PCP Characterization of NP: Proof chapter 1

1. PCP Characterization of NP: Proof chapter 1

2. Current Status • Starting Point:Gap-QS[O(n),,2/||] is NP-hard for every finite field ||>2. • Goal: For every <1, there exists c>0, s.t Gap-QS[O(1),,2/||] is NP-hard for ||=2clogn. Coming up: Roadmap…

3. Gap-QS[O(n),,2||-1] Sum Check quadratic equations, whose dependencies are linear. Gap-QScons[O(1),,2||-1] quadratic equations of constant size with consistency assumptions Consistent Reader conjunctions of constant number of quadratic equations, whose dependencies are constant. quadratic equations of constant size. Gap-QS*[O(1),O(1),,||-] Hadamard Gap-QS[O(1),,2||-1]

4. Today's road Putting Quadratic Terms Aside

5. First Path Putting Quadratic Terms Aside:

6. Restricted Linear Equations Def: (Gap-LSres[D,,]) Instance: • a set of n linear equations (polynomials) over . Each equation depends on at most D variables. • a restriction: xij=xixj for every 0i,jDn. Problem:to distinguish between: There is an assignment satisfying the restriction and all the equations. No more than an  fraction of the equations can be satisfied simultaneously when the restriction is satisfied.

7. Gap-QS Reduces to Gap-LSres Claim: Gap-QS[D,,] reduces to Gap-LSres[D,,]. Proof:Replace every quadratic expression xixj with xij. Make sure that: • Yes instances of Gap-QS[D,,] are transformed into Yes instances of Gap-LSres[D,,]. • No instances of Gap-QS[D,,] are transformed into No instances of Gap-LSres[D,,]. • The reduction is efficient.

8. Consistency Assumptions Solvability with a consistency promise:For any Solvability problem S, let S denote the same problem, but in which only assignments satisfying some promise are legal, and where that promise has the following structure: • <M1,r1>, <M2, r2> … : each Mi is a 1-1 mapping of d to the set of variables and ri is some degree (namely, associates some variables of the system each with a point in d) • the only assignments considered are those where for each Mi there exists a degree-ri polynomial fi, where every variable Mi(x) is assigned the value f(x)

9. Gap-LSres*Reduces to Gap-QS+* Claim: Gap-LSres+[D,,] reduces to Gap-QS+[D+3,,(r·+1)·||-1] (When ||D is at most polynomial in the size of the input). Proof: Given an instance of Gap-LSres[D,,],generate the set of all linear combinations of pairs: one linear polynomial and one restriction.

10. The Promise There exists a degree-r polynomial P s.t. for every 0i,jDn,xij=P(i,j). There exists a degree-r/2 polynomial Q s.t. for every 0iDn,xi=Q(i). The probability the restriction is not satisfied, but for some 0i,jDn,xij=xixjis at most r/||.

11. Correctness of the Reduction • If the original system had a common solution, so does the new system. • Otherwise, fix an assignment and observe a linear combination. • the probability the two polynomials are satisfied  r||-1· • if not both of the polynomials are satisfied, the probability the combination is satisfied  ||-1 • The total error probability is (r·+1)·||-1. 

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

13. l’1(xn,x2n)=0  l’2(x1,x2)=0 l’3(xm)=0 l’2(x12,x234)=0 l’4(xn)=0 l’5(x1)=0  l’4(x1)=0 ... l’m(x3,x4,x5,xn+1)=0  l’n(x1)=0 l1(x1,...,xn) = 0 l2(x1,...,xn) = 0 ... ln(x1,...,xn) = 0 quadratic restrictions quadratic restrictions 1 Left To Do To reduce Gap-LSres[O(n),,2||-1] to Gap-LSres+*[O(1),O(1),,K||-1] for some constant ||>>K>1. consistency assumptions

14. p1(xn,x2n)=0  p2(x1,x2)=0 p3(xn)=0 p2(x12,x234)=0 ... pn(x3,x4,x5,xn+1)=0 consistency assumptions 2 Left To Do To reduce Gap-QS+*[O(1),O(1),,K||-1] to Gap-QS*[O(1),O(1),,K’||-1] for some constants ||>>K’>K>1. q1(xn,x2n)=0  q2(x1,x2)=0 q3(xm)=0 qn(x12,x234)=0 q4(xn)=0 q5(x1)=0  q4(x1)=0 ... qm(x3,x4,x5,xn+1)=0  qn(x1)=0