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Unformatted text preview: MATH 239 Winter 2009 Assignment 5 Solutions [23 marks] 1. Prove that the following two graphs are not isomorphic. Solution. [3 marks] First step: note that each graph has only one vertex of degree 3. (In the first graph, it is B, and in the second, V.) In the first graph, the vertex of degree 3 is contained in exactly one 3cycle, with just two of its neighbours (G and D). In the second graph, the vertex of degree 3 is in two different 3cycles, VUW and VZW, including all three of its neighbours. So the graphs cannot be isomorphic. For a bit more explanation: note that if an isomorphism existed from the first graph to the second, it would have to map B to V, since they are the only vertices of degree 3. It also must map each vertex in a 3cycle with B to a vertex in a 3cycle with V. This becomes impossible because the number of such vertices is not the same in each of the two graphs, but the isomorphism needs to be a bijection. An alternative answer can be based on assuming the result of the following Question 2, and showing that the complements of the two graphs are not isomorphic. The complements have less edges so it is easier to see what is going on. 2. The complement of a graph G is the graph G such that V ( G ) = V ( G ) and for all u,v ∈ V ( G ), uv ∈ E ( G ) iff uv / ∈ E ( G ). (a) Prove that G is isomorphic to H if and only if G is isomorphic to H . Solution. [4 marks] If G is isomorphic to H then there exists an isomorphism φ from V ( G ) to V ( H ) with uv ∈ E ( G ) iff φ ( u ) φ ( v ) ∈ E ( H ). Given such an isomorphism, we have uv ∈ E ( G ) iff uv / ∈ E ( G ) (by definition of complement) iff φ ( u ) φ ( v ) / ∈ E ( H ) (by properties of isomorphism) iff φ ( u ) φ ( v ) ∈ E ( H ). Thus φ is an isomorphism from G to H ....
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This note was uploaded on 08/09/2009 for the course MATH 239 taught by Professor M.pei during the Spring '09 term at Waterloo.
 Spring '09
 M.PEI
 Math, Combinatorics

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