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MATH 2200 Homework #9 Solutions MATH 2200 Homework #9 Solutions Thomas Goller April 16, 2013 General Comments: Thank you for your hard work! • Please make your drawings of graphs BIG and neat! • The singular form of “vertices” is “vertex”. Think of this as analogous to “matrices” • and “matrix”. The correct spelling of “bipartite” is “bipartite”. • If you got a “2”, then please check in with me so I can give you tips on what to work • on for LC #5. I probably had trouble understanding your proofs. I only had time to grade some of the questions. Compare your other answers with the • solutions in this document. Many of the exercises have multiple “right” answers, so your answer doesn’t have to match mine to be fully correct. In case of doubt, come talk to me. Exercise 7.3.10 n Claim. Let n Z 2. The complete graph Kn has exactly edges. ∈ ≥ 2 Proof #1. Kn has n vertices and exactly one edge between! "every pair of distinct vertices. n n Since there are 2 pairs of distinct vertices, Kn has 2 edges. Proof #2. Since! ev" ery vertex in Kn is adjacent to ev!er"y other vertex, the degree of every vertex is n 1. By Exercise 7.1.9, − 2 E = deg(v) = (n 1) = n(n 1), · | | − − v V v V #∈ #∈ so n(n 1) n E = − = . | | 2 2 $ % 1 Exercise 7.3.15 Claim. Let G be a graph. Then G is simple if and only if G has no cycles of length 1 or 2. Proof. A loop together with its endpoint is a cycle of length 1. A pair of multiple edges together with their endpoints are a cycle of length 2. Thus G has no loops and no multiple edges if and only if it has no cycles of length 1 or 2. Exercise 7.3.20 Claim. Let m, n Z 0. The complete bipartite graph Km,n has exactly mn edges. ∈ ≥ Proof. Since K is complete bipartite, we can write V = V V , where V = m, V = n, m,n 1 # 2 | 1| | 2| and Km,n has exactly one edge between every vertex in V1 and every vertex in V2. Thus there is exactly one edge for each choice of a vertex in V1 and a vertex in V2, and by the product rule there are V V = mn ways to choose a vertex in V and a vertex in V . | 1| · | 2| 1 2 Exercise 7.3.21 Claim (a). Let H be a subgraph of a graph G. If G is simple, then H is simple. Proof. Since G has no loops or multiple edges, neither does H. Claim (b). Let H be a subgraph of a graph G. If G is a cycle, then H is a cycle. This claim is false! Counterexample: The subgraph of the cycle is not a cycle. • • Claim. Let H be a subgraph of a graph G. If G is bipartite, then H is bipartite. Proof. Since G = (V, E, φ) is bipartite, we can write V = V V , where every edge in G has 1 # 2 one endpoint in V1 and the other endpoint in V2. Then V = (V V ) (V V ). H 1 ∩ H # 2 ∩ H If e E , then also e E, so e has one endpoint in V and the other endpoint in V . Since ∈ H ∈ 1 2 both endpoints of e are in VH , we see that e has one endpoint in V1 VH and the other endpoint in V V . Thus H is bipartite. ∩ 2 ∩ H Claim. If G is complete bipartite, then H is complete bipartite. This claim is false! Counterexample: Let • • G = K1,2 = and H = • • • • 2 Exercise 7.4.4 Claim. Let G be a bipartite graph. Then every walk in G that starts and ends at the same vertex contains an even number of edges. Proof. Since G = (V, E, φ) is bipartite, we can write V = V V , such that every edge in E 1 # 2 has one endpoint in V1 and the other endpoint in V2. Suppose (v0, e1, v1, e2, . , vn 1, en, v0) − is a walk in G that starts and ends at the same vertex v0. Since every edge has one endpoint in V1 and the other in V2, the vertices in the walk alternate between V1 and V2. Since the walk starts and ends at the same vertex, the number of vertices in the walk must be odd, namely n + 1 is odd. Thus n, which is the number of edges in the walk, is even. Exercise 7.4.10 Claim (a). Let G be a graph. If G is simple, then G is connected. This claim is false! Counterexample: G = . • • Claim (b). Let G = (V, E, φ) be a graph. If G is complete, then G is connected. Proof. Let v, w V . If v = w, then (v) is a walk in G from v to w. If v = w, then since G is complete, ther∈e is e E such that φ(e) = v, w . Then (v, e, w) is a wa% lk in G from v to w. So G is connected. ∈ { } Claim (c). If G = (V, E, φ) is a cycle of length n, then G is connected. Proof. Since G is a cycle of length n, we can write V = v , . , v and E = e , . , e , { 1 n} { 1 n} such that φ(e1) = vn, v1 and φ(ei) = vi 1, vi for all 2 i n. Let vi, vj V . We may − assume i j. Then{ } { } ≤ ≤ ∈ ≤ (vi, ei+1, vi+1, . , ej, vj) is a walk in G from vi to vj. So G is connected. Claim (d). Let G be a graph. If G is bipartite, then G is connected. This claim is false! Counterexample: Let G = K = . 2,0 • • Claim (e). Let G be a graph. If G is complete bipartite, then G is connected. This claim is false! Counterexample: Let G = K = . 2,0 • • 3 Claim (f). Let G = (V, E, φ) be a graph. If G is the complete bipartite graph Km,n for m, n 1, then G is connected. ≥ Proof. Since G is complete bipartite, we can write V = V1 V2, where V1 = m, V2 = n, and G has exactly one edge between each vertex in V and ea# ch vertex in| V| . Let v|, w| V . 1 2 ∈ If v V1 and w V2, then there is e E such that φ(e) = v, w , and (v, e, w) is a walk in G fro∈m v to w. If∈ instead v, w V , t∈hen choose any u V {. Ther} e are e , e E such that ∈ 1 ∈ 2 1 2 ∈ φ(e1) = v, u and φ(e2) = w, u , and (v, e1, u, e2, w) is a walk in G from v to w. Similar argumen{ts wo}rk when v V{ and}w V or when v, w V . Thus G is connected. ∈ 2 ∈ 1 ∈ 2 Exercise 7.4.11 Claim (a). Let H and H be connected subgraphs of a graph G. Then H H is connected. 1 2 1 ∪ 2 v w v w This claim is false! Counterexample: Let G = , H1 = , and H2 = . Then H and H are connected, but H H = G is not co• nnect•ed. • • 1 2 1 ∪ 2 Claim (b). Let H and H be connected subgraphs of a graph G. Then H H is connected. 1 2 1 ∩ 2 This claim is false! Counterexample: Let G = , H1 = , and H2 = . Then H H = is not connected• . • • • • • 1 ∩ 2 • • 4.
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