An Upper Bound for the Number of Different Positions of the Fully-Colored Magic120Cell The following is an upper bound for the number of permutations of the fully-colored Magic120Cell. To establish it is the exact answer (which I am virtually certain of), a number of algorithms must be found, which I will describe in my explanation. Here is the upper bound: (600!/2)*(1200!/2)*(720!/2)*((2^720)/2)*((6^1200)/2)*((12^600)/3) The exact value was computed using the free computer algebra system Yacas; here it is in its full glory: 23435018363697222779126210606140343600982219866708667227704291465940007 37743198001537086016413748065359228217622633869330769129523601891497799 90823414733250819032377663096727895392891107724676361939174468537213471 84699260131924584724938945790242680862147295113762851571432130901040238 96149551266842769465158629370618815995041314328829732432057176063611161 23422302770133676753359134856348612503635674252607065815753807941966366 98057536512196715919594180779891338303538085006270891583549467992567391 85180535778985103137974951114346934416286264525322532242698044327455362 45594013789336900464999769975314632465421639791307831594564938014806846 43170816415770010483963217284920963354265992129473309221874222731561178 16542353296798264919536869373254412130565886195935986667368898349834998 21329543130389608025077440970771216857898162084209764122888617411552472 35969553038560987694512525780640891845410615026444483377179855326132898 75234261959552618641928258383934632570287455387991780347467121010722113 61928836844443707162412473785444967682885281547729595860180055748863425 82987146883285105106538113133816701062677558383952546932927579065352378 01699938857635611816907866063280477566711511600651402621287007177419657 47137395706297269591116929204261763967322064643743204180740840609622274 50477533328851963152796037024975768039218238701900252954269938177351575 02677389416404108346187189424180896325665818763923753999882873138580846 77228966566309226326618668840915289587325249776450099751298794240127365 82338830099863762586940626070992713912334648392737597361950653986530252 30327207836333881250393819594360829900032177090494274930448392889865527 16614065052187986459391365691818487107035801789611790557625739664732160 65851938727927023709296602229334790711981400149187774330090686038188693 91261606235286494542181170865114851849953373717543236061723434256780261 70547107650201457180103595669061215322965129698879686174938686442023883 00748340309101390837462739717879861015966407725537701315760595302551716 71986415959726651855198833270638521668923163972072488967633862968020464 59857154591489337994834077319666200102594473324166310981151194815089205 57151182894539191982468293894472660275087710846277134524865818266133054 42334380903271185037626999712118120957184637997454373033536708899691932 67133888402013848180589600375369501541798982850283307728992349369110105 46520467826426004880473115209607600645972315535718172987982451474844986 38207939550826131991321502334364118044702920268720341762396367094618866 13506333873119893045317942691097910138170606688929064386560230196639558 14850013029742851991570001253742052023463664478943640221712458729805575 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92219730863040706516534661225522594916193135982360752630702024083643694 98309370698498093084824000430278614244379613737044583505562135097992678 05260822568001712636170958894341958120566294218192269192589063843887982 76399691863692115560126140715610109114003149849449412445547918396053785 42102203322863117771608181202015567452798664995683183676667131261081040 30610697346947842941118989099929505001072885907143020380705671971970771 74641197649400613289762440747999594711818677478380093322693390443497615 06790858102509872412149022901579978879595015323736540446504645407248271 24442974862512599608887589752218559193144931596281284315382618742792620 66616881593787942961115669105927538622586908510205223079160329890976613 24318437454227007437361053657522104635365530904941109094426111379946913 71854373062622155659107585797616686931874970164036381491924856162327083 87215983278489287122517383893445550987886905214362677925682743059318092 90715982378923238819174377671246115948128333247228319128499096287090364 06418925007261761200623236257957747401814104819201322380783299932882694 97705069648412898606682892058216414535137703172325349226036435233500060 08811101917211049364489819890827973553466812312700794247970136249599713 68830975248367892523082396738072274831267273049793679458450960225575330 90840325055925127369491405078011569600959829055592354981900265212992888 74537523085955044911868546931655826676111114140917917721449373043059908 24075969774780698659600903247322382509271117981454345778343923721701140 34040387142730946291948768518544291460594918104272439297270660195239204 69851212038726474485921192066725395225842350618752505691550098017532445 29742915483006071654290990776376332377597123229369363319211034520828156 16383626599751692734054125142693424208441259140739967321942103460390485 73512549204538199361441602981588922796564372729802637150967463996220269 92509662606254579651749991204772662937610983604733514590588466763484779 75336521786978901110936704729127455396942554264720541493172351367586278 52118009553781736752460941012653895714556808971968820220233708185524289 26324734529251413367934964381909880343066993726638347012446562279909471 00665871099287936575368913555297252102692185719691751527961183922455231 72371787084227168597930766375481134730976264840880937562376002100356262 35107696623982892463214959186113390887996406714662349935032809366747705 65928739057221107528446966775483155772142995330294320084551917275407602 87830218813852897349462068161826723496382625465167461901840049431850524 89018407136301332421978685188429040584176333209422917640992438642326814 48471988797114061727140678598275664209888253021405519815632168746508451 06268660985414101246860290152701992710734632469139060273152361598118124 26751417487110100479881455904096718779749277515897027333976945734065218 13427043475236100473238801150143720068671986307968791851735260237830993 59283951655340545557118534217560207955199404424096337283839953943573882 77230454238664055061747285720327136114251359554586129278357158728391085 51846977682603655222729137145829356583863818649600000000000000000000000 00000000000000000000000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000 What a number! Here is an outline of how I derived this upper bound: First of all, we can count that Magic120Cell has 120 1-colored immobile center pieces, 720 2-colored pieces (120*12/2), 1200 3-colored pieces (120*30/3), and 600 4-colored pieces (120*20/4); each piece being uniquely colored. There are three types of rotations to consider when examining how Magic120Cell can be scrambled (or solved!) When a dodecahedral face rotates 72 degrees about a line connecting the midpoints of two opposite pentagonal faces, there are two 5-cycles of the 2-colored pieces, six 5-cycles of the 3-colored pieces, and four 5-cycles of the 4-colored pieces. Since 5-cycles are even permutations, all permutations of each type of piece, including facelets, will be even when this rotation is used. A dodecahedral face can also rotate 180 degrees about a line connecting the midpoints of two opposite edges. When this happens, there are six 2-cycles of the 2-colored pieces, 14 2-cycles of the 3-colored pieces, and ten 2-cycles of the 4-colored pieces. 2-cycles are odd permutations, but note that each type of piece performs an even number of 2-cycles, making the total for each type of piece even. Hence, once again, all permutations of each type of piece, including facelets, will be even when this rotation is used. Finally, a dodecahedral face can be rotated 120 degrees about two opposite verticies. For this rotation, there are four 3-cycles of 2-colored pieces, ten 3-cycles of 3-colored pieces, and six 3-cycles of 4-colored pieces. Since 3-cycles are even permutations, all permutations of each type of piece, including facelets, will be even for this rotation. Thus we see that no matter what type of rotation is used, an even permutation of pieces and facelets will occur for each type of piece. Therefore, the 720 2-coloreds can be permuted 720!/2 ways, dividing by 2 because of the even parity. Similarly, the 3-coloreds and 4-coloreds can be permuted 1200!/2 and 600!/2 ways, respectively. Multiplying these three terms together, we obtain an upper bound for the number of ways the pieces can be permuted without regards to orientation: (600!/2)*(1200!/2)*(720!/2) To show this number is exact, we will have to find 3 algorithms: one that performs a 3-cycle of any three 2-coloreds without affecting any other pieces, a 3-cycle of any three 3-coloreds without affecting any other pieces, and a 3-cycle of any three 4-coloreds without affecting any other pieces. These three algorithms, when combined with each other and conjugates (setup moves), can produce any possible permutation of the pieces. Now for the orientations. Since the facelets also undergo even permutations, all orientations will be limited to even permutations of facelets. Therefore 719 2-coloreds can be oriented in any of 2 ways each, but the last will be determined by the others because of the even parity. This results in (2^720)/2 or 2^719 ways of orienting the 2-coloreds. To show this number is exact, we must find an algorithm that flips two 2-colored pieces without affecting the others. For the 3-coloreds and 4-coloreds, I followed closely the methods of Keane and Kamack in their paper, "The Rubik Tesseract", modifying their arguments as neccesary to apply them to Magic120Cell. Any 3-colored piece can be oriented in 6 different ways. (not three, because in four dimensions we can reflect 3-colored pieces as well as twist them!) Notice that a twist (a 3-cycle of the facelets on that piece) is an even permutation, while a reflection (a 2-cycle of two of the facelets on that piece) is an odd permutation. Since the total parity of all of the 3-coloreds must be even, the first 1199 3-coloreds can be oriented in 6 ways each, while the last can be oriented in only 3. (If the first 1199 3-coloreds total to an even permutation, the last 3-colored must be one of the 3 even twists, while if they total to an odd permutation, the last 3-colored must be one of the 3 odd reflections) This gives a total of (6^1200)/2 or (6^1199)*3 ways of orienting the 3-colored pieces. To show this number is exact, we must find an algorithm that twists one 3-colored piece without affecting the others, and an algorithm that reflects two 3-colored pieces without affecting the others. Finally, the toughest part. The orientation of the 4-coloreds required me to generalize the group theory based solution for the corners of the 3^4 cube by Keane and Kamack to Magic120Cell. In their paper, Keane and Kamack first describe that there are 24 permutations of the facelets of a 4-colored piece, comprising the S4 group (The symmetric group on four letters.) They describe orientations using cycle notation of the four faces, labeled a, b, c, and d. The 24 different orientations can be broken down into four crosses, (ab)(cd), (ac)(bd), (ad)(bc), and I, the identity; eight 3-cycles, called twists; six 2-cycles, and six 4-cycles. It turns out that for n-colored pieces in n-dimensional space, only even permutations of the facelets can occur, because the odd permutations are n-dimensional mirror images (This rule also applies to the 3-dimensional Rubik's Cube, and to Magic120Cell.) Thus, 2-cycles and 4-cycles cannot occur in 4-space because they are odd permutations. Hence, each 4-colored piece can only be oriented in 12 ways. The even permutations are all possible, and form the alternating group A4. Keane and Kamack continue by observing that the crosses are a normal subgroup of the alternating group they call N. So, N = [I, (ab)(cd), (ac)(bd), (ad)(bc)] The cosets of N consist of the twists, which they call S and Z: S = [(abc), (adb), (acd), (bdc)] Z = [(acb), (abd), (adc), (bcd)] They then note that the sets N, S, and Z form the quotient-group of A4 by N, in which N acts as the identity: A4/N = [N, S, Z] We can then see that the group multiplication table is: N S Z N N S Z S S Z N Z Z N S From this table, we can see that this quotient-group is isomorphic to the group of residue classes, mod 3. This means that we can assign the number 0 to N, the number 1 to S, and the number -1 to Z, and adding these numbers mod 3 is the same as taking the product of elements of these three subgroups. Notice that this entire argument applies equally well to Magic120Cell as to the tesseract. Now the only thing left to show is that the sum of the orientations of the 4-coloreds (counting 0 for an orientation in N, etc.) mod 3 is always the same, whether to the tesseract or Magic120Cell. The orientations can be defined by assigning, to each 4-colored piece, a letter to each facelet and each position of each facelet. Then each orientation can be described by a 4-letter string (e.g. ABCD) relative to the position it is occupying. When pieces or cubies rotate in a cycle, their facelets undergo n seperate cycles if they have n facelets. The important thing is that there are always n disjoint cycles. In the tesseract, every corner rotation boils down to four 4-cycles of facelets for each cycle of four cubies. In Magic120Cell, it is the same except it is four 5-cycles. If we can show that in cycles of any length, the sum of the orientations of the pieces does not change, we will have proved this for both the tesseract and Magic120Cell. Consider four 2-cycles: ABCD 1 ABCD 2 Each row represents a 4-colored piece. The actual 4-cycles are vertical in direction. For example: ABCD 1 CDAB 2 This means that facelet A on piece 1 goes where facelet C on piece 2 was, etc. In this example, piece 1 performed an N-twist. Now notice that since we are dealing with 4-cycles, the facelets of piece 2 must return to the original positions of the facelets of piece 1. Therefore, piece 2 also performed an N-twist. It can be checked that if piece 1 performs a Z-twist, piece 2 performs an S-twist, and if piece 1 performs an S-twist, piece 2 performs a Z-twist. Therefore, the sum of the values does not change, and equals zero. (N=0, Z=-1, S=1) Now we can do a proof by induction to show that four cycles of any length has this property. Assume that 4 k-cycles always sum to zero: ABCD 1 . . . ABCD k (not neccesarily this orientation) We can think of this as a single k-cycle of the pieces (1 2 ... k). Now consider adding a piece, to produce the cycle (1 2 ... k (k+1)). The piece k now replaces piece k+1, (and hence may change the type of twist it performs) and piece k+1 replaces piece 1. Notice that we can write (1 2 ... k (k+1)) as a product of the two cycles (1 2 ... k)(1 (k+1)). Now by induction, the elements of the first cycle sum to zero. But notice that the second cycle also sums to zero by induction. Thus, the cycle (1 2 ... (k+1)) sums to zero as well. Therefore, no matter what the length of the cycles, the sum of the values of the orientations equals zero. This means it is true for both the tesseract and Magic120Cell. So now we know the sums of the orientations, mod 3, always equal zero in Magic120Cell. Since an N-twist is 0, we can have an isolated N-twist without affecting any other pieces. The value S - Z must therefore be congruent to 0, mod 3. This means that the first 599 4-colored pieces can each be in any of 12 orientations. If the value of orientations up to that point is 0, the remaining value must be an N-twist. If it is 1, the remaining value must be a Z-twist, and if it is -1, the remaining value must be an S-twist. In each case there are four possible orientations left for the last 4-colored piece. Therefore, the upper bound for the orientations of the 4-colored pieces is (12^600)/3 or (12^599)*4. To prove this is exact, we must find algorithms that show that any N-twist can be performed without affecting the rest of the pieces, and algorithms showing that any Z-twist can be performed along with any S-twist without affecting any other pieces. Keane and Kamack were able to find these 20 algorithms, all generated from a single pair of twists. This technique can probably also be applied to Magic120Cell. When we multiply these figures with the ones for permutations, we arrive at the final answer. I believe that before long, all of the algorithms neccesary to make this figure exact will be found. I wish to thank Roice Nelson with the deepest gratitude for creating Magic120Cell, which without, I would have never had the inspiration to perform this calculation. -David Smith Update: Roice has discovered 3 algorithms that perform 3-cycles on the 2-colored, 3-colored, and 4-colored pieces. They are actually less restrictive than the ones that I said would be required. The algorithm that cycles the 2-coloreds scrambles the 3-coloreds and 4-coloreds, the one that cycles the 3-coloreds scrambles the 4-coloreds and preserves the 2-coloreds, and the one that cycles the 4-coloreds preserve the rest of the pieces. Therefore, as long as they are applied in that order, any possible permutation of pieces can be achieved. I am sure that before long the same will also be true for orientations!