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

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

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

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

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

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)

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

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

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

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.

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 | § |

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…]

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)

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

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)

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.

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) ).

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.

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

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

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

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

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

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

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

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

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

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

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

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)

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.

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

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.

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

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’)”

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!

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

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

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

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

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

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

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

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

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

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

Sunflower vs. Manifold 47 B A k

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

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

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

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