Characteristics of Eulerian Graph

Theorem
A finite graph is Eulerian if and only if it is connected and each vertex is even.

Note that the definition of graph here includes:
 * Simple graph;
 * Loop-graph;
 * Multigraph;
 * Loop-multigraph.

Sufficient Condition
First, we prove that any Eulerian graph is connected and has only even vertices.

Suppose that $$G$$ is Eulerian.

Then $$G$$ contains an Eulerian circuit, that is, a circuit that uses each vertex and passes through each edge exactly once.

Since a circuit must be connected, $$G$$ is connected.

Beginning at a vertex $$v$$, follow the Eulerian circuit through the graph.

As the circuit passes through each vertex, it uses two edges: one going to the vertex and another leaving.

Each edge is used exactly once, so each of the vertices must be even. Since the circuit must also end at $$v$$, so $$v$$ is also even.

Necessary Condition
To prove the converse, suppose that a graph $$G$$ is connected and its vertices all have even degree.

If there is more than one vertex in the graph, then each vertex must have degree greater than $$0$$.

Begin at a vertex $$v$$.

Since the graph is connected, there must be an edge $$\{v, v_1\}$$ for some vertex $$v_1 \ne v$$.

Since $$v_1$$ has even degree greater than $$0$$, there is an edge $$\{v_1,v_2\}$$.

These two edges make a trail from $$v$$ to $$v_2$$.

Continue this trail, leaving each vertex on an edge that was not previously used, until returning to $$v$$.

Call the circuit formed by this process $$C_1$$.

If $$C_1$$ covers all the edges of $$G$$, then we are done.

Otherwise, remove $$C_1$$ from the graph, leaving the graph $$G_0$$.

The remaining vertices are still even, and since $$G$$ is connected there is some vertex $$u$$ in both $$G_0$$ and $$C_1$$.

Repeat the same process as before, beginning at $$u$$.

The new circuit, $$C_2$$, can be added to $$C_1$$ by starting at $$v$$, moving along $$C_1$$ to $$u$$, travelling around $$C_2$$ back to $$u$$ and then along the remainder of $$C_1$$ back to $$v$$.

Repeat this process, adding each new circuit found to create a larger circuit.

Since $$G$$ is finite, this process must end at some point, and the resulting circuit will be an Eulerian circuit.