There is just one way to go from XXXXOOOO to XOXOXOXO in five moves if *three* adjacent coins are moved at each turn. How?

Could someone please clarify what it means to "move" three adjacent coins? I can think of two meanings since the problem doesn’t specify.

Does it mean that I have a long strip of squares that are all place holders for where the coins are, and when I move 3 coins I leave a gap in the original arrangement? For example, if the coins are initially in spots 1-8, and then I take coins 2-4 and move the to the right end, do I now have X _ _ _ OOOXXX? So now I’m using spots 1-11 (but with 2-4 empty).

Or does it mean that I always have only 8 spots, and it’s just that when I move 3 coins I can insert them anywhere (pushing some coins aside) and the gap I left closes up? For example, if I had XXXXOOOO and I want to move coins 2-4 in between 7 and 8, I would have XOOOXXXO. Is that a legal move?

I have a solution for the second way, but not the third. But this solution is different than DJ’s (who seems to have used the second movement definition I described), which doesn’t seem to fit with the "There is just one way…" statement in the problem. Also, even if the movement follows the first definition, I don’t see how there could be just one way, since it is symmetric. Any solution that you can find should have a "mirror" equivalent (so you move Xs instead of Os, and left and right is switched). I wonder if that made sense.

Clarification is appreciated!

If you are curious about the solution I found, here it is (using DJs notation as well)
XXXXOOOO
XOOOXXXO
XOXXOOOX
XOXOXOOX
OXOXOXOX
XOXOXOXO

Posted by nikki on 2004-10-26 20:30:55