13 Aug '07 11:06>
Originally posted by David113True, but such positions would not be reachable on an unscrewdrivered cube anyway. So let's restate the question to make it more interesting:
Such a move does not exist. If it existed, the group of permutations of the cube would be cyclic and thus Abelian. However, it's easy to find pairs of moves which don't commute. QED
Is there a series of moves that will cycle through every single position which can be reached from the starting position?
The answer to that question is trivially yes: move to position 1; back to the starting position; move to position 2; back again; and so on. Such a series would, of course, be unfathomably huge, and useless in practice. Ok, question 2:
Is there a series of moves that will cycle through every single reachable position only once?
The answer to that is probably no, but I'm not about to even try to prove that. The most interesting variation on the question is probably this:
There are any number of series of moves that will cycle through every single position which can be reached from the starting position. How long is the shortest of these? What is it?
And no, I'm not going to have a shot at that, either. It is going to be at least as long as the number of reachable positions anyway, so that's your lower bound right there.