Complexity ©D.Moshkovits 1 Hardness of Approximation

Complexity©D.Moshkovits

1

Hardness of Approximation


2

Introduction

• Objectives:– To show several approximation

problems are NP-hard• Overview:

– Reminder: How to show inapproximability?

– Probabilistic Checkable Proofs– Hardness of approximation for clique


3

Promise Problems

• Sometimes you can promise something about the input

• It doesn’t matter what you say for inputs that do not satisfy the promise

I know my graph has clique of size n/4! Does it have a clique of size

n/2?


4

Promise Problems & Approximation

• We’ll see promise problems of a certain type, called gap problems, can be utilized to prove hardness of approximation.


5

Optimization Problems

Consider an optimization problem P:

instances: x1,x2,x3,…

optimization measure

feasible solutions

all graphs

Example:

all cliques in that graph

the clique’s size (max)


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Each Instance Has an Optimal Solution

OPTx1 x2 x3x4


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Approximation (Max Version)

OPTxi


8

How To Show Hardness of Approximation?

Hardness of distinguishing far off instances Hardness of approximation

OPT

A B

gap

xi


9

Gap Problems (Max Version)

• Instance: …

• Problem: to distinguish between the following two cases:

The maximal solution B

The maximal solution ≤ A


10

Formally:

Claim: If the [A,B]-gap version of a problem

is NP-hard, then that problem is NP-hard to

approximate to within factor B/A.


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Formally:

Proof (for maximization): Suppose there is an approximation algorithm that outputs C≤C* so that C*/C ≤ B/A

A proper distinguisher:* If CA, return ‘YES’* Otherwise return ‘NO’


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Proof

Since C ≥ C*·A/B, (1) If C* > B (the correct answer is

‘YES’), then necessarilyC ≥ C*·A/B > B·A/B = A

(we answer ‘YES’)

(2) If C*≤A (the correct answer is ‘NO’), then necessarily C≤C*≤A(we answer ‘NO’).


13

Translating To Decision Problems To Prove Hardness

Optimization Problems Approximation Problems

Threshold Problems Gap Problems

Is the size of the max

clique > ½n?

Is the size of the max

clique > ¾n or < ¼n?


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Idea

• We’ve shown “standard” problems are NP-hard by reductions from 3SAT.

• We want to prove gap-problems are NP-hard,

• Why won’t we prove some canonical gap-problem is NP-hard and reduce from it?

• If a reduction reduces one gap-problem to another we refer to it as gap-preserving


15

Gap-3SAT[]

Instance: a set of 3-clauses {c1,…,cm} over variables v1,…,vn.

Problem: to distinguish between the following two cases:

There exists an assignment that satisfies all clauses.No assignment can satisfy more than 7/8+ of the clauses.


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Gap-3SAT: Example

( x1 x2 x3 )

( x1 x2 x2 )

( x1 x2 x3 )

( x1 x2 x2 )

(x1 x2 x3 )

( x3 x3 x3 )

= { x1 F ; x2 T ; x3 F }satisfies 5/6 of the clauses


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Why 7/8?

Claim: For any set of clauses with exactly three independent literals,

there always exists an assignment that satisfies at least 7/8 clauses.


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The Probabilistic Method

Proof: Consider a random assignment.

x1 x2 x3 xn

. . .


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1. Find the Expectation

Let Yi be the random variable indicating the outcome of the i-th clause.

For any 1im,

F F F F

F F T T

F T F T

F T T T

T F F T

T F T T

T T F T

T T T T

87

181

0YE i 87

181

0YE i


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1. Find the Expectation

The number of clauses satisfied is a random variable Y=Yi.

By the linearity of the expectation:

E[Y] = E[ Yi] = E[Yi] = 7/8m


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2. Conclude Existence

Thus, there exists an assignment which satisfies at least the expected number of clauses.


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PCP (Without Proof)

Theorem (PCP): For any >0,

Gap-3SAT[] is NP-hard.

This is tight! Gap-3SAT[0] is polynomial time

decidable


23

Why Is It Called PCP? (Probabilistically Checkable Proofs)

3SAT has a polynomial membership proof checkable in polynomial time.

)(...)( 42146121 xxxxxx n

x1

x2

x3

x4

x5

x6

x7

x8

xn-3

xn-2

xn-1

xn

. . .

My formula is satisfiable!

Prove it!

This assignment satisfies it!


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…Now our verifier has to check the assignment satisfies all clauses…


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But gap-3SAT also has a polynomial membership proof checkable in polynomial time.

)(...)( 42146121 xxxxxx n

x1

x2

x3

x4

x5

x6

x7

x8

xn-3

xn-2

xn-1

xn

. . .

My formula is satisfiable!

Prove it!

This assignment satisfies it!


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And for gap-3SAT the verifier would be right with high probability, even if he

(1)picks at random a constant number of clauses and

(2)checks only them

In a NO instance of gap-3SAT, 1/8

of the clauses are not satisfied!


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• Since gap-3SAT is NP-hard, All NP problems have probabilistically checkable proofs.


28

Gap Preserving Reductions

P1 P2

•YES

•don’t care

•NO

• YES

• don’t care

• NO


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Hardness of Approximation

• Do the reductions we’ve seen also work for the gap versions (i.e approximation preserving)?

• We’ll revisit the CLIQUE example.


30

CLIQUE Construction

.

.

.

a part for each

clause

a vertex for each literal

edge indicates consistency: one is not the

negation of the other


31

Cliques & Truth-Assignments

• A Clique CV corresponds to the assignment A:V{T,F} s.t C A()=T.

• An edge between two vertices implies the corresponding literals can be both assigned T.

• Thus each clique corresponds to a satisfying truth-assignment.

.

.

.


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Gap Preservation

• If there is an assignment that satisfies all clauses, there is a clique of size m.

• If there is a clique of size m (for some 0<<1) there is an assignment that satisfies at least of the clauses.


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Gap-CLIQUE (Ver1)

The following problem is NP-hard for any >0:

Instance: a graph G=(V,E) composed of m independent sets of size 3.

Problem: to distinguish between:

There’s a clique of size m=|V|/3

Every clique is of size at most (7/8+)m


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Corollary

Theorem: for any >0,CLIQUE is hard to approximate

within a factor of 1/(7/8+)


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Can We Do Better?

• The bigger the gap is, the better the hardness result.

• We’ll see an improved result for CLIQUE.


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.

.

.

...

...

Amplification

A part for every k clauses

vertex for each satisfying

assignment to the k clauses

edge indicates

consistency

Given an instance of the Gap-3SAT problem and a constant k (to be determined later):


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Boolean Assignments

• Each clause has at most 7 satisfying assignments.

• Thus k clauses have at most 7k satisfying assignments.

F F F F

F F T T

F T F T

F T T T

T F F T

T F T T

T T F T

T T T T


38

Consistency

• Two assignments are inconsistent, if they give the same variable different truth-values.

x y z

x y z w

F F T

x w y F T T


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The Graph G=<V, E>

• Given = {C1, …, Cm} over variables y1, …, yn denote Y(C1, …, Ck) the set of variables which appear in C1, …, Ck

• Vertices

• Edges between every two consistent assignments

1 1 1k k kV C, ,C ,A A : Y(C, ,C ) T,F sat C, ,C 1 1 1k k kV C, ,C ,A A : Y(C, ,C ) T,F sat C, ,C

1 2 1 11 2 1 2E C ,A , C ,A y Y(C ) Y(C ),A (y) A (y) 1 2 1 11 2 1 2E C ,A , C ,A y Y(C ) Y(C ),A (y) A (y)


40

Cliques & Assignments

Observation: A clique on of the parts corresponds to an assignment which satisfies all relevant clauses.

.

.

.

...

...

.

.

.

...

...


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Correctness (1)

• If there is a satisfying assignment, then picking the corresponding assignment in each of the parts yields a clique of size

k

m

k

m

read: “m choose k”

i.e. m!/k!(m-k)!

k

m

k

m


42

Observation

Fix an assignment. If 1/8 of the clauses are false, then only (7/8)k of the sets of k clauses are satisfiable.


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Correctness (2)

• For any 0<<1, set k so (7/8+)k < • If there is a clique with

representatives in ≥ of the parts• There is an assignment satisfying ≥

fraction of the k-tuples of clauses• Ruling out the NO case, in which no

assignment satisfies more than 1/8- of the clauses.


44

Gap-CLIQUE (Ver2)

The following problem is NP-hard for any 0<<1:

Instance: a graph G=(V,E) composed of m independent sets of size r.

Problem: to distinguish between:

There’s a clique of size m = |V|/r

Every clique is of size at most m


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Corollary

Theorem: MAX-CLIQUE is NP-hard to approximate to within any constant factor.


46

Chromatic Number

• Instance: a graph G=(V,E).• Problem: To minimize , so that

there exists a function f:V{1,…, }, for which

(u,v)E f(u)f(v)


47

Chromatic Number


48

Chromatic NumberObservation:

Each color class is an

independent set


49

Clique Cover Number (CCN)

• Instance: a graph G=(V,E).• Problem: To minimize , so that

there exists a function f:V{1,…, }, for which

(u,v)E f(u)=f(v)


50

Clique Cover Number (CCN)


51

Observation

Claim: The CCN problem on graph G is the CHROMATIC-NUMBER problem of the complement graph Gc.


52

Reduction Idea

.

.

.

CLIQUE CCN

.

.

.

q

same under cyclic shift

clique preserving

m G G’


53

Correctness

• Given such transformation:– MAX-CLIQUE(G) = m CCN(G’) = q– MAX-CLIQUE(G) < m CCN(G’) > q/


54

Transformation

T:V[q]

for any v1,v2,v3,v4,v5,v6,

T(v1)+T(v2)+T(v3) T(v4)+T(v5)+T(v6) (mod q)

{v1,v2,v3}={v4,v5,v6}T is unique for triplets


55

Observations

• Such T is unique for pairs and for single vertices as well:

• If T(x)+T(u)=T(v)+T(w) (mod q), then {x,u}={v,w}

• If T(x)=T(y) (mod q), then x=y


56

Using the Transformation

0 1 2 3 4 … (q-1)

vi

vj

T(vi)=1

T(vj)=4

CLIQUE

CCN


57

Completing the CCN Graph Construction

T(s)

T(t)

(s,t)ECLIQUE

(T(s),T(t))ECCN


58

Completing the CCN Graph Construction

T(s)

T(t)

Close the set of edges under shift:

For every (x,y)E,

if x’-y’=x-y (mod q), then (x’,y’)E


59

Edge Origin Unique

T(s)

T(t)

First Observation: This edge comes

only from (s,t)


60

Triangle Consistency

Second Observation: A

triangle only comes from a triangle


61

Clique Preservation

Corollary: {c1,…,ck} is a clique in the CCN graph

iff {T(c1),…,T(ck)} is a clique in the CLIQUE graph.


62

What Remains?

• It remains to show how to construct the transformation T in polynomial time.


63

feasible values

Greedy Constructionv6

v6

v2

v2

v1

v1

v5

v5v

3

v3

v4

v4

vertices we determined

forbidden values


64

Greedy Construction - Analysis

At most values are ruled out totally, so for q=n5 the greedy construction works.

Corollary: There exists a polynomial time algorithm which constructs a triplet unique transformation with q=n5

5

n


65

Corollaries

Theorem: CCN is NP-hard to approximate within any constant factor.

Theorem: CHROMATIC-NUMBER is NP-hard to approximate within any constant factor.


66

Summary

• We saw how to show hardness of approximation and explained the concept of gap problems.

• We presented the PCP theorem, stating that 3SAT is hard to approximate within some constant factor.


67

Summary

• We saw that some of the reductions we know were approximation preserving.

• That was the case for the 3SATpCLIQUE reduction.


68

Summary

• However, that reduction gave us a weak result for CLIQUE,

• So we showed how to amplify it.


69

Summary

• Then we introduced a new problem, called CHROMATIC-NUMBER.

• We reduced gap-CLIQUE to its gap version, showing it was in fact NP-hard to approximate.

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Complexity ©D.Moshkovits 1 Hardness of Approximation