罗密欧与朱丽叶在森林般的区域相遇
Romeo and Juliet Meeting in Forest Like Regions
摘要：
与对手会合的博弈是一个由两个参与者在图上进行的博弈：协调者和分割者。辅导员有两个座席，而分割者有一个由$k \ge 1$座席组成的团队。当引导者的代理人的初始位置固定时，分割者可以选择他的代理人的初始位置。然后，他们轮流将他们的代理移动到相邻的顶点（或保持不变），协调者的目标是将她的代理带到同一个顶点，而分割者的目标是阻止它。感兴趣的计算问题是确定协调者是否有一个获胜的策略来对抗具有$k$代理的分割者。Fomin，Golovach和Thilikos [WG，2021]介绍了这个游戏，并证明了它是PSPACE-硬和co-W[2]-硬的，参数是代理的数量。这种困难自然促使问题的结构参数化。作者证明，它承认FPT算法时，参数的模块宽度和允许轮数。然而，他们从其他结构参数的角度来看问题的复杂性。特别是，他们明确地问，这个问题是否允许FPT或XP算法相对于输入图的树宽。我们回答这个问题的否定，并表明，Renaissance是合作NP-硬，即使是图的恒定树宽。进一步，我们证明了当以反馈顶点集数和代理数为参数时，该问题是co-W[1]-hard的，而当以顶点覆盖数和代理数为参数时，该问题不可能存在多项式核.补充这些硬度的结果，我们表明，雷诺数是FPT参数化时的顶点覆盖数和解决方案的大小。最后，对于树宽至多为2且网格数为2的图，我们证明了该问题可以在多项式时间内求解。
英文摘要：
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.
机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。
