TITLE: Solving Rubik's cube and other permutation puzzles
QUESTION [25 upvotes]: I've seen two questions on solving the Rubik's cube but none of the answers have given a complete solution using mainly mathematical techniques. Furthermore, I've not seen a good explanation of general techniques for solving permutation puzzles in general, including those of the Rubik's cube family, without involving any memorized move sequences. So I've decided to write up my method below as an example of how group theory can be applied to solving such puzzles.

REPLY [5 votes]: May I add that if anyone wondered about the maximum number of 3-cycles required to solve/generate any even permutation of any $n\times n\times n$ Rubik's cube orbit of pieces (ignoring corner and middle edge orientations), I get that it takes a maximum of:

12 3-cycles to solve every non-fixed center and wing edge orbit (one at a time).


The exact average number of 3-cycles needed to solve all $24!/2$ even permutation cases of 24 objects is $7272437897/669278610$ (about 10.87).

For middle edges (12 objects), a maximum of 6 3-cycles is required.


The exact average number of 3-cycles to handle all $12!/2$ even permutations is 34757/6930 which is about 5.02.

For corners (8 objects), a maximum 4 3-cycles is required.


The exact average number of 3-cycles to handle all 8!/2 even permutations is 649/210 which is about 3.09.


All of these calculations include the solved case (which obviously is an even permutation and is included in the k!/2 quantities).
For everyone's possible curiosity, this means that it takes a maximum of this number or an average of this number of 3-cycles to solve the $n\times n\times x\times n$ cube. (Just substitute an integer which corresponds to the cube size for k at the right of the input in Wolfram Alpha.)
That is, the maximum number of 3-cycles function to solve the $n\times n\times n$ supercube (where we ignore the twists of fixed centers on the odd $n\times n\times n$ supercube, and non-fixed centers are unique) is simply $3n^2  - 6n + \frac{{3\cos \left( {n\pi } \right) + 5}}{2}$.
Suppose that we use 3-cycles which are 10 moves in length.  This means that it should take about (8)(10) = 80 moves to solve the 3x3x3 on average using this approach (where the "solved" state will be one in which corners and middle edges will most likely be unoriented), and about (10)(26635) = 266,350 moves on average to solve the 100x100x100.  We of course apply at most Floor[n/2] slice quarter turns to get the nxnxn supercube into the nxnxn supercube commutator subgroup, and thus we add this term to our total number of estimated moves.
Clearly since these calculations are for the $n\times n\times n$ supercube, the actual maximum number of required 3-cycles to solve the regular 6 colored $n\times n\times n$ cube is most likely less, although doing such a calculation would require a much more involved brute force search than the one I did for these supercube calculations.