Gillat Koljoint work with Ran Raz

Locally Testable Codes Analogues to the

Unique Games Conjecture Do Not Exist

Summary• The Unique Games Conjecture (UGC) is an important

open problem in the study of PCPs• It conjectures the existence of PCPs with special

properties• Known PCP constructions are based on Locally

Testable Codes (LTCs) with analogues properties• We show that LTCs with properties analogues to the

UGC do not exist• Thus, show limitations of some of the current PCP

constructions techniques

The PCP Theorem

The PCP Theorem• A unbounded prover wants to convince a poly-time

verifier that SAT, by supplying a proof• The verifier wants to only read constant number of

symbols from the proof

• PCP Thm [BFL,FGLSS,AS,ALMSS ‘92]: This can be done!‐ Completeness: SAT proof accepted whp‐ Soundness: SAT proof rejected whp

The PCP Theorem

Probabilistically Checkable Proof p

i j

(2 queries)

1. Toss coins to get locations i and j2. Query pi and pj

3. Using pi and pj, decide if to accept

b q p myp wyr u t

Verifier

The Unique Games Conjecture

Why is the UGC Interesting?• Almost all hardness of approximation results rely on

the PCP Theorem• Yet, for many fundamental problems, optimal

hardness results are still not know• The UGC is a strengthening of the PCP Theorem

shown to imply many improved hardness resultsMax-Cut [MOO ‘05, KKMO ‘07], Vertex-Cover [KR ‘08], CSPs [Rag ‘08],…

Unique Tests• The UGC deals with verifiers V that read 2 locations

and only make unique tests:i,j queried by V permutation ij: s.t.

V accepts iff ij(pi) = pj

• That is, after reading location i, there exists a unique value for location j that makes V accept (and vice versa)

The Unique Games Conjecture

Unique Games Conjecture [Khot ‘02]:,s > 0 consts (const size depends on ,s) s.t. V checking proofs for “SAT” over by only performing unique tests Completeness 1-: SAT proof accepted wp ≥ 1- Soundness s: SAT proof accepted wp < s

Parallel Repetition Theorem [Raz ‘98]: Such a verifier exists when uniqueness is relaxed to projection

Locally Testable Codes

Error Correcting Codes• Hamming Distance:

‐ dist(u,w) = frac of coordinates u and w disagree on‐ agree(u,w) = frac of coordinates u and w agree on

• Error Correcting Code: C n

• Relative Distance: C has relative distance 1- if u w C, dist(u,w) ≥ 1-

equiv. agree(u,w) High relative distance Good error correcting ability

Locally Testable CodesLocally Testable Code: A code C with a tester (prob algo) that checks if a given word v is in C by only reading a constant number of locations Completeness 1-: vC accept wp ≥ 1- Soundness s: dist(v,C) > 1/3 accept wp < s

equiv. accept wp s uC, agree(u,v) 2/3

Low Soundness LTCs• Soundness (review): dist(v,C) > 1- = 1/3 accept wp < s• Observation: s cannot be lower than #queries

s is proportional to : Can only expect low accept prob (small s) for words that are far from the code (small )

• Soundness (generalized): Let s():(0,1)[0,1] be arbitrary (monotone) function

dist(v,C) > 1- accept wp < s() equiv. accept wp ≥ s() uC, agree(u,v)

PCPs and LTCs• Both PCP verifiers and LTC testers test if a given string is

“close” to being “good” (good = valid proof /codeword) by reading only a constant number of locations in it

• Known PCP constructions are based on LTCs with analogues properties

“LTCs Analogues to the UGC”?• (,,s)-LTC: , > 0, s:(0,1)[0,1]

Relative distance 1- (codewords agree frac) Completeness 1- (codewords accepted wp 1-) Soundness s() (dist > 1- accept wp < s())

• The UGC requires a low-error PCP with unique tests

• Uniqueness: A Unique LTC is an LTC with unique tests• Low-error: In known PCPs, the error originates from the

completeness, soundness, and distance of the LTC usedThus, we would have wanted:

> 0 const, LTC with , < and s() < for some

Our Results

Our Result

Theorem (Main):Let n, , s:(0,1)[0,1] be arbitrary (monotone)Assume s() 10-5 for some fixed Denote c1 = 10-102 and c2 = 1010||/

Let C n be an (,,s)-unique LTC. If , c1 then |C| c2

• I.e., fixing s fixes a const c1, s.t. and cannot both be smaller than c1, unless C is of const size

• Some Tightness: = {a, b, c, …}, C = {an, bn, cn, …}. C is a unique-LTC with ==0 (test: vi = vj), and |C|=||

Proof

Constraint Graphs• Proof by way of contradiction:

Let C be such a unique LTC with tester T • T can be viewed as a constraint graph G

‐ Vertex set = [n]‐ There exist an edge (i,j) if T may query locations (i,j) ‐ The edge (i,j) is associated with ij

• A word v satisfies the edge (i,j) if ij(vi) = vj

Step 1 (Main): Decompose GDecompose G to small connected components by removing only a small number of edges (obtain G*)• Each connected component of G* contains n vertices • G* contains 210-4e edges (e = #edges in G)

n vertices

210-4e edges

G*G

Step 2: Constructing a “Bad” Word• Set k 1/ constant• Partition the connected components of G* to k sets, each

containing n/k vertices (components of G* are small) • Let v* be “balanced” hybrid of any k different codewords

(|C| large), agreeing with each on one of the k parts of G*

G*

• v* is far from the code: ‐ v* is a hybrid of codewords‐ Codewords disagree on most coordinates (relative dist) ‐ v* cannot agree with either on many coordinates

• v* is accepted with non-negligible prob: ‐ On every component of G*, v* agrees with a codeword‐ On this component, v* only violates the edges violated

by the codeword‐ v* satisfies most of the edges in G* (Completeness)‐ v* satisfies many edges (G* contains many edges)

v* violates soundness!

v* Violates Soundness

Graph Decomposition(Main)

Decomposition (Review)

n vertices

210-4e edges

G*G

Decompose G to small connected components by removing only a small number of edges (obtain G*)• Each connected component of G* contains n vertices • G* contains 210-4e edges (e = #edges in G)

Decomposition Algo: First Attempt• Decomposition Algorithm:

Repeat‐ Select two new codewords u w‐ Disconnect A, the set of coordinates uand w agree on

• A is small: |A| n (relative distance)

• What about the numberof removed edges?

GA = {i: ui = wi}

How Many Edges Removed?

A = {i: ui = wi}

iijjG

• Observation: Each removed edge (i,j) violates either uor w• Proof: Assume iA, and (i,j) satisfied by both uand w.

Then, uj = ij (ui) = ij(wi) = wj jA • Conclusion: 2e edges were removed (Completeness)

Is 2 Good Enough? No!

We still may be removing too many edges:• |A| n /|| (assume C is a random code)

To decompose G, repeat || times

• Each iteration removes up to 2 frac of the edges Algo removes up to 2|| frac of the edges

• Recall that || may be much larger than 1/ All edges may be removed!

Cutting Down Expenses• Observation (review): Each removed edge violates uor w• Denote: Ev = set of edges violated by the word v

Ev∩A = edges in Ev with end-point in A

• Observation’: We only need to remove edges in Eu∩A and Ew∩A!

Assume A is a random set of size n, and G is regular• v, frac of the edges in Ev are in Ev∩A • Thus, each iteration removes 2 frac of the edges

But A = agree(u,w) Is Not Random• Fix u. Since there are many codewords, u cannot agree

with all on roughly the same set of coordinates• Thus, random selection of w yields “random enough” A

Most of the proof is devoted to showing that…

Thank You!