1. PCP Characterization of NP:
Proof chapter 1

2. PCP Proof Map In previous lectures: 3SAT Solvability QS
Clauses to polynomials
Solvability
Introducing new variables QS
Error correcting codes
Gap-QS[O(n),,2||-1]

3. PCP Proof Map In following lectures: Gap-QS[O(n),,2||-1]
Sum Check
quadratic equations of constant size with consistency assumptions
Gap-QScons[O(1),,2||-1]
Consistent Reader
conjunctions of constant number of quadratic equations, whose dependencies are constant.
Gap-QS*[O(1),O(1),,||-]
Error correcting codes
Gap-QS[O(1),,2||-1]

4. Today's road
The sum check lemma

5. Definitions
Def: Given a finite field  and a positive parameter d, we define the corresponding domain as
Fi={ xk | kd }.
The variables in the domain range over .
Def: An assignment f:d to a domain is said to be feasible if it’s a degree-r polynomial.
Def: An assignment f:d to a domain is said to be good if it’s a degree-s polynomial.
(sr and d are some global constants.)

6. Definitions YES NO Def: (Gap-QScons[D,,])
Instance: A set of domains F1,...,Fk and n quadratic equations over . Each equation depends on at most D variables, some of them belong to some domains.
Problem: to distinguish between:
There is a good assignment satisfying all the equations.
No more than an  fraction of the equations
can be satisfied simultaneously by a feasible assignment.
YES NO

7. Definitions x1 2 + 2 x2 + x3 + ... + 3 xn = 0 equation x1 x3 xn
promise
Variables belong to some domains.
The promise is that the values to a domain’s variables form a low-degree polynomial
(3,3,3,1)
(0,0,1,1)
(1,0,1,2)
domain

8. The Sum-Check Lemma Lemma (Sum-Check):
Gap-QS[O(n),,2/||] is efficiently reducible to Gap-QScons[O(1),,2/||].

9. Overview We precede the proof by a general scheme:
Our starting point is the gap-QS instance, and 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
The consistency will be checked later on in the proof utilizing the efficient consistent-readers we have seen
Our test 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)

10. Representing a Quadratic-Polynomial
Given a quadratic-polynomial P, over variables Yi, let us write the value of P in a certain point in the space as follows:
A is an assignment
to the variables
( (i,j) is the coefficient of the monomial yiyj )
Let us convert the polynomial to linear form: let’s assume a set of variables yij, i,j [1..m], with the intention that A(yij) = A(yi) · A(yj), and the special case where A(yii) = A(yi) which lets us write:

11. Representing a Quadratic-Polynomial
Next, we associate each variable yij with some point xHd. Define the following one-to-one function:
Notice that:
As a consequence we can define:
For a value in xHd without a source define:

12. Representing a Quadratic-Polynomial
Using the new definitions we can write:
Where  , A are functions:

13. Low Degree Extension (LDE)
Def: (low degree extension): Let  : Hd  H be a string (where H is some finite field).
Given a finite field F, which is a superset of H, we define a low degree extension of  to F as a polynomial LDE : Fd  F which satisfies:
LDE agrees with  on Hd (extension).
The degree-bound of LDE is |H| in each variable (low degree).

14. Using the LDE Let ƒ be a low-degree-extension of  · A:
Notice that f is define by all  · A

15. Using the LDE We therefore can write:
Notice that 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 total degree 2r.

16. What’s ahead
We show next a test that uses a small number of variables:
For any assignment for which some variables corresponds to a function ƒ of degree 2r, the test 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.
Later on we will replace the assumption that ƒ is a low-degree-function by evaluating that single point accessed with an efficient consistent-reader for ƒ

17. Partial Sums
For any j[0..d] define:
That is, Sumƒ is the function that does not vary on the first j variables, and sums over all points for which the rest of the variables are all in H
Proposition: Sumƒ is of degree 2rd
Proof:
Immediate since ƒ is of degree 2r and Sumƒ is the linear combination of d degree-r functions

18. Partial Sums Proposition: For every a1, .., ad   and any j[0..d] :
Proof:
Homework...

19. The Sum-Check Test
Now we can assume Sumƒ to be of degree 2r (this is the consistency assumption – to be verified later on with a consistent reader) and verify property 2, namely that for j=0, Sumƒ gives the appropriate sum of values of ƒ:
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  ; one which 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])

20. The Sum-Check Test
The above test already drastically reduce the number of variables each linear-sum accesses to O(d |H|)
However, we have introduces a consistency assumption (that the functions f are low degree)
We shall now reduce the number of variables accessed to a constant O(1) (and strengthen the consistency assumption on the way) using the exact same method.

21. The Sum-Check Test (repeated)
Define the following function using the Sumƒ function defined previously, for a certain (a1, .., ad):
Using this notation, recall the sum-check test was to verify that for a certain (a1, .., ad):
Define now a new function:

22. The Sum-Check Test (repeated)
Claim: If for a certain (a1, .., ad):
Then for a random uniform (i1, .., id):
Proof:
Homework…
Note, from a function f with O(Hd) variables we’ve received a function T with only O(rd) variables.

23. The Sum-Check Test (repeated)
Claim: If for a certain (a1, .., ad):
Then for a random uniform (i1, .., id):
Proof:
Homework…
Using this fact we can now devise another local test that relies on less variables…