To answer the question about a graph $G$ where both the graph $G$ and its complement $\overline{G}$ are Eulerian, we need to first understand the definition of an Eulerian graph.

Definition of an Eulerian Graph:

A graph is Eulerian if it contains a closed Eulerian circuit, which means a cycle that visits every edge of the graph exactly once and returns to the starting point. For a graph to be Eulerian:

• Every vertex must have an even degree.

Eulerian complement graph $\overline{G}$:

The complement $\overline{G}$ of a graph $G$ has the same set of vertices, but its edges consist of all the edges that are not in $G$. Therefore, for both $G$ and $\overline{G}$ to be Eulerian:

• The degree of every vertex in $G$ must be even.
• The degree of every vertex in $\overline{G}$ must also be even.

Example of such a graph $G$:

A complete graph with 4 vertices, denoted as $K_4$, serves as a perfect example. Let’s break it down:

• Graph $G = K_4$:

  • A complete graph with 4 vertices has all possible edges between the 4 vertices. Each vertex in $K_4$ has degree 3 (since it is connected to the other 3 vertices).
  • For $G$ to be Eulerian, each vertex should have an even degree. However, this does not work for $K_4$ because its vertices have odd degrees.

• Graph $G$ = $K_4 \setminus E$: This notation represents removing some of the edges, and we adjust it to make degrees even. A simple example can be a disjoint union of two $K_2$. So, 4 vertices (with degrees that are calculated for combining.