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This document delves into the intriguing intersection of direct-product testing, parallel repetitions in games, and the isoperimetric problem across various dimensions. We analyze the quest for minimizing surface area while given a volume, drawing connections to optimal shapes such as spheres and complex tilings. In addition, the discussion extends to probabilistic constructions and theoretical frameworks that push the boundaries of understanding in mathematical physics, chemistry, and engineering, highlighting significant historical contributions and open questions in the realm of complexity and coding theory.
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Direct-Product testing Parallel Repetitions And Foams Avi Wigderson IAS
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-Phelan1994 Even better
Our Problem Minimum surface area of body tiling Rd with period Zd ? Volume=1 d=2 area: Choe’89: Optimal! >4 4
[Kindler,O’Donnell, Rao,Wigderson] Bounds in d dimensions ≤OPT≤ ≤ 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 y 1 Bound does not depend on d
Spine Surface blocking all cycles that wrap around Torus
Probabilistic construction of spine Step 1 Randomly construct B in [0,1)d , which in expectation satisfies B 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 elimination 2) 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 Game G Draw j [m] at random Xij Yij Alice Bob αj βj Check if αj +βj = bj Pr [YES] ≤ 1-δ 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
Hardness amplification byparallel repetition Game Gk Draw j1,j2,…jk [m] at random Xij1Xij2 XijkYij1Yij2 Yijk Alice Bob αj1αj2 αjkβj1βj2 βjk Check if αjt +βjt = bjt t [k] Pr[YES] ≤ (1-δ2)k [Raz,Holenstein,Rao] Pr[YES] ≥ (1-δ2)k 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 [Feige-Kindler-O’Donnell] Spherical Cubes X [Raz] [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…XijkYij1…Yijk Alice Bob αj1…αjkβj1…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] NewTest’ with Pr[YES] ≤ (1-δ)√k
Direct-product testingPart of - local testing of codes- property testing- discrete rigidity / stabilityRelated to- local decoding of codes- Yao’s XOR lemma
Direct Product: Definition • For f : U R, the k -wise direct productfk : Uk Rk is fk (x1,…, xk) = ( f(x1), …, f(xk) ). [Impagliazzo’02, Trevisan’03]: DP Code TT (fk)is DP Encoding ofTT ( 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 anyfk • (2’) [Inverse Thm] Pr [Test accepts C ] > • Cfk on > t() 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 .99 Dinur-Reingold‘062 .99 Dinur-Goldenberg‘082 1/kα Dinur-Goldenberg’082 1/k Impagliazzo-Kabanets-W‘083 exp(-kα) Impagliazzo-Kabanets-W‘08* 2 1/kα / *Derandomization
V-Test[GS00,FK00,DR06,DG08] Pick two random k-setsS1 = (B1,A), S2 = (A,B2) withm = k1/2common elements A. Check if C(S1)A = C(S2)A [DG08]: IfV-Testaccepts with probability ² > 1/kα, then there is g : U R s.t. C ¼gkon at least ²/4fraction of k-sets. [IKW09]: Derandomize [DG08]: V-Testfails for ²<1/k B1 B2 S1 S2 A
Z-Test Pick three random k-setsS1 =(B1, A1), S2=(A1,B2), S3=(B2, A2) with|A1| = |A2| = m = k1/2. Check if C(S1)A1= C(S2)A1andC(S2)B2 = C(S3)B2 Theorem [IKW09]: IfZ-Testaccepts with probability² > exp(-kα), then there is g : U R s.t. C ¼gkon at least ²/4fraction of k-sets. B1 A1 S1 S2 B2 A2 S3
Proof steps • 1. Pr [ Test accepts C ] > structure • 2. Structure local agreement • 3. local agreement global agreement • Agreement: there is g : U R • s.t. C ¼gkon at least • ²/4fraction of k-sets.
Flowers, cores, petals Flower: 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 for DP-decoding. Symmetry arguments! B1 B B2 A A B3 B5 B4
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 Consare • almost perfectly consistent. B1 B B2 A A B3 B5 B4
V-Test: Harmony Almost all B1 = (E,D1) andB2 = (E,D2) in Cons (with |E|=|A|) satisfy C(A, B1)E C(A, B2)E Proof: Symmetry between A and E (few errors in AuE) Chernoff: ²¼ exp(-kα) D1 B E Implication: Restricted to Cons, an approx V-Test on E accepts almost surely: Unique Decode! A D2 A E
Harmony ) Local DP Main 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 D1 Intuition: g = g(A,B) is the unique (approximate) decoding of C on Cons(A,B) B E D2 A A Idea: Symmetry arguments. Largness guarantees that random selections are near-uniform. x B’ Challenge: Our analysis gets stuck in²¼ exp(-√k) Can one get ²¼ exp(-k)??
Local DP structure across Uk Field of flowers (Ai,Bi) For each, gi s.t C(S)¼gik (S) if S2 Cons(Ai,Bi) Global g? B2 B1 B3 Bi B A3 A2 A1 Ai A A A A A A
From local DP to global DPQ: How to “glue” local solutions ?A: If a typical S has two disjoint large, harmonious A’s² > 1/kα high probability (2 queries) [DG]² > exp(-kα)Z-test (3 queries) [IKW]
DerandomizationDP code whose length ispoly (|U|), instead of |U|k
m-subsets A Inclusion graphs are Samplers Most lemmas analyze sampling properties of U k-subsets elements S Cons x Subsets: Chernoff bounds – exponential error Subspaces: Chebychev bounds – polynomial error
Derandomized DP Test Derandomized DP: U=(Fq)d Encodefk (S), S subspaceof 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!
Summary Spherical 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 S1 S2 A