Locally Testable Codes Analogues to the Unique Games Conjecture Do Not Exist

1. Locally Testable Codes Analogues to the Unique Games Conjecture Do Not Exist Gillat Kol joint work with Ran Raz

2. 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

3. The PCP Theorem

4. 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

5. The PCP Theorem i j q p y m b p r y w u t Verifier Probabilistically Checkable Proofp • (2 queries) • Toss coins to getlocations i and j • Query pi and pj • Using pi and pj, decide if to accept

6. The Unique Games Conjecture

7. 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 results Max-Cut [MOO ‘05, KKMO ‘07], Vertex-Cover [KR ‘08], CSPs [Rag ‘08], …

8. 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 locationj thatmakes V accept (and vice versa)

9. The Unique Games Conjecture Unique Games Conjecture [Khot ‘02]: ,s > 0consts  (const size depends on ,s) s.t. V checking proofs for “SAT” over by onlyperformingunique 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

10. Locally Testable Codes

11. 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

12. Locally Testable Codes Locally Testable Code: A code C with a tester (probalgo) that checks if a given word v is in C by only reading a constant number of locations Completeness 1-: vCaccept wp≥ 1- Soundness s: dist(v,C) > 1/3  acceptwp< s equiv.acceptwp s uC, agree(u,v)  2/3

13. Low Soundness LTCs Soundness (review):dist(v,C) > 1- = 1/3  acceptwp< 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- acceptwp< s() equiv. acceptwp≥ s() uC, agree(u,v)  

14. PCPs and LTCs Both PCPverifiers 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

15. “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: AUnique LTC is an LTC with unique tests Low-error: Inknown PCPs, the error originates from the completeness, soundness, and distance of the LTC used Thus, we would have wanted:  > 0 const, LTC with , < ands() < for some

16. Our Results

17. Our Result Theorem (Main): Let n, , s:(0,1)[0,1] be arbitrary(monotone) Assume s()  10-5for some fixed  Denote c1 = 10-102and c2 = 1010||/ LetC nbe an (,,s)-unique LTC. If, c1then|C|  c2 • I.e., fixing s fixes a const c1, s.t. and cannot both be smaller thanc1, 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|=||

18. Proof

19. Constraint Graphs Proof by way of contradiction: Let C be such a unique LTCwith 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 vsatisfies the edge (i,j)if ij(vi) = vj

20. Step 1 (Main): Decompose 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  210-4e edges (e = #edges in G) G* G  n vertices  210-4e edges

21. 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 ofG*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*

22. 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 everycomponentof 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!

23. Graph Decomposition(Main)

24. Decomposition (Review) 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  210-4e edges (e = #edges in G) G* G  n vertices  210-4e edges

25. Decomposition Algo: First Attempt Decomposition Algorithm: Repeat Select two new codewords u w Disconnect A, the set of coordinates uandw agree on Ais small:|A|n  (relative distance) What about the number of removed edges? G A = {i: ui = wi}

26. How Many Edges Removed? ij j i • Observation:Each removed edge (i,j) violates either uorw • Proof:Assume iA,and (i,j) satisfied by both uandw. Then, uj = ij (ui) = ij(wi) = wj jA • Conclusion: 2e edges were removed (Completeness) A = {i: ui = wi} G

27. 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!

28. Cutting Down Expenses Observation (review):Each removed edge violates uorw Denote:Ev= set of edges violated by the word v Ev∩A= edges in Evwith end-point in A Observation’: We only need to remove edges in Eu∩Aand Ew∩A! Assume Ais a random set of size n,andGis regular v,  frac of the edges in Ev are in Ev∩A Thus, each iteration removes  2frac of the edges

29. ButA= 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…

30. Thank You!