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2. 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

3. 3 Using consistent readers

4. 4 Consistent Readers Remove Consistency Assumptions: Concept: Readers are plugged into equations, and ensure (w.h.p) the domain variables are only assigned values corresponding to some  - permissible polynomial. Def: A function f is called a  -permissible polynomial (with respect to an assignment A) if at least 1-  of the points x are assigned the value of a dimension d and degree r polynomial.

5. 5 Consistent Readers: Scheme Local Readers Local Reader... Local Test Evaluator Representation Consistent Reader variable... Q: How do the consistency gadgets we’ve seen fit into this scheme?

6. 6 The Desired Properties of A Consistent Reader Efficiency: There is only a polynomial number of local readers. Efficiency: There is only a polynomial number of local readers. Low Dependency: Each local reader depends on a constant number of representation variables. Low Dependency: Each local reader depends on a constant number of representation variables. Linearity: Each local test is a conjunction of linear equations over representation variables. Each evaluator is a linear combination of such. Linearity: Each local test is a conjunction of linear equations over representation variables. Each evaluator is a linear combination of such. Range: All variables should range over . Range: All variables should range over .

7. 7 The Desired Properties of A Consistent Reader Completeness: All local tests accept every assignment to the variables corresponding to some  -permissible polynomial. Moreover, every evaluator must return the evaluations of that polynomial. Completeness: All local tests accept every assignment to the variables corresponding to some  -permissible polynomial. Moreover, every evaluator must return the evaluations of that polynomial. Soundness: At most an  fraction of the tests can be satisfied when no  -permissible polynomial exists. Soundness: At most an  fraction of the tests can be satisfied when no  -permissible polynomial exists.

8. 8 The Consistent Reader We Have Representation: One variable for every cube (affine subspace) of dimension k+2, containing the k points corresponding to variables from this domain appearing in equations. (Values of the variables range over all degree-r, dimension k+2 polynomials) One variable for every cube (affine subspace) of dimension k+2, containing the k points corresponding to variables from this domain appearing in equations. (Values of the variables range over all degree-r, dimension k+2 polynomials) One variable for every point x  d. (Values of the variables range over  ). One variable for every point x  d. (Values of the variables range over  ).

9. 9 The Consistent Reader We Have Consistent-Reader: One local-reader for every cube-variable C and a point-variable y  C, which One local-reader for every cube-variable C and a point-variable y  C, which rejects if A’s value for C restricted to y disagrees with A’s value on y, rejects if A’s value for C restricted to y disagrees with A’s value on y, otherwise: returns A’s values on C restricted to x 1,...,x k. otherwise: returns A’s values on C restricted to x 1,...,x k.

10. 10 Does This Consistent Reader Suffice? Implicitly, our consistent reader relies itself on consistency assumptions. Implicitly, our consistent reader relies itself on consistency assumptions. This will be resolved in the proof of the Composition-Recursion lemma, assuring us there exists a satisfactory consistent reader. This will be resolved in the proof of the Composition-Recursion lemma, assuring us there exists a satisfactory consistent reader. In the meantime, let us assume that, and continue with the PCP proof. In the meantime, let us assume that, and continue with the PCP proof. You’d might want to take a better look at the desired properties and the construction first.desired propertiesthe construction

11. 11 Gap-QS cons Reduces To Gap- QS* The Construction: Let p be one of the equations. Let p be one of the equations. Assume there are k domains F 1,...,F k (associated with k consistent readers). Assume there are k domains F 1,...,F k (associated with k consistent readers). Generate a conjunction for every choice of one local reader of each consistent reader. Generate a conjunction for every choice of one local reader of each consistent reader. Each conjunction has the following format: Each conjunction has the following format: the original equation, where each domain variable is replaced by its evaluation  local test 1...   local test k Note the uniformity of the construction: Each equation contributes the exact same number of conjunctions.

12. 12 Correctness: Completeness Let A denote an instance of Gap-QS cons [O(1), ,2/|  |]. Let A denote an instance of Gap-QS cons [O(1), ,2/|  |]. Let B denote the corresponding instance of Gap-QS*[O(1),O(1), ,|  | -  ]. Let B denote the corresponding instance of Gap-QS*[O(1),O(1), ,|  | -  ]. Completeness: a good satisfying assignment for A can be easily extended into a satisfying assignment for B. Completeness: a good satisfying assignment for A can be easily extended into a satisfying assignment for B.

13. 13 How Are We Going To Prove Soundness? equations conjunctions We’ll show that for any assignment, more than few satisfied conjunctions imply more than few satisfied equations even under the consistency assumption. satisfied “contributes” we’re only interested in equations, for which this fraction is big. Such can be satisfied when  - permissible polynomials are assigned to their domains. Since there’re few permissible polynomials, even a random choice should satisfy enough equations.

14. 14 Correctness: Soundness Suppose there exists an assignment which satisfies  >|  | -  of B’s conjunctions. Suppose there exists an assignment which satisfies  >|  | -  of B’s conjunctions. Let p be an equation which more than k  of its conjunctions are satisfiable. Let p be an equation which more than k  of its conjunctions are satisfiable. At least  -k  of the equations satisfy this. At least  -k  of the equations satisfy this.

15. 15 Correctness: Soundness For every reader, at most  of the local tests err. For every reader, at most  of the local tests err. Overall at most k  of all local tests err. Overall at most k  of all local tests err. Hence there exists a satisfied conjunction with no erroneous tests. Hence there exists a satisfied conjunction with no erroneous tests.   satisfying assignment for p which assigns every domain a  -permissible polynomial.   satisfying assignment for p which assigns every domain a  -permissible polynomial.

16. 16 Auxiliary Lemma: Not Many Permissible Polynomials Lemma: For any big enough  (  2 >4rd|  | -1 ), there are less than 2  -1  -permissible polynomials. Proof: Suppose there were 2  -1  -permissible polynomials. Suppose there were 2  -1  -permissible polynomials. Each agrees with every other polynomial on at most rd|  | -1 of the points. Each agrees with every other polynomial on at most rd|  | -1 of the points.

17. 17 Auxiliary Lemma: Not Many Permissible Polynomials Overall on at most a 2  -1 rd|  | -1 fraction there exists some polynomial it agrees with. Overall on at most a 2  -1 rd|  | -1 fraction there exists some polynomial it agrees with. Which is less than ½  by the choice of . Which is less than ½  by the choice of . But a  -permissible polynomial agrees with the assignment on at least  of the points. But a  -permissible polynomial agrees with the assignment on at least  of the points.

18. 18 Auxiliary Lemma: Not Many Permissible Polynomials Thus, for each polynomial for more than ½  of the points it’s the only one agreeing with the assignment on them. Thus, for each polynomial for more than ½  of the points it’s the only one agreeing with the assignment on them. Contradiction! Recall that by our assumption there are as many as 2  -1 polynomials.  Contradiction! Recall that by our assumption there are as many as 2  -1 polynomials. 

19. 19 Proof of Soundness Continued If we assign each domain a random  - permissible polynomial, p should be satisfied with probability at least O(  k ). If we assign each domain a random  - permissible polynomial, p should be satisfied with probability at least O(  k ). Hence the expected number of satisfied equations is at least O((  -k  )  k ). Hence the expected number of satisfied equations is at least O((  -k  )  k ). Thus at least one assignment satisfies that many equations. Thus at least one assignment satisfies that many equations. For an appropriate choice of the parameter , this number is greater than 2/|  |.  For an appropriate choice of the parameter , this number is greater than 2/|  |. 

20. 20 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...