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Optimal Patrolling of High Priority Segments While Visiting the Unit Interval with a Set of Mobile Robots

Published 2 Oct 2019 in cs.DC | (1910.01250v1)

Abstract: Consider a region that requires to be protected from unauthorized penetrations. The border of the region, modeled as a unit line segment, consists of high priority segments that require the highest level of protection separated by low priority segments that require to be visited infinitely often. We study the problem of patrolling the border with a set of $k$ robots. The goal is to obtain a strategy that minimizes the maximum idle time (the time that a point is left unattended) of the high priority points while visiting the low priority points infinitely often. We use the concept of single lid cover (segments of fixed length) where each high priority point is covered with at least one lid, and then we extend it to strong double-lid cover where each high priority point is covered with at least two lids, and the unit line segment is fully covered. Let $\lambda_{k-1}$ be the minimum lid length that accepts a single $\lambda_{k-1}$-lid cover with $k-1$ lids and $\Lambda_{2k}$ be the minimum lid length that accepts a strong double $\Lambda_{2k}$-lid cover with $2k$ lids. We show that $2\min(\Lambda_{2k}, \lambda_{k-1})$ is the lower bound of the idle time when the max speed of the robots is one. To compute $\Lambda_{2k}$ and $\lambda_{k-1}$, we present an algorithm with time complexity $O(\max(k, n)\log{n})$ where $n$ is the number of high priority sections. For the upper bound, first we present a strategy with idle time $\lambda_{k-1}$ where one robot covers the unit line, and the remaining robots cover the lids of a single $\lambda_{k-1}$-lid cover with $k-1$ lids. Then, we present a simple strategy with idle time $3\Lambda_{2k}$ that splits the unit line into not-disjoint $k$ segments of equal length that robots synchronously cover. Then, we present a complex strategy that split the unit line into $k$ non-disjoint segments that robots asynchronously cover.

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