The mathematics of tavern puzzles

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 $n\leq 4$.

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Although, someone has shown that the Borromean rings cannot be circles, which is one reason I find this interesting. – Brian Rushton Dec 12 '13 at 21:36
OK. Sorry to beat this to death, but I want to make sure I understand. For this question, reflections in R^3 are disallowed for identification purposes. If I have that wrong, please correct me. – The Masked Avenger Dec 12 '13 at 22:09
Reflections are not allowed, just like vampires. – Brian Rushton Dec 12 '13 at 22:11
The following may help: "The theorem stating Borromean rings to be impossible with flat circles is proved rigorously in the article "Borromean circles are impossible," Amer. Math. Monthly, 98 (1991) 340-341, by B.Lindström and H.-O. Zetterström." That is a quote from the wedsite: popmath.org.uk/sculpmath/pagesm/borings.html. – Tim Porter Dec 13 '13 at 6:32
Was it perhaps this arxiv paper? arxiv.org/abs/1208.6545 – j.c. Dec 13 '13 at 13:00

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.