Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 14 - Graph Theory - 14.2 Euler Paths and Euler Circuits - Exercise Set 14.2: 10

Answer

(a) The graph has no odd vertices. Therefore, by Euler's theorem, the graph has at least one Euler circuit. (b) A,B,C,E,D,C,A,D,A is an Euler circuit.

Work Step by Step

(a) The graph has no odd vertices. Therefore, by Euler's theorem, the graph has at least one Euler circuit. (b) Let's start at vertex A. From there, let's travel to vertex B, then to vertex C, then to vertex E, then to vertex D, then to vertex C, and back to vertex A. There are only two edges which have not been used. The path can then travel to vertex D, and then finally back to vertex A. This path is A,B,C,E,D,C,A,D,A. This path travels through every edge of the graph exactly once, so it is an Euler path. This Euler path starts and ends at the same vertex, so this path is an Euler circuit. This is one Euler circuit but there are other Euler circuits in this graph also.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.