罗密欧和朱丽叶能相遇吗？图上有对手的Renminbi博弈
Can Romeo and Juliet Meet? Or Rendezvous Games with Adversaries on Graphs
摘要：
我们引入了与对手的会合博弈。在这个游戏中，两个玩家{ \sl Facilitator}和{ \sl Disruptor}在一个图上互相对抗。Facilitator有两个代理，Disruptor有一组$k$代理，位于图的某些顶点。他们轮流将他们的代理移动到相邻的顶点（或停留）。协调者赢，如果他的代理人在图的某个顶点相遇。破坏者的目标是阻止协调者的代理人会合。我们的兴趣是决定调解人是否能赢。一般来说，这个问题是PSPACE困难的，当用$k$参数化时，它是co-W[2]困难的。此外，即使是游戏的变体，我们问主持人是否可以确保他的代理人在$τ$步骤中的会议对于$τ=2$来说已经是共同NP完全的。另一方面，对于弦图和无$P_5$图，我们证明了该问题在多项式时间内是可解的。这些算法利用了一个有趣的关系的游戏和最小顶点削减在某些图类。最后，我们证明了这个问题是固定参数的易处理的参数图的邻域多样性和$τ$。
英文摘要：
We introduce the rendezvous game with adversaries. In this game, two players, {\sl Facilitator} and {\sl Disruptor}, play against each other on a graph. Facilitator has two agents, and Disruptor has a team of $k$ agents located in some vertices of the graph. They take turns in moving their agents to adjacent vertices (or staying). Facilitator wins if his agents meet in some vertex of the graph. The goal of Disruptor is to prevent the rendezvous of Facilitator's agents. Our interest is to decide whether Facilitator can win. It appears that, in general, the problem is PSPACE-hard and, when parameterized by $k$, co-W[2]-hard. Moreover, even the game's variant where we ask whether Facilitator can ensure the meeting of his agents within $τ$ steps is co-NP-complete already for $τ=2$. On the other hand, for chordal and $P_5$-free graphs, we prove that the problem is solvable in polynomial time. These algorithms exploit an interesting relation of the game and minimum vertex cuts in certain graph classes. Finally, we show that the problem is fixed-parameter tractable parameterized by both the graph's neighborhood diversity and $τ$.
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。
