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Unformatted text preview: Section 13.1 363 we have V3 = V2 + 1, while E3 2: E2 + 1. Hence, E2  V2 ~ E3  V3 and so EI  VI < E3  V3. Thus, 9 and its sub graph K provide another counterexample, but this contradicts the minimality of VI  V2 since VI  V3 < VI  V2. (b) Say 9 is not planar. Then 9 has a subgraph 1i with VI vertices and EI edges which is homeomor phic to K3 ,3 or K 5 For K3,3, E  V = 9  6 = 3. For K 5 , E  V = 10  5 = 5. By Exercises 12 and 13, we conclude that EI  Vi = 3 or 5, but we are given that EI  VI ~ 2, contradicting part (a). (c) The graph need not be planar if E = V + 3 since K 3 ,3 has six vertices and nine edges. 15. [BB] Yes. An example is shown to the right. The graphs are homeomorphic since the one on the right is obtainable from the other by adding a vertex of degree 2. GE) 0 0 0 A B A B 16. Doubtlessly Poland's most famous mathematician, Kazimierz Kuratowski was born in 1896 in Warsaw where he died on the 18th of June, 1980. Renowned for his lectures, Kuratowski's research was where he died on the 18th of June, 1980....
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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.
 Summer '10
 any
 Graph Theory

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