[a_e_k]

I assume that's only possible in the general case for an odd number of vertices?

A fully connected graph with an even number of vertices will have odd degrees for all of them, and so can't have an Eulerian path.

[phalf]

I could be mixing things up here, it's been too long ago. I think this was about a directed graph, so the number of edges is actually N^2 and your walk roughly of that length.

[hgsgm]

/r/2IsNotEven