You are given 12 balls, 12 identical except one weighs slightly more or slightly less than the others.

Here's a statement of the 12 ball problem. I recently got hooked by the 12 ball problem (defined shortly), and while I found some solutions and discussion, you left me without a satisfying solution: one that really got to the meat of the issue and explained what's going on. Kth weighing putting those whose kth letter is A in pan AĪnd those whose kth letter is B in pan B. The non-constant sequences of N letters from A,B,C whoseįirst change is A-to-B or B-to-C or C-to-A, and at the (3^N - 3)/2 coins in the same way, by labelling them with all N weighings can be used to locate a possible fake from among Otherwise,įor just one of the two pans the sequence is among the 12 above,Īnd names the fake coin, whose weight is Above or Below the proper If this sequence is CCC, then there's no fake coin. So the weighings determine a sequence of three of these letters. Now in a given weighing, a pan will either end up in theĬanonical position (C) that it assumes when the pans are balanced, or Problem first, and then explain the generalization.ĪAB ABA ABB ABC BBC BCA BCB BCC CAA CAB CAC CCAĢnd weighings put AAB CAA CAB CAC in pan A, ABA ABB ABC BBC in pan B. It was found by Dyson and Lyness in 1946 Here's the solution to the n-weighing version of this problem that I just thought I'd let you know I found an even better, more general solution to the 12 ball problem mentioned yesterday, generalized to (3^N-3)/2 balls where N = the number of weighings.

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