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标题: 科克曼女生问题 [打印本页]

作者: Jenny    时间: 2005-4-12 01:31
标题: 科克曼女生问题

一个小组7个人,要安排他们做连续7天的值日,每天一组,每组三个人。要求一周下来,每个小组成员都要与小组内其他所有成员同在一组一次(且一次)做值日。按:这是著名科克曼女生问题最简单的特例。这个问题英国人科克曼于1850年提出的,原题是:一位女老师安排15个女生连续7天的散步,每天安排五组,每组3个人,要求7天下来,每个人与其他14个人都在一组一次且一次散步。这道题一般化后,原题就成了一个特例。我国数学家陆家羲于20世纪60年代独立解决了科克曼女生问题。
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作者: Harbin    时间: 2005-4-12 20:19
标题: 回复:科克曼女生问题

1, 2, 3, 4, 5, 6, 7www.ddhw.com
 
123, 145, 167, 246, 257, 347, 356.
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作者: fzy    时间: 2005-4-12 20:22
标题: An interesting solution

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 ) 
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作者: 乱弹    时间: 2005-4-12 21:44
标题: 妙!

  妙!





作者: fzy    时间: 2005-4-13 00:57
标题: 回复:An interesting solution

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}www.ddhw.com
{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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