Vertex cover might be hard to approximate

within 2 - ε

Subhash Khot, Oded Regev

Slides by: Ofer Neiman

Given a graph G(V,E), VC(G) is the minimal set of vertices touching every edge.

Vertex Cover

•Assuming only P≠NP the best known hardness factor is 1.36.

•2-approximation is easy.

Independent set

Given a Graph G(V,E), IS(G) is the maximal set of vertices without edges between them.

We saw it’s the complement of Vertex Cover.

Bottom Line

With certain assumption, it is NP-hard to distinguish between:

• IS(G)>½-2ε

• IS(G)<δ

VC is NP-hard to approximate within 2-ε’

VC(G)≤½+ε

VC(G)≥1-δ

Overview

Unique gamesconjecture

Unique labelcover

Strong labelcover

Graphconstruction

CompletenessSoundness

Unique games conjecture

Unique: answer of

one prover determines

the answer of the other.

Prover 2

Prover 1

verifier

Acceptance map},{: falsetrueRYRX

Probabilistically choose

q1

q2

YXqq 2,1Strategy p1:X→R

Strategy p2:Y→R

a1

a2

Unique Games Conjecture

• There’s a permutation for every pair of questions: πxy:R→R.

• verifier accepts iff πxy(p1(x))=p2(y).

NP-hard to decide whether•Success probability >1-ς

•Success probability <γ

Unique Label Cover

An assignment A:XUY→R that satisfies the maximal weight of constraints.

x1

xn

x2

y1

y2

ym

Labeled from R

Labeled from R

πxy

constraintsΠxy:R→R

One to oneAnd onto! wxy

Weights wxy≥0

Complete bi-partite graph

Notations

• A constraint πxy is satisfied if πxy)A(x))=A(y).

• w(x) = Yy

xyw

Xx

AA xww )(

•wA(x) is the weight of the satisfied x-constraints.

• is the total weight of satisfied constraints.

Unique game=unique-LC

Unique-LC conjecture:

For any ς,γ>0 there is constant |R| such that it is NP-hard to distinguish between:

1,

yx

xyw

1. There is an assignment A, such that wA>1-ς.

2. For all assignments A, wA<γ.

For a unique-LC with label set R such that

Why 1-ς??

• The gap of (γ,1) is easy:• There’s a linear algorithm determining whether a Unique-

LC can have all its tests satisfied.

x1

xn

x2

y1

y2

ym

}1,2…,R{

πxyx1 y1

y2x2

xn ym

Is this enough??

New hardness results:• Min-2SAT-Deletion.• Not-All-Equal predicate on 3 variables.• a hardness for vertex cover.2

Closing the gap requires stronger tools….

Strong Label Cover

x1

xn

x2

y1

y2

ym

Labeled from R

Labeled from R

πxy

constraintsΠxy:R→R

One to oneAnd onto!

No weights!

Degree d

d left degree bi-partite graph

Theorem

Assuming the unique games conjecture:

For any ς,γ>0 there are constants d,|R| such that it is NP-hard to distinguish between:

1. There is an assignment in which at least 1-ς fraction of the X vertices have all of their tests satisfied.

2. No assignment satisfies more than γ fraction of the tests.

For a Strong-LC with label set R and left degree d.

Road mapUnique-LC

Unique-Lc+ Weights adjustements

Strong-LC

Unique-Lc+ Weights rounding

Lemma 1

For any ς,γ,β>0 there is constant |R| such that it is NP-hard to distinguish between:

1. There is an assignment A such that 1-β fraction of the X vertices have wA(x)>1-ς.

2. For any assignment A at most β fraction of the X vertices have wA(x)>γ.

XxFor a Unique-LC with label set R and the property that for all w(x)=1.

Construction

• Given (X’,Y’,Π’,W’) with parameters ς’,γ’ to be chosen later, let l be a large constant.

)('|'|)( xwXlxk

x1

x2

Xk(x)

x yxy' xyw'

xy')('

'

xw

w xy

y

Proof

• Fact 2: (l-1)|X’|≤|X|≤l|X’|

• Fact 1: w(x)=1

Completeness

• Assume there’s an assignment A’ for which w’A≥1-ς’.

• Define A(xi)=A’(x), A(y)=A’(y).

• Objective: 1-β fraction of the X vertices have wA(x)>1-ς.

Step 1: '21

XxwXx

A

Step 2: for at least X vertices

'21 '21 xwA

Soundness

• Assume by contradiction:• There is an assignment A for which β fraction of the X

vertices have wA(x)>γ.

Objective: build an assignment A’(x)=A(x1) for which w’A>γ’.

w.l.o.g: A(x1)=A(x2)=…=A(xk(x))

claim

'

1'

XxA l

lxw

Lemma 2

• We can change the unique-LC from lemma 1 a little…

• For a large integer α=O(Y) and some integer i.

wxy iwxy

Proof of Strong-LC

So far we have (X’,Y’,Π’,W’) with parameters ς’,β’,γ’>0 to be chosen later, and a constant |R| such that:

•w’(x)=1.

•wxy=i/α

Construction

x1

xn

x2

y1

y2

ym

x1

Complete bi-partite graph

d

ixyi

wxk1

')(

π’xy w’xy

d-left regular

π’xy

X,(y1,y2,…,yd)

x1

x2

Xk(x)

y1

y2

yd

The new graph

xx

xx

xy1

y2

ym

xx

xx

x

xx

xx

x

d-Left regulat

•For each we create a large (constant) set of vertices.

•Y=Y’.

'Xx

The new graph

• Weightless.

d

d

Yyxy

d

Yyy

d

ixy ww

dd

i

''),...,( 11

•d-left regular.

•|X|=αd|X’|, since:

Completeness• Assume A’ is an assignment such that 1-β’ fraction of

the X’ vertices have w’A(x)>1-ς’.

•For each denote Yx to be the set of vertices such that π’xy is satisfied.

'XxYy

•We define A(xi)=A’(x) if xi was created from x.

A(y)=A’(y).

Objective: an assignment A such that for 1-ς of X vertices all the test are satisfied.

Completeness

x

y1

y2

ym

yes

no

yes

xi

y1

y2

yd

xi connected to Yx only

dxd Yyy ),...,( 1

•For each having w’A(x)>1-ς’, there are at least

vertices connected only to Yx

Counting…

How many good vertices are there???

dd )'1( 'Xx

XX ddd )'1()'1(')'1()'1(

•The total number is:

•A’(y)=A(y).

•A’(x)=i, where is the label maximizing wA’(x).

Soundness

• Assume no assignment A’ to the unique-LC has more than β’ fraction of the X’ vertices with wA’(x)>γ’.

Ri

•Assume we have A: an assignment to the Strong-LC, we will build A’ as follows:

Objective: show that no assignment satisfy more than γ

Observations

• For 1-β’ fraction of the X’ vertices, after fixing labels for Y’ any assignment will satisfy at most γ’ test weight.

1)(')(:,:,...,1 jxydx yAijiyyZj

• For any such x, let

• A vertex created from a tuple in Zx will have at most 1 satisfied test.

There are many such vertices…

Counting• 1-β’ of the x in X’ create at least

good vertices (one successful test at most).

21'1 ddd d

β’ fraction of |X’|: d satisfied edges.

fraction1 satisfied edge

fractiond sat edges

)1(2 )21)('1(

X

End of part 1

• The total fraction of tests satisfied by A is at most γ.

Part 2: The Graph(X,Y,Ē,Π)

x1

xn

x2

y1

y2

ym

Degree d

•There is A such that 1-ς of the X vertices have all tests satisfied

•For all A at most γ tests satisfied

ς,γ>0 chosen later

w(IS(G)) > ½ -2ε

w(IS(G)) < δ

G(V,E)

Construction

x1

xn

x2

x1 x1

xn

x2

x1 R2

R2

R2

B[x1]

B[x2]

B[xn]

RFFxxB :,][

Construction

• For each define• Fix p=½-ε

RF FRFp ppF /1

FX

Fxweight p1

,

RFXx

Fxweight,

1,

Construction

x1

y

πx1 y

x2

π x 2y

x2

x1 F

G

R2

R2

B[x1]

B[x2]

Edge exists between all pairs for whom πx1y)F)∩πx2y(G)=Ø

Completeness

XX 10

• X0 will be the set of vertices for which all the tests are satisfied.

FxAXxFxIS )(,:),( 0

x IS(x)A(x)

• Let

• Assume by contradiction: EGxFx ,,, 21

x1

y

πx1 y

x2

π x 2y

ISGxFx ,,, 21

021, Xxx yAxAxA yxyx 21 21

GxAFxA 21 , GF yxyx 21

Weight?

• Weight(IS(x)) = p = ½-ε

•Weight(IS) = (1-ς)(½-ε) ≥ ½-2ε

,For all

0Xx

Soundness

For all A at most γ tests satisfied w(IS(G))<δ

IFxRFFx ,,|F

Assume by contradiction that there is independent set Iof weight δ.

2:* xxX p F

Definitions

XX 2*

Monotone families

• A family F is called monotone if and

implies .

FFF'F

'FF

F(x) is a monotone family:

Reminder: EGxFxGF yxyx ,,, 2121

, For all IFxIFx ',, 'FF

Idea

hxLXx ][1:*

xX

X*L[x]

Lemma: • there is a constant h such that for all

21 21xLxL yxyx

• for all*

21, Xxx

x1

x2

X*

X

L[x1]

L[x2]

yx1

yx2

y

Assuming lemma

• Let be the set a vertices having a test with some

.

• For any choose that there is a test πx(y)y *Yy

*XxYY *

*)( Xyx

x1 y1

y2x2

xn ym

X* Y*

L[y]=πx(y)y(L[x(y)])

End of proof

Claim• For all we have

**,: YyXxxy

yLxLxy

A(x), A(y) are randomly chosen from L[x], L[y]

With probability at least we have πxy(A(x))=A(y)21

h

The expected fraction of satisfied tests: 22h

Choosing yields contradiction22h