Hardness of Approximating

Multicut

S. Chawla, R. Krauthgamer, R. Kumar, Y. Rabani, D. Sivakumar

(2005)

Presented by Adin Rosenberg

Multicut

Input: An undirected graph G=(V,E), where |V|=n k pairs of vertices {si,ti}i=1,…,k, called demand pairs Optional: a cost function c on E

Goal: A multicut: a subset of edges M, whose removal

disconnects all of the demand pairs. Of course, minimize c(M) (or |M| if c isn’t defined)

Multicut – an example

What are we going to prove?

Assuming the Unique Games conjecture is true, Multicut is NP-hard to approximate within any constant factor L

How are we going to prove this?

We will show a reduction from a UG instance to a Multicut instance.

Unique Games

Input: A bipartite graph G=(Q,EQ) Each side p=1,2 contains n=|Q|/2 vertices (or questions)

labeled q1p, q2

p, …, qnp

Each edge (qi1,qj

2) (called a question edge) is associated with a bijection bij:[d]→[d]

Each edge (qi1,qj

2) has a nonnegative (normalized) weight wij

Unique Games (cont.)

A solution is an answer 1≤Aip≤d for each question

qip

A solution satisfies an edge (qi1,qj

2) if the answers Ai

1 and Aj2 agree, i.e. Aj

2=bij(Ai1)

Goal: Find a solution with maximum value (total weight of

satisfied edges)

Unique Games Conjecture [Khot 2002]

For every η,δ>0 there exists d=d(η,δ) such that it is NP-hard to determine whether a unique 2-prover game with answer set of size d has a value of: at least (1- η), or at most δ

A Little About Hypercubes

A d-dimensional hypercube is a graph G=(V,E) where V={0,1}d and there is an edge between two vertices if they differ in exactly one coordinate.

An edge (u,v) is called a dimension-a edge if u and v differ in coordinate a.

A dimension-a cut is the set of dimension-a edges.

The antipodal of a vertex u is the vertex which differs from u in every coordinate.

A Little About Hypercubes

(0,0,0) (1,0,0)

(0,1,0) (1,1,0)

(0,0,1) (1,0,1)

(0,1,1) (1,1,1)

Dimension-1 edges

The Reduction from Unique Games to Multicut

For every vertex qip construct a d-dimensional

hypercube Cip. Let the edges of these 2n cubes

(called hypercube edges) have cost 1. For each question edge (qi

1,qj2) extend bij to a

bijection b’ij:{0,1}d→{0,1}d defined by

Connect each vertex with using and edge (called a cross edge) with cost wijΛ, where Λ=n/η.

Define the demand pairs to be the antipodal pairs.

),...,,()(')()2()1( 111 dbbbij

ijijijuuub u

1iCu 2)(' jij Cub

The Reduction from Unique Games to Multicut

w11

w23

w11Λ

w23Λ

1 1

11

The Yes Instance

Claim: If there is a solution A for the unique 2-prover game with value of at least 1-η, then there exists a multicut M for the Multicut instance such that c(M) ≤ 2d+1n

Proof: Construct the following multicut M: For every answer Ai

p take the dimension- Aip cut in

cube Cip.

For every edge (qi1,qj

2) that the solution A doesn’t satisfy, take all the cross edges between Ci

1 and Cj

2.

The Yes Instance (cont.)

Removing M disconnects all the demand pairs: For every vertex v in Ci

p, define f(v) to be the Aip-th

coordinate of v. For every edge (u,v) left, f(u)≠f(v)

The cost of M is at most 2d+1n: Let S be the set of question edges not satisfied by the

solution A. nnwnMc dndd

Sqqij

dd

ji

1

),(

1 222222)(21

A Little More About Hypercubes

For a function f on the vertices of a hypercube, define Ia

f to be the fraction of dimension-a edges (u,v) for which f(u) ≠ f(v).

For a cutset M in a hypercube, define IaM to

be the fraction of dimension-a edges that belong to M.

Observe that |M| = 2d-1ΣaIaM

And now some lemmas…

Lemma 1

Let M be a cutset in a hypercube, and let g be the function labeling each vertex with the index of the connected component it belongs to. Then Ia

M≥Iag.

Proof: M contains every edge (u,v) for which g(u) ≠ g(v)

Lemma 2

Let M be a cutset in a hypercube H. Suppose M disconnects at least a β fraction

of the antipodal pairs in H. Then for every x>0, if ΣaIa

M ≤ βx then there exists a dimension a* such that Ia*

M ≥ 2-6x/27

Lemma 3

Let f,g be two function on the vertices of a hypercube.

If f(v) = g(v) for all but a β fraction of the vertices v, then for every dimension a we have |Ia

f – Iag| ≤ 2β.

The No Instance

Claim: There exists a constant c such that if the Multicut instance has a cutset of cost at most 2dnL (where L=c(log(1/(η+δ))) ) whose removal disconnects α≥7/8 fraction of the demand pairs, then there exists a solution for the unique 2-prover game whose value is larger than δ.

The No Instance (cont.)

Proof: Let M we such a cutset for the Multicut instance. Let Ia

p,i be the influence of M for each cube Cip.

Construct a randomized solution A for the unique 2-prover game instance.

For each vertex qip, we choose Ai

p to be the answer a with probability Ia

p,i / Σa Iap,i.

The expected value of A is at least δ, and therefore there exist a solution with such a value.

The No Instance (cont.)

Bound the probability of the following “bad” events (for a choice of the question edge (qi

1,qj2) ):

E1 – fewer than half the demand pairs in Ci1 are

disconnected in G \ M E2 – M contains more than 2d+2L hypercube edges in Ci

1.

E3 – M contains more than 2d+2L hypercube edges in Cj2.

E4 – M contains more than 2d / 296L+7 cross edges between Ci

1 and Cj2.

All “bad” events do not occur with probability of at least 1/8.

The No Instance (cont.)

Assuming none of the “bad” events occur: There exists a dimension a* s.t. Ia*

1,I ≥ 2-96L/27 (according to Lemma 2)

Ibij(a*)2,j ≥ Ia*

1,i – 2-96L-6 ≥ 2-96L/54 (Lemma 3)

The expected value of A is

)2(8

1

*)](*,Pr[)](Pr[

962,2

,2*)(

,1

,1*

2112

L

a

ja

jab

a

ia

ia

ijjiiijj

LI

I

I

I

abAaAAbA

ij

ji

Liijjij LAbAw

,

96212 )2()](Pr[

What have we seen?

If the unique game has a value greater than 1-η, then the Multicut instance has a cutset M which disconnects all of the demand pairs with cost c(M) ≤ 2d+1n

If the unique game has a value less than δ, then the disconnected 7/8 of the demand pairs in the Multicut instance costs (at least) 2dLn

Therefore, the Unique Games Conjecture implies that it is NP-hard to approximate Multicut within a factor of any L>0.

Proof of Lemma 2

ΣaIaM ≤ βx means Ia*

M ≥ 2-6x/27 for some a* Proof:

Convert M to a two-sided cut M’ Define a Boolean function f according to the

connected components of M’ Use KKL’s lemma: ΣaIa

f/α + Σa(Iaf)4/3 ≥ 2p*logα/α

(where f is a Boolean function on a hypercube and p ≤ ½ is the balance of f)

Proof of Lemma 3

If f(v) = g(v) for all but a β fraction of the vertices v, then for every dimension a we have |Ia

f – Iag| ≤ 2β.

Proof: For all but at most β2d of the edges, we have

f(u)=g(u) and f(v)=g(v). Therefore, only a β2d/2d-1=2β fraction of edges

can contribute to the difference between Iaf and

Iag.

Credits to…

The authors of the paper for giving me what to talk about.

Kahn, Kalai and Linial for saving us the Fourier analysis.

Sarai for taking care of the lights. And of course, thank you all for listening…