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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

{\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

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.


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.

© 1996-9 Eric W. Weisstein