direct-product testing parallel repetitions and foams avi wigderson ias
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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
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