# An Eulerian Graph without an Eulerian Circuit?

I want to make sure I understand the definition of an Eulerian Graph / Trail / Cycle / Circut.

I'm reading Schaum's Discrete Mathematics Outline which gives the following definitions:

A graph G is called an Eulerian Graph if there exists a closed traversable trail, called an Eulerian trail.

It then goes on to say that:

A finite connected graph is Eulerian if and only if each vertex has even degree.

I then read the following in Wikipedia:

Euler proved that a necessary condition for the existence of Eulerian circuits is that all vertices in the graph have an even degree

...not every Eulerian graph possesses an Eulerian cycle

So which one is right? Can a graph be Eulerian, having only vertices of even degree, yet not contain an Eulerian circuit?

Also, isn't there a distinction between an Eulerian trail and an Eulerian Cycle / Circuit?

• There is a distinction between an Eulerian trail/path and an Eulerian circuit, but note that the Schaum’s Outline definition specified a closed Eulerian trail $-$ which would be a circuit. – Brian M. Scott Jan 26 '12 at 9:08
• For the question about Eulerian graphs, note that Wikipedia also says: ‘The term Eulerian graph has two common meanings in graph theory. One meaning is a graph with an Eulerian circuit, and the other is a graph with every vertex of even degree.’ When they say that not every Eulerian graph possesses an Eulerian cycle, they’re using the second definition and thinking of graphs that are not connected. – Brian M. Scott Jan 26 '12 at 9:10
• So the only case of a graph whose vertices are all even but doesn't contain an Eulerian circuit would be disconnected graphs, correct? – Robert S. Barnes Jan 26 '12 at 9:19
• That’s correct. They would be unions of Eulerian graphs in the stronger sense. – Brian M. Scott Jan 26 '12 at 9:21
• I edited the Wikipedia definition to be clearer about the two possible senses of the term. It's now a bit redundant, but that's better than being confusing. – Ilmari Karonen Jan 26 '12 at 13:20