Lower Bounds for Geometric Diameter Problems Herv´ e Fournier University of Versailles St-Quentin en Yvelines Antoine Vigneron INRA Jouy-en-Josas Lower Bounds for Geometric Diameter Problems – p.1/40
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 . Lower Bounds for Geometric Diameter Problems – p.2/40
Previous work Lower Bounds for Geometric Diameter Problems – p.3/40
The diameter problem P INPUT: a set P of n points in R d . OUTPUT: diam( P ) := max { d( x, y ) | x, y ∈ P } . Lower Bounds for Geometric Diameter Problems – p.4/40
The diameter problem P p p ′ diam( P ) = d( p, p ′ ) . Lower Bounds for Geometric Diameter Problems – p.5/40
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 Lower Bounds for Geometric Diameter Problems – p.6/40
Observation P p p ′ P lies in the intersection of the two balls with radius d( p, p ′ ) centered at p and p ′ . Lower Bounds for Geometric Diameter Problems – p.7/40
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 . Lower Bounds for Geometric Diameter Problems – p.8/40
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 . Lower Bounds for Geometric Diameter Problems – p.9/40
Finding the antipodal pairs The rotating calipers technique. ℓ ′ ℓ Lower Bounds for Geometric Diameter Problems – p.10/40
Finding the antipodal pairs The rotating calipers technique. ℓ ′ ℓ Lower Bounds for Geometric Diameter Problems – p.11/40
Finding the antipodal pairs The rotating calipers technique. Lower Bounds for Geometric Diameter Problems – p.12/40
Finding the antipodal pairs The rotating calipers technique. Lower Bounds for Geometric Diameter Problems – p.13/40
Finding the antipodal pairs The rotating calipers technique. Lower Bounds for Geometric Diameter Problems – p.14/40
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. Lower Bounds for Geometric Diameter Problems – p.15/40
3D-diameter problem Randomized O ( n log n ) time algorithm (Clarkson and Shor, 1988). Randomized incremental construction of an intersection of balls and decimation. Deterministic O ( n log n ) time algorithm (E. Ramos, 2000). These two algorithms compute the diameter of an n -point set in R 3 . Can we compute the diameter of a convex 3 D -polytope in linear time? No, we give an Ω( n log n ) lower bound. Lower Bounds for Geometric Diameter Problems – p.16/40
Model of computation Lower Bounds for Geometric Diameter Problems – p.17/40
Real-RAM Real Random Access Machine. Each registers stores a real number. Access to registers in unit time. Arithmetic operation (+ , − , × , / ) in unit time. Lower Bounds for Geometric Diameter Problems – p.18/40
Algebraic computation tree: definition x = ( x 1 , x 2 , . . . , x n ) ∈ R n . Input: ¯ Output: YES or NO It is a binary tree with 3 types of nodes Leaves: YES or NO Degree-1 nodes: computation nodes. Perform { + , − , × , /, √·} on two operands. An operand is a real constant, some input number x i , or the value obtained by computation nodes that are ancestors of the current node. Degree-2 nodes: branching nodes. Compares with 0 the value obtained at a computation node that is an ancestor of the current node. Lower Bounds for Geometric Diameter Problems – p.19/40
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 ∈ S , it reaches a leaf labeled NO. ∀ ( x 1 , . . . , x n ) / 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 1 (Ben-Or) Any ACT that decides S has depth Ω(log( number of connected components of S )) . Lower Bounds for Geometric Diameter Problems – p.20/40
Lower bound for 3D convex polytopes Lower Bounds for Geometric Diameter Problems – p.21/40
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. Example: P is given in a doubly-connected edge list. 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. Lower Bounds for Geometric Diameter Problems – p.22/40
Polytopes P ( ¯ β ) The family of polytopes is parametrized by ¯ β ∈ R 2 n − 1 . When n is fixed, only the 2 n − 1 blue points change with ¯ β . Lower Bounds for Geometric Diameter Problems – p.23/40
Polytopes P ( ¯ β ) The family of polytopes is parametrized by ¯ β ∈ R 2 n − 1 . When n is fixed, only the 2 n − 1 blue points change with ¯ β . Lower Bounds for Geometric Diameter Problems – p.24/40
Notation Example where n = 3 . c 1 2 c 1 b 2 ( β 2 ) a 3 − 3 a 2 a 1 a 0 a − 1 b 0 b − 2 ( β − 2 ) c − 1 a − 2 2 c − 1 a − 3 1 c − 1 0 c − 1 − 1 c − 1 − 2 c − 1 − 3 Lower Bounds for Geometric Diameter Problems – p.25/40
Notation a := ( 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 , c 1 − n , c 1 − n +1 , . . . , c 1 n − 1 ) . ¯ C := { c − 1 − n , c − 1 − n +1 , . . . , c − 1 n − 1 , c 1 − n , c 1 − n +1 , . . . , c 1 n − 1 } . P (¯ β ) := CH( A ∪ B (¯ β ) ∪ C ) . Lower Bounds for Geometric Diameter Problems – p.26/40
Point sets A and C z r ≃ 1 ψ c 1 2 c 1 1 2 ϕ − ψ c 1 0 a 3 c 1 y − 1 c 1 a 2 c 1 − 2 − 3 a 1 α x a 0 = O γ c − 1 a − 1 2 a − 2 c − 1 1 c − 1 a − 3 0 c − 1 − 1 c − 1 − 2 c − 1 − 3 Lower Bounds for Geometric Diameter Problems – p.27/40
Point b j ( β j ) z b j ( β ) y 1 2 β (cos( jψ ) , sin( jψ ) , 0) jψ O x (1 , 0 , 0) The blue region is parallel to Oxz . β ∈ [ − α, α ] Lower Bounds for Geometric Diameter Problems – p.28/40
Coordinates of points in A , B ( ¯ β ) and C 1 2 (1 − cos( iγ )) a i := 0 1 2 sin( iγ ) i + 1 �� � � r cos ψ 2 c s i + 1 �� � � i := r sin ψ 2 1 2 sα cos( jψ ) − 1 2 (1 − cos β ) b j ( β ) := sin( jψ ) 1 2 sin( β ) Lower Bounds for Geometric Diameter Problems – p.29/40
Parameters f r 1 2 ψ g h e t t r 1 ϕ = 1 / 4 n α is small. ψ = ϕ/n γ = α/n Lower Bounds for Geometric Diameter Problems – p.30/40
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 2 n connected components. a 3 < 1 � 1 a 2 < 1 a 1 a 0 a − 1 b j ( β ) a − 2 a − 3 Lower Bounds for Geometric Diameter Problems – p.31/40
Proof Proof of Lemma 1: Calculations, until the second-order terms. 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 )) . Lower Bounds for Geometric Diameter Problems – p.32/40
Proof Definitions: a, ¯ b (¯ c ) | ¯ β ∈ [ − α, α ] 2 n − 1 } S n = { (¯ β ) , ¯ β ∈ [ − α, α ] 2 n − 1 and diam( P (¯ a, ¯ b (¯ c ) | ¯ E n = { (¯ β ) , ¯ β )) < 1 } Notice that E n ⊂ S n ⊂ R 24 n . Lemma 4 The set S n can be decided by an ACT with depth O ( n ) . 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 (2 n ) 2 n − 1 connected components. Apply Ben-Or’s bound (Theorem 1). Lower Bounds for Geometric Diameter Problems – p.33/40
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