1

H • AA • P

R • P

D • R

N • O

E • X

S • I

S • M

• A

O • T

F • I• O

• N

Goal:

• Show some approximation problems NP-hard

Plan:

• How to show inapproximability?• Probabilistically Checkable P

roofs (PCP)

• CLIQUE, COLORING

2

3

Promise Problems – Ilustrated

*

• Must accept

Good inputs

• Must reject

Bad inputs

• Whatever

Other inputs

EG

Is there an assignment satisfying 90% of a

3SAT, assuming

one can satisfy 80%?

Gap problems? A special

case

• Best solution according to optimization parameter

Objective

Optimization

4

• Find a solution within some factor of optimal

Approximation

problem

CLIQUE

3SAT

S.C.

min/max

max

max

min

instance

graph

3CNF

family

solution

clique

assign.

cover

parameter

|clique|

#clause

|cover|

optimal

algorithm

• C, efficient h-approximation for A resolves gap-A[C, hC]:

Observation:

<CYes

>hCNo

• Parameter is

In optimal Solution

GAP

Gap-A[C,hC] (minimization)

• Whatever

GAP

5accept if <hC

• 3-CNF

Instance:

• Maximum, over assignments, of fraction of satisfiable clauses:

Gap Problem: [7/8+, 1]

6

Gap-3SAT

Yes

• =1

No

• < 7/8+

EG( 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 clauses

Why 7/8?

• E3CNF (clauses: exactly 3 independent literals) assignment that satisfies ≥7/8 of clauses

Claim:

Proof:

7

7/8

• How many does an assignment satisfy expectedly?E

• clause Ci, let Yi be a 0/1 variable: “is Ci satisfied?” Yi

• Yi’s expectation: E[Yi] = 7/8E Yi

• E[ Yi] = E[Yi] = m7/8 (m = #clauses)E

• assignment with at least that number satisfied

• >0, Gap-3SAT[7/8+,1] is NP-hard

Theorem (PCP):

•Cuius rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet

Proof:

8

PCP Characterization of NPi.e.

LNP, Karp reduction f to

3CNF:xLf(x)3SAT

xLmax satisfy <7/8+ of f(x)

Approximating max-3SAT to within which factor is thus proven NP-hard?

Verify witness

• Choose only O(1) bits to read (randomly)

9

Probabilistic Checking of Proofs

_ _

_ _

_ -

_ _

_ _

a a

b a

b -

-

InputW

ork

Witness

for gap-3SAT: pick O(1) clauses

Instance:

• G=(V,E) (of m IS’s each of size 3)

Gap Problem:[(7/8+)/3 ,1/3]

• Parameter: fractional size of largest CLIQUE:

Yes

• ≥ 1/3

No

• < (7/8+)/310

Gap-CLIQUE

Theorem:

• Gap-CLIQUE [(7/8+)/3,1/3] is NP-hard

Corollary:

• It is NP-hard to approximate MAX-CLIQUE to within 7/8+

How does one prove such an NP-hardness theorem?Reduce from gap-3SAT?

Gap Preserving Reduction

*

*

f

GAP

• Whatever11

3SAT

CLIQUE

G,m

12

SAT p CLIQUE

1 vertex for 1 occurrence• within triplets. =.inconsistency non-edge

m= number of clausesA clique of size l an assignment satisfying ≥l clauses

13

Color CoordinationParty

• A set of partygoers; each pair of friends agree on possible pairs of colors

Solution: MAXV

• Invite some partygoers --- set colors s.t. all pairs OK

Optimize:

• How many invited

Solution: MAXE

• Assign a color to every partygoer

Optimize:

• Number of OK pairs

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Constraints` GraphCSG Input:

• U = (V, E, , ) - : EP[2]

Solution: MAXV

• A: V{} s.t. (u,v)E: A(u),A(v) (A(u),A(v)) (u,v)

Optimize:

• Pr[A(u)]

Solution: MAXE

• A: V

Optimize:

• Pr[(A(u),A(v))(u,v)]

EGCLIQUEIS3SATV.C.(G)

EGMAX-CUT(G)

Instance:

• U=(V,E, [k], )

Gap Problem:[, ]

• Fractional size of largest all-consistent assignment:

Yes

• ≥

No

• < 15

gapV-kCSG[, ]

Observe:

• GapV-3CSG[(7/8+),1] is NP-hard

Theorem:

• >0 k=k()gap-IS[/k,1/k]NP-hard

How does one prove kCSG is NP-hard?Reduce from gap-3SAT?

3SAT is a ‘special case’

Proof: via kCSG

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CSGClaim [CSGIS]:

• gapV-kCSG[, 1] L gap-IS[/k, 1/k]

Proof:

• V’=V,• ij ((u,i),(u,j))E’• (i,j)(u,v)} ((u,i),(v,j))E’•I = (u, A(u))Yes

•A(u) = i | (u, i) INo

1 in each clique

u there is ≤1 such i

GapV-3CSG[(7/8+),1] is NP-hard

gapV-kCSG[, 1] L gapV-klCSG[l, 1]

gapV-kCSG[, 1] L gap-IS[/k, 1/k]

Whatever constraints, I can always invite 1% of partygoers

In that case, I can do 99%, whatever the constraints are!

How?

‘Invent’ a guest for every set of l real guests

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Party on

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AmplifyTheorem:

• gapV-kCSG[, 1] L gapV-klCSG[l, 1]

Proof:

• V’=Vl, ’=l

• E’: (u’, v’) s.t. uu’,vv’ s.t. u=v or (u, v)E

• ’ forbids: A(u’)u A(v’)u or (A(u’)u, A(v’)v)(u, v)

•Natural, consistent assignmentYes

•Color uu' by A(u’)u if any such u’V’ is colored. u’V’ is colored only if all uu’ are coloredNo

GapV-3CSG[(7/8+),1] is NP-hard

gapV-kCSG[, 1] L gapV-klCSG[l, 1]

gapV-kCSG[, 1] L gap-IS[/k, 1/k]

Constraints depend only on difference I can invite of partygoers

In that case, I can do the same for general constraints

And what does it get you?

Show approximate coloring of a graph is NP-hard

19

Settle differences

20

If colors are (mod q)

constraint (u, v) depends only on A(u)-A(v) (mod q)

Many symmetric colorings

21

qCSGqCSG Instance:

• U = (V, E, , ) : EP[2] is define by

: E P{[q]} thus (u, v) = {(i,j) | i-j mod q (u, v)}

Theorem:

• gapV-kCSG[, 1] L gapV-qCSG[, 1]q = (nk)5

P{[q]} = all subsets of {0..q-1}

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q-ColoringCorollary:

• gap-[q, q/] is NP-hard

Proof:

• Apply same CSGIS reduction

• Consistent A d Ad(v)=A(v)+d (mod q) consistentYes

•vertexes partitioned into IS’s of fractional size /q – how many?No

q disjoint IS covering all

Lemma:• For q=|X|5, can efficiently

construct T:X[q] s.t. x1,x2,x3,x4,x5,x6,T(x1)+T(x2)+T(x3) T(x4)+T(x5)+T(x6) (mod q) {x1,x2,x3}={x4,x5,x6}Proof:

• Incrementally assign values to members of U

23

Triplets` Unique T

Corollary:

• T is unique for pairs and for single vertexes as well

Proof:

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

• T(x)T(y) (mod q) x=y

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Proof:

• q=(|V|||)5

• (u, v) = {T(u,i)-T(v,j) mod q | (i,j)(u,v)}

Pairs: assuming A’ colors vertexes u,v

• !d s.t. A’(u)=T(u,i)+d & A’(v)=T(v,j)+d for (i,j)(u,v)

Triplets: assuming A’ colors vertexes u,v,w

• dA’(u,v)=dA’(v,w) as A’(u)-A’(v) + A’(v)-A’(w) + A’(w)-A’(u)0 (mod q)

General: assuming A’ any partial coloring

• If dA’ is not everywhere the same, inconsistent triplet (u,v,w)

qCSG

denote: dA’(u,v) = that unique d

•for good A: A’(u) = T(u, A(u)) goodYes•for good A’: !d A(u)=T-1(u,A’(u)-d) goodNo

Assume a complete graph – add trivial constraints

Question:

• Is it in P or NP-hard?

Constraints

• Colors` permutations

In optimal Solution

• Pr[(A(u),A(v))(u,v)]<No

>1-Yes

GAP

Gap-UG[, 1-] .

GAP

• Whatever

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Note:

• gap-Max-Cut[1-1.01, 1-] NP-hard

<1-No

>1-Yes

Optimal cut

• Pr[A(u)A(v)]GAP

Gap-MC[1-, 1-] .

GAP

• Whatever

26

WWindex

27

PCP PCP Theorem

Clique SAT

3SAT Vertex Cover

Coloring

CNF

NP-Hard

Interactive Proof System