Hardness of Approximation

Complexity1

Hardness of Approximation

Complexity2

Introduction• Objectives:

– To show several approximation problems are NP-hard

• Overview:– Reminder: How to show

inapproximability?– Probabilistic Checkable Proofs– Hardness of approximation for clique

Complexity3

Optimization ProblemsConsider 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)

Complexity4

Each Instance Has an Optimal Solution

OPTx1 x2 x3x4

Complexity5

Approximation (Max Version)

OPTxi

Complexity6

How To Show Hardness of Approximation?

Hardness of distinguishing far off instances Hardness of approximation

OPTA B

gap

xi

Complexity7

Gap Problems (Max Version)

• Instance: …

• Problem: to distinguish between the following two cases:

The maximal solution B

The maximal solution ≤ A

Complexity8

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

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

approximate within factor B/A.

Complexity9

Formally:Proof: Suppose there is an

approximation algorithm that outputs C so that C/C*≤B/A

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

Complexity10

ProofSince C*≥AC/B, (1) If C>B (we answer ‘YES’), then

necessarily C*>A (the correct answer cannot be ‘NO’).

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

Complexity11

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 approximation-preserving

Complexity12

Gap-3SAT[]Instance: a set of clauses {c1,…,cm}

over variables v1,…,vn.Problem: to distinguish between the

following two cases:There exists an assignment which

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

Complexity13

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

Complexity14

Why 7/8?Claim: For any set of clauses with

exactly three independent literals, there always exists an assignment

which satisfies at least 7/8.

Complexity15

The Probabilistic MethodProof: Consider a random

assignment.

x1 x2 x3 xn

. . .

Complexity16

1. Find the ExpectationLet Yi be the random variable

indicating the outcome of the i-th clause.

For any 1im, E[Yi]=0·1/8+1·7/8=7/8

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

Complexity17

2. Conclude ExistenceExpectedly, the number of clauses

satisfied is 7/8m.

Thus, there exists an assignment which satisfies at least that many.

Complexity18

PCP (Without Proof)Theorem (PCP): For any >0,

Gap-3SAT[] is NP-hard.

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

decidable

Complexity19

Approximation Preservation

A B

•YES

•don’t care

•NO

• YES

• don’t care

• NO

Complexity20

Hardness of Approximation

• Do the reductions we’ve seen also work for the gap versions?

• We’ll revisit the CLIQUE example.

Complexity21

CLIQUE Construction

.

.

.

a part for each

clause

a vertex for each literal

edge indicates

consistency

Complexity22

Approximation Preservation

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

• If there is a clique of size (7/8+)m, there is an assignment which satisfies more than 7/8+ of the clauses.

Complexity23

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

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

Complexity24

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

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

Complexity25

Amplification• The bigger the gap is, the better

the hardness result.• We’ll see how a gap can be

amplified.

Complexity26

.

.

.

...

...

Amplification

A part for every k vertices

vertex for each Boolean

assignment

edge indicates

consistency

Given an instance of the Gap-CLIQUE problem and a constant k:

Complexity27

Boolean assignments• A Boolean assignment over k vertices

{v1,…,vk} is a function A:{v1,…,vk}{0,1}.

• Think about it as if it indicates whether each vertex belongs to the clique.

Complexity28

Good Assignments

Complexity29

Consistency• Two assignments are inconsistent,

when they give the same vertex different truth-values.

. . .n

. . .. . .

Complexity30

Consistency• They are also inconsistent, if they

both assign 1 to two vertices not connected by an edge.

non-edge

Complexity31

Correctness

Complexity32

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

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

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

Complexity33

Chromatic Number

Complexity34

Chromatic NumberObservation: Each color group is an

independent set

Complexity35

Clique Cover Number (CCN)

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

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

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

Complexity36

Clique Cover Number (CCN)

Complexity37

Reduction Idea

.

.

.

CLIQUE CCN

.

.

.

q

• cyclic shift-morphic• clique preserving

m...

Complexity38

Correctness

Complexity39

TransformationT: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

Complexity40

Observations• Such T is unique for pairs and for

single vertices as well:• If T(x)+T(u)=T(v)+T(w), then

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

Complexity41

feasible values

Greedy Constructionv6

v2

v1

v5v

3

v4

vertices we determined

forbidden values

Complexity42

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

5n

Complexity43

Using the Transformation

0 1 2 3 4 … (q-1)

vi

vj

T(vi)=1T(vj)=4

CLIQUE

CCN

Complexity44

Completing the CCN Graph Construction

T(s)

T(t)

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

Complexity45

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

Complexity46

Max Clique of G-clique and G-ccn

• Lemma:Max-Clique(G-clique) = Max-Clique(G-CCN)

• Corollary: – MAX-clique(G-clique) = m CCN(G-ccn)=q– MAX-clqiue(G-clique) < m CCN(G-ccn)>

q

Complexity47

Edge Origin Unique

T(s)

T(t)

First Observation: This edge comes

only from (s,t)

Complexity48

Triangle Consistency

Second Observation: A

triangle only come from a triangle

Complexity49

Clique PreservationCorollary:

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

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

Complexity50

Summary• We’ve seen how to show hardness of

approximation results in general, • and even proven several such using the

PCP theorem:– CLIQUE– CHROMATIC NUMBER