Neeldhara Misra, M. Mulpuri, P. Tale, Gaurav Viramgami
{"title":"罗密欧与朱丽叶在森林般的地区相遇","authors":"Neeldhara Misra, M. Mulpuri, P. Tale, Gaurav Viramgami","doi":"10.48550/arXiv.2210.02582","DOIUrl":null,"url":null,"abstract":"The game of rendezvous with adversaries is a game on a graph played by two players: Facilitator and Divider. Facilitator has two agents and Divider has a team of $k \\ge 1$ agents. While the initial positions of Facilitator's agents are fixed, Divider gets to select the initial positions of his agents. Then, they take turns to move their agents to adjacent vertices (or stay put) with Facilitator's goal to bring both her agents at same vertex and Divider's goal to prevent it. The computational question of interest is to determine if Facilitator has a winning strategy against Divider with $k$ agents. Fomin, Golovach, and Thilikos [WG, 2021] introduced this game and proved that it is PSPACE-hard and co-W[2]-hard parameterized by the number of agents. This hardness naturally motivates the structural parameterization of the problem. The authors proved that it admits an FPT algorithm when parameterized by the modular width and the number of allowed rounds. However, they left open the complexity of the problem from the perspective of other structural parameters. In particular, they explicitly asked whether the problem admits an FPT or XP-algorithm with respect to the treewidth of the input graph. We answer this question in the negative and show that Rendezvous is co-NP-hard even for graphs of constant treewidth. Further, we show that the problem is co-W[1]-hard when parameterized by the feedback vertex set number and the number of agents, and is unlikely to admit a polynomial kernel when parameterized by the vertex cover number and the number of agents. Complementing these hardness results, we show that the Rendezvous is FPT when parameterized by both the vertex cover number and the solution size. Finally, for graphs of treewidth at most two and girds, we show that the problem can be solved in polynomial time.","PeriodicalId":175000,"journal":{"name":"Foundations of Software Technology and Theoretical Computer Science","volume":"64 1","pages":"0"},"PeriodicalIF":0.0000,"publicationDate":"2022-10-05","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"0","resultStr":"{\"title\":\"Romeo and Juliet Meeting in Forest Like Regions\",\"authors\":\"Neeldhara Misra, M. Mulpuri, P. Tale, Gaurav Viramgami\",\"doi\":\"10.48550/arXiv.2210.02582\",\"DOIUrl\":null,\"url\":null,\"abstract\":\"The game of rendezvous with adversaries is a game on a graph played by two players: Facilitator and Divider. Facilitator has two agents and Divider has a team of $k \\\\ge 1$ agents. While the initial positions of Facilitator's agents are fixed, Divider gets to select the initial positions of his agents. Then, they take turns to move their agents to adjacent vertices (or stay put) with Facilitator's goal to bring both her agents at same vertex and Divider's goal to prevent it. The computational question of interest is to determine if Facilitator has a winning strategy against Divider with $k$ agents. Fomin, Golovach, and Thilikos [WG, 2021] introduced this game and proved that it is PSPACE-hard and co-W[2]-hard parameterized by the number of agents. This hardness naturally motivates the structural parameterization of the problem. The authors proved that it admits an FPT algorithm when parameterized by the modular width and the number of allowed rounds. However, they left open the complexity of the problem from the perspective of other structural parameters. In particular, they explicitly asked whether the problem admits an FPT or XP-algorithm with respect to the treewidth of the input graph. We answer this question in the negative and show that Rendezvous is co-NP-hard even for graphs of constant treewidth. Further, we show that the problem is co-W[1]-hard when parameterized by the feedback vertex set number and the number of agents, and is unlikely to admit a polynomial kernel when parameterized by the vertex cover number and the number of agents. Complementing these hardness results, we show that the Rendezvous is FPT when parameterized by both the vertex cover number and the solution size. 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引用次数: 0
摘要
与对手会合的游戏是一个由两名玩家玩的游戏:促进者和分割者。Facilitator有两个代理,而Divider有一个由1个代理组成的团队。当Facilitator的代理的初始位置固定时,Divider可以选择其代理的初始位置。然后,他们轮流将他们的代理移动到相邻的顶点(或停留在原地),Facilitator的目标是将她的两个代理都带到同一个顶点,而分隔者的目标是防止它。感兴趣的计算问题是确定Facilitator是否具有对抗具有$k$代理的Divider的获胜策略。Fomin, Golovach, and Thilikos [WG, 2021]引入了这个博弈,并证明了它是PSPACE-hard和co-W[2]-hard(通过agent数量参数化)。这种硬度自然激发了问题的结构参数化。通过模宽度和允许轮数的参数化,证明了该算法的可行性。然而,从其他结构参数的角度来看,他们没有揭示问题的复杂性。特别是,他们明确地询问了该问题是否允许使用FPT或xp算法来考虑输入图的树宽。我们以否定的方式回答了这个问题,并证明了即使对于恒定树宽的图,交会也是共同np困难的。进一步,我们证明了当用反馈顶点集数和智能体数参数化时,问题是co-W[1]-hard,当用顶点覆盖数和智能体数参数化时,问题不太可能存在多项式核。补充这些硬度结果,我们表明,当顶点覆盖数和解大小都参数化时,交会是FPT的。最后,对于树宽最多为2的图和网格图,我们证明了该问题可以在多项式时间内解决。
The game of rendezvous with adversaries is a game on a graph played by two players: Facilitator and Divider. Facilitator has two agents and Divider has a team of $k \ge 1$ agents. While the initial positions of Facilitator's agents are fixed, Divider gets to select the initial positions of his agents. Then, they take turns to move their agents to adjacent vertices (or stay put) with Facilitator's goal to bring both her agents at same vertex and Divider's goal to prevent it. The computational question of interest is to determine if Facilitator has a winning strategy against Divider with $k$ agents. Fomin, Golovach, and Thilikos [WG, 2021] introduced this game and proved that it is PSPACE-hard and co-W[2]-hard parameterized by the number of agents. This hardness naturally motivates the structural parameterization of the problem. The authors proved that it admits an FPT algorithm when parameterized by the modular width and the number of allowed rounds. However, they left open the complexity of the problem from the perspective of other structural parameters. In particular, they explicitly asked whether the problem admits an FPT or XP-algorithm with respect to the treewidth of the input graph. We answer this question in the negative and show that Rendezvous is co-NP-hard even for graphs of constant treewidth. Further, we show that the problem is co-W[1]-hard when parameterized by the feedback vertex set number and the number of agents, and is unlikely to admit a polynomial kernel when parameterized by the vertex cover number and the number of agents. Complementing these hardness results, we show that the Rendezvous is FPT when parameterized by both the vertex cover number and the solution size. Finally, for graphs of treewidth at most two and girds, we show that the problem can be solved in polynomial time.