Tight Integrality Gaps for Lov á sz-Schrijver LP relaxations of Vertex Cover

1. Tight Integrality Gaps for Lovász-Schrijver LP relaxations of Vertex Cover Grant Schoenebeck Luca Trevisan Madhur Tulsiani UC Berkeley

2. Integer Optimum Optimum of Program Minimum Vertex Cover • For G = (V,E) find the smallest subset of vertices containing at least one endpoint of every edge. • Integrality Gap = MaxG = 2 – o(1) for both SDP (Lovász -function) Minimize u2Vz0¢zu k zuk 1 8u 2 V, k z0k= 1 (z0 - zu)¢(z0 - zv) = 0 8 (u,v) 2 E k zuk 2 = z0¢zu LP Minimize u2V xu xu2 [0,1] 8 u 2V xu + xv¸ 1 8 (u,v) 2 E

3. Integrality Gaps with more constraints • For the complete graph (Kn) on n vertices: LP Optimum = n/2; Integer Optimum = n-1 Integrality Gap = 2 – 2/n • What if we add the constraint xu+xv+xw¸ 2 for every triangle (u,v,w) in G? Performance ratio for Kn is 3/2, but the integrality gap still remains 2-o(1). • What if add constraints analogous to the one above for every odd cycle? What if we add xu + xv + xw + xz 3 for every clique of size 4? Size 5? • One needs to prove integrality gaps from scratch every time new constraints are added.

4. Automatically generating “natural” constraints • LS/LS+ hierarchies define define “cut operators” applied to (convex) solution space. • Operators can be iteratively applied to generate tighter LP/SDP relaxations. • Relaxation obtained by r cuts (rounds) solvable in nO(r) time. • Constant number of rounds produce most known LP/SDP relaxations. • Lower bounds against these hierarchies imply lower bounds for large class of LP/SDP relaxations (ABL’02)

5. LS/LS+ lower bounds for Vertex Cover LS 2 -  (log n) rounds ABL’02, ABLT’06 (log2 n) rounds Tourlakis’06 3/2 -  2 -  (n) rounds STT’07 LS+ 2 -  1 round GK’98 2 -  Charikar’02 1 round + triangle inequality 7/6 -  1 round (random 3XOR) FO’06 (n) rounds STT’07 7/6 -  (√(logn/loglogn)) 2 -  GMPT’07

6. The Lovász-Schrijver Hierarchy • Goal: Only allow convex combinations of 0/1 solutions. Probability distributions! y = (y1, y2, …, yn)  S Marginal distribution! Ask for conditionals. 2/3 1 0 1 = 1/3 x + 1/3 x + 1/3 x 1 1 1 1 2/3 1 1 0 1/2 1 = 2/3 x + 1/3 x 1 1 0 1 u y(u), y( u) y = (yu) y(u) + (1-yu) y( u)

7. The Lovász-Schrijver Hierarchy u y(u), y( u) y = (yu) y(u) + (1-yu) y( u) y1y(1) y2y(2)  yny(n) Conditions we can check : y  S is “good” if  Y s.t. • Y = YT • Diagonal(Y) = y • Yi/yi  S, (y –Yi)/(1-yi)  S • Y is p.s.d. Y = LS LS+ Yij = Pr[i=1 ^ j=1]

8. Prover-Adversary Game Showing y  LSr(VC) can be viewed as “almost” a 2-player game y  LSr(VC) u (0 < yu < 1) y(u) ,y( u) LSr-1(VC) y(u) ,v  Prover Adversary

9. 1 1/2 2/3 1 ? 1 1/2 ? 2/3 1 2/3 1 2/3 1 1/2 ? ? 2/3 1 1 1/2 0 1/2 1 1 2/3 1 1 1 0 2/3 Prover-Adversary Game (contd…) 2/3 x 1/3 x Ha! 1/2 x 1/2 x

10. The Graphs [ABLT’06] • Pick one at random: Gn,p with np = O(1) • Removing O(n1/2) edges ) no cycles of size O(log n) (locally like trees) • All subgraphs on at most n vertices are sparse |ES| < (1+)|S| • |VC| > (1-)n • (1/2+, …, 1/2+) survives (n) rounds ) gap of 2-O()

11. The game begins… V v = 1 ½ + 

12. 0 0 ½+ 1/2 1 ½+ 1 1 1/2 0 1 1-4 1-8 8 4 O(1/ log(1/)) The “splash” 1 + ½ = (½-) = ½ + (½+) Like a tree: Tree: 0 o.w. 1 w.p. 2/(1/2+) After ½+ v Before

13. Idea 1: Cheat while you can Cycles after O(log n) rounds Easy Doable Conditionals exist for trees Girth is essentially the limitation for earlier works

14. Idea 2: Actively simplify the solution • If y =  ciy(i) and each y(i) survives r rounds, then so does y. • Express solution at every step as a convex combination of “nice” solutions with splashes being far apart. • “Process” solution to obtain 0/1 values in regions with complicated distributions.

15. Making things simpler(?) • [ABLT’06]: If every subgraph is sparse (|ES| < (1+)|S|), then (½+, …, ½+) is a combination of 0/1 solutions. • [STT’07]: If yi + yj¸ 1+2 for all (i,j)2 E not both 0/1, then y is a convex combination of (0,1, ½+) solutions. 0/1 ½+ STT ABLT

16. But wait… that was wrong! ½+ Processing fixes region to arbitrary 0/1 values. Inconsistensies on boundary. 0 Solution: Create splashes around boundary! But what if they intersect again? Wasn’t that the whole problem to begin with!

17. Idea 3: Fix things aggressively • Recursively include short paths between points in the region to be processed. • No short paths between two points on boundary ) splashes at boundary are far apart.

18. And this finally works! • Splash • Aggressively simplify • Splash on boundaries Continue

19. How many rounds does this work for? • The region to be processed should never get bigger than n (only small subgraphs are sparse). • Splashes affect only constant number of vertices at each round. • Need to show that “aggressive simplification” does not fix too many vertices.

20. The cost of aggression • We always add paths - one more edge than vertices. • Adversary fixes trees – one more vertex than edges. • If too many more paths than trees, then subgraph becomes too dense (violates sparsity). • Set up potential function, show that size of processed region is O(r) after r rounds ) can survive (n) rounds.

21. Conclusions/Open Problems • 2- gap for (n) rounds of LS+. • Similar results for Sherali-Adams/Lassere hierarchies? • LS+ results for other problems? • Sparsest Cut • Unique Games CMM’07: 2- for (n) rounds of SA

22. Thank You Questions?