a one-move solution? (not chess moves...)

Originally posted by David113
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

True, but such positions would not be reachable on an unscrewdrivered cube anyway. So let's restate the question to make it more interesting:

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.

Richard

Originally posted by Shallow Blue
True, but such positions would not be reachable on an unscrewdrivered cube anyway. So let's restate the question to make it more interesting:

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 th ...[text shortened]... the number of reachable positions anyway, so that's your lower bound right there.

Richard

what position is not reachable on an unscrewable cube?

david113 had exactly the answer i was looking for. as he said, the permutations on the cube form a mathematical group. the "moves", as i attempted to define them, are permutations. if there was a single move, or permutation, that cycled through every single possible (unscrewable) position of the cube only once then it would generate every permutation too (as every permutation takes the solved cube an puts it into a unique position). but this would imply that the group is cyclic (by defintion) and so abelian*. however, it is obviously not abelian (twist the top 90 deg then twist the right side 90 deg gives a different position from first twisting the right side 90 deg then the top 90 deg) and thus we have a contradiction.

also, the last question you asked was what i was asking in reply to david113's answer. once you realise the parallel with group theory the answer isn't too hard to come by.

*abelian is ab=ba. cyclic groups are abelian as, a^n=p, a^m=q (by definition). p.q=(a^n).(a^m)=a^(n+m)=a^(m+n)=(a^m).(a^n)=q.p

Originally posted by Shallow Blue
True, but such positions would not be reachable on an unscrewdrivered cube anyway. So let's restate the question to make it more interesting:

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 th the number of reachable positions anyway, so that's your lower bound right there.

Richard

repeat post.

Originally posted by Shallow Blue
True, but such positions would not be reachable on an unscrewdrivered cube anyway. So let's restate the question to make it more interesting:

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 th the number of reachable positions anyway, so that's your lower bound right there.

Richard

second repeat post. my browser was being wierd...