1 / 31

Sum Check

Sum Check. Where Quadratic-Polynomials get slim. Introduction. Our starting point is a gap-QS instance [HPS]. We need to decrease (to constant) the number of variables each quadratic-polynomial depends on.

bevan
Télécharger la présentation

Sum Check

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Sum Check Where Quadratic-Polynomials get slim

  2. Introduction • Our starting point is a gap-QSinstance [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.

  3. Introduction • 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) .

  4. Representing a 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) mand A(yi) means the assignment of ai to yi.

  5. Polynomials are hard, linear forms are 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 writeP(A) =i,j[1..m] (i,j) · A(yij) .

  6. Checking Sum over an LDE Next, we associate each pair ij with some xHdP(A) =i1, …, idH (i1, …, id) · A(i1, …, id) . Let ƒ: below-degree-extension of  · A ƒ = 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, .., idH ƒ(i1, …, id) .

  7. 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 ofƒ. 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.

  8. Partial Sums For any j[0..d] letSumƒ(j, a1,..,ad)=ij+1, .., idH 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) .

  9. Properties of Sumƒ Proposition:For every a1, .., ad   and any j[0..d] , • Sumƒ ( d, a1, .., ad ) = ƒ(a1, .., ad) . • Sumƒ ( 0, a1, .., ad ) = i1, .., idH ƒ(i1, …, id) . • Sumƒ (j, a1,..,ad)= iHSumƒ(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.

  10. The Sum-Check Test Representation:One variable [j , a1, .., ad ]for everya1, .., 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]) = iH A([j+1,a1,..,aj,i,aj+2,..,ad]) .

  11. 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.

  12. Embedding 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).

  13. Example: P(x) = X12 + X25 Y1 = X, Y2 = X3, Y3 = X9 X12 = Y3Y2, X25 = Y32Y22Y1 Pe(Y1,Y2,Y3) = Y3Y2 + Y32Y22Y1 [Note that the degree of Pe is much smaller then that of P, However it is defined over 3 rather than over ].

  14. Embedding-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,…,Ytwhereas 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).

  15. Encoding an LDF • Given an [r,t]-LDF P over t we construct its proper extension Pe: • Each term (Xi)jin 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).

  16. The degree of the Proper-Extension • 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.

  17. Linearizing-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].

  18. Linearizing-Extension • Take for example the LDF: P(x,z) = 5x2z3 + 2xz + 3x • We replace it by Pe(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 assign Yx2z3 = x2z3, Yxz = xz and Yx = x then Pe(Yx2z3,Yxz,Yx) = P(x,z)

  19. Linearizing-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.

  20. Show that as M has only d coordinates, we must have |S| = (r+r t)  d .

  21. Encoding 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. • Peis 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.

  22. The Composition-Recursion Consistent Reader • 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.

  23. The Composition-Recursion Consistent Reader • 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.

  24. Ft The manifold representing the hyper-cube in the higher dimension space The hyper-cube (affine subspace) Embedding Map t < d Fd

  25. The Composition-Recursion Consistent Reader • 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.

  26. A different map for every different affine-subspace Map 2 Map 1 Map 3 Map 8785 …

  27. Contraction - 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).

  28. Embedding Extension maps Embedding Extension maps O(1) … Linearization maps

  29. The Composition-Recursion Consistent Reader • 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.

  30. The Composition-Recursion Consistent Reader • 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/]

  31. Summary • We started with [HPS]: gap-QS[ n, , 2||-1 ] is NP-hard As long as ||  ncfor 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 that QS[ O(1), , 2/ ] is NP-hard,as long as log||  logn for any constant < 1.

More Related