math239_f16_a6_sln - Math 239 Fall 2016 Assignment 6...

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Math 239 Fall 2016 Assignment 6 Solutions 1. { 6 marks } Consider the following three graphs. G H 2 1 2 3 4 5 6 7 8 9 N I O G L M W E C H 1 N I O G L M W E C Among the graphs H 1 and H 2 , one of them is isomorphic to the graph G , and the other is not. For the one that is isomorphic, give an isomorphism (no justification required). For the one that is not isomorphic, give a proof that it is not isomorphic to G . Solution. Graph G is not isomorphic to graph H 1 . There are two vertices of degree 4 in G , which are 3 and 7. In addition, vertex 6 is adjacent to both 3 and 7. There are also two vertices of degree 4 in H 1 , which are L and O. So 3 and 7 are mapped to L and O in some way. However, L and O do not have any common neighbours in H 1 , so it is not possible to map 6 to a vertex in H 1 so that it is adjacent to both L and O. Therefore, no isomorphism exists. Graph G is isomorphic to graph H 2 , we give an isomorphism f : V ( G ) V ( H 2 ): v 1 2 3 4 5 6 7 8 9 f ( v ) W E L C O M I N G 2. (a) { 2 marks } Draw a 3-regular graph with 6 vertices, and one with 8 vertices. Solution. (b) { 4 marks } Prove that any 3-regular graph must have even number of vertices, and the number of edges is divisible by 3. Solution. In a 3-regular graph, every vertex has odd degree. The corollary of the Handshaking Lemma tells us that there are even number of odd-degree vertices in every graph. Hence the number of vertices must be even.
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