Date: 4 May 1982 00:15-EDT From: Alan Bawden Subject: The Archives To: CUBE-LOVERS at MIT-MC Those of you who look through the archives of old Cube-Lovers mail will notice that I have split off a new section of the archive. The mail now lives in: MC:ALAN;CUBE MAIL0 ;oldest mail in foward order MC:ALAN;CUBE MAIL1 ;next oldest mail in foward order MC:ALAN;CUBE MAIL2 ;more of same MC:ALAN;CUBE MAIL3 ;still more of same MC:ALAN;CUBE MAIL ;recent mail in reverse order (I have also corrected a problem with the tabs in MAIL0 so that the diagrams contained therein are readable once again.)  Date: 10 May 1982 0004-EDT From: ROBG at MIT-DMS (Rob F. Griffiths) To: cube-lovers at MIT-MC Subject: Games magazine... Message-id: <[MIT-DMS].231410> There is a conteset in the cureent issue of Games magazine called Air Race. Its basically modeled after the type of puzzle where you try to run through once and rack up as many points as possible without retracing your steps. I am wondering if nay of you out there subscribeto Games, and are trying the contest? What types of scores are you achieving? Any info appreciated. (I would like to try to find out if my scores are higher/lower than average) Thanks; -Rob.  Date: Thursday, 13 May 1982, 14:48-EDT From: Bernard S Greenberg Subject: 4 x 4 x 4 To: alan at MIT-MC, dcp at MIT-MC, cube-lovers at MIT-MC 4 x 4 x 4's are here. Games People Play on Mass Ave sell's 'em at $15.00 apiece, called "Rubik's Revenge". I have two here (at Symbolics). No progress yet, other than the obvious corner moves.  Date: 13 May 1982 23:44-EDT From: David C. Plummer Subject: 4 x 4 x 4 To: BSG at SCRC-TENEX cc: ALAN at MIT-MC, DCP at MIT-MC, CUBE-LOVERS at MIT-MC Indeed, I have one also. I thought about solving it for a couple of months now. Never wrote anything down, I think I just have a working knowledge of the types of transforms needed. It took me about 1.5 hours to solve it. I thought I had ALL the necessary tools, but there is still one tool that alludes me. It happens half the time, and when I get it, I try to mess it up just enough to allow it to be solved again. As ACW pointed out, most intuition about edges and centers is WRONG. The first thing I tried to the original solved cube was the extended Pons. Much to my dismay, it didn't work. So if you think you know how to extend your 3x3x3 transforms to the 4x4x4, put your fingers where your mouth is!!  Date: 14 May 1982 0413-edt From: Ronald B. Harvey Subject: 4 x 4 x 4 To: cube-lovers @ mit-mc I just got mine (two) today. I had a friend who was in Boston on a business trip get them (to Phoenix) for me. They are also available at The Name of the Game in Faneuil Hall-Market Square also for $15 (14.99). I spent a few hours playing this evening, wondering what tools others have come up with. I seem to be able to get three faces which share a vertex into perfect shape, and then kind of falter. Oh well, it will have to wait until after work tomorrow (Friday). - Ron  Date: 14 May 1982 14:35-EDT From: Alan Bawden Subject: cube talk To: CUBE-LOVERS at MIT-MC Date: 14 May 1982 13:53-EDT From: Alias for WGD Sender: ___100 To: ALAN cc: CWH Re: cube talk There is a talk at Northeastern on the group of the cube this Monday by Ramshaw. I don't recall the details wrt to time and place but there is a notice on the seminar bulletin board on the 2nd floor of building 2 (last corridor).  Date: 14 May 1982 21:46-EDT From: Alan Bawden Subject: cube talk To: BIL at MIT-MC cc: CUBE-LOVERS at MIT-MC, CWH at MIT-MC Well, I looked for the announcement to get the details, but I couldn't find it. The closest thing I could find was an announcement for a conference this summer in Canada on finite groups (hot stuff these days). Perhaps someone else can supply details?  Date: 16 May 1982 21:24-EDT From: Richard Pavelle Subject: 4 x 4 x 4 = C^4 To: CUBE-LOVERS at MIT-MC cc: RP at MIT-MC I played with C^4 and I conjecture that the tools from C^3 are sufficient to solve it. I have not completely finished it but I think it is just a matter of time. One transformation I use repeatedly, in generic terms, is S = top 180, 2nd slice up or down, top 180, 2nd slice down or up. This is the verticle slice which is facing you. This is just the move in C^3 to move 3 edges in a plane whereas in C^4 the outcome is far more complicated. The steps for C^4 are then 1) Align the centers with a map. Some use of S is necessary. 2) Do all corners as in C^3. 3) Use S repeatedly to position the edges and this is very laborious. 4) Use the C^3 edge flip (Rubik's transformation) to finish it off. After several hours of C^4 I find C^3 looks like a toy.  Date: 17 May 1982 04:07 edt From: RHarvey.Multics at MIT-MULTICS Subject: Re: 4 x 4 x 4 = C^4 To: cube-lovers at MIT-MC In-Reply-To: Message of 17 May 1982 00:11 edt from Alan Bawden I have also been using the tools described by Richard Pavelle as S to do things to edges on the C^4. However, there is nothing that I have found that will flip to edges in place (that is, two halves of a C^3). Also, for aligning centers, I have been doing it AFTER edges. The tools I use do not really map into C^3, and I have not yet mastered them to the point wwhere I can finish them off. I sit and stare at the damn thing a lot. This behavior was considered strange even at the science fiction convention that I attended this weekend. Speaking of strange, what kind of reactions have people been getting from folks who see one for the first time? I have gotten the equivalent of double-takes from ones who do notice a difference. Most, however, tend to say "Oh, I have a friend who can do that in X minutes/seconds", in which case you know that they never learned (or probably even played) to solve. - Ron  Date: 17 May 1982 11:39:22 EDT (Monday) From: Bernie Cosell Subject: solving the 4x4x4x4 cube To: cube-lovers at mit-mc Cc: frye at BBN-UNIX I've managed to solve C4 in a way similar to those mentioned, but using a different transform. I solve C3 a little differently than most folk because I use only one (!) transform to do ALL of the edge work. On C3 the result of this transform is (looking at the edges only): a a from b c to ~d b d ~c Using it and its inverse and a few simple conjugates of it I can do all the work that is necessary for the edges. For me, at least, it has the twin advantages of 1) being easy to remember (since it is fairly short and there is only one of it), and 2) it has such bounded consequences that it is easy to fix a cube without requiring a lot of planning (in the picture above, nothing that is not shown changes: no other edges, no corners). Anyhow, since that little move is a favorite of mine, I tried it on C4. On C3 it comes in two flavors (the move and its inverse, or as it turns out, the right-handed and left-handed versions). On C4 it comes in four versions: the move and its inverse, but each in a `left central slice' and `right central slice' version. Now for the fun part: First off I started ignoring the centers and I noticed that the move (lets call it `M') only moves around edges in a single plane. As I tried to figure out what the damn thing did I discovered that it is a move of order 5!!!. I find it truly hard to plan out what happens when five cubes move around a little orbit, but I'm getting better at predicting it. The result is: from: to: a b a b c d ~d e e f ~f g g h c h With some pain I have been able to use ONLY M (and conjugates and powers of it, of course) to get all the edges in place. Then I looked at M a little more to see what it did with centers. This one is NOT planar, unfortunately, but is simple enough to be useful: only two sets of centers are affected. If you are doing M on the top (to get the above edge transform), only the top centers and the rear centers are affected. And what happens is that there are two disjoint three-cycles each involving two of the top center cubes and one of the rear center cubes. thus, you can easily use conjugates of M to move cubes, one or two at a time, into place on all of the faces around any particular top. Voila: done! An amusing thing about M: since its edge permutation is of order 5 and the center permutation is of order 3, the damn thing is an order 15 move. /Bernie  Date: 17 May 1982 17:30-EDT From: Allan C. Wechsler Subject: Reactions to 4^3. To: CUBE-LOVERS at MIT-AI Bernie and I have divided the double-takers into three major classes: 1. "Oh, a four-by-four cube!" 2. "Oh, a four-sided cube!" (and the best of all) 3. "Oh, a four-dimensional cube!" One of the contributors to this list inadvertently made error #3 by calling the 4x4x4 a "C^4". Which brings me to an interesting question. What would a four-dimensional cube really be like? Let's just start with a 2x2x2x2. It would have sixteen hypercubies, all of the "corner" type. Each hypercubie presents to the outside world four three-dimensional hyperstickers. The 2^4 has eight three-dimensional hyperfaces, presumably each its own color. I like the idea of using black and violet along with the traditional red, yellow, orange, blue, green, and white. We can call the hyperfaces Back, Front, Up, Down, Left, Right, In, and Out. In is across the puzzle from Out. I doesn't touch O, but does touch all the other hyperfaces. The 2^3 can be seen as two square slabs stuck together. Of course, they aren't really "square" since they have thickness. A move consists of rotating one of these squares with respect to the other. Similarly, the 2^4 is two cubical hyperslabs stuck together. Of course, they aren't really cubical since they have hyperthickness. A move consists of rotating one of these cubes with respect to the other. While the two slices of a 2^3 can only have four relative positions, the hyperslices of a 2^4 can have twenty-four different alignments. There are twenty-four corresponding twists. One is the null twist, which consists of just sitting and looking at the thing. (In this case, this is not such a trivial operation!) Then there are six different quarter-twists, as opposed to two in the 3D case. Then there are, get this, eight "third-twists", which when repeated three times bring the slice home. These have no analog in the 3D puzzle. The remaining nine twists are half-twists, three of one kind and six of another. As in the three-dimensional case, the half-twists and third-twists are all products of quarter-twists. If we regard one of the sixteen hypercubies as fixed (without loss of generality, if you'll believe that) then there are twenty-four different quarter-twists in all. Twelve of these are inverses of the other twelve, but selecting the twelve "clockwise" ones is a lot harder than it is in the three-dimensional case. My intuition fails me. I haven't tried to apply the Furst-Hopcroft-Luks algorithm to this monster. At most there are 6^15*15! = 4.7*10^11*1.3*10^12 = (very roughly) 6*10^22. I suspect that some kind of parity, trinity, or quaternity argument will reduce this by a factor of two, three, or four. Yours with a headache, --- Allan  Date: 17 May 1982 2218-edt From: Ronald B. Harvey Subject: 4x4x4 solution To: cube-lovers @ mit-mc Amazing! I log in after working on my cube for an hour or so tonight in order to announce my solution and I find another solution! As previously stated, I use S to put edges into place. This is done to totality after I have put all of the corners together. Now comes the hard part - all of the centers. The tools that I have to work on centers work on 4 or eight center cubies on opposite faces, plus, on the 4-cubie swap version, two pair of edge cubies get swapped. After having a few edges swap (these are now treated like 3x3x3 edges...), I usually replace these so I can undo the complex conjugations more reliably. Now that I can solve it, I going to search for tools that do finer things, like double swaps for centers or edges without munging the other centers or edges. BTW, I have had the tools all weekend - it has just taken this long to get the proper perspective in order to set up the proper conjugations... Has anyone solved the construction problem? Is Ideal's version similar to Plummer's design of last December? Who DID do Ideal's version? Any truth to the rumor (that I just started) that the reason that it has taken Ideal so long to actually start selling the things was that they were working on a solution booklet all this time? (They DO offer one) - Ron  Date: 18 May 1982 09:12-EDT From: Richard Pavelle Subject: C^4 To: CUBE-LOVERS at MIT-MC I solved it after about 10 hours. The very tricky part, as I mentioned in my last message, is to move the edges about in a plane. I still do not have a move which moves 3 in one slice (nicely), rather 2 in one slice and one in the adjacent slice. This makes the process very time consuming. I also believe centers first is probably the way to approach it although it is just a feeling. I think that moving only some of the center cubies will require very complicated transformations. I would guess this cube requires about 10 times more moves than C^3. Does anyone else care to speculate? I will be surprised if anyone will ever solve C^4 regularly in under 10 minutes. Two questions: 1) Any comments about cube-lube. This cube is massive and needs some. 2) I heard there are about 10^50 configurations, true?  Date: 19 May 1982 0107-edt From: Ronald B. Harvey Subject: 4^3 Cube To: cube-lovers @ mit-mc In response to Richard Pavelle's message, there are about 7.4*10^45 color combinations. See "mc:alan;cube 4x4x4" for more details on the subject (especially the derivations) by Dan Hoey. I bought a solution book from my favorite bookstore today. The title is "The Winning Solution to Rubik's Revenge". It is a sequel to "A Winning Solution to Rubik's Cube" by the same author (Minh Thai), who is billed as the U.S. National Champion of the Rubik's Cube-A-Thon. I assume the Revenge book is not REALLY a winning solution... at least not yet. His method is to put all corners together, go for two opposite centers, then the edges of said centers. Next he goes for the remaining edges (numbering eight), and then the last four centers. He also goes into many patterns, and has developed a notation which I have not found immediately obvious. I haven't looked at the book more than to scan it yet, however. He does list a few pretty-patterns, none of which are checkerboards of any sort. On a different subject, Paul Schauble accidentally found out how to take apart the 4^3 today - twist an outer layer about 30 degrees so that an edge of the twisted face is directly over the edge cube that now forms a corner on the rest of the cube. Pop out the cube from the outer layer, twist the face again so that the popped cube's partner is in a similar position for popping, and then pop it. The corners now come out fairly easily. Unlike Plummer's design, the insides of this beastie is a sphere with grooves running along it to make a kind of universal joint. The center cubies ride in these grooves and hold all of the other pieces in. We only took out about a half-dozen cubies because the cube was NOT in an initialized state, but closer to a pretty pattern. I was definitely not interested in getting it ALL apart in order to get it back together correctly. (we did it correctly the first try!). Unlike the 3^3 cube, the 4^3 does not seem to come apart easily after the first few are out. I seem to have gone on for a bit longer than I intended. I will study the pamphlet (published by Dell/Banbury by the way) and report in more detail on notation later. I just noticed that in the section on Cubology, the author lists the number of "arrangements is something in excess of 3.7*10^45". Since Hoey's number is larger, I guess the statement is correct. - Ron  Date: 21 May 1982 1439-EDT From: SWG at MIT-DMS (S. W. Galley) To: cube-lovers at MIT-MC Subject: Newsweek 4/19 (in case you missed it) Message-id: <[MIT-DMS].232584> "Rubikmania: Lots Of New Twists" Newsweek, 19 April 1982, pp. 16f It has been hurled out of moving buses, dumped into trash mashers and pounded to bits with blunt instruments. With more than 43 quintillion possible arrangements, Rubik's Cube puzzle may be the most infuriating plaything ever marketed, as well as one of the most popular. Seven years after Hungarian architecture Prof. Erno Rubik constructed the first cube to help his students understand three-dimensional objects, world sales have passed the 30 million mark. "It's phenomenal," says a vice president for the Ideal Toy Corp., which makes the plastic cube under an agreement with a Hungarian manufacturing company. "Every month we pinch ourselves and say it won't last, but the cube is still selling like nothing else." Ideal has capitalized on the cube's success with a number of spin-offs. There's Rubik's Revenge, which has sixteen tiles on a side, instead of nine; Rubik's Pocket Cube, a simpler version intended for children; Rubik's World, a globe made of 26 sections that twist apart; Rubik's Game, a three-dimensional pegboard, and Rubik's Race, a two-player game in which the multicolored tiles must duplicate various patterns. True masochists might also want to try something called the Calendar Cube---which requires twiddling the tiles every day to form the correct date. Rubikmania has also spread to the publishing industry. Ideal's solution booklet and another one written by a 13-year-old London schoolboy ("You Can Do the Cube") are both big sellers. "The Simple Solution to Rubik's Cube," which runs to 64 pages, has sold 7 million copies; it is the fastest-selling title in the history of Bantam Books. There are even books for people who are fed up with the craze. Among them: "Not Another Cube Book," "You Can Kick the Cube" and "101 Uses for a Dead Cube." Rubik receives about 5 percent of the puzzle profits, making him perhaps the only self-made millionaire in Hungary. "With the money I earn, I can afford to buy myself a new Fiat every two days," he jokes. "A little Fiat." He has taken a leave of absence from his teaching post to help Ideal organize a world cube-twisting championship in Budapest this spring. To qualify for the contest, you must be able to solve the puzzle in less than one minute---which eliminates Rubik himself. It takes him at least twice that long.  Date: 21 May 1982 1710-PDT From: ISAACS at SRI-KL Subject: frustration and Alexander's Star To: cube-lovers at MIT-MC Aauugghh - no one I've been able to find in the Bay ARea has the 4^3. If anyone is flying in from Boston, BRING SOME! Several other of Ideals new stuff is out here - including the game "Rubik's Race", some of the Puzzle pens, and the 2^3. The latter uses the sliding disks on an internal sphere mechanism, by the way. If one comes apart, it seems difficult (but not impossible) to get back together. Anyway, the only really worthwhile new Ideal thing is Alexander's Star, which is a worthy addition to the sliding axis puzzle field. It is a Great Dodecahedron in form, and each "star", that protrudes from a pentagonal face, rotates (5 positions). Each face (the pentagons, not the stars) is monochromatic in the solved state. It is colored with 6 colors, opposite faces the same. (Look in a decent geometry book for a picture of a great dodecahedron. It's one of the 4 Kepler-Poinsot regular concave polyhedra, this one having 12 pentagonal faces interpenatrating each other in a star-like manner). It's not too hard to solve, the main difficulty is figuring where each piece goes. (Each moving piece is the triangular wedge which can be found between the points of a stellated dodecahedron, which "turn it into" a great dodecahedron). If anyone knows where to get the 4^3 out here, or is coming visiting and can bring one, my phone number is (415)326-7788, if you can't get at a terminal. My work number is (415)497-2577. -- Stan Isaacs -------  Date: 22 May 1982 19:13-EDT From: Richard Pavelle Subject: C^4 To: CUBE-LOVERS at MIT-MC I take it back. There is more to C^4 than I first thought. I encountered a final move today and do not have the tools and perhaps this is what DCP meant in his message of May 14th. The final configuration requires flipping only ONE pair of edges. Can anyone explain this difference between C^3 and C^4?  Date: 22 May 1982 22:53:41 EDT (Saturday) From: Bernie Cosell Subject: A missing transformation for C^4 To: cube-lovers at mc I take back my statement about solving the cube, also: I do have one transform that is the product of an order-3 centers-only cycle and an order-5 edges-only cycle, and I can use that transform to do very nearly everything...except for one maneuver. If the normal layout of the edges on one face is: a b h c g d f e Then, when I try to solve the cube, once in a while I end up with one of the following two equivalent forms: ~a ~b c b h c or h a g d g d f e f e I have not yet fully determined exactly which of the two identical edges of each color is where. In particular, I strongly suspect that the first form is really `~b ~a' (that is, if you think of the two edges together as a single C^3 edge, then that edge was flipped), while I suspect that the second form is correctly rendered. I have found that if I do sensible things to the cube, I can mostly work until the cows come home and not fix the thing, but if I screw up a transform and then have to recover from a somewhat scrambled mess, more often than not it comes out OK when I get the cube back in shape. sigh. right now I'm working on a group-theoretic analysis to try to get some idea for what type of move or conjugate I have to be looking for in order to make the thing work out. /Bernie  Date: 23 May 1982 16:15:19 EDT (Sunday) From: Winston Edmond Subject: Re: A missing transformation for C^4 In-Reply-to: Your message of 23 May 1982 01:48 EDT To: Alan Bawden Cc: Cube-Lovers at MIT-MC, Edmond at BBN-UNIX, Mann at BBN-UNIX The 4x4x4 cube does indeed have a "parity" problem. It may be described approximately thusly: Look at one face of the cube. Number the cubies across one edge as 1, 2, 3, 4. It can be shown that an edge cubie of type two can be either in a type 2 position unflipped or in a type 3 position flipped. Next, assuming that the top three planes (or slices) are correct, that the corners of the last face are correct, and that the proper color is up for all of the edge cubies, then the edge cubies in group 2 and those in group 3 either have the same "parity" or opposite parity. The parity groups for the edge cubies are defined as a a b d b c c d and any of the twelve variations of each obtainable by one or more rotations of three elements at a time. (Each letter represents a different color.) When solving the cube, the cubies of group 2 and group 3 will either end up with the same or opposite parity. If the parity is the same, the cube can be solved straightforwardly. (I assume people have discovered the transforms that invert the parity of both groups at the same time.) However, when the parity is opposing, there is only one "transform" that will correct the problem. People accustomed to thinking of a useful transform as one which performs some limited rearrangement of cubies while leaving everything basically unchanged will find that none of their transforms is sufficient to solve the problem. Credit goes to Bill Mann for discovering the "transform" that solves the problem. I leave it up to him to describe the solution if he wishes to. -WBE  Date: 24 May 1982 09:33-EDT From: Richard Pavelle Subject: Ideal and the C^4 To: CUBE-LOVERS at MIT-MC I just spoke to Ideal Toy. They are aware of the edge flip problem and give a tool to do it in their published solution. However, it requires 32 moves and they ask us to provide them with a shorter one if found. They also say they are unable to do a complete checkerboard of any kind on C^4 and would like to know if we can do it.  Date: 24 May 1982 10:56-EDT From: David C. Plummer Subject: Ideal and the C^4 To: RP at MIT-MC cc: CUBE-LOVERS at MIT-MC I have done a complete search on paper for a complete checkboard. It cannot be done for any cobe of even side. The restrictions are in the corners. I would like somebody to double check this though. The standard problems to run into are: You need two of one type of corner, or You have to rotate exactly one corner, which is impossible. There are a couple other amazing things I found. As it is known, any two edges can be exchanged (or appear to be exchaged because of center arbitrariness). It is ALSO possible to (appear to) exchange two corners, for about the same reason. This is impossible on the 3^3 becuase it requires the exchange of two edges. But in the 4^3 there are two cubies per 3^3 edge. Therefore, we just do a double exchange, which does not violate any parity arguments. Combining these two moves, you can flip a 1x1x4 edge. Happy revenge.  Date: 24 May 1982 08:47 PDT From: Hoffman at PARC-MAXC Subject: Re: frustration and Alexander's Star In-reply-to: ISAACS at SRI-KL's message of 21 May 1982 1710-PDT To: Cube-Lovers at MIT-MC I shared your frustration. I'm in L.A., and, upon calling Ideal's office here, I learned that they are following a rigid schedule of TV ads first before introducing the toy in any area. Here in L.A. (and probably in the Bay Area), the TV ads are not scheduled until mid-June! Curiously, Minh Thai's solution book is already in some bookstores here. Anyway, I wound up calling Games People Play in Cambridge, and mail-ordering one ($15 + $2 handling + UPS mailing cost) using a credit card. A two-minute transcontinental call before 8 am PDT is about 40 cents. I hope to have my Rubik's Revenge in a day or so now. --Rodney Hoffman  Date: 24 May 1982 19:43-EDT From: Martin Minow To: CUBE-LOVERS at MIT-AI Please add me to the mailing list, using the address decvax!minow at berkeley (are back issues available?) The 4x4 Rubic's revenge has just come out. Does anybody have a good nomenclature for positions and moves yet? Thanks very much. Martin Minow decvax!minow  Date: 24 May 1982 17:19-EDT From: Allan C. Wechsler Subject: |4^3| = 1.7*10^55 To: CUBE-LOVERS at MIT-AI The corner group of 4^3 is exactly like that of the 3^3. It has 8!*3^7 elements. Now to calculate the order of the whole group, we have to find the order of the corner stabilizer group. That's the group of moves that leave the corners fixed. SO let's start by thinking about the edge group of the corner stabilizer. I personally have a tool that exchanges two edge cubies without moving corners. Since the edge group is transitive, I can exchange any two edges without moving corners. You cannot flip any edge of the 4^3 without moving it. So there's no edge-flippage in the 3^3 sense on the 4^3. That means that the edge group in the corner stabilizer has order 24!. Now all there is left is compute the order of the edge AND corner stabilizer. The inside twists (slices) are center-even. The outside twists (faces) are center-odd, but they are also corner-odd, and so if we want to bring the corners home we have to make an even number of outside twists. That means that all the center-permutations in the edge-and-corner stabilizer are even. But how many of these can be achieved? I have a tool that exchanges two pairs of centers. I think (but I haven't yet proved) that the center group is 4-transitive, so that ANY two pairs of centers can be exchanged. That means that all the even permutations of centers can be done without perturbing edges and corners. Hence the size of the edge-and-corner stabilizer is 24!/2. As for supergroupiness, Dave Plummer just pointed out to me that once you know a center's position, you know its orientation, since center cubies always keep one corner pointing to the middle of the face. So 4^3 has no supergroup. So the order of the 4^3 group is 8!*3^7*24!*24!/2. In numbers, if you insist, |4^3|= 16,972,688,908,618,238,933,770,849,245,964,147,960,401,887,232,000,000,000 or about 1.7*10^55. Now that number is a little deceptive, because it includes whole-cube rotations. The 4^3 has no nice fixed reference frame like the 3^3 has. If you don't want to count whole-cube rotations you have to divide by 24, to get 707,195,371,192,426,622,240,452,051,915,172,831,683,411,968,000,000,000 or about 7.1*10^53. Finally, we have to face the fact that the center cubies are in six nonintradistinguishable sets of four. (All the edge cubies are distinguishable by color or orientation). So we have to divide our last result by 4!^6/2. That leaves 7,401,196,841,564,901,869,874,093,974,498,574,336,000,000,000 or about 7.4*10^45 distinguishable color patterns. Remember that these do not form a group. I leave it as an exercise for Hoey and Saxe to find a lower bound for the diameter of the group. ---Wechsler  Date: 24 May 1982 22:54:03 EDT (Monday) From: Bernie Cosell Subject: The missing C^4 transform found To: cube-lovers at mc I have to admit to not quite yet fully understanding all of the parity issues involved, however I have managed to apply some VERY SIMPLE logic to the mysterious transform and since it worked I'd like to share the insight with you-all. As I mentioned in an earlier note I have been trying to do some analysis to try to sneak up on the missing transform. Mostly I've been unsuccessful at fully identifying the parity issues (my understanding is basically at the same not-so-useful level as wbe's is: I kind of know that some moves seem to come in pairs and there are two `classes' of edge cubes and if you interchange one from each class they both flip... but that mostly didn't help me see what had to be done to make the `missing transform' happen, nor would it let me look at a scrambled cube (or even a nearly-done cube) and guess which parity class it was in). However, I did make one observation: all of my normal transforms contained an EVEN number of quarter-twists. I fumbled around a bit to try to find a limited-change transform that used an ODD number of twists and mostly I couldn't. So I decided to simply take the `simplest' odd-twist transform I could: a single twist! IT WORKS!! As far as I can tell it must be a slice twist (but ANY one), a face twist doesn't do it. If I get the cube edges all solved except for a pair that must be flipped, I simply make a SINGLE quarter slice and then re-solve the cube from there. Since all of my normal solving maneuvers are EVEN, when I get the thing solved again I will have preserved the ODD parity of the configuration, and poof! the edges are solved. Now to start on the search for pretty patterns and elegant (and short) operators... /Bernie  Date: 27 May 1982 00:08 edt From: Senft.Multics at MIT-MULTICS (HVN 341-7244) Subject: Cube-lovers To: Cube-Lovers at MIT-MC cc: Senft.Multics at MIT-MULTICS (outgoing.sv) Acknowledge-To: Senft.Multics at MIT-MULTICS Please add "Senft.Multics@PCO-Multics" -at MIT-Multics to the cubelovers list.  Date: 28 May 1982 1149-EDT From: SWG at MIT-DMS (S. W. Galley) To: cube-lovers at MIT-MC Subject: Moleculon v. Ideal Message-id: <[MIT-DMS].233206> Cambridge firm seeks $60m in Rubik's Cube suit By Christy George Special to The [Boston] Globe [5/27/82] Moleculon Research Corp., a Cambridge chemical, research and development firm, yesterday filed a $60 million patent infringement lawsuit against the Delaware-based Ideal Toy Corp., which manufactures and markets the highly-lucrative puzzle known as Rubik's Cube. Moleculon holds a 1972 US patent for a similar mathematical cube puzzle invented by Dr. Larry Nichols of Moleculon. The prototype of the Nichols Cube is a 2x2x2 cube held together by magnets, while the original Rubik's Cube is a 3x3x3 cube held together mechanically. However, the 1972 Moleculon patent also lists specifications for a mechanically-constructed cube as well for cubes of other varying sizes. Ideal has no US patent for Rubik's Cube, although its inventor, Erno Rubik, was issued a Hungarian patent in 1978. In its complaint, Moleculon alleges that Ideal "willfully and maliciously" continued to manufacture and sell Rubik's Cube despite having been notified a year ago that Moleculon holds the only valid US patent on the invention. According to Moleculon's president, Dr. Arthur Obermayer, the Cambridge firm unsuccessfully tried to interest Ideal in its cube puzzle in 1969. In 1970, the firm applied for a patent at the suggestion of other national toy manufacturers, who also were uninterested in marketing the puzzle. Lawyers for Ideal would not comment on the pending suit because, they said, they have not yet received the complaint. But Ideal's general counsel, Samuel Cohen, dismissed the notion that Ideal might have stolen the idea from Moleculon. "It's ridiculous on its face," Cohen said, "we're paying heavy sums to the Hungarians for use of their patent. It would be stupid to do that if we had it in our pocket all the time." Obermayer says the company didn't realize how similar Rubik's Cube was to the Nichols Cube until the phenomenal success of the product began to make headlines. Rubik's Cube was first marketed worldwide in 1980. Because of the puzzle's complexity, inventor Nichols wasn't surprised that toy firms were unimpressed. "I'm delighted people enjoy solving the puzzle," Nichols said, "but I'm also frustrated that neither I nor Moleculon has seen any financial reward all these years." Obermayer is confident of success in the lawsuit, citing the fact that three prominent patent law firms have agreed to collaborate on Moleculon's behalf on a contingency basis. Moleculon is asking for $20 million in estimated lost royalties and $40 million in damages, as well as interest on other costs. Even more could be at stake in terms of future royalties. According to an article in Newsweek magazine last month, annual world sales for the basic 3x3x3 Rubik's Cube, which retails for between $7 and $8 in the Boston area, have exceeded the $30 million mark. And the company also markets a range of spin-off products, including cubes in three other sizes and types, a Rubik's Cube solution booklet, a Rubik's Cube world globe and a Rubik's Cube board game. "Sales are just incredible," said Carol Monica, owner of the Cambridge store, The Games People Play. Monica said she already has sold more than a hundred of Ideal's latest spinoff, a 4x4x4 verison of Rubik's Cube which costs $16 and went on the market only three weeks ago. Ironically, in an action pending before the International Trade Commission, Ideal claims it is besieged by countless "knockoffs" of Rubik's Cube by unscrupulous foreign competitors. The Delaware company is asking the commission to squelch international imports of similar mathematical puzzles.  Date: 30 May 1982 04:52:01-PDT From: decvax!duke!uok!mwm at Berkeley Date-Sent: Wed May 26 19:09:15 1982 To: duke!decvax!ucbvax!cube-lovers@ai Subj: Diameter of the group I just saw a letter on the list that talked about finding the diameter of the group for 4^3. This implies (to me, anyway) that somebody out there knows something about the diameter of the group for 3^3. Since I just started getting mail from the list, I missed any discussion that may have happened on that topic, and would appreciate it if someone would send me the apporpriate information, or tell me which archive the discussion is in. Thanx in advance, mike meyer  Date: 30 May 1982 10:52:24-PDT From: decvax!duke!uok!mwm at Berkeley Date-Sent: Wed May 26 19:09:15 1982 To: duke!decvax!ucbvax!cube-lovers@ai Subj: Diameter of the group I just saw a letter on the list that talked about finding the diameter of the group for 4^3. This implies (to me, anyway) that somebody out there knows something about the diameter of the group for 3^3. Since I just started getting mail from the list, I missed any discussion that may have happened on that topic, and would appreciate it if someone would send me the apporpriate information, or tell me which archive the discussion is in. Thanx in advance, mike meyer  Date: 30 May 1982 16:30-EDT From: Alan Bawden Subject: God's number To: CUBE-LOVERS at MIT-MC OK, it's been some time since I pointed out where I keep archives and things... Old cube-lovers mail is archived in the following places: MC:ALAN;CUBE MAIL0 ;oldest mail in forward order MC:ALAN;CUBE MAIL1 ;next oldest mail in forward order MC:ALAN;CUBE MAIL2 ;more of same MC:ALAN;CUBE MAIL3 ;still more of same MC:ALAN;CUBE MAIL ;recent mail in reverse order (Files can be FTP'd from MIT-MC without an account.) In addition, I have the following two excerpts from the archives sitting on my directory since they contain some of the more asked-for material: MC:ALAN;CUBE 4X4X4 ;Contains some pre-release speculations on the 4x4x4 ;cube. Some are out of date, but it contains the only ;analysis this list has seen of the 4x4x4 group, I ;believe. MC:ALAN;CUBE S&LM ;While most of the speculation about the diameter of ;the 3x3x3 group is scattered randomly through the ;archives, this file contains the single message with ;the highest content. Hoey and Saxe's message on ;Symmetry and Local Maxima. To briefly remind you all of ALL that we know about the diameter of the 3x3x3 group (refered to as "God's number" in many of our discussions): We know that God's number is greater or equal to 21 quarter twists. (See Hoey's message of January 9 1981: "The Supergroup -- Part 2 ..." in MAIL1 for a good explanation of this, as well as some other interesting bounds.) We know that God's number is greater or equal to 18 half twists. (See Singmaster.) We know that God's number is less than or equal to 52 half twists. (See Singmaster again, this is Thistlethwaite's algorithm of several years ago. I'll bet it's been improved upon by now. There is a persistent rumor that he was trying for 41.) We have never bothered to figure out an upper bound on God's number in quarter twists ("Q"s). It must be less than 104 Qs because of the half twist result, but we could probably do better than that if we took the trouble to understand Thistlethwaite's algorithm. Proofs of these numbers, and a great deal of other discussion can be found by sifting through the archives (unfortunately they are spread all throught the files). I would urge people to sift through the archives before starting any new discussions on the subject.  Date: 1 June 1982 2220-EDT (Tuesday) From: Dan Hoey at CMU-10A To: Cube-Lovers at MIT-MC Subject: Lower bounds for the 4x4x4 Message-Id: <01Jun82 222052 DH51@CMU-10A> Well, since you insist, here are my lower bounds for the 4^3. For the "colored" 4^3, where only the color pattern matters, some positions require at least 41 qtw to solve. For the "marked" 4^3, where center facets of the same color are distinguished so as to force a unique home position for each, some positions require at least 48 qtw. The proof is in MC:ALAN;CUBE4 LB and is about 5K characters long. A qtw of the 4^3 is either a quarter-twist of a face relative to the rest, or a quarter-twist of half of the puzzle relative to the other half. Note that this makes a slice twist into two moves. I like this metric because it is consistent with our conventions for the 3^3. One of these days I'll explain why I like those conventions.  Date: 1 June 1982 2222-EDT (Tuesday) From: Dan Hoey at CMU-10A To: Cube-Lovers at MIT-MC Subject: Lower bounds for the 4x4x4 (Long) Message-Id: <01Jun82 222230 DH51@CMU-10A> Lower bounds for solving the 4^3 puzzle In order to prevent counting positions which arise from whole-cube moves, I do not move the DLB corner. So the generators I use are U1 U1' R1 R1' F1 F1' U2 U2' R2 R2' F2 F2' U3 U3' R3 R3' F3 F3' where the digit indicates how many layers are being moved and the prime indicates a counterclockwise quarter-twist. Let us break up any process into "syllables", where a syllable is a maximal nonempty string of generators with the same letter. Note that the order of the generators within a syllable is irrelevant. We may make a syllable canonical by simplifying so that 1. A generator and its inverse do not both appear, 2. Clockwise generators appear at most twice, 3. Counterclockwise generators appear at most once, and 4. Generators appear in numerical order. For each letter there are 63 canonical syllables. Omitting the letter, they are: 1,1',2,2',3,3' Six of length one 11,12,12',13,13',1'2,1'2',1'3,1'3',22, 23,23',2'3,2'3',33 Fifteen of length two 112,112',113,113',122,123,123',12'3, 12'3',133,1'22,1'23,1'23',1'2'3,1'2'3', 1'33,223,223',233,2'33 Twenty of length three 1122,1123,1123',112'3,112'3',1133,1223, 1223',1233,12'33,1'223,1'223',1'233, 1'2'33,2233 Fifteen of length four 11223,11223',11233,112'33,12233,1'2233 Six of length five 112233 One of length six (Exercise for the reader: There is a reason these are binomial coefficients. Why did we skip a row?) The number of canonical syllable strings containing N generators is P(-N) = 0 if -N < 0, P(0) = 1, P(N) = 6 C(N-1) P(N-1) + 15 C(N-2) P(N-2) + 20 C(N-3) P(N-3) + 15 C(N-4) P(N-4) + 6 C(N-5) P(N-5) + C(N-6) P(N-6) if N > 0, where C(N) is the number of ways of choosing a new letter after N generators: C(-N) = 0 if -N < 0, C(0) = 3, C(N) = 2 if N > 0. Evaluating this recurrence yields P( 0)= 1 P(13)<1.338E15 P(26)<1.404E30 P(39)<1.472E45 P( 1)= 18 P(14)<1.914E16 P(27)<2.007E31 P(40)<2.105E46 P( 2)= 261 P(15)<2.737E17 P(28)<2.871E32 P(41)<3.011E47 P( 3)= 3732 P(16)<3.915E18 P(29)<4.106E33 P(42)<4.307E48 P( 4)= 53379 P(17)<5.599E19 P(30)<5.873E34 P(43)<6.160E49 P( 5)= 763506 P(18)<8.009E20 P(31)<8.400E35 P(44)<8.811E50 P( 6)=10920771 P(19)<1.146E22 P(32)<1.202E37 P(45)<1.261E52 P( 7)<1.563E 8 P(20)<1.639E23 P(33)<1.719E38 P(46)<1.803E53 P( 8)<2.235E 9 P(21)<2.344E24 P(34)<2.459E39 P(47)<2.579E54 P( 9)<3.196E10 P(22)<3.353E25 P(35)<3.516E40 P(48)<3.688E55 P(10)<4.572E11 P(23)<4.795E26 P(36)<5.029E41 P(49)<5.275E56 P(11)<6.539E12 P(24)<6.858E27 P(37)<7.194E42 P(50)<7.545E57 P(12)<9.352E13 P(25)<9.810E28 P(38)<1.029E44 P(51)<1.080E59 The number of positions exactly N qtw from SOLVED is at most P(N), so the number of positions within N qtw of SOLVED is at most P(0)+P(1)+...+P(N). Since there are 7.07E53 marked and 7.40E45 colored positions, this would give us lower bounds of 40 qtw for the marked cube and 47 qtw for the colored cube. But we can improve these lower bounds by one each, using the parity hack I described some time ago. Every qtw is an odd permutation of the corner cubies, so the number of positions with even corner permutations within 2N+1 qtw of SOLVED is at most P(0)+P(2)+...+P(2N) and the number of positions with odd corner permutations within 2N+2 qtw of solved is at most P(1)+P(3)+...+P(2N+1). There are 3.70E45 colored positions with odd corner permutations, and fewer than 1.583E45 odd-length canonical processes with length at most 39. So some odd colored positions require at least 41 qtw to solve. There are 3.53E53 marked positions with even corner permutations, and fewer than 1.939E53 even-length canonical processes with length at most 46. So some even marked positions require at least 48 qtw to solve. Directions for future hacks Note that we could also distinguish between positions with even or odd edge permutations. The recurrence gets hairier, but my analysis of that problem indicates that the numbers get very close very quick, so no luck. We could divide the positions into buckets based on other quotients of the group. For instance, count the number of positions with each of the 729 possible corner orientations. This should be feasible, but probably no help for the 4^3 puzzle. For the 3^3, where there are 2187 corner buckets, some of which stay empty for a fair number of moves (I seem to recall 8), and our current lower bound is small, I think there is a chance of improvement. Any God's numerologist willing to hack the good hack?  Date: 2 June 1982 10:33-EDT From: Richard Pavelle Subject: Moleculon/Ideal, cont. To: CUBE-LOVERS at MIT-MC I spoke to a person at Moleculon and was given the U.S. patent number, 3655201. I sent for a copy and will report again when I have read it. It will likely be several weeks before I receive it so if someone is near DC and feels motivated to get the patent sooner, please do so. I will also check out the Patent Gazette and see if there is anything of interest there.  Date: 4 Jun 1982 1006-EDT From: PDL at MIT-XX (P. David Lebling) Subject: Re: Moleculon/Ideal, cont. To: ALAN at MIT-MC, Cube-Lovers at MIT-MC In-Reply-To: Your message of 3-Jun-82 1933-EDT There was a report on the Moleculon suit on WCVB, including a closeup of the patent's picture page. The picture is of a 2x2 cube with the cubies held together by magnets (the patent shows how the magnets are to be arranged so the object will hold together). The inventor of the "Nicholls Cube" was shown working it. It doesn't turn; you pull it apart, turn it, and then stick it back together. He never quite claimed to have made a 3x3 version, and carefully tried to fudge the distinction between his 2x2 frob and Rubik's 3x3. My impression was that the guy is a sleazo trying to get an out-of-court settlement from Ideal for $100k or so. Dave -------  Date: 4 Jun 1982 1508-PDT From: ISAACS at SRI-KL Subject: 4^3 To: cube-lovers at MIT-MC Well, I finally got one (UPS from Boston). I agree with the preceding messages; basically, I could solve most of it with 3^3 techniques, and work out a couple of variations to get edge halves together; then if a 2-edge monoflip or exchange is left over, do an odd slice move and recover. Some of the tools are quite long, however; does anyone have a short edge flip or exchange, or a short half-center exchange? -------  Date: 4 Jun 1982 1531-PDT From: ISAACS at SRI-KL Subject: Winning Ways To: cube-lovers at MIT-MC The book "Winning Ways" by Berlekamp, Conway, and Guy is finally out. (A puzzlists' dream - to get 4^3 one day, and Winning Ways the next.) It seems, on quick scanning, to be all rumor had it to be. It's sort of a sequel to Conway's "On Numbers and Games", and contains a more understandable of his Nimbers (Surreal Numbers), with lots of examples applied to game analysis. There are chapters on many puzzles, including one on Rubik's Cube. Also included are sliding block puzzles, Instant Insanity, Life, Sprouts, Polyominoes and Polyiamonds, Wire puzzles, Chinese Rings, peg jumping, etc, etc. An almost encyclopaedic collection of puzzles and games (of certain types) from a mathematical standpoint. With plenty of humor and puns. The only problem is its price - about $20 for each of the two volumes, in paperback. (When the hardback comes out later this year, it's supposed to be about $60 per volume.) The 4 sections are "Spade Work" and "Change of Hearts" in the first volume (on "Games in General"), and "Games in Clubs" and "Solitaire Diamonds" (puzzles) in Vol. 2. ("Games in Particular). Academic Press. Don't miss it. --- Stan -------  Date: 5 June 1982 01:07-EDT From: Alan Bawden Subject: 4x4x4 mechanics To: CUBE-LOVERS at MIT-MC I had a small accident the other day. I dropped Dave Plummer's 4x4x4 cube and broke one of the center cubies (sorry about that Dave). This gave me an opportunity to closely examine the insides of the beastie. I can't possibly describe it through the mail (I can barely describe it in person), but there is an interesting problem raised by the insides: Inside of a 4x4x4 cube is a 57th piece. It is not permanently connected to any of the cubies. (The center cubies are free to slide back and forth in slots cut in the center piece. The rest of the cubies are held in by the centers.) It is impossible to determine by examining the outside of a cube exactly what orientation the center piece has. However, it IS deterministic how the center piece will move under a certain twist. When twisting a face, the center piece stays fixed with respect to the other 3 layers of the cube. When doing an "equatorial" twist, the center piece can follow only one of the halves of the cube (as determined by it's orientation). (I believe I have just constrained things enough so that you can figure out exactly how the thing moves on your own.) Thus there is an even larger permutation group to the 4x4x4 (beyond the supergroup problem where the identities of the center cubies are considered) that includes the center piece. Call this the "hypergroup". And since the center piece has a 3 element symmetry group there is another group beyond that ("superhypergroup"?) that takes that into account. Now the first question to consider about the hypergroup is: Is it really larger than the original group or supergroup? In other words: When you have solved the 4x4x4, does the center piece necessarily return to its original position? How about if you solve the cube in the hypergroup? A problem with this problem, is that you cannot learn how to manipulate the orientation of the center piece without taking your cube apart to look at the thing. (I DON'T recommend that. My experience with Plummer's cube has taught me that those center pieces are fragile.) Anybody have any insights into the problem? -Alan  Date: 5 June 1982 08:34-EDT From: Martin Minow Subject: A notation and a simple edge/center process To: CUBE-LOVERS at MIT-AI I have been using the following notation for the 4x4 cube: Name the faces L, R, F, B, U, D. This follows the 3x3 notation. The inner edges are then l, r, f, b, u, d. The front, upper, right quadrant consists of one corner cube, FUR, two edge cubes, FUr and FRu, and one center cube, fur. Moves follow 3x3 notation: F moves the front face clockwise, f moves the front slice clockwise. F' moves counterclockwise. ** SPOILER WARNING ** SPOILER WARNING ** SPOILER WARNING ** The remainder of this message describes two processes for the 4x4 cube. ** SPOILER WARNING ** SPOILER WARNING ** SPOILER WARNING ** The following process rotates three edge cubes. It preserves parity. (FUl UBl URb) l' RUR'U' L URU'R' The following process rotates three center cubes. It can be used to move one center cube between two faces. (Ful Ulb Urb) l' rur'u' l uru'r' While the center process is sufficient, the edge process will not organize all edges. It can easily be extended to a process that moves (FUl to UBl) but that process messes up centers, corners, and other edges. It also does not do 4-way moves. So the bloody thing still takes an hour. Martin Minow decvax!minow@berkeley  Date: 5 Jun 1982 10:44:47-PDT From: decvax!minow at Berkeley To: ucbvax!cube-lovers@mit-ai Subject: Rubic's Domino I found one (for $11.00) at Games People Play in Cambridge. Made in Hungary, not by Ideal. Haven't solved it yet. The notes say you should be able to solve it in 15 minutes. It is a 3x3x2 construction; one face white; one black. Each cube is numbered as on a domino. The object is to get the numbers and colors arrainged. Martin Minow decvax!minow at berkeley  Date: 5 Jun 1982 2332-EDT From: Joseph A. Bowbeer Subject: 4x4x4 solution book To: cube-lovers at MIT-MC Jeff Adams (MIT math graduate student) and another person have coauthored a 4x4x4 solution book which should be on the shelves shortly. I'll try to get more information. The 3x3x3 solution posters which are on sale at the MIT math undergraduate office, among other places, were designed by these two cubists, also. -------  Date: 6 Jun 1982 17:11:14-PDT From: decvax!minow at Berkeley To: ucbvax!cube-lovers@MIT-AI Note: this is a resubmission of my note of Saturday June 6 which was bit by typo's. Thanks to RP@MIT-MC for pointing them out. I have been using the following notation for the 4x4 cube: Name the faces L, R, F, B, U, D. This follows the 3x3 notation. The inner edges are then l, r, f, b, u, d. The front, upper, right quadrant consists of one corner cube, FUR, two edge cubes, FUr and FRu, and one center cube, fur. Note that the Ideal book uses Lx etc for the mid-slices. Moves follow 3x3 notation: F moves the front face clockwise, f moves the front slice clockwise. F' moves counterclockwise. ** SPOILER WARNING ** SPOILER WARNING ** SPOILER WARNING ** The remainder of this message describes two processes for the 4x4 cube. ** SPOILER WARNING ** SPOILER WARNING ** SPOILER WARNING ** The following process rotates three edge cubes. It preserves parity. (FUl UBl URb) l' RUR'U' l URU'R' The following process rotates three center cubes. It can be used to move one center cube between two faces. (Ful Ulb Urb) l' rUr'U' l UrU'r' While the center process is sufficient, the edge process will not organize all edges. It can easily be extended to a process that moves (FUl to UBl) but that process messes up centers, corners, and other edges. It also does not do 4-way moves. So the bloody thing still takes an hour. Martin Minow decvax!minow@berkeley  Date: 7 Jun 1982 15:34:28 EDT (Monday) From: Roger Frye Subject: 4^3 breakage, notation, and solution To: Cube-Lovers at MIT-MC Cc: decvax!minow at BERKELEY, ISAACS at SRI-KL, frye at BBN-UNIX About broken 4^3 center cubies: The center cubies take all the stress of holding in the edges and corners. The post which breaks is 4mm square compared to the 16mm square cubie. When I called Games People Play about my broken cube, another cube had just been returned with the same problem. So I traded my broken cubie for an unbroken one and picked up a spare. About Minow notation for 4^3: Name the faces L, R, F, B, U, D. This follows the 3x3 notation. The inner edges are then l, r, f, b, u, d. The front, upper, right quadrant consists of one corner cube, FUR, two edge cubes, FUr and FRu, and one center cube, fur. [decvax!minow at Berkeley] Isn't there still a mistake here? I think of "fur" as being a spot buried in the cube. If you mean the four cubies in the upper right quadrant of the front face, then the name of the center cubie would be "Fur". If you mean the front, upper, right octant then the two other center cubies are "Urf" and Rfu". About 4^3 processes: ** SPOILER WARNING ** SPOILER WARNING ** SPOILER WARNING ** My process to exchange three edge cubies is a U face commutator: (RDf URf UFl) = R'd'R U' R'dR U I think of my other 4^3 process as exchanging two half centers: (Rfu Flu Rbd) \ (Rub Fur Rdf) = R2 u' R2 u (RBd FRd BRu RFu FLu) / You can get lost trying to read the permutations. Just think of it as injecting the two upper cubies from the front face into the right face in exchange for the two upper cubies from the right face. This is a move ISAACS requested. My solution strategy: ** SPOILER WARNING ** SPOILER WARNING ** SPOILER WARNING ** 1) Solve the top center using random methods. 2) Solve the corners using 3^3 moves. (See \\Jeff Conquers the Cube//.) 3) Solve all centers with my center process. 4) Solve all edges with my edge process. 5) If a 2-edge monoflip or exchange is left over: 5a) In 2^3 mode, turn top and rotate three corners back; i.e. Uu (Rr)2 (Bb)2 Rr Ff R'r' (Bb)2 Rr F'f' Rr. 5b) In 3^3 mode, undo 5a; i.e. U' L2 B2 L' F' L B2 L F L. 5c) Repeat steps 3 and 4. -Roger Frye  Date: 7 Jun 1982 1606-PDT From: Dolata at SUMEX-AIM Subject: I quit! To: cube-lovers at MIT-AI I have decided to give up cubing for my sanity! Therfor, I have 2 ideal cubes in good tight condition for sale. They go to the highest bidder, assuming the bid is sufficiently above postage to make it worthwhile. Dan (ex-cubeist) Dolata -------  Date: 8 June 1982 19:34-EDT From: Alan Bawden Subject: [DANIEL: FYI. [dave: Rubic's Cube World Championship]] To: CUBE-LOVERS at MIT-MC Just for the record: Date: 8 June 1982 11:44-EDT From: Daniel Weise Re: FYI. [dave: Rubic's Cube World Championship] Date: 7 June 1982 2054-PDT (Monday) From: dave at UCLA-Security (David Butterfield) Re: Rubic's Cube World Championship Ming Thai, the Vietnamese student who I met a couple of weeks ago, won the World Rubic's Cube Contest in Budapest on June 5. His time was around 23 seconds. Very impressive time.  Date: 9 Jun 1982 1001-PDT From: ISAACS at SRI-KL Subject: S&M and other moves To: cube-lovers at MIT-MC BACKGROUND On the 3^3, many people use variants of a 3-cycle of edges, more-or-less as follows: let Rs = Right slice = R'L + roll cube so front is up T (for tri-cycle?) = Rs U2 Rs' U2 = (UF,DF,UB) to cycle 3 edges on a slice Some people use it in the form R2 D' - T - D R2 = (UF,UR,UB) to get 3 on a face, without flips, and some use Rs' U - T - U' Rs = Rs' U Rs U2 Rs' U Rs = (UF,LU,RU) for 3 on a face, with flips, since it saves a move due to U2 U' = U. END BACKGROUND Now, this move, when transferred to the 4^3, seems to be the basis of both Richard Pavelle's "S" move of 16-May, and Bernie Cosell's "M" move of 17-May (Bernie using a left slice version of the third form). The move Roger Frye describes on 7-June is also the same as this (re-oriented to a different face), and so is the "quite long" tool I use. To move only edges requires S3, and to move only centers requires S5!!! I still would like a "nice" (preferably short) center pair move. SPOILER SPOILER SPOILER We have now several 3-cycles: (UBl,UFr,LDf) = U2 f' D f - U2 f' D' f (from Minh Thai's book) (UFl,BUl,RUb) = l' - R U R' U' - l - U R U' R' (Minow, 6-June) (RDf,URf,UFl) = R' d R - U' - R' d R - U (Frye, 7-June) Note that all of these have corresponding 3-cycles of centers, by simply substituting a slice for its' corresponding edge in each of the moves. For instance, in Thai, substitute d for D; in Minow, r for R, etc. There is also a version which cycles an edge an center together, by rotating the appropriate face and slice together. In fact, it looks as though any simple move on the 4^3 will have a left and right version, a forward and back version, and a slice and edge version or two, along with combinations of these. Is there any way to put these into a canonical form so we can recognize related moves? What about canonical form on the 3^3? One more spoiler: to switch opposite centers (all 4 cubies): (F,B) = (r2 U2 l2 U2)3. END SPOILER END SPOILER END SPOILER Can the notation be usefully extended so as to talk about UB as the pair of upper back edges, F as the front 4 center cubies, and Fu as the upper two of F, etc. It might make notating the permutations easier. Is there any notation to make various repeats easier; for instance, some expansion of "squared" to indicate "primed" (commutator) type repeat, or a repeat with all quarter twists in the opposite direction. Maybe ( )2 means repeat; ( )i2 means repeat with sub-parts "primed", and ( )-2 repeats with qtws opposite. The purpose is to try to make it more obvious what each move is really doing, and to be able to compare moves easier. Has anybody given any thought to notation the 5^3? By the way, if Bill Mann is listening, would you describe your transform (mentioned by Edmond on 23-May)? --- Stan -------  Date: 10 Jun 1982 0940-PDT From: ISAACS at SRI-KL Subject: correction To: cube-lovers at MIT-MC At least 2 errors appeared in my previous message: In Minh Thai's 3-cycle, substitute "r" for "f" to do the permutation reported. In the Frye 3-cycle, the first "d" should be "d'". I thought I had a half center move: (r2 u2)2 = (Fur,Bur) but when repeated, I got (r2 u2)4 = (Fdr,Bur), so I realized that what I really had was the 6-cycle of centers: (r2 u2)2 = (Ful,Fdr,Bul,Bdl,Fur,Bur) (I think). Oh, well. ---- Stan -------  Date: 14 Jun 1982 11:17:40 EDT (Monday) From: Bernie Cosell Subject: Magic Domino To: cube-lovers at mc The magic Domino (available, at least, at Games People Play here in Cambridge) is a delightful little puzzle. I would guess that it is the easiest of the `Rubik-type' puzzles to date -- I think it is even easier than the 2x2x2. I solve it with two basic moves: one fixes the corner (but screws up the edges), and then the other fixes up the edges. An interesting sidelight on the domino is the ability to interchange ONE pair of edges. The trick here (for reasons that I don't understand yet) is that the supergroup rotation seems to invert the parity of the edges. That is: if you have the whole domino solved and then re-solve the thing with both faces rotated by a quarter turn around the center (5) cube. You'll discover that you can't do this hardly at all on just one face, since you wont be able to get the corners fixed. But if you rotate both the white and black faces, then the corner parity is OK, BUT.... the edge parity has been flipped and so you have done a single interchange of two edges. /Bernie  Date: 14 Jun 1982 16:10:19-PDT From: decvax!minow at Berkeley To: ucbvax!cube-lovers@mit-ai Subject: Some simple edge processes for the 4x4 cube. ** ** ** ** SPOILER WARNING ** ** ** ** This message describes some edge processes for the 4x4 Rubic's Cube. My collegue, Marty Hiller, came up with two simple processes which she uses to order edges on the 4x4 cube. They are sufficient to solve all edge permutations: (FUl UBl FDr DBr) (r U2)^4 r (FUl UFr UBl BUr FDr) (r U2 r' U2)^3 The ^n notation means repeat the process n times. Note that these processes probably modify the center cubes as well but this is invisible if you have already solved the center. Martin Minow  Date: 15 June 1982 1045-EDT (Tuesday) From: Dan Hoey at CMU-10A To: Cube-Lovers at MIT-MC Subject: Patterns for Rubik's Revenge Message-Id: <15Jun82 104503 DH51@CMU-10A> CHECKERBOARDS I confirm Dave Plummer's result that it is impossible to make a checkerboard on all faces of a cube of even side. Proving this for the 2^3 is sufficient, and implies the corollary I sent last year that it is impossible to make S or Z or Zig-Zag patterns on all sides of the 3^3. This time I will outline my analysis of the problem. What we are asking for in each case is a pattern in which every face has matching diagonally opposite corner facets and contrasting adjacent corner facets. First, consider any position in which diagonally opposite corner facets match on each face. If we connect every pair of corner cubies that have diagonally opposite corner facets we get two tetrahedrons of corner cubies. A quick examination will show that the four cubies in such a tetrahedron must either be 1) Not all from the same cube, 2) Four copies of the same cubie, or 3) Four different cubies from the same cube, in the correct position and orientation relative to each other. Assuming case 3, we may place one tetrahedron in the home position and consider how the other tetrahedron has been rigidly rotated with respect to the first. There are 12 rotations of the tetrahedron. One is the identity, three are 180-degree rotations about the midpoints of two opposite edges, and six are 120-degree rotations about a vertex. In the identity every face's adjacent corner facets fail to contrast. The 180-degree rotations have the corners like the 3^3 Zig-zag pattern, where two faces have noncontrasting corners. The 120-degree rotations would make checkerboards, but violate the corner twist invariant. Approximate checkerboards can be made from the 180-degree rotation and from the 120-degree rotation with one corner twisted. In the 180-degree approximation, two faces have two wrong facets each. In the 120-degree approximation, three faces have one wrong facet each. Both are achievable with the 2^3 and 4^3, and I think these are as close as you can get on any even-sided cube. SPOTS Spot patterns of the 4^3 are those which have the pattern X X X X X Y Y X X Y Y X X X X X on all nonblank faces. There are a lot of them: every permutation of the centers is possible. There are thus 6! = 720 spot patterns. We may cut this number down by identifying positions that are M-conjugates (recolorings) of each other. As long as I am listing the permutations by conjugacy class, I may as well break them down by recoloring type. Last July I asked a question about the possible rearrangements of colors on the cube. I worked on the solution long enough to find that given a ``standard'' cube, there are five kinds of recoloring up to M-conjugacy. Identity -- The standard coloring. Reflection -- Identity in a mirror. Swap -- Identity with two adjacent colors exchanged. Wrench -- Identity with three adjacent colors cycled. Befuddler -- Wrench in a mirror. The Identity and Reflection are unique colorings. There are twelve Swaps and eight each of the Wrench and Befuddler recolorings. Each recoloring corresponds to 24 face permutations, achievable by whole-cube moves. Here, then, are the spot patterns. Columns correspond to the number of spots in the pattern. The row groupings show the kind of cube coloring the spot pattern comes from. The patterns are given as a permutation of faces: (...XY...) and (Y...X) both mean that face X has a spot colored Y, as shown at the beginning of this message. The number of permutations in each conjugacy class is also given. Number of spots 0 2 3 4 5 6 Coloring -------------------------------------------------------------------- Identity :8 (BUFD):6 (BUL)(FDR):8 (BF)(UD):3 (BF)(UL)(DR):6 -------------------------------------------------------------------- Reflection (BF):3 (BU)(FD):6 (BULFDR):8 (BF)(ULDR):6 (BF)(UD)(LR):1 -------------------------------------------------------------------- Swap (BU):12 (BFU):24 (BFUD):12 (BUFDL):48 (BFULDR):48 (BF)(UL):12 (BF)(UDL):24 (BF)(UDLR):12 (BU)(FDL):48 (BU)(FDLR):48 -------------------------------------------------------------------- Wrench (BUL):16 (BUFL):24 (BFUDL):48 (BFULRD):24 (BU)(FLR):48 (BUL)(FRD):8 (BU)(FLDR):24 -------------------------------------------------------------------- Befuddler (BFUL):48 (BFULD):48 (BFUDLR):16 (BU)(FL):24 (BUFLDR):24 (BFU)(DLR):24 (BU)(FL)(DR):8 -------------------------------------------------------------------- The answer: Fifteen six-spot patterns, six five-spots, eight four-spots, two three-spots, two two-spots, and one no-spot. CROSSES What cross patterns are possible on the 4^3? We must first ask what a cross pattern on the 4^3 should look like. I consider the following two kinds of cross. Thick Cross Thin Cross X O O X X O X X O O O O O O O O O O O O X O X X X O O X X O X X Every thick cross pattern is a rigid rotation or reflection of the edge and face center pieces with respect to the corners. This is just like the 3^3 case except that the 3^3 has face centers that are fixed relative to each other, and so does not allow reflec- tions. So in addition to the thick versions of Plummer's Cross and Cristman's Cross, there are three new crosses. Thick Pons Cross Thick Fliptwist Cross Thick Interlaced Cross U D D U U L L U D D D D B U U B L L L L D D D D U U U U L L L L U D D U U U U U U L L U B U U B L R R L F B B F R L L R L D D L F B B F R U U R R R R R B B B B L L L L U L L U D D D D B B B B U U U U R R R R B B B B L L L L L L L L D D D D B B B B U U U U L R R L F B B F R L L R L L L L L D D L F B B F R U U R U L L U D U U D D R R D U U U U L F F L F D D F R B B R R R R R U U U U F F F F D D D D B B B B R R R R D U U D F F F F D D D D B B B B D R R D L F F L F D D F R B B R B F F B B F F B F F F F D R R D F F F F F F F F R R R R F F F F B F F B R R R R B F F B D R R D For thin crosses, we first examine those in which the arms of the crosses meet at the edges. Again the figure facets are rigidly rotated and reflected with respect to the ground. This time cubie conservation becomes an issue, because of the impossibility of flipping an edge cubie, so there are only three such thin crosses. Thin Pons Cross Thin Plummer Cross Thin Interlaced Cross U U D U U U R U U U D U B B U B U U R U D D D D B B U B R R R R U U D U U U U U U U R U B B U B L L R L F F B F R L R R L L B L F F U F R F R R R R R R B B B B L L L L U U L U B B B B U U U U F F F F L L R L F F B F R L R R U U L U L L B L F F U F R F R R L L R L F F B F R L R R L L L L L L B L F F U F R F R R U U L U D D U D D D L D U U U U L L F L F F D F R B R R L L L L D D U D F F F F D D D D B B B B D D L D D D U D L L F L F F D F R B R R D D L D L L F L F F D F R B R R B B F B B B D B B B F B D D R D B B D B F F F F R R R R D D D D B B F B D D R D B B D B D D R D When we relax the constraint that thin cross arms must meet at the edges, the figure is no longer rigidly transformed with respect to the ground. Indeed, we might expect that adjacent crosses whose arms do not meet might have colors that are opposite on the cube. I carried out a long examination of the cases, and found that this does not happen. In fact, only one new color permutation arises. Thin Fliptwist Cross U U B U B B B B U U B U U U B U L R L L F F U F R L R R B D B B L R L L F F U F L L L L B D B B R R R R U U U U R L R R D D D D L R L L F F U F R L R R B D B B D D F D D D F D F F F F D D F D The only other thin cross patterns in which not all crosses meet at the edges are 43 versions of the Thin Pons Cross (modulo my missing a case or two in the analysis).  Date: 24 June 1982 16:28-EDT From: Richard Pavelle Subject: Moleculon vs Ideal, cont. To: CUBE-LOVERS at MIT-MC I have read the patent of Larry Nichols who assigned it to Moleculon and it looks to me like they have a very strong case against Ideal. One should keep in mind, however, that more than 70% of patent infringement cases go against the patentee. The basic drawings and descriptions in the patent deal with the 2^3 held together (non-rigidly) by magnets. Nichols discusses the possibility of fixing the cubies rigidly in his description in a manner not unlike Rubik. But the fact that he does not mention this aspect in his claims may be his undoing. He mainly stresses the magnetic attachment. I would not be surprised if the PTO told him that the rigid attachment would comprise another implementation and he wished to avoid the extra expense (I am speaking from experience here). He discusses the puzzle aspects and some of the higher order cubes and non-cubes we have seen on the market. In conclusion, I doubt he will get his 60M from Ideal but I think he will get a non-trivial percentage and a continuing royalty.  Date: 25 Jun 1982 0916-PDT From: ISAACS at SRI-KL Subject: Scientific American To: CUBE-LOVERS at MIT-MC The July issue is out, with an update on Cubes by Hofstadter. Don't miss it. Also, the first Rubik's Cube Newsletter from Ideal came. It's an 8 page glossy, mostly about Cube contests and Ideal's new puzzles. So far, mostly an Ideal advertising sheet. I got a letter from Ideal which said that "Revenge" won't hit California until August! Adam Alexander is doing a promotional tour for Ideal with his "Alexander's Star". He said it is not out in the East yet - we get the star, East coast gets the revenge, and the meet somewhere in the middle. He lives in N.Y., was (is) a mathematician, and has lots of Cube-like inventions and models, mostly built out of cardboard so far. -- Stan -------  Date: 25 June 1982 1225-EDT (Friday) From: Dan Hoey at CMU-10A To: Cube-Lovers at MIT-MC Subject: New Scientific American article Message-Id: <25Jun82 122513 DH51@CMU-10A> Date: 6 August 1980 16:52 edt From: Greenberg.Multics at MIT-Multics Subject: Re: 'cube lovers digest' I am really interested in when the _ Gardner is going to get his act together and publish THE cubing article-- each month for the last n I have eagerly taken the cover of SciAm and been disappointed. This is clearly THE mathematical game (since the inception of SciAm, with the possible exception of Conway's LIFE), and I wonder what he's waiting for. About a week ago, I actually dreamt that I opened the SciAm wrapper and found the Cube on the cover, introducing a WHOLE ISSUE about it (Social implications, Ancient Cubing, Cubing in the Soviet Union, etc...) Well, there finally was a Metamagical Themas column on the cube in March '81. Unfortunately, that article was mostly old hat to the readers of Cube-Lovers. The July '82 issue, just out, is a different story. Hofstadter discusses about twenty different ``Cube'' puzzles, based on all the regular polyhedra (and the tesseract), several arrangements of twist axes and several coloring schemes. It even mentions cubing in the Soviet Union! Stan Isaacs is the Cube-Lovers representative for the new article. At last I can believe that Hofstadter's column will replace Gardner's.  Date: 29 Jun 1982 2232-PDT From: ISAACS at SRI-KL Subject: Twisting Cube To: CUBE-LOVERS at MIT-MC Take a cube(3x3x3) and re-color it in two colors, as follows: Make 2 opposite faces with color 1 in the center, and color 2 on the rest. Call them "dots". Take another pair of opposite faces and make the 4 corners color 2 and the cross color 1. Call them "pluses". Finally take the last pair and make an "H" with color 1 and the 2 edges NOT adjacent to "dots" color 2. Call them "H's". In this cube, every edge is the same, so is each corner, and the centers. The result is a cube where you only have to solve orientation problems, and never need to position any cubie. However, it's not quite the entire orientation sub-cube - if you flip all 12 edges, you can't tell the difference. Can anybody come up with a coloring (2 or 3 colors - the centers could all be colored with a different color from the edges and corners. In fact, I guess you could use 6 colors; the only necessity is for all edges to be the same, and all corners the same. )... come up with a coloring which uses the complete twisting sub-group. By the way, an elegant solution to the "edge-only" cube is to recolor an Octahedron Cube so it is vertex centered. Much nicer than peeling all the corners of a regular cube. --- Stan -------  Date: 6 Jul 1982 2318-PDT From: Zellich at OFFICE-3 (Rich Zellich) Subject: Mailing-list for "List of lists" update notices To: All mailing-lists: cc: ZELLICH For those of you not previously aware of it, I maintain a master list of ARPANET mailing-lists/digests/discussion groups (currently 756 lines or ~29,000 characters) on OFFICE-3 in file: INTEREST-GROUPS.TXT For ARPANET users, OFFICE-3 supports the net-standard ANONYMOUS login within FTP, with any password. To keep people up to date on the large number of such lists, I have established a mailing list for list-of-lists \update notices/. I do not propose to send copies of the list itself to the world at large, but for those ARPANET users who seriously intend to FTP the updated versions when updated, I will send a brief notice that a new version is available. For those counterparts at internet sites who maintain or redistribute copies for their own networks (DECNet, Xerox, etc.) and can't reach the master by ARPANET FTP, I will send out the complete new file. I do \not/ intend to send file copies to individual users, either ARPANET or internet; our system is fairly heavily loaded, and we can't afford it. There is no particular pattern to the update frequency of INTEREST- GROUPS.TXT; I will occasionally receive a burst of new mailing-lists or perhaps a single change of address for a host or mailing-list coordinator, and then have a long period with no changes. To get on the list, send requests to ZELLICH@OFFICE-3, \not/ to the mailing-list this message appears in. Cheers, Rich -------  Date: Friday, 16 July 1982, 10:20-EDT From: Clark M. Baker Subject: Supplement to the Oxford English Dictionary To: cube-lovers at MIT-AI contains Rubic's Cube, described as an international craze of 1981 and invented in 1975 by Erno Rubik of Hungary.  Date: 28 Jul 82 22:20:52-PST (Wed) From: Scott.uci at UDel-Relay To: cube-lovers.uci at UDel-Relay Subject: the 4x4 cube (a.k.a. Rubik's Revenge) Via: UCI; 29 Jul 82 5:43-EDT Has anyone seen a 4x4 cube anywhere in Orange County, or even L.A.? I'd appreciate any leads on where/how to get one. I saw on (a friend's) in Santa Barbara, but he got it in Boston. I've also seen solution books (e.g. Crown Books). But the object itself is elusive. If you like the 3x3, I highly recommend the 4x4. If you have a scrambled one, I can solve it for you. -- Scott Huddleston  Date: 4 Aug 1982 08:41 PDT From: Mendelson.es at PARC-MAXC Subject: Supercube To: Cube-Lovers at MIT-MC cc: Mendelson.es I have been patiently(?) waiting for the 4^3 Supercube (Rubik's Revenge) to show up in the stores in the Los Angeles area. To my knowledge it has not done so. Does anyone know where it can be obtained in Los Angeles? If it cannot, is there anyone out there who will volunteer to send me one, all costs to be borne by me? Thanks for your responses. Jerry Mendelson  Date: Friday, 6 August 1982, 18:51-EDT From: Jonathan L. Handel Subject: Reagan cubed To: cube-lovers at MIT-MC, bsg at SCRC-TENEX Cc: jlh at scrc-tenex From The Nation, August 7-14, page 103 (a continuing column of Reagan's slip ups, called There He Goes Again: Reagan's Reign of Error): Reagan: Elizabeth Drew wrote in the June 21 New Yorker that in a White House meeting, the President lauded the inventor of Rubik's Cube as exemplifying the virtues of American free enterprise. Truth: Erno Rubik, the inventor of the cube, is a Hungarian professor living in Communist Budapest.  Date: 6 August 1982 19:51-EDT From: Allan C. Wechsler Subject: Invisible group of the 4^3 To: CUBE-LOVERS at MIT-AI The 4^3 doesn't have a supergroup in the sense of the 3^3 -- the orientations of the ceter cubies are determined by their positions. However, there is one fairly natural adjunct group that people might try thinking about and solving. A 4^3 shows 24 center cubies, 24 edge cubies, and eight corner cubies. But if it were really a solid cube chopped up by parallel slices, it would have eight more cubies buried inside. Call them stomach cubies. The eight stomach cubies form a 2^3 buried in the 4^3. They move when you twist slices. Can people come up with tools to frob the stomach cubies without disturbing the visible cubies? What is the order of the adjunct group? --- Allan  Date: 7 Aug 1982 1446-PDT From: ISAACS at SRI-KL Subject: 4^3 Supergroup To: CUBE-LOVERS at MIT-MC Of course there is a supergroup on the 4^3. Just number the 4 centers to see it. I think you can exchange any 2 pairs of centers. -- Stan -------  Date: 7 Aug 1982 1501-PDT From: ISAACS at SRI-KL Subject: POLYCUBE (on IBM PC) To: cube-lovers at MIT-MC I got the POLYCUBE program mentioned in the latest Scientific American cube article, which has from 1x1x1 up to 7x7x7 cubes on the IBM Personal Computer. I found it disappointing. It doesn't show the back of the cube, making solving it VERY difficult. It should also allow user-defined shorthand, so one could build macros (or simply define better notation). The notation is good for the general case, but hard for the 3^3 case and down - it is a general X-Y-Z notation, R or L direction, 1-n layer. Thus RX1 is our "R"; "ZL1" = U', etc. The colors are pretty. You can save a cube on disk if you haven't finished solving it, but only one. Why doesn't someone design and write a general group-theory puzzle simulation program. Draw any pattern (2 or 3 dimensions) on a screen, associate it with a matrix, name some permutations in the matrix for moves, and you should have any conceivable (drawable) rotating axis puzzle modeled. -- Stan -------  Date: 7 August 1982 20:02-EDT From: Yekta Gursel Subject: 4^3 is available in LA area in some of the CHESS and GAMES stores... To: CUBE-LOVERS at MIT-MC A friend of mine just got his for about $11. Apparently they were having an opening sale. I got mine a month ago from a friend in Boston. YEKTA@MC  Date: 9 August 1982 0737-EDT (Monday) From: Dan Hoey at CMU-10A To: Cube-Lovers at MIT-MC Subject: Invisible Revenge Message-Id: <09Aug82 073740 DH51@CMU-10A> When you have seen Rubik's Revenge, have you seen everything? Maybe not. Supergroups There is some confusion about the meaning of the term Supergroup for the larger cubes. There are two issues at stake: (1) In the 4^3 and larger cubes, there are pieces that may be permuted but are colored the same. Are positions that differ only in the permutation of identically- colored facets to be considered distinct? (2) In all odd-sized cubes, each face has a center facet that may be twisted. Are positions that differ only in the twist of face centers to be considered distinct? If the answer to (1) is `no', the puzzle is not a group, but a collection of cosets. I find this interesting only in that it is understood by the masses, and then the answer to (2) is `no'. If the answer to both (1) and (2) is `yes', we have what I would call the Supergroup. I mark my cubes so that I can distinguish patterns in this group. I discussed a way of doing this for the 3^3 in my message of 9 January 81 0551-EST. For Rubik's Revenge, I use a similar procedure, except now two spots must be cut out on each face: +-----------------------+ | | | | | | | *|* | | |-----+-----+-----+-----| | | *|* | | | | | *|* | |-----+-----+-----+-----| | | | *|* | | | | | | |-----+-----+-----+-----| | | | | | | | | | | +-----------------------+ Also, I now arrange the faces T-symmetrically, with the spots toward the girdle. Jim Saxe first brought it to my attention that we may answer `yes' to (1) and `no' to (2). We get a group that ignores face center orientation. This is probably what people mean when they say there is no Supergroup for Rubik's Revenge: This is the group of the puzzle, and the Supergroup (as I have defined it) is the same for the 4^3. But in the 5^3 this group is distinct both from the Supergroup and from the Color Cosets. The 57th Piece Alan Bawden, in his message of 5 June 82, despaired of explaining the mechanical workings of Rubik's Revenge. I will now rush in. If you take your Rubik's Revenge apart (as described in Ronald B. Harvey's message of 19 May 82, not recommended by Alan Bawden), you will find that the cubies ride around on a sphere: the mysterious 57th piece. Only the face centers are connected to the sphere; they have flanges that hold the edges in place, and the centers and edges have flanges that hold the corners in place. The tricky part is the linkage between the sphere and the face center cubies. The four center cubies of a face have extensions that together form a mushroom-shaped plug. Each plug extends from the center of the face inward and is cut in quarters lengthwise. There are six sockets on the sphere in which these plugs will fit and may rotate, but may not be pulled out. This is sufficient to implement a puzzle isomorphic to the 3^3, namely the 4^3 where we allow only face twists. The other necessity for a 4^3 is to be able to twist the six center slices, which we name after their adjacent faces. To accomplish this, there are grooves in the sphere that form three orthogonal great circles. Each groove has the cross section of half of a socket, and includes the four half-sockets that correspond to a center slice. When we twist one of those center slices, the half-plugs formed by pairs of face centers in that slice ride around the sphere in the grooves. Of course there is an adjacent center slice; when it is moved, it takes the sphere with it. The reason the grooves cannot have the cross-section of a socket is that then, when we twisted half the cube with respect to the other, the sphere might turn forty-five degrees, preventing center slice twists along the other great circles. When you twist the U center-slice of your cube, does the sphere move or stay fixed? To find out without taking the cube apart, hold the cube by the D center slice and repeatedly twist the U center slice clockwise. Don't touch the U or D faces while doing this. Mostly, the U face will turn and the D face will stay fixed, because the cubie-cubie friction is greater than the cubie-sphere friction. Eventually, however, you will either see the U face lag behind the U center slice, or the D face move to follow the U center slice. If the U face lags, the sphere is not moving; if the D face moves, the sphere is moving. I will take this opportunity to mention another feature of the interior sphere: It has screws in it. I took my screws out, but the sphere didn't come apart. Then I put them back in, on the `don't screw with it' principle. Perhaps they are there so Rubik's Revenge won't float? This issue was raised by Tom Davis (12 August 80) back when people were interested in solving the 3^3 underwater. The last I heard, Richard Pavelle (25 July 1980) was able to solve the cube with only five gulps of air. I imagine some of the 30-second whiz kids can solve Rubik's Cube while completely submerged. But Rubik's Revenge? Don't hold your breath. The Mechanical Invisible Group Alan Bawden's message posed an interesting question about the 57th piece, which I will state somewhat differently. Suppose we paint the sockets of the sphere according to the colors of the face centers that inhabit them. Then we mix up Rubik's Revenge and solve it. Must the sockets still match their face centers? The answer is no. In fact, the sphere may be in any of the twenty-four positions consistent with it having once matched the face centers. To show this, we will show how to perform a ninety-degree whole-cube move of the outside without moving the sphere. This is equivalent to turning the sphere ninety degrees, and we can repeat the procedure and its conjugates to move the sphere to any of its twenty-four positions. In order to twist Rubik's Revenge without moving the sphere, we place the cube in a position such that the U, F, and R center slices do not move the sphere and restrict ourselves to the moves: U1 Clockwise quarter twist of the U face U2 Clockwise quarter twist of the U half (face and center slice together) D1 Clockwise quarter twist of the D face U1',U2',D1' Counterclockwise quarter twists R1,R2,L1,R1',R2',L1' Likewise for R F1,F2,B1,F1',F2',B1' Likewise for F The move is actually fairly simple. The tricky part is moving the L center slice cubies without moving the sphere. To do this, remember the eight-flip X = (R1 L1 U1 D1 F1 B1)^2 from the 3^3. This exchanges the L and R center slices in the 4^3, allowing us to cycle the L center slice cubies in the R center slice. R2 L1' X R2 R1' X is a sphere-fixing whole-cube move taking 24 qtw after cancellation. The Theoretical Invisible Group So much for tawdry reality. As Allan C. Wechsler pointed out on 6 August 82, we can imagine a 4^3 puzzle that contains a 2^3 on the inside. If we solve the outside, must the inside be solved? The process shown above for the 57th piece implies not, for that process performs the RL antislice on the 2^3 with respect to the outside. Can any move of the 2^3 be accomplished? Elementary group theory says no, for odd permutations of the Rubik's Revenge edge cubies are also odd permutations of the 2^3's cubies. I ran the Furst, Hopcroft, and Luks algorithm on the problem. It turns out that the permutation parity is the only restriction on what can be done with the 2^3. Thus if we solve the outside, the inside may be in any one of the (8! 3^7)/2 positions obtainable in an even number of quarter twists on a 2^3 puzzle. This in particular includes all whole-cube moves of the 2^3. Unfortunately, I don't know any simple processes for whole-cube moves or for turning two adjacent faces.  Date: 11 Aug 1982 2204-PDT From: ISAACS at SRI-KL Subject: Tsukuda's Square To: CUBE-LOVERS at MIT-MC Just got a new group-theory puzzle, called Tsukuda's Square. It's sort of like the 15/16 puzzle, but harder. It consists of a 4x4 matrix in the center with the numbers from 1 to 16. On the left side are 4 plungers, one for each row. At the top is one plunger, which pushes down columns 2, 3, and 4, all at once. Both the top and the side plungers push the rows or columns over 1 square; releasing them causes the same row(s) or columns to slide back. When the top plunger is pushed down, the number 1 row plunger cannot be pushed because of interference; that is the only interference. A typical move, for instance, would be to push in the #2 row plunger, push the top, release the #2, release the top. This would change 1 2 3 4 5 6 7 8 (only looking at the top 2 rows) to 1 3 4 8 2 5 6 7. You can push 2 or 3 (or even4) side plungers at once. Pushing an adjacent pair is even useful in solving this thing. At any rate, it has the normal even parity problem - that is, you cannot exchange a single pair, but you can exchange 2 pairs, or permute three squares. Because of the limited moves, it is not nearly so simple as it looks. I can permute 3 in one specific position, and it takes me 12 moves to do so (where a move is a push of 1 or more side plungers, then push the top, then release them, or the reverse [push the top first]). Can anyone develop some more efficient algorithms? -- Stan -------  Date: 12 Aug 1982 08:49 PDT From: Mendelson.es at PARC-MAXC Subject: Re: Tsukuda's Square In-reply-to: ALAN's message of 12 August 1982 02:38-EDT To: ISAACS at SRI-KL cc: Cube-Lovers at MIT-MC Sorry, I don't have any answers for you. I'm strictly a hacker and not a theorist, but I always find puzzles such as the one you describe to be a lot of fun. I would like to suggest that whenever a new one such as this shows up the announcement of its arrival includes a reference to where it can be obtained. Jerry Mendelson  Date: 12 Aug 1982 1129-PDT From: Isaacs at SRI-KL Subject: re: Mendelson To: cube-lovers at MIT-MC In the south Bay Area, Tsakuda's Square can be bought at Tex's Toys at the San Antonio Shopping Center. They also, at last report (about 2 days ago) had Rubik's Revenge -- Stan -------  Date: 12 Aug 1982 1725-CDT From: Clive Dawson Subject: Rubik's Revenge To: cube-lovers at MIT-MC For those of you tracking the progress of Ideal's Rubik's Revenge into stores across the country, it has made it to Texas. Walgreen's drugstores here in Austin have put them on sale for $9.99, which is the best price I've heard of so far. They say the regular price after the sale ends will be $12.99, but stickers on the boxes show $15.99. -------  Date: 14 August 1982 17:23-EDT From: Alan Bawden Subject: [Hoffarth.wbst: Rubik's Revenge] To: CUBE-LOVERS at MIT-MC Date: 13 Aug 1982 10:41 EDT From: Hoffarth.wbst at PARC-MAXC To: Alan Bawden Re: Rubik's Revenge Is it best to send messages to you for forwarding? Last week Gold Circle, Rochester, N.Y. area, started advertising Ideal's Rubik's Revenge for $9.99.  Date: 14 August 1982 17:27-EDT From: Alan Bawden Subject: Where to send your messages. To: Hoffarth.wbst at PARC-MAXC cc: CUBE-LOVERS at MIT-MC Date: 13 Aug 1982 10:41 EDT From: Hoffarth.wbst at PARC-MAXC Is it best to send messages to you for forwarding? No it isn't. It is best to mail your messages directly to Cube-Lovers@MIT-MC, otherwise I have to forward your messages there myself before I can make a digest.  Mail-from: SU-NET host SU-SHASTA rcvd at 16-Aug-82 1204-PDT Date: Monday, 16 Aug 1982 12:04-PDT To: cube-lovers at Mit-mc Subject: Cube Mechanics From: Tom Davis I finally found a 4x4x4 cube a couple of days ago, and have a couple of interesting observations. Forgive me if I repeat anything said so far, but I have been ignoring everything on this list having to do with the 4^3 cube for fear of any sort of spoilers. Using my 3^3 knowledge, I found it fairly easy to get it almost solved. Half the time, however, I got it to the state where everything was solved except that two adjacent edge cubies were flipped. I finally convinced myself by means of a somewhat involved (and probably fallacious) "proof" that I would have to exchange them before I could solve the cube. My first observation is simply a trivial proof of that fact that I discovered immediately after I took the cube apart for the first time to see what was inside -- it is mechanically impossible to put the cube back together with the cubies flipped (but not exchanged). Some similar parity-type arguments can be made about possible configurations of the center cubies. What is interesting is that this presents a new method of proving things about configurations -- if one can dream up a mechanical model of a cube with different guts, it may be obvious that some sorts of things are impossible. The cube simply has to behave the same way externally. I wonder if there are nice ways to look at the various parity-trinity features of the three-cube by looking at it using a different model of the internal mechanics. My second observation is that although a 5^3 and a 6^3 may someday appear on the market, the 7^3 will be pretty tricky to build. When one of the faces of a 7^3 is turned 45 degrees, the corner will lie completely outside the original cube. Any mechanical linkage will be complicated indeed. Maybe a cube could be built with little microprocessors inside each cubie controlling little arms and hooks to grab adjacent cubie faces ... -- Tom Davis  Date: 17 Aug 1982 2103-PDT From: ISAACS at SRI-KL Subject: RUBIK's Magazine To: CUBE-LOVERS at MIT-MC I just got my first issue of "RUBIK'S", International Game Magazine, subtitled "Logic & Fantasy in Space 1/82". This is a magazine published in Hungary, Editor in Chief: Erno Rubik. "Responsible Editor": Norbert Siklosi. It contains "The Order of Disorder", by Peter Gnadig, likening the Cube to Entropy; a sketch of Rubik; the words to the song Mr. Rubik, "Games and Mathematics" by Gerzson Keri, articles on competitions, clubs, fans; on the company that manufactures the Cube, book reviews (Bosserts' book and Singmasters "Notes"), puzzles, advertisements, and more. I haven't finished reading all of it yet, but it seems to be an attempt at a serious cube magazine, a little stilted because of translation from the Hungarian. It's a quarterly, single issues are US $2.00, I guess the yearly subscription is $8.00. The address: Lapkiado Vallalat Kereskedelmi Iroda H-1906 Budapest P.O.B. 223 HUNGARY The Chorus and first verse of "Mr. Rubik" (out in England, by the Barron Knights Mr. Rubik Rubik Rubik Is your Cube from outer space? Mr. Rubik Rubik Rubik He got three sides then lost his place Mr. Rubik Rubik Rubik He just twists your cube all day This ain't my idea of child's play. Being the kind of guy I am, I told him I would try To help him solve the secrets of the cube that made him cry Well that was thirty days ago and half a million moves My wife's black and blue 'cos I keep dreaming she's a cube. ----Stan -------  Date: 18 Aug 1982 2226-PDT From: ISAACS at SRI-KL Subject: KERI's Article To: CUBE-LOVERS at MIT-MC The "Games and Mathematics" article in "Rubik's" Magazine (mentioned yesterday) asks an interestion question: how can you characterize the random coloring on a cube in order to determine if the cube is 1) solvable by twisting, or 2) solvable by dismantling and reassembling. The obvious criteria are 6 colors, 9 of each, 4 on edges, 4 on corners, 1 on a center, no 2 facies of a cubie the same color. For case 2, Keri claims you need 4 more tests. For instance, he gives test 1: Given 1 corner with colors A, B, and C, let the other 3 colors be a, b, and c. Then you can't have a capital and small of the same letter on one corner, and the 8 corners are exactly the 8 combinations. What are the other tests? Are 4 really necessary? What are the tests for case 1? By the way, he (from some other article) classifies the 3 unscrambling methods as follows: 1) Chemical unscrambling: repaint the sides. 2) Physical unscrambling: dismantle and reassemble 3) Mechanical (or mathematical): normal way, by twisting. --- Stan -------  Date: 20 August 1982 0242-EDT From: James.Saxe at CMU-10A (C410JS30) To: Cube-Lovers at MIT-MC Subject: Rubik's Revenge problem--deep & shallow hypermoves Message-Id: <20Aug82 024233 JS30@CMU-10A> Consider all manipulations of Rubik's revenge as consisting of two sorts of moves, namely (1) shallow moves, which turn an outer layer with respect to the remaining three layers, and (2) deep moves, which turn an outer layer and the adjacent inner layer with respect to the remaining two layers. [For the purposes of this problem, we will regard a manipulation that turns only an inner layer--resulting, for example, in faces that look like XXXX OOOO XXXX XXXX when applied to a solved cube--as consisting of two moves, one deep and one shallow, in opposing directions.] If only shallow moves are permitted, the 4x4x4 simulates a 3x3x3. If only deep moves are permitted, the 4x4x4 simulates a 2x2x2. Define a shallow (deep) hypermove as an arbitrary sequence of shallow (resp. deep) moves. My question is: What is the maximum number of hypermoves required to solve the 4x4x4? Notice that the answer to this question may depend on whether or not one considers identically-colored face centers to be distinct (as Hoey points out, the puzzle is not a group if identically-colored face centers are not distinguished) and on whether or not one worries about the positions of the eight hypothetical stomach cubies. Also, if the minimal number of hypermoves is odd, then it might be important to start with one class of move. That is, it is plausible that sequences of the form SDSDS may be sufficient while sequences of the form DSDSD may not. Jim Saxe  Date: 23 August 1982 1623-EDT (Monday) From: Dan Hoey at CMU-10A To: Cube-Lovers at MIT-MC Subject: Hypermove lower bounds CC: James Saxe at CMU-10A Message-Id: <23Aug82 162336 DH51@CMU-10A> There is an easy lower bound on the number of hypermoves needed to solve Rubik's Revenge. If we distinguish like-colored face centers, let us fix the BL center of the D face, permitting only the shallow moves B1, F1, U1, U3, L1, R1, and their inverses, and the deep moves F2, U2, R2, and their inverses. Let us compute the number of hypermoves needed to solve just the face centers of Rubik's Revenge. A shallow hypermove can achieve SF = 4^6 = 4096 different face center positions. A deep hypermove can achieve DF = 7! 3^6 = 3674160 different face center positions. So in four hypermoves, at most 1 + (SF + DF) + 2 SF DF + (SF + DF) SF DF + 2 SF DF SF DF = 453,021,789,719,303,692,337 face center positions can be achieved. Since this is fewer than the 23! = 25,852,016,738,884,976,640,000 face center positions of Rubik's Revenge, some face-center positions will require at least five hypermoves. If like-colored face centers are not distinguished, the best lower bound I can find using this method is three hypermoves. If stomach cubies are considered, I think both bounds increase by one, since only deep moves can touch them. It seems strange that this method relies only on the face center solution. Similar arguments about edges are not as good, because so many edge positions are achievable using shallow hypermoves. Corners are practically irrelevant, since they can be fixed using only shallow hypermoves. With respect to the question of odd sequences of hypermoves, Jim Saxe mentions that ``it is plausible that sequences of the form SDSDS may be sufficient while sequences of the form DSDSD may not.'' I would like to add the further plausibility that both types may be sufficient, while neither may suffice alone.  Date: 31 Aug 1982 1022-PDT From: ISAACS at SRI-KL Subject: re:hypermove lower bounds To: Hoey at CMU-10A cc: cube-lovers at MIT-MC, isaacs at SRI-KL Dan: Could you explain to me how you got the formula of you message of 23 Aug. I can see more-or-less what you're doing, but I haven't been able to parse the formula. Also, I haven't seen your notation before. I take it that "U3" is equivalent to "D1'", except hold the D1 in place. How do you represent half twists? Only by two quarters, or is there a shorthand? From a group theory perspective, is it easier to talk about hypermoves than slice moves? Will that also be true on the 5^3, 6^3, etc? From a solving perspective, it seems clumsy. -- Stan -------  Date: 2 September 1982 0755-EDT (Thursday) From: Dan Hoey at CMU-10A To: ISAACS at SRI-KL Subject: Re: Hypermove Lower Bounds CC: Cube-Lovers at MIT-MC In-Reply-To: ISAACS@SRI-KL's message of 31 Aug 82 12:22-EST Message-Id: <02Sep82 075524 DH51@CMU-10A> Date: 31 Aug 1982 1022-PDT From: ISAACS at SRI-KL Could you explain to me how you got the formula of your message of 23 Aug. There are SF different shallow hypermoves and DF different deep hypermoves, and two similar hypermoves in succession can be collapsed into one. The formula expresses the number of alternating sequences of length at most four, which is the sum of the ``Number'' column below. Length Type Number 0 1 1 S SF 1 D DF 2 SD SF * DF 2 DS SF * DF 3 SDS SF^2 * DF 3 DSD SF * DF^2 4 SDSD SF^2 * DF^2 4 DSDS SF^2 * DF^2 I can see more-or-less what you're doing, but I haven't been able to parse the formula. Well, I did leave out the multiplication signs. Also, I haven't seen your notation before. I take it that "U3" is equivalent to "D1'", except hold the D1 in place. Right. I used this notation in "Lower Bounds for the 4x4x4" on 2 June and "Invisible Revenge" on 9 August. How do you represent half twists? Only by two quarters, or is there a shorthand? There is U3^2, not much of a shorthand. For the U slice move, I hinted at U21'. From a group theory perspective, is it easier to talk about hypermoves than slice moves? Hypermoves are a curiosity that Jim Saxe dreamed up. Any sequence of depth 1 (or 3) moves is a single hypermove, as is any sequence of depth 2 moves. I assume you mean to ask whether it's easier to talk the way I usually talk, in terms of what I will call "twist moves" to distinguish them from "slice moves". The question boils down to what set of generators (moves) you want to use when counting the length of a process. This topic was endlessly rehashed in 1980 when people were trying to decide whether to call a half-twist a single move or two. Jim Saxe nearly sent a message in 1980 about using only two generators to solve Rubik's cube. [As I recall, computer failure trashed the message and he never retyped it.] Certainly we can all do slices and half-twists. The question is how many moves to charge for such an operation. The richer the set of generators, the fewer the number of moves, but the more complex the explanation of the generators. I use the "quarter-twist" convention for the 3^3. The generators are 90-degree rotations of faces. This seems natural, because it is the minimal set that satisfies the following criteria. 1. Every possible cube position can be created using these generators, up to whole-cube moves. This is a basic criterion. 2. The inverse of every generator is a generator. This is necessary so that we have a metric. 3. Any position that can be reached by performing part of a generator is a generator. This criterion ties the mechanical operations used in the cube to the permutation group. Otherwise we could have generators like FUF and F'U' and perform F with their composition. Charging two moves for F in that circumstance is somewhat bizarre. 4. Every M-conjugate of a generator is a generator. This is an aesthetic consideration. We could leave out the D and D' twists and still solve the 3^3, but that breaks up the symmetry of the puzzle. Why do I want the generator set to be minimal? Well, we could make it maximal, but then we would have ``over 3 billion'' generators. What I am looking for is a canonical set, and minimality seems like the best way of choosing among metrics. Thus we exclude slice moves as generators because they are not necessary. For the 3^3, this set is particularly fortunate, because the converse of criterion 4 holds: Every two generators are M-conjugate. This allows us to identify some local maxima without long computations (14 December 1980) and to tighten lower bounds using parity principles (9 January 1981). Will that also be true on the 5^3, 6^3, etc? I would just as soon stick with a compatible metric. This is not to say that there cannot be abbreviations for these moves, simply that for the sake of asking ``how many moves does this take'' we count the number of quarter-twists. We unfortunately don't have the converse of criterion 4 for cubes larger than 3^3. For the 4^3, for instance, there are two flavors of move: deep and shallow. Dave Plummer (26 September 1981) described certain positions of the 4^3 as local maxima, but I have convinced him that we cannot demonstrate the truth of that assertion using known techniques other than exhaustive search. My note of 2 June was able to use only one kind of parity in the lower bound argument. Both problems are due to the lack of the converse. From a solving perspective, it seems clumsy. I'll agree that the generators are few in this scheme, but it is possible to generate macros. For instance, consider describing a slice move on the 3^3 cube. Singmaster uses the notation Fs to denote the F1B1' = F12'3 slice move. We can now also talk about the F12' slice move, which is how everyone actually does the move. I think this is much easier to remember than Allan Wechsler's IJK notation (introduced 18 July 1980) or Doug Landauer's HPS notation (27 August 1980) for dealing with whole-cube moves. I'm still not too happy about the state of RubikSong, but I think it's a matter of human engineering, and I like to stick with the mathematics.  Date: 9 Sep 1982 1453-PDT From: ISAACS at SRI-KL Subject: IMPOSSI-BALL To: CUBE-LOVERS at MIT-MC If you live in the SF Bay area, BESTs now has the Impossi-ball, as well as most of the other better group-theory puzzles at fairly reasonable prices. Rubik's Revenge is $9.00. The Impossi-ball was in the Scientific American column; its the icosahedron-like ball with 2 each of 6 colors (as Alexanders Star). If fact, I think it is equivalent to the star. It doesn't look as good, but the mechanism is more clever; what's more, you can remove one face and use it as a sliding block puzzle. Has anybody seen the other new puzzles from the column? -- Stan -------  Date: 15 September 1982 12:36-EDT From: Richard Pavelle To: CUBE-LOVERS at MIT-MC Ideal is sponsoring another cubathon (3x3) on October 9 at the South Shore Plaza in Braintree. Prizes will be awarded in various categories but unless you are under 40 seconds you will not be good enough. I understand that Ideal has sold about 22 million cubes but their world-wide rights have not helped too much. They estimate that 3 times that number have been sold by illegal competition.  Date: 17 Sep 1982 0918-PDT From: Dave Dyer Subject: Cubes in L.A. To: cube-lovers at MIT-MC I picked up my 4x4x4 at Toys R Us for $10. They also have a good selection of 2x2x2, 3x3x3 "cubes" in various shapes and other cube inspired puzzles. -------  Date: 22 Sep 1982 1425-PDT From: ISAACS at SRI-KL Subject: OMNI cubes To: CUBE-LOVERS at MIT-MC The current OMNI (Oct.) has mention of various "cubes" from Meffert, in the Games section. Doesn't say much, but has nice pictures. The main one of interest is the Megaminx, a cube-like cutting of a dodecahedron, so each face has 5 corners, 5 edges, and one fixed center (which only rotates, but does not move relative to the others), exactly as in the cube. Scot Morris claims the Cube has 10^19 combinations, the Revenge has 10^48, and that the Megaminx has 10^69. Is this correct? Meffert claims the Megaminx is harder than the Revenge, but I would bet it is easier, in the sense that cube moves should move over easier, and do closer to the same things as before. -- Stan -------  Date: 23 September 1982 1206-EDT (Thursday) From: Dan Hoey at CMU-10A To: Cube-Lovers at MIT-MC Subject: MegaMinx and orientation theory In-Reply-To: ISAACS's message of 22 Sep 1982 1425-PDT Message-Id: <23Sep82 120644 DH51@CMU-10A> Funny you should ask about the positions of the MegaMinx. I analyzed this puzzle with the FHL algorithm. It was the longest run of that algorithm that I have heard of, something like 20 CPU hours. Necessity was the mother of checkpointing. The standard MegaMinx has 20 corners and 30 edges. Both the edge and corner permutation parities must be even. Corner and edge orientation parities are zero (mod 3) and (mod 2), respectively. These are the only invariants, so there are twenty-four orbits and (20! 3^20 30! 2^30)/24 ~ 1.007E68 positions. In the SuperMegaMinx, a 1/5 twist of a single face center is possible, so there are (20! 3^20 30! 2^30 5^12)/24 ~ 2.458E76 positions. I think the MegaMinx is going to be easier than Rubik's Revenge, not because there is more carryover from the cube, but because there is more of the puzzle that is not affected by a single generator. Computing a lower bound that takes account of generators that commute is going to be difficult, however: there are triples of commuting generators, and we can find commuting triples {A, B, C} and {A, B, D} such that C and D do not commute. A curious question: What do we mean by the corner orientation of the MegaMinx when the corner permutation is not the identity? On the cube, we took the corner colortabs on two opposite faces of a solved cube and marked both the tab and its position. Then in a scrambled cube we counted the position of each marked corner colortab relative to the marked position on its cubie. This procedure doesn't work on the dodecahedron: where should we mark the tabs on those corner cubies that are not on the two opposite faces? It was a year before I realized that the choice of placing the marks on opposite faces of the cube was arbitrary. The marked tabs and positions can be chosen any way that gives us one marked tab and one marked position per cubie, and has zero orientation parity in a solved cube. It is customary to enforce the second criterion by requiring that marked positions be the positions of marked tabs in a solved cube. The same argument holds for edges, and both generalize to the MegaMinx. Why is there orientation parity? When we twist an n-gon face, the edge cubies move in an n-cycle and their colortabs move in two n-cycles. We call this an ``untwisted cycle''. But we could conceive (see below) of a puzzle where the edge colortabs would move in a single 2n-cycle. This would be a ``twisted cycle'', and would violate orientation parity. Similar arguments hold for corner cubies, whose kn tabs must move in k n-cycles rather than k/d nd-cycles, for d a divisor of k (note that k=4 for the icosahedron (but the icoshahedron has two corner tab orbits, so quite a different argument applies)). To summarize, any puzzle that moves pieces in untwisted cycles must preserve orientation parity. I wonder if some argument of this type can be made for the tesseract. The argument depends on the orientation group being Abelian (it has in fact been cyclic in our examples), and at least some of the hypercubies have nonabelian orientation groups. Perhaps we have to use Abelian quotients of the orientation groups. Has anyone seen the paper on the tesseract that was mentioned by Hofstatder? Retreating to three dimensions, let's consider a variant on Rubik's cube. Suppose you had a gear (heh) embedded in the URF and BLD corners, such that whenever an edge cubie passed by one of these corners (e.g. from UR to UF by twisting U) it would flip (go from UR to FU). This would violate EOP on every quarter-twist! Of course, you know what positions would be possible in such a puzzle. So now let's consider gears in the FU, BL, and RD edges that twist corner cubies as they go by. I don't know about this puzzle, but I suspect that there are only four orbits.  Date: 26 September 1982 22:12-EDT From: Martin Minow Subject: Solving Rubic's Cube To: CUBE-LOVERS at MIT-AI The fall catalog of the Cambridge Center for Adult Education had a course (8 1-hour sessions) on solving Rubic's Cube. It was apparently cancelled for lack of interest. Martin Minow decvax!minow @ Berkeley  Date: 28 Sep 1982 2102-PDT From: ISAACS at SRI-KL Subject: Meffert Catalogue To: cube-lovers at MIT-MC I recieved a copy of the Meffert "Pyraminx" catalogue a couple of days ago. The only solid axis-rotation puzzles not already mentioned in the Scientific American article or the OMNI article are the Pyriminx Magic Triangle, Pentagon, and Hexagon, 3 puzzles, all prisms of the named shape, each side face with 9 squares, and the top and bottom the appropriate n-gon with corners, edges, and a center. Ie, the Magic Hexagon has a hexagon on top and bottom, with 6 corners and 6 edges. These should be similar to a cube, except, of course, the side faces cannot make quarter turns. Also in the catalogue are a junior and senior "Magic Barrell", a 3-layered cylinder with 2 (jr) or 3 (sr) arcs cut down from top to bottom, which look as if they allow half twists; a Double Pyraminx, an octagon with 4 triangles per side, which I would assume twists about the 4 central hexagons, and is mechanically similar to the Skewb. Also shown are some textured pyraminxes, the Pyraminx Star, with small tetrahedra on several Pyraminx faces (it looks strange, but is equivalent to the Pyraminx, I assume); some bead puzzles (a version of the Hungarian Rings, another related to the Orb), some sliding block puzzles, a group of tetrahedron dissections (into 2, 3, 4, 5, and 6 pieces),and a couple of jigsaw-like puzzles. It's quite a catalogue, but he says most of the items are not yet in production! More waiting. -- Stan -------  Date: 28 Sep 1982 2118-PDT From: ISAACS at SRI-KL Subject: Cubic Curcular and Scientific American To: cube-lovers at MIT-MC Ole Jacobson is visiting, and he brought the new issue of David Singmasters Cubic Circular. It is labeled Issues 3&4; neither of us ever recieved issue 2. Did anyone else? This issue contains more anecdotes, reports on contests, new cube products (try combining pieces from octagonal prism-cubes and regular cubes in various ways..someone even made "Siamese Cubes", gluing parts of 2 cubes together), some processes for the Revenge, some information on the 5^3 (a prototype exists by 21-year-old Gaston Saint- Pierre of Quebec; Singmaster claims 2.83E74 patterns), an analysis and theorems about magic polyhedra in general, more pretty patterns, a magic disc, and a table of the 73 possible orders of elements with how many elements are in each order (average order is 122, median is 67.3, and the order with the most elements is 60, with 4.6e18 elements or about 10.6%). Also, this months Scientific American has the cube again in Hofstadters' "Metamagical Themas", this time as an example of creativity via variations on a theme. Enough writing for tonight. -- Stan -------  Date: 29 Sep 1982 18:16:54 EST (Wednesday) From: Mike Meyer Subject: Re: Meffert Catalogue & Cubic Curcular and Scientific American In-Reply-to: Your message of 29 Sep 1982 18:48 EDT To: Alan Bawden Cc: Cube-Lovers at MIT-MC I have issue number 2 of the cubic circular, and have had it since late apr. If you are interested, I can forward a summary. mike  Date: 22 Oct 1982 19:09-EDT From: Dan Hoey at CMU-10A Subject: The 2x2x2x2 magic tesseract To: Cube-Lovers at MIT-MC Allan Wechsler's message of 17 May 1982 contains some interesting comments on the four-dimensional hyper-cube, or tesseract. I will expand on them, and offer a correction. The tesseract has eight cubical sides, labeled Back, Front, Up, Down, Left, Right, Out, and In. Each side may be twisted in any of the twenty-four ways that a cube may be rotated in three-space. Since these twenty-four twists are generated by repeated application of the six quarter-twists of the cube, I consider a move to be a single quarter-twist of one of the cubical sides. I have picked three of the quarter-twists of the Out side to be the ``clockwise'' twists, given as Of, Ou, and Or below. Given the constraint that clockwise twists must be conjugates of each other with respect to the movement group of the tesseract in four-space, the remaining clockwise quarter-twists are determined. In the following list, the upper-case letter denotes the side to be twisted, and the quarter-twist is displayed as a permutation on the (square) faces of that (cubical) side. Of=(URDL) Ou=(RFLB) Or=(FUBD) If=(RULD) Iu=(FRBL) Ir=(UFDB) Ro=(UFDB) Rf=(OUID) Ru=(FOBI) Lo=(FUBD) Lf=(UODI) Lu=(OFIB) Ur=(OFIB) Uo=(FRBL) Uf=(ROLI) Dr=(FOBI) Do=(RFLB) Df=(ORIL) Fu=(ORIL) Fr=(UODI) Fo=(RULD) Bu=(ROLI) Br=(OUID) Bo=(URDL) These twists have the satisfying property that when two different twists move an edge from position E1 to position E2, then one of the twists is clockwise and the other counterclockwise. For instance, both the Dr and the Fr' twists move an edge from FID to FOD. Another property that mimics the three dimensional cube is that clockwise twists on opposite sides are reversed: The action of Of on the O side is the inverse of the action of If on the I side. To see how the table above is constructed we must describe the movement group of the tesseract (the group of whole-tesseract moves in four-space). I look at it as operating on quadruples VWXY of mutually adjacent sides. To see if VWXY->V'W'X'Y' is in the group, replace all occurrences of B, D, L, and I with F, U, R, and O, respectively. The resulting permutation must have the same parity as the number of replacements performed. Thus FLOD->UROB is in the group, because we perform three replacements to form FROU->UROF, an odd permutation. To tell whether a quarter-twist is clockwise or not, take the side V to be twisted, two consecutive letters WX from the permutation, and a fourth, orthogonal letter Y from {F, U, R, O}. If VWXY->OURF is in the movement group of the tesseract, then we have the clockwise quarter-twist Vy, otherwise the counterclockwise quarter-twist Vy'. For instance, if we twist the U side as (LFRB), then VWXY=ULFO->OURF is in the movement group (one replacement creates the four-cycle URFO->OURF), so we have the clockwise twist Uo. Let us now examine the reachable configurations of the corners of the tesseract. Every quarter-twist moves eight corners in two four-cycles, so only even permutations of the corners are achievable. The orientations of the corners are more complex. If we move corner VWXY to V'W'X'Y', then VWXY->V'W'X'Y' must be in the movement group of the tesseract. Thus only half of the twenty-four permutations of {V', W', X', Y'} are achievable, because of the permutation parity constraint. To define the orientation of the corners, we label the sides of each corner and each corner position with the letters VWXY, and read the orientation of a cubie as the letters it has in the V, W, X, and Y sides of its position. It is important here to obey the the permutation parity constraint when doing the labelling, so that each cubie may be placed in the home (VWXY) orientation in any position. For instance, one possible labelling is as follows, where each column refers to a corner: V F F F F F F F F B B B B B B B B W U U U U D D D D U U U U D D D D X R O L I O L I R O L I R R O L I Y O L I R R O L I R O L I O L I R Thus if the FURO corner (column 1) is in the FLUO position (column 2), then its orientation is VXYW. Any orientation that is an even permutation of VWXY is possible. The group of orientations, A4, is the same as the movement group of the tetrahedron (with vertices labeled VWXY) in three-space. As I suggested in my message of 23 September 1982, this orientation group is not Abelian, so the orientation of the last corner is not completely determined by the orientations of the other fifteen. To see what is determined, let us look at the tetrahedron with vertices at half of the corners of a three-dimensional cube, say the FUR, FDL, BUL, and BDR corners. As I reported on 15 June 1982, the twelve movements of the tetrahedron consist of the identity, three 180-degree rotations, and eight 120-degree rotations. The June message also mentions that if those corner cubies are moved as a unit, preserving their positions and orientations relative to each other, then the 180-degree rotations are achievable on the 3^3 (they are the corners of the Zig-Zag pattern) but the 120-degree rotations violate the corner twist invariant. Of course, four of them perform a net clockwise twist, and four of them perform a net counterclockwise twist. Define the twist of a tetrahedron movement to be the net clockwise twist it applies to the corners of the cube. Thus the twist of the movements of a tetrahedron VWXY is given by the following table. Twist 0: VWXY, WVYX, XYVW, YXWV Twist 1: VXYW, WYXV, XVWY, YWVX Twist 2: VYWX, WXVY, XWYV, YVXW By reasoning about the actions on corners of the cube, it is clear that the twist of the product of two movements is the sum of their twist, modulo three. Thus the twist group is an Abelian quotient of A4, isomorphic to the cyclic group on three elements. Since the orientation group of the tesseract corners is also A4, we may use the twist group to construct an orientation invariant of the corners of the tesseract. As described in the September message, each qtw moves the corners in untwisted cycles, so the sum of the twists of the orientations of the corners must be zero, modulo 3. I ran the Furst, Hopcroft, and Luks algorithm on the 2^4 tesseract and found that this is the only invariant of corner orientation. Therefore, the number of reachable positions of the tesseract is (15! / 2) (12^15 / 3) ~ 3.358 x 10^18. This is larger than Allan Wechsler's upper bound because he thought there were only six orientations of each corner.  Date: 25 Oct 1982 0846-PDT From: ISAACS at SRI-KL Subject: megaminx; octahedron To: cube-lovers at MIT-MC The Megaminx is now on sale in the San Francisco Bay Area. It is, as pictured, a dodecahedron with each face twistable in fifths, and containing 5 corners, 5 edges, and a center. Instead of 12 different colors it seems to have 10, with the red and yellow duplicated; that means you can have a parity problem at the end if you exchange the duplicate edges. Solving seems to be pretty straightforward, except new edge moves must be developed - the "slice" moves aren't very effective. I also found an octahedron much more analogous to the cube then the one with the 9 triangular faces (and independent vertices). ON this one there is a triangular center, 3 diamond shaped corners, and longish edges on each face. The centers are equivalent to the corners of a cube, but monochromatic; the corners of the octahedron are equivalent to the centers of a cube, but have 4 colors. You can solve it with supergroup cube moves, but it can be solved more efficiently, I think, by doing the corners first (matching up the colors), then the edges, then the centers. Needed are moves that move edges without twisting corners (do we have any good corner moves that don't rotate centers on a cube?) One final puzzle that has come out - it's called Inversion, and it's a sliding block cube. There are 19 identical cubies, each colored half red and half blue (3 faces each) and arranged around the edges of a cube with one extra empty space. They are held in place by a Rubiks-like mechanism through the centers of the big cube. Thus each edge of the big cube has 3 of the little cubes, or 2 of them plus a space. The little cubes are each in one of the 8 possible orientations; One orientation is represented by one cubie, 3 orientations by 2 cubies each, and 4 orientations by 3 cubies each. The idea is to slide the cubies around the edges of the big cube so that the outside is all red or all blue, or some other regular pattern; Inverting from (say) red to blue means sliding all cubies to more-or-less the diagonal opposite positions. -- Stan -------  Date: 1 Nov 1982 0902-PST From: ISAACS at SRI-KL Subject: Re: megaminx; octahedron To: BSG@SCRC-TENEX at MIT-MC cc: cube-lovers at MIT-MC, Isaacs at SRI-KL In-Reply-To: Your message of 29-Oct-82 0922-PDT Yes, I solved it. It didn't take much work at all - mainly deciding which cube moves transfered well. The only really diferent moves involved switching edges on the last face, or flipping them. Both had fairly easy solutions. I do one face first, and then work upwards from there. There is a 7 move sequence on the cube, which moves an edge from the upper face to a middle edge, which transfers nicely to the Megaminx. Corners just work, without any special algorithms. The 4 move commutator F R' F' R moves to the Megaminx and will move corners around. Of course, none of the slice moves work, so new edge moves are needed. We also need an easy way to move centers, and to move stars around, to make pretty patterns. By the way, the one two I got, and several a friend got in L.A., all had 10 colors, with red and yellow duplicated. However, last Friday I saw ones with 12 colors (I think - they were in boxes), with dark green and brown as the new colors. I don't know if the 10 colors are a mistake or intentional. The presence of a duplicate edge piece makes the solution slightly more difficult. -- Stan -------  Date: 5 Nov 1982 16:54:01-EST From: acw at scrc-vixen To: rp at scrc-vixen Subject: "The Cube Lovers at MIT" Cc: CUBE-LOVERS@mit-oz The article could be salvaged without damaging it too much. Cube-Lovers could be described as a group based at MIT which communicates via electronic mail. All mention of ARPA and even the word "network" should be eschewed, and the nationwide nature of Cube-Lovers should be downplayed. The rest of the article can stand, as far as I am concerned. Although I think the style is a little dippy, I'm sure he doesn't want my advice about it. --- Allan  Date: 11 Nov 1982 1421-PST From: Carolyn Tajnai Subject: Pentominoes To: cube-lovers at MIT-MC My husband, Joe, has requested a set of pentominoes on his Christmas wish list. Don Woods suggested I sndmsg to cube-lovers for pointers. Where can I buy a set? Help! Carolyn (please snd answers, solutions, suggestions to CSD.Tajnai@SU-SCORE) -------  CEL@MIT-ML 11/23/82 16:00:18 Re: Pentominoes request To: Cube-lovers at MIT-MC Hexominoes are easier to find than pentominoes. If you treat four of the basic trianglular pieces of the Snake puzzle as a unit square, a 24-piece Snake can be configured to make any hexomino. Just buy enough snakes. If you want to break the rivets on the snake, you can make pentominoes by removing four pieces. In fact, I wish there were a modular Snake consisting of snap-together pieces so that different length Snakes could be realized more simply.  Date: 29 November 1982 22:51-EST From: Alan Bawden Subject: [YEKTA: "Pentominoes"...] To: CUBE-LOVERS at MIT-MC Date: 24 November 1982 19:57-EST From: Yekta Gursel To: ALAN, CSD.Tajnai at SU-SCORE Re: "Pentominoes"... GABRIEL toy company markets a nice set of pentominoes under the name "HEXED". They are the makers of the puzzles "Hi-Q" and "PYTHAGORAS". The latter is a nice set of "tangrams" if you are interested in them. You should be able to get any of these in any well stocked toy store ( like Toys 'R Us ). If you are interested in getting solutions for them, me ( Yekta Gursel ) and a friend of mine ( Douglas Macdonald ) wrote a program on a microcomputer to solve any pentomino problem. The description of the program appeared in November 1979 issue of BYTE. It is written for a COMMODORE PET, but can be adopted toi any microcomputer relatively easily. If you do not have access to a microcomputer we can print the solutions out for you just for the cost of paper and postage ( a few dollars ). If you are interested please let me know. My network address is YEKTA@MIT-MC. [ ALAN: If anybody asks about pentominoes, you can refer them to me... Thanks!] Date: 25 November 1982 05:12-EST From: Yekta Gursel To: ALAN Re: Addition to my pentominoes message... Sorry, I accidentally sent the pentominoes message to you instead of the CUBE-LOVERS. What I wanted to add is that one can also get a set of hexaminoes, though this is a little bit harder. They are marketed in England under the name "Spear's Multipuzzle". The set comes with a few extra pieces, but nevertheless contains a full set of hexaminoes. The quality is not as good as Gabriel's "HEXED".  Date: 30 Nov 1982 18:16-EST From: hoey Subject: Pentominoes To: CSD.TAJNAI at SU-SCORE, Cube-Lovers at MIT-MC Message-Id: <82/11/30 1816.433@NRL-AIC> Origin: NRL-AIC A friend of mine has a set of pentominoes called ``Quintillions'', available from Kadon Enterprises, Inc. 1227 Lorene, Suite 16 Pasadena MD 21122 Basic Quintillions is a set containing the twelve planar pentominoes for $29.00. Super Quintillions is a set containing the nonplanar pentominoes for $39.00. (The literature says there are eighteen pieces, but I think there are only seventeen nonplanar pentominoes. Perhaps they include some extra piece so you can make a 3x5x6 rectangular prism) The basic set is a high-quality puzzle, each piece laser-cut from a single piece of wood, with a velour box. If you want to make your own, you can glue little cubes together. Here are patterns for the planar pentominoes: X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X For the nonplanar pieces, let U, D, and B stand for cubes on the upper level, lower level, and both levels. The pieces I know of are as follows, where an asterisk denotes a piece that can be reflected to form a new piece: * * * * * * B D D D B D D D B D U B D D B D D B D D D B D D B D B D D B D D D D D B B D B D D D  Date: 30 Nov 1982 18:16-EST From: hoey at CMU-10A Subject: Pentominoes To: CSD.TAJNAI at SU-SCORE, Cube-Lovers at MIT-MC A friend of mine has a set of pentominoes called ``Quintillions'', available from Kadon Enterprises, Inc. 1227 Lorene, Suite 16 Pasadena MD 21122 Basic Quintillions is a set containing the twelve planar pentominoes for $29.00. Super Quintillions is a set containing the nonplanar pentominoes for $39.00. (The literature says there are eighteen pieces, but I think there are only seventeen nonplanar pentominoes. Perhaps they include some extra piece so you can make a 3x5x6 rectangular prism) The basic set is a high-quality puzzle, each piece laser-cut from a single piece of wood, with a velour box. If you want to make your own, you can glue little cubes together. Here are patterns for the planar pentominoes: X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X For the nonplanar pieces, let U, D, and B stand for cubes on the upper level, lower level, and both levels. The pieces I know of are as follows, where an asterisk denotes a piece that can be reflected to form a new piece: * * * * * * B D D D B D D D B D U B D D B D D B D D D B D D B D B D D B D D D D D B B D B D D D  Date: 2 December 1982 06:14-EST From: Richard Pavelle To: CUBE-LOVERS at MIT-MC The times for REVENGE are now under 90 seconds!!  Date: 2 December 1982 06:42-EST From: Richard Pavelle Subject: material for Ripleys' To: CUBE-LOVERS at MIT-MC The following are true: The times for REVENGE are now under 90 seconds! The one-handed record for the 3x3x3 is 89 seconds!! The time for the 3x3x3 with two feet is 2 minutes and 50 seconds!!! Does any have times for no hands and no feet?  Mail-From: CMUFTP host CMU-EE-AMPERE received by CMU-10A at 3-Dec-82 03:24:38-EST Date: 3 Dec 1982 03:19:57-EST From: Phil-Servita-H at CMU-EE-AMPERE at CMU-10A To: Cube-Lovers@mc@a Subject: puzzle search does anybody out there know where i can get either Megaminx (otherwise known as Pyraminx Ultimate) or the Pyraminx Cube (known as the Skewb to all you Scientific American readers) ? Ive been trying for months. Any help will be greatly appreciated. -phil ( pxs@cmu-ee-ampere@cmu-10a )  Date: 3 Dec 1982 1820-PST From: ISAACS at SRI-KL Subject: Megaminx To: CUBE-LOVERS at MIT-MC Megaminx, under that name, should be available at your local stores (you don't say where you live). It is out in the San Francisco Bay Area, and L.A., I know, including in such chains as GEMCO and TOYS R US. It is distributed by TOMY Corp. I have not seen the Skewb anywhere, except in the hands of a couple of people that got it from Meffert personally; he did not send me one by mail, though I asked. He did send his Double Pyramid, though, a nice version of an octahedron, cut as Rubik's Cube, e.g. with centers. (The other octahedron, the "Star Cube", has no centers and a lot of extra corners.) -- Stan ------- Mail-From: CMUFTP host CMU-EE-AMPERE received by CMU-10A at 11-Dec-82 11:15:41-EST Date: 11 Dec 1982 11:08:22-EST From: Phil-Servita-H at CMU-EE-AMPERE at CMU-10A To: cube-lovers@mc@a Subject: 4x4x4 competition can you solve the 4x4x4? can you solve it fast? dont you wish that SOMEBODY would hold a 4x4x4 competition? if you answered yes to any of those questions, then read on. i have been in contact with Ideal/Gabriel about this topic. this is all that they want: people. if enough people seem interested in a 4x4x4 competition, they may hold one sometime between spring and fall '83. so here's what you do: if you are interested in a 4x4x4 competition, mail your response to: pxs@cmu-ee-ampere@cmu-10a Cc: meister@ccc@mc, ps0k@topsf@cmuc@cmu-10a ill keep a list of responses and send it to Bob Weissman at Ideal. -phil