Direct-Product testingParallel Repetitions

And Foams

Avi WigdersonIAS

Parallel Repetition of Games and Periodic Foams

Isoperimetric problem: Minimize surface area given volume.

One bubble. Best solution: Sphere

Many bubbles Isoperimetric problem: Minimize surface area given volume.

Why? Physics, Chemistry, Engineering, Math… Best solution?: Consider R3 Lord Kelvin 1873 Optimal… Wearie-Phelan 1994 Even better

Our Problem

Minimum surface area of body tiling Rd with period Zd ?Volume=1

d=2 area:

4>4Choe’89:Optimal!

Bounds in d dimensions

≤ OPT ≤

[Kindler,O’Donn[Kindler,O’Donnell,ell, Rao,Wigderson]Rao,Wigderson] ≤OPT≤

“Spherical Cubes” exist!Probabilistic construction!(simpler analysis [Alon-Klartag])

OPEN: Explicit?

Randomized Rounding

Round points in Rd to points in Zd

such that for every x,y

1.

2.

x y1

Bound does not depend on d

Spine

TorusSurface blocking allcycles that wrap around

Probabilistic construction of spine

Step 1

Randomly construct B in [0,1)d , which in expectation satisfies

BB

Step 2

Sample independent translations of B until [0,1)d is covered, adding new boundaries to spine.

PCPs & Linear equations over GF(2)

m linear equations: Az = b in n variables: z1,z2,…,zn

Given (A,b)1) Does there exist z satisfying all m equations? Easy – Gaussian elimination2) Does there exist z satisfying ≥ .9m equations? NP-hard 3) Does there exist z satisfying ≥ .5m equations? Easy – YES!

[Hastad] δ>0, it is NP-hard to distinguish (A,b) which are not (½+δ)-satisfiable, from those (1-δ)-satisfiable!

Linear equations as Games

2n variables: X1,X2,…,Xn, Y1,Y2,…,Yn

m linear equations:Xi1 + Yi1 = b1

Xi2 + Yi2 = b2

…..

Xim + Yim = bm

Promise: no setting of the Xi,Yi satisfy more than (1-δ)m of all equations

Game G

Draw j [m] at random

Xij Yij Alice Bob

αj βj

Check if αj + βj = bj

Pr [YES] ≤ 1-δ

Hardness amplification byparallel repetition

2n variables: X1,X2,…,Xn, Y1,Y2,…,Yn

m linear equations:Xi1 + Yi1 = b1

Xi2 + Yi2 = b2

…..

Xim + Yim = bm

Promise: no setting of the Xi,Yi satisfy more than (1-δ)m of all equations

Game Gk

Draw j1,j2,…jk [m] at random

Xij1Xij2 Xijk Yij1Yij2 Yijk Alice Bob

αj1αj2 αjk βj1βj2 βjk

Check if αjt + βjt = bjt t [k]

Pr[YES] ≤ (1-δ2)k

[Raz,Holenstein,Rao] Pr[YES] ≥ (1-δ2)k

[Feige-Kindler-O’Donnell] Spherical Cubes

[Raz]X[KORW]Spherical Cubes

Hardness amplification byother means?Xi1 + Yi1 = b1

Xi2 + Yi2 = b2

…..

Xim + Yim = bm

Promise: no setting of the Xi,Yi satisfy more than (1-δ)m of all equations

Amplification

Xij1… Xijk Yij1… Yijk

Alice Bobαj1… αjk βj1… βjk

Test: αjt + βjt = bjt t ?

Pr[YES] ≤ (1-δ2)k

[Raz,Holenstein,Rao] Pr[YES] ≥ (1-δ2)k [Raz]

Major open question: Is there Test’ s.t. Pr[YES] ≤ (1-δ)k ?[Khot] Unique games conjecture

Idea: force each player to answer consistently - e.g. make Alice commit to one assignment of Xi’s

[Impagliazzo-Kabanets-W] New Test’ with Pr[YES] ≤ (1-δ)√k

Direct-product testing

Part of - local testing of codes- property testing- discrete rigidity / stability

Related to- local decoding of codes- Yao’s XOR lemma

Direct Product: Definition

For f : U R, the k -wise direct product fk : Uk Rk is

fk (x1,…, xk) = ( f(x1), …, f(xk) ).

[Impagliazzo’02, Trevisan’03]: DP Code

TT ( fk ) is DP Encoding of TT ( f )

Rate and distance of DP Code are “bad”, but the code is still very useful in Complexity …

Direct-Product Testing

Given an oracle C : Uk Rk

Test makes few queries to C, and(1) Accept if C = fk.(2) Reject if C is “far away” from any fk

(2’) [Inverse Thm] Pr [ Test accepts C ] > C fk on > () of inputs.

- Minimize # queries e.g. 2, 3,.. ? - Analyze small e.g. < 1/k , < exp(-k) ? - Reduce rate/Derandomize e.g.|C| = poly (|U|) ?

DP Testing History

Given an oracle C : Uk Rk, is C¼ gk ? #queries acc prob Goldreich-Safra’00* 20 .99Dinur-Reingold‘06 2 .99Dinur-Goldenberg‘08 2 1/kα

Dinur-Goldenberg’08 2 1/kImpagliazzo-Kabanets-W‘08 3 exp(-kα)Impagliazzo-Kabanets-W‘08* 2 1/kα

/

*Derandomization

Consistency tests

V-Test [GS00,FK00,DR06,DG08] Pick two random k-sets S1 = (B1,A), S2 =

(A,B2) with m = k1/2 common elements A.

Check if C(S1)A = C(S2)A

B1 B2

A

[DG08]: If V-Test accepts withprobability ² > 1/kα, then there is g : U Rs.t. C ¼ gk on at least ²/4 fraction of k-sets.

[IKW09]: Derandomize

[DG08]: V-Test fails for ²<1/k

S1 S2

Z-Test Pick three random k-sets S1 =(B1, A1),

S2=(A1,B2), S3=(B2, A2) with |A1| = |A2| = m =

k1/2.

Check if C(S1)A1= C(S2)A1

and C(S2)B2 = C(S3)B2

Theorem [IKW09]:

If Z-Test accepts withprobability ² > exp(-kα), then there is g : U Rs.t. C ¼ gk on at least ²/4 fraction of k-sets.

B1

B2

A1

A2

S1S2

S3

Proof Ideas

Proof steps

1. Pr [ Test accepts C ] > structure

2. Structure local agreement

3. local agreement global agreement

Agreement: there is g : U Rs.t. C ¼ gk on at least ²/4 fraction of k-sets.

Flowers, cores, petalsFlower: determined by S=(A,B)

Core: A

Core values: α=C(A,B)A

Petals: ConsA,B =

{ (A,B’) | C(A,B’)A =α }

In a flower, all petals agree on core values!

[IJKW08]: Flower analysis forDP-decoding. Symmetry arguments!

B

B4

AA B2

B3

B1

B5

V-Test ) Structure (similar to [FK, DG])

Suppose V-Test accepts with probability ².

ConsA,B = { (A,B’) | C(A,B’)A = C(A,B)A }

• Largeness: Many (²/2) flowers (A,B) have many (²/2) petals ConsA,B

• Harmony: In every large flower, almost all pairs of overlapping sets in Cons are almost perfectly consistent.

B

B4

AA B2

B3

B1

B5

V-Test: HarmonyAlmost all B1 = (E,D1) and B2 = (E,D2) in

Cons(with |E|=|A|) satisfy C(A, B1)E C(A, B2)E

B

D2

D1

AE

Proof: Symmetry between A and E (few errors in AuE )Chernoff: ² ¼ exp(-kα) E

A€

∪

Implication: Restricted to Cons, an approxV-Test on E accepts almost surely: Unique Decode!

Harmony ) Local DPMain Lemma: Assume (A,B) is harmonious. Define

g(x) = Plurality { C(A,B’)x | B’2 Cons & x 2 B’ }

Then C(A,B’)B’ ¼ gk (B’), for almost all B’ 2 Cons

B

AAD2

D1

E

Intuition: g = g(A,B) isthe unique (approximate) decoding of C on Cons(A,B)

B’x

Idea: Symmetry arguments.Largness guarantees thatrandom selections are near-uniform.Challenge: Our analysis gets stuck in ² ¼ exp(-√k)

Can one get ² ¼ exp(-k) ??

Local DP structure across Uk

Field of flowers (Ai,Bi)

For each, gi s.tC(S) ¼ gi

k (S) ifS2 Cons(Ai,Bi)

Global g?

B2

AA2

Bi

AAi

B

AA

B3

AA3B1

AA1

From local DP to global DP

Q: How to “glue” local solutions ?

A: If a typical S has two disjoint large, harmonious A’s

² > 1/kα high probability (2 queries) [DG]

² > exp(-kα) Z-test (3 queries) [IKW]

Derandomization

DP code whose length is

poly (|U|), instead of |U|k

Inclusion graphs are Inclusion graphs are SamplersSamplers

Most lemmas analyze sampling properties

m-subsets

A

Subsets: Chernoff bounds – exponential error

Subspaces: Chebychev bounds – polynomial error

Cons

S

k-subsets

x

elements of U

Derandomized DP Test Derandomized DP: U=(Fq)d Encode fk (S), S subspace of const dimension (as [IJKW08] )

Theorem (Derandomized V-Test): If derandomized V-Test accepts C with probability ² > poly(1/k), then there is a function g : U R such that C (S) ¼ gk (S) on poly(²) of subspaces S.

Corollary: Polynomial rate testable DP-code with [DG] parameters!

SummarySpherical cubes exist Power of consistency

Counterexample [DG]

For every x 2 U pick a random gx: U R

For every k-subset S pick a random x(S) 2 S

Define C(S) = gx(S)(S)

C(S1)A=C(S2)A “iff” x(S1)=x(S2)

V-test passes with high prob:

² = Pr[C(S1)A=C(S2)A] ~ m/k2

No global g if ² < 1/k2

B1 B2

AS1 S2