Date: 12 JUL 1980 1342-EDTFrom: ALAN at MIT-MC (Alan Bawden)To: CUBE-LOVERS at MIT-MCWelcome to the cube-lovers mailing list. I don't know what we will betalking about, but another mailing list cannot hurt (too much).Mail to cube-lovers@mc or cube-hackers@mc whatever strikes your fancy.

Date: 15 July 1980 01:41-EDTFrom: Jef Poskanzer Subject: Complaints about :CUBE program.To: CUBE-HACKERS at MIT-MCThis is a real-neat program, BUT:In solving the cube, the description it gives for the action it isabout to perform can be ambiguous, I think. For example, can't "TurnTOP center to BOTTOM" be performed in two non-equivalent ways?I don't know what the display format is on color LISP machines, but onmy terminal is sucks. How about something like this:+--------+ / YL YL YL \ / \ / YL YL YL \ / \ / YL YL YL \ ,-+------------------+-. ,-' | | `-. ,-' GR | RD RD RD | BL `-.+-' GR | | BL `-+----------+| GR | | BL | OR OR OR || | | | || GR GR GR | RD RD RD | BL BL BL | OR OR OR || | | | || GR | | BL | OR OR OR |+-. GR | | BL ,-+----------+ `-. GR | RD RD RD | BL ,-' `-. | | ,-' `-+------------------+-' \ WH WH WH / \ / \ WH WH WH / \ / \ WH WH WH /+--------+Maybe add FRONT, BACK, LEFT, RIGHT, TOP, BOTTOM labels; maybe compressit vertically so it fits on 24-line screens; maybe three-char abbrevsinstead of two; maybe the back face should be shown reversed (i.e.from the inside of the cube looking out) to facilitate mentalmanipulations; the basic format is better, though, don't you agree?---Jef

Date: 15 JUL 1980 1413-EDTFrom: ALAN at MIT-MC (Alan Bawden)To: CUBE-HACKERS at MIT-MCSince we have this mailing list I am tempted to put on record aninteresting property of the cube that some of you might not be awareof:The cube has a degree of freedom that is rarely considered. The sixcenter faces which are the pivots about which rotations are performeddo not always return to their original orentation. In other words, ifyou were to paint little arrows on the center square of each face of asolved cube, randomized the cube, and then solved it, you might wellfind that the arrows no longer pointed in the same directions as theydid before. (Now you have to get them back. And you thought you hadalready solved the damn thing!)This extended cube problem has been solved independently twice to myknowledge (not by me, I cheated and learned someone else's solution).The property was first shown to me by Spencer Love. Does anybody knowif anybody else ever discovered it independently?Without considering this property we already knew that there were43252003274489856000 = (8! * 3^8 * 12! * 2^12)/12 differentrearrangements of the cube (all those numbers are obvious except the12 in the denominator). Considering this property raises that numberto 88580102706155225088000 = (8! * 3^8 * 12! * 2^12 * 4^6)/24 (the 24being just as hard to explain as the 12 was).

CMB@MIT-ML 07/15/80 15:06:58To: cube-lovers at MIT-MCI had known about the center faces being turnable for a while, and have a cubethat I have marked up so I could work on solving the center face problem. Ingeneral, I solve the cube except for the center faces (since that was what Ialready knew how to do), and then have three transforms: one that will turnany center face 180 degrees and leave everything else unchanged, one that willturn one center face 90 in one direction, and an adjacent center face 90 inthe other direction, and the last turns two adjacent center faces 90 in thesame direction. My transforms are rather long and rep(it)*ous.

Date: 15 JUL 1980 2158-EDTFrom: ALAN at MIT-MC (Alan Bawden)To: CMB at MIT-MLCC: CUBE-HACKERS at MIT-MCThe last two transforms you describe sound similar to the two Ilearned. Mine are also rather repititous. Perhaps it is the casethat the two configurations are very distant using the obvious metric:smallest number of twists from one to the other.I wonder if anyone knows very much about the nature of that metricanyway? I understand that it is known that no two points are morethan 94 (or is it 93?) twists apart (disregarding the extendedproblem). I don't know if that number is actually attained, or if itis only the currently known upper bound based on the best algorithm.(Or perhaps there isn't an algorithm that good yet, just a proof ofthe fact.)I believe that you and ACW and I once did the math to show thatwhatever that longest distance is, it has to be greater than somethingaround 30, and for the extended cube problem it must be even biggerthan that, so since I can do the transformations you speek of in about28 (I think) moves, those must not be most distant points.

Date: 15 July 1980 2245-edtFrom: Bernard S. Greenberg Subject: Cube minimaTo: CUBE-LOVERS at MIT-MCI believe the 94 number comes from Singmaster's book. UNFORTUNATELYSingmaster doesnt seem to know what the word "canonical" means,and 180-degree twists count as single moves "too" in his religion.This makes his number kind of worthless.

ED@MIT-ML 07/15/80 23:21:07 Re: Singmeister who?To: cube-lovers at MIT-MCI've never heard of Singmaster's Book. Is it in the Old Testamentor the New? Is it prophetic, historical or lackadaisical?

Date: 16 July 1980 09:35 edtFrom: Greenberg.Multics at MIT-MulticsSubject: SingmasterTo: cube-lovers at MIT-MCDavid Singmaster is a prof at an English university whosename escapes me, who publishes a 40-page pamphlet on cubingfor about $5.00. It took me months to get; I have it for xeroxingin my posession. It contains many transforms, including manywierd solve-the-edge-cubes-first solutions. As I havepointed out, his noncanonical move counting is a problem.

ACW@MIT-AI 07/16/80 15:08:09 Re: CubespeakTo: CUBE-HACKERS at MIT-MCI would like to see some talk about a good language fordescribing cube manipulations. I know Bernie has one thathe swears by, but I would rather not see the discussionSTART with everybody giving their favorite cubespraak...this can get confusing, dogmatic, counterproductive, nobodylistens, etc. So let's keep the "My language is better thanyour language" to a minimum at first, and see what desideratapeople consider as basic.My first contribution is: I think that turning the cube over,rotating it, etc., without performing any twists -- that is,all the things that you could do just as well to a solid block ofwood -- SHOULD be considered manipulations in their own right.This includes performing a mirror reversal. Why? So thatmanipulations that only differ in starting orientation will nothave representations that look completely different. ---Wechsler

Date: 16 JUL 1980 2051-EDTFrom: RP at MIT-MC (Richard Pavelle)To: CUBE-LOVERS at MIT-MCA simple transformation with a pretty resulting design:Let CS = center slice which faces you and isparallel to the axis of your body.Then CS up 90, rotate cube 90 in either direction sothat the CS turns in the direcion of rotation.Performing these two operations 3 more times givesthe desired result.

Date: 16 Jul 1980 2130-PDT (Wednesday)From: Lauren at UCLA-SECURITY (Lauren Weinstein)Subject: confusionTo: CUBE-HACKERS at MCI am relatively new to the whole idea of cubing, in fact I do not yeteven own a cube. I am a littel confused as to what the state of theart is in solving an actual randomized Rubik's Cube. Are thesealgorithms so complex that they are only usable with computer aid, or atleast pencil and paper? Or are they such that, handed a random cube,you can just sit there and (given sufficient time) put it right?--Lauren---------

Date: 17 July 1980 01:22-EDTFrom: Alan Bawden Subject: confusionTo: Lauren at UCLA-SECURITYcc: CUBE-HACKERS at MIT-MCGiven a randomized cube it can take about 5 to 10 minutes to set itstraight with no aid whatsoever. Pencil and paper or a computer canbe a great deal of help when one is first learning to solve the cube(I used both), but I know of no one who uses such aids once they havelearned how. A method of solution that required some computationalaid to perform (some hairy calculation based on the currentconfiguration of the cube, resulting in a single 259 twist sequencethat brings it immediately back to the solved state) is notinconceivable, but most people have solutions composed of short, easilycomprehended steps.Can anyone tell me who this Rubik character is? His name appears tobe attached to the new American version of what some of us once knewas the "Hungarian Cube". Is Rubik the Hungarian who invented it? Hashe done anything else? I heard this rumor that there was a 4x4x4 cubeout there somewhere, anybody else heard about it?

Date: 17 JUL 1980 0846-EDTFrom: JURGEN at MIT-MC (Jon David Callas)To: CUBE-LOVERS at MIT-MCYeah, sounds great, I always loved linear Alg.I'm not really COMPLETELY sure what you're doing, butcount me in. It sounds like I might learn something.Thanx,Jurgen@mc

Date: 17 July 1980 09:40 edtFrom: Greenberg.Multics at MIT-MulticsSubject: Re: confusionTo: Alan Bawden cc: Lauren at UCLA-Security, CUBE-HACKERS at MIT-MCIn-Reply-To: Message of 17 July 1980 01:23 edt from Alan BawdenFor the record, Erno^H" Rubik is the hungarian teacher ofarchitecture and design at some Budapest equivalent_high_school, who invented the Cube. Apparently he has a solution,but it is not a particularly good one. I would also like to add toALAN's note that given my method or ALAN's method, I have neverseen either take longer than 5 minutes (I promise clockedsolutions in under 4 and have taken much less) and Singmasterhas heard of solutions in 2 minutes, but I find this difficult to believe.I think I can also assert that most of the certified Cubemeistersdid NOT use a computer in solving it, although it can be of great help.For the record for the purposes of this list, my algorithmis implemented in the :CUBE program that runs on all ITS's.

Date: 17 JUL 1980 1042-EDTFrom: RP at MIT-MC (Richard Pavelle)Subject: confusion but simplicityTo: Lauren at UCLA-SECURITYCC: CUBE-HACKERS at MIT-MCI do not use a computer but I found it useful in keeping track ofmoves and testing various transformations. I can fix the cubenow in 5 minutes or less. In fact, I have won cubes atstores by challenging the owner to afreeby if I can do it under 10 or 15 minutes. My transformations arevery easy to remember because after one face is complete I need only 4to fix the cube. The transformations are positioning the corners, topplingthem, fixing edges and then toppling edges. These are not designed forspeed but for simplicity. I have taught some people to solve the cube thisway with a few hours of practice. To do it in 5 minutes, however, requiresa few more transformations.

Date: 17 JUL 1980 1114-EDTFrom: ACW at MIT-MC (Allan C. Wechsler)Subject: Short Introductory SpeechTo: CUBE-LOVERS at MIT-MCNobody else has made this kind of flame yet, so Iguess I will.Welcome to CUBE-LOVERS. We are devotees of a certainmathematical puzzle variously called the HungarianCube, the Magic Cube, and Rubik's Cube. It is a hardpuzzle. Very intelligent people often take weeks tolearn to solve it. Once they have learned, though, theycan solve it in a few minutes.The puzzle embodies mathematical sophistication andmechanical ingenuity in a pleasing and intriguingsynthesis. I have forgotten the Hungarian inventor'sname, but we should learn it: this person deservesour profound respect.For those who have not yet become Cube Solvers: youcan only solve the Cube for the first time ONCE. Afterthat, although there are a lot of problems to thinkabout, the initial challenge is gone. So, in the wordsof Mr. Duffey: SPOILER WARNING! SPOILER WARNING!Messages to this list will often deal with particularsolution techniques. If you haven't solved the cube yet,and want to do it on your own, reading these messagesmay ruin your fun.If there is any demand, I am willing to hack up a shortintroduction to Group Theory for Cubans. Group Theorygives us a mathematical language for talking about thecube.Also, if anyone out there knows Hungarian, there are somepamphlets we need to translate. Cubans should informeveryone on the list of any written material they know of,so that we can compile a bibliography.Happy cubing, ---Wechsler

Date: 17 JUL 1980 1420-EDTFrom: RP at MIT-MC (Richard Pavelle)Subject: the cross designTo: CUBE-LOVERS at MIT-MCDoes anyone know a transformation (or series of) which will producethe following pattern on each face where X and O are two different colors.|X O X||O O O||X O X|I think this may not be possible which brings me the question whichis how one can say whether a particular arrangement is or isnot possible.

Date: 17 July 1980 14:38 edtFrom: Greenberg.Multics at MIT-MulticsSubject: Re: the cross designTo: RP at MIT-MC (Richard Pavelle)cc: CUBE-LOVERS at MIT-MCIn-Reply-To: Message of 17 July 1980 14:20 edt from Richard PavelleThere are two such sets of patterns known, one with threesets of paired colors (the Christman cross) and onewith two triplets of colors (the Plumer cross).The Plummer cross is achieved by two orthogonal applicationsof the transform to the Christman cross. The transforms arefairly long and hairy, and I hesitate a little before attemptingto describe them, but will if people want.

Date: 17 JUL 1980 1635-EDTFrom: ALAN at MIT-MC (Alan Bawden)Subject: the cross designTo: RP at MIT-MCCC: CUBE-HACKERS at MIT-MCYes, the cross design is possible. I will let BSG try and describe itif people want. It is fairly straightfoward to tell if a position isreachable. I am thinking of onlining the proof that there are 12equivalence classes of cubes (I think I have a fairly simplifiedversion), and if I do it should contain an algorithm to tell if aposition is reachable.

Yekta@MC (Sent by ___117) 07/17/80 16:45:00 Re: Checker board pattern...To: cube-lovers at MIT-MCIs the pattern in which evry face of the cube is a checkerboardof two colors possible?? ( I have not played with the cube at all,so my apologies if it is too trivial to get... )

Date: 17 July 1980 17:33 edtFrom: Greenberg.Multics at MIT-MulticsSubject: Re: Checker board pattern...To: Yekta at MIT-MC (Sent by ___117)cc: cube-lovers at MIT-MCIn-Reply-To: Message of 17 July 1980 16:45 edt from Sent by ___117Yes. This is absolutely trivial. Rotate each face 180. Si x totalonehunred-eighty degree twists.

Date: 17 Jul 1980 1358-PDT (Thursday)From: Mike at UCLA-SECURITY (Michael Urban)Subject: ConfusionTo: laurenCC: cube-hackers at mit-mc Actually, dealing with the cube is a learn-as-you-go experience.The appeal of the cube, which makes it superior to Soma, or "InstantInsanity", etc, is that you actually have to analyze what's going onin order to approach the solution. For example, I began by learninghow to put one face right; this required certain simple transformationsthat were useful later. As you go, you develop your own heuristicsfor moving the faces you need around without messing up what you'vedone so far. I can do the whole thing from an arbitrary startingposition in around 20 minutes; I'm still not very adept at movingcorners around. Even after you have solved it, there are still things to dowith the cube, including improving your personal algorithm, as wellas creating nifty patterns, etc. A Worthy Toy. According to "Games & Puzzles" magazine, Rubik is the Hungarianfellow that devised this evil little time-stealer. Thearticle also impies a 2x2 version is in the works, which is even harderto understand mechanically than the 3x3 version. How the heck IS the thing put together, anyhow?Mike-------

Date: 17 July 1980 17:38 edtFrom: Greenberg.Multics at MIT-MulticsSubject: Re: the cross designTo: ALAN at MIT-MC (Alan Bawden)cc: RP at MIT-MC, CUBE-HACKERS at MIT-MCIn-Reply-To: Message of 17 July 1980 16:35 edt from Alan BawdenThere is a wonderful alogorithm for ANY pattern, which constitutesan existence proof. Given that you know how to "solve cubes", youcan achieve a givenpattern if it is achievable simply by "solving" to that position,which may in fact be faster than some set of arcane transforms.Before the "neat" algorithms for the Cruces Plummeri et Christmaniwere discovered, they were achieved by cubemeisters only by"solving" to that position, lambda-binding the target state(lessee, this guy wants to go here, etc.) to the desired pattern.I will burn some neurocomputrons tonight to describe thealgorithms for the Crosses.Incidentallly, if you try to solve for some pattern and come toa roadblock of the form "this cant come here because its here"(we need two of these cubies ( a local jargonfor the little cubes)), or a parity/trinity argument, you haveproven that you can't achieve it.

Date: 17 July 1980 17:33 edtFrom: Greenberg.Multics at MIT-MulticsSubject: Re: Checker board pattern...To: Yekta at MIT-MC (Sent by ___117)cc: cube-lovers at MIT-MCIn-Reply-To: Message of 17 July 1980 16:45 edt from Sent by ___117Yes. This is absolutely trivial. Rotate each face 180. Si x totalonehunred-eighty degree twists.

Date: 17 July 1980 21:18 edtFrom: Greenberg.Multics at MIT-MulticsSubject: Re: ConfusionTo: Mike at UCLA-Security (Michael Urban)cc: lauren at UCLA-Security, cube-hackers at MIT-MCIn-Reply-To: Message of 17 July 1980 18:12 edt from Michael UrbanThe easiest way to see how the thing is put together is to take it apart.If you turn the "top" plane 45 degrees, you can pry out thetop edge cube (any of the 4) fairly easily, and it becomes clearhow the devilish thing is put together.In three lines or less, all the center-of-face pieces are on a6-armed, solid, rigid cross. They can spin, but that's it.The EDGE pieces (i.e., not the corneres) have a rectangularprojecction from their non-visible edge which gets them stuckbehind whichever two center-of-face pieces they're currentlyvisiting. While not rotating, it's stuck behind two. whilerotating it (a edge piece) is stuc behind the oneon its rotating face, and a groove that all edge pieceshave on their sides, in this case on the edge pieces of thenext plane back. The corner pieces have little cube-likeprojections on their invisible corners that basicallywedge in behind the edge pieces, which are stuck as desribedabove. No magnets, wires, universaljoints or rubber bands.IF youshould decide to take one apart, be SURE to put ittogether SOLVED to ensure solvability.

Date: 17 July 1980 2228-edtFrom: Bernard S. Greenberg Subject: The Higher CrossesTo: CUBE-LOVERS at MIT-MCFor the amusement of the experienced cubemeister, and theeducation of those desirous of learning the Art, I have hereproduced a Description of the Methods, as I have used,for the production of the Cruces Plummeri & Christmani.These are the elegant configigurations of lid Crossesof which we spoke earlier today.**********************************************************************I herein describe the algorithms for construction ofthe higher (Plummer, Christman) crosses from a solved position.From an unsolved position, it is faster to "solve" directly tothe desired configuration; by following the steps below,the eager cubist may learn exactly what these configurations are.Of the words and phrases I use:I call the faces front, back, right, left, top, bottom. A facehas 9 cubies, viz., 4 corner cubies, 4 edge cubies, and itscenter cubies. Separating 2 opposite faces, is a "centerslice", being of 4 center cubies and 4 edge cubies. As Ihold the cube, I call three center slices: floor-parallel,body-parallel, body-slicing. For instance, the body-paralleland body-slicing centerslices meet in the front face.I name "double-swap" the transform which is performed as follows: Double-swap (front, back) ;parameter-faces Turn body-slicing centerslice 180. Turn bottom face 180. Turn body-slicing centerslice 180. Turn bottom face 180.Observe well what it has done, viz. swapped the two cubies of theturned centerslice on the front with those of the back. Youwill use it as needed during the following shenanigans:----------------------------------------------------------------------To achieve Christman's (DPC at MIT-MC) Cross, the simpler of the two:Rotate the body-slicing centerslice 180. Rotate the floor-parallelcenterslice 90 either way (your choice).Stare hard at what you have. The CORNER CUBIES and CENTER CUBIESare in their final position for the Crux Christmani; all furtherhacking will be simply to move the EDGE CUBIES, IN PAIRS, intoplace. To achieve ANY Crux Plummeri or Crux Christmani configuration,learn how to do the initial rotations (see below for the CP)so that you get the center cubies to corners you want, and hack fromthere.I will now describe the edge-cubie moves for the CC given thatthe centerslices have been aligned to orient the center cubiesas needed: Among the six faces you now have, find one of the two thathave a solid stripe between two sides of the same color, i.e., x y x x y x x y xand align it like so, so that the stripe is vertical, and thisface is the front.Note that the edge cubies of the y y y stripe want to be exchangedwith the two x-showing edge cubies, i.e., x x x y y y x x x(Remember that the goal is x y x/y y y/x y x)You can tell that they want to be inthe horiz. positions by theirnon-showing faces, which you will observe match the center-cubieson the right and left sides.To do this: 1. Perform doubleswap on front-back. 2. Rotate the FRONT so that when you do (3), the two cubies we just moved to the back will come to such place so that when we undo this step (see 4), they will be in the right place. This will be either 90 deg. left or right. 3. Perform doubleswap on front-back. 4. Undo step 2, i.e., turn FRONT 90 deg the "other" way.Whehter you blew (2) or not, you will now find you have (x x x/y y y/x x x) on front. If you understood 2 and DIDNTblow it, you will have the sides of the y y edge cubies matchingthe side centers (if you blew it, doubleswaps on the side facescan fix you up).You will see the floor-parallel centerslice begin to form a band.We will now finish that band. The two appropriate cubies (to goin the two rear positions of the floor-parallel centerslice arenow on the front plane, the x x cubies of the last step. Notethat a simple doubleswap on front-back would move them to theback face, but the WRONG two places on the back face. Easy.So, turn the back face 90 degrees and do the doubleswap,and unturn the back. Choose which 90 such that these twocubies wind up in the right place.You will now find you have solid bands and solid crossesgalore. The front and back should have solid crosses, and thefloor-parallel slice should now be a solid band.Look at the top of the cube. Make it the front.Orient it so that it is (a b a/c c c/a b a). Do a front-backdoubleswap, and now look at the remaining face pair wehavent been thinking about. Do the appropriate doubleswap onthem to get solid crosses, and then you should have the Crux Christmani.Study well what you have: three pairs of alternated crosses.----------------------------------------------------------------------The Crux Plummeri (after DCP at MIT-MC who first came up with it,altho by solving-to) is exactly equal to doing the entireabove transformation twice, at 90 degrees. The following, however,is a direct route from solved that is more intuitive.Take the cube, turn the body-slicing centerslice up 90 deg.Turn the floor-parallel centerslice 90 deg clockwise as seen fromthe top. Note well the configuration of corners versus centers;it is the final one. Note that you will have two tripletsof trebly-interleaved colors: that is the characteristic of the CP.Look at the TOP or BOTTOM. Let's say the TOP. Make it the front.Orient it so that you see x y x x y x x y xOnly the top or bottom look like this; this is what you have toremember to look for after aligning centers to taste.We're gonna rotate the y y y band into the horiz position.Do this exactly as for the CC above, producing (x x x/ y y y/x x x)Next goal is again to complete the solid band of the floor-parallelcenterslice by doubleswapping front/back so thatthe x x edgecubies,w hich would complete that band, go tothe back. Of course, we must temporarily rotate theback during this doubleswap, so that they go to the sidepositions ofthe back when swapped. Do so, completing thesolid color-band of the floor-parallel slice.Now consider the top and bottom. You note that exactly one appropriatedoubleswap between top and bottom would give us solidcrosses on both. Do it.Take what had been the top just now, and call that the front.Note that there are solid crosses on front and back, and thebody-parallel plane is correct and complete.Think about the front: it looks like a b a b b b a b aAlthough it looks right fromt the front, the two vertical b-edgecubies want to be the two horizontal b-edge cubies, as a cursoryinspectionof the top bottom and sides of the cube will show.This is true of the back, as well.Tofix up the FRONT do this: 1. Doubleswap front/back 2. Rotate the FRONT (temporarily) 90 degrees sothat the twovertical b-edge cubies are gonna come to the right place, 3. And doubleswap front/back 4. Undo 2. 5. Doubleswap front/back.Now you see all is right save the back. It wants the samething done to it. Do it for it; Do this same thingjust doNe in the last 5 steps for the back (viewing it as thetemporary front).It is done. Consider it.An exquisite variant ont he CP is obtained by taking on of thetrebly-bound sides and rotating the centers via the well-knowncenter-cubie rotating algorithm. As the centers are rotatedleft or right, either a sextuple checkerboard or a stunningtriply-rotated canon of centers , edges, and corners appears.The checkerboard is amusing insofar as it appears to anovice cubist to be the Pons Asinorum 6tuple checkerboardmade by 6 twists (described earlier today), but cannot befixed (solved, or produced) without the consummate hairof the CP that only true cubemeisters can execute.The application of the Pons Asinorum checkerboard transformto the CP (as well as the CC) produces interesting andsuprising results.----------------------------------------------------------------------The Higher Crosses are fascinating insofar as they appear to bevery simple edge-cube hacks, but are in fact quite "far"from home; the CP being exactly twice as "hairy" (far)as the CC (discovered by ALAN) is in itself a source ofwonderment.

Date: 17 JUL 1980 2245-EDTFrom: ALAN at MIT-MC (Alan Bawden)Subject: The Higher CrossesTo: Greenberg at MIT-MULTICSCC: CUBE-HACKERS at MIT-MC Date: 17 July 1980 2228-edt From: Bernard S. Greenberg ... The Higher Crosses are fascinating insofar as they appear to be very simple edge-cube hacks, but are in fact quite "far" from home; the CP being exactly twice as "hairy" (far) as the CC (discovered by ALAN) is in itself a source of wonderment.I wish I could really say something was "exactly twice as far" fromhome as something else. Unfortunately, as I complained before, themetric by which one measures the distance of one configuration fromanother is not well enough understood to be able to make claims likethis. It might well be that the CP is less than twice as far as theCC, all we can really be sure is that it is not any further than that.

Date: 17 July 1980 22:52 edtFrom: Greenberg.Multics at MIT-MulticsSubject: Postscript to aboveTo: cube-hackers at MIT-MCI should have noted in the above flamage(btw, will all of those of us who are paid by our respective employersto hack this lunacy please let me know at once)that there is room for opitimization and lookahead in the finaldoubleswaps in correcting the top/bottom, and thedoubleswap preceding it, but I do not do this, so thatI might better keep track of what I am doing.

Date: 17 July 1980 22:54 edtFrom: Greenberg.Multics at MIT-MulticsSubject: Bug in aboveTo: cube-hackers at MIT-MCIn the terminOlogy section, note that thebody-slcing and FLOOR-PARALLEL centerslices meet in the front face,not the body-parallel and body-slicing as given.

ED@MIT-AI 07/18/80 00:12:53 Re: Patterns, designs &c.To: cube-lovers at MIT-MCTo jump the gun slightly on the group-theoretic explanation,any sequence of rotations of any number of faces can bethought of as an atomic "transformation" for the purposes ofgroup theory. One of the precepts of this theory is that anysuch transformation, repeated often enough, will return thecube to the original state. For instance, given the transform"rotate top ccw 90", 4 iterations suffice to return the cubeto the original state.Mike Speciner, a fellow Camexian, claims that no transformationcan be created that requires more than 216 (=6^3) iterations toreturn to the virgin state. (He doesn't yet have a cube, buthas been stealing his daughter's blocks and modeling the cubewith them.)Where does this number come from, and is it true?I have been playing with various transforms, and have foundat least one reasonably trivial one that requires the 216iterations: rotate a face 90, then turn the cube 90 and repeat.The transform here is 4 twists in a band around the axis ofcube rotation. The patterns generated in the process are interesting,too, though none of them are as unique as the cruciform or center-face patterns.

Date: 18 JUL 1980 0205-EDTFrom: ALAN at MIT-MC (Alan Bawden)Subject: Patterns, designs &c.To: ED at MIT-AICC: CUBE-HACKERS at MIT-MCEr, perhaps I don't understand the move you describe, but in any case,by my calculations it would take 1260 repetitions to come back homedoing that one (1260=2^2*3^2*5*11). It is certainly the case that1260 > 216 so that number must be wrong (I could be miscalculating,but I don't see how). 1260 is rather similar in appearance to 216,perhaps you spazzed somehow?

Date: 18 JUL 1980 0351-EDTFrom: ALAN at MIT-MC (Alan Bawden)Subject: 1260To: ED at MIT-MCCC: CUBE-HACKERS at MIT-MCI now have a proof that no element of the cube group can be of ordergreater than 1260. Since you have so thoughfully provided me with anelement of order 1260, I must conclude that this element is indeed themaximum, as you claimed (but where did the number 216=6^3 come from?).My proof contains no nice derivation of the number 1260, you will bedissapointed to see where it comes from, it is just all that is leftafter a number of cases have been eliminated.Perhaps someone can devise a "nice" proof of this fact. Is this factin the literature? (Bernie?)

Date: 18 JUL 1980 0932-EDTFrom: RP at MIT-MC (Richard Pavelle)To: CUBE-LOVERS at MIT-MCCC: ZIM at MIT-MCThe file which contains all cube-lovers mail is ALAN;CUBE MAIL on MC.

ACW@MIT-AI 07/19/80 01:42:56 Re: Where to get them in the Boston Area, Cube Language.To: CUBE-LOVERS at MIT-MC"Games People Play" on Mass Ave, near the fork where Mt. AuburnSt. branches off, carred them the last I knew. According torumor, the domestic product ("Rubik's Cube"), selling forabout $7, is mechanically inferior to the Hungarian import,costing $15. I don't know how everyone feels about money,but to me, not having to fight with the thing would be worththe extra $8.Bernie's explanation of how to achieve the Plummer and ChristmanCrosses is a prime example of why we need a cube language.Since no one has proposed anything, I will jump into the fray.Center the cube at the origin of a 3-d coordinate system. Theaxes of the coordinate system protrude from the centers of the faces.Make a hitch-hiker's gesture with your right hand and point the thumbalong the X axis. Imagine rotating the WHOLE CUBE one quarter turnin the direction your curled fingers are pointing. I call this operation"I". (The X axis is the horizontal axis that does not skewer you.)If the cube was lying on a table, "I" would roll it toward you.Now point up, along the Y axis, with your thumb. A quarter-rotationin the direction your curled fingers point is the operation I call "J".The Z axis goes right through your belly. A quarter turn around itI call "K". Actually, K=IIIJI.To simplify things a little, we define I'=III, J'=JJJ, and K'=KKK.In general, M' is M done backwards. If we call the do-nothingmanipulation "1", then MM'=M'M=1.For my own nefarious reasons I define "H" as the operation (unachievablein real life) of REFLECTING the cube right-to-left through the YZ plane.We note H'=H.Twisting the front (Z=1) face 90 degrees counter-clockwise I call"T".One more piece of notation: For any manipulations M and N, I writeM'NM as N[M], reading this as if N were a function: "N of M".Note M[1]=M.Examples:T[I] means "Twist the top face"T[II]=T[JJ] means "Twist the back face"T[I'] means "Twist the bottom face"T'[J] means "Twist the left face CLOCKWISE"T[I] T'[I'] J' means "Rotate the floor-parallel center-slice aquarter turn counter-clockwise as seen from above."Note that (MN)'=N'M', and (N[M])'=N'[M].Also notice that (MN)[P] = M[P] N[P].To do the Pons Asinorum checkerboard:Set Q= (TT)[J] (TT)[ZJ] "Half turn body-slicing center-slice."Then the Pons is Q Q[J] Q[K].Does anybody see what I'm getting at or am I a lone, madgenius?Set Q= T[J] T[J'].Then (Q Q[J])^3 is quite pretty. ---Wechsler

Date: 19 JUL 1980 0249-EDTFrom: ALAN at MIT-MC (Alan Bawden)Subject: 1260 vs. 2520To: ED at MIT-MC, CUBE-HACKERS at MIT-MCSorry folks, there was a slight error in my last message about themaximum order of any element in the cube group. To understand why,let us examine the transformation that ED described. I like to labelthe cube this way: A0 AE A1 AB AX AD A3 AC A2B0 BA B3 C3 CA C2 D2 DA D1 E1 EA E0BE BX BC CB CX CD DC DX DE ED EX EBB7 BF B4 C4 CF C5 D5 DF D6 E6 EF E7 F4 FC F5 FB FX FD F7 FE F6(I won't bother to describe the features of this labeling, and I willpresume that the correspondence between this unfolded labeling and anactual 3D cube is obvious.)If I understand ED's directions this is the way the cube looks afterapplying his transformation: A3 AB A0 AC AX AE D1 DA D6C3 CA E1 A1 AD F6 E6 EA E0 B0 BA B3CB CX BF FB DX DE ED EX EB BE BX BCC2 EF E7 B7 CF B4 C4 DC C5 D5 DF D2 F7 FC F4 FE FX CD A2 FD F5Note that the center faces BX, CX, DX and EX have all moved. My proofthat the maximum order of any element is 1260 assumes that thesecenter faces remain fixed. There is nothing wrong with thatassumption, it simply relects the intuition that picking a cube up andputting it down again on a different face doesn't change theconfiguration at all. If we decide that the orentation of the cubeDOES matter, then we get transformations like ED's here. The groupwe are dealing with is also 24 times larger than before. And my proofnow shows that no element has an order greater than 2520 (twice as big).Could it be that ED's transformation is not actually a maximal one?Could there be one with higher order? Or can my proof be tightened upsome, so that even in the larger group the maximum order remains thesame?To answer these questions (hopefully) I present a transformationthat I claim has order 2520: D2 AB A0 DC AX AE D5 AD A1C2 CD C5 F5 DA D1 E1 EA E0 B0 BA A2CA CX CF FC DX DE ED EX EB BE BX ACC3 CB C4 F4 DF D6 E6 EF E7 B7 BF A3 B4 FD F6 BC FX FE B3 FB F7(This transformation, like ED's, is not a member of the usual group,but the larger one where the center faces are allowed to move.)It is easy to check that this permutation has order 2520, the realquestion is whether you can get to this configuration from a solvedcube. I haven't actually tried to do that, but I am fairly certainthat it is possible. If anyone would like to try it, and report to metheir success or failure, I would be very grateful.

Date: 19 JUL 1980 0346-EDTFrom: ALAN at MIT-MC (Alan Bawden)Subject: OOPSTo: ACW at MIT-AICC: CUBE-LOVERS at MIT-MCLet me suggest the following corrections to your cube languagemessage. If any of these are not actual errors, but are actuallymisunderstandings on my part, then I apologize for causing undueconfusion, but I thought it would be helpfull to people trying tounderstand you system to have these corrected as soom as possible. ACW@MIT-AI 07/19/80 01:42:56 ... One more piece of notation: For any manipulations M and N, I write M'NM as N[M], reading this as if N were a function: "N of M".I think you must intend N[M] to mean M N M' . ... T[I] T'[I'] J' means "Rotate the floor-parallel center-slice aquarter turn counter-clockwise as seen from above."T[I] and T'[I'] don't touch the floor-parallel center-slice at all andJ' rotates it clockwise (as seen from above), my guess is that thedescription should simply read: "Rotate the floor-parallelcenter-slice a quarter turn clockwise as seen from above." ... Set Q= (TT)[J] (TT)[ZJ] "Half turn body-slicing center-slice."This must be a typo or some discarded notation. My guess is that thisshould read Q= (TT)[J] (TT)[J']. I don't think that the descriptionhere ("Half turn body-slicing center-slice") fits, but it works inproducing the pattern anyway. If you had said Q= (TT)[J] (TT)[J'] IIthen you could get the pattern and fit the description too!

Date: 19 Jul 1980 11:27 PDTFrom: McKeeman.PA at PARC-MAXCSubject: Re: Where to get them in the Boston Area, Cube Language.In-reply-to: Your message of 07/19/80 01:42:56To: ACW@MIT-AIcc: CUBE-LOVERS at MIT-MC, Lynn.ES, Horning, SturgisYour proposal for a language leads me to suspect that you have not seen:NOTES ON THE 'MAGIC CUBE'David SingmasterMathematical Sciences and ComputingPolytechnic of the South BankLondon, SE1 0AA, Englandwhich you can get by sending him $3. The fourth printing runs 36 pages. It isdefinitely worth the money.Cube quality: I have handled about a dozen Hungarian import versions of thecube with enormous variation in quality. They were all the colortab-on-blackvariety. They sell out here in the discount stores like K-Mart for less than $9. Ihave also seen a colortab-on-white version which is not substantially differentin quality from the Hungarian versions. My advice is to pick and choose as bestyou can from the ones on the shelf, regardless of what kind you get.As for your language proposal, I like the RLUDFB abbreviations given bySingmaster and generally used in Europe to your T plus whole cube moves.They mean rotate the (Right, Left, Up, Down, Front, Back) surface 90oclockwise. I also use rludfb for the inverses of them. However since(fortunately) these two notations have a null intersection, there is nodisadvantage in mixing them. For consistency, however, I suggest ijk forcounterclockwise moves, and IJK for clockwise ones. This does little violence tothe quaternion origins where lower case is generally used anyway. The ' forinverse notation is fine; in fact I frequently prefer R' to r when I write downalgorithms. The mappings then are, substituting f for your T:F = f'R = F[J]r = f[J] = R'L = F[j]l = f[j] = L'U = F[i]u = f[i] = U'D = F[I]d = f[I] = D'B = F[ii]b = f[ii] = B'It also solves the problem that I was having of saying IJK in English since thereis no way to say it in "Singmaster".I didn't follow your Pons Asinorum checkerboard solution because I didn'tunderstand the Z in Q, but in the mixed notation I think it is:rrllffbbuuddor, if Q = rrll; QjQkQYour "quite pretty" is (lrfb)^3 = (lr (lr)[j])^3I propose the following grammer for RCML (Rubik Cube ManipulationLanguage):algorithm ::= definition* move*definition ::= letter = algorithm ;move ::= letter | move' | move^digit | move[algorithm] | (algorithm)digit ::= 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9where the letters RLUDFBrludfbIJKijk are predefined as noted above.--Bill

Date: 19 July 1980 1455-edtFrom: Bernard S. Greenberg Subject: General remarksTo: CUBE-LOVERS at MIT-MCWhile watching these various "linguae cubisticae" make the rounds,I note one categoric deficiency, whose perception may indeed bedue to the idiosyncracies of my own meta-algorithms.In specific, my algorithms (including my 200-line procedure of theother day) include DECISIONS of the form "find a face that hasa pattern like-so" or "look around for the target cubie, andcategorize its place like so". Now clearly, a generalsolution algorithm must include some expression of such decisions,although I find Singmaster's procedures deficient in this regard(he tends to say "Use enough of FOO transform to get all the..."often). However, the generation of specific patterns from anothercanonical state NEED not involve "decisions", although most of myprocedures (and hence the lisp-macro language of :cube) do.I offer that looking for certain patterns during the course ofeven one of these pretty-pattern-generations leads to morerememberable (^= memorable) algorithms than even a well-subroutinized set ofabsolute moves.On the Physical Reality of cubes:I have seen three species of the genus Cubi, the original Hungarians(C. Hungaricus), which is black faced,cubes sold by Ideal (C. Americanus), the AmericanBlack-faced cube, and the American Whiteface (C. Albus).The last-mentioned was first shown to me by the lady in theCambridge cube-store, who called me at home to make me aware ofthem when they first appeared (she says they are done bysome Friend of Rubik in Virginia).C. Americanus seems a good deal lighter in weight and build thanC. Hungaricus. While C. Hungaricus takes two to three weeks ofconstant use before they are loose enough for Concert performances,C. Americanus can be turned with one hand, yea, one fingerwhile held by the rest of the hand WHEN NEW. C. Americanido not seem to bind at all; the Hungarians not only bind, butdecompose and bind more. I confess a certain sentimental attachmentto C. Hungaricus, on which I mastered the Art. The stiff,clean solidity of a new Hungarian is indeed reassuring, but speedof movment and non-binding is something else. I have not subjectedAmericanus to truly extensive use, and I do not know howit ages, but so far, they seem to show less CHANGE per time thanthe Hungarians, and are usable from the start. I think theywill be a win.Of the American White-faced Cube, I have little to say; C. Albusis a curio item, and has nothing to recommend it over either of theothers.

Date: 19 July 1980 1524-edtFrom: Bernard S. Greenberg Subject: :cube featureTo: CUBE-LOVERS at MIT-MCOk, everybody wants a way to tell :CUBE "I have a cube likeso and so, solve it". The reason I have avoided doing this(other than laze) is that you may specify a non-reachablestate. (Duplicate cubies, or purple faces, etc. are easyto check for). I am not convinced that there is anybetter or faster check for an inconsistent/unreachablecube than trying to solve it and blowing out: lestwe find such a check (other possibility: run the program oncesilently and once loudly; if the first time fails(no, i will not make it list all cubes it cant solve)give an error), all internal breakpoints becomeuser errors.The second issue is what is the best languageto specify a configuration in? Comments?

Date: 19 JUL 1980 1548-EDTFrom: ALAN at MIT-MC (Alan Bawden)To: Greenberg at MIT-MULTICSCC: CUBE-HACKERS at MIT-MC Date: 19 July 1980 1524-edt From: Bernard S. Greenberg ... I am not convinced that there is any better or faster check for an inconsistent/unreachable cube than trying to solve it and blowing out: ...You mean I haven't convinced you of that yet? Show me how your cubeis represented and I will write one for you. What would be so badabout blowing out anyway? What happens when you hand a person animpossible cube? his algorithm "blows out" eventually! There is noshame in being tricked into trying to solve a bad cube.

Date: 19 July 1980 15:53 edtFrom: Greenberg.Multics at MIT-MulticsTo: ALAN at MIT-MC (Alan Bawden)cc: CUBE-HACKERS at MIT-MCIn-Reply-To: Message of 19 July 1980 15:48 edt from Alan BawdenI certainly believe you have shown me sufficient conditions forinconsistency, but until now I wa not aware that you claimedthat they were necessary...

Date: 19 July 1980 16:31-EDTFrom: Ed Schwalenberg Subject: Nope, I guess we don't understandTo: ALAN at MIT-MCcc: CUBE-LOVERS at MIT-MCFirst of all, I was sitting here inventing a GOOD cube notation,when McKeeman's message about his notation came in. I beginto fear that ACW was right, and we will all die before agreeingon a notation. I dislike all proposed notations so far, for thefollowing reasons:ALAN's labeling of the cube for describing a POSITION is excellent;however it is not useful for describing transforms, which areoperations and not positions.ACW's language I find baffling, principally because it lacksverbosity. I would much rather have a notation like[doubleswap] than F[X,[Y']] since it is descriptive. An analogywith Life is useful here; I think that the adoption of nameslike "traffic lights" is far more usable in the long run thanany of [the-oscillator-of-period-two-composed-of-three-blots-placed-orthogonally-adjacent] or | . ... | . | .or "if you put 3 blots in a row, the center dot remains alive foreverwhile the two outer dots appear first horizontally then vertically".Creating names like "The Christman Cross" is fun, and makes forinteresting wordplay, even if you don't resort to Latin. So myproposed language is Augmented English, which has the great featureof being able to put in comments (a feature notably absent in theothers).I urge people to describe the transform and its result in anymessage describing a nifty transform or pattern (provided both areknown, of course!). RP's pretty pattern may not be what I think itis, since "pretty" is not a good description. I propose that thisbe called "swapping-centers-in-triplets" (procedural notation) orTwelve Squares (which is not a movie by Mel Brooks, but the positionalnotation).Bernie's comment about decision-making I think is important: itseems to me that cubemeisters do in fact approach the problem witha set of "subroutines", which are defined to NOT containconditional branches. Doubleswap and checkerboard-all-faces areexamples of subroutines. Conditional branching is generally simplythe matter of selecting an orientation of the cube before applyinga transformation, "setting up" the subroutine if you will. I thinkthat in the case of generating patterns from a solved cube,branches are unnecessary but helpful: rather than say "doubleswapthen rotate the cube 90 clockwise in Z and doubleswap again"saying "doubleswap, then doubleswap the remaining solid face" muchmore clearly indicates what is going on. (the examples above arespurious).I herein announce two patterns I have independently invented; I donot know if they are elsewhere available. The first is calledTen Squares by analogy with Twelve Squares, since it is highly related.Twelve Squares causes the 3 centercubies of three mutually adjacentfaces to move to an adjacent face; three iterations of "swapping-centers-in-triplets" suffice to return to the solved state. TenSquares, on the other hand, is a configuration wherein two oppositefaces are solid, while the other 2 sets of opposing faces each possessthe centercubie belonging to its opposing face. This is created byfirst swapping-centers-in-triplets, then swapping-centers-in-tripletsagain, only with the cube rotated 90 degrees away from you. Note thatthis results in the final centerslice rotation of the first transformand the first centerslice rotation of the second effectively combininginto a single 180-degree centerslice rotation. To resolve the cube,simply do swapping-centers-in-triplets without regard to the orientationof the cube, then you are back to the trivially soluble Twelve Squares.The second I call Laughter, after the use of the string \/\/ in comlinksto signify laughter. It leaves the top and bottom faces solid, whilecausing pairs of opposing faces to have a diagonal stripe of theopposing color. (I propose that the color of the face opposite a givenface be called the complementary color; my cube has complementary pairsof red-white, orange-yellow and blue-green.) To create Laughter, selecta top-bottom pair, which will remain solid. Rotate the left and rightsides clockwise (in "opposite" directions as viewed from the top, thusresulting initially in the top being 3 differently-colored stripes). Icall this "splaying". Then rotate the cube 90 degrees while preservingthe top/bottom orientation (i.e., rotate it about the Z axis.). Sixiterations of splay/rotate suffice. Laughing the cube again solves it.Laughing the cube, then Frowning it (same as laughing, only rotate thefaces anticlockwise) results in Four Crosses: 2 complementary pairsof crosses with the top and bottom solid.

Date: 22 Jul 1980 1211-PDT (Tuesday)From: Mike at UCLA-SECURITY (Michael Urban)Subject: Yet another checkerboardTo: cube-lovers at mit-mc Nobody has mentioned (or discovered?) the fully-interlockedcheckerboard pattern obtained by applying the Pons Asinorumtransformation to the Plummer's Cross as suggestedby Greenberg during his description thereof, and then performinga few triple-center swaps (unfortunately, I was playing rather randomlywith the 3c swaps at the time and must leave them as an "exercise forthe reader"). The six colors are interlocked in pairs of AB, BC, CD, DE,EF, and FA on each checkerboard. Quite mysterious.Mike-------

Date: 23 Jul 1980 4:10 am PDT (Wednesday)From: Woods at PARC-MAXCSubject: Xerox cube-loversTo: Cube-Lovers at MIT-MCThis message is of (possible) interest only to Xerox members of Cube-Lovers, butthere's no easy way for me to send it only to them; apologies to the rest of you.I've created a Laurel-format message file containing the Cube-Lovers mail (fromthe archive at MC) up through approximately last weekend. The message is inchronological order (instead of reverse chronological the way MIT's stupid mailerdefaults) and has otherwise been cleaned up slightly (e.g., to get rid of lines justbarely too wide). If you're interested in looking through this archive, it's in[maxc]Cube.mail. It'll probably be around for several days at least.-- Don.

Date: 23 JUL 1980 1044-EDTFrom: RP at MIT-MC (Richard Pavelle)To: CUBE-LOVERS at MIT-MCThere is a short article on the cube written by Singmaster. Thereference is Mathematical Intelligencer, Vol.2, #1, pp.29, 1979.

Date: 23 Jul 1980 1640-PDT (Wednesday)From: Mike at UCLA-SECURITY (Michael Urban)Subject: ClarificationTo: cube-lovers at mit-mc My description of the checkerboard pattern was clearlyinadequate, due to haste and the fact that my cube was "borrowed"and scrambled before I had a chance to see precisely what wasgoing on. The pattern in question is formed from Crux Plummeri byapplying the "Pons Asinorum" transformation; from theresulting almost-checkerboard, a simple "12-squares" transformationwill provide the 6 checkerboards; it isn't hard to inspectthe almost-checkerboard to find a trio of center cubies torotate. The resulting checkerboards are a completely interlockedset. In the following unfolded cube, A/B means that the centerand corners are color A, and the edges are color B. A/BC/A B/D D/E F/C E/F I don't THINK anyone has mentioned this pattern; the checkerboardpattern mentioned by Greenberg consists of two cycles of three colorseach, something like A/BB/C C/A D/E E/F F/D although I may have the handedness wrong.Mike-------

Date: 23 Jul 1980 5:23 pm PDT (Wednesday)Sender: Woods at PARC-MAXCFrom: Woods at PARC-MAXC, Boyce at PARC-MAXCSubject: 216 vs 1260 vs . . .To: Cube-Lovers at MIT-MCRegarding finding a sequence of twists that requires a maximal number ofrepetitions to restore the original state, it is indeed true that 1260 is themaximum, but the sequence suggested by ED (on July 18) is not it. (Hethought its period was 216; ALAN thought it was 1260. It's actually 315.)This assumes that we don't consider the orientation of the center-face cubiesnor the orientation of the cube as a whole.First off, note that the transformation given by ED was ill-specified, since hedidn't say which way to rotate the cube relative to the rotations of the faces.In Singmaster's notation the two operations would be, say, LBRF and LFRB.It turns out these are each period-315, though it's not immediately obviousthat the two should be at all similar. Having done (LBRF)^1, you'll findthere are 7 edge cubies that have cycled through each other's positions, andthe other 5 edge cubies have done likewise, 5 corner cubies have rotated 120degrees in place, and the other 3 corner cubies have cycled through eachother's positions and one of them has rotated. Hence, if we do (LBRF) amultiple of 9 times, the corner cubies will be back where they started, and ifwe do it a multiple of 35 times, the edge cubies will be restored, so if we do(LBRF)^315, all of the cubies will be back to the solved state.Perhaps the reason ALAN thought this was period 1260 was because it takes1260 twists to restore the original cube. But since the transformation that weare repeating is a sequence of four twists, it actually has period 315. If youcount total twists, you can obviously get "identity" sequences with lengthsmuch greater than 1260. Or perhaps ALAN was considering the transformationto be LJ (twist left face, rotate cube about vertical axis). This indeed hasperiod 1260, but it "cheats" in that we're not supposed to consider reorientingthe entire cube as a legitimate operation. If we do, then it is again possibleto get periods greater than 1260.To get a period-1260 transformation, we need to observe whence the periodarises. If we do a sequence of twists, and then look at the cycles of the form"face X moved to where face Y used to be, face Y moved to where face Z was,. . . , face moved to where face X was", we take the least commonmultiple of the lengths of those cycles. If we repeat the sequence of twiststhat many times, the faces will moved around those cycles an integral numberof times and end up where they started; if we stop short of that manyrepetitions, at least one of the cycles will be left stuck in the middle.(Apologies to any readers who are familiar with group theory and are boredby this verbosity.)To get a period of 1260, we need to have a cycle of 7 edge cubies, anothercycle of 2 edge cubies where one of them is also rotated (so that the fourfaces on those cubies form a single period-4 cycle), a cycle of 5 corner cubies,and a cycle of 3 corner cubies where again one of them is rotated (forminga face-cycle of length 9). If, instead of the 2 edge cubies, we could get 4edge cubies cycling with one of them rotated, that would be a period-8 cycleand the overall transformation would have period 2520, but due to the factthat it's impossible to swap exactly two cubies it turns out you can't get thatcombination of subcycles.Based on this analysis, it's not hard to construct a period-1260 sequence,although this one is almost certainly not the shortest such (it's 24 twists):F U F^2 u r f b l r f D^2 F b R U (R^2 F^2)^3 u r B

Date: 23 Jul 1980 1757-PDTFrom: DDYERSubject: impossible cubesTo: cubr-lovers at MIT-MCRemailed-date: 23 Jul 1980 1758-PDTRemailed-from: Dave Dyer Remailed-to: cube-lovers at MIT-MC According to Singmaster, there are 12 disjoint orbits (his term) of cubepositions. The article I read doesn't give a derivation, but the basicidea is clear. It should be possible to characterize the orbit of a givenposition as a function of the transformations that would have had tobe done on each cubie ( as if it could be moved alone ) to bring itto its proposed position. Given such a charaterization, one coulddeclare positions fed to :CUBE illegal without trying to solve theproblem. This relates to my thoughts on the problem of inversion of an unknowntransformation. Suppose you are given a Cube that is N unit movesfrom home, determine the correct sequence to get it there. I thinka representation of the Cube in terms of combined tranformations onthe cubies is the place to start.-------

Date: 24 JUL 1980 2115-EDTFrom: RP at MIT-MC (Richard Pavelle)To: CUBE-LOVERS at MIT-MCMany of you have seen this message before. However, since the mailinglist of CUBISTS is growing rapidly I shall repeat this on occasion. A file of past cube mail is on ALAN;CUBE MAIL on MC.

Date: 25 JUL 1980 0845-EDTFrom: ZIM at MIT-MC (Mark Zimmermann)Subject: "blindfold" cube?To: CUBE-HACKERS at MIT-MCHow long does it typically take for one to be able to work with onlya mental image of the cube? (As in"blindfold" chess, soma, pentominoes....)

Date: 25 JUL 1980 0906-EDTFrom: RP at MIT-MC (Richard Pavelle)Subject: "blindfold" cube?To: ZIM at MIT-MCCC: CUBE-LOVERS at MIT-MC Date: 25 JUL 1980 0845-EDT From: ZIM at MIT-MC (Mark Zimmermann) How long does it typically take for one to be able to work with only a mental image of the cube? (As in"blindfold" chess, soma, pentominoes.I do not know of anyone who can do it blindfolded (I cannot). However,a few days ago mine fell into a swimming pool and I did it underwater withfive gulps of air.

Date: 25 July 1980 10:53 edtFrom: Greenberg.Multics at MIT-MulticsSubject: Re: "blindfold" cube?To: RP at MIT-MC (Richard Pavelle)cc: ZIM at MIT-MC, CUBE-LOVERS at MIT-MCIn-Reply-To: Message of 25 July 1980 09:06 edt from Richard PavelleThis interchange very elegantly points out the difference betweenthe Mathematician's view of the Cube and theHacker(System Programmer-type)'s view. While an algorithmthat looked at the initial state and categorized it("Perform the folliwng 152 steps for configuration106xy205a (left-handed) and it will be solved") would thrill mathematicians, practicalcube-solving ALGORITHMS require iteration, conditionals,subroutines, and other program-like techniques. Ineed, manynon-cognoscenti have watched me solve the cube, noting thatI look at it only a very small percent of the time(seeing which subroutine to invoke, as you will, )but they don't know that, and conclude that I "solve cubesbasically without looking". The whole notion of algorithmiccubism is based upon "I will do this hairy thing and it willhave this desired effect, and I need not thinK about theintermediate states, unless, god forbid, the phone rings, etc."To contain a cube map and intermediate states of transformsin one's head woud, in my mind, involve a greater skillthan blindfold chess; it is not a normal human-memorycapability, although doubtless institutiOnalized freakswith pathological intelligence/visulization problems(in the positive direction) exist who could conceivable do such a thing.

Date: 26 Jul 1980 1315-PDTFrom: Davis at OFFICE-3Subject: Some (perhaps well-known) TransformationsTo: cube-lovers at MIT-MCI have just recently joined the cube-hackers mailing list, but I have read overall the old mail. I suspect, however, that some of what I have to say will bewell-known already; on the other hand, I suspect that some of it is new.When I first got my cube, I had the advantage and disadvantage of working in avacuum. Hence, my original solution was more complicated than it should havebeen, but I did go about it another way, and discovered a few interesting factsalong the way. Having only seen one cube, I foolishly assumed that the colorpattern was the same on all cubes, and my notation for a move was simply acolor. A R (or Red) move meant that you look at the red face, and turn it 90degrees courter-clockwise. I wrote a little hack based on that notation, andsince all my notes are in that form, I will use it here. It should be trivialenough to convert it to the Left, Right, ... form. For reference, here is mycolor pattern in the "solved" condition: GGG GGG GGG OOO RRR YYY WWW OOO RRR YYY WWW OOO RRR YYY WWW BBB BBB BBBWhen I purchased my cube, the store was out of all but the demonstration model,so I never got a chance to mess around with a virgin cube, and I approached theproblem as follows: My program would simply take moves and print out thecondition after so many moves. It would also, given a sequence of moves, findthe order of the move as a group element. I then tried a number of patterns,some from my experience with the cube, and some completely at random, and foundthe orders of those moves. If the order came out to be, say, 90, then I woulddo 30 and 45 moves to see what the cube looked like after that many moves.These half- and third- way patterns were often fairly simple, and after tryingaout about 20 or 30 likely candidates, I generated most of the primitivetransformations I used to solve the cube originally. I also generated (as halfway positions) a large number of nice patterns, which are pretty to look at,but not of much use for solving a cube.In particular, laughter, mentioned by ALAN, is gotten by: 3(OYRW)I find 3(RYYR) a useful solving transformation, as well as 3(RYRRRYYY) and3(RYYYRRRY) (these last two being commutators). I have not yet gotten a lookat any of Singmaster's stuff, but these last two may be well known. Believeeit or not, one of the transformations I used to solvee the cube originally was45(RGWY), which has the net effect (after 180 moves) of flipping the R-Y andG-Y side cubies in place. I'm glad I found a better one than that.In fact, if you haul out any book on group theory, it is interesting to findgroup equations, and plug in random cube moves as the group elements to seewhat happens. Needless to say, things which group theorists find interestingusually tend to do interesting things to the cube.One other thing I discovered which also does interesting things to the cube isthe following transformation: Leave the body-slicing center slice in place,and rotate the other two sides, one toward and one away from you. Then rotatethe top and bottom 180 degrees, and rotate the sides back. This transformationalso gives a cube configuration which looks like something in the center-slicegroup, but is not. -- Tom Davis-------

Date: 26 July 1980 22:25 edtFrom: Greenberg.Multics at MIT-MulticsSubject: Re: Some (perhaps well-known) TransformationsTo: Davis at OFFICE-3cc: cube-lovers at MIT-MCIn-Reply-To: Message of 26 July 1980 16:15 edt from DavisOut of curiosity (I am responsible for :cube), I'd be interestedin seeing your code. Is it someplace accessible? -Bernie Greenberg

Date: 27 Jul 1980 1004-PDTFrom: Davis at OFFICE-3Subject: Re: Some (perhaps well-known) TransformationsTo: Greenberg.Multics at MIT-MULTICScc: cube-lovers at MIT-MC, DAVISIn response to your message sent 26 July 1980 22:25 edtHi, You're welcome to look at the code, although it is hardly fit forhuman consumption. I wrote it essentially in one sitting, and stoppedas soon as I solved the cube. There are some poorly thought outdesign features, etc. Anyway, you can get it at SAIL. It is writtenin PASSCAL, and is called MCUBE.PAS[1,TRD]. I scribbled in a fewcomments this morning, so maybe there is enough information tofigure out how to use it. Unfortunately I am at home on a typewriterterminal, and I refuse to try to edit without a display. Good luck. By the way, I have never used :cube, and have no idea exactly whatit does. Is there some documentation associated with it somewhere? Thanks, -- Tom DAvis-------

Date: 31 JUL 1980 1006-EDTFrom: RP at MIT-MC (Richard Pavelle)To: CUBE-LOVERS at MIT-MCIS IT POSSIBLE?I just got a note from Martin Gardner who plans an article about thecube in Sci.Am. sometime. But he claims that a person namesTHISTLETHWAITE has proved a restoration procedure of 50 movesand believes he can reduce it to 41.Gardner also says that a 2x3x3 solid is selling in Hungary, alsoby Rubic.

Date: 31 July 1980 10:52 edtFrom: Greenberg.Multics at MIT-MulticsTo: RP at MIT-MC (Richard Pavelle)cc: CUBE-LOVERS at MIT-MCIn-Reply-To: Message of 31 July 1980 10:06 edt from Richard PavelleOK, your'e on the spot..... you'd better produce these artifactsor we're all not gonna let you log out....

Date: 31 July 1980 13:06-EDTFrom: Alan Bawden To: RP at MIT-MCcc: CUBE-HACKERS at MIT-MC Date: 31 JUL 1980 1006-EDT From: RP at MIT-MC (Richard Pavelle) IS IT POSSIBLE?The Singmaster notes claim that Thistlethwaite had an 85 twistalgorithm in an addenda dated November 30, 1979. I presume that sincethen Thistlethwaite has continued to cube-hack, so why not 50 (or even41)?It should be noted that Singmaster insists on counting a 180 twist asONE twist, so I presume that the 85 number is measured that way. Howis Gardner counting?It is certainly possible. If you count twists Singmaster's way, youcan show that there are positions at least 18 twists away from home.There is nothing to suggest that this might not in fact be themaximum. So there might be room for Thistlethwaite to lower hisnumber all the way to 18!(If you count 180 twists as TWO twists, then a similar proof showsthat there are positions 21 twists away from home. In a past messageI reported that some of us had proved the existence of positions asfar away from home as around 30. I believe that the reasoning thatled to such a high number was incorrect. (Although I cannot provethat there AREN'T positions that far away, I now believe that I havenever seen a proof that there ARE.))

Date: 31 Jul 1980 10:34 am PDT (Thursday)From: Woods at PARC-MAXCSubject: Re: 180 degree twistsIn-reply-to: ALAN's message of 31 July 1980 13:06-EDTTo: CUBE-HACKERS at MIT-MCIt appears that Singmaster, Thistlethwaite, and just about all the cubehackers I know at and around Stanford consider anything you can dowith one wrist motion to be a single twist. Since this gives a moreaccurate measure of how complicated a sequence is, I'm happy with it.Why do you folks at MIT insist that you're right and the world is wrong?(I admit it complicates the notation. My own cube notation uses two charsper twist, one being the face and the other being the direction: left-arrowfor counterclockwise, right-arrow for clockwise, down-arrow for 180 -- isn'textended ASCII wonderful?)-- Don.

Date: 31 July 1980 1413-EDT (Thursday)From: Sandeep.Johar at CMU-10ASubject: cube group theory.To: cube-lovers at mit-mcMessage-ID: In one of the previous letters on the subject of cubingthere was a letter from some one offering to put together a pieceon group theory and cubing, I for one would certainly be interested inseeing it. When I took my cube apart and tried to put it togetherI wondered as to how many ways there are of putting it togetherso that the cube is unsolvable, i.e. how many equivalence classesare there which are 'wrong'.Any one worked it out, I would but group theory is unfortunately notmy strongest subject.Sandeep

ACW@MIT-AI 07/31/80 14:28:38To: cube-lovers at MIT-MCYeah, we have a prejudice against regarding180-degree twists as atomic. I understandyour feeling that a 180-degree twist isintuitively a single operation.Many of the cube-hackers at MIT becameinterested in the mathematical aspects ofthe cube, and the preference for countingquarter twists arose from this (admittedlyrather Spartan) mathematical viewpoint.When the cube first appeared, the mathematiciansamong us instantly exclaimed, with greatdelight, "Wow, here we have a group, whoseelements are possible manipulations of thecube, and whose binary operation consistsof following one manipulation with another."We immediately got interested in group-theory questions like, "What is the orderof this group?" "Does it have any interestingsubgroups?" and, in general "What kind ofobject is this group? Does understandingit help us solve the cube better?"There are several common ways of representinggroups. One is as a subgroup of a permutationgroup. This doesn't really help in the case ofthe Hungarian Cube, because it is too closeto what the cube really is: few new facts orinsights are revealed. Another way is withgenerators and relations. This means, to lista few basic group elements from which the wholegroup may be derived by multiplying them together.We soon figured out (along with hundreds of othermathematically-inclined cube-hackers) that thewhole group of possible manipulations could begenerated from six elements: the quarter-twistsof each of the six faces. This observationlater turned out to be crucial in calculatingthe order (number of possible states) of the group.Hence our predilection for counting quarter-turns.The half-turns were already accounted for, andwe thought of them as two juxtaposed quarter-turns.I guess some of us believe that the mathematicalstructure of the cube group is built on quarter-turns. Those whose delight in the cube is notmathematical will not agree: after all,a half-twist is as easy as a quarter-twistto perform. But you will miss things likethe fact that many useful manipulationsare 8, 12, or 24 quarter-turns long. If youcount half-turns, you get a whole spectrumof random move counts, thus missing somefundamental (and as yet little-understood)kinship between these manipulations.Of course, if you are not interested in suchthings, any measure of complexity (why notcount equator twists? why not penalize forcounter-clockwise twists, since they aremarginally harder for right-handed people todo?) will suffice. ---Wechsler

Date: 31 JUL 1980 1431-EDTFrom: RP at MIT-MC (Richard Pavelle)To: CUBE-LOVERS at MIT-MCSorry to spoil your day, but.....As you may know Ideal Toy has international rights to the manufactureand distribution of the cube. I just spoke to a fellow from Idealwho is writing up a solution booklet for them and he told me thefollowing:1) When Ideal first bought the rights (and by the way Rubik gets nothing- his "institute" gets it all) they discussed various national promotion schemes. One of them was to offer 1000K bucks to the first person to solve it less than 1/2 hour- that right 1 million.2) One of the first stores in NY to carry it put adverts in the paper offering $50 to anyone who could solve 1 face in 1/2 hour. They lost $3,000 before they ended the offer the next day.3) A fourth grade girl from Florida solved the cube in 3 weeks and can now do it in under 5 minutes.4) The 2x3x3 solid does exist and will be marketed by Ideal later this year. However, instead of colors it has "dominoe" like faces, whatever that means.

Date: 31 Jul 1980 16:44 PDTSender: McKeeman.PA at PARC-MAXCSubject: Re: The shortest solution?In-reply-to: ALAN's message of 31 July 1980 13:06-EDTTo: Alan Bawden From: (Bill) McKeemancc: CUBE-HACKERS at MIT-MC, Lynn.ESA lower bound on the number of twists can be derived as follows: There are4.3*10^19 distinct reachable arrangments of the cube. Suppose the moves arerestricted to the (more than sufficient) set RLFBUD. Then there are at most sixindependent choices at each step and the number of reachable places is boundedby 6^n. That gives6^25 < 4.3*10^19 < 6^26,or 26 moves as the (probably unachievable) minimum. If all single-hand-motiontwists, R RR RRR L LL .... DDD are allowed, there are 18 choices, giving18^15 < 4.3*10^19 < 18^16,or 16 moves as a minimum. This isn't very interesting since Singmaster hasexamples 18 twists away. If the orientation of the center squares is alsoconsidered, then the combinatoric is 8.8*10^22, and the minima are, respectively,30 and 19.

Date: 31 Jul 1980 5:13 pm PDT (Thursday)From: Woods at PARC-MAXCSubject: Re: The shortest solution?In-reply-to: McKeeman's message of 31 Jul 1980 16:44 PDTTo: CUBE-HACKERS at MIT-MCYou can do better than that for a lower bound! Say you consider allsingle-hand-motion twists to be okay. Then yes, there are 18 such,but there's no point in twisting the same face twice consecutively, soafter the first twist the tree branching factor is only 15. In fact, there'sno point in twisting a face twice if the only intervening twist was doneon the opposite face; if we look at the operations of the form "twist faceX thusly and the opposite face thusly", there are 45 initial such operations,and 30 at each succeeding branch, but since some branches now representtwo twists and some only one twist, it's much harder to compute theminimum depth of the tree.-- Don.

Date: 31 JUL 1980 2159-EDTFrom: ALAN at MIT-MC (Alan Bawden)Subject: The shortest solution?To: McKeeman at PARC-MAXCCC: CUBE-HACKERS at MIT-MC Date: 31 Jul 1980 16:44 PDT Sender: McKeeman.PA at PARC-MAXC In-reply-to: ALAN's message of 31 July 1980 13:06-EDT From: (Bill) McKeeman A lower bound on the number of twists can be derived as follows: There are 4.3*10^19 distinct reachable arrangments of the cube. Suppose the moves are restricted to the (more than sufficient) set RLFBUD. Then there are at most six independent choices at each step and the number of reachable places is bounded by 6^n. That gives 6^25 < 4.3*10^19 < 6^26, or 26 moves as the (probably unachievable) minimum.This is not an improvement on my result. I (and the rest of the cubehackers I know) consider a unit move to be a 90 degree twist in EITHERdirection. You are only considering CLOCKWISE 90 degree twists.Let me point out that if we were to count twists your way, we wouldno longer have a metric. Both the quarter twist method and Singmaster'smethod result in a measure of distance that is a true metric. Date: 31 Jul 1980 5:13 pm PDT (Thursday) From: Woods at PARC-MAXC Subject: Re: The shortest solution? In-reply-to: McKeeman's message of 31 Jul 1980 16:44 PDT To: CUBE-HACKERS at MIT-MC You can do better than that for a lower bound! Say you consider all single-hand-motion twists to be okay. Then yes, there are 18 such, but there's no point in twisting the same face twice consecutively, so after the first twist the tree branching factor is only 15. In fact, there's no point in twisting a face twice if the only intervening twist was done on the opposite face; if we look at the operations of the form "twist face X thusly and the opposite face thusly", there are 45 initial such operations, and 30 at each succeeding branch, but since some branches now represent two twists and some only one twist, it's much harder to compute the minimum depth of the tree.-- Don.Singmaster's notes are aware of these factors, that is how he improveson the 16 count computed by McKeeman to arrive at 18. Similarly Iused the same factors to improve on the 12^n argument for quartertwists (which gives 19 as a lower bound) to arrive at the number 21.I also compute 24 as the quarter twist lower bound for the extendedcube (considering orentations of the center faces).

Date: 1 Aug 1980 11:53 PDTFrom: Lynn.ES at PARC-MAXCSubject: Re: cube group theory.In-reply-to: Sandeep.Johar's message of 31 July 1980 1413-EDT (Thursday), To: Sandeep.Johar at CMU-10Acc: cube-lovers at mit-mcSingmaster ("Notes on the 'Magic Cube'", p. 12 of edition 4) gives 12 classes or"orbits" of assemblies, each one such that the other 11 cannot be reached withoutdisassembling the cube. To quote Singmaster, "This also means that if wereassemble the cube at random, there is only a 1/12 chance of being able to getback to START," and, "It is advisable to reassemble in the starting pattern."Singmaster's notes have several pages of relevant group theory for thoseinterested./Don Lynn

Date: 1 Aug 1980 12:20 PDTFrom: Lynn.ES at PARC-MAXCSubject: Re: Sorry to spoil your day, but.....In-reply-to: RP's message of 31 JUL 1980 1431-EDTTo: RP at MIT-MC (Richard Pavelle)cc: CUBE-LOVERS at MIT-MC2) Another TRUE story of rubiking:The May Company department stores in the Los Angeles area held contests atfour of their stores when they started carrying the cube, a couple of monthsago. The object was to solve one face in three minutes with their scrambledcube. Reward was a 50 buck gift certificate. They allowed about 6 people everyfifteen minutes to enter for two days. The store where I won (and my 10 yearold daughter, my wife, and three of my next door neighbors) had 21 winners. Iassume the other three stores had similar numbers. They also gave away lots ofcube tee shirts for "good tries".I got my cube the night before the contest. But it took me another month tosolve the whole thing. I might have worked faster at it if the megabuck prizehad been offered. But then you all would have also.4) Singmaster (edition 4, p. 34) reports on the Rubik domino (2x3x3 cubies), andsays each cubie has spots like a domino (1 up to 9). They are to be lined up innumeric order. Top and bottom (the 3x3 faces) are two different colors.Unfortunately, Singmaster laments, magic square type patterns (all directions addto same number) are not possible, since all edges are even, all corners odd. Themechanics of the device are said to be "more complicated than for the MagicCube."/Don Lynn

Date: 1 Aug 1980 15:47 PDTFrom: McKeeman at PARC-MAXCSubject: Re: The shortest solution?In-reply-to: ALAN's message of 31 JUL 1980 2159-EDTTo: ALAN at MIT-MC (Alan Bawden)cc: McKeeman, CUBE-HACKERS at MIT-MCAlan,Sorry I missed your earlier words of wisdom on the subject. Anyway, I aminterested in how you show any of these is, or is not, a metric. Can youforward same?Thanks, Bill

Date: 2 August 1980 01:55-EDTFrom: Alan Bawden Subject: a metric for the cube group.To: CUBE-HACKERS at MIT-MC, McKeeman at PARC-MAXCFirst off, a metric is a function (call it D) that assigns anon-negitive number to every pair of points in some set. This numberis to be thought off as the distance between those two points. Thefunction must satisfy the following:For all a, b and c1) D(a,b) >= 02) D(a,b) = D(b,a)3) D(a,b) = 0 if and only if a = b4) D(a,b) + D(b,c) >= D(a,c)(Number 4 is usually called the "triangle inequality". It is theconstraint that most makes D act like a distance, and not somethingrandom.)We wish to construct a metric on the set of all attainable cubeconfigurations. So from now on, those lower case letters willrepresent cube configurations.Now we have recently been talking a lot about methods of counting thenumber of "twists" that it takes to perform certain manipulations ofthe cube. We have been looking for a function (call it T) thatassigns a non-negitive integer to each manipulation. I claim that itis obvious that any such function should satisfy the following:For all M and N1) T(M) >= 03) T(M) = 0 if and only if M = I (I is the identity manipulation)4) T(M) + T(N) >= T(MN)(We adopt the convention of using upper case letters to representmanipulations. Also we shall use M' to denote the inverse manipulationfrom M.)Now manipulations can be applied to configurations to yeild otherconfigurations. We use aM to denote the configuration resulting fromapplyint the manipulation M to the configuration a. (Note that(aM)N = a(MN), so we may omit the parens and simply write aMN.)Now how may we use our twist measuring function T to obtain a metricon the configurations? Again I think it is obvious that we wish therelationship D(a,aN) = T(N) to be true for all configurations a, andall manipulations N.It is easy to show that given that D(a,aN) = T(N), metric propertynumber 1 is equivalent to twist measure property number 1. Similarlyfor numbers 3 and 4. But what about metric property number 2?Well, if T(N) = D(a,aN), and D(a,aN) = D(aN,a) (property 2!), anda = aNN', then we have that T(N) = D(aN,aNN') = T(N'). So the missingproperty of twist measures must be that T(N) = T(N').So this means that if we agree that T(L) = 1, and we like metrics(how can we use words like "distance" unless we have a metric?), thenT(LLL) = T(L') = T(L) = 1. We can argue about T(LL) some other time!

Date: 2 Aug 1980 12:26 PDTFrom: McKeeman at PARC-MAXCSubject: Re: a metric for the cube group.In-reply-to: ALAN's message of 2 August 1980 01:55-EDT (yawn)To: Alan Bawden cc: CUBE-HACKERS at MIT-MCAlan, I enjoyed your presentation. I am convinced that counting the RLFBUDmanipulations will not give a metric. I do not, however, see an easy way tocompute twists T(M). I think one gets a metric only if one takes the minimumover some set of manipulations. That is, take a set, AM, of atomic movesincluding their inverses, let AM* be the strings of AM, and |M| be the lengthof M in AM*. ThenD(a, b) = min |M| such that a = bMdefines a metric. D(a,b) would sometimes be undefined if AM did not generatethe whole group. The recent discussion on shortest solutions is in fact about themaximum of such a T(M) for all M in some AM*.Bill

Date: 3 August 1980 02:20-EDTFrom: Alan Bawden Subject: more on metricsTo: McKeeman at PARC-MAXCcc: CUBE-HACKERS at MIT-MCYes, it is true that the four conditions I gave for twist measuresdon't guarentee that the function will behave anything like the kindof complexity measure we are looking for. I was only trying to showhow some of the properties you might expect of a twist measure couldbe used to generate a metric, so I didn't actually need strong enoughproperties to ensure reasonable twist measures. The additional propertyI have been using to assure reasonability is the following:5) For all M, if T(M) > 1, then there exists an N such that0 < T(N) < T(M) and T(N) + T(N'M) = T(M).Note that N'M has the property that 0 < T(N'M) < T(M) (easy to show)so the two manipulations N and N'M are both "simpler" than M. We canthus easily show that any manipulation M can be expressed as theproduct of T(M) twists (where a twist is defined as a manipulationsuch that T(N) = 1).

Date: 4 Aug 1980 1156-PDTFrom: Dave Dyer Subject: cube permutationsTo: cube-lovers at MIT-MC I'm not quite satisfied with the numbers that have been quotedfor various cube groups. I believe they are probably correct,but I haven't seen anything resembling a satisfactory sketchof a proof.Number of Reachable Positions: The standard calculation runs like : we have 8 corners that canbe in all 8! arrangements, and 12 sides that can be in all 12!arrangements, except that the total permutation must be even.( I'll talk about orientations later) I can accept this figure as an upper bound, but can anyone demonstratethat all the positions not ruled out can actually be reached? The onlyway I can think of is an actual construction, which means findinga generator for the group (for instance of corner cube positions) andshowing that its order is 8!. This is somewhat less than elegant,and requires an unspecified bit of magic to 'find' a generatorfor the group. I also don't like the recourse to geometric arguments ( the cornersand sides can't be interchanged ) but I am willing to accept it.Number of Reachable orientations: A similar line of reasoning is used to argue the number of reachableorientations : the amount of twist on all corners and (independantly)on all sides is a multiple of 360 degrees. I haven't seen a demonstrationthat all the not-forbidden orientations are actually reachable.Finally, I haven't seen a demonstration that the orientation and permutationsubgroups are indepentant, that is, that you can get to an arbitrarilyselected location and an arbitrarily selected orientation at the sametime. This assumption is the basis for Singmasters claim that thereare 12 orbits of cubes : ( even-spacial-permutation X even-side-orientationX one-of-three-corner-orientation = 2 X 2 X 3 = 12 )-------

Date: 4 August 1980 20:36-EDTFrom: Alan Bawden Subject: cube permutationsTo: DDYER at USC-ISIBcc: CUBE-LOVERS at MIT-MCThe part of the proof that shows that you can actually reach all theconfigurations in a particular equivalence class is not particularlyelegant. Basically, you have to appeal to the details of a particularcube solving algorithm.For example, I have a tool that "flips" two edge cubies in place,without desturbing anything else. This tool shows that I can orientthe edge cubies in ANY even permutation of the edge cubie faces. Thefact that I cannot obtain any odd permutations is a result of the factthat a quarter twist is itself an even permutation of the edge cubiefaces.Most people can examine their own cube solving tools and see that infact, they are capable of obtaining all the configurations notforbidden by the familiar constraints.

Date: 4 Aug 1980 2204-PDTFrom: Davis at OFFICE-3Subject: Cube PermutationsTo: ddyer at USC-ISIBcc: cube-lovers at MIT-MCThe following two transformations are useful in demonstrating that all of theclaimed elements of an equivalence class of positions can be reached. Usingthe RLFBUP notation, the transformation RB'R'B'U'BU reverses two of the cornercubies and leaves all other corner cubies in place (It does, however, shufflearound the side cubies, and does some random twists to the corners). If theabove transformation is repeated four times in a row, everything is leftexactly fixed, except that three corner cubies are each rotated one-third of aturn. By then performing the inverse of this operation on two of the threecorners which were turned and a new corner, it is not hard to see that any twocorners can be the only ones moved, and that they are each rotated one-third ofa turn in opposite directions.If we look at just the corners alone, and ignore in-place rotation, since wecan exchange any adjacent pair, we can obviously get to all permutations of thecorners. A similar argument can be made to show that all the edge cubies canbe arranged arbitrarily (permuted arbitrarily, that is). An easytransformation rotates three of them (among themselves) on a face, and since weare also allowed to rotate the face, it is easy to generate a transposition ofany pair.Unfortunately, the two operations described above are not independent. If wejust look at the blocks and label them ignoring color, a primitive (one-quarterturn) transformation moves four corners into four corners, and four edge cubiesinto four edge cubies. If this is viewed as a member of a permutation group,it is obviously even (the set being permuted is all the movable cubes). Thus,at least half of the positions are impossible. If we ignore the corner cubies,and look at the colors of the edge cubies, every primitive rotation rotates thefour front colors, and the four colors around the outside, again, an evenpermutation. Since one of the above used coloring and no corner cubes, and theother did not, there must be at least a factor of 4 impossible positions. Amuch more complicated argument shows the necessity of a factor of three in theset of impossibles. (Does anyone know of a simple way to see this? I just didthe obvious thing of defining "standard" orientations of every cube in everycorner, and showed that all the primitive transformations caused the totalrotation away from standard to be a multiple of 2 pi.)Using the transformation which flips any two in place, and the two discussedabove, it is not hard to see that the factor is at most and at least 12.Boy, It sure is hard to prove things on a computer. On myy notes withdiagrams and all, this is perfectly clear, but I get confused trying to read myonline proof. I hope that someone can make some sense out of it.-------

Date: 6 AUG 1980 0939-EDTFrom: JURGEN at MIT-MC (Jon David Callas)Subject: Random Notes on CubismTo: CUBE-LOVERS at MIT-MCCC: JURGEN at MIT-MCI am beginning to feel competant enough to jump into the fray, so ...1) language: I guess that RLUDFB has become the de facto lg. for cubing.(by the way, how IS 'RLUDFB' pronounced?) Is this to include the IJK rotationsthat ACW described on 7-19? They are very handy, but I know that there is a desire tokeep the lq. brief. Also what (if anything is being done with the centerslices that Bernie used while describing the higher crosses? Given RLUDFB+IJKnotation, they are easy algorithms (subroutines?) that can be described as:FPC {floor-parallel center} := udKBPC {body-|| center} := fbIBSC {body-slicing center} := rlJwithout IJK, they're much more complex. Perhaps this is a case for usingRLUDFB & IJK.2) I have been playing a little with the group of rotations, twists and so forth. I thought that it mighht be very wieldy until I remembered that Don Woods has given us a transform with period 1260. (Is it just myimagination or is 1260 important? it seems like a familiar number) This means(by la Grange's thm ) that the order of this group is some integral multipleof 1260 (ugh!) That eliminates hand-enumerating (at least by me) it. Howmuch does anyone else know about this, and are thre any interesting subgroups?It also seems that this thing might be related in some way to the dihedral gps.but I'm not sure.3)Somewhat connected with (2) above, it would be nice if we compiled"Famous Transforms I Have Known" or something to that effect. Not only wouldit help in finding my desired sbgp, but it would help keep some of thefolklore out of cubing (while there is nothing wrong with folklore,there are problems when falsities (especially VERY inobvious ones) creepinto the body of lore. Folklore also does tend to cause duplication ofeffort and that most of us have probably been wasting time re-inventingthe Pons Ansinorum (which, by the way, IS 'rrllffbbuudd', isn't it?)Since I'm suggesting this nonesense, I guess it's only fair tovolunteer to do this compendium (unless someone else is dying to).4) Would it be possible to modify :cube to work on aprinting tty? I really don't care if it produces large volumes of paper.I am avoiding implementing my own form of a Cube program on myAPPLE ][ (see my notes a