1. Lecture 52 Section 11.2 Wed, Apr 26, 2006
Paths and Circuits
Lecture 52
Section 11.2
Wed, Apr 26, 2006

2. The Seven Bridges of Königsberg
In the city of Königsberg, two branches of the Pregel River came together, with an island at their junction.

3. The Seven Bridges of Königsberg
There were seven bridges crossing the river at various places.

4. The Seven Bridges of Königsberg
The challenge was to start at one point, cross each bridge exactly once, and return to the starting point. ?

5. Euler’s Solution
Euler abstracted the bridges as a graph with four vertices and seven edges.

6. Euler’s Solution
Each vertex represents a land mass and each edge represents a bridge.
North Shore
South Shore
Island
Peninsula

7. Walks and Paths
A walk from vertex v to vertex w is a finite alternating sequence of adjacent vertices and edges from v to w:
v0 e1 v1 e2 … en – 1 vn – 1 en vn,
where v0 = v and vn = w.
A path from v to w is a walk that does not repeat any edge.

8. Walks and Paths
A simple path is a path that does not repeat any vertices.
A closed walk is a walk that starts and ends at the same vertex.
A circuit is a closed path.
A simple circuit is a circuit that does not repeat any vertex.

9. Synopsis walk = from A to B, no restrictions.
path = walk, no repeated edge.
closed = from A to A.
circuit = closed walk.
simple = no repeated vertex.

10. Euler Circuits
An Euler circuit is a circuit that contains every vertex and every edge of the graph.
The problem of the Seven Bridges of Königsberg is to find an Euler circuit.

11. Connected Graphs
A graph is connected if, for every pair of vertices v and w, there is a walk from v to w.
A connected component of a graph is a maximal connected subgraph.

12. Euler’s Solution
Theorem: A graph has an Euler circuit if and only if it is connected and every vertex has even degree.
Thus, an Euler circuit over the Seven Bridges of Königsberg does not exist.

13. The Two Bridges of Ashland
At Randolph-Macon College, they have been trying to solve the Two Bridges of Ashland problem for decades. ?
King’s Dominion
RMC
I-95

14. Proof Proof (): Suppose a graph G has an Euler circuit. Let v  V(G).
Then as we travel the circuit, each time we pass through v, we “use up” two of the edges incident to v.
When we finish the circuit, we have used all the edges incident to v.

15. Proof Thus, v had an even number of edges.
Obviously, G must be connected.

16. Proof
Proof ():
Now suppose that G is connected and that every vertex of G has even degree.
Choose a vertex v at which to begin.
deg(v) > 0 since G is connected, so follow one of the edges incident to v.
Let w be the next vertex.
We used one of w’s edges to get there.

17. Proof
w has even degree, so there is at least one more edge available that we can follow.
This happens at every vertex that we visit.
Thus, the circuit is forced to terminate only when we return to the starting vertex v.
This procedure alone does not necessarily produce an Euler circuit.

18. Proof Suppose there are edges that were not used.
Follow the original circuit until a vertex is reached that is incident to one of the unused edges.
Apply the original procedure to produce a circuit that starts and ends at this vertex.
“Splice” it into the original circuit.

19. Proof
Continue in this way, splicing circuits into the existing circuit, until there are no unused edges remaining.
The result is an Euler circuit.

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61. Euler Paths
Theorem: A graph G has an Euler path from v to w if G is connected, v and w have odd degree, and all other vertices have even degree.
Proof:
Add an edge from v to w.
Then the graph has an Euler circuit.
Remove the new edge from the circuit.

62. Hamiltonian Circuits
A Hamiltonian circuit is a simple circuit that includes every vertex of the graph.
The Traveling Salesman Problem seeks a Hamiltonian circuit of minimal length.

63. Hamiltonian Circuits
Theorem: If a graph G has a nontrivial Hamiltonian circuit, then G has a subgraph H such that
V(H) = V(G).
H is connected.
|E(H)| = |V(G)|.
deg(v) = 2 for all v  V(H).
These conditions are necessary, but not sufficient.

64. Hamiltonian Circuits
The following graph does not have a Hamiltonian circuit.