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

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”. 2

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 ) 9 ¼, P(V ¼ accepts)=1. - Soundness: φ not sat ) 8 ¼, P(V ¼ accepts)≤ε. 3

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

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

6. Projection Tests ? A B 6 Proof partitioned into two: A, B. 1.Verifier queries (a,b) where a  A, b  B. 2.Projection f a,b : § A  § B  {  } 3.Verifier checks f a,b (  ( a) ) =  ( b)

7. Main Parameters of PCP | φ | = n. #Queries q. Error ε. Size s. (Randomness r; s  q ¢ 2 r ). Alphabet §. (Answer size log| § |). size queries alphabet 7 Note: ε ≥ 1/| § | q

8. Initial Parameters 8 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=n 1+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]. 9

10. Two Queries 10 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 + . 11 PCP Thm q = 2 (projection tests) error ε size s alphabet § SAT reduces to approximating 3SAT within 7/8 + ε Ω(1) on inputs of size s∙2 | § |

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

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

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) 14 downside Large polynomial size, only constant error

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

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

17. Implication to 3SAT 17 PCP Thm q = 2 (projection tests) error ε size s alphabet § SAT reduces to approximating 3SAT within 7/8 + ε Ω(1) on inputs of size s∙2 | § | 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=n 1+o(1) ).

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

19. More Applications to Hardness 1.3LIN. NP-hard to approximate within ½+o(1) under almost-linear reductions. [Håstad’97] 2.Amortized query complexity (q/log(1/ε)) 1+o(1). [Samorodnitsky-Trevisan’00] 3.Free bit complexity (f/log(1/ε)) o(1). [Samorodnitsky-Trevisan’00] … 19

20. The Construction 1.Construction with large alphabet | § |≥ n ω(1). 2.Reduce alphabet to log| § |=poly(1/ ε ). 20

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]. 21

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  →0 for the algebraic construction. Techniques: algebraic and combinatorial.... 22

23. The Construction 1.Algebraic Construction – Difficulty in composition with two queries 2.Combinatorial transformations on algebraic construction 3.Composition with two queries 23

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

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

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

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 27

28. Approach: Simulate V With Two Provers 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. 28 AB Protocol will guarantee that provers answer according to ¼ that is independent of k queries

29. Low Degree Extension 29 Def: low degree extension of ¼ is the m-variate polynomial p over F of degree at most (|H|-1) in each var s.t. p(x)= ¼ (x) for every x 2 H m. Low Degree: d=m ¢ (|H|-1) <<|F|. H F FmFm 1· · ·s |H| m = s ¼

30. Algebraic Construction 30 FmFm k 1. Pick random r. Let V’s queries be i 1,…,i k. 2. Pass a random degree-k manifold S through the k points. Pick random x 2 S. 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 i 1,…,i k satisfy V & A,B’s answers consistent. 1. Pick random r. Let V’s queries be i 1,…,i k. 2. Pass a random degree-k manifold S through the k points. Pick random x 2 S. 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 i 1,…,i k satisfy V & A,B’s answers consistent. S x S “p |S ” “p(x)” x AB S={(q 1 (t 1 …t a ),…,q m (t 1 …t a ))|t 1 …t a 2 F}, deg q i · k, a=O(1)

31. AB Use low degree testing to argue B evals poly. Manifold Vs. Point 31 k S x S “p |S ” “p(x)” x Assume B always returns p(x) Main point: If A answers q  p |S, then q(x)  p |S (x) on most x 2 S. FmFm 1. Pick random r. Let V’s queries be i 1,…,i k. 2. Pass a random degree-k manifold S through the k points. Pick random x 2 S. 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 i 1,…,i k satisfy V & A,B’s answers consistent. 1. Pick random r. Let V’s queries be i 1,…,i k. 2. Pass a random degree-k manifold S through the k points. Pick random x 2 S. 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 i 1,…,i k satisfy V & A,B’s answers consistent.

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

33. Composition with Three Provers 33 x S “p |S ” “p(x)” A B A.A A.B Evaluate p |S on x, i 1,…,i k Answers of A.A, A.B of length polylogarithmic in length of A’s answer = polylog(polylogs)<<logn.

34. Technicality Polynomial p |S of degree kd in O(1) variables Polynomial p S of degree O(logkd) in O(logkd) variables: x i,j =x i 2 j 34 k S x k x

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

36. Composition with Three Provers 36 x “p(x)” B A.A A.B Evaluate p S on x, i 1,…,i k k x x' S’ x' S’ “p S|s’ ” “p S (x’)” Problem: Three Provers!

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

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

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

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

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

42. Right Degree Reduction Prover B receives h b,i i, and supposedly answers question b in original game G. Prover A and verifier are as in G. 42 A..................

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

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

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

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

47. Sunflower vs. Manifold 47 B A k

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

49. Question to Prover A Pick sunflower for the manifold, for all D manifolds. 49 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 50

51. The Main Ideas Use the recursive structure of the algebraic construction Change the game Combinatorial transformations: right degree reduction, sunflowers Composition 51