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Evacuating equilateral triangles and squares in the face-to-face model
Computational Geometry ( IF 0.4 ) Pub Date : 2020-02-19 , DOI: 10.1016/j.comgeo.2020.101624
Huda Chuangpishit , Saeed Mehrabi , Lata Narayanan , Jaroslav Opatrny

Consider k robots initially located at a point inside a region T. Each robot can move anywhere in T independently of the other robots with maximum speed one. The goal of the robots is to evacuate T through an exit at an unknown location on the boundary of T. The objective is to minimize the evacuation time, which is defined as the time the last robot reaches the exit. We consider the face-to-face communication model for the robots: a robot can communicate with another robot only when they meet in T.

In this paper, we give upper and lower bounds for the face-to-face evacuation time by k robots that are initially located at the centroid of a unit-sided equilateral triangle or square. For the case of a triangle with k=2 robots, we give a lower bound of 1+2/32.154, and an algorithm with upper bound of 2.3367 on the worst-case evacuation time. We show that for any k, any algorithm for evacuating k2 robots requires at least 3 time. This bound is asymptotically optimal, as we show that even a straightforward strategy of evacuation by k robots gives an upper bound of 3+3/k. For k=3 and 4, we give better algorithms with evacuation times of 2.0887 and 1.9816, respectively. For the case of the square and k=2, we give an algorithm with evacuation time of 3.4645 and show that any algorithm requires time at least 3.118 to evacuate in the worst-case. Moreover, for k=3, and 4, we give algorithms with evacuation times 3.1786 and 2.6646, respectively. The algorithms given for k=3 and 4 for evacuation in the triangle or the square can be easily generalized for larger values of k.



中文翻译:

撤消面对面模型中的等边三角形和正方形

考虑最初位于区域T内某个点的k个机器人。每个机器人可以独立于其他机器人以最大速度移动到T中的任何位置。机器人的目的是为了疏散ŧ在边界上的一个未知地点通过出口牛逼。目的是最小化疏散时间疏散时间定义为最后一个机器人到达出口的时间。我们考虑了机器人的面对面交流模型:一个机器人只有在T相遇时才能与另一个机器人通信。

在本文中,我们给出了k个机器人面对面疏散时间的上限和下限,这些机器人最初位于单位面等边三角形或正方形的质心处。对于三角形的情况ķ=2 机器人,我们给 1个+2/32.154,以及在最坏情况下的疏散时间上限为2.3367的算法。我们证明对于任何k,任何用于疏散的算法ķ2 机器人至少需要 3时间。这个边界是渐近最优的,因为我们表明,即使使用k个机器人进行疏散的简单策略也会给出3+3/ķ。对于ķ=3和4,我们给出了更好的算法,疏散时间分别为2.0887和1.9816。对于正方形和ķ=2,我们给出了疏散时间为3.4645的算法,并表明在最坏的情况下,任何算法都需要至少3.118的时间进行疏散。而且,对于ķ=3,和4,我们分别给出了疏散时间为3.1786和2.6646的算法。给出的算法ķ=3对于较大的k值,可以很容易地概括出用于疏散在三角形或正方形中的2和4 。

更新日期:2020-02-19
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