TITLE: The mathematics of tavern puzzles
QUESTION [12 upvotes]: I remember seeing a paper on the arxiv this year (which I cannot now find Edit: This paper: http://arxiv.org/abs/1208.6545, found by j.c.) proposing to study the linkage of rigid bodies such as tavern or carnival puzzles. However, they only worked out one simple example (like showing that two circles of the same diameter can't pass through each other).
This is an interesting idea which I'd like to see more about. I know that such ideas have been studied in symplectic geometry (e.g. the symplectic camel).
It would help to examine a relatively simple case. So my question is:
Given $n$ circles of equal radius in 3-space, how many distinct linkages are there? I.e. how many configuration are there which cannot be made the same by isotopies of 3-space which restrict to a family of isometries on the circles?
I would also accept a solution for $n\leq 4$.

REPLY [2 votes]: A general useful theory probably does not exist. The reason is that even in dimension $n=2$, and even if all but one of the the bodies are rectangles, the problem is PSPACE-hard. That is because a popular board game Rush Hour is PSPACE-complete. This a result of Flake and Baum.
By taking a product with an interval, and enclosing the puzzle in a box, one can of course transfer the result to three (and also higher) dimensions.