Inapproximability of MAX-CUT

Khot,Kindler,Mossel and O’Donnell

Moshe Ben Nehemia June 05

Main Result

It is NP-Hard problem to approximate MAX-CUT to within a factor

is the approximation ratio achieved by the algorithm of Goemans & Williamson.

The result follows from: 1. Unique Games conjecture 2. Majority is Stablest Theorem

GW878567.0GW

Hardness of Approximation

History: Bellare & Goldreich & Sudan :It is NP

Hard to approximate MAX-CUT within factor higher than 83/84

Hasted improved the result to 16/17 Today: closing the gap…

Introduction

MAX-CUT: Given a Graph G =(V,E), find a partition C=(V1, V2) that

maximize:

Unique Label Cover: Given a bi-partite graph with left side vertices- V ,right side W, and edges- E each edge have a constraint bijection The goal: assign each vertex a label which satisfy the

constraint.

][][:, MMwv

)())((, vlabelwlabelwv

EVVe

ew)( 21

)(

Unique Games Conjecture: For any there exist a constant Such that it is NP-hard to distinguish whether the

Unique Label Cover problem with label set in size M has optimum at least or at most

0, ),( MM

Some defintions Let be an arbitrary boolean

function The influence of xi on f

Let x be a uniformly random string in :E[X]=0 and form y by flipping each bit with prob The noise stability of f for a noise rate is:

)],,,,...(),...,([Pr)( 1111}1,1{

niiinx

i xxxxxfxxffInfn

n}1,1{

)]()([Pr , yfxfyx

}1,1{}1,1{: nf

The Correlation between x,y is define to be: E[XY] = 2 Pr[X=Y]-1

)]()([)( , yfxffS yxE

Let x be a uniformly random string in y be -correlated copy :i.e. pick each bit independently s.t. The noise correlation of f with parameter is:

n}1,1{

][ ii yxE

Result[60’] :

arccos1))((lim 2 MajoritySn

Fix then for any there is a small enough s.t. if is any function satisfying :

Then:

)1,0[ 00),(

}1,1{}1,1{: nf

ni

f

..1.2

0)(.1

E

)( fInfi

arccos1)( 2fS

The Majority is Stablest Theorem

On the Geometry of MAX-CUT

The Goemans-Williamson algorithm: Embedding the graph in the unit sphere of Rn : The embedding is selected s.t. this sum is

maximize

A cut in G is obtained by choosing a random

hyperplane through the origin . And this sum bounds from above the size of the

maximal cut

Evu

vu xx),(

21

21 ,

On the Geometry of MAX-CUT

The probability that vertics u,v lie on opposite sides of the cut is:

So the expected weight is

),arccos( ji vv

ji

jiji vvwWE ),arccos(1

][ ,

On the Geometry of MAX-CUT

So to get:

Set the approximation ratio to:

})(arccos

{21

21

]1,1[min

GW

OPTWE ][

Reminder

The Long Code: The codeword encoding the message is by the truth table of the “dictator”

function:

][ni

}{:)( iXiLC

Technical Background

The Bonami Beckner operator

Proposition: Let and then:

SS

S SffT )(ˆ)(

R nf }1,1{: ]1,1[

][

2)(ˆ)(,)(nS

S sffTffS

Technical Background

Proposition:Let then for every

Proof:Define:

And :

And using the Parseval identity we get the proposition

R nf }1,1{: ][ni

Si

i SffInf 2)(ˆ)(

2

22)()( )()()( xffInfxf iiixfxf

i

SiS

Si Sfxf:

)(ˆ)(

Technical Background

Let and let

The k-degree influence of coordinate i on f is defined by:

Proposition: The “Majority is Stablest” Theorem remains true if we change the assumption to ')( fInf k

i

R nf }1,1{: ][ni

kSSi

ki SffInf 2)(ˆ)(

)( fInfi

Reverse version of the “Majority is Stablest” Fix then for any there is a small enough s.t. if is any function satisfying :

Then:

]0,1( 00),(

}1,1{}1,1{: nf

ni ..1 )( fInf ki

arccos1)( 2fS

Reverse version of the “Majority is Stablest”

Proof: Take such f, and define:

Now g holds:

And now apply the original Theorem

SoddS

Sfxfxf

xg

)(ˆ2

))()(()(

)()()(

)()(.2

0)(.1

gSgSfS

fInfgInf

gE

ii

Reduction from Unique LC to MAX-CUT

Notations: denote the string and xy the coordinatewise product of x and y

x ),...,( )()1( nxx

Lemma 1: Completeness If ULC have OPT then MAX-CUT have

cut Lemma 2: Soundness If ULC have OPT then MAX-CUT have cut

at most

1c 212

1

s /)(arccos

Reduction from Unique LC to MAX-CUT

Unique Label Cover

W V

'

vW’

w

MAX-CUT

j

J’ i

-{1,1}M

Reduction from Unique LC to MAX-CUT

The Reduction: Pick a vertex at random and 2 of its

neighbors: Let and be the constrains for

those edges Let f,g be the supposed Long Codes of the labels Pick at random Pick by choosing each coordinate

independently to be 1 with probability and -1 with prob. Edge in Cut iff

VvWww ',

wv, ',' wv

Mx }1,1{M}1,1{

2121

2121

))'(()( xgxf

Reduction from Unique LC to MAX-CUT

Completeness Assume that the LC instance has a

labeling which satisfies fraction of the edges.

now encode these labels via Long Code with prob both the edges are

satisfied by the labeling Denote the label of v,w,w’ by i,j,j’

1

21

Reduction from Unique LC to MAX-CUT

Completeness note that: Now f,g are the Long Codes of j,j’, so: The two bits are unequal iff and that happens with prob.

hence the completeness :

)'(')( jij

1' j

'')'('

)(

))'((

)(

jijj

ij

xxxg

xxxf

2121

)()21( 21

21

Reduction from Unique LC to MAX-CUT

Soundness – The Proof Strategy if the max-cut bigger than we’ll

be able to “decode” the “Long Code“ and create a labeling which satisfy significant fraction of the edges in the LC problem, and get a contradiction by choosing small enough.

/arccos

Reduction from Unique LC to MAX-CUT

)]()()()(ˆ)(ˆ[

)()(

)]()'()()(ˆ)(ˆ[

')'('][',

)(,2

121

)(

'][',

',2

121

SSMSS

Sx

SS

SMSS

SSx

xxSgSfE

xx

xxSgSfE

)])'(()([]Pr[ 21

21

,,',,

xgxfaccept Exwwv

From the Fourier Transform:

)()'( )'('' xx SS

Reduction from Unique LC to MAX-CUT

The expectation over x vanishes unless and then s,s’ have the same size. Because:

We got:

Because of for at least v in V (“good” v) We have

)'(')( ss

)(ˆ)(ˆ

)]([

)'(')(:',21

21

''

SgSf

E

SSSS

S

SSS

/arccos]Pr[accept2/

21

'

121

21 /)(arccos)]'('(ˆ[)]((ˆ[ SgESfE

wwS

S

Reduction from Unique LC to MAX-CUT

)(arccos1)])((ˆ[(

/)(arccos)])((ˆ[(

221

221

21

21

SfE

SfE

wS

S

wS

S

S

S

w

w

M

Sh

SfESh

xfExh

h

arccos1)(ˆ

)]((ˆ[)(ˆ

)]([)(

]1,1[}1,1{:

22

1

Define

Now:

Reduction from Unique LC to MAX-CUT

Now ,from the “Majority is stablest” theorem: We conclude that h has at least one coordinate j s.t. label the vertex v with j

)(hInf kj

)]([]))((ˆ[))]((ˆ[)(ˆ)(

212121 fInfESfESfESh k

jw

kSSj

kSSj

ww

kSSj

And we have:

Reduction from Unique LC to MAX-CUT

From the above equation we have that for at least fraction of neighbors w of v we have

Define

And so,

Because we got that

2/)()(1

fInf k

j

}2/)(:][{][ fInfMiwCand ki

kfInfi

ki )( /2][ kwCand

2

][)(1 wCandj

Reduction from Unique LC to MAX-CUT

Now ,if we label each vertex w in W by random element from Cand[w], then among the “good” vertices v

at least satisfied. or among the edges , and that yields the contradiction

k22

k8

2