From cube-lovers-errors@curry.epilogue.com Tue May 7 15:11:29 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id PAA02528 for ; Tue, 7 May 1996 15:11:28 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Date: Tue, 7 May 1996 17:15:39 +0200 Message-Id: <199605071515.RAA16897@mailsvr> X-Sender: geohelm@mailsvr.pt.lu X-Mailer: Windows Eudora Version 1.4.4 Mime-Version: 1.0 Content-Type: text/plain; charset="us-ascii" To: cube-lovers@ai.mit.edu From: Georges Helm Subject: cube solutions prior to 1980 Some time ago someone asked a question about early cube solutions. Unfortunately I don't find the corresponding mail, but here is what I have found. I have solutions by the following people which were all written in 1979: Angevine, Beasley, Cairns/Griffiths, Dauphin, Howlett, Jackson (3-D), Johnson, Maddison, Singmaster, Sweenen and Truran. At http://ourworld.compuserve.com/homepages/Georges_Helm/cubbib.htm other info on solutions can be found. Georges Helm geohelm@pt.lu http://www.geocities.com/Athens/2715 http://ourworld.compuserve.com/homepages/Georges_Helm ------------------------------------- Phone: ++352-503896 (answer machine) ++352-38019 (office) ++352-021 19 13 13 (GSM) Fax: ++352-38535 ------------------------------------- From cube-lovers-errors@curry.epilogue.com Wed May 8 02:22:29 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id CAA04036 for ; Wed, 8 May 1996 02:22:28 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Sender: s2394459@csc.cs.technion.ac.il Message-Id: <31903CF4.3470@cs.technion.ac.il> Date: Wed, 08 May 1996 09:19:32 +0300 From: Rubin Shai X-Mailer: Mozilla 2.01 (X11; I; SunOS 5.5 sun4m) Mime-Version: 1.0 To: Cube-Lovers@ai.mit.edu Cc: s2394459@cs.technion.ac.il Subject: Rubik's cube Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Hi I'm looking for solutions to the 2X2X2 cube. I need solution that put the cubiks ONE AFTER THE OTHER. Is anyone can help? Thanks Shai Rubin From cube-lovers-errors@curry.epilogue.com Fri May 10 16:55:11 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id QAA11643 for ; Fri, 10 May 1996 16:55:10 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Message-Id: <31932827.7790@cytex.com> Date: Fri, 10 May 1996 04:27:35 -0700 From: "Michael B. Parker" Reply-To: mbparker@cytex.com Organization: CYTEX CORPORATION X-Mailer: Mozilla 3.0b3 (Win95; I) Mime-Version: 1.0 Newsgroups: rec.puzzles,geometry.puzzles,rec.games.abstract,comp.ai.games,rec.games.design,rec.games.misc,oc.general,la.general To: PuzzleParty@cytex.com, Cube-Lovers@ai.mit.edu, www-designer@cytex.com, 506maple-residents@cytex.com, mitacas@cytex.com, Pierre Wuu , "Julie S. Peterson" Cc: Wei-Hwa Huang Subject: Puzzle Party THIS SATURDAY, 7pm, Cal Tech Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit #?????????????????# ? PUZZLE PARTY 5! ? #?????????????????# HEY, LA! PUZZLE PARTY 5 IS AT *CALTECH*!, and is hosted by the International Puzzle Champion (and CalTech Junior) Wei-Wwa Huang! Wei-Hwa was recently featured in the LA Times for leading the US Puzzle Team to victory in the 1996 International Puzzle Chapionships. So show and share the brain teasers and mechanical puzzles you have, and the mental games you know, and discover dozens of new ones, and new tricks! (Or, if you're still puzzleless, *this is the place* to get clued in!) Plenty of snacks, refreshments, and good conversation provided. WHEN: Saturday, 1996 May 11, 7pm until the wee hours of the morning... WHERE: Winnett Student Center, 1200 San Pasqual, Pasadena, CA Caltech is located in a rectangle bordered on the north by Del Mar Blvd., south by California Blvd., west by Wilson Avenue, east by Hill Avenue. Winnett is the small building right in the middle of campus. Parking near Winnett is limited, so try to find a local parking space and walk to Caltech. There is some parking near the northwest and southeast of Caltech, but they can be quite full. Signs will be posted on and around major entrances. Directions: From 210 fwy: exit south on either: (1) Hill Avenue: Caltech will be on your right after 2 to 3 miles; or (2) Lake Avenue: Turn left on Del Mar after 2 miles, then right on Wilson. Caltech will be on your left. From 110 north: continue until in Pasadena. Turn right at California. Caltech will be on your left after 3 miles. COST: $10 Non-MITCSC Members without puzzles $ 8 MITCSC Members without puzzles $ 6 Non-MITCSC Members with puzzles $ 4 MITCSC Members with puzzles Free Caltech Students (with Student ID) [Hey, it's a condition of them letting us use the room...] Please RSVP (with the number of puzzles you'll be bringing) so we know how many people (and puzzles) to expect. RSVP: Mike Parker, MIT '89 - mbparker@cytex.com, 800-MBPARKER xLIVE, xFAXX (if you get lost, call 800-MBPARKER xLIVE -- will have cell phone) HOST: Wei-Hwa Huang, CIT '97 - whuang@cco.caltech.edu, 818-395-1599 PS: for the latest Puzzle Party updates, just tune in to http://www.cytex.com/~mitcsc/ ----------------------------------------------------------------------------- Michael B. Parker, MIT '89 CYTEX CORP. President http://www.cytex.com/~mbparker/ email mbparker@cytex.com, direct voice 714-639-6436, fax 714-639-5381 CYTEX CORPORATION, ** WE PUT YOUR COMPANY ON THE INTERNET ** 506 N. Maplewood St., Orange, CA 92667-6917 * 1-800-33CYTEX (332-9839) Dial 800#, then enter extension (pin): SALES(7253), TECH(8324), FAXX(3299) World-Wide-Web http://www.cytex.com/ * email info@cytex.com (r19960229) From cube-lovers-errors@curry.epilogue.com Sat May 11 00:25:22 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id AAA12417 for ; Sat, 11 May 1996 00:25:22 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Message-Id: <199605102323.TAA15582@mail-e2b-service.gnn.com> X-Mailer: GNNmessenger 1.3 Mime-Version: 1.0 Content-Type: text/plain; charset="us-ascii" Date: Fri, 10 May 1996 18:24:19 From: Sean Brewer To: cube-lovers@ai.mit.edu Subject: Rubik's Revenge My father has been looking for the sixteen sided Rubik's Revenge for years now. If you have any idea where I can get one for him please let me know. Thanks for any help you can give. From cube-lovers-errors@curry.epilogue.com Sat May 11 04:40:17 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id EAA12883 for ; Sat, 11 May 1996 04:40:16 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Date: Sat, 11 May 96 01:52:57 +0400 (EST) From: "Robert P. Munafo" Reply-To: "Robert P. Munafo" Message-Id: <7003%mrob.uucp@ursa-major.spdcc.com> To: cube-lovers@ai.mit.edu Subject: The barcode Dale Newfield wrote: > [...] > Charlottesville, VA 22901-2708 > |..|.|..|.||.|..||......||..|.||...|||...|..|.||....|.|..|..|| > (No, this barcode is not necessary, but I figured this would be a good > place to ask: "Has anyone figured out what information is encoded in > this, or how it is encoded?" :-) It's your zip code. The Post Office recognizes the zipcode with some sort of OCR (Optical Character Recognition) then prints a barcode on the envelope so that simpler machines can sort the mail later in its delivery path. Sometimes the sender prints the barcodes (if your mail is part of a large mailing, like junk mail, magazines, tax forms etc.). This is from the URL http://www.advanstar.com/autoidnews/barcofaq.txt > POSTNET symbols are different from other symbologies because the > individual bar height alternates rather than the bar width. Each > number is represented by a pattern of five bars. A single tall bar is > used for the start and stop bars. > > Each symbol includes a check digit defined as the single digit that > must be added to the sum of all the digits to make the total the next > multiple of 10. For example, 98116's check digit is 5 because: > 9+8+1+1+6=25 and 25 + 5 = 30. > > POSTNET can be used for 5-digit, 9-digit ZIP+4, and the new 11-digit > Delivery Point Barcode. They are often used in conjunction with one > of the three FIM bars (Facing Identification Marks) which are found > on the upper right corner of a mail piece like Business Reply Mail. The encoding is as follows: ||... ...|| ..|.| ..||. .|..| .|.|. .||.. |...| |..|. |.|.. 0 1 2 3 4 5 6 7 8 9 -- Robert P. Munafo UUCP: ...!harvard!spdcc!mrob!cube CUBE-LOVERS Account Internet: cube%mrob.uucp@spdcc.com From cube-lovers-errors@curry.epilogue.com Sat May 11 17:40:28 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id RAA15973 for ; Sat, 11 May 1996 17:40:28 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com From: 100021.1617@compuserve.com Date: 11 May 96 10:15:57 EDT To: CUBE-LOVERS@ai.mit.edu, Sean Brewer Subject: Re: Rubik's Revenge Message-Id: <960511141556_100021.1617_EHV69-2@CompuServe.COM> > My father has been looking for the sixteen sided Rubik's Revenge > for years now. If you have any idea where I can get one for him > please let me know. Thanks for any help you can give. Me too. Somebody posted this information some time ago in the cube-lovers list... It's a shop with Rubik-related puzzles, among others, including #130 Rubik's Revenge (4x4x4) #131 Professor's Cuve (5x5x5) They sell on the Internet, too. They also have a complete list of distribuitors worldwide, including *Game Preserve 222 D Street, Suite 4 Davis, CA 95616 USA Tel.: (916) 753 42 63 *Star Magic 4026 24th Street San Francisco, CA 94114 USA Greetings, Alvaro From cube-lovers-errors@curry.epilogue.com Sat May 11 17:53:19 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id RAA15991 for ; Sat, 11 May 1996 17:53:18 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Date: Sat, 11 May 1996 17:22:02 +0100 Message-Id: <9605111622.AA25415@mecmdb.me.ic.ac.uk> X-Sender: ars2@mecmdb.me.ic.ac.uk Mime-Version: 1.0 Content-Type: text/plain; charset="us-ascii" To: cube-lovers@ai.mit.edu From: The Unofficial Thermodynamics Fan Club of the UK Subject: Re: Rubik's Revenge X-Mailer: > My father has been looking for the sixteen sided Rubik's Revenge >for years now. If you have any idea where I can get one for him >please let me know. Thanks for any help you can give. > > I always thought that the Rubik's cube principle could only be applied to Polyhedrons of sides totalling 4, 6, 8, 12 or 20 as these are the only ones that could be made from only one shape of side. If it is a 12 sided Rubik's polyhedron, there is one called the MegaMinx made by Uwe Meffret in Hong Kong. This should be available around the world as it is still in production. Meffret's Company is known as: IDI, P.O.Box 24455, Aberdeen, Hong Kong. Tel: 852-2518-3080 Fax: 852-2518-3282 (I haven't found anything on the Net for his Company, but I'd be interested to learn....) Another possibility is that it could be a 'Cut down' Rubik Cube, but I haven't seen any of this type in years. Andy, (The Artist Currently Known as the Unofficial Thermodynamics Fan Club of The U.K.) From cube-lovers-errors@curry.epilogue.com Mon May 27 19:46:08 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id TAA28038 for ; Mon, 27 May 1996 19:46:07 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Date: Thu, 23 May 1996 12:53:27 -0500 (EST) From: Jerry Bryan Subject: Compact Cube Representation for Shamir and Otherwise To: Cube-Lovers Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Content-Transfer-Encoding: 7BIT I said I wasn't going to write again about Shamir's method until I had a working program. Well, I don't have a working program yet but this is only indirectly about Shamir. Rather, it is about how we might represent the cube compactly in a way that is easy to work with. We would like a compact representation that is usable by Shamir's method. But more importantly, we would like a compact representation that is easily usable for forming compositions in general. The compact representation I will describe is useful in a number of contexts, not just Shamir's method. My standard model is an S24 x S24 model, modeling the corner and edge facelets separately and not modeling the face centers. At one byte per facelet, this representation requires 48 bytes per position without packing. David Moews has described a wreath product representation (e.g., 23 Feb 1996) which requires 40 bytes without packing. There are 8 bytes to describe the position of each corner cubie, and 8 more bytes to describe the twist of each corner cubie. Similarly, there are 12 bytes to describe the position of each edge cubie, and 12 more bytes to describe the flip of each edge cubie. This representation has the virtue of being 8 bytes smaller than the S24 X S24 representation, while still being easy to work with and manipulate. For my very large searches, I always used a supplement representation for the external files. That is, I only stored one facelet from each of the 8 corner cubies and one facelet from each of the 12 corner cubies for a total of 20 bytes unpacked. (I also packed the 20 bytes into 13 bytes to use less tape, but that is not important for this particular story.) However, I found the supplement representation awkward to manipulate, so I always expanded the supplement representation to a full S24 x S24 representation inside the program. None of my programs were more than a few K (not a few Meg, just a few K because the storage was external), so the extra few bytes were a non-issue. But now that I want to implement Shamir, my programs will be very large. Therefore, I wanted to figure out how to manipulate the supplement representation directly. The representation itself is not new, but the technique to manipulate it is. Here is what I have come up with. I think it is applicable to Shamir programs and non-Shamir programs alike. I will use the corners as an example. Similar comments would apply to the edges. My standard supplement for the corners is the Front facelets and the Back facelets. The way I number the facelets, these are facelets 1 through 4 for the Front and 21 through 24 for the Back. In the vector notation we have been talking about in this thread, the supplement of the identity is [1,2,3,4,21,22,23,24]. 1 is mapped to 1, 2 is mapped to 2, 3 is mapped to 3, and 4 is mapped to 4. However, 5 is not mapped to 21. Rather, 21 is mapped to 21, 22 is mapped to 22, etc.. You have to think of the last 4 indexes as being offset by 16 because 16 of the facelets are left out. From this vector, we can reconstruct the fact that 5 is mapped to 5, 6 is mapped to 6, etc. based on which facelets are part of which cubies. Composition of these supplement vectors can be hard or easy depending on what we are trying to do. Suppose X is a permutation on the corners represented by an 8 byte supplement vector and q is a quarter-turn on the corners represented by a 24 byte permutation vector. Then, the calculation of Xq more or less "just works", and the composition is an 8 byte supplement vector. For some kinds of things you have to worry a little bit about the offset of the last 4 indexes, but the computer coding is basically very straightforward. The code even runs faster than the code for composing two 24 byte permutation vectors. But suppose for some reason we need to form qX instead of Xq. The q vector will map into values that simply aren't there in the X vector. The programming symptom will be an out-of-bounds subscript. It doesn't help to use two supplement vectors. If X and Y are both supplement vectors, then neither the product XY nor the product YX can be formed. The same problem occurs anytime a supplement vector is pre-multiplied, no matter whether it is pre-multiplied by another supplement vector or whether it is pre-multiplied by a full-length permutation vector. With some searches you can probably get by with only post-multiplying supplement vectors by full-length permutation vectors. I think you could form a breadth first search tree that way by always post-multiplying by full-length vectors q in Q. But I always want to form M-conjugates m'Xm, so I have to be able to pre-multiply. Here is how to do it with supplement vectors. As I said, my old programs expand an 8 byte supplement vector for the corners into a 24 byte permutation vectors on the corners when a position is read from a file into memory. Two special 24 byte vectors are used in the process. One of the 24 byte vectors defines which facelet is to the right of each other facelet on the corner cubies, and the other of the 24 byte vectors defines which facelet is to the left of each other facelet on the corner cubies. So the supplement is expanded by mapping each of the 8 bytes in the supplement into itself, and in addition by mapping each of the 8 bytes into its respective right and left. These "right of" and "left of" vectors can be identified with the permutations which twist each corner cubie right and left, respectively. These permutations are not in the Start orbit. But we can nonetheless observe that both of them commute with every other permutation. I am focusing this example on the corners, but my old programs also have to expand a 12 byte supplement vector for the edges into a 24 byte permutation vector. The vector which accomplishes this mapping defines for each edge facelet the other facelet which is on the same edge cubie. This permutation can be identified with Superflip, and Superflip also commutes with every other permutation. The center of G consists of the identity plus Superflip. These permutations fix the corners and either fix or flip the edges. But the center of the constructable group consists of fixing or flipping the edges composed with fixing or twisting right or twisting left the corners. So there are six positions in the center of the constructable group, and it is precisely these six permutations which are required to make composition of supplement vectors work. I normally write an M-conjugate in E-mail just as m'Xm. But for this example, let me write it as (i)m'Xm, where i is the argument of the permutation and where i runs from 1 to 24 for the corners. The trick to make composition of supplements work is going to be to write the permutation as something like (i)m'k'Xkm, where k is not really a permutation. Rather, it is some magic to be defined below. The trick is not just for M-conjugation. It is for pre-multiplication in general. The Xm part of m'Xm is not a problem; it is the m'X part which is a problem. Similarly, to multiply supplement X (or full-length vector X) by supplement Y, the k trick would be Xk'Yk, which we could group as X(k'Yk) for emphasis. As with M-conjugation, I will make the argument explicit and write (i)Xk'Yk. But just what is this k? For the corners, we define k[1] as the identity, k[2] as twist all corners right, and k[3] as twist all corners left. We also define a 24 byte vector j which defines which corner facelets are in the supplement (a value of 1), right of the supplement (a value of 2), or left of the supplement (a value of 3). j is a function, but is not a permutation. With my particular numbering scheme and choice of supplement, j looks something like [1,1,1,1,2,3,2,3,......3,2,3,2,1,1,1,1]. That is, only the first four and last four facelets are in the supplement. The j vector is used to index k. For the edges we would define k[1] as the identity and k[2] as Superflip. An M-conjugate would then be calculated as (i)m' k[j[t]]' X k[j[t]] m for i in 1..24 and where t=(i)m'. Effectively, k' maps (i)m' into the supplement so that X operates only on the supplement, and k undoes (untwists and/or unflips) whatever k' does. However, the k-conjugation must be applied on a facelet by facelet basis. k[1] might be used for one facelet, k[2] for another facelet, and k[3] for still another. Nonetheless, since each of the k's is in the center of the constructable group, we have X=k'Xk for all X, irrespective of which k is used for a particular facelet. It is not strictly necessary, but this procedure would be slightly simpler if the facelets were renumbered. That is, renumber the supplement 1 to 8 for the corners and 1 to 12 for the edges. It is easy to see how to construct the tree required by Shamir's method using this supplement representation. The supplement representation does not reduce the number of potential branches out of each node. But it does reduce the number of levels of nodes. I plan to have the branching for the first 8 levels of my tree be controlled by the supplement for the corners, and the branching for the next 12 levels of my tree be controlled by the supplement for the edges. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) jbryan@pstcc.cc.tn.us Pellissippi State (423) 539-7127 10915 Hardin Valley Road (423) 694-6435 (fax) P.O. Box 22990 Knoxville, TN 37933-0990 From cube-lovers-errors@curry.epilogue.com Tue May 28 14:04:51 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id OAA29949 for ; Tue, 28 May 1996 14:04:51 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Message-Id: <9605281411.AA29962@sarofim-sun.MIT.EDU> To: cube-lovers@ai.mit.edu Subject: Square 1 Combinations Date: Tue, 28 May 1996 10:11:46 EDT From: "Michael C. Masonjones" Hi, I'm new to the group, but I have read the entire archive. I noticed rather little work done on Square 1. It seems to me that this puzzle deserves a closer look for finding God's algorithm. Mike Reid's calculations notwithstanding (archive 17), I have found that the problem can be reduced by at least a factor of 400 if we just get rid of combinations that result from trivial face turns, and if we note that the Start position has a degeneracy of 16. (One center slice is assumed fixed - another factor of 2 is tempting but not possible) Mike's calcualtion for the number of states would reduce to: (2*(1/6)*(9/2)+2*(28/3)*3+(35/4)*(35/4))*2*8!*8!=435891456000 combinations. Divide by the start degeneracy, multiply by 2 storage bits per state, and you get a storage requirement of 6.81GB. This seems very close to being doable. Maybe in another 10 years, I can do this project on my PC, if no one has done it yet. On another note, when I signed up, I mentioned to Alan that I must be crazy enough to join this group since I have a five foot mockup of a rubik's type puzzle as my coffee table. He thought its description might be of general interest. Skip the rest of this paragraph if you couldn't care less about its origins. I built it for Caltech's ditch day event Maybe you have heard of it. That's where all the seniors leave for the day with their room locked only with a puzzle of some sort, and the object is for the undergraduates to get into the room by solving it (with a couple of clues, of course). Anyway, being as it was that I had a mechanical engineer roommate... The rest is history, and I now have a five foot diameter puzzle coffee table. OK, a description. The puzzle is a three centered version of the Puzzler, widely available in the last few years in puzzle/game specialty stores. The differences being that it is colored so that the maximum number of combinatins are possible (including the supergroup of distinguishing face centers). For those of you who have not seen the Puzzler, and thus have no frame of reference, consider one vertex of a cube and it's surrounding faces. 7 vertices, 9 edges. Faces can undergo 4 quarter turns. Extrapolate to the Megaminx and you again get one central vertex for a total of 10 vertices and 12 edges. Faces can undergo 1/5 turns. Extrapolate again to six sided faces, and you get a flat puzzle with one central vertex for a total of 13 vertices, and 15 edges. Faces can undergo 1/6 turns. So it is basically the group for a' cube' with hexagonal faces. The extra face over the Puzzler also serves to remove the significant parity constraints on the edge pieces. (compare group to group of regular cube). You, too, can make a smaller version of the Ditch Day puzzle at home. The advantage of the flat puzzle is that it is easily constructed. I built the 6 inch diameter prototype with poster board, lamination, magic markers, and an easily machined smooth pressboard frame. You only need to drill three 3 inch holes. The rest is trimming. Oh yeah, you will need a plexiglass faceplate to keep the pieces in too. Cutting out and gluing together the poster board to make sufficiently thick pieces was the hardest part. Number of combinations = (13!*15!*3^13*2^15*6^3)/24 = 3.83E33 Difficulty is comparable with Megaminx. Happy cubing. This is already too long. mikem. Mike Masonjones. From cube-lovers-errors@curry.epilogue.com Tue May 28 18:56:50 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id SAA00500 for ; Tue, 28 May 1996 18:56:50 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Date: Tue, 28 May 1996 13:18:51 -0700 From: "Jason K. Werner" Message-Id: <9605281318.ZM12960@neuhelp.corp.sgi.com> In-Reply-To: "Michael C. Masonjones" "Square 1 Combinations" (May 28, 10:11) References: <9605281411.AA29962@sarofim-sun.MIT.EDU> X-Face: 6]L85m[]|?5>dL9qI]8j>PPk/:]fF4Ma`5O&VJU)U.6"lo:gX{D`?bNqWl~),bS~`rrB5+P d=NQ_[sXE*#|;SZ)PanGF^&Q-Ch[[|Q)Pgx%ts.JdPJ,3bwU84qc^s2q"sH{l9+g]$cD&a"?S]PQ)F b~4}Y93=ZOimDi_J^(lR;OLeN^W\]/&!v8S=~8Qw'HJ.ksu:R/!iV:WiExaWEXw!v$&hyp[mC X-Mailer: Z-Mail-SGI (3.2S.2 10apr95 MediaMail) To: cube-lovers@ai.mit.edu, "Michael C. Masonjones" Subject: Re: Square 1 Combinations Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii On May 28, 10:11, Michael C. Masonjones wrote: > Subject: Square 1 Combinations ..... > On another note, when I signed up, I mentioned to Alan that I must be > crazy enough to join this group since I have a five foot mockup of a > rubik's type puzzle as my coffee table. He thought its description > might be of general interest. Skip the rest of this paragraph if you > couldn't care less about its origins. I built it for Caltech's ditch day event > Maybe you have heard of it. That's where all the seniors leave for the > day with their room locked only with a puzzle of some sort, and the > object is for the undergraduates to get into the room by solving it > (with a couple of clues, of course). Anyway, being as it was that I > had a mechanical engineer roommate... The rest is history, and I now > have a five foot diameter puzzle coffee table. ..... Speaking of oversized Rubik puzzles... A good friend of mine built an oversized, fully functional Rubik's Magic about 3 years ago. She painted all of the artwork that went inbetween the plastic squares, cut out all the grooves, and used a heavy duty grade of fishing wire to connect all the pieces. We kind of thrashed my Rubik's Magic to see how many wires were used in the Magic and all the paths they took. I _think_ each square was 1'X1', so that would have made the puzzle 2'X4'. It's fun to play with, but only if you have the stamina; it's heavy! :) -Jason -- Jason K. Werner, Silicon Graphics U.S. Field Operations I/S Sys Admin mrhip@corp.sgi.com, 415-933-6397 "I will choose free will".....Neil Peart "These go to eleven".....Nigel Tufnel *********************** THIS IS A FREE SPEECH ZONE ************************ In defiance of the Communications Decency Act, I refuse to self-censor the content of my e-mail, my online postings, and my Web pages. I urge other Constitutionally-protected Americans to declare their online communications FREE SPEECH ZONES and to fight any attempts at regulating, censoring and "dumbing down" the Internet. The Net is not TV and radio! Let's keep it that way. http://www.eff.org/blueribbon.html mrhip@corp.sgi.com **************** I SUPPORT THE EFF'S BLUE RIBBON CAMPAIGN ***************** From cube-lovers-errors@curry.epilogue.com Tue May 28 23:35:28 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id XAA01092 for ; Tue, 28 May 1996 23:35:27 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Message-Id: <31ABA343.23B@erols.com> Date: Tue, 28 May 1996 21:07:15 -0400 From: Charlie Dickman Reply-To: charlied@erols.com X-Mailer: Mozilla 2.01 (Macintosh; U; 68K) Mime-Version: 1.0 To: Cube-Lovers@ai.mit.edu Subject: A 4-dimensional Cube Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit I have a document that describes a model of a 4-dimensional Rubik's Cube (3x3x3x3) and a program that implements the model. The document is a stand-alone that executes on a Macintosh and the implementation of the model runs on a Mac as well. This paper/program is based on an unpublished paper by Harry Kamack and Tom Keene that was referenced in Hofstader's '82 SA column. I'll be more than happy to share them with any interested parties - let me know your interest. Anyone with a site on the Web who would like me to upload the files please let me know. The document is 305K bytes, the program is 197K bytes. Charlie Dickman charlied@erols.com From cube-lovers-errors@curry.epilogue.com Wed May 29 01:37:34 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id BAA01350 for ; Wed, 29 May 1996 01:37:34 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Date: Wed, 29 May 1996 00:54:08 -0400 From: AirWong@aol.com Message-Id: <960529005406_544579496@emout14.mail.aol.com> To: CUBE-LOVERS@ai.mit.edu Subject: ULTIMATE Rubik's cube? I'm not sure if this has been discussed or not, but I was asked the following question, and I am not sure of thie answer. Is there an ULTIMATE Rubik's cube that, if an algorithm for it was known, it would contain an algorthm for ANY Rubik's cube? I guessed that it was four, but I'm not so sure about it. Aaron Wong AirWong@AOL.com From cube-lovers-errors@curry.epilogue.com Wed May 29 15:25:56 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id PAA02702 for ; Wed, 29 May 1996 15:25:56 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Date: Wed, 29 May 1996 10:56:15 +0100 Message-Id: <9605290956.AA05172@mecmdb.me.ic.ac.uk> X-Sender: ars2@mecmdb.me.ic.ac.uk Mime-Version: 1.0 Content-Type: text/plain; charset="us-ascii" To: CUBE-LOVERS@ai.mit.edu From: "We Love Stress Analysis." Subject: Ultimate Rubik's Cube. (how to make a 4x4x4) X-Mailer: I was thinking about something similar to this a while back. I asked (as did many others) about 4x4x4s, but soon after I sent that E-mail I realised that I was holding one with a few extra pieces. If we can consider a 3x3x3 Rubik's cube as a 2x2x2 with mid-edges and centres, in effect we can reverse this and ignore the centres and mid-edges of a 3x3x3 thus making it into a 2x2x2. Anyone who wanted a 2x2x2 could just pull the stickers from the central columns and rows of a 3x3x3. As this makes those pieces indistinguishable, they are no longer part of the puzzle. the relationship between 4x4x4 and 3x3x3 is slightly harder, but the above is true for 4x4x4 and 2x2x2 (but if ANYONE tries that with a 4x4x4, I'll hit them!!!!!!!). A more common cube than the 4x4x4 is the 5x5x5 (which is still in production, c/o Uwe Meffret). This can be transformed from a 5x5x5 into a 4x4x4 by removing the central lines of stickers. It can also be transformed into a 3x3x3 (why anyone would want to.................) by removing columns & rows 2&4 from each side, and 2x2x2 (I won't bother saying it.........) by removing columns and rows 2,3,&4 from each side. Higher orders of cubes aren't in production, but apparently do exist in cyberspace, These would display similar properties. The pattern is simple: smaller cubes with odd number of pieces per side can be incorporated with other pieces to form larger cubes with odd number of pieces per side. etc. etc. etc. 2x2x2 + (extras) = 3x3x3 2x2x2 + (extras) = 4x4x4 3x3x3 + 4x4x4 + (extras) = 5x5x5 I'm sure there is some mathematical proof to what I am trying to say, but I'm no mathematician. It might start off: Pieces Jump(from last) used before: 2x2x2 8 8 0 3x3x3 26 18 8 4x4x4 56 30 8 5x5x5 98 42 74 6x6x6 152 54 ??? I would say that the ultimate Rubik cube was in fact the 2x2x2 because it features in all the solutions. However, based on this logic, the 8086 is the ultimate P.C. as any P.C. can run 8086 software, but an 8086 can't run 80386 software................ I hope this E-mail hasn't been too scatter brained................... Andy. Fact: did you know that British Airways has more Super-Sonic flying time than any air force in the world?? From cube-lovers-errors@curry.epilogue.com Wed May 29 15:27:35 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id PAA02706 for ; Wed, 29 May 1996 15:27:34 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com X-Sender: mag@void.ncsa.uiuc.edu Message-Id: Mime-Version: 1.0 Content-Type: text/plain; charset="us-ascii" Date: Wed, 29 May 1996 09:59:23 -0600 To: cube-lovers@ai.mit.edu From: Tom Magliery Subject: Re: Square 1 Combinations At 10:11 AM 5/28/96, Michael C. Masonjones wrote: >On another note, when I signed up, I mentioned to Alan that I must be >crazy enough to join this group since I have a five foot mockup of a >rubik's type puzzle as my coffee table. Neato! It's long been a carpentry dream of mine to build a giant functioning 3x3x3 Rubik's Cube. (Just how giant, who knows, but I've usually envisioned about 1-foot cubies.) Has anyone ever written up plans or built a giant cube? mag -- .---o Tom Magliery, Research Programmer .---o `-O-. NCSA, 605 E. Springfield (217) 333-3198 `-O-. o---' Champaign, IL 61820 O- mag@ncsa.uiuc.edu o---' From cube-lovers-errors@curry.epilogue.com Wed May 29 20:41:19 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id UAA03160 for ; Wed, 29 May 1996 20:41:18 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Date: Wed, 29 May 1996 17:08:37 -0400 (EDT) From: Nicholas Bodley To: cube-lovers@ai.mit.edu Subject: Another subscriber Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Thanks to Alta Vista and Rubik as a keyword, I did an all-nighter over the holiday weekend, and discovered you folks. I haven't yet downloaded the archives, but will do my best to be a good Netizen before posting "real" queries. The first Cube I saw must have been pre-Ideal; I was riding home on the West Side IRT local in NYC, and noticed someone sitting at the other end of the car manipulating a puzzle that looked so unbelievable (mechanically) that I really wondered whether my perceptions had gone haywire from too many consecutive late bedtimes and regular mornings. It was *some* sleep debt! I have little doubt that it was a Cube. When I got mine, I came quite close to solving it by sheer persistence and brute force (probably about 4 cubies out of position); beginner's luck! When Meffert was mentioned in (Martin Gardner's col.?) in Sci. Am., I wrote away for his catalog, which I'm just about sure I still have. I bought a "5" from him, and sent another check for more items; never received them. He said Customs must have confiscated them; Customs never notified me. At the time, I could afford $112 or so. I consider it lost; I hope Meffert used it to good advantage. (This would have been around 1987.) It was fascinating to see that he apparently is active once more. I've seen "5"s for sale again within the past year or so, I think at The Compleat Gamester (?) in Waltham, and also The Games People Play in Cambridge, which has moved (not too far) about a year ago. Does *everybody* know there's a ball inside Rubik's Revenge? I'm at least as much of a gadget-hound as a puzzle-solver; I have a decent collection. I get a real bang out of dismantling group-theory puzzles to see how they're built; almost all can be disassembled, although (as most people probably know) the "2" (Pocket Cube) is quite hard both to disassemble and to reassemble. I have the Hungarian Globe, which is truly impossible to dismantle, IMO. (I haven't dared to scramble it!) This one has printed metal surfaces attached to a plastic structure; the "tiles" take paths like the grooves in the ball inside the "4" (R.R.). I hope I might be forgiven for posting one question that has been paining me-- I'd dearly love to know the answer! Is it true that a physical prototype of the "6" (6 X 6 X 6) has been constructed; if so, could anyone tell me the approximate date(s) of messages that discuss it? I would not want anyone to do lots of searching on my behalf, but just a recollection would be welcome. I'm also very curious about the mechanism for a "7"; it seems to me that locking pins (or the equivalent) would be necessary. I really wonder whether the mechanical design can be practical. I'm also a mechanical calculator (See Erez Kaplan's pages on the Web, in particular) and also mechanical analog computer enthusiast. Paradise was being a Navy fire control tech. who correctly diagnosed a loose screw inside the Mk. 1A main battery computer on a destroyer; it took three weeks to repair. The Master Technician scheduled things well; it happened just before the ship went in for its every-3-year yard overhaul. I expect to be enjoying this List! NB Nicholas Bodley Autodidact & Polymath |*| Keep smiling! It makes | Waltham, Mass. Electronic Technician |*| people wonder what | nbodley@tiac.net Amateur musician |*| you have been up to. | -------------------------------------------------------------------------* From cube-lovers-errors@curry.epilogue.com Thu May 30 18:15:45 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id SAA05355 for ; Thu, 30 May 1996 18:15:45 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Date: Thu, 30 May 1996 15:23:58 -0400 From: der Mouse Message-Id: <199605301923.PAA18428@Collatz.McRCIM.McGill.EDU> To: cube-lovers@ai.mit.edu Subject: Re: Another subscriber > I hope I might be forgiven for posting one question that has been > paining me-- I'd dearly love to know the answer! Is it true that a > physical prototype of the "6" (6 X 6 X 6) has been constructed; [...] > I'm also very curious about the mechanism for a "7"; it seems to me > that locking pins (or the equivalent) would be necessary. I really > wonder whether the mechanical design can be practical. In my opinion mechanical designs for the 7 and above will have to be fundamentally different from those for the 6 and below, because that's the point at which the "buried" corner of a corner cubie extends past the surface of the face during a face turn and thus it's not possible to build the thing as rigid pieces connected to a central mechanism, at least not without cutting away part of some face-center cubies. (Specifically, that buried corner is at sqrt(2)*(.5-1/N) from the center, taking the cube side as 1 and N as the order of the cube. The face is at .5 from the center. The former becomes greater than the latter at about N=6.83...not that non-integer N make physical sense.) This is not to say that a 7 is impossible, just that it will have to be rather drastically different - somehow, when a turn is started, the corner cubie will have to be mechnically locked to the rest of the face that's turning with it. I can easily enough imagine possible mechanisms, but coming up with one simple enough to mass-produce at a price people are likely to be willing to pay would be a major challenge. On the other hand, a straightforward locking mechanism could probably be put together by a good watchmaking shop at no more than the price of a high-end watch. A few collectors might go for it, especially since the result - particularly if made out of metal - would feel much better than the plastic-on-plastic feel of most cubes. der Mouse mouse@collatz.mcrcim.mcgill.edu From cube-lovers-errors@curry.epilogue.com Thu May 30 18:15:15 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id SAA05351 for ; Thu, 30 May 1996 18:15:14 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com To: Cube-Lovers@AI.MIT.EDU From: Wei-Hwa Huang Subject: Re: Another subscriber Date: 30 May 1996 05:53:22 GMT Organization: California Institute of Technology, Pasadena Lines: 31 Message-Id: <4ojd4i$g2o@gap.cco.caltech.edu> References: Nntp-Posting-Host: accord.cco.caltech.edu X-Newsreader: NN version 6.5.0 #12 (NOV) As a first aside, I'd like to mention that a nice project was to get a standard 3x3x3 cube into a standard American spaghetti sauce jar. Makes a good conversation piece, though hard to scramble quickly... Nicholas Bodley writes: > I'm at least as much of a gadget-hound as a puzzle-solver; I have a >decent collection. I get a real bang out of dismantling group-theory >puzzles to see how they're built; almost all can be disassembled, although >(as most people probably know) the "2" (Pocket Cube) is quite hard both to >disassemble and to reassemble. I have the Hungarian Globe, which is truly >impossible to dismantle, IMO. (I haven't dared to scramble it!) This one >has printed metal surfaces attached to a plastic structure; the "tiles" >take paths like the grooves in the ball inside the "4" (R.R.). I feel compelled to mention that there's a small company in Taiwan which makes two variants of the Hungarian Globe that are harder. One variant allows for a move that turns the 9 pieces on one side; the other variant allows for the 5 pieces at every intersection to be rotated. Let me dig out the address... International Puzzles and Games Fl. 3 No. 192 Chung Ching N. Rd. Sec 2 Taipei Taiwan, Republic of China Tel: 886-2-5532575 Fax: 886-2-5536757 -- Wei-Hwa Huang, whuang@cco.caltech.edu, http://www.ugcs.caltech.edu/~whuang/ --------------------------------------------------------------------------- Caught Porfiry, Raskolnikov sung his swan Sonia when he went Dounia to Siberia. From cube-lovers-errors@curry.epilogue.com Fri May 31 02:25:14 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id CAA06390 for ; Fri, 31 May 1996 02:25:14 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Message-Id: <31AE492B.6FC7@erols.com> Date: Thu, 30 May 1996 21:19:39 -0400 From: Charlie Dickman Reply-To: charlied@erols.com X-Mailer: Mozilla 2.01 (Macintosh; U; 68K) Mime-Version: 1.0 To: Cube-Lovers Subject: Re: An Ultimate Cube References: <31ADF6C2.28E3@is.ge.com> Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit On the subject of an ultimate cube... There can not possibly be an ultimate cube just like (because) there is no "ultimate", ie., largest, integer. But if you can solve the (N+1)x(N+1)x(N+1) cube then you can surely solve the NxNxN cube. You would simply reduce the (N+1)x(N+1)x(N+1) permutation to one that is in the NxNxN group and continue from there like MBW did when looking for God's algorithm in the 3x3x3 group. The group of an NxNxN cube is a proper subgroup of an (N+1)x(N+1)x(N+1) cube. For example, the 2x2x2 cube group is the 3x3x3 group minus the edge moves and the center cubie orientation moves - that is, as Singmaster pointed out, it is just the corners of the 3x3x3 cube. Adding the 3rd cut added 2 additional types of cubies to the 2x2x2 cube, the edges and the centers, and along with them came the edge moves (to form the group of the 3x3x3 cube) and the center orientations (to form the 3x3x3 super-group). The edge moves alone are a proper subgroup of the cube group and the cube group is a proper subgroup of the super-group. A similar situation occurs when you go from the 3x3x3 cube to the 4x4x4 cube. If you constrain the cube so that the central 2 slices can not be moved independently of one another then the 2 central edge pieces act exactly like the edges of a 3x3x3 cube and the 4 face center pieces act exactly like the face centers of the 3x3x3 cube. When the central slices are allowed to move independently of one another permutations are added to the 3x3x3 group and super-group to make up the 4x4x4 group and super-group. Thus the 3x3x3 groups are proper subgroups of the 4x4x4 groups. The pattern continues as the value of N increases with the N+1 group being larger than the N group and properly containing the N group. So the answer is no, there is no ultimate cube. From cube-lovers-errors@curry.epilogue.com Fri May 31 02:24:27 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id CAA06383 for ; Fri, 31 May 1996 02:24:26 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Message-Id: <9605310001.AA29475@SSD.intel.com> Cc: cube-lovers@ai.mit.edu Subject: realizing 7x7x7 or larger cubes In-Reply-To: Your message of "Thu, 30 May 96 15:23:58 EDT." <199605301923.PAA18428@Collatz.McRCIM.McGill.EDU> Date: Thu, 30 May 96 17:00:59 -0700 From: Scott Huddleston >In my opinion mechanical designs for the 7 and above will have to be >fundamentally different from those for the 6 and below, because that's >the point at which the "buried" corner of a corner cubie extends past >the surface of the face during a face turn and thus it's not possible >to build the thing as rigid pieces connected to a central mechanism, at >least not without cutting away part of some face-center cubies. One solution to this dilemma is to let some of the "cubies" become "brickies" (i.e., rectangular bricks instead of cubes). In this approach, there's no limit in principle on N to how large an NxNxN puzzle you could build with the standard mechanism. There is, of course, the lower limit you just described to how small the corner cubies could become. From cube-lovers-errors@curry.epilogue.com Fri May 31 16:05:37 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id QAA07784 for ; Fri, 31 May 1996 16:05:37 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Message-Id: <199605311941.PAA17615@zaphod.caz.ny.us> X-Mailer: exmh version 1.6.2 7/18/95 To: charlied@erols.com Cc: Cube-Lovers Reply-To: bmbuck@acsu.buffalo.edu Subject: Re: An Ultimate Cube In-Reply-To: Your message of "Thu, 30 May 1996 21:19:39 EDT." <31AE492B.6FC7@erols.com> Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii Date: Fri, 31 May 1996 15:41:34 -0400 From: Buddha Buck > On the subject of an ultimate cube... > > There can not possibly be an ultimate cube just like (because) there is > no "ultimate", ie., largest, integer. But if you can solve the > (N+1)x(N+1)x(N+1) cube then you can surely solve the NxNxN cube. You > would simply reduce the (N+1)x(N+1)x(N+1) permutation to one that is in > the NxNxN group and continue from there like MBW did when looking for > God's algorithm in the 3x3x3 group. I would not necessarily agree with the assertion that if one can solve an N^3 cube, one can solve an (N-1)^3 cube. Your construction, of solving the N^3 cube into a subgroup that is homeomorphic to the (N-1)^3 cube, assumes that one already knows how to solve a (N-1)^3 cube. > > The group of an NxNxN cube is a proper subgroup of an (N+1)x(N+1)x(N+1) > cube. For example, the 2x2x2 cube group is the 3x3x3 group minus the > edge moves and the center cubie orientation moves - that is, as > Singmaster pointed out, it is just the corners of the 3x3x3 cube. Adding > the 3rd cut added 2 additional types of cubies to the 2x2x2 cube, the > edges and the centers, and along with them came the edge moves (to form > the group of the 3x3x3 cube) and the center orientations (to form the > 3x3x3 super-group). The edge moves alone are a proper subgroup of the > cube group and the cube group is a proper subgroup of the super-group. True, and I will conceed that if you know how to solve a 3x3x3, you can solve a 2x2x2. > > A similar situation occurs when you go from the 3x3x3 cube to the 4x4x4 > cube. If you constrain the cube so that the central 2 slices can not be > moved independently of one another then the 2 central edge pieces act > exactly like the edges of a 3x3x3 cube and the 4 face center pieces act > exactly like the face centers of the 3x3x3 cube. When the central slices > are allowed to move independently of one another permutations are added > to the 3x3x3 group and super-group to make up the 4x4x4 group and > super-group. Thus the 3x3x3 groups are proper subgroups of the 4x4x4 > groups. Yes, the 3x3x3 groups are proper subgroups (or, probably more accurately, homeomorphic to proper subgroups) of the 4x4x4 groups, but that doesn't mean that knowing how to solve the 4x4x4 allows one to solve the 3x3x3. For instance, I can solve a 4x4x4. However, my solution to the 4x4x4 involves slice moves that don't exist on a 3x3x3 cube, through all stages of my solution, including the final stage. I cannot directly apply my 4x4x4 solution to a 3x3x3 cube. (I can to the 2x2x2 cube, since the techniques for solving the corners are applicable to cubes of all order N). If my solution for solving the 4x4x4 involved reducing it to the subgroup of the 4x4x4 generated by face turns only, then yes, I could directly solve a 3x3x3 by the methods I use for a 4x4x4, but I don't. > The pattern continues as the value of N increases with the N+1 group > being larger than the N group and properly containing the N group. So > the answer is no, there is no ultimate cube. The question originally asked (by Aaron Wong) was "Is there an ULTIMATE Rubik's cube that, if an algorithm for it was known, it would contain an algorthm for ANY Rubik's cube?" There might be no answer to the general question of if -any- algorithm was known for the U^3 cube, than an algorithm could be derived for any N^3 cube. For instance, few here would argue the assertion that if you can solve a 3x3x3, you can solve a 2x2x2, but from the discriptions I've heard of it, I wonder how well Thistlewaite's algorithm would work on a 2x2x2 cube. A Thistlewaite type algorithm for a (2N)^3 cube might very well reduce the (2N)^3 cube to the subgroup that is equivilant to a 2^3 cube in its final stages. Such an algorithm would be totally unsuited for solving a (2M+1)^3 cube, because there would be no way to reduce that to a 2^3 cube. (In general, I would guess that any algorithm for an n^3 cube that involved reducing it to an m^3 cube, where n = km, would be unsuited for solving a l^3 cube, where l does not have n or m as a factor). However, I think the question can be divided into two parts, if we look at it differently (requiring the existance of an algorithm with the stated property for order U^3 cubes, rather than requiring that all algorithms for order U^3) cubes have the stated property): First, is there a general algorithm that can be used to solve cubes of all orders? I think the answer is "yes". Second, what is the smallest order U^3 cube requiring a complete description of the algorithm? I think the answer is U=5. My current solution for the 4^3 cube is very closely related to my current solution for the 3^3. There are only minor changes in one stage, major changes in another, (both to deal with the split edge pieces) and the addition of a completely new stage to handle the centers, which aren't in the 3^3 at all. Transforming this algorithm to the 5^5 and higher is relatively easy, once I have the 3^3 and 4^4 down. All the important components of the two lower order solutions are needed for the 5^5, and nothing really new is added. The same goes for the higher orders. The tedium of solving increases, but not the real difficulty. I have been thinking (but haven't done much yet) of writing a collection of web pages describing my general solution (at least, for the 2^3, 3^3, and 4^4 cubes). -- Buddha Buck bmbuck@acsu.buffalo.edu "She was infatuated with their male prostitutes, whose members were like those of donkeys and whose seed came in floods like that of stallions." -- Ezekiel 23:20 From cube-lovers-errors@curry.epilogue.com Fri May 31 16:05:07 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id QAA07780 for ; Fri, 31 May 1996 16:05:06 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com To: Cube-Lovers@AI.MIT.EDU From: Wei-Hwa Huang Subject: Re: Another subscriber Date: 31 May 1996 07:30:49 GMT Organization: California Institute of Technology, Pasadena Lines: 19 Message-Id: <4om779$aip@gap.cco.caltech.edu> References: Nntp-Posting-Host: accord.cco.caltech.edu X-Newsreader: NN version 6.5.0 #12 (NOV) der Mouse writes: >In my opinion mechanical designs for the 7 and above will have to be >fundamentally different from those for the 6 and below, because that's >the point at which the "buried" corner of a corner cubie extends past >the surface of the face during a face turn and thus it's not possible >to build the thing as rigid pieces connected to a central mechanism, at >least not without cutting away part of some face-center cubies. >(Specifically, that buried corner is at sqrt(2)*(.5-1/N) from the >center, taking the cube side as 1 and N as the order of the cube. The >face is at .5 from the center. The former becomes greater than the >latter at about N=6.83...not that non-integer N make physical sense.) There's a really simple solution to this. Just don't make the 7 slices evenly spaced. -- Wei-Hwa Huang, whuang@cco.caltech.edu, http://www.ugcs.caltech.edu/~whuang/ --------------------------------------------------------------------------- Caught Porfiry, Raskolnikov sung his swan Sonia when he went Dounia to Siberia. From cube-lovers-errors@curry.epilogue.com Sat Jun 1 00:18:13 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id AAA08704 for ; Sat, 1 Jun 1996 00:18:13 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Message-Id: <2.2.32.19960601020111.009ebee0@greatdane.cisco.com> X-Sender: ronnie@greatdane.cisco.com X-Mailer: Windows Eudora Pro Version 2.2 (32) Mime-Version: 1.0 Content-Type: text/plain; charset="us-ascii" Date: Fri, 31 May 1996 19:01:11 -0700 To: Scott Huddleston From: "Ronnie B. Kon" Subject: Re: realizing 7x7x7 or larger cubes Cc: cube-lovers@ai.mit.edu At 05:00 PM 5/30/96 -0700, Scott Huddleston wrote: > >>In my opinion mechanical designs for the 7 and above will have to be >>fundamentally different from those for the 6 and below, because that's >>the point at which the "buried" corner of a corner cubie extends past >>the surface of the face during a face turn and thus it's not possible >>to build the thing as rigid pieces connected to a central mechanism, at >>least not without cutting away part of some face-center cubies. > >One solution to this dilemma is to let some of the "cubies" become >"brickies" (i.e., rectangular bricks instead of cubes). In this approach, >there's no limit in principle on N to how large an NxNxN puzzle you could >build with the standard mechanism. There is, of course, the lower limit >you just described to how small the corner cubies could become. I've had this dream of making cubies which attach (via bars or perhaps electromagnets) to their neighbors, with the smarts to detect the torque of a turn and release until the turn has been completed. You could then sell corner cubies, edge cubies, face cubies, and internal cubies one-at-a-time and people could build their own puzzles as large as they wanted. I'll buy enough for an order 10 cube if anyone cares to make this. :-) Ronnie From cube-lovers-errors@curry.epilogue.com Sat Jun 1 00:17:48 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id AAA08700 for ; Sat, 1 Jun 1996 00:17:47 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Date: Fri, 31 May 1996 21:27:05 -0400 (EDT) From: Nicholas Bodley To: Wei-Hwa Huang Cc: Cube-Lovers@ai.mit.edu Subject: Cube in a jar In-Reply-To: <4ojd4i$g2o@gap.cco.caltech.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII On 30 May 1996, Wei-Hwa Huang wrote: > As a first aside, I'd like to mention that a nice project was to get a > standard 3x3x3 cube into a standard American spaghetti sauce jar. Makes > a good conversation piece, though hard to scramble quickly... When one does this, is it OK to dismantle the Cube, and then reassemble it within the jar? (I assume not). {Snips} > Wei-Hwa Huang, whuang@cco.caltech.edu, http://www.ugcs.caltech.edu/~whuang/ > Regards to all, NB Nicholas Bodley Autodidact & Polymath |*| Keep smiling! It makes | Waltham, Mass. Electronic Technician |*| people wonder what | nbodley@tiac.net Amateur musician |*| you have been up to. | -------------------------------------------------------------------------* From cube-lovers-errors@curry.epilogue.com Sat Jun 1 01:40:03 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id BAA08874 for ; Sat, 1 Jun 1996 01:40:01 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Date: Sat, 1 Jun 1996 00:34:50 -0400 (EDT) From: Nicholas Bodley To: der Mouse Cc: cube-lovers@ai.mit.edu Subject: Locking mechanism for a 7^3 or larger (some thoughts) (fairly long) In-Reply-To: <199605301923.PAA18428@Collatz.McRCIM.McGill.EDU> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Some years ago, I think I remember reading (in Douglas Hofstadter's second (?) "major" book, as I think of them) that someone had designed a mechanism for a 7^3. When I started this message, I thought I'd offer a description of such a mechanism to the members of this list, but then realized that the problem is even harder than I thought. I hope these thoughts are not a waste of bitspace; let me know if so! Evidently, the design Hofstadter (probably) alluded to is not generally known to this List's members. Perhaps he could enlighten us. Also, one wonders whether Erno Rubik has contemplated the mechanical design for a 7^3. Edge cubies could be retained by schemes such as are now used, but only until they are moved out "into midair"; then they have one or two rubbing surfaces completely exposed, so to speak. The mechanism for retaining them is probably closely related to that for retaining corner cubies when out of alignment. Once the scheme for retaining corners is worked out, then a minor variant might be what's needed for the edges. (Or possibly conversely...) (Ideally, one would not want to prohibit an inner slice from being rotated all by itself; less-clever locking schemes might require that no more than one plane in the whole Cube be sheared at once.) For a first try, the corners would be retained by three locking pins to hold them in alignment with their neighbors. However, the problem is to retract one of the three pins before shearing an outside layer with respect to its neighbor. Until the pin is retracted, you can't shear the layer by much! Furthermore, that surface has to be locked once more when the corner is realigned, and this must happen automatically, reliably, and quickly. IMHO, it takes someone like Mr. Rubik to invent such a mechanism, and it might be one of the most difficult to invent yet. In the next few paragraphs, I get into speculative electronic/mechanical engineering, but try not to presume specialized knowledge any more than necessary. If one allows an electronic/mechanical scheme, cost balloons, and one has the disgusting thought of a battery to power things. Sliding contacts to distribute power from one central battery are bound to be unreliable unless carefully maintained; they would be a pain. They also would constitute an interesting problem in their own right: Can power be distributed without short circuits caused 1) by intermediate physical relative positions, and 2) by any possible configuration of a Cube? Keep in mind that two paths for current are needed, and they must never intersect. This is an interesting problem in topology, if I'm thinking correctly. A related problem is to ask whether every cubie could always have power connected to it. The thought of a small battery inside each corner and each edge cubie (or every other one) is even more painful. But, if there were power inside, here's what could be done. The locking pins would have a certain amount of "give", to permit limited shear movement. They could be mounted so they could tilt slightly (say, 10 degrees max.) against spring tension, and their mating sockets could be narrow "funnels", wide end at the surface. Some sort of sensor would detect misalignment well before the limit of "give" was reached. Misalignment would cause internal electronics to apply a pulse to a coil to retract the locking pin. (The pin should have minimal friction; a polished surface and a Teflon-lined mating hole should do.) Once the pin was retracted, the electronics would ignore other misalignments. (Otherwise, one could simply push against a cubie in "midair" and detach it!) The pin would be kept both extended and retracted magnetically, by a remanent alloy that stays magnetized, but which can have its magnetization reversed by the flux from a coil. Extending the pin would be done by pulsing the coil with the opposite polarity. (The principle is closely-related to pulsed magnetic latching relays, which do not require continuous coil power in either of their states.) As has been implied, once the cubie is realigned, a second (reverse-polarity) pulse re-extends the locking pin. Pulsed operation should give acceptable battery life; the misalignment sensor might be somewhat of a challenge to engineer, but not a major problem. One could consider mechanical contacts; with electronics, they wouldn't need to be kept scrupulously clean, only clean enough to permit a milliampere or so to flow. (Contamination becomes a factor if you make the electronics too sensitive.) The electronics seems quite straightforward, and by today's standards, quite simple and entirely practical. The locking-pin mechanism would be the most costly, more than likely; it would have to be custom-built, and might cost $2 US apiece in 10,000 lots, perhaps more. I can imagine many people-weeks of development to create a decently-reliable locking-pin design. It's essentially a miniature solenoid. Battery access would be via a screw-threaded cover with a slot that fits the edge of a coin; the Cube would look distinctive. The clicking sound of the retaining pins would be interesting! It's not immediately obvious (to me) how the pins and mating sockets should be arrayed; their number must be minimal. Edge cubies would need four locks apiece, while corners would need three apiece. The mind wants sleep, so I think I'll give this mad message a quick proofread and post it. --------------------- On Thu, 30 May 1996, der Mouse wrote: {Snips} > > that locking pins (or the equivalent) would be necessary. I really {Snips} > > On the other hand, a straightforward locking mechanism could probably > be put together by a good watchmaking shop at no more than the price of > mouse@collatz.mcrcim.mcgill.edu Best regards to all, NB Nicholas Bodley Autodidact & Polymath |*| Keep smiling! It makes | Waltham, Mass. Electronic Technician |*| people wonder what | nbodley@tiac.net Amateur musician |*| you have been up to. | -------------------------------------------------------------------------* From cube-lovers-errors@curry.epilogue.com Sat Jun 1 01:40:27 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id BAA08881 for ; Sat, 1 Jun 1996 01:40:26 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Date: Sat, 1 Jun 1996 01:01:26 -0400 (EDT) From: Nicholas Bodley To: Wei-Hwa Huang Cc: Cube-Lovers@ai.mit.edu Subject: Re: Another subscriber In-Reply-To: <4om779$aip@gap.cco.caltech.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII On 31 May 1996, Wei-Hwa Huang wrote: {Snips} > der Mouse writes: > >In my opinion mechanical designs for the 7 and above will have to be > >fundamentally different from those for the 6 and below, because that's > There's a really simple solution to this. Just don't make the 7 slices > evenly spaced. > -- > Wei-Hwa Huang, whuang@cco.caltech.edu, http://www.ugcs.caltech.edu/~whuang/ _____________________ Large corner cubies, and rectangular edge pieces to match, when seen face-on, right? Has anyone worked out the innards? Perhaps a further extension of the scheme used for the 5^3? (If so, the retaining "foot" on a corner cubie would have a truly wondrous shape! The "foot" of a 5^3 is quite impressive.) I should have read this earlier; it might have saved bitspace from my mad electronic/mechanical scheme! :) It would also be quieter... My best to all, NB Nicholas Bodley Autodidact & Polymath |*| Keep smiling! It makes | Waltham, Mass. Electronic Technician |*| people wonder what | nbodley@tiac.net Amateur musician |*| you have been up to. | -------------------------------------------------------------------------* From cube-lovers-errors@curry.epilogue.com Sat Jun 1 16:37:55 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id QAA12114 for ; Sat, 1 Jun 1996 16:37:54 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com To: Cube-Lovers@AI.MIT.EDU From: Wei-Hwa Huang Subject: Re: realizing 7x7x7 or larger cubes Date: 1 Jun 1996 09:22:10 GMT Organization: California Institute of Technology, Pasadena Lines: 32 Message-Id: <4op242$5mh@gap.cco.caltech.edu> References: Nntp-Posting-Host: accord.cco.caltech.edu X-Newsreader: NN version 6.5.0 #12 (NOV) First, as a comment on the other thread, I think it is safe to say: If one can solve an (2n+1)^3 cube, then one can solve a (2n)^3 cube. "Ronnie B. Kon" writes: >I've had this dream of making cubies which attach (via bars or perhaps >electromagnets) to their neighbors, with the smarts to detect the torque of >a turn and release until the turn has been completed. You could then sell >corner cubies, edge cubies, face cubies, and internal cubies one-at-a-time >and people could build their own puzzles as large as they wanted. It would certainly require a very creative design for the corners; your description seems to say that in the stable state the corners are not attached by anything! Perhaps corner cubies could be equipped with buttons that had to be depressed before a face would turn? For instance, imagine a cube with three faces that have a button on the middle, each one triggering a bar on the opposite side. When a button is pressed, the bar retracts into the cube. This would make a workable corner cube, although it would be a bit awkward to press the face that you wanted to turn! As another aside, I don't understand the rationale behind the canonical 4x4x4 design. It would seem to me that it's better to have two rings of grooves in each dimension, so that the face pieces could have "fatter" legs and not break off as easily. -- Wei-Hwa Huang, whuang@cco.caltech.edu, http://www.ugcs.caltech.edu/~whuang/ --------------------------------------------------------------------------- Caught Porfiry, Raskolnikov sung his swan Sonia when he went Dounia to Siberia. From cube-lovers-errors@curry.epilogue.com Sat Jun 1 16:43:04 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id QAA12140 for ; Sat, 1 Jun 1996 16:43:03 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com To: Cube-Lovers@AI.MIT.EDU From: Wei-Hwa Huang Subject: Yet another (silly) idea on realizing the 7x7x7 Date: 1 Jun 1996 20:38:23 GMT Organization: California Institute of Technology, Pasadena Lines: 8 Message-Id: <4oq9nv$l8m@gap.cco.caltech.edu> Nntp-Posting-Host: accord.cco.caltech.edu X-Newsreader: NN version 6.5.0 #12 (NOV) Hey, if the corners are going to fall off, let them! After all, anyone who actually bothers to buy a 7x7x7 should know how to solve the corners... :) -- Wei-Hwa Huang, whuang@cco.caltech.edu, http://www.ugcs.caltech.edu/~whuang/ --------------------------------------------------------------------------- Caught Porfiry, Raskolnikov sung his swan Sonia when he went Dounia to Siberia. From cube-lovers-errors@curry.epilogue.com Sun Jun 2 04:12:01 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id EAA14161 for ; Sun, 2 Jun 1996 04:12:01 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Date: Sun, 2 Jun 96 00:17:03 EDT From: hoey@aic.nrl.navy.mil Message-Id: <9606020417.AA04623@sun13.aic.nrl.navy.mil> To: Jerry Bryan Cc: Cube-Lovers Subject: Re: Compact Cube Representation for Shamir and Otherwise I'm not sure this is so interesting to all of cube-lovers; e-mail me if you have opinions pro or con. Jerry writes of the standard S24 x S24 model, which uses 48 bytes per position without packing. He also has a "supplement" representation that uses one facelet from each edge and corner, for 20 bytes. He packs them into 13 bytes on tape. The way I did it the last time I worked on brute force was to pack eight twelve-bit fields: The orientations in two twelve-bit fields (2^11 and 3^7), The edge permutation in four twelve-bit fields, each of three base-12 digits (12^3), and The corner permutation in two twelve-bit fields, each of four base-8 digits (8^4). Unpacking the fields can be done with native arithmetic or table lookup. In the latter case, it is better to use 12*11*10 instead of 12^3 and 8*7*6*5 instead of 8^3. Also, postmultiplying by a fixed permutation can be done with table lookup without unpacking. I used this feature for twelve permutations of particular interest. I am somewhat rusty on the implications of using this representation in conjunction with Shamir's algorithm. I think it provides an ordering of the permutations that enables at least an approximation to the random access you need, then you unpack it and do a better job. Dan Hoey Hoey@AIC.NRL.Navy.Mil From cube-lovers-errors@curry.epilogue.com Tue Jun 4 14:07:52 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id OAA04174 for ; Tue, 4 Jun 1996 14:07:51 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Date: Tue, 4 Jun 1996 08:35:17 -0400 (EDT) From: Nicholas Bodley To: Wei-Hwa Huang Cc: Cube-Lovers@ai.mit.edu Subject: Fragile parts in 4^3 In-Reply-To: <4op242$5mh@gap.cco.caltech.edu> Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII On 1 Jun 1996, Wei-Hwa Huang wrote: {Mostly snipped} > As another aside, I don't understand the rationale behind the canonical > 4x4x4 design. It would seem to me that it's better to have two rings of > grooves in each dimension, so that the face pieces could have "fatter" > legs and not break off as easily. > > Wei-Hwa Huang, whuang@cco.caltech.edu, http://www.ugcs.caltech.edu/~whuang/ It probably isn't necessary for the legs to be so thin; the mechanical engineer probably had optimistic estimates of the likely forces and the strength of the particular polymer used. The latter isn't, by any means, cheap stuff. Wider legs might still meet the constraints that ensure the Cube not fall apart. (Sorry for a slow reply.) Regards to all, NB Nicholas Bodley Autodidact & Polymath |*| Keep smiling! It makes | Waltham, Mass. Electronic Technician |*| people wonder what | nbodley@tiac.net Amateur musician |*| you have been up to. | -------------------------------------------------------------------------* From cube-lovers-errors@curry.epilogue.com Tue Jun 4 18:29:55 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id SAA04761 for ; Tue, 4 Jun 1996 18:29:54 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Date: Tue, 4 Jun 1996 18:08:15 -0400 From: Jim Mahoney Message-Id: <199606042208.SAA13307@ marlboro.edu> To: Cube-Lovers Subject: A essay on the NxNxN Cube : counting positions and solving it Thoughts on the NxNxN Cube -------------------------- Lately I've seen a few comments here on the NxNxN cube, how to build one, and how what an algorithm to solve one might look like. I have no idea how the inside mechanics should work, nor have I made significant progress thinking about how to find God's Algorithm for these guys, but I did work out a recipe for solving the NxNxN, way back in 1981, and worked out the number of possible positions. Perhaps given those questions this is the right time to post a synopsis. My approach is rather pedestrian, and I wave my hands quite a lot, but I'm confident that an an arbitrarily large cube can be unscrambled with the method I describe below. But first, since this is the first time I've posted to this discussion group, I guess I should say who I am. I wrote my undergraduate thesis on Rubik's Cube fifteen years ago, as part of a math/physics degree. Ironically, even though I was at MIT and this mailing list is based there, too, I never knew of it until this year, when I started (slowly) reading through the archives of this mailing list. There's a lot there, and not all of it is exactly quick reading. These days I teach physics and astronomy at Marlboro College, where I have occasionally taught an "ideas of group theory" course based on puzzles like TopSpin and Rubik's Cube, designed to show mostly non-math majors why group theory is so pretty. I see from previous posts that other people have taught similar courses. Anyway, what follows is one way of thinking about the NxNxN. Much of this isn't new, but perhaps it hasn't been said quite this way, and will therefore be worth saying. Hope it isn't too long-winded, and that I don't offend the folks who've posted similar things already by not referencing them; as I said, I've swallowed some but not all of the 1500 or so posts to this mailing list. ====================================================================== (I) The Cube Itself ================================================= ====================================================================== First, I will envision the NxNxN Cube (the whole thing) as a solid, transparent array of N^3 cubies (the pieces its made of), most of which are 'hidden' on the inside. (For example, this would imply that the the 3x3x3 cube has a single hidden cubie at the center.) All the real mechanical 3x3x3, 4x4x4, 5x5x5 Cubes that I've seen only have cubies on the outside, but if you can put back all N^3 cubies in the one I'm describing then you can certainly do the real ones. (In Dan Hoey's notation, I believe that this means I treat the Cube as the G+C group, where G is generated by the outer slice rotations, and C is the rotations of the entire thing. I am not including spatial inversions, because I'm physicall obtainable positions here. And while I agree that G is what I see on the 3x3x3, I think that what I describe here is a more general and elegant approach to the NxNxN.) Second, let a 'move' be a rotation of any plane of cubies, including the interior slices. There are 3N of these planes, and each has NxN cubies in it. Since I'm not going to try to count moves here, it doesn't matter whether you consider a half turn as one move or two quarter turns. The slices are numbered from 1 to N, so that rotating the 1-slice or N-slice is a rotation of an outside face, while the (N+1)/2 slice is through the center of an odd Cube. (I chose this convention rather than numbering at zero in the center because it lets me talk about the N'th-layer, as defined below, and the N'th-slice in the same breath without getting myself confused.) Third, I imagine that each of face of each cubie has a distinct color, which I will take as usual to be (Front, Back, Up, Down, Left, Right), and that the unique 'solved' position has all N^3 cubies in the same orientation, with their colors all aligned. ====================================================================== (II) Layers, Orbits, and Types ===================================== ====================================================================== Now I would like to see how many different kinds of cubies there are, where they live, and how they behave. The first thing to notice is that there are two distinct kinds of NxNxN Cubes, depending on whether N is even or odd. For N odd, there is a single cubie at the center (which I will call the 1-layer), surrounded by the cubies on the outside of the 3x3x3 (which I will call the 3-layer), which in turn are surrounded by those cubies on the outside of the 5x5x5 Cube (the 5-layer), and so on until I reach the outermost N-layer. When N is even, the innermost layer is 2x2x2, which is surrounded by a 4x4x4 "4-layer", and so on. Thus the entire Cube is made up of disjoint layers which are either all odd or all even. Moreover, it is easy to see that the cubies on a given layer always stay on that layer; the allowed rotations cycle cubies within a layer but never between layers. Next, I will define any complete set of cubies that can move into each other's position as an "orbit." (This name is at least suggestive of the group theory notion of a closed sequence of elements.) For example, the 8 corner cubies on the 3x3x3 Cube form one orbit since any one of those cubies can be put in any of those eight positions. Likewise, the 12 edge cubies on the 3x3x3 form another orbit. Finally, distinct orbits which have similar properties will be called members of the same "type." For example, the 4x4x4 Cube has an orbit of eight outer corners on the 4-layer, and a second orbit of eight corner cubies on the inside, in the 2-layer. Although these sets of cubies are in distinct orbits, they are both "Corner" types. (I know that my notion of what exactly "similar properties" means is vague here, but I think the general idea is clear.) One approach to solving the cube, then, is to identify each kind of type - it turns out there aren't very many - and find some method of manipulating the cubies in an orbit of that type without disturbing any other the rest of the Cube. I'll explain one way to do this further down, after listing the different types. ====================================================================== (III) The Eight Types ============================================== ====================================================================== Without further ado, here they are. Name What: ------ ------ Central The unique cubie in the center of Cubes with N odd. Corner Corners in each layer. In each layer there is 1 corner orbit consisting of 8 cubies, each of which can be in 3 orientations in each of 8 positions. (8 positions x 3 orientations = 24 total.) However, while all 8! position rearrangements are permissible, all rotations are not; as is well known, only 1/3 of them are. One way to see this is to define "twist" state as (0,1,2) for each orientation of a cubie at a corner, and to notice that the sum of all these states isn't changed by a single move. This means that you cannot turn just one corner in place. Edge-Single The ones like the outer edges on the 3x3x3. In each odd layer there is 1 of these orbits, consisting of 12 cubies, each of which can be in 2 states. (12 positions x 2 orientations = 24 total.) All 12! placements are accessible, but again only some of the flips; you cannot turn just one edge. Face-Center The cubies like the centers of the 3x3x3 face. In each odd layer there is on of these orbits, which has 6 cubies each of which can be in 4 rotation states. (12 places x 4 states each = 24 total.) This time all rotations are possible; however, the cubies can only move in space as a rigid whole, and therefore there are 24 different positions for these cubies, which are completely determined by the orientation of the central cubie. Edge-Double, Face-Corner, Face-Edge, Face-Offset Each of these orbits consists of exactly 24 cubies, as shown in the pictures below. There are in general many of each of these orbits in each layer, as given by the formulae (simple geometry and counting - see the diagrams) in the table below. *None* of these cubies in these orbits can "flip" or "twist" in place like the Corners and Single Edges do; in every case there are exactly 24 cubies which implies that there must be only one orientation at each possible position. Another way to see this is to draw in an orientation on each cubie of a given orbit, with arrows, and then show that no possible move changes the positions of the arrows. Here's a summary of the specs for each type. Note that the "number of positions" given is for both only one parity, that is, for both an even or odd number of quarter turns, and ignoring the all other orbits. The number of positions of the whole Cube is *not* a simple product of all these numbers; the parities of different orbits must agree. More on this later. As usual, I use "!" and "^" for factorial and "raise-to-the-power-of", i.e. 8!=8*7*6*5*4*3*2*1 and 3^7=3*3*3*3*3*3*3. - Types -- ( n = which layer ) ------------------------------------ Name # of # of orbits per layer. # of positions per orbit cubies (n odd) (n even) (both even/odd parity) ------ ------ ----------- ------- --------------------- n=1 Central 1 1 0 24 n>1 Corner 8 1 1 (3^7) 8! Edge-Single 12 1 0 (2^11) 12! Face-Center 6 1 0 (4^6) n>3 Edge-Double 24 (n-3)/2 (n-2)/2 24! Face-Corner 24 (n-3)/2 (n-2)/2 24! Face-Edge 24 (n-3)/2 0 24! n>5 Face-Offset 24 (n-3)(n-5)/4 (n-2)(n-4)/4 24! --------------------------------------------------------------------- It's also convenient to define "h" and "H" such that h = n/2 (n even); H = N/2 (N even) h = (n-1)/2 (n odd); H = (N-1)/2 (N odd) which makes the counting a bit easier. "h" stands for "half", and is the number of the slice just before the center slice, if there is a center slice. With this "h", the expressions for the number of orbits per layer are much simpler, namely Name # of orbits per layer ----------- ---------------- Double Edge h-1 Face-Corner h-1 Face-Edge h-1 Face-Offset (h-1)(h-2) And now for the pictures. This is much easier to visualize in 3D with real drawings, but I'll do what I can with ASCII. The smallest layer n that contains all the distinct types (except the central cubie, of course) is n=7, so I've drawn in one outside (n=7) plane of a 7x7x7 Cube below and sketched in where they live. You can either think of this as the outer-most layer of an N=7 Cube, or part of an inner n=7 slice of a larger Cube. The slices (rows and columns in the pictures) can be numbered either left to right or right to left, so when I refer to the "n-slice" I also mean the "(N+1-n)-slice" where N=(size of entire Cube)=7 here, and 1<=n<=N is a particular slice. I also note to the right of each picture which slice rotations can disturb the cubies in that orbit, and whether a quarter turn of that kind of move gives an even or odd permutation of the cubies. ("even" or "odd" refers to how many pairwise swaps it takes to get that permutation. A cycle of 2 cubies is odd, a cycle of 3 cubies is even, and a cycle of 4 cubies is odd. For example, it takes an even number of moves to corner number 1 to corner 2's place, corner 2 to 3's place, and corner 3 to 1's place.) These parities will be discussed further in the next section. Where there is more than one possible orbit I have used the labels "p" and "q" to specify which one is shown. The letter "H" (described above) is in this case (with N=7) H = (N-1)/2 = 3. 7 6 5 4 3 2 1 1 2 3 4 5 6 7 ----------------------- | 7 1| C . . . . . C n-Corner | 6 2| . . . . . . . ( N >= n >= H ) | 5 3| . . . . . . . Moved By Parity | ------- ------ 4 4| . . . . . . . | n-slice odd 3 5| . . . . . . . | 2 6| . . . . . . . | 1 7| C . . . . . C 1 2 3 4 5 6 7 ----------------------- | 1| . . . ES . . . n-Edge-Single | 2| . . . . . . . ( N >= n >= H ) | 3| . . . . . . . Moved By Parity | -------- ------ 4| ES . . . . . ES | n-slice odd 5| . . . . . . . | (H+1)-slice odd 6| . . . . . . . | (Note that H+1 is the slice 7| . . . ES . . . through the center.) 1 2 3 4 5 6 7 ----------------------- | 1| . . . . . . . n-Face-Center | 2| . . . . . . . Moved By Parity | -------- ------ 3| . . . . . . . center (H+1)-slice odd | 4| . . . FC . . . | Rotated By 5| . . . . . . . --------- | n-slice "odd" 6| . . . . . . . | 7| . . . . . . . 1 2 3 4 5 6 7 ----------------------- | 1| . ED . . . ED . n-p-Edge-Double | 2| ED . . . . . ED ( 1 < p <= H; p=2 shown here.) | 3| . . . . . . . Moved By Parity | -------- ------ 4| . . . . . . . n-slice even (2 4-cycles) | p-slice odd (1 4-cycle) 5| . . . . . . . | (The "p-slice" referred to here 6| ED . . . . . ED and in the next figures cuts into the | Cube in the 3rd dimension not shown 7| . ED . . . ED . into the paper. The "n-slice" move turns this diagram by 90 degrees.) 1 2 3 4 5 6 7 ----------------------- | 1| . . . . . . . n-p-Face-Corner | 2| . FC . . . FC . ( 1 < p <= H; p=2 shown here.) | 3| . . . . . . . Moved By Parity | -------- ------ 4| . . . . . . . n-slice odd (4 cubies move) | p-slice even (8 cubies move) 5| . . . . . . . | 6| . FC . . . FC . | 7| . . . . . . . 1 2 3 4 5 6 7 ----------------------- | 1| . . . . . . . n-p-Face-Edge | 2| . . . FE . . . ( 1 < p <= H; p=2 shown here.) | 3| . . . . . . . Moved By Parity | -------- ------ 4| . FE . . . FE . n-slice odd (4) | p-slice odd (4) 5| . . . . . . . (H+1)-slice even (8) | 6| . . . FE . . . | 7| . . . . . . . 1 2 3 4 5 6 7 ----------------------- | 1| . . . . . . . n-p-q-Face-Offset | 2| . . FO . x . . ( 1 < p <= H; p=2 shown here; | 1 < q <= H; q=3 shown here; 3| . x . . . FO . p not equal to q. ) | 4| . . . . . . . Moved By Parity | -------- ------ 5| . FO . . . x . n-slice odd | p-slice odd 6| . . x . FO . . q-slice odd | 7| . . . . . . . The x's are *not* part of this orbit, but mark a distinct mirror-image orbit. There are no moves which will bring one of the FO's to one of the x's. Every cubie on any size NxNxN Cube fits one of these patterns. For large values of N, nearly all the cubies are in Face-Offset orbits. ====================================================================== (IV) Parity and the Total Number of NxNxN Positions ================== ====================================================================== I'd guess most of you who read this will already understand parity considerations, but let me run through it quickly anyway. The basic idea is that any permutation of a group of symbols can be broken into a sequential set of pair exchanges, and the number of these pair exchanges, even or odd, determines the parity of the permutation, even or odd. Thus if ABCDE is re-arranged into BAECD, then this rearrangement is odd because it requires three pair swaps, and three is odd: (1) swap AB to BA in the original, (2) swap D and E to get the D at the end, (3) swap the E and C. There are other swap sequences, but all are odd. Any slice rotation which cycles four cubies ABCD into BCDA puts those cubies into an odd permutation since it would require three pair swaps (AB, AC, AD) to change one into the other. The point all this is that different orbits must have consistent parities. The best known Cube example is that on the 3x3x3 cube, a quarter turn of any outside slice changes the parity of *both* corners and edges; therefore, positions which have the corners and edges in different parities are impossible, and therefore one cannot exchange two corners without exchanging two edges somewhere, too. On the NxNxN things are a bit messier. Any given slice rotation will change the parity of some orbits, and leave many others unchanged. Most choices of arbitrary placements of cubies won't be a possible cube position, not only because of the "twist" of the corners and "flip" of the edges (described above briefly) but also because the parity of all the orbits must be consistent. Here's one way to do it. Since each slice n>1 moves exactly one Corner orbit, and since the n=1 slice moves the Central cubie, I can use the N Corner/Central orbits to *define* the parity of each slice. Then the parity of all other orbits is fixed, and each has available exactly one-half (only one parity) of its total number of positions as given in the table Another way of doing the same thing is to count the only one parity, that is 1/2 of *all* the orbits, and then multiply by 2^N as the number of ways to choose the parity on each of N slices. So I can now calculate the total number of available positions T(N) of the NxNxN cube by multiplying the number of possible positions of each orbit, taking into account how many orbits there are in each layer, over all N layers, and keeping the parity in agreement. (Note that here I *am* including a rotation of the entire cube as a new "position". To take out this factor, divide what follows by 24. Note also that I'm distinguishing between different rotations of the face centers, which is also a bit different from what is usually done.) The counting is a pretty straightforward. Using the initials of the types as abbreviations, the total number of each type of orbit is: N odd: #1 = 1 #C = (N-1)/2 #ES = (N-1)/2 #FC = (N-1)/2 #ED = Sum n= (3, 5, 7, 9, ..., N) of { (n-3)/2 } = (N-1)(N-3)/8 #FC = same as #ED #FE = same as #ED #FO = Sum n= (3, 5, 7, 9, ..., N) of { (n-3)(n-5)/4 } = (N-1)(N-3)(N-5)/24 N even: #1 = 0 #C = N/2 #ES = 0 #FC = 0 #ED = Sum n= (2, 4, 6, 8, ..., N) of { (n-2)/2 } = (N)(N-2)/8 #FC = same as #ED #FE = 0 #FO = Sum n= (2, 4, 6, 8, ..., N) of { (n-2)(n-4)/4 } = (N)(N-2)(N-4)/24 And so --------- | T(N) = Total Positions of NxNxN cube, all orientations, | all N^3 cubies | | = 2^N (24/2)^#1 (3^7 8!/2)^#C (2^11 12!/2)^#ES (4^6/2)^#FC | * (24!/2)^(#ED+#FC+#FE+#FO) ----------------------------- which simplifies to either T(N odd)= 24 [3^7 8! 2^10 12! 4^6/2]^((N-1)/2) [24!/2]^( (N-1)(N-3)(N+4)/24 ) = 24 [ 8.85801e22 ]^((N-1)/2) [3.102242e23]^((N-1)(N-3)(N+4)/24) = (44.9) (0.0561)^N (9.52)^(N^3) or T(N even) = [3^7 8!]^(N/2) [24!/2]^( N(N-2)(N+2)/24 ) = (8.817984e7)^N (3.102242e23)^(N(N-2)(N+2)/24) = (1.14)^N (9.52)^(N^3) which is an awful lot of positions no matter how you look at it. Note that usually the number of 3x3x3 positions is given without the factor of 24 (spacial rotations) or the 4^6/2 (face center rotations), which leaves (3^7 8! 2^10 12!) = 4.3e19. For large N, T(N) is dominated by the (24!/2)^(N^3/24) term, which is about 9.524^(N^3), which implies that for very big cubes, each of the N^3 cubies acts as if it has nearly 10 independent places it can be. (Oh, and the notation 1.23e4 means 1.23 10^4 = 1230.) ===================================================================== (V) Solving It ===================================================== ===================================================================== Here's where the handwaving really gets going. What I describe here is closer to an outline of a method than a real algorithm, but you can probably fill in the details yourself. The basic idea is to use a general-purpose idea for cycling three cubies of any given orbit without disturbing any other part of the Cube. A small variation on this same theme can be used to twist two corners or single edges in the place, too. If I can do this for every type, then all I have to do to solve the whole cube is the following. -- An NxNxN Recipe ------------------------------------------ (A) If N is odd, orient the Central cubie correctly, and at the same time, turn each Face-Center so that it is aligned with the Central cubie. (If you can't see the orientations of the Face-Centers, as is usually the case on the typical 3x3x3, then just skip that step and move on.) (B) For each layer, examine the parity of corresponding Corner orbit. If its parity is odd, make one arbitrary 1/4 turn rotation on that layer; otherwise, don't move it. At this point all the Corner orbits have even parity, and therefore *all* the orbits have even parity. (C) And finally, I have these nested loops: (i) Starting at the innermost layer and working outward, (ii) on each orbit in that layer, (iii) 1. Restore each cubie of that orbit to its proper position with the 3-cycle technique described below. Since all the orbits already have even parity, these even-move combinations are enough to restore everything to their proper places. 2. If the current orbit is a Corner or Edge-Single, then once the cubies are in the right places apply the "twist" and "flip" operations described below to orient them correctly. That's it. Now all I have to do is describe three tricks, (1) how to cycle three cubies on any given type, (2) how to twist two corners in opposite directions, and (3) how to flip two Edge-Singles, all without disturbing anything else. All these tricks are well known, I think. And there are certainly many, many other tricks; however, these are the simplest that I know of that can be generalized to any size Cube. Moreover, they have the nice property that you can actually "think" your way through them without actually needing to memorize a long sequence of moves. ===================================================================== (VI) How to Cycle Three Cubies ===================================== ===================================================================== The basic idea is to find a move sequence that will (1) take a chosen cubie off from its "hot seat" on a chosen slice *without* (here's the trick) disturbing any other cubie on that slice. The rest of the cube can be completely scrambled by this operation. Then (2) rotate the chosen slice, (3) undo step (1), putting the original cubie back into its original slice and undo whatever changes were made to the other cubies, and (4) undo step 2. The sequence always of the form A R A' R' where "A" is step 1, "R" is a rotation of a single slice, and the ' mark means, as usual, the inverse operation. Here's a detailed example, using the Corner orbit of a 3x3x3 cube, with the top layer as the "chosen slice" and the cubie marked "1" in the unfolded sketch of a cube below as the focus of attention. In eight moves the cubies in locations 1, 2, and 3 will trade places. The starting position: U a - 1 - 2 - d - (a,1,2,d,e,3,g,h) are a Corner orbit. | L | F | R | B e - 3 - g - h - (U, D, L, R, F, B) are the possible D clockwise rotations. (1) Get "b" off the chosen slice, without disturbing any other cubie on that slice. Replace it with the cubie that you want to put in its place. e - a - 2 - d - -> L -> | | | | 3 - 1 - g - h - e - a - 2 - d - -> D -> | | | | h - 3 - 1 - g - a - 3 - 2 - d - -> L' -> | | | | After L D L' e - h - 1 - g - The top layer was (a,b,c,d); now it is (a,f,c,d). "b" has been taken off the top slice, and "f" is in its place. (2) Rotate the chosen slice to place a new cubie in the hot seat. 3 - 2 - d - a - -> U -> | | | | After (L D L') U e - h - 1 - g - (3) Undo step 1, which pops the chosen cubie "b" back to its original slice, *and* (here's the key part), restore (nearly) all other cubies to their original locations, since none of the disturbed ones were on the slice that rotated in step (2). 3 - 1 - d - a - -> L D' L' -> | | | | After (L D L') U (L D' L') e - 2 - g - h - (4) Undo step 2, restoring the chosen slice back to its original position. a - 3 - 1 - d - -> U' -> | | | | After (L D L') U (L D' L') U' e - 2 - g - h - So the move sequence to cycle corners (1,2,3) is simply (L D L') U (L D' L') U' (reading left to right). With a few extra moves before this sequence (which should be undone afterwards) to arrange the cubies which should be moved into the places which are actually modified by this operation (or a similar one), this trick and its variations can be used to put back all 8 corners into their proper places. And with a bit of exploration, this same idea can be used to cycle three cubies of any type, in any orbit, on any layer, without disturbing anything else. For the Edge-Singles on the 3x3x3, for example, to bring an edge off the top slice without disturbing anything else on top, step (1) can be S D S', where "S" vertical is a rotation of a center slice. Or it could be F H F', where "H" is a horizontal rotation of of center slice parallel to the top and bottom; either works. I have actually tried this for all eight of the types of orbits, and it does indeed work. Yes, I know this is pure handwaving, but this essay is already long enough, and it really is pretty straightforward once you get the idea. ===================================================================== (VII) Turning Corners and Flipping Edges =========================== ===================================================================== I don't think I need to say too much about this, because basically the same tricks that work on the 3x3x3 Cube will work on the orbits in the NxNxN. My usual approach is to find a sequence that will bring a corner or single-edge cubie out of its position, and then back with a turn or flip, *without* changing any cubie on on slice. Call that entire operation "A". Then just like before, A R A' R' where "R" is a rotation of the slice which was left (nearly) unchanged will restore all the parts of the Cube which were messed up by A, and leave only two corners or two edges turned or flipped. For example, on the 3x3x3, this sequence will turn two corners. [(L DD L') (F' DD F)] U [(F DD F') (L' DD L)] U' The stuff in the brackets brings a corner cubie off the top (up) slice, and brings it back with a twist. If "H" is a clockwise (as viewed from the top) quarter turn on the horizontal center slice of the 3x3x3 (the plane parallel to the top and bottom), then this similar sequence will flip two edges. [ (L HH L') U' (F' HH F) U ] U [ U' (F HH F') U (L' HH L) ] U' Again, the moves in the brackets bring one of the 3x3x3 edge cubies off the top layer, and bring it back with a twist. One can also combine several different 3-cycles from section VI to twist and flip the corners and edges. ===================================================================== (VII) Comments =================================================== ===================================================================== Well, that turned out to be a lot longer than I'd planned. If anyone has actually bothered to read down this far, I hope it was worthwile. I think that when all is said in done, the 3x3x3 is by far the most interesting of the sizes. All the new types of orbits on the larger Cubes are fairly boring, actually, since none of the cubies can be flipped or turned in place the way that the 3x3x3 corners and edges can. And I confess that I like how the only cubie on the 3x3x3 that you can't see - the one that I like to imagine is hidden in the center - is the only one that's completely specified by the locations of the other orbits, namely by the positions of the face centers. When I was first working out this stuff, back in 1981, I built a 7x7x7 Cube out of colored dice. None of them were "stuck" to the others; it was just stack of dice. Manipulating it was pure hell, but I could usually squeeze a layer and carefully turn and put it back. One slip and I had dice all over the room. I have a few other ideas kicking around in the back of my head, but they'll have to wait for another time, and another note. Regards, Dr. Jim Mahoney mahoney@marlboro.edu Physics & Astronomy Marlboro College, Marlboro, VT 05344 From cube-lovers-errors@curry.epilogue.com Wed Jun 5 02:48:31 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id CAA05776 for ; Wed, 5 Jun 1996 02:48:30 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Date: Wed, 5 Jun 1996 09:33:41 +0300 (IDT) From: Rubin Shai X-Sender: s2394459@csc To: Cube-Lovers@ai.mit.edu Subject: Computer representation to the cube. Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Hi all I'm looking for an easy to implementation / easy to debug / easy to print / and most important chip to manipulate (in computer time) for the 3X3X3 cube. Does anyone heard of somthing? Shai From cube-lovers-errors@curry.epilogue.com Wed Jun 5 15:28:47 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id PAA07399 for ; Wed, 5 Jun 1996 15:28:46 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Date: Wed, 05 Jun 1996 10:22:31 -0500 (EST) From: Jerry Bryan Subject: Re: A essay on the NxNxN Cube : counting positions and solving it In-Reply-To: <199606042208.SAA13307@marlboro.edu> To: Cube-Lovers Message-Id: Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Content-Transfer-Encoding: 7BIT On Tue, 4 Jun 1996, Jim Mahoney wrote: It's going to take a while to absorb your whole note, but I do have a couple of quick comments/questions. > Next, I will define any complete set of cubies that can move into each > other's position as an "orbit." (This name is at least suggestive of > the group theory notion of a closed sequence of elements.) For > example, the 8 corner cubies on the 3x3x3 Cube form one orbit since > any one of those cubies can be put in any of those eight positions. > Likewise, the 12 edge cubies on the 3x3x3 form another orbit. This has been discussed before on Cube-Lovers, but I am still puzzled or curious about the usage of the word "orbit". Your definition is consistent with the usage advocated by Martin Schoenert on Cube-Lovers. For example, Martin talked about the corner orbit, the edge orbit, and the face center orbit of the 3x3x3. (I suppose for completeness, we should include in this list of orbits the orbit for the invisible center of the whole 3x3x3 cube.) David Singmaster, on the other hand, has always talked about the twelve orbits of the constructable group of the 3x3x3, where orbits are defined in terms of twists, flips, and parity. Depending on what you mean by "closed sequence of elements", your definition may be consistent with Singmaster's usage as well. That is, Singmaster's orbits are certainly closed. However, Martin says that Singmaster's orbits should be called cosets. Secondly, if my understanding of your model is correct, you are treating positions as distinct which cannot be distinguished with normal coloring of a physical cube (even an imaginary physical cube for large N). The issue appears as early as the 4x4x4, and persists for larger values of N. I don't necessarily disagree with your treatment. Indeed, it makes the cube theory tenable. Otherwise, your model tends to become a coset model rather than a group model. But I wondered if my understanding of your model is correct? There are several implications of how you treat visibly indistinguishable positions. For example, it impacts your counts of how many positions there are. For another example, it impacts your solutions (e.g., "invisible" incorrect parity on the 4x4x4. "Invisible" bad parity can also occur on the 3x3x3 if you remove the face center color tabs. A slice move will give the edges and corners opposite parity that is not visible.) Perhaps you could discuss these issues with respect to your model. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) jbryan@pstcc.cc.tn.us Pellissippi State (423) 539-7127 10915 Hardin Valley Road (423) 694-6435 (fax) P.O. Box 22990 Knoxville, TN 37933-0990 From cube-lovers-errors@curry.epilogue.com Wed Jun 5 15:27:15 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id PAA07391 for ; Wed, 5 Jun 1996 15:27:14 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Date: Wed, 5 Jun 1996 09:30:21 +0100 Message-Id: <96060509302127@glam.ac.uk> From: VANESSA PARADIS WANTS ME To: CUBE-LOVERS@ai.mit.edu X-Vms-To: RUBIKCUBE To have a bit more challange when doing the cube, complete it so that each horizontal slice, is 1 turn (quarter of a full circle) out of place. Therefore, top and bottom face are one colour, but all side faces contain 3 colours. Boy, is it hard! By the way, I can do the cube in 1 minute 26 seconds. How does that compare with everyone else! (P.S. Please be fair, I`ve only been doing it for 6 weeks or so!) Chris CMAGGS@GLAM.AC.UK From cube-lovers-errors@curry.epilogue.com Wed Jun 5 15:27:57 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id PAA07395 for ; Wed, 5 Jun 1996 15:27:56 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Date: Wed, 5 Jun 1996 09:53:36 -0400 Message-Id: <199606051353.JAA21055@chara.BBN.COM> From: Allan Wechsler To: s2394459@cs.technion.ac.il Cc: Cube-Lovers@ai.mit.edu In-Reply-To: (message from Rubin Shai on Wed, 5 Jun 1996 09:33:41 +0300 (IDT)) Subject: Re: Computer representation to the cube. Reply-To: awechsle@bbn.com Date: Wed, 5 Jun 1996 09:33:41 +0300 (IDT) From: Rubin Shai Hi all I'm looking for an easy to implementation / easy to debug / easy to print / and most important chip to manipulate (in computer time) for the 3X3X3 cube. Does anyone heard of somthing? Shai Ani lo y'khol lavin et-ha-anglit shelkha. Are you looking for a program? An algorithm? A piece of hardware? A circuit diagram? And I apologize in advance if I got your gender wrong in my wretched Hebrew. -A From cube-lovers-errors@curry.epilogue.com Wed Jun 5 19:50:32 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id TAA07826 for ; Wed, 5 Jun 1996 19:50:31 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Message-Id: Mime-Version: 1.0 Content-Type: text/plain; charset="us-ascii" Date: Wed, 5 Jun 1996 18:02:49 -0500 To: CUBE-LOVERS@ai.mit.edu From: Kristin Looney Subject: fastest hands in the midwest... > By the way, I can do the cube in 1 minute 26 seconds. > How does that compare with everyone else! 37.72 won me the midwest championship, my best official time was 35.30 seconds which placed me 5th in the country. I think it was 1981. Now? I don't get timed very often, but it's still usually under a minute. I guess it is like riding a bicycle. Anyone else on this list from those contest days? Minh Thai - are you out there? How about Jeff Verasono? or David P. Conrady? I've often wondered what that crazy guy with the bright maroon hair ended up doing with his life... Kristin (used to be Wunderlich) Looney kristin@tsi-telsys.com From cube-lovers-errors@curry.epilogue.com Wed Jun 5 19:50:57 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id TAA07832 for ; Wed, 5 Jun 1996 19:50:56 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Date: Wed, 5 Jun 96 18:32:29 EDT From: hoey@aic.nrl.navy.mil Message-Id: <9606052232.AA22039@sun34.aic.nrl.navy.mil> To: Nicholas Bodley , Wei-Hwa Huang Cc: Cube-Lovers@ai.mit.edu Subject: Fragile parts in 4^3 On 1 Jun 1996, Wei-Hwa Huang wrote: {Mostly snipped} > As another aside, I don't understand the rationale behind the canonical > 4x4x4 design. It would seem to me that it's better to have two rings of > grooves in each dimension, so that the face pieces could have "fatter" > legs and not break off as easily. If the center pieces had one leg each (instead of a 1/4-leg) you would have _one_ groove around each equator (instead of _half_ a groove). Remember, it's important that the inner sphere stay in sync with at least one of the sets of face centers so that after you've finished the turn you will be able to turn in an orthogonal direction. I don't know how that would work with the turns of the face. You might need a switch that looks kind of like the following where two equators meet: I I * * * I * * ============O===============O======== * I * I * I * * I * * I * I * I * O O I * * I * * * I where the legs live at the "O" positions when a turn is not in progress. But this looks dangerous to me; I think there is a lot of potential for derailment. Dan Hoey Hoey@AIC.NRL.Navy.Mil From cube-lovers-errors@curry.epilogue.com Wed Jun 5 19:51:48 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id TAA07836 for ; Wed, 5 Jun 1996 19:51:47 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Date: Wed, 5 Jun 96 18:54:44 EDT From: hoey@aic.nrl.navy.mil Message-Id: <9606052254.AA22046@sun34.aic.nrl.navy.mil> To: Jim Mahoney Cc: Cube-Lovers Subject: Re: A essay on the NxNxN Cube : counting positions and solving it > All the > real mechanical 3x3x3, 4x4x4, 5x5x5 Cubes that I've seen only have > cubies on the outside, but if you can put back all N^3 cubies in the > one I'm describing then you can certainly do the real ones. > (In Dan Hoey's notation, I believe that this means I treat the Cube as > the G+C group, where G is generated by the outer slice rotations, and > C is the rotations of the entire thing.... Actually, the distinction between G and G+C is that in the latter we draw a distinction between cubes that differ by a whole-cube move as different. When we take account of the internal cubies I call it the "Theoretical Invisible cube", described in my Invisible Revenge article 9 August 1982. A solution method is given in Eidswick, J. A., "Cubelike Puzzles -- What Are They and How Do You Solve Them?", 'American Mathematical Monthly', Vol. 93, #3, March 1986, pp. 157-176. that is pretty much like yours, I think. As for counting the positions, I haven't got around to checking the numbers in "Groups of the larger cubes", 24 Jun 1987. You might want to see how they compare to yours. Dan Hoey Hoey@AIC.NRL.Navy.Mil From cube-lovers-errors@curry.epilogue.com Wed Jun 5 19:52:19 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id TAA07842 for ; Wed, 5 Jun 1996 19:52:17 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Date: Wed, 5 Jun 96 19:27:12 EDT From: hoey@aic.nrl.navy.mil Message-Id: <9606052327.AA22049@sun34.aic.nrl.navy.mil> To: Jerry Bryan Cc: Cube-Lovers Subject: Re: A essay on the NxNxN Cube : counting positions and solving it Jerry remarks: > For example, Martin talked about the corner orbit, the edge orbit, and the > face center orbit of the 3x3x3.... > David Singmaster, on the other hand, has always talked about the twelve > orbits of the constructable group of the 3x3x3, where orbits are defined > in terms of twists, flips, and parity.... When a group G has a representation as permutations of a set X, the orbits are the equivalence classes of X induced by x~y if a (x)g=y for some g in G. But these orbits will be different depending on the representation, and in particular depending on X. If we represent the Rubik group as the usual permutations on cubies and facies, the orbits are corners, edges, etc. as Martin uses. I agree this is the usual kind of orbit to talk about. If we represent the Rubik group as permutations on itself (I think it's called the right regular representation) you get one orbit. This is always true of the right regular representation, since for any f, g in G, let h=f'g, and we have (f)h = g, so f~g. But consider the constructible group C, the set of positions you can get by taking the cube apart and putting it back together. We can extend the right regular representation to a representation on C. In this case, there are twelve orbits of mutually accessible positions. This is Singmaster's usage. They are indeed the cosets of C/G, as with any subgroup of a larger group. But the fact that we usually do not consider the group structure of C (as in taking products of reassemblies) militates against calling them cosets, so I can understand why Singmaster might prefer orbits. But we have to remember to disambiguate which kind of orbit we are talking about. Dan Hoey Hoey@AIC.NRL.Navy.Mil From cube-lovers-errors@curry.epilogue.com Wed Jun 5 22:17:39 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id WAA08111 for ; Wed, 5 Jun 1996 22:17:38 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Date: Wed, 5 Jun 1996 16:47:13 -0400 Message-Id: <199606052047.QAA21223@chara.BBN.COM> From: Allan Wechsler To: cmaggs@glam.ac.uk Cc: CUBE-LOVERS@ai.mit.edu In-Reply-To: <96060509302127@glam.ac.uk> (message from VANESSA PARADIS WANTS ME on Wed, 5 Jun 1996 09:30:21 +0100) Reply-To: awechsle@bbn.com Date: Wed, 5 Jun 1996 09:30:21 +0100 From: VANESSA PARADIS WANTS ME To have a bit more challange when doing the cube, complete it so that each horizontal slice, is 1 turn (quarter of a full circle) out of place. Therefore, top and bottom face are one colour, but all side faces contain 3 colours. I'm not sure I understand this modified goal. Isn't this achieved by solving the cube aas usual, and then giving the top and bottom faces a clockwise quarter twist? Then the top and bottom are solid, and the sides are tricolor horizantal stripes. Even if I haven't understood the goal position, solving for any achievable position is not in principle harder than solving for any other. There might be perceptual problems, but surely these would go away after a little practice, no matter what the goal configuration. -A From cube-lovers-errors@curry.epilogue.com Wed Jun 5 22:18:25 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id WAA08115 for ; Wed, 5 Jun 1996 22:18:24 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com Message-Id: <199606060031.UAA16994@orbit.flnet.com> From: Chris and Kori Pelley To: CUBE-LOVERS@ai.mit.edu Subject: RE: Contest days Date: Wed, 5 Jun 1996 20:31:12 -0400 X-Msmail-Priority: Normal X-Priority: 3 X-Mailer: Microsoft Internet Mail 4.70.1080 Mime-Version: 1.0 Content-Type: text/plain; charset=US-ASCII Content-Transfer-Encoding: 7bit >Anyone else on this list from those contest days? Minh Thai - are you >out there? How about Jeff Verasono? or David P. Conrady? I've often >wondered what that crazy guy with the bright maroon hair ended up >doing with his life... I was in the contests. First place in Peoria, IL with 48 seconds or so. Then I won 5th place in Chicago with 47ish seconds, 3rd in St. Louis with 46 seconds. I was only 13 at the time and it was just a thrill to be there with so many other avid cube-solvers. I still have my official Ideal "Cubists Do It Faster" T-shirts. From the St. Louis contest, which was held in a large outdoor mall, I managed to come away with one of the stage props which was a giant cardboard cube about 3 feet per side. I used it as a table for a few years, then it finally collapsed. Chris Pelley ck1@flnet.com From cube-lovers-errors@curry.epilogue.com Wed Jun 5 22:19:02 1996 Return-Path: cube-lovers-errors@curry.epilogue.com Received: from curry.epilogue.com (localhost [127.0.0.1]) by curry.epilogue.com (8.6.12/8.6.12) with SMTP id WAA08119 for ; Wed, 5 Jun 1996 22:19:01 -0400 Precedence: bulk Errors-To: cube-lovers-errors@curry.epilogue.com To: Cube-Lovers@AI.MIT.EDU From: Wei-Hwa Huang Subject: Re: fastest hands in the midwest... Date: 6 Jun 1996 02:06:31 GMT Organization: California Institute of Technology, Pasadena Lines: 14 Message-Id: <4p5ef7$cvj@gap.cco.caltech.edu> References: Nntp-Posting-Host: accord.cco.caltech.edu X-Newsreader: NN version 6.5.0 #12 (NOV) Kristin Looney writes: >Anyone else on this list from those contest days? Minh Thai - are you >out there? How about Jeff Verasono? or David P. Conrady? I've often >wondered what that crazy guy with the bright maroon hair ended up >doing with his life... I talked with Minh Thai some t