Ma/CS 6b Class 12: Graphs and Matrices

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1 Ma/CS 6b Class 2: Graphs and Matrices 3 3 v 5 v 4 v By Adam Sheffer Non-simple Graphs In this class we allow graphs to be nonsimple. We allow parallel edges, but not loops.

2 Incidence Matrix Consider a graph G = V, E. We order the vertices as V = v,,, v n and the edges as E = e, e 2,, e m The incidence matrix of G is an n m matrix M. The cell M ij contains if v i is an endpoint of e j, and otherwise. v v 5 e 4 v 4 e e 2 e 3 e 5 e 6 Playing with an Incidence Matrix Let M be the incidence matrix of a graph G = V, E, and let B = MM T. What is the value of B ii? The degree of v i in G. What is the value of B ij for i j? The number of edges between v i and v j. M = e v 5 e 4 v 4 v e 2 e 3 e 5 e 6 2

3 Adjacency Matrix A = Consider a graph G = V, E. We order the vertices as V = v,,, v n. The adjacency matrix of G is a symmetric n n matrix A. The cell A ij contains the number of edges between v i and v j. v 3 3 v 5 v 4 A Connection Problem. Let G = V, E be a graph with incidence matrix M and adjacency matrix A. Express MM T using A. Answer. This is a V V matrix. MM T ii is the degree of v i. MM T ij for i j is number of edges between v i and v j. Let D be a diagonal matrix with D ii being the degree of v i. We have MM T = A + D. 3

4 Example A = 3 3 D = v M = MM T = = A + D Multiplying by s Let n denote the n n matrix with in each of its cells. Problem. Let G = V, E be a graph with adjacency matrix A. Describe the values in the cells of B = A V. Answer. It is a V V matrix. The column vectors of B are identical. The i th element of each column is the degree of v i. 4

5 Playing with an Adjacency Matrix Let A be the adjacency matrix of a simple graph G = V, E, and let M = A 2. What is the value of M ii? The degree of v i in G. What is the value of M ij for i j? The number of vertices that are adjacent to both v i and v j. A = v 5 v 4 v What about M 3? Let A be the adjacency matrix of a simple graph G = V, E. For i j: A ij tells us if there is an edge v i, v j E. A 2 ij tells us how many vertices are adjacent to both v i and v j. What is A 3 ij? It is the number of paths of length three between v i and v j. In fact, A 2 ij is the number of paths of length two between v i and v j. 5

6 The Meaning of M k Theorem. Let A be the adjacency matrix of a (not necessarily simple) graph G = V, E. Then A k is the number of ij (not necessarily simple) paths between v i and v j. An Example a b d c A = A 2 = AA = = A 3 = A 2 A = =

7 The Meaning of M k Theorem. Let M be the incidence matrix of a (not necessarily simple) graph G = V, E. Then M k is the number ij of (not necessarily simple) paths between v i and v j. Proof. By induction on k. Induction basis. Easy to see for k = and k = 2. The Induction Step We have A k = A k A. That is V A k ij = Ak im A mj m= By the induction hypothesis, A k is the im number of paths of length k between v i and v m. A mj is the number of edges between v m, v j. 7

8 The Induction Step (cont.) V A k ij = Ak im A mj m= For a fixed m, A k A im mj is the number of paths from v i to v j of length k with v m as their penultimate vertex. Thus, summing over every m V results in the number of paths from v i to v j of length k. Which Classic English Rock Band is More Sciency? 8

9 Computing the Number of Paths of Length k Problem. Consider a graph G = V, E, two vertices s, t V, and an integer k >. Describe an algorithm for finding the number of paths of length k between s and t. Let A be the adjacency matrix of G. We need to compute A k, which involves k matrix multiplication. Matrix Multiplication: A Brief History We wish to multiply two n n matrices. Computing one cell requires about n multiplications and additions. So computing an entire matrix takes cn 3 (for some constant c). In 969 Strassen found an improved algorithm with a running time of cn In 987, Coppersmith and Winograd obtained an improved cn After over 2 years, in 2, Stothers obtained cn Williams immediately improved to cn In 24, Le Gall improved to cn

10 A More Efficient Algorithm To compute A k, we do not need k matrix multiplications. If k is a power of 2: A 2 = AA, A 4 = A 2 A 2,, A k = A k/2 A k/2. Only log 2 k multiplications! If k is not a power of 2: We again compute A, A 2, A 4,, A k. We can then obtain A k by multiplying a subset of those. For example, A 57 = A 32 A 6 A 8 A. At most 2log k multiplications! Connectivity and Matrices Problem. Let G = V, E be a graph with an adjacency matrix A. Use M = I + A + A 2 + A A V to tell whether G is connected. Answer. G is connected iff every cell of A is positive. The main diagonal of A is positive due to I. A cell A ij for i j contains the number of paths between v i and v j of length at most k. If the graph is connected, such paths exist between every two vertices.

11 And Now with Colors Problem. Consider a graph G = V, E, two vertices s, t V, and an integer k >. Moreover, every edge is colored either red or blue. Describe an algorithm for finding the number of paths of length k between s and t that have an even number of blue edges. Solution We define two sets of matrices: Cell ij of the matrix E m contains the number of paths of length m between v i an v j using an even number of blue edges. Cell ij of the matrix O m contains the number of paths of length m between v i an v j using an odd number of blue edges. What are E and O ()? E is the adjacency matrix of G after removing the blue edges. Similarly for O after removing the red edges.

12 Solution (cont.) We wish to compute E k. We know how to find E and O. How can we compute E m and O (m) using E m and O (m )? E m = E m E + O (m ) O (). E m E is the number of paths of ij length m between v i and v j with an even number of blue edges and whose last edge is red. Similarly for O m O last edges of the paths is blue. ij except that the Solution (cont.) How can we similarly compute O m using E m and O m? O m = E m O + O (m ) E (). Concluding the solution. We compute about 2k matrices. Each computation involves two matrix multiplications and one addition. 2

13 Identical Graphs? Are the following two graphs identical? v v Possible answers: No, since in one v has degree 2 and in the other degree. Yes, because they have exactly the same structure. Graph Isomorphism An isomorphism from a graph G = V, E to a graph G = V, E is a bijection f: V V such that for every u, v V, E has the same number of edges between u and v as E has between f u and f v. We say that G and G are isomorphic if there is an isomorphism from G to G. In our example, the graphs are isomorphic (f v = u 2, f = u, f = u 3 ). v u 2 u 3 u 3

14 Uniqueness? Question. If two graphs are isomorphic, can there be more than one isomorphism from one to the other? Yes! a w, b z, c x, d y. a y, b z, c x, d w. a b w x c d y z Isomorphisms and Adjacency Matrices What can we say about adjacency matrices of isomorphic graphs? c a b d w y z x 4

15 Answer Two graphs G, G are isomorphic if and only if the adjacency matrix of G is obtained by permuting the rows and columns of the adjacency matrix of G. (The same permutation should apply both to the rows and to the column)., 2 4, 4 3, 3 2: Incidence Matrix of a Directed Graph Consider a directed graph G = V, E. We order the vertices as V = v,,, v n and the edges as E = e, e 2,, e m The incidence matrix of G is an n m matrix M. The cell M ij contains - if e j is entering v i, and if e j is leaving v i. v v 5 e 4 v 4 e e 2 e 3 e 5 e 6 5

16 The End: Queen Ph.D. in astrophysics Degree in Biology Electronics engineer Freddie Mercury 6

Ma/CS 6b Class 20: Spectral Graph Theory

Ma/CS 6b Class 20: Spectral Graph Theory By Adam Sheffer Recall: Parity of a Permutation S n the set of permutations of 1,2,, n. A permutation σ S n is even if it can be written as a composition of an