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

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

4. 4 The sum check lemma

5. 5 Definitions Def: Given a finite field  and a positive parameter d, we define the corresponding domain as F i ={ x k | 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. 6 Definitions Def: (Gap-QS cons [D, ,  ]) Instance: A set of domains F 1,...,F k and n quadratic equations over . Each equation depends on at most D variables, some of them belong to certain 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.

7. 7 Definitions x 1 2 + 2 x 2 + x 3 +... + 3 x n = 0 x1x1 xnxn (1,0,1,2) (0,0,1,1) x3x3 (3,3,3,1) equation domain promise Variables belong to certain domains. The promise is that the values to a domain’s variables form a low-degree polynomial

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

9. 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 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 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 (chapter 2) utilizing the efficient consistent-readers we have seen The consistency will be checked later on in the proof (chapter 2) 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) 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. 10 Representing a Quadratic-Polynomial Given a quadratic-polynomial P, over variables Y i, let us write the value of P in a certain point in the space as follows: Let us convert the polynomial to linear form: let’s assume a set of variables y ij, i,j  [1..m], with the intention that A(y ij ) = A(y i ) · A(y j ), and the special case where A(y ii ) = A(y i ) which lets us write: A is an assignment to the variables (  (i,j) is the coefficient of the monomial y i y j )

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

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

13. 13 Low Degree Extension (LDE) Def: (low degree extension): Let  : H d  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  : F d  F which satisfies:  LDE  agrees with  on H d (extension).  The degree-bound of LDE  is |H| in each variable (low degree).

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

15. 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. 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 H d equals a given value. Each local-test accesses much smaller number than |H d | 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. 17 Partial Sums For any j  [0..d] define: That is, Sum ƒ 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: Sum ƒ is of degree 2rd Proof: Immediate since ƒ is of degree 2r and Sum ƒ is the linear combination of d degree-r functions

18. 18 Partial Sums Proposition: For every a 1,.., a d   and any j  [0..d] : Proof:Homework...

19. 19 The Sum-Check Test Now we can assume Sum ƒ 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, Sum ƒ gives the appropriate sum of values of ƒ: Representation: One variable  [j, a 1,.., a d ] for every a 1,.., a d   and j  [0..d] Supposedly assigned Sum ƒ (j, a 1,.., a d ) (hence ranging over  ) Test: One local-test for every a 1,.., a d   ; one which accepts an assignment A if for every j  [0..d]: A(  [j,a 1,..,a d ]) =  i  H A(  [j+1,a 1,..,a j,i,a j+2,..,a d ])

20. 20 The Sum-Check Test Analysis Define the following function using the Sum ƒ function defined previously, for a certain (a 1,.., a d ): Using this notation, the sum-check test is to verify that for a certain (a 1,.., a d ): Define now a new function:

21. 21 The Sum-Check Test Analysis Claim: If for a certain (a 1,.., a d ): Proof:Homework… Then for a random uniform (i 1,.., i d ): Hence, the sum-check test is w.h.p a suitable replacement for the original local test and relies on less variables.

22. 22 The Sum-Check Test Analysis The above test already reduces the number of variables each local test accesses from O(H d ) to O(d |H|). The above test already reduces the number of variables each local test accesses from O(H d ) to O(d |H|). However, we have introduces a consistency assumption (that the functions f are low degree) 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. 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.

23. 23 The Sum-Check Test made linear Recall out goal was to verify that for a certain (a 1,.., a d ): Now we can apply the Hadamard code on and obtain all linear combinations: (the number of combinations is: F d ) The new test is to pick randomly three combinations and verify linearity:

24. 24 The Sum-Check Test made linear Completeness: if the sum check test is positive, then all of the delta are zero, and the linear test succeeds with probability 1. On the other hand, if the sum check test fails then as we saw: And the test fails with high probability. Notice that the number of variables per local test was reduced to 3 !