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Sum CheckSum Check

Where Quadratic-Polynomials get slim

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IntroductionIntroduction

Our starting point is a gap-QS instance [HPS]. We need to decrease (to constant) the number

of variables each quadratic-polynomial depends on.

We will add variables to those of the original gap-QS instance, to check consistency, and replace each polynomial with some new ones.

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IntroductionIntroduction

We utilize the efficient consistent-reader above.

Our test thus assumes the values for some preset sets of variables to correspond to the point-evaluation of a low-degree polynomial (an assumption to be removed by plugging in the consistent reader) .

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Representing aRepresenting a Quadratic-Polynomial Quadratic-Polynomial

Given: A polynomial P,

We write: P(A) = i,j[1..m] (i,j) · A(yi) · A(yj)((i,j) is the coefficient of the monomial yiyj ).

Note that P(A) means estimating P at a point A = (a1,…,am) m and A(yi) means the

assignment of ai to yi.

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PolynomialsPolynomials are hard, are hard, linear forms linear forms are easyare easy

Assume A(yij) = A(yi) · A(yj), for special the variables yij, i,j [1..m], and also A(y00) = 1

we can then write

P(A) = i,j[1..m] (i,j) · A(yij) .

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Checking Sum over an Checking Sum over an LDELDE

Next, we associate each pair ij with some xHd

P(A) = i1, …, idH (i1, …, id) · A(i1, …, id) .

Let ƒ: be low-degree-extension of · Aƒ = LDE() · LDE(A) .

LDE of both and A is of degree |H|-1 in each variable, hence of total degree r = d(|H|-1), which makes ƒ of degree 2r.

We therefore can write:

P(A) = i1, .., idH ƒ(i1, …, id) .

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We show next a test that, for any assignment for which some variables corresponds to a function ƒ: of degree 2r, verifies the sum of values of ƒ over Hd equals a given value.

Each local-test accesses much smaller number than |Hd| of representation variables, and a single value of ƒ.

We will then replace the assumption that ƒ is a low-degree-function by evaluating that single point accessed with an efficient consistent-reader for ƒ.

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Partial SumsPartial Sums

For any j[0..d] let

Sumƒ(j, a1,..,ad)=ij+1, .., idH ƒ(a1,..,aj,ij+1,..,id) .

That is, Sumƒ is the function that ignores all indices larger than j, and instead sums over all points for which these indices are all in H.

Proposition:Since ƒ is of degree 2r, Sumƒ is of degree 2rd (being the combination of d degree-r functions) .

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Properties of Properties of SumSumƒƒ

Proposition:For every a1, .., ad and any j[0..d] ,

Sumƒ ( d, a1, .., ad ) = ƒ(a1, .., ad) .

Sumƒ ( 0, a1, .., ad ) = i1, .., idH ƒ(i1, …, id) .

Sumƒ (j, a1,..,ad)= iHSumƒ(j+1,a1,..,aj,i,..,ad) .

Now we can assume Sumƒ to be of degree 2r (and later plug in a consistent-reader) and verify property 2, namely that for j=0, Sumƒ gives the appropriate sum of values of ƒ.

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The Sum-Check TestThe Sum-Check Test

Representation:One variable [j , a1, .., ad ] for every

a1, .., ad and j[0..d] Supposedly assigned Sumƒ (j, a1, .., ad )(hence ranging over ) .

Test:One local-test for every a1, .., ad that accepts an assignment A if for every j[0..d] ,

A([j,a1,..,ad]) = iH A([j+1,a1,..,aj,i,aj+2,..,ad]) .

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How good is it ?How good is it ?

The above test already drastically reduces the number of variables each linear-sum accesses to O(d |H|), nevertheless, we would like to decrease it to constant.

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Embedding ExtensionEmbedding Extension

As we've seen, representing an assignment by an LDF might not be enough.

We want a technique of lowering the degree of the LDF (even at the price of moderately increasing the dimension).

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Example:Example:

P(x) = X12 + X25

Y1 = X, Y2 = X3, Y3 = X9

X12 = Y3Y2, X25 = Y32Y2

2Y1

Pe(Y1,Y2,Y3) = Y3Y2 + Y32Y2

2Y1

[Note that the degree of Pe is much smaller then that of P, However it is defined over 3 rather than over ].

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Embedding-MapEmbedding-Map

Def: (Embedding-Map) Let d,t and r be natural numbers, d > t (r is the degree of the LDF that should be encoded). The embedding-map for parameters (t,r,d) is a manifold M: t d of degree bl-1, as defined shortly. Let l = d/t, and let b = r1/l.

Given X = (X1,…,Xt) t, M(X) is a vector Y d, structured by concatenating of sub vectors Y1,…,Yt whereas Yi = (Y(i,0),…Y (i,l-1)) = ((xi),(xi)b,(xi)b2,…,(xi)bl-1)

M(X) = Y = (Y1,…Yt,0,…0) d, (Y is padded with zeros if necessary, so it has d coordinates).

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Encoding an LDFEncoding an LDF

Given an [r,t]-LDF P over t we construct its proper extension Pe: Each term (Xi)j in P is represented by low-degree

monomial m(i,j): d over the coordinates (Y(i,0),…Y

(i,l-1)) which correspond to powers of xi.

m(i,j) =(Y) = (Y(i,0)) 0(Y(i,1)) 1…Y (i,l-1)) l-1

Wheres 1 2…l-1 is the base b representation of j, that is k=0…l-1 k = j, 0 k < b for all k.

Note that Pe is not just an encoding of P, but is actually an extension of the function that P induces on M(t).

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The degree of the Proper-The degree of the Proper-ExtensionExtension

The degree of Pe in each variable is no more than b-1.

Pe has d variables, therefore

deg(Pe) d(b-1) < db.This roughly equals drt/d, which is much

smaller than r ( = deg(P) ) as long as d is not very large.

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Linearizing-ExtensionLinearizing-Extension

If the parameters r and t (of an [r,t]-LDF) are small enough, namely (r +

r t) < d it is

possible to transform this [r,t]-LDF to a linear LDF over d.

This technique, adopted from [ALM+92], is similar to the embedding-extension technique.

This is the (more) formal representation os what we saw in Polynomials are hard, linear forms are easy].

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Linearizing-ExtensionLinearizing-Extension

Take for example the LDF: P(x,z) = 5x2z3 + 2xz + 3x

We replace it byPe(Yx2z3,Yxz,Yx) = 5Yx2z3 + 2Yxz + 3Yx

We now have three coordinates (instead of two) but linear function.

Pe Encodes P in the sense that if we assignYx2z3 = x2z3, Yxz = xz and Yx = x thenPe(Yx2z3,Yxz,Yx) = P(x,z)

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Linearizing-MapLinearizing-Map

For the general case:Def: Let d,t and r be natural numbers, Let S = {m1,…ms} be the set of monomials

of degree r over t variables. The linearizing-map for parameters (t,r,d) is

function M: t d defined by M(x1,…xt) = (Ym1,…,Yms,0,…,0) where Ymi = mi(x1,…,xt)

(padding with zeros to have d coordinates).M(Ft) is therefore a manifold in Fd of degree r.

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Show that as M has only d coordinates, we must have |S| = (r +

r t) d .

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Encoding and DecodingEncoding and Decoding

Consider an [r,t]-LDF P over t. The proper-extension of P is a linear LDF over d such that Pe

o M = P To construct Pe, we write P as a linear

combination of monomials P = i=1…s i mi.

Pe is then defined by Pe(Ym1,…,Yms) = i=1…s i Ymi. The other direction (Decoding from a linear LDF

over d to an LDF over t) is left to the reader.

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The Composition-The Composition-Recursion Consistent Recursion Consistent ReaderReader

We now have sufficient tools to construct the final consistent reader, which will access only constant number of variables.

We do so by recursively applying the (hyper)-cube-vs.-point reader upon itself.

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The Composition-The Composition-Recursion Consistent Recursion Consistent ReaderReader

The consistent-reader asks for a value of a (hyper-)cube (supposedly a LDF ) P. It then bounds it to a point x and compare the result with the value on x.

This will be done by “compressing” this LDF (using the embedding-extension) and sending it to another consistent reader.

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Embedding M

ap

Ft

Fd

t < d

The hyper-cube (affine subspace)

The manifold representing the hyper-cube in the higher dimension space

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The Composition-The Composition-Recursion Consistent Recursion Consistent ReaderReader

Every (hyper-)cube induces a different embedding-map, followed by a different consistent-reader. We will therefore get a tree-like structure.

Of course this tree-like structure should be constructed in polynomial time, and be of polynomial size. It will follow from the height (the recursion depth) being bounded by a constant.

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Map

1

Map

2 Map 3Map 8785

…

A different map for every different affine-subspace

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Contraction - ExpansionContraction - Expansion

For every cube (affine subspace) we apply embedding-extension.

After several applications of this contraction - expansion procedure, we get an LDF of degree small enough for the application of linearizing - extension.

(When we say “we get an LDF”, we mean – if the original function was a LDF).

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Embedding

Extension maps

Embedding

Extension maps

Linearization maps

…

O(1)

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The Composition-The Composition-Recursion Consistent Recursion Consistent ReaderReader

We’ll then have a representation of the original LDF by linear functions.

Which means that a (hyper)-cube variable (in the new space) can be represented by a constant number of scalars.

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The Composition-The Composition-Recursion Consistent Recursion Consistent ReaderReader

Each cube-vs.-point reader adds a constant number of scalars (3).

The depth of the recursion was constant.Hence the composition recurtion

constisten reader access only constant number of scalar.

Thus proving gap-QS PCP[O(1), , 2/]

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SummarySummary

We started with [HPS]: gap-QS[ n, , 2||-1 ] is NP-hard As long as || nc for some c>0.

We recursively applied the (hyper-)cube-vs.point consistent reader with the Embedding-Extension and the Linearizaion-Extension techniques to construct the CR-Consistent-reader which access only constant numer of variables.

Thus we proved thatQS[ O(1), , 2/ ] is NP-hard,

as long as log|| logn for any constant < 1.