: Subject: Re: Unique antipode of edges only In-Reply-To: Message of 12/06/93 at 10:45:00 from mark.longridge@canrem.com On 12/06/93 at 10:45:00 mark.longridge@canrem.com said: >-> I was somewhat startled to see the unique antipode of the 3x3x3 edges >-> in the quarter-turn metric. Do you know what pattern that is? >-> >-> Dan >It's got to be all edges flipped in place. I had to stare at my picture for a couple of minutes to be sure, but yes it is. How did you know? >I would like to see the process generating the position! This is doable, but it is a little harder said than done. My "data base" is just a simple flat file with the states and the level associated with each state. In the case of the 2x2x2, the file is about 625K, and I have programs to search it readily. If you use the file in "Solver mode", my "Solver program" just generates all successors of the current node, finds each successor in the data base (it is a simple binary search, the file is sorted), chooses one at level N-1 (there is always at least one), and makes that the new current node. It stops when N=0. I have a "Solver program" for the "corners plus centers of the 3x3x3" as well, but again the data base is small. It is actually the original 625K file for the 2x2x2 case, plus three additional 625K files. This simple file structure was chosen to keep the file small. There are no pointers, trees, or processes stored in the data base. The "edges of the 3x3x3 without centers" is a little tougher. Early in the project, I generated a data base for the first few levels (six or seven, I think), and I have a "Solver program" that will work up to that level. However, the full "edges of the 3x3x3 without centers" is a 4.2 gigabyte file on tape, so it is hard to process. Also, the size of the equivalence classes is not in the data base, only the level. I have to calculate the size of each equivalence class, and it is an expensive calculation. I made a pass at the file and calculated the number of equivalence classes (took 125 hours on a mainframe), but I only saved a summary. I did not save the number of equivalence classes for each state. I found the antipodal by looking for level 15, since I knew there was only one occurrence, and since the level was in the data base. I did save the summaries by tape, so I should only have to look on two tapes to find the other two equivalence classes which have 24 elements. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU If you don't have time to do it right today, what makes you think you are going to have time to do it over again tomorrow? From mouse@collatz.mcrcim.mcgill.edu Tue Dec 7 07:38:26 1993 Return-Path: Received: from Collatz.McRCIM.McGill.EDU by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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 cube-lovers@ai.mit.edu; Tue, 7 Dec 1993 07:38:09 -0500 Date: Tue, 7 Dec 1993 07:38:09 -0500 From: der Mouse Message-Id: <199312071238.HAA16886@Collatz.McRCIM.McGill.EDU> To: cube-lovers@ai.mit.edu Subject: Re: Unique Antipodal of the 3x3x3 Edges > In answer to the question by Dan Hoey, I printed out the unique > antipodal of the 3x3x3 edges [...]. > It is really quite extraordinary and wonderful. [...]. Without > further ado: Someone else remarks that it's "got to be all edges flipped in place", and Jerry Bryan remarks that it is. > *6* *6* > 6*6 3*4 > *6* *1* > *2* *5* > 2*2 3*4 > *2* *2* > *3**1**4* *1**1**1* > 3*31*14*4 5*23*42*5 > *3**1**4* *6**6**6* > *5* *2* > 5*5 3*4 > *5* *5* I disagree. Look at the 1-2 edge. If it's "flipped in place", then since it appears to be fixed, the cube must flip around it. But then the four 3 faces would be where the 4 faces actually are. No, it's more complicated than just all-edges-flipped. "[Q]uite extraordinary and wonderful" it is. der Mouse mouse@collatz.mcrcim.mcgill.edu From ccw@eql12.caltech.edu Tue Dec 7 08:25:59 1993 Return-Path: Received: from EQL12.Caltech.Edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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) Message-Id: <931206185340.20400b26@EQL12.Caltech.Edu> Subject: Re: Unique antipode of edges only In-Reply-To: Your message <9312070232.AA10000@life.ai.mit.edu> dated 6-Dec-1993 To: BRYAN%WVNVM.WVNET.EDU%WVNVM.WVNET.EDU@mitvma.mit.edu, cube-lovers@ai.mit.edu On 12/06/93 at 10:45:00 mark.longridge@canrem.com said: >-> I was somewhat startled to see the unique antipode of the 3x3x3 edges >-> in the quarter-turn metric. Do you know what pattern that is? >-> >-> Dan >It's got to be all edges flipped in place. Unfortunately, this is wishfull thinking. This antipode is 15 qtw from Home, an odd distance. All edges flipped is an even distance from Home in the qtw metric. Looking at Jerry Bryan's pictures, I see 5 two edge swaps. > > *6* *6* > 6*6 3*4 > *6* *1* > *2* *5* > 2*2 3*4 > *2* *2* > *3**1**4* *1**1**1* > 3*31*14*4 5*23*42*5 > *3**1**4* *6**6**6* > *5* *2* > 5*5 3*4 > *5* *5* > > Start Antipodal > If we assume face 1 is F, I get (FU) (BD) (FD,BU) (FL,LU) (FR, RU) (LD,BL) (RD,BR) Is the 1152 number the result of factoring out the 24 spatial rotations and 2 reflections of the centers? Are there any estimates of how many distinct sequences actually generate this Antipodal Class? Ideally, it would be interesting to have a total list of these sequences. From formail.TCPBRIDGE.FS3.FAA1.ERICM%smte@formail.formation.com Tue Dec 7 09:06:46 1993 Return-Path: Received: from uu3.psi.com by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA28712; Tue, 7 Dec 93 09:06:46 EST Received: from formail.formation.com by uu3.psi.com (5.65b/4.0.071791-PSI/PSINet) via SMTP; id AA20412 for CUBE-LOVERS@AI.MIT.EDU; Tue, 7 Dec 93 09:06:37 -0500 Received: by formail.formation.com (4.1/SMI-4.1) id AA19052; Tue, 7 Dec 93 09:01:04 EST Message-Id: <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 cube-variants can be purchased? Thanks. From andyl@harlequin.com Tue Dec 7 10:33:00 1993 Return-Path: Received: from hilly.harlequin.com by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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 Message-Id: <6474.199312071537@phaedrus.harlequin.com> To: Alan@lcs.mit.edu Cc: BRYAN@wvnvm.bitnet, Cube-Lovers@ai.mit.edu In-Reply-To: Alan Bawden's message of Mon, 6 Dec 93 20:16:26 -0500 <6Dec1993.195513.Alan@LCS.MIT.EDU> Subject: Unique Antipodal of the 3x3x3 Edges Date: Mon, 6 Dec 93 20:16:26 -0500 From: Alan Bawden 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 Return-Path: Received: from scumby.clipper.ingr.com by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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) Message-Id: <199312071611.AA18849@scumby.clipper.ingr.com> Subject: Re: cube availability To: cube-lovers@ai.mit.edu, formail.TCPBRIDGE.FS3.FAA1.ERICM%smte@formail.formation.com (Moyer Eric) Date: Tue, 7 Dec 93 8:11:26 PST In-Reply-To: <9312071401.AA19052@formail.formation.com>; from "Moyer, Eric" at Dec 7, 93 9:10 am X-Mailer: ELM [version 07.00.00.00 (2.3 PL11)] >After reading the recent archives I rushed out to find a square1 >and fell in love all over again, Congratulation. >only this time, I'm armed. Watch out, he is dangerous. :-) >I was somewhat amazed to find, however, that not a single other >cube puzzle was available at ToysRUs or at any of the stores I >tried first. Instead of toy stores, I'd try game stores. I don't know if you'll be able to find a Square-1 or not. It has been a couple of years since they come out. >came across the address for Uwe >M'effert Novelties, Princewell (Far East), Ltd., P.O. Box 31008, >Causeway Bay, Hong Kong. Does anyone know if this company still >exists? I believe that this company has not been around for several years. I did buy a couple of their "cubes" about ten years ago. But at some point they seemed to have disapeared. Too bad they had some unique variations. >Additionally, does anyone know of any mail order company >where cubes and cube-variants can be purchased? Thanks. Yes. It seems that just recently Ishi Press has made some of the cube-variants available. I saw a 5x5x5 cube in a game store recently from Ishi Press. Game stores are a good place to look since Ishi has long been providing products for the games; Go and Shogi. Note that there is even an email address! Ishi Press International 76 Bona Ventura San Jose Ca 95134 (408)944-9900 fax - (408)944-9110 email - ishius@ishius.com Have fun, Tom Granvold tom@clipper.ingr.com From senya@rainbow.ldgo.columbia.edu Tue Dec 7 11:15:30 1993 Return-Path: Received: from lamont.ldgo.columbia.edu (ldgo.columbia.edu) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA04949; Tue, 7 Dec 93 11:15:30 EST Received: from rainbow.ldgo.columbia.edu by lamont.ldgo.columbia.edu (4.1/SMI-3.2) 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:cube-lovers@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) Message-Id: <9312071614.AA29505@rainbow.ldgo.columbia.edu> To: cube-lovers@ai.mit.edu Subject: Needed: Basic elements of solving Rubic Cube: Could you gentelmen suggest me the place where I can find the basic (elementary) combinations to solve the cube? Like the sequence to swap two "internal" side's boxes while not changing the positions of all other "internal" boxes but probably only rotating them? Or could you tell me about the method to rotate some of edge elements without swapping them? ___________________________________________________________________________ ____ ______ __ __ ______ Semyon Basin, / ___\ /\ ___\ /\ \ /\ \ / ____ \ Lamont-Doherty Earth Observatory, /\ \/_/ \/\ \/_/ \/\ \_\/\ \ /\ \___\ \ Route 9W, \/\ \ \/\ _\ \/\ ____ \\/\__ __ \ Palisades, NY 10964 \/\ \____\/\ \/___\/\ \/_/\ \\/_/ /_/\ \ \/\_____\\/\_____\\/\_\ \/\_\ /\_\ \/\_\ Internet:senya@rainbow.ldgo.columbia.edu \/_/_/_/ \/_/_/_/ \/_/ \/_/ \/_/ \/_/ ________________________________________________________________________________ From dseal@armltd.co.uk Tue Dec 7 12:02:56 1993 Return-Path: Received: from eros.britain.eu.net by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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) cube-lovers@ai.mit.edu Subject: Re: Unique antipode of edges only Message-Id: <2D04ABBF@dseal> > Someone else remarks that it's "got to be all edges flipped in place", > and Jerry Bryan remarks that it is. > > > *6* *6* > > 6*6 3*4 > > *6* *1* > > *2* *5* > > 2*2 3*4 > > *2* *2* > > *3**1**4* *1**1**1* > > 3*31*14*4 5*23*42*5 > > *3**1**4* *6**6**6* > > *5* *2* > > 5*5 3*4 > > *5* *5* > > I disagree. Look at the 1-2 edge. If it's "flipped in place", then > since it appears to be fixed, the cube must flip around it. But then > the four 3 faces would be where the 4 faces actually are. No, it's > more complicated than just all-edges-flipped. > > "[Q]uite extraordinary and wonderful" it is. It is in fact the position arrived at by flipping all edges in place, *then* reflecting the entire configuration. I believe this also tells us what the other two equivalence classes with just 24 elements are: they are the results of doing each of these two operations separately. David Seal dseal@armltd.co.uk From andyl@harlequin.com Tue Dec 7 12:24:30 1993 Return-Path: Received: from hilly.harlequin.com by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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 Message-Id: <12292.199312071729@phaedrus.harlequin.com> To: ccw@eql12.caltech.edu Cc: BRYAN%WVNVM.WVNET.EDU%WVNVM.WVNET.EDU@mitvma.mit.edu, cube-lovers@ai.mit.edu In-Reply-To: Chris Worrell's message of Mon, 6 Dec 93 19:13:20 PST <931206185340.20400b26@EQL12.Caltech.Edu> Subject: Unique antipode of edges only Unfortunately, this is wishfull thinking. This antipode is 15 qtw from Home, an odd distance. All edges flipped is an even distance from Home in the qtw metric. Looking at Jerry Bryan's pictures, I see 5 two edge swaps. > > *6* *6* > 6*6 3*4 > *6* *1* > *2* *5* > 2*2 3*4 > *2* *2* > *3**1**4* *1**1**1* > 3*31*14*4 5*23*42*5 > *3**1**4* *6**6**6* > *5* *2* > 5*5 3*4 > *5* *5* > > Start Antipodal > The antipodal position is an interesting one. If you take the antipodal position, and flip all the edges, you get: *5* 5*5 *5* *1* 1*1 *1* *3**2**4* 3*32*24*4 *3**2**4* *6* 6*6 *6* Antipodal with edges flipped. This looks like a rotation of the solved state at first glance, since all the faces on a given side of the cube are the same color. But look again! This is not the solved state of the original cube, but of the mirror image cube. If you added in the centers or the corners, there would be no way to add them to make this a solved state. Andy Latto andyl@harlequin.com From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Tue Dec 7 13:42:56 1993 Return-Path: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU> Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA13489; Tue, 7 Dec 93 13:42:56 EST Message-Id: <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 X-Acknowledge-To: Date: Tue, 7 Dec 1993 13:40:20 EST From: "Jerry Bryan" To: "Cube Lovers List" Subject: Re: Unique antipode of edges only In-Reply-To: Message of 12/07/93 at 12:29:32 from andyl@harlequin.com On 12/07/93 at 12:29:32 Andy Latto said: >The antipodal position is an interesting one. If you take the antipodal >position, and flip all the edges, you get: > *5* > 5*5 > *5* > *1* > 1*1 > *1* > *3**2**4* > 3*32*24*4 > *3**2**4* > *6* > 6*6 > *6* >Antipodal with edges flipped. >This looks like a rotation of the solved state at first glance, since >all the faces on a given side of the cube are the same color. But >look again! This is not the solved state of the original cube, but >of the mirror image cube. If you added in the centers or the corners, >there would be no way to add them to make this a solved state. Indeed. I spoke too quickly when I said the antipodal was simply Start with the edges flipped. I stared at it, flipped the edges in my mind, and it "looked" solved, so I assumed it was Start. I am not yet for sure what they look like, but of the other two states with order-24 equivalence classes, one is at level 9 and the other is at level 12. Since the only one at an even level is at level 12, I am assuming that will be the one which is Start with the edges all flipped. The one at level 9 will probably be the mirror image of Start. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU If you don't have time to do it right today, what makes you think you are going to have time to do it over again tomorrow? From hoey@aic.nrl.navy.mil Tue Dec 7 20:13:23 1993 Return-Path: Received: from Sun0.AIC.NRL.Navy.Mil by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA05026; Tue, 7 Dec 93 20:13:23 EST Received: by Sun0.AIC.NRL.Navy.Mil (4.1/SMI-4.0) 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) Message-Id: <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 All-Edges-Flipped composed with a mirror reflection of Start. Begging the question: *which* mirror reflection? The answer is, it doesn't matter: since these are the edges of a cube without centers, all reflections are the same position. As long as we get to choose which reflection, the canonical one would be the central reflection. When composed with All-Edges- Flipped, it makes the following antipode. (I think using BFTDLR notation instead of 123456 makes these diagrams a lot easier to read). + T + + F + T T R L + T + + B + + L + + F + + R + + D + + D + + D + L L F F R R => F B R L B F + L + + F + + R + + T + + T + + T + + D + + B + D D R L + D + + F + + B + + T + B B R L + B + + D + > I am not yet for sure what they look like, but of the other two states > with order-24 equivalence classes, one is at level 9 and the other > is at level 12. Since the only one at an even level is at level 12, > I am assuming that will be the one which is Start with the edges all > flipped. The one at level 9 will probably be the mirror image of Start. If an order-24 equivalence class means what I think it does, I'm pretty sure those two states have to be Mirror-Start and All-Edges- Flipped, being the only sufficiently symmetric positions. But as to their depth, the parity argument (which Chris Worrell also cited) is not valid here. Remember that the cube has no face centers, so the position is not changed by rotating the assemblage of edges in space (i.e., with respect to the absent face centers). But a quarter-turn of the cube in space is an odd permutation of the edges. So permuta- tion parity is not an intrinsic property of edge positions. We can show that there is no sort of parity here by explicitly constructing an odd cycle. Just use a process that would permute the edges of a cube with faces as (FR,FT,FL,FD)(BR,BT,BL,BD)(RT,TL,LD,DR). This has to be an odd process, but it's an identity on the faceless cube. My (very cheap) guess for where we will find the other two M-symmetric positions is opposite to Jerry Bryan's. On a cube with faces, the central reflection of the edges with respect to the faces is Pons Asinorum, which has the easy 12-qt tight lower bound we've seen before (or if not, you can of course get it from me with email). I'm guessing that this bound happens to be tight on the cube without faces, as well. But I have no proof of this guess, and I'm very grateful we won't have to settle for guesses for very long. Dan Hoey Hoey@AIC.NRL.Navy.Mil From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Wed Dec 8 10:04:52 1993 Return-Path: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU> Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA02610; Wed, 8 Dec 93 10:04:52 EST Message-Id: <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 X-Acknowledge-To: Date: Wed, 8 Dec 1993 10:02:15 EST From: "Jerry Bryan" To: "Cube Lovers List" Subject: Re: Unique antipode of edges only In-Reply-To: Message of 12/07/93 at 20:13:08 from hoey@aic.nrl.navy.mil On 12/07/93 at 20:13:08 hoey@aic.nrl.navy.mil said: >My (very cheap) guess for where we will find the other two M-symmetric >positions is opposite to Jerry Bryan's. On a cube with faces, the >central reflection of the edges with respect to the faces is Pons >Asinorum, which has the easy 12-qt tight lower bound we've seen before >(or if not, you can of course get it from me with email). I'm >guessing that this bound happens to be tight on the cube without >faces, as well. But I have no proof of this guess, and I'm very >grateful we won't have to settle for guesses for very long. >Dan Hoey >Hoey@AIC.NRL.Navy.Mil Dan Hoey is correct. Mirror-Image-of-Start is at level 12. Edges-Flipped is at level 9. Mirror-Image-of-Start-and-Edges-Flipped is at level 15. And, of course, Start is at Level 0. This exhausts the list of configurations with order-24 symmetry. I am still thinking about the easiest way to extract sequences of operators from my data base. I gather from Dan's comments that a 12-qt operator is known for Mirror-Image-of-Start. Are operators known for the other two cases? This is going to be sufficiently time-consuming that I don't want to try to find operators that are already known. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU If you don't have time to do it right today, what makes you think you are going to have time to do it over again tomorrow? From senya@rainbow.ldgo.columbia.edu Wed Dec 8 12:21:51 1993 Return-Path: Received: from lamont.ldgo.columbia.edu (ldgo.columbia.edu) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA12094; Wed, 8 Dec 93 12:21:51 EST Received: from rainbow.ldgo.columbia.edu by lamont.ldgo.columbia.edu (4.1/SMI-3.2) 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:cube-lovers@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) Message-Id: <9312081720.AA20676@rainbow.ldgo.columbia.edu> To: cube-lovers@ai.mit.edu Subscribe me please From @mail.uunet.ca:mark.longridge@canrem.com Wed Dec 8 14:21:29 1993 Return-Path: <@mail.uunet.ca:mark.longridge@canrem.com> Received: from mail.uunet.ca (uunet.ca) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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/SMI-4.1) id AA09857; Wed, 8 Dec 93 13:50:53 EST Received: by canrem.com (PCB-UUCP 1.1f) id 18D8B1; Wed, 8 Dec 93 13:41:19 -0400 To: cube-lovers@life.ai.mit.edu Reply-To: CRSO.Cube@canrem.com Sender: CRSO.Cube@canrem.com Subject: Antipodal Edge Position From: mark.longridge@canrem.com (Mark Longridge) Message-Id: <60.600.5834.0C18D8B1@canrem.com> Date: Wed, 8 Dec 1993 12:33:00 -0500 Organization: CRS Online (Toronto, Ontario) >>It's got to be all edges flipped in place. Oops. Well I figured if all edges flipped was one of the hardest know cube states that in the case of edges-only it would be the antipode. I'm now sure (I think) that it is really: all edges flipped + 4 X (with the 4 X on sides F, R, B, L which should match Dan's diagram) Hmmmm, I don't know if this is a standard form of representation, but this picture looks like a folded out cube: + T + + F + T T R L + T + + B + + L + + F + + R + + D + + D + + D + L L F F R R => F B R L B F + L + + F + + R + + T + + T + + T + + D + + B + D D R L + D + + F + + B + + T + B B ---------> R L + B + | + D + | + D + In my program I would have L R on the screen for the bottom face. + T + The idea is you are always looking at a cube face head-on (just to clarify the difference in diagrams). More quotes for Jerry Bryan: >The "edges of the 3x3x3 without centers" is a little tougher. Early >in the project, I generated a data base for the first few levels >(six or seven, I think), and I have a "Solver program" that will >work up to that level. However, the full "edges of the 3x3x3 without >centers" is a 4.2 gigabyte file on tape, so it is hard to process. >Also, the size of the equivalence classes is not in the data base, >only the level. I have to calculate the size of each equivalence >class, and it is an expensive calculation. > >I made a pass at the >file and calculated the number of equivalence classes (took >125 hours on a mainframe), but I only saved a summary. I did not >save the number of equivalence classes for each state. I found >the antipodal by looking for level 15, since I knew there was >only one occurrence, and since the level was in the data base. > >I am not yet for sure what they look like, but of the other two states >with order-24 equivalence classes, one is at level 9 and the other >is at level 12. Since the only one at an even level is at level 12, >I am assuming that will be the one which is Start with the edges all >flipped. The one at level 9 will probably be the mirror image of Start. I'd still like to see the process for all-edges-flipped (not caring about the centres or corners). So "level" is the number of moves required to solve the position? That means edges flipped in place can be done in 12 qtw. From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Wed Dec 8 14:42:11 1993 Return-Path: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU> Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA19673; Wed, 8 Dec 93 14:42:11 EST Message-Id: <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 X-Acknowledge-To: Date: Wed, 8 Dec 1993 14:08:15 EST From: "Jerry Bryan" To: "Cube Lovers List" Subject: 1152-fold Symmetry and 24-fold Symmetry I guess it's time to try to explain what I mean by 1152-fold symmetry and 24-fold symmetry. Let me start with two or three very simple ideas. First, consider two equally colored and oriented cubes at Start. To one of them, apply F, and to the other one apply R. The obvious solution to the first one is then F' and the obvious solution to the second on is then R'. But take both cubes and toss them through the air to someone else, so that the spatial orientation is lost. Almost anyone would solve either cube by finding the one face that was twisted clockwise and simply twisting it counter-clockwise. No distinction between F and R would be made, and it would be "obvious" to any reasonable person that the cubes were equivalent. As a slightly more formal application of this idea, consider again Start to which R has been applied. We could rotate the whole cube in space using Singmaster's script-U operation. That is, grasp the Up (top) of the cube and turn the whole cube in space clockwise. Now, the cube looks like F has been applied rather than R, and the solution looks like F' rather than R'. If we applied F', the cube would be solved, but the colors would be oriented wrong. We could restore the colors by script-U'. Thus, (script-U F' script U') is exactly the same thing as R' (we are just using conjugates in a very simple way). Continuing in this vein, take any two equally colored and oriented cubes at Start. To one of them, apply some long sequence of permutations P. To the second one, apply (script-U P script-U'). At this point, the two cubes are definitely not "equal" in some sense -- you could clearly tell them apart. Yet, they are clearly "equivalent" in some sense, because if P' is a solution to the first cube, then (script-U P' script-U') is a solution to the second one. Furthermore, it should be obvious that it is not really necessary to use the (script-U script-U') conjugate on the second cube. Rather we can think of some rotation as having been performed on P to give Q, and then of Q as having been performed on Start, so that the same rotation that was applied to P could be applied to P' to give Q', and Q' is equivalent to (script-U P' script-U'). If I can wax sophomorically philosophical for a minute, I tend to think of there being two kinds of permutations in mathematics. The first is the "permutations and combinations" kind of thing you run into in probability and statistics. The second is the permutation operator kind of thing you run into in abstract algebra or group theory. With this kind of thinking, the cube itself represents the first kind of permutation, where the cube is an object being operated on, and the twists of the cube are the second kind of permutation, where the twists are permutation operators and are doing the operating. Well, at some deep level, the two kinds of permutations are very much the same thing, so it is very natural to think of operating on (rotating, in this particular case) a permutation P, where P itself is an operator. I need one more simple idea. Again, think of a cube in Start, and think of Singmaster's script-U operator. We can (informally) write script-U = (Front --> Left --> Back --> Right --> Front). But suppose the cube is colored as Font=Red, Left=White, Back=Orange, Right=Blue). We could just as well write script-U = (Red --> White --> Orange --> Blue --> Red). It looks as if for any fixed coloring, we can freely interchange Singmaster's notation with a notation based on colors. But we can't really. For example, colored as I described it above, script-F is equivalent to script-Red. Either is the same as grasping the front of the cube and rotating the whole cube clockwise. But first perform script-U. Now, script-F is the same as script-Blue). The Front/Back/Up/Down/Left/Right notation is fixed in space, but the color notation is not. Now, we try to put all this together. Once again, consider two equally colored and equally oriented cubes in space, and apply F to the first one and (R script-U) to the second one. Both cubes can now be described as "Start with the front twisted clockwise by 90 degrees), but the colors are not the same. They are clearly equivalent, but under my internal computer model for the cube, they are not equal. Furthermore, no amount of additional application of Singmaster's whole cube "script" operators will make them equal. The only thing that will make them equal will be to rotate the colors. The exact same thing applies to reflections. Consider two equally colored and oriented cubes in Start, and apply F to one and F' to the other. The cubes are equivalent but not equal. Hold up the cube to which F' has been applied to a mirror. The mirror-image now has F applied instead of F', but the colors are wrong (they have been reflected). To make the cubes equal, it is necessary to reflect the colors of the mirror-image. Hence, my program generates equivalence classes by applying a cube rotation, a color rotation, a cube reflection, and a color reflection. There are 24 cube rotations and 24 color rotations (one of each is the identity), and any cube rotation can occur with any color rotation. There are 2 cube reflections and 2 color reflections (one each is the identity), but the cube reflection identity must occur with the color reflection identity and vice versa. Thus, there are (in general) 24x24x2 elements in each equivalence class. I only store one element of each equivalence class in my data base. Some of the equivalence classes have fewer than 24x24x2 elements, namely those for which the cube configuration inherently has a high degree of symmetry. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU If you don't have time to do it right today, what makes you think you are going to have time to do it over again tomorrow? From @mail.uunet.ca:mark.longridge@canrem.com Wed Dec 8 15:31:43 1993 Return-Path: <@mail.uunet.ca:mark.longridge@canrem.com> Received: from mail.uunet.ca (uunet.ca) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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/SMI-4.1) id AA16889; Wed, 8 Dec 93 15:00:21 EST Received: by canrem.com (PCB-UUCP 1.1f) id 18D8D1; Wed, 8 Dec 93 14:56:34 -0400 To: cube-lovers@life.ai.mit.edu Reply-To: CRSO.Cube@canrem.com Sender: CRSO.Cube@canrem.com Subject: More corrections From: mark.longridge@canrem.com (Mark Longridge) Message-Id: <60.601.5834.0C18D8D1@canrem.com> Date: Wed, 8 Dec 1993 13:52:00 -0500 Organization: CRS Online (Toronto, Ontario) Mark gaffs again: > I'm now sure (I think) that it is really: > > all edges flipped + 4 X > (with the 4 X on sides F, R, B, L which should match Dan's diagram) * sign * No, I see I entered the position into my program wrong. A central reflection of the edges with respect to the faces is simply 6 X or checkerboard order 2, solvable in 12 qtw or 6 htw. So the edges-only antipode is: all-edges-flipped + 6 X. Jerry Byran quote: >Dan Hoey is correct. Mirror-Image-of-Start is at level 12. >Edges-Flipped is at level 9. Mirror-Image-of-Start-and-Edges-Flipped >is at level 15. And, of course, Start is at Level 0. This exhausts >the list of configurations with order-24 symmetry. Ok, only 9 qtw.... it's got to play havoc with corners. I got it now. * Hmmm, what are all the possible orders of symmetry? * Also I note my "Symmetry Level" is the opposite of Jerry's Order-N symmetry: > If we define "symmetry level" as the number of distinct patterns >generated by rotating the cube through it's 24 different orientations in >space then most known antipodes are symmetry level 6. Thus the lower the >number the higher the level of symmetry. The least symmetric positions >have level 24, and this is very common. The most symmetric positions >have level 1, the two positions START and 6 X order 2. Of course all-edges-flipped I never included, as at the time I was looking at the square's group. ------------------------------- As a small postfix to my cyclic decomposition article, I found the following patterns. I'm fond of pattern 16 myself. I am looking for CD-type processes for 6 X order 3 and 6 X order 6. I find when I am physical cubing (as opposed to computer cubing or old fashioned mental cubing!) it really helps having a CD-type process memory-wise. Memorizing the computer generated processes is like memorizing prime numbers. p161 Mark's Pattern 16 (F1 R1 L1 B1) ^3 + F2 B2 D2 F2 B2 T2 (18) p162 2 X, 4 H full (F1 T2 B1) ^4 (12) p163 4 ARM Full (F2 T1 B2) ^4 + T1 D3 (14) p164 4 Y's Rotated (F1 T2 D2) ^6 + F1 (19) p165 2 Swap, 4 H full (F1 L2 T2 R2 B1) ^2 + L2 R2 T2 D2 (14) p166 2 H adj swap (F1 L2 T2 R2 B1) ^2 + L2 T2 R2 D2 L2 T2 (16) No doubt these are compressible and hence not as efficient, but if you consider ease of execution.... -> Mark <- From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Thu Dec 9 03:38:45 1993 Return-Path: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU> Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA21581; Thu, 9 Dec 93 03:38:45 EST Message-Id: <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 X-Acknowledge-To: 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 q-turns. The size of the equivalence class is 48. + B + R L + F + + D + L R + U + + B + + F + + B + U D R L D U + F + + B + + F + + U + L R + D + = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU If you don't have time to do it right today, what makes you think you are going to have time to do it over again tomorrow? From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Thu Dec 9 04:19:46 1993 Return-Path: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU> Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA22353; Thu, 9 Dec 93 04:19:46 EST Message-Id: <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 X-Acknowledge-To: Date: Wed, 8 Dec 1993 23:16:50 EST From: "Jerry Bryan" To: "Cube Lovers List" Subject: Pretty Pattern (again) Pretty pattern, 8 q-turns from Start, size of equivalence class is 48. + B + F F + B + + U + D D + U + + L + + F + + R + R R B B L L + L + + F + + R + + D + U U + D + = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU If you don't have time to do it right today, what makes you think you are going to have time to do it over again tomorrow? From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Thu Dec 9 04:56:35 1993 Return-Path: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU> Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA22759; Thu, 9 Dec 93 04:56:35 EST Message-Id: <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 X-Acknowledge-To: Date: Wed, 8 Dec 1993 23:39:35 EST From: "Jerry Bryan" To: "Cube Lovers List" Subject: That's all the 48-Fold Symmetries The two patterns I just posted are the only two for which the size of the equivalence class is 48. The next size up is 72, and there are twelve patterns whose equivalence class size is 72. Things become less interesting as the equivalence class size increases because they are less symmetrical overall. Also, there are more patterns. Finally, these things take a good deal of time to chase down. Therefore, I am going to stop chasing down "pretty patterns" for now unless somebody is just dying to see the patterns whose equivalence class size is 72. Finally, nobody really answered an earlier question, but is the following true: 1) Mirror-Image-of-Start is 12 q-turns from Start, and a sequence is known (Dan Hoey sent me a well-known sequence), 2) All-Edges-Flipped is 9 q-turns from Start, and a sequence is known, and 3) Mirror-Image-of-Start-and-All-Edges-Flipped (the antipodal of Start) is 15 q-turns from Start, and a sequence is *not* known? If this is true, then I will start working on the 15 move sequence. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU If you don't have time to do it right today, what makes you think you are going to have time to do it over again tomorrow? From mouse@collatz.mcrcim.mcgill.edu Fri Dec 10 09:16:34 1993 Return-Path: Received: from Collatz.McRCIM.McGill.EDU by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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 Message-Id: <199312101416.JAA04504@Collatz.McRCIM.McGill.EDU> To: senya@rainbow.ldgo.columbia.edu Subject: Re: Needed: Basic elements of solving Rubic Cube: Cc: cube-lovers@ai.mit.edu > Subject: Needed: Basic elements of solving Rubic Cube: (Does anyone happen to know whether Ern Rubik knows what's happened to his name? I don't mean just here, but how common a term it's become. And I apologize for using instead of what I think is the proper long accent, but Latin-1 doesn't have the proper one....) > Could you gentelmen suggest me the place where I can find the basic > (elementary) combinations to solve the cube? There are no *the* basic combinations. I would guess there are as many different sets of combinations as there are cube solvers. > Like the sequence to swap two "internal" side's boxes while not > changing the positions of all other "internal" boxes but probably > only rotating them? > Or could you tell me about the method to rotate some of edge elements > without swapping them? Well, for what it's worth, when I solve a cube, I do it as follows. (Slice turns: if I write FB, I mean turn the F-B slice in the direction one would turn F for an F turn, similarly for other slice turns: turn the slice as if it were carried along by the turn named by the first letter. Thus LR and RL are inverses. Is there some standard representation for slice turns?) - Solve one layer ad-hoc. (This refers to a layer of cubies, not just one face of the cube.) I often don't worry about edge flips at this stage. Some simple operators I use: To get corners in place: F D' F, or F' D F, depending on corner orientation. To get edges in place: If the cubie is on the D face, FB/BF/RL/LR, D/D', inverse of the slice turn. If it's on the middle layer, F/B/R/L, UD/DU, inverse of face turn. - Turn the cube so the solved face is L. Solve what then becomes the R-L slice layer with a combination of R2 U2 R2 U2 R2 U2, to move cubies around within the slice layer, and either of two sequences to move cubies between the R layer and the slice layer: R2 D R' B2 R2 B2 R2 B2 R' D' FB D R' B2 R2 B2 R2 B2 R' D' BF (The first one is a sequence that normally ends with R2, but since the R layer is scratch at this point I often don't bother.) These are, of course, interspersed with R, R2, and R' turns to get edges in the correct places for them. At this point you will have two layers solved, except possibly for some flipped edges. Next, corners of the "scratch" layer. Check them for placement, ignoring orientation. They can be: 1) All in place. This is the easy case. :-) 2) Two swapped. Make one quarter-turn to reach case 3. (They can't be diagonal, they must be adjacent - or some joker has taken your cube apart.) 3) One in place, other three permuted. 4) Two pairs, each swapped. If the swaps are diagonal, turn the layer a half-turn to reach case 1. In case 3 or 4, the corners can be put in place by holding the cube with the unsolved layer as U and repeating L F L' F' L F L' F' L F L' F' twice, turning U so as to place appropriate pairs of cubies in the UFL and UBL corners. To orient the corners correctly, hold the cube with the unsolved layer as F and use R B2 R' U' B2 U and its inverse U' B2 U R B2 R' with a turn of the F face in between; this will allow you to twist the corners into correct orientation. All that remains at this point is positioning the edges on the last layer, and possibly some edge flips. To position the edges, I normally use (with U as the unsolved layer) R2 D R' B2 R2 B2 R2 B2 R' D' R2 FB D R' B2 R2 B2 R2 B2 R' D' BF R2 U2 R2 U2 R2 U2 with appropriate turns of U in between, swapping the FR and BR edges repeatedly as auxiliaries while swapping pairs of edges on U to get them in place. (The difference between the first two sequences is that the first one swaps UB and UR, the second UB and RU.) Edge flips are all that's left at this point. Judicious choice of which of the two sequences above can often drastically reduce the work to be done here, but there's often some left anyway. The basic sequence I use for this is RL U RL U RL U RL U, which flips four edge cubies in-place: UB, UL, DB, and DF. (A similar sequence U RL U RL U RL U RL is similar but flips UR instead of UL; this can be thought of as U X U', where X is the first-given sequence.) My use of this sequence is usually ad-hoc; sequences such as X F X F' will let you flip arbitrary pairs of edges. Presto! You have a solved cube. :-) In practice, I often take shortcuts; for example, if X represents the R B2 R' U' B2 U sequence, then X B2 X B2 X B2 will twist three corners on B.... der Mouse mouse@collatz.mcrcim.mcgill.edu From hoey@aic.nrl.navy.mil Mon Dec 13 22:31:38 1993 Return-Path: Received: from Sun0.AIC.NRL.Navy.Mil by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA02377; Mon, 13 Dec 93 22:31:38 EST Received: by Sun0.AIC.NRL.Navy.Mil (4.1/SMI-4.0) 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) Message-Id: <9312140331.AA25974@Sun0.AIC.NRL.Navy.Mil> To: Cube-Lovers@ai.mit.edu Subject: Symmetry I suppose it's time for a few observations on symmetry. After all, tomorrow is the thirteenth anniversary of "Symmetry and Local Maxima." As Jerry Bryan notes, we can perform the "R" turn by rotating the cube to put the R face in front, performing "F", and undoing the rotation. But we can also perform "R'" by reflecting the cube in a left-to-right mirror, performing "L", and undoing the reflection. Thus conjugation can be extended to use the 48-element group of rotations and reflections, which we call M. In the absence of face centers, there is another kind of reduction that takes account of the 24 possible positions of the resulting collection of edges in space. So two positions X and Y are considered equivalent if X = m' Y m c where m is a rotation or reflection in M, and c is a rotation. My understanding of Jerry Bryan's method is that he combines "m c" into a single rotation or reflection, and factors out the reflection on both sides. It seems to me that what he calls a a "color rotation" is premultiplication, while a "cube rotation" is postmultiplication. [I am somewhat uncertain of this, because it doesn't explain how there can be a 1252-element symmetry group when face centers are present, so perhaps I'm missing something crucial.] But I think we are at least conceptually better off dealing with M when dealing with conjugation, because it takes account of the essential similarity between clockwise and anticlockwise turns. Alan Bawden mentioned back in 1980 that certain positions with sufficient symmetry were local maxima (in terms of distance from start), on the grounds that any clockwise or anticlockwise turn gives us essentially the same position. Jim Saxe and I formalized the notion in a paper entitled "Symmetry and Local Maxima" that we posted on 14 December 1980. [You can find it in /pub/cube-lovers/cube-mail-1.Z on FTP.AI.MIT.Edu]. We had some hope that some of these local maxima might turn out to be global maxima. My hopes for that have been somewhat low in recent years. That is perhaps my best excuse for not noticing immediately that the single global maximum for the edge group turns out to be one of these symmetric local maxima. In fact, all four of the positions with 24-element equivalence classes appear in the list of M-symmetric positions. The paper on Symmetry and Local Maxima also catalogues the positions that have 48-element equivalence classes and 72-element equivalence classes. The The former are the H-symmetric positions, "Six-H" and "Six-H with all edges flipped". The latter are the twelve T-symmetric positions. For T-symmetry, the set of flipped edges may be any of {none, girdle-edges, off-girdle-edges, or all}; the set of edges exchanged with their antipodes may be any of the four as well. But if we choose "none" or "all" for all both choices we get one of the four M-symmetric positions with 24-element equivalence classes, so only twelve of the sixteen possibilities have 72-element equivalence classes. With regard to the edge cube, I should mention that no one has mentioned a 9 QT process for the all-flip nor a 15 QT process for the pons-asinorum-all-flip. Of course, the latter would be somewhat more interesting, being the longest optimal sequence. Dan Hoey Hoey@AIC.NRL.Navy.Mil From avm@bgerug51.bitnet Tue Dec 14 02:42:21 1993 Return-Path: Received: from mserv.rug.ac.be ([157.193.40.37]) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA09383; Tue, 14 Dec 93 02:42:21 EST Received: by mserv.rug.ac.be id AA28828 (5.65c/IDA-1.4.4 for cube-lovers@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: Anne-Mie Vandermeeren - RUG Subject: Unsunscribe rob@bgerug51.bitnet To: cube-lovers@ai.mit.edu Message-Id: <01H6GP6445K0000WTQ@BGERUG51.BITNET> X-Envelope-To: cube-lovers@ai.mit.edu X-Vms-To: IN%"cube-lovers@ai.mit.edu" Hi, Please remove rob@bgerug51.bitnet from your mailing list cube-Lovers Thanks, Anne-Mie Vandermeeren Postmaster for BGERUG51 From anandrao@hk.super.net Tue Dec 14 03:18:49 1993 Return-Path: Received: from hk.super.net by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA10381; Tue, 14 Dec 93 03:18:49 EST Received: by hk.super.net id AA15191 (5.65c/IDA-1.4.4 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 Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII I managed to pick up all four Tangle puzzles in an obscure shop in Jakarta, Indonesia. The puzzles are similar, except that the extra(25th) piece is different in each. The solutions for each puzzle are very different and I could not see any pattern. I solved all 4 using 'intelligent brute force', i.e. made the search as efficient as I could. But the 10*10 puzzle seems intractable. The 5*5 could be solved using a 486DX2-66 PC in about 20 minutes. The 10*10 will take several months using my algorithm. Does anyone have a more intelligent, or a more brute method? Once this puzzle has been put in the public domain, we MUST find a solution. So, any ideas are welcome! Thanks From dn1l+@andrew.cmu.edu Tue Dec 14 11:29:57 1993 Return-Path: Received: from po4.andrew.cmu.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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 Cube-Lovers@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) Message-Id: Date: Tue, 14 Dec 1993 11:28:41 -0500 (EST) From: "Dale I. Newfield" To: Cube Lovers Subject: Re: Tangle Cc: In-Reply-To: Could you explain what your algorithm was? I have one of the puzzles, number 4, I believe, and spent a large amount of time trying to find a solution that was not trial and error. I could not. The algorithm that I used to have the computer solve it for me was to fill the 5x5 in the following manner, recursively, returning when no possible pieces fit. 1 2 4 7 11 / / / / / / / / 3 5 8 12 16 / / / / / / / / 6 9 13 17 20 / / / / / / / / 10 14 18 21 23 / / / / / / / / 15 19 22 24 25 (wrapping at the edges to keep incrementing properly) I did that because given any pieces diagonal from one another, there are at most two pieces that can fill the gap (line up with both correctly). (When the four colors are different, there are two tiles When there is a single repeated color, there is one tile When there are 2 pairs of colors there is no tile And in all these cases, if the tile(s) was already used, or didn't exist, that is the bottom of that branch of the search tree) Is this better or worse than the algorithm you used? Has anyone found a non-brute-force solution scheme? -Dale From hoch@chem.wisc.edu Tue Dec 14 12:13:15 1993 Return-Path: Received: from ernie.chem.wisc.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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 Message-Id: <9312141713.AA19022@ernie.chem.wisc.edu> To: cube-lovers@ai.mit.edu Subject: List removal Please remove hoch@pigggy.chem.wisc.edu from your cube-lovers list. Thanks. From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Tue Dec 14 21:23:57 1993 Return-Path: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU> Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA24330; Tue, 14 Dec 93 21:23:57 EST Message-Id: <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 X-Acknowledge-To: Date: Tue, 14 Dec 1993 20:50:51 EST From: "Jerry Bryan" To: "Cube Lovers List" Subject: Re: Symmetry In-Reply-To: Message of 12/13/93 at 22:31:31 from hoey@aic.nrl.navy.mil On 12/13/93 at 22:31:31 hoey@aic.nrl.navy.mil said: >In the absence of face centers, there is another kind of reduction >that takes account of the 24 possible positions of the resulting >collection of edges in space. So two positions X and Y are considered >equivalent if > X = m' Y m c >where m is a rotation or reflection in M, and c is a rotation. >My understanding of Jerry Bryan's method is that he combines "m c" >into a single rotation or reflection, and factors out the reflection >on both sides. It seems to me that what he calls a "color rotation" >is premultiplication, while a "cube rotation" is postmultiplication. >[I am somewhat uncertain of this, because it doesn't explain how there >can be a 1252-element symmetry group when face centers are present, so ^^^^ should be 1152 >perhaps I'm missing something crucial.] I just reread "Symmetry and Local Maxima". Let's see if I can make some sense of this. I believe "pre-multiplication" and "post-multiplication" are correct. In my computer model, the corner facelets are simply numbered from 1 to 24, and any configuration of the corners is an order-24 row vector. The rotation and reflection operators are also order-24 row vectors, again with each cell simply containing a number from 1 to 24. In almost anybody's programming language you would copy an order-24 row vector with something like For i = 1 to 24 B(i) = A(i) Well, if P is a rotation operator, you could perform a rotation two ways. I guess one is pre-multiplication and one is post-multiplication. 1) For i = 1 to 24 B(i) = A(P(i)) 2) For i = 1 to 24 B(i) = P(A(i)) (As an aside, this illustrates the question I raised in my previous post about "which is the operator and which is the thing being operated on?" Is P operating on A, or is A operating on P?) In fact, if what I am doing is properly called pre- and post- multiplication, then I am doing both as a part of a single, composite operator. I.e., For i = 1 to 24 B(i) = P(A(P(i))) More completely, there are 24 rotations, P1 through P24, so the actual loop looks something like For j = 1 to 24 for k = 1 to 24 for i = 1 to 24 Bj,k(i) = Pj(A(Pk(i))) Finally, if Q is a reflection (actually, if Q1 is the identity and Q2 is the reflection), then we have For j = 1 to 24 for k = 1 to 24 for m = 1 to 2 for i = 1 to 24 Bj,k,m(i) = Qm(Pj(A(Qm(Pk(i))))) I believe this loop calculates Dan Hoey's M. In my data base, I store the minimum of Bj,k,m over j = 1 to 24, k = 1 to 24, and m = 1 to 2. I tend to call the minimum of Bj,k,m a canonical form. I am not sure if that is the best terminology. The minimal element is not any simpler than any other. It is just that I need a function to choose an element from a set, and picking the minimal element seems very natural. Any other element would do as well, provided I could always be sure of picking the same element. Also, my criterion for equivalence is slightly different (but isomorphic, I think) than the one described by Dan Hoey. Suppose A and B are two cubes. Rather than mapping A to B or B to A in M, I map both A and B to their respective canonical forms. A and B are equivalent if their respective canonical forms are equal. I hasten to add that the actual loop in the program is a bit more complex than the one shown above. The one above would be far too slow. The actual loop is several hundred times faster. Now, as to the centers. I still sometimes have a certain doubt about the centers. They are fixed, so how can you reduce the problem (i.e., increase the size of the equivalence classes) by both rotating the cube and rotating the colors (by both pre- and post-multiplication)? In my computer model for the centers, I simply number center facelets from 1 to 6, and the centers are stored as an order-6 row vector. The centers are disjoint from the corners (as well as from the edges), so there is no problem in numbering one set of objects from 1 to 24 and another from 1 to 6. I define a set of 24 rotation operators P* on the centers, corresponding to the 24 rotation operators P on the corners, and a set of 2 reflection operators Q* on the centers, corresponding to the 2 reflection operators Q on the corners. Then, if C is an order-6 row vector representing the centers, I calculate Dj,k,m = Q*m(P*j(C(Q*m(P*k)))) anytime I calculate Bj,k,m = Qm(Pj(A(Qm(Pk)))). (Read the asterisks above as superscripts. I am not intending the multiplication operation which the asterisk denotes in many programming languages.) Hence, I rotate and reflect the centers right along with the corners. But there are only 24 distinct states for the centers, and each can occur with any canonical form for the corners. Hence, the "corners plus centers" data base is exactly 24 times larger than the "2x2x2" data base. My model for the cube seems to start out 24 times larger than everybody else's. However, by storing only the canonical form for each equivalence class, and since most of the equivalence classes have 1152 elements, my data base seems to end up about 48 times smaller than everybody else's. This fact seems to remain true, even when the "fixed centers" are added in. I am not sure if this answers Dan's question about my model with centers added. Effectively, I am using a "fixed corners" representation of the cube, and rotating the centers. Each equivalence class for the corners under M has (up to) 1152 elements, and each equivalence class for the centers under M has only 24 elements. But it doesn't seem to matter. (Up to) 48 different configurations of the corners within M share each configuration of the centers. Since I am in this deep, let me finish explaining certain details of my model. I don't really store all 24 elements of each row vector. I really just store 8. That is, I store the facelets for the front and back face. The other 16 facelets can be reconstructed from the first 8. In effect, storing a number from 1 to 24 stores both the location of each cubie and its twist. Finally, I really, really only store 7 elements. In the canonical form, the first element is always 1, so there is no reason to store it. Thus, a data base record for the 2x2x2 looks like CCCCCCC,L where the CCCCCCC are the seven elements representing the canonical form, and L is the corresponding level. When you add the centers, I started out with notion that the order-6 row vector for center only has 24 possible states. Thus, it can be encoded as a number from 1 to 24. This lead to the following CCCCCCC,L,R where CCCCCCC and L are as before, and R is an index encoding the orientation of the centers. But this can be improved upon even further. With my model for the corners plus centers, each distinct value of CCCCCCC will occur exactly 24 times, and each distinct value of CCCCCCC is already represented in my data base for the 2x2x2. Hence, I can have the exact same number of (longer) records, and encode the corners of the 3x3x3 as CCCCCCC,L,L1 L2 L3 .... L23 L24 where CCCCCCC is as before, L is the level of CCCCCCC in the 2x2x2, and L1 through L24 are the levels of CCCCCCC in the corners of the 3x3x3 when the index of the position of the centers with respect to the corners is 1 through 24, respectively. Hence, my data base for the corners of the 3x3x3 has the same number of records as the data base for the 2x2x2, and is physically only four times larger. >With regard to the edge cube, I should mention that no one has >mentioned a 9 QT process for the all-flip nor a 15 QT process for the >pons-asinorum-all-flip. Of course, the latter would be somewhat more >interesting, being the longest optimal sequence. I will work on these two cases, but it will take some time. My model is very good at storing a great many states of the cube very compactly, but it does not encode processes at all. I will have to extract the processes by hand. This is quite easy in my data bases for the 2x2x2 and corners of 3x3x3. But it is quite hard for the edges because the data base is 4.2 gigabytes. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU If you don't have time to do it right today, what makes you think you are going to have time to do it over again tomorrow? From Don.Woods@eng.sun.com Wed Dec 15 06:04:15 1993 Return-Path: Received: from Sun.COM by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA07720; Wed, 15 Dec 93 06:04:15 EST Received: from Eng.Sun.COM (zigzag.Eng.Sun.COM) by Sun.COM (4.1/SMI-4.1) id AA22455; Tue, 14 Dec 93 14:48:16 PST Received: from colossal.Eng.Sun.COM by Eng.Sun.COM (4.1/SMI-4.1) id AA21191; Tue, 14 Dec 93 14:47:01 PST Received: by colossal.Eng.Sun.COM (5.0/SMI-SVR4) 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) Message-Id: <9312142248.AA22891@colossal.Eng.Sun.COM> To: Cube-Lovers@ai.mit.edu Subject: Re: Tangle X-Sun-Charset: US-ASCII Content-Length: 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 cube-lovers on April 27, 1992. I own only #1 myself and so cannot confirm or deny the accuracy of the colors. But the basic idea applies, given that each puzzle (a) has the same pattern of ropes on all pieces and (b) has each permutation of colors exactly once except for one permutation which appears twice. Solving the 10x10 is another kettle of fish, and I haven't tried it. I do have a program that solves the 5x5 in about 45 seconds on a SparcStation II, but I haven't looked into how much longer it would take on the 10x10. "Dale I. Newfield" writes: > Could you explain what your algorithm was? > Has anyone found a non-brute-force solution scheme? My solution was brute-force. I posted to cube-lovers (again, in April '92) asking if anyone had found a more logical approach to the puzzle, but got no affirmative responses. Dale's method is a little inefficient in the order in which it tries tiles. Mine used the sequence: Dale's used: 1 3 5 7 9 1 2 4 7 11 2 4 6 8 10 3 5 8 12 16 11 12 13 14 15 6 9 13 17 20 16 17 18 19 20 10 14 18 21 23 21 22 23 24 25 15 19 22 24 25 The first three tiles in our two methods are equally constrained, but the next seven in Dale's methods are constrained along 1-2-1-1-2-2-1 edges, while mine are constrained along 2-1-2-1-2-1-2 edges. So I suspect my search tree gets trimmed a bit more quickly. Another way in which the search can be made more efficient is in finding the pieces to try in each position. For each pair of colors that can appear along an edge, my program precomputes a table listing all tiles that can match that pair of colors, including how to rotate the tiles. -- Don. From @mail.uunet.ca:mark.longridge@canrem.com Wed Dec 15 11:07:13 1993 Return-Path: <@mail.uunet.ca:mark.longridge@canrem.com> Received: from mail.uunet.ca (uunet.ca) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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/SMI-4.1) id AA18531; Wed, 15 Dec 93 10:50:45 EST Received: by canrem.com (PCB-UUCP 1.1f) id 18E5EA; Wed, 15 Dec 93 10:46:24 -0400 To: cube-lovers@life.ai.mit.edu Reply-To: CRSO.Cube@canrem.com Sender: CRSO.Cube@canrem.com Subject: Lib of Congress From: mark.longridge@canrem.com (Mark Longridge) Message-Id: <60.618.5834.0C18E5EA@canrem.com> Date: Wed, 15 Dec 1993 09:38:00 -0500 Organization: CRS Online (Toronto, Ontario) I Pulled this list from the Library of Congress. The hungarian book I've never seen before (#2), and #3 and #4 are new to me. "Zen of Cubing" may be interesting, has anyone read this one? ITEMS 1-4 OF 11 SET 15: BRIEF DISPLAY FILE: LOCI (DESCENDING ORDER) 1. 86-8699: Buvos kocka. English. Rubik's cubic compendium / Oxford ; New York : Oxford University Press, 1987. xi, 225 p. : ill. (some col.) ; 23 cm. LC CALL NUMBER: QA491 .B8813 1987 2. 85-109601: Mezei, Andras. Magyar kocka, avagy, Meg mindig ilyen gazdagok vagyunk? / Budapest : Magveto, c1984. 473 p. : ill. ; 21 cm. LC CALL NUMBER: QA491 .M49 1984 3. 82-72610: Feder, Happy Jack. Zen of cubing : in search of the seventh side / 1st ed. South Bend, Ind. : And Books, c1982. 100 p. : ill. ; 21 cm. LC CALL NUMBER: PN6231.R78 F42 1982 4. 82-3755: O'Grady, Miles. You can kick the cube] : the cube hater's handbook / New York, N.Y. : Penguin Books, 1982. p. cm. NOT IN LC COLLECTION 5. 82-1264: Bandelow, Christoph. Inside Rubik's cube and beyond / Boston : Birkhauser, c1982. 120, [5] p., [6] leaves of plates : ill. (some col.) ; 23 cm. LC CALL NUMBER: QA491 .B2613 1982 6. 81-85850: Schlafly, Roger. The complete cube book / Chicago : Regnery Gateway, c1982. vi, 51 p. : ill. ; 21 cm. LC CALL NUMBER: QA491 .S34 1982 7. 81-81556: Taylor, Don. Mastering Rubik's cube : the solution to the 20th century's most amazing puzzle / 1st American ed. New York : Holt, Rinehart and Winston, 1981, c1980. 31 p. : ill. ; 22 cm. LC CALL NUMBER: QA491 .T38 1981 8. 81-21650: Varasano, Jeffrey, 1966- Conquer the cube in 45 seconds / New York : Bell Pub. Co. : Distributed by Crown Publishers, c1981. 48 p. : ill. ; 21 cm. NOT IN LC COLLECTION 9. 81-16682: Varasano, Jeffrey, 1966- Jeff conquers the cube in 45 seconds : and you can too] / New York : Stein and Day, 1981. p.cm. NOT IN LC COLLECTION 10. 81-12525: Frey, Alexander H. Handbook of cubik math / Hillside, N.J. : Enslow Publishers, c1982. viii, 193 p. : ill. ; 23 cm. LC CALL NUMBER: QA491 .F73 1982 11. 80-27751: Singmaster, David. Notes on Rubik's magic cube / Hillside, N.J. : Enslow Publishers, 1981. vi, 73 p. : ill.; 24 cm. LC CALL NUMBER: QA491 .S58 1981 More to follow -> Mark <- From @mail.uunet.ca:mark.longridge@canrem.com Wed Dec 15 12:23:33 1993 Return-Path: <@mail.uunet.ca:mark.longridge@canrem.com> Received: from mail.uunet.ca (uunet.ca) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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/SMI-4.1) id AA18542; Wed, 15 Dec 93 10:50:49 EST Received: by canrem.com (PCB-UUCP 1.1f) id 18E5EB; Wed, 15 Dec 93 10:46:25 -0400 To: cube-lovers@life.ai.mit.edu Reply-To: CRSO.Cube@canrem.com Sender: CRSO.Cube@canrem.com Subject: 6 X order 3 From: mark.longridge@canrem.com (Mark Longridge) Message-Id: <60.619.5834.0C18E5EB@canrem.com> Date: Wed, 15 Dec 1993 09:42:00 -0500 Organization: CRS Online (Toronto, Ontario) A few messages back I mentioned a cyclicly decomposable process for the pattern 6 X order 3. Success! Those familar with Christoph Bandelow's "Inside Rubik's Cube and Beyond" will recognize the notation, but for those who don't: Mr is the middle slice adjacent to face R Mu is the middle slice adjacent to face U (or T) Mf is the middle slice adjacent to face F Thus Mr1 rotates the middle slice in the same direction as r1, etc. ...fairly intuitive. The 28 slice moves are rather lengthy, but one can follow the progression to 6 X order 3 easily. Before the discovery of process p1b, memorization and execution of this pattern was difficult. By memorization I don't mean retention for days or weeks or even months as I wanted a CD-type process with which I could always reconstruct it in my head. Perhaps this could be improved upon, nevertheless now the checkerboard order 3 is easy to execute and easy to remember! (rotates edges 120 degrees around the FTR corner and BDL corner) p1b alternate method 2 (Mr2 D3 Mr2 U1) ^3 TOP becomes LEFT (28s) (Mr2 D3 Mr2 U1) ^3 LEFT becomes TOP Mr3 Mt1 Mr1 Mt3 From @mail.uunet.ca:mark.longridge@canrem.com Wed Dec 15 13:04:34 1993 Return-Path: <@mail.uunet.ca:mark.longridge@canrem.com> Received: from mail.uunet.ca (uunet.ca) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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/SMI-4.1) id AA20478; Wed, 15 Dec 93 11:00:22 EST Received: by canrem.com (PCB-UUCP 1.1f) id 18E5F0; Wed, 15 Dec 93 10:58:09 -0400 To: cube-lovers@life.ai.mit.edu Reply-To: CRSO.Cube@canrem.com Sender: CRSO.Cube@canrem.com Subject: Part 2 From: mark.longridge@canrem.com (Mark Longridge) Message-Id: <60.620.5834.0C18E5F0@canrem.com> Date: Wed, 15 Dec 1993 09:58:00 -0500 Organization: CRS Online (Toronto, Ontario) Hmmmm, a little inconsistent (sp?) with the notation there. I usually only use U when quoting the processes of others. I'm going to try tackling the 1152-fold symmetry idea later tonight. When looking for CD-type processes I find it helps to think in terms of distinct states you pass through in approaching the goal state. Sort of like factoring a composite pattern into simpler ones you add together. Rotating the cube in space definitely helps too. -> Mark From dn1l+@andrew.cmu.edu Wed Dec 15 13:17:11 1993 Return-Path: Received: from po4.andrew.cmu.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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) Message-Id: Date: Wed, 15 Dec 1993 13:15:45 -0500 (EST) From: "Dale I. Newfield" To: cube-lovers@ai.mit.edu Subject: Re: Description of Tangle, Part 2 Cc: don.woods@eng.sun.com, acw@riverside.scrc.symbolics.com In-Reply-To: <920425084746.2bc000e4@EQL.Caltech.Edu> Just to make sure everyone knows what we are talking about, here is a message from the archives: Excerpts from mail: 25-Apr-92 Description of Tangle, Part 2 by Chris Worrell@eql.caltec > Annotating Don.Woods diagram (which is in the correct orientation) > 2 3 > --------------------- > | @ # | > | @ # | > 1 |$$ @ # %%%%| 4 > | $ @ %#% | > | $ @ %% # | > | $ %@ # | > | $ %% @@# | > | %%% #@@ | > 4 |%%%% $ # @@@| 2 > | $ # | > | $ # | > --------------------- > 1 3 > > The duplicate piece in each tangle is: > 1 2 3 4 > Tangle 1 Blue Red Yellow Green > Tangle 2 Yellow Blue Green Red > Tangle 3 Green Yellow Blue Red > Tangle 4 Red Green Yellow Blue > > All 4 Tangles are the same puzzle, just colored differently. > Each has all 24 color permutations, plus a duplicate. I had kind of hoped that the connectivity on the different puzzles was different, instead of just the colors. (Actually, the sequence I sent before was slightly wrong--here is the one I actually used. Using Don's format) >Don used the sequence: Dale used: > > 1 3 5 7 9 1 2 6 10 15 > 2 4 6 8 10 3 4 7 11 16 > 11 12 13 14 15 5 8 12 17 20 > 16 17 18 19 20 9 13 18 21 23 > 21 22 23 24 25 14 19 22 24 25 But yes, Don's fillpattern still gets more constraints in earlier--here is the number of constraints at each step Don's: 0 1 1 2 1 2 1 2 1 2 1 2 2 2 2 1 2 2 2 2 1 2 2 2 2 Mine: 0 1 1 2 1 1 2 2 1 1 2 2 2 1 1 2 2 2 2 2 2 2 2 2 2 As you can see, I had my 1's clustered more toward the beginning, which is non-optimal. Assuming that there is only a change in color(and not in connectivity), as was posted by Chris in april of 92, I would think modifying code to attempt the 10x10 would be fairly simple...(seeing as my code went poof sometime last year, when a disk crashed(not that it was complicated))...wanna try? (Thanks for the pointers to the Apr 92 discussion) I agree with the concensus expressed in the archives that this puzzle is inherently "not that great" because no non-brute-force method has been found/seems to exist. -Dale From anandrao@hk.super.net Wed Dec 15 20:14:26 1993 Return-Path: Received: from hk.super.net by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA17907; Wed, 15 Dec 93 20:14:26 EST Received: by hk.super.net id AA22615 (5.65c/IDA-1.4.4 for Cube-Lovers@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: Cube-Lovers@ai.mit.edu In-Reply-To: <9312142248.AA22891@colossal.Eng.Sun.COM> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII 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 cube-lovers on April > 27, 1992. I own only #1 myself and so cannot confirm or deny the accuracy > of the colors. But the basic idea applies, given that each puzzle (a) has > the same pattern of ropes on all pieces and (b) has each permutation of > colors exactly once except for one permutation which appears twice. > > Solving the 10x10 is another kettle of fish, and I haven't tried it. I do > have a program that solves the 5x5 in about 45 seconds on a SparcStation II, > but I haven't looked into how much longer it would take on the 10x10. > > "Dale I. Newfield" writes: > > Could you explain what your algorithm was? > > Has anyone found a non-brute-force solution scheme? > > My solution was brute-force. I posted to cube-lovers (again, in April '92) > asking if anyone had found a more logical approach to the puzzle, but got no > affirmative responses. > > Dale's method is a little inefficient in the order in which it tries tiles. > Mine used the sequence: Dale's used: > > 1 3 5 7 9 1 2 4 7 11 > > 2 4 6 8 10 3 5 8 12 16 > > 11 12 13 14 15 6 9 13 17 20 > > 16 17 18 19 20 10 14 18 21 23 > > 21 22 23 24 25 15 19 22 24 25 > > The first three tiles in our two methods are equally constrained, but the > next seven in Dale's methods are constrained along 1-2-1-1-2-2-1 edges, > while mine are constrained along 2-1-2-1-2-1-2 edges. So I suspect my > search tree gets trimmed a bit more quickly. Another way in which the > search can be made more efficient is in finding the pieces to try in each > position. For each pair of colors that can appear along an edge, my program > precomputes a table listing all tiles that can match that pair of colors, > including how to rotate the tiles. > > -- Don. > > From anandrao@hk.super.net Wed Dec 15 20:27:19 1993 Return-Path: Received: from hk.super.net by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA18806; Wed, 15 Dec 93 20:27:19 EST Received: by hk.super.net id AA23068 (5.65c/IDA-1.4.4 for cube-lovers@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: cube-lovers@ai.mit.edu, don.woods@eng.sun.com, acw@riverside.scrc.symbolics.com In-Reply-To: Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Just because no non-brute-force method has been found, does not make this puzzle any less intersting. As we have been told that there is a solution, it is exciting to search for one, even by brute force methods. The real challenge is to find a brute-force method with sufficient insight to solve the problem within a reasonable time-frame. All the algorithms so far are exponential. We may never find a linear algorithm for this problem. The idea is to find one algorithm that can be used in actual practice. We can then bury this puzzle into the archives, for the next generation to pick up! > > (Thanks for the pointers to the Apr 92 discussion) > I agree with the concensus expressed in the archives that this puzzle is > inherently "not that great" because no non-brute-force method has been > found/seems to exist. > > -Dale > Is this the reason why Rubik has gone into hiding? I haven't seen any puzzle from him after this set of 4 released in 1990/1991. I tried to contact the Hong Kong office of Matchbox which gets Rubik's puzzles in China, but they have closed shop. Matchbox UK said that they have discontinued this line. If anyone has found another source for Rubik's puzzles, or discovered anyone else who has taken the responsibility of giving us sleepless nights, please let me know! From Don.Woods@eng.sun.com Wed Dec 15 20:39:18 1993 Return-Path: Received: from Sun.COM by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA19011; Wed, 15 Dec 93 20:39:18 EST Received: from Eng.Sun.COM (zigzag.Eng.Sun.COM) by Sun.COM (4.1/SMI-4.1) id AA12769; Wed, 15 Dec 93 17:39:17 PST Received: from colossal.Eng.Sun.COM by Eng.Sun.COM (4.1/SMI-4.1) id AA06793; Wed, 15 Dec 93 17:38:04 PST Received: by colossal.Eng.Sun.COM (5.0/SMI-SVR4) 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) Message-Id: <9312160139.AA26306@colossal.Eng.Sun.COM> To: cube-lovers@ai.mit.edu Subject: Re: Description of Tangle, Part 2 X-Sun-Charset: US-ASCII Content-Length: 897 > Is this the reason why Rubik has gone into hiding? I haven't seen any > puzzle from him after this set of 4 released in 1990/1991. Hm, didn't "Square-1" come out later than the Tangles? 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 Return-Path: Received: from charon.cwi.nl by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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/CWI-Amsterdam); Thu, 16 Dec 1993 03:03:19 +0100 Received: by boring.cwi.nl id AA10975 (4.1/2.10/CWI-Amsterdam); Thu, 16 Dec 93 03:03:19 +0100 Date: Thu, 16 Dec 93 03:03:19 +0100 From: Dik.Winter@cwi.nl Message-Id: <9312160203.AA10975.dik@boring.cwi.nl> To: anandrao@hk.super.net Subject: Re: Description of Tangle, Part 2 Cc: cube-lovers@ai.mit.edu My memory may be extremely faulty of course, but was there not more than one single solution for the 5x5? (Not unprecedented, I have one puzzle that promises a single solution but there are hundreds.) And, is there a solution for the 10x10? I seem to remember that there was (or I had) a convincing argument that such a thing did not exist. I should go through the lesser used parts of my memory one of these days. dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland home: bovenover 215, 1025 jn amsterdam, nederland; e-mail: dik@cwi.nl From dik@cwi.nl Wed Dec 15 21:40:59 1993 Return-Path: Received: from charon.cwi.nl by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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/CWI-Amsterdam); Thu, 16 Dec 1993 03:40:43 +0100 Received: by boring.cwi.nl id AA11474 (4.1/2.10/CWI-Amsterdam); Thu, 16 Dec 93 03:40:42 +0100 Date: Thu, 16 Dec 93 03:40:42 +0100 From: Dik.Winter@cwi.nl Message-Id: <9312160240.AA11474.dik@boring.cwi.nl> To: Don.Woods@eng.sun.com Subject: Re: Description of Tangle, Part 2 Cc: cube-lovers@ai.mit.edu > > Is this the reason why Rubik has gone into hiding? I haven't seen any > > puzzle from him after this set of 4 released in 1990/1991. > > Hm, didn't "Square-1" come out later than the Tangles? Square-1 is not by Rubik. But he came this year with two new puzzles (at least, they are in his name). Rubik's Maze and Rubik's Hat. In the first there are 6 connected cubes with a black/yellow pattern on them. The cubes can turn around each other fairly freely. The purpose is to get a 1x2x3 where there is a single black continuous line along the cubes. Not very difficult, interesting. Rubik's Hat is in the form of a hat with six rings on it. You can look trough it (and through the rings by implication). By turning rings you see more or less rabbits. The purpose is to see a rabbit in every position. I think the puzzle is based on light polarization, with different polarizations coming through the segments of the rings. dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland home: bovenover 215, 1025 jn amsterdam, nederland; e-mail: dik@cwi.nl From anandrao@hk.super.net Thu Dec 16 01:12:12 1993 Return-Path: Received: from hk.super.net by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA29768; Thu, 16 Dec 93 01:12:12 EST Received: by hk.super.net id AA08737 (5.65c/IDA-1.4.4 for cube-lovers@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: cube-lovers@ai.mit.edu In-Reply-To: <9312160203.AA10975.dik@boring.cwi.nl> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII 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 Return-Path: Received: from hk.super.net by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA29860; Thu, 16 Dec 93 01:19:38 EST Received: by hk.super.net id AA09218 (5.65c/IDA-1.4.4 for cube-lovers@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: cube-lovers@ai.mit.edu In-Reply-To: <9312160139.AA26306@colossal.Eng.Sun.COM> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII > > 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 Return-Path: Received: from hilly.harlequin.com by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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 Message-Id: <21332.199312161538@phaedrus.harlequin.com> To: Don.Woods@eng.sun.com Cc: cube-lovers@ai.mit.edu In-Reply-To: Don Woods's message of Wed, 15 Dec 93 17:39:20 PST <9312160139.AA26306@colossal.Eng.Sun.COM> Subject: Description of Tangle, Part 2 Date: Wed, 15 Dec 93 17:39:20 PST From: Don.Woods@eng.sun.com (Don Woods) X-Sun-Charset: US-ASCII Content-Length: 897 > Is this the reason why Rubik has gone into hiding? I haven't seen any > puzzle from him after this set of 4 released in 1990/1991. Hm, didn't "Square-1" come out later than the Tangles? Did Rubik have anything to do with Square-1? In any case, it's a great puzzle, and I recommend it to anyone on the list who hasn't tried it. While there's a group structure lurking here as usual, this is the only puzzle I've seen where the set of attainable positions is not a subgroup. This means lots of the usual ways of thinking about puzzles like this (e.g. conjugation) don't always work, which makes it quite challenging. Andy Latto andyl@harlequin.com From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Thu Dec 16 17:11:29 1993 Return-Path: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU> Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA21529; Thu, 16 Dec 93 17:11:29 EST Message-Id: <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 X-Acknowledge-To: 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 non-cubing programming examples. When I first started computer cubing, I was struck by the fact that (at least with my model of the cube), the computer code to implement a permutation operation looked exactly like the computer code to translate between various character codes. For example, I have had frequent occasion to translate between ASCII and EBCDIC (in both directions). The code to translate between the ASCII string X and the EBCDIC string Y is something like for i = 1 to n Y(i) = T(X(i)) where T is the translate table. To make this clear by an example, the ASCII code for the letter A is decimal 33 and the EBCDIC code for the letter A is decimal 193. Hence, the 33-rd position of T contains decimal 193, and the 193-rd position of T' contains 33. Beyond this simple little loop above, many (if not most) programming languages have a function (often called TRANSLATE or TRANSFORM) which does exactly the same thing. There are also hardware architectures which implement the TRANSLATE in hardware. For example, you might have something like Y = TRANSLATE(X,T) where X is the string to be translated and T is the translate table. X and T are clearly not interchangeable as input to the TRANSLATE function. However, (and repeating myself) I think there is an essential duality between X and T. For example, consider what would happen if you reversed the role of X and T as follows. Let X be the hexadecimal string 010201020301020403. Then, Y = TRANSLATE(X,' ABC') would yield the string ' A AB ACB'. Such a role reversal for the "permutation operator" and "permutation operatee" can be a very powerful programming technique. For example, I have used it to redistribute data, creating well-formatted print lines or well-formatted display screens (text mode) with one fell swoop (with only a single invocation of the TRANSLATE function). I am going to continue reading, but perhaps I could pose a question to Dan Hoey anyway: is reversing the role of X and T in the TRANSLATE function above essentially the same thing as switching between pre-multiplication and post-multiplication? = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU If you don't have time to do it right today, what makes you think you are going to have time to do it over again tomorrow? From anandrao@hk.super.net Thu Dec 16 20:14:55 1993 Return-Path: Received: from hk.super.net by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA00360; Thu, 16 Dec 93 20:14:55 EST Received: by hk.super.net id AA28316 (5.65c/IDA-1.4.4 for cube-lovers@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, cube-lovers@ai.mit.edu In-Reply-To: <9312160240.AA11474.dik@boring.cwi.nl> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII On Thu, 16 Dec 1993 Dik.Winter@cwi.nl wrote: > Square-1 is not by Rubik. But he came this year with two new puzzles (at > least, they are in his name). Rubik's Maze and Rubik's Hat. > > In the first there are 6 connected cubes with a black/yellow pattern on > them. The cubes can turn around each other fairly freely. The purpose > is to get a 1x2x3 where there is a single black continuous line along > the cubes. Not very difficult, interesting. > > Rubik's Hat is in the form of a hat with six rings on it. You can look > trough it (and through the rings by implication). By turning rings you > see more or less rabbits. The purpose is to see a rabbit in every position. > I think the puzzle is based on light polarization, with different > polarizations coming through the segments of the rings. > dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland > home: bovenover 215, 1025 jn amsterdam, nederland; e-mail: dik@cwi.nl Where can we get these puzzles from? Do you know of anyone who can take credit card orders and mail? From dik@cwi.nl Thu Dec 16 20:22:01 1993 Return-Path: Received: from charon.cwi.nl by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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/CWI-Amsterdam); Fri, 17 Dec 1993 02:21:55 +0100 Received: by boring.cwi.nl id AA15294 (4.1/2.10/CWI-Amsterdam); Fri, 17 Dec 93 02:21:54 +0100 Date: Fri, 17 Dec 93 02:21:54 +0100 From: Dik.Winter@cwi.nl Message-Id: <9312170121.AA15294.dik@boring.cwi.nl> To: anandrao@hk.super.net Subject: Re: Description of Tangle, Part 2 Cc: Don.Woods@eng.sun.com, cube-lovers@ai.mit.edu I would not know sources for Rubik's Maze and Rubik's Hat. They are on sale in the local shops here. I have looked, the distributer is no longer Matchbox but Parker, so that would imply availability in the US I think. dik -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland home: bovenover 215, 1025 jn amsterdam, nederland; e-mail: dik@cwi.nl From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Fri Dec 17 00:56:31 1993 Return-Path: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU> Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA11504; Fri, 17 Dec 93 00:56:31 EST Message-Id: <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 X-Acknowledge-To: 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 Mirror-Image- of-Edges-Flipped. The antipode is 15 qturns from Start. Also, I have a complete data base of equivalence classes in the edge group documenting the distance from Start for each configuration of the edges. It seems to me that given these two facts, some additional distances can be determined. For example, it is possible to determine the distance from any configuration to Mirror-Image-of-Edges-Flipped. Let Z be a sequence of operators that converts Start to Mirror-Image-of-Edges-Flipped, and let A be any configuration of the edges. Then apply Z' to A, look up the result in the data base of distances from Start, and that will be the distance from A to Mirror-Image-of-Edges-Flipped. The reason is quite simple. Let P be a sequence which takes Z'(A) to Start. Then, Z'PZ takes A to Mirror-Image-of-Edges-Flipped. This is a very nice use of conjugates. Another consequence of this result is the following: suppose you began with Mirror-Image-of-Edges-Flipped and performed a breadth-first exhaustive search. Start would be antipodal, and the number of nodes at each level of the tree would be identical to the existing tree which begins at Start. In addition, all of the above applies to Mirror-Image-of-Start and Edges-Flipped with respect to each other. They are mutually antipodal, and are 15 qturns apart. A tree built with either at the root would have exactly the same number of nodes at each level as the existing tree with Start at the root. Finally, the distance of any configuration from Mirror-Image-of-Start or Edges-Flipped can be determined. Let Y be a sequence of operators which converts Start to Mirror-Image-of-Start, and let X be a sequence of operators that converts Start to Edges-Flipped. Let A be any cube. Then, the distance of A from Mirror-Image-of-Start is the same as the distance of Y'(A) from Start, and the distance of A from Edges-Flipped is the same as the distance of X'(A) from Start. I have the sensation in describing this that the Edge group is square, with Start and Mirror-Image-of-Edges-Flipped 180 degrees apart, and Mirror-Image-of-Start and Edges-Flipped at the other two corners of the square. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU If you don't have time to do it right today, what makes you think you are going to have time to do it over again tomorrow? From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Fri Dec 17 11:24:53 1993 Return-Path: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU> Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA27504; Fri, 17 Dec 93 11:24:53 EST Message-Id: <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 X-Acknowledge-To: 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 Cube-Lovers for some time: How many positions are there in Rubik's >Cube? We know from Ideal that the number is somewhat over three >billion. Most cube lovers will tell you a number of about 43 >quintillion. But I really don't see why we should count twelve >distinct positions at one quarter-twist from solved--all twelve are >essentially the same position. So the question, suitably rephrased, is >of the number of positions that are distinct up to conjugacy in M, the >48-element symmetry group of the cube. I think this is an interesting >question, but I don't see any particularly easy way of answering it. >My best guess is that it involves a case-by-case analysis of the 98 >subgroups of M, or at least the 33 conjugacy classes of those >subgroups. In ``Symmetry and Local Maxima'', Jim Saxe and I examined >five of the classes, which we called M, C, AM, H, and T. > >Even finding the numbers for the pocket cube is a little tricky. If we >limit ourselves to symmetry in S, I believe the pocket cube has 2 >positions with a six-element symmetry group, 160 positions with a >three-element symmetry group, 3882 positions with a two-element >symmetry group, and 3670116 positions with a one-element symmetry >group, for 613062 positions distinct up to S-conjugacy. But the >numbers for M-conjugacy are still elusive; I am not even sure how to >deal with factoring out whole-cube moves in the analysis. I hope to >find time to write a program for it. > >I expanded my pocket cube program to deal with the corner group of >Rubik's cube. This group is 24 times as large as the group of the >pocket cube, having 3^7 * 8! = 88179840 elements. The number of >elements P(N) and local maxima L(N) at each (quarter-twist) distance N >from solved are given below. > > N P(N) L(N) > 0 1 0 > 1 12 0 > 2 114 0 > 3 924 0 > 4 6539 0 > 5 39528 0 > 6 199926 114 > 7 806136 600 > 8 2761740 17916 > 9 8656152 10200 > 10 22334112 35040 > 11 32420448 818112 > 12 18780864 9654240 > 13 2166720 2127264 > 14 6624 6624 > >The alert reader will notice that rows 10 through 14 contain values >exactly 24 times as large as those for the pocket cube. This is not >surprising, given that the groups are identical except for the position >of the entire assembly in space, and each generator of the corner cube >is identical to the inverse of the corresponding generator for the >opposite face except for the whole-cube position. Thus when solving a >corner-cube position at 10 qtw or more from solved, it can be solved as >a pocket cube, making the choice between opposite faces in such a way >that the whole-cube position comes out right with no extra moves. > I wish to propose an answer to Dan's question. I will propose an approximation then (hopefully) the exact answer. The approximation is simply 4.3*(10^19) / 1152, or about 3.7*(10^16). 1152=24*24*2, and is based on my version of Dan's M symmetry group. I remain convinced that my version of M is isomorphic to Dan's, but the subject deserves some more thought and discussion. But we can do better. We already know (under my version of M) how many equivalence classes there are for the corner group (namely, 77,802). But each of the equivalence classes for the corners can be rotated 24 ways with respect to the centers, so we have 77,802*24. We also already know (under my version of M) how many equivalence classes there are for the edge group (namely 851,625,008). But each of the equivalence classes for the edges can be rotated 24 ways with respect to the centers, so we have 851,625,008*24. Hence, we have (77,802*24) * (851,625,008*24) = 38,164,682,230,511,620 This figure is gratifyingly close to 3.7*(10^16), and I believe it is the correct answer to Dan's question. It is slightly larger than the approximation because some of the equivalence classes have fewer than 1152 elements, and consequently there are a few more equivalence classes than the approximation suggests. However, the alert reader should have noticed a problem. Why did I not divide by 2 to take into account the fact that odd edge permutations can only occur with odd corner permutations and vice versa? Actually, I did, but the division by 2 cancelled. The reason it canceled is slightly tricky. Also, remember that we are talking about equivalence classes, not specific cube configurations. Any equivalence class has both even and odd members, depending on how the members are rotated. Hence, any corner equivalence class can be matched up with any edge equivalence class, assuming the rotations are compatible. But you still have to worry about "dividing by 2", as follows. Let G be the number of states of the whole cube without M, namely the 4.3*(10^19) figure, and similarly let C be the number of states of the corners without M and let E be the number of states of the edges without M. Then, we have the trivial relation G = C * E / 2. Here, the division by 2 does properly reflect the odd/even parity of the corners vs. the edges. Let Gm = G / (24*24*2), Cm = C / (24*24*2), and Em = E / (24*24*2). Hence, G = Gm * (24*24*2), C = Cm * (24*24*2), and E = Em * (24*24*2). What I have available (approximately) is Cm and Em, and what I want is Gm. Hence, Gm = G / (24*24*2) Gm = (C * E / 2) / (24*24*2) Gm = ((Cm * (24*24*2)) * (Em * (24*24*2)) / 2) / (24*24*2) Gm = (Cm*24) * (Em*24) Therefore, I replace Cm by the real figure for the number of corner equivalence classes, replace Em by the real figure for the number of equivalence classes, and Gm becomes the real figure for the total states of the cube. The "division by 2" is in the formula, but it is invisible because of all the cancellations. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU If you don't have time to do it right today, what makes you think you are going to have time to do it over again tomorrow? From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Fri Dec 17 14:25:29 1993 Return-Path: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU> Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA08679; Fri, 17 Dec 93 14:25:29 EST Message-Id: <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 X-Acknowledge-To: Date: Fri, 17 Dec 1993 14:22:34 EST From: "Jerry Bryan" To: "Cube Lovers List" Subject: Re: Size of the Cube Group In-Reply-To: Message of 12/17/93 at 11:21:51 from , BRYAN%WVNVM.BITNET@mitvma.mit.edu On 12/17/93 at 11:21:51 Jerry Bryan said: >However, the alert reader should have noticed a problem. Why did I >not divide by 2 to take into account the fact that odd edge >permutations can only occur with odd corner permutations and vice >versa? Actually, I did, but the division by 2 cancelled. The reason >it canceled is slightly tricky. Also, remember that we are talking >about equivalence classes, not specific cube configurations. Any >equivalence class has both even and odd members, depending on how ^^^^^^^^^^^^^^^^^^^^^^^^^ >the members are rotated. Hence, any corner equivalence class can be >matched up with any edge equivalence class, assuming the rotations >are compatible. But you still have to worry about "dividing by 2", >as follows. It is pretty bad when you have to followup with corrections to your own posts. I hurried to complete the previous post before lunch, and just didn't think clearly enough -- till I had time to think *during* lunch. Let's try this again. A qturn of the whole cube (a 90 degree rotation of the whole cube) is odd. However, if you think of a qturn rotation of the whole cube as disjoint between edges and corners, a qturn rotation of the corners is even, and a qturn rotation of the edges is odd. Hence, for any equivalence class of the corners under M, either the whole equivalence class is even, or the whole equivalence class is odd. For any equivalence class of the edges under M, half of the equivalence class is even and half is odd. Thus, any equivalence class of the corners can occur with any equivalence class of the edges, but with only half the members of the edge equivalence class -- namely those with the same parity. I believe my calculations were correct, but a piece of the justification was not. I hope I am not still missing something. You do have to "divide by 2", and my calculations do indeed "divide by 2" as previously described, but the parity of edges vs. the parity of corners was incorrect in the previous post. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU If you don't have time to do it right today, what makes you think you are going to have time to do it over again tomorrow? From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Sat Dec 18 17:08:38 1993 Return-Path: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU> Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA00538; Sat, 18 Dec 93 17:08:38 EST Message-Id: <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 X-Acknowledge-To: 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) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU If you don't have time to do it right today, what makes you think you are going to have time to do it over again tomorrow? From xirion!jandr@relay.nl.net Mon Dec 20 06:35:54 1993 Return-Path: Received: from sun4nl.NL.net by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA00411; Mon, 20 Dec 93 06:35:54 EST Received: from xirion by sun4nl.NL.net via EUnet id AA05737 (5.65b/CWI-3.3); Mon, 20 Dec 1993 10:39:28 +0100 Received: by xirion.xirion.nl id AA03876 (5.61/UK-2.1); Mon, 20 Dec 93 10:38:37 +0100 From: Jan de Ruiter Date: Mon, 20 Dec 93 10:38:37 +0100 Message-Id: <3876.9312200938@xirion.xirion.nl> X-Organization: Xirion Unix Software & Consultancy bv Burgemeester Verderlaan 15 X 3454 PE De Meern The Netherlands X-Phone: +31 3406 61990 X-Fax: +31 3406 61981 To: cube-lovers@ai.mit.edu To: cube-lovers@ai.mit.edu Subject: Re: Search order of Tangle I saw the discussion of Dale and Don about the search order (fillpattern) for rubiks tangle come by, and wondered why they both missed an even better search order (the best?): Don: Dale: Jan: Equivalent to: 1 3 5 7 9 1 2 6 10 15 1 2 5 10 17 17 16 15 14 13 2 4 6 8 10 3 4 7 11 16 3 4 6 11 18 18 5 4 3 12 11 12 13 14 15 5 8 12 17 20 7 8 9 12 19 19 6 1 2 11 16 17 18 19 20 9 13 18 21 23 13 14 15 16 20 20 7 8 9 10 21 22 23 24 25 14 19 22 24 25 21 22 23 24 25 21 22 23 24 25 The number of constraints is illustrative: don: 0 1 1 2 1 2 1 2 1 2 1 2 2 2 2 1 2 2 2 2 1 2 2 2 2 dale: 0 1 1 2 1 1 2 2 1 1 2 2 2 1 1 2 2 2 2 2 2 2 2 2 2 jan: 0 1 1 2 1 2 1 2 2 1 2 2 1 2 2 2 1 2 2 2 1 2 2 2 2 I disliked the irregularity in both don and dales search orders, and in search for a more regular order, I found this one, which is better. It is readily extendible to the 10 by 10 tangle. - Jan D. de Ruiter From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Mon Dec 20 06:43:01 1993 Return-Path: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU> Received: from by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AB00195; Mon, 20 Dec 93 06:43:01 EST Message-Id: <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 X-Acknowledge-To: 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) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU If you don't have time to do it right today, what makes you think you are going to have time to do it over again tomorrow? From xirion!jandr@relay.nl.net Mon Dec 20 07:31:08 1993 Return-Path: Received: from sun4nl.NL.net by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA04598; Mon, 20 Dec 93 07:31:08 EST Received: from xirion by sun4nl.NL.net via EUnet id AA15711 (5.65b/CWI-3.3); Mon, 20 Dec 1993 13:31:05 +0100 Received: by xirion.xirion.nl id AA04506 (5.61/UK-2.1); Mon, 20 Dec 93 13:30:27 +0100 From: Jan de Ruiter Date: Mon, 20 Dec 93 13:30:27 +0100 Message-Id: <4506.9312201230@xirion.xirion.nl> X-Organization: Xirion Unix Software & Consultancy bv Burgemeester Verderlaan 15 X 3454 PE De Meern The Netherlands X-Phone: +31 3406 61990 X-Fax: +31 3406 61981 To: cube-lovers@life.ai.mit.edu To: cube-lovers@life.ai.mit.edu Subject: Re: Search order of Tangle I saw the discussion of Dale and Don about the search order (fillpattern) for rubiks tangle come by, and wondered why they both missed an even better search order (the best?): Don: Dale: Jan: Equivalent to: 1 3 5 7 9 1 2 6 10 15 1 2 5 10 17 17 16 15 14 13 2 4 6 8 10 3 4 7 11 16 3 4 6 11 18 18 5 4 3 12 11 12 13 14 15 5 8 12 17 20 7 8 9 12 19 19 6 1 2 11 16 17 18 19 20 9 13 18 21 23 13 14 15 16 20 20 7 8 9 10 21 22 23 24 25 14 19 22 24 25 21 22 23 24 25 21 22 23 24 25 The number of constraints is illustrative: don: 0 1 1 2 1 2 1 2 1 2 1 2 2 2 2 1 2 2 2 2 1 2 2 2 2 dale: 0 1 1 2 1 1 2 2 1 1 2 2 2 1 1 2 2 2 2 2 2 2 2 2 2 jan: 0 1 1 2 1 2 1 2 2 1 2 2 1 2 2 2 1 2 2 2 1 2 2 2 2 I disliked the irregularity in both don and dales search orders, and in search for a more regular order, I found this one, which is better. It is readily extendible to the 10 by 10 tangle. - Jan D. de Ruiter From @cannon.ecf.toronto.edu:malone@ecf.toronto.edu Mon Dec 20 14:17:33 1993 Return-Path: <@cannon.ecf.toronto.edu:malone@ecf.toronto.edu> Received: from cannon.ecf.toronto.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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: cube-lovers@ai.mit.edu Subject: Please remove ... Message-Id: <93Dec20.141719edt.7382@cannon.ecf.toronto.edu> Date: Mon, 20 Dec 1993 14:17:10 -0500 Please remove malone@ecf.toronto.edu from the cube-lovers list. Thanks Matt From Don.Woods@eng.sun.com Mon Dec 20 19:21:46 1993 Return-Path: Received: from Sun.COM by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA10335; Mon, 20 Dec 93 19:21:46 EST Received: from Eng.Sun.COM (zigzag.Eng.Sun.COM) by Sun.COM (4.1/SMI-4.1) id AA04625; Mon, 20 Dec 93 16:21:45 PST Received: from colossal.Eng.Sun.COM by Eng.Sun.COM (4.1/SMI-4.1) id AA17679; Mon, 20 Dec 93 16:20:26 PST Received: by colossal.Eng.Sun.COM (5.0/SMI-SVR4) 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) Message-Id: <9312210021.AA10356@colossal.Eng.Sun.COM> To: cube-lovers@ai.mit.edu Subject: Re: Search order of Tangle Cc: jandr@xirion.nl X-Sun-Charset: US-ASCII Content-Length: 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 Return-Path: Received: from hk.super.net by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA13155; Mon, 20 Dec 93 20:16:04 EST Received: by hk.super.net id AA23740 (5.65c/IDA-1.4.4 for cube-lovers@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: cube-lovers@ai.mit.edu In-Reply-To: <3876.9312200938@xirion.xirion.nl> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Your concept is theoretically extendable to the 10*10 tangle, but even with this optimisation the puzzle would take a long time to solve. How long do you take for the 5*5 Tangle on your computer? On Mon, 20 Dec 1993, Jan de Ruiter wrote: > To: cube-lovers@ai.mit.edu > Subject: Re: Search order of Tangle > > I saw the discussion of Dale and Don about the search order > (fillpattern) for rubiks tangle come by, and wondered why they both > missed an even better search order (the best?): > > Don: Dale: Jan: Equivalent to: > 1 3 5 7 9 1 2 6 10 15 1 2 5 10 17 17 16 15 14 13 > 2 4 6 8 10 3 4 7 11 16 3 4 6 11 18 18 5 4 3 12 > 11 12 13 14 15 5 8 12 17 20 7 8 9 12 19 19 6 1 2 11 > 16 17 18 19 20 9 13 18 21 23 13 14 15 16 20 20 7 8 9 10 > 21 22 23 24 25 14 19 22 24 25 21 22 23 24 25 21 22 23 24 25 > > The number of constraints is illustrative: > don: 0 1 1 2 1 2 1 2 1 2 1 2 2 2 2 1 2 2 2 2 1 2 2 2 2 > dale: 0 1 1 2 1 1 2 2 1 1 2 2 2 1 1 2 2 2 2 2 2 2 2 2 2 > jan: 0 1 1 2 1 2 1 2 2 1 2 2 1 2 2 2 1 2 2 2 1 2 2 2 2 > > I disliked the irregularity in both don and dales search orders, and > in search for a more regular order, I found this one, which is better. > It is readily extendible to the 10 by 10 tangle. > > - Jan D. de Ruiter From pbeck@pica.army.mil Tue Dec 21 00:24:43 1993 Return-Path: Received: from COR6.PICA.ARMY.MIL by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA22931; Tue, 21 Dec 93 00:24:43 EST Date: Mon, 20 Dec 93 8:33:55 EST From: Peter Beck (BATDD) To: Cube-Lovers@ai.mit.edu Cc: pbeck@pica.army.mil Subject: test Message-Id: <9312200833.aa09624@COR6.PICA.ARMY.MIL> i am having trouble posting. please excuse this message From xirion!jandr@relay.nl.net Tue Dec 21 02:45:28 1993 Return-Path: Received: from sun4nl.NL.net by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA24967; Tue, 21 Dec 93 02:45:28 EST Received: from xirion by sun4nl.NL.net via EUnet id AA28145 (5.65b/CWI-3.3); Tue, 21 Dec 1993 08:45:07 +0100 Received: by xirion.xirion.nl id AA00997 (5.61/UK-2.1); Tue, 21 Dec 93 08:43:59 +0100 From: Jan de Ruiter Date: Tue, 21 Dec 93 08:43:59 +0100 Message-Id: <997.9312210743@xirion.xirion.nl> X-Organization: Xirion Unix Software & Consultancy bv Burgemeester Verderlaan 15 X 3454 PE De Meern The Netherlands X-Phone: +31 3406 61990 X-Fax: +31 3406 61981 To: anandrao@hk.super.net, cube-lovers@ai.mit.edu Subject: Re: Rubiks tangle To: anandrao@hk.super.net Cc: cube-lovers@ai.mit.edu >Your concept is theoretically extendable to the 10*10 tangle, but even >with this optimisation the puzzle would take a long time to solve. How >long do you take for the 5*5 Tangle on your computer? I am sorry to say I haven't implemented the search yet. The 5x5 is solved, so that isn't that interesting anymore; the 10x10 has such a huge search space, that it will need a very efficient algorithm and/or clever representation. I just haven't decided on the representation yet. I did decide on the search order though. - Jan D. de Ruiter From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Tue Dec 21 09:27:36 1993 Return-Path: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU> Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA01929; Tue, 21 Dec 93 09:27:36 EST Message-Id: <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 X-Acknowledge-To: Date: Tue, 21 Dec 1993 08:54:12 EST From: "Jerry Bryan" To: "Cube Lovers List" Subject: 9 Qturn Sequence for All-Edges-Flipped in the Edge Group RUB DRB LDB (the spaces are for readability only). Remember that this is for the "edges without corners and without centers" case. Hence, the edges are all flipped and are all properly configured with respect to each other, but they are not flipped "in place" with respect to a fixed coordinate system of centers. They are rotated with respect a fixed coordinate system of centers. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU If you don't have time to do it right today, what makes you think you are going to have time to do it over again tomorrow? From pbeck@pica.army.mil Tue Dec 21 18:32:39 1993 Return-Path: Received: from COR6.PICA.ARMY.MIL by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA14683; Tue, 21 Dec 93 18:32:39 EST Date: Tue, 21 Dec 93 7:49:23 EST From: Peter Beck (BATDD) To: Cube-Lovers@ai.mit.edu Cc: pbeck@pica.army.mil Subject: puzzle party Message-Id: <9312210749.aa08453@COR6.PICA.ARMY.MIL> ROBERT HOLBROOK 11837 LINDEN CHAPEL ROAD CLARKSVILLE, MD 21029 410-531-6135 IS Planning a puzzle party for FEB 19,20 1994 at his home. IF YOU ARE INTERESTED PLEASE CONTACT BOB directly. Clarksville is 1/2 way between DC and Baltimore. VENUE is low keyed with trading, buying, selling and TALKING. THE FUTURE IS PUZZLING, BUT CUBING IS FOREVER !!! From pbeck@pica.army.mil Tue Dec 21 19:05:28 1993 Return-Path: Received: from COR6.PICA.ARMY.MIL by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA15488; Tue, 21 Dec 93 19:05:28 EST Date: Tue, 21 Dec 93 7:51:39 EST From: Peter Beck (BATDD) To: cube-lovers@ai.mit.edu Cc: pbeck@pica.army.mil Subject: puzzle party Message-Id: <9312210751.aa09222@COR6.PICA.ARMY.MIL> ROBERT HOLBROOK 11837 LINDEN CHAPEL ROAD CLARKSVILLE, MD 21029 410-531-6135 IS Planning a puzzle party for FEB 19,20 1994 at his home. IF YOU ARE INTERESTED PLEASE CONTACT BOB directly. Clarksville is 1/2 way between DC and Baltimore. VENUE is low keyed with trading, buying, selling and TALKING. THE FUTURE IS PUZZLING, BUT CUBING IS FOREVER !!! From hoey@aic.nrl.navy.mil Wed Dec 22 13:58:45 1993 Received: from Sun0.AIC.NRL.Navy.Mil by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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/SMI-4.0) id AA20803; Wed, 22 Dec 93 13:58:43 EST Return-Path: 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 Message-Id: <9312221858.AA08479@sun1.aic.nrl.navy.mil> To: Cube-Lovers@ai.mit.edu Subject: The 4^3 and 3^4 Rubik puzzles Organization: Naval Research Laboratory, Washington, DC [ Cube-Lovers, There has recently been a discussion on Usenet group rec.puzzles about some cube topics. There were a few pieces of new information, such as that you can now get Ishi's 5^3 cubes in a lot of places (I got mine in Learningsmith's) for about $35. Here's a message I sent that's relevant to some Cube-Lovers topics. By the way, I'm still working through Jerry Bryan's articles on his brute-force program and his approach to symmetry. I hope to get a reply out soon. ] eric@gsb002.cs.ualberta.ca (Holleman Eric) wrote: > By the way, I found the Revenge somewhat easier than the Cube, and I > don't think that it was because of my familiarity with the earlier > puzzle. x87bennett@gw.wmich.edu (Joe) wrote: > From my experience, if you can solve a Rubik's Revenge, you can > solve the Cube very easily. Once you get each of the middle 2 cubes > on each edge to match, and all 4 center cubes on each face to match, > it works exactly like a Rubik's cube. and alan@saturn.cs.swin.oz.au (Alan Christiansen) wrote: > I have both. I solved both. The 4x4x4 is a superset of the 3x3x3. > ie by fixing all the face centres and then pairing all edges you are > left with a 3x3x3 cube, except that when you have solved this 3x3x3 > there may be a single pair of edges flipped. This is impossible > on a real 3x3x3. Fixing this requires a middle layer to be rotated > 1/4 revolution and then all the bits put back. > I cant see how it can be [any] easier than a 3x3x3. I, too, found the 3^3 easier than the 4^3. But I can imagine ways in which a solver could find the 4^3 easier. Let us first consider a 4^3 with the faces fixed, the edges together, and the correct simulated edge flip parity. I would solve this as if it were a 3^3, and a lot of people do. But another solver might find it easier to take advantage of the extra moves that are not possible on a 3^3. To take a concrete example, it could be that the solver has a hard time with flipping edges by pairs, as is needed to solve the 3^3. On the 4^3 you can flip one edge at a time. So the solver would find the 4^3 position easier than the corresponding position on a 3^3. If the solver finds this so much easier that it overcomes the difficulty of putting the faces and edges together--or in fact puts the faces and edges together in the course of solving the corners and the edge positions--then the 4^3 could be easier. It depends on the solution procedure. alan@saturn.cs.swin.oz.au (Alan Christiansen) continues: > ANyway the real reason I am writing this is that I have written > a cube simulator. > It can simulate 3x3x3 4x4x4 5x5x5 .... cubes. > I am working on 4x4x4x4 cube simulation. This is interesting, as there is more than one way to model the four-dimensional cube problem. Consider the 3^4 cube. It has eight hyper-faces, each in the shape of a cube. One model of this puzzle is that you could turn any face of any hyper-face as if it were a face of a 3^3 Rubik's cube. In a second model, you cannot move part of a hyper-face, but can turn each hyper-face as if it were a solid cube in space. A third model allows either kind of move. These models are different from each other. The second model permits the face centers of the hyper-faces to move around, whereas in the first model only edges and corners move. In the first model, odd permutations of corners are possible, which is not true in the second model. Of course, the third model is the closure of the first two. According to Hofstatder's column reprinted in _Metamagical_Themas_, there is an unpublished 1982 manuscript by H J Kamack and T R Keane entitled ``The Rubik Tesseract''. They calculated the size of the group of the 3^4 puzzle, but I don't know which model was used. Alan Christiansen indicates he has gone directly to the 4^4 puzzle. I don't know which model he plans, or if the models become more similar with the extra possibilities inherent in the larger cube. I don't even know whether he plans to figure out how big the groups are or whether they are identical. Dan Hoey Hoey@AIC.NRL.Navy.Mil From xirion!jandr@relay.nl.net Thu Dec 23 02:48:30 1993 Return-Path: Received: from sun4nl.NL.net by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA08367; Thu, 23 Dec 93 02:48:30 EST Received: from xirion by sun4nl.NL.net via EUnet id AA04747 (5.65b/CWI-3.3); Thu, 23 Dec 1993 08:48:27 +0100 Received: by xirion.xirion.nl id AA04326 (5.61/UK-2.1); Thu, 23 Dec 93 08:47:44 +0100 From: Jan de Ruiter Date: Thu, 23 Dec 93 08:47:44 +0100 Message-Id: <4326.9312230747@xirion.xirion.nl> X-Organization: Xirion Unix Software & Consultancy bv Burgemeester Verderlaan 15 X 3454 PE De Meern The Netherlands X-Phone: +31 3406 61990 X-Fax: +31 3406 61981 To: cube-lovers@ai.mit.edu Subject: Re: Rubiks tangle To: anandrao@hk.super.net Cc: cube-lovers@ai.mit.edu >Your concept is theoretically extendable to the 10*10 tangle, but even >with this optimisation the puzzle would take a long time to solve. How >long do you take for the 5*5 Tangle on your computer? Your question prompted me to actually write the program, and to squeeze as much efficiency from the program as I could. You wrote on december the 14th, your program took about 20 minutes on a 486DX2-66, Don Woods writes on the same date, that his program takes 45 seconds on a SparcStation II, And now I am proud to present my timing: trrrrr (drum roll) 7 seconds on a Compacq Deskpro 386/33. (and still only brute force!) Now I am ready to try the 10x10. Some thoughts in the mean time: If the algorithm treats the duplicate pieces just as ordinary pieces, i.e. as different, this will cause the program to find 4 solutions for the 5x5 where only 2 exist (by exchanging the duplicate pieces). This factor of 2 may not be dramatical, but if the same algorithm tries the 10x10, then for every 1 solution that exists, the program will find (5!)^4 x (4!)^20 identical versions (combinations of duplicate exchanges). My program views duplicate pieces as one, which may be placed several times. So for some position X a piece with duplicates will only be tried once. Don Woods writes: >Regarding solving the Tangle, I forgot one other minor optimisation: >When my program is picking a corner piece other than the first, it >requires that the piece "number" be less than or equal to that of the >first corner. I.e., it refuses to search for solutions that are >rotations of other solutions. My program prevents finding rotations of solutions, by excluding the rotations of just one piece. The list of possibilities to try on any position includes this one piece just once, and every other piece four times. You can choose any piece for this, except the duplicated one. Regrettably this approach works only for the 5x5: the 10x10 will probably have to use Don Woods method. - Jan D. de Ruiter From anandrao@hk.super.net Thu Dec 23 04:29:34 1993 Return-Path: Received: from hk.super.net by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA10668; Thu, 23 Dec 93 04:29:34 EST Received: by hk.super.net id AA16480 (5.65c/IDA-1.4.4 for cube-lovers@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: cube-lovers@ai.mit.edu In-Reply-To: <4326.9312230747@xirion.xirion.nl> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII 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 Return-Path: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU> Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA20316; Sat, 25 Dec 93 22:48:40 EST Message-Id: <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 X-Acknowledge-To: 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) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU If you don't have time to do it right today, what makes you think you are going to have time to do it over again tomorrow? From @mail.uunet.ca:mark.longridge@canrem.com Mon Dec 27 02:33:51 1993 Return-Path: <@mail.uunet.ca:mark.longridge@canrem.com> Received: from mail.uunet.ca (uunet.ca) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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/SMI-4.1) id AA18888; Mon, 27 Dec 93 02:00:05 EST Received: by canrem.com (PCB-UUCP 1.1f) id 18F5BD; Mon, 27 Dec 93 01:58:20 -0400 To: cube-lovers@life.ai.mit.edu Reply-To: CRSO.Cube@canrem.com Sender: CRSO.Cube@canrem.com Subject: Local Maxima Revisited From: mark.longridge@canrem.com (Mark Longridge) Message-Id: <60.656.5834.0C18F5BD@canrem.com> Date: Mon, 27 Dec 1993 00:58:00 -0500 Organization: CRS Online (Toronto, Ontario) Some thoughts on Local Maxima ----------------------------- I have verified that the position I call "6 H order 2 type 2" is a local maximum. p175 6 H order 2 type 2 T2 B2 L2 T2 D2 L2 F2 T2 (8) A rare example of a pattern with symmetry level 2, perhaps even the only one of this type, and also the most symmetric of the 6 H patterns. Nothing new here, as this was noted formerly by David Singmaster in one of the Cubic Circulars and by Jim Saxe & Dan Hoey in the archives. Somewhat more interesting is the conclusion that the pattern 4 H order 2, or H's on the F,R,B,L faces (oriented like the letter H) is also a local maximum, at least in the square's group. p160 4 H order 2 Type 2 B2 D2 (L2 R2 F2) ^2 T2 F2 (10) From the archives: > We include a description of 71 local maxima, which we believe > to be all of the local maxima that can be proven using known > techniques other than exhaustive search. Oh well, I used an exhaustive search. p160 is 10 moves long in the htw metric, and each of the moves ( T2, D2, F2, B2, L2, R2 ) all bring one to a position requiring nine 180 degree twists, thusly.... 4 H + T2 = L2, R2, F2 B2, T2, L2 R2, B2, F2 (9) 4 H + D2 = L2, R2, F2 B2, D2, L2 R2, B2, F2 (9) 4 H + F2 = B2, D2, L2 R2, F2, L2 R2, F2, T2 (9) 4 H + B2 = F2, D2, L2 R2, F2, L2 R2, F2, T2 (9) 4 H + L2 = R2, D2, L2 F2, B2, L2 F2, B2, T2 (9) 4 H + R2 = L2, D2, L2 F2, B2, L2, F2, B2, T2 (9) ---------------------------------------------------------------------- I did discover an interesting property of the "Cube in a cube" pattern I didn't notice before. p7a Cube in a cube U2 F2 R2 U3 L2 D1 (B1 R3) ^3 B1 D3 L2 U3 (15) Let's say you are entertaining some cube guests at a cube party and the topic is (cube) patterns. Your guests are impressed with the efficiency of the well-memorized process. You would like to go on to the next pattern but you don't quite remember how the inverse goes. No problem! Rotate the whole cube so TOP becomes BACK then BACK becomes DOWN, and finally FRONT becomes RIGHT. Simply repeat the process p7a and your reputation as a cube expert is saved. ;-> This same idea works for the 6 X order 3 pattern as well. And now for an unsymmetric local maximum!! (Just kidding) -> Mark <- From @mail.uunet.ca:mark.longridge@canrem.com Mon Dec 27 12:31:47 1993 Return-Path: <@mail.uunet.ca:mark.longridge@canrem.com> Received: from mail.uunet.ca (uunet.ca) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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/SMI-4.1) id AA26935; Mon, 27 Dec 93 12:30:32 EST Received: by canrem.com (PCB-UUCP 1.1f) id 18F603; Mon, 27 Dec 93 12:23:56 -0400 To: cube-lovers@life.ai.mit.edu Reply-To: CRSO.Cube@canrem.com Sender: CRSO.Cube@canrem.com Subject: Cube Rotations From: mark.longridge@canrem.com (Mark Longridge) Message-Id: <60.659.5834.0C18F603@canrem.com> In-Reply-To: <19166.199312271606@phaedrus.harlequin.com> Date: Mon, 27 Dec 1993 11:12:00 -0500 Organization: CRS Online (Toronto, Ontario) -> goes. No problem! Rotate the whole cube so TOP becomes BACK then -> BACK becomes DOWN, and finally FRONT becomes RIGHT. Simply repeat -> the process p7a and your reputation as a cube expert is saved. ;-> -> -> The faces FRONT and BACK are opposite each other. After your -> rotation, they become RIGHT and DOWN, which are not opposite each -> other. This would certainly establish a reputation for you, but if -> you did it with my cube, it might not be the sort of reputation you -> wanted to have :-) -> Andy Latto -> andyl@harlequin.com Perhaps my description of the rotations was unclear... Rotate the entire cube so that TOP -> DOWN FRONT -> LEFT Ok, before I meant rotate the cube in space in 3 steps so that the TOP face becomes BACK, then the face that is the BACK at this point becomes DOWN, and the face that is the FRONT at this point becomes the RIGHT. The reason I used this type of description is because there are multiple ways for the TOP to become the DOWN face.... TOP becomes BACK becomes DOWN and TOP becomes RIGHT becomes DOWN and TOP becomes LEFT becomes DOWN etc... Perhaps it is better to use the form old FACE A -> new FACE A old FACE B -> new FACE B Where the faces A & B are adjacent. Mark Email: mark.longridge@canrem.com ....wait a second, I don't think faces A & B have to be adjacent for the rotation to be unambiguous. Any 2 faces should do! From hoey@aic.nrl.navy.mil Mon Dec 27 17:52:28 1993 Received: from Sun0.AIC.NRL.Navy.Mil by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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/SMI-4.0) id AA08057; Mon, 27 Dec 93 17:52:17 EST Return-Path: 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 Message-Id: <9312272252.AA22049@sun30.aic.nrl.navy.mil> To: Cube-Lovers@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 order-24 row > vector. The rotation and reflection operators are also order-24 row > vectors, again with each cell simply containing a number from 1 to > 24. That is the most usual way of doing it, but it's important to specify what you represent by those vectors. When I do it, I number the corner facelet locations from 1 to 24, and these locations retain their numbers through manipulations of the cube. I use a vector A to specify a position in which the facelet whose home location is i has been moved to location A(i), for each i. I use a vector P to specify the transformation that moves the facelet in location i to location A(i), for each i. I'll assume you're doing the same, though you could, for instance, be representing the inverse of the operators, or the locations from which the facelets originate. Note that a position is represented by the same vector that represents the transformation that takes SOLVED to that position. > Well, if P is a rotation operator, you could perform a rotation > two ways. I guess one is pre-multiplication and one is > post-multiplication. > 1) For i = 1 to 24 B(i) = A(P(i)) I would write this as B = P A, and say that A is premultiplied by P, or equivalently that P is postmultiplied by A. In a general group, we could have B = P A where the multiplication is not considered to be the composition of permutations. But it turns out we can restrict our attention to permutation groups without loss of generality. For instance, when we are dealing with the supergroup, we can consider the orientation of a face center to be a permutation of the corners of the face center. > 2) For i = 1 to 24 B(i) = P(A(i)) Here B = A P, A is postmultiplied by P, and P is premultiplied by A. (Note that the operator or position name appears in the reverse order from the prefix format. Algebraists sometimes avoid this by writing (i)B = ((i)A)P. I kid you not.) > (As an aside, this illustrates the question I raised in my previous > post about "which is the operator and which is the thing being > operated on?" Is P operating on A, or is A operating on P?) Well, the answer is ``both''. I agree it's easy to get confused, which is why proofs are a good idea. > Finally, if Q is a reflection (actually, if Q1 is the identity and > Q2 is the reflection), then we have > For j = 1 to 24 for k = 1 to 24 for m = 1 to 2 > for i = 1 to 24 Bj,k,m(i) = Qm(Pj(A(Qm(Pk(i))))) > I believe this loop calculates Dan Hoey's M. On the the theory that proofs are a good idea, let's see what this loop calculates. I'm going to put brackets around the subscripts. Then I'll substitute "R" for "Q", because I use Q for the set of quarter-turns of faces. Furthermore, I'll use "C" instead of "P", because the P[j] are just the elements of C, the group of cube rotations. So you are computing B[j,k,m] = C[k] R[m] A C[j] R[m] (1) for j in {1,...,24}, k in {1,...,24}, and m in {1,2}. Now every member of M (the group of cube rotations and reflections) has a unique representation as M[n] = C[k] R[m]. Let us define Cind() and Rind() as the functions for which M[n]=C[Cind(M[n])] R[Rind(M[n])]. So we can write (1) as B[j,k,m] = M[n] A M'[n] (M[n] C[j] R[Rind(M[n])]) Note that (M[n] C[j] R[Rind(M(n))] must be an element of C. So B is a set of elements of the form M[n] A M'[n] C[o]. To see that we have all such elements, first observe that (M[n]' C[o] R[Rind(M[n])]') is an element of C, say C[j]. So equation (1) includes: C[Cind(M[n])] R[Rind(M[n])] A C[j] R[Rind(M[n])] = M[n] A (M[n]' C[o] R[Rind(M[n])]') R[Rind(M[n])] = M[n] A M'[n] C[o]. Thus the set of all B[j,k,m] is the set of all M[n] A M'[n] C[o]. Or in English, that's the set of all M-conjugates of A, operated on by all whole-cube rotations. > In my data base, I store the minimum of Bj,k,m over j = 1 to 24, > k = 1 to 24, and m = 1 to 2. I tend to call the minimum of Bj,k,m a > canonical form. I am not sure if that is the best terminology. The > minimal element is not any simpler than any other. It is just that > I need a function to choose an element from a set, and picking the > minimal element seems very natural. Any other element would do as > well, provided I could always be sure of picking the same element. It's pretty common terminology. You might be slightly better off calling it a ``representative element,'' as that connotes that the element is ordinary except in that it represents the equivalence class (like representatives in the U.S. Congress). > Also, my criterion for equivalence is slightly > different (but isomorphic, I think) than the one described by > Dan Hoey. Suppose A and B are two cubes. > Rather than mapping A to B or B to A in M, I map both A and B > to their respective canonical forms. A and B are equivalent if > their respective canonical forms are equal. This is straightforward once we show that M-conjugacy is an equivalence relation, and B[j,k,m] is an equivalence class. If A ~ Representative[A] = Representative[B] ~ B, then by transitivity A ~ B. Conversely, if A ~ B, then Class[A] = Class[B], and therefore Representative[A] = Representative[B]. This shows that the criteria are equivalent. > Now, as to the centers. I still sometimes have a certain doubt > about the centers. They are fixed, so how can you reduce the > problem (i.e., increase the size of the equivalence classes) > by both rotating the cube and rotating the colors (by both pre- > and post-multiplication)? What you have done is to increase the size of the whole cube problem by a factor of 24, by dealing with all rotations of the cube, and the equivalence classes expand by the same factor, from 48 to 1152. This has allowed you to calculate something like M-conjugacy classes for cube problems that lack face centers. But the size of the equivalence classes doesn't shrink the problem for cubes that have face centers. You could have just calculated M-conjugates and got the same answer. > I am not sure if this answers Dan's question about my model > with centers added. It's clear now. I hadn't realized you were rotating the cube in space when the face centers were present. I expected that to be a wasted effort. But I am impressed by the way it allows you to shrink the database by storing positions together that differ only by whole-cube moves of the face centers. I think it should be possible to shrink the database without the effort, though. In your message of Thu, 16 Dec 1993 15:36:58 EST, on the ``Duality of Operators and Operatees'': > I have mentioned several times my discomfort about "an operator" as > opposed to "the thing being operated on" when it comes to groups. I > am never quite sure just which of the two it is that people are > talking about, even (or especially) when I am listening to myself > talk. It is hard to keep it straight. Sometimes we all get it wrong. The best way to avoid errors, as far as possible, is to avoid such language and talk about group multiplication. But then we have to explain what is going on with the cube, so we get caught into talking about operators again. It's a discomfort that must be endured. > The code to translate between the ASCII string X and > the EBCDIC string Y is something like > for i = 1 to n Y(i) = T(X(i)) > where T is the translate table. Yes, or Y = X T as above. > I am going to continue reading, but perhaps I could pose a question to > Dan Hoey anyway: is reversing the role of X and T in the TRANSLATE > function above essentially the same thing as switching between > pre-multiplication and post-multiplication? Yes. Dan Hoey Hoey@AIC.NRL.Navy.Mil From dik@cwi.nl Mon Dec 27 18:43:04 1993 Return-Path: Received: from charon.cwi.nl by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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/CWI-Amsterdam); Tue, 28 Dec 1993 00:43:03 +0100 Received: by boring.cwi.nl id AA25571 (4.1/2.10/CWI-Amsterdam); Tue, 28 Dec 93 00:43:02 +0100 Date: Tue, 28 Dec 93 00:43:02 +0100 From: Dik.Winter@cwi.nl Message-Id: <9312272343.AA25571.dik@boring.cwi.nl> To: Cube-Lovers@life.ai.mit.edu Subject: Re: Group theory basics (Re: Symmetry) One additional remark: > > Well, if P is a rotation operator, you could perform a rotation > > two ways. I guess one is pre-multiplication and one is > > post-multiplication. > > 1) For i = 1 to 24 B(i) = A(P(i)) > I would write this as B = P A, and say that A is premultiplied by P, > or equivalently that P is postmultiplied by A. There is quite a bit of confusion about this. When permutation groups are considered; even text-books do not agree. When A and P are permutations you can find both that P A means: apply P first, A next, but also: apply A first, P next. (The first meaning comes from the pure group theorists, the second meaning more from the algebra inclined.) Sorry to confuse the issue, but when I read such texts I have always to think hard to get at the intended meaning. I think the functional notation is much clearer and leads to less confusion. Of course, doing notations for cube rotations the group theorists notation is applied, but when doing abstract operations... From hoey@aic.nrl.navy.mil Tue Dec 28 14:17:02 1993 Received: from Sun0.AIC.NRL.Navy.Mil by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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/SMI-4.0) id AA19858; Tue, 28 Dec 93 14:16:50 EST Return-Path: 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 Message-Id: <9312281916.AA25640@sun30.aic.nrl.navy.mil> To: Cube-Lovers@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 brute-force search has shown that d(I,PE)=15, and he notes that conjugation by E shows us that d(P,E)=15 as well. He concludes: > I have the sensation in describing this that the Edge group is > square, with Start and Mirror-Image-of-Edges-Flipped 180 degrees > apart, and Mirror-Image-of-Start and Edges-Flipped at the other > two corners of the square. Well, it's not quite a square, since d(I,P)=12 and d(I,E)=9, according to Jerry's message of Wed, 8 Dec 1993 10:02:15 EST. Conjugation will similarly show that d(E,PE)=12 and d(P,PE)=9. So we are dealing with a rectangle. The sides of the rectangle are 9 and 12, and the diagonal is 15: a most fortuitous set of numbers, in that we can actually embed such a rectangle in the Euclidean plane! We can map the positions of the edge group to 4-tuples of distances. For any position X, let f(X)=(d(I,X), d(E,X), d(P,X), d(PE,X)). If f(X)=(a,b,c,d), then conjugation shows us that f(X E)=(b,a,d,c), f(X P)=(c,d,a,b), and F(X PE)=(d,c,b,a). So the set of quadruples has the symmetries of the rectangle. We know f(I)=(0,9,12,15). What is more, the earlier results on symmetry show us that I is at a local maximum distance from E, P, and PE. So, letting I1 be the unique (up to M-conjugacy) position adjacent to I, we have F(I1)=(1,8,11,14). (This destroys Euclidean embeddability.) An analogous result holds for the unique neighbor of each corner of the rectangle. We also have Jerry's results of Wed, 8 Dec 1993 22:41:28 EST and 23:16:50 EST that H (the 6-H pattern) and HE=H E are at distances 8 and 13 from start, respectively. Since H is an M-conjugate of P H, this gives us f(H)=(8,13,8,13). [Note: there are two distinct M-conjugates of H, call them H and Hbar. This distinction is important when we compose permutations: H H = I, but H Hbar = P. So we have to be careful when conflating M-conjugates.] We can by symmetry find f(H1)=(7,12,7,12) for H's unique neighbor H1. What quadruples are possible? If f(X)=(a,b,c,d), and X is not one of the eight corners and neighbors, we have max(2,9-b,12-c,15-d) <= a <= min(14,9+b,12+c) with constraints on b, c, and d from symmetry. A quick hack tells me there are 7836 such quadruples. I wonder how many of them are realized? If it's fairly few, I would like to see a diagram of quadruples, with lines between those quadruples that represent adjacent positions (adjacent quadruples differ by at most one in each coordinate). Maybe with the number of positions for each quadruple, too. I have an idea that such a diagram might tell us something about the problem, or at least look pretty. Dan Hoey Hoey@AIC.NRL.Navy.Mil From hoey@aic.nrl.navy.mil Tue Dec 28 18:42:22 1993 Received: from Sun0.AIC.NRL.Navy.Mil by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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/SMI-4.0) id AA06105; Tue, 28 Dec 93 18:40:52 EST Return-Path: 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 Message-Id: <9312282340.AA25691@sun30.aic.nrl.navy.mil> To: Cube-Lovers@ai.mit.edu Subject: Re: Cube Rotations Cc: CRSO.Cube@canrem.com mark.longridge@canrem.com (Mark Longridge) writes: > Perhaps my description of the rotations was unclear... Yes. > ...Perhaps it is better to use the form > old FACE A -> new FACE A > old FACE B -> new FACE B > Where the faces A & B are adjacent. That will serve to uniquely identify a rotation, but it's somewhat verbose. Worse, it does not suffice to uniquely identify a symmetry from the group of rotations and reflections, M. I find it's far more informative to identify a rotation or reflection as a permutation of the faces, in cycle format. There are only ten kinds: Even rotations: I=Identity (1), (FRT)(BLD)=120-degree rotation (8), (FB)(RL)=180-degree orthogonal rotation (3). Odd rotations: (FRBL)=90-degree rotation (6), (FB)(TR)(DL)=180-degree diagonal rotation (6). Even reflections: (FR)(BL)=diagonal reflection (6), (FRBL)(TD)=90-degree glide reflection (6), Odd reflections: (FB)=orthogonal reflection (3), (FRTBLD)=60-degree glide reflection (8), (FB)(RL)(TD)=central reflection (1). In case it isn't clear, the cycle notation for (e.g.) a 120-degree rotation (FTL)(BDR) means that the F, T, L, B, D, and R faces move to the T, L, F, D, R, and B, locations, respectively. The only thing I'm afraid of with this notation is that someone will think I'm describing a magic-cube process rather than a whole-cube move. So when you say Top->Down, Front->Left, I would say (TD)(FL)(BR) for the 180-degree diagonal rotation, to distinguish it from (TD)(FLBR) the 90-degree glide reflection. > ....wait a second, I don't think faces A & B have to be > adjacent for the rotation to be unambiguous. Any 2 faces > should do! No, you're back to your original bogosity. Knowing the destinations of two opposite faces doesn't give you any more information than knowing the destination of one (unless you go breaking the axles). Dan Hoey Hoey@AIC.NRL.Navy.Mil From hoey@aic.nrl.navy.mil Wed Dec 29 17:43:47 1993 Return-Path: Received: from Sun0.AIC.NRL.Navy.Mil by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA12501; Wed, 29 Dec 93 17:43:47 EST Received: by Sun0.AIC.NRL.Navy.Mil (4.1/SMI-4.0) 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) Message-Id: <9312292243.AA05575@Sun0.AIC.NRL.Navy.Mil> To: Cube-Lovers@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 4-tuples of distances f(X)=(d(I,X), d(E,X), d(P,X), d(PE,X)). I wrote: ?? If f(X)=(a,b,c,d), then conjugation shows us that ?? ?? f(X E)=(b,a,d,c), f(X P)=(c,d,a,b), and F(X PE)=(d,c,b,a). ?? ?? So the set of quadruples has the symmetries of the rectangle. ?? The first sentence is incorrect, though the argument as a whole is reparable. First, I'll do what I should have done yesterday, and define the distance function d(X,Y). We want the minimum length process Z such that X Z = Y. But premultiplying both sides by X', we have Z = X' Y. So I define d(X,Y)=Length(X' Y). From the properties of the length function (Length(I)=0, Length(X)=Length(X'), and Length(X Y)<=Length(X) + Length(Y)) we can conclude that d(X,Y) is a metric. Suppose f(X)=(a,b,c,d). I claim f(E X)=(b,a,d,c), f(P X)=(c,d,a,b), and F(PE X)=(d,c,b,a). Proof: To show f(E X)=(b,a,d,c), first observe that I=I', E=E', and P E = E P. d(I,E X) = Length(I' E X) = Length(E' X) = d(E,X), so d(E,E X) = d(I, E E X) = d(I,X); d(P,E X) = Length(P' E X) = Length((PE)' X) = d(PE,X) so d(PE,E X)=d(P,E E X)=d(P,X). To show that f(P X)=(c,d,a,b), exchange P and E in the above argument. To show that f(PE X)=(d,c,b,a), use both occurrences of the argument. QED. So the idea of yesterday's message is correct, but I had X E, X P, and X PE instead of E X, P X, and PE X, respectively. I would show you a counterexample to yesterday's formulation, but it turns out there is none. I claim that f(X,E)=f(E,X), f(X,P)=f(P,X), and f(X,PE)=f(PE,X). Proof: Recall that E commutes with every element of the Rubik cube group, so f(X E)=f(E X). It turns out that ``up to M-conjugacy'', P commutes with every element of the edge group as well. For P performs a mirror-reflection of the edges, and so can be regarded as an element of M acting on the edge group. So P' X P = Xbar is an M-conjugate of X, and X P = P Xbar. Since Length(X) agrees on M-conjugates, so does d(X,Y), and so f(X), so f(X P)=f(P Xbar) = f(P X). Finally, f(X PE) = f(X P E) = f(E X P) = f(P E X) = f(PE X). QED. So it turns out it that the statement about f was true. But I am no less embarrassed for asserting it, for I had no reason to think it would be true. It's only rescued by the surprising commutativity of the Pons Asinorum. Finally, I would like to note something that I nearly included in yesterday's message, but yanked when I decided it was false: f(X')=f(X). Now I'll prove it: Proof: For W among {I,E,P,PE}, we have X W = W Xbar, for Xbar an M-conjugate of X. So d(X,W)=Length(X'W)=Length(W'Xbar')=Length(W'X')=d(W,X'). QED. Dan Hoey Hoey@AIC.NRL.Navy.Mil From Don.Woods@eng.sun.com Sun Jan 2 20:10:16 1994 Return-Path: Received: from Sun.COM by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA12259; Sun, 2 Jan 94 20:10:16 EST Received: from Eng.Sun.COM (zigzag.Eng.Sun.COM) by Sun.COM (4.1/SMI-4.1) id AA11819; Sun, 2 Jan 94 17:10:12 PST Received: from colossal.Eng.Sun.COM by Eng.Sun.COM (4.1/SMI-4.1) id AA03228; Sun, 2 Jan 94 17:08:39 PST Received: by colossal.Eng.Sun.COM (5.0/SMI-SVR4) 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) Message-Id: <9401030110.AA13229@colossal.Eng.Sun.COM> To: cube-lovers@ai.mit.edu Subject: 10x10 Tangle Content-Length: 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 Cube-Lovers re which tiles are duplicated in the four 5x5s; I have no way of checking whether that info was correct. Has anyone out there ever heard definitely that someone has found a solution to the 10x10? Is it possible that the makers of Tangle (Matchbox, using Rubik's name under license) merely claimed that such a solution exists, without actually verifying it? (Seems pretty sleazy if so, but then, having Tangles 2-4 be merely color permutations of #1 is pretty weak in the first place.) -- Don. From dik@cwi.nl Sun Jan 2 21:52:23 1994 Return-Path: Received: from charon.cwi.nl by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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/CWI-Amsterdam); Mon, 3 Jan 1994 03:52:21 +0100 Received: by boring.cwi.nl id AA07139 (4.1/2.10/CWI-Amsterdam); Mon, 3 Jan 94 03:52:20 +0100 Date: Mon, 3 Jan 94 03:52:20 +0100 From: Dik.Winter@cwi.nl Message-Id: <9401030252.AA07139.dik@boring.cwi.nl> To: Don.Woods@eng.sun.com, cube-lovers@ai.mit.edu Subject: Re: 10x10 Tangle > Has anyone out there ever heard definitely that someone has found a > solution to the 10x10? As I wrote before, I have embedded in my memory that there is an easy argument that the 10x10 is *not* solvable. I do not know whether I found it myself (and ever did mail it to other people) or whether I found it somewhere on the net; it is a long time ago. When I find the time I will do a check. (I know very sure that I have had a program running at that time but that I abandoned the search because it would be fruitless.) > Is it possible that the makers of Tangle (Matchbox, > using Rubik's name under license) merely claimed that such a solution > exists, without actually verifying it? Yes, very probable. You should never trust the number of solutions the manufacturers give. Sometimes it is much more, in this case it is less. An actual example is a puzzle that consists of of nine rings (eh, this is from memory, I do not have access to the puzzle at this time). Five rings contain digits; three rings contain operators; one ring contains equal signs. All in four positions around the rings. The idea is to create correct sums (like 5 + 1 - 4 + 1 = 3) on all four positions of the rings. The claim was that there was only a single solution. Actually there are many. If there is interest I can hunt down the rings and describe them in more detail. (An interesting detail is that my father was the first to find the puzzle; he had correct solutions like: 1 + 3 : 2 + 1 = 3. He was a physicist. The accomanying leaflet did not give details about operator priorities. Hence it actually makes two puzzles; one with regards to priorities, the other just going left to right.) > (Seems pretty sleazy if so, > but then, having Tangles 2-4 be merely color permutations of #1 is > pretty weak in the first place.) Indeed, the mass manufacturers are sleazy. Cheers. I will mail when I find back the argument disallowing 10x10. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924098 home: bovenover 215, 1025 jn amsterdam, nederland; e-mail: dik@cwi.nl From xirion!jandr@relay.nl.net Mon Jan 3 02:29:28 1994 Return-Path: Received: from sun4nl.NL.net by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA21848; Mon, 3 Jan 94 02:29:28 EST Received: from xirion by sun4nl.NL.net via EUnet id AA18380 (5.65b/CWI-3.3); Mon, 3 Jan 1994 08:29:22 +0100 Received: by xirion.xirion.nl id AA22110 (5.61/UK-2.1); Mon, 3 Jan 94 08:29:37 +0100 From: Jan de Ruiter Date: Mon, 3 Jan 94 08:29:37 +0100 Message-Id: <22110.9401030729@xirion.xirion.nl> X-Organization: Xirion Unix Software & Consultancy bv Burgemeester Verderlaan 15 X 3454 PE De Meern The Netherlands X-Phone: +31 3406 61990 X-Fax: +31 3406 61981 To: cube-lovers@ai.mit.edu To: Don.Woods@eng.sun.com, cube-lovers@ai.mit.edu Subject: RE: 10x10 Tangle >Hm. Well, I split up the 10x10 Tangle exhaustive search and ran it on >several machines over Christmas break, getting the 90 days of compute >time done in about 10. > >And turned up no solutions. My program is a bit faster, but as I have less machines at my disposal and I started a bit later, my programs are still running. Up until now they did not produce a solution either. I am starting to get worried. >There could of course be a bug in my program, but the same code with >minor changes finds the same solutions as others have found for the 5x5. The same goes for me. > I did doublecheck that the 100 tiles matched the info >posted to Cube-Lovers re which tiles are duplicated in the four 5x5s; >I have no way of checking whether that info was correct. I have the puzzles myself, and checked the info in the message from Dale I Newfield (15 Dec 1993), which quotes the archives. I can assure you: those are indeed the duplicate pieces. > >Has anyone out there ever heard definitely that someone has found a >solution to the 10x10? Is it possible that the makers of Tangle (Matchbox, >using Rubik's name under license) merely claimed that such a solution >exists, without actually verifying it? (Seems pretty sleazy if so, >but then, having Tangles 2-4 be merely color permutations of #1 is >pretty weak in the first place.) > I thought about that too, but considered that the choice for precisely those four duplicate pieces could be dictated by the desire to have a solution for the 10x10. >I also tried adding some extra tiles for the 10x10, and it began finding >solutions okay. Question: did you add pieces at random, or did you add more duplicate pieces? In the latter case you may have found the duplications that should have been made to get a solvable 10x10. That in turn would show there could not exist an argument disallowing 10x10 (as claimed by Dik Winter), unless that argument is based on the particular colouring of the four duplicated pieces... -- Jan From Don.Woods@eng.sun.com Mon Jan 3 05:45:40 1994 Return-Path: Received: from Sun.COM by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA23952; Mon, 3 Jan 94 05:45:40 EST Received: from Eng.Sun.COM (zigzag.Eng.Sun.COM) by Sun.COM (4.1/SMI-4.1) id AA10025; Mon, 3 Jan 94 02:45:24 PST Received: from colossal.Eng.Sun.COM by Eng.Sun.COM (4.1/SMI-4.1) id AA09005; Mon, 3 Jan 94 02:43:43 PST Received: by colossal.Eng.Sun.COM (5.0/SMI-SVR4) 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) Message-Id: <9401031045.AA14295@colossal.Eng.Sun.COM> To: cube-lovers@ai.mit.edu, jandr@xirion.nl Subject: Re: 10x10 Tangle Content-Length: 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 Return-Path: Received: from sun4nl.NL.net by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA26254; Mon, 3 Jan 94 08:17:10 EST Received: from xirion by sun4nl.NL.net via EUnet id AA06106 (5.65b/CWI-3.3); Mon, 3 Jan 1994 14:17:08 +0100 Received: by xirion.xirion.nl id AA22788 (5.61/UK-2.1); Mon, 3 Jan 94 14:16:52 +0100 From: Jan de Ruiter Date: Mon, 3 Jan 94 14:16:52 +0100 Message-Id: <22788.9401031316@xirion.xirion.nl> X-Organization: Xirion Unix Software & Consultancy bv Burgemeester Verderlaan 15 X 3454 PE De Meern The Netherlands X-Phone: +31 3406 61990 X-Fax: +31 3406 61981 To: cube-lovers@ai.mit.edu To: Don.Woods@Eng.Sun.COM Subject: Re: 10x10 Tangle >Incidentally, I would be interested in seeing your program. (And am >willing to send you mine.) I'm always willing to learn something about >how to make combinatorial searches more efficient. Will be sent separately (and yes, I would like to see yours too!) >It's a shame, really. I'll bet that it would be possible to come up with >four Tangles that (a) really are different instead of being simple color >permutations of each other, (b) each have a unique solution (not counting >rotations) instead of two, and (c) can be combined to form a 10x10 that has >a unique solution. Well, strike the "unique" from (c) and I'd make the bet; >but with the "unique" I certainly wouldn't bet against it! When you limit yourself to 4 ropes with 4 colours, you always get 24 pieces, and when you want to build a puzzle of 25 pieces, you will have to duplicate one, which causes (a). Using 5 colours instead, creates a set of 120 pieces, from which you could probably pick 25 pieces (without duplication) which would satisfy both (a) and (b), and probably 4 such sets could be found to satisfy (c) as well, but such a puzzle would be less attractive, because the choice of 25 from 120 is somewhat arbitrary, and a puzzler would probably be more inclined to use all 120 pieces... Of course it is all a matter of taste. -- Jan From Don.Woods@eng.sun.com Mon Jan 3 17:11:06 1994 Return-Path: Received: from Sun.COM by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA21618; Mon, 3 Jan 94 17:11:06 EST Received: from Eng.Sun.COM (zigzag.Eng.Sun.COM) by Sun.COM (4.1/SMI-4.1) id AA09153; Mon, 3 Jan 94 14:10:57 PST Received: from colossal.Eng.Sun.COM by Eng.Sun.COM (4.1/SMI-4.1) id AA19324; Mon, 3 Jan 94 14:09:21 PST Received: by colossal.Eng.Sun.COM (5.0/SMI-SVR4) 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) Message-Id: <9401032211.AA15405@colossal.Eng.Sun.COM> To: cube-lovers@ai.mit.edu, jandr@xirion.nl Subject: Re: 10x10 Tangle X-Sun-Charset: US-ASCII Content-Length: 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 Return-Path: Received: from Sun.COM by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA28614; Mon, 3 Jan 94 19:50:59 EST Received: from Eng.Sun.COM (zigzag.Eng.Sun.COM) by Sun.COM (4.1/SMI-4.1) id AA24247; Mon, 3 Jan 94 16:50:55 PST Received: from colossal.Eng.Sun.COM by Eng.Sun.COM (4.1/SMI-4.1) id AA22695; Mon, 3 Jan 94 16:49:20 PST Received: by colossal.Eng.Sun.COM (5.0/SMI-SVR4) 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) Message-Id: <9401040051.AA16535@colossal.Eng.Sun.COM> To: cube-lovers@ai.mit.edu Subject: tangled cube X-Sun-Charset: US-ASCII Content-Length: 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 red-blue, the adjacent tile on that face touching the same edge will also touch the edge with red- blue. I see no obvious reason why the solution should have this property, but it does. Solution below. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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 Return-Path: Received: from abernathy.tenet.edu (Kay-Abernathy.tenet.edu) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA28829; Mon, 3 Jan 94 20:04:38 EST Received: by abernathy.tenet.edu id AA16334 (5.65c/IDA-1.4.4 for CUBE-LOVERS@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: CUBE-LOVERS@life.ai.mit.edu Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII WOULD YOU BE WILLING TO SHARE WHAT YOU KNOW WITH PRE-COLLEGE STUDENTS AND TEACHERS BY ELECTRONIC MAIL? Recent estimates indicate that there are now more than 300,000 classroom teachers from primary, middle, and secondary schools who hold accounts on the Internet. This makes a very special kind of learning available to them: one which directly involves subject matter experts communicating with students and teachers about their specialties, via electronic mail. With support from the Texas Center for Educational Technology, we (at the University of Texas at Austin) have piloted and are now expanding an Internet-based service (the "Electronic Emissary") that brings together pre-college students, their teachers, and subject matter experts (SMEs) electronically, helping them to create telecomputing exchanges centered around the students' learning in the SMEs' disciplines. For example, * A class studying South America could learn about recent global environmental research results from a scientist who studies rainforest deforestation in Brazil. * A class studying geometry might "talk" electronically with Euclid, who is actually a mathematics professor. * A class studying the future of education might converse with an emerging technologies specialist from California's Silicon Valley. * A class studying American History might electronically interview Harry Truman, who is really a curator with the National Archives. * A class exploring the rapidly-changing governmental structures that are emerging in what was once the Soviet Union might correspond with a group of graduate political science students at a university in the CIS. * Or, a class reading _Huckleberry Finn_ might correspond with an African-American studies scholar about the repercussions resulting from the enacting of the Emancipation Proclamation. In successive phases of the project, increasing numbers of SMEs or SME groups are needed to correspond regularly (approximately 4 times per week) with primary, middle school, or secondary students and their teachers (1 SME or expert group per class, study group, or "special student"). Each electronic exchange will begin with approximately 2 weeks of project planning via electronic mail between the SMEs and the teachers. Communications with students will begin on mutually convenient dates, and will continue for previously-arranged periods of time, usually between 2 and 10 weeks. Subject matter expert volunteers are sought in all disciplines, but there is immediate need for SMEs with expertise in: ~ gravity and satellite motion ~ heat transfer ~ Hitler's rise to power during World War II ~ the Indian Wars (1870's & 1880's) ~ 20th century fragmentation due to weapons of war, especially the atom bomb ~ Maya Angelou (and other women in literature) ~ Native American literature, specifically Cherokee ~ George Orwell's _Animal Farm_ & Russian revolutions ~ personal finance ~ geometry ==> If you would like to find out more about ==> participating in this project, please send ==> electronic mail to Judi Harris, jbharris@tenet.edu. ==> Please include your name, institution, and areas of ==> expertise. ==> PLEASE RESPOND ASAP; teacher-SME pairs in the ==> specific areas requested above will be formed on ==> 1/12/93. From xirion!jandr@relay.nl.net Tue Jan 4 02:30:20 1994 Return-Path: Received: from sun4nl.NL.net by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA10206; Tue, 4 Jan 94 02:30:20 EST Received: from xirion by sun4nl.NL.net via EUnet id AA24141 (5.65b/CWI-3.3); Tue, 4 Jan 1994 08:30:19 +0100 Received: by xirion.xirion.nl id AA23801 (5.61/UK-2.1); Tue, 4 Jan 94 08:30:56 +0100 From: Jan de Ruiter Date: Tue, 4 Jan 94 08:30:56 +0100 Message-Id: <23801.9401040730@xirion.xirion.nl> X-Organization: Xirion Unix Software & Consultancy bv Burgemeester Verderlaan 15 X 3454 PE De Meern The Netherlands X-Phone: +31 3406 61990 X-Fax: +31 3406 61981 To: cube-lovers@ai.mit.edu To: Don.Woods@eng.sun.com, cube-lovers@ai.mit.edu Subject: RE tangle cube >Well, a pleasant surprise! It _is_ possible to take a set of 24 distinct >tiles from any Rubik's Tangle, and use them to tile the surface of a 2x2x2 >cube such that all touching ropes match. And the solution is unique! > >I'll include the solution below, after some blank lines to avoid spoiling it >for anyone who wants to try solving the puzzle without seeing the answer... You may have noticed it yourself, but the solution you promised was missing from your message. But I take your word for it that you found it, because (sorry to spoil your scoop) a solution for the tangle-cube as you described was published before in CFF (Cubism For Fun) the periodical of the NKC (Nederlandse Kubus Club = Dutch Cubist Club). Contrary to what the name suggests members are not solely interested in cubes. Membership to that club is open to anyone interested in puzzles like these and highly recommended! The periodical CFF is published in English, and appears three or four times a year. Further information can be obtained via gm@phys.uva.nl -- Jan. From mouse@collatz.mcrcim.mcgill.edu Tue Jan 4 07:25:33 1994 Return-Path: Received: from Collatz.McRCIM.McGill.EDU ([132.206.78.1]) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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 Message-Id: <199401041220.HAA11684@Collatz.McRCIM.McGill.EDU> To: jandr@xirion.nl Cc: cube-lovers@ai.mit.edu >> Well, a pleasant surprise! It _is_ possible to [tile a 2x2x2 cube >> with distinct Rubik's Tangle pieces, uniquely] >> I'll include the solution below, after some blank lines to avoid >> spoiling it for anyone who wants to try solving the puzzle without >> seeing the answer... > You may have noticed it yourself, but the solution you promised was > missing from your message. Not from the copy I got - though the lines in question weren't blank; they each contained a dot. Perhaps someone's mailer isn't doing the hidden-dot algorithm correctly? der Mouse mouse@collatz.mcrcim.mcgill.edu From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Tue Jan 4 11:12:37 1994 Return-Path: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU> Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA23149; Tue, 4 Jan 94 11:12:37 EST Message-Id: <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 X-Acknowledge-To: 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 quarter-turns, by definition you cannot rotate the cube in space as a single operation. Yet, a simple quarter-turn sequence such as RL' does rotate the 2x2x2 cube because it is faceless. Is Start of the 2x2x2 operated on by RL' solved? If so, you can argue that the 2x2x2 has 24 Starts. Most people would not. They would argue that there is only one Start, and that 2x2x2 cubes that differed only by a rotation are equivalent. 3. Combining #1 and #2, I *think* that most people would argue that Start and Pons Asinorum on the Edge Cube are not equivalent, but that simple rotations of the 2x2x2 are equivalent. If I am correct about "most people", why? Is a rotation symmetry intrinsically a stronger or weaker symmetry than a reflection symmetry? 4. When I was first posting my results about the Edge Group, and particularly when it first began to sink in what the four equivalence classes with only 24 elements really were, I had a moment of panic. Since Start and Pons Asinorum differ only by a simple reflection, why had not my version of M-conjugation declared them to be equivalent? (I speak of "my version of M-conjugation", but the question is no different if you look at Dan Hoey's original M-conjugation). I think I know the answer, but I will leave the problem as an exercise for the student. Furthermore, I think the answer to #4 is really the same as the answer to #3. 5. What is a reflection, really? Here is an exercise to illustrate the question. Take two identically colored and oriented 3x3x3 cubes. On one, perform F and on the other perform F'. Examine the two cubes, plus their images in a mirror. Why are there four distinct cubes rather than only two? At one level of abstraction, the answer is simple. Of the four, one is not reflected, one is pre-reflected, one is post-reflected, and one is both pre- and post-reflected. Is this a sufficient answer, or is there something deeper? At this point, I can't help but note Martin Gardner's famous mirror question in Scientific American many years ago: why does a mirror reverse left and right but not up and down? 6. I found Dan Hoey's postings about the four special states of the Edge Group to be delightful. Some of the results were based on a computer search of the group, for example the fact that f(I)=(0,9,12,15) could only reasonably be determined from a computer search. However, the thought occurred to me that most of Dan's results were independent of the computer search, and I was curious precisely which results would stand without the search? For example, if we identified the group as being rectangular, would we be led to saying which of the four special states were diagonally opposed without the computer search? Without the search, I might be tempted to say that Start and Pons Asinorum were diagonally opposed. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU If you don't have time to do it right today, what makes you think you are going to have time to do it over again tomorrow? From mouse@collatz.mcrcim.mcgill.edu Tue Jan 4 13:48:23 1994 Return-Path: Received: from Collatz.McRCIM.McGill.EDU by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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 cube-lovers@ai.mit.edu; Tue, 4 Jan 1994 13:48:03 -0500 Date: Tue, 4 Jan 1994 13:48:03 -0500 From: der Mouse Message-Id: <199401041848.NAA12863@Collatz.McRCIM.McGill.EDU> To: cube-lovers@ai.mit.edu Subject: Re: Which is the Real Start? > The net is so wonderful about answering questions, here are a few > more: I am hardly more than a dilettante of the Cube, but I can perhaps offer a few suggestions, since this seems to me to be largely psychology and 3-D geometry rather than Cube group theory. > 1. Take a standard 3x3x3 Rubik's cube, and remove the corner and > center labels to make an Edges Cube. [...] Scramble the cube. > Give it to a cubemeister to solve. How will the cubemeister know > if the cube is solved? In other words, how will the cubemeister > distinguish Start from Pons Asinorum? > One answer is that the cubemeister cannot. [...] Another answer > is that either one is Start -- that there are two Starts. Obviously, it is correct to state that without some a priori knowledge of the cube's coloring, the cubemeister cannot tell. As for whether you call them both Start: > However, if you like this answer, and if you identify the identity > with Start, you are in the disquieting situation of having a group > with two distinct identities (grin!). Not at all. All you have to do is consider the group elements to be equivalence classes under not only whole-cube rotation but also reflection. If you take a(n ordinary) cube and rotate the whole thing a quarter-turn, the result is not essentially different from the original - most programs and virtually all humans would consider them "the same". Taking the stand that Start and P.A. are the same on the Edge Cube means only that on the Edge Cube you consider a single group element to consist of not only a position and all those reachable by whole-cube rotations, but also those reachable by reflections as well. The group-theoretic identity is then neither Start nor Pons Asinorum, but rather the equivalence class containing both those (and 46 other elements). > 2. [...] Just what is the 2x2x2 cube? Or more correctly, how do you > know when it is solved? When you have achieved any of the 24 elements of the class that we lump together as Start. > With any size of cube, if you restrict yourself to quarter-turns, > by definition you cannot rotate the cube in space as a single > operation. I'd argue the 1x1x1 breaks this statement :-) What's more, it's not clear what "quarter-turn" includes: it usually doesn't include slice turns on the 3-Cube, but on the 4-Cube and higher, they must of necessity be included. > Yet, a simple quarter-turn sequence such as RL' does rotate the > 2x2x2 cube because it is faceless. Is Start of the 2x2x2 operated > on by RL' solved? Yes, I would say so. I would hope most people would. > If so, you can argue that the 2x2x2 has 24 Starts. Most people > would not. They would argue that there is only one Start, and > that 2x2x2 cubes that differed only by a rotation are equivalent. Right. I would say that RL' produces a cube that is precisely as solved as that produced by RR' is - that on the 2x2x2, R and L are in some sense the same thing. My position would be that there is only one Start on the 2-Cube, and it is an equivalence class with 24 members. > 3. Combining #1 and #2, I *think* that most people would argue that > Start and Pons Asinorum on the Edge Cube are not equivalent, but > that simple rotations of the 2x2x2 are equivalent. If I am > correct about "most people", why? I would say that Start and Pons Asinorum on the Edge Cube can be looked at as mathematically equivalent (though they need not be, if you choose) but are not intuitively equivalent. Physical objects generally cannot be turned into reflected versions of themselves; they normally *can* be turned into rotated versions of themselves. Thus, rotations "feel" equivalent, but reflections don't. > 4. [...] Since Start and Pons Asinorum differ only by a simple > reflection, why had not my version of M-conjugation declared them > to be equivalent? I'm too lazy to answer this; I no longer have the messages describing exactly what your M-conjugation operation is online. Presumably, you implemented some intuitively-reasonable operation, and it produced identical results for rotations but not reflections. > 5. What is a reflection, really? Ouch. Mathematically, this is easy enough: given a center of reflection P in Cartesian 3-space, one computes the reflection of a point p as P+(P-p). All the things one thinks of as reflections can be represented as this operation compounded with rotation and/or translation. > Here is an exercise to illustrate the question. Take two > identically colored and oriented 3x3x3 cubes. On one, perform F > and on the other perform F'. Examine the two cubes, plus their > images in a mirror. Why are there four distinct cubes rather than > only two? There are certainly four cubes - or at least four cube images. For there to be only two distinct cubes, one would have to identify some of them with one another. However, the only operations (on the cube as a whole) that will allow identifying two of them are (1) reflection and (2) recoloring. If your mathematical treatment considers reflections or recolorings to be equivalent, then mathematically, there are only two distinct cubes. Neither of these operations "feels" trivial, though, so the four cubes all "feel" distinct. > At this point, I can't help but note Martin Gardner's famous > mirror question in Scientific American many years ago: why does a > mirror reverse left and right but not up and down? (rot13 for those who would rather think about this for a while.) Nf abgrq va jungrire vg jnf V ernq gung dhrfgvba va, vg qbrfa'g - vg erirefrf sebag-gb-onpx (jurer "sebag" naq "onpx" ner qrsvarq va grezf bs gur fhesnpr qbvat gur ersyrpgvat). Jul guvf *nccrnef* gb nzbhag gb erirefvat yrsg naq evtug vf n zber vagrerfgvat dhrfgvba, naq vg nzbhagf gb nfxvat jung xvaqf bs fcngvny bcrengvbaf jr cresbez jvgubhg abgvpvat (pbafvqrevat gurz abbcf). Va gur pnfr bs n ersyrpgvba bs n crefba, gur bcrengvba jr'er cresbezvat jvgubhg abgvpvat vf gung bs znccvat crefba-vzntr bagb frys-vzntr ol ebgngvba, fb nf gb (1) znc urnq bagb urnq naq srrg bagb srrg naq (2) znc obql-sebag gb obql-sebag naq obql-onpx gb obql-onpx. Ersyrpgvba, pbzcbhaqrq jvgu guvf ebgngvba, *qbrf* erirefr yrsg-gb-evtug. > 6. [...] I'm not qualified to comment. der Mouse mouse@collatz.mcrcim.mcgill.edu From hoey@aic.nrl.navy.mil Tue Jan 4 19:05:27 1994 Return-Path: Received: from Sun0.AIC.NRL.Navy.Mil by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA17502; Tue, 4 Jan 94 19:05:27 EST Received: by Sun0.AIC.NRL.Navy.Mil (4.1/SMI-4.0) 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) Message-Id: <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 M-conjugacy classes of Rubik's cube positions. Having calculated the number of conjugacy classes of the corner and edge groups separately, his idea was to combine these to calculate the number of conjugacy classes of the entire group. Eventually, he withdrew the calculation, after realizing that he had not found enough information to determine the answer. This article is about how we might calculate the answer from separate searches of the edge and corner groups. My first idea is to formalize the concept of symmetry in the conjugacy classes that Jerry used in his searches. Recall that the conjugacy class of a position X is defined to be the set of all positions m'Xmc, where m is an element of M, the 48-element symmetry group of the cube, and c is an element of C, the 24-element subgroup of M consisting of rigid rotations of the cube in space. The reason some positions have more symmetry than others is that for some positions, there are nontrivial elements m and c such that m'Xmc=X. The way in which this arises can be formalized as a kind of symmetry group of X. For an edge-group position X, let CSymm(X) be the set of all f in M such that X'f'Xf is an element of C. First, I'll claim that CSymm(X) is a subgroup of M [see proof 1, below]. Second, I note that CSymm(X) is the set of all m in M such that there exists c in C with m'Xmc=X [proof: m'Xmc=X iff X'm'Xm=c']. Third, I'll claim that if m'Xmc=Y, then CSymm(X) and CSymm(Y) are conjugate subgroups of M [proof 2]. So when Jerry says that a position X has order-N symmetry, he is saying that CSymm(X) has 1152/N elements. But the identity of CSymm(X) has more information than just its size, and I believe this information is crucial if we are to combine symmetry groups. It looks to me as if it would be sufficient to record the conjugacy class of CSymm(X), and there are only 33 possibilities. Now the usual symmetry group of X, Symm(X), is defined to be the group consisting of all f in M such that X'f'Xf=I [or, equivalently, Xf=fX. Symm(X) is the largest group such that X is Symm(X)-symmetric, in the sense of the Symmetry and Local Maxima article]. The first step in combining the corner and edge sets is to calculate the symmetry groups of the rotations of a position X, AllSymms(X)={Symm(Xc) : c in C}. This corresponds to computing the symmetry groups of the edges-and- centers group from the symmetry groups of the edges group. I suspect there is a way of computing this from CSymm(X), but I do not know it. I am not even sure that AllSymms(X) is determined by CSymm(X). One useful experiment would be to calculate CSymm(X) and AllSymms(Xc) for all elements of the corner group and see what the correspondence is. Barring an ability to calculate AllSymms(X) from CSymm(X), we could calculate AllSymms(X) directly. This involves a great number of calculations, though: 24 symmetry group calculations for each element of the edge group. My first thought was to try to split the problem up further, to deal with the group of permutations separately from the group of orientations. But I abandoned this when I realized there is a problem that shows up when we try this with the corner group. The permutation of the corners that takes each cubie to its antipode is clearly M-symmetric, and no matter how we decide to measure orientation, there is a way to perform this permutation leaving the cubies in their `home' orientation. But there is no way to compose the two together in an M-symmetric way. I suspect the same problem arises in the edge group. But there may be some help from the edge search available in calcu- lating AllSymms(X). For take a position Y in the edges-and-centers group; Y is also a rotated position in the edges group, so Y=m'Xmc for some X in Jerry Bryan's list. So for f in Symm(Y), Y'f'Yf=I is an element of C, so f is in CSymm(Y). This says that Symm(Y) is a subgroup of CSymm(Y), which is a conjugate of CSymm(X). So if Symm(Y) is nontrivial, then CSymm(X) will also be nontrivial. So to find the symmetry groups of the edges-and-centers group we need only look at those positions that have nontrivial groups in Jerry's list (i.e. less than order-1152 symmetry), as all the others will have Symm(Y)=I. So, Jerry, do you have the data on how many positions of the edge group have less than order-1152 symmetry, and which positions those are? So, on to finding the symmetry groups of the Rubik's group positions. We need to calculate Symm(X) for every element X of the edges-and-cen- ters group and Symm(Y) for every element Y of the corners-and-centers group, while keeping track of the permutation parity of X and Y. (The permutation parity will be constant over each Symm(X), Symm(Y)). The symmetry groups in the Rubik's group will be the intersections of symmetry groups of edge and corner positions of the same parity. We need not keep track of the particular positions here, only the numbers for each parity and each (conjugacy class of) symmetry group. I have a program that could produce a table easily enough. Recently, I took a look in Paul B. Yale's _Geometry_and_Symmetry_ and it looks like this is the sort of problem we could use the Polya-Burnside theorem on. Unfortunately, I don't understand it yet, and it looks like the number of cases here might be too large to conveniently carry out by hand. So it would help to go after this problem computationally. The rest of this article has the proofs for the claims I mentioned in the second paragraph. ================================================================ Proof 1: Suppose f, g are elements of CSymm(X); it suffices to show that f'g is an element of CSymm(X). X'(f'g)'X(f'g)=X(g'f)X(f'g) =X'g'(Xgg'ff'X')fXf'g =(X'g'Xg) g'f (f'X'fX) f'g, =(X'g'Xg) (f'g)' (X'f'Xf)' (f'g). Since we assumed f, g in CSymm(X), X'g'Xg and X'f'Xf must be in C. (f'g)' and (f'g) are elements of M that are either both in C or neither. So the product is in C, so f'g is in CSymm(X). Therefore CSymm(X) is a group, QED. Proof 2: Suppose Y=m'Xmc. First let f be an element of CSymm(X), so that X'f'Xf is in C. I will first show that m'fm is an element of CSymm(Y). Y'(m'fm)'Y(m'fm)=(m'Xmc)'(m'fm)'(m'Xmc)(m'fm) =(c'm'X'm)(m'f'm)(m'Xmc)(m'fm) =c'm'(X'f'X)(mcm'fm) =c'm'(X'f'Xf)(f'mcm'fm) All of which are elements of M, with an even number in C. Therefore the expression is in C, so m'fm is in CSymm(Y). Now let g be an element of CSymm(Y), so that Y'g'Yg is in C. Let f=mgm', so f is an element of M such that m'fm is in CSymm(Y). I will show that f is an element of CSymm(X): X'f'Xf=(mc)(mc)'X'(mm')f'(mm')Xf(f'mcm'fm)(f'mcm'fm)' =(mc)(m'Xmc)'(m'fm)'(m'Xmc)(m'fm)(f'mcm'fm)' =(mc)Y'(m'fm)'Y(m'fm)(f'mcm'fm)' =(mc)Y'g'Yg(f'mcm'fm)', which is in C, so f is in CSymm(X). I've shown that for every element f of CSymm(X), m'fm is an element of CSymm(Y), and that every element of CSymm(Y) is m'fm for some f in CSymm(X). Therefore CSymm(Y)=m' CSymm(X) m, QED. ================================================================ Dan Hoey Hoey@AIC.NRL.Navy.Mil From hoey@aic.nrl.navy.mil Tue Jan 4 21:36:19 1994 Return-Path: Received: from Sun0.AIC.NRL.Navy.Mil by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA23585; Tue, 4 Jan 94 21:36:19 EST Received: by Sun0.AIC.NRL.Navy.Mil (4.1/SMI-4.0) 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) Message-Id: <9401050236.AA07661@Sun0.AIC.NRL.Navy.Mil> To: "Jerry Bryan" , Cube-Lovers@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 less-symmetric group in which one of the edges never moves, or 2) a larger group in which we distinguish positions that differ by rigid motions of the cube, or 3) a non-group in which we consider cosets--equivalence classes of group #2, where group elements that differ by rigid motions are equivalent. You have got a lot of mileage out of working with group #2 to save duplication among symmetries, then reducing to non-group #3. But what you lose is the group structure of the object you are studying. Instead, you have to work in the large group and then deduce information about the cosets. All in all, though, I'm very glad of it, for the lost symmetries of group #1 were sorely missed. For most of the other questions, mouse@collatz.mcrcim.mcgill.edu provides satisfactory answers. However, strictly speaking we should not call an equivalence class to be a group element (unless it is a coset of a normal subgroup, and neither C nor M is normal in the large group). I'll admit I've also abused the term when considering distances in the ``edge group'', as if all 24 rotations of a position were the same element of some group. But when we start dealing with the distinction between fixed and movable cubes I think we need to start being more careful. [ mouse also mentions that quarter-turn ``usually doesn't include slice turns on the 3-Cube, but on the 4-Cube and higher, they must of necessity be included.'' I'll take that as an argument for eccentric slabism: a QT rotates any 1xNxN slab except a central slab of an odd-edged cube. As opposed to cutism, where a QT consists of a rotation of part of the cube with respect to the other. ] Other questions: > ...since Start and Pons Asinorum differ only by a simple > reflection, why had not my version of M-conjugation declared them > to be equivalent? Your versino treats positions X,Y for which m'Xmc=Y (m in M, c in C) as equivalent. If you instead determine when m'Xmn=Y (m,n in M) you would find them equivalent. This is equivalent to changing the loop in your version of M-conjugacy. > For j = 1 to 24 for k = 1 to 24 for m = 1 to 2 > for i = 1 to 24 Bj,k,m(i) = Qm(Pj(A(Qm(Pk(i))))) so that the two occurrences of Qm need not be the same. > (I speak of "my version of M-conjugation", but the question is no > different if you look at Dan Hoey's original M-conjugation). No, I didn't use M-conjugation except for a cube with a fixed orientation in space [or equivalently, with face centers]. So in the original concept of M-conjugation that Jim Saxe and I put together, Start and Pons Asinorum don't just differ by a reflection. > I found Dan Hoey's postings about the four special states of the > Edge Group to be delightful.... However, [without the results on > distances] if we identified the group as being rectangular, would we > be led to saying which of the four special states were diagonally > opposed without the computer search? Without the search, I might be > tempted to say that Start and Pons Asinorum were diagonally opposed. Well, really the `group' is in the shape of a sphenoid, a word I learned yesterday for a tetrahedron whose three pairs of opposite edges are equal. [Or equivalently, a tetrahedron whose edges are face diagonals of a rectangular prism.] But it might be more accurate to consider it as a large ball of string with a bunch of symmetries. Calling it a rectangle or sphenoid may lead us to ignore the structure that is not representable in Euclidean space. Dan Hoey Hoey@AIC.NRL.Navy.Mil From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Thu Jan 6 04:11:20 1994 Return-Path: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU> Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA29330; Thu, 6 Jan 94 04:11:20 EST Message-Id: <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 X-Acknowledge-To: 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 M-conjugation with a whole cube rotation, Dan Hoey inserted square brackets to make his exposition more readable. And therein lies a story. As it turns out, I had composed most of my original note at work, and I had used square brackets. Actually, I had used square brackets to delimit the indexes for the rotation and reflection operations, and I had used parenthesis to delimit the indexes for the individual cells in the row vectors. I wanted to make a distinction between the two kinds of indexing. Dan avoids the necessity for a distinction by simply not detailing the indexing of the individual cells. Anyway, I completed the note at home. Much to my dismay, all of my square brackets had disappeared! I pretty much understand the problem. My E-mail system is an IBM mainframe which uses EBCDIC as its basic code. EBCDIC does not deal very well with square brackets. There are at least two "standards" for encoding square brackets in EBCDIC. There are any number of ways to access an IBM mainframe, but the native terminal support is 3270 terminals using EBCDIC. Both at home and at work, I use a PC running TN3270 to access our mainframe. TN3270 is a 3270 version of TELNET. However, the TN3270 I use at work is considerably different from the TN3270 I use at home. One TN3270 implements one of the square bracket standards, and the other TN3270 implements the other. For similar reasons, mail gateways often have difficulties with square brackets. They may have to translate EBCDIC to ASCII or ASCII to EBCDIC, and it is difficult to know how best to set up the translate tables. My experience is that some gateways get it "right" and others get it "wrong". I therefore had a great fear that if I posted my note with square brackets, that the square brackets might appear as gibberish to at least some of you. Thus, I sort of temporized and faked the subscripts with upper and lower case letters (e.g., Bk to mean B-sub-k), omitting square brackets entirely. It is probably no accident that old programming languages such as FORTRAN and COBOL use parentheses for indexing. These languages originated in the 50's. At that time, the dominant character code was BCD, which did not include square brackets. EBCDIC is just extended BCD, and the original EBCDIC did not include square brackets, either. Square brackets are a latter day addition to EBCDIC, and the implementation of square brackets in EBCDIC is inconsistent. ASCII has always included square brackets. "Modern" languages (say, starting in the 70's) such as Pascal and C (and their descendents) grew up in the ASCII world, and tend to use square brackets for indexing and parentheses for function arguments. FORTRAN compilers to this day have difficulty figuring out with things like Y=X(I) or Y=F(X) -- which are functions and which are subscripted arrays. Also, Pascal and C tend not to co-exist very well in the EBCDIC world because of these kinds of character set difficulties. There are several other characters with similar difficulties. For example, if G is the cube group, you might want to refer to the size of the cube group as |G|. But the delimiting vertical bars can be very different between EBCDIC and ASCII. Finally, FORTRAN used ** for exponentiation. More modern languages tend to use some sort of up-arrow or carat. But such characters don't translate well between EBCDIC and ASCII. For example, if I write |G| = 4.3 * 10^19, it is highly problematic whether the character between the 10 and the 19 which I am using to express exponentiation will make any sense on your particular system. For whatever it is worth, here are my home and work versions of square brackets: home left square bracket [[[[[[[ x'AD' home right square bracket ]]]]]]] x'BD' work left square bracket x'BA' work right square bracket x'BB' = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU If you don't have time to do it right today, what makes you think you are going to have time to do it over again tomorrow? From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Thu Jan 6 13:52:03 1994 Return-Path: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU> Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA24385; Thu, 6 Jan 94 13:52:03 EST Message-Id: <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 X-Acknowledge-To: 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 B-equivalent if B(X)=B(Y). Note that the B function calculates the "canonical form" or the "representative element" of an equivalence class under M-conjugation and rotation of the cube in space. Dan Hoey already gave a thorough analysis of this situation. I would like to provide an alternative analysis which I hope to be equivalent to Dan's. My analysis will be fairly informal as compared to Dan's. Here are a couple of preliminaries. First, multiplication of permutations is generally not commutative. For example, if X is any cube, then it is generally not the case that m'Xm=mXm', where m is in M, the set of all cube rotations composed with reflections. However, we can calculate all M-conjugates of X as B[n]=m[n]' X m[n] for n in {1 to 48}, or we can calculate all M-conjugates of X as B[n]=m[n] X m[n]'. Either way, B will be the same set of cubes. It will be in a different order, but it will be the same set. The reason is that the set of all m[n] is the same as the set of all m[n]', namely just M, but m[n] is in a different order than m[n]'. This means that as long as we are calculating all M-conjugates as opposed to a specific M-conjugate, we can sort of "violate" the normal rules about multiplication commutivity. Second, if X is any cube, consider the set of all rotations of X, namely B[i] = X c[i] for i in {1 to 24}, and where c[i] is in C, the set of all cube rotations. Having generated the set of all rotations of X, we can rotate as many times as we wish, for example B[j] = X c[i] c[j] for i in {1 to 24} and j in {1 to 24}, or even B[k] = X c[i] c[j] c[k] for i in {1 to 24}, j in {1 to 24}, and k in {1 to 24}. No matter how many times we multiply, B will be the same set, it will just be in a different order. Conversely, if we have any number of adjacent rotations in the multiplication, we can eliminate all but one rotation, and B will be the same set, and again will just be in a different order. With the preliminaries out of the way, we note that the set of all M-conjugates of X is generated as B[i]=m[i]' X m[i] for i in {1 to 48}. But we can also generate the same set in a different order as B[i]=m[i] X m[i]' for i in {1 to 48}. We can decompose M and calculate all M-conjugates as B[i,j]=c[i] r[j] X r[j]' c[i]' for i in {1 to 24} and j in {1 to 2}. But r[1]'=r[1] (r[1] is the identity) and r[2]'=r[2] (the reflection is its own complement). So we have B[i,j]=c[i] r[j] X r[j] c[i]' for i in {1 to 24} and j in {1 to 2}. The set of all c[i] is the same as the set of all c[i]' (just in a different order), so we define i' as the index for which c[i']=c[i]'. Hence the calculation of an M-conjugate can be written as B[i,j]=c[i] r[j] X r[j] c[i'] for i in {1 to 24} and j in {1 to 2}. Finally, we wish to multiply the M-conjugate by the set of all rotations, so we have B[i,j,k]=c[i] r[j] X r[j] c[i'] c[k] for i in {1 to 24}, j in {1 to 2}, and k in {1 to 24}. But as we noted in our second preliminary note, we can collapse multiple rotations into one, and we have B[i,j,k]=c[i] r[j] X r[j] c[k], and B will be the set of all M-conjugates of X multiplied by all rotations. I guess I am overusing the letter "B" a bit, because the "B" function is simply the minimum of the "B" matrix. But in any case, we have shown that the "B" loop in my program is simply calculating the set of all M-conjugates multiplied by all rotations. This is the exact result already proven by Dan Hoey, but I found the above derivation a little easier to follow. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU If you don't have time to do it right today, what makes you think you are going to have time to do it over again tomorrow? From pbeck@pica.army.mil Thu Jan 6 14:22:12 1994 Return-Path: Received: from COR6.PICA.ARMY.MIL by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA26466; Thu, 6 Jan 94 14:22:12 EST Date: Thu, 6 Jan 94 14:06:54 EST From: Peter Beck (BATDD) To: Cube-Lovers@ai.mit.edu Subject: Mickey's Challenge Message-Id: <9401061406.aa23113@COR6.PICA.ARMY.MIL> NEW PUZZLE "MICKEY'S CHALLENGE" is at your Disney store now, price $10. This is a legal MACHBALL, ie, a spherical SKEWB. It comes with a solution book. Christoph Bandelow (a longer time cuber) wrote the solution. I haven't bought one or it played with it yet. GOOD PUZZLING pete beck pbeck@pica.army.mil From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Thu Jan 6 14:31:34 1994 Return-Path: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU> Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA26764; Thu, 6 Jan 94 14:31:34 EST Message-Id: <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 X-Acknowledge-To: 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) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU If you don't have time to do it right today, what makes you think you are going to have time to do it over again tomorrow? From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Fri Jan 7 10:35:45 1994 Return-Path: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU> Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA09588; Fri, 7 Jan 94 10:35:45 EST Message-Id: <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 X-Acknowledge-To: 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 M-conjugate classes for G. In this case, |G\M| is approximately 48 times smaller than |G|. I believe that when Dan Hoey asked in 1984 the question "how big is G, really?", that he was really asking how big is G\M, and that he was asking for the approximation to be resolved to an exact number. Let GC\M be the set of M-conjugate classes for GC. In this case, |GC\M| is approximately 48 times smaller than |GC|. Let GE\M be the set of M-conjugate classes for GE. In this case, |GE\M| is approximately 48 times smaller than |GE|. Recall that B is the function which calculates the canonical form for a cube under the composed operations of M-conjugation plus rotation. My programs calculate equivalence classes under B. Let G\B be the set of B-classes for G. Let GC\B be the set of B-classes for GC. Let GE\B be the set of B-classes for GE. So far, my programs have built complete search trees for GC\B and GE\B. Let Gx denote any of G, GC, and GE. Then, we have Gx\B=(Gx\C)\M=(Gx\M)\C. In English, we can decompose B into a multiplication by C and M (in either order). Also, note that Gx\C=(Gx\C)\C=((Gx\C)\C)\C=.... Similarly, (G\M)\C=((G\M)\C)\C=.... In English, having reduced once by C, we can reduce again by C as many times as we wish, but we simply get the same set back again each time. This notation can help us address the question of whether B actually accomplishes a "times 48" or a "times 1152" reduction in the size of the cube. If we are dealing with Gx, then Gx\B is a "times 1152" reduction. However, information is lost. For example, consider GC and GC\B. GC is "corners plus centers", and B-reduction of GC removes the centers and calculates M-conjugates of the corners. But you really don't have the same problem any more because the centers are gone. If on the other hand we are dealing with Gx\C, then (Gx\C)\B is a "times 48" reduction. All we have really done is calculate M-conjugates. The reduction by the C that is composed into B is duplicate effort which accomplishes nothing. I have come to realize that my program for the 2x2x2 actually models GC (corners with centers without edges) rather than GC\C (corners without centers without edges). My program does not explicitly encode the centers. However, it encodes all eight corner cubies, and when it makes qturns, any of the eight cubies can move. Hence, rotational information is encoded, even if the centers themselves are not explicitly encoded. If I wanted to model GC\C, I would have had to either model only seven of the cubies, or else modeled all eight but moved only seven of them. Since what I really wanted was (GC\C)\M, and since what I had was GC, I had to invent this funny B thing, where GC\B=(GC\C)\M. If I had been clever enough to model GC\C in the first place, I never would have had to invent B. Similar comments apply to my model for the edges. To convince yourself that eight corner cubies model GC and seven corner cubies model GC\C, just think about calculating |GC| and |GC\C|. For |GC|, there are eight ways to pick the first cubie, seven ways to pick the second cubie, and so forth yielding the familiar |GC|=8!*(3^8)/3. For |GC\C|, we let one of the cubies be fixed, then there are seven ways to pick the second cubie, and so forth yielding |GC\C|=7!*(3^7)/3, and |GC| = |GC\C| * 24. Hence, the "corners of the 3x3x3" problem is 24 times larger than the "2x2x2" problem. I will discuss the "times 24" reduction that is accomplished by reducing by C in a followup note. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU If you don't have time to do it right today, what makes you think you are going to have time to do it over again tomorrow? From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Sat Jan 8 10:21:12 1994 Return-Path: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU> Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA26746; Sat, 8 Jan 94 10:21:12 EST Message-Id: <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 X-Acknowledge-To: 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 B-conjugacy classes of X is the set of all m'Xmc for all m in M and all c in C. We denote this set as BClass(X). B is the function B(X)=min(BClass(X)). Note that we could have defined BClass(X) equivalently as the set of all mXm'c, or as the set of all cm'Xm, or as the set of all cmXm'. It is in general not the case that m'Xmc = mXm'c = cm'Xm = cmXm' for any fixed value of m and c. (Quite the contrary!). However, when we say "the set of all...", the four ways of generating BClass(X) become equivalent. This is the justification for the assertion in a previous note that Gx\B = (Gx\M)\C = (Gx\C)\M. Two cubes X and Y are B-equivalent if BClass(X) = BClass(Y). Equivalently, two cubes X and Y are B-equivalent if B(X) = B(Y). |X| is the length of X (the distance of X from Start). We have |B(X)| = |X| for centerless cubes, but it is generally not the case that |B(X)| = |X| for cubes with centers. In fact, let X and Y be cubes with centers such that B(X)=B(Y). It is not necessarily the case that |X| = |Y|. For example, consider the set GC of cubes with corners with centers without edges. We have B(RL')=B(I), but |RL'|=2 and |I|=0. |BClass(X)| is the number of elements in BClass(X). If |BClass(X)| = N, then X is said to have order-N symmetry. (I sincerely regret ever using this terminology. As has been noted on the list, it seems "backwards" somehow. But given that this usage exists, the value 1152/N is generally more useful than the value N.) We note the following: 1. B(X) is a cube. 2. BClass(X) is a set of cubes. 3. B(B(X)) = B(X) 4. BClass(B(X)) = BClass(X). 5. Both X and B(X) are in BClass(X). = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU If you don't have time to do it right today, what makes you think you are going to have time to do it over again tomorrow? From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Sat Jan 8 11:25:43 1994 Return-Path: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU> Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA28874; Sat, 8 Jan 94 11:25:43 EST Message-Id: <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 X-Acknowledge-To: 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 M-conjugate classes of G, which I think in some sense is the "real" size of G. My proposal draws *very* heavily on Dan's note. My first (incorrect) proposal was based on the following idea. By computer search, we already have a database of GC\B and GE\B, the corners and edges of G, respectively, reduced by B-conjugacy. Hence, we know |GC\B| and |GE\B|. For each X in GC\B and each Y in GE\B, we calculate BClass(X) and BClass(Y). Then, we can combine BClass(X) and BClass(Y) in all legal ways (minding parity considerations). Call the combinations BClass(X) * BClass(Y). For any fixed X and Y, BClass(X) * BCLass(Y) is a set of cubes in G. Hence we can calculate (BClass(X) * BClass(Y))\M and |(BClass(X) * BClass(Y))\M|. My idea then was just to sum |(BClass(X) * BClass(Y))\M| over all values of X and Y to calculate |G\M|. And we know how many X's and Y's there are! But, alas, |(BClass(X) * BClass(Y))\M| is not the same across all X's and Y's because, well because of symmetry. All X's and Y's are not equally symmetrical. I was assuming that |(BClass(X) * BClass(Y))\M| was constant, and of course it is not. My next (incorrect) proposal was never posted to the list. It was a slight improvement on the first idea. We have a data base of all X's in GC\B and of all Y's in GE\B. For each X in GC\B and for each Y in GE\B, we know |BClass(X)| and |BClass(Y)|. (Actually, we don't. I have to calculate it. I have done so, and I have posted summaries of those calculations. However, I did not store the order of the equivalence classes in the data base. I kick myself for not doing so, but this is a minor problem, so let us continue). There are only 10 distinct values for |BClass(X)| and for |BClass(Y)|, namely 24, 48, 72, 96, 144, 192, 288, 384, 576, and 1152. (By the way, I have never figured out why it is *exactly* the same set of values for both the corners and for the edges. It is easy to see why it is approximately the same set of values, but the structure of the corners is enough different from the structure of the edges that I see no obvious reason the set of values should be exactly the same in both cases.) Let GC[m] be the set of all X for which |BClass(X)| = m and let GE[n] be the set of all Y for which |BClass(Y)| = n. Hence, GC\B is partitioned into GC[m]\B for m=24,48..., and GE\B is partitioned into GE[n]\B for n=24,48,... Now, we form the sets BClass(X)[m] * BClass(Y)[n] for all X in GC[m]\B and for all Y in GE[n]\B, and for all legal values of m and n. There will be 100 such sets. For any fixed X, Y, m and n, BClass(X)[m] * BCLass(Y)[n] is a set of cubes in G. Hence we can calculate (BClass(X)[m] * BClass(Y)[n])\M and |(BClass(X)[m] * BClass(Y)[n])\M|. My idea then was just to sum |(BClass(X)[m] * BClass(Y)[n])\M| over all X in GC[m]\B and over all Y in GE[n]\B to calculate |(BClass(X)[m] * BClass(Y)[n])\M|. We know how many X's there are in GC[m]\B and we know how many Y's there are in GE[n]\B, so the calculation seemed possible. I then intended to sum again over all m and over all n, and I would be done. But, alas, in order to perform the sum over all X and all Y, I needed a theorem which I couldn't prove and which I now believe is not true anyway. I needed to be able to prove that for a fixed m and n, that |(BClass(X)[m] * BClass(Y)[n]| had the same value for all X in GC[m]\B and all GE[n]\B. For a while I thought it was true, but right now I can't think of any reason why it should be. But perhaps Dan Hoey comes to the rescue with his CSymm function. I still need a theorem which I cannot (yet) prove, but I believe it is true. If it can be proven, my basic overall scheme can be rescued. In my second proposal, I used the values of all possible BClass sizes as indexes -- 24, 48, 72... It would perhaps be more convenient to make these sizes a set {24, 48, 72, ...}, and to think of the indexes m and n taking on the values from 1 to 10, where the values from 1 to 10 index the set {24, 48, 72, ....}. With this understanding, all the above results are valid, and the indexing is more convenient. We can now say that GC\B is partitioned into GC[1]\B, GC[2]\B, ... through GC\B[10] and similarly for GE\B. Unfortunately, using the B-equivalence class sizes to partition GC\B and GE\B did not permit us perform the calculations we wanted to perform. However, suppose we partition GC\B and GE\B a different way, namely using CSymm. Suppose, for each X in GC\B and for each Y in GE\B, we calculate CSymm(X) and CSymm(Y). (We would have to do this by computer). CSymm(X) and CSymm(Y) are sets, but there are only a (relatively) small number of such sets. Let each distinct value CSymm(X) and CSymm(Y) be mapped to an index. We can call such a mapping function CSind, and we can calculate CSind(CSymm(X)) and CSind(CSymm(Y)). Actually, there is no reason not to define CSind in such a way that the domain is the set of X's and Y's, so that we can calculate CSind(X) and CSind(Y). Now, we use m=CSind(X) and n=CSind(Y) to form a partition of GC\B and GE\B. All our results from before are valid. The only issue is, can we now perform the sum? In order to perform the sum, we need the following to be true: For a fixed m and n, |BClass(X)[m] * BClass(Y)[n]| is constant for all X in GC[m]\B and all Y in GE\[n]\B, where GC[m] is the set of all X in GC such that CSind(X)=m and GE[n] is the set of all Y in GE such that CSind(Y)=n. It really seems true to me, and I shall strive to prove it. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU If you don't have time to do it right today, what makes you think you are going to have time to do it over again tomorrow? From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Sat Jan 8 18:47:56 1994 Return-Path: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU> Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA13806; Sat, 8 Jan 94 18:47:56 EST Message-Id: <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 X-Acknowledge-To: 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) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU If you don't have time to do it right today, what makes you think you are going to have time to do it over again tomorrow? From CPELLEY@delphi.com Mon Jan 10 17:31:46 1994 Return-Path: Received: from bos1a.delphi.com (delphi.com) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA11105; Mon, 10 Jan 94 17:31:46 EST Received: from delphi.com by delphi.com (PMDF V4.2-11 #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: Cube-Lovers@ai.mit.edu Message-Id: <01H7INM9NNEQ91X4TE@delphi.com> X-Vms-To: INTERNET"Cube-Lovers@ai.mit.edu" Mime-Version: 1.0 Content-Type: TEXT/PLAIN; CHARSET=US-ASCII Content-Transfer-Encoding: 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 Return-Path: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU> Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA26716; Mon, 10 Jan 94 23:08:35 EST Message-Id: <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 X-Acknowledge-To: 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 B-conjugate classes are of order 1152, the calculations are trivial. About 99.9923% of the edge conjugate classes and about 96.924% of the corner conjugate classes are of order 1152, so we can calculate the correct number of M-conjugates of G for a very large percentage of the cases. Consider some fixed X in GC\B and some fixed Y in GE\B where |BClass(X)|=1152 and |BClass(Y)|=1152. Form BClass(X) * BClass(Y). Now, |BClass(X) * BClass(Y)| = |BClass(X)| * |BClass(Y)| / 2 = 1152 * 1152 / 2. (The division by 2 takes care of parity). Finally, form (BClass(X) * BClass(Y))\M, and we have |(BClass(X) * BClass(Y))\M| = |BClass(X) * BClass(Y)| / 48 = (1152 * 1152) / 2 / 48 = 13,824. We know the number of BClasses of GC of order 1152 from computer search (namely 75,392), and we know the number of BCLasses of GE of order 1152 from computer search (namely 851,493,140). Hence, for the special case of both BClasses being of order 1152, we have the total number of elements of G\M being 851,493,140 * 75,392 * 13,824 = 887,442,335,689,605,120. We can derive similar results if only one of BCLass(X) and BClass(Y) are of order 1152. For example, there are 4 elements of GE\B for which |BClass(Y)|=24. Choose such a Y, and choose X in GC\B such that |BClass(X)|=1152. Form BClass(X) * BClass(Y). It will be the case that |BClass(X) * BClass(Y)| = 1152 * 24 / 2 = 13,824. Hence, |(BClass(X) * BClass(Y))\M| = 13,824/48 = 288. There are 75,392 values of X for which |BClass(X)|=1152, 4 values of Y for which |BCLass(Y)|=24, and hence there are 75,392 * 4 * 288 = 86,851,584 elements of G\M of the form BClass(X) * BClass(Y) for which |BCLass(X)| = 1152 and |BClass(Y)| = 24. There are nineteen cases in all for which at least one of BClass(X) and BCLass(Y) are of order 1152, and this note calculates only two of the nineteen. Completing the other seventeen would be trivial but tedious. However, a total solution to the problem will require coming up with some way to deal with the cases where neither |BClass(X)|=1152 nor |BClass(Y)|=1152. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU If you don't have time to do it right today, what makes you think you are going to have time to do it over again tomorrow? From tonyd@earwax.pd.uwa.edu.au Tue Jan 11 02:06:32 1994 Return-Path: Received: from earwax.pd.uwa.edu.au by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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 Message-Id: <199401110709.PAA20165@earwax.pd.uwa.edu.au> Mime-Version: 1.0 Content-Type: text/plain; charset="us-ascii" Date: Tue, 11 Jan 1994 15:08:16 +0800 To: Cube-Lovers@ai.mit.edu From: tonyd@earwax.pd.uwa.edu.au Subject: Rubik chaos? On sci.nonlinear... In article <1994Jan5.120409@oxygen.aps1.anl.gov> Thomas D. Orth, orth@oxygen.aps1.anl.gov writes: >A friend of mine has written a few papers on the subject of >the Rubik's Cube Group, and the elements of Chaos within >it, or Pseudo-chaos as she calls it. The papers are being submitted to the journal CHAOS. cheers, Tony From @mail.uunet.ca:mark.longridge@canrem.com Tue Jan 11 16:07:24 1994 Return-Path: <@mail.uunet.ca:mark.longridge@canrem.com> Received: from mail.uunet.ca (uunet.ca) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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/SMI-4.1) id AA09812; Tue, 11 Jan 94 13:52:09 EST Received: by canrem.com (PCB-UUCP 1.1f) id 190BA3; Tue, 11 Jan 94 13:41:58 -0400 To: cube-lovers@life.ai.mit.edu Reply-To: CRSO.Cube@canrem.com Sender: CRSO.Cube@canrem.com Subject: Andras Mezei's Book From: mark.longridge@canrem.com (Mark Longridge) Message-Id: <60.692.5834.0C190BA3@canrem.com> Date: Tue, 11 Jan 1994 12:18:00 -0500 Organization: CRS Online (Toronto, Ontario) A while back I reported on the list of cube books available at the Library of Congress. At the time, I did not realize the significance of #2: 2. 85-109601: Mezei, Andras. Magyar kocka, avagy, Meg mindig ilyen gazdagok vagyunk? / Budapest : Magveto, c1984. 473 p. : ill. ; 21 cm. Digging through some old magazines I reread the cube article in the March 1986 issue of "Discover" magazine. In this issue John Tierney talks to Rubik himself. The article itself is excellent, showing pictures of Rubik's first wooden prototype, and having discussions on the Golden Age of the Cube when only Rubik had access to his invention. I learned that Andras Mezei (a Budapest writer) wrote a book and a play called "The Hungarian Cube". This chronicles the debacle that occured when the Hungarians attempted to expand their operations to meet the huge demand, rather than farm the work to other factories. Andras writes: "Everyone made money on the cube except the Hungarians". Does anyone on Cube-Lovers have this book? Judging by the size of the book, and the fact that it's illustrated, I think it would be a worthy addition to any cubist's library, even more so if there exists an english translation. If not, I feel encouraged enough to get an English-Hungarian dictionary and read it anyway! -> Mark From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Wed Jan 12 23:46:28 1994 Return-Path: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU> Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA20019; Wed, 12 Jan 94 23:46:28 EST Message-Id: <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 X-Acknowledge-To: 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 B-conjugate class for corners) B - |BClass(Y)| for Y in GE\B (order of B-conjugate class for edges) C - |GC[A]\B| (number of B-conjugate classes of size A for corners) D - |GE[B]\B| (number of B-conjugate classes of size B for edges) E - |BClass(X) * BClass(Y)| = |BClass(X)| * |BClass(Y)| / 2 F - |BClass(X) * BClass(Y)| * |GC[A]\B| * |GE[B]\B| / 48 (number of M-conjugates of |GC[A]\B * GE[B]\B) A B C D E F 1152 24 75,392 4 13,824 86,851,584 1152 48 75,392 2 27,648 86,851,584 1152 72 75,392 12 41,472 781,664,256 1152 96 75,392 16 55,296 1,389,625,344 1152 144 75,392 110 82,944 14,330,511,360 1152 192 75,392 70 110,592 12,159,221,760 1152 288 75,392 1,544 165,888 402,296,537,088 1152 384 75,392 1,252 221,184 434,952,732,672 1152 576 75,392 128,858 331,776 67,149,128,466,432 1152 1152 75,392 851,493,140 663,552 887,442,335,689,605,120 24 1152 1 851,493,140 13,824 245,230,024,320 48 1152 1 851,493,140 27,648 490,460,048,640 72 1152 3 851,493,140 41,472 2,207,070,218,880 96 1152 1 851,493,140 55,296 980,920,097,280 144 1152 14 851,493,140 82,944 20,599,322,042,880 192 1152 15 851,493,140 110,592 29,427,602,918,400 288 1152 135 851,493,140 165,888 397,272,639,398,400 384 1152 32 851,493,140 221,184 125,557,772,451,840 576 1152 2,208 851,493,140 331,776 12,995,229,448,765,440 Total 901,082,361,368,033,280 Note that we have covered over 99.99 percent of edge positions (which are combined with all corner positions), and over 96.9 percent of the remaining corner positions (which are combined with all edge positions). Hence, we have covered about 99.99969 percent of all positions. However, that last fraction of a percent is going to be devilishly difficult. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU If you don't have time to do it right today, what makes you think you are going to have time to do it over again tomorrow? From @mail.uunet.ca:mark.longridge@canrem.com Thu Jan 13 05:03:10 1994 Return-Path: <@mail.uunet.ca:mark.longridge@canrem.com> Received: from mail.uunet.ca (uunet.ca) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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/SMI-4.1) id AA23090; Wed, 12 Jan 94 18:30:28 EST Received: by canrem.com (PCB-UUCP 1.1f) id 190EF0; Wed, 12 Jan 94 18:20:47 -0400 To: cube-lovers@life.ai.mit.edu Reply-To: CRSO.Cube@canrem.com Sender: CRSO.Cube@canrem.com Subject: 4x4x4 Cube moves From: mark.longridge@canrem.com (Mark Longridge) Message-Id: <60.694.5834.0C190EF0@canrem.com> Date: Wed, 12 Jan 1994 17:07:00 -0500 Organization: CRS Online (Toronto, Ontario) Some comments on flipping a single pair of edges on the 4x4x4 cube: Singmaster notation on the 4x4x4 (same notation as Jeffery Adams) -------------------------------- L left face l inner left slice r inner right slice R right face F front face f inner front slice b inner back slice B back face U up face u inner up slice d inner down slice D down face So L1 would be turn the left face 90 degrees clockwise and l1 would be turn the inner left slice 90 degrees clockwise. I'll use the suffix "2" to be for 180 degree turns and the suffix "3" to be for 270 degree turns clockwise or 90 degree turns counterclockwise. This is the shortest sequence I found for flipping 2 adjacent edges on the 4x4x4 cube (LD pair): (r3 D3) ^3 + (r1 D1) ^4 + Rr3 D3 R1 D1 r3 D3 R3 D1 R1 D3 Note the use of Rr to represent both the turns R face & r inner slice. Counting slice turns the sequence is 25 turns, or 24 "hyper moves". This sequence moves some centre pieces around. However, on checking David Singmaster's Cubic Circular, in issues 5 & 6, Autumn & Winter 1982 there is a shorter process on page 15, (UB pair): r2 D2 l3 D1 R3 U1 R3 U3 l3 U1 R1 U3 l1 R1 D1 r2 This process, although more difficult to memorize, is only 16 slice moves. It also disturbs centre pieces, although in a simpler way. I always solve the centre pieces last on the 4x4x4. Hope this helps! -> Mark Email: mark.longridge@canrem.com From Mikko.Haapanen@otol.fi Thu Jan 13 09:19:35 1994 Return-Path: Received: from lassie.eunet.fi by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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/IDA-1.5 for ); Thu, 13 Jan 1994 16:19:16 +0200 Received: from rhea.otol.fi by sulo.otol.fi with SMTP (PP) id <01542-0@sulo.otol.fi>; Thu, 13 Jan 1994 16:19:11 +0200 Received: from otol.fi by rhea.otol.fi id <26762-0@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" Reply-To: "M. Haapanen" Subject: Re: 4x4x4 Cube moves To: cube-lovers@life.ai.mit.edu In-Reply-To: <60.694.5834.0C190EF0@canrem.com> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; CHARSET=US-ASCII > ... > 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 Return-Path: Received: from holonet.net (giskard.holonet.net) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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 Message-Id: <199401131737.JAA25594@holonet.net> To: cube-lovers@life.ai.mit.edu From: ishius@ishius.com (Ishi Press International) Subject: Skewb, 5x5x5 cubes available I've been watching all the discussion here, and I thought some people on this list might appreciate knowing that 5x5x5 Rubik's Cubes and the Skewb are available from Ishi Press International. We also have hundreds of other mechanical puzzles. If you would like to be on our puzzle e-mail list, write us. If you would like a color catalog of our puzzles, send us your snail-mail address. I apologize for intruding with a commercial message, but it did seem to me that at least a few people on this list would like to get their hands on some of these puzzles, and I don't know of any other distributor for these two items. Anton Dovydaitis Customer Support ======================================================================== Ishi Press International 800/859-2086 voice, 408/944-9110 FAX 76 Bonaventura Drive ishius@ishius.com The Americas San Jose, CA 95134 ishi@cix.compulink.co.uk Europe From CPELLEY@delphi.com Thu Jan 13 23:25:54 1994 Return-Path: Received: from bos2a.delphi.com (delphi.com) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA15306; Thu, 13 Jan 94 23:25:54 EST Received: from delphi.com by delphi.com (PMDF V4.2-11 #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: cube-lovers@life.ai.mit.edu Message-Id: <01H7NGQCR7WI99DX0P@delphi.com> X-Vms-To: INTERNET"cube-lovers@life.ai.mit.edu" Mime-Version: 1.0 Content-Type: TEXT/PLAIN; CHARSET=US-ASCII Content-Transfer-Encoding: 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 Return-Path: Received: from holonet.net (giskard.holonet.net) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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 Message-Id: <199401141913.LAA17654@holonet.net> To: Cube-Lovers@ai.mit.edu From: ishius@ishius.com (Ishi Press International) Sender: ishius@ishius.com (Unverified) Subject: 4x4x4 and 5x5x5 cubes. I've been getting a lot of requests for 4x4x4 cubes, and we're looking into getting them. However, I have a couple questions. 1) Why are 4x4x4 cubes so interesting? Do the additional symmetries make for interesting questions, are they more fun, or easier to solve? 2) It appears to me that if you know how to solve the 3x3x3 Rubik's cube, then you can easily solve the 5x5x5 rubiks (i.e., the solution is derivative). For example, you can treat the inner 3x3 faces of the 5x5x5 as a single 3x3x3 cube. Alternately, you can treat the edges/faces along with the the middle three slices combined into a single slice as its own 3x3x3 cube, and this would not really disturb the "inner face" 3x3x3 cube. Is this really so, or am I missing something? Is the 5x5x5 cube simply the group product of two 3x3x3 cubes and one or two sub-groups of a 3x3x3, or is it more complex than that? How does this relate to the 4x4x4? I do have a Bachelor's degree in mathematics and am familiar with abstract algebra. I appreciate any light you can shed on these questions. I would like to be able to converse intelligently about the cubes; that is why I subscribed to this list. Anton Dovydaitis ======================================================================== Ishi Press International 800/859-2086 voice, 408/944-9110 FAX 76 Bonaventura Drive ishius@ishius.com The Americas San Jose, CA 95134 ishi@cix.compulink.co.uk Europe From mouse@collatz.mcrcim.mcgill.edu Fri Jan 14 15:45:34 1994 Return-Path: Received: from Collatz.McRCIM.McGill.EDU by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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 cube-lovers@ai.mit.edu; Fri, 14 Jan 1994 15:44:48 -0500 Date: Fri, 14 Jan 1994 15:44:48 -0500 From: der Mouse Message-Id: <199401142044.PAA00960@Collatz.McRCIM.McGill.EDU> To: cube-lovers@ai.mit.edu Subject: Re: 4x4x4 and 5x5x5 cubes. > I've been getting a lot of requests for 4x4x4 cubes, and we're > looking into getting them. I'd (probably) buy one - I don't know what's become of the one I had. Depends on price, of course, but I found the 5-Cube price acceptable. > 1) Why are 4x4x4 cubes so interesting? Do the additional symmetries > make for interesting questions, are they more fun, or easier to > solve? I doubt it. If one ignores the center slice in each dimension on a 5x5x5, one has a 4x4x4. I think it's the completist instinct any collector has. :-) Now what I'd *really* like is something topologically equivalent to a 2x2x2x2 Cube. Obviously a 2x2x2x2 Cube can't really be built, but it should be possible to build something topologically equivalent. (A 3x3x3x3 would be nice too, but perhaps too difficult.) The hard part is designing an emulation that has some aesthetic appeal; it's easy enough to write a program that lets you permute appropriate overlapping 4-cycles of objects without any intuitively-obvious structure. > 2) It appears to me that if you know how to solve the 3x3x3 Rubik's > cube, then you can easily solve the 5x5x5 rubiks (i.e., the > solution is derivative). No, not really. If you can do the 3-Cube *and the 4-Cube*, then the 5-Cube has no new challenges to offer (nor, I believe, does any size). But the 4-Cube and 5-Cube do have challenges the 3-Cube doesn't, namely edge cubies and face-center cubies. All the 3-Cube experience in the world won't help you if you get your 5-Cube solved except for two edge cubies which are swapped. (Or rather, general Cube-type-puzzle experience will help - for example, how to use conjugates - but 3-Cube-specific experience won't.) > For example, you can treat the inner 3x3 faces of the 5x5x5 as a > single 3x3x3 cube. Alternately, you can treat the edges/faces > along with the the middle three slices combined into a single > slice as its own 3x3x3 cube, and this would not really disturb the > "inner face" 3x3x3 cube. Is this really so, or am I missing > something? You're missing something, but not much. :-) As you say, there are two ways of emulating a 3-Cube on the 5-Cube, namely to paste slices 2-1-2 along each dimension and to paste them 1-3-1. (I hope that's not too abbreviated to be comprehensible - I mean, along each dimension, paste the 5-Cube slices together into three groups, taking two, then one, then two slices, or one, three, and one.) However, it is entirely possible to scramble the 5-Cube in ways that cannot be solved by treating the 5-Cube as virtual 3-Cubes, except in the trivial sense that any 5-Cube turn can be viewed as one or more turns of appropriately-chosen virtual 3-Cubes. For example, I can swap two edge cubies (and also permute center cubies in invisible ways); alternatively, I can permute the face cubies so that the 2-1-2 virtual 3-Cube has two identical corner v-cubies. der Mouse mouse@collatz.mcrcim.mcgill.edu From ncramer@bbn.com Fri Jan 14 17:05:26 1994 Return-Path: Received: from LABS-N.BBN.COM by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA05939; Fri, 14 Jan 94 17:05:26 EST Message-Id: <9401142205.AA05939@life.ai.mit.edu> Date: Fri, 14 Jan 94 7:47:37 EST From: Nichael Cramer To: Ishi Press International Cc: cube-lovers@life.ai.mit.edu Subject: Re: Skewb, 5x5x5 cubes available >Date: Thu, 13 Jan 1994 09:37:53 -0800 >Message-Id: <199401131737.JAA25594@holonet.net> >From: Ishi Press International >Subject: Skewb, 5x5x5 cubes available > >I've been watching all the discussion here, and I thought some people on this >list might appreciate knowing that 5x5x5 Rubik's Cubes and the Skewb are >available from Ishi Press International. We also have hundreds of other >mechanical puzzles. > >If you would like to be on our puzzle e-mail list, write us. If you would >like a color catalog of our puzzles, send us your snail-mail address. Anton Yes to both the above, please. e-mail : ncramer@bbn.com land-mail: Nichael Cramer 123 B Spring St Cambridge MA 02141 Thanks much Nichael From @mail.uunet.ca:mark.longridge@canrem.com Fri Jan 14 23:47:23 1994 Return-Path: <@mail.uunet.ca:mark.longridge@canrem.com> Received: from mail.uunet.ca (uunet.ca) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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/SMI-4.1) id AA19059; Fri, 14 Jan 94 23:21:00 EST Received: by canrem.com (PCB-UUCP 1.1f) id 191357; Fri, 14 Jan 94 23:16:58 -0400 To: cube-lovers@life.ai.mit.edu Reply-To: CRSO.Cube@canrem.com Sender: CRSO.Cube@canrem.com Subject: Higher Order Cubes, correction From: mark.longridge@canrem.com (Mark Longridge) Message-Id: <60.704.5834.0C191357@canrem.com> Date: Fri, 14 Jan 1994 22:10:00 -0500 Organization: CRS Online (Toronto, Ontario) -> (fm2 + u2) ^2 + (fm2 + lm2) ^2 (corrects the centres) This should be: (fm2 + u2) ^2 + (fm2 + l2) ^2 -> Mark From @mail.uunet.ca:mark.longridge@CANREM.COM Sat Jan 15 00:52:48 1994 Return-Path: <@mail.uunet.ca:mark.longridge@CANREM.COM> Received: from mail.uunet.ca (uunet.ca) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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/SMI-4.1) id AA19041; Fri, 14 Jan 94 23:20:55 EST Received: by canrem.com (PCB-UUCP 1.1f) id 191356; Fri, 14 Jan 94 23:16:57 -0400 To: cube-lovers@life.ai.mit.edu Reply-To: CRSO.Cube@canrem.com Sender: CRSO.Cube@canrem.com Subject: Higher Order Cubes From: mark.longridge@canrem.com (Mark Longridge) Message-Id: <60.703.5834.0C191356@canrem.com> Date: Fri, 14 Jan 1994 21:59:00 -0500 Organization: CRS Online (Toronto, Ontario) Anton Dovydaitis writes: >I've been getting a lot of requests for 4x4x4 cubes, and we're >looking into getting them. However, I have a couple questions. > >1) Why are 4x4x4 cubes so interesting? Do the additional symmetries >make for interesting questions, are they more fun, or easier to >solve? A well lubed 4x4x4 cube is still relatively easy to physically manipulate. As der Mouse suggests, it is arguably the largest interesting cube from a solver's point of view. Once one starts actually twisting with a 5x5x5 cube, the physical problems become more severe, e.g. the stickers come off easier, turning the slice you want to is more of a challenge, etc. In the virtual realm of computer cubing the patterns you can create are more elaborate, although I find in practice that finding pretty patterns on the 5x5x5 can become wearisome due to fact there are 9 centre pieces per side! >2) It appears to me that if you know how to solve the 3x3x3 Rubik's > cube, then you can easily solve the 5x5x5 rubiks (i.e., the > solution is derivative). For example, you can treat the inner 3x3 > faces of the 5x5x5 as a single 3x3x3 cube. Using the 4x4x4 cube we can produce a single exchange of centres and an exchange of edge pairs, and we can invert a single edge pair. Thus we can construct all the impossible 3x3x3 patterns except those involving a twist of a single corner! That is why I think the 4x4x4 cube is a good cube to have. The individual centre cubies can naturally wander all over the cube, and on the 3x3x3 cube they are fixed. In the case of the 5x5x5 cube, lots of the 3x3x3 knowledge does help. When dealing with the 5x5x5's middlemost slice (let's call one such slice "fm" for middlemost front slice) some of the 3x3x3 move sequences will move the appropriate edges, but now these sequences will also move centre pieces, specifically the ones which have no counterpart on the 3^3 and 4^3. To solve cubes 4x4x4 and greater requires new sequences to efficiently move centre cubies at will, and in the case of the 5^3 there really is no standard language to interchange move sequences and label individual cubies. I find having a letter as a mnemonic helps, so I'll suggest the following as an extension of Singmaster's 4x4x4 notation for the 5x5x5 cube: L left face l inner left slice lm left middlemost slice R right face r inner right slice rm right middlemost slice F front face f inner front slice fm front middlemost slice B back face b inner back slice bm back middlemost slice U up face u inner up slice um up middlemost slice D down face d inner down slice dm down middlemost slice Again, we follow the alphabetic component by a number to signify the rotation (1 = 90, 2 = 180, 3 = 270 or -90) This is overkill, and we can dispense with rm, bm and dm. Thus we could flip 2 middlemost edges at FD and BD with: (fm1 D1) ^3 + fm1 D2 + (fm1 D1) ^3 + fm1 (disturbs some centres) followed by: (fm2 + u2) ^2 + (fm2 + lm2) ^2 (corrects the centres) I believe this is correct, and I will double-check on my physical 5x5x5 at home. Definitely 5^3 cubing is a sport for the specialist ;-> -> Mark Email: mark.longridge@canrem.com From ncramer@bbn.com Sat Jan 15 09:22:26 1994 Return-Path: Received: from LABS-N.BBN.COM by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA03599; Sat, 15 Jan 94 09:22:26 EST Message-Id: <9401151422.AA03599@life.ai.mit.edu> Date: Sat, 15 Jan 94 9:12:37 EST From: Nichael Cramer To: CRSO.Cube@canrem.com Cc: cube-lovers@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 "brand-named" cubes, not cheap twiz-o knock-offs. On the other hand all of the 5X's I've owned have been _very_ easy to turn without any special customization. Except for the tendency (as Mark mentions) for the stickers to come off of one of them, they're consistently more comfortable to the hand than any of the 3X's I've owned. Nichael ncramer@bbn.com -- Captain and left quarter guard, BBN Calvinball Team From mouse@collatz.mcrcim.mcgill.edu Sat Jan 15 12:33:29 1994 Return-Path: Received: from Collatz.McRCIM.McGill.EDU by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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 cube-lovers@ai.mit.edu; Sat, 15 Jan 1994 12:33:21 -0500 Date: Sat, 15 Jan 1994 12:33:21 -0500 From: der Mouse Message-Id: <199401151733.MAA02409@Collatz.McRCIM.McGill.EDU> To: cube-lovers@ai.mit.edu Subject: Re: Higher Order Cubes >> A well lubed 4x4x4 cube is still relatively easy to physically >> manipulate. As der Mouse suggests, it is arguably the largest >> interesting cube from a solver's point of view. Probably true; it's the largest cube that actually offers new challenges. However, bigger cubes are better in that they offer more variety for making pretty patterns. :-) >> Once one starts actually twisting with a 5x5x5 cube, the physical >> problems become more severe, e.g. the stickers come off easier, >> turning the slice you want to is more of a challenge, etc. > This is interesting, because it's almost exactly the opposite of my > experience. > The problem seems to be the difference between the internal > mechanisms of the odd- and even- ordered cubes. This brings up an interesting point. Perhaps it would be possible to build a 4-Cube that was internally a 5-Cube but for which the middle slice was not actually visible on the surface? Or a 2-Cube that's internally a 3-Cube? I wonder if it might make for smoother-turning cubes. > [O]f the four 4X's I've owned, only one was really useable; two were > so stiff they were very difficult to turn (even with lubrication) and > one was so loose that it never lasted more than about 20 minutes > before dissolving into a pile of cubelets [...]. I have owned only one 4-Cube, and it's been long enough since I knew where it was that I don't recall how easy it was to turn. I now have two 3-Cubes and a 5-Cube. One of the 3-Cubes is a joy to turn; it's lubed enough that it turns readily and easily, even when the turn has a good deal of skew to correct, but it's not so loose that it turns when I don't want it to. (The other 3-Cube is (a) missing one center cubie face and (b) much more difficult to turn.) The 5-Cube (one of the recent Ishi Press cubes, btw) is mechanically quite good, though the orange stickers did tend to come off (no other color did, and contact cement worked just fine for putting them back on). Not as good as my good 3-Cube, though. I've wondered whether it would be possible to build higher-order cubes. The corners of the 5-Cube still catch by a respectable amount as the face turns, but by little enough that it makes me wonder if the 6-Cube or 7-Cube is actually feasible. (Oh, for a really good force-reflecting dataglove...then such a thing could be done virtually with no problem at all!) der Mouse mouse@collatz.mcrcim.mcgill.edu From @mail.uunet.ca:mark.longridge@canrem.com Sat Jan 15 17:03:53 1994 Return-Path: <@mail.uunet.ca:mark.longridge@canrem.com> Received: from mail.uunet.ca (uunet.ca) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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/SMI-4.1) id AA29781; Sat, 15 Jan 94 17:00:39 EST Received: by canrem.com (PCB-UUCP 1.1f) id 1914B9; Sat, 15 Jan 94 16:53:33 -0400 To: cube-lovers@life.ai.mit.edu Reply-To: CRSO.Cube@canrem.com Sender: CRSO.Cube@canrem.com Subject: 4x4x4 & 5x5x5 processes From: mark.longridge@canrem.com (Mark Longridge) Message-Id: <60.710.5834.0C1914B9@canrem.com> Date: Sat, 15 Jan 1994 15:44:00 -0500 Organization: CRS Online (Toronto, Ontario) Here are a couple of processes for larger cubes, plus the requested edge pair flip without disturbing centres (p2), as well as a minor correction for the 5x5x5 process: 4x4x4 processes (measured in slice moves) --------------- p1 Flip LD edge pair (r3 D3) ^3 + (r1 D1) ^4 + Rr3 D3 R1 D1 r3 D3 R3 (disturbs centres) D1 R1 D3 (25) p2 Flip UB edge pair r2 D2 l3 D1 R3 U1 R3 U3 l3 U1 R1 U3 l1 R1 D1 (retain centre positions ) + U2 r1 (u2 r2 l2) ^2 + r3 U2 r2 (26) 5x5x5 processes (measured in slice moves) --------------- Flip 2 middlemost edges at FD and BD with: p1 (fm1 D1) ^3 + fm1 D2 + (fm1 D1) ^3 + fm1 (disturbs some centres) followed by: D1 + (fm2 u2) ^2 + (fm2 l2) ^2 + D3 (corrects the centres) (25) -> Mark Email: mark.longridge@canrem.com From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Mon Jan 17 09:09:39 1994 Return-Path: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU> Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA27830; Mon, 17 Jan 94 09:09:39 EST Message-Id: <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 X-Acknowledge-To: Date: Mon, 17 Jan 1994 09:06:59 EST From: "Jerry Bryan" To: "Cube Lovers List" Subject: Number of M-Conjugate Classes for GC\M On 4 December 1993, I posted results from a breadth-first exhaustive search of GC\M, the corners of the 3x3x3, reduced by M-conjugation. The posting included a summary of how many conjugate classes there were at each level of the search tree (i.e., distance from Start). It occurred to me that I had not also posted a summary of M-conjugates for the corners of the 3x3x3 by the size of the conjugate classes. I searched my records, only to discover that I had never calculated such sizes. If I had, I probably would have been forced to analyze properly the distinction between M-conjugation and B-conjugation, because B-conjugation makes no sense for the corners of the 3x3x3. B-conjugation *can* be performed for the corners of the 3x3x3, but you end up with the 2x2x2 instead because B-conjugation effectively removes the centers. Anyway, I have now calculated M-conjugate class sizes for GC\M via computer search, and here are the results. M-Class Number Size of Classes 1 1 2 1 3 3 4 1 6 34 8 33 12 301 16 104 24 9064 48 1832428 Total 1841970 Notice that with M-conjugation, the maximum class is size is 48, rather than 1152 as it is with B-conjugation. Hence, my posting of 4 December 1993 incorrectly identified the results as being for "1152 fold symmetry". The results are correct, but they should be labeled as being for "48 fold symmetry", i.e., for M-conjugation rather than for B-conjugation. In calculating M-conjugate class sizes for GC\M, I did not "start from scratch". Rather, I used the existing results for B-conjugate classes as a base. In the case of B-conjugate classes of order 1152, no calculations are required. Each such B-class can simply be partitioned into 24 M-classes of order 48. Hence, I had to perform calculations for less than 4% of the B-classes. Here is a summary matrix, showing for each B-class size the number of each M-class size which are derived. M-Class Size 1 2 3 4 6 8 12 16 24 48 Total 24 1 0 1 0 2 1 0 0 0 0 5 B-Class 48 0 1 0 0 1 2 2 0 0 0 6 Size 72 0 0 2 0 11 0 2 0 5 0 20 96 0 0 0 1 0 1 3 0 2 0 7 144 0 0 0 0 20 0 42 0 30 14 106 192 0 0 0 0 0 29 0 8 73 16 126 288 0 0 0 0 0 0 252 0 682 406 1340 384 0 0 0 0 0 0 0 96 0 224 320 576 0 0 0 0 0 0 0 0 8272 22360 30632 1152 0 0 0 0 0 0 0 0 0 1809408 1809408 Total 1 1 3 1 34 33 301 104 9064 1832428 1841970 The first row of the matrix exemplifies the process of calculating M-Class sizes from B-Class sizes. In the case of corners, there is only one B-class of order 24, namely Start. The 24 elements of the B-class are the 24 elements of the form Ic, where c is in C, the 24 rotations of the cube. Under B-conjugation, these 24 elements are equivalent (i.e., in a centerless cube such as the 2x2x2, the 24 rotations of I are indistinguishable). But in a cube with centers, such as the corners of the 3x3x3, the 24 elements are not equivalent. For example, the M-class of order 1 is {I}. One of the M-classes of order 6 is {FB', UD', RL', LR', BF', DU'}. The M-class of order 3 is {FFB'B', RRL'L', UUD'D'}. That's as many as I can do in my head, but I think the pattern is clear. M-classes are a partition of the B-classes. In the case of B-classes of order 1152, the partition is regular -- i.e., you get exactly 24 M-classes of order 48. However, all partitions are not regular. In the partition of the B-class of I which we just discussed, there is 1 M-class of order 1, 1 M-class of order 3, 2 M-classes of order 6, and 1 M-class of order 8, for a total of 24 M-classes. Many other partitions are not regular, as well. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU If you don't have time to do it right today, what makes you think you are going to have time to do it over again tomorrow? From dseal@armltd.co.uk Mon Jan 17 14:14:08 1994 Return-Path: Received: from eros.britain.eu.net by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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) cube-lovers@ai.mit.edu Subject: Re: Higher Order Cubes Message-Id: <2D3AD5C5@dseal> In-Reply-To: <199401151733.MAA02409@Collatz.McRCIM.McGill.EDU> > This brings up an interesting point. Perhaps it would be possible to > build a 4-Cube that was internally a 5-Cube but for which the middle > slice was not actually visible on the surface? Or a 2-Cube that's > internally a 3-Cube? I wonder if it might make for smoother-turning > cubes. Yes, I think you could build such a 4-Cube. Likewise, you could build a 2-Cube as a 3-Cube with invisible middle slices. But I don't believe you'd want one: it could get completely jammed much too easily. The reason: If you take a 3-Cube and rotate its left and right slices 45 degrees each, you cannot rotate any of its other faces. This isn't surprising, since you don't expect to be able to perform one rotation halfway through another. If its middle slices were hidden, however, it would *appear* to be a 2-Cube which is not halfway through a rotation, and the fact that you couldn't move any faces but the left and right ones would be surprising - and undesirable. Unfortunately, I believe such a situation could probably arise quite easily. If you were to take the 2-Cube concerned and rotate its right face 90 degrees relative to its left face, you're going to be OK if the hidden middle layer rotates 0 or 90 degrees relative to the left face, but not OK if it rotates any other amount. I suspect most mechanisms would be more liable to rotate it an intermediate amount! There may be a way out, though. If you can anchor the place where the three axes meet to one of the corner cubelets in some way, the problem is solved: if the "anchor cubelet" is in the left face, then the hidden layer will rotate 0 decrees; if it is in the right face, then 90 degrees. David Seal dseal@armltd.co.uk From mouse@collatz.mcrcim.mcgill.edu Mon Jan 17 16:23:05 1994 Return-Path: Received: from Collatz.McRCIM.McGill.EDU by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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 cube-lovers@ai.mit.edu; Mon, 17 Jan 1994 16:22:50 -0500 Date: Mon, 17 Jan 1994 16:22:50 -0500 From: der Mouse Message-Id: <199401172122.QAA05806@Collatz.McRCIM.McGill.EDU> To: cube-lovers@ai.mit.edu Subject: Re: Higher Order Cubes >> Perhaps it would be possible to build a 4-Cube that was internally a >> 5-Cube but for which the middle slice was not actually visible on >> the surface? Or a 2-Cube that's internally a 3-Cube? > Yes, I think you could build such a 4-Cube. Likewise, you could build > a 2-Cube as a 3-Cube with invisible middle slices. But I don't > believe you'd want one: it could get completely jammed much too > easily. > The reason: If you take a 3-Cube and rotate its left and right slices > 45 degrees each, you cannot rotate any of its other faces. Duh, yeah; that never occurred to me. > There may be a way out, though. If you can anchor the place where the > three axes meet to one of the corner cubelets in some way, the > problem is solved: [...]. Yes. I think this may be possible, too...consider a normal 3-Cube, and restrict yourself to R, U, and F turns. Then ignore the center and edge cubies - the ones that get invisibilized. You're left with a 2-Cube. Three edge cubies never move with respect to the center cubies or the corner cubie they surround; glue those together. Presto! The same treatment is not possible for making a 4-Cube out of a 5-Cube, but an alternative occurs to me, that I *think* will work for higher cubes: key three of the (invisible) center cubies to the center six-pronged piece, so that they can't turn. Then half the face turns will cause the invisible center slice to turn with them; non-face slices (which don't exist on the 2/3-Cube) work normally. I notice with this construction for (say) a 4-Cube, the puzzle core turns whenever certain face slices do. With the 4-Cube I owned (and presumably still own, if I could find it), the puzzle core turns whenever certain next-to-center slices do. I suspect the latter would make for a smoother-turning puzzle. Perhaps someone will someday build a 5-Cube-turned-4-Cube and this can be determined. In the (IMO unlikely) event I originated any of the above ideas, I hereby place it/them in the public domain. Go wild, Ishi Press. :-) der Mouse mouse@collatz.mcrcim.mcgill.edu From Mikko.Haapanen@otol.fi Wed Jan 19 15:09:31 1994 Return-Path: Received: from lassie.eunet.fi by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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/IDA-1.5 for ); Wed, 19 Jan 1994 22:07:44 +0200 Received: from rhea.otol.fi by sulo.otol.fi with SMTP (PP) id <27853-0@sulo.otol.fi>; Wed, 19 Jan 1994 22:07:41 +0200 Received: from otol.fi by rhea.otol.fi id <25938-0@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: cube-lovers@ai.mit.edu In-Reply-To: <60.710.5834.0C1914B9@canrem.com> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII 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 Return-Path: Received: from Sun0.AIC.NRL.Navy.Mil by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA17275; Fri, 21 Jan 94 18:32:30 EST Received: by Sun0.AIC.NRL.Navy.Mil (4.1/SMI-4.0) 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) Message-Id: <9401212332.AA13137@Sun0.AIC.NRL.Navy.Mil> To: Cube-Lovers@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 24-element sets, each of which is an equivalence class when two positions are considered equivalent when they differ by their position with respect to the corners. The classes are called the cosets of C in G, GC, and GE, respectively. > Let G\M be the set of M-conjugate classes for G..... > Let GC\M be the set of M-conjugate classes for GC.... > Let GE\M be the set of M-conjugate classes for GE.... The partition of a group into conjugacy classes is not at all the same as the partition into cosets. So I would prefer to use different symbology, like "\" for conjugacy and "/" for cosets, but.... > Recall that B is the function which calculates the canonical form > for a cube under the composed operations of M-conjugation plus > rotation. My programs calculate equivalence classes under B. > Let G\B be the set of B-classes for G [ and likewise for GE, GC ]. Well, if you are using "\" for a generic partition into equivalence classes, then we should really do something like G\Conj(M) for partitions into conjugacy classes. At least then you can say G/C=G\Cosets(C). > Then, we have Gx\B=(Gx\C)\M=(Gx\M)\C. In English, we can decompose > B into a multiplication by C and M (in either order). No, that's _multiplication_ by C and _conjugation_ by M. A good example of why it's important not to use confusing symbols. M and C are not at all treated the same, except inasmuch as they are used to induce partitions into equivalence classes. Say instead that Gx\B = (Gx/C)\Conj(M) = (Gx\Conj(M))/C. > If I wanted to model GC\C, I would have had to either model only > seven of the cubies, or else modeled all eight but moved only seven > of them. Since what I really wanted was (GC\C)\M, and since what I > had was GC, I had to invent this funny B thing, where GC\B=(GC\C)\M. > If I had been clever enough to model GC\C in the first place, I > never would have had to invent B. Similar comments apply to my > model for the edges. Well, the part about moving only seven (corner) cubies is the approach that's been taken before on this list to deal with cubes that don't have face centers. It has the advantage that the object being treated is a group. But the problem is that the group is no longer cubically symmetrical (in some vague sense). This led me, at least, to lose track of the structure that would allow analysis of M-conjugacy. So I have to admire your tackling GE as a whole, instead of trying to stick to GE/C. At first blush, it looks like GE/C is 24 times smaller. But since GE/C\Conj(M) is almost 48 times smaller still, it's important to work in GE at least enough to be able to use the conjugation. Which is beside the point that I'm actually very interested in the structure of Gx/Conj(M) itself. And that is what I was really getting at in 1984 when I asked about how many positions there really are. Dan Hoey Hoey@AIC.NRL.Navy.Mil From walace@ntiaa.embrapa.ansp.br Fri Jan 21 21:35:11 1994 Return-Path: Received: from fpsp.fapesp.br by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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: cube-lovers@life.ai.mit.edu Message-Id: <4CD9C4A5C000C949@fpsp.fapesp.br> X-Envelope-To: cube-lovers@ai.ai.mit.edu Dear reader, Due to the wide spectrum of people that might be interested in the subjects to be discussed during the III International Conference on Systems Integration, we decided to post this call for papers in your mailing list. We encourage you to participate in this event as a paper author. The paper arrival deadline is March 3, 1994. Please forward this message to whoever you think it might be of interest and we appreciate your effort to post it. Thanks, Prof. Fuad Gattaz Sobrinho Conference Chairman ----------------------------------------------------------------- Call for Papers The Third International Conference for Systems Integration Sao Paulo City - Brazil July 30th - August 6th, 1994 ----------------------------------------------------------------- The Integration of Society for the Social, Economical, Scientific and Technological Development. This conference focuses on the integration of technologies, processes and systems, and the development of mechanisms and tools enabling solutions to complex multi-disciplinary problems dealing with agriculture, housing, telecommunications, financing and business, public services, education and software. The conference will provide an international and interdisciplinary forum in which researchers, educators, managers, practitioners and politicians, involved within the production process, can share novel research and development, education, production, trading, management and political experiences. Papers should deal with recent effort in theory, design, implementation, methodology, technics, tools and experiences of integration. Topics to be addressed include, but are not limited to: Technical and Scientific Aspects: - Integration, Modeling, Characterization and Automation of Process and Systems - Reengineering and Simplification of Processes - Computational Environments and Software Factories for Engineerind, Design, Manufacturing and System Development - Rol of Human Engineering in Integration - Experiences within National or Continental Software Projects - The Implication of Systems Integration for Manpower Skills - Quality Control and Certification in Organizational and Process Integration. Social, Political and Economical Aspects: - Experiences in Modeling, Development, Evolution and Integration of Enterprises - Experiences in Management and Identification of Value-Add Chains within Agriculture, Housing, Telecommunications, Financing and Business, Public Services, Education and Software - Public Policies and City Management - Management of Multi-dimensional Integration. Infrastructure Aspects: - Qualified Information Resources - Education and Training - Science and Technology - Enterprise Development. Information and Instructions for Authors: All papers must be in English or Portuguese, typed in double spaced format, and may not exceed 6,000 words. Each submission should provide a cover page containing author(s), affiliation(s), complete address(es), identification of principal author, and telephone number. Also include SIX copies of complete text with a title and abstract. Notice of acceptance will be mailed to the principal author(s) by March 15, 1994. If accepted, the author(s) will prepare the final manuscript, in English, in time for inclusion in the conference proceedings and will present the paper at the conference; otherwise, the author(s) will incur a page charge. Authors of accepted papers must sign a copyright release form. The proceedings will be published by the IEEE Computer Society Press. Send SIX copies of your paper(s) to: Prof. Peter A. Ng IIISis - USA Office - New Jersey Institute of Technology University Heights Newark, NJ 07102 USA For Further Information, Contact: Prof. Peter A. Ng Prof. Fuad Gattaz Sobrinho Fone:(1) (201) 596-3387 OR Phone:(55)(192) 41-4504 Fax: (1)(201) 596-5777 Fax: (55)(192) 41-3098 Email: ng_p@vienna.njit.edu Email: iiisis@ccvax.unicamp.br ------------------------------------------------------------------- >>>>>>>>>> Paper Arrival Deadline: March 3rd, 1994 <<<<<<<<<<<<<<<< ------------------------------------------------------------------- CONFERENCE COMMITTEE Conference Chair Fuad Gattaz Sobrinho IIISis Program Chair Peter A. Ng NJIT Finance & Business Co-Chair Alcir A. Calliari Banco do Brasil Agriculture Co-Chair Ney B. Araujo ABAG European Co-Chair Herbert Weber University of Dortmund Pac!fic Co-Chair Fumihiko Kamijo IPA Middle East Co-Chair Asuman Dogac METU South America Co-Chair Julio C. S. P. Leite PUC/RJ North America Co-Chair Bruce Berra Syracuse University Tutorials Co-Chairs Oscar Ivan Palma Pacheco EMBRAPA Murat M. Tanik SMU Organization Co-Chairs Rita de Cassia A. Marchiore IIISis Carole Poth NJIT Steering Committee Chair Peter A. Ng NJIT Honorary Advisors Raymond T. Yeh C. V. Ramamoorthy Laurence C. Seifert Honorary Conference Chair Irma Rossetto Passoni Sc&Tech, Info. and Comm. Comission of Brazilian Congress. Sponsored by IIISis - International Institute for Systems Integration, BB - Banco do Brasil, TELEBRAS, FINEP, CNPq, FBB, with colaboration of NJIT, SUCESU, EMBRAPA, ABAG, ACM e IEEE Computer Society. Instituto Internacional de Integracao de Sistemas - IIISis - Brazil. From hoey@aic.nrl.navy.mil Mon Jan 24 19:15:23 1994 Received: from Sun0.AIC.NRL.Navy.Mil by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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/SMI-4.0) id AA07954; Mon, 24 Jan 94 19:15:16 EST Return-Path: 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 Message-Id: <9401250015.AA05746@sun13.aic.nrl.navy.mil> To: Cube-Lovers@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 B-conjugacy classes of X is > the set of all m'Xmc for all m in M and all c in C. We denote > this set as BClass(X). B is the function B(X)=min(BClass(X)). That's a little unfortunate--I'd prefer to use B(X) for the equivalence class, and min(B(X))--or repr(B(X))--for the canonical representative. That's because the representative is not the important thing here, it's just a convenient way to represent (!) the class in a computer. > Note that we could have defined BClass(X) equivalently as the set of > all mXm'c, or as the set of all cm'Xm, or as the set of all > cmXm'.... This is the justification for the assertion in a previous > note that Gx\B = (Gx\M)\C = (Gx\C)\M. Not quite. The justification for (Gx\Conj(M))/C = (Gx/C)\Conj(M) is that instead of m'Xmc we could choose m'Xcm, a possibility you didn't list. In his message of "Sat, 8 Jan 1994 10:52:22 EST", Jerry continues with discussion on combining conjugacy classes. We've exchanged some private email on the subject material, but in case anyone on the list is following this stuff.... > There are only 10 distinct values for |BClass(X)| and for > |BClass(Y)|, namely 24, 48, 72, 96, 144, 192, 288, 384, 576, and > 1152. (By the way, I have never figured out why it is *exactly* the > same set of values for both the corners and for the edges. It is > easy to see why it is approximately the same set of values.... I'm not sure what kind of approximation you mean, but certainly those ten values are all that are possible: Proof: For if m1,m2 are in the same coset of M/CSymm(X), then (m1 m2') is in CSymm(X), so X' (m1 m2')' X (m1 m2') = c0 in C so m1' X m1 = m2' X c0 m2. It's then clear that { m1' X m1 c : c in C } = { m2' X m2 c : c in C } (*) are equal 24-element sets. The same manipulation in reverse shows that if (*) holds for some m1,m2 in M, then m1 and m2 are in the same coset of M/CSymm(X). So |BClass(X)|=24 |M/CSymm(X)|. |M/CSymm(X)| must be a divisor of |M|=48, QED. It wouldn't have been all that surprising to see one of the possible sizes of |CSymm(X)| fail to appear as a symmetry group of the corners or edges, but it's not surprising that they all do, either. > [For the original approach] I needed to be able to prove that for a > fixed m and n, that |(BClass(X)[m] * BClass(Y)[n]| had the same > value for all X in GC[m]\B and all GE[n]\B. That is to say, that the sizes |CSymm(X)| and |CSymm(Y)| might determine {|Symm(X*Y)|} for X in GC\B, Y in GE\B, and so (X*Y) in G\Conj(M). It doesn't, but the situation is even worse. Jerry goes on to suppose that perhaps CSymm(X) and CSymm(Y) themselves might determine {|Symm(X*Y)|}, and even that isn't true. I've discovered this by a computer search of GC\B. (A search of GE\B is in progress, but for the current result we can take Y=I in GE\B). I have found that AllSymms(X) is not determined, even up to subgroup sizes, by CSymm(X). According to the search, the following are the only positions of GC\B for which |CSymm(X)|=16. X1 X2 X3 +---+ +---+ +---+ |F F| |B F| |F B| |B B| |F B| |F B| +---+---+---+ +---+---+---+ +---+---+---+ |R R|D D|L L| |L L|D T|R R| |L L|D T|R R| |R R|T T|L L| |L L|T D|R R| |L L|D T|R R| +---+---+---+ +---+---+---+ +---+---+---+ |B B| |F B| |B F| |F F| |B F| |B F| +---+ +---+ +---+ |T T| |T D| |T D| |D D| |D T| |T D| +---+ +---+ +---+ Coincidentally, I have been (privately) calling the CSymm(Xi) subgroups the "X subgroups" of M, an X subgroup being the subgroup that maps an orthogonal axis of the cube (in the above examples, the L-R axis) to itself. X1 is a notable position, in that each corner has been swapped with its opposite corner. Symm(X1) is an X subgroup as well, and there is another X subgroup in AllSymms(X1). There is, however, no 16-element subgroup in AllSymms(X2) or AllSymms(X3). (We have seen X2 before: it is the corners of the Laughter (or 4/) position). In fact, my program says that AllSymms(X1) contains two occurrences of 16-element X subgroups, two occurrences of the 8-element HX subgroup, two occurrences of 8-element R subgroups, two occurrences of 8-element S subgroups, eight occurrences of 4-element HS subgroups, and eight occurrences of the 2-element HV subgroup. AllSymms(X2) and AllSymms(X3) each contain two occurrences of 8-element CX subgroups, two occurrences of 8-element AX subgroups, two occurrences of 8-element P subgroups, two occurrences of 8-element Q subgroups, eight occurrences of 2-element ES subgroups, and eight occurrences of 2-element HW subgroups. The names of these groups are part of a taxonomy of the subgroups of M I've developed, which I won't go into just now. But the point that I find surprising here is that AllSymms(X1) and AllSymms(X2) are completely disjoint. While that can't happen all the time (smaller CSymm() groups have many occurrences of the one-element "I" subgroup) I think the tendency to disjointness is too pronounced to be simple anti-coincidence. Dan Hoey Hoey@AIC.NRL.Navy.Mil From cuf@aol.com Thu Feb 10 23:09:33 1994 Return-Path: Received: from mailgate.prod.aol.net by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers 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 X-Mailer: America Online Mailer Sender: "cuf" Message-Id: <9402102217.tn53025@aol.com> To: cube-lovers@life.ai.mit.edu Date: Thu, 10 Feb 94 22:17:05 EST Subject: Computer & Health The Computer User Family (CUF) is concerned about the health problem associated with computers. Video Display Terminals, emit UV and ELF radiation and may cause cancer, immune system irregularities, miscarriages and eye fatigue. Computer noise from fans, disk and CD drives is also becoming a source of anxiety, stress and general discomfort . We usually don't realize how loud our computers are: 50dB and more. These problems should be dealt with and add-ons should be provided for present computers to avoid putting us at risks. Some safe screens and quiet power supplies are coming out but they are marginal and prices are prohibitive. Meanwhile the general guidelines for the users are: 1. Position yourself approximately 22 inches to 28 inches (arm's length) from the screen and four feet from the sides and rear of other terminals. 2. Eliminate sources of glare and lower light levels in the room. Don't sit facing a bright window. If necessary, use screen hoods, glare shields over the screen or wear anti-UV/anti-glare glasses. 3. Put a noise absorbing mat under your computer. Pull your computer away from the wall or any hard surface that reflects noise and vibration back to you. 4. Rest occasionally during periods of intense concentration. Closing your eyes helps. 5. Turn off the VDT when not in use. From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Sun Feb 13 16:59:40 1994 Return-Path: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU> Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA17813; Sun, 13 Feb 94 16:59:40 EST Message-Id: <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 X-Acknowledge-To: Date: Sun, 13 Feb 1994 16:59:22 EST From: "Jerry Bryan" To: "Cube Lovers List" Subject: Re: Some Proposed Terminology In-Reply-To: Message of 01/21/94 at 18:32:15 from hoey@AIC.NRL.Navy.Mil On 01/21/94 at 18:32:15 hoey@AIC.NRL.Navy.Mil said: >I welcome Jerry Bryan's efforts to >improve the terminology of the groups associated with Rubik's cube. >But there is some additional clarification I think is necessary. >> Let G\C be the corners with edges without centers group. I intend >> for the notation to indicate G reduced by C, where C is the rotation >> group for the cube.... >> Let GC\C be the corners without edges without centers group.... >> Let GE\C be the edges without corners without centers group.... >First, these are not, strictly speaking, groups. Well, you can make >them groups, by defining what the group operation is. But I don't >know any way of doing that without losing the symmetrical nature of >the problem. >Second, I would suggest that G/C, GC/C, and GE/C are more standard >names for these objects. The elements are nominally 24-element sets, >each of which is an equivalence class when two positions are >considered equivalent when they differ by their position with respect >to the corners. The classes are called the cosets of C in G, GC, and >GE, respectively. Dan Hoey's criticism's are quite valid. I will attempt to repair the damage as follows: 1) accept the Gx/C notation in lieu of Gx\C, 2) define an operation within Gx/C such that Gx/C is a group, and 3) use Gx/C as a model for cubes without centers in such a way that the symmetrical nature of the problem is retained. Let C be the set of twenty-four whole cube rotations of the cube, and let G be the standard 3x3x3 cube group. We observe that if X is a cube in G, then c'Xc is also a cube in G for every c in C. We could call this operation C-conjugancy. However, there is seldom (if ever) any reason to speak of C-conjugancy. That is, C is just a subset of M, the set of forty-eight whole cube rotations and reflections. Indeed, C is half of M, and the other half of M is the reflection of C. Hence, M-conjugancy of the form m'Xm is more powerful than C-conjugancy, and there is normally no reason to speak of C-conjugancy. I only bring it up to emphasize that if X is in G, then c'Xc is in G. On the other hand, if I understand correctly the model most people use for G, elements of the form Xc or cX are not in G except for the trivial case where c=I. The problem is that C is considered to move the centers, but G is generated by Q, the set of quarter-turns of the faces, and Q does not move the centers. For example, there is a c in C such that F=c'Rc, but there is not a c in C such that F=Rc or F=cR. And indeed, neither Rc nor cR are in G at all unless c=I. As we said, G is generated as G=, where Q is the set of quarter-turns Q={F,B,U,D,L,R,F',B',U',D',L',R'}. Elements of Q move the corners and edges, but Q is the identity on the centers. C, on the other hand, is generally considered to move the centers. Hence, the group generated as is a supergroup of G, and there are elements of the supergroup which are not in G. (This supergroup, by the way, is not The Supergroup. The Supergroup is generated by Q alone, but with orientations of the (otherwise fixed) centers considered.) Therefore, our first order of business is to make C into a sub-group of G. We observe that since the elements of Q are the identity on the centers, the primary function of the centers is to provide a frame of reference. But we can provide a frame of reference without the centers actually being there. For example, consider the group GC consisting of cube centers and corners. You can model this group by removing the edge labels from a physical cube. Establish the cube at Start and perform RL'. The corners will be rotated forward, and will be positioned properly with respect to each other, but the cube is clearly not solved. You can tell that the cube is not at Start because the corners are not aligned properly with the centers. Now, do the same thing except remove both the edge and center labels. If you perform RL' at Start, the cube "looks" solved but rotated forward. However, we can adopt the convention that the cube is solved only if the Up color is Up, the Front color is Front, etc. With this convention in place, RL' is clearly seen not to be solved; it is two moves from Start. The convention provides the fixed frame of reference. Furthermore, RL' (which is in GC) is equal to an element of C, and indeed all elements of C are in GC, as are all elements of the form Xc or cX for c in C and X in GC. Hence, we have =. Similar comments apply to GE, the group of edges and centers, except that processes composed from elements of Q to accomplish rotations in C are not quite so short in GE as they are in GC. G, the full 3x3x3 cube group consisting of corners, edges, and centers is a bit more difficult. The problem is that if X is in G, then objects of the form Xc or cX are in G only if c is even. Twelve elements of C are even and twelve are odd. Indeed, C[even] is a sub-group of C, but C[odd] is not. We will deal with this situation (as circumstances require) in two different ways. One is simply to restrict ourselves to C[even] when dealing with G. The other is to define a new group we will call GS. In our model for G in which the centers are implied by a frame of reference convention rather than by actual physical centers, we can easily add slice moves to the standard face moves. If the centers were physically present, then the slice moves would move the centers, but without the physical centers there is no problem. If S is the set of slice moves, then GS is generated as . GS is essentially G with parity restrictions removed. Hence we observe that |G|=|GC|*|GE|/2, |GS|=|GC|*|GE|, and |GS|=|G|*2. Also, if X is in G or in GS, then elements of the form cX or Xc are in GS for all c in C. In those occasions where we are willing to think of GS rather than G, we can use C rather than C[even]. At this point, we can say that GS/C, G/C[even], GC/C, and GE/C are cosets of C in GS, C[even] in G, C in GC, and C in GE, respectively. To be a little more conformant with standard coset notation, we will write cube elements as lower case letters for the remainder of this note, and hence for a particular cube x a coset of C is denoted as Cx={y: y=cx} or xC={y: y=xc}. Now, we propose a group operator for the cosets: Cx Cy = C(xy) and xC yC = (xy)C. Showing that we have a group is easy. I originally included a proof in this note, but there is a proof in Chapter 8 of Frey and Singmaster's _Handbook of Cubik Math_. Hence, I will defer to their proof instead. According to Frey and Singmaster, G/C is called the factor group of C in G, or the quotient group of G by C. Of most significance to us right now is the fact that the identity of the factor group is Ci or iC, where i is the identity of G. But Ci or iC is just C. Hence, the identity of the factor group is C. This justifies our identification of G/C with a centerless cube. In English, it means that we can rotate a centerless cube in space without changing anything. I think this would comply with most people's intuitive sense of what it means for a cube to be centerless. Finally, as to whether this model retains the "symmetrical nature of the problem", I will have to leave that as an open question, depending on precisely what we mean by "symmetrical". It seems to me that this model does a better job of being "symmetrical" than a model which includes only seven corner cubies or only eleven edge cubies, but maybe not. What does "symmetrical" mean when it comes to centerless cubes? = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU If you don't have time to do it right today, what makes you think you are going to have time to do it over again tomorrow? From anandrao@hk.super.net Thu Feb 17 02:18:45 1994 Return-Path: Received: from hk.super.net by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA29723; Thu, 17 Feb 94 02:18:45 EST Received: by hk.super.net id AA16490 (5.65c/IDA-1.4.4 for Cube-Lovers@ai.mit.edu); Wed, 2 Feb 1994 10:13:07 +0800 Date: Wed, 2 Feb 1994 10:10:28 +0800 (HKT) From: "Mr. Anand Rao" Subject: Re: Mickey's Challenge To: Peter Beck Cc: Cube-Lovers@ai.mit.edu In-Reply-To: <9401061406.aa23113@COR6.PICA.ARMY.MIL> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII I recently picked up this puzzle on a trip to Houston, TX. It is really sad that even though this puzzle is made in China by a Hong Kong company, I had to go to the US to get it! It is not available for sale here in Hong Kong. Many thanks for your tip. If any one knows about any other interesting puzzles, they are welcome to contribute! Cheers! On Thu, 6 Jan 1994, Peter Beck wrote: > > NEW PUZZLE "MICKEY'S CHALLENGE" is at your Disney > store now, price $10. > > This is a legal MACHBALL, ie, a spherical > SKEWB. It comes with a solution book. > Christoph Bandelow (a longer time cuber) > wrote the solution. > I haven't bought one or it played with it > yet. > > GOOD PUZZLING > > pete beck > > pbeck@pica.army.mil > From xirion!jandr@relay.nl.net Fri Feb 18 08:45:09 1994 Return-Path: Received: from sun4nl.NL.net by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA17847; Fri, 18 Feb 94 08:45:09 EST Received: from xirion by sun4nl.NL.net via EUnet id AA26061 (5.65b/CWI-3.3); Fri, 18 Feb 1994 14:45:06 +0100 Received: by xirion.xirion.nl id AA02038 (5.61/UK-2.1); Fri, 18 Feb 94 14:43:52 +0100 From: Jan de Ruiter Date: Fri, 18 Feb 94 14:43:52 +0100 Message-Id: <2038.9402181343@xirion.xirion.nl> X-Organization: Xirion Unix Software & Consultancy bv Burgemeester Verderlaan 15 X 3454 PE De Meern The Netherlands X-Phone: +31 3406 61990 X-Fax: +31 3406 61981 To: cube-lovers@life.ai.mit.edu To: cube-lovers@life.ai.mit.edu Subject: Re: 10x10 Tangle Sorry about not reporting this earlier, but my search for solutions for Rubiks Tangle 10x10 confirms the finding of Don Woods: no solutions! Dik Winter writes: >As I wrote before, I have embedded in my memory that there is an easy >argument that the 10x10 is *not* solvable. I do not know whether I >found it myself (and ever did mail it to other people) or whether I >found it somewhere on the net; it is a long time ago. When I find the >time I will do a check. (I know very sure that I have had a program >running at that time but that I abandoned the search because it would >be fruitless.) I am beginning to get real curious about that 'easy argument'. Does this argument depend on the particular choice for the four duplicated pieces or not? If it does, there could exist a choice that does allow a solution, and we could re-define the puzzle as follows: find which four pieces to duplicate in order to find solutions for the 10x10. If the number of solutions varies depending on the choice, you could even add a restriction: find which four pieces to duplicate in order to find a set which has the minimum number of solutions for the 10x10. But if the easy argument does NOT depend on the choice, i.e.: any choice would lead to no solutions, then the above puzzles would be senseless as well. So if anyone at all knows this argument, please tell us and solve the mystery. Jan From hochberg@gnumath.rutgers.edu Fri Feb 18 14:36:47 1994 Return-Path: Received: from gnumath.rutgers.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA16297; Fri, 18 Feb 94 14:36:47 EST Received: by gnumath.rutgers.edu (5.59/SMI4.0/RU1.5/3.08) id AA26578; Fri, 18 Feb 94 14:36:45 EST Date: Fri, 18 Feb 94 14:36:45 EST From: hochberg@gnumath.rutgers.edu (Rob.) Message-Id: <9402181936.AA26578@gnumath.rutgers.edu> To: Cube-Lovers@ai.mit.edu Subject: Add a name, please... Could you add edgemstr@orange.cc.utexas.edu to the cube lovers list? Thank you kindly. Rob.