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The P C P starting point

Quadratic Solvability. The P C P starting point. Overview. In this lecture we’ll present the Quadratic Solvability problem. We’ll see this problem is closely related to PCP . And even use it to prove a (very very weak...) PCP characterization of NP. Example:  =Z 2 ; D=1 y = 0

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The P C P starting point

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  1. Quadratic Solvability The PCP starting point

  2. Overview • In this lecture we’ll present the Quadratic Solvability problem. • We’ll see this problem is closely related to PCP. • And even use it to prove a (very very weak...) PCP characterization of NP.

  3. Example: =Z2; D=1 y = 0 x2 + x = 0 x2 + 1 = 0 Quadratic Solvability or equally: a set of dimension D total-degree 2 polynomials Definition (QS[D, ]): Instance: a set of n quadratic equations over  with at most D variables each. Problem: to find if there is a common solution.  0  1 1  1

  4. Solvability A generalization of this problem is the following: Definition (Solvability[D, ]): Instance: a set of n polynomials over  with at most D variables. Each polynomial has degree-bound n in each one of the variables. Problem: to find if there is a common root.

  5. Solvability is Reducible to QS: Proof Idea w2 w2 y2 x2 + x2 t + tlz + z + 1 = 0 w3 w1 w1 = y2 w2 = x2 w3 = tl the parameters (D,) don’t change (assuming D>2)! Could we use the same “trick” to show Solvability is reducible to Linear Solvability?

  6. 3-SAT For completeness we provide the definition of 3-SAT: Definition (3-SAT): Instance: a 3CNF formula. Problem: to decide if this formula is satisfiable. (123)...(m/3-2m/3-1m/3) where each literal i{xj,xj}1jn It is well known that 3-SAT is NP-Complete.

  7. 3-SAT is Reducible to Solvability Given an instance of 3-SAT, use the following transformation on each clause: Tr[ xi ] = 1 - xi Tr[ xi ] = xi Tr[ ( i  i+1  i+2 ) ] = Tr[ i ] * Tr[ i+1 ] * Tr[ i+2 ] The corresponding instance of Solvability is the set of all resulting polynomials. For the time being, assume the variables are only assigned {0,1}

  8. 3-SAT is Reducible to Solvability: Removing the Assumption In order to remove the assumption we need to add the equation xi * ( 1 - xi ) = 0 for every variable xi. This concludes the description of a reduction from 3SAT to Solvability[O(1),] for any field . What is the maximal dependency?

  9. QS is NP-hard Proof: by the two above reductions. 3-SAT Solvability QS

  10. Arithmization • To translate 3-SAT to Solvability we used the idea of aritmization. • The simple trick is widely used in PCP proofs, as well as in other fields.

  11. Gap-QS Definition (Gap-QS[D, ,]): Instance: a set of n quadratic equations over  with at most D variables each. Problem: to distinguish between: There is a common solution No more than an  fraction of the equations can be satisfied simultaneously. YES NO

  12. Gap-QS and PCP the variables of the input system Gap-QS[D,,] Reminder: LPCP[D,V, ] if there is an efficient algorithm, which for any input x, producesa set of efficient Boolean functions over variables of range 2V,each depending on at mostD variables. xL iff there exits an assignment to the variables, which satisfies all the functions xL iff no assignment can satisfy more than an -fraction of the functions. Gap-QS[D,,] PCP[D,log||,] quadratic equations system For each quadratic polynomial pi(x1,...,xD), add the Boolean function i(a1,...,aD)pi(a1,...,aD)=0 values in 

  13. Proving PCP Characterizations of NP through Gap-QS • Therefore, every language which is efficiently reducible to Gap-QS[D,,]is also in PCP[D,log||,]. • Thus, proving Gap-QS[D,,] is NP-hard, also proves the PCP[D,log||,] characterization of NP. • And indeed our goal henceforth will be proving Gap-QS[D, ,] is NP-hard for the best D,  and  we can.

  14. degree-2 polynomials p1 p2 p3 . . . pn i 650 . . . 0 any assignment Some Gap-QS is NP-hard Proof: by reduction from QS[O(1),]. Proof Idea: Observe an instance of QS[O(1),]: we need an efficientdegree-preserving transformation on the polynomials which induces a trans. E on the evaluations s.t.: 1) E(0n)=0m 2) v0n, (E(v),0m) is big. there might be a lot of zeroes p1’p2’p3’. . . pm’ not many zeroes 024 . . . 3

  15. c1 . . . cm p p1 p2 ... pn c11 . . . c1m . . . . . . . . . cn1 . . .cnm p•c1 . . . p•cm  = Multiplication by a Matrix Preserves the Degree polynomials poly-time, if m=nc inner product a linear combination of polynomials scalars

  16. c1 . . . cm  e1 e2 ... en c11 . . . c1m . . . . . . . . . cn1 . . .cnm •c1 . . . •cm  = How Does a Multiplication Affect the Evaluations Vector? the values of the polynomials under some assignment the values of the new polynomials under the same assignment a zero vector if =0n

  17. Suitable Matrices • A matrix Anxm which satisfies for every vu,(vA,uA)1- is a linear code. • Note, that this is completely equivalent to saying Anxm satisfies for every v0n,(vA,0m)1-. • That’s because (vA,uA)=((v-u)A,0m).

  18. What’s Ahead • We proceed with several examples for linear codes: • Reed-Solomon code • Random matrix • And finally even a code which suits our needs... the “generic -code” from the Encodings lecture.

  19. Using Reed-Solomon Codes • Define the matrix as follows: That’s really Lagrange’s formula in disguise... • One can prove that for any 0i||-1, (vA)i is P(i), where P is the unique degree n-1 univariate polynomial, for which P(i)=vi for all 0in-1. • Therefore for any v the fraction of zeroes in vA is bounded by (n-1)/||. using multivariate polynomials we can even get =O(logn/||)

  20. for any 1||-1 A Random Matrix Should Do Lemma: A random matrix Anxm satisfies w.h.p. v0nn,|{i : (vA)i = 0}| / m < 2||-1 Proof: Let v0nn. • 1im PrAnxm[ (vA)i = 0 ] = ||-1 The inner product of v and a random vector is random. •  |{i : (vA)i = 0}| (denoted Xv) is a binomial r.v with parameters m and ||-1. • By the Chernoff bound, Pr[ Xv  2m||-1 ]  2e-m/4||.

  21. A Random Matrix Should Do Every v0n disqualifies at most 2e-m/4|| of the matrices nxm At most 2||ne-m/4|| of the matrices are disqualified That is, Pr[ v0n: Xv/m  2||-1 ]  2||ne-m/4||. For m=O(n||log||), the claim holds. 

  22. p2k pk-1 Deterministic Construction Assume =Zp. Let k=logpn+1. (Assume w.l.o.g kN) Let Zpk be the dimension k extension field of . associate each column with a pair(x,y)ZpkZpk associate each row with 1ipk-1 <xi,y>

  23. Analysis degree-pk-1 polynomial, denoted G(x) • For any vn, for any (x,y)ZpkZpk, • The number of zeroes in vA where v0n x,y: G(x)=0 + x,y: G(x)0  <G(x),y>=0 • And thus the fraction of zeroes 

  24. Summary of the Reduction Given an instance {p1,...,pn} for QS[O(1),], find a matrix A which satisfies v0n|{i : (vA)i = 0}| /m < 2||-1 {p1,...,pn}  QS[O(1),] iff {p1A,...,pnA}  Gap-QS[O(n),,2||-1] !!

  25. Hitting the Road This proves a PCP characterization with D=O(n) (hardly a “local” test...). Eventually we’ll prove a characterization with D=O(1) ([DFKRS]) using the results presented here as our starting point.

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