Discrete Mathematics with Graph Theory (3rd Edition) 251

Discrete Mathematics with Graph Theory (3rd Edition) 251 -...

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Section 9.2 11. (a) Such a coloring is shown at the right. (b) Suppose the cat and the mouse are each at black positions, as they are initially, and it is the cat's move. The cat must move to a white position, so she doesn't have the mouse since the mouse is at a black position. 249 CAT Since every vertex in the graph is joined to at least two others, the mouse can always move to a vertex where the cat isn't. Now both animals are at white vertices and it's the cat's tum to move. The cat moves to a black vertex where the mouse isn't (the mouse is at a white position). Again, the mouse can move to a (necessarily black) position different from the cat's. And so on. Exercises 9.2 V2
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Unformatted text preview: 1. [BB] Here is one possibility. v'6 v , V5 V4 Vl V2 Vl V2 Vl V2 G------E) \ 0 0 /., o V3 V3 2. o V3 o edges; 0, 0, 0 1 edge; 1, 1,0 1 edge; 1, 1,0 1 edge; 1, 1, 0 Vl V2 Vl V2 Vl V2 \: v: \1., 2 edges; 2, 1, 1 2 edges; 2, 1, 1 2 edges; 2, 1, 1 3 edges; 2, 2,2 The question doesn't make sense for pseudographs because we could just keep adding edges indefi-nitely. 3. [BB] o 4. (a) There are six vertices, 12 edges and the degree sequence is 6, 5, 4, 4, 3, 2. (b) Proposition 9.2.5: L: deg Vi = 6 + 5 + 4 + 4 + 3 + 2 = 2(12) = 21t"l-Corollary 9.2.6: There are two odd vertices (of degrees 3 and 5), two being even. 5. [BB] 10 edges. This is JC 5 , the complete graph on five vertices....
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.

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