From cube-lovers-errors@oolong.camellia.org Mon Jun 9 19:04:20 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id TAA09819; Mon, 9 Jun 1997 19:04:20 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Message-Id: <3.0.1.32.19970609160106.00ade4a0@sdgmail.ncsa.uiuc.edu> X-Sender: mag@sdgmail.ncsa.uiuc.edu X-Mailer: Windows Eudora Pro Version 3.0.1 (32) Date: Mon, 09 Jun 1997 16:01:06 -0500 To: cube-lovers@ai.mit.edu From: Tom Magliery Subject: Re: Designations for the cubes (proposal) In-Reply-To: Mime-Version: 1.0 Content-Type: text/plain; charset="us-ascii" At 11:31 AM 6/8/97 -0400, Nicholas Bodley unabashedly said: > > Peter (Reitan, I think; sorry) (who is not Karen) brought up the >clumsiness of such designations as "5X5X5". I find these downright clumsy >to type (although the Caps Lock key helps). In my private world, I simply >refer to the Pocket Cube as "[the] two", the original Rubik's as "[the] >three", Revenge as "[the] four", and the biggest available as "[the] >five". > > I think that provided we understand that we are referring to the >well-known family of true cubes, it should be OK simply to refer to "the >three", for instance. Granted, these names require more keystrokes, but >numerals should be OK, as in "the 3". I have another suggestion, which might be slightly less likely to require explanation to a newcomer. I know how to *pronounce* it, but I'm not sure how I would recommend *spelling* it. (Considerations include terseness, ease of typing -- which is of course not the same thing!, and likeliness to be mispronounced by a reader.) The pronunciation is three-bye, four-bye, five-bye, ... Possible spellings include: 3by, 4by, 5by, ... 3-by, 4-by, 5-by, ... 3x, 4x, 5x, ... three-by, four-by, five-by, ... mag -- .---o Tom Magliery, Research Programmer (217) 333-3198 .---o `-O-. NCSA, 605 E. Springfield O- mag@ncsa.uiuc.edu `-O-. o---' Champaign, IL 61820 http://sdg.ncsa.uiuc.edu/~mag/ o---' From cube-lovers-errors@oolong.camellia.org Mon Jun 9 19:03:57 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id TAA09815; Mon, 9 Jun 1997 19:03:57 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Message-ID: <339C56DB.5809@hrz1.hrz.th-darmstadt.de> Date: Mon, 09 Jun 1997 21:17:47 +0200 From: Herbert Kociemba X-Mailer: Mozilla 3.0 (Win95; I) MIME-Version: 1.0 To: cube-lovers@ai.mit.edu Subject: Re: Detailed explanation of two phase algorithm References: <970608193131.21411978@iccgcc.cle.ab.com> Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit SCHMIDTG@iccgcc.cle.ab.com wrote: > > And if we want to show that all depth one nodes will be pruned when > we are at some search depth d where 1 < d < h[0] we would need to show > that: > > 1.9 1 + h[1] > h[0] > Why do you say 1 < d < h[0] and not d = 1? What I wanted to show is, that unter the assumption 2.0 D < h(0) all depth-one nodes will be pruned. As you correctly stated before, for pruning we need 1.8 d + h(d) > D and in the case d=1 this means 1.9a 1 + h(1) > D, which is different from 1.9, because of 2.0 . But 1.9a can be shown easily: In my last message, I tried to explane that 2.1 |h(n-1)-h(n)| <=1, I try to explain it once more in other words. A node at depth n is generated from a node at depth n-1 by applying a single face-turn on it. And as I told, h is defined by h(x,y,z):=max{h1(x,y),h2(x,z),h3(y,z)}, where for example h1(x,y) is the length of the shortest maneuver sequence which transforms (x,y,z) to (x0,y0,z') for any z' (this means the z-coordinate is ignored). And this length can maximal change by one when applying a single move. The same holds for h2(x,z) and h3(y,z). For this reason, h(x,y,z) also can change maximal by one, which implies 2.1 . In the case n=1, from 2.1 follows h(0) <= 1 + h(1), and because of 2.0 we have D < h(0) <= 1 + h(1), which proves 1.9a . > (1) > / \ > (2) (3*) cost = .9 > / > (4*) cost = .7 > > Suppose nodes 3 and nodes 4 were both solutions. Even though node 4 > has a lower cost, phase1 would find node 3 to be our first solution > whereas IDA* wouldn't. I don't think we are far away from each other. Of course, the phase1 (or phase2) algorithm does not claim to be an universal IDA* for any sort of problem. But for a special problem like the cube you can simplify the general IDA* and the simplified algorithm will be equivalent to the one I developed for phase1. Best regards, Herbert From cube-lovers-errors@oolong.camellia.org Mon Jun 9 21:30:42 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id VAA10040; Mon, 9 Jun 1997 21:30:41 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org From: bandecbv@mailhost.rz.ruhr-uni-bochum.de Message-Id: <199706100041.UAA11455@life.ai.mit.edu> Comments: Authenticated sender is To: mgirard@videotron.ca Date: Tue, 10 Jun 1997 02:38:52 +0000 MIME-Version: 1.0 Content-type: text/plain; charset=US-ASCII Content-transfer-encoding: 7BIT Subject: Magic Cubes CC: cube-lovers@ai.mit.edu Priority: normal X-mailer: Pegasus Mail for Windows (v2.42a) Mathieu Girard wrote: >I am in a quest to buy both 4x4x4 Rubik's Revenge and also 5x5x5 >cube.. but i just can't find any! Regrettably the 4x4x4 cubes seem to be sold out everywhere in the world. With the 5x5x5 Magic Cubes (in Japan once sold under the name Professor's Cube), the situation is much better: They are still available from me (and as far as I know nowhere else now). Since they will probably never be produced again (the production cost is too high and the general interest too low), they will also be sold out soon. I shall inform the cube-lovers when this is the case. My price of the 5x5x5 cube is 40 DM or 24 USD plus postage. I send my free mail order catalog (containing also many other twisting puzzles like the Magic Dodecahedron , the Skewb, the Pyraminx, Mickey's Chellenge and several books and details how to order) to every cube-lover requesting it and providing a postal address. A few days ago, Joe McGarity complained bitterly about his 5x5x5 cube which fell apart. Fortunately, I did not encounter this problem before, and Joe is not in my files so he has probably not bought his cube from me. On the other hand you will destroy every twisting puzzle by twisting it with force without sufficient aligning the layers before every single move. Since the 5x5x5 cube contains 98 visible little cubies compared to the 26 of the 3x3x3 (and 92 compared to 20 if we only count the freely floating ones), one should accept that it requires a little bit more care. Joe McGarity also mentioned that some orange stickers sometimes do not behave according to there name. I have to admit that this sometimes also happens with my 5x5x5 cubes. Furtunately it happens only to the orange stickers and it can be repaired easily by warm pressure or - better - some glue. Christoph Christoph Bandelow mailto:Christoph.Bandelow@rz.ruhr-uni-bochum.de From cube-lovers-errors@oolong.camellia.org Tue Jun 10 15:15:53 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id PAA11724; Tue, 10 Jun 1997 15:15:53 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Date: Tue, 10 Jun 1997 08:16:53 -0400 (EDT) From: Nicholas Bodley To: Tom Magliery cc: cube-lovers@ai.mit.edu Subject: Re: Designations for the cubes (proposal) In-Reply-To: <3.0.1.32.19970609160106.00ade4a0@sdgmail.ncsa.uiuc.edu> Message-ID: MIME-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII I like Tom's version; it is concise, distinctive enough not to be confused (however, I haven't studied all the various math. notations going around), and easy to say and type. Of the ones you gave, I prefer such a form as "3-by". Thanks! |* Nicholas Bodley *|* Electronic Technician {*} Autodidact & Polymath |* Waltham, Mass. *|* ----------------------------------------------- |* nbodley@tiac.net *|* When the year 2000 begins, we'll celebrate |* Amateur musician *|* the 2000th anniversary of the year 1 B.C.E. -------------------------------------------------------------------------- From cube-lovers-errors@oolong.camellia.org Tue Jun 10 15:16:08 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id PAA11728; Tue, 10 Jun 1997 15:16:08 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Date: Tue, 10 Jun 1997 13:07:15 -0400 (Eastern Daylight Time) From: Jerry Bryan Subject: Re: Some Face Turn Numbers In-reply-to: To: Cube-Lovers Message-id: MIME-version: 1.0 Content-type: TEXT/PLAIN; charset=US-ASCII X-X-Sender: jbryan@PSTCC6.pstcc.cc.tn.us On Thu, 5 Jun 1997, Jerry Bryan wrote: > The following table gives the > best known results for face turns. The results through depth 7 have been > calculated (my message of 19 July 1994). The rest are based on Dan Hoey's > recursion formula PH[n] = 6*2*PH[n-1] + 9*2*PH[n-2] for n>2, where PH[n] > is the number of face turns which are n moves from Start Rats, here is a little correction. I think my meaning was clear from the overall context of the note, but Dan's formula is an upper bound, so it should read PH[n] <= 6*2*PH[n-1] + 9*2*PH[n-2] for n>2. For depth 0 through 7, my table provided exact values. As I hope was clear from the context, my table included upper bounds rather than exact values for depths greater than 7. My apologies. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) jbryan@pstcc.cc.tn.us Pellissippi State (423) 539-7198 10915 Hardin Valley Road (423) 694-6435 (fax) P.O. Box 22990 Knoxville, TN 37933-0990 From cube-lovers-errors@oolong.camellia.org Tue Jun 10 15:15:37 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id PAA11720; Tue, 10 Jun 1997 15:15:37 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Message-ID: <339CB8B7.1B8D@snowcrest.net> Date: Mon, 09 Jun 1997 19:15:19 -0700 From: Joe McGarity Reply-To: joemcg3@snowcrest.net X-Mailer: Mozilla 3.01Gold (Win95; I) MIME-Version: 1.0 To: "Mailing List, Rubik's Cube" Subject: 5x cubes Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Yes, the cube I purchased was not from Cristoph Bandelow. He is off the hook. It was purchased in San Francisco at the Game Gallery. Although I am considering buying my next one from him. And no, I don't force it either. I treated it with kindness and love, yet it betrayed me. From cube-lovers-errors@oolong.camellia.org Wed Jun 11 00:38:42 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id AAA12658; Wed, 11 Jun 1997 00:38:41 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org From: SCHMIDTG@iccgcc.cle.ab.com Date: Wed, 11 Jun 1997 0:35:55 -0400 (EDT) To: cube-lovers@ai.mit.edu Message-Id: <970611003555.21417ec3@iccgcc.cle.ab.com> Subject: Re: Detailed explanation of two phase algorithm Herbert Kociemba wrote: >SCHMIDTG@iccgcc.cle.ab.com wrote: >> >> And if we want to show that all depth one nodes will be pruned when >> we are at some search depth d where 1 < d < h[0] we would need to show >> that: >> >> 1.9 1 + h[1] > h[0] >> > >Why do you say 1 < d < h[0] and not d = 1? Oops, I think that should have been 'D' and not 'd'. >[...slightly different restatement of earlier proof omitted...] After examining this once again, I have now satisfied myself that it is correct. It's just that for some reason, I seem to find the result rather counter-intuitive. But that makes the result all the more interesting. So I think this may yet me another case where the phase1 algorithm differs slightly from IDA*, but the difference is not significant since, in this case, one is able to prove a special property of the heuristic that demonstrates that the number of nodes explored by the two algorithms is comparable. At this point, I think we can wind down this thread, (I do hope others on this list have found it interesting) and I will still continue to think of possible ideas for improving the algorithm. I do have one last question regarding the pruning tables. While the three tables used in phase1 are clear, I do not recall reading a description of the tables that are used in phase2. I examined Dik Winter's program and he seems to have a few more "maximum move" (i.e. "mm" tables) than I expected, namely: phase1 ------ mm_twists[] mm_flips[] mm_choices[] /* and the following "mixed" tables */ mm_tf[][] /* twist & flip */ mm_tc[][] /* twist & choice */ mm_fc[][] /* flip & choice */ phase2 ------ mm_eperms[] /* edge perms */ mm_cperms[] /* corner perms */ mm_sperms[] /* slice orderliness */ /* "mixed" tables follow */ mm_cs[][] /* corner perms & slice orderliness */ mm_es[][] /* edge perms and slice orderliness */ Are you using the same tables? Or are the "mixed" tables ones that Dik added to the algorithm? It appears that Dik was able to use them because he had a machine with more memory at his disposal than your 1MB Atari ST. His program can be built with or without the "mixed" tables and is 11MB with them. He also mentions that the small program finds a reasonable solution in 30 minutes whereas the large program finds it in only a few seconds. I have also been studying his code to try to understand how he generates these tables. He does not seem to be using breadth-first-search to fill in these tables as Korf does. I will be interested in looking at your new program when it becomes available. Thanks again for your patience. Best regards, -- Greg From cube-lovers-errors@oolong.camellia.org Wed Jun 11 14:05:23 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id OAA14044; Wed, 11 Jun 1997 14:05:23 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Date: Wed, 11 Jun 1997 08:39:17 -0700 From: "Jason K. Werner" Message-Id: <9706110839.ZM926@isdn-rubik.corp.sgi.com> In-Reply-To: Joe McGarity "5x cubes" (Jun 9, 19:15) References: <339CB8B7.1B8D@snowcrest.net> Reply-to: "Jason K. Werner" X-Mailer: Z-Mail-SGI (3.2S.3 08feb96 MediaMail) To: cube-lovers@ai.mit.edu, Mark Pilloff , Mathieu Girard , joemcg3@snowcrest.net Subject: Re: 5x cubes Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii Here's the place where I got my 5x cube, which has been extremely solid and nearly indestructible: Game Gallery 2855 Stevens Creek Blvd. Santa Clara, CA 95050 USA 408-241-4263 408-241-5945 FAX http://www.gamegallery.com From cube-lovers-errors@oolong.camellia.org Wed Jun 11 14:04:50 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id OAA14037; Wed, 11 Jun 1997 14:04:50 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Date: Wed, 11 Jun 1997 08:32:27 -0400 (EDT) From: der Mouse Message-Id: <199706111232.IAA00315@Twig.Rodents.Montreal.QC.CA> To: cube-lovers@ai.mit.edu Cc: Mathieu Girard Subject: Re: ... MIME-Version: 1.0 Content-Type: text/plain; charset=iso-8859-1 Content-Transfer-Encoding: quoted-printable X-MIME-Autoconverted: from 8bit to quoted-printable by life.ai.mit.edu id IAA16376 > I am in a quest to buy both 4x4x4 Rubik's Revenge and also 5x5x5 > cube.. but i just can't find any! > I live in Qu=E9bec, Canada, Where? Montr=E9al, or Qu=E9bec City, or what? I'm in Montr=E9al; I got = my 5-Cube at Valet de Coeur, on the west side of St-Denis, somewhere a bit south of Mont-Royal. I don't know whether they still have them; this _was_ back in 1993 (December 15, according to my records). > By the way.. if it is not asking too much... could u please give me > an aproximate of the prices of thoses cubes... in canadian or us > dollar... I paid $45.01, including tax, for my 5-Cube. But as I remarked above, that _was_ three and a half years ago. der Mouse mouse@rodents.montreal.qc.ca 7D C8 61 52 5D E7 2D 39 4E F1 31 3E E8 B3 27 4B From cube-lovers-errors@oolong.camellia.org Wed Jun 11 16:11:56 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id QAA14368; Wed, 11 Jun 1997 16:11:56 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org To: Cube-Lovers@AI.MIT.EDU From: Wei-Hwa Huang Subject: Re: 5x5x5 Stuctural Integrtity Date: 11 Jun 1997 19:01:30 GMT Organization: California Institute of Technology, Pasadena Message-ID: <5nmsma$t63@gap.cco.caltech.edu> References: NNTP-Posting-Host: blend.ugcs.caltech.edu X-Newsreader: NN version 6.5.0 #2 (NOV) David Litwin writes: > The orange sticker problem seems to be with all of them. Mine had >some small documentation with it mentioning that putting a piece of paper >on the orange side and ironing it a bit will help fix the stickers on the >cube. It worked well for me and I've not had any problems with them >anymore. I must say that the first sticker I lost on my 5x5x5 was red. (And I don't know where it went! ARGH!!!) -- Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/ ------------------------------------------------------------------------------- Inspiration strikes suddenly, so be prepared to defend yourself. From cube-lovers-errors@oolong.camellia.org Wed Jun 11 16:11:24 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id QAA14364; Wed, 11 Jun 1997 16:11:24 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Message-ID: <339EF3BF.766D@hrz1.hrz.th-darmstadt.de> Date: Wed, 11 Jun 1997 20:51:44 +0200 From: Herbert Kociemba X-Mailer: Mozilla 3.0 (Win95; I) MIME-Version: 1.0 To: cube-lovers@ai.mit.edu Subject: Re: Detailed explanation of two phase algorithm References: <970611003555.21417ec3@iccgcc.cle.ab.com> Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit SCHMIDTG@iccgcc.cle.ab.com wrote: > I do have one last question regarding the pruning tables. While > the three tables used in phase1 are clear, I do not recall reading > a description of the tables that are used in phase2. In phase2, the state of the cube also is described by a triple (x,y,z), in this case 0<=x<8! describes a permutation of the 8 corners, 0<=y<8! describes a permutation of the 8 UD-slice edges and 0<=z<4! describes a permutation of the middleslice edges. Because the overall permutation must be even, only half of the triples belong to physical cubes. We could correct this, by defining the z coordinate to describe one of the 12 possibilities for the locations of two middleslice edges - the other two edges will then be corrected automatic. But there are good reasons not to do so (which I think is not necessary to explain here). > I have also been studying his code to try to understand how he generates > these tables. He does not seem to be using breadth-first-search to > fill in these tables as Korf does. > I only use the "mixed" tables. How to generate the tables is quite obvious and though I don't know how Dik does it I'm sure it is similar: 1. On initialisation set all elements of the table to 0xf (we use four bits per entry), only the element belonging to (x0,y0,z0) is set to 0. Set L=0, n_done=1, n_old=1 (n_done denotes the number of valid tableentries). 2. Check all elements of the table one after the other. If an entry is 0xf, do nothing. If the entry is L, compute the the 18 possible "child nodes" and check, if the corresponding tableentry is 0xf. Only in this case set it to L+1 and increment n_done. 3. Check if n_done=n_old. In this case we are ready. Else increment L, set n_old=n_done and goto 2. > I will be interested in looking at your new program when it becomes > available. I'm writing too much to this mailing list and do not work at my windows-help! The program itself is ready. I did a two hours run on each of Rich Korfs 10 random cubes on a Pentium133 with 16MB RAM and the result were really pleasing: The generated maneuver lenghts were on the average less than 1 move away from Rich Korfs optimal solutions (exactly: 9 moves more for the 10 cubes). Best regards, Herbert From cube-lovers-errors@oolong.camellia.org Wed Jun 11 20:44:06 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id UAA15754; Wed, 11 Jun 1997 20:44:06 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org From: SCHMIDTG@iccgcc.cle.ab.com Date: Wed, 11 Jun 1997 20:42:11 -0400 (EDT) To: cube-lovers@ai.mit.edu Message-Id: <970611204211.2141971d@iccgcc.cle.ab.com> Subject: Re: Detailed explanation of two phase algorithm Herber Kociemba wrote: [...much additional detailed explanation deleted...] Thank again, I found this information helpful, especially when my only other option is to examine code in great detail in order to extract out the general principles. >> I will be interested in looking at your new program when it becomes >> available. > >I'm writing too much to this mailing list and do not work at my >windows-help! Sorry about that. I'll stop with my questions. In fact, no need to even answer this response! I'm sure your program will be well worth the wait :). > The program itself is ready. I did a two hours run on each >of Rich Korfs 10 random cubes on a Pentium133 with 16MB RAM and the >result were really pleasing: The generated maneuver lenghts were on the >average less than 1 move away from Rich Korfs optimal solutions >(exactly: 9 moves more for the 10 cubes). Very impressive. And if you perform some longer runs and find optimal solutions, please be sure to let us know the run times. Best regards, -- Greg From cube-lovers-errors@oolong.camellia.org Thu Jun 12 13:23:55 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id NAA17772; Thu, 12 Jun 1997 13:23:54 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Date: Thu, 12 Jun 1997 01:03:11 -0400 (EDT) From: Nicholas Bodley To: Wei-Hwa Huang cc: Cube-Lovers@ai.mit.edu Subject: Re: 5x5x5 Structural Integrtity (Stickers) In-Reply-To: <5nmsma$t63@gap.cco.caltech.edu> Message-ID: MIME-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII One cure might be to take a solved cube (more convenient...) and remove all the red stickers; carefully clean off the adhesive, perhaps with 99% isopropyl alcohol, and paint the surfaces neatly with the type of paint used for plastic model kits. I have also thought of removing the stickers, cleaning all the adhesive off both the stickers and the cubies, and then reattaching the stickers with a different type of adhesive. These are just ideas, and I hope no source of trouble. My best to all, |* Nicholas Bodley *|* Electronic Technician {*} Autodidact & Polymath |* Waltham, Mass. *|* ----------------------------------------------- |* nbodley@tiac.net *|* When the year 2000 begins, we'll celebrate |* Amateur musician *|* the 2000th anniversary of the year 1 B.C.E. -------------------------------------------------------------------------- From cube-lovers-errors@oolong.camellia.org Wed Jun 18 16:19:15 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id QAA04432; Wed, 18 Jun 1997 16:19:15 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org To: cube-lovers@ai.mit.edu Date: Wed, 18 Jun 1997 15:05:54 -0500 Subject: Square One Message-ID: <19970618.150557.11350.0.shaggy34@juno.com> X-Mailer: Juno 1.38 X-Juno-Line-Breaks: 0-2 From: Josh D Weaver Does anyone know how to solve one of those "Square One" puzzles? Josh From cube-lovers-errors@oolong.camellia.org Wed Jun 18 18:43:38 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id SAA04738; Wed, 18 Jun 1997 18:43:37 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Message-ID: <33A84C0E.94@hrz1.hrz.th-darmstadt.de> Date: Wed, 18 Jun 1997 22:58:54 +0200 From: Herbert Kociemba X-Mailer: Mozilla 3.0 (Win95; I) MIME-Version: 1.0 To: cube-lovers@ai.mit.edu Subject: Windows95 program now available Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit The Windows95 program which implements my algorithm to solve Rubik's Cube is now availabe at http://home.t-online.de/home/kociemba/cube.htm It not only solves Rubik's cube, but also does a few other nice things... Herbert [ Moderator's note: This program is also available in the Cube-Lovers Archive. See: ftp://ftp.ai.mit.edu/pub/cube-lovers/contrib/cubexp10.zip - Alan ] From cube-lovers-errors@oolong.camellia.org Thu Jun 19 01:16:47 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id BAA05551; Thu, 19 Jun 1997 01:16:47 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org From: Benjamin Wong To: Josh D Weaver Date: Thu, 19 Jun 1997 11:37:43 +1000 (EST) X-Sender: chi@pipe02.orchestra.cse.unsw.EDU.AU cc: cube-lovers@ai.mit.edu Subject: Re: Square One In-Reply-To: <19970618.150557.11350.0.shaggy34@juno.com> Message-ID: On Wed, 18 Jun 1997, Josh D Weaver wrote: ._@_.Does anyone know how to solve one of those "Square One" puzzles? ._@_. http://www.cfar.umd.edu/~arensb/Square1/ is the only page on the net (that i can find) which describe how to solve square 1 however, either i can not follow instruction, or error in it's instructuion i just can not solve it with their algorimthm I bought square 1 mess them up, only manage to solve them 2 times. (beginner luck) but the page does not help very much. ._@_.Josh ._@_. o------------------------------------------------------o |Error: Reality.sys Corrupt? Reboot Universe [Y,N,Q] | +---------------o--------------------------------------o | Benjamin Wong | E-mail: chi@cse.unsw.edu.au | | | or benjaminwong@hotmail.com | | | http://www.cse.unsw.edu.au/~chi | o---------------o--------------------------------------o |=A1u=C2=E5=A5=CD=A1I=BD=D0=B0=DD=A1y=BA=B5=BF=DF=B2=B4=A1z=AA=BA=A6=A8=A6]=ACO=AC=C6=BB=F2=A1H=A1v | |Quick Quiz: Describe Universe ? Give Three Example. | o------------------------------------------------------o From cube-lovers-errors@oolong.camellia.org Thu Jun 19 12:31:32 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id MAA06811; Thu, 19 Jun 1997 12:31:31 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Message-ID: <33A8DBC4.2A8F@snowcrest.net> Date: Thu, 19 Jun 1997 00:12:04 -0700 From: Joe McGarity Reply-To: joemcg3@snowcrest.net X-Mailer: Mozilla 3.01Gold (Win95; I) MIME-Version: 1.0 To: Josh D Weaver CC: "Mailing List, Rubik's Cube" Subject: Re: Square One References: <19970618.150557.11350.0.shaggy34@juno.com> Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit After three months of agony and wondering if Square One was actually the puzzle box from Clive Barker's Hellraiser, I managed to come up with a solution that covered all the bases, i.e. worked every time. I have never tried to write it out in a step by step form however. I will try to cover the basics. First the puzzle looks like a Rubik's Cube when solved in that it is a cube with a solid color on each face, but the similarity ends there. Square One more closely resembles the Orb, Masterball and Smart Alex in the ways that it moves. If you can solve any of those you will be a step closer to the Square One. I see the square one as nearly identicle to the Smart Alex. The shapes of the pieces are different, but they move as a disk divided into sectors (exactly like the Masterball). There are six pieces on each face if you count the small sectors as half pieces. Count the pieces and you will see what I mean. The little ones are half the the size of the angle of the big ones. The idea then for me was to get the little ones paired up like the picture in the instruction booklet. Once they were paired correctly I could solve it just like the Masterball or Smart Alex making it look like it did when it was new in the package (remember it came in a slightly scrambled state with instructions on how to solve it from there in about six moves). Then I could just follow the booklet for the final part. Like I said it took three months and scores of note paper to finally get it. When I did, the walls opened up and the Cenobites took me away, but it was worth it. I hope I haven't caused more confusion. It is difficult to describe without having one in my hands to show you. This is just a sketchy overview of how I solve it. If I get a chance to document this solution I will send you a copy, but it probably won't be for a while. I'm sure that someone has a better solution and I'd be interested in seeing what others have come up with. My solution takes about twenty minutes to do and there must be a faster way. The ones I have trouble with are the Sqewb and the Alexander's Star. Anybody got a good solution for any of these? Joe McGarity From cube-lovers-errors@oolong.camellia.org Thu Jun 19 12:32:15 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id MAA06816; Thu, 19 Jun 1997 12:32:15 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Message-ID: From: "joyner.david" To: "'Josh D Weaver'" Cc: "'cube-lovers@ai.mit.edu'" Subject: RE: Square One Date: Thu, 19 Jun 1997 08:06:58 -0400 X-Mailer: Microsoft Exchange Server Internet Mail Connector Version 4.0.994.63 MIME-Version: 1.0 Content-Type: text/plain; charset="us-ascii" Content-Transfer-Encoding: 7bit >---------- >From: Josh D Weaver[SMTP:shaggy34@juno.com] >Sent: Wednesday, June 18, 1997 4:05 PM >To: cube-lovers@ai.mit.edu >Subject: Square One > >Does anyone know how to solve one of those "Square One" puzzles? There's a paper on my web page which indirectly explains how. (It's actually a math paper written with a student of mine explaining the group theory of the puzzle.) What's useful are some of the moves which we give. If you can't print it out (It's a dvi file) I'll mail it to you if you give me your postal address. http://www.nadn.navy.mil/MathDept/wdj/rubik.html The idea, if I remember, is 1. get into a square form, 2. use the special moves we give (moves which permute 3 pieces only and leave the others alone, for example) to solve the puzzle as one solves the Rubik's cube. - David Joyner > >Josh > > From cube-lovers-errors@oolong.camellia.org Thu Jun 19 12:32:39 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id MAA06820; Thu, 19 Jun 1997 12:32:39 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Date: Thu, 19 Jun 1997 09:19:01 -0400 Message-Id: <199706191319.JAA27307@bso.newvision.com> From: Carl Woolf To: chi@cse.unsw.edu.au CC: shaggy34@juno.com, cube-lovers@ai.mit.edu In-reply-to: (message from Benjamin Wong on Thu, 19 Jun 1997 11:37:43 +1000 (EST)) Subject: Re: Square One Square One is a great puzzle! I think there is an instruction booklet, published in Massachusetts or thereabouts, and available from Puzzlets (mgreen@puzzletts.com). I developed a set of techniques that let me solve the thing, but I haven't worked my notes into a form intelligible by other humans (or by me on a bad day). -- -- Carl ----------------------------------------------------- Business: woolf@newvision.com Personal: woolf@ccs.neu.edu http://www.ccs.neu.edu/home/woolf From cube-lovers-errors@oolong.camellia.org Thu Jun 19 21:55:51 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id VAA07813; Thu, 19 Jun 1997 21:55:51 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Message-Id: <199706200151.AA15092@world.std.com> To: "cube-lovers@ai.mit.edu" Subject: Square One Solution Date: Thu, 19 Jun 97 21:53:07 -0500 From: Mike Masonjones X-Mailer: E-Mail Connection v2.5.03 Cube Lovers, This is long, but complete solutions are long. I even left quite a bit of the boring and obvious stuff out, and it still took me 3 hours to write it. Any errors, let me know. I hope this satisfies. Apparently I have the least extracurricular life of all of you, since I know how to solve Square One much quicker than any reports I've seen here. Quite a dubious honor, I suspect. Anyway, here goes my solution, which with a little practice, guarantees a solution within 75-100 seconds for someone who can do The Cube in about 55-60 seconds (assuming there's a correlation in hand speed from puzzle to puzzle). The solution is completely my own, as is the notation. Sorry if it offends anyone. 1. Start by getting to the cube state for the top and bottom faces. Ignore the middle slice til the very end. This can be most efficiently done by memorizing a table. A pretty good (but not error-free) one is on the only square-one web site in existence that I know of. Sorry, you'll have to find the site yourself with a browser, since I can't look it up right now. I have a scheme written down somewhere that is quite a bit easier to memorize, but why should I take the fun away from any of you looking for the solution yourselves. If there is a big response to this letter, then I will dig out my easy table. Tables are difficult to memorize, so I usually just try to get to a six pointed star on one side and all the little wedges plus the remaining 2 big wedges on the other. It is easy to get to one of the five possible states that result, thus requiring memorization of only 5 solutions to get back to a cube. This method takes about 5-10 seconds longer, an average, than the table technique. Using a notation where L represents a large wedgie, and S a small wedgie, the five possible states can be written as: 1) bottom = LLLLLL, top = LLSSSSSSSS 2) bottom = LLLLLL, top = LSLSSSSSSS 3) bottom = LLLLLL, top = LSSLSSSSSS 4) bottom = LLLLLL, top = LSSSLSSSSS 5) bottom = LLLLLL, top = LSSSSLSSSS For cases 1),3),5), rotate the top face so that it will be sliced symmetrically between the two L pieces when the center is flipped. The next move in each of these cases involves moving the top face one way and the bottom face the other, when looking from the front. (Front will be the term used from now on to denote the end nearest you of the central cut through which flipping occurs (a 180 turn of one half of the cube)). After a flip, cases 1) and 5) should give two barrel shapes (LLSSLLSS), top and bottom. You should aim in case 3) for two tomahawk (LLSSLSLS) shapes. Any self- respecting cubist should be able to get home from here. Cases 2 and 4 are a little more complicated. For both cases align the top so that the left half of the top face reads, going clockwise, SLSSS. Flip right side. Now rotate the bottom so that when you flip with the right hand , the top will read SlSSSSSLL starting from the front and going clockwise. Now rotate the top 1/12 turn counterclockwise and the bottom so it reads LLLLSSL going clockwise from the front and flip again. Now you're in an easy state to get home from (LLLLSSSS on top and LLSLSSLS on bottom). 2. Now that you're in a cube state top and bottom, get all the wedgies on their correct side (top and bottom face all the same color, respectively). This is very straightforward and intuitive. I usually start with one large wedgie, and sequentially put in one at a time next to it going around a face until you get down to one S wedgie stuck on the wrong side. Sometimes it is easier to do LSL on one half of the top, and then do LSL on the other, and then putting in the second to last S between the groups. Now position the top face so that the Odd S wedgie (O) is positioned as LSLOLSLS going clockwise from the front. Put the bottom odd wedgie in front with the bottom square skewed from the top (bottom should read LSLSLSLO going clockwise from the front). Now do FT4B1FT-4B-1FT4B1FT-3F, where F = flip with right hand, Bx = turn bottom face clockwise x/12 of a turn, B-x = same thing counterclockwise, Tx, T-x mean similar things. 3. Now get L's positioned. Case 1. No L's are correctly adjacent to each other. Position top and bottom (top = LSLSLSLS, bottom = SLSLSLSL, each going clockwise from front). Now go FB3FT-3B-3FT3F, turn the whole puzzle 180 degrees so that the back of the central cut is now the front, and repeat the move. Case 2. Two sets of adjacent pairs are out of whack, one on top, and one on bottom. Do the move for case 1 once, with the components of the pairs in question all nearest the front. Case 3. Only one adjacent pair correct. Position the top so that the correctly adjacent pair (denoted as A) is positioned as ASASLSLS, and the bottom reads LSLSLSLS (same conventions as before). Now do FB-3FB3FB-3FB3F. Case 4. Only one pair incorrect. Position the top (with the incorrect pair) so that the correctly adjacent pair is positioned as LSASASLS, and the bottom reads LSLSLSLS. Do the move in Case 3 twice with a T3 between instances. Case 5. One side is OK, the other has no correct adjacent pairs. Bad side = top. top = LSLSLSLS, bottom = LSLSLSLS. Do the move in Case 3 twice with T6 between instances. 4. Now check for parity. With the L's in place it is easy to identify whether you need to change the parity of the system. It should take an even number of switches to right the S's at this stage. A cycle of three is even, since it would take two switches to fix it. A cycle of two or four is odd. If the overall parity is odd, do the following: starting with top = LSLSLSLS, bottom = LSLSLSLS, go FT3B3FT1B2FT2B2FT- 2FT2B2FT3B2FT-3B-3T-3B-1FT-2B-2F This may not be the optimum way, but it preserves the corners, and it's easy to remember the path. (Try it) 5. Place the S's (they should already be on their correct face). The most useful moves are the below: All permutations of S's can be solved with application of a maximum of three of these short moves in sequence, combined with the appropriate turns in between to set things up. Move 1. Start with top = LOLSLSLO, bottom = LOLSLSLO, where O = pairs that will be switched on a given side. Do FT-3FT1B1FT2B-1F. Repeated twice with a T3 between makes a three-cycle on the top side. Move 2. top = LSLOLSLO, bottom = LSLOLSLO, O definition same as Move 1. Do FT1B1FT6FT-1B-1F. There may be quicker solutions than applying these moves for a 4-cycle/2- cycle combination or a 4-cycle/4-cycle combination, where you have to apply 3 moves in succession. I'd like to hear about suggestions. I haven't investigated it too much since these modes come up so rarely. 6. Fix middle slice. If square shaped but wrong, do FT6B6F. Otherwise, position the bad half on the right, and do FB6FB6F. Congratulations, you have a solved Square One. Happy cubing. Mike Masonjones From cube-lovers-errors@oolong.camellia.org Thu Jun 19 21:55:37 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id VAA07809; Thu, 19 Jun 1997 21:55:36 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Message-Id: <2.2.32.19970619180050.0068b11c@uclink4.berkeley.edu> X-Sender: mdp1@uclink4.berkeley.edu X-Mailer: Windows Eudora Pro Version 2.2 (32) Mime-Version: 1.0 Content-Type: text/plain; charset="us-ascii" Date: Thu, 19 Jun 1997 11:00:50 -0700 To: cube-lovers@ai.mit.edu From: Mark Pilloff Subject: Re: Square One At 12:12 AM 6/19/97 -0700, J.M. wrote: >The ones I have trouble with are the >Sqewb and the Alexander's Star. Anybody got a good solution for any of >these? > >Joe McGarity I finally came up with a solution to the Alexander Star last year. I haven't ever written out all of the details, but here are some helpful hints. First of all, the star is almost identical to the Megaminx (aka, magic dodecahedron, etc.) with all of the corners pieces removed. The only reason I say "almost" is that on the star, every individual piece is doubly degenerate. This sometimes leads to a problem wherein using the Megaminx moves seems to leads to an insoluble position. The trick in this case is two swap two of the degenerate pieces while disturbing as little of the rest of the star as possible. This has always worked perfectly for me. As for the rest of the star, I usually find that I can solve most of the star (except for the uppermost regions) just by inspection. From there, very slight modifications of Rubik's cube manipulations are useful. It's worthwhile to note that locally, the star (and the megaminx) are identical to the cube (except, perhaps, for the missing corners on the star). Thus, cube moves which only affect small portions of the cube will often be successful on the star or megaminx. In any event, I'm not going to write out explicit moves because I believe solutions to the megaminx are floating on the net as well, but I hope this is somewhat helpful. If there is really strong demand for explicit solutions, I'll see what I can do about that. Good luck, Mark ************************************ ** Mark D. Pilloff ** ** mdp1@uclink4.berkeley.edu ** ************************************ From cube-lovers-errors@oolong.camellia.org Fri Jun 20 11:36:32 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id LAA09262; Fri, 20 Jun 1997 11:36:32 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Date: Fri, 20 Jun 1997 07:05:50 +0200 (MET DST) Message-Id: <1.5.4.16.19970620070547.2e6f908a@mailsvr.pt.lu> X-Sender: geohelm@mailsvr.pt.lu X-Mailer: Windows Eudora Light Version 1.5.4 (16) Mime-Version: 1.0 Content-Type: text/plain; charset="us-ascii" To: Carl Woolf From: Georges Helm Subject: Re: Square One Cc: cube-lovers@ai.mit.edu At 09:19 19/06/1997 -0400, you wrote: >Square One is a great puzzle! > >I think there is an instruction booklet, published in Massachusetts or >thereabouts, and available from Puzzlets (mgreen@puzzletts.com). I >developed a set of techniques that let me solve the thing, but I >haven't worked my notes into a form intelligible by other humans (or >by me on a bad day). > There is a solution in Puzzle World. Check out details at http://ourworld.compuserve.com/homepages/Georges_Helm/cubeold.htm Georges geohelm@pt.lu http://ourworld.compuserve.com/homepages/Georges_Helm http://www.geocities.com/Athens/2715 From cube-lovers-errors@oolong.camellia.org Fri Jun 20 11:36:18 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id LAA09258; Fri, 20 Jun 1997 11:36:17 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Date: Fri, 20 Jun 1997 07:03:46 +0200 (MET DST) Message-Id: <1.5.4.16.19970620070348.2e6f1bf6@mailsvr.pt.lu> X-Sender: geohelm@mailsvr.pt.lu X-Mailer: Windows Eudora Light Version 1.5.4 (16) Mime-Version: 1.0 Content-Type: text/plain; charset="us-ascii" To: joemcg3@snowcrest.net From: Georges Helm Subject: Re: Square One Cc: cube-lovers@ai.mit.edu At 00:12 19/06/1997 -0700, you wrote: >The ones I have trouble with are the >Sqewb and the Alexander's Star. Anybody got a good solution for any of >these? I have solutions for both. Check out my list of solutions at http://ourworld.compuserve.com/homepages/Georges_Helm/cubeold.htm Alexander's book on his star Flettermann has a good solution for the skewb (but I realize I don't have him listed) I can send copies, though Georges geohelm@pt.lu http://ourworld.compuserve.com/homepages/Georges_Helm http://www.geocities.com/Athens/2715 From cube-lovers-errors@oolong.camellia.org Sat Jun 21 14:03:33 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id OAA13787; Sat, 21 Jun 1997 14:03:33 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Message-Id: <199706201853.TAA09093@mail.iol.ie> From: Goyra To: cube-lovers@ai.mit.edu Subject: Re: Square One Date: Thu, 19 Jun 1997 19:00:58 +0100 X-MSMail-Priority: Normal X-Priority: 3 X-Mailer: Microsoft Internet Mail 4.70.1161 MIME-Version: 1.0 Content-Type: text/plain; charset=ISO-8859-1 Content-Transfer-Encoding: 7bit > >Does anyone know how to solve one of those "Square One" puzzles? Can anyone tell me what this looks like so I can put up a Java version? David Byrden From cube-lovers-errors@oolong.camellia.org Mon Jun 23 21:05:29 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id VAA03450; Mon, 23 Jun 1997 21:05:28 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Date: Mon, 23 Jun 1997 18:33:40 -0400 From: "Jonathan R. Ferro" Message-Id: <199706232233.SAA51996@knave.ece.cmu.edu> Organization: Electrical and Computer Engineering, CMU X-Disclaimer: This disclaimer is not required by Leader Kibo. This article does not necessarily represent the opinions of Leader Kibo. Have a nice day! X-Exclaimer: Yow! To: cube-lovers@ai.mit.edu Subject: An art project... http://www.wunderland.com/EBooks/Window/Pages/SUTW-JD.html -- Jon From cube-lovers-errors@oolong.camellia.org Tue Jun 24 20:45:27 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id UAA06941; Tue, 24 Jun 1997 20:45:27 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Message-ID: <33B0668C.97B@ibm.net> Date: Tue, 24 Jun 1997 17:30:04 -0700 From: Jin "Time Traveler" Kim Organization: The Fourth Dimension X-Mailer: Mozilla 3.01Gold (Win95; I) MIME-Version: 1.0 To: cube-lovers@ai.mit.edu Subject: Re: An art project... References: <199706232233.SAA51996@knave.ece.cmu.edu> Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Jonathan R. Ferro wrote: > > http://www.wunderland.com/EBooks/Window/Pages/SUTW-JD.html > > -- Jon Very impressive. How many cubes and were they altered in any way except turning them? -- Jin "Time Traveler" Kim chrono@ibm.net VGL Costa Mesa From cube-lovers-errors@oolong.camellia.org Wed Jun 25 13:29:10 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id NAA08775; Wed, 25 Jun 1997 13:29:10 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Date: Wed, 25 Jun 1997 18:11:31 BST From: David Singmaster Computing & Maths South Bank Univ To: cube-lovers@ai.mit.edu Message-ID: <009B651B.AC285B40.328@vax.sbu.ac.uk> Subject: Pilloff's query about 5^3; Korf's method. I've been away and have just seen email for late May and early June. Pilloff asks about getting the parity of the edges correct on the 5^3. As he notes, the commutators cannot solve this. Examining the basic moves, one sees that the rotation of a inner face changes the parity of the edges while conserving the parity of some pieces, but not of others. However, the pieces whose parity is not conserved are the pieces next to the centre and there are four apparently identical copies of these, so one can simulate an exchange by a 3-cycle with two pieces the same. Hence one wants to apply a rotation of an inner face. What one does is to move the two edges into the same inner face. Then rotate the face. Then make a 3-cycle of the edges. This produces an exchange of edges - and rather messes up the centre pieces. Then put things back. The edges go A B C D to D A B C to B A C D. Because this is relatively easy to do, but messes up the centres, I normally do the edges and corners first and then put centres in place. Re: Korf. Someone has said it reminds them of alpha-beta pruning. It reminds me of branch and bound search. Both are older names for the general process of using information about the remainder of the problem to estimate the number of steps for a solution via a partial solution. Going back to Pilloff's query, I have several methods in my files for exchanging two edges without moving anything else (I think, but that seems to contradict what I said about parities??) DAVID SINGMASTER, Professor of Mathematics and Metagrobologist School of Computing, Information Systems and Mathematics Southbank University, London, SE1 0AA, UK. Tel: 0171-815 7411; fax: 0171-815 7499; email: zingmast or David.Singmaster @sbu.ac.uk From cube-lovers-errors@oolong.camellia.org Wed Jun 25 13:28:50 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id NAA08771; Wed, 25 Jun 1997 13:28:50 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Message-Id: Mime-Version: 1.0 Content-Type: text/plain; charset="us-ascii" Date: Wed, 25 Jun 1997 09:53:21 -0500 To: cube-lovers@ai.mit.edu From: Kristin Looney Subject: Re: An art project... Cc: jake@wunderland.com >> http://www.wunderland.com/EBooks/Window/Pages/SUTW-JD.html > > Very impressive. How many cubes and were they altered in any > way except turning them? 100 cubes, scrambled - not altered in any other way. This is, I believe, the sixth cube sculpture that Jacob has done in the window of my gameroom. A seventh is currently in progress. Previous sculptures include: "Rubik's Cube", "Merry Xmas" with a picture of a Christmas tree, a bizarre (and not too successful) abstract thingamabob, a pacman with several ghosts, and the Apple logo. I think the pacman is probably my favorite, it's a hard choice. They are all very much worth a look... I have pictures of them, and I will encourage Jake to put them on his web page for all to see. The one he is currently working on is, believe it or not, TWO SIDED. Unlike most which he just sorta fiddles with while playing games at my gameroom table, this one was carefully planned out in advance with graph paper. -K. 5th fastest hands in the nation (at least back in 1981) kristin@wunderland.com www.wunderland.com/kristin ------------------------------------------------------------- "I'm really angry that I, a superior human being in every way, have less money than my neighbor, who's wife I would love to nail, if only I weren't so busy sleeping and eating pork chops." -- George "Cannonball" Carlin, on the 7 deadly sins From cube-lovers-errors@oolong.camellia.org Wed Jun 25 17:18:25 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id RAA09126; Wed, 25 Jun 1997 17:18:25 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Date: Wed, 25 Jun 1997 22:14:32 BST From: David Singmaster Computing & Maths South Bank Univ To: cube-lovers@ai.mit.edu CC: notation@vax.sbu.ac.uk, for@vax.sbu.ac.uk, 4^3@vax.sbu.ac.uk, and@vax.sbu.ac.uk, 5^3@vax.sbu.ac.uk Message-ID: <009B653D.9EE57540.331@vax.sbu.ac.uk> Subject: notation for 4^3 and 5^3 David Barr has described some moves for the 5^3 using his own notation. In the early 1980s, I extended my 3^3 notation to the 4^3 and this can be used on the 5^3 as well. I described this in my Cubic Circular but perhaps it would be useful to give it here. I will describe the situation for the 4^3. On the 5^3, we don't ever have to make a slice move of just a central layer. Further, a combination of 4^3 and 3^3 processes will solve the 5^3, so we don't really need to label the central layers. Consider the four layers from L to R. I denote the inner layers by l and r. So the four layers are: LlrR. Similarly we have four layers: FfbB and UudD. To describe a piece requires more effort than before, but each piec lies in three layers and we can describe the piece by these layers. E.g. FUR is a corner piece; FUr is an edge piece, lying on the FU edge - but there are two of these and they are distinguished as being FUr and FUl (I usually give the layers in clockwise order, but it is not essential and there are times when it is more informative to use the other order.); Fur is a face-centre piece, the one in the upper right corner of the inner four cells of the F face. If you have been paying attention, you will ask about fur. This is one of the body-centre cells, invisible to you unless you make a transparent cube! Using the standard notation of [F, R] = FRF'R', we find a number of easy 3-cycles. [[F,R],L] = (FLU,ULB,RFU) [[F,R],l] = (FlU,UlB,DfR) (I've copied this from my Circular, but I wonder if it's right as I thought there'd be some symmetry with the preceding??) [[F,r],l] = (Flu,Ulb,Drb) In theory, these and a careful consideration of parity are sufficient to solve the 4^3 and the 5^3. However, the parity problem is a bit awkward. In my original approach, I got the corners in place and then all edges except leaving the four edges along the FU and BU edges. Examine the parity of these carefully. If they are in an odd permutation, apply r^2 D^2 l' D^2 r^2 which 4-cycles these four edges and moves some centres. Once the parity is corrected, there is little difficulty restoring the rest of the cube. For the 5^3, once you have paired up the edges, one can solve the central edges by treating the 5^3 as a 3^3 with fat slices. To correct a single pair of edges, one can use the following. rrDDl'DDrr rrD'RR [[R,U],l'] RRDrr = rrDDl'DR'UR'U'l'URU'lRDrr = (UBl,UBr) (Ful,Ubl,Bdl,Ufr) (Fdl,Ufl,Bul,Ubr). This messes up some centres, but they are not too hard to restore. Indeed applying rrUUr (uurrll)^2 r'UUrr wil correct the F and B centres disturbed by the above, leaving a 180 degree rotation of the four U centres. After I had developed the notation and solution, a Peter Lees pointed out an unexpected feature. The exchange of upper and lower case letters is a duality. The dual of URF is urf, while the dual of URf is urF. This gives us a Pricniple of Duality: The dual of a sequence of moves is the same process on the dual pieces. E.g., we had [[F,R],l] = (FUl,UBl,DRf), so [[f,r],L] = (fuL,ubL,drF). This duality allows one to transfer a number of 3^3 processes to 4^3 processes and to solve the invisible interior part of the cube! By always moving an outer layer with its inner layer, one is obviously simulating the 2^3. However, if one always moves, say R and l together, one is also simulating the 2^3 in eight copies! Ooops, one wants to move R and l' together. If one moves R and l together, I think you get eight versions of the 2^3, but each is a reflection of its neighbours! If you are tired of thinking about God's Algorithm on the 3^3, try the 4^3. I'm not even sure how to count moves. E.g., to do r, one normally does Rr and then R', so does r count as one move or two? Likewise, does Rr count as one move or two? Enough for now. DAVID SINGMASTER, Professor of Mathematics and Metagrobologist School of Computing, Information Systems and Mathematics Southbank University, London, SE1 0AA, UK. Tel: 0171-815 7411; fax: 0171-815 7499; email: zingmast or David.Singmaster @sbu.ac.uk From cube-lovers-errors@oolong.camellia.org Wed Jun 25 18:29:12 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id SAA09246; Wed, 25 Jun 1997 18:29:12 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Date: Wed, 25 Jun 1997 23:28:16 BST From: David Singmaster Computing & Maths South Bank Univ To: cube-lovers@ai.mit.edu Message-ID: <009B6547.EBD77100.328@vax.sbu.ac.uk> Subject: 4^3 and 5^3 I've just seen a comment about the 5^3 saying the writer had problems with the four pieces at distance 1 from the center. My approach treated both these and the pieces ate distance 2 from the center in the same way. We know that the commutator of two slice moves on the 3^3 produces two 3-cycles of the centres. Applying this idea to the 4^3, we find that [f,r] gives two 3-cycles of central pieces, one on each face. By turning one face and then inverting, we get a 3-cycle of central pieces, two being in one face. E.g. [[r,b],U] = (Fur,Ubr,Ubl). A similar result holds if we combine any two inner moves, so we can replace the b above by a central slice on the 5^3 to obtain a 3-cycle of the pieces at distance 1 from the central piece, while the process directly gives us a 3-cycle of the pieces at distance 2 from the central piece. Although tedious, these moves mean that once one has the corners and edges in place, the rest of the problem is easy - though very tedious - it generally took me about an hour to do the 5^3, assuming I could get the corners and edges correct without making too many mistakes. DAVID SINGMASTER, Professor of Mathematics and Metagrobologist School of Computing, Information Systems and Mathematics Southbank University, London, SE1 0AA, UK. Tel: 0171-815 7411; fax: 0171-815 7499; email: zingmast or David.Singmaster @sbu.ac.uk From cube-lovers-errors@oolong.camellia.org Wed Jun 25 23:51:13 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id XAA00509; Wed, 25 Jun 1997 23:51:13 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Date: Thu, 26 Jun 1997 04:44:18 BST From: David Singmaster Computing & Maths South Bank Univ To: cube-lovers@ai.mit.edu Message-ID: <009B6574.123497C0.321@vax.sbu.ac.uk> Subject: names for cubes What's wrong with 2^3, 3^3, 4^3, 5^3, pronounced 2 cube, 3 cube, 4 cube, 5 cube. If you have superscripts available, you can use them instead of the uparrows. DAVID SINGMASTER, Professor of Mathematics and Metagrobologist School of Computing, Information Systems and Mathematics Southbank University, London, SE1 0AA, UK. Tel: 0171-815 7411; fax: 0171-815 7499; email: zingmast or David.Singmaster @sbu.ac.uk From cube-lovers-errors@oolong.camellia.org Thu Jun 26 15:40:18 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id PAA06801; Thu, 26 Jun 1997 15:40:17 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Date: Thu, 26 Jun 1997 07:44:43 -0400 (EDT) From: Nicholas Bodley To: David Singmaster Computing & Maths South Bank Univ cc: cube-lovers@ai.mit.edu, notation@vax.sbu.ac.uk, for@vax.sbu.ac.uk, 4^3@vax.sbu.ac.uk, and@vax.sbu.ac.uk, 5^3@vax.sbu.ac.uk Subject: Hidden cubies; Spaceball In-Reply-To: <009B653D.9EE57540.331@vax.sbu.ac.uk> Message-ID: MIME-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Reading David Singmaster's recent posts, it seems almost obvious to me that (for the ambitious, which I'm definitely not!), computer simulations that are unhindered by real-world mechanical constraints and opacity can permit manipulation of the normally-hidden (and physically-nonexistent) internal cubies. It's very likely that a new collection of maneuvers would need to be developed for this. I haven't thought much about coloring the internal faces of the outer cubies... Incidentally, if this weren't the cube-lovers' List, I would have split that first sentence into a few shorter ones. Sorry if "cubie" is not the most-preferred term; should be no great problem. On another topic, it seems to me that an ideal device for controlling a computer-simulated Cube (or other similar puzzle) would be the Spaceball, a ball that you can grip. It senses torque around all three mutually- orthogonal axes, as well as "translational" force along those axes. It's not a consumer item; not sure it's still being made. I'm reasonably sure of the tradename. It was/is used with workstations. "Spaceball" sounds much like the name of a puzzle. (I expect some astute reader to tell me that the MIT Media Lab did just this thing 5 years ago!) My best to all, |* Nicholas Bodley *|* Electronic Technician {*} Autodidact & Polymath |* Waltham, Mass. *|* ----------------------------------------------- |* nbodley@tiac.net *|* When the year 2000 begins, we'll celebrate |* Amateur musician *|* the 2000th anniversary of the year 1 B.C.E. -------------------------------------------------------------------------- From cube-lovers-errors@oolong.camellia.org Thu Jun 26 16:35:00 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id QAA06927; Thu, 26 Jun 1997 16:35:00 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Message-ID: <33B2D235.3A77@ibm.net> Date: Thu, 26 Jun 1997 13:33:57 -0700 From: Jin "Time Traveler" Kim Organization: The Fourth Dimension X-Mailer: Mozilla 3.01Gold (Win95; I) MIME-Version: 1.0 To: Nicholas Bodley CC: cube-lovers@ai.mit.edu Subject: Re: Hidden cubies; Spaceball References: Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Nicholas Bodley wrote: > > On another topic, it seems to me that an ideal device for controlling a > computer-simulated Cube (or other similar puzzle) would be the Spaceball, > a ball that you can grip. It senses torque around all three mutually- > orthogonal axes, as well as "translational" force along those axes. It's > not a consumer item; not sure it's still being made. I'm reasonably sure > of the tradename. It was/is used with workstations. "Spaceball" sounds > much like the name of a puzzle. > > (I expect some astute reader to tell me that the MIT Media Lab did just > this thing 5 years ago!) > > My best to all, > Actually, the Spaceball that you talk about is still in existence of sorts. I have three Spaceballs. Actually, they were known as the Spacetec Spaceball Avengers. Those are no longer produced. They've been replaced by the newer model, the SpaceOrb 360. Not to get too far off subject, but the SpaceOrb is used by some people to play Quake. You can't beat a good Mouse and Keyboard for Quake, but the SpaceOrb's multiple axes of movement does allow for some interesting possibilities. Due to its 3D nature, I think the SpaceOrb would be a natural extension for the solving of 3d puzzles in graphical environments. -- Jin "Time Traveler" Kim chrono@ibm.net VGL Costa Mesa From cube-lovers-errors@oolong.camellia.org Thu Jun 26 16:55:53 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id QAA06984; Thu, 26 Jun 1997 16:55:52 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Date: Thu, 26 Jun 1997 13:53:47 -0700 From: "Jason K. Werner" Message-Id: <9706261353.ZM3850@neuhelp.corp.sgi.com> In-Reply-To: Nicholas Bodley "Hidden cubies; Spaceball" (Jun 26, 7:44) References: Reply-to: "Jason K. Werner" X-Mailer: Z-Mail-SGI (3.2S.2 10apr95 MediaMail) To: Nicholas Bodley , cube-lovers@ai.mit.edu Subject: Re: Hidden cubies; Spaceball Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii On Jun 26, 7:44, Nicholas Bodley wrote: > Subject: Hidden cubies; Spaceball > On another topic, it seems to me that an ideal device for controlling a > computer-simulated Cube (or other similar puzzle) would be the Spaceball, > a ball that you can grip. It senses torque around all three mutually- > orthogonal axes, as well as "translational" force along those axes. It's > not a consumer item; not sure it's still being made. I'm reasonably sure > of the tradename. It was/is used with workstations. "Spaceball" sounds > much like the name of a puzzle. In case anyone is interested: http://www.spacetec.com/ http://www.spaceorb.com/ -Jason -- Jason K. Werner Email: mrhip@sgi.com Systems Administrator Phone: 415-933-6397 USFO I/S Technical Services Fax: 415-932-6397 Silicon Graphics, Inc. Pager: 415-317-4084, mrhip_p@sgi.com "Winning is a habit"-Vince Lombardi;"These go to eleven"-Nigel Tufnel From cube-lovers-errors@oolong.camellia.org Fri Jun 27 16:39:01 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id QAA09468; Fri, 27 Jun 1997 16:39:01 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Date: Fri, 27 Jun 1997 16:28:04 -0400 From: Corey Folkerts Subject: First corners, then sides To: Cube-Lovers@ai.mit.edu Message-ID: <199706271628_MC2-1961-C7B8@compuserve.com> During the last 5 months or so I have been fiddling around with a Rubik's Cube that I got for Toys R Us. I have become pretty good at solving it with the layers method (90 seconds is my record so far.) My question is this : Is the method of solving the cube corners first and then sides any faster (once one becomes good at it) then solving it by layers? If so, could someone please reply with a description of that method. I don't know the names of any fancy moves, so if possible please use F B U D L R for the faces during specific moves and expain the other parts of the strategy clearly. I realize this is asking alot. Thanks in advance to anyone who replies! Corey Folkerts From cube-lovers-errors@oolong.camellia.org Sat Jun 28 14:03:46 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id OAA13053; Sat, 28 Jun 1997 14:03:46 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Date: Sat, 28 Jun 1997 05:50:29 -0400 (EDT) From: Jiri Fridrich X-Sender: fridrich@bingsun2 To: Corey Folkerts cc: Cube-Lovers@ai.mit.edu Subject: Re: First corners, then sides In-Reply-To: <199706271628_MC2-1961-C7B8@compuserve.com> Message-ID: MIME-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII It appears that both systems are good if worked out into sufficient detail. As far as I know, the fastest speed cubists can achieve an average of 16-17 seconds irrespective! of whether they use corners-edges or by-slices systems. You can look at a system which I have developed some time ago and which enables me to solve the cube in 17 sec. on average. It is described in detail here: http://ssie.binghamton.edu/~jirif/. I also recommend the section on speed cubing tips. Good luck! Jiri On Fri, 27 Jun 1997, Corey Folkerts wrote: > > > During the last 5 months or so I have been fiddling around with a > Rubik's Cube that I got for Toys R Us. I have become pretty good at solving > it with the layers method (90 seconds is my record so far.) My question is > this : Is the method of solving the cube corners first and then sides any > faster (once one becomes good at it) then solving it by layers? If so, > could someone please reply with a description of that method. I don't know > the names of any fancy moves, so if possible please use F B U D L R for > the faces during specific moves and expain the other parts of the strategy > clearly. I realize this is asking alot. Thanks in advance to anyone who > replies! > > Corey Folkerts > > ********************************************************************** | Jiri FRIDRICH, Research Associate, Dept. of Systems Science and | | Industrial Engineering, Center for Intelligent Systems, SUNY | | Binghamton, Binghamton, NY 13902-6000, Tel.: (607) 797-4660, | | Fax: (607) 777-2577, E-mail: fridrich@binghamton.edu | | http://ssie.binghamton.edu/~jirif/jiri.html | ********************************************************************** ...................................................................... Remember, the less insight into a problem, the simpler it seems to be! ---------------------------------------------------------------------- From cube-lovers-errors@oolong.camellia.org Sun Jun 29 22:09:51 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (oolong.camellia.org [206.119.96.100]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id WAA16484; Sun, 29 Jun 1997 22:09:51 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Message-Id: <199706292239.CAA27387@telecom.lek.ru> From: Alex Joukov To: Cube-Lovers@ai.mit.edu Subject: Newcomer: Algoritms. How to solv U-slice if D-slice and UD-slice have been solvd yet? Date: Mon, 30 Jun 1997 03:11:05 +0400 X-MSMail-Priority: Normal X-Priority: 3 X-Mailer: Microsoft Internet Mail 4.70.1155 MIME-Version: 1.0 Content-Type: text/plain; charset=KOI8-R Content-Transfer-Encoding: 7bit Dear Cube-Lovers, How to solv U-slice if D-slice and UD-slice have been solvd yet? I remember that in about 1994 when I was a child I used 3 or 5 algoritms (with mirrow variants). But now I don't remember the way. I read cube-lovers arhives (from #0) and step by step find out some algoritms. But before I will be able to solve Cube I lose a lot of interesting in archive messages. I just can't try a lot not having solved Cube! Please help me. A need just "1. U2DFTU'RD' 2. UR'D2L'D2 3..." without comments. Or, may be somebody have created such type FAQ which is accesible for ftp? lllykob@telecom.lek.ru Sasha From cube-lovers-errors@oolong.camellia.org Mon Jun 30 15:09:03 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id PAA18257; Mon, 30 Jun 1997 15:09:02 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Date: Mon, 30 Jun 1997 14:08:40 -0400 (Eastern Daylight Time) From: Jerry Bryan Subject: Inverses of Local Maxima To: cube-lovers@ai.mit.edu Reply-to: Jerry Bryan Message-id: MIME-version: 1.0 Content-type: TEXT/PLAIN; charset=US-ASCII X-X-Sender: jbryan@PSTCC6.pstcc.cc.tn.us One of the oldest unsolved problems on Cube-Lovers (aside from God's Algorithm itself) has to do with inverses of local maxima. It seems obvious that the inverse of a local maximum also ought to be a local maximum. But is it necessarily so? In Symmetry and Local Maxima, Jim Saxe and Dan Hoey suggest that it may not be. Their example is UFF, which can end with F or F' because we can write it as UF'F'. But the inverse is F'F'U', which can only end with U' Hence, there are very simple positions where the number of q-turns with which the position can end is different than the number of q-turns with with the inverse of the position can end. If the same thing should happen with a local maximum, then the inverse would not be a local maximum. On the other hand, for all known local maxima in G, the inverse is also a local maximum. What are we to think? I have some small progress. I can report that for the corners-only group, there are local maxima for which the inverse is not a local maximum. The results were obtained with my new Shamir program. For each position x, we define E(x) to be the set of all quarter-turns with which a minimal process for the position can end. As an example, if x=UFF, then E(x)={F,F'}. E(x) is a subset of Q, the set of twelve quarter-turns, or equivalently it is an element of P(Q), the power set of Q. As such, it is conveniently represented in my program as a bit-string of twelve bits. In this notation, we would say that a position x is a local maximum if E(x)=Q or if |E(x)|=12. We also define S(x) to be the set of all quarter-turns with which a minimal process for a position can start. In this notation, for x=UFF we would say that |S(x)|=1 and |E(x)|=2. So the general question for local maxima becomes the following: if |E(x)|=12, does it necessarily follow that |S(x)|=12? My program calculates S(x) and E(x) as follows. Any breadth-first search may be characterized as calculating products of the form z=xy for suitable choices of x and y. Most typically, x comes from Q[n], the set of all quarter-turns of length n, and y comes from Q[1], the set of all quarter-turns of length 1. But in my more general Shamir program, x comes from Q[m] and y comes from Q[n] to form products of length m+m. In any case, S(z) is the union of S(x) over all x which can be a part of a product which produces z, and E(z) is the union of E(y) over all y which can be a part of a product which produce z. For each q in Q, we initialize with S(q)=E(q)={q} and go from there. Here is a portion of a printout from my program. |x| |E(x)| |S(x)| M-Conjugacy Positions Classes 0 0 0 1 1 1 1 1 1 12 2 1 1 2 96 2 2 2 3 18 3 1 1 12 576 3 1 2 3 96 3 2 1 3 96 3 2 2 4 96 3 3 3 2 60 As you can see, the effect pointed out by Saxe and Hoey first shows up three moves from Start, where there are six positions unique up to M-conjugacy where |S(x)| is not equal to |E(x)|. (Actually, three of the six positions are just the inverses of the other three.) The first local maxima are six moves from Start in the corners-only group. |x| |E(x)| |S(x)| M-Conjugacy Positions Classes 6 12 12 8 114 As you can see, there are 114 local maxima of which 8 are unique up to M-conjugacy. However, for all 8 of them, the inverse is also a local maximum so we discover nothing new. The new discovery occurs 7 moves from Start. |x| |E(x)| |S(x)| M-Conjugacy Positions Classes 7 12 8 4 120 7 12 10 3 144 7 12 12 14 336 As you can see, there are 21 local maximu unique up to M-conjugacy. For 14 of them, the inverse is also a local maximum. But for the other 7, the inverse is not a local maximum. In 4 cases, we have |S(x)|=8, and in 3 cases we have |S(x)|=10. Here follow summaries for local maximum up to a distance of 11 moves from Start. |x| |E(x)| |S(x)| M-Conjugacy Positions Classes 8 12 6 14 576 8 12 8 12 576 8 12 10 86 4128 8 12 11 13 624 8 12 12 272 12012 9 12 4 26 1152 9 12 6 31 1344 9 12 8 24 1152 9 12 10 14 576 9 12 12 131 5976 10 12 2 14 576 10 12 4 88 4032 10 12 6 218 10368 10 12 8 144 6336 10 12 10 168 8064 10 12 12 140 5664 11 12 4 384 18432 11 12 6 2687 128688 11 12 8 5550 264192 11 12 10 5014 240576 11 12 12 3617 166224 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) jbryan@pstcc.cc.tn.us Pellissippi State (423) 539-7198 10915 Hardin Valley Road (423) 694-6435 (fax) P.O. Box 22990 Knoxville, TN 37933-0990 From cube-lovers-errors@oolong.camellia.org Mon Jun 30 19:21:04 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id TAA18780; Mon, 30 Jun 1997 19:21:03 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Message-ID: <19970630230531.25512.rocketmail@send1.rocketmail.com> Date: Mon, 30 Jun 1997 16:05:31 -0700 (PDT) From: Bill Webster Subject: Hi To: cube-lovers@ai.mit.edu MIME-Version: 1.0 Content-Type: text/plain; charset=us-ascii Hi, I first encountered the cube sometime in the early 80s when the fad hit Australia. Solving it absorbed all my spare time for two weeks along with several wads of A4 and $15 for a second cube which I used to develop what I called 'sequences'. My experiments on the second cube were conducted with great care, as I had not yet discovered that Satan's Algorithm could be had for the price of a small screwdriver. I never became a speed-freak, or even employed operators outside my own meagre discoveries, but could solve the cube comfortably inside three minutes. I solved corners, then edges because the first reasonable sequences I discovered were edge-disrupting corner operators. My operators were short and scant so my method incurred a lot of short term memory overhead, manipulating faces into susceptible positions, applying the sequence, then inverting the prior manipulation. A friend of mine aquired a cube at about the same time and much to my chagrin, solved it in less than a day, without paper, without explicitly developing any operators. In fact, he couldn't give a satisfactory account of exactly how he'd solved it. He may have just got *extremely* lucky and stumbled on something close to START, but I don't think so. I handed him a scrambled cube a couple of weeks later and he was quite taken aback - he wasn't going through all that again, he'd done it hadn't he?. I always felt that my own solution was somewhat contrived after witnessing this feat. Does anyone else have examples of GestaltCube? I have coded a C++ class which represents cube states and operators and which includes methods to manipulate the cube. I would like to implement overloaded C++ operators in a manner which is consistent with (and perhaps extends) the appropriate mathematical notation *if this is feasible*. Is there anyone out there familiar with both grammars and willing to make suggestions? I am aware and prepared to accept that the use of some (C++) operators may introduce inefficiencies in the form of temporary objects created during expression evaluation - such operators will not be used in time critical code. I have been using a freeware ray-tracer, POV-Ray to produce 'photo-realist' cube images. I intend to extend my software to export animation scripts, so that I can produce (externally rendered) animated solutions. These take forever to trace, so their value is aesthetic rather than practical. I have been experimenting with solid gold cubes inlaid with coloured marble etc., but still prefer the platonic form. I have the POV source for a static cube if anyone is interested. The POV team are true heroes - details... "The internet home of POV-Ray is reachable on the World Wide Web via the address http://www.povray.org and via ftp as ftp.povray.org." "POV-Ray can be used under MS-DOS, Windows 3.x, Windows for Workgroups 3.11, Windows 95, Windows NT, Apple Macintosh 68k, Power PC, Commodore Amiga, Linux, UNIX and other platforms." Regards, Bill Webster _____________________________________________________________________ Sent by RocketMail. Get your free e-mail at http://www.rocketmail.com From cube-lovers-errors@oolong.camellia.org Mon Jun 30 21:31:40 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id VAA19022; Mon, 30 Jun 1997 21:31:39 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Date: Mon, 30 Jun 1997 20:01:16 -0400 (EDT) From: Jerry Bryan Subject: Re: Inverses of Local Maxima In-reply-to: To: cube-lovers@ai.mit.edu Message-id: MIME-version: 1.0 Content-type: TEXT/PLAIN; charset=US-ASCII On Mon, 30 Jun 1997, Jerry Bryan wrote: > > I have some small progress. I can report that for the corners-only > group, there are local maxima for which the inverse is not a local > maximum. > There is a minor interesting point that might be added. When we find a local maximum x for which |S(x)|<12, we can form a new, longer local maximum qx for suitable q in Q. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) jbryan@pstcc.cc.tn.us Pellissippi State (423) 539-7198 10915 Hardin Valley Road (423) 694-6435 (fax) P.O. Box 22990 Knoxville, TN 37933-0990 From cube-lovers-errors@oolong.camellia.org Mon Jun 30 21:57:52 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id VAA19066; Mon, 30 Jun 1997 21:57:52 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Message-Id: <199707010132.VAA26778@life.ai.mit.edu> Date: Mon, 30 Jun 1997 21:37:39 -0400 From: michael reid To: cube-lovers@ai.mit.edu, jbryan@pstcc.cc.tn.us Subject: example of a local maximum whose inverse is not a local maximum jerry bryan asks if the inverse of a local maximum is necessarily a local maximum. the following example shows that this need not be the case. the interesting "six-two-one" pattern is produced by the sequence B U2 F2 R U' R' B' R' U F2 U2 (15q) this position has six symmetries, generated by the cube rotation C_UFR and central reflection. therefore we also have the maneuvers L F2 R2 U F' U' L' U' F R2 F2 D R2 U2 F R' F' D' F' R U2 R2 F' D2 B2 L' D L F L D' B2 D2 R' B2 L2 D' B D R D B' L2 B2 U' L2 D2 B' L B U B L' D2 L2 for the same position. it is not hard to check (by computer) that these are minimal maneuvers. note that for each quarter turn, we have a maneuver that ends with that quarter turn. thus, from this position, any quarter turn brings us closer to start, so our position is a local maximum. consider now the inverse position; it is produced by U2 F2 U' R B R U R' F2 U2 B' (15q) it is not hard to check (by computer) that applying the quarter turn B' to this moves us further from start (16q), so this position is not locally maximal. note that this is already in the archives; i first reported it on april 20, 1995 in my message "correction and an interesting example" mike From cube-lovers-errors@oolong.camellia.org Sat Jul 5 20:21:49 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id UAA00582; Sat, 5 Jul 1997 20:21:48 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Message-Id: <199707052054.QAA25375@life.ai.mit.edu> Date: Sat, 5 Jul 1997 16:58:20 -0400 From: michael reid To: cube-lovers@ai.mit.edu Subject: optimal cube solver the recent work by rich korf on finding optimal solutions has prompted me to try my hand at writing an optimal cube solving program. so far, i've done this for the face turn metric. a description of my program follows. let H denote the intermediate subgroup which we've seen before. we'll use distances to this intermediate subgroup for our pruning tables (or "pattern databases"). calculating these distances involves doing a breadth first search on the coset space H \ G , and storing these distances in memory. (i've written this as a right coset space, rather than a left coset space.) this search has been done several times, by dik winter and by myself. some review. positions in H are characterized by the following. corners cannot change orientation; their U or D facelet remains on the U or D face. similar, edges cannot change orientation. furthermore, the four U-D slice edges remain in the U-D slice. therefore, cosets in H \ G are described by triples (c, e, l), where c denotes corner orientation, e denotes edge orientation and l denotes the location of the four U-D slice edges. there are 3^7 = 2187 possible corner orientations, 2^11 = 2048 possible edge orientations / 12 \ and \ 4 / = 495 possible U-D slice edge locations. all combinations are possible, so there are 2187 * 2048 * 495 = 2217093120 cosets. since this is too many configurations to store in memory, we use symmetry to to reduce this number. there are 16 symmetries of the cube that preserve the U-D axis, and therefore the intermediate subgroup H. rather than store all the cosets, we'll just store one of each up to symmetry. actually, this is slightly more complicated than necessary; instead, we could just divide the corner coordinate by symmetry. this is what i did in my message of january 7, 1995. however, i encountered a pitfall along the way. i discovered (very late in the development stage) the need for very large transformation tables. although i continued with the same approach at that time, i gave two options for overcoming this problem: > i) only use the 8 symmetries that preserve my choice of > 12 edge facelets. > > ii) combine the two coordinates edge and location into a single > coordinate and divide this coordinate by the 16 symmetries. of these, clearly the second is the better choice, since it utilizes more symmetry. this new edge coordinate has 2048 * 495 = 1013760 possibilities. up to symmetry, there are 64430 possibilities. we need room for 64430 * 2187 = 140908410 cosets in memory. for each of these, we store its distance to the identity coset. this is an integer between 0 and 12 (inclusive), so each is stored in half a byte. thus the whole table requires 67 megabytes. essentially, what we're doing here is changing coordinates from (c, e, l) to (c, e', s), where e' is our new edge coordinate, and s is a symmetry coordinate. some cosets have multiple coordinates in this new system, but that causes no harm. a breadth first search of this space takes under 11 minutes. the increase in speed is partially due to a more powerful computer, and partially due to switching to "backward searching" (or "bidirectional search") at the optimal time. we'll also use distances to the intermediate subgroups and . we don't need to store additional coset spaces, since we can derive that information from our first coset space. note that the cube rotation C_UFR takes the subgroup to the subgroup . therefore it transforms the first coset space into the second coset space. furthermore, it preserves distances, so the one pattern database suffices for all three applications. an attractive feature of this approach is that it uses the 16 symmetries to reduce the size of the pattern database, and then uses the remaining symmetry of the cube in applying it in different orientations. these are the only pruning tables my program currently uses. note that they cannot "see the entire group". specifically, let H_0 = , H_1 = , H_2 = , and let T denote the intersection of these three subgroups. for a given position, the three distances to these subgroups depend only upon the corresponding coset in T \ G . thus T might be thought of as a "target subgroup". this target subgroup T is interesting. it consists of those positions that "look like" they're in the "square group" , i.e. F and B colors mix only with each other, and similarly for R and L , and also for U and D. however, this is strictly larger than the square group; it contains the square group as a subgroup of index 6. the searching is done in the way that korf describes as "IDA*" (or at least the "ID" part of that terminology). we traverse the tree of all sequences of length 1, hoping to find a solution. that generally fails, so we continue to sequences of length 2, and so forth, until a solution is found. the "A*" part of the algorithm is to use the pruning tables to avoid searching large parts of the tree that are guaranteed not to bear fruit. in his paper, korf uses the expected value of his heuristic functions to get an estimate of how effective they are at pruning the search tree. actually, he should subtract 1 from this expected value, since we must generate (at least partially) the top node of a subtree that gets pruned. this is only a rough estimate; getting a more precise figure is a delicate matter which i won't address here. korf reports an expected value of 8.878. i generated 10 million random cubes (i did not use the long sequence of random twists method) and got an expected value of 9.941. my program generates slightly more than 500000 nodes per second. korf generates them at 700000 per second, so i've got more overhead per node. however, it generates many fewer nodes, since it prunes the search tree more efficiently. i solved korf's ten random cubes, and found all minimal solutions, rather than stopping at the first. this entailed one complete search through length 16f, three through length 17f and six through length 18f. the position at distance 16f has a unique minimal solution, as do the three positions at distance 17f. of the six positions at distance 18f, one has a unique minimal solution, one has 3 minimal solutions, two have 4 minimal solutions and two have 6 minimal solutions. the total run time for these was just under 198 hours. korf estimates 4000 hours for the same search, so on these positions, my program is twenty times as fast. my computer has a 200 MHz pentium pro processor, and is configured with 128 megabytes of RAM. i'd expect a similar increase in performance for most positions, but not all. for example, positions inside the target subgroup T run very slowly, as do positions very close to it. hopefully, most of these are close enough to start, so that searches don't have to go very deep. i suspect that there are probably also positions that give korf's program difficulty. as you can see, i've made only minor modifications to korf's method. the only differences are: 1. use different pattern databases that allow more efficient pruning. 2. apply the same pattern database in multiple orientations. 3. allow a target subgroup larger than just the identity. it's clear that more experimentation is needed with different pattern databases. for any subgroup K of G , we could consider distances to that subgroup. it seems likely that we want small subgroups, so that the average distance is large. for this reason, using symmetry to reduce the size of the database is an important tool. i encourage others to experiment with different subgroups. more results to come ... mike From cube-lovers-errors@oolong.camellia.org Mon Jul 7 02:22:55 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id CAA04972; Mon, 7 Jul 1997 02:22:54 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Message-Id: <199707070459.AAA17039@life.ai.mit.edu> Date: Mon, 7 Jul 1997 01:04:35 -0400 From: michael reid To: cube-lovers@ai.mit.edu Subject: symmetry reductions for superflip a simple counting argument shows that some cube positions are at least 18 face turns from start, and thus the diameter of the cube group is at least 18f. in january 1995, i showed, by exhaustive search, that the position "superflip" is exactly 20 face turns from start. therefore the diameter is at least 20f. this gave the first improvement to the lower bound obtained by the counting argument. the searching method i used at the time was my version of kociemba's algorithm. although my symmetry reductions fit together quite well with kociemba's algorithm, this might not be the most appropriate searching method to use for this purpose. (i guess i could have hacked it not to bother looking for solutions longer than 19f. i don't remember why i didn't do this.) my new optimal solving program can do an exhaustive search in much less time. the symmetry reductions are similar, but much simpler. i will try be more coherent this time with my explanation, hopefully without being overbearing. the first thing to note is that dik winter found a maneuver for superflip in 20f: F B U2 R F2 R2 B2 U' D F U2 R' L' U B2 D R2 U B2 U (20f) therefore our concern is with searching for maneuvers of length at most 19f. there are three ways to transform a maneuver for superflip to get another such maneuver, which do not change its length: 1. we may conjugate the maneuver by any symmetry of the cube. 2. we may cyclically shift the maneuver; i.e. replace sequence_1 sequence_2 by sequence_2 sequence_1 3. we may replace the maneuver by its inverse. (in fact, we won't use 3 here, but it might be helpful elsewhere.) our first result is proposition 1. any maneuver for superflip in 19f contains a 180 degree face turn. proof. if the proposition were false, then superflip would be an odd number of quarter turns from start, contradiction. qed. the relevance of this proposition is proposition 2. suppose that a maneuver for superflip contains a 180 degree face turn. then it can be transformed, using the above tranformations, into a maneuver that begins with U R2. proof. we first claim that the maneuver has two consecutive "syllables" such that the first contains a 90 degree face turn and the second contains a 180 degree face turn. a "syllable" is a sequence of one or two face turns along the same axis; e.g. U D2. by hypothesis, the maneuver has a syllable that contains a half turn. if the claim is false, then the preceding syllable contains no 90 degree turns, and therefore consists only of half turns. but then the syllable before that contains only half turns, by the same reasoning. continuing in this way, we see that every syllable consists only of half turns. therefore we have a maneuver for superflip consisting only of half turns. this is a contradiction, so the claim is true. now, since the individual face turns within a syllable commute, we may suppose that the maneuver has a 90 degree face turn followed by a 180 degree face turn, which are along different axes, and thus are adjacent faces. now we may conjugate by an appropriate symmetry of the cube to suppose that these turns are U R2. finally, we may cyclically shift the maneuver so that these are the first two turns. qed. proposition 3. suppose that superflip is exactly 19 face turns from start. then applying the sequence U R2 to it brings us 2 face turns closer to start, i.e. 17f from start. proof. apply proposition 1 and proposition 2. qed. we now know how to handle the case that superflip's distance from start is exactly 19f. if the distance is less than 19f, we use the following proposition 4. under any circumstances, applying the sequence U R2 to superflip brings us at least 1f closer to start. proof. a minimal maneuver for superflip must contain a 90 degree twist, and we may suppose that the next face turned is an adjacent one. by cyclically shifting the maneuver, we may bring these two turns to the beginning. furthermore, by symmetry, we may suppose that the first turn is U and the second is some twist of the R face. now by applying U to superflip, we've moved 1f closer to start, and applying R2 to this doesn't move us any further from start, since it either combines with, or cancels the next turn in the minimal maneuver. qed. putting this all together, we get our desired result. proposition 5. suppose that superflip is within 19f of start. then the position superflip U R2 is within 17f of start. proof. this is just combining props 3 and 4. qed. i don't claim that these are the best reductions possible. they suffice for our purposes. i tested the position superflip U R2 (i.e. the position obtained by first doing superflip, and then doing the sequence U R2) with my optimal solver. my program took 2 hours and 40 minutes to exhaustively search this position through 17 face turns (not including about 11 minutes to generate all the lookup tables). there were no solutions. thus superflip is exactly 20 face turns from start. when i did the search in january 1995, the run time was 6 days. so we see quite a bit of improvement. mike From cube-lovers-errors@oolong.camellia.org Tue Jul 8 00:16:07 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id AAA07853; Tue, 8 Jul 1997 00:16:06 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Message-Id: <199707080413.AAA00423@life.ai.mit.edu> Date: Tue, 8 Jul 1997 00:18:18 -0400 From: michael reid To: cube-lovers@ai.mit.edu Subject: superfliptwist requires 20 face turns i can now show that the pattern "superfliptwist" is exactly 20 face turns from start. this position was proposed as a likely antipode of start by cubologist christoph bandelow. in the german edition of his book "einfuerung in die cubologie" he offered a prize for the shortest maneuver for this pattern. the prize was collected by rainer aus dem spring, who found a maneuver in 22 face turns. much later, a maneuver of length 20f was found by herbert kociemba: D F2 U' B2 R2 B2 R2 L B' D' F D2 F B2 U F' L R U2 F' (20f) as one of the first applications of his ingenious searching algorithm. i'll try not to be so verbose with my symmetry reductions this time. first note that "superfliptwist" does not describe a unique position of the cube; there are two possible orientations. in this context, i use the term "position" to refer to one of the 43252003274489856000 possible configurations, and the term "pattern" to refer to an equivalence class of positions under symmetries of the cube. (this concept has been discussed by dan hoey and jerry bryan as the "real size of cube space" i.e. the number of patterns.) the following two facts are easily verified: * superfliptwist commutes with the square of each face turn. * it does not commute with 90 degree slices (e.g. U D') or 90 degree antislices (e.g. U D), however, if A is a 90 degree slice or antislice, then A superfliptwist A^(-1) is also superfliptwist, but in the other orientation. these facts lead to the importance of the following proposition. superfliptwist is not in the subgroup generated by slices and antislices. (note that this group contains all squares of face turns.) proof. we may ignore the corners and just show that all edges cannot be flipped in this subgroup. to do this, we choose dominant facelets on the 12 edges as follows: choose the U or D facelet of the edges in the R-L slice, the R or L facelet of the edges in the F-B slice and the F or B facelet of the edges in the U-D slice. now we may define the flip of an edge that is not in its correct location. all edges start in the correct orientation. a 90 degree slice or antislice along the U-D axis changes the orientation of all eight edges in the F-B slices and R-L slices. similarly, a 90 degree slice or antislice along the F-B or R-L axis flips all edges in two different slices. within this subgroup, either all edges in a given slice are flipped, or none are flipped, and furthermore, the number of the three slices with flipped edges is even, i.e. 0 or 2. however, superfliptwist has all three slices with flipped edges, so it is not in this subgroup. qed. now consider the first syllable of a minimal maneuver for superfliptwist. ("syllable" was defined in my previous message.) if this is a single 180 degree turn, then we may cyclically shift this to the end of the maneuver. similarly, a slice squared may also be shifted to the end of the maneuver. furthermore, 90 degree slices and or antislices may also be shifted to the end of the maneuver, with only the mild effect of changing which orientation of superfliptwist we're doing. from the proposition, we eventually find a syllable which is not of these types, and is therefore of type U or D2 U. in the case of D2 U , we may shift the D2 to the back of the maneuver, so we may suppose that the first face turn is U . furthermore, by conjugating by the cube rotation C_U, if necessary, we may suppose that our maneuver solves our preferred orientation of superflip. the second face turn is in a different syllable, so it is an adjacent face. conjugating by C_U2, if necessary, brings this face to either R or F. therefore we may suppose that the first two face turns are one of the six sequences U R , U R2 , U R', U F , U F2 or U F' . to show that superfliptwist is not within 19f of start, i tested the six patterns obtained by applying these sequences to it. it took my program 7.5 hours to exhaustively search all of these through 17f. (these positions ran a bit faster than most of the others i've tested. this is partly because superfliptwist is 15 face turns from my "target" subgroup, so larger parts of the search tree are pruned.) no solutions were found, so superfliptwist requires 20 face turns. i also let the first situation run partially through depth 18f. in about 4 and a half hours, it found a solution which yields U R F' B U' D' F U' D F L F' L' U R D F U R L (20f, 20q) this is automatically minimal in the quarter turn metric! mike From cube-lovers-errors@oolong.camellia.org Tue Jul 8 17:33:24 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id RAA09883; Tue, 8 Jul 1997 17:33:24 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Message-Id: <199707082039.QAA04704@life.ai.mit.edu> Date: Tue, 8 Jul 1997 16:43:50 -0400 From: michael reid To: cube-lovers@ai.mit.edu Subject: composition of superflip and pons asinorum the position which is the composition of superflip and pons asinorum is exactly 19 face turns from start. some time ago, jerry bryan found that this position is exactly 20 quarter turns from start, and he gave all minimal maneuvers, up to symmetry. one of these is 19 face turns long: B' D' L' F' D' F' B U F' B R2 L U D' F L U R D (19f, 20q) symmetry reductions for this position are much simpler (but not nearly as good) as for superflip and/or superfliptwist. if the first face turn is a 90 degree turn, then by symmetry, we may suppose it is U . if the first face turn is a 180 degree turn, then we may suppose it is U2 . i tested the two positions obtained by applying these possible initial turns. my program took about 6 and a half hours to exhaustively search these through 17 face turns. no solutions were found, and therefore the original position is more than 18 face turns from start. i realize that this is not nearly as satisfying as obtaining all minimal maneuvers. that will take about 13 times as long, but is feasible with my current program. mike From cube-lovers-errors@oolong.camellia.org Thu Jul 10 00:56:16 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id AAA12818; Thu, 10 Jul 1997 00:56:16 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Message-Id: <199707100450.AAA14097@life.ai.mit.edu> Date: Thu, 10 Jul 1997 00:54:58 -0400 From: michael reid To: cube-lovers@ai.mit.edu Subject: all minimal maneuvers for superflip (face turn metric) i can now give all minimal maneuvers for superflip in the face turn metric. recall that there are three operations we may apply to any maneuver for superflip which give another maneuver of the same length: 1. we may conjugate the maneuver by any symmetry of the cube. 2. we may cyclically shift the maneuver; i.e. replace sequence_1 sequence_2 by sequence_2 sequence_1 3. we may replace the maneuver by its inverse. my original search (january 1995) for superflip in 19 face turns was divided into 16 cases. since i used my (unhacked) version of kociemba's algorithm, the search through each case produced maneuvers for superflip, and 8 of these cases found maneuvers of length 20f. i previously reported that these were each equivalent to dik winter's maneuver, using the three operations above. however, i was mistaken about this; there were two different maneuvers which differ only very slightly. to facilitate an exhaustive search through 20f, i'll use a result of a previous search. proposition. any maneuver for superflip in 20f contains a 180 degree face turn. proof. otherwise the maneuver would be 20 quarter turns long. however, i did an exhaustive search through 20q and found no maneuvers. qed. (in fact, this quarter turn result was later improved by jerry bryan, who showed that superflip is not within 22q of start, and therefore is exactly 24q from start.) now the symmetry reductions show that we may take the first two face turns to be U R2 . my program exhaustively searched the position superflip U R2 through 18f. it took 35 hours, and found 30 maneuvers, which came in two different types: U R2 F B R B2 R U2 L B2 R U' D' R2 F R' L B2 U2 F2 (20f) U R2 F B R B2 R U2 L B2 R U' D' R2 F D2 B2 U2 R' L (20f) note that these are the same except for the last 5 face turns. (this gives the relation R' L B2 U2 F2 R L' U2 B2 D2 = identity; alternatively, the same sequence produces (f++)(d++) in the supergroup.) from this, we can count the exact number of 20f sequences for superflip. both of the above may be cyclically shifted in 23 different ways. we get 23 different ways, instead of 20, because there are three separate pairs of consecutive twists of opposite faces. we'd consider sequence_1 U D sequence_2 and sequence_1 D U sequence_2 to be the same, but we wouldn't consider U sequence D and D sequence U to be the same. yet cyclic shifting of these last two produces the same maneuver. we can also conjugate by any of the 48 symmetries of the cube, and we can also invert any of the maneuvers. all these operations produce different maneuvers, so we get a total of 2 * 23 * 48 * 2 = 4416 different maneuvers. by counting, the number of different sequences of length <= 19f is about 82 times as many positions the cube has. thus a position has, on average, 82 maneuvers of length <= 19f, although superflip has 0. the number of different sequences of length 20f is about 1016 times the number of positions, so a position has, on average, 1016 different maneuvers of length 20f. superflip has more than 4 times that many. here are the 30 solutions my program found for superflip U R2. hopefully i haven't made any mistakes this time. they should all be equivalent to one of the two listed above. U R2 F U2 F2 D2 R' L U R2 F' B' R D2 L F2 R D2 R D (20f) U R2 F B R B2 R U2 L B2 R U' D' R2 F R' L B2 U2 F2 (20f) U R2 F B R B2 R U2 L B2 R U' D' R2 F D2 B2 U2 R' L (20f) U R2 F R' L F2 D2 B2 U R2 F' B' R D2 L F2 R D2 R D (20f) U R2 F2 L2 F D2 R L D R2 D F2 U R2 D F' B' D2 L D' (20f) U R2 F2 L2 F' U2 R' L' U' R2 U' F2 D' R2 U' F B U2 L' D' (20f) U R2 F2 L2 B D2 R' L' D F2 U R2 D F2 D F B D2 R D' (20f) U R2 F2 L2 B' U2 R L U' F2 D' R2 U' F2 U' F' B' U2 R' D' (20f) U R2 F' U2 B2 D2 R L' D' R2 F' B' R' B2 R' D2 L' B2 R' D (20f) U R2 F' B' R D2 L F2 R D2 R U D R2 F U2 F2 D2 R' L (20f) U R2 F' B' R D2 L F2 R D2 R U D R2 F R' L F2 D2 B2 (20f) U R2 F' R L' B2 D2 F2 D' R2 F' B' R' B2 R' D2 L' B2 R' D (20f) U R2 U B2 D R2 U F' B' U2 L F2 R2 B2 U' D B U2 R L (20f) U R2 U B2 D R2 U F' B' U2 L U' D R2 B2 L2 B U2 R L (20f) U R2 U R L U2 F L2 F2 R2 U' D L U2 F' B' U R2 D F2 (20f) U R2 U R L U2 F U' D F2 R2 B2 L U2 F' B' U R2 D F2 (20f) U R2 U2 L2 F' B R F2 U' D' F L2 B U2 F L2 F R L F2 (20f) U R2 B R' L B2 U2 F2 D R2 F' B' R U2 L B2 R U2 R D (20f) U R2 B D2 B2 U2 R' L D R2 F' B' R U2 L B2 R U2 R D (20f) U R2 B' R L' F2 U2 B2 U' R2 F' B' R' F2 R' U2 L' F2 R' D (20f) U R2 B' D2 F2 U2 R L' U' R2 F' B' R' F2 R' U2 L' F2 R' D (20f) U R2 D F2 U R2 U R L U2 F L2 F2 R2 U' D L U2 F' B' (20f) U R2 D F2 U R2 U R L U2 F U' D F2 R2 B2 L U2 F' B' (20f) U R2 D F2 U R' L' U2 B L2 F2 R2 U' D R U2 F B U F2 (20f) U R2 D F2 U R' L' U2 B U' D F2 R2 B2 R U2 F B U F2 (20f) U R2 D F2 D F B D2 R U D' R2 F2 L2 B D2 R' L' D F2 (20f) U R2 D F2 D F B D2 R B2 R2 F2 U D' B D2 R' L' D F2 (20f) U R2 D F' B' D2 L U D' R2 F2 L2 F D2 R L D R2 D F2 (20f) U R2 D F' B' D2 L B2 R2 F2 U D' F D2 R L D R2 D F2 (20f) U R2 D2 L2 F B' L B2 U D B D2 B L2 F D2 B R' L' B2 (20f) mike From cube-lovers-errors@oolong.camellia.org Sun Jul 13 19:35:33 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id TAA06922; Sun, 13 Jul 1997 19:35:33 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Message-Id: <199707132333.TAA15870@life.ai.mit.edu> Date: Sun, 13 Jul 1997 19:37:58 -0400 From: michael reid To: cube-lovers@ai.mit.edu Subject: minimal maneuvers for composition of superflip and pons asinorum i've finished calculating all minimal maneuvers (in the face turn metric) for the composition of superflip and pons asinorum. in my search for maneuvers of length <= 18f for this position, i used symmetry to show that we may suppose that the first face turn is either U or U2 . in fact, there is more symmetry available, and this time i will use it. in the case beginning with U , there are four symmetries, generated by the cube rotation C_U . using these, we may suppose that the second face turn is one of D , D2 , D' , R , R2 or R' . in the case beginning with U2 , there are eight symmetries. these are generated by the cube rotation C_U and reflection through the left-right plane. using these, we may assume that the second face turn is one of D , D2 , R or R2 . we can reduce these cases somewhat further. the cases beginning with U D2 and with U2 D are equivalent, so only one needs to be seacrhed. the cases beginning with U D' and with U2 D2 can also be eliminated. both U D' and U2 D2 commute with both pons asinorum and with superflip, so we may cyclically shift these turns to the end of the maneuver. our position cannot be achieved only using these "slice" turns, so we'll always be able to cyclically shift until we do not begin with a slice turn. (alternatively, note that any maneuver of length 19f , or any odd length cannot consist only of slice turns!) that leaves seven cases to search. my program took just less than one day to search all through 17f. it found 26 maneuvers, 16 for the case beginning with U D . however, this case has 8 symmetries, so there are just 2 different maneuvers, each in 8 different orientations. this leaves 12 different maneuvers, which come in 6 pairs of inverses. they are: U R F D R U' D L' U' D F' B2 R L' D' F' L' B' R' (19f) U D F R L' F B' L D2 R L F' B' U' L2 F B' U2 L' (19f) U D F' B' L' U2 F' B L2 U' R' L' F' U' D F' B D' L2 (19f) U2 R F U F B' L' D' F B' L B R L' U D2 B' R' U2 (19f) U2 R F U2 D' R' L F' L' F B' U L F B' D' B' R' U2 (19f) U2 R U2 D2 R U' L' U B R F2 U' D B' R' F' D B' L2 (19f) and their inverses. the first of these is the maneuver found by jerry bryan. it's also the only of these that is 20 quarter turns long, which is consistent with his findings. mike From cube-lovers-errors@oolong.camellia.org Mon Jul 14 12:37:17 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id MAA08816; Mon, 14 Jul 1997 12:37:16 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Message-Id: <3.0.1.32.19970714113921.006aa4a4@pop.tiac.net> X-Sender: kangelli@pop.tiac.net X-Mailer: Windows Eudora Light Version 3.0.1 (32) Date: Mon, 14 Jul 1997 11:39:21 -0400 To: cube-lovers@ai.mit.edu From: karen angelli Subject: hockey puck puzzle Mime-Version: 1.0 Content-Type: text/plain; charset="us-ascii" I recently purchased a hockey puck puzzle from Whole Systems Design, who also brought us the Masterball. Since I received it, I've played with for only a couple of hours without beginning any systematic attempt at solving it. My initial impression is that, although it looks like is should be very easy, it may actually be pretty hare. Or, it's one of those puzzles that is so easy that I'm thinking too hard and can't see the answer. I've never heard anyone in the cube-lovers discuss the puzzle, and I wondered if any of you had any impressions of it. The puzzle has the shape, size and feel of a regulation hockey puck, and is divided into twelve wedges. It's basically a flattened masterball in which only the lines of longitude twist, and the lines of latitude do not. There are several designs, with various degrees of difficulty and redundancy. For example, on some, there are printed hockey players, Maple leafs, or American flags. On my version, the wedges are numbered consecutively from one through twelve. The pristine version is with the numbers lined up in consecutive order. I'm not really interested in learning a solution from anybody, but I would be interested in comments about whether you think the puzzle is harder or easier than it looks. YOu can see pictures of the hockey puck puzzle or order them at www.wsd.com/HockeyPuck/home. 'e-ya later, Pete. From cube-lovers-errors@oolong.camellia.org Mon Jul 14 14:02:22 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id OAA08973; Mon, 14 Jul 1997 14:02:22 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Date: Mon, 14 Jul 1997 13:59:16 -0400 (EDT) Message-Id: <199707141759.NAA00043@spork.bbn.com> From: Allan Wechsler To: karen angelli Cc: cube-lovers@ai.mit.edu Subject: hockey puck puzzle In-Reply-To: <3.0.1.32.19970714113921.006aa4a4@pop.tiac.net> References: <3.0.1.32.19970714113921.006aa4a4@pop.tiac.net> Please give a clearer description of the puck. The photo gives only a slight clue about how many pieces there are, and how they are arranged. Are there twelve pieces, or twenty-four, or more? -A From cube-lovers-errors@oolong.camellia.org Tue Jul 15 13:18:50 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id NAA10963; Tue, 15 Jul 1997 13:18:49 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Date: Tue, 15 Jul 1997 17:57:11 +0200 From: Rob Hegge Subject: Description of hockey puck puzzle To: cube-lovers@ai.mit.edu Message-id: <9707151557.AA06449@sumatra.mp.tudelft.nl> Content-transfer-encoding: 7BIT X-Sun-Charset: US-ASCII The hockey puck puzzle is a flat disk with a diameter of about 9 cm or 3.5 inches and a thickness of about 2.5 cm or 1 inch. It basically consists of a circle (the hart) and a "ring" surrounding the circle. The circle is cut into two equal halves like "(|)". The two halves are connected so that you can turn one half upside down, while holding the other half. The ring is cut (from front to back) into 12 equal wedges, each of which is attached to the circle by a dovetail so that the ring with the wedges can be moved around the circle. One can also flip six wedges including one half of the circle around so that afterwards those 6 wedges and the half circle face backwards. Thus the puzzle is similar to a puzzle called saturn (which has only 8 wedges ?). The type of moves reminds me of moves possible on square-1. In the puzzle I own the 12 wedges on the front are numbered from 1 to 12 and on the back with the letters of "hockeypuzzle", while the left half circle contains the letters "pu" and right half circle the letters "ck" as shown below. I do not have it here so this was straight from memory. front: back: 12 1 c k 11 2 o e 10 | 3 h | y | pu|ck 9 | 4 p | e 8 5 u l 7 6 z z The three "|" denote the cut through the circle. A flip as described above would give for instance 12 k c 1 11 e o 2 10 | y h | 3 |ck pu| 9 | e p | 4 8 l u 5 7 z z 6 while then a clockwise turn of the ring for one wedge would give: 11 12 1 2 10 k c 3 9 | e o | 4 |ck pu| 8 | y h | 5 7 e p 6 z l u z For a rotational puzzle it is not that difficult. Rob From cube-lovers-errors@oolong.camellia.org Tue Jul 15 13:18:23 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id NAA10958; Tue, 15 Jul 1997 13:18:23 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Date: Tue, 15 Jul 1997 09:41:28 -0400 (EDT) Message-Id: <199707151341.JAA00944@spork.bbn.com> From: Allan Wechsler To: jandr@xirion.nl Cc: Allan Wechsler , cube-lovers@ai.mit.edu Subject: Re: hockey puck puzzle In-Reply-To: <9707151143.AA27610@la1.apd.dec.com> References: <199707141759.NAA00043@spork.bbn.com> <9707151143.AA27610@la1.apd.dec.com> [Jan de Ruiter:] The puzzle contains edge pieces and center pieces, all with the same thickness as the puck. There are always two center pieces which together form the inner circle. In this case there are 12 edge pieces which together form the outer ring. Simpeler pucks might contain less than 12 edge pieces, but always an even number. The possible moves are: 1. rotate one center piece with half of the edge pieces 180 degrees relative to the other center piece and the other half of the edge pieces. 2. rotate the edge pieces around the center pieces, always multiples of 30 degrees, or 1/12 of a circle (or more if there are less edge pieces) I know the puck with 6 edge pieces is near trivial to solve. I haven't tried the other ones yet. Jan de Ruiter Thanks for the description -- Pete ("Karen Angelli") provided an identical one in a private reply. But this is still incomplete. Are the obverse and reverse of the individual pieces distinguishable? Suppose I manage to flip every other edge piece over in place (not sure this is possible). Does it then look solved? Or do the two sides have different colors or a distinguishing mark or something? I haven't tried it, but I can't imagine that this puzzle would be very difficult. -A From cube-lovers-errors@oolong.camellia.org Tue Jul 15 13:17:34 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id NAA10953; Tue, 15 Jul 1997 13:17:33 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Message-Id: <9707151143.AA27610@la1.apd.dec.com> Subject: Re: hockey puck puzzle To: Allan Wechsler Date: Tue, 15 Jul 1997 13:42:54 +0200 (CETS) From: Jan de Ruiter Cc: cube-lovers@ai.mit.edu In-Reply-To: <199707141759.NAA00043@spork.bbn.com> from "Allan Wechsler" at Jul 14, 97 01:59:16 pm Reply-To: jandr@xirion.nl X-Mailer: ELM [version 2.4 PL23] Content-Type: text > > Please give a clearer description of the puck. The photo gives only a > slight clue about how many pieces there are, and how they are > arranged. Are there twelve pieces, or twenty-four, or more? > > -A > The puzzle contains edge pieces and center pieces, all with the same thickness as the puck. There are always two center pieces which together form the inner circle. In this case there are 12 edge pieces which together form the outer ring. Simpeler pucks might contain less than 12 edge pieces, but always an even number. The possible moves are: 1. rotate one center piece with half of the edge pieces 180 degrees relative to the other center piece and the other half of the edge pieces. 2. rotate the edge pieces around the center pieces, always multiples of 30 degrees, or 1/12 of a circle (or more if there are less edge pieces) I know the puck with 6 edge pieces is near trivial to solve. I haven't tried the other ones yet. Jan de Ruiter From cube-lovers-errors@oolong.camellia.org Wed Jul 16 09:37:39 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id JAA12856; Wed, 16 Jul 1997 09:37:38 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Message-ID: From: "joyner.david" To: "'cube-lovers@ai.mit.edu'" Subject: RE: hockey puck puzzle Date: Wed, 16 Jul 1997 07:53:21 -0400 X-Mailer: Microsoft Exchange Server Internet Mail Connector Version 4.0.994.63 MIME-Version: 1.0 Content-Type: text/plain; charset="us-ascii" Content-Transfer-Encoding: 7bit >---------- >From: Jan de Ruiter[SMTP:jandr@apd.dec.com] >Sent: Tuesday, July 15, 1997 7:42 AM >To: Allan Wechsler >Cc: cube-lovers@ai.mit.edu >Subject: Re: hockey puck puzzle > >> >> Please give a clearer description of the puck. The photo gives only a >> slight clue about how many pieces there are, and how they are >> arranged. Are there twelve pieces, or twenty-four, or more? Spencer Robinson, a former student, and I have written a WWW page which sketches an easy but rather inefficient solution of the 12 piece hockey puck puzzle: http://www.nadn.navy.mil/MathDept/wdj/mball/puck.htm Have fun! - David Joyner >> >> -A >> > From cube-lovers-errors@oolong.camellia.org Wed Jul 16 10:18:34 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id KAA12915; Wed, 16 Jul 1997 10:18:33 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Date: Wed, 16 Jul 1997 09:44:21 -0400 (Eastern Daylight Time) From: Jerry Bryan Subject: No Local Maxima 11q from Start To: cube-lovers@ai.mit.edu Message-id: MIME-version: 1.0 Content-type: TEXT/PLAIN; charset=US-ASCII X-X-Sender: jbryan@pstcc6.pstcc.cc.tn.us My new Shamir program has now generated the entire search tree for the standard cube group G up to 11q from Start. This search was accomplished once before using my old tape spinning programs, so there is limited new information. One good result is that all the numbers match between the two programs. The matching results were obtained using different programs, implementing different algorithms, in different programming languages, on different hardware platforms, and under a different operating systems. So I feel pretty good about the numbers. They have been posted before, so I won't post them again. With problems this big, it is always good to have some sort of independent verification because it is impossible to verify anything by hand. Another interesting result is in fact new. The old program was only able to determine local maxima up to 10q from Start while calculating the 11q tree. The new program is able to determine local maxima up to the same distance from Start it is searching. There are no local maxima 11q from Start. I find this result somewhat surprising, since there are four local maxima (unique up to M-conjugacy) 10q from Start. The new program did confirm the previously known 10q local maxima, but failed to find any 11q local maxima. In its search for local maxima, for each position x the p