Simple Extractors for all Min-Entropies

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Simple Extractors for all Min-Entropies. R.Shaltiel and C.Umans. Definitions. Def (min-entropy): The min-entropy of a random variable X over {0, 1} n is defined as:

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## Simple Extractors for all Min-Entropies

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Simple Extractors

for all Min-Entropies

R.Shaltiel and C.Umans

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)

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)

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.
Extractors

High

Min-Entropy distribution

Uniform-distribution seed

2t

2n

2m

E

Close to uniform output

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:

Proof

Now define the following hybrid distributions:

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:

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!

Proof

And fis indeed a next bit predictor:

Q.E.D.

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:

(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

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)

h ~ n1/2

j ~ lgn

m ~ desired entropy

Basic Example – Reconstruction Function

Random line

“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

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

Problems with Safra, Ta-Shma, Zukerman
• Curse of dimensionality - too many lines!Solution: generator matrix.
Next-q-it List-Predictor

f is allowed to output a small list of l possible next elements

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

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).

(“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.

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

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:

• 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
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)

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:

Extension Fields - Example

And multiplication:

And now modulo the irreducible polynomial:

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:

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:

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) .

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

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)
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):

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

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)

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

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)

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

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

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

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:

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).

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 ).
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
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)

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

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

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)!

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.

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.

From q-ary extractors to (regular) extractors

The last theorem allows using theorem 1 for  = O(e/log*m) , and implies a (k,e) extractor with seed length t=O(log n) and output length m=k/(log n)O(1)

Extractor  PRG
• Identify:
• string x{0,1}log n with the
• function x:{0,1}log n{0,1} by setting x(i)=xi
• Denote by S(x) the size of the smallest circuit computing function x

Def (PRG): an -PRG for size s is a function G:{0,1}t{0,1}m with the following property: 1im and all function f:{0,1}i-1{0,1}i with size s circuits,

Pr[f(G(Ut)1...i-1)=G(Ut)i]  ½ + /m

This imply:

for all size s-O(1) circuits C

|Pr[C(G(Ut))=1] – Pr[C(Um)=1]| 

q-ary PRG

Def (q-ary PRG): Let F be the field with q elements. A -q-ary PRG for size s is a function G:{0,1}tFm with the following property: 1im and all function f:Fi-1F(-2) with size s circuits,

Pr[j f(G(Ut)1...i-1)j=G(Ut)i]  

Fact: O()-q-ary PRG for size s can be transformed into (regular) m-PRG for size not much smaller than s

Note: Gx(j) corresponds to using our q-ary extractor construction with the “successor function” Amj

The Construction

We show: x is hard  at least one Gx(j) is a q-ary PRG

Plan for building a PRG Gx:{0,1}t {0,1}m:

• use a hard function x:{0,1}log n {0,1}
• let z be the low-degree extension of x
• obtain l “candidate” PRGs, where l=d(log q / log m) as follows:For 0j<l define Gx(j):{0,1}d log q Fm byGx(j)(v) = z(A1mjv)  z(A2mjv) ... z(AMmjv)where A is a generator of Fd\{0}
Getting into Details

Note F’d is a subset of Fd

think of Fd as both a vector space and the extension field of F

perhaps we should just say: immediate from the correspondence between the cyclic group GF(qd) and Fd\{0} ??? otherwise in details we may say:

Proof:

• There exists a natural correspondence between Fd and GF(qd), and between F’d and GF(hd),
• GF(qd) is cyclic of order qd-1, i.e. there exists a generator g
• gp generates the unique subgroup of order hd-1, the multiplicative group of GF(hd).
• A and A’ are the linear transforms corresponding to g and gp respectively.

Let F’ be a subfield of F of size h

Lemma: there exist invertible dd matrices A and A’ with entries from F which satisfy:

•  vFd s.t. v0, {Aiv}i=Fd\{0}
•  vF’d s.t. v0, {A’iv}i=F’d\{0}
• A’=Ap for p=(qd-1)/(hd-1)
• A and A’ can be found in time qO(d)

since hd>n, there are enough “slots” to embed all x in a d dimensional cube of size hd

• and since A’ generates F’d\{0}, indeed x is embedded in a d dimensional cube of size hd
• Note h denotes the degree in individual variables, and the total degree is at most hd
• The computation of z from x can be done in poly(n,qd)=qO(d) time
• require hd>n
• Define z as followsz(A’i1)=x(i), where 1 is the all 1 vector (low degree extension).
• Recall: For 0j<l define Gx(j):{0,1}d log q Fm byGx(j)(v) = z(A1mjv)  z(A2mjv) ... z(AMmjv

Theorem (PRG main): for every n,d, and h satisfying hd>n, at least one of Gx(j) is an -q-ary PRG for size (-4 h d2 log2q). Furthermore, all the Gx(j)s are computable in time poly(qd,n) with oracle access to x.

Extension Field

Def: if F is a subset of E, then we say that E is an extension field of F.

Lemma: let

• E be an extension field of F,
• f(x) be a polynomial over F(i.e.f(x)F[X]),
• cE,

then f(x)f(c) is an homomorphism of F[X] into E.

Construction of the Galois Field GF(qd)

Thm: let p(x) be irreducible in F[X], then there exists E, an extension field of F, where there exists a root of p(x).

Proof Sketch:

• add a  (a new element)to F. is to be a root of p(x).
• In F[] (polynomials with variable )
Example:
• F=reals
• p(x)=x2+1