Automorphisms of S 6 and the Colored Cubes Puzzle
Ethan Berkove, David Cervantes Nava, Daniel Condon, Rachel Katz
AAUTOMORPHISMS OF S AND THE COLORED CUBESPUZZLE
ETHAN BERKOVE, DAVID CERVANTES NAVA, DANIEL CONDON, AND RACHEL KATZ
Abstract.
Given a palette of six colors, a colored cube is a cube where eachface is colored with exactly one color and each color appears on some face.Starting with an arbitrary collection of unit length colored cubes, one can tryto arrange a subset of the collection into a n × n × n cube where each face isa single color. This is the Colored Cubes Puzzle . In this paper, we determineminimum size sets of cubes required to complete an n × n × n cube’s frame ,its corners and edges. We answer this problem for all n , and in particularshow that for n ≥ S action. In addition to the problem simplification this actionprovides, it also gives another way to visualize the outer automorphism of S . Introduction
Given a palette of six colors, a “colored cube” is one where each face of thecube is one color and all six colors appear on some face. A pleasant combinatorialargument shows there are exactly 30 distinct colored cubes, which can be used ina number of interesting puzzles. For example:(1) Select one colored cube. Find seven other distinct colored cubes and as-semble all eight into a 2 × × × × n arbitrary colored cubes and determines when it is possibleto construct an n × n × n cube where each face is the same color. This is the formal“Colored Cubes Puzzle.” The following is the main theorem from [2]. Date : November 5, 2018.2000
Mathematics Subject Classification.
Key words and phrases.
Discrete geometry, Cube stacking puzzle, Outer automorphism of S .This project was funded by NSF grant DMS-1063070. We also gratefully acknowledge supportfrom the Provost Office at Lafayette College. a r X i v : . [ m a t h . C O ] M a r ETHAN BERKOVE, DAVID CERVANTES NAVA, DANIEL CONDON, AND RACHEL KATZ
Theorem 1.1.
Let n > . Given n arbitrary colored cubes, it is always possibleto solve the Colored Cubes Puzzle. As noted in [2], there is a variation of this problem that is a better indicator of itsdifficulty. In particular, the cube varieties that make up the ( n − × ( n − × ( n − n × n × n cube can be arbitrary. The same is true for the 6( n − cubes in the interior of the six faces, since all six colors are represented on everycolored cube. The only cubes that cannot be chosen arbitrarily are those that lie onthe edges, which we call the frame of the n × n × n cube. Therefore, once a cube’sframe has been constructed, the Colored Cubes problem is solved. The focus ofthis paper is Conjecture 5.4 in [2], which posits that for n sufficiently large, whenthere are enough cubes to fill in a frame then one can actually construct it. Ourmain results are: Theorem A . Any set of 24 cubes is sufficient to build a 2 × × × × Theorem B . When n >
3, given an arbitrary collection of 12 n −
16 colored cubes,one can always build a frame.An important component of a number of our proofs involves the obvious S action on the cubes induced by color permutation. There is an arrangement ofthe collection of 30 colored cubes–the tableau–where the action has particularlynice properties. This action allows us to simplify many our arguments, and playsa central role is a number of proofs in this paper. As a bonus, the S actionon the tableau provides another concrete demonstration of the action of an outerautomorphism of S .In the next section, we introduce notation and a formal statement of the ColoredCubes problem. In the third section, we describe the tableau, give some of itsproperties, and discuss its associated S action. In the last three sections we usethe S action and other results to prove our main theorems. We conclude withsome open questions.It is a pleasure to thank Liz McMahon and Gary Gordon for many helpfulconversations during the construction of these arguments, and for comments andsuggestions which improved the overall quality of this paper.2. Definitions and Problem Statement
To distinguish between different cubes, we set up an equivalence and say thatcubes that have the same coloring up to isometry of R are of the same variety . Asmentioned in the introduction, there are 30 distinct varieties of colored cubes. A solution to the puzzle is an arrangement of n cubes that forms an n × n × n cubewhere each face is one color. A corner solution is a set of eight cubes that formsthe corners of a solution. We say that an n × n × n solution cube is modeled aftera colored cube if the solution and its model are of the same variety. We note thata solution and its corner solution are also always of the same variety.We also track relative positions of face colors on the cube. The unordered pairof colors that are opposite each other on a cube form an opposite pair . Similarly,the unordered pair of colors on faces that share an edge of a cube are an adjacentpair . We call the three colors on the faces which meet at the corner of cube a cornertriple . Since order matters in this last case, we read the colors clockwise around thecorner. Furthermore, two corner triples are equivalent if one is a cyclic permutation UTOMORPHISMS OF S AND THE COLORED CUBES PUZZLE 3 of the other. That is, (2 , , ∼ (6 , , (cid:54)∼ (6 , , mirror image corner triples if they contain the same colors but are not equivalent. Note that ona given colored cube, every possible unordered pair of distinct colors appears eitheras an adjacent pair or an opposite pair. So every colored cube contains 12 of the15 possible adjacent pairs, and 8 of the 40 possible corner triples.In general, we will denote a generic cube by c and its mirror image by c ∗ ; c and c ∗ together form a pair of mirror cubes . We note that mirror cubes have the sameopposite and adjacent pairs, but have mirror image corner triples. We have thefollowing useful characterization. Lemma 2.1.
The varieties c and c ∗ are related through the exchange of an oppositepair. Therefore, the varieties c and c ∗ are the only two with the same opposite pairs.Proof. This is directly seen by visualizing a mirror placed parallel to some face ofa cube c . For the second part, iterate the mirror procedure. (cid:3) As mentioned in the introduction, the Colored Cubes puzzle is solved once theframe is complete. In an n × n × n cube, this consists of 12( n −
2) + 8 = 12 n − n cubes. Since the number of cubes in theframe grows linearly with n , starting with n cubes is generous. In fact, since thereare only 30 distinct varieties of colored cubes, when 12 n − < n , the pigeonholeprinciple guarantees a solution for the frame using just one variety of cube. Thishappens for n ≥
19. We therefore focus our attention towards determining the valueof the frame function fr( n ), which gives the size of the smallest collection of cubeswhere one is guaranteed to be able to build a frame, regardless of the componentcube varieties. Knowing the value of fr( n ) gets to the heart of the Colored Cubespuzzle, as it is the best possible result.3. The cube tableau and automorphisms of S We describe an arrangement of the 30 colored cube varieties, the tableau, whichwill aid in the proofs that follow. This tableau, which can be found in the appendix,is an arrangement of the 30 cube varieties in a 6 × S action on the tableau that arises from permuting the setof six face colors. Furthermore, the action provides a way to visualize the outerautomorphism of S . Using terminology that goes back to Sylvester [13], call an ETHAN BERKOVE, DAVID CERVANTES NAVA, DANIEL CONDON, AND RACHEL KATZ unordered pair of six colors a duad and a collection of three duads a syntheme . Acollection of five synthemes such that each of the 15 duads appears exactly once isa pentad . Duads, synthemes, and pentads correspond to opposite pairs, the threeopposite pairs in a cube, and the collection of such pairs in each row and columnof the tableau, respectively.
Lemma 3.1.
A permutation of the color palette induces a permutation that sendsrows to rows and columns to columns. Once the permutation is determined on thetop row of the tableau, the action on the rest of the tableau is uniquely determined.Proof.
A color permutation of the palette takes pentads to pentads, so it is sufficientto show that a row isn’t taken to a column. Since all permutations in S aregenerated by transpositions, we will show that applying a transposition sends rowsto rows. Let ( a , a ) be the transposition. There is precisely one cube per rowwith the duad { a , a } as part of its syntheme, so call it cube c . By Lemma 2.1,exchanging colors for c has the geometric effect of sending it to its mirror image c ∗ .That means that c ’s row pentad is sent either to the row pentad of c ∗ or the columnpentad of c ∗ . Take another cube d in c ’s row with syntheme { a a }{ a a }{ a a } .Applying ( a , a ) to d yields the syntheme { a a }{ a a }{ a a } . Since d ∗ is part of c ∗ ’s column pentad and already contains the duad { a a } , ( a , a ) must send d to c ∗ ’s row pentad. Finally, since the tableau has mirror symmetry across the diagonal,once the action on the rows is determined, so is the action on the columns. (cid:3) Lemma 3.2.
Consider the set of permutations of the color palette that fix a dis-tinguished variety in the top row of the tableau. This set is isomorphic to S , andit acts faithfully on the non-distinguished cubes in the top row.Proof. The stabilizer of the distinguished cube is its group of direct isometriesin R , which is S , and by Lemma 3.1, any palette permutation that fixes a thedistinguished variety sends its row to itself. A simple check shows that at least oneelement of the stabilizer acts non-trivially on the tableau. Further, if a permutationfixes every cube in the first row, by Lemma 3.1 it fixes every cube in the tableauand must be in the tableau stabilizer, which is a normal subgroup of S . The onlypossible non-trivial normal subgroup, A , is clearly too large to be this stabilizer.We conclude the action is faithful. (cid:3) Corollary 3.3.
Let σ be the map that takes α ∈ S to its action on the six pentadsvia permutation of the color palette. Then σ is an outer automorphism of S .Proof. Given a transposition ( a , a ), since every row and column in the tableaucontains exactly one cube with the duad { a , a } , each of these cubes gets sentto a different row, the row containing its mirror image. Therefore, the effect of( a , a ) on the tableau is to swap three pairs of rows. This map isn’t an innerautomorphism, since it doesn’t preserve cycle structure. (cid:3) The construction of the tableau and its relation to automorphisms of S werepreviously known. J. H. Conway used the tableau to provide a complete answer toPuzzle 2 in the Introduction [9]. In Conway’s notation, tableau rows are labeled A through F and tableau columns are labeled a through f . The cubes in the tableauthen have the following “coordinates.” UTOMORPHISMS OF S AND THE COLORED CUBES PUZZLE 5
Ab Ac Ad Ae AfBa Bc Bd Be BfCa Cb Cd Ce CfDa Db Dc De DfEa Eb Ec Ed EfFa Fb Fc Fc FeIn this nomenclature, cubes Xy and Yx form a mirror pair.In addition, P. Cameron mentions the connection between the tableau and au-tomorphisms of S in a WordPress blog [3]. The tableau provides a particularlynice demonstration of the action of the outer automorphism of S , and we believeit deserves to be better known.There are additional relationships between varieties in the rows and columnswhich we will use in subsequent sections. Before we state these results we recalltwo lemmas from [2]. Lemma 3.4. [2, Lemma 2.6]
Two colored cubes share exactly nine, ten, or twelveadjacent pairs according to whether they share exactly zero, one, or three oppositepairs, respectively.
Lemma 3.5. [2, Lemma 2.7]
Given two varieties: (1)
If they have no opposite pairs in common, then they share zero or twocorner triples. (2)
If they have one opposite pair in common, then they share exactly two cornertriples. (3)
If they have three opposite pairs in common, then they share zero or eightcorner triples.
The proof of Lemma 3.5 follows from the description of how a variety c is relatedto the 29 other cube varieties, which we summarize here. There are eight varietiesthat arise from choosing a corner triple of c and cyclically permuting its colorsclockwise and counterclockwise. These varieties share two corner triples but noopposite pairs with c . Their mirror images form eight new varieties that shareno corner triples and no opposite pairs with c . There are twelve varieties thatcome from exchanging an adjacent pair on c . These varieties all share two adjacentcorners and one opposite pair with c . The last variety is c ∗ , which shares threeopposite pairs and no corner triples with c . This characterization also providesadditional structure to the tableau, as well as a means for its construction. Corollary 3.6.
Fix a distinguished variety c in position Ab in the tableau. Then (1) The variety Ba is c ∗ . (2) The eight varieties related to variety c by cyclically permuting of a cornerfollowed by a mirror image are in column b and row A. (3) The eight varieties related to variety c by cyclically permuting of a cornerare in column a and row B. (4) The twelve varieties related to variety c by edge flipping are in columns c,d, and e, and rows C and D, and E.Proof. Since the tableau has mirror symmetry, c ∗ is in position Ba. Column band row A are pentads containing variety c . From the characterization in Lemma3.5 and the discussion afterwards, these eight varieties must be formed from c by cyclically permuting a corner followed by a mirror image. By checking case, ETHAN BERKOVE, DAVID CERVANTES NAVA, DANIEL CONDON, AND RACHEL KATZ one finds that the varieties that are formed using a clockwise cyclic permutationconstitute one pentad, and the varieties that are formed using a counterclockwisecyclic permutation form the other. Using mirror symmetry, column a and row Bare formed from c by cyclically permuting a corner. That leaves the last twelvevarieties in the claimed positions. (cid:3) Corollary 3.7.
Let Xy be an arbitrary variety in the tableau. Then the varietiesthat share no corners with variety Xy are precisely variety Yx, the varieties incolumn y, and the varieties in row X.Proof.
The variety Yx is the mirror image of Xy, so shares no corner triples withit. The statement about the row and column follows from Lemmas 3.6 and 3.1once we note that there is a permutation of the color palette that moves Xy to thedistinguished position. (cid:3) Building a × × solution: fr(2) = 24In [2] it was (incorrectly) conjectured that fr(2) = 23. Using properties of thetableau from Section 3, one can quickly construct a counterexample. We firstdescribe a way to keep track of the cubes in an arbitrary set using the tableau. Aset may contain many cubes of one variety; if variety Xy occurs k times in the set,put k into Xy’s position in the tableau. We call k the repetition number of thevariety, and note that if k ≥
8, then there is a corner solution modeled after Xy. Ifvariety Xy is not part of the set, put a dot in its position in the tableau.Consider the following collection of 23 cubes. Within the five cubes of one rowpentad, take seven copies of three varieties and one copy of the other two. ByLemma 3.1 there is a color automorphism that moves one variety with multiplicityseven to position Ab in the tableau. By Lemma 3.2 there is another automorphismthat puts the row A into the following form:7 7 7 1 1This collection cannot be used to build any variety in the top row of the tableausince no two varieties in this row share corners. One can’t build a variety Xa incolumn a, since Lemma 3.5 implies that the other varieties in row A can contributeat most two corners to the frame, variety Xa shares no corners with variety Ax,and at least one variety only appears once in the collection. In a similar way, avariety Xy in rows b through f shares no corners with variety Ay. The remainingfour cube varieties are insufficient to construct a corner solutions.We claim that fr(2) = 24. Our argument uses partitions of the set of 24 cubesinto k subsets, where the size of a subset determines a variety’s repetition number.We write our partitions in decreasing order as like 54 . This tells us that theset of 24 cubes is divided into eight distinct varieties, one variety with repetitionnumber 5, three with repetition number 4, etc. Once we fix a partition, we stillneed to assign varieties to the repetition numbers, and with a set of 30 possibilitiesthe number of ways to do this is quite large. For example, the number of cases wehave to check to ensure that the partition 54 of 24 cubes always has a solutionis about 1 . × . Instead of checking them all, we determine conditions on thestructure of the partition which will always yield a solution. This is the focus ofthe following computer calculations and lemmas. Lemma 4.1.
Any collection which contains ten varieties always has a corner so-lution.
UTOMORPHISMS OF S AND THE COLORED CUBES PUZZLE 7
Proof.
The proof is by computer search, written in
Mathematica . One builds the cube-corner bipartite graph, where one set of vertices is the set of 30 possiblecolored cubes and the other set is the 40 possible corners. Edges in the cube-cornergraph connect varieties in the first vertex set with their corners in the second.The algorithm proceeds by choosing a subset of ten varieties from the 30 possible.Rather than work with the entire bipartite graph each time, for each subset thealgorithm loops through all 30 possible corner solutions, building the subgraph ofthe cube-corner graph where one set of vertices is the subset of ten varieties, andother set is the set of eight corners from the variety of the potential corner solution.
Mathematica then determines a maximal matching, where a matching of size eightmeans the distinguished corner solution can be assembled from the subset. (cid:3)
At first glance, Lemma 4.1’s proof requires checking (cid:0) (cid:1) ≈ . × cases, butwe can use the tableau to reduce this number some. Given ten distinct varieties inthe tableau, at least two are in the same row. This row can be moved to the topof the tableau using the S action, and furthermore we can assume that the twovarieties are Ab and Ac. This leaves (cid:0) (cid:1) ≈ . × cases, about one tenth of theprevious number. This shorter computation still requires almost four hours of CPUtime using a 2.70 GHz Intel Core i7-3740QM CPU with a 64-bit operating systemand 8 GB of memory.The result in Lemma 4.1 implies that it is sufficient to consider subsets of 24cubes comprised of between four and nine different varieties. This results in 354partitions to consider. We use the next lemma, proven by computer search, toshow that the vast majority of these contain a subset that forms a corner solution,regardless of the varieties which appear. Lemma 4.2.
Given the following sets of cubes, one can always construct a cornersolution. •
18 cubes consisting of two varieties with repetition number and any fourother varieties. •
16 cubes consisting of two varieties with repetition number and two vari-eties with repetition number . •
19 cubes consisting of two varieties with repetition number and three va-rieties with repetition number . •
16 cubes consisting of three varieties with repetition number and two va-rieties with repetition number . •
18 cubes consisting of six varieties with repetition number . •
14 cubes consisting of seven varieties with repetition number . We note that the corresponding Lemma 4.1 in [2] erroneously claimed that onecan always construct a corner solution from two varieties with repetition number4 and four with repetition number 2. The correct statement is above, that threevarieties with repetition number 4 and two with repetition number 2 suffices. In-terested readers can contact the first author for copies of the code that was used toprove Lemmas 4.1 and 4.2.Lemma 4.2 immediately implies the existence of a corner solution in all but 22of the partitions. (One partition for which the lemma does not apply, for example,is 753 .) The next two lemmas provide tools to handle many of these remainingcases. ETHAN BERKOVE, DAVID CERVANTES NAVA, DANIEL CONDON, AND RACHEL KATZ
Lemma 4.3.
Given a set of eight cubes consisting of four varieties from any pentad,each with repetition number , one can always construct a corner solution.Proof. Taking mirror images if necessary, we can assume the pentad is a row pentad.Then by Lemmas 3.1 and 3.2 we can assume that the varieties are in the top row(row A) and in columns b through e of the tableau, like so:2 2 2 2 · By Corollary 3.7, each of these varieties shares exactly two corners with varietiesBf, Cf, Df, or Ef. Furthermore, the corners are distinct, since the top row is apentad. So there are at least four possible corner solutions. (cid:3)
Lemma 4.4.
Given a decreasing partition ( a , a , . . . , a n ) with a ≥ , if a = 7 and a + n ≥ , then there is a subset of cubes that forms a corner solution. Thesame is true if a = 6 and a + n ≥ .Proof. For simplicity, we will refer to the variety with repetition number a i asvariety i . By using the lemmas from Section 3, we can assume that variety 1,which appears 7 times, is in position Ab in the tableau. Then by Corollary 3.7,there is automatically a solution unless the remaining cube varieties are amongthe nine in row A, column b, and the mirror variety Ba. Therefore, variety 2 iseither in the same pentad as c , or is c ∗ . By using mirror symmetry and/or a colorautomorphism, we may assume that variety 2 is in position Ac or Ba as in thediagrams below. Boxes represent the cube varieties that don’t automatically resultin a corner solution as noted in Corollary 3.7. a a (cid:3) (cid:3) (cid:3)(cid:3) · · · ·· (cid:3) · · ·· (cid:3) · · ·· (cid:3) · · ·· (cid:3) · · · Case Ac a (cid:3) (cid:3) (cid:3) (cid:3) a · · · ·· (cid:3) · · ·· (cid:3) · · ·· (cid:3) · · ·· (cid:3) · · · Case Ba
Case Ac : Variety 2 shares two corners with variety 1 ∗ and all the varieties in thecolumn’s boxed positions. We claim that if k ≤ k distinct corners of a corner solution modeledafter variety 2. Since there are only five slots in the top row of the tableau, we canalways build a corner solution when a + ( n − ≥ Case Ba : Here, variety 2 shares two corners with every boxed variety, so a +( n − ≥ a + n ≥ a = 6, then one variety with repetition number 1 can be in the dottedpositions in the tableau without yielding a corner solution. Now there are 7 distinct UTOMORPHISMS OF S AND THE COLORED CUBES PUZZLE 9 corners available to build a corner solution modeled after variety 1, and we’re inthe prior case. (cid:3)
We apply Lemma 4.4 to the 22 partitions, and end up with eight remaining casesto consider, listed by increasing number of varieties:75 , , , , , , , These cases can all be handled by ad hoc methods, although broadly speaking thetechniques are similar to those in the proof of Lemma 4.4. One places variety 1 is inposition Ab. Although variety 2 can be in either position Ac or Ba, we usually onlyneed to check position Ac, since that case, as in Lemma 4.4, is the most involved.Also, Lemma 4.3 implies that if there are four varieties in the same pentad withrepetition number greater than 1 then there is a solution, so we avoid that situationtoo. Keeping these observations in mind, we sketch arguments for the three mostsubtle of the remaining cases. The other five cases are similar, but easier.(1) The partition 743 : Let a = 7 and a = 4 as in the Case Ac. UsingLemma 4.3, we can assume that there is one variety with repetition number3 in the row pentad and the other two varieties with repetition number 3are in the column pentad, say Cb and Db. Variety 2 also shares two cornerswith Cb and Db, and since the varieties in the column form a pentad thesecorners are distinct. That is, these two varieties contribute four corners toa corner solution modeled after variety 2.(2) The partition 73 : This case is similar to the one above, although thereare essentially two cases to consider, representatives of which are shownbelow. The final variety that occurs once can be in either boxed position.7 3 3 1 (cid:3) · · · ·· · · ·· · · ·· · · ·· (cid:3) · · · Case 1 7 3 3 1 (cid:3) · · · ·· · · ·· · · ·· · · ·· (cid:3) · · · Case 2In both cases, variety 2 in position Ac has repetition number 3, and varietyBa has repetition number 2 or 3. In both cases, the three varieties in thesame column as variety 1 are part of a pentad. Therefore, those threevarieties can contribute 5 corners towards a corner solution modeled aftervariety 2.(3) The partition 5 : Variety 2 has repetition number 5, and there mustbe at least three varieties in the column pentad of variety 1. These threecontribute three distinct corners to a corner solution modeled after variety2.Since there are corner solutions for all possible partitions we have the main resultof this section and the first part of Theorem A. Theorem 4.5 (Theorem A, part 1) . Given any set of colored cubes, there isalways a subset from which one can construct a corner solution. Consequently, fr(2) = 24 . Building a × × solution: fr(3) = 24In contrast to the 2 × × n ) for n > n = 2. In order to determine the way toplace cubes into the interior of the frame, we need to know a bit more about howcubes varieties are related to each each. This is the subject of the next few results,which describe how to construct partial frames given cubes of a particular type,how to place cubes into edges of the frame, and how cubes that share an oppositepair are related. We use these results both in this section and in Section 6. Lemma 5.1.
Given a corner solution and k ( n − cubes that share k edge pairswith the corner solution, then it is possible to place all k ( n − cubes into the n -frame.Proof. In light of Lemma 3.4, k = 9 ,
10, or 12. The result is clear when k = 12.If k <
12, assume that in the construction of the frame, fewer than k edges arecomplete. Since any single cube shares at least k edges with the corner solution, itcan be fit into the cube. This process can always be repeated when fewer than k edges are complete in the frame, which requires k ( n −
2) cubes. (cid:3)
The next results examine the situation where one has a collection of cubes thatall share one opposite pair.
Lemma 5.2.
There are exactly six distinct cube varieties that share any oppositepair. Given any three of the six, all adjacent pairs on one variety can be found oneither one or both of the remaining two varieties.Proof.
Once one opposite pair has been determined, there are three ways to par-tition the remaining four colors into two opposite pairs, and there are two mirrorvarieties, c and c ∗ , for each set of three opposite pairs.For the second statement in the lemma, if two of the varieties are mirror images,the result follows since these varieties have the same adjacent pairs by Lemma 3.4.Otherwise, Lemma 3.4 implies that two distinct varieties share ten adjacent pairs,so the two together have 24 −
10 = 14 distinct adjacent pairs. Since all the cubesshare one opposite pair, these represent all possibilities. (cid:3)
Lemma 5.3.
Fix c , one of the six cube varieties that share one opposite pair. Thevariety c shares no corners with c ∗ , and shares exactly two unique corners with eachof the other four varieties.Proof. Assume that the opposite pair are the colors 5 and 6, and that the colorsaround the girth of c are given by the cyclic permutation (1234). The cyclic per-mutations for the other five varieties are (2134), (1324), (1243), (4231), and (4321).In the girth, the first four share exactly one adjacent pair with c , the edges { , } , { , } , { , } , and { , } respectively. These give rise to a pair of corners that match c ’s. The last cube is c ∗ . (cid:3) Corollary 5.4.
Take four copies of c and two copies each of two other varietiesthat share the same opposite pair. If the two other varieties are not c ∗ , then theeight cubes can be assembled into a corner solution modeled on c . There is a similarresult using six copies of c and two cubes that are not of the same variety as c ∗ . UTOMORPHISMS OF S AND THE COLORED CUBES PUZZLE 11
When a cube and its mirror image occur with large repetition number we cansay even more.
Lemma 5.5.
Let S be a set consisting of seven cubes each of varieties c and c ∗ along with one cube of any other variety. Then S has a subset of eight cubes thatforms a corner solution modeled on either variety c or variety c ∗ .Proof. In the tableau, assume that c is the distinguished variety. By Corollary 3.7,any variety that isn’t c or c ∗ shares two corners with either c or c ∗ . (cid:3) Lemma 5.6. [2, Lemma 3.2]
It is always possible to find a corner solution given15 cubes that share one opposite pair.Proof.
A sketch is as follows: Find the variety, c , that occurs with largest repetitionnumber in the set of 15. If | c | = 7, then a set of seven copies each of c and c ∗ doesnot have a subset that forms a corner solution, but the set formed by adding anyother variety does by Lemma 5.5. The cases of smaller | c | are similar, and havelower thresholds. (cid:3) We showed that fr(2) = 24 in Section 4. Since 20 cubes are required to build the3 × × ≥
24. We now show that this inequality issharp, which is the second part of Theorem A.We base our proof on the technique used in Theorem 3.4 of [2].
Theorem 5.7 (Theorem A, part 2) . The frame of a × × puzzle can alwaysbe completed given arbitary cubes, so fr(3) = 24 .Proof. Given 24 cubes, there is always a 2 × × S . We know fromLemma 5.1 that the cubes from S can be fit into at least nine out of the twelveslots in the 3 × × e , through e , and let c i denote the number of cubes out of the 16 remaining that have an adjacent pair e i . We note that c i ≥ i is a sufficient condition for a solution to the frame–pick anycube that has adjacent pair e as its representative in the frame and continue theprocess in ascending order. If this cannot be done, then there must be a largestindex j with c j < j .We bound the total number of adjacent pairs two ways. On the low end, each ofthe 16 cubes in S shares at least nine adjacent pairs with the solution by Lemma3.4. On the other hand, the j edges e , . . . , e j occur no more than j − − j edges e j +1 , . . . , e occur as many as 16 times. We have theinequalities 16 × ≤ (cid:88) i =1 c i ≤ ( j )( j −
1) + (12 − j )(16) . When we solve this quadratic inequality over the integers, we find that j ≤ j ≥
15. The latter case is impossible as j ≤
12. We conclude that if wecannot complete the frame, it is because at most three edges could not be matched.Although this is consistent with the results of Lemma 5.1, we now know exactlyhow we might fail to have a solution.
Case 1: e = 0. In this case, there is one adjacent pair, say { , } , that is oppositein all 16 cubes in S . By Lemma 5.6, there is a subset of eight cubes that forms acorner solution, and this corner solution must have { , } as an opposite pair. Werepeat the enumerating process above with this new corner solution. Each cubeshares at least ten adjacent pairs with the corner solution by Lemma 3.4, so16 × ≤ (cid:88) i =1 c i ≤ ( j )( j −
1) + (12 − j )(16) . This implies that j ≤
2. If there is no solution, then there are two new possibil-ities. The first is that c = 0. In this case, there is another pair { x, y } , which isopposite on all 16 cubes. Since the cubes in S already have { , } as an oppositepair, they must have the same three opposite faces. That is, the cubes consistof a variety c and its mirror variety c ∗ , so they all have the same adjacent pairs.At least eight of these cubes are of the same variety and form a corner solution.Using Lemma 5.1, we can place the eight cubes not in S into the 3 × × S now complete the solution.The second possibility is that c = c = 1. Furthermore, we can assume thatadjacent pairs e and e are on the same cube, since if the adjacent pairs are ondifferent cubes then each can be used as its representative in the frame. As in theprevious case, there are at least 15 cubes that share the same opposite pairs andare of exactly two varieties. Pick eight identical cubes for the corner solution andproceed as before. Case 2: e = e = 1. We can again assume that the adjacent pairs e and e arefrom the same cube. In fact, we can also assume that the pairs e and e have nocolors in common. For say e and e are the adjacent pairs { x, y } and { w, z } . Then { x, y } and { w, z } must be opposite pairs on the remaining 15 cubes of S, which canonly happen if x , y , w , and z are distinct. Therefore, these 15 cubes have the samethree opposite pairs. We now proceed as in Case 1. Case 3: e = e = e = 2. If these adjacent pairs are from three cubes then wecan pick representatives for the three edges and complete the 3 × × S that have the same opposite pairs. If there are eight of one variety we can completethe frame as above. Otherwise, if there are seven each of a variety c and its mirrorimage c ∗ , then take a cube of another variety from S and apply Lemma 5.5 to builda corner solution modeled on c or c ∗ . Start filling in the edges of the frame usingthe cubes of varieties other than c and c ∗ , then use varieties c and c ∗ to completethe frame. (cid:3) Building a n × n × n solution: fr( n ) = 12 n − for n ≥ n ≥ n × n × n frame was always possible. Unfortunately, that does not work in our setting, sincein [2] it was assumed that there were roughly n cubes available to build the frame,whereas here we assume that the number only grows linearly with n . Theorem 6.1 (Theorem B) . If n ≥ , then fr( n ) = 12 n − . UTOMORPHISMS OF S AND THE COLORED CUBES PUZZLE 13
Proof.
We proceed using induction. For the base case, we note that the 4-framecontains 32 cubes. By Theorem 5.7, any collection of 24 cubes contains a 3 × × n − × ( n − × ( n − n − n -frame is twelve cubes, onefor each edge. If the 12 cubes can not be inserted to extend the frame, we willshow that it is possible to construct another frame using a different corner solution.Referencing Lemma 5.1, we see that we will be able to fit at least nine of the twelvecubes into the frame. We proceed by cases. Case 1: exactly eleven edges of the n -frame are complete . Assume thatthe missing adjacent pair on the frame is { , } , and that, in a worst case, theunplaced cube shares only nine edges with the corner solution. If a cube in oneof the corresponding nine completed edges of the frame contains the adjacent pair { , } , move it to the unfilled position and put the unused cube in its place tocomplete the puzzle. If such a swap is not possible, then { , } must be an oppositepair in all 9( n −
2) cubes as well as in the unused cube.Next, assume there is a cube c among the 2( n −
2) cubes on the other two non- { , } frame edges that has { , } as a (hidden) adjacent pair. Denote c ’s adjacentpair contribution to the frame by { x, y } . If there is another cube in the nine otheredges that also has an adjacent pair { x, y } , use it to replace c and put the originalunplaced cube into its slot. If not, on each of the 9( n −
2) cubes colors x and y must form an opposite pair. Since { , } is already an opposite pair on these cubes,neither x nor y can be colors 1 or 2.By Lemma 3.5, it follows that all 9( n −
2) cubes are of two mirror varieties, andhave the same adjacent pairs. Since n ≥
4, we can pick eight of one variety for acorner solution. Start building the frame using the cubes from the two edges andthe unfinished edge. This is possible by Lemma 5.1. Complete the frame with theremainder of the 9( n −
2) cubes, any of which can be used for any edge.The last possibility is that there is indeed no cube among the 2( n −
2) on theother two edges that has a { , } adjacent pair, implying that { , } is an oppositepair on at least 11( n −
2) cubes and the unplaced cube. Denote this set of 11 n − S , and the remaining n − T . Now 11 n − ≥ n ≥
4. Using Lemma 5.2, partition S into three subsets, each consisting of avariety and its mirror image (which have the same adjacent pairs). One of thesesubsets will have at least eight elements, so at least one cube variety, say c , hasmultiplicity 4. Assume first that there are at least two cubes in each of the othertwo subsets. Then by Corollary 5.4, these can be assembled into a corner solutionmodeled on c . We note that (cid:24) n − (cid:25) − ≥ n −
2) for n ≥ , so after building the corner solution, there are enough cubes with the same adjacentpairs as c to complete any two edges. Use Lemmas 3.4 and 5.1 to fill in as muchas possible of ten edges of the n -frame using cubes from the other subsets and T .Then the last (cid:100) n − (cid:101) − n -frame.Next, assume that there is only one cube in one of the subsets of S . Move thatcube to T ; now S consists of four varieties in two subsets. Since 11 n − ≥
22 for n ≥
4, there is at least one subset with 11 cubes and six of some variety, say c . Asabove, (cid:24) n − (cid:25) − ≥ n −
2) for n ≥ . As long as there are two cubes in the other subset, we can apply Corollary 5.4 tobuild a corner solution modeled on c . The rest of the construction is as before.Finally, if there is no more than one cube in the other subset, then 11 n −
23 cubesare one of two varieties. Use the one that appears most frequently for the cornersolution, and the construction of the n -frame is straightforward. Case 2: exactly ten edges of the n -frame are complete . There are two cubesof the twelve that remain to be placed. Note that since each cube cannot be placedinto the edges, the colors that are adjacent in the unfilled places of the frame areopposite pairs in the two cubes. This also implies that the two unfilled adjacentpairs do not share a common color. In a worst case the two cubes share the samenine edges with the corner solution. If some cube in one of the nine correspondingedges of the frame also has one of the missing adjacent pairs, then move it to theunfilled position and put one of the two unused cubes in its place. We are now inthe situation in Case 1, so there is a solution.If such a swap cannot be done, then none of the 9( n −
2) cubes from the completededges nor the two used cubes have the two unfilled adjacent pairs in the cornersolution. These 9 n −
16 cubes must therefore share the same two, and hence three,opposite pairs. Call the set of these cubes S , and the remaining 3 n − T . We note that each cube in S is one of two varieties of mirrorcubes, and all the cubes in S share the same adjacent pairs. Since 9 n − ≥ n ≥
4, at least ten of the 9 n −
16 cubes are identical cubes. Pick eight of theseto make the corner solution, then start filling in the edges of the frame using cubesfrom the set T . Since | T | < n − T can beplaced into the new frame. The cubes from S share the same adjacent pairs as thecorner solution, so they can be used to complete the ( n + 1)-frame. Case 3: exactly nine edges of the n -frame are complete . There are threecubes of the twelve that remain to be placed. Since none of the three fit into theexisting frame, by Lemma 3.4 all three share the same nine adjacent pairs withthe n -frame. As in the prior cases, if some cube in the corresponding edges of theframe can be used to complete an unfilled edge, swap it out and put one of thethree unused cubes in its place. We are now in Case 2, so there is a solution.If no such swap is possible, then the 9( n −
2) cubes from the completed edgesin the frame and three unused cubes share the same three opposite pairs. We callthe set of these 9 n −
15 cubes S , the remaining 3 n − T , and proceed as in Case 2. (cid:3) Open Questions and Final Remarks
Although MacMahon’s original questions are now nearly 100 years old, they arestill generating fruitful problems. In this section we describe a number of openquestions for the interested reader to pursue. We start with the tableau, whoseassociated S action made it very useful in reducing the number and type of cases weneeded to consider in this paper. We believe that there is additional structure in the UTOMORPHISMS OF S AND THE COLORED CUBES PUZZLE 15 tableau still to be discovered that would further reduce the amount of computationrequired to complete the arguments. This motivates our first question.
Problem 7.1.
Refine the analysis of the S action on the tableau, and determinewhich of the results in Lemma 4.2 follow from this finer understanding. The Colored Cubes Puzzle whose solution is in this paper is just one member of alarger family of related puzzles. A pretty generalization in the spirit of MacMahon’sProblem 2 in the introduction is determine the minimum number of cubes requiredto solve the 3 × × n × n × n version of this problem would be very challenging,but even asymptotic bounds on the number of cubes would be interesting.Another way to generalize the problem is by changing the number of colors. Forexample: Problem 7.2.
For n > , determine g ( n, k ) , the minimum number of cubes coloredwith k colors required to solve the n × n × n Colored Cubes Problem.
There are two variations of this problem, depending on whether k < k > k <
6, one might start by assuming a regularity condition, and say that acube is k-colored if each cube face has a single color and all k colors appear on atleast one face of the cube. In this case, a solution to the frame implies a solutionfor the puzzle, and g ( n, k ) is the same as the function fr( n, k ), the analog of fr( n ).The authors of this paper have completed the calculations for k = 2 and k = 3[1]. The cases of n = 4 and n = 5 are more challenging, in large part because ofthe large number of distinct cubes. The total number of distinct cubes up to rigidrotation can be determined using a Polya counting argument, and are given in thefollowing table for k ≤ k >
6, it is no longer possible for all colors to appear on each cube,although we can still apply the regularity condition that no color appear more thanonce on a face of any cube. We note, however, that the successful construction of aframe no longer implies that the rest of the n × n × n cube can be completed. Weexpect that g ( n, k ) > fr( n, k ) for k > n (probably n = 2 or3!). In addition, we also believe that for fixed n , g ( n, k ) − fr( n, k ) should increasewith k , the number of colors. This leads into the third problem. Problem 7.3.
Determine asymptotic bounds, both upper and lower, on the sizesof fr( n, k ) and/or g ( n, k ) . For all values of k , there are analogous problems that arise when the regularitycondition is dropped. For k <
6, this means that some colors might not appear orcertain (or any) cubes. For k ≥
6, this means that a color may appear more thanonce on a cube.Finally, a number of people have analyzed the complexity of puzzles like theColored Cubes Puzzle. A well-known example is Instant Insanity R (cid:13) , a 4-coloredpuzzle whose elegant graph theoretic solution is presented in many introductorytexts on combinatorics (see [4], for example). Robertson and Munro showed in[12] that the solution to a generalization of the Instant Insanity Puzzle with n cubes and n colors is NP-complete. A more recent work by Demaine, et. al. [5] studied variations of Instant Insanity with several types of prisms; some of thesepuzzles have solutions that are NP-complete, others can be solved in polynomialtime. There is an analogous problem for the Colored Cubes Puzzle. Problem 7.4.
Determine the complexity of solving the n × n × n problem with n − n -colored cubes. Towards the end of the writing of this paper, we became aware of an arXivpreprint “On a Generalization of the Eight Blocks to Madness Puzzle” by KazuyaHaraguchi [6], where, among other things, Haraguchi determines the value of fr(2).Section 4 of this work and the preprint share a number of results, like Lemma 4.1and the example at the beginning of Section 4 of a collection of 23 cubes withouta corner solution. The proofs in [6] differ from ours, and rely on encoding a subsetof cubes into an associated multigraph. The existence of corner solutions can thenbe determined by counting the number of tree components in the graph. Thearguments are clever, and we recommend the article to the interested reader. Somearguments in [6] also utilize the tableau, although to a lesser extent than in thispaper.
UTOMORPHISMS OF S AND THE COLORED CUBES PUZZLE 17
Appendix: The Cube Tableau61254 3 { }{ }{ } { }{ }{ } { }{ }{ } { }{ }{ } { }{ }{ } { }{ }{ } { }{ }{ } { }{ }{ } { }{ }{ } { }{ }{ } { }{ }{ } { }{ }{ } { }{ }{ } { }{ }{ } { }{ }{ } { }{ }{ } { }{ }{ } { }{ }{ } { }{ }{ } { }{ }{ } { }{ }{ } { }{ }{ } { }{ }{ } { }{ }{ } { }{ }{ } { }{ }{ } { }{ }{ } { }{ }{ } { }{ }{ } { }{ }{ } References [1] E. Berkove, D Condon, D. Cervantes-Nava, and R. Katz,
The colored cubes problem with and colors , in progress.[2] E. Berkove, J. Hummon B. Kogut, and J. VanSickle, An analysis of the (colored cubes) puzzle , Discrete Mathematics
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Graphs as mathematical models , Prindle, Weber & Schmidt, Inc., 1977.[5] E. Demaine, M. Demaine, S. Eisenstat, T. Morgan, and R. Uehara,
Variations on InstantInsanity , Lecture notes in comput. sci. 8066.[6] K. Haraguchi,
On a generalization of “Eight Blocks to Madness puzzle” . arXiv:1408.3696[cs.DM] .[7] O. H˝older, Bildung zusammengestzter gruppen , Math. Ann. (1895), 321–422 (German).[8] A. Jebasingh and A. Simoson, Platonic solid insanity , Congressus Numeratium .[9] J˝urgen K˝oller, MacMahon’s coloured cubes , 2014. , accessed: February 15, 2015.[10] P. A. MacMahon,
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NP-completeness, puzzles and games , Utilitas Mathematica (1978), 99–116.[13] J.J. Sylvester, Xliv. Elementary researches in the analysis of combinatorial aggregation ,Philosophical Magazine Series 3 (1844), no. 159, 285–296, available at http://dx.doi.org/10.1080/14786444408644856 .[14] R. Vakil, B Howard, J. Millson, and A. Snowden, A new description of the outer automor-phism of S , and the invariants of six points in projective space , Philosophical Magazine (2008), 1296–1303. Mathematics Department, Lafayette College, Quad Drive, Easton, PA 18042, USA
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