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Avoider-Enforcer games

机译:避让游戏

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摘要

Let p and q be positive integers and let R be any hypergraph. In a (p, q, H) Avoider-Enforcer game two players, called Avoider and Enforcer, take turns selecting previously unclaimed vertices of R. Avoider selects p vertices per move and Enforcer selects q vertices per move. Avoider loses if he claims all the vertices of some hyperedge of H; otherwise Enforcer loses. We prove a sufficient condition for Avoider to win the (p, q, R) game. We then use this condition to show that Enforcer can win the (1, q) perfect matching game on K-2n for every q <= cn /log n for an appropriate constant c, and the (1, q) Hamilton cycle game on K, for every q <= cn log log log log n/log n log log log n for an appropriate constant c. We also determine exactly those values of q for which Enforcer can win the (1, q) connectivity game on K-n. This result is quite surprising as it substantially differs from its Maker-Breaker analog. Our method extends easily to improve a result of Lu [X. Lu, A note on biased and non-biased games, Discrete Appl. Math. 60 (1995) 285-291], regarding forcing an opponent to pack many pairwise edge disjoint spanning trees in his graph. (c) 2006 Elsevier Inc. All rights reserved.
机译:令p和q为正整数,令R为任何超图。在(p,q,H)避让者—执行者游戏中,两个名为回避者和执行者的玩家轮流选择以前未领取的R顶点。回避者每步选择p个顶点,而Enforcer则选择每步q个顶点。如果避让者声称H的某些超边的所有顶点,他就会输掉;否则Enforcer将会失败。我们证明了避免者赢得(p,q,R)游戏的充分条件。然后,我们使用此条件表明,对于适当的常数c,对于每个q <= cn / log n,Enforcer都能在K-2n上赢得(1,q)完美匹配游戏,并且在(-1,q)Hamilton循环游戏上K,对于每个q <= cn日志常数log log n / log n日志常数n的日志log n。我们还准确确定Enforcer可以在K-n上赢得(1,q)连通性游戏的q值。这个结果与它的Maker-Breaker类似物有很大的不同,因此非常令人惊讶。我们的方法易于扩展以改善Lu [X. Lu,关于有偏和无偏游戏的说明,Discrete Appl。数学。 60(1995)285-291],关于强迫对手在其图表中打包许多成对的边缘不相交的生成树。 (c)2006 Elsevier Inc.保留所有权利。

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