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CS225: Pseudorandomness Prof. Salil Vadhan
Lecture 13: More Connections with Expanders
March 22, 2007
Based on scribe notes by Adam Kirsch and Alexandr Andoni.
1 Lossless Condensers vs. Expanders
Last time we saw the notion of a k →ε k′^ condenser Con : { 0 , 1 }n^ × { 0 , 1 }d^ → { 0 , 1 }m^ , where for every k-source X on { 0 , 1 }n, Con(X, Ud) is ε-close to some k′-source. We can define the entropy loss of a condenser to be = k′^ − (k + d). As we have discussed, an extractor (i.e. m = k′) must have ≥ 2 log(1/ε) − O(1), whereas if we allow m to be larger than k′^ (specifically, m ≥ k′^ + log(1/ε) + O(1)), then it is possible for a condenser to be lossless (i.e. have ` = 0).
As we have seen for extractors in Lecture Notes 11, every function Con : { 0 , 1 }n^ ×{ 0 , 1 }d^ → { 0 , 1 }m can be viewed as a bipartite multigraph G with N = 2n^ left vertices, left-degree D = 2d, and M = 2m^ right-vertices where the y’th neighbor of left-vertex x is Con(x, y). Generalizing what we showed for extractors, the condenser property implies a vertex-expansion property of the graph G. Specifically, if we define a (= K, A) vertex expander to be one in which sets of size exactly K expand by a factor of A, then we have:
Lemma 1 If Con : { 0 , 1 }n^ × { 0 , 1 }d^ → { 0 , 1 }m^ is a k →ε k′^ condenser, then the corresponding bipartite graph is a (= K, A) vertex expander for A = (1 − ε) · K ′/K = (1 − ε) · D/L, where K = 2k, K′^ = 2k
′ , D = 2d, L = 2, and = k′^ − (k + d) is the entropy loss of the Con.
Thus, if L ≤ D, the expansion factor A is in fact bigger than 1 (but the case A < 1 is still interesting and nontrivial, because the graphs are unbalanced). In general, the vertex expansion property is weaker than the property of being a condenser — for example, when m = k′, it corresponds to a disperser rather than an extractor. However, for the special case of lossless condensers (i.e. L = 1), it turns out that the two properties are equivalent.
Lemma 2 A function Con : { 0 , 1 }n^ × { 0 , 1 }d^ → { 0 , 1 }m^ is a k →ε k + d (lossless) condenser if and only if the corresponding bipartite graph is a (= K, A) vertex expander for A = (1 − ε) · D, where K = 2k^ ∈ N, K′^ = 2k′ , and D = 2d.
Proof: The ‘only if’ direction follows from Lemma 1, so we only prove the ‘if’ direction. Assume that the bipartite graph corresponding to Con is a (= K, (1 − ε) · D) vertex expander. To show that Con is a lossless condenser, it suffices to show that Con(X, Ud) is ε-close to a k′-source for every flat k-source X. Let S be the support of X, which is of size K. By the vertex expansion of the graph, |N (S)| ≥ (1 − ε) · DK. Since there are only DK edges leaving S, we can make all of these edges lead to distinct vertices by shifting an ε fraction of them. Let T ⊂ [M ] be the set of KD vertices hit after this shifting. Then Con(X, Ud) is ε-close to the uniform distribution on T , which is a (k + d)-source.
Thus, the lossless condenser that we assumed in the previous lecture follows immediately from the following expander:
Theorem 3 For every constant α > 0 , every N ∈ N, K ≤ N , and ε > 0 , there is an explicit (K, (1−ε)D) expander with N left-vertices, M right-vertices, left-degree D = O((log N )(log K)/)1+1/α and M ≤ D^2 · K1+α. Moreover, D is a power of 2.
We will construct this expander after Spring Break, using ideas based on list-decodable error- correcting codes.
Note that the kind of expander given by this theorem can be used for data structure application on Problem Set 3 — storing a K/2 subset of S with M = K 1+α^ · polylog(N ) bits in such a way that membership can be probabilistically tested by reading only 1 bit of the data structure. (An efficient solution to this application actually requires more than the fact that the graph is explicit in the usual sense, but also that there are efficient algorithms for finding all left-vertices having a δ fraction neighbors in a set T ⊆ [M ] of right vertices, but it turns out that the expanders we will construct turn out to have this property.)
The connection of Lemma 2 is also useful in the reverse direction. Indeed, as we saw last time, there are explicit extractors with entropy loss ` = O(log(1/ε)) and seed length d = O(log k · log(n/ε)). By applying an almost pairwise-independent hash function, these can be converted into lossless condensers with the same (polylogarithmic) seed length and optimal output length m = k + d + log(1/ε) + O(1). These correspond to expanders with expansion (1 − ε) · D whose degree is quasipolynomial in (log N )/ε, but whose right-hand side is of the optimal size M = O(KD/ε).
Open Problem 4 Construct highly unbalanced bipartite expanders with degree D = poly(log N ), expansion (1 − ε) · D for an arbitrarily small constant ε > 0 , and M = O(KD) right-hand vertices.
Such a construction would give lossless condensers whose output is of extremely high min-entropy (k′^ = m−O(1)), and thus we could get extractors that extract all the min-entropy by then applying the high min-entropy extractor based on spectral expanders.
2 Block-Source Extraction vs. the Zig-Zag Product
Recall the block-source extraction method presented last time. We define Ext′^ : { 0 , 1 }n^1 +n^2 × { 0 , 1 }d^2 → { 0 , 1 }m^1 by Ext′((x 1 , x 2 ), y 2 ) = Ext 1 (x 1 , Ext 2 (x 2 , y 2 )). (Here we consider the special case that m 2 = d 1.
Viewing the extractors as bipartite graphs, the left-vertex set is [N 1 ] × [N 2 ] and the left-degree is D 2. A random step from a vertex (x 1 , x 2 ) ∈ [N 1 ] × [N 2 ], corresponds to taking a random step from x 2 in G 2 to obtain a right-hand vertex y 1 ∈ { 0 , 1 }m^2 , which we view as an edge label y for G 1. We then move to the y’th neighbor of x 1.
This is just like the first two steps of the zig-zag graph product. Why do we need a third step in the zig-zag product? It is because of the slightly different goals in the two setting. In a (spectral) expander, we consider an arbitrary initial distribution that does not have too much (Renyi) entropy, and need to add entropy to it. In a block-source extractor, our initial distribution is constrained
