The PCP Theorem via gap amplification

Irit DinurHebrew University

The PCP Theorem [AroraSafra, AroraLundMotwaniSudanSzegedy, 1992]

PProb.rob.CCheckable.heckable.PProofroof

SAT Instance: SAT Instance:

VerifierVerifier

If sat() = 1 then 9proof, Pr[Ver accepts] = 1 If sat() < 1 then 8proof, Pr[Ver accepts] < ½

The PCP Theorem[AroraSafra, AroraLundMotwaniSudanSzegedy, 1992]

variablesvariables

VV11 VV22 VV33 VVnn…

PCP Thm <--> reduction from SAT to gap-CSP Given a constraint graph G, it is NP hard to decide

between1. gap(G)=02. gap(G)>

xx11 xx22

xx44

xx55

xxnn

xx33

VV 11

VV 22

This talk

New proof for the PCP Theorem: Given a constraint graph G, it is NP-hard to

decide between1. gap(G) = 02. gap(G) >

Based on: gap amplification, inspired by Reingold’s SL=L proof

Also: “very” short PCPs

step 0: Constraint Graph SAT is NP-hard

Given a constraint graph G, it is NP-hard to decide if gap(G)=0 or gap(G) > 0

proof: reduction from 3coloring. ={1,2,3}, inequality constraints on edges. Clearly, G is 3-colorable iff gap(G)=0.

PCP Thm: Given a constraint graph G, it is NP-hard to decide if gap(G)=0 or gap(G) >

Basic Plan

• Start with a constraint graph G (from 3coloring)

• G G1 G2 … Gk= final output of reduction

• Main Thm: gap(Gi+1) ¸ 2 ¢ gap(Gi) (if not already too large)

• size(Gi+1) = const ¢ size(Gi), degree, alphabet, expansion all remain the same.

• Conclusion: NP hard to distinguish between gap(Gk)=0 and gap(Gk)> (constant)

Main Step

Standard Standard transformations: making

transformations: making G a regular constant

G a regular constant degree expander, w/ self-

degree expander, w/ self-loopsloops

Composition with Composition with P P = a a “constant-size” PCP. “constant-size” PCP.

PP can be as inefficient can be as inefficient as possibleas possible

Gi Gi+1 : Gi+1 = ( prep(Gi) )t ² P

1. Preprocess G2. Raise to power t3. Compose with P = constant size PCP

Key step: G Gt, multiplies the gap by t; Keeps size linear !

poweringpowering

Powering a constraint graph

Vertices: same Edges: length-t paths (=powering of adj. matrix) Alphabet: dt reflecting “opinions” about neighbors Constraints: check everything you can!

uu vv

Powering a constraint graph

Vertices: same Edges: length-t paths (=powering of adj. matrix) Alphabet: dt reflecting “opinions” about neighbors Constraints: check everything you can!

uu vv

Observations:1. New Degree = dt

2. New Size = O(size) (#edges is multiplied by dt-1)

3. If gap(G)=0 then gap(Gt)=04. Alphabet increases from to dt

Amplification Lemma: gap(Gt) ¸ t ¢ gap(G)

Amplification Lemma: gap(Gt) > t ¢ gap(G)

Intuition:Spread the information inconsistencieswill be detected more often

uu vv

Assumption: G is d-regular d=O(1), expander, w self-loops

Given A:V dt“best”

extract a:V

by most popular value in a random t/2 step walk

vv

Amplification Lemma: gap(Gt) > t ¢ gap(G)

Given A:V dt “best”

extract a:V

by most popular value in a random t/2 step walk

uu

Extracting a:V

vv

Given A:V dt“best”

extract a :V

and consider F = { edges rejecting a }

Note: F/E ¸ gap(G)

vvuu

Extracting a:V

Amplification Lemma: gap(Gt) > t ¢ gap(G)

Relate fraction of rejecting paths to fraction of rejecting edges ( = F/E )

vvuu

Two Definitions

= (v0,v1,…,u,v,…,vt); j = (vj-1,vj)

Definition: the j-th edge strikes if 1. |j – t/2| < t

2. (u,v)2F, i.e., (u,v) rejects a(u), a(v)

3. A(v0) agrees with a(u) on u &

A (vt) agrees with a(v) on v .

Definition: N() = # edges that strike . 0 · N() < 2t If N()>0 then rejects, so gap(Gt) ¸ Pr[N()>0]

vv00

vvuu vvtt

jj

We will prove: Pr[N()>0] > t ¢ F/E

Lemma 1: E[N] > t ¢ F/E ¢ const(d, )

Intuition: Assuming N() is always 0 or 1, Pr[N>0] = E[N]

Lemma 2: E[N2] < t ¢ F/E ¢ const(d, )

Standard: Pr[N>0] ¸ (E[N])2/E(N2)

pf: E[N2|N>0]¢Pr[N>0]2¸(E[N|N>0])2¢Pr[N>0]2

Pr[N>0] > (t ¢ F/E )2 / (t ¢ F/E) = t ¢ F/E

gap(Ggap(Gtt) ) ¸̧ ¸ ¸ gap(G)gap(G)

Lemma 1: E[N] = t ¢ F/E

Ni() = indicator for event “the i-th edge strikes ” N =

i2JNi where J = { i : |i-t/2|< t }

Claim: if i 2 J E[Ni] ¼ 1/2 ¢ F/E

can be chosen by the following process:1. Select a random edge (u,v)2E, and let i = (u,v).2. Select a random i-1 step path from u3. Select a random t-i step path from v

Clearly, Pr[i 2 F] = F/E What is the probability that A (v0) agrees with a(u) and A (vt)

agrees with a(v) ?

vv00

vvuu vvtt

i-1i-1 t-it-i

Claim: if i 2 J E[Ni] ¼ 1/2 ¢ F/E chosen by :

1. Select a random edge (u,v)2E, and let i = (u,v).2. Select a random i-1 step path from u3. Select a random t-i step path from v

i-1 = t/2 walk from u reaches v0 for which A(v0) thinks a(u) of u, with prob. ¸ 1/.

i 2 J: roughly the same !! (because of self-loops)

vv00

vvuu vvtt

i-1i-1 t-it-it/2t/2

Back to E[N]

Fix i2J. Select by the following process:1. Select a random edge (u,v), and let i = (u,v).

2. Select a random i-1 step path from u3. Select a random t-i step path from v

1. Pr[i 2 F] = F/E

2. Pr[A(v0) agrees with a on u | (u,v) ] > 1/2

3. Pr[A(vt) agrees with a on v | (v0,…,u,v) ] > 1/2

E[Ni] = Pr[Ni=1] > F/E ¢ 1/2 ¢ const

so E[N] = i2JE[Ni] > t ¢ F/E ¢ const QED

vv00

vvuu vvtt

i-1i-1 t-it-i

We will prove: Pr[N()>0] > t ¢ F/E

Lemma 1: E[N] > t ¢ F/E ¢ const(d, )

Lemma 2: E[N2] < t ¢ F/E ¢ const(d, )

read: “most striked paths see · a constant number of striking edges”

By Pr[N>0] > (E[N])2 / E[N2]

Pr[N>0] > (t ¢ F/E )2 / (t ¢ F/E) = t ¢ F/E

gap(Ggap(Gtt) ) ¸̧¸ ¸ gap(G)gap(G)

Lemma 2: Upper bounding E[N2]

Observe: N() · # middle intersections of with F

Claim: if G=(V,E) is an expander, and F½E any (small) fixed set of edges, then E[(N’)2] < t¢F/E¢(t¢F/E+const)

proof-sketch: Compute i<jE[N’iN’j]. Conditioned on i 2 F, the expected # remaining steps in F is still · constant.

The full inductive step

Gi Gi+1 : Gi+1 = ( prep(Gi) )t ² P

1. Preprocess G2. Raise to power t3. Compose with P = constant size PCP

Preprocessing

G H=prep(G) s.t. H is d-regular, d=O(1) H is an expander, has self-loops.

maintain

size(H) = O(size(G)) gap(G) ¼ gap(H), i.e.,

1. gap(G) = 0 gap(H) = 02. gap(G)/const · gap(H)

Add expander edgesAdd expander edges

Add self-loopsAdd self-loops

[PY][PY] Blow up every vertex

Blow up every vertex

u into a cloud of deg(u)

u into a cloud of deg(u)

vertices, and inter connect

vertices, and inter connect

them via an expander.

them via an expander.

Reducing dt to

Consider the constraints {C1,…,Cn} (and forget the graph structure)

For each i, we replace Ci by {cij} = constraints over smaller alphabet .

P = algorithm that takes C to {cj}, cj over s. t. If C is “satisfiable”, then gap({cj})=0 If C is “unsatisfiable”, then gap({cj}) >

Composition Lemma: [BGHSV, DR]The system C’ = [iP(Ci) has gap(C’) ¼ gap(C)

CC11 CC22 CC33 CC44 CCnn……

cc1111 cc1212 cc1313 cc1414 cc1515

PP

ccn1n1 ccn2n2 ccn3n3 ccn4n4 ccn5n5

PP

Assignment-testers [DR] / PCPPs [BGHSV]

Composition

If P is any AT / PCPP then this composition works. P can be

Hadamard-based Longcode-based found via exhaustive search (existence must be

ensured, though)

P’s running time only affects constants.

Summary: Main theorem

Gi Gi+1 : Gi+1 = ( prep(Gi) )t ² P gap(Gi+1) > 2¢gap(Gi) and other params stay same

1. G [, ]2. G prep(G) [, /const]

3. G Gt [dt, t¢/const]4. G G ² P [, t¢/const’] = [,2]

G=G0 G1 G2 … Gk= final output of reduction After k=log n steps,

If gap(G0) = 0 then gap(Gk)=0

If gap(G0) > 0 then gap(Gk) > const

Application: short PCPs

…[PS, HS, GS, BSVW, BGHSV, BS] [BS’05]: NP µ PCP1,1-1/polylog[ log (n¢polylog ), O(1) ] There is a reduction taking constraint graph G to G’ such that

|G’| = |G|¢ polylog |G| If gap(G)=0 then gap(G’)=0 If gap(G)>0 then gap(G’)> 1/polylog|G|

Applying our main step loglog|G| times on G’, we get a new constraint graph G’’ such that

If gap(G) = 0 then gap(G’’)=0 If gap(G) > 0 then gap(G’’) > const

i.e., NP µ PCP1,1/2[ log (n¢polylog ), O(1) ]

final remarks

Main point: gradual amplification

Compare to Raz’s parallel-repetition thm

Q: get the gap up to 1-o(1)