I have an assignment to count all nonequivalent Eulerian trails in the famous kindergarten graph:
I think that there are $12$ distinct trails in total, but I have reached this conclusion by drawing them all. On top of that, I think that I can reason like this:
The graph has 2 odd vertices, so the trail has to start from one of them and end in the other one. If I start from vertex $A$, I can go in $3$ different routes and each one then splits one more time to $2$ other. So that is $3 \times 2 = 6$ different trails. And another $6$ if I start from $B$, using the same reasoning.
My question is - is my reasoning correct? And is there any general algorithm for counting Eulerian trails in undirected graphs? I know it exists for directed graphs.
Thanks in advance!