From @mail.uunet.ca:mark.longridge@canrem.com Mon Sep 4 23:11:05 1995 Return-Path: <@mail.uunet.ca:mark.longridge@canrem.com> Received: from seraph.uunet.ca (uunet.ca) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA03764; Mon, 4 Sep 95 23:11:05 EDT Received: from portnoy.canrem.com ([198.133.42.17]) by mail.uunet.ca with SMTP id <210113-1>; Mon, 4 Sep 1995 23:13:08 -0400 Received: from canrem.com by portnoy.canrem.com (4.1/SMI-4.1) id AA07431; Mon, 4 Sep 95 23:06:36 EDT Received: by canrem.com (PCB-UUCP 1.1f) id 1F3009; Mon, 4 Sep 95 23:01:18 -0500 To: cube-lovers@life.ai.mit.edu Reply-To: CRSO.Cube@canrem.com Sender: CRSO.Cube@canrem.com Subject: Dino Cube From: mark.longridge@canrem.com (Mark Longridge) Message-Id: <60.1225.5834.0C1F3009@canrem.com> Date: Mon, 4 Sep 1995 23:41:00 -0400 Organization: CRS Online (Toronto, Ontario) # Here are a few Dino cube calculations. The calculations for the # cube with an X cut on each of the 6 sides, assuming period 3 # rotations of 3 edges (there are 8 of these, one for each corner) # The Dino cube has 12! /2 = 239,500,800 essential states # Fixing one edge gives the Dino cube a fixed orientation # in space and gives 11! /2 = 19,958,400 combinations # It has less combinations then the standard pyraminx, but more # than the 2x2x2 Rubik's Pocket cube. # The Dino cube has 12 edges which can not flip, observed by Rubik # himself back in 1982 (re: Rubik's Logic & Fantasy in Space.) # Dino cube has trivial centre dino := Group( (1,24,7) (2,23,5), (2,12,22) (4,11,24), (4,19,10) (3,17,12), (3,5,20) (1,6,19), (13,21,11) (14,22,9), (14,8,23) (16,7,21), (16,18,6) (15,20,8), (15,9,17) (13,10,18) );; From BRYAN@wvnvm.wvnet.edu Sat Sep 9 09:52:21 1995 Return-Path: Received: from WVNVM.WVNET.EDU by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA24673; Sat, 9 Sep 95 09:52:21 EDT Received: from WVNVM.WVNET.EDU by WVNVM.WVNET.EDU (IBM VM SMTP V2R3) with BSMTP id 1908; Sat, 09 Sep 95 09:51:57 EDT Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU (LMail V1.2a/1.8a) with BSMTP id 1333; Sat, 9 Sep 1995 09:51:57 -0400 Message-Id: Date: Sat, 9 Sep 1995 09:51:56 -0400 (EDT) From: "Jerry Bryan" To: "Cube Lovers List" Subject: A Proposal for a More General Definition of Symmetry Our normal definition of symmetry for Rubik's cube is based ultimately on the 48 symmetries of the standard math book wire model of a cube, and the 48 symmetries were discovered long before Rubik's cube was ever dreamed of. This note is based on the conviction that these 48 symmetries do not really capture all that we might think of as "symmetry" when we think of Rubik's cube. This note has the further purpose to propose a more general definition of symmetry for Rubik's cube. I want to start with a couple of really basic concepts. I think every reader of this list knows what a permutation is, namely it is a one-to-one onto function on a set. In the case of a finite set as we have with Rubik, a function on a set is one-to-one if and only if it onto, so we can sometimes get by with speaking only of one-to-one or by speaking only of onto. But what is a symmetry? A very standard definition is something like "the set of all rigid motions that transform a given geometric figure onto itself" (James and James Mathematics Dictionary). Another way to say it is that the transformation preserves the figure. Working with that definition, a symmetry almost inevitably may be interpreted as a permutation. With simple geometric figures, the permutation would usually be described as being a permutation on Euclidean n-space -- 2-space for a square or circle, etc., and 3-space for a cube or sphere, etc. Hence, we might think of a symmetry as being a special kind of permutation, namely one that preserves a geometric figure in Euclidean n-space. I have had a difficult time finding books that address the issue of symmetry vs. permutation to my satisfaction. It is very hard to think of a symmetry abstractly enough that it doesn't simply turn into a permutation right before your eyes. Paul Yale's "Geometry and Symmetry" doesn't really seem to define a symmetry (it sort of assumes you know what one is), but it does describe the relationship between a symmetry and a permutation. I would paraphrase as follows. Label your geometric figure in some fashion -- e.g., label the edges, label the axes, label the vertices, or label *something*. Then, there is a homomorphism between the set of symmetries and the corresponding set of permutations on the labels. But I repeat that it is hard for me to conceive of the set of symmetries in a sufficiently abstract fashion that the symmetries themselves aren't already permutations on *some* set or other. So it seems to me that Yale could just as well be talking about homomorphisms between one set of permutations and another set of permutations as talking about homomorphisms between symmetries and permutations. A couple of quick additional points, and then I will go on: 1) since we are talking about homomorphisms, it is obvious that both the set of symmetries and the set of permutations to which they map are groups, and 2) most homomorphisms between symmetries and permutations turn out in fact to be isomorphisms. This latter observation gives added weight to the notion that symmetries are just a special kind of permutation. Given all that has been said so far, we could informally say that the normal definition of a symmetry is that it is a permutation that preserves a geometric figure. Our more general definition will simply be that a symmetry is a permutation that preserves some property. If we were sufficiently liberal in our notion of "preserving some property", then most any permutation could be interpreted as a symmetry. We will not be quite that liberal by the time we are done, but we will be more liberal than would be permitted by the standard 48 math book symmetries of the cube. But what property of Rubik's cube should we try to preserve if we want a more general definition of symmetry than the normal one? I wish to motivate our definition of that property in several steps. The standard Rubik's cube definition of symmetry for a position X is Symm(X) is the set of all m in M such that X=m'Xm, or equivalently such that mX=Xm. M is the set of 48 permutations on the Rubik's cube corresponding to the 48 symmetries of a cube. Write a position Z as Z=XY, where X is the permutation on the corners and Y is the permutation on the edges. We have Symm(Z)=Symm(XY)=Symm(X) intersect Symm(Y). For example, we could have Symm(X)=M, Symm(Y)=I, and Symm(Z)=Symm(XY)=I. Such a position would look very "symmetrical" because the corners would be fixed (or "solved"), although the edges would be scrambled. Most people would consider such a position to be more "symmetrical" than one where both the corners and edges were scrambled, although we would have Symm(Z)=I in either case. A couple of points before proceeding: 1) From an information theory point of view, Symm(X) and Symm(Y) separately contain more information than does Symm(XY). There is an obvious loss of information when we calculate Symm(X) intersect Symm(Y). This is a strong indication that Symm(XY) does not tell us everything we might like to know about the symmetry of a position. 2) The set of positions Z=XY for which Symm(X)=M forms a group (as does the set of positions for which Symm(Y)=M). This anticipates where we are headed, namely that group membership is the property that we should seek to preserve in a more general definition of symmetry. A third (and equivalent) definition for Symm(X) is that Symm(X) is the set of all m in M such that X'm'Xm=I. Most readers will recognize X'm'Xm as a commutator. Per Dan Hoey, we can generalize and define CSymm(X) to be the set of all m in M such that X'm'Xm is in C, the set of 24 rotations of the cube. For example, if we have Z=XY as before, then CSymm(X)=M means that the corners are positioned properly with respect to each other, although they might be rotated with respect to the fixed face centers. Such a position would look fairly "symmetrical", even to a non-cubemeister, even though we might have Symm(Z)=I. Again, we have the set of all positions for which CSymm(X)=M forms a group. Similarly, the set of all positions for which CSymm(Y)=M forms a group, and the set of all positions for which CSymm(Z)=CSymm(XY)=M forms a group. Recall that there are 98 subgroups of M. For each subgroup K of M, there is a corresponding subgroup of G consisting of all the K-symmetric positions. So would could just as well define symmetry in terms of these 98 subgroups of G. But there are far more than 98 subgroups of G. (We don't know how many, and I doubt than even GAP could tell us). Why not simply define symmetry in G in terms of subgroup membership in G? The symmetry of a position X is then the set of all subgroups of H of G for which X is in H. And a symmetry operation (in the sense that a symmetry is a permutation that preserves something) is an operation that preserves subgroup membership. That pretty much completes my proposal, but I have a few closing remarks. 1) The proposed general definition of symmetry is analogous to the Thistlethwaite algorithm for solving the cube. Typical cube solutions gradually solve more and more of the cube. The "more and more of the cube" that gets solved can be characterized as a sequence of nested subgroups. Thistlethwaite reversed the process and created a sequence of nested subgroups which in turn solves more and more of the cube. Similarly, the standard definition of symmetry implies a set of 98 subgroups of G. We reverse the process and let all the subgroups of G define symmetry instead. 2) The proposed general definition of symmetry has the virtue that it includes the standard definition as a special case, since the 98 K-symmetric subgroups of G are in fact subgroups of G. 3) The proposed general definition of symmetry has the virtue that there is only one position that is "completely symmetric", namely Start itself (the identity permutation). The standard definition of symmetry has four positions which are "completely symmetric", which to me is an unsatisfactory state of affairs. (Recall that we have Symm(X)=M for Start, Pons Asinorum, Superflip, and the composition of Pons and Superflip. I am still bummed out that this is the case while at the same time only Start and Superflip are in the center of G. This suggests that Superflip is "more symmetric" than Pons. I wonder if such a suggestion would be supported by my proposed general definition 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 From morabito@omni.voicenet.com Mon Sep 11 20:48:15 1995 Return-Path: Received: from omni.voicenet.com by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA08218; Mon, 11 Sep 95 20:48:15 EDT Received: from cherryhill24.voicenet.com by omni.voicenet.com (5.x/SMI-SVR4) id AA03223; Mon, 11 Sep 1995 20:47:41 -0400 Date: Mon, 11 Sep 1995 20:47:40 -0400 Message-Id: <9509120047.AA03223@omni.voicenet.com> X-Sender: morabito@omni.voicenet.com Mime-Version: 1.0 Content-Type: text/plain; charset="us-ascii" To: cube-lovers@ai.mit.edu From: morabito@omni.voicenet.com (Morabito) Subject: submission X-Mailer: Hello. I would like to be part of your mail group. Please send me the letters at morabito@omni.voicenet.com. Thanks. From @mail.uunet.ca:mark.longridge@canrem.com Wed Sep 13 02:15:29 1995 Return-Path: <@mail.uunet.ca:mark.longridge@canrem.com> Received: from seraph.uunet.ca (uunet.ca) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA01159; Wed, 13 Sep 95 02:15:29 EDT Received: from portnoy.canrem.com ([198.133.42.17]) by mail.uunet.ca with SMTP id <210967-2>; Wed, 13 Sep 1995 02:17:54 -0400 Received: from canrem.com by portnoy.canrem.com (4.1/SMI-4.1) id AA14615; Wed, 13 Sep 95 02:11:18 EDT Received: by canrem.com (PCB-UUCP 1.1f) id 1F4769; Wed, 13 Sep 95 01:47:39 -0500 To: cube-lovers@life.ai.mit.edu Reply-To: CRSO.Cube@canrem.com Sender: CRSO.Cube@canrem.com Subject: Alexander's Star From: mark.longridge@canrem.com (Mark Longridge) Message-Id: <60.1229.5834.0C1F4769@canrem.com> Date: Wed, 13 Sep 1995 02:21:00 -0400 Organization: CRS Online (Toronto, Ontario) > From: pbeck@pica.army.mil (Peter Beck) > Subject: ranking > > noticed you did not include > square-1 or alexanders star > > thanks for the compilation Very little literature exists on Alexander's Star. Ideal Toy published a solution booklet, and Adam Alexander (the puzzle's inventor) wrote a book, and David Singmaster wrote his analysis in one of his Cubic Circulars. Dr. Singmaster also mentions Alexander's Star in Rubik's Cubik Compendium in a chapter about variations on Rubik's cube. The Ideal booklet mentions that Alexander's Star has 24 start positions, and Dr. Singmaster states there are 12 start positions, each one with it's own mirror reflection, again giving 24 start positions in total. Alexander's Star is akin to an "edge-only" Dodecahedron (Megaminx) without centres. It has 30 edges, that is 15 pairs of distinct 2 coloured edges. To calculate the number of permutations of N objects with N1 objects which are like, N2 objects which are like ... Nrth objects which are like we use the formula: N! -------------------- N1! * N2! * ... Nr! In the case of Alexander's Star it is a bit more complicated because we must contend with edge flips and no centres, however I believe the correct calculation is.... 30! / 2^15 * 2^29 / 60 = 72,431,714,252,715,638,411,621,302,272,000,000 approx = 7.2 * 10^34 I think the term for this is approximately 72 decillion. (decillion = 10^33) Here is a bit of an explanation.... There are 30 total objects, with 15 different pairs of 2 like edges. 29 edges may be flipped in any fashion but the 30th edge is forced. The Great Dodecahedron has 60 orientations in space. Here is another way to calculate the same number... Pick one edge and lock it in position and orientation. We still have 29 edges which can move in position and orientation. There are still 28 edges which can be flipped freely, the 29th edge being forced. We still have 15 different pairs of 2 like edges. 29! / 2^15 * 2^28 = 7.2 * 10^34 (approx) To be completely clear the calculation is really 29!/2 / 2^15/2 * 2^28 ....because only half of the 29! arrangements are possible and only half of the 2^15 arrangements are possible but both the /2's cancel. On the Great Cosmic Ranking List this puzzle has less combinations than Rubik's Revenge, but more than the VIP sphere. I know of no Square 1 calculations whatsoever, but would be interested in seeing anyone's calculations. -> Mark <- From dlitwin@geoworks.com Thu Sep 14 23:45:03 1995 Return-Path: Received: from quark.geoworks.com ([198.211.201.100]) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA20748; Thu, 14 Sep 95 23:45:03 EDT Received: from radium.geoworks.com by quark.geoworks.com (4.1/SMI-4.0) id AA15388; Thu, 14 Sep 95 20:40:52 PDT Date: Thu, 14 Sep 95 20:40:52 PDT From: dlitwin@geoworks.com (David Litwin) Message-Id: <9509150340.AA15388@quark.geoworks.com> To: cube-lovers@life.ai.mit.edu Subject: Alexander's Star While we are on the subject of the Alexander's star, I have never been entirely satisfied with the arrangment of its solution. I've noticed that most people who look at it don't even know it is solved and I have to explain that the solution is that all the stickers of pieces laying in a common plane around the points are the same color. Hard to say, but I can point it out to people. To this end I have spent some time trying to find a more satisfying solution, one more visually clear and simple. I've only come up with one alternative that I consider reasonable, and it isn't as pure as I would like. This solution involves grouping colors in the depressions of the star. The main problem lies in the fact that the edges come together in groups of three, but there are 10 stickers of each color so at some point having all the depressions of the star a single color breaks down. For this reason I choose one color (White is my preference) to be an exception and have it remain in the original configuration, i.e. all in two parallel planes. With the rest of the colors, I group them in groups of five: three in one depression, and two in an adjacent depression with the third color of this second depression being one of a white. The result of this is a set of interlocking "diamonds" that group visually because the are of the same color. Unrolled, the star would look like this (this should just fit on a display of 80 columns): /|\ /|\ /|\ /|\ /|\ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / Y_|_Y \ / B_|_B \ / O_|_O \ / R_|_R \ / G_|_G \ / _-' W `-_ \ / _-' W `-_ \ / _-' W `-_ \ / _-' W `-_ \ / _-' W `-_ \ _/-'_________`-\_/-'_________`-\_/-'_________`-\_/-'_________`-\_/-'_________`-\ |\-,_ _,-/|\-,_ _,-/|\-,_ _,-/|\-,_ _,-/|\-,_ _,-/ | \ `-_Y_-' / | \ `-_B_-' / | \ `-_O_-' / | \ `-_R_-' / | \ `-_G_-' / | \ Y " Y / | \ B " B / | \ O " O / | \ R " R / | \ G " G / | \ | / | \ | / | \ | / | \ | / | \ | / |_O \ | / R_|_R \ | / G_|_G \ | / Y_|_Y \ | / B_|_B \ | / O_ O `-_ \ | / _-' R `-_ \ | / _-' G `-_ \ | / _-' Y `-_ \ | / _-' B `-_ \ | / _-' _____`-\|/-'_________`-\|/-'_________`-\|/-'_________`-\|/-'_________`-\|/-'____ _,-/ \-,_ _,-/ \-,_ _,-/ \-,_ _,-/ \-,_ _,-/ \-,_ W_-' / \ `-_W_-' / \ `-_W_-' / \ `-_W_-' / \ `-_W_-' / \ `-_ " O / \ R " R / \ G " G / \ Y " Y / \ B " B / \ O | / \ | / \ | / \ | / \ | / \ | / \ | / \ | / \ | / \ | / \ | / \ | / \ | / \ | / \ | / \ |/ \|/ \|/ \|/ \|/ \ I haven't found a nice way of having more solid depressions than this. Has anyone else found any nice solutions? Dave Litwin From @mail.uunet.ca:mark.longridge@canrem.com Thu Sep 21 02:25:16 1995 Return-Path: <@mail.uunet.ca:mark.longridge@canrem.com> Received: from seraph.uunet.ca (uunet.ca) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA05649; Thu, 21 Sep 95 02:25:16 EDT Received: from portnoy.canrem.com ([198.133.42.17]) by mail.uunet.ca with SMTP id <249737-1>; Thu, 21 Sep 1995 02:17:23 -0400 Received: from canrem.com by portnoy.canrem.com (4.1/SMI-4.1) id AA23248; Thu, 21 Sep 95 02:10:41 EDT Received: by canrem.com (PCB-UUCP 1.1f) id 1F5B3D; Thu, 21 Sep 95 02:05:43 -0500 To: cube-lovers@life.ai.mit.edu Reply-To: CRSO.Cube@canrem.com Sender: CRSO.Cube@canrem.com Subject: VIP Sphere & Masterball From: mark.longridge@canrem.com (Mark Longridge) Message-Id: <60.1234.5834.0C1F5B3D@canrem.com> Date: Thu, 21 Sep 1995 02:56:00 -0400 Organization: CRS Online (Toronto, Ontario) # Calculations on the VIP sphere # 32 pieces with all distinct colours # Sphere where each of the 2 hemispheres rotate 180 degrees # and the 4 rows (of 8 pieces each) can slide around # the circumference # # # Size (vip) = 437,763,136,697,395,052,544,000,000 # No fixed Orientation # approx. = 4.4 * 10^26 or 437 septillion! # trivial centre vip := Group( (1,2,3,4,5,6,7,8), (9,10,11,12,13,14,15,16), (17,18,19,20,21,22,23,24), (25,26,27,28,29,30,31,32) , (1,28)(2,27)(3,26)(4,25)(9,20)(10,19)(11,18)(12,17), (5,32)(6,31)(7,30)(8,29)(13,24)(14,23)(15,22)(16,21) );; Within the 2 orbits of 16 pieces any exchange is possible. One orbit is "polar" and the other is "equatorial". (28,1) in vip; true Thus on the VIP Sphere a single 2-cycle is legal, although I know of no simple process as yet. The original calculation by Dr. Singmaster was (16!)^2, and I have confirmed his result with GAP. I have also played with the Masterball somewhat. This puzzle is awful! Just how accurately does this thing have to be lined up to turn it? It locked up several times on me when I tried to randomize it. It is the single most difficult puzzle to turn I have ever encountered, save the Equator puzzle only. My first thoughts on calculating the number of positions on the Masterball was it was the same as the VIP sphere divided by 2^16 but I'm not sure. I can't use GAP in the case of the Masterball (rainbow edition in this case) to verify this because of the identical pieces. The booklet which came with the Masterball refers to some number like 350 quadrillion but there are more zeroes than the american quadrillion, and I get a totally different number anyways. Thoughts anyone?? -> Mark <- From VinylM@aol.com Fri Sep 22 02:42:39 1995 Return-Path: Received: from emout05.mail.aol.com by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA18328; Fri, 22 Sep 95 02:42:39 EDT Received: by emout05.mail.aol.com (8.6.12/8.6.12) id CAA29847 for Cube-Lovers@ai.mit.edu; Fri, 22 Sep 1995 02:42:39 -0400 Date: Fri, 22 Sep 1995 02:42:39 -0400 From: VinylM@aol.com Message-Id: <950922024237_105790716@emout05.mail.aol.com> To: Cube-Lovers@ai.mit.edu Subject: Rubik's Revenge I am currently looking for a solution to Rubik's Revenge (the 4 X 4 cube) any ideas? Aron Siegel 404-321-0445 From boland@sci.kun.nl Sat Sep 23 18:31:59 1995 Return-Path: Received: from wn1.sci.kun.nl by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA21232; Sat, 23 Sep 95 18:31:59 EDT Received: from canteclaer.sci.kun.nl by wn1.sci.kun.nl via canteclaer.sci.kun.nl [131.174.132.34] with SMTP id AAA29847 (8.6.10/2.13) for ; Sun, 24 Sep 1995 00:31:56 +0200 Message-Id: <199509232231.AAA29847@wn1.sci.kun.nl> To: cube-lovers@ai.mit.edu Subject: Order problems Date: Sun, 24 Sep 95 00:31:55 +0200 From: Michiel Boland Hello all, Had a great time reading the archives. What I haven't found there are order problems: what is the shortest (in terms of quarter turns or half- and quarter turns, whatever you prefer) transformation of the cube with a given order? Here is a list that my good old PC produced this afternoon. I hope some of you find this interesting. :) A couple of notes on the list: "Len" is the length of the transformation in terms of quarter twists. You will notice that I listed two transforms with order 3: one is minimal wrt quarter-turn metric, the other wrt half-turn metric. A notable absentee is number 11. I suspect that (U.R.F2B.D')2 is the shortest possible with order 11, but my comp just isn't fast enough to confirm this. Note that (U.R.F2B.D')2 yields an 11-cycle on the edges (see also Mark Longridge's mail from 15 Jul 1994.) (I use dots to maintain readability; personally, I do not like the U1F2L3 notation, but that's just a matter of taste :) Order Len 1 0 2 2 U2 3 6 U.R.U'D'R.D. 3 8 U2R2U2R2 4 1 U. 5 4 U.R.U.R' 6 4 U2R2 7 4 U.R.U'F. 8 4 U.R2D. 9 4 U.R.F2 10 4 U'R.U.F. 11 ? ???????????? 12 4 U.R.F.D' 14 6 U'R.U.R'F.D. 15 6 U.R2U.R2 16 5 U.R.U'F.D. 18 5 U.R.U'R'F. 20 5 U.R.U'L2 21 6 U2R.U2F. 22 6 U.R.F2B.D' 24 4 U.R2D' 28 4 U.R.U'L. 30 3 U.R2 33 4 U.R.F'D' 35 6 U2R.U2L' 36 4 U2R'F' 40 5 U.R.U2L. 42 6 U.R2U2R' 44 4 U'R.F'D. 45 4 U.R.U.L. 48 5 U2R.U.F. 55 6 U.R.F'U'B'L. 56 5 U2R.F'D. 60 3 U.R'F' 63 2 U.R' 66 6 U.R.U.F2L' 70 6 U.R'U.R.F.R' 72 4 U.R.U.F' 77 4 U.R'F'L' 80 3 U'R'F' 84 3 U.R.F. 90 3 U.R.D. 99 6 U.R2F.L2 105 2 U.R. 110 8 U.R.U2R'F.R.L' 112 6 U.R'U.F'R.D. 120 4 U.R.F.L' 126 4 U'R.F'L' 132 4 U.R.F'L. 140 4 U.R'U.F' 144 5 U.R'F'D2 154 6 U.R.U.F.L.D' 165 6 U.R'U.F2L' 168 4 U.R.D2 180 3 U.R.D' 198 6 U2R.F.D2 210 4 U.R'D.L' 231 4 U.R.F'D. 240 5 U'R.F'L2 252 4 U.R.F.L. 280 5 U'R'U'F.L' 315 4 U.R.D.L. 330 6 U2R.F'D'L' 336 6 U.R.U.F.D2 360 3 U.R.F' 420 4 U.R.D.L' 462 6 U'R.F'D2L' 495 6 U.R2U.F'L' 504 5 U.R2F.L' 630 6 U'R'U'F'L2 720 6 U'R'U'F'D2 840 5 U2R'F'D. 990 6 U'R'U'F'L.D. 1260 6 U.R'U.F'D2 -- Michiel Boland University of Nijmegen The Netherlands From news@nntp-server.caltech.edu Sun Sep 24 09:54:40 1995 Return-Path: Received: from chamber.cco.caltech.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA15598; Sun, 24 Sep 95 09:54:40 EDT Received: from gap.cco.caltech.edu by chamber.cco.caltech.edu with ESMTP (8.6.12/DEI:4.41) id GAA19960; Sun, 24 Sep 1995 06:54:38 -0700 Received: by gap.cco.caltech.edu (8.6.7/DEI:4.41) id GAA27751; Sun, 24 Sep 1995 06:54:35 -0700 To: mlist-cube-lovers@nntp-server.caltech.edu Path: whuang From: whuang@cco.caltech.edu (Wei-Hwa Huang) Newsgroups: mlist.cube-lovers Subject: Re: Alexander's Star Date: 24 Sep 1995 13:54:34 GMT Organization: California Institute of Technology, Pasadena Lines: 42 Message-Id: <443nuq$r35@gap.cco.caltech.edu> References: <9509150340.AA15388@quark.geoworks.com> Nntp-Posting-Host: accord.cco.caltech.edu X-Newsreader: NN version 6.5.0 #12 (NOV) dlitwin@geoworks.com (David Litwin) writes: > While we are on the subject of the Alexander's star, I have never >been entirely satisfied with the arrangment of its solution. I've noticed >that most people who look at it don't even know it is solved and I have to >explain that the solution is that all the stickers of pieces laying in a >common plane around the points are the same color. Hard to say, but I can >point it out to people. > To this end I have spent some time trying to find a more satisfying >solution, one more visually clear and simple. I've only come up with one >alternative that I consider reasonable, and it isn't as pure as I would >like. > This solution involves grouping colors in the depressions of the >star. The main problem lies in the fact that the edges come together in >groups of three, but there are 10 stickers of each color so at some point >having all the depressions of the star a single color breaks down. For >this reason I choose one color (White is my preference) to be an exception >and have it remain in the original configuration, i.e. all in two parallel >planes. With the rest of the colors, I group them in groups of five: three >in one depression, and two in an adjacent depression with the third color >of this second depression being one of a white. The result of this is a >set of interlocking "diamonds" that group visually because the are of the >same color. Unrolled, the star would look like this (this should just fit >on a display of 80 columns): > I haven't found a nice way of having more solid depressions than this. > Has anyone else found any nice solutions? > Dave Litwin I came up with this solution independently. I also liked on that picked two opposite depressions on the star and made sure they contained one of each color. That allowed me to fill each of the other 18 depressions with homogenous colors. -- -- Wei-Hwa Huang (whuang@cco.caltech.edu) Homepage (under construction): http://www.ugcs.caltech.edu/~whuang/ Microsoft: small and limp. From boland@sci.kun.nl Sun Sep 24 12:19:36 1995 Return-Path: Received: from wn1.sci.kun.nl by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA21113; Sun, 24 Sep 95 12:19:36 EDT Received: from canteclaer.sci.kun.nl by wn1.sci.kun.nl via canteclaer.sci.kun.nl [131.174.132.34] with SMTP id RAA13678 (8.6.10/2.13) for ; Sun, 24 Sep 1995 17:19:34 +0100 Message-Id: <199509241619.RAA13678@wn1.sci.kun.nl> To: cube-lovers@ai.mit.edu Subject: Re: Order problems In-Reply-To: Your message of "Sun, 24 Sep 95 00:31:55 +0100." <199509232231.AAA29847@wn1.sci.kun.nl> Date: Sun, 24 Sep 95 17:19:32 +0100 From: Michiel Boland Sorry to follow up on my own post, but I messed up a bit :) > 12 4 U.R.F.D' There is also U2R2F' which has less face turns but more quarter turns. > 70 6 U.R'U.R.F.R' This transform has in fact order 140! The correct ones are: 70 6 U.R.U'R.F.B' 70 7 U2R'U2F'L' > 110 8 U.R.U2R'F.R.L' This should be U'R'U.R.F.D'B'L' (8 qt) I converted my order-searching program to C and am now running it on a Sun - expecting results on order 11 soon. -- Michiel Boland University of Nijmegen The Netherlands From BRYAN@wvnvm.wvnet.edu Sun Sep 24 18:36:02 1995 Return-Path: Received: from WVNVM.WVNET.EDU by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA07039; Sun, 24 Sep 95 18:36:02 EDT Received: from WVNVM.WVNET.EDU by WVNVM.WVNET.EDU (IBM VM SMTP V2R3) with BSMTP id 1007; Sun, 24 Sep 95 18:35:36 EDT Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU (LMail V1.2a/1.8a) with BSMTP id 5196; Sun, 24 Sep 1995 18:35:37 -0400 Message-Id: Date: Sun, 24 Sep 1995 18:35:36 -0400 (EDT) From: "Jerry Bryan" To: "Michiel Boland" , "Cube Lovers List" Subject: Re: Order problems In-Reply-To: Message of 09/24/95 at 00:31:55 from boland@sci.kun.nl On 09/24/95 at 00:31:55 Michiel Boland said: >Hello all, >Had a great time reading the archives. What I haven't found >there are order problems: what is the shortest (in terms of >quarter turns or half- and quarter turns, whatever you prefer) >transformation of the cube with a given order? I would be curious to hear how you are doing your search. It is trivial to see how to calculate the order of a particular position. However, it is not obvious to me how to find a position of a particular order. I hope it is not the case that it is in the archives and I just haven't seen it. I would guess that you are building a search tree of length 0, length 1, length 2, etc. as has been done many times before, and calculating the order of each position as you encounter it. You could then easily build a table of shortest positions of each order, provided the order appeared in your search. I would further guess that you have searched down to about level 6. However, if that is how you are doing it, I don't see how you could have proved that the shortest position of order 110 is of length 8. I don't see how a PC program could have searched to level 8 in just a little while. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU From boland@sci.kun.nl Sun Sep 24 19:44:45 1995 Return-Path: Received: from wn1.sci.kun.nl by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA10253; Sun, 24 Sep 95 19:44:45 EDT Received: from canteclaer.sci.kun.nl by wn1.sci.kun.nl via canteclaer.sci.kun.nl [131.174.132.34] with SMTP id AAA21239 (8.6.10/2.13) for ; Mon, 25 Sep 1995 00:44:43 +0100 Message-Id: <199509242344.AAA21239@wn1.sci.kun.nl> To: "Cube Lovers List" Subject: Re: Order problems In-Reply-To: Your message of "Sun, 24 Sep 95 18:35:36 -0400." Date: Mon, 25 Sep 95 00:44:40 +0100 From: Michiel Boland Jerry wrote: >I would be curious to hear how you are doing your search. It is >trivial to see how to calculate the order of a particular >position. However, it is not obvious to me how to find a >position of a particular order. I hope it is not the case that >it is in the archives and I just haven't seen it. I use a simple brute-force method, that is, I compute the order of each transform and the number of quarter turns. If there is already a transform with that order & number of qt, I forget all about it and go to the next transform. I have the C source available for anyone who wants to peek at it. Note that (almost) all transforms start with UxRx, since you have to twist two adjacent faces in order to get something with order other than 1, 2 or 4. That saves a bit of time. On my PC, i finished all transforms of length 6 (quarter- and half turns), and did some of length 7. Fortunately, as I mentioned earlier, I managed to get it to work on a somewhat faster machine, and am now waiting for the results of that. I am searching all transforms of 10 quarter turns or less. A process with order 110 must have an even number of quarter turns, since the permutation on the edges has to be even (the only possibility is to have an 11-cycle on the edges, which is an even permutation). Therefore, since there are no processes with order 110 with 6 or less quarter turns, this proves that the one of length 8 is indeed the shortest possible. Cheers, -- Michiel Boland University of Nijmegen The Netherlands From @mail.uunet.ca:mark.longridge@canrem.com Sun Sep 24 22:25:13 1995 Return-Path: <@mail.uunet.ca:mark.longridge@canrem.com> Received: from seraph.uunet.ca (uunet.ca) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA16210; Sun, 24 Sep 95 22:25:13 EDT Received: from portnoy.canrem.com ([198.133.42.17]) by mail.uunet.ca with SMTP id <250313-1>; Sun, 24 Sep 1995 22:27:40 -0400 Received: from canrem.com by portnoy.canrem.com (4.1/SMI-4.1) id AA03338; Sun, 24 Sep 95 22:20:55 EDT Received: by canrem.com (PCB-UUCP 1.1f) id 1F62D8; Sun, 24 Sep 95 22:13:27 -0500 To: cube-lovers@life.ai.mit.edu Reply-To: CRSO.Cube@canrem.com Sender: CRSO.Cube@canrem.com Subject: Least Commutative Element From: mark.longridge@canrem.com (Mark Longridge) Message-Id: <60.1242.5834.0C1F62D8@canrem.com> Date: Sun, 24 Sep 1995 22:56:00 -0400 Organization: CRS Online (Toronto, Ontario) # The 2nd most commutative element of the cube group # It is the position of 7 clockwise and 1 counter-clockwise twists # This commutes with 1 out of every 8 elements of the cube commuter := ( 1,35, 9)( 3,27,33)( 6,11,17)( 8,19,25)(24,43,30)(32,48,38) (14,40,46)(16,22,41);; # The least commutative element of the cube group ( I think! ) # This commutes with 1 out of every 450,541,700,775,936,000 # or approximately 1 out of every 4.5 * 10^17 patterns least := ( 1, 6,32,19, 3,41,24,46)( 2, 4,13,42,29, 7,31,15,39,37,26,21) ( 5,28,34,10,20,23,36,18,45,44,47,12)( 8,33,16,30,40, 9,17,38) (11,48,25,27,22,43,14,35);; > after thinking about it, i realized that > > corners: (8) edges: (12) > > commutes with even fewer elements. again, elements with > this cycle structure split into two conjugacy classes. > > mike With GAP we must deal with permutations of cube facelets, and that is why the permutation 'least' has 3 sets of 8 numbers and 2 sets of 12 numbers. Moreover, as I'm sure Mike will appreciate, the least commutative element I've found is has a 8-cycle of corners and a 12-cycle of edges. Size (Centralizer (cube, least)); 96 -> Mark <- From boland@sci.kun.nl Mon Sep 25 04:19:19 1995 Return-Path: Received: from wn1.sci.kun.nl by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA24902; Mon, 25 Sep 95 04:19:19 EDT Received: from canteclaer.sci.kun.nl by wn1.sci.kun.nl via canteclaer.sci.kun.nl [131.174.132.34] with SMTP id JAA28924 (8.6.10/2.13) for ; Mon, 25 Sep 1995 09:19:18 +0100 Message-Id: <199509250819.JAA28924@wn1.sci.kun.nl> To: cube-lovers@ai.mit.edu Subject: 11-cycle Date: Mon, 25 Sep 95 09:19:17 +0100 From: Michiel Boland U.R.U.F.R'L.U'R'F'D' This completes the order list. :) -- Michiel Boland University of Nijmegen The Netherlands From mouse@collatz.mcrcim.mcgill.edu Mon Sep 25 07:07:11 1995 Return-Path: Received: from Collatz.McRCIM.McGill.EDU by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA27224; Mon, 25 Sep 95 07:07:11 EDT Received: (root@localhost) by 3544 on Collatz.McRCIM.McGill.EDU (8.6.12 Mouse 1.0) id HAA03544; Mon, 25 Sep 1995 07:06:43 -0400 Date: Mon, 25 Sep 1995 07:06:43 -0400 From: der Mouse Message-Id: <199509251106.HAA03544@Collatz.McRCIM.McGill.EDU> To: boland@sci.kun.nl Subject: Re: Order problems Cc: cube-lovers@ai.mit.edu >> I would be curious to hear how you are doing your search. [...] > I use a simple brute-force method, that is, I compute the order of > each transform and the number of quarter turns. If there is already > a transform with that order & number of qt, I forget all about it and > go to the next transform. This sounds to me as though you're assuming that all transforms with a given order are equivalent as far as deriving further transforms of other orders go. That is, if you find that a given transform X of length L has order N, it sounds as though you're assuming that there is no need to store any other transforms of length L and order N. I'm not convinced this is justified. If you've found X of (say) length L and order N, and then find a different Y of length L and order N, I can't see any justification for the assumption that you can prune the entire subtree below Y, because if the cycle decompsition of Y is different from that of X, they may behave entirely differently when followed by more twists, even though they have the same order. der Mouse mouse@collatz.mcrcim.mcgill.edu From boland@sci.kun.nl Mon Sep 25 07:12:23 1995 Return-Path: Received: from wn1.sci.kun.nl by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA27309; Mon, 25 Sep 95 07:12:23 EDT Received: from canteclaer.sci.kun.nl by wn1.sci.kun.nl via canteclaer.sci.kun.nl [131.174.132.34] with SMTP id MAA05350 (8.6.10/2.13) for ; Mon, 25 Sep 1995 12:12:22 +0100 Message-Id: <199509251112.MAA05350@wn1.sci.kun.nl> To: cube-lovers@ai.mit.edu Subject: Re: Order problems In-Reply-To: Your message of "Mon, 25 Sep 95 07:06:43 -0400." <199509251106.HAA03544@Collatz.McRCIM.McGill.EDU> Date: Mon, 25 Sep 95 12:12:20 +0100 From: Michiel Boland >If you've found X of (say) length L and >order N, and then find a different Y of length L and order N, I can't >see any justification for the assumption that you can prune the entire >subtree below Y [...] I do not prune the search tree. If I say I "forget" about Y, then I do not mean "forget" all transforms that start with Y. That would be a bad thing, of course. -- Michiel Boland University of Nijmegen The Netherlands From mreid@ptc.com Wed Sep 27 17:28:34 1995 Return-Path: Received: from ptc.com (poster.ptc.com) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA08352; Wed, 27 Sep 95 17:28:34 EDT Received: from ducie.ptc.com by ptc.com (5.x/SMI-SVR4-NN) id AA17019; Wed, 27 Sep 1995 17:24:55 -0400 Message-Id: <9509272124.AA17019@ptc.com> Received: by ducie.ptc.com (1.38.193.4/16.2.nn) id AA04376; Wed, 27 Sep 1995 17:50:00 -0400 Date: Wed, 27 Sep 1995 17:50:00 -0400 From: michael reid To: cube-lovers@ai.mit.edu Subject: number of positions of square1 mark says > I know of no Square 1 calculations whatsoever, but would be > interested in seeing anyone's calculations. i also haven't seen any figures for square1 (although i guess that ideal toy corp.'s generic "more than three billion" might apply). but it's not hard to derive some figures. i guess there might be some question about what constitutes a "position". i think it's reasonable to consider only configurations where all three axes are free to turn. note that up to rotation, there are 29 different shapes for the top face. they occur as 19 symmetric shapes and 5 mirror image pairs. these are grouped into five different classes according to the number of 60 degree pieces ("corners"?), which i'll call doubles. for each shape, we also count the number of rotationally distinct orientations, as well as the number of orientations where the locations of the doubles do not block the half-turn axis. description # rotational # orientations name of shape symmetries # orientations which allow half-turn type 6-0 A 222222 6 2 1 type 5-2 B 2222211 1 12 6 C 2222121 1 12 4 D 2221221 1 12 2 type 4-4 E 22221111 1 12 6 F 22122111 1 12 4 G 22112211 2 6 4 H 22212111 1 12 4 H' 22211121 1 12 4 I 22211211 1 12 6 J 22121211 1 12 6 J' 22112121 1 12 6 K 22121121 1 12 4 L 21212121 4 3 2 type 3-6 M 222111111 1 12 6 N 221211111 1 12 6 N' 221111121 1 12 6 O 221121111 1 12 8 O' 221111211 1 12 8 P 221112111 1 12 6 Q 212121111 1 12 8 R 212112111 1 12 6 R' 212111211 1 12 6 S 211211211 3 4 2 type 2-8 T 2211111111 1 12 8 U 2121111111 1 12 8 V 2112111111 1 12 8 W 2111211111 1 12 8 X 2111121111 2 6 5 the top and bottom faces have complementary type, (i.e. are 6-0 and 2-8, 5-2 and 3-6, 4-4 and 4-4, 3-6 and 5-2, or 2-8 and 6-0). type total # valid orientations 6-0 1 5-2 12 4-4 46 3-6 62 2-8 37 thus we have 1 * 37 + 12 * 62 + 46 * 46 + 62 * 12 + 37 * 1 = 3678 valid possibilities of the doubles. each permutation of the doubles and singles is possible, and the middle layer has two orientations. any combination of these is possible. therefore we get a final count of 3678 * 2 * 8! * 8! = 11958666854400 positions. mike From BRYAN@wvnvm.wvnet.edu Thu Oct 5 21:01:18 1995 Return-Path: Received: from WVNVM.WVNET.EDU by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA12445; Thu, 5 Oct 95 21:01:18 EDT Received: from WVNVM.WVNET.EDU by WVNVM.WVNET.EDU (IBM VM SMTP V2R3) with BSMTP id 0286; Wed, 04 Oct 95 10:26:46 EDT Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU (LMail V1.2a/1.8a) with BSMTP id 0750; Wed, 4 Oct 1995 10:26:43 -0400 Message-Id: Date: Wed, 4 Oct 1995 10:26:41 -0400 (EDT) From: "Jerry Bryan" To: "Cube Lovers List" Subject: Question It is well known that if we define G= for the twelve quarter turns q in Q, we can also generate G as G=, leaving out B and B'. Leaving out any other quarter turn would do as well, but I am going to stick to leaving out B for illustrative purposes. However, when one of the quarter turns is left out, the length of most positions will change. In particular, we will no longer have |B|=1. My reading of the archives indicates that we do not know what the length of B would be in this situation, nor what a minimal process for B would be. I am going to take a crack at this problem via exhaustive search. But I like to use representative elements of conjugacy classes in my searches, and I don't think I can do so in this situation. For full-blown searches of G, I use M-conjugacy classes. For subsets and/or restrictions of G, I use appropriate subsets and/or restrictions of M. But I don't think I can use conjugacy classes at all for this problem. The group is still G, even though lengths have changed, so no subset and/or restriction of M is appropriate. But when G is generated as , we do not necessarily have |X|=|m'Xm| for all m in M. Am I missing something obvious? I don't think so, but in the meantime I am going to have to start the search without conjugacy classes. Bummer. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU From bagleyd@source.asset.com Fri Oct 6 16:34:56 1995 Return-Path: Received: from source.asset.com by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA00393; Fri, 6 Oct 95 16:34:56 EDT Received: by source.asset.com (AIX 3.2/UCB 5.64/4.03) id AA51717; Fri, 6 Oct 1995 16:37:09 -0400 Date: Fri, 6 Oct 1995 16:37:09 -0400 From: bagleyd@source.asset.com (David A. Bagley) Message-Id: <9510062037.AA51717@source.asset.com> To: cube-lovers@life.ai.mit.edu Subject: pyraminx-like puzzles Hi I have a question, I hope this makes sence. ;) On a "nxnxn" tetrahedron with period 2 or period 3 turning or a "nxnxn" octahedron with period 3 or period 4 turning, can the orientation of any of the center triangles change when the puzzle is solved? If so, where does this start to happen. I know from "experience" that this is not true on a pyraminx. A Pyraminx, according to me, has period 3 turning where n = 3. n is the number of triangles along an edge not counting corners of triangles. Center triangles are those triangles that are not on an edge (except for maybe the corners of the triangle). A face of a 2x2x2 has one center triangle /\ /__\ /\C /\ /__\/__\ INSERT MODE A face of a 3x3x3 has 3 center triangles /\ /__\ /\C /\ /__\/__\ /\C /\C /\ /__\/__\/__\ A face of a 4x4x4 has 7 center triangles I would like to know this because it will help me in the design of my puzzles which I am in the process now of converting to Motif (you will still be able to run them if you just have X). I hear that a freely distributable Lesstif Widget set will be out soon. (I am also considering a Windows version and am getting ready to convert them). Estimated completion time for the motif programs: 2 weeks. Cheers, --__--------------------------------------------------------------- / \ \ / David A. Bagley \ | \ \ / bagleyd@source.asset.com | | \//\ Some days are better than other days. | | / \ \ -- A short lived character of Blake's 7 | \ / \_\puzzles Available at: ftp.x.org/contrib/games/puzzles / ------------------------------------------------------------------- From BRYAN@wvnvm.wvnet.edu Fri Oct 6 16:39:08 1995 Return-Path: Received: from WVNVM.WVNET.EDU by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA01095; Fri, 6 Oct 95 16:39:08 EDT Received: from WVNVM.WVNET.EDU by WVNVM.WVNET.EDU (IBM VM SMTP V2R3) with BSMTP id 8544; Fri, 06 Oct 95 09:25:11 EDT Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU (LMail V1.2a/1.8a) with BSMTP id 2139; Fri, 6 Oct 1995 09:25:12 -0400 Message-Id: Date: Fri, 6 Oct 1995 09:25:11 -0400 (EDT) From: "Jerry Bryan" To: "Cube Lovers List" Subject: Preliminary Search Results for The first step in finding |B| in is to build a little Start-rooted data base using only the ten generators F, F', U, U', etc. I was able to search eight levels deep with a "quick and dirty" search. To search deeper would take a good bit longer. I hope that eight levels will be enough to calculate |B|. By constructing a companion B-rooted data base, I should be able to test up to 15 levels deep. (B is odd, so eight levels deep for each half-depth search only gets you 15 levels total instead of 16.) I haven't started the search for B yet, but I thought the results so far might be of minor interest in their own right. As I indicated before, these are actual cube positions. I haven't figured out any way to apply conjugacy classes to this problem. I find it a little puzzling that the branching factor is not monotonically decreasing (c.f., level 3 to level 4). Level Number of Local Branching Positions Max Factor 0 1 0 1 10 0 10.000 2 77 0 7.700 3 584 0 7.584 4 4,434 0 7.592 5 33,664 0 7.592 6 255,320 0 7.584 7 1,933,936 7.575 8 14,635,503 7.568 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU From alan@curry.epilogue.com Sat Oct 7 06:36:21 1995 Return-Path: Received: from curry.epilogue.com by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA29078; Sat, 7 Oct 95 06:36:21 EDT Received: (from alan@localhost) by curry.epilogue.com (8.6.8/8.6.6) id GAA02153; Sat, 7 Oct 1995 06:36:19 -0400 Date: Sat, 7 Oct 1995 06:36:19 -0400 Message-Id: <7Oct1995.062936.Alan@LCS.MIT.EDU> From: Alan Bawden Sender: Cube-Lovers-Request@ai.mit.edu To: Cube-Lovers@ai.mit.edu In-Reply-To: "Samantha Jerrings, President, International Students\ Association, Eastern Division"'s message of Sat, 7 Oct 1995 01:06:45 -0500 Subject: ===>> FREE 1 yr. Magazine Sub sent worldwide- 300+ Popular USA Titles If you're wondering why you got the fllowing piece of mail: Date: Sat, 7 Oct 1995 01:06:45 -0500 From: "Samantha Jerrings, President, International Students\ Association, Eastern Division" Hi fellow 'netters, My name is Samantha Jerrings and I recently started using a magazine subscription club in the USA that has a FREE 1 yr. magazine subscription deal with your first paid order- and I have been very pleased with them. ... it's because you're on Cube-Lovers. Needless to say, I'll be protesting this misuse of our mailing list. Please do not respond to -me- about this message. And I would certainly encourage you not to order any magazines from these slime-balls either! From mark.longridge@canrem.com Sun Oct 8 00:06:06 1995 Return-Path: Received: from itchy.crso.com (itchy.canrem.com) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA08342; Sun, 8 Oct 95 00:06:06 EDT Received: by canrem.com (PCB-UUCP 1.1f) id 1F8131; Sat, 7 Oct 95 23:56:06 -0500 To: cube-lovers@life.ai.mit.edu Reply-To: CRSO.Cube@canrem.com Sender: CRSO.Cube@canrem.com Subject: Antislice Correction From: mark.longridge@canrem.com (Mark Longridge) Message-Id: <60.1248.5834.0C1F8131@canrem.com> Date: Sat, 7 Oct 95 23:53:00 -0500 Organization: CRS Online (Toronto, Ontario) I'll start with a small correction: I wrote on Mon, 3 Jul 1995 14:53:00 > Patterns in the Anti-Slice Group > -------------------------------- > > p4 8 flip (Op sides) (R1 L1 U1 D1 F1 B1) ^2 (12) > p10a pons asinorum (L3 R1 U3 D1)^3 (12) > p16a 4 cross order 2 F1 B1 U1 D1 L2 R2 U1 D1 F1 B1 U2 D2 (12) > p17 4 diagonal (F1 B1 R1 L1) ^3 (12) > p18a 4 diagonal,2 cross (F1 B1 R3 L3) ^3 (12) > p22 2 DOT, 2 Stripe R1 L1 U2 D2 R3 L3 (6) > p64a 4 Z F1 B1 L3 R3 F1 B1 L1 R1 F3 B3 L1 R1 (12) > p143 Pinwheels F1 B1 L1 R1 F3 B3 U3 D3 L1 R1 U1 D1 (12) > p175a 6 H order 2 U3 D3 L3 R3 F2 B2 U2 D2 L3 R3 U1 D1 (12) > p198a 2 X, 4 Diag no C L1 R1 F1 B1 L3 R3 F3 B3 L1 R1 F1 B1 (12) > p201 Pinwheels + Pons L1 R1 F3 B3 L1 R1 U3 D3 F1 B1 U3 D3 (12) > > p201 is a quite interesting position. The reference to p201 is incorrect. It should read: p175a is a quite interesting position. ^^^^^ > The square's group equivalent is no shorter in q turns: > > p175 6 H order 2 type 2 U2 B2 L2 U2 D2 L2 F2 U2 (8) > > Note that p201 = |{m'Xm}|=2 and |Symm(X)|=24. Someone also edited my original entry and changed all the T's to U's which is more consistent and standard. In addition..... Dan Hoey wrote on Fri, 28 Oct 94 11:38:15 EDT > But there's another reason. Remember the annoying feature that the > color assignments to faces were never standardized? The first cube I > bought had red opposite yellow, blue opposite white, and orange > opposite green (I think). Even though in later days most cubes are > manufactured with opposite faces ``differing by yellow''--red opposite > orange, blue opposite green, and yellow opposite white--there does not > seem to be a standard for the handedness of the coloring. This has > long been a problem on cube-lovers, where everyone ....... There was a standard in the cube contests. The original Ideal colour arrangement was the tournament standard in the U.S. and Canada. Top=White, Down=Blue, Left=Red, Right=Orange, Front=Yellow, Back=Green -> Mark <- From mark.longridge@canrem.com Sun Oct 8 00:27:13 1995 Return-Path: Received: from itchy.crso.com (itchy.canrem.com) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA08706; Sun, 8 Oct 95 00:27:13 EDT Received: by canrem.com (PCB-UUCP 1.1f) id 1F8132; Sat, 7 Oct 95 23:56:06 -0500 To: cube-lovers@life.ai.mit.edu Reply-To: CRSO.Cube@canrem.com Sender: CRSO.Cube@canrem.com Subject: Using 5 Generators From: mark.longridge@canrem.com (Mark Longridge) Message-Id: <60.1249.5834.0C1F8132@canrem.com> Date: Sat, 7 Oct 95 23:54:00 -0500 Organization: CRS Online (Toronto, Ontario) From: "Jerry Bryan" Subject: Question > It is well known that if we define G= for the twelve quarter turns > q in Q, we can also generate G as G=, leaving out B and B'. > Leaving out any other quarter turn would do as well, but I > am going to stick to leaving out B for illustrative purposes. > > However, when one of the quarter turns is left out, the length of most > positions will change. In particular, we will no longer have |B|=1. > My reading of the archives indicates that we do not know what the > length of B would be in this situation, nor what a minimal process > for B would be. This problem was solved by David Benson in Oct. 1979, who was one of the earliest cube pioneers. Dr. Singmaster reports on this in his 2nd Addendum of "Notes". Let A = R1 L3 F2 B2 R1 L3, then AUA = D1 AUA = R1 L3 F2 B2 R1 L3 U1 R1 L3 F2 B2 R1 L3 (17 q, 13 q+h) Perhaps Jerry will find something shorter. -> Mark <- From mark.longridge@canrem.com Sun Oct 8 00:27:14 1995 Return-Path: Received: from itchy.crso.com (itchy.canrem.com) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA08707; Sun, 8 Oct 95 00:27:14 EDT Received: by canrem.com (PCB-UUCP 1.1f) id 1F8133; Sat, 7 Oct 95 23:56:06 -0500 To: cube-lovers@life.ai.mit.edu Reply-To: CRSO.Cube@canrem.com Sender: CRSO.Cube@canrem.com Subject: Picture Cubes From: mark.longridge@canrem.com (Mark Longridge) Message-Id: <60.1250.5834.0C1F8133@canrem.com> Date: Sat, 7 Oct 95 23:55:00 -0500 Organization: CRS Online (Toronto, Ontario) David Singmaster (ZINGMAST@VAX.SBU.AC.UK) writes on Wed Sep 13 11:50:09 1995: ------------------------------------------------ I think the ranking is not quite fair because the puzzles are of very different types. E.g. the 15 puzzle has nearly as many patterns as the 3^3 but no one would claim it was anywhere near as difficult. Indeed the Babylon Tower has 36! = 3.72 x 10^41 basic positions. One can divide by some small value such as 2 or 6 or perhaps more, depending on what one considers the same position. This puts it between 3 and 4 in your list, but it is not a difficult puzzle, except that it is hard to see the gradations of the colors! Indeed, the commercial 7 x 7 'fifteen puzzles' have 49! =6.08 x 10^62 basic patterns - again one has to divide by something, in this case 2. This falls between 2 and 3 in your list, but again it is hardly a difficult puzzle. So I think you are comparing puzzles which are of such different type that the number of patterns is not a fair comparison of their difficulties.I would group them in three (or perhaps 2) types. Rubik Cube, etc. Fifteen Puzzles, etc. in the plane. Cylindrical Puzzles - barrels, etc. ------------------------------------------------ I quite agree. One of my reasons for making that list was to simply rank all the puzzles by number of combinations only, to show the feasibility (or lack of) for a brute force search to find God's Algorithm. The major drawback, as you point out, is that difficulty in solving is not only a function of the number of combinations. Dr. Singmaster continues: ------------------------ Re your Case E. Almost all the picture cubes have all four orientations distinct on the face centres - both those with nine little pictures on each face and those with a big picture spread over all nine facelets. These are actually pretty common. ------------------------------------------------- Case E, that is cases that have only a fraction of the total possible number of combinations for a Rubik's picture cube, are unfortunately well represented in my own cube collection. The following cubes are all in Case E: Rubik's Calendar Cube, Rubik's Cube 4th Dimension, Rubik's World, Blind Man's Cube (from Germany), Royal Wedding Cube (with Charles & Di). Although I don't doubt that, over all, these cases are exceptional. In the case of Rubik's World there are 3 blank centre pieces, and in the Royal Wedding Cube only 2 opposite faces can show all 4 possible orientations. Name Combinations Inventor 8. Picture Cube (3x3x3) (E) 8.8*10^22 Erno Rubik, Dan Hoey 9. Calendar Cube (3x3x3)(F) 4.4*10^22 Marvin Silbermintz 10. Rubik's Cube 4th Dim.(D) 1.1*10^22 Erno Rubik 11 Rubik's World (G) 2.7*10^21 Erno Rubik 12. Royal Wedding Cube 6.9*10^20 Unknown 13. Rubik's Cube (3x3x3) 4.3*10^19 Erno Rubik -> Mark <- From alan@curry.epilogue.com Sun Oct 8 03:58:52 1995 Return-Path: Received: from curry.epilogue.com by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA11503; Sun, 8 Oct 95 03:58:52 EDT Received: (from alan@localhost) by curry.epilogue.com (8.6.8/8.6.6) id DAA00533; Sun, 8 Oct 1995 03:58:50 -0400 Date: Sun, 8 Oct 1995 03:58:50 -0400 Message-Id: <8Oct1995.033935.Alan@LCS.MIT.EDU> From: Alan Bawden Sender: Cube-Lovers-Request@ai.mit.edu To: Cube-Lovers@ai.mit.edu In-Reply-To: "Samantha Jerrings, President, Association of International\ Students, Eastern Division"'s message of Sun, 8 Oct 1995 00:00:17 -0500 Subject: ===>> FREE 1 yr. Magazine Sub sent worldwide- 300+ Popular USA Titles Date: Sun, 8 Oct 1995 00:00:17 -0500 From: "Samantha Jerrings, President, Association of International\ Students, Eastern Division" ... Alas, there is little I can currently do about this, beyond complaining to various postmaster addresses (which I have done). As the Internet becomes more of a capitalist free-for-all, even a moderately small mailing list like Cube-Lovers (less than 200 members) becomes a target for things like this. The only real solution, given that it is no longer possible to rely on people's courtesy, is to go for some kind of moderation. And in fact, I am developing some software to moderate Cube-Lovers (and another mailing list I maintain). When I'm done, the only changes you all should notice are: (1) no more junk mail and (2) a slightly longer turnaround time when you submit a legitimate message. Until I've got that working all you can do is ignore these losers. From tdb@delta1.deltanet.com Sun Oct 8 12:53:11 1995 Received: from delta1.deltanet.com by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA25937; Sun, 8 Oct 95 12:53:11 EDT Received: by delta1.deltanet.com (5.65/1.2-eef) id AA16056; Sun, 8 Oct 95 09:53:09 -0700 Return-Path: From: tdb@delta1.deltanet.com (Tom D. Baccanti) Newsgroups: cube-lovers To: cube-lovers@life.ai.mit.edu Subject: Intro and question Date: Sun, 08 Oct 1995 09:33:01 -0700 Message-Id: <90/dwor+BYIO085yn@delta1.deltanet.com> X-Newsreader: Yarn 0.85 with YES 0.20 Editor by QEDIT Lines: 23 Briefly, I am a old time cube fan and I am blind. I have been blind for about 9 years now due to complications from diabetes. I recently encountered a "Braille Or Tactile Rubik's cube" available from Japan. I finally received my cube from there after much expense and effort but it was worth it. Describing the cube: it has designs on each side that can easily be differentiated from the other. ie; circles, plus, dotted circle, smooth, six dotsand textured side. I am now able to solve all but the last bottom slice from memory but I would like to ask if anyone has a xolution that I can read that would make sense to me? I wish to solve this and then move on to the other puzzles I have read on this list mentioned. I will have to mark them tactilly also. Thanks for the time, Tom Baccanti -- Every crowd has a silver lining. Tom D. Baccanti /** Reach me at Internet: TDB@Deltanet.com TDB@crl.com /* PGP key from: pgp-public-keys@pgp.mit.edu get key id: 8EB942B1 From laz@smartlink.net Sun Oct 8 15:32:13 1995 Return-Path: Received: from warp10.smartlink.net (smartlink.net) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AB02106; Sun, 8 Oct 95 15:32:13 EDT Received: from alumina.smartlink.net by warp10.smartlink.net(8.6.12/SMARTLINK-1.0) with id MAA12980 ESMTP for on Sun, 8 Oct 1995 12:33:33 -0700 Received: (from laz@localhost) by alumina.smartlink.net (8.6.11/8.6.9) id MAA00115; Sun, 8 Oct 1995 12:32:00 -0700 Date: Sun, 8 Oct 1995 12:32:00 -0700 Message-Id: <199510081932.MAA00115@alumina.smartlink.net> From: Laszlo Takacs To: Cube-Lovers-Request@ai.mit.edu Cc: Cube-Lovers@ai.mit.edu In-Reply-To: <8Oct1995.033935.Alan@LCS.MIT.EDU> (message from Alan Bawden on Sun, 8 Oct 1995 03:58:50 -0400) Subject: Re: ===>> FREE 1 yr. Magazine Sub sent worldwide- 300+ Popular USA Titles Reply-To: laz@smartlink.net |Until I've got that working all you can do is ignore these losers. It that really all? Why not have everyone return the mail 100 fold? From hazard@niksula.hut.fi Mon Oct 9 06:19:55 1995 Return-Path: Received: from nukkekoti.cs.hut.fi by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA29409; Mon, 9 Oct 95 06:19:55 EDT Received: from neppari.cs.hut.fi (hazard@neppari.cs.hut.fi [130.233.40.139]) by nukkekoti.cs.hut.fi (8.6.12/8.6.11) with ESMTP id MAA17196 for ; Mon, 9 Oct 1995 12:19:40 +0200 Received: (hazard@localhost) by neppari.cs.hut.fi (8.6.12/8.6.10) id MAA00485; Mon, 9 Oct 1995 12:19:38 +0200 Date: Mon, 9 Oct 1995 12:19:37 +0200 (EET) From: Mikko Haapanen X-Sender: hazard@neppari.cs.hut.fi To: cube-lovers@ai.mit.edu Subject: Re: ===>> FREE 1 yr. Magazine Sub sent worldwide- 300+ ... In-Reply-To: <199510081932.MAA00115@alumina.smartlink.net> Message-Id: Content-Conversion: prohibited Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=ISO-8859-1 Content-Transfer-Encoding: QUOTED-PRINTABLE On Sun, 8 Oct 1995, Laszlo Takacs wrote: > |Until I've got that working all you can do is ignore these losers. > It that really all? Why not have everyone return the mail 100 fold? This is dirty but i think it's efficient! ------Mikko Haapanen------hazard@niksula.hut.fi--------- Jos tahto siirt=E4=E4 vuoren, niin tahdon siirt=E4=E4 tahdon Vuorta siirt=E4m=E4=E4n, ett=E4 vuoren sis=E4=E4n n=E4=E4n -P.Hanhiniemi -------------------------------------------------------- From ronnie@cisco.com Mon Oct 9 14:18:38 1995 Return-Path: Received: from hubbub.cisco.com by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA21011; Mon, 9 Oct 95 14:18:38 EDT Received: from madhatter.cisco.com (ronnie-ss10.cisco.com [171.69.61.22]) by hubbub.cisco.com (8.6.12/CISCO.GATE.1.1) with ESMTP id LAA26693; Mon, 9 Oct 1995 11:18:35 -0700 Received: from cisco.com (localhost.cisco.com [127.0.0.1]) by madhatter.cisco.com (8.6.8+c/CISCO.WS.1.1) with ESMTP id LAA00588; Mon, 9 Oct 1995 11:18:34 -0700 Message-Id: <199510091818.LAA00588@madhatter.cisco.com> To: laz@smartlink.net Cc: Cube-Lovers-Request@ai.mit.edu, Cube-Lovers@ai.mit.edu Subject: Re: ===>> FREE 1 yr. Magazine Sub sent worldwide- 300+ Popular USA Titles In-Reply-To: Your message of "Sun, 08 Oct 1995 12:32:00 PDT." <199510081932.MAA00115@alumina.smartlink.net> Date: Mon, 09 Oct 1995 11:18:32 -0700 From: "Ronnie B. Kon" > |Until I've got that working all you can do is ignore these losers. > > It that really all? Why not have everyone return the mail 100 fold? Note that there is an e-mail address buried in the message which is different from the sending e-mail address. The sending address is a forgery, probably an innocent person. Ronnie From alan@curry.epilogue.com Mon Oct 9 15:07:15 1995 Return-Path: Received: from curry.epilogue.com by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA23048; Mon, 9 Oct 95 15:07:15 EDT Received: (from alan@localhost) by curry.epilogue.com (8.6.8/8.6.6) id PAA04970; Mon, 9 Oct 1995 15:07:13 -0400 Date: Mon, 9 Oct 1995 15:07:13 -0400 Message-Id: <9Oct1995.140201.Alan@LCS.MIT.EDU> From: Alan Bawden Sender: Cube-Lovers-Request@ai.mit.edu To: Cube-Lovers@ai.mit.edu In-Reply-To: Mikko Haapanen's message of Mon, 9 Oct 1995 12:19:37 +0200 (EET) Subject: ===>> FREE 1 yr. Magazine Sub sent worldwide- 300+ ... Date: Mon, 9 Oct 1995 12:19:37 +0200 (EET) From: Mikko Haapanen To: cube-lovers@ai.mit.edu On Sun, 8 Oct 1995, Laszlo Takacs wrote: > |Until I've got that working all you can do is ignore these losers. > It that really all? Why not have everyone return the mail 100 fold? This is dirty but i think it's efficient! Sigh. I was hoping I wouldn't have to say this. PLEASE DO -NOT- CARRY ON A CONVERSATION ABOUT HOW TO DEAL WITH JUNK MAIL ON CUBE-LOVERS! From aaweint@io.org Mon Oct 9 15:35:38 1995 Return-Path: Received: from io.org by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA24768; Mon, 9 Oct 95 15:35:38 EDT Received: from shemp04.slip.yorku.ca (shemp04.slip.yorku.ca [130.63.122.53]) by io.org (8.6.12/8.6.12) with SMTP id PAA23524 for ; Mon, 9 Oct 1995 15:35:27 -0400 Date: Mon, 9 Oct 1995 15:35:27 -0400 Message-Id: <199510091935.PAA23524@io.org> X-Sender: aaweint@io.org X-Mailer: Windows Eudora Version 1.4.4 Mime-Version: 1.0 Content-Type: text/plain; charset="us-ascii" To: Cube-Lovers@ai.mit.edu From: aaweint@io.org (Aaron Weintraub) Subject: Rubik's Revenge, where? I'm looking fora Rubik's Revenge puzzle. Do places still sell them? If not, is there a place where I can mail order one from? Thanks for any information on this. Aaron aaweint@io.org From BRYAN@wvnvm.wvnet.edu Mon Oct 9 19:48:30 1995 Return-Path: Received: from WVNVM.WVNET.EDU by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA08164; Mon, 9 Oct 95 19:48:30 EDT Received: from WVNVM.WVNET.EDU by WVNVM.WVNET.EDU (IBM VM SMTP V2R3) with BSMTP id 2565; Mon, 09 Oct 95 08:58:40 EDT Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU (LMail V1.2a/1.8a) with BSMTP id 7680; Mon, 9 Oct 1995 08:58:40 -0400 Message-Id: Date: Mon, 9 Oct 1995 08:58:39 -0400 (EDT) From: "Jerry Bryan" To: Subject: Re: Antislice Correction In-Reply-To: Message of 10/07/95 at 23:53:00 from mark.longridge@canrem.com On 10/07/95 at 23:53:00 mark.longridge@canrem.com said: >Someone also edited my original entry and changed all the T's to U's >which is more consistent and standard. I agree that U's are more "consistent and standard", but for reasons that have been discussed several times, T's would probably be better. However, the effort to switch from U's to T's has not been very successful. Personally, I have *tried* to switch to T's, and it just doesn't feel right. So I have to share the blame for sabotaging the switch to T's. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU From BRYAN@wvnvm.wvnet.edu Wed Oct 11 18:10:11 1995 Return-Path: Received: from WVNVM.WVNET.EDU by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA24526; Wed, 11 Oct 95 18:10:11 EDT Received: from WVNVM.WVNET.EDU by WVNVM.WVNET.EDU (IBM VM SMTP V2R3) with BSMTP id 4678; Wed, 11 Oct 95 13:50:52 EDT Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU (LMail V1.2a/1.8a) with BSMTP id 7014; Wed, 11 Oct 1995 13:50:53 -0400 Message-Id: Date: Wed, 11 Oct 1995 13:50:51 -0400 (EDT) From: "Jerry Bryan" To: Subject: Re: Using 5 Generators In-Reply-To: Message of 10/07/95 at 23:54:00 from mark.longridge@canrem.com On 10/07/95 at 23:54:00 mark.longridge@canrem.com said: >This problem was solved by David Benson in Oct. 1979, who was one of >the earliest cube pioneers. Dr. Singmaster reports on this in his >2nd Addendum of "Notes". >Let A = R1 L3 F2 B2 R1 L3, then AUA = D1 >AUA = R1 L3 F2 B2 R1 L3 U1 R1 L3 F2 B2 R1 L3 (17 q, 13 q+h) >Perhaps Jerry will find something shorter. Well, without seeing the original article, I am not sure if I would agree that the problem was solved or not back in 1979. By that I mean that the problem I was proposing to solve was finding a minimal solution. I don't know if the original article claimed that 17q was minimal. I can now confirm that 17q is indeed a minimal solution. With my data base of positions through level 8, I was able to perform half-depth searches to confirm that there are no solutions through 15q. Given the existence of a 17q solution, that is sufficient to show that 17q is minimal. But just to be sure, I extended my data base of positions through level 9, and with my extended data base I was able to find several solutions of length 17q. I am not quite sure what we should count as a unique solution. But I can report that my search found 16 unique (*not* unique up to conjugacy) half-way positions. I use the term "half-way" advisedly. The "half-way" positions are 9q from Start and 8q from B or vice versa. I guess you could say that the vice versa gives you a total of 32=16+16 half-way positions, but the whole concept of "half-way" is pretty slippery in this case anyway. Just because we know that 17q is minimal does not mean that we know that 13 q+h is minimal. I have not done any searches of the q+h case. With my extended data base, the results of the search with five generators are as follows: Level Number of Local Branching Positions Max Factor 0 1 0 1 10 0 10.000 2 77 0 7.700 3 584 0 7.584 4 4,434 0 7.592 5 33,664 0 7.592 6 255,320 0 7.584 7 1,933,936 0 7.575 8 14,635,503 7.568 9 110,685,344 7.562 One more thing, and perhaps this particular problem can be put to rest. I have mentioned several times that I could not figure out how to use conjugacy for this particular problem. Well now I have, although it is too late for doing the search. You certainly cannot use M-conjugacy, but you can use a subgroup of M. The subgroup includes four rotations and four reflections. The four rotations are i, b, bb, and bbb, where we use lower case letters to simulate Frey and Singmaster's script notation for rotations. For example, b is the whole cube rotation consisting of grasping the Back face and rotating the whole cube (not just the Back face) clockwise by 90 degrees. For the reflections, we use Dan Hoey's "permutations of face centers" notation. The four reflections are (UL)(DR), (UR)(DL), (UD), and (LR). To tie the two notations together, we could write the rotations as i=(), b=(ULDR), bb=(UD)(RL), and bbb=(URDL). These eight rotations and reflections form the subgroup of M which is called Q2 in Dan's taxonomy of the 98 subgroups of M. Hence, we could have used Q2-conjugacy, which would have reduced the size of the problem by about eight times. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU From hoey@aic.nrl.navy.mil Wed Oct 11 22:56:23 1995 Received: from Sun0.AIC.NRL.Navy.Mil by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA12649; Wed, 11 Oct 95 22:56:23 EDT Received: from sun13.aic.nrl.navy.mil by Sun0.AIC.NRL.Navy.Mil (4.1/SMI-4.0) id AA15062; Wed, 11 Oct 95 22:56:16 EDT Return-Path: Received: by sun13.aic.nrl.navy.mil; Wed, 11 Oct 95 22:56:15 EDT Date: Wed, 11 Oct 95 22:56:15 EDT From: hoey@aic.nrl.navy.mil Message-Id: <9510120256.AA04061@sun13.aic.nrl.navy.mil> To: "Jerry Bryan" , Cube-Lovers@life.ai.mit.edu Subject: Re: Using 5 Generators I certainly agree that confirming minimality is an important result. Thanks, Jerry. > But I can report that my search found 16 unique (*not* unique up to > conjugacy) half-way positions. I use the term "half-way" advisedly. > The "half-way" positions are 9q from Start and 8q from B or vice > versa. I guess you could say that the vice versa gives you a total > of 32=16+16 half-way positions, but the whole concept of "half-way" > is pretty slippery in this case anyway. If I understand this, there are 16 positions at 9q from Start and 8q from B, and there are 16 other positions at 8q from Start 9q from B. Is each of the first bunch adjacent to exactly one of the other? And vice versa? It would be good to get them reduced by Q2-conjugacy, as well. [in Q2] > The four rotations are i, b, bb, and bbb, where we use lower case > letters to simulate Frey and Singmaster's script notation for rotations. > For example, b is the whole cube rotation consisting of grasping > the Back face and rotating the whole cube (not just the Back face) > clockwise by 90 degrees. The reflections are similarly rrv, rrbv, ttv, and ttbv, where t and r are the whole-cube rotations by the Top and Right faces, and v is the central inversion. Dan Hoey Hoey@AIC.NRL.Navy.Mil From preux@lil.univ-littoral.fr Thu Oct 12 10:48:47 1995 Return-Path: Received: from lilserv.univ-lille1.fr by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA10654; Thu, 12 Oct 95 10:48:47 EDT Received: from elgon.univ-littoral.fr (elgon.univ-littoral.fr [194.57.179.17]) by lilserv.univ-lille1.fr (8.7/jtpda-5.1) with SMTP id PAA10611 for ; Thu, 12 Oct 1995 15:46:38 +0100 (MET) Message-Id: <199510121446.PAA10611@lilserv.univ-lille1.fr> Received: by elgon.univ-littoral.fr Thu, 12 Oct 1995 14:45:45 GMT Date: Thu, 12 Oct 1995 14:45:45 GMT From: preux@lil.univ-littoral.fr (Preux Philippe) To: Cube-Lovers@life.ai.mit.edu Subject: I am in search of Thistlewaite's algorithm Hi, As long as I am a new comer to this mailing list, I will briefly introduce myself. As a computer science researcher, I am working on evolutionary algorithms trying to assess their ability to solve combinatorial optimization problems. One of my old dream has been to solve the Rubik's cube (for the moment, the very basic 3x3x3 version) with this kind of algorithms. I have heard about the Thistlewaite's algorithm which is able to solve the problem in less than 50 or so moves. I have also heard about a publication of D. Singmaster that, among other things, describes this algorithm. Thus, I am looking for information about this algorithm: how does it work precisely? I wonder whether this algorithm, either a description or an implementation of it, is available somewhere on the net. I would also be interested in having a copy of D. Singmaster's report (either via ftp or paper). Can someone help me? Thanks a lot in any case. Philippe -- From SHV6937@ocvaxa.cc.oberlin.edu Thu Oct 12 17:22:08 1995 Return-Path: Received: from OCVAXA.CC.OBERLIN.EDU by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA06733; Thu, 12 Oct 95 17:22:08 EDT Received: from OCVAXA.CC.OBERLIN.EDU by OCVAXA.CC.OBERLIN.EDU (PMDF V5.0-4 #7710) id <01HWCZLQUZM800TL3Z@OCVAXA.CC.OBERLIN.EDU> for Cube-Lovers@life.ai.mit.edu; Thu, 12 Oct 1995 17:17:22 -0400 (EDT) Date: Thu, 12 Oct 1995 17:17:21 -0400 (EDT) From: Huy Vo Subject: Unsubscribe me ... To: Cube-Lovers@life.ai.mit.edu Message-Id: <01HWCZLR02V600TL3Z@OCVAXA.CC.OBERLIN.EDU> X-Vms-To: IN%"Cube-Lovers@life.ai.mit.edu" Mime-Version: 1.0 Content-Type: TEXT/PLAIN; CHARSET=US-ASCII Content-Transfer-Encoding: 7BIT Please unsubscribe me from the listserv. ( shv6937@ocvaxa.oberlin.edu ) I apologize for the occupation of your disk space but the system I am on is being changed. -- huy From BRYAN@wvnvm.wvnet.edu Fri Oct 13 15:46:51 1995 Return-Path: Received: from WVNVM.WVNET.EDU by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA08454; Fri, 13 Oct 95 15:46:51 EDT Received: from WVNVM.WVNET.EDU by WVNVM.WVNET.EDU (IBM VM SMTP V2R3) with BSMTP id 1264; Fri, 13 Oct 95 11:47:44 EDT Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU (LMail V1.2a/1.8a) with BSMTP id 2160; Fri, 13 Oct 1995 11:47:44 -0400 Message-Id: Date: Fri, 13 Oct 1995 11:47:43 -0400 (EDT) From: "Jerry Bryan" To: "Dan Hoey" , Subject: Re: Using 5 Generators In-Reply-To: Message of 10/11/95 at 22:56:15 from hoey@aic.nrl.navy.mil On 10/11/95 at 22:56:15 hoey@aic.nrl.navy.mil said: >> But I can report that my search found 16 unique (*not* unique up to >> conjugacy) half-way positions. I use the term "half-way" advisedly. >> The "half-way" positions are 9q from Start and 8q from B or vice >> versa. I guess you could say that the vice versa gives you a total >> of 32=16+16 half-way positions, but the whole concept of "half-way" >> is pretty slippery in this case anyway. >If I understand this, there are 16 positions at 9q from Start and 8q >from B, and there are 16 other positions at 8q from Start 9q from B. >Is each of the first bunch adjacent to exactly one of the other? And >vice versa? It would be good to get them reduced by Q2-conjugacy, as >well. I don't think I can answer your questions without further analysis, and I don't have much time to devote to the problem. But let me clarify as follows. First, the search looked as follows: Distance from Distance from Total Number of Start B Distance Matching Positions 0 1 1 0 1 2 3 0 2 3 5 0 3 4 7 0 4 5 9 0 5 6 11 0 6 7 13 0 7 8 15 0 8 9 17 16 There is a certain arbitrariness in at least two respects. For one example, to test for a total distance of 11, you could just as well use distances from Start and B respectively of 4 and 7 instead of 5 and 6. For another example, the Start-rooted tree and the B-rooted tree have identical structures, so the first two columns could be reversed. Indeed, you could get the B-rooted tree simply by pre-multiplying the Start-rooted tree by B. (This reminds me of one of my most foolish errors on Cube-Lovers. For search trees where the nodes are conjugacy classes (or representatives of conjugacy classes), the tree looks different depending on which class or representative is at the root. But when the nodes are all the positions, then the tree is essentially the same in all cases, just pre-multiplied by the root. I once claimed the tree structure depended on the root for trees containing all positions, confusing that situation with the situation for trees of conjugacy classes. Arrg!) Hence, I am not especially comfortable talking about "half-way" positions. But continuing anyway, denote the 16 positions which are 8 moves from Start and 9 moves from B as X_i for i in 1..16. Then, the 16 positions which are 8 moves from B and 9 moves from Start are B(X_i) for i in 1..16. A solution to the problem would then look something like (X_j)Y(X_k)', where Y is in Q-{B,B'}. But I don't think we can say a priori that there is no Z in Q-{B,B'} and no X_m such that (X_j)Z(X_m)' is also a solution (Z not equal Y and X_m not equal X_k). I think to analyze the problem properly you would have to take the positions X_i for i in 1..16 and the positions B(X_j) for j in 1..16 and match up each X_i with each B(X_j) to see which ones differ by a quarter turn. Each X_i is going to match up with at least one B(X_j) and vice versa, but there might be more than 16 matches overall. Reduction by Q2-conjugacy is important, but I don't think it would tell you how many solutions there are that you would really want to consider to be unique. Recall the solution of Pons Asinorum by half-depth searches. There are 5 positions unique up to M-conjugacy which are 6q from Start and 6q from Pons. But most people would consider that there are only two minimal solutions to Pons that are really different. The trouble is that 4 of the 5 half-way positions for the Pons are in the middle of a sub-process which commutes. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU From BRYAN@wvnvm.wvnet.edu Fri Oct 13 16:40:29 1995 Return-Path: Received: from WVNVM.WVNET.EDU by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA12120; Fri, 13 Oct 95 16:40:29 EDT Received: from WVNVM.WVNET.EDU by WVNVM.WVNET.EDU (IBM VM SMTP V2R3) with BSMTP id 4011; Fri, 13 Oct 95 15:01:31 EDT Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU (LMail V1.2a/1.8a) with BSMTP id 2466; Fri, 13 Oct 1995 15:01:32 -0400 Message-Id: Date: Fri, 13 Oct 1995 15:01:31 -0400 (EDT) From: "Jerry Bryan" To: "Dan Hoey" , Subject: Re: Using 5 Generators In-Reply-To: Message of 10/11/95 at 22:56:15 from hoey@aic.nrl.navy.mil On 10/11/95 at 22:56:15 hoey@aic.nrl.navy.mil said: >[in Q2] >> The four rotations are i, b, bb, and bbb, where we use lower case >> letters to simulate Frey and Singmaster's script notation for rotations. >> For example, b is the whole cube rotation consisting of grasping >> the Back face and rotating the whole cube (not just the Back face) >> clockwise by 90 degrees. >The reflections are similarly rrv, rrbv, ttv, and ttbv, where t and r >are the whole-cube rotations by the Top and Right faces, and v is the >central inversion. I am not quite sure why it took me so long to figure out that Q2-conjugation was the appropriate symmetry when generating G without B and B'. With the wisdom if hindsight, it is perhaps because the Q2 subgroup of M could be described as fixing the B and F face centers, and I was thinking only in terms of fixing the B face center. Obviously, you cannot fix the B face center without also fixing the F face center. Thus, it is clear that Q2-conjugation is also the appropriate symmetry when generating G without F and F'. Q1 and Q3 are subgroups of M which are conjugate with Q2 in Dan's taxonomy of subgroups of M. Q1 fixes the T and D face centers, and would be the appropriate symmetry when generating G without T and T', or without D and D'. Q3 fixes the R and L face centers, and would be the appropriate symmetry when generating G without R and R', or without L and L'. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU From BRYAN@wvnvm.wvnet.edu Tue Oct 17 12:48:14 1995 Return-Path: Received: from WVNVM.WVNET.EDU by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA23661; Tue, 17 Oct 95 12:48:14 EDT Received: from WVNVM.WVNET.EDU by WVNVM.WVNET.EDU (IBM VM SMTP V2R3) with BSMTP id 5797; Mon, 16 Oct 95 23:08:03 EDT Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU (LMail V1.2a/1.8a) with BSMTP id 1260; Mon, 16 Oct 1995 23:08:03 -0400 Message-Id: Date: Mon, 16 Oct 1995 23:08:02 -0400 (EDT) From: "Jerry Bryan" To: "Cube Lovers List" Subject: Correctness of Large Searches When I was in graduate school in another life many years ago, one of the big issues was program correctness, and whether programs could be proven to be correct. As I recall, the short answer is that non-trivial programs cannot be proven to be correct. I haven't really followed the issue since then, but I doubt that the answer has changed. Also, many mathematicians are suspicious or unaccepting of proofs or other results which involve large computer searches. For example, the proof of the Four Color Map Theorem is a famous result involving a large search which is held in certain suspicion. For one thing, the referees for the paper were not able to reproduce the results of the computer search (not enough computing power). For another thing, the referees were neither able to confirm that the programs were error free (who can?), nor that some sort of computer error (software, hardware, or procedural) did not occur during the running of the programs (c.f., the famous Pentium floating point divide error). Two factors cause me to raise the question of program correctness at this time. One is simply that some of the searches we do are so large, how do we know that the answers are correct? The other is that I have found an error in one of my (smaller) searches that I need to report. Those of you who actually fight through the details of my longer, more boring posts will recall that there are essentially three different models I use from time to time. For cubes without centers, I usually use representative elements of sets of the form {m'Xmc}. For cubes with centers, I use either representative elements of sets of the form {m'Xm}, or else I use representatives of the form Repr{m'Xmc}*C to simulate Repr{m'Xm}. The latter model is very compact for cases where complete searches can be accomplished For the case of the 3x3x3 cube, corners only, with centers, q-turns only, I recently discovered the following anomaly. Note in particular level 8 and below of the search. 3x3x3 Corners Only -- Qturns Distance Repr{m'Xmc}*C Repr{m'Xm} from Model Model Start 0 1 1 1 1 1 2 5 5 3 24 24 4 149 149 5 850 850 6 4257 4257 7 16937 16937 8 57800 57848 9 180639 180787 10 466052 466220 11 676790 676786 12 392558 392342 13 45744 45600 14 163 163 Total 1841970 1841970 The 3x3x3 corners only case has been searched a number of times by a number of people, all of whom got the same results. However, these results all were for every position, including those which are M-conjugate. My searches are for conjugacy classes only, and therefore I have nobody with whom I can compare results directly. The only way I can compare results is to expand the conjugacy classes, and regrettably I did not do so when I first calculated the corners only case. For a variety of reasons to be detailed below, I believe the {m'Xm} results above are correct and that the {m'Xmc}*C results are incorrect. But at the same time, I do not believe there is an error per se in the {m'Xmc}*C model. Let's see if we can make some sense of this. When I first discovered the anomaly, my reaction was that the {m'Xm} model is simpler, and Occum's Razor suggested that the error was in the {m'Xmc}*C model. Also, when I expanded the conjugacy classes for the Repr{m'Xm} model, the results matched what everybody else had posted. Hence, I went back to the {m'Xmc}*C program, which I hadn't looked at in years. After many hours of reviewing the model, and many more hours of reviewing the program itself, I not unsurprisingly found no errors. Furthermore, I was no longer convinced that Occum's Razor still applied. The {m'Xmc}*C model is "more complicated" in some ways, but on the other hand, every cell of storage for the data base can be addressed directly. There is no sorting, merging, and matching of huge files as there is with the {m'Xm} model. So I ran the {m'Xmc}*C program again. Very surprisingly, this time it produced different answers, and the answers were the same as for the {m'Xm} model! So what's going on? I don't know. I am presently a bureaucrat (cubing is a hobby), but in previous lives I have been a technical support person. In that role, I have found numerous problems that caused programs to produce incorrect results, even though the program was "correct" (whatever that means). I have found hardware errors vaguely similar to the Pentium error, operating system errors, compiler errors (especially with optimizing compilers), and subroutine library errors. So my best guess is that something of this nature caused the {m'Xmc}*C program to produce incorrect results one time and correct results another time. Of course, an uninitialized variable or pointer can cause similar symptoms, but I have not been able to find anything like that in my program. The program is written in Turbo-Pascal for a PC using DOS. I have the exact same compiler now I have had for ten years or so. The Pascal source code is unchanged from when I ran it before. The first time I ran the program, I ran it under DOS, or maybe in the DOS box in an early version of OS/2 (can't remember for sure), and it ran on a 286 or an early 386 (can't remember for sure). This time, it ran in the DOS box under OS/2 Warp on a 486/50. So lots of things have changed. Furthermore, both times I ran it, I used the VDISK facility to cache all the disk files in memory, and OS/2 Warp surely has a newer VDISK than whatever I was using before. I even wondered if the data base was correct the first time, but maybe I had a bad counting program analyzing the data base. But I still had the original data base and counted it again. It really was wrong. So the mystery of the incorrect results is not solved. In light of the above discussion, I thought it might be appropriate to summarize some background about my larger searches. I want to indicate which ones of them seem pretty well verified, and which ones of them might benefit from further study. - I believe this one is ok. I ran the q turn case with and without conjugacy classes. Upon expanding the conjugacy classes, the results matched the results without conjugacy classes. Also, the results matched results posted by Mark Longridge as far as they went (although my anomaly with corners-only suggests that such matching doesn't prove very much). I ran the q+h turn case only with conjugacy classes, and expanded then to get the results without conjugacy classes. The model was Repr{w'Xw} (W-conjugacy instead of M-conjugacy), with much sorting, merging, and matching of external files. edges only This ran for about a year, so there is potential (without face for error. The model was Repr{m'Xmc}. The centers) runs to create neighbors were performed primarily on two 486's and a 386, using a mainframe as a file server. A UNIX station was added for the last few months. All sorts, merges, and matches were run on the mainframe. The whole process was driven by REXX scripts on the PC's and UNIX stations (we run REXX on all our UNIX boxes). There were actually more lines of code for the scripts than for the programs. The scripts moved files back and forth with FTP, and contained code to detect and correct errors with the FTP's. The problem has only been run once, so there is no independent verification. edges only This was run twice, once using the {m'Xm} model, (with face and once using the Repr{m'Xm}*C model. The centers) answers matched through level 11. They did not match at level 12. I have only posted through level 11 to the list. I have not had time to find the error. I do not expect to find the error in any of the programs. It took about a year to get to level 11, running on the mainframe. I expect the error to be somewhere in a script, probably failing to recover from some error condition like a system failure. These things run around the clock, and inevitably get caught in system failures from time to time. Although I was not able to do so, this is probably the largest cubing problem where we can conceive of running the problem to completion using today's technology. Had I been able to do run the problem to completion, the Repr(m'Xm)*C model would have served as verification for the "without face centers" case because the minimum of each row of the solution matrix with centers serves as the solution for the without centers case. Whole cube These are the most important runs. The problem (edges and corners has been run twice using the {m'Xm} model, although with face centers) you might not wish to count it as twice. The difference is that the edges were the major sort in one run, and the corners were the major sort in the other. The two runs did match, which is a good (but not perfect indication) that there was no error (e.g., such as the scripts failing to include something in in a sort or merge that should have been included). It would still be nice to have verification by someone else. I was able to run up through level 11. I did not try level 12 because it was just too big. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU From bagleyd@source.asset.com Tue Oct 17 13:56:03 1995 Return-Path: Received: from source.asset.com by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA27953; Tue, 17 Oct 95 13:56:03 EDT Received: by source.asset.com (AIX 3.2/UCB 5.64/4.03) id AA47533; Tue, 17 Oct 1995 14:01:53 -0400 Date: Tue, 17 Oct 1995 14:01:53 -0400 From: bagleyd@source.asset.com (David A. Bagley) Message-Id: <9510171801.AA47533@source.asset.com> To: cube-lovers@life.ai.mit.edu Subject: Motif puzzles Hi I converted my puzzles at ftp.x.org in /contrib/games/puzzles to use Motif. By default, the puzzles will compile and link with just the X-Windows includes and libs. If there is demand, I could supply the Motif versions statically linked for SunOS4.1.3. The Motif versions have no increased functionality but may be easier to use. Cheers, --__--------------------------------------------------------------- / \ \ / David A. Bagley \ | \ \ / bagleyd@source.asset.com | | \//\ Some days are better than other days. | | / \ \ -- A short lived character of Blake's 7 | \ / \_\puzzles Available at: ftp.x.org/contrib/games/puzzles / ------------------------------------------------------------------- From BRYAN@wvnvm.wvnet.edu Tue Oct 17 14:20:20 1995 Return-Path: Received: from WVNVM.WVNET.EDU by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA00357; Tue, 17 Oct 95 14:20:20 EDT Received: from WVNVM.WVNET.EDU by WVNVM.WVNET.EDU (IBM VM SMTP V2R3) with BSMTP id 4701; Tue, 17 Oct 95 14:19:53 EDT Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU (LMail V1.2a/1.8a) with BSMTP id 3051; Tue, 17 Oct 1995 14:19:53 -0400 Message-Id: Date: Tue, 17 Oct 1995 14:19:52 -0400 (EDT) From: "Jerry Bryan" To: Subject: Re: I am in search of Thistlewaite's algorithm In-Reply-To: Message of 10/12/95 at 14:45:45 from preux@lil.univ-littoral.fr On 10/12/95 at 14:45:45 preux@lil.univ-littoral.fr said: >As long as I am a new comer to this mailing list, I will briefly >introduce myself. As a computer science researcher, I am working on >evolutionary algorithms trying to assess their ability to solve >combinatorial optimization problems. One of my old dream has been to >solve the Rubik's cube (for the moment, the very basic 3x3x3 version) >with this kind of algorithms. I have heard about the Thistlethwaite's >algorithm which is able to solve the problem in less than 50 or so >moves. I have also heard about a publication of D. Singmaster that, among >other things, describes this algorithm. Thus, I am looking for >information about this algorithm: how does it work precisely? I wonder >whether this algorithm, either a description or an implementation of >it, is available somewhere on the net. I would also be interested in >having a copy of D. Singmaster's report (either via ftp or paper). I don't know if you received any private replies or not, but I will take a crack at this one. There are two places I have personally read about Thistlethwaite's algorithm. One is Douglas Hofstadter's article in Scientific American in March of 1981. The article is reprinted in Hofstadter's "Metamagical Themas". The other is Frey and Singmaster's "Handbook of Cubik Math". There are probably other sources as well, but some of them (e.g., Singmaster's Cubik Circulars) may be hard to come by at your local library. Any "by hand" solution to the Cube generally involves something like "corners first, then edges" (or vice versa), or "top layer, then middle layer, and finally the bottom layer", or (usually) some combination or variation of these themes. Any such theme has states where it is visually obvious that the cube is becoming more and more solved. These plateau states generally have the attribute that they define a subgroup of G. For example, the set of states where the top layer is solved is a subgroup of G. The subgroups defined by the plateau states form a nested sequence of subgroups as the Cube becomes more and more solved. However, progress is not monotonic. You almost always have to give up some of your progress temporarily on the way to the next plateau. Thistlethwaite's algorithm reverses the role of the plateau states and the subgroups. Instead of plateau states defining nested subgroups, he has nested subgroups defining plateau states. In fact, his nested subgroups are arranged in such a way that once a particular subgroup is achieved, there is no loss of progress on your way to the next subgroup. Also, the plateau states achieved by reaching the next subgroup are generally not visually obvious, and indeed it is not visually obvious that any progress at all is being made until the cube is almost solved. The nested subgroups given by Frey and Singmaster are as follows. I believe that other nested subgroups have been used as well. H0==G H1= H2= H3= H4= I do not recall seeing any practitioners of Thistlethwaite's algorithm posting to Cube-Lovers. However, there are several practitioner's of an algorithm called Kociemba's algorithm who contribute to Cube-Lovers. Kociemba's algorithm is a close cousin to Thistlethwaite's, but does not depend on pre-searching as does Thistlethwaite's. Also, I believe that the best Kociemba programs now utilize only two nested subgroups, and are able to achieve results far better than those achieved by Thistlethwaite. As I understand it, Kociemba's algorithm differs from Thistlethwaite's in two primary respects. Kociemba uses searching, and Thistlethwaite uses tables (what I called "pre-searching" in the previous paragraph). Also, Thistlethwaite simply hooks the nested subgroups together and stops when they are hooked, whereas Kociemba continues searching to see if improvements are possible by merging the nested subgroups at their boundaries. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU From BRYAN@wvnvm.wvnet.edu Tue Oct 17 19:45:07 1995 Return-Path: Received: from WVNVM.WVNET.EDU by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA20990; Tue, 17 Oct 95 19:45:07 EDT Received: from WVNVM.WVNET.EDU by WVNVM.WVNET.EDU (IBM VM SMTP V2R3) with BSMTP id 9305; Tue, 17 Oct 95 19:44:40 EDT Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU (LMail V1.2a/1.8a) with BSMTP id 8479; Tue, 17 Oct 1995 19:44:41 -0400 Message-Id: Date: Tue, 17 Oct 1995 19:44:40 -0400 (EDT) From: "Jerry Bryan" To: "Cube Lovers List" Subject: Positions 8q from Start, 9q from B, Five Generators Much analysis is not done, and I have little time. However, I am able to print out the positions very easily. DDD FFD RRR DBD RBB UBR DRD RRD RRR BUB UUF UUU UUB UUB UUU BBB UUR UBU LLL UUU RRR BBB LLU FDL BBB LDL FFF LLB DFL BRR FLB LFU FRR LLB LFL FRF LLL UUU RRR RRR BLU FFL BLB LRL FFF FFF DDL DBD FDF DDL DDD FFF DDB DDD 1. 2. 3. ---------------------------------------------------- UUF LLL FFF UBF LBD BBL RBR LLL LLU DUD UUU UUR RUF UUU UUR FFF UBU UUR FBU LLL DDB FFF RLR BRB FFF RRD BBB LLL DFD RRR FLF RFR BRR LLF RFD BRF LLL DDB RRR FFF RUR BBB RUD BRB ULL BBU DBD LDL BDU DDD BDD BFU DDD DDD 4. 5. 6. ---------------------------------------------------- DBF ULU ULL DBF UBU BBL LDF UUU UFF BFD FFF BUU BUD FUF RUU BBL FFF RUU UUU RRB DRR LLL DDD RRR RFF DRB LLL LLL UFU RRR LLB RFU BRR LLF DFR BRF LLL UUB DBR LLL DDD RRR RUF DRR BBB FFR BBB LDD FDL BDD BDD FDU BDB FDD 7. 8. 9. ---------------------------------------------------- LBL FFF FFF FBD FBF FBF FDD FFF FFF DFB DDD DDD DUB UUU UUU DBB DDD DDD LLL BUU RRR LLL BBB RRR LRL BBB RLR LLL UFU RRR LLL BFB RRR LLL BFB RRR BUU RRR DBF LLL BBB RRR LRL BBB RLR FFF UUU UUU FDL DDD DDD UDU UUU UUU 10. 11. 12. ---------------------------------------------------- DRR FFF FFF RBB FBF FBF FFF FFF FFF UUU UUU UUU UUB UUU DUD RUR UUU UUU RRD BLB UDL RLR BBB LRL RLR BBB LRL FLB UFL FRR RLR BFB LRL RLR BFB LRL BBB ULL FFF RLR BBB LRL RLR BBB LRL LDD DDD DDD LDD DDD UDU LDD DDD DDD 13. 14. 15. ---------------------------------------------------- FUR FBU FBU DUF RUF LFF LBU BDD RRR LLL DFD RRR LLU BLL DDD RBB UDB UFB 16. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU From mark.longridge@canrem.com Wed Oct 18 01:38:43 1995 Return-Path: Received: from itchy.crso.com (itchy.canrem.com) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA12454; Wed, 18 Oct 95 01:38:43 EDT Received: by canrem.com (PCB-UUCP 1.1f) id 1F9828; Wed, 18 Oct 95 01:28:18 -0500 To: cube-lovers@life.ai.mit.edu Reply-To: CRSO.Cube@canrem.com Sender: CRSO.Cube@canrem.com Subject: Dino Cubes Revisited From: mark.longridge@canrem.com (Mark Longridge) Message-Id: <60.1251.5834.0C1F9828@canrem.com> Date: Wed, 18 Oct 95 01:17:00 -0500 Organization: CRS Online (Toronto, Ontario) As I mentioned recently: > The original Ideal colour arrangement was the tournament standard in > the U.S. and Canada. > > Top =White, Down =Blue > Left =Red, Right=Orange > Front=Yellow, Back =Green Dan Hoey mentions: > manufactured with opposite faces ``differing by yellow''--red opposite > orange, blue opposite green, and yellow opposite white-- The ``differing by yellow'' colouring was pretty common on those "Wonderful Puzzlers". The only difference in original Ideal cubes colouring and the Wonderful Puzzlers colouring is that Blue and Yellow are transposed. I just the 3 Dino Cubes I ordered from Gametrends. More new cubes! There are 3 different kinds: A) Each of the 6 faces of the cube is a unique solid colour. B) Each of the 6 faces of the cube is a unique solid colour with four dinosaurs (six different dinos, 1 kind for each face) C) Each of the 4 tetrads is a unique colour. The Dino colouring is: (Just like the puzzlers) TOP =White, DOWN =Yellow LEFT =Red, RIGHT=Orange FRONT=Blue, BACK =Green The packaging is a small white cardboard box devoid of any trademarks. The name on the box is "Dinosaur Rubik's Cube". All the regulars on cube-lovers will make mincemeat of this puzzle, and the 4 colour version is easier still! The mechanism is quite good (nice and smooth) and it is a good puzzle to have. The 6 colour version has 19,958,400 combinations. The 4 colour version has only 42,600 combinations. -> Mark <- From BRYAN@wvnvm.wvnet.edu Wed Oct 18 20:56:25 1995 Return-Path: Received: from WVNVM.WVNET.EDU by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA09250; Wed, 18 Oct 95 20:56:25 EDT Received: from WVNVM.WVNET.EDU by WVNVM.WVNET.EDU (IBM VM SMTP V2R3) with BSMTP id 5392; Wed, 18 Oct 95 20:56:03 EDT Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU (LMail V1.2a/1.8a) with BSMTP id 1451; Wed, 18 Oct 1995 20:56:04 -0400 Message-Id: Date: Wed, 18 Oct 1995 20:56:03 -0400 (EDT) From: "Jerry Bryan" To: "Cube Lovers List" Subject: Re: Positions 8q from Start, 9q from B, Five Generators In-Reply-To: Message of 10/17/95 at 19:44:40 from BRYAN@wvnvm.wvnet.edu I can add a bit of additional information. The 16 positions 8q from Start and 9q from B can be reduced to 4 positions unique up to Q2-conjugacy. As I have discussed before, it is still difficult to claim that the 4 positions are really "different" without further analysis because of the possibility that the positions are variations within a commuting subsequence of moves. I don't really have a Q2-conjugacy program. It would be easy to make one, but I don't have time so I used my M-conjugacy program. Recall that Q2={i,b,bb,bbb,rrv,rrbv,ttv,ttbv}, where b, r, and t are whole cube rotations of the Back, Right, and Top faces, respectively, and v is the central inversion. For 12 of the 16 positions X the program reports Symm(X)={i}, which is to say m'Xm is not equal X for any m in M except the identity. Obviously, the same is true for all m in Q2 since Q2 is a subgroup of M. We have |Q2|=8, so |{m'Xm | m in Q2}=6. Therefore, the 12 positions for which Symm(X)={i} form two Q2-conjugacy classes. Using the M-conjugacy program for the other 4 positions is trickier, but only slightly so. For the other 4 positions, the M-conjugacy program reports Symm(X)=HX, where HX={i,bb,rr,tt,v,bbv,rrv,ttv}. But HX is not a subgroup of Q2, and what we need is sort of "Symm(X) with respect to Q2", which I will call Symm(X/Q2). (A better notation is probably available). It is easy to see that Symm(X/Q2)=(Symm(X) intersect Q2), and we have (HX intersect Q2)={i,ttv,bb,rrv}. This subgroup is called HQ2 in Dan's taxonomy. We have |Q2|=8 and |HQ2|=4, so |{m'Xm | m in HQ2}|=2 when Symm(X/Q2)=HQ2. Therefore, the last 4 positions form two Q2-conjugacy classes. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU From geohelm@pt.lu Thu Oct 19 03:20:44 1995 Return-Path: Received: from menvax.restena.lu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA28589; Thu, 19 Oct 95 03:20:44 EDT Date: Thu, 19 Oct 95 03:20:44 EDT Received: from telinf1.pt.lu by menvax.restena.lu with SMTP; Thu, 19 Oct 1995 8:20:41 +0100 (MET) Received: from slip8.pt.lu by telinf1.pt.lu id aa11995; 19 Oct 95 8:20 CET X-Sender: geohelm@mailsvr.pt.lu X-Mailer: Windows Eudora Version 1.4.4 Mime-Version: 1.0 Content-Type: text/plain; charset="us-ascii" To: cube-lovers@ai.mit.edu From: Georges Helm Subject: Thisthlethwaite's algorithm Message-Id: <9510190820.aa11995@telinf1.pt.lu> There are two papers by Morwen Thistlethwaite on this subject, both dating from 1980. I got these copies from David Singmaster after having tried unsuccessfully to get them from Morwen himself. I have since then made many copies of these notes for many people all around the world. I will continue to do this as long as I feel comfortable about it. So if there is still anybody out there who wants them, please e-mail me directly at geohelm@pt.lu or at georges.helm@comnet.eo.lu Georges Helm From boland@sci.kun.nl Thu Oct 19 10:21:21 1995 Return-Path: Received: from wn1.sci.kun.nl by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA12514; Thu, 19 Oct 95 10:21:21 EDT Received: from canteclaer.sci.kun.nl by wn1.sci.kun.nl via canteclaer.sci.kun.nl [131.174.132.34] with SMTP id PAA06003 (8.6.10/2.14) for ; Thu, 19 Oct 1995 15:21:19 +0100 Message-Id: <199510191421.PAA06003@wn1.sci.kun.nl> To: cube-lovers@ai.mit.edu Subject: 3-cycle on edges in group Date: Thu, 19 Oct 95 15:21:18 +0100 From: Michiel Boland URURU R'U'R'U'R' -- Michiel Boland University of Nijmegen The Netherlands From bagleyd@source.asset.com Thu Oct 19 13:27:59 1995 Return-Path: Received: from source.asset.com by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA24422; Thu, 19 Oct 95 13:27:59 EDT Received: by source.asset.com (AIX 3.2/UCB 5.64/4.03) id AA40695; Thu, 19 Oct 1995 13:16:06 -0400 Date: Thu, 19 Oct 1995 13:16:06 -0400 From: bagleyd@source.asset.com (David A. Bagley) Message-Id: <9510191716.AA40695@source.asset.com> To: cube-lovers@life.ai.mit.edu Subject: pyraminx-like puzzles (again) Hi Recently I asked the question: > I have a question, I hope this makes sence. ;) On a "nxnxn" > tetrahedron with period 2 or period 3 turning or a "nxnxn" octahedron with > period 3 or period 4 turning, can the orientation of any of the center > triangles change when the puzzle is solved? If so, where does this > start to happen. I know from "experience" that this is not true on > a pyraminx. Well, if you believe proof by example on a simulated puzzle, then Tetrahedron period 2 turning: never happens Tetrahedron period 3 turning: starts when n=4 with center triangle Octahedron period 3 turning: starts when n=4 with center triangle Octahedron period 4 turning: starts with n=4 with center triangle New versions of my pyraminx and octahedron puzzles will be out soon to reflect this. Cheers, --__--------------------------------------------------------------- / \ \ / David A. Bagley \ | \ \ / bagleyd@source.asset.com | | \//\ Some days are better than other days. | | / \ \ -- A short lived character of Blake's 7 | \ / \_\puzzles Available at: ftp.x.org/contrib/games/puzzles / ------------------------------------------------------------------- From boland@sci.kun.nl Thu Oct 19 21:41:58 1995 Return-Path: Received: from wn1.sci.kun.nl by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA25492; Thu, 19 Oct 95 21:41:58 EDT Received: from canteclaer.sci.kun.nl by wn1.sci.kun.nl via canteclaer.sci.kun.nl [131.174.132.34] with SMTP id CAA03659 (8.6.10/2.14) for ; Fri, 20 Oct 1995 02:41:58 +0100 Message-Id: <199510200141.CAA03659@wn1.sci.kun.nl> To: cube-lovers@ai.mit.edu Subject: Embedding G in a symmetrical group Date: Fri, 20 Oct 95 02:41:55 +0100 From: Michiel Boland It is clear that the group G of the cube (the one with 4.3252x10^19 elements) can be embedded in a symmetrical group, e.g. S_48, since each move of the cube can be seen as a permutation of 48 objects. Hence, there is a smallest number n such that G can be embedded in S_n. I'm curious to find out what this number is. It can be shown with some counting arguments that n>=32 (I'm happy to write these down but it's nicer if you thought about this first). I would be surprised if n=32 but you never know. -- Michiel Boland University of Nijmegen The Netherlands From mark.longridge@canrem.com Fri Oct 20 02:32:34 1995 Return-Path: Received: from itchy.crso.com (itchy.canrem.com) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA05992; Fri, 20 Oct 95 02:32:34 EDT Received: by canrem.com (PCB-UUCP 1.1f) id 1F9CFB; Fri, 20 Oct 95 02:24:41 -0500 To: cube-lovers@life.ai.mit.edu Reply-To: CRSO.Cube@canrem.com Sender: CRSO.Cube@canrem.com Subject: Cube Verification From: mark.longridge@canrem.com (Mark Longridge) Message-Id: <60.1255.5834.0C1F9CFB@canrem.com> Date: Fri, 20 Oct 95 02:22:00 -0500 Organization: CRS Online (Toronto, Ontario) In Jerry's message from Date: Mon, 16 Oct 1995 23:08:02 -0400 (EDT) Subject: The Correctness of Large Seaches > In light of the above discussion, I thought it might be appropriate > to summarize some background about my larger searches. I want to > indicate which ones of them seem pretty well verified, and which ones > of them might benefit from further study. > > - I believe this one is ok. I ran the q turn case > with and without conjugacy classes. Upon > expanding the conjugacy classes, the results > matched the results without conjugacy classes. > Also, the results matched results posted by > Mark Longridge as far as they went (although > my anomaly with corners-only suggests that > such matching doesn't prove very much). Well, I did get up to 12 q turns deep ;-) Good enough for 2 half deep searches... But there is another possible verification method by counting the number of even positions and odd positions and totalling them. Analysis of < U, R > group on 3x3x3 cube by Parity -------------------------------------------------- Even Positions Odd Positions -------------- ------------- 0 1 1 4 2 10 3 24 4 58 5 140 6 338 7 816 8 1,970 9 4,756 10 11,448 11 27,448 12 65,260 13 154,192 14 360,692 15 827,540 16 1,851,345 17 3,968,840 18 7,891,990 19 13,659,821 20 18,471,682 21 16,586,822 22 8,039,455 23 1,511,110 24 47,351 25 87 ---------- ---------- 36,741,600 36,741,600 This is almost Cube Philosophy... how can we be certain about the true nature of God's Algorithm? How can we be certain our cube databases are completely accurate? I suppose it is not really a big problem as long as the various cube programs all agree, and a human observer executes such processes on a real cube. -> Mark <- From din5w@server.cs.virginia.edu Fri Oct 20 06:24:11 1995 Received: from virginia.edu (uvaarpa.Virginia.EDU) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA10327; Fri, 20 Oct 95 06:24:11 EDT Received: from server.cs.virginia.edu by uvaarpa.virginia.edu id aa26937; 19 Oct 95 22:57 EDT Received: from cobra.cs.Virginia.EDU (cobra-fo.cs.Virginia.EDU) by uvacs.cs.virginia.edu (4.1/5.1.UVA) id AA07239; Thu, 19 Oct 95 22:57:16 EDT Posted-Date: Thu, 19 Oct 1995 22:57:15 -0400 (EDT) Return-Path: Received: by cobra.cs.Virginia.EDU (5.x/SMI-2.0) id AA24433; Thu, 19 Oct 1995 22:57:15 -0400 Date: Thu, 19 Oct 1995 22:57:15 -0400 (EDT) From: Dale Newfield X-Sender: din5w@cobra.cs.Virginia.EDU Reply-To: DNewfield@virginia.edu To: cube-lovers@ai.mit.edu Subject: Re: Embedding G in a symmetrical group In-Reply-To: <199510200141.CAA03659@wn1.sci.kun.nl> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII On Fri, 20 Oct 1995, Michiel Boland wrote: > It is clear that the group G of the cube (the one with > 4.3252x10^19 elements) can be embedded in a symmetrical group, e.g. > S_48, since each move of the cube can be seen as a permutation of 48 > objects. Um...If I were a better net.person, I'd look up which version of the cube has that number of elements, but wouldn't it be correct to say that each move of the cube is a permutation of the pieces of the cube, i.e. the 26 cubies? (Or even, depending on which cube-model you are using(This is what I should have looked up), if you ignore center cubie orientation, the 20 cubies?) If that logic holds, then the largest possible S_n would be S_20, much less than the 32 that you claim is minimal... ...I think I'm just confused--can you alleviate that problem? -Dale Newfield From hazard@niksula.hut.fi Fri Oct 20 07:00:31 1995 Return-Path: Received: from nukkekoti.cs.hut.fi by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA10664; Fri, 20 Oct 95 07:00:31 EDT Received: from ummagumma.tky.hut.fi (hazard@ummagumma.tky.hut.fi [130.233.33.120]) by nukkekoti.cs.hut.fi (8.6.12/8.6.11) with SMTP id NAA14906 for ; Fri, 20 Oct 1995 13:00:30 +0200 Date: Fri, 20 Oct 1995 13:00:30 +0200 Message-Id: <199510201100.NAA14906@nukkekoti.cs.hut.fi> X-Sender: hazard@pop.niksula.cs.hut.fi X-Mailer: Windows Eudora Pro Version 2.1.2 Mime-Version: 1.0 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: 8bit To: cube-lovers@ai.mit.edu From: Mikko Haapanen Subject: Old question about 2 adj edges Hello! I want to ask if somebody can tell me how to flip 2 adj. edges (and nothing else) in 4x4x4 cube? I have my own formula to do this but i cannot write it down because i have to discover it every time ;) and i turn whole cube so many times during it. I remember 39 or 40 turns have been the shortest way i've seen. What i need is that formula in BFUDLR (or something) language. I have another question about this mailing list. I have seen many lists similar to cube-lovers in my .nersrc file. Can i find cube-lovers list somewhere with my newsreader? Thanks. ------Mikko Haapanen------hazard@niksula.hut.fi--------- Jos tahto siirtää vuoren, niin tahdon siirtää tahdon Vuorta siirtämään, että vuoren sisään nään -P.Hanhiniemi -------------------------------------------------------- From BRYAN@wvnvm.wvnet.edu Fri Oct 20 09:02:16 1995 Return-Path: Received: from WVNVM.WVNET.EDU by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA15552; Fri, 20 Oct 95 09:02:16 EDT Received: from WVNVM.WVNET.EDU by WVNVM.WVNET.EDU (IBM VM SMTP V2R3) with BSMTP id 6200; Fri, 20 Oct 95 08:38:42 EDT Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU (LMail V1.2a/1.8a) with BSMTP id 4649; Fri, 20 Oct 1995 08:38:42 -0400 Message-Id: Date: Fri, 20 Oct 1995 08:38:41 -0400 (EDT) From: "Jerry Bryan"