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Classification of the symmetries and antisymmetries of Rubik's cube
Submitted by Herbert Kociemba on Sat, 10/05/2013 - 16:38.
In 2005, Mike Godfrey and me computed the number of of essentially different cubes regarding the 48 symmetries of the cube (group M) and the inversion, see here for details.
We used the Lemma of Burnside to find this number. Since then I wondered if it would be possible to confirm this number by explicitly analyzing all possible symmetries/antisymmetries of the cube.
M has 98 subgroups, up to conjugacy there are 33 different subgroups of M and hence 33 different types of symmetries ignoring the inversion. Including the inversion we have 420 subgroups, up to conjugacy there are 131 different subgroups which give us 131 essentially different kinds of symmetry/antisymmetry.
The group we use to deal antisymmetries can be thought as the direct product M x C2 of M with the cyclic group C2 ={1,a}. This group has 96 elements. We define a group action of M x C2 on the set of all cubes by
For any cube y, the stabilizer subgroup Stab(y) defines the symmetry/antisymmetry of the cube. Stab(y) contains all elements of M x C2 which leave y fixed. It is not difficult to characterize the subgroups of M x C2. They are closely related to the subgroups of M. There are three different kind of subgroups:
The most interesting column is the column "Number of cubes mod M x C2 with exactly this symmetry". To get these numbers in a first step I computed for each of the 4.3*10^19 cubes to which of the 420 possible stabilizer subgroups they belong. Because egdes and corners can be dealt separately and because the stabilizer subgroup of a fully defined cube is the intersection of the stabilizers of the corners and edges, this computation can be done in a couple of hours.
Let's take for example the case 3.39 C3{S6}. There are 4 subgroups of type C3{S6}. The computation showed that each of these subgroups is the stabilizer subgroup of 1456 different cubes. In fact it would mean that something went wrong if the count would be different for the 4 conjugate subgroups. The union U of all these cubes has 4*1456 elements. Since the orbit of any element y in U is completely in U, we can compute the number of orbits in U by dividing 4*1456 by the orbit size. When we sum up the corresponding numbers of all 131 subgroups we exactly get 450541810590509978 in accordance with the result of 2005. A task not yet solved is to compute the shortest maneuver for a cube which gives an example for each of the 131 possible symmetries/antisymmetries classes. Even more demanding is a distance table for each of these subgroups. Silviu Radu managed to do this in the case of the subgroups of M, I wonder if someone is able to manage this for for the subgroups of M x C2. |
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