1. A Survey of Mathematics with Applications
Tenth Edition
Chapter 13
Graph Theory
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2. Section 13.2 Euler Paths and Euler Circuits

3. What You Will Learn
Upon completion of this section, you will be able to:
Solve problems involving Euler paths and Euler circuits.

4. Euler Path
An Euler path is a path that passes through each edge of a graph exactly one time.

5. Euler Circuit
An Euler circuit is a circuit that passes through each edge of a graph exactly one time.

6. Euler Path Versus Euler Circuit (1 of 2)
The difference between an Euler path and an Euler circuit is that an Euler circuit must start and end at the same vertex.

7. Euler Path Versus Euler Circuit (2 of 2)
Euler Path D, E, B, C, A, B, D, C, E
Euler Circuit D, E, B, C, A, B, D, C, E, F, D

8. Euler’s Theorem (1 of 2)
For a connected graph, the following statements are true:
A graph with no odd vertices (all even vertices) has at least one Euler path, which is also an Euler circuit. An Euler circuit can be started at any vertex and it will end at the same vertex.

9. Euler’s Theorem (2 of 2)
A graph with exactly two odd vertices has at least one Euler path but no Euler circuits. Each Euler path must begin at one of the two odd vertices, and it will end at the other odd vertex.
A graph with more than two odd vertices has neither an Euler path nor an Euler circuit.

10. Example 3: Solving the Königsberg Bridge Problem (1 of 3)
Could a walk be taken through Königsberg during which each bridge is crossed exactly one time?

11. Example 3: Solving the Königsberg Bridge Problem (2 of 3)
Solution Here’s a representation of the problem: vertices are the land, edges are the bridges.

12. Example 3: Solving the Königsberg Bridge Problem (3 of 3)
Solution Does an Euler path exist? Four odd vertices: A, B, C, D So, according to item 3 of Euler’s Theorem, no Euler path exists.

13. Fleury’s Algorithm (1 of 4)
To determine an Euler path or an Euler circuit
Use Euler’s theorem to determine whether an Euler path or an Euler circuit exists. If one exists, proceed with steps 2-5.

14. Fleury’s Algorithm (2 of 4)
If the graph has no odd vertices (therefore has an Euler circuit, which is also an Euler path), choose any vertex as the starting point. If the graph has exactly two odd vertices (therefore has only an Euler path), choose one of the two odd vertices as the starting point.

15. Fleury’s Algorithm (3 of 4)
Begin to trace edges as you move through the graph. Number the edges as you trace them. Since you can’t trace any edges twice in Euler paths and Euler circuits, once an edge is traced consider it “invisible.”

16. Fleury’s Algorithm (4 of 4)
When faced with a choice of edges to trace, if possible, choose an edge that is not a bridge (i.e., don’t create a disconnected graph with your choice of edges).
Continue until each edge of the entire graph has been traced once.

17. Example 6: Crime Stoppers Problem (1 of 9)
On the next slide is a representation of the Country Oaks subdivision of homes. The Country Oaks Neighborhood Association is planning to organize a crime stopper group in which residents take turns walking through the neighborhood with cell phones to report any suspicious activity to the police.

18. Example 6: Crime Stoppers Problem (2 of 9)

19. Example 6: Crime Stoppers Problem (3 of 9)
a) Can the residents of Country Oaks start at one intersection (or vertex) and walk each street block (or edge) in the neighborhood exactly once and return to the intersection where they started? b) If yes, determine a circuit that could be followed to accomplish their walk.

20. Example 6: Crime Stoppers Problem (4 of 9)
Solution a) Does a Euler circuit exist?
There are no odd vertices.
By item 1 of Euler’s theorem, there is at least one Euler circuit.

21. Example 6: Crime Stoppers Problem (5 of 9)
Solution Start at A. Choose A B or A E, neither is a bridge. Choose A B.

22. Example 6: Crime Stoppers Problem (6 of 9)
Solution Continue to trace from vertex to vertex around the outside of the graph. Notice that no edge chosen is a bridge.

23. Example 6: Crime Stoppers Problem (7 of 9)
Solution At vertex E. E A is a bridge, so choose either E B or E I. Choose E B.
Now from vertex B we must choose B F.

24. Example 6: Crime Stoppers Problem (8 of 9)
Solution
From vertex F. F I is a bridge, so we must choose either F C or F I. Choose F C.
From vertex C, our only choice is C G.

25. Example 6: Crime Stoppers Problem (9 of 9)
Solution
From vertex G, G J is a bridge, so we must choose G G.
Back at vertex G, and from now on, there is only one choice at each vertex.