CMPT 881 - Pseudorandomness : Solutions to Problem Set 2

(a) In this question you are asked to show that randomness extraction is possible only from sources that are statistically close to sources with high min-entropy; thus, high minentropy is both sufficient and necessary for randomness extraction. More formally, let X be any distribution over {0, 1}n and let Ext : {0, 1}n × {0, 1}d → {0, 1}m. Suppose that Ext(X,Ud) is ǫ-close to the uniform distribution Um. Prove that X is O(ǫ)-close to some k-source X ′ where k ≥ m− d− 1. (Hint: Consider the set A of strings z ∈ {0, 1}m that get assigned probability more than 2 ∗ 2−m by Ext(X,Ud). Argue that the set A gets the probability at most 2ǫ with respect to the distribution Ext(X,Ud). Fix the seed y ∈ {0, 1}d so that Pr[Ext(X, y) ∈ A] ≤ 2ǫ. Argue that for every x such that Ext(x, y) 6∈ A, we have Pr[X = x] ≤ 2−(m−d−1). Conclude that there exists a k-source X ′ such that ∆(X,X ′) ≤ 2ǫ.) Solution: Following the hint, we get by the definition of extractor that Pr[Ext(X,Ud) ∈ A] ≤ μ(A) + ǫ, where μ(A) = |A|/2m is the density of the set A. On the other hand, by the definition of A, Pr[Ext(X,Ud) ∈ A] ≥ |A|2 ∗ 2 −m = 2μ(A). Combining these two inequalities we get μ(A) ≤ ǫ and hence Pr[Ext(X,Ud) ∈ A] ≤ 2ǫ. By averaging, there exists a string y such that Pr[Ext(X, y) ∈ A] ≤ 2ǫ. Let x be any string such that z = Ext(x, y) 6∈ A. If Pr[X = x] > 2−(m−d−1), then the pair (x, y) gets the probability greater than 2−(m−d−1) ∗ 2d = 2 ∗ 2−m according to (X,Ud). This means that z = Ext(x, y) gets probability greater than 2 ∗ 2−m according to Ext(X,Ud), and hence, by the definition of A, z must be in A, which contradicts our earlier choice of z 6∈ A. Thus, for every such string x, we must have Pr[X = x] ≤ 2−(m−d−1). Finally, define the distribution X ′ to be an arbitrary flat k-source for k = m − d − 1. We have ∆(X,X ′) = Pr[X ∈ T ] − Pr[X ′ ∈ T ] ≤ Pr[X ∈ T ] for T = {z | Pr[X = z] > Pr[X ′ = z]} (this is easy to verify using the definition of the statistical distance ∆(X,X ′) = maxT {Pr[X ∈ T ]− Pr[X ′ ∈ T ]}). Now, observe that in our case, T ⊂ {x | Ext(x, y) ∈ A}. Hence, Pr[X ∈ T ] ≤ Pr[Ext(X, y) ∈ A] ≤ 2ǫ, and so, ∆(X,X ′) ≤ 2ǫ. (b) Show that every k-source X over {0, 1}n, for large k, can be viewed as a block source X = Y Z. More precisely, let X = Y Z be an (n − ∆)-source, for some ∆, where Y is a distribution over l-bit strings for any l ≤ n, and Z is a distribution over strings of length m = n − l. Prove that Y is a (l −∆)-source. Prove also that, for every ǫ > 0, with probability at least (1− ǫ) over the choice of y according to the distribution Y , the conditional distribution Z|Y=y is an (m−∆− log(1/ǫ))-source. Solution: For any y ∈ {0, 1}l, we have Pr[Y = y] = ∑ z∈{0,1}m Pr[X = yz] ≤ 2m2−(n−∆) = 2−(l−∆), where the last inequality is due to the condition that X is an (n−∆)-source. Thus, we know that Y is an (l−∆)-source.