# Euler circuit with ten edges, two vertices of degree 2 and the rest of vertices with the same degree

Can an Euler circuit with ten edges, two vertices of degree 2 and the rest of vertices with the same degree exist?

My approach is that it can't exist since for any Euler circuit you need that every vertex has a pair degree, since one edge will leave the vertex and at the end the other one will enter the vertex. So if you have 2 vertices of degree 2, you have used 4 of the ten edges possible. You need to arrange 6 edges taking into consideration their degree must be pair, so you have 3 vertices with degree 2 but if you connect them to the other 2 vertices, you end up with vertices with different degrees.

The two green vertices have degree $$2$$, whilst the rest has $$4$$, and there are $$10$$ edges in total.
Note that for an Euler path to exist, you only need the vertices to have even degree. They don't have to be of degree exactly $$2$$.
• The sum of the degrees of all vertices is equal to twice the number of edges. So if we have two vertices of degree $2$, and $k$ vertices of degree $n$, we would need $4 + kn = 20$, i.e. $kn = 16$. Thus, $n$ can only be $2,4,8,16$ with $k=16/n$. – glowstonetrees Jun 12 '20 at 11:22
• $n=16$, $k=1$ is clearly not possible. $n=2$ corresponds to just having $10$ vertices joined together in a circle. It should also be obvious that the only configuration for $n=8$, $k=2$ is \begin{align} & \equiv \\ 0 = 0 & \equiv 0 = 0 \end{align} where the $0$ represent vertices. – glowstonetrees Jun 12 '20 at 11:27
• Finally, we have the $n=k=4$ case as shown above, hence there are $3$ such graphs in total. – glowstonetrees Jun 12 '20 at 11:28