Discrete Mathematics with Graph Theory (3rd Edition) 255

# Discrete Mathematics with Graph Theory (3rd Edition) 255 -...

This preview shows page 1. Sign up to view the full content.

Section 9.2 253 18. (a) [BB] No such graph exists. The sum of the degrees of the vertices is an odd number, 17, which is impossible. (b) No such graph exists: 100(101) 2 + 2 + 2 + 3 + 4 + . .. + 100 = 1 + (1 + 2 + 3 + . .. + 100) = 3 + 2 is odd. (c) [BB] Impossible. A vertex of degree 5 in a graph with six vertices must be adjacent to all other vertices. Two vertices of degree 5 means all other vertices have degree at least 2, but the given degree sequence contains a 1. (d) 0-0 0-0 0-0 (e) Not possible. Since there is a vertex of degree 5 there must be at least six vertices in the graph. (0 Not possible. The vertex of degree 5 would be joined to all other vertices. But now, the degrees of the two vertices of degree 1 have been accounted for, and it is not possible to have a vertex of degree 4. (g) Not possible. The vertices of degree 6 use up all the degrees 1, 1, 2, 2, 2, leaving a vertex of degree 4 impossible. (h) Yes. 19. No. Since there are five vertices, we would have to have at least three edges in order that every vertex
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: be incident with at least one of them, but then some vertex is incident with two edges, contradicting no adjacent edges. 20. (a) [BB] E degvi = 5(4) + 2(2) = 24 = 21£1, so there are 12 edges. (b) E deg Vi = 4(3) + 2(4) + 2(5) = 30 = 21£1, so there are 15 edges. 21. (a) [BB] This is not bipartite because it contains a triangle. (b) This is bipartite with bipartition sets indicated R and W on the graph at the left below. R W R W R BJ W R w~w R W R (c) [BB] This is bipartite with bipartition sets indicated R and W on the graph at the right above. (d) This is not bipartite. Consider the vertices labeled 1, 2, 3, 4, 5 in the picture at the left below. Vertices 1 and 2 are would have to lie in different bipartition sets and, thus, 1 and 3 would lie in the same set. Since 3 and 4, and 4 and 5 lie in different sets, 3 and 5 lie in the same set. But this...
View Full Document

## 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.

Ask a homework question - tutors are online