Abstract
AbstractGiven a simple polygon P of n vertices, the watchman route problem asks for a shortest (closed) route inside P such that each point in the interior of P can be seen from at least one point along the route. We present a simple, linear-time algorithm for computing a watchman route of length at most 2 times that of the shortest watchman route. The best known algorithm for computing a shortest watchman route takes O(n 4 log n) time, which is too complicated to be suitable in practice.This paper also involves an optimal O(n) time algorithm for computing the set of so-called essential cuts, which are the line segments inside the polygon P such that any route visiting them is a watchman route. It solves an intriguing open problem by improving the previous O(n log n) time result, and is thus of interest in its own right.KeywordsShort PathLine SegmentIntersection PointSimple PolygonLeft EndpointThese keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.
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