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Hardness Result for MAX-3SAT

This lecture is given by:

Limor Ben Efraim

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3Sat CNF formula: a formula of n variables (xi) given by m clauses (Cj), each clause contains exactly 3 literals.

Max-3Sat: Given: 3Sat CNF formula.Goal: Find an assignment x that maximize the number of satisfied clauses.

Hastard (1997), Khot(2002): For any constant > 0 , it is NP Hard to distinguish whether a MAX-3SAT instance is satisfiable or there is no assignment that satisfies ⅞+ fraction of the clauses.

Fact: Any random assignment satisfies 7/8 from the clauses.

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Max-3Lin-2: Given: a system of linear equations over Z2, exactly

3 variables in each equation.Goal: Find an assignment that maximize the

number of satisfied equations.We saw

MAX-3Lin-2Gap(½+,1-)

MAX-3SatGap(⅞+,1-)

4 gadget

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Label Cover - Reminder

Each vertex V is a set of u variables.

Each vertex W is a set of u clauses

When an assignment to LC satisfies the edge (V,W)?

If satisfies W, and (V) is a restriction of (W).

Bipartite graph

… …

Constraints Functions

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type-0 block A set of Tu clauses and u variables.

type-1 block A set of (T+1)u clauses.

0 is the family of all type-0 blocks

1 is the family of all type-1 blocks

Given W 2 1 (V 2 0):MW (MV) – the set of all satisfying assignments to W (V).

Given W 2 1 (V 2 0), how many satisfying assignments there are ?

Answer: At most 7(T+1)u (2u7Tu) values.

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When a type-0 block V is a sub block of type-1 block W ?

V,W:MW ! MV is the operation of taking a sub assignment.

If we can replace u clauses {ci| i=1,2,…u} in W by u variables {xi| i=1,2,…,u} in V such that the variable xi is in the clause ci for 1 · i · u.

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Label Cover +

(V,W) 2 E(LG+) if V is a sub-block of W.

Each vertex V is in 0

Each vertex W is in

When an assignment to LC+ satisfies the edge (V,W)?

If satisfies both V and W, and V,W((W))=(V).

Bipartite graph

… …

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Theorem: It is NP Hard to distinguish between the following two cases:

YES: There is an assignment that satisfies every edge in the graph

NO: No assignment can satisfies more that 2-(u)

of the edges

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Lemma: W 2 1. Let ,’ 2 MW. If V is a random sub-block of W then

PrV [V,W()=V,W(’)] · 1/T

Proof: ,’ differ at least on one clause. For a choice of a random sub-block V, one replaces at random u clauses out of (T+1)u clauses in W. With probability · 1/T each different clause is replaced.

Corollary: W 2 1. Let 0 µ MW and 2 . If V is a random sub block of W thenPrV [8 ’ 2 , ’ , V,W() V,W(’)] ¸ 1-||/T

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The Smoothness Lemma: For any set 0 µ MW

Proof:

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Label Cover + with smoothness property.

3Sat CNF formula

Our Plan

If there is a satisfiable assignment to the Label Cover+

The 3Sat CNF formula is satisfiable.

If the Label Cover+ is 2-(u) satisfiable

The 3Sat CNF formula is · ⅞+ 8 satisfiable.

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Long Code

FV is the set of all functions f:MV! {-1,1}.

FW is the set of all functions f:MW! {-1,1}.

Long code of an assignment x 2 MV is the mapping A:FV ! {-1,1} where A(f)=f(x).

Size: 22|V|

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Building…

VW

… …

AW

AV

We will replace each vertex W (V) in a set of boolean variables, a variable for each bit of AW (AV), the long code of W (V). (W,f) ! XW,f. (V,f) ! XV,f.

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Building - Continue

What are the clauses ?

To answer this, we define a test for each (W,V) 2 E(LC+)

V is a sub block of W

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The test

Pick a block W 2 1

Pick a random sub-block V of W.

Let = V,W

Let A,B be the supposed long codes of supposed satisfying assignment to the blocks V,W resp.

Pick a function f:MV ! {-1,1} with the uniform probability.

Pick a function g:MW ! {-1,1} with the uniform probability.

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Accept unless A(f)=B(g)=B(h)=1.

Define a function h:MW ! {-1,1} independently 8 y 2 MW

if f((y))=1 then h(y)=-g(y)if f((y)=-1 then:

Equivalent: Accept if the clause XV,f Ç XW,g Ç XW,h is satisfiable

{0,1} !{1,-1}x)=(-1)x

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Completeness

This test has perfect completeness. If f(y|V)=1, by definition one of g(y),h(y) will be -1

B(g)=-1 or B(h)=-1.

If f(y|V)=-1, we have A(f)=-1.

How many clauses we got ?

Polynomially in n for constant u !!!

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Fourier Analysis

Reminder: FV is the set of all functions f:MV! {-1,1}.

Orthonormal basis to FV is:

(f)= x 2 f(x) 8 µ {-1,1}V. v=|V|.

The inner product of 2 functions A,B is

(A,B)=2-2v f 2 FV A(f)B(f)

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Fourier Analysis-Continue

Lemma: For any f,g 2 FV and , µ {-1,1}V:

. (fg)=(f)(g)

2. (f)(f)=M (f).

Lemma:

1. Ef[(f)]=0 8 µ {-1,1}V , ;.

2. Ef[A(f)]=0

Parseval’s Formula:

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Soundness

The acceptance criteria can be written as:

Ef,g,h[ 1-⅛ (1+A(f)) (1+B(g)) (1+B(h)) ] =

⅞ - ⅛ [ Eg,h(B(g)B(h)) + Ef,g,h ( A(f)B(g)B(h) ) ]

We will show that each term · O()

For the rest of the proof fix T=

(f,g),(f,h) are independent

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Eg,h,[B(g)B(h)]

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sx is the number of y 2 s.t. y|V=x

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Pr[sx=1] ¸ 1-

24e(x) · 1-

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Ef,g,h,[A(f)B(g)B(h)]

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Lemma: For any ,

Proof: the left size is equal to

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Cauchy-Schwartz inequality

E[X]2 · E[X2]

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Cauchy-Schwartz inequality

Goal: to see that this is bounded by O()

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Reminder

Label Cover+ 3Sat CNF Formula

Given assignment to the 3Sat-CNF formula

We can find an assignment to the Label Cover+

Goal: to see that if the assignment satisfies ¸ ⅞+O( of the clausesThen we can find an assignment that satisfies ¸ 2-(u) of the edges.

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The Folding Mechanism

Goal: To make sure that A(f)=-A(-f) 8 f.

Action: Given A: FU ! {-1,1}, define A’: for every pair (f,-f) selecting one of (f,-f).

IF: f is selected (A’(f),A’(-f))=(A(f),-A(f))IF: -f is selected (A’(f),A’(-f))=(-A(-f),A(-f))

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We will assume all our long codes are of the folding mechanism !!!

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We will create an assignment to Label Cover+ based on

By the folding lemma , ;

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For the choicePrevious theoremon Label Cover+

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Summary

Label Cover +Gap(2-(u),1)

Long Code + Testing

Max-3SatGap(⅞+,1)

FIN