Every cube <vertex> has <degree of a vertex>[degree] three, so $Q_3$ is \b[not Eulerian]; it even has eight odd-<degree of a vertex>[degree] <vertices>, so has no open <Eulerian trail>. After the four diagonals are added every <vertex> has <degree of a vertex>[degree] four, and the <graph> remains <connected>. Thus $Q'_3$ \b[is Eulerian], by the circuit criterion proved above.

Solution

Past exam of the Probability, statistics and options II course of the University of Oxford / 2008 / R / 2 / iv / a / Solution

 Solution (source code)