1.Quiz 5 due tomorrow afternoon in E309 from 1pm to 4pm.

2.Homework grades will be based on ten graded homework assignments (dropping the lowest one).

3.Final Exam will be on Thursday April 30, 12:30pm to 2:30pm.

4.Review problems (for the final) will be available before Saturday evening (Check the News and

Announcements)

5.Office hour cancelled this afternoon.

Path and Connectivity (9.4 and 9.5)

Not a connected graph because there is no path between v1 and v7.

Proof:

Because the graph is connected, there is a path between u and v. Throw out multiple edges to make the path simple.

No cut edge or cut vertex

In G1 and G2, every edge is a cut edge.

In the union, no edge is a cut edge.

The vertex e is a cut vertex in all graphs.

Mr is the r-th power of M (using matrix multiplication).

An Euler circuit in a graph G is a simple circuit containing every edge of G. An Euler path in G is a simple path containing every edge of G.

a

d

b

e

c

a b

c d e

a b

d c

e

a b

d c

e

There are Euler circuits. For example, a, e, d, c, b, a.

a

d

b

e

cNo Euler circuit.

a b

c d e

No Euler circuit but has Euler paths.

a, c, d, a b, d, e, b

Konigsberg Bridge Problem

The problem was to find a walk through the city that would cross each bridge once and only once.

Konigsberg is a city (in Prussia) with 7 bridges.

Problem: Find an Euler path or circuit in this graph.

Theorem: A connected multigraph with at least two vertices has an Euler circuit if and only if each of its vertices has even degree.

Theorem: A connected multigraph has an Euler path but not an Euler circuit if and only if its has exactly two vertices of odd degree.

a b

d c

e

a

d

b

e

c

a b

c d e

Theorem: A connected multigraph with at least two vertices has an Euler circuit if and only if each of its vertices has even degree.

Theorem: A connected multigraph has an Euler path but not an Euler circuit if and only if its has exactly two vertices of odd degree.

Proofs are on Page 635 in the textbook.

Necessary Condition: If G has Euler circuit, then every vertex has even degree.

Sufficient Condition: Reading assignment.

Starting vertexany other vertex

One edge in and one edge out.

first edge

last edge

Suppose there is an Euler circuit

All vertices have even degree.

One edge in and one edge out.

any other vertex

Suppose there is an Euler path

All vertices except two have even degree.

first edge

Starting vertex

last edge

Ending vertex

That’s all Folks !Have a great summer !!