1. List Decoding Using the XOR Lemma Luca Trevisan U.C. Berkeley

2. Yao’s XOR Lemma If f:[N] -> {0,1} is weakly hard on average –every circuit of size s computes f correctly on at most 1-  fraction of inputs Define f +k :[N] k ->{0,1} as f +k (x1,…,xk) = f(x1)+…+f(xk) mod 2 Then f +k is very hard on average –every circuit of size ~s computes f correctly on at most ~1/2+ (1-2  k  fraction of inputs

3. Concatenation Lemma Essentially equivalent to XOR Lemma If f:[N] -> {0,1} is weakly hard on average –circuits of size s succed on at most 1-  fraction of inputs Define f k :[N] k ->{0,1} k as f k (x1,…,xk) = f(x1),…,f(xk) Then f k is very hard on average –every circuit of size ~s computes f correctly on at most ~(1-  k fraction of inputs

4. This Talk We observe: – any black-box proof of Concatenation Lemma or XOR Lemma gives way of converting ECCs with weak unique-decoding algorithms into ECCs with strong list-decoding algorithms Better codes if the Concatenation Lemma –Holds even if x1,…,xk not fully independent –Proof uses limited non-uniformity

5. This Talk We give ‘almost uniform’ version of Impagliazzo’s Concatenation Lemma for pairwise independent x1,…,xk We get codes with quadratic encoding length and quasi-linear time list-decoding (no polynomials in the construction) The uniform version of Impagliazzo’s result also gives a weak uniform version of O’Donnell’s amplification of hardness within NP (work in progress)

6. Concatenation Lemma A black-box proof argues: Let f:[N]->{0,1}, and define –F(x1,…,xk)=f(x1),…,f(xk) Let G have agreement  with F There is a circuit that computes f on 1-  fraction of inputs and –makes poly(1/ ,1/  ) oracle queries to G –has size poly(log N, 1/ , 1/ , k) (  can be as small as ~(1-  ) k )

7. Error-Correcting Code Let C:{0,1} m ->{0,1} N be ECC with decoding algorithm that corrects up to  N errors Define C’:{0,1} m ->({0,1} k ) N k –Given message M, compute C(M) –C’(M) has an entry for each k entries of C(M) –C’(M)[x1,…,xk]=C(M)[x1],C(M)[x2],…,C(M)[xk]

8. Decoding Given a corrupted G that has agreement  with C’(M), think of – C(M) as f –C’(M) as F –G as A Enumerate all exp(log N, 1/ ,1/ ,k) circuits having oracle access to A. At least one defines string having agreement 1-  with C  Apply unique decoding algorithm for C() to each string in list List will contain M

9. How to Improve Encoding length is N k –Shorter encoding length if x1,…,xk not fully independent –Impagliazzo proves concatenation lemma for pairwise independent x1,…,xk. Proof inherently non-uniform List size and decoding time are quasi-polynomial –Shorter list / faster decoding if reduction is uniform –Levin’s and GNW’s proofs are uniform, but need full independence of x1,…,xk We give almost uniform proof of pairwise- independent concatenation lemma –Quadratic encoding length –Quasi-Linear decoding time

10. Impagliazzo’s Proof 1.If f is hard to solve on more than 1-  fraction of inputs. –Then there is a set H containing  fraction of inputs and f is hard to solve on more than ½+  fraction of H 2.If f is hard to solve on more than ½+  fraction of set H of density  –then F(a,b)=f(a+b),…,f(a+kb) hard to solve on more than O(1/  k) fraction of inputs

11. 1 st Part: Hard-Core Sets Thm: –If f cannot be solved on (1-  ) fraction of inputs with circuits of size s, –Then there is set H of size  N such that f cannot be solved on (½+  ) fraction of H with circuits of size s*poly( ,  ) Equivalently: –If for every H of size  N there is circuit of size s that solves f on (½+  ) fraction of H –Then there is circuit of size s*poly(1/ ,1/  ) that solves f on (1-  ) fraction of inputs

12. Impagliazzo’s Proof Set/function Game. At step i: –Player A produces a set H i of size  N –Player B produces a function g i that agrees with f i on ½ +  fraction of H i Player A wins at step t if g(x)=maj{g 1 (x),…,g t (x)} agrees with f on 1-  fraction of inputs Thm [Imp95]: there is a winning strategy for A that suceeds in poly(1/ ,1/  ) steps

13. Impagliazzo’s Proof If for every H there is circuit of size s and agreement ½+  with f on H –Let Player A play Impagliazzo’s strategy –Let Player B always reply with a circuit size s Construct size s*poly(1/ ,1/  ) circuit that computes f on 1-  fraction of inputs

14. Uniform Version Thm: suppose there is distrib C over circuits such that for every H of size  N –Pr C~C [ Pr x~H [C(x)=f(x)]> ½ +  ] >  Then there is distrib C’ over circuits such that –Pr C~C’ [ Pr x [C(x)=f(x)]> 1 -  ] >  poly(1/ ,1/  ) Proof: pick t=poly(1/ ,1/  ) ckts from C and take majority. There is at least probability  t that it works.

15. 2 nd Part: Pairwise Independence Suppose f is (1-  )-hard, but algorithm A computes –F(a,b)=f(a+b),f(a+2b),…,f(a+kb) On  ~1/  k fraction of inputs Let H be set of size  k. –[Imp95]: can compute f on ½+  ’ frac of H. Let A(a,b)= A1(a,b),A2(a,b),…,Ak(a,b) Define 0/1 random variables Z1,…,Zk –Zi=1 iff Ai(a,b)=f(a+ib) AND a+ib is in H

16. Studying the Zi A(a,b)= A1(a,b),A2(a,b),…,Ak(a,b) Zi=1 iff Ai(a,b)=f(a+ib) AND a+ib is in H Note: E[Zi]= Pr[Ai(a,b)=f(a+ib)|a+ib in H]  Suppose E[Zi]>  (1/2 +  ’) or E[Zi] <  (1/2-  ’) Then easy to solve f on H better than ½ Remains to consider case all E[Zi] ~  /2

17. Studying the Zi A(a,b)= A1(a,b),A2(a,b),…,Ak(a,b) Zi=1 iff Ai(a,b)=f(a+ib) AND a+ib is in H E[Zi] ~  /2 for all i Suppose the Zi are almost pairwise independent Then sum of Zi concentrated around k  /2 Number of a+ib in H concentrated around k  Impossible for A to have noticeable prob of being correct on all inputs

18. Studying the Zi A(a,b)= A1(a,b),A2(a,b),…,Ak(a,b) Zi=1 iff Ai(a,b)=f(a+ib) AND a+ib is in H There are i,j such that E[ZiZj] >   /4 +  ’’ Equivalent: Pr[ Ai(a,b)=f(a+ib) AND Aj(a,b)=f(a+jb)]> ¼+  ’’ conditioned on a+ib and a+jb are in H Equivalent: pick x,y in H Pr[ Ai(a,b)=f(x) AND Aj(a,b)=f(y)]> ¼+  ’’ where a,b such that a+ib=x and a+jb=y

19. Using the Dependency Pick x,y in H Pr[ Ai(a,b)=f(x) AND Aj(a,b)=f(y)]> ¼+  ’’ where a,b such that a+ib=x and a+jb=y Then one of: –Pr[ Ai(a,b)=f(x)]>1/2 +2  ’’/3 –Pr[ Aj(a,b)=f(y)]>1/2 +2  ’’/3 –Pr[ Ai(a,b) XOR Aj(a,b) XOR f(y)=f(x)]>1/2 +2  ’’/3 In each case we get circuit for f on H

20. Uniform Version Let f be a function and A be an algorithm that computes –F(a,b)=f(a+b),f(a+2b),…,f(a+kb) on  ~1/  k fraction of inputs Then can define distribution of circuits such that for each set H there is prob at least poly( , ,1/k) that circuit computes f on H on more than ½+  ’ inputs Proof: replace non-uniformity in Impagliazzo’s argument with random choices.

21. Everything Together Suppose function f and algorithm A are so that –F(a,b)=f(a+b),f(a+2b),…,f(a+kb) agrees with A on  ~1/  k fraction of inputs Then can sample distribution of circuits such that there is prob. exp(1/ ,1/ ,k) of sampling a circuit that agrees with f on 1-  frac of inputs Also can produce list of size exp(1/ ,1/ ,k) that contains whp circuit  -close to f

22. Coding Application From –binary error-correcting code with codewords of length N and unique decoding algorithm for  fraction of errors Error-correcting code with –alphabet of size 2 k, –codewords of length N 2 –list decoding up to 1-O(1/  k) errors, with list of size exp(1/ ,k) –implicit representation of list is computed in polylog N time. Explicit representation in Npolylog N time.

23. Comparison to Previous Work Sudan’97: –linear encoding length + –quasi-linear encoding time + –polynomial list-decoding time – –list is polynomial in 1/  + Feng’99, Alenkkhnovich’02: –improve to quasi-linear decoding time = Guruswami-Indyk’01-02 –Even better rates but quadratic decoding time +/- –Do not use polynomials =

24. Possible Improvement Prove a Concatenation Lemma for almost pairwise independent inputs. As in BSVW’03, let F t be vector space and S be small bias space, then consider –F(a,b)=f(a+b),f(a+2b),…,f(a+kb) Where a ranges in F t and b ranges in S Whole argument works out up to the point –Pr[f(a+ib)=A i (a,b) XOR A j (a,b) XOR f(a+jb)]>½

25. Other Applications O’Donnell proves a (non-uniform) amplification of hardness result in NP using Impagliazzo’s hard-core sets. We can prove: –Let f be NP function, can construct f’ such that –If f’ has BPP algorithm that works on ½+  fraction of inputs –Then f has BPP algorithm that produces exp(1/ ,1/  ) circuits, one of them solves f on 1-  fraction of inputs

26. How to Choose the Circuit Suppose every problem in NP has BPP algorithm that works on ¾+  ’ fraction of inputs Let f be NP function, C1,…,Cl circuits such that one of them solves f on 1-  fraction of inputs Define –F(x1,…,xt)=g(f(x1),…,f(xt)) –Di(x1,…,xt)=g(Ci(x1),…,Ci(xt)) Where g() is noise-sensitive function

27. How To Choose The Circuit Define –F(x1,…,xt)=g(f(x1),…,f(xt)) –Di(x1,…,xt)=g(Ci(x1),…,Ci(xt)) If Ci  -close to f then Di  t-close to F If Ci  ’-far from f then Di (½+  ’’)-far from F We have algorithm to compute F on > ¾ of inputs, can distinguish two cases

28. Uniform Result Suppose for every NP problem there is BPP algorithm that works on 1-1/(log n) c fraction of inputs Then for every NP problem there is BPP algorithm that works on ¾+1/(log n) c fraction of inputs (Proof not completely written up)

29. Conclusions ``Everything's got a moral, if only you can find it.’' (Lewis Carroll, Alice's Adventures in Wonderland)

30. Conclusions An information-theoretic interpretation of black- box concatenation lemmas –Conversion of weaker to stronger error-correcting codes In coding theory –Suggests a new technique for list-decoding –What is it? In complexity theory –Emphasis on two aspects of Concatenation Lemma proofs: non-uniformity and derandomization