University of Toronto Department of Mathematics

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1 University of Toronto Department of Mathematics MAT332H1F, Graph Theory Midterm, October 21, 2014 Instructor: Kasra Rafi First Last Student Number Instructions: No aids allowed. Write solutions on the space provided. To receive full credit you must show all your work. If you run out of room for an answer, continue on the back of the page. This exam has 6 questions, for a total of 50 points. Problem # Grade Bonus Total

2 1. (8 points) Define the following terms and expressions: (a) bipartite A graph G is bipartite if the vertex set of G can be partitioned into two sets X and Y so that there every edge in G has one end in X and one end in Y. (b) induced subgraph And induces subgraph of G is a subgraph of G that is obtained from G by vertex deletion only. (c) isomorphic Two graphs G(V, E) and G (V, E ) are isomorphic if there are bisections θ : V V and η : E E so that x and y are ends points of an edge e E if and only of θ(x) and θ(y) are end points of η(e). (d) equivalence relation An equivalence relation elements of a set A is a relation that is (a) reflexive: a A, a a. (b) symmetric: a, b A, a b = b a. (c) transitive: a, b, c A, a b and b c = a c. Page 2 of 7 Please go on to the next page...

3 2. (12 points) Answer true of false. Justify your answer with an argument or a counter example. (a) False Every graph that is edge transitive is also vertex transitive. For example P 2, the path of length 2, is edge transitive but not vertex transitive. (b) False Every subgraph of a bipartite graph is bipartite. The graph with one vertex is the subgraph of every graph but it is not bi-partite. (Note: this is assuming that the sets X and Y in the definition of a bi-partite graphs are always non-empty which, we decided in class, is a matter of convention. The points were given to all students.) (c) True All spanning trees of a graph G have the same number of edges. Every spanning tree in a graph with n vertices has n 1 edges. (d) False Every tournament has a unique directed Hamilton path. For example, a tournament between 3 teams {A, B, C} where A beats B, B beats C and C beats A. Both A, B, C and B, C, A are Hamilton paths, Page 3 of 7 Please go on to the next page...

4 3. (10 points) Prove the following theorem from the book. Theorem 4.7. A graph is bipartite if and only if it contains no odd cycle. See the textbook. Page 4 of 7 Please go on to the next page...

5 4. (10 points) Which of these graphs are isomorphic? Either provide an isomorphism or prove they are not isomorphic. v 2 v 3 w 2 w 3 w 1 w 4 w 5 v 4 v 5 v 1 b 2 b 3 a 2 a 3 a 1 b 1 a 4 a 5 b 4 b 5 x 4 x 3 y 1 y y 0 2 y 3 x 2 x 1 x 5 x 6 They are all different. For example, we can count the number of 4 cycles in these graphs. The first one has no 4 cycles. The second one has five 4 cycles. The last one has two 4 cycles. Page 5 of 7 Please go on to the next page...

6 5. (10 points) By counting the number of branchings whose root lies in the m set of K m,n, show that t(k m,n ) = m n 1 n m 1. Let X be the m set and Y be the n set of K m,n. Every vertex of a branching, except for the root, has the in-degree one. Hence, a branching in K m,n has n arcs going from X to Y and (m 1) arcs going from Y to X. There are mn possibilities for an arc from X to Y. For the second arc we have m(n 1) possibilities since one vertex in Y has already an incoming arc. We can proceed this way; for the ith arcs we have m(n i) possibilities. Hence, there are a total of m n n! choices. But choosing these arcs in a different order would be equivalent. Which means every selection has been counted n! times. Therefore, there m n possibilities for the arcs going from X to Y. Similarly, we can select arcs going from Y to X. But no arc can end in the root. Hence, the first arc has n(m 1) possibilities, the second arcs and n(m 2) possibilities and the ith arc has n(m i) possibilities. The total number is n m 1 (m 1)!. But every selection has been counted (m 1)! times. So there are n m 1 possible choices for arcs going from Y to X. Thus, the number of possible branching of K m,n is m n n m 1. But each spanning tree of K m,n has m associated branching where the root is in the m set. Therefore, t(k m,n ) = mn n m 1 m = mn 1 n m 1. Page 6 of 7 Please go on to the next page...

7 6. (Bonus Problem) Recall that a finite projective plane is a geometric configuration of points and lines in which: (i) any two points lie on exactly one line, (ii) any two lines meet in exactly one point, (iii) there are four points no three of which lie on a line. (This serves only to exclude trivial configurations.) Prove that, in a projective plane, there is a number k so that every line contains k vertices and every vertex has k lines passing through it. The main observation is that if a point X is not on a line l then the number of lines passing through X is the same as the number point on l. Let P, Q, R, S be the points so that no 3 are collinear. Let k be the number of points in the line P Q. Then every point not in P Q (for example R and S) has k lines passing through them. We need to show points on P Q also have this property. Let X be a point on P Q, say X Q. Then the lines passing through X is the same as points on any line not containing X, say P R, which the same as the number of lines passing through S which we have shown is equal to k. We have shown every points has k lines going through it. But every line l has a a point X that is not on it and the number of points on l is the same as the number of line passing through X which we have shown to be k. Page 7 of 7 End of exam.