1, 2, 3, 4, 5, 6, 7 123, 145, 167, 246, 257, 347, 356. |
Consider the classic NIM game: Given three piles of stones with 3, 5, and 7 in each pile. The players take any number of stones, but from one pile only. There are 7 winning states when none of the three piles is empty: {1,2,3}, {1,4,5}, {1,6,7}, {2,4,6}, {2,5,7}, {3,4,7}, {3,5,6}. These 7 groups also give a solution to this special problem. In general, a solution exists if and only if N = 1 or 3 mod 6. ( http://mathworld.wolfram.com/SteinerTripleSystem.html ) |
This method works for N = any 2^k - 1. For 15, ie Kirkman's original problem, the solution is {1,2,3}, {1,4,5}, {1,6,7}, {1,8,9}, {1,10,11}, {1,12,13}, {1,14,15} {2,4,6}, {2,5,7}, {2,8,10}, {2,9,11}, {2,12,14}, {2,13,15} {3,4,7}, {3,5,6}, {3,8,11}, {3,9,10}, {3,12,15}, {3,13,14} {4,8,12}, {4,9,13}, {4,10,14}, {4,11,15} {5,8,13}, {5,9,12}, {5,10,15}, {5,11,14} {6,8,14}, {6,9,15}, {6,10,12}, {6,11,13} {7,8,15}, {7,9,14}, {7,10,13}, {7,11,12} I do not know whether this solution satisfies Kirkman's other condition: Divide the 35 triples into 7 groups, so that every girl appears in each group exactly once. |
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