[Math] How does an Eulerian path have vertices with odd degree

graph theory

Shouldn't a Eulerian path have vertices having only an even degree of paths? One leading in and one leading out?
How can a vertex with three paths (odd degree) become a part of Eulerian path since you would have to cross the vertex twice to follow all three paths?

Best Answer

An Eulerian path need not be a circuit. If it’s not a circuit, it has two endpoints, and they must have odd degree. In the case of the graph shown below, the endpoint labelled $F$ has degree $3$, while that labelled $S$ has degree $1$:

                      S    F
                      o----o----o  
                           |    |  
                           o----o
Related Question