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Two Query PCP with Sub-constant Error

Dana MoshkovitzPrinceton University

Ran RazWeizmann Institute

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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”.

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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: φ not sat ) 8¼, P(V¼ accepts)≤ε.

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Hardness of Approximation

[FGLSS91,ALMSS92…]: PCP Theorems approximation

problems are NP-hard.

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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

Projection Tests

?

?

A

B

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Proof partitioned into two: A, B.1. Verifier queries (a,b) where

aA, bB.2. Projection fa,b:§A§B {}3. Verifier checks fa,b((a)) = (b)

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Main Parameters of PCP

|φ| = n.• #Queries q.• Error ε. • Size s. (Randomness r; sq¢2r).• Alphabet §. (Answer size log|§|).

size

queriesalphabet

Note: ε ≥ 1/|§|q

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Initial Parameters

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

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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].

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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.

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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∙ |§|

General Paradigm for Hardness of Approximation

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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∙ |§|

Many hardness of

approx. results [BGS95,H97,ST00,DS02,

ABHK05…]

General Paradigm for Hardness of Approximation

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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∙ |§|

Many hardness of

approx. results [BGS95,H97,ST00,DS02,

ABHK05…]

Parallel Repetition Theorem (Raz94)

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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

General Paradigm for Hardness of Approximation

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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∙ |§|

Many hardness of

approx. results [BGS95,H97,ST00,DS02,

ABHK05…]

Parallel Repetition Theorem (Raz94) Our Work

= constant= nc

→ 0 = n1+o(1)

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Our WorkThm: 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)β for some β>0. • Constant alphabet size for constant error.

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Implication to 3SAT

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∙ |§|

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)).

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Results Under Stronger Assumptions

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∙ |§|

Previous result: Unless NPµTIME(nloglogn) cannot efficiently approximate 3SAT on inputs of size N within 7/8 + 1/(logN) for some constant >0.

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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]

…

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The Construction

1. Construction with large alphabet |§|≥ nω(1).

2. Reduce alphabet to log|§|=poly(1/ε).

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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].

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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.

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The Construction

1. Algebraic Construction– Difficulty in composition with two queries

2. Combinatorial transformations on algebraic construction

3. Composition with two queries

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The Construction

Simplifications: - Polynomial size/logarithmic randomness.- Polynomial alphabet.

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Two-Prover Game

A B

a b

¼(a) ¼(b)

projection test

φ sat

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1. The Algebraic Construction

Two query PCP with sub-constant error, but super-polynomial alphabet

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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

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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.

A B

Protocol will guarantee that provers answer according to ¼ that is independent of k queries

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Low Degree Extension

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 x2Hm.

Low Degree: d=m¢(|H|-1) <<|F|.

H

F

Fm

1 · · ·s

|H|m = s ¼

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Algebraic Construction

Fm

k

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.

S

xS “p|S” “p(x)”

x

A B

S={(q1(t1…ta),…,qm(t1…ta))|t1…ta 2 F}, deg qi·k, a=O(1)

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A BUse low degree testing to argue B evals poly.

Manifold Vs. Point

k

S

xS “p|S” “p(x)”

x

Assume B always

returns p(x)

Main point: If A answers qp|S, then q(x)p|S(x) on most x2S.

Fm 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.

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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).

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Composition with Three Provers

xS “p|S” “p(x)”

A B

A.A A.BEvaluate p|S on x, i1,…,ik

Answers of A.A, A.B of length polylogarithmic in length of A’s answer = polylog(polylogs)<<logn.

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Technicality

Polynomial p|S of degree kd in O(1) variables

Polynomial pS of degree O(logkd) in O(logkd) variables: xi,j=xi

2j

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Composition with Three Provers

x“p(x)”

B

A.A A.BEvaluate pS on x, i1,…,ik

(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.

kx

x'S’

x'S’ “pS|s’” “pS(x’)”

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Composition with Three Provers

x“p(x)”

B

A.A A.BEvaluate pS on x, i1,…,ik

kx

x'S’

x'S’ “pS|s’” “pS(x’)”

Problem: Three Provers!

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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.

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2. Combinatorial Transformations

Changing the Manifold vs. Point game, so both provers know S,x

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Manifold vs. Point GraphA

B

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Possible questions to A

Possible questions to B

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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.

A

B

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Right Degree Reduction

• 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).

A

B

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i

j

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Right Degree Reduction

• Prover B receives hb,ii, and supposedly answers question b in original game G.

• Prover A and verifier are as in G.

A

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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.

A B

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Sunflowers

Outcome: Both provers know the manifold!

Downside: length of A’s answer is ≥ poly(1/ε).

A B

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Making Both Provers Know The Point

Perform right degree reductionFinal verifier:• Pick sunflower and manifold• Send Prover A the sunflower• Send Prover B the manifold

and D neighboring centers• Check Sunflower vs. Manifold

A B

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3. Composition with Two Provers

For the Sunflower vs. Manifold game

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Sunflower vs. Manifold

BA

k

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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 on k+D points.

BA

kk

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Question to Prover A

Pick sunflower for the manifold,for all D manifolds.

BA

kk

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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

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The Main Ideas

• Use the recursive structure of the algebraic construction

• Change the game• Combinatorial transformations: right degree

reduction, sunflowers• Composition