11

The The PPCCPP starting point starting point

22

OverviewOverview

In this lecture we’ll present the In this lecture we’ll present the Quadratic SolvabilityQuadratic Solvability problem. problem.

We’ll see this problem is closely We’ll see this problem is closely related to related to PCPPCP..

And even use it to prove a (very very And even use it to prove a (very very weak...) weak...) PCPPCP characterization of characterization of NPNP..

33

Quadratic SolvabilityQuadratic Solvability

Def: (QS[D, Def: (QS[D, ]):]): InstanceInstance:: a set of a set of nn quadratic equations over quadratic equations over

with at most with at most DD variables each. variables each.ProblemProblem:: to find if there is a common solution. to find if there is a common solution.

or equally: a set of dimension Dtotal-degree 2 polynomials

Example: =Z2; D=1

y = 0

x 2 + x = 0

x 2 + 1 = 0

1 1

1

0

44

SolvabilitySolvability

A generalization of this problem:A generalization of this problem:

Def: (Solvability[D, Def: (Solvability[D, ]):]):

InstanceInstance:: a set of a set of n n polynomials over polynomials over with with at most at most DD variables. Each polynomial has variables. Each polynomial has degree-bound degree-bound n n in each one of the in each one of the variables. variables.

ProblemProblem:: to find if there is a common root. to find if there is a common root.

55

Solvability is Reducible to Solvability is Reducible to QS:QS:

y2 x2 + x2 t + tlz + z + 1 = 0w1

w1 = y2

w2

w2 = x2

w3

w3 = tl

w2

Could we use the same “trick” to show Solvability is reducible

to Linear Solvability?

the parameters (D,) don’t

change (assuming

D>2)!

66

QS is NP-hardQS is NP-hardLet us prove that QS is NP-hard by reducing Let us prove that QS is NP-hard by reducing 3-SAT 3-SAT

to it:to it:

(123)...(m/3-2m/3-1m/3) where each literal i{xj,xj}1jn

Def: (3-SAT):Def: (3-SAT): InstanceInstance:: a a 3CNF3CNF formula. formula. ProblemProblem:: to decide if this formula is satisfiable. to decide if this formula is satisfiable.

77

Given an instance of Given an instance of 3-SAT3-SAT, use the following , use the following transformation on each clause:transformation on each clause:

QS is NP-hardQS is NP-hard

xxii 1-x1-xii

xxii xxii

( ( ii i+1i+1 i+2i+2 ) ) Tr[ Tr[ ii ] * Tr[ ] * Tr[ i+1i+1 ] * Tr[ ] * Tr[ i+2i+2 ] ]

The corresponding instance of The corresponding instance of SolvabilitySolvability is the set is the set of all resulting polynomials (which, assuming the of all resulting polynomials (which, assuming the variables are only assigned Boolean values, is variables are only assigned Boolean values, is equivalent)equivalent)

88

QS is NP-hardQS is NP-hard

In order to remove the assumption we need to add In order to remove the assumption we need to add the equation for every variable the equation for every variable xxii::

xxi i * ( 1 - x* ( 1 - xi i ) = 0) = 0

This concludes the description of a reduction from This concludes the description of a reduction from 3SAT3SAT to to Solvability[O(1),Solvability[O(1),]] for any field for any field ..

What is the maximal degree of the resulting equations ?

99

QS is NP-hardQS is NP-hardAccording to the two previous reductions:According to the two previous reductions:

1010

Gap-QSGap-QS

Def: (Gap-QS[D, Def: (Gap-QS[D, ,,]):]):

InstanceInstance:: a set of a set of nn quadratic equations over quadratic equations over with with at most at most DD variables each. variables each.

ProblemProblem:: to distinguish between the following two to distinguish between the following two cases:cases:

There is a common solutionThere is a common solution

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

1111

Def:Def:

LLPCP[PCP[DD,,VV, , ]] if there is a polynomial time if there is a polynomial time algorithm, which for any input algorithm, which for any input xx, produces, produces a set a set of efficient Boolean functions over variables of of efficient Boolean functions over variables of range range 22VV,, each depending on at mosteach depending on at most DD variables variables so that:so that:

xxLL iff there exits an assignment to the variables, iff there exits an assignment to the variables, which satisfies all the functionswhich satisfies all the functions

xxLL iff no assignment can satisfy more than an iff no assignment can satisfy more than an --fraction of the functions.fraction of the functions.

Gap-QS and PCPGap-QS and PCP

Gap-QS[Gap-QS[DD,,,,]] PCP[ PCP[DD,,log|log|||,,]]

Gap-QS[D,,]

quadratic equations system

For each quadratic polynomial pi(x1,...,xD), add the Boolean

function i(a1,...,aD)pi(a1,...,aD)=0the variables

of the input system

values in

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Gap-QS and PCPGap-QS and PCP

Therefore, every language which is Therefore, every language which is efficiently reducible to Gap-QS[efficiently reducible to Gap-QS[DD,,,,] is ] is also in also in PCP[PCP[DD,,log|log|||,,]]..

Thus, proving Gap-QS[Thus, proving Gap-QS[DD,,,,] is ] is NP-hardNP-hard, , also proves the also proves the PCP[PCP[DD,,log|log|||,,] ] characterization of characterization of NPNP..

And indeed our goal henceforth will be And indeed our goal henceforth will be proving Gap-QS[proving Gap-QS[DD,,,,] is ] is NP-hardNP-hard for the for the best best DD, , and and we can. we can.

1313

Gap-QS[Gap-QS[nn,,,,2/|2/||| ]] is NP-hard is NP-hard

Proof:Proof: by reduction from by reduction from QS[O(1),QS[O(1),]]

p1 p2 p3 . . . pn

degree-2 polynomials Instance of Instance of QS[O(1),QS[O(1),]:]:

Satisfying assignment : Satisfying assignment : i

00 00 0 . . .0 . . . 0000 00 0 . . .0 . . . 00

Non-satisfying assignment : Non-satisfying assignment :

j 00 33 7 . . .7 . . . 0000 33 7 . . .7 . . . 00

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Gap-QS[Gap-QS[O(1)O(1),,]] is NP-hard is NP-hardIn order to have a gap we need an efficient degree-In order to have a gap we need an efficient degree-

preserving transformation on the polynomials so preserving transformation on the polynomials so that any non-satisfying assignment results in few that any non-satisfying assignment results in few satisfied polynomials:satisfied polynomials:p1 p2 p3 . . . pn

degree-2 polynomials

p1’ p2’ p3’ . . . pm’

Transformation:Transformation:

Non-Non-satisfying satisfying assignment assignment

: : j

00 22 4 . . . 34 . . . 300 22 4 . . . 34 . . . 3

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Gap-QS[Gap-QS[O(1)O(1),,]] is NP-hard is NP-hard

For such an efficient degree-preserving For such an efficient degree-preserving transformation transformation EE it must hold that: it must hold that:

n m

n m

1)E(0 ) 0

2) v 0 . (E(v),0 ) 1 2/

Thus Thus EE is an error correcting code ! is an error correcting code !

We shall now see examples of degree-preserving transformations We shall now see examples of degree-preserving transformations which are also error correcting codes:which are also error correcting codes:

1616

The linear transformation: The linear transformation: multiplication by a matrixmultiplication by a matrix

p1 p2 ... pn

c11 . . . c1m

. . .

. . .

. . .

cn1 . . . cnm

p

cc1 . . . 1 . . . ccmm

p•c1 . . . p•cm

=

scalarsscalars

polynomialspolynomials

inner productinner product

a linear combination of a linear combination of polynomialspolynomials

poly-time, if poly-time, if m=nm=ncc

1717

The linear transformation: The linear transformation: multiplication by a matrixmultiplication by a matrix

e1 e2 ... en

c11 . . . c1m

. . .

. . .

. . .

cn1 . . . cnm

c1 . . . cm

•c1 . . . •cm

=

the values of the polynomials under some the values of the polynomials under some assignmentassignment

the values of the new polynomials under the values of the new polynomials under the same assignmentthe same assignment

a zero vector if =0n

1818

What’s AheadWhat’s Ahead

We proceed with several examples for linear We proceed with several examples for linear error correcting codes:error correcting codes: Reed-Solomon codeReed-Solomon code Random matrixRandom matrix And finally even a code which suits our needs...And finally even a code which suits our needs...

1919

Using Reed-Solomon Codes Using Reed-Solomon Codes Define the matrix as follows:Define the matrix as follows:

nik

nikij ki

kja

0

0

)(

)(

using multivariate polynomials we can even

get =O(logn/||)

That’s really Lagrange’s formula

in disguise...

One can prove that for any One can prove that for any 00ii|||-1|-1, , (vA)(vA)ii is is P(i)P(i), , where where PP is the unique degree is the unique degree n-1n-1 univariate univariate polynomial, for which polynomial, for which P(i)=vP(i)=vii for all for all 00iin-1n-1..

Therefore for any Therefore for any vv the fraction of zeroes in the fraction of zeroes in vAvA is is bounded by bounded by (n-1)/|(n-1)/|||..

2020

Using a Random MatrixUsing a Random Matrix

Lem:Lem: A random matrix A random matrix AAnxm nxm satisfies w.h.p:satisfies w.h.p:

The fraction of zeros in the output vector

12

}0)(|{.0

Fm

Aviv in

2121

Using a Random MatrixUsing a Random MatrixProof:Proof: (by the probabilistic method) (by the probabilistic method)

Let Let vv00nnnn. .

Because the inner product of Because the inner product of vv and and

a random vector is random:a random vector is random:

1]0)[(Pr.1

FAvmi iA

Hence, Hence, |{i : (vA)|{i : (vA)ii = 0}| = 0}| (denoted (denoted XXvv) is a ) is a binomial random variable with parameters binomial random variable with parameters mm and and ||||-1-1..For this random variable, we can compute the For this random variable, we can compute the

probability probability Pr[ XPr[ Xvv 2m| 2m|||-1 -1 ] ] (the probability that (the probability that the fraction of zeros exceedsthe fraction of zeros exceeds 2| 2|||-1 -1 ))

2222

Using a Random MatrixUsing a Random Matrix

The Chernoff bound:The Chernoff bound:

For a binomial random variable with parameters For a binomial random variable with parameters mm and and ||||-1-1::

1

2

|F|4

mδ

1v 2eδ||X

m

1Pr

F

Hence:Hence:

1|F|4

m

1v 2e||2XPr

Fm

2323

Using a Random MatrixUsing a Random Matrix

Overall, the number of different vectors Overall, the number of different vectors vv is is | |||nn

Hence, according to the union bound, we can Hence, according to the union bound, we can multiply the previous probability by the multiply the previous probability by the number of different vectors number of different vectors vv to obtain a bound to obtain a bound on the probability :on the probability :

1|F|4

m

11v

n 2e||||2.X0vPr

FFm

And this probability is smaller then And this probability is smaller then 11 for: for: m=O(n|m=O(n||log||log||)|). Hence, for such . Hence, for such mm, a random , a random matrix satisfies the lemma with positive matrix satisfies the lemma with positive probability. probability.

The union bound:The probability for a union of events isSmaller then or equal to the sum of

Their probabilities

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Deterministic Construction Deterministic Construction

Define a random matrix Define a random matrix AAnxm nxm ::

Assume Assume =Z=Zpp. Let . Let k=logk=logppn+1n+1. (Assume w.l.o.g . (Assume w.l.o.g kkNN))

Let Let ZZppkk be the dimension be the dimension kk extension field of extension field of ..

Associate each row with Associate each row with 11iippk-1k-1

Hence, Hence, n=pn=pk-1k-1

Associate each column with a pair Associate each column with a pair (x,y)(x,y)ZZpp

kkZZppkk

Hence, Hence, m=pm=p2k2k

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Deterministic Construction Deterministic Construction

pk-1

p2k

<xi,y>

And define And define A(i,(x,y))=A(i,(x,y))= <x<xii,y>,y> (inner (inner product)product)

2626

AnalysisAnalysis For any vector For any vector vvnn, for any column , for any column (x,y)(x,y)ZZpp

kkZZppkk,,

1k1k p

1i

ii

p

1i

iiy)(x, y,xvy,xv(vA)

p2

p

p)p(ppp2k

1k1kkk1k

And thus the fraction of zeroes And thus the fraction of zeroes

The number of zeroes in The number of zeroes in vAvA where where vv00nn

x,y: G(x)=0x,y: G(x)0 <G(x),y>=0

k1k pp 1k1kk p)p(p +

2727

Summary of the ReductionSummary of the Reduction

Given an instance Given an instance {p{p11,...,p,...,pnn}} for for QS[O(1),QS[O(1),]],,

We found a matrix We found a matrix AA which satisfies which satisfies

vv00nn |{i : (vA)|{i : (vA)ii = 0}| /m < 2| = 0}| /m < 2|||-1-1

!!

Hence:Hence:{p{p11,...,p,...,pnn} } QS[O(1), QS[O(1),] ]

If and only if:If and only if: {p{p11A,...,pA,...,pnnA} A} Gap-QS[O(n), Gap-QS[O(n),,2|,2|||-1-1]]

This proves This proves Gap-QS[O(n),Gap-QS[O(n),,2|,2||-1]|-1] is NP-hardis NP-hard

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Hitting the RoadHitting the Road

This proves a This proves a PCPPCP characterization with characterization with D=O(n)D=O(n) (hardly (hardly a “local” test...).a “local” test...).

Eventually we’ll prove a characterization with Eventually we’ll prove a characterization with D=O(1)D=O(1) (([DFKRS][DFKRS]) using the results presented here as our ) using the results presented here as our starting point.starting point.