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A Linear Round Lower Bound for Lovasz-Schrijver SDP relaxations of Vertex Cover

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

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A Linear Round Lower Bound for Lovasz-Schrijver SDP relaxations of Vertex Cover

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  1. A Linear Round Lower Bound for Lovasz-Schrijver SDP 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 LP Minimize u2V xu xu2 [0,1] 8 u 2V xu + xv¸ 1 8 (u,v) 2 E SDP (Lovasz -function) Minimize u2Vz0¢zu k zuk 1 8u 2 V, k z0k= 1 (z0 - zu)¢(z0 - zv) = 0 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.

  5. y y1z(1) y2z(2)  ynz(n) The Lovasz-Schrijver Hierarchy • Goal: Only allow convex combinations of 0/1 solutions. Probability distributions! y = (y1, y2, …, yn)  S Marginal distribution! Ask for conditionals. z(n) = (z(n)1, z(n)2, …, 1) w(n) = (w(2)1, w(n)2, …, 0) ynz(n)+ (1-yn)w(n) = y z(1) = (1, z(1)2, …, z(1)n) w(1) = (0, w(1)2, …, w(1)n) y1z(1)+ (1-y1)w(1) = y z(2) = (z(2)1, 1, …, z(2)n) w(2) = (w(2)1, 0, …, w(2)n) y2z(2)+ (1-y2)w(2) = y … • Y = YT • Diagonal(Y) = y • Yi/yi  S, (y –Yi)/(1-yi)  S • Y is p.s.d. LS Yij = Pr[i=1 ^ j=1] LS+ Y =

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

  7. 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

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

  9. 001 010 110 000 011 101 111 100 The Graphs • Random 3XOR formula (m = cn clauses) (FO’06) • Show solutions for Independent Set (y1, …, yn)  LSr(VC)  (1-y1, …, 1-yn)  LSr(IS) x1 + x2 + x3 = 1 x3 + x4 + x5 = 0 1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4

  10. 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)

  11. 1/2 1/2 1/2 1 1/2 The conditional distributions 1/4 0 1/4 1 1/4 1/4 A 1/4 0 1/4 1/4 1/4 0 B A B ? Expansion prevents propogation of effects from conditioning.

  12. 0 0 0 0 1/3 0 0 1 1 0 0 1/3 1/3 1 0 0 The conditional distributions (contd…) • Conditional distribution for yi = 0 is a convex combination of distributions for yj = 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. = 1/3 + 1/3 + 1/3

  13. 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 = iy(i), then y(i) S  i y S. • Express y as uniform distribution over consistent assignments to clauses in B. B

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

  15. B 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. 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

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

  17. But the title says LS+! • Still need to show matrix Y at each round is p.s.d. • A matrix Y Rm  m is p.s.d. iff v1, …, vm Rm such that Yij = vivj  i,j • Need to exhibit a vector for every vertex with above property (also shows symmetry).

  18. 001 00 010 111 100 11 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. vi has  yi in the coordinate corresponding to the classes contained and 0 elsewhere. One extra coordinate = yi in vi. x1 + x2 + x3 = 1 x3 + x4 = 0 Classes: (x1), (x2), (x3 & x4) vi vj= yiyj(1 + nagree – ndisagree) 0 if contradict = yi yj no shared vars 2yi yj shared agree (¼, ¼, ¼, ¼) (¼, -¼, ¼, -¼) (½, 0,0, - ½) (¼, ¼, ¼, ¼) (¼, ¼, -¼, -¼) (½, 0,0, ½)

  19. 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

  20. Thank You Questions?

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