Abstract

Let $B$ be a point robot in the plane, whose path is constrained to have curvature of at most 1, and let $\Omega$ be a set of polygonal obstacles with n vertices. We study the collision-free, optimal path-planning problem for B. Given a parameter $\varepsilon$, we present an $O( (n^2/ \eps^4) \log n)$-time algorithm for computing a collision-free, curvature-constrained path between two given positions, whose length is at most $(1+\varepsilon)$ times the length of an optimal path, provided it is robust. (Roughly speaking, a path is robust if it remains collision-free even if certain positions on the path are perturbed). Our algorithm thus runs significantly faster than the previously best known algorithm by Jacobs and Canny whose running time is $O( (\frac{n+L}{\eps^2})^2 + n^2 (\frac{n+L}{\eps^2}) \log n)$, where L is the total edge length of the obstacles. More importantly, the running time of our algorithm does not depend on the size of obstacles. The path returned by this algorithm is not necessarily robust. We present an $O((n^{2.5}/ \eps^4) \log n)$-time algorithm that returns a robust path whose length is at most $(1+\varepsilon)$ times the length of an optimal path, provided it is robust. We also give a stronger characterization of curvature-constrained shortest paths, which, apart from being crucial for our algorithm, is interesting in its own right. Roughly speaking, we prove that, except in some special cases, a shortest path touches obstacles at points that have a visible vertex nearby.

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