Discrete Mathematics with Graph Theory (3rd Edition) 266

Discrete Mathematics with Graph Theory (3rd Edition) 266 -...

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

View Full Document Right Arrow Icon
264 4. (a) [BB] (b) No Eulerian circuit. The graph is not connected. (c) 0 49 3 1 5 8 6 7 (d) 12 ~ 5 10 3 1 6 9 ~ 7 8 (0 (e) No Eulerian circuit. There are (four) vertices of odd degree. 5. [BB] This definition does not work. In the graph to the right, the sequence el e2e3 of edges fits Gerard's criteria, but does not define a walk. 6. With the proposed definition, an Eulerian graph would no longer necessarily be connected, as the example shows. 7. (a) [BB] Yes, there is because A and B are the only vertices of odd degree. (b) No. C has even degree. 8. (a) [BB] No. In the graph representing the modified Konigsberg Bridge Problem there are two vertices of odd degree. o Solutions to Exercises (b) [BB] This question asks about the possibility of an Eulerian trail and there is one, between the two
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: vertices A and B of odd degree. One possibility is ACBDADCAB. 9. The graph depicting the situation is shown at the right. (a) This question asks whether the graph is Eulerian. It is not: there are vertices of odd degree, A, for instance. B~ C (b) This question asks whether the graph has an Eulerian trail. It does, between A and F, since these are the unique vertices of odd degree in a connected graph. One possible trail is ABDBCDCECFEDFADF. 10. [BB] Yes. In this case, both 9 and 1{ must be cycles. 11. The resulting graph has an Eulerian trail (between VI and V2) but no Eulerian circuit. This is because, in the new graph, VI and V2 are the only vertices of odd degree....
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