Discrete Mathematics with Graph Theory (3rd Edition) 258

Discrete Mathematics with Graph Theory (3rd Edition) 258 -...

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256 Solutions to Exercises Case 2: There is no vertex of degree 12. In this case, there is a vertex v of degree 11. Then v must be joined to at least two of the vertices of degree one. These vertices cannot be connected to any other vertex, so the other two vertices must have degree at most 10. If one had degree 10, then it would be connected to the third vertex of degree one, and the final vertex would be of degree at most 9, which is impossible because 11 + 10 + 9 = 30 < 31. Otherwise, the other two vertices have degrees at most 9 and then the three vertices have degree sum ~ 11 + 9 + 9 = 29 < 31, again a contradiction. Exercises 9.3 1. [BB] (i) and (ii) are not isomorphic because (i) has five edges and (ii) has four. (i) and (iii) are isomorphic, as shown by the labeling.
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Unformatted text preview: A BAD ~ll C DB C (i) and (iii) ~~G C D D (i) and (v) (i) and (iv) are not isomorphic because (iv) has a vertex of degree 1 and (i) does not. (Also, (iv) has only four edges.) (i) and (v) are isomorphic, as shown by the labeling. (ii) and (iii) are not isomorphic because (ii) has four edges and (iii) has five edges. (ii) and (iv) are not isomorphic because (iv) has a vertex of degree 1 and (ii) does not. (ii) and (v) are not isomorphic because (v) has five edges and (ii) has four edges. (iii) and (iv) are not isomorphic because (iv) has a vertex of degree 1 and (iii) does not. (iii) and (v) are isomorphic, as shown by the labeling. i1A~G B C D (iv) and (v) are not isomorphic because (iv) has a vertex of degree 1 [or because (v) has five edges]. 2. (a) (b)...
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