Spy-Game on Graphs

Nathann Cohen, Mathieu Hilaire, Nícolas A. Martins, N. Nisse, S. Pérennes
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引用次数: 8

Abstract

We define and study the following two-player game on a graph G. Let k in N^*. A set of k guards is occupying some vertices of G while one spy is standing at some node. At each turn, first the spy may move along at most s edges, where s in N^* is his speed. Then, each guard may move along one edge. The spy and the guards may occupy same vertices. The spy has to escape the surveillance of the guards, i.e., must reach a vertex at distance more than d in N (a predefined distance) from every guard. Can the spy win against k guards? Similarly, what is the minimum distance d such that k guards may ensure that at least one of them remains at distance at most d from the spy? This game generalizes two well-studied games: Cops and robber games (when s=1) and Eternal Dominating Set (when s is unbounded). We consider the computational complexity of the problem, showing that it is NP-hard and that it is PSPACE-hard in DAGs. Then, we establish tight tradeoffs between the number of guards and the required distance d when G is a path or a cycle. Our main result is that there exists beta>0 such that Omega(n^{1+beta}) guards are required to win in any n*n grid.
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图上的间谍游戏
我们定义并研究图g上的二人博弈,设k为N^*。一组k个守卫占据了G的一些顶点,而一个间谍站在某个节点上。在每个回合中,首先间谍最多可以沿着s条边移动,其中N^*中的s是他的速度。然后,每个守卫可能沿着一个边缘移动。间谍和守卫可能占据相同的顶点。间谍必须逃脱守卫的监视,也就是说,必须到达距离每个守卫的距离大于N中的d(预定义的距离)的顶点。间谍能战胜k个卫兵吗?同样地,k个守卫能保证至少有一个守卫与间谍的距离不超过d的最小距离是多少?这款游戏概括了两种被广泛研究的游戏:Cops and robbers游戏(当s=1时)和Eternal domination Set(当s无界时)。我们考虑了问题的计算复杂性,表明它是np困难的,在dag中是pspace困难的。然后,当G是一条路径或一个循环时,我们建立了保护数量和所需距离d之间的紧密权衡。我们的主要结果是存在β >0,使得需要Omega(n^{1+ β})守卫才能在任何n*n网格中获胜。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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