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Almost Tight Bound for a Single Cell in an Arrangement of Convex Polyhedra in R 3 Esther Ezra Tel-Aviv University
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A single cell of an arrangement of convex polyhedra Input: = {P 1, …, P k } a collection of k convex polyhedra in 3-space with n facets in total. A( ) : The arrangement induced by . The problem What is the maximal number of vertices/edges/faces that form the boundary of a single cell of A( ) ? Combinatorial complexity
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Motivation: Translational motion planning Input Robot R, a set A = {A 1, …, A k } of k disjoint obstacles. The free space The set of all legal placements of R. R does not intersect any of the obstacles in A The workspace collision
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The configuration space The robot R is mapped to a point. Each obstacle A i is mapped to the set: P i = { (x,y,z) : R(x,y,z) A i } = A i (-R(0,0,0)) A point p in P i corresponds to an illegal placement of R and vice versa. The forbidden placements of R The Minkowski sum The expanded obstacle
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The free space The free space is An algorithm that constructs the union? Not efficient when the complexity of the whole union is high (cubic).
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Restriction: A single component of the free space A single component of The subset of all placements reachable from a given initial free placement of R via a collision-free motion.
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Restatement: A single component in the complement of the union Input = {P 1, …, P k } a collection of k convex polyhedra in 3-space with n facets in total. The problem What is the maximal number of vertices/edges/faces that form the boundary of a single component of ? It is sufficient to bound the number of intersection vertices Minkowski sum of a convex obstacle with a convex part of -R A single cell of A( )
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Single (bounded) cell in 2D
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The unbounded cell in 3D Ω(nk) vertices Ω(k 2 ) vertices Can be modified to Ω(nk (k)) vertices
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Previous results 1.R 2 : Aronov & Sharir 1997. Θ(n (k)). 2.R 3 : Aronov & Sharir 1990. O(n 7/3 log n). 3.R d : Aronov & Sharir 1994. O(n d-1 log n). 4.R 3 : Halperin & Sharir 1995. O(n 2+ ), > 0. 5.R d : Basu 2003. O(n d-1+ ), > 0. 1-4: Comparable algorithmic bounds. The case of convex polyhedra in R 3 : Use [Aronov & Sharir 1994] O(n 2 log n). This bound does not depend on k. Many components Simply-shaped regions Curved simply-shaped regions
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Our result The combinatorial complexity of a single cell of A( ) is O(nk 1+ ), > 0. We use a variant of the technique of [Halperin & Sharir 1995]. We present a deterministic algorithm that constructs a single cell in O(nk 1+ log 2 n) time, > 0. The bound depends on the number k of polyhedra Crucial: The input regions are of constant description complexity
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Classification of the intersection vertices Outer vertex: The intersection of an edge of a polyhedron with a facet of another polyhedron. Overall number: O(nk). Inner vertex: The intersection of three facets of three distinct polyhedra. Overall number: O(nk 2 ). u
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The combinatorial complexity of the unbounded cell How many inner vertices are on the unbounded cell of A( ) ?
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Analysis: Exposed convex chains Not meeting any polyhedra After the removal of P’: 4 steps Classify each vertex v by: How long can we freely go from v when alternating out-of/into the unbounded cell. 1 step
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Analysis: Continue We trace this way Exposed convex chains. Number of steps = length of the chain V (j) ( ) – the number of inner vertices of the unbounded cell of A( ) with j steps. V (0) ( ) bounds the overall number of inner vertices of the unbounded cell. 5 steps
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The overall complexity of exposed chains Exposed chains of length 4 Use recurrence: V (j) ( ) V (j+1) ( ) Exposed chains of length 4 or 5 Lemma: The number of vertices on exposed chains of length 5 is O(nk). The number of vertices on exposed closed chains (of length 4) is O(nk). Multiply by O(k ). This is the only interesting case.
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Solving the recurrence V (j) ( ) = O(nk 1+ ), > 0, 0 j 4 The combinatorial complexity of a single cell of A( ) is O(nk 1+ ), > 0.
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Thank you
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The charging scheme: Case (2)
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Exposed chains of length 5 M=F_1 P_2 ’=M P’ =M P_3 ’’ ’’
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Special quadrilateral
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Special vertex
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Combinatorial complexity. Union of polyhedra in R 3 Input: = {P 1, …, P k } a collection of k polyhedra in 3-space with n facets in total. The problem What is the maximal number of vertices/edges/faces that form the boundary of the (complement of) the union? Trivial upper bound: O(n 3 ). Lower bound: Ω(n 3 ), for non-convex polyhedra. An algorithm that constructs the union: Not efficient.
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The combinatorial problem: Convex polyhedra Motion planning [Aronov, Sharir 1997] is a set of convex polyhedra that arise in the case of convex translating robot R Minkowski sums of (-R) and the obstacles: O(nk log k) Lower bound: (nk (k)) Construction time: O(nk log k log n) The general problem [Aronov, Sharir, Tagansky 1997] is a set of convex polyhedra : O(k 3 + nk log k) Lower bound: (k 3 + nk (k)) Construction time: O(k 3 + nk log k log n) Cannot be applied when R is non-convex.
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The combinatorial problem: Non-convex polyhedra is a set of general polyhedra : Θ(n 3 ). k = O(1) Also holds in translational motion planning problem. Not necessarily convex.
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