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Equivalence of Search Capability Among Mobile Guards with Various Visibilities

机译:具有各种可见性的移动警卫中搜索能力的等价性

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Given a polygonal region with n vertices, a group of searchers with vision are trying to find an intruder inside the region. Can the searchers find the intruder or can the intruder evade searchers' detection for ever? It is likely that the answer depends on the visibility of the searchers, but we present quite a general result against it. We assume that the searchers always form a simple polygonal chain within the polygon such that the first searcher moves along the boundary of the polygon and any two consecutive searchers along the chain are always mutually visible. Two types of visibility of the searchers are considered: on the one extreme every searcher has 360° vision - called an ∞-searcher; on the other extreme every searcher has one-ray vision - called a 1-searcher. We show that if any polygon is searchable by a chain of ∞-searchers it is also searchable by a chain of 1-searchers consisting of the same number of searchers as the oo-searchers. Our proof uses simple simulation techniques. The proof is also interesting from an algorithmic point of view because it allows an O(n2)-time algorithm for finding the minimum number of 1-searchers (and thus ∞-searchers) required to search a polygon [9]. No polynomial-time algorithm for a chain of multiple oo-searchers was known before, even for a chain of two ∞-searchers.
机译:给定具有N个顶点的多边形区域,一个具有愿景的搜索者正在尝试在该地区内找到入侵者。搜索人可以找到入侵者或者入侵者逃避搜索者的检测吗?答案很可能取决于搜索者的可见性,但我们呈现出相当一般的结果。我们假设搜索者始终在多边形内形成一个简单的多边形链,使得第一搜索者沿着多边形的边界移动,并且沿着链条的任何两个连续搜索者始终相互可见。考虑了搜索者的两种类型的可视性:在一个极端的每个搜索者上有360°视力 - 称为∞搜索者;另一个极端,每个搜索者都有一个射线视觉 - 称为1-searcher。我们表明,如果任何多边形是由∞搜索链中可搜索的,它也可以由一个由与OO搜索者组成的1次搜索者链中搜索。我们的证据采用简单的仿真技术。证明也是从算法的角度感兴趣的,因为它允许O(n2)-time算法查找搜索多边形所需的最小1-搜索(以及因此∞搜索者)[9]。之前,不知道多个oo-Searcher链的多项式时间算法,即使是两个∞搜索者的链。

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