1. Almost Tight Bound for a Single Cell in an Arrangement of Convex Polyhedra in R 3 Esther Ezra Tel-Aviv University

2. 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

3. 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

4. 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

5. 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).

6. 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.

7. 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(  )

8. Single (bounded) cell in 2D

9. The unbounded cell in 3D Ω(nk) vertices Ω(k 2 ) vertices Can be modified to Ω(nk  (k)) vertices

10. 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

11. 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

12. 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

13. The combinatorial complexity of the unbounded cell How many inner vertices are on the unbounded cell of A(  ) ?

14. 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

15. 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  

16. 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. 

17. 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.

18. Thank you

19. The charging scheme: Case (2)   

20. Exposed chains of length  5 M=F_1  P_2  ’=M  P’  =M  P_3  ’’ ’’

21. Special quadrilateral 

22. Special vertex

23. 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.

24. 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.

25. 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.