Abstract

Given 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. In this paper, we present a simple, linear-time algorithm for computing a watchman route of length at most two 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.

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