Originally Posted by

**CarlBrannen**
Of course the vast majority of initial conditions will be solved long before the algorithm indicates: Since the algorithm has thousands of moves, it has to, on average cover all the states thousands of times.

I.e. I'm assuming that the Hamiltonian circuit works in that if we call the algorithm A, then A raised to the J power is not equal to A for any J less than the number of cube positions. But the number of (not different) positions you've actually traversed in doing this is MxA^N where M is the number of 1/4 turns in the algorithm.

## Bookmarks