The Rubik's Clock has 12^14 or 1,283,918,464,548,864 positions. Even if we solve a million positions a second, we'd need more than 41 years to evaluate them all. Luckily, we can both reduce the number of positions to individually consider using symmetry, and we can solve them much faster than that. In total it took about three days on a couple of desktops to calculate the following distance distribution:

0 1 1 330 2 51,651 3 4,947,912 4 317,141,342 5 14,054,473,232 6 428,862,722,294 7 8,621,633,953,202 8 101,600,180,118,726 9 528,107,928,328,516 10 613,251,601,892,918 11 31,893,880,879,492 12 39,248 -- --------------------- 1,283,918,464,548,864No wonder distance-12 positions were so hard to find!

The subgroup we used held the front cross fixed, but permitted all possible movements of the back cross and corner clocks; this subgroup has size 12^9 or 5,159,780,352. By Langrange's theorem, this means there were 12^14/12^9 or 12^5 (248,832) cosets to consider. This is precisely the number of positions of the front cross. Each coset, defined by one position of the front cross, consists of the set of all possible positions of the corners and back cross. Our coset solver will optimally solve all five billion positions in one execution.

But a quarter-million cosets is a lot. An equivalence relation can be defined between different front cross positions based on symmetry and modular arithmetic. From symmetry, we know that we can relate a position with its left-right mirror image; the distance distribution of the coset for a particular front cross position is identical to that of its mirror image. Similarly, we can take 90 degree rotations of the entire clock into account. Finally, there is an isomorphism between positions when you multiply each clock by any integer relatively prime to the number of hours, this gives another potential reduction. When only considering the front cross, this means for a particular front cross position, there can be up to 31 other positions that are equivalent (in terms of their distance distribution in the coset due to a 1:1 mapping among elements of the two cosets). It's a fairly trivial Perl program to compute representative front cross elements for each of these equivalence classes, and there turns out to be exactly 9,906 of these. So instead of solving 248,832 unique cosets, we only need solve 9,906.

Unlike most twisty puzzles, the moves on the clock commute; they can be executed in any order. Another way to say this is that the group is Abelian. This makes solving the puzzle, and analysis of the group, much easier. For the Rubik's clocks, moves can be partitioned into two sets---those that affect the front cross, and those that affect the back cross. In our case, our coset solver considers all possible ways to solve the front cross, with respect to the impact on the corner clocks, completely ignoring any moves that affect the back clocks. For a given coset (defined by a particular front cross position), this generates a set of distances for all positions with a solved front cross, and a particular setting of the corner clocks. There are only 12^4 or 20,736 possible corner settings so a depth-first or breadth-first search guided by a cross-solving pruning table can quickly enumerate all of these. Our coset solver uses iterated depth-first search.

Once we have distances for all permutations of the corners with a solved front cross (and no impact on the back cross), we can then consider only moves that affect the back cross, to enumerate all optimal solutions to positions in the coset. Again, since the group is Abelian, we only need consider each move once, scanning the distance table (with five billion entries) and extending it based on executions of that move. Once we've done this for all moves, our distance table is complete and ready to sum.

Some tricky bit hacks can be used to handle the inner loop of the table extension. The source code I used is here for your perusal; I plan to document it as a literate program soon. In sum, the program took less than three minutes a coset to run, using one core; I was able to run one of these programs on each core over the 9,906 cosets and sum the results to get the table above.

But how do we know the program is free of bugs? To help test the program, I made it general enough to work for any value of hours (the number of positions of each clock) from 2 to 12. I then explored the Rubik's Clock state space for all of these sizes using this coset program, and also wrote simpler, slower programs that enumerated the state space using other methods. I was able to exhaustively verify the full distances for sizes 2 and 3, and was able to confirm the lower distance values for the remaining sizes. In addition, I used the OptClock program to validate that the distance-12 positions found actually take 12 moves to solve.

The 39,248 distance-12 positions are given here; they are all in OptClock format, which gives the clock settings of the front, row by row, followed by the clock settings of the back cross only, row by row. Surprisingly, there are distance-12 positions with as many as six clocks already solved; there are distance-12 positions with only four distict values on the clocks.