arXivDaily arXiv每日学术速递 周一至周五更新
arXiv周末暂无论文更新,休息一下吧,周末愉快~~

环上的警察与致命强盗:一种分布式视角

Cops and Lethal Robber on Rings: A Distributed Perspective

Amanpreet Singh Saini, Ashish Saxena, Kaushik Mondal

arXiv 2609.06388首次发表:更新:

发表机构

Indian Institute of Technology Ropar(印度技术研究所罗帕尔分校)

机构由 AI 辅助整理,请以论文原文为准。

AI 中文总结

本文在分布式设置中引入可杀死警察的致命强盗,定义警察与致命强盗问题,证明环上n名警察不足,并提出需n+⌊log n⌋+4名警察的算法。

AI 中文摘要

警察与强盗博弈在顺序设置中已被广泛研究,其主要目标是捕获强盗。捕获强盗意味着至少一名警察与强盗在某一时刻位于同一顶点。该博弈存在多种变体,包括警察的目标是包围强盗的变体。包围强盗意味着在强盗位置的每个相邻顶点上至少有一名警察。在本文中,我们在分布式设置中引入该博弈,同时赋予强盗更强的能力,即它可以杀死警察。具体而言,在我们的模型中,强盗在奇数轮移动且具有无限速度,警察在偶数轮移动,如果一名或多名警察移动到强盗当前所在的顶点,这些警察将全部被杀。我们称这种强盗为“致命强盗”。这也将我们的工作与入侵者捕获和黑洞搜索问题联系起来,因为引入了既动态又致命的实体,据我们所知,这种设置尚未被研究。在这项工作中,我们引入了致命强盗,在分布式设置中定义了“警察与致命强盗”问题,并在大小为$n$的静态环上对其进行了研究。我们证明了即使所有警察从同一顶点出发,$n$名警察也是不够的,并提供了一种从任意初始配置开始的算法,该算法在最坏情况下需要$n+\rfloor\log n \rfloor+4$名警察。

英文摘要

The Cops and Robber game is extensively studied in the sequential setting where the main goal is to capture the robber. Capturing the robber means at least one cop, and the robber will be at the same vertex together at some time. There are several variants, including variants where the goal of the cops is to surround the robber. Surrounding the robber means there is at least one cop in each of the neighboring vertices of the robber's position. In this paper, we introduce it in the distributed setting while empowering the robber by saying it can even kill cops. Specifically, in our model, the robber moves in odd rounds and has unbounded speed, cops move in even rounds, and if one or more cops move into a vertex where the robber is currently residing, all these cops get killed. We call this {\it lethal robber}. This also connects our work to the Intruder Capture and Black Hole Search problems by introducing an entity which is dynamic as well as lethal, a setting that, to the best of our knowledge, has not been studied. In this work, we introduce the lethal robber, define the {\it cops and lethal robber} problem in the distributed setting and study it on a static ring of size $n$. We prove $n$ cops are not enough, even if all start from the same vertex, and provide an algorithm starting from an arbitrary initial configuration that requires $n+\lfloor\log n \rfloor+4$ cops in the worst case.

论文原文

arXiv 摘要页 · PDF 原文 · HTML 原文

↑