sum check n.
Skip this Video
Loading SlideShow in 5 Seconds..
Sum Check PowerPoint Presentation
Download Presentation
Sum Check

Loading in 2 Seconds...

play fullscreen
1 / 9

Sum Check - PowerPoint PPT Presentation

  • Uploaded on

Sum Check. Where Quadratic-Polynomial get slim. Scheme. We precede the proof by a general scheme: Our starting point is the gap-QS instance [HPS] We need to decrease (to constant) the number of variables each quadratic-polynomial depends on

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
Download Presentation

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.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
sum check

Sum Check

Where Quadratic-Polynomial get slim


We precede the proof by a general scheme:

  • Our starting point is the 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 many new ones
  • We utilize the efficient consistent-reader above
  • Our test thus assumes the values for some preset sets of variables to corresponds to the point-evaluation of a low-degree polynomial (an assumption to be removed by plugging in the consistent reader)
representing a quadratic polynomial
Representing a Quadratic-Polynomial

Given a quadratic-polynomial P, over variables Yi, let us write P as follows:P(A) =i,j[1..m] (i,j) · A(yi) · A(yj)( (i,j) is the coefficient of the monomial yiyj )

Since we know how to deal with linear forms, let’s assume a set of variables yij, i,j [1..m],with the intention that A(yij) = A(yi) · A(yj), and have a special variable y11 so A(y11) = 1which lets us writeP(A) =i,j[1..m] (i,j) · A(yij)

checking sum over an ldf
Checking Sum over an LDF

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)

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

partial sums
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)

properties of sum
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

the sum check test
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])

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