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Unformatted text preview: MATH 239 ASSIGNMENT 6 Draft Solutions 33 marks total 1. 5 marks (a) Let G be a graph in which exactly two of the vertices u,v have odd degree. Prove that G contains a path from u to v . Solution: Suppose to the contrary that there is no path from u to v . This means that u and v are in different components H u and H v of G . The components H u and H v would each be graphs with exactly one vertex of odd degree. This contradicts the handshake theorem (Corollary 4.1.8), which states there must be an even number of vertices of odd degree. Thus there must be a path from u to v . (b) Let G be a graph where every vertex has degree 4 or 6. Prove that G cannot have a bridge. Solution: Suppose G contains a bridge e = { u,v } . Let H be the connected component of G that contains e . Then by lemma 4.4.6 the graph H e contains two components, and u and v are in different components . Since u and v have even degree in H , then they have odd degree in H e ; the remaining vertices still have the same degree as in H (either 4 or 6, both even). Part a) implies that there must be a path from u to v in H e , which contradicts the fact that u and v are in different components of H e ; thus G does not contain a bridge. 2. 12 marks Let G n be the graph with vertices equal to the binary strings of length n 2 with an even number of 1s, and two vertices are joined if they differ in exactly two consecutive positions. E.g. 0110011 and 0101011 are adjacent vertices in G 7 since they differ in positions 3 and 4....
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 Spring '09
 M.PEI
 Math, Combinatorics

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