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2 Overview of the Previous Lecture Gap-QS[O(n), ,2|  | -1 ] Gap-QS[O(1), ,2|  | -1 ] QS[O(1),  ] Solvability[O(1),  ] 3-SAT This will imply a strong PCP characterization of NP ??

3 Aim To reduce Gap-QS[O(n), ,2|  | -1 ] to Gap-QS[O(1), ,2|  | -1 ]. p 1 (x 1,...,x n ) = 0 p 2 (x 1,...,x n ) = 0... p n (x 1,...,x n ) = 0 q 1 (x 16,x 21,x 32 ) = 0 q 2 (x 13,x 26 ) = 0 q 3 (x 1,x n,x 5n ) = 0... q m (x 3,x 4,x 5,x n+1 ) = 0 each equation may depend on many variables each equation depends only on a constant number of variables

4 New Version Definition (Gap-QS*[D, ,  ]): Instance: a set of n conjunctions of constant number of quadratic equations (polynomials) over . Each equation depends on at most D variables. Problem: to distinguish between: There is an assignment satisfying all the conjunctions. No more than an  fraction of the conjunctions can be satisfied simultaneously.

5 Conjunctions Rather Than Just Equations An example for a NO instance of Gap-QS*[1,Z 2,½] x = 0  y = 0 x = 0  y = 1 x = 1  y = 0 Nevertheless, we can satisfy more than a half of the equations!! Henceforth, we’ll assume the number of equations in all the conjunctions is the same. Is this a restriction?

6 Relaxation Claim: Gap-QS*[O(1), ,2|  | -1 ] reduces to Gap-QS[O(1), ,3|  | -1 ] (When |  | is at most polynomial in the size of the input). Proof: Given an instance of Gap- QS*[O(1), ,2|  | -1 ], replace each conjunction with all linear combinations of its polynomials. Make sure that: 1) The number of linear combinations over  is polynomial. 2) The dependency of each linear combination is constant.

7 Correctness of the Reduction If the original system had a common solution, so does the new system. Otherwise, fix an assignment to the variables of the system and observe the two instances:

8 Analysis  2|  | -1 fraction of unsatisfied conjunctions fraction of satisfied conjunctions polynomials originating from the blue set polynomials originating from the pink set all satisfied  2|  | -1 fraction of satisfied polynomials originating from unsatifiable conjunctions  |  | -1

9 Relaxation Yes instance of Gap-QS*[O(1), ,2|  | -1 ] are transformed into Yes instances of Gap- QS[O(1), ,3|  | -1 ]. No instance of Gap-QS*[O(1), ,2|  | -1 ] are transformed into No instances of Gap- QS[O(1), ,3|  | -1 ]. The construction is efficient when |  | is at most polynomial in the size of the input. What proves the claim. 

10 Amplification Claim: Gap-QS[O(1), ,3|  | -1 ] reduces to Gap-QS[O(1), ,2|  | -1 ] (When |  |>3 is at most polynomial in the size of the input). Proof: Given an instance of Gap- QS[O(1), ,3|  | -1 ], generate the set of all linear combinations of N polynomials. Make sure that: 1) The number of linear combinations over  is polynomial. 2) The dependency of each linear combination is constant. A constant to be determined later.

11 Correctness of the Reduction Again if the original system had a common solution, so does the new system. Otherwise, fix an assignment and observe a linear combination. –the probability all the N polynomials are satisfied  (3|  | -1 ) N –if not all the polynomials are satisfied, the probability the combination is satisfied  |  | -1 –When |  |>3 for N=4 we get the desired error probability. 

12 What Have We Done So Far? Gap-QS[O(n), ,2|  | -1 ] Gap-QS[O(1), ,2|  | -1 ] Gap-QS*[O(1), ,2|  | -1 ] Gap-QS[O(1), ,3|  | -1 ] QS[O(1),  ] Solvability[O(1),  ] 3-SAT This will imply a strong PCP characterization of NP ??

13 New Aim To reduce Gap-QS[O(n), ,2|  | -1 ] to Gap-QS*[O(1), ,2|  | -1 ]. p 1 (x 1,...,x n ) = 0 p 2 (x 1,...,x n ) = 0... p n (x 1,...,x n ) = 0 q 1 (x n,x 2n )=0  q 2 (x 1,x 2 )=0 q 3 (x 13 )=0  q 2 (x 12,x 234 )=0 q 4 (x n )=0  q 5 (x 1 )=0  q 4 (x 1 )=0... q m (x 3,x 4,x 5,x n+1 )=0  q n (x 1 )=0 each equation may depend on many variables each conjunction is composed of O(1) equations, which depend only on O(1) variables

14 Rewriting the Equations Every quadratic polynomial P(x 1,...,x D ) can be written as for some series of coefficients {c ij }, where (1) a ij =x i x j for any 1  i  j  D (2) a i0 =x i for any 1  i  D (3) a 00 =1 depends on the assignment depends on the polynomial

15 Representing Polynomials As Sums For any given point x  d we can associate each a ij with a point in H d (H is some finite field, d=O(log |H| D)). Now the evaluation of P in this point can be written as

16 Representing Polynomials As Sums The evaluation of every quadratic polynomial at a point can be written as where f P (h)=LDE c P (h)·LDE A (h). Its total degree is at most 2·d·(|H|-1)

17 Partial Sums Define: Sum f can be thought of as a polynomial of total degree at most 2·d 2 ·(|H|-1)

18 Verifying the Polynomial Zeroes