1 Replacement Product and Zig-Zag Product

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1 Stanford University CS59: raph Partitioning and Expanders Handout 7 Luca Trevisan March, 0 Lecture 7 In which we define and analyze the zig-zag graph product. Replacement Product and Zig-Zag Product In the previous lecture, we claimed it is possible to combine a d-regular graph on vertices and a -regular graph on N vertices to obtain a d -regular graph on N vertices which is a good expander if the two starting graphs are. Let the two starting graphs be denoted by H and respectively. Then, the resulting graph, called the zig-zag product of the two graphs is denoted by z H. Using λ() to denote the eigenvalue with the second-largest absolute value for a graph, we claimed that if λ(h) b and λ() a, then λ( z H) a + b + b. In this lecture we shall describe the construction for the zig-zag product and prove this claim. Replacement product of two graphs We first describe a simpler product for a small d-regular graph on vertices (denoted by H) and a large -regular graph on N vertices (denoted by ). Assume that for each vertex of, there is some ordering on its neighbors. Then we construct the replacement product (see figure) r H as follows: Replace each vertex of with a copy of H (henceforth called a cloud). For v V (), i V (H), let (v, i) denote the i th vertex in the v th cloud. Let (u, v) E() be such that v is the i-th neighbor of u and u is the j-th neighbor of v. Then ((u, i), (v, j)) E( r H). Also, if (i, j) E(H), then u V () ((u, i), (u, j)) E( r H). Note that the replacement product constructed as above has N vertices and is (d + )-regular.

2 B B B C A E B C F r H A E A A A E E E B C F F F C F C r H Zig-zag product of two graphs iven two graphs and H as above, the zig-zag product z H is constructed as follows (see figure): The vertex set V ( z H) is the same as in the case of the replacement product. ((u, i), (v, j)) E( z H) if there exist l and k such that ((u, i)(u, l), ((u, l), (v, k)) and ((v, k), (v, j)) are in E( r H) i.e. (v, j) can be reached from (u, i) by taking a step in the first cloud, then a step between the clouds and then a step in the second cloud (hence the name!).

3 B B B C A E B C F z H A E A A A E E E B C F F F C F C z H It is easy to see that the zig-zag product is a d -regular graph on N vertices. Let M R ([N] []) ([N] []) be the normalized adjacency matrix of z H. Using the fact that each edge in r H is made up of three steps in r H, we can write M as BAB, where { 0 if u v B[(u, i), (v, j)] = M H [i, j] if u = v And A[(u, i), (v, j)] = if u is the j-th neighbor of v and v is the i-th neighbor of u, and A[(u, i), (v, j)] = 0 otherwise. Note that A is the adjacency matrix for a matching and is hence a permutation matrix. Preliminaries on Matrix Norms Recall that, instead of bounding λ, we will bound the following parameter (thus proving a stronger result). efinition Let M be the normalized adjacency matrix of a graph = (V, E), and λ... λ n be its eigenvalues with multiplicities. Then we use the notation λ(m) := max i=,...,n { λ i } = max{λ, λ n }

4 The parameter λ has the following equivalent characterizations. Fact λ(m) = M x max x R V {0},x x = max x R v,x, x = Mx Another equivalent characterization, which will be useful in several contexts, can be given using the following matrix norm. efinition (Spectral Norm) The spectral norm of a matrix M R n n is defined as M = max x R V, x = M x If M is symmetric with eigenvalues λ,..., λ n, then the spectral norm is max i λ i. Note that M is indeed a norm, that is, for every two square real matrices A, B we have A + B A + B and for every matrix A and scalar α we have αa = α A. In addition, it has the following useful property: Fact For every two matrices A, B R n n we have Proof: For every vector x we have AB A B ABx A Bx A B x where the first inequality is due to the fact that Az A z for every vector z, and the second inequality is due to the fact that Bx B x. So we have min x R n,x 0 We can use the spectral norm to provide another characterization of the parameter λ(m) of the normalized adjacency matrix of a graph. Lemma 5 Let be a regular graph and M R n n matrix. Then be its normalized adjacency λ(m) = M n J where J is the matrix with a in each entry.

5 Proof: Let λ = λ λ n be the eigenvalues of M and v = n, v,..., v n a corresponding system of orthonormal eigenvector. Then we can write Noting that v v T = J, we have n M = λ v v T + + λ n v n v T n M n J = 0 v v T + n λ i v i vi T and so v,..., v n is also a system of eigenvectors for M J, with corresponding n eigenvalues 0, λ,..., λ n, meaning that i= M n J = max{0, λ,..., λ n } = λ(m) The above lemma has several applications. It states that, according to a certain definition of distance, when a graph is a good expander then it is close to a clique. (The matrix J is the normalized adjacency matrix of a clique with self-loops.) The n proof of several results about expanders is based on noticing that the result is trivial for cliques, and then on approximating the given expander by a clique using the above lemma. We need one more definition before we can continue with the analysis of the zig-zag graph product. efinition 6 (Tensor Product) Let A R N N and B R be two matrices. Then A B R N N is a matrix whose rows and columns are indexed by pairs (u, i) [N] [] such that (A B) (u,i),(v,j) = A u,v B i,j For example I M is a block-diagonal matrix in which every block is a copy of M. 5 Analysis of the Zig-Zag Product Suppose that and H are identical cliques with self-loops, that is, are both n-regular graphs with self-loops. Then the zig-zag product of and H is well-defined, because the degree of is equal to the number of vertices of H. The resulting graph z H is a n -regular graph with n vertices, and an inspection of the definitions reveals that z H is indeed a clique (with self-loops) with n vertices. 5

6 The intuition for our analysis is that we want to show that the zig-zag graph product preserves distances measured in the matrix norm, and so if is close (in matrix norm) to a clique and H is close to a clique, then z H is close to the zig-zag product of two cliques, that is, to a clique. (Strictly speaking, what we just said does not make sense, because we cannot take the zig-zag product of the clique that is close to and of the clique that H is close to, because they do not have the right degree and number of vertices. The proof, however, follows quite closely this intuition.) Theorem 7 If λ(m ) = a and λ(m H ) = b, then λ( z H) a + b + b Proof: Let M be the normalized adjacency matrix of z H, and let x be a unit vector such that x and λ(m) = Mx Recall that we defined a decomposition M = BAB where A is a permutation matrix, and B = I M H. Let us write E := M H J, then B = I J + I E. Let us call J := I J and Ē := I E. First, we argue that the matrix norm of Ē is small. Take any vector z R N and write is as z = (z,..., z N ), where, for each u [N], z u is the -dimensional restriction of z to the coordinates in the cloud of u. Then (I E)z = u Ez u u E z u = E z and so we have I E E b Then we have BAB = ( J + Ē)A( J + Ē) = JA J + JAĒ + ĒA J + ĒAĀ and so, using the triangle inequality and the property of the matrix norm, we have BABx JA Jx + ĒA J + JAĒ + ĒAĒ where ĒA J Ē A J Ē b 6

7 JAĒ J A Ē Ē b ĒAĒ Ē A Ē Ē b It remains to prove that JA Jx a. If we let A = M be the adjacency matrix of, then we can see that That is, ( JA J) (u,i),(v,j) = (A ) u,v = (M ) u,v = (M J) (u,i),(v,j) JA J = M J Finally, we write x = (x,..., x N ), where x u is the -dimensional vector of entries corresponding to the cloud of u, we call y u := i x u(i)/, and we note that, by Cauchy-Schwarz: y = u ( i ) x u,i u ( i ) ( i x u,i ) = x The final calculation is: ( JA Jx = M ) J x = ( ) (M ) u,v x u,i u,i v,j = ( ) (M ) u,v y u u,i v = ( ) (M ) u,v y u u v = M y a y a x 7