Lecture 8: Construction of Expanders

Transcription

1 Spectral Graph Theory and Applications WS 0/0 Lecture 8: Construction of Expanders Lecturer: Thomas Sauerwald & He Sun In this lecture we study the explicit constructions of expander graphs. Although we can construct expanders probabilistically, this does not suffice for many applications. One applications of expander graphs is for reducing the randomness complexity of algorithms (cmp. Lecture 6), thus constructing the graph itself randomly does not sere this purpose. Sometimes we may een need expanders of exponential size. In this case, we cannot store the whole description of the graph. Hence we need a more explicit ersion of expanders. We call a family of expander graphs explicitly constructible if the construction satisfies the following properties:. We can construct the whole graph in time poly(n), where n is the number of ertices in a graph.. For any ertex and integer i {,, d}, we can find the i-th neighbor of in time poly(log n, log d).. For any ertices and u, we can determine if they are adjacent in time poly(log n). Let G = (V, E) be a d-regular graph. For each ertex V, we label the edges adjacent to and let [i] be the i-th edge of. Define a rotation map Rot G : V [d] V [d] by Rot G (u, i) = (, j) where is the i-th neighbor of u and u is the j-th neighbor of. Example: For the graph shown in Figure, we hae Rot(a, ) = (h, ). Figure : Rotation Map

2 Lecture 8: Construction of Expanders Graph Powering We know that if G is an (N, d, λ)-graph, then G k, the k-th powering of the adjacency matrix of G, represents an (N, d k, λ k )-graph. From the rotation map s perspectie, if G is a d-regular graph with rotation map Rot G, then the k-th powering of G is a d k -regular graph whose rotation map is gien by Rot G k( 0, (a,, a k )) = ( k, (b k,, b )), where the alues b,, b k and k are computed ia the rule ( i, b i ) = Rot G ( i, a i ). Replacement Product For a D-regular graph G with N ertices and a d-regular graph H with D ertices, the replacement product, denoted as G r H, is a (d + )-regular graph with N D ertices. Each ertex in G is replaced by a graph H, called a cloud. Moreoer, Rot G r H ((u, k), i) = ((, l), j) if and only if u = and Rot H (k, i) = (l, j), or i = j = d + and Rot G (u, k) = (, l). Figure : Replacement Product Some comments: Replacement product depends on the arbitrary labels for the ertices and the edges. Replacement product reduces the (relatie) ertex degree without losing the connectiity. For seeral problems in graph theory it can be shown that it suffices to sole them for graphs obtained by the Replacement product. Zig-Zag Product Based on rotation maps, the zig-zag product is defined as follows. Definition 8.. [RVW00] Let G be a D-regular graph on [N] with rotation map Rot G and H be a d-regular graph on [D] with rotation map Rot H. Then their zig-zag product G z H is defined to be the d -regular graph on [N] [D] whose rotation map Rot G z H is as follows:. Let (a, i ) = Rot H (a, i). Let (w, b ) = Rot G (, a ). Let (b, j ) = Rot H (b, j) 4. Output ((w, b), (j, i )) as the alue of Rot G z H ((, a), (i, j)). Example: Let G and H be two graphs shown in Figure. Then Rot G z H ((C, z), (, )) = ((F, z), (, )).

3 Lecture 8: Construction of Expanders A F B E C D (C,z) Figure : An example of the zig-zag product u Figure 4: Intuition behind the zig-zag product The intuition behind the zig-zag product is shown in Figure 4. The zig-zag product corresponds to -step walks on the replacement product graph, where the first and the last steps are along the inner-cloud edges and the middle step is along an inter-cloud edge, and each ertex in the cloud corresponds to an edge starting from the ertex which the cloud represents. Theorem 8.. Suppose that G is an (N, d, λ )-expander and H is a (d, d, λ )-expander. Then G z H is an (N d, d, f(λ, λ ))-expander, where f(λ, λ ) λ + λ + λ. Proof. The number of ertices and degree of G z H are obtained directly from the definition of the zig-zag product and we only need to analyze the spectral expansion of G z H. Let M be the normalized adjacency matrix of G z H. It suffices to show that for any α N d, α R N d, it holds that Mα, α f(λ, λ ) α, α. Let α R N d with the property that α N d. For any ertex [N ], define α R d by (α ) k = α k. Also, let C : R N d R N be a linear mapping such that (Cα) = d k= α k. Then we can express α as α = e α. [N ]

4 Lecture 8: Construction of Expanders 4 Let α = α + α where α d. Then α = ( ) e α + ( ) e α := α + α. That is, α is uniform within any gien cloud and can be expressed as α = Cα d d. Let A and B be the normalized adjacency matrices of G and H, respectiely. Let B = I N B and à be the permutation matrix corresponding to Rot G, i. e. an N d N d matrix where { if RotG (u, i) = Rot G (, j) à (u,i)(,j) = 0 otherwise. Then M = Bà B. Since B is real symmetric, we hae Mα, α = Bà Bα, α = à Bα, Bα. On the other hand, we hae Bα = B(α + α ) = α + Bα. Thus ( Mα, α = à α + Bα ), (α + Bα ) = Ãα, α + Ãα, Bα + à Bα, α + à Bα, Bα and Mα, α Ãα, α + Ãα Bα + à Bα α + à Bα Bα = Ãα, α + α Bα + Bα, () where the last equality holds as à is a permutation and Ãx = x for any x NN d. Notice that ( ) Bα = B e α = e Bα = Bα λ α λ α. () So we only need to bound Ãα, α. Ãα, α = Ãα, Cα d /d = CÃα, Cα /d = ACα, Cα /d

5 Lecture 8: Construction of Expanders 5 and Ãα, α λ Cα, Cα /d = λ Cα d, Cα d /d = λ α, α α = λ. () Combining equation () and equation (), we hae α Mα, α λ α + λ α + λ α. By taking p = α α and q = α α, we hae p + q =. Therefore Mα, α α, α λ p + λ pq + λ q λ + λ + λ, which completes the proof. Some comments on the zig-zag product. The edge labels in G z H are just pairs of edges labeled in the small graph. By taking a product of a large graph with a small graph, the resulting graph inherits (roughly) its size from the large one, its degree from the small one and its expansion properties from both. This was the key to creating arbitrary large graphs with bounded degree. 4 Construction of Expanders In this section we use the zig-zag product to construct expander graphs. Theorem 8.. Let H be a (d 4, d, λ 0 ) graph for some λ 0 /5. Define G = H and G t+ = G t z H for t. Then for all t, G t is a (d 4t, d, λ)-expander with λ /5. Proof. We proe the theorem by induction. When t =, it is straightforward to see that G is a (d 4, d, λ 0 )-expander. Assume that G t is a (d 4(t ), d, λ)-expander for λ /5. By Definition 8., the number of ertices in G t is d 4t. So it suffices to show the spectral expansion of G t. By Theorem 8., the spectral expansion of G t is λ(g t ) λ(g t ) + λ(h) + λ(h ) ( ) = = 5. Remark. search. Since H is a graph of constant-size, we can find it in constant time by brute-force Generalization.

6 Lecture 8: Construction of Expanders 6 References [RVW00] Omer Reingold, Salil P. Vadhan, and Ai Wigderson. Entropy waes, the zig-zag graph product, and new constant-degree expanders and extractors. In FOCS, pages, 000.