Inapproximability of Hypergraph Vertex-Cover. A k-uniform hypergraph H= : V – a set of vertices E - a collection of k-element subsets of V Example: k=3.

Inapproximability Inapproximability

ofof

Hypergraph Vertex-CoverHypergraph Vertex-Cover

A k-uniform hypergraph H=<V,E>:

V – a set of vertices

E - a collection of k-element subsets of V

Example: k=3

The Vertex-Cover problem (Ek-Vertex-Cover):

Given a k-uniform hypergraph, find the smallest subset of vertices that intersects every hyperedge.

The matching decision problem is NP-Complete

We will show that the problem is hard to approximate

within a factor of 2

k

(The actual known bound is )1k

We will show this by using a reduction to the

Independent Set Problem

Both results by Dinur, Guruswami, Khot, and Regev (2004)

Independent Set :

Find the maximal set of vertices which doesn’t contain an edge.

This problem is VC’s complement:

is the minimal VC is the maximal ISC V \V C

Proof:

– a vertex cover.C V

Suppose that isn’t an independent set .

Then there is an edge e whose vertices are all in .

\V C

\V CSo e isn’t represented in C – a contradiction!

So all of e‘s vertices are in S – a contradiction!

- an independent set.S VSuppose that isn’t a vertex cover .

Then there is an edge e which isn’t represented in

\V S

\V S

So, if we have a maximal Independent Set, its complement

is a minimal Vertex Cover!

Given a 4-uniform hypergraph G , for every ,

It is NP-hard to distinguish between the following:

, 0

( ) 1/ 2IS G

( )IS G

( ) 1/ 2VC G

( ) 1VC G

VC is hard to approximate within

( ) 2VC G

We will see:

Lecture’s outline:

• Intersecting families and a helpful lemma

• Reduction from Label Cover to IS

• Proof of completeness and soundness

Intersecting Families

[ ] {1,..., }n n[ ]2 { | [ ]}n A A n

Denote

A family is called pairwise intersecting

if for every 2 sets ,

[ ]2 nF

1 2,A A F 1 2A A

What is the maximal size of a pairwise intersecting family?

Of every pair we can only choose one set ,A A

12n

Is this bound tight?

i

All sets such that i AAFor odd nAll sets such thatA | | / 2A n

n

For a bias parameter and a set ,

the weight of a set is

0 1p R

A R

| | | \ |: (1 )R A R Ap p p

The weight of a family is 2RF

( ) : ( )p pA F

F A

When clear from the context, we will omit the R

What is the weight of the family ?

iIt’s the probability that i appears in a given set: p

What is the weight of the family ?

nDepends on p

If - nearly 01

2p If - nearly 1

1

2p If -

1

2p 1

2

3. For any , all sets such that [ ]2 nA

| [ 2 ] |A t j t j

0j

A family is called pairwise t-intersecting

if for every 2 sets ,

[ ]2 nF

1 2,A A F 1 2A A t

Examples:

1. All sets such that [ ]2 nA [ ]t A [t]

2. All sets such that [ ]2 nA | [ 2] | 1A t t t+2

(i,j)-shift on a family:

replacing j with i in all sets

such that and

A F

,j A i A ( \{ }) { }A j i F

Left-shifted family: a family which is invariant with respect to

(i,j)-shift for any 1 i j n

By iterating the (i,j)-shift for all we eventually get a

left-shifted family S(F)

1 i j n

Lemma 1:

Left-shifting a family maintains its size and the size of the sets

In the family.

If F is a pairwise t-intersecting family, then so is S(F).

Lemma 2:

Let be a left-shifted pairwise t-intersecting family.

Then, for every , there exists a with

[ ]2 nF

A F 0j

| [ 2 ] |A t j t j

Proof

Lemma 3:

For arbitrary , let .

Then, there exists s.t. for any

pairwise t-intersecting Family , .

, 0 1

2p

[ ]2 nF ( )p F

( , 2, )t t

Meaning, for any ,

a family of non-negligible weight ( )

can’t be pairwise t-intersecting for a large enough t

1

2p

( )p F

Correctness of Lemma3:

For every , there exists a with ( )A S F 0j | [ 2 ] |A t j t j

the probability that such a j exists for a random set

chosen according to . ( ( ))p S F

p

if F is pairwise t-intersecting then so is S(F) , and

( ) ( ( ))p pF S F

Pr[| [ 2 ] | ]A t j t j Look at

For some 0j

Since , the prob. of a set having a big intersection

with a big section ( [t+2j] ) is very small.

1

2p

So any pairwise t –intersecting family can’t be heavy for a

large enough t.

Label Cover Constraints - Functions

fromto YR ZR

Variable Set

Y Over

YR

Variable Set

Z Over

ZR

The graph is bi-regular

Goal : find a labeling for X and Y that satisfies the maximal fraction of constraints

a constraint is satisfied if ( ( )) ( )y z y z

(1), OY ZR R R

ynzm

1 2y z

y2

y1 z1

z2

Theorem (PCP theorem + Raz’s parallel Repetition Theorem):

There exists a universal constant such that for every

(large) constant R it is NP-hard to distinguish between

the following two cases:

No: no labeling can satisfy more than

of the constraints .

1

R

Yes: There is a labeling

such that every constraint is satisfied by :

for every

: ,Y ZY R Z R

, ( ( )) ( )y z y z y z

0

We will use the PCP theorem to prove a factor hardness

for E4-Vertex-Cover, or in other words:

2

Given a 4-uniform hypergraph G , for every ,

it is NP-hard to distinguish between the following cases:

, 0

( ) 1/ 2IS G

( )IS G

We start with a Label Cover instance as described

- A local set of constraints over variables whose

respective ranges are

Y Z

,Y ZR R

Parameters: (from Lemma 3)( , 2, )2

t t

( - the constant from the PCP theorem)2

1/2( )

tR

The Reduction

From the Label Cover instance, we construct a weighted 4-uniform hypergraph G

The weight of each vertex: 1

2

1( , ) : ( )

| |y A A

Y

G’s Vertices: 2 YRY

yn

y2

y1

1 2 YRy

2 2 YRy

2 YRny

1[ ]V y

2[ ]V y

[ ]nV y

G’s Hyperedges:

iff there’s no and such that1 1 2r A A

1 1 2 2( ) ( )y z y zr r 2 1 2r B B

1 2 YRy

2 2 YRy

1 1,y A1 2,y A

2 2,y B

2 1,y B

1y z y1

z2y z

y2

For each pair of constraints sharing a

common z

Completeness: if there’s a labeling satisfying all constraints

then ( ) 1/ 2IS G

Proof: Assume a satisfying labeling : Y ZY Z R R

The following set is an independent set of G:

{ , [ ] | , ( ) }I y A V y y Y y A

We need to show : there’s no edge which consists solely of

vertices of the above form

y y,a a

ai

ajak

y1,a aiaj

ak

Suppose that an edge like that exists:

y1,a ajal

y2,bbi

bj y2,b bk

y1 a

y2 b

But we have:

1 2( ) ( )y z y za b 1 2 1 2,a A A b B B

By the definition of the edges, this is a contradiction!

What is the Independent Set’s weight?

=Pr choosing a vertex out of this IS, when picked according to A’s weight

1

2

Soundness: if no assignment satisfies more than

of the constraints in the Label Cover instance then ( )IS G 1/ R

The same as: if then there exists an assignment

that satisfies more than a fraction

of the constraints in the Label Cover instance.

( )IS G : Y ZY Z R R 1/ R

We need to prove:

Proof: Let be an independent set of size at least .S V

Look at , the set of all variables y for which the

weight of in is at least .

'Y Y

[ ]S V y [ ]V y / 2

1[ ]V y

2[ ]V y

[ ]nV y

SS

yn

y2

y1

Y’Y’

/ 2

/ 2

What is the size of ? 'Y

If less than of the blocks have an intersection

smaller than , S’s size can’t be at least

/ 2

| ' | | | / 2Y Y

/ 2

For each , define .'y Y { 2 | , }yR

yF A y A S

y

y

y

y Fy

So, we have: for each , 1

2

( )2yF

'y Y

We saw (L3), that for , a family of weight

can’t be pairwise t-intersecting for our choice of t ( )

1

2p

2

( , 2, )2

t t

Therefore, there must exist s.t. 1 2| |A A t 1 2, yA A A

We shall call the core of assignments for y.1 2( )B y A A

Intuitively, for any y for which V[y] has a large

intersection with S, we will match B(y) as potential values.

We will translate the cores into an assignment satisfying

More than a fraction of the constraints. 1/ R

For every and

such that are constraints, we must have

z Z 1 2 1 2, ',y y Y y y

1 2,y z y z

1 1 2 2( ( )) ( ( ))y z y zB y B y

y1,A1 y1,

y2, y2,

A2

C1 C2

An edge in the IS – a contradiction!

Otherwise…

Let denote the set of all Z variables

That participate in some constraint with some .

'Z Z

'y Y

ynzm

1 2y z

y2

y1 z1

z2Y’Y’ Z’Z’

We associate each with an arbitrary

for which there exists

'z Z 'y Y

y z

1y z y1

z2y z

y2

Y’Y’ Z’Z’

Denote ( ) : ( ( ))y z ZB z B y R

We have matched each variable a set of values B(y)

We have matched each variable a set of values B(z)

We will now use these values to define a labeling

Which will satisfy a sufficient number of constraints

'z Z

'y Y

We define a random assignment :

For each we independently select a random value from B(y)'y Y

For each we independently select a random value from B(z)'z Z

For the rest of the variables we assign any

arbitrary value

( \ ') ( \ ')Y Y Z Z

ynzm

y2

y1 z1

z2

B(y1)

B(yn)

B(z2)

RZ

RZ

Ry

We now show that for every , each constraint

is satisfied by with probability

'y Y

2

1

t

We saw: '( ( )) ( ( '))y z y zB y B y ( ( )) ( )y z B y B z

so there is at least one value s.t.( )ya B y ( ) ( )y z ya B z

Look at for somey z 'y Y

Assume that z is associated with some ' 'y Y

y z y

z

y’

y z

ory zy’=

For every , and so'y Y | ( ) |B y t | ( ) |B z t

So there is at least prob. of having2

1

t( ( )) ( )y z y z

Pr[ ( ) ] 1/iy a t

y z y z

B(y)ai

aj

ak

B(z)

bl

b( )ia

Pr[ ( ) ( )] 1/iz a t

Variables from Y’ participate in of the constraints| ' |

| |

Y

Y

And we are done!

There exists some assignment that meets the expectation.

satisfies of the constraints2

1

2t R

We chose:2

1/2( )

tR

22( )

tR

so the expected number of local constraints satisfied by is

2

| ' | 1| |

| |

Y

Y t 2

| |2t

We saw:

| ' |/ 2

| |

Y

Y

How many constraints are satisfied?

The reduction’s complexity:

The number of vertices in the hypergraph is at most:

(1)

| | 2ORY

And for any k?

We prove for any even k as follows:

Pairwise t-intersecting K/2-wise t-intersecting

1

2p 1

1/ 2

pk

And build a k-uniform hypergraph.

To prove for any odd k, we add a new vertex to each

edge in a (k-1)-uniform hypergraph.

Obviously, that doesn’t change the size of the vertex cover

And for an unweighted Hypergraph?

It is proved possible to turn a weighted Hypergraph into

an unweighted one, in polynomial time.

Lemma 2:

Let be a left-shifted s-wise t-intersecting family.

Then, for every , there exists a with

[ ]2 nF

A F 0j

| [ ] | ( 1)A t js t s j

Proof:

For two sets A, B s.t. |A|=|B|=l,

1 1{ ,..., },1 ...l lA a a a a n

1 1{ ,..., },1 ...l lB b b b b n

We say that if for all A B i ia b 1,...,i l

We first prove: For , if then also A B [ ] \n A F [ ] \n B F

Proof by induction:

for i=0,…,l ,let 1 1[ ] \{ ,..., , ,..., }i i i lF n a a b b[ ] \lF n A and therefore lF F

We now show that implies that : iF F 1iF F

If then , and the claim holds.i ia b 1i iF F

Otherwise, and hence i ia b i ib F

F is left-shifted and =>i ia F 1\{ } { }i i i iF b a F F

Since , we are done. 0 [ ] \F n B

Assume, by contradiction, that there exists

such that for all ,

A F

| [ ] | ( 1)A t sj t s j 0j

Let , and 1 2{ , ,..., }lC c c c[ ] \C n A 1 2 ... lc c c

A has at least i “holes” in [ ( 1) ]t i s

Therefore, ( 1)ic t i s

A has at least one “hole” for each s elements from t to n

Therefore, ( ) / 1l n t s

Since F is s-wise t-intersecting, the size of each set must be

greater than t, and therefore ( )l n t

0,..., 1k s For , define the set

by

,1 ,2 ,{ , ,..., }k k k lB b b b

, min{ ( 1) , ( )}k ib t k i s n l i

Since are all integers, 1 2 ... lc c c n ( )ic n l i

And so, or in other words, 0,i ic b0C B

( )l n t

So , and therefore,k ib t [ ] \ [ 1]in B t

Notice that: ( 1)t t k i s

( )t n l i

We saw: ( 1)ic t i s

Since , we have, 1,k i k ib b 0 1 1... sC B B B

We saw: For , if then also A B [ ] \n A F [ ] \n B F

A contradiction!

0 1 1... sC B B B

0 0[ ] \ [ 1] ( \ ) ... ( \ ) [ 1]in B t n B n B t But

F is s-wise t-intersecting

Therefore, since , for all , 0,..., 1k s [ ] \ in B F

We get: 0 1( \ ) ... ( \ )sn B n B t

[ ] \n C F

Lemma 2 is proved: For every , there exists a with

A F 0j

| [ ] | ( 1)A t js t s j