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 Ernö 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 ö 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-T