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Cop-Robber Guarding Game with Cycle Robber Region

机译:带有强盗区域的警察强盗守卫游戏

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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:强盗玩家和警察玩家在图G上以顶点集的二分{R,C}玩警察强盗保护游戏。强盗玩家通过在R的一个顶点上放置一个强盗(她的棋子)来开始游戏,接着是警察角色,后者在C的某些顶点上放置一组警察(她的棋子)。两名玩家轮流移动他们的棋子到G中的相邻顶点。警察玩家将警察移到C内,以防止强盗玩家将强盗移动到C中的任何顶点。强盗玩家只要轮到将强盗移动到某个顶点就获胜。在C中没有警察,否则警察将获胜。问题是找到允许警察玩家采用获胜策略的最小警察人数。已经表明,如果R诱导路径,则该问题可以多项式解决,而如果R诱导树木,则该问题是NP完全的。当R诱导一个循环时,它是否可解是开放的。本文肯定地回答了这个问题。

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