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Unformatted text preview: Section 9.2 255 28. [BB] Let n be the number of vertices. Since all vertices have degree;:::: 3, it follows that 'E deg Vi ;:::: 3n. But 'E deg Vi = 21E I = 70, so 70 ;:::: 3n and 23! ;:::: n. Since n must be an integer, the largest possible number is 23. 29. We know that mlVI ::; 'E degvi ::; MIVI. But 'E degvi = 21EI. SO the result follows. 30. [BB] 'E deg Vi = kiVI. But also, 'E deg Vi = 21EI. Therefore, 21EI = klVI and so k divides 21EI. But k is odd, so k 1 lEI 31. Suppose 9 has vertex set V and edge set E. Since 'EVEV deg V = 21EI, we have dn = 34, so d = 1,2,17,34 and n = 34,17,2,1, respectively. Clearly the cases d = 34, n = 1 and d = 17, n = 2 are impossible. The case d = 1, n = 34 defines a graph with 17 components each of which is a single edge. The case d = 2, n = 17 arises when each (connected)component of 9 is a cycle. 32. [BB] There are n vertices and n possible degrees for the vertices; namely, 0,1,2, ... , n - 1. If, however, we have a vertex of degree 0, then it is not possible to have another vertex of degree...
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- Summer '10
- Graph Theory