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Cop-Robber Guarding Game with Cycle Robber Region

机译:Cop-robber守卫游戏与循环强盗区域

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A cop-robber guarding game is played by the robber-player and the cop-player on a graph G with a bipartition {R, C} of the vertex set. The robber-player starts the game by placing a robber (her pawn) on a vertex in R, followed by the cop-player who places a set of cops (her pawns) on some vertices in C. The two players take turns in moving their pawns to adjacent vertices in G. The cop-player moves the cops within C to prevent the robber-player from moving the robber to any vertex in C. The robber-player wins if it gets a turn to move the robber onto a vertex in C on which no cop situates, and the cop-player wins otherwise. The problem is to find the minimum number of cops that admit a winning strategy to the cop-player. It has been shown that the problem is polynomially solvable if R induces a path, whereas it is NP-complete if R induces a tree. It was open whether it is solvable or not when R induces a cycle. This paper answers the question affirmatively.
机译:由强盗播放器和COP-Player在Traph G上使用顶点{r,c}的poc-player和cop-player播放了一个警察守卫游戏。抢劫者队员通过将抢劫者(她的典当)放在R中的顶点上,然后在C的一些顶点上放置一组警察(她的典当)的Cop-Player来开始游戏。两名球员轮流转动他们的典当在G的邻近顶点.Cop-Player将COPS移动到C内,以防止强盗播放器将强盗将强盗移动到C中的任何顶点。如果它变得转动将强盗移动到顶点上,强盗播放器赢在C没有警察的情况下,否则会赢得COP-Player。问题是找到对COP播放器承认获胜策略的最小警察数量。已经表明,如果r引起路径,则问题是多项式可溶性的,而如果r诱导树,则为np-treate。它是打开的,无论是可解决的吗?当r引起一个循环时。本文肯定地回答了这个问题。

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