Two Query PCP with Sub-constant Error

1. Two Query PCP with Sub-constant Error Dana Moshkovitz Princeton University Ran Raz Weizmann Institute

2. Probabilistically Checkable Proofs The PCP Theorem (...,AS92,ALMSS92,…): “Any proof can be transformed into a proof that can be checked probabilistically by reading only a constant number of proof symbols”.

3. The PCP Theorem There is an efficient probabilistic verifier V for verifying the satisfiability of φ, that uses O(log|φ|) random bits to make O(1) queries to proof . - Completeness:φ sat )9¼, P(V¼ accepts)=1. - Soundness:φnot sat ) 8¼, P(V¼ accepts)≤ε.

4. Hardness of Approximation [FGLSS91,ALMSS92…]: PCP Theorems  approximation problems are NP-hard.

5. In This Work PCP Verifier: • Makes two queries to the proof • Makes projection test on queries • Has error ε→0 Many applications in hardness of approximation

6. Projection Tests A Proof partitioned into two: A, B. Verifier queries (a,b) where aA, bB. Projection fa,b:§A§B  {} Verifier checks fa,b((a)) = (b) B

7. Main Parameters of PCP |φ| = n. • #Queriesq. • Errorε. • Sizes. (Randomnessr; sq¢2r). • Alphabet§. (Answer sizelog|§|). Note:ε ≥ 1/|§|q size alphabet queries

8. Initial Parameters The PCP Theorem (AS92,ALMSS92): PCP verifier: • q = O(1) • ε = ½ • s = poly(n) • |§| = O(1)

9. Previous Work on PCP • Almost-linear size s=n1+o(1)[GS02,BSVW03,BGHSV04,BS05,D05,MR07]. • Record:s=n polylog n[BS05,D05]. • Sub-constant error ε→0 [AS97,RS97,DFKRS99,MR07]. • Record:ε=2-(logn)1-α for any α>0[DFKRS99].

10. Two Queries • Importantly: all results for error ε→0 were for q>2. • Folklore: can obtain error ε→0 and q=3. • Our focus:q=2 (projection tests) and error ε→0.

11. Hardness of Approximation Theorem (Håstad97): For any constant >0,3SAT is NP-hard to approximate within ⅞ + . • PCP Thm • q = 2 (projection tests) • error ε • size s • alphabet § SAT reduces to approximating 3SAT within 7/8 + εΩ(1)on inputs of size s∙2|§|

12. General Paradigm for Hardness of Approximation • Many hardness of approx. results [BGS95,H97,ST00,DS02,ABHK05…] SAT reduces to approximating 3SAT within 7/8 + εΩ(1)on inputs of size s∙2|§| • PCP Thm • q = 2 (projection tests) • error ε • size s • alphabet §

13. General Paradigm for Hardness of Approximation Parallel Repetition Theorem (Raz94) • Many hardness of approx. results [BGS95,H97,ST00,DS02,ABHK05…] SAT reduces to approximating 3SAT within 7/8 + εΩ(1) on inputs of size s∙2|§| • PCP Thm • q = 2 (projection tests) • error ε • size s • alphabet §

14. Parallel Repetition Parallel Repetition PCP (Raz94): For any ε>0, PCP verifier with error ε: • q = 2 (projection tests) • s = n£(log1/ε) • log|§| = £(log1/ε) downside Large polynomial size, only constant error

15. General Paradigm for Hardness of Approximation Our Work Parallel Repetition Theorem (Raz94) • Many hardness of approx. results [BGS95,H97,ST00,DS02,ABHK05…] SAT reduces to approximating 3SAT within 7/8 + εΩ(1) on inputs of size s∙2|§| • PCP Thm • q = 2 (projection tests) • error ε • size s • alphabet § = constant → 0 = n1+o(1) = nc

16. Our Work Thm: For any ε>0, PCP verifier with error ε: • q = 2 (projection tests) • s = n1+o(1) poly(1/ε) • log|§| = poly(1/ε) downside • Remarks: • Sub-constant error: ε = 1/(logn)βfor some β>0. • Constant alphabet size for constant error.

17. Implication to 3SAT • PCP Thm • q = 2 (projection tests) • error ε • size s • alphabet § Our Result:NP-hard to approximate 3SAT on inputs of size N within 7/8 + 1/(loglogN)for some constant >0(almost-linear blow-up N=n1+o(1)). SAT reduces to approximating 3SAT within 7/8 + εΩ(1)on inputs of size s∙2|§|

18. Results Under Stronger Assumptions • PCP Thm • q = 2 (projection tests) • error ε • size s • alphabet § Previous result:Unless NPµTIME(nloglogn) cannot efficiently approximate 3SAT on inputs of size N within 7/8 + 1/(logN)for some constant >0. SAT reduces to approximating 3SAT within 7/8 + εΩ(1)on inputs of size s∙2|§|

19. More Applications to Hardness • 3LIN. NP-hard to approximate within ½+o(1) under almost-linear reductions. [Håstad’97] • Amortized query complexity (q/log(1/ε)) 1+o(1). [Samorodnitsky-Trevisan’00] • Free bit complexity (f/log(1/ε)) o(1). [Samorodnitsky-Trevisan’00] …

20. The Construction • Construction with large alphabet |§|≥ nω(1). • Reduce alphabet to log|§|=poly(1/ε).

21. Construction with Large Alphabet • Algebraic construction based on low degree testing of sub-constant error [AS97,RS97…]. • To get almost-linear size: • Use sub-constant error low degree test of almost linear size [MR06]. • For the PCP construction, use idea from [MR07].

22. Alphabet Reduction • Alphabet reduction in PCP via composition. • [AS92,…,DR04,BGHSV04]:existing techniques either yield q>2 or  ≥ ½. • The heart of our work:composition with q=2 and →0for the algebraic construction. • Techniques:algebraic and combinatorial. ...

23. The Construction • Algebraic Construction • Difficulty in composition with two queries • Combinatorial transformations on algebraic construction • Composition with two queries

24. The Construction Simplifications: • Polynomial size/logarithmic randomness. • Polynomial alphabet.

25. Two-Prover Game φ sat B A a b projection test ¼(a) ¼(b)

26. 1. The Algebraic Construction Two query PCP with sub-constant error, but super-polynomial alphabet

27. Starting Point Sequential repetition PCP Verifier V for SAT: • Randomness complexity: O(logn+log1/ε) • Queries: k=£(log 1/ε) queries • Size: s=poly(n) • Alphabet: {0,1} • Error: ε

28. Approach: Simulate V With Two Provers Protocol will guarantee that provers answer according to ¼ that is independent of k queries Will show a two prover protocol: • Provers should decide on proof ¼ for V. • For random r, provers should answer V’s k queries according to ¼. • Simulate V. A B

29. Low Degree Extension |H|m = s ¼ Fm 1 · · · s H F • Def:low degree extensionof ¼ is the m-variatepolynomial p over Fof degree at most (|H|-1) in each vars.t. p(x)=¼(x) for every x2Hm. • Low Degree: d=m¢(|H|-1) <<|F|.

30. Algebraic Construction A B “p|S” “p(x)” S x • 1. Pick random r. Let V’s queries be i1,…,ik. • 2. Pass a random degree-k manifold S through the k points. Pick random x2S. • 3. Ask A what is the restriction of p to S. Ask B what is p(x). • 4. Check A answers low degree poly & evals to i1,…,ik satisfy V & A,B’s answers consistent. Fm k x S S={(q1(t1…ta),…,qm(t1…ta))|t1…ta2 F}, deg qi·k, a=O(1)

31. Manifold Vs. Point Assume B always returns p(x) A B Use low degree testing to argue Bevals poly. “p|S” “p(x)” S x • 1. Pick random r. Let V’s queries be i1,…,ik. • 2. Pass a random degree-k manifold S through the k points. Pick random x2S. • 3. Ask A what is the restriction of p to S. Ask B what is p(x). • 4. Check A answers low degree poly & evals to i1,…,ik satisfy V & A,B’s answers consistent. Fm k x S Main point: If A answers qp|S, then q(x)p|S(x) on most x2S.

32. Large Alphabet #symbols to represent ProverA‘s answer ¸kd¸ ω(logn). • Standard parameter setting: m,|H|= logs/loglogs. • For almost-linear size: m = (logs)1- ,|H|=2(logs).

33. Composition with Three Provers B A Answers of A.A, A.B of length polylogarithmic in length of A’s answer = polylog(polylogs)<<logn. Evaluate p|Son x, i1,…,ik A.A A.B “p(x)” “p|S” x S

34. Technicality Polynomial p|S of degree kd in O(1) variables Polynomial pS of degree O(logkd) in O(logkd) variables:xi,j=xi2j k k x S x

35. Composition with Three Provers B Evaluate pS on x, i1,…,ik A.A A.B “p(x)” x “pS|s’” “pS(x’)” S’ x' • (a) Pass a random degree-k+1 sub-manifold S’ through the k+1 points. Pick random x’2S’. • (b) Ask A.A what is the restriction of pS to S’. Ask A.B what is pS(x’). • (c) Check A.A answers low degree poly & A.A,A.B’s answers consistent. S’ x' k x

36. Composition with Three Provers B Evaluate pS on x, i1,…,ik A.A A.B “p(x)” x “pS|s’” “pS(x’)” S’ x' S’ x' k x Problem: Three Provers!

37. Composition for TwoProvers?? • The Idea:change the Manifold vs. Point game, such that both provers know bothS, x. * Provers will also get more information to confuse them.

38. 2. Combinatorial Transformations Changing the Manifold vs. Point game, so both provers know S,x

39. Manifold vs. Point Graph A B Possible questions to A . . . . . . Possible questions to B

40. Right Degree Reduction A Reduce the degree of B vertices to D=poly(1/ε). I.e., given proverB’s point, there are only D possibilities to proverA’s manifold. Remarks: • Uses expanders • Relies (only!) on projection; no left degree reduction. • Optimality: D¸1/ε. B . . . . . .

41. Right Degree Reduction A • Replace each B vertex of degree N with N new vertices hb,ii, i2[N]. • Expander H=([N],[N],EH) of degree D. • If a = i’th neighbor of b and (i,j)2EH then put (a,hb,ji). B . . . j i . . . . . .

42. Right Degree Reduction A • ProverB receives hb,ii, and supposedly answers question b in original game G. • ProverA and verifier are as in G. . . . . . . . . .

43. Sunflowers A B Sunflowers verifier: • Pick manifold and point • Ask ProverB about manifold. • Ask ProverA about all D neighbors of point. • Check all of ProverA’s answers (including consistency on point)& A,B answer same on manifold. . . . . . . . . . . . .

44. Sunflowers A B Outcome: Bothprovers know the manifold! Downside: length of A’s answer is ≥poly(1/ε). . . . . . .

45. Making Both Provers Know The Point A B Perform right degree reduction Final verifier: • Pick sunflower and manifold • Send ProverA the sunflower • Send ProverB the manifold and D neighboring centers • Check Sunflower vs. Manifold . . . . . .

46. 3. Composition with Two Provers For the Sunflower vs. Manifold game

47. Sunflower vs. Manifold B A k

48. Sunflower vs. Manifold Gamefor Prover B’s Manifold • In the inner Sunflower vs. Manifold game, provers agree on poly for B’s manifold and evaluate it onk+D points. B A k k

49. Question to Prover A Pick sunflower for the manifold, for allD manifolds. B A k k

50. The Full Construction • Almost-linear size. Different parameter setting, almost-linear size low degree test [MR06], idea from [MR07]. • Small alphabet. Note: cannot store a field element. Solution: composition with Hadamard construction