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2 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  BUT it remains to prove the composition- recursion lemma...

3 Using error correcting codes

4 Conjunctions of Equations Definition (Gap-QS*[D 1,D 2, ,  ]): Instance: a set of n conjunctions of D 1 quadratic equations (polynomials) over . Each equation depends on at most D 2 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 of Equations An example instance of Gap-QS*[2,1,Z 2,½]: Notice that 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 The reduction Claim: Gap-QS*[D 1,D 2, ,  ] reduces to Gap- QS[D 1 ·D 2, ,  +|  | -1 ] (as long as |  | D 1 is at most polynomial). Proof: Given an instance of Gap-QS*[D 1,D 2, ,  ], replace each conjunction with all linear combinations of its polynomials. (i.e. apply Hadamard code to conjuncts)

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

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

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

10 Amplification Claim: For any constant C, Gap-QS[D, ,  ] reduces to Gap-QS[C·D, ,  C +|  | -1 ] (When |  | is at most polynomial in the size of the input). Proof: Given an instance of Gap-QS[D, ,  ], generate the set of all linear combinations of C polynomials. By an argument similar to the former, the claim holds. 

11 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  BUT it remains to prove the composition- recursion lemma...