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Overview of the Previous Overview of the Previous LectureLecture

Gap-QS[O(n),,2||-1]

Gap-QS[O(1),,2||-1]

QS[O(1),]

Solvability[O(1),] 3-SAT

This will imply a strong PCP

characterization of NP

??

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AimAim

To reduce Gap-QS[O(n),,2||-1] to

Gap-QS[O(1),,2||-1].

p1(x1,...,xn) = 0

p2(x1,...,xn) = 0

...

pn(x1,...,xn) = 0

q1(x16,x21,x32) = 0

q2(x13,x26) = 0

q3(x1,xn,x5n) = 0

...

qm(x3,x4,x5,xn+1) = 0

each equation may depend on many variables

each equation depends only on a constant number of variables

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New VersionNew Version

Definition (Gap-QS*[D, ,]): Instance: a set of n conjunctions of constant

number of quadratic equations (polynomials) over . Each equation depends on at most D variables.

Problem: to distinguish between:

There is an assignment satisfying all the conjunctions.

No more than an fraction of the conjunctions can be satisfied simultaneously.

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Conjunctions Rather Than Just Conjunctions Rather Than Just EquationsEquationsAn example for a NO instance of Gap-QS*[1,Z2,½]

x = 0 y = 0

x = 0 y = 1

x = 1 y = 0

Nevertheless, we can satisfy more than a half of the

equations!!

Henceforth, we’ll assume the number of equations in all the

conjunctions is the same. Is this a restriction?

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RelaxationRelaxation

Claim: Gap-QS*[O(1),,2||-1] reduces to Gap-QS[O(1),,3||-1] (When || is at most polynomial in the size of the input).

Proof: Given an instance of Gap-QS*[O(1),,2||-1], replace each conjunction with all linear combinations of its polynomials.

Make sure that:1) The number of linear combinations over is polynomial.2) The dependency of each linear combination is constant.

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Correctness of the ReductionCorrectness of the Reduction

If the original system had a common solution, so does the new system.

Otherwise, fix an assignment to the variables of the system and observe the two instances:

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AnalysisAnalysis

2||-1

fraction of

unsatisfied

conjunctions

fraction of satisfied

conjunctions

polynomials originating

from the blue set

polynomials originating from

the pink set

all satisfied

2||-1

fraction of satisfied polynomials originating

from unsatifiable conjunctions ||-1

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RelaxationRelaxation

Yes instance of Gap-QS*[O(1),,2||-1] are transformed into Yes instances of Gap-QS[O(1),,3||-1].

No instance of Gap-QS*[O(1),,2||-1] are transformed into No instances of Gap-QS[O(1),,3||-1].

The construction is efficient when || is at most polynomial in the size of the input.

What proves the claim.

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AmplificationAmplification

Claim: Gap-QS[O(1),,3||-1] reduces to Gap-QS[O(1),,2||-1] (When ||>3 is at most polynomial in the size of the input).

Proof: Given an instance of Gap-QS[O(1),,3||-1], generate the set of all linear combinations of N polynomials.

Make sure that:1) The number of linear combinations over is polynomial.2) The dependency of each linear combination is constant.

A constant to be determined

later.

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Correctness of the ReductionCorrectness of the Reduction

Again if the original system had a common solution, so does the new system.

Otherwise, fix an assignment and observe a linear combination. – the probability all the N polynomials are

satisfied (3||-1)N

– if not all the polynomials are satisfied, the probability the combination is satisfied ||-1

– When ||>3 for N=4 we get the desired error probability.

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What Have We Done So Far?What Have We Done So Far?

Gap-QS[O(n),,2||-1]

Gap-QS[O(1),,2||-1]

Gap-QS*[O(1),,2||-1]

Gap-QS[O(1),,3||-1]

QS[O(1),]

Solvability[O(1),] 3-SAT

This will imply a strong PCP

characterization of NP ??

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New AimNew Aim

To reduce Gap-QS[O(n),,2||-1] to Gap-QS*[O(1),,2||-1].

p1(x1,...,xn) = 0

p2(x1,...,xn) = 0

...

pn(x1,...,xn) = 0

q1(xn,x2n)=0 q2(x1,x2)=0

q3(x13)=0 q2(x12,x234)=0

q4(xn)=0 q5(x1)=0 q4(x1)=0

...

qm(x3,x4,x5,xn+1)=0 qn(x1)=0

each equation may depend on many variables

each conjunction is composed of O(1) equations, which depend only on O(1) variables

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Rewriting the EquationsRewriting the Equations

Every quadratic polynomial P(x1,...,xD) can be written as

for some series of coefficients {cij}, where

(1) aij=xixj for any 1ijD

(2) ai0=xi for any 1iD

(3) a00=1

Dji ,0

ijij acdepends on the assignment

depends on the polynomial

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Representing Polynomials As Representing Polynomials As SumsSums For any given point xd we can associate

each aij with a point in Hd (H is some finite field, d=O(log|H|D)).

Now the evaluation of P in this point can be written as

A(h)(h)cdHh

P

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Representing Polynomials As Representing Polynomials As SumsSumsThe evaluation of every quadratic polynomial at

a point can be written as

where fP(h)=LDEcP(h)·LDEA(h).

dHh

P(h)f

Its total degree is at most 2·d·(|H|-1)

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Partial SumsPartial Sums

Define:

Hh

d1jj1Hh

d1fd1j

)h,...,h,a,...,f(a...)a,...,a(j,Sum

Sumf can be thought of as a polynomial of total degree at most

2·d2·(|H|-1)

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Verifying the Polynomial ZeroesVerifying the Polynomial Zeroes