一个组合对策问题的解
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Solution to a Combinatorial Game
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    摘要:

    本文求解如下的组合对策问题:设有一堆棋子,总数N是奇数,甲乙两人轮流取子,每人每次可取一颗、二颗,最多可取 s 颗,但不能不取,直至取完后分别来数甲乙两人所取棋子的总数,总数为奇数者获胜。站在甲的立场上考虑获胜的策略,文中解决了如下两个问题:(Ⅰ)总数N应是什么样的奇数,甲才有获胜策略;(II)当 N 一定时,甲应采取什么样的策略取子,才能获胜。

    Abstract:

    Let there be a heap of beans with total number N(odd) and an integer s,we difine a 2-person game Γ(N,s) as follows: The first player P takes some beans from the heap, at least one bean and at most s beans. Player two,PII now picks some beans under the same constraint. The play then reverts to PII and continues in the same way until all beans have been removed. The player with odd number of beans at hand wins. In this paper,We completely solve the game. At first,we give the winning strategy (if exists) for P. Theorem Let r be the number of beans P leaving to PII at any step,and q be the number p have at the end of the step. Then p will win if r=1 or 0(mod(2s+2)) if q is odd ,and r=s+1 or s+2(mod(2s+2)) if q is even when s is odd and r=1 or 0 (mod(s+2)) if q is odd,and r=s+1(mod(s+2)) if q is even when s is even. The theorem can be proved by induction. As a by-product of the theorem we have Corollary Γ(n,s) is a win for PII iff n=S十1 (mod(2s+2)) when s is odd and n=s+1 (mod(s+2)) when s is even.

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黄振高,黄受安.一个组合对策问题的解. Solution to a Combinatorial Game[J].国防科技大学学报,1990,12(3):15-20.

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  • 收稿日期:1989-08-16
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  • 在线发布日期: 2015-07-04
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