Baguenaudier

A Puzzle involving disentangling a set of rings from a looped double rod (also called Chinese Rings). The minimum number of moves needed for $n$ rings is

$$\cases{ {\textstyle{1\over 3}}(2^{n+1}-2) & $n$\ even\cr {\textstyle{1\over 3}}(2^{n+1}-1) & $n$\ odd.\cr}$$

By simultaneously moving the two end rings, the number of moves can be reduced to

$$\cases{ 2^{n-1}-1 & $n$\ even\cr 2^{n-1} & $n$\ odd.\cr}$$

The solution of the baguenaudier is intimately related to the theory of Gray Codes.

References

Dubrovsky, V. ``Nesting Puzzles, Part II: Chinese Rings Produce a Chinese Monster.'' Quantum 6, 61-65 (Mar.) and 58-59 (Apr.), 1996.

Gardner, M. ``The Binary Gray Code.'' In Knotted Doughnuts and Other Mathematical Entertainments. New York: W. H. Freeman, pp. 15-17, 1986.

Kraitchik, M. ``Chinese Rings.'' §3.12.3 in Mathematical Recreations. New York: W. W. Norton, pp. 89-91, 1942.

Steinhaus, H. Mathematical Snapshots, 3rd American ed. New York: Oxford University Press, p. 268, 1983.