two query pcp with sub-constant error dana moshkovitz princeton university ran raz weizmann...
Post on 18-Dec-2015
216 Views
Preview:
TRANSCRIPT
1
Two Query PCP with Sub-constant Error
Dana MoshkovitzPrinceton University
Ran RazWeizmann 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: φ not 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
Projection Tests
?
?
A
B
6
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)
7
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
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∙ |§|
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 + εΩ(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 + εΩ(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)
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 + 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)
16
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.
17
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)).
18
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.
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]
…
20
The Construction
1. Construction with large alphabet |§|≥ nω(1).
2. 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 →0 for the algebraic construction.• Techniques: algebraic and combinatorial.
23
The Construction
1. Algebraic Construction– Difficulty in composition with two queries
2. Combinatorial transformations on algebraic construction
3. Composition with two queries
24
The Construction
Simplifications: - Polynomial size/logarithmic randomness.- Polynomial alphabet.
25
Two-Prover Game
A B
a b
¼(a) ¼(b)
projection test
φ sat
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
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
29
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
30
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)
31
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.
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).
33
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.
34
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
35
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’)”
36
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!
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.
38
2. Combinatorial Transformations
Changing the Manifold vs. Point game, so both provers know S,x
39
Manifold vs. Point GraphA
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.
A
B
.
.
.
.
.
.
41
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
.
.
.
.
.
.
.
.
.
i
j
42
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
.
.
.
.
.
.
.
.
.
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.
A B
.
.
.
.
.
.
.
.
.
.
.
.
44
Sunflowers
Outcome: Both provers know the manifold!
Downside: length of A’s answer is ≥ poly(1/ε).
A B
.
.
.
.
.
.
45
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
.
.
.
.
.
.
46
3. Composition with Two Provers
For the Sunflower vs. Manifold game
47
Sunflower vs. Manifold
BA
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 on k+D points.
BA
kk
49
Question to Prover A
Pick sunflower for the manifold,for all D manifolds.
BA
kk
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
51
The Main Ideas
• Use the recursive structure of the algebraic construction
• Change the game• Combinatorial transformations: right degree
reduction, sunflowers• Composition
top related