# Jakub Stepo´s solutions to the two Skewb Star Competition problems

For those who don´t know about this competition, please refer to my post of 14 June 2019 entitled "Skewb Star Special Challenge/Competition, with Special Prize", and to my Winner Announcement post of 29 August 2019, in which, as you can see, it was stated that the Special Prize had been awarded but that the challenge itself was to remain open until the New Year; well, the New Year has now arrived, so please find below Jakub Stepo´s solutions to the two competition problems:

Question 1
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Let’s say that the cube is solved and ﬁxed in position. We have to ﬁnd out which positions are permissible while having solved cube.

First we take care of permutations. The octahedron has 6 pieces in total and the cube has 8. But notice that all moves move a cube’s piece to a face-diagonally opposite piece. This means that each piece can reach only 4 positions; hence, on the cube, there are 2 orbits of 4 pieces. We expect permutation groups of our pieces to be S_8 (octahedron), S_4 and S_4 (cube) or their normal subgroups (when a permutation can be applied to any elements, it forms a normal subgroup).*

During a move, a 3-cycle of the octahedron’s pieces emerges, which is an even permutation; it also makes a 3-cycle of the cubes’ pieces. Normal subgroups of S_8 and S_4 with the lowest order containing these permutations are A_8 and A_4, respectively. Since A_8 is simple, the choice of a particular permutation on the cube cannot limit the group of all possible permutations on the octahedron (it would have to be a coset of A_8 and some of its nontrivial normal subgroups). So while the cube is solved, the octahedron can achieve all permutations in A_8, i. e. the even ones.

To show that it is correct, we need to ﬁnd an algorithm which can generate all even permutations while leaving the cube unchanged.

A move always makes a 3-cycle of pieces of only one orbit of the cube, so it is suﬃcient to show our algorithm is possible while permuting only pieces of one orbit. We now have to permute the pieces on one orbit from solved position to solved position so that the octahedron is not solved in the end (here, “the solved position” refers to the starting state). The simplest nontrivial way to move pieces back to their positions is as such: Meanwhile, on the octahedron, the pieces are permuted like this – they form a 4-cycle and a 2cycle: It could be possible that the orientation of the cube’s pieces would be wrong – however, their orientation is of order 3, so by performing this sequence of moves thrice, the cube is surely solved. This only has the eﬀect of reversing the 4-cycle, and by performing the whole tripled sequence thrice, the position is back as shown; this means that we can make a 4-cycle + 2-cycle permutation like on the picture on the octahedron alone. This is an even permutation which can be used to make all even permutations on the octahedron (two composed such permutations can be used to make a 3-cycle, two of which can make a 3-cycle like the one created after a move, which can generate all even permutations). Therefore, all even permutations of the octahedron are possible while the cube is solved.

Now on to orientations. Notice that if we split the octahedron’s faces in two sets of four, each set consisting of faces adjacent only by corners, after a move, all stickers of all pieces stay in their respective set; this means that the octahedron’s pieces have only two orientations, and their stickers are in two orbits. A rotation (incorrect orientation) of a piece creates 2 2-cycles of the stickers on it; however, these 2-cycles lie in diﬀerent orbits, so such an orientation is an odd permutation of the stickers in one orbit. We observe that a move is an even permutation of both orbits of the stickers (it makes 2 3-cycles in each orbit), so the permutation of the octahedron’s stickers in each orbit has to be even. This is equivalent to saying that all positions in which the number of incorrectly orientated pieces is even are possible.

So ﬁnally, to test the truthfulness of this, we have to show that there exists an algorithm which orientates two of the octahedron’s pieces incorrectly while leaving the cube intact (rotation of two pieces incorrectly can generate all even numbers of incorrectly orientated pieces). There indeed exists a simple (but repetitive) algorithm: If we label the octahedron’s faces as F, R, L, U, B, Br, Bl and D, the algorithm (F’ R F R’)6 rotates 4 pieces (FLUBl, FRUBr, FRLD and RBBrD). Now we just rotate the cube and do it again in a way that exactly three of incorrectly orientated pieces will be back solved, so the ﬁnal algorithm looks like (F’ R F R’)6 (F’ U F U’)6 and it rotates pieces RBBrD and UBBrBl. Therefore, all orientations which have an even number of incorrectly orientated pieces of the octahedron are possible when the cube is solved.

Only what is left now is ﬁnding the solutions. We just list all possible rotations of the octahedron and their properties; if the permutation is not even or the total orientation is not 0 (or even the orientations are impossible), the rotation is impossible. Here is the list: Only possible rotations are patently 1-fold, 2-fold (corner) and 3-fold; in total, 1 + 3 + 8 = 12 solutions.

Therefore, we have proved that there are exactly 12 solutions to the puzzle Skewb Star.

*Strictly speaking, there are some more technicalities to deal with, for example the groups could be groups of symmetries of the octahedron/tetrahedron and their respective normal subgroups, but the argument ultimately does hold.

Question 2
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It would be useful to know which faces a given face of an octahedron is projected to after the rotation.

First, consider all of the octahedron’s rotations (even the impossible ones). Obviously, among them, a given face is projected on every of the octahedron’s faces. There are 3 ways to rotate the octahedron without changing the position of a certain face (in our case, the face the given face lands on). This means that, all of the rotations considered, a given face would get projected on each face of the octahedron three times.

Of course, not all of the rotations are possible. But notice that all the impossible ones have impossible orientations (denoted by me as ‘–’ in the table above) of the pieces. This means with the possible rotations, all rotations of the octahedron that have possible orientations of the pieces occur. As we have seen, the possible rotations of the pieces mean that the stickers of the pieces stay in their orbit, both orbits consisting of faces adjacent only by corners. This signiﬁes that, among the possible rotations, a given face can be projected only on faces in its orbit. As all of such projections (rotations), are possible, a given face gets projected exactly three times on each member of its orbit. If we consider a face of the cube of the colour A, only one octahedron’s piece has stickers adjacent to it. This piece has 2 stickers per each orbit.

The A-coloured face of the octahedron will be projected on each of the 2 stickers of whichever orbit it is in exactly thrice among all rotations, so for this face, 6 colour-matchings happen in total. This can be said for all of the cube’s faces, which are 6 in total, so the total number of colour-matchings is 6 × 6 = 36.

Therefore, we have proved that, no matter how are the faces coloured, if the octahedron contains the same colours as the cube, exactly 36 colour-matchings occur.