290 likes | 547 Views
Expanders. Presented by Alon Levin 29.10.2006. Expander - Intuitive Definition. Expander graph is an undirected graph: Without “bottlenecks” W ith high connectivity With a “large” minimal cut With “large” edge expansion. Expanders and their applications. Computational Complexity Theory
E N D
Expanders Presented by Alon Levin 29.10.2006
Expander - Intuitive Definition Expander graph is an undirected graph: • Without “bottlenecks” • With high connectivity • With a “large” minimal cut • With “large” edge expansion
Expanders and their applications • Computational Complexity Theory • Economical Robust Networks (Computer Network, Phone Network) • Construction of Hash Functions • Error Correcting Codes • Extractors • Pseudorandom Generators • Sorting Networks
Expander – Definition via Edge Expansion • The edge expansion of a graph G=(V,E) is : • Theorem: Gn are d-regular. • {Gn } is an explicit expander family.
Expanders – Definition via Spectral Gap • We shall discuss undirected d-regular graphs from now and on • We shall adopt the notion of A = A(G) as G’s adjacency matrix • Since A is a real symmetric nn matrix it has n real eigenvalues: • We will denote = max( |2|, |n|) • The spectral gapis defined as d- .
Expanders – Definition via Spectral Gap • A d-regular graph G is an (n, d, )-expander if = max { |i|:1<i≤n } = max { |2(G)|, |n(G)| } and d> • If G is an (n, d, )-expander then 2(G)/2d ≤d- ≤ 2(G) large expansion ~ large spectral gap! * We will prove the inequality right after we show a rather helpful characterization of 2(G)
Reminder – Euclidean Scalar Product and Norm Cauchy–Schwarz inequality : Proof:
Reminder – Triangle Inequality The triangle inequality: Proof:
Rayleigh Quotient • The second largest eigenvalue of a real symmetric matrix can be computed this way: Proof: Let be an orthonormal basis of A’s eigenvectors, where the ith vector corresponds to eigenvalue i(A). “≤”: If we let then we would get |2| and |n|, so the maximum is at least .
Rayleigh Quotient • (Proof continues): “≥”: Denote an arbitrary x as and since a1=0. Now and since that: =max{|i|:1<i≤n}
Spectral gap and edge expansion Now we are ready to prove the following lemma: • if G is an (n, d, )-expander then d- ≤ 2(G) Proof:G=(V,E), |V|=n. Let SV, |S|≤n/2. We shall set a vector x similar to S’s indicator, but so that <x, >=0 and so by the Rayleigh quotient we have So, we will define:
Spectral gap and edge expansion • As we can easily see, x is orthogonal to since • We can also notice the equalities: • ;
X1 - Xi - xn 1 0 0 1 0 1 0 0 0 1 0 Spectral gap and edge expansion • By A’s definition, , and since A is symmetric:
Spectral gap and edge expansion Edges originating in S Cut in half due to edges double count Cut edges
Spectral gap and edge expansion • Which implies since by our assumption , because
Regular Graph Union • Here we shall prove a simple lemma: If G is a d-regular graph over the vertex set V, and H is a d’-regular graph over the same vertices, then G’ = GH = (V,E(G) E(H)) is a d+d’-regular graph such that (G’) (G) + (H) Proof: Take x s.t. Then, Rayleigh quotient AG’=AG+ AH Since it’s a multigraph (edge set is a multiset)
F The Final Expander Lemma • Let G=(V,E) be an (n, d, )-expander and FE. Then the probability that a random walk which starts on an edge from F will pass on an edge from F on it’s tth step is bounded by • Motivation: Showing that a random walk on a good expander (large spectral gap) behaves similarly to independent choice of random edges
The Final Expander Lemma • Proof: x is a vertices distribution vector s.t. xv=Pr[Our walk starts at vertex v] • The probability to reach vertex u at certain point is , thus , where x is the current distribution and x’ is the new one. By the definition of A=A(G) we can write x’=Ax/d. We denote Ã=A/d, so x’=Ãx. • After i steps, the distribution is x’=Ãix. Pr[u is reached from v]=1/d Since G is d-regular
The Final Expander Lemma • P is the probability that an edge from F will be traversed on the tth step. • Let yw be the number of edges from F incident on w divided by d. • Then,
The Final Expander Lemma • Calculation of initial x: first step is picking an edge from F and then one of it’s vertices, so for vertex v it’s dyv is the number of edges from F incident on v
The Final Expander Lemma • G is d-regular, thus each row in à sums to one. • If is the uniform distribution on G, i.e. , then . Since x is a probability function, it can be decomposed into , where since x and are probability functions: • Then,
The Final Expander Lemma • When G is d-regular, each row in à sums to one, and thus if then . • A more intuitive way of seeing this, other than algebra, is considering a random walk with uniform distribution on a d-regular graph:
The Final Expander Lemma Linearity of scalar product • Hence,
The Final Expander Lemma Cauchy-Schwartz inequality (Ã)=(A)/d + Rayleigh quotient t-1 times
The Final Expander Lemma • Since xi are positive, • Maximum is achieved when all edges incident to v are in F, and in that case , so:
The Lubotzky-Phillips-Sarnak Expander • Take a prime p, let V=Zp{}. Define 0-1= and connect every vertex x to: • x+1 • x-1 • x-1 • It’s a 3-regular graph with <3
The Margulis/Gaber-Galil Expanders • Take V=ZnZn, so |V|=n2. Given v=(x,y)V connect it to the following vertices: • (x+2y,y) • (x,2x+y) • (x,2x+y+1) • (x+2y+1,y) (all operations are done modulo n) • This is an 8-regular graph with =52<8, so it’s spectral gap is about 0.93
The End • Questions?