Approximation algorithms for shortest path motion planning

This paper gives approximation algorithms of solving the following motion planning problem: Given a set of polyhedral obstacles and points <italic>s</italic> and <italic>t</italic>, find a shortest path from <italic>s</italic> to <italic>t</italic> that avoids the obstacles. The paths found by the algorithms are piecewise linear, and the length of a path is the sum of the lengths of the line segments making up the path. Approximation algorithms will be given for versions of this problem in the plane and in three-dimensional space. The algorithms return an <italic>ε</italic>-short path, that is, a path with length within (1 + <italic>ε</italic>) of shortest. Let <italic>n</italic> be the total number of faces of the polyhedral obstacles, and <italic>ε</italic> a given value satisfying <italic>&Ogr;</italic> < <italic>ε</italic> ≤ π. The algorithm for the planar case requires <italic>&Ogr;</italic>(<italic>n</italic> log <italic>n</italic>)/<italic>ε</italic> time to build a data structure of size <italic>&Ogr;</italic>(<italic>n</italic>/<italic>ε</italic>). Given points <italic>s</italic> and <italic>t</italic>, and <italic>ε</italic>-short path from <italic>s</italic> to <italic>t</italic> can be found with the use of the data structure in time <italic>&Ogr;</italic>(<italic>n</italic>/<italic>ε</italic> + <italic>n</italic> log <italic>n</italic>). The data structure is associated with a new variety of Voronoi diagram. Given obstacles <italic>S</italic> ⊂ <italic>&Egr;</italic><supscrpt>3</supscrpt> and points <italic>s</italic>, <italic>t</italic> ε <italic>E</italic><supscrpt>3</supscrpt>, an <italic>ε</italic>-short path between <italic>s</italic> and <italic>t</italic> can be found in <italic>&Ogr;</italic>(<italic>n</italic><supscrpt>2</supscrpt>λ(<italic>n</italic>) log(<italic>n</italic>/<italic>ε</italic>)/<italic>ε</italic><supscrpt>4</supscrpt> + <italic>n</italic><supscrpt>2</supscrpt> log<italic>np</italic> log(<italic>n</italic> log<italic>p</italic>)) time, where <italic>p</italic> is the ratio of the length of the longest obstacle edge to the distance between <italic>s</italic> to <italic>t</italic>. The function λ(<italic>n</italic>) = α(<italic>n</italic>)<supscrpt><italic>&Ogr;</italic>(α(<italic>n</italic>)<supscrpt><italic>&Ogr;</italic>(1)</supscrpt>)</supscrpt>, where the α(<italic>n</italic>) is a form of inverse of Ackermann's function. For log(1/<italic>ε</italic>) and log <italic>p</italic> that are <italic>&Ogr;</italic>(log <italic>n</italic>), this bound is <italic>&Ogr;</italic>(log <italic>n</italic><supscrpt>2</supscrpt>(<italic>n</italic>)λ(<italic>n</italic>)/<italic>ε</italic><supscrpt>4</supscrpt>).