ORourkeCH8

# Orourkech8

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Unformatted text preview: UMass Lowell Computer Science 91.504 Advanced Algorithms Computational Geometry Prof. Karen Daniels Spring, 2010 O'Rourke Chapter 8 Motion Planning Chapter 8 Motion Planning Shortest Paths (robot = point) Moving a Disk Translating a Convex Polygon Moving a Ladder Robot Arm Motion Separability Problem Specification s t Assume fixed environment of impenetrable objects e.g. Polygons or polyhedra Robot: movable object at initial position e.g. point robot starting at s Goal: Plan motion to final position (e.g. t) avoiding robot penetration of objects Sliding contact is acceptable Questions: Does there exist a "free" (collision-avoiding) path? How to construct such a path? How to find shortest such path? Shortest Paths s t Reduce possibilities to finite list! Shortest path segment endpoint vertex of obstacle (consider s,t as "point obstacles) Shortest path is subpath of visibility graph of vertices of obstacle polygons visibility graph requires (n2) time Algorithm: DIJKSTRA'S ALGORITHM T {s} while t not in T do Find edge e in (G \ T) that augments T to reach a node x whose distance from s is minimum T T + {e} Assume: - Polygonal obstacles have total of n vertices - Points s, t are outside obstacles - Points s, t are degenerate polygons - Obstacles are disjoint Moving a Disk Revised Goal: find any path to t not necessarily shortest path Shrink disk to a point Grow obstacles by disk radius Form union of grown obstacles If t, s in same component of plane, there is a free path t s t find it by modifying visibility graph to include circular arcs from grown obstacles s O(n2 lg n) Why Does it Work? Minkowski Sum: vector sum for point sets A B = {a + b | a A, b B} a+b a b a1 a4 a 3 a2 a4+b3 a3+b2 b2 A B Simple Case b3 A B a2+b1 b1 Convex/Convex Case Why Does it Work? (continued) (a5 + b4 ), (a5 + b3 ) (a3 + b4 ), (a4 + b4 ) a4 a5 a3 a1 a2 a5+b4 a4+b4 a3+b4 a5+b3 a2+b3 b4 b3 b2 b1 a3+b2 a3+b1 a2+b2 A B (a3 + b1 ), (a3 + b2 ) (a2 + b2 ), (a2 + b3 ) Polygonal Nonconvex/Nonconvex Case Minkowski Sum: Some Properties Shape is translationally invariant ( A + t ) B = ( A B) + t = A ( B + t ) Commutative A B = B A Union formulation A B = aA,bB {a + b} = {a + B} = {b + A} aA bB When A, B convex, sum is convex Minkowski Sum: Properties (continued) TRANSLATIONAL INTERSECTION ( B + t ) A iff t ( A (- B )) Why?? t a b Consider Simple Case: a=A, b=B b+t = a t = a -b Note: -b = b rotated by Mi...
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