Tight Integrality Gaps for Lovász-Schrijver LP

relaxations of Vertex Cover

Grant Schoenebeck Luca Trevisan Madhur Tulsiani

UC Berkeley

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

Minimum Vertex Cover

SDP (Lovász -function)

Minimize u2V z0¢zu

k zuk 1 8 u 2 V, k z0k 1

(z0 - zu)¢(z0 - zv) = 0 8 (u,v) 2 E

k zuk 2 = z0¢zu

LP

Minimize u2V xu

xu 2 [0,1] 8 u 2 V

xu + xv ¸ 1 8 (u,v) 2 E

Integer Optimum

Optimum of Program

Integrality Gaps with more constraints For the complete graph (Kn) on n vertices:

LP Optimum = n/2; Integer Optimum = n-1Integrality 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.

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)

LS/LS+ lower bounds for Vertex Cover

2 - (√(logn/loglogn))GMPT’07

7/6 - (n) roundsSTT’07

7/6 - 1 round (random 3XOR)FO’06

2 - 1 round + triangle inequalityCharikar’02

2 - 1 roundGK’98

2 - n) roundsSTT’07

3/2 - (log2 n) roundsTourlakis’06

2 - (log n) roundsABL’02, ABLT’06

LS

LS+

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

12/3

1

10

0

11

1

11

= 1/3 x ++ 1/3 x1/3 x

1

10

1/2

11

= 2/3 x + 1/3 x

u y(u) , y( u) y = (yu) y(u) + (1-yu) y( u)

2/3 1/3 2/3

2/3 2/3 2/3

2/3 2/3 1

The Lovász-Schrijver Hierarchy

y1y(1)

y2y(2)

yny(n)

Y = 1. Y = YT

2. Diagonal(Y) = y

3. Yi/yi S, (y –Yi)/(1-yi) S

4. Y is p.s.d.Yij = Pr[i=1 ^ j=1]

LSLS+

u y(u) , y( u) y = (yu) y(u) + (1-yu) y( u)

Conditions we can check : y S is “good” if Y s.t.

Prover-Adversary Game

y LSr(VC)

u (0 < yu < 1)

y(u) , y( u) LSr-1(VC)

y(u) , v

Prover Adversary

Showing y LSr(VC) can be viewed as “almost” a 2-player game

Prover-Adversary Game (contd…)

2/3 x

2/3

2/3

2/3

2/32/3

2/3 2/3

0

1

1 1

1 1

1

1/2

1/2

1/21/2

1/2

1

1

0 1

1 1

1

?

1 ?

? ?

1/3 x

1/2 x1/2 x

Ha!

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(

The game begins…

½ + V v = 1

The “splash”

½+v

Before

After1

4

1-4

8

1-8

1/2

1/2

0

1

1

0

= ½ + ½ Tree:

½+

½+

0

1

1

1 w.p. 2/(1/2+)

= (½- + (½+Like a tree:

0o.w.

O(1/ log(1/))

Idea 1: Cheat while you can

Easy

Doable

Cycles after O(log n) rounds

Girth is essentially the limitation for earlier works

Conditionals exist for trees

Idea 2: Actively simplify the solution If y = ci y(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.

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.

STT

0/1

½+

ABLT

But wait… that was wrong!½+

Processing fixes region to arbitrary 0/1 values. Inconsistensies on boundary.

Solution: Create splashes around boundary!

But what if they intersect again? Wasn’t that the whole problem to begin with!

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.

And this finally works!1. Splash

2. Aggressively simplify

3. Splash on boundaries

Continue

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.

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.

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

Thank You

Questions?