Lower Bounds for Geometric Diameter Problems

Transcription

1 Lower Bounds for Geometric Diameter Problems p. 1/48 Lower Bounds for Geometric Diameter Problems Hervé Fournier University of Versailles St-Quentin en Yvelines Antoine Vigneron INRA Jouy-en-Josas

2 Lower Bounds for Geometric Diameter Problems p. 2/48 Outline Review of previous work on the 2D and 3D diameter problems. Ω(n log n) lower bound for computing the diameter of a 3D convex polytope. Reduction from Hopcroft s problem to the diameter problem for point sets in R 7.

3 Previous work Lower Bounds for Geometric Diameter Problems p. 3/48

4 Lower Bounds for Geometric Diameter Problems p. 4/48 The diameter problem P INPUT: a set P of n points in R d. OUTPUT: diam(p ) := max{d(x, y) x, y P }.

5 Lower Bounds for Geometric Diameter Problems p. 5/48 The diameter problem p P p diam(p ) = d(p, p ).

6 Lower Bounds for Geometric Diameter Problems p. 6/48 Decision problem We will give lower bounds for the decision problem associated with the diameter problem. INPUT: a set P of n points in R d. OUTPUT: YES if diam(p ) < 1 NO if diam(p ) 1

7 Lower Bounds for Geometric Diameter Problems p. 7/48 Observation P p p P lies in the intersection of the two balls with radius d(p, p ) centered at p and p.

8 Lower Bounds for Geometric Diameter Problems p. 8/48 The diameter problem l l P p p P lies between two parallel hyperplanes through p and p. We say that (p, p ) is an antipodal pair.

9 Lower Bounds for Geometric Diameter Problems p. 9/48 The diameter problem l l P p p Any antipodal pair (and therefore any diametral pair) lies on the convex hull CH(P ) of P.

10 Lower Bounds for Geometric Diameter Problems p. 10/48 Finding the antipodal pairs The rotating calipers technique. l l

11 Lower Bounds for Geometric Diameter Problems p. 11/48 Finding the antipodal pairs The rotating calipers technique. l l

12 Lower Bounds for Geometric Diameter Problems p. 12/48 Finding the antipodal pairs The rotating calipers technique.

13 Lower Bounds for Geometric Diameter Problems p. 13/48 Finding the antipodal pairs The rotating calipers technique.

14 Lower Bounds for Geometric Diameter Problems p. 14/48 Finding the antipodal pairs The rotating calipers technique.

15 Lower Bounds for Geometric Diameter Problems p. 15/48 Computing the diameter of a 2D-point set Compute the convex hull CH(P ) of P. O(n log n) time. Find all the antipodal pairs on CH(P ). There are at most n such pairs in non degenerate cases. O(n) time using the rotating calipers technique. Find the diametral pairs among the antipodal pairs. O(n) time by brute force. Conclusion: The diameter of a 2D-point set can be found in O(n log n) time The diameter of a convex polygon can be found in O(n) time.

16 Lower Bounds for Geometric Diameter Problems p. 16/48 Diameter in R 3 and higher dimensions Randomized O(n log n) time algorithm in R 3 (Clarkson and Shor, 1988). Randomized incremental construction of an intersection of balls and decimation. Deterministic O(n log n) time algorithm in R 3 (E. Ramos, 2000). In R d, algorithm in n 2 2/( d/2 +1) log O(1) n (Matoušek and Schwartzkopf, 1995).

17 Lower Bounds for Geometric Diameter Problems p. 17/48 Lower bound on the diameter Ω(n log n) lower bound in R 2. Reduction from Set Disjointness. Given A, B R, decide if A B =. Ã B

18 Lower Bounds for Geometric Diameter Problems p. 18/48 Lower bound on the diameter Ω(n log n) lower bound in R 2. Reduction from Set Disjointness. Given A, B R, decide if A B =. Ã B

19 Lower Bounds for Geometric Diameter Problems p. 19/48 Diameter of a polytope The diameter of a convex polygon in R 2 can be found in O(n) time. Can we compute the diameter of a convex 3D-polytope in linear time? No, we give an Ω(n log n) lower bound.

20 Model of computation Lower Bounds for Geometric Diameter Problems p. 20/48

21 Lower Bounds for Geometric Diameter Problems p. 21/48 Real-RAM Real Random Access Machine. Each registers stores a real number. Access to registers in unit time. Arithmetic operation (+,,, /) in unit time.

22 Lower Bounds for Geometric Diameter Problems p. 22/48 Algebraic computation tree Input: x = (x 1, x 2,..., x n ) R n. Output: YES or NO It is a dag with 3 types of nodes Computation nodes: a real constant, some input number x i, or an operation {+,,, /, } performed on ancestors of the current node. Branching nodes arranged in a tree: compares with 0 the value obtained at a computation node that is an ancestor of the current node. Leaves: YES or NO

23 Lower Bounds for Geometric Diameter Problems p. 23/48 Algebraic computation tree: example x y z + no > 0 yes + + no > 0 yes no > 0 yes N Y Y N

24 Lower Bounds for Geometric Diameter Problems p. 24/48 Algebraic computation tree (ACT) We say that an ACT decides S R n if (x 1,..., x n ) S, it reaches a leaf labeled YES, and (x 1,..., x n ) / S, it reaches a leaf labeled NO. The ACT model is stronger than the real RAM model. To get a lower bound on the worst-case running time of a real-ram that decides S, it suffices to have a lower bound on the depth of all the ACTs that decide S Theorem (Ben-Or). Any ACT that decides S has depth Ω(log(number of connected components of S)).

25 Lower bound for 3D convex polytopes Lower Bounds for Geometric Diameter Problems p. 25/48

26 Lower Bounds for Geometric Diameter Problems p. 26/48 Problem statement We are given a convex 3-polytope P with n vertices. P is given by the coordinates of its vertices and its combinatorial structure: All the inclusion relations between its vertices, edges and faces. The cyclic ordering of the edges of each face. Remark: the combinatorial structure has size O(n). Problem: we want to decide whether diam(p ) < 1. We show an Ω(n log n) lower bound. Our approach: We define a family of convex polytopes. We show that the sub-family with diameter < 1 has n Ω(n) connected components. We apply Ben-Or s bound.

27 Lower Bounds for Geometric Diameter Problems p. 27/48 Polytopes P ( β) The family of polytopes is parametrized by β R 2n 1. When n is fixed, only the 2n 1 blue points change with β.

28 Lower Bounds for Geometric Diameter Problems p. 28/48 Polytopes P ( β) The family of polytopes is parametrized by β R 2n 1. When n is fixed, only the 2n 1 blue points change with β.

29 Lower Bounds for Geometric Diameter Problems p. 29/48 Notation Example where n = 3. c 1 2 a 3 c 1 3 b 2 (β 2 ) a 2 a 1 a 0 a 1 a 2 a 3 b 2 (β 2 ) b 0 c 1 1 c 1 2 c 1 3 c 1 2 c 1 1 c 1 0

30 Lower Bounds for Geometric Diameter Problems p. 30/48 Notation ā := (a n, a n+1..., a n ). A := {a n, a n+1,..., a n }. β := (β n+1,..., β n 1 ). b( β) := (b n+1 (β n+1 ),..., b n 1 (β n 1 )). B( β) := {b n+1 (β n+1 ),..., b n 1 (β n 1 )}. c := (c 1 n, c 1 n+1,..., c 1 n 1, c1 n, c 1 n+1,..., c1 n 1 ). C := {c 1 n, c 1 n+1,..., c 1 n 1, c1 n, c 1 n+1,..., c1 n 1 }. P ( β) := CH(A B( β) C).

31 Lower Bounds for Geometric Diameter Problems p. 31/48 Point sets A and C z r 1 ψ c 1 2 2ϕ ψ c 1 1 a 0 = O a 3 a 2 a 1 γ a 1 a 2 a 3 α y c 1 3 c 1 3 c 1 2 c 1 2 c 1 1 c 1 1 c 1 0 c 1 0 c 1 1 c 1 2 x

32 Lower Bounds for Geometric Diameter Problems p. 32/48 Point b j (β j ) z y 1 2 β b j (β) O jψ (cos(jψ), sin(jψ), 0) (1, 0, 0) x The blue region is parallel to Oxz. β [ α, α]

33 Lower Bounds for Geometric Diameter Problems p. 33/48 Coordinates of points in A, B( β) and C a i := c s i := b j (β) := 1 2 (1 cos(iγ)) sin(iγ) r cos (( i + 1 ) ) 2 ψ r sin (( i + 1 ) ) 2 ψ 1 2 sα cos(jψ) 1 2 (1 cos β) sin(jψ) 1 2 sin(β)

34 Lower Bounds for Geometric Diameter Problems p. 34/48 Parameters f e ϕ = 1/4n α is small. ψ = ϕ/n γ = α/n r 1 ψ 2 g r t t 1 h

35 Lower Bounds for Geometric Diameter Problems p. 35/48 Proof Notation: diam(e, F ) := max{d(e, f) (e, f) E F }. Lemma 1. The set {b j (β) β [ α, α] and diam(a, {b j (β)}) < 1} has at least 2n connected components. a 1 a 2 a 2 a 1 a 0 a 3 < 1 1 < 1 b j (β) a 3

36 Proof Proof of Lemma 1: Calculations, until the second-order terms. 1, , , ,15-0,1-0,05 0 0,05 0,1 0,15 Lower Bounds for Geometric Diameter Problems p. 36/48

37 Lower Bounds for Geometric Diameter Problems p. 37/48 Proof Lemma 2. The combinatorial structure of CH(A B( β) C) is independent of β. We denote P ( β) = CH(A B( β) C). Lemma 3. diam(a B( β) C) = diam(a, B( β)).

38 Lower Bounds for Geometric Diameter Problems p. 38/48 Proof Definitions: S n = {(ā, b( β), c) β [ α, α] 2n 1 } E n = {(ā, b( β), c) β [ α, α] 2n 1 and diam(p ( β)) < 1} Notice that E n S n R 24n. Restriction to S n is easy. Lemma 4. The set S n can be decided by an ACT with depth O(n). Decinding E n over S n is hard. Lemma 5. Any ACT that decides E n has depth Ω(n log n). Proof: By lemmas 1 and 3, E n has at least (2n) 2n 1 connected components. Apply Ben-Or s bound.

39 Lower Bounds for Geometric Diameter Problems p. 39/48 End of the proof Theorem. Assume that an algebraic computation tree T n decides whether the diameter of a 3-polytope is smaller than 1. Then T n has depth Ω(n log n). Let d n be the depth of T n. The computation tree T n can be transformed into T n of depth d n + O(n) which decides E n over S n. Thus T n has depth Ω(n log n) and the same holds for T n.

40 Lower Bounds for Geometric Diameter Problems p. 40/48 Related work (Chazelle) The convex hull of two 3-polytopes can be computed in linear time. (Chazelle et al.) It is not known whether the convex hull of a subset of the vertices of a 3-polytope can be computed in linear time. (Chazelle et al.) However, we can compute in linear time the Delaunay triangulation of a subset of the vertices of a Delaunay triangulation.

41 Diameter is harder than Hopcroft s problem Lower Bounds for Geometric Diameter Problems p. 41/48

42 Lower Bounds for Geometric Diameter Problems p. 42/48 Hopcroft s problem P is a set of n points in R 2. L is a set of n lines in R 2. Problem: decide whether (p, l) P L : p l.

43 Lower Bounds for Geometric Diameter Problems p. 43/48 Hopcroft s problem P is a set of n points in R 2. L is a set of n lines in R 2. Problem: decide whether (p, l) P L : p l.

44 Lower Bounds for Geometric Diameter Problems p. 44/48 Complexity of Hopcroft s problem An o(n 4/3 log n) algorithm is known. (Matoušek). No o(n 4/3 ) algorithm is known. Erickson gave an Ω(n 4/3 ) lower bound in a weaker model. Partitioning algorithms, based on a divide-and-conquer approach.

45 Lower Bounds for Geometric Diameter Problems p. 45/48 From Hopcroft s problem to Diameter We give a linear-time reduction from Hopcroft s problem to the diameter problem in R 7. Known upper bound: n 1.6 log O(1) n. We first give a reduction to the red-blue diameter problem in R 6 : compute diam(e, F ) when E and F are n-point sets in R 6.

46 Lower Bounds for Geometric Diameter Problems p. 46/48 Proof θ(x, y, z) := 1 x 2 + y 2 + z 2 (x2, y 2, z 2, 2xy, 2yz, 2zx). Note that θ(x, y, z) = 1. For 1 i n p i = (x i, y i, 1) l i = (u i, v i, w i ) is the line l i : u i x + v i y + w i = 0. Let p i := θ(p i) and l j = θ(l j). We get p i l j 2 = p i 2 + l j 2 2 < p i, l j > = 2 2 < p i, l j > 2 p i 2 l j 2

47 Lower Bounds for Geometric Diameter Problems p. 47/48 Proof Note that p i l j iff < p i, l j >= 0. Thus, there exists i, j such that p i l j if and only if diam(θ(p ), θ(l)) = 2. θ(p ) and θ(l) are n-point sets in R 6. Similarly, we can get a reduction from Hopcroft s problem to the diameter problem in R 7, using this linearization: θ(x, y, z) := ( 1 x 2 + y 2 + z 2 (x2, y 2, z 2, 2xy, 2yz, 2zx), ±1 )

48 Lower Bounds for Geometric Diameter Problems p. 48/48 Related work The red-blue diameter in R 4 can be computed in O(n 4/3 polylog n) (Matoušek and Scharzkopf). It would be interesting to get a reduction from Hopcroft s problem. Erickson gave reduction from Hopfcroft problem to other computational geometry problems. Ray shooting in polyhedral terrains, Halfspace emptyness in R 5 are harder than Hopcroft s problem.