Clarkson s Algorithm for Violator Spaces. Yves Brise, ETH Zürich, CCCG Joint work with Bernd Gärtner

Size: px

Start display at page:

Download "Clarkson s Algorithm for Violator Spaces. Yves Brise, ETH Zürich, CCCG Joint work with Bernd Gärtner"

Angela Fleming
5 years ago
Views:

1 Clarkson s Algorithm for Violator Spaces Yves Brise, ETH Zürich, CCCG Joint work with Bernd Gärtner

2 A Hierarchy of Optimization Frameworks Linear Programming Halfspaces (constraints), Optimization direction Optimal point LP-type Problems, Sharir, Welzl (1992) Set of constraints H, Total order on 2 H, Value function ϕ :2 H R Optimal subset of H Violator Spaces, Skovron (2002) Set of constraints H, Violator mapping V :2 H 2 H B H minimal and V (B) =

3 Smallest Enclosing Ball Given a set H of n points in R d, find the smallest enclosing ball. Consider R H. Basis: A minimal set of points that have the same smallest enclosing ball. R Violators: Points that lie outside of the smallest enclosing ball of R. H

4 Definition Violator Space The pair (H, V ) is a violator space if H is a finite set, and V :2 H 2 H satisfies the following conditions. Consistency: G H, V (G) G = Locality: F G H, if V (F ) G = then V (F )=V(G) Basis: A basis of G H is a minimal subset B of G, s.t. V (B) =V (G). The combinatorial dimension of (H, V ), denoted by dim(h, V ), is the size of the largest basis in (H, V ).

5 Sampling Idea choose R H u.a.r.; B basis(r); while (V (B) = ) { choose h V (B); B basis(b {h}); } return B;

6 Clarkson s Algorithm Main Idea: Sampling! But do it the right way. while (V (B) = ) { choose h V (B); B basis(b {h}); } return B; 1. Don t discard too many violators. German Algorithm 2. Adjust weights and do resampling in every round. Swiss Algorithm

7 Overview Clarkson (H, V ), H = n, dim(h, V )=δ German Algorithm (δ + 1) (H, V ), n = O(δ n) Brute (H, V ), O(δ ln n ) Swiss Force H = O(δ 2 ) Algorithm Las Vegas Algorithms for Linear and Integer Programming When the Dimension is Small, Clarkson, Initially developed for Linear Programming.

8 Swiss Algorithm µ h 1, h H; 1 do { choose R H according to µ; B basis(r); µ h 2µ h, h V (B); } until (V (B) = ) return B; µ h is the multiplicity of node h

9 Simplified Clarkson µ h 1, h H; while (V (B) = ) { choose R H according to µ; B basis(r); µ h 2µ h, h V (B); } return B; if µ(v (B)) 1 2d µ(h) then µ h 2µ h, h V (B); Not intuitive! Only there for the analysis.

10 Overview Clarkson (H, V ), H = n, dim(h, V )=δ German Algorithm (δ + 1) (H, V ), n = O(δ n) Brute (H, V ), O(δ ln n ) Swiss Force H = O(δ 2 ) Algorithm Theorem: Let (H, V ) be a violator space, H = n, dim(h, V )=δ. We can compute a basis of (H, V ) using an expected number of O(δ 2 n + f (δ)) unit cost operations.

11 Result Overview Gärtner, Welzl, 2001 Brise, Gärtner, 2009 German Swiss Brute Force LP Simplified Simplified Subexponential LP-type Simplified Simplified Subexponential VS Simplified Simplified Exponential Kalai, Matousek, Sharir, Welzl, 1992 Gärtner, 1995

12 Thank You For Your Attention! Yves Brise,

13 Runtime 2nd Stage Lemma: Let (H, V ) be a violator space, H = n, dim(h, V )=δ, B any basis of (H, V ), and k N. Then after kδ rounds of Stage II, 2 k E[µ (kδ) (B)] < ne k/2. By choosing k large enough we derive a contradiction. Proof: (lower bound) By a simple Lemma we have that B contains at least one violator of every round. After kδ rounds there must be an element in B that has doubled its weight at least k times, because B δ.

14 Runtime 2nd Stage Proof: (upper bound) E[µ () (H)] = E[µ (0) (H)] + i=1 E[ (i) (H)], where (i) is the increment in multiplicity in round i. i=1 i=1 E[ (i) (H)] = E[ (i) (H) µ (i 1) (H) =t] Pr[µ (i 1) (H) =t] t=0 Sampling Lemma: E[ (i) (H) µ (i 1) (H) =t] δ t r r +1

15 Runtime 2nd Stage Calculating, we get a recursive equation: E[µ () (H)] n + δ r +1 i=1 E[µ (i 1) (H)] δr r +1 Finally, plugging in r =2δ 2, we get E(µ () (H)) (1 + δ r ) n ( δ )kδ n < ne k/2 This concludes the proof.

16 Open Problems

17 Clarkson Dimension # rounds / points in R d Average of 10^3 trials Lemma: Let (H, V )bea violator space, dim(h, V )=δ. Stage 1 runs for at most δ + 1 rounds dimension d unit cube unit ball simplex

18 Clarkson Dimension Average of 10^3 trials Size of G / points in R d 22,000 18,400 14,800 11,200 7,600 Theorem: Let (H, V )bea violator space, dim(h, V )=δ, H = n, and G (δ) the working set of Stage 1 after δ rounds. Then, E[ G (δ) ] 2(δ + 1) n/2. 4, dimension d unit cube unit ball simplex

19 Cyclic VS, dim=2? The generalized linear complementarity problem with P-matrices gives rise to possibly cyclic unique sink orientations (USO) of hypercube grids. 123 GLCP Grid USO Violator Space Lemma (Gärtner, Morris, Rüst, 2005): 2-dimensional cyclic grid USO Grid USO cannot be a model for cyclic, non-degenerate violator spaces of dim = 2.

20 Other Open Problems Subexponential analysis for violator spaces Number of violator spaces Unified Clarkson algorithm

21 Junk

22 Sampling Lemma Let P be a set of size n, ϕ :2 P R, and R P, r = R. Violators: V (R) := {p P\R ϕ(r {p}) = ϕ(r)} Extreme: X (R) := {p R ϕ(r\{p}) = ϕ(r)} Observation: p violates R p is extreme in R {p} Lemma (Gärtner, Welzl, 2001): For 0 r n, E[ V (R) ] n r = E [ X (R) ] r+1

23 Sampling Lemma (proof) n r E[ V (R) ] = R ( S r) = = = R ( S r) Q ( S r+1) n r +1 s S\R s S\R s Q [s violates R] [s is extreme in R {s}] [s is extreme in Q] E[ X (R) ]

24 2nd Stage Calculation E[µ () (H)] n + = n + i=1 i=1 dr r +1 = n + d t=0 r +1 d t r r +1 Pr[µ(i 1) (H) =t] d r +1 t=0 t=0 t Pr[µ (i 1) (H) =t] Pr[µ (i 1) (H) =t] i=1 E[µ (i 1) (H)] dr r +1

25 History In 1992, the first subexponential algorithms for Linear Programming are developed by Matousek, Sharir, and Welzl, and independently by Kalai. In 1996, Sharir and Welzl extend the machinery to LP-type problems, which are a generalization of Linear Programming. In 2002, Skovron introduces Violator Spaces. Subsequently, it is showed by Gärtner, Matousek, Rüst, and Skovron that a lot of the machinery carries over. In LP-type there is a value function ϕ, which is used to define violators and extreme points. It turns out that ϕ is not necessary. The problem defines the violator mapping V. Extreme: X (R) := {p R V (R\{p}) = V (R)}

26 Smallest Enclosing Ball Nondegeneracy: Less than d + 2 points on any ball. P Regularity: Every smallest ball has exactly the same number of points on the boundary.

27 Definition Violator Space The pair (H, V ) is a violator space if H is a finite set, and V :2 H 2 H satisfies the following conditions. Consistency: G H, V (G) G = Locality: F G H, if V (F ) G = then V (F )=V(G) s Not everything is local! Consider the diameter problem: Given P, find the p D r F = {q, s} G = {p, q, s} Euclidean diameter of P. q = V (F ) = V (G) ={r}

28 Cyclic Violator Spaces f g G B(G) V (G) f, g, h f f h g g f h h g f, g f h f, h h g g, h g f f, g, h f, g, h h A point is locally smaller than another if it is further clockwise w.r.t. the center (measure the acute angle).

29 Some Examples Linear Programming Polytope Distance OPT s c OPT dim = d dim = d +1 s Largest Ellipsoid in Polytope Rectilinear 3-Center Problem dim = 43 dim d 2 Sharir, Welzl, 1996

30 German Algorithm choose R H u.a.r.; B basis(r); while (V (B) = ) { R R V (R); B basis(r); } return B;

31 Combinatorial Dimension Consider the violator space (H, V ) Basis: A basis of G H is a minimal subset B of G, s.t. V (B) =V (G). The combinatorial dimension of (H, V ), denoted by dim(h, V ), is the size of the largest basis in (H, V ). Smallest Enclosing Ball: It is well known that d + 1 points in R d uniquely determine a ball. Therefore dim d + 1.

Linear Programming Randomization and Abstract Frameworks

appeared in Proc. 13th Ann. Symp. on Theoret. Aspects of Comput. Sci. Lecture Notes in Computer Science 1046 (1996) 669-687. Linear Programming Randomization and Abstract Frameworks Bernd Gärtner and Emo