1. A Linear Round Lower Bound for Lovasz-Schrijver SDP relaxations of Vertex Cover Grant Schoenebeck Luca Trevisan Madhur Tulsiani UC Berkeley

2. For G = (V,E) find the smallest subset of vertices containing at least one endpoint of every edge. Integrality Gap = Max G = 2 – o(1) for both Minimum Vertex Cover SDP (Lovasz -function) Minimize  u2V z 0 ¢z u k z u k  1 8 u 2 V, k z 0 k  1 (z 0 - z u )¢(z 0 - z v ) = 0 8 (u,v) 2 E LP Minimize  u2V x u x u 2 [0,1] 8 u 2 V x u + x v ¸ 1 8 (u,v) 2 E Integer Optimum Optimum of Program

3. Integrality Gaps with more constraints For the complete graph (K n ) on n vertices: LP Optimum = n/2; Integer Optimum = n-1 Integrality Gap = 2 – 2/n What if we add the constraint x u +x v +x w ¸ 2 for every triangle (u,v,w) in G? Performance ratio for K n 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 x u + x v + x w + x z  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 n O(r) time. Constant number of rounds produce most known LP/SDP relaxations.

5. From polytopes to cones Allow scaling of solutions to convert solution set to a cone. For Vertex Cover: Minimize  u2V y u y 2 R n+1 0  y u  y 0  u  V y u + y v  y 0  (u,v)  E y 0 = 1 Define a cone VC(G)

6. The Lovasz-Schrijver Hierarchy  Goal: Only allow convex combinations of 0/1 solutions. Probability distributions! y = (1, y 1, y 2, …, y n )  K z (1) = (1, 1, z (1) 2, …, z (1) n ) w (1) = (1, 0, w (1) 2, …, w (1) n ) y 1  z (1) + (1-y 1 )  w (1) = y z (n) = (1, z (n) 1, z (n) 2, …, 1) w (n) = (1, w (2) 1, w (n) 2, …, 0) y n  z (n) + (1-y n )  w (n) = y z (2) = (1, z (2) 1, 1, …, z (2) n ) w (2) = (1, w (2) 1, 0, …, w (2) n ) y 2  z (2) + (1-y 2 )  w (2) = y … y y 1  z (1) y 2  z (2)  y n  z (n) Y = 1.Y = Y T 2.Diagonal(Y) = y 3.Y i  K, y - Y i  K  i 4.Y is p.s.d. Y ij = Pr[i=1 ^ j=1] LS LS + Marginal distribution! Ask for conditionals.

7. The Lovasz-Schrijver Hierarchy  Goal: Only allow convex combinations of 0/1 solutions. Probability distributions! y = (y 1, y 2, …, y n )  S z (1) = (1, z (1) 2, …, z (1) n ) w (1) = (0, w (1) 2, …, w (1) n ) y 1  z (1) + (1-y 1 )  w (1) = y z (n) = (z (n) 1, z (n) 2, …, 1) w (n) = (w (2) 1, w (n) 2, …, 0) y n  z (n) + (1-y n )  w (n) = y z (2) = (z (2) 1, 1, …, z (2) n ) w (2) = (w (2) 1, 0, …, w (2) n ) y 2  z (2) + (1-y 2 )  w (2) = y … y y 1  z (1) y 2  z (2)  y n  z (n) Y = 1.Y = Y T 2.Diagonal(Y) = y 3.Y i /y i  S, (y –Y i )/(1-y i )  S 4.Y is p.s.d. Y ij = Pr[i=1 ^ j=1] LS LS + Marginal distribution! Ask for conditionals.

8. Prover-Adversary Game y  LS r (VC) u (0 < y u < 1) z (u), w (u)  LS r-1 (VC) z (u), v ProverAdversary Showing y  LS r (VC) can be viewed as “almost” a 2-player game 

9. Prover-Adversary Game (contd…) 2/3 x 2/3 0 1 1 1 11 1 1/2 1 1 0 1 11 1 ? 1 ? ?? 1/3 x 1/2 x Ha!

10. Work on LS + for Vertex Cover k-1-   (n) rounds (VC in k-uniform hypergraphs) AAT’05 2 -  (√(logn/loglogn)) GMPT’06 7/6 -   (n) rounds This paper 7/6 -  1 round (random 3XOR)FO’06 2 -  1 round + triangle inequalityCharikar’02 2 -  1 roundGK’98

11. The Graphs Random 3XOR formula (m = cn clauses) (FO’06) Show solutions for Independent Set (y 1, …, y n )  LS r (VC)  (1-y 1, …, 1-y n )  LS r (IS) x 1 + x 2 + x 3 = 1 001 100111 010 x 3 + x 4 + x 5 = 0 101 110 011 000 1/4

12. Properties of random 3XOR formulas Any assignment satisfies at most (1/2 +  )m clauses. For VC: Integer optimum  4m - (1/2+  )m = (7/2-  )m (IS  assignment, vertices included  clauses satisfied) Fractional optimum = 4m – ¼  4m = 3m Integrality gap  7/6 -  Two clauses share at most one variable (constant probability). k clauses involve at least 1.9k variables for k  n (AAT’05, other works in proof complexity) (implies no small unsatisfiable subset)

13. The conditional distributions Expansion prevents propogation of effects from conditioning. ? B A A B 1/4 0 0 0 1 1/2 1

14. The conditional distributions (contd…) Conditional distribution for y i = 0 is a convex combination of distributions for y j = 1 for other j’s. All values are 0, 1/4, 1/2 or 1 and neighbors of 1 are 0. Solutions are fractional independent sets. Need to maintain expansion even after variables are assigned. 0 1/3 0 0 1 0 0 0 0 1 0 1 0 0 = + 1/3  1/3  + 1/3 

15. Maintaining the expansion Problem: Adversary fixes variables – may cause loss of expansion. Solution: We fix more variables - all variables in a maximal non-expanding subset of clauses, say B. Adversary fixes t (= 1 or 2) clauses, we fix |B| (k, 1.9)  (k – t – |B|,1.9) If y =  i y (i), then y (i)  S  i  y  S. Express y as uniform distribution over consistent assignments to clauses in B. B

16. Maintaining the expansion (contd…) If B is consistent, k clauses & l vars  2 l-k solutions. For C  B (with 3 variables), if every assignment to C is consistent with B, every assignment appears in 2 k-l-2 solutions (solutions form affine subspace). B consistent with any assignment to C as contradictions have very low expansion. 1/4 2 l-k-2

17. Assigning clauses consistently Let S  B be minimal unsatisfiable subset after fixing an assignment to C  B. S involves at most 1.5|S| variables as each must occur twice. B Adversary : 1 (or 2) clauses, at most 3 (or 4) variables C : 1 clause, at most 3 variables S : |S| clauses, at most 1.5|S| variables 1.5|S| + 3 + 3  1.9 (|S| + 2)  |S|  5

18. How many rounds? Start with (  n, 1.95)-expansion. At i th round, adversary fixes set S i, we fix T i. 1.95(  i |S i | + |T i |)  #(fixed vars)  3  i |S i | + 1.9  i |T i | Stop after r rounds if  n -  i (|S i | + |T i |)  4   n – 44r  4

19. But the title says LS + ! Still need to show matrix Y at each round is p.s.d. A matrix Y  R m  m is p.s.d. iff  v 1, …, v m  R m such that Y ij = v i  v j  i,j Need to exhibit a vector for every vertex with above property (also shows symmetry).

20. The vector solutions (FO’06) Divide variables into equivalence classes: 3-variable eqn  different classes 2-variable eqn  same class One coordinate for each class. v i has  y i in the coordinate corresponding to the classes contained and 0 elsewhere. One extra coordinate = y i in v i. 001 100111 010 x 1 + x 2 + x 3 = 1 00 11 x 3 + x 4 = 0 Classes: (x 1 ), (x 2 ), (x 3 & x 4 ) v i  v j = y i y j (1 + n agree – n disagree ) 0 if contradict = y i y j no shared vars 2y i y j shared agree (¼, ¼, -¼, -¼)(¼, ¼, ¼, ¼) (¼, -¼, ¼, -¼)(¼, ¼, ¼, ¼) (½, 0,0, - ½) (½, 0,0, ½)

21. Conclusions/Open Problems 2-  gap for  (n) rounds. Reduction from CSPs  LS + rounds  partial assignments. 2-  gap using other CSPs? Interpretation as probability distributions still does not give a natural interpretation of the vector solutions. Other natural ways of looking at the hierarchy? Similar results for Sherali-Adams hierarchy? LS + results for other problems? Sparsest Cut Khot’s SDP for Unique Games

22. Thank You Questions?