Simple Extractors for all Min-Entropies

1. Simple Extractors for all Min-Entropies R.Shaltiel and C.Umans

2. Definitions Def (min-entropy): The min-entropy of a random variable X over {0, 1}n is defined as: Thus a random variable X has min-entropy at least k if Pr[X=x]≤2-k for all x. The maximum possible min-entropy for such a R.V. is n Def (statistical distance): Two distributions on a domain D are e-close if the probabilities they give to any AD differ by at most e (namely, using norm 1)

3. Weak random source n E Random string m Seed t Definitions Def (extractor): A (k,e)-extractor is a functionE:{0,1}n  {0,1}t{0,1}ms.t. for any R.V. X with min-entropy ≥kE(X,Ut) is e-close to Um(where Um denotes the uniform distribution over {0,1}m)

4. Weak random source n E Random string m Seed t Parameters The relevant parameters are: • min entropy of the weak random source input – k. Relevant values log(n) k  n(the seed length is t ≥ log(n), hence useless to consider lower min entropy). • seed lengtht ≥ log(n) . • Quality of the output: e. • Size of the output m=f(k). The optimum is m=k.

5. Extractors High Min-Entropy distribution Uniform-distribution seed 2t 2n 2m E Close to uniform output

6. Next Bit Predictors Claim: to prove E is an extractor, it suffices to prove that for all 0<i<m+1 and all predictorsf:{0,1}i-1{0,1} Proof: Assume E is not an extractor; then exists a distribution s.t. X s.t. E(X,Ut) is note-close to Um, that is:

7. Proof Now define the following hybrid distributions:

8. Proof Summing the probabilities for the event corresponding to the set A for all distributions yields: And because |∑ai|≤ ∑|ai| there exists an index 0<i<m+1 for which:

9. The Predictor We now define a function f:{0,1}i-1{0,1} that can predict the i’th bit with probability at least ½+e/m (“a next bit predictor”): The function f uniformly and independently draws the bits yi,…,ym and outputs: Note: the above definition is not constructive, as A is not known!

10. Proof And fis indeed a next bit predictor: Q.E.D.

11. Basic Example – Safra, Ta-Shma, Zukerman Construction: • Let BC:F{0,1}s be a (inefficient) binary-code • Given • x, a weak random source, interpreted as a polynomial x:F2F and • s, a seed, interpreted as a random point (a,b), and an index j to a binary code. • Def:

12. (a,b+1) (a,b+m) x(a,b+m) x(a,b) x(a,b+1) 001 001 110 110 000 000 101 101 110 110 Basic Example – Illustration of Construction • x  x, s = ((a,b), 2) • E(x,s)=01001 (a,b) (inefficient) binary code

13. Basic Example – Proof Sketch • Assume, by way of contradiction:exists a next bit predicator function f. • Next, show a reconstructionfunction R • Conclude, a contradiction!(to the min-entropy assumption of X)

14. h ~ n1/2 j ~ lgn m ~ desired entropy Basic Example – Reconstruction Function Random line “advice” “Few” red points: a=mjO(h) Repeat using the new points, until all Fd is evaluated List decoding by the predictor f Resolve into one value on the line

15. Counting Argument For Y X, let (Y)=yYPr[y] (“the weight of Y”) Let R:{0,1}a{0,1}n, s.t. Prx~X[z R(z)=x] 1/2 • (for a uniform X, |R(S)|  |X|/2 ) • For an arbitrary distribution X, (R(S)) (X)/2 • Let X ~ min-entropy  k, • then (R(S))2a-k(there are at most 2a strings in R(S), and xX Pr[x]  2-k) • and therefore k  a - log2(1/2)(1 = (X)  (R(S)) 22a-k 2-1  a-k hence k  a+1) 2nX R(S) R 2aS

16. Problems with Safra, Ta-Shma, Zukerman • Curse of dimensionality - too many lines!Solution: generator matrix.

17. Next-q-it List-Predictor f is allowed to output a small list of l possible next elements

18. q-ary Extractor Def: Let F be a field with q elements. A (k, l)q-ary extractor is a functionE:{0,1}n  {0,1}tFms.t. for all R.V. X with min-entropy ≥k and all 0<i<m and all list-predictors f:Fi-1Fl

19. Generator Matrix Def: Define the generator matrix for the vector space Fd as a matrix Ad×d, s.t. for any non-zero vector vFd: (that is, any vector 0≠vFd multiplied by all powers of A generates the entire vector space Fd except for 0) Lemma: Such a generator matrix exists and can be found in time qO(d).

20. Note that for such a polynomial, the number of coefficients is exactly: (“choosing where to put d-1 bars between h-1 balls”) Construction • Let F be a field with q elements, • Let Fd be a vector space over F. • Let h be the smallest integer s.t. • For x {0,1}n, let xdenote the unique d-variate polynomial of total degree h-1 whose coefficients are specified by x.

21. x(Aiv) Amv x(v) x(Amv) Aiv Fd v v Aiv Amv Construction • The definition of the q-ary extractor: E:{0,1}n {0,1}d log qFm seed, interpreted as a vector v Fd Generator matrix

22. Main Theorem Thm: For any n,q,d and h as previously defined, E is a (k, l)q-ary extractor if: Alternatively, E is a (k, l)q-ary extractor if:

23. What’s Ahead • Proving existence of a generator matrix • How the counting argument works • The reconstruction paradigm • Basic example – Safra, Ta-Shma, Zukerman • Proof of the main theorem • From extractors to PRGs

24. Extension Fields A field F2 is called an extension of another field F if F is contained in F2 as a subfield. Thm: For every power pk (p prime, k>0) there is a unique (up to isomorphism) finite field containing pk elements. These fields are denoted GF(pk).All finite fields’ cardinality have that form. Def: A polynomial is called irreducible in GF(p) if it does not factor over GF(p) Thm: Let f(x) be an irreducible polynomial of degree k over GF(p). The finite field GF(pk) can be constructed using the set of degree k-1 polynomials over Zp, with addition and multiplication carried out modulo f(x)

25. Extension Fields - Example Construct GF(25) as follows: Let the irreducible polynomial be: Represent every k degree polynomial as a vector of k+1 coefficient: Addition over this field:

26. Extension Fields - Example And multiplication: And now modulo the irreducible polynomial:

27. Generator Matrix – Existence Proof Denote by GF*(qd) the multiplicative group of the Galois Field GF(qd). This multiplicative group of the Galois Field is cyclic, and thus has a generator g: Let jbe the natural isomorphism between the Galois Field GF(qd) and the vector space Fd, which matches a polynomial with its vector of coefficients:

28. Generator Matrix – Existence Proof Now define the generator matrix A of Fd as the linear transformation that corresponds to multiplication by the generator in GF*(qd) : A is a linear transformation because of the distributive property of both the vector space and the field GF(qd), according to the isomorphism properties:

29. Generator Matrix – Existence Proof It remains to show that the generator matrix A of Fd can be found in time qO(d). And indeed: • The Galois Field GF(qd) can be constructed in time qO(d) using an irreducible polynomial of degree d over the field Zq (and such a polynomial can also be found in time qO(d) by exhaustive search). • The generator of GF(qd) can be found in time qO(d) by exhaustive search • Using the generator, for any basis of Fd, one can construct d independent equations so as to find the linear transformation A.This linear equation system is also solvable in time qO(d) .

30. “Reconstruction Proof Paradigm” Proof sketch: For a certain R.V. X with min-entropy at least k, assume a function f that violates the properties of a q-ary extractor, construct another function, R :{0,1}a{0,1}n, the “reconstruction function”. This function, using f as a procedure, has the property that: Applying the “counting argument”, this is a contradiction to the assumption that X has min-entropy at least k

31. Proof Sketch • Let X be a random variable with min-entropy at least k • Assume, by way of contradiction:exists a next bit predicator function f. • Next, show a reconstructionfunction R • Conclude, a contradiction!(to the min-entropy assumption of X)

32. Main Lemma Lemma: Let n,q,d,h be as in the main theorem. There exists a probabilistic function R:{0,1}a{0,1}n with a = O(mhd logq) such that for every x on which: The following holds (the probability is over the random coins of R):

33. The Reconstruction Function (R) • Task: allow many strings x in the support of X to be reconstructed from very short advice strings. • Outlines: • Use f in a sequence of prediction steps to evaluate z on all points of Fd,. • Interpolate to recover coefficients of z, • which gives x Next We Show: there exists a sequence of prediction stepsthat works for manyx in the support of X and requires few advice strings

34. Curves • Let r=Q(d), • Pick random vectors and values • 2r random points y1,…,y2rFd, and • 2r values t1,…,t2rF, and • Define degree 2r-1 polynomials p1,p2 • p1:FFd defined by p1(ti)=yi, i=1,..,2r. • p2:FFd defined by p2(ti)=Ayi, i=1,..,r, and p2(ti)=yi, i=r+1,..,2r. • Define vector sets P1={p1(z)}zF and P2={p2(z)}zF • i>0 define P2i+1=AP2i-1 and P2i+2=AP2i({Pi}, the sequence of prediction stepsare low-degree curves in Fd, chosen using the coin tosses of R)

35. Ai*(y2) A(y2) A2(y2) A3(y2) A2(y1) Ai*(y1) A(y1) A3(y1) A2(yr) Ai*(yr) A(yr) A3(yr) Ai*(y2) Ai*(yr+1) Ai*(y1) Ai*(y2r) Ai*(yr) Amv A(y2) A2(y2) A2(yr+1) A(yr+1) A(y1) A2(y1) A2(y2r) A(y2r) A(yr) A(yr) Amv y2 Aiv yr+1 Aiv y1 y2r yr A2(yr+1)) yr+1 A(yr+1)) Ai*-1(yr+1)) v t1 t2 tr tr+1 A(y2r) Ai*-1(y2r) A2(y2r) y2r t2r v Curves Fd F

36. Simple Observations • A is non-singular linear-transform, hence i • Pi is 2r-wise independent collection of points • Pi and Pi+1 intersect at r random points • z|Pi is a univariate polynomial of degree at most 2hr. • Given evaluation of z on Av,A2v,…,Amv, we may use the predictor function f to predict z(Am+1v) to within l values. • We needadvice string: 2hr coefficients of z|Pi for i=1,…,m. (length: at most mhr log q ≤ a)

37. A(y2) A2(y2) A3(y2) Ai*(y2) Ai*(y1) A2(y1) A(y1) A3(y1) Ai*(yr) A(yr) A2(yr) A3(yr) Ai*(y2) Ai*(yr+1) Ai*(y1) Ai*(y2r) Ai*(yr) Amv A2(y2) A(y2) A2(yr+1) A(yr+1) A(y1) A2(y1) A(y2r) A2(y2r) A(yr) A(yr) y2 Aiv yr+1 y1 y2r yr A(yr+1)) yr+1 A2(yr+1)) Ai*-1(yr+1)) v t1 t2 tr tr+1 y2r A2(y2r) Ai*-1(y2r) A(y2r) t2r Using N.B.P. Cannot resolve into one value! Fd F

38. Ai*+1(y2) A2(y2) A(y2) Ai*(y2) A3(y2) Ai*+1(y1) Ai*+1(yr) Ai*(y1) A2(y1) A(y1) A3(y1) A3(yr) Ai*(yr) A2(yr) A(yr) Ai*(y2) Ai*(yr+1) Ai*(y1) Ai*(y2r) Ai*(yr) Amv A2(y2) A(y2) A(yr+1) A2(yr+1) A(y1) A2(y1) A(y2r) A2(y2r) A(yr) A(yr) y2 Aiv yr+1 y1 y2r yr yr+1 Ai*-1(yr+1)) A(yr+1)) A2(yr+1)) v t1 t2 tr tr+1 Ai*-1(y2r) A(y2r) y2r A2(y2r) t2r Using N.B.P. Can resolve into one value using the second curve! Fd F

39. yr+1 y2r Ai*+1(y2) Ai*(y2) A3(y2) A(y2) A2(y2) Ai*+1(y1) Ai*+1(yr) A(y1) Ai*(y1) A3(y1) A2(y1) A(yr) Ai*(yr) A2(yr) A3(yr) Ai*(y2) Ai*(yr+1) Ai*(y1) Ai*(y2r) Ai*(yr) Amv A(y2) A2(y2) A(yr+1) A2(yr+1) A2(y1) A(y1) A(y2r) A2(y2r) A(yr) A(yr) y2 Aiv yr+1 y1 y2r yr yr+1 A(yr+1)) Ai*-1(yr+1)) A2(yr+1)) v t1 t2 tr tr+1 y2r A2(y2r) Ai*-1(y2r) A(y2r) t2r Using N.B.P. Can resolve into one value using the second curve! Fd F

40. Main Lemma Proof Cont. • Claim: with probability at least 1-1/8qd over the coins tosses of R: • Proof: We use the following tail bound: Let t>4 be an even integer, and X1,…,Xn be t-wise independent R.V. with values in [0,1]. Let X=Xi, =E[X], and A>0. Then:

41. Main Lemma Proof Cont. • According to the next bit predictor, the probability for successful prediction is at least 1/2√l. • In the i’th iteration we make q predictions (as many points as there are on the curve). • Using the tail bounds provides the result. Q.E.D (of the claim). Main Lemma Proof (cont.): Therefore, w.h.p. there are at least q/4√l evaluations points of Pithat agree with the degree 2hr polynomial on the i’th curve (out of a total of at most lq).

42. Main Lemma Proof Cont. • A list decoding bound: given n distinct pairs (xi,yi) in field F and Parameters k and d, with k>(2dn)1/2, There are at most 2n/k degree d polynomials g such that g(xi)=yi for at least k pairs. Furthermore, a list of all such polynomials can be computed in time poly(n,log|F|). • Using this bound and the previous claim, at most 8l3/2degree 2rh polynomials agree on this number of points (q/4√l ).

43. Lemma Proof Cont. • Now, • Pi intersect Pi-1 at r random positions, and • we know the evaluation of z at the points in Pi-1 • Two degree 2rh polynomials can agree on at most 2rh/q fraction of their points, • So the probability that an “incorrect” polynomial among our candidates agrees on all r random points in at most

44. Main Lemma Proof Cont. • So, with probability at leastwe learn points Pi successfully. • After 2qd prediction steps, we have learned z on Fd\{0} (since A is a generator of Fd\{0}) • by the union bound, the probability that every step of the reconstruction is successful is at least ½. Q.E.D (main lemma)

45. Proof of Main Theorem Cont. • First, • By averaging argument: • Therefore, there must be a fixing of the coins of R, such that:

46. Ai*+1(y2) A2(y2) A(y2) Ai*(y2) A3(y2) Ai*+1(y1) Ai*+1(yr) Ai*(y1) A2(y1) A(y1) A3(y1) A3(yr) Ai*(yr) A2(yr) A(yr) Ai*(y2) Ai*(yr+1) Ai*(y1) Ai*(y2r) Ai*(yr) Amv A2(y2) A(y2) A(yr+1) A2(yr+1) A(y1) A2(y1) A(y2r) A2(y2r) A(yr) A(yr) y2 Aiv yr+1 y1 y2r yr yr+1 Ai*-1(yr+1)) A(yr+1)) A2(yr+1)) v t1 t2 tr tr+1 Ai*-1(y2r) A(y2r) y2r A2(y2r) t2r Using N.B.P. – Take 2 Unse N.B.P over all points in F, so that we get enough ”good evaluation” Fd F

47. Proof of Main Theorem Cont. • According to the counting argument, this implies that: • Recall that r=Q(d). • A contradiction to the parameter choice: Q.E.D (main theorem)!

49. From q-ary extractors to (regular) extractors The simple technique - using error correcting codes: Lemma: Let F be a field with q elements. Let C:{0,1}k=log(q){0,1}n be a binary error correcting code with distance at least 0.5-O(2) . If E: {0,1}n *{0,1}t ->Fm is a (k,O(r)) q-ary extractor, then E’: {0,1}n *{0,1}t+log(n) ->Fm defined by: Is a (k,rm) binary extractor.

50. From q-ary extractors to (regular) extractors A more complex transformation from q-ary extractors to binary extractors achieves the following parameters: Thm: Let F be a field with q<2m elements. There is a polynomial time computable function: Such that for any (k,r) q-ary extractor E, E’(x;(y,j))=B(E(x;y),j) is a (k,r log*m) binary extractor.