hihi. all - let's start by stating the problem in a way that can be solved the graph certainly needs to be finite and connected in addition, you also have to limit the number of times each edge can be retraced, or else it is trivial to draw any connected finite graph with a continuous path, whether or not there are odd order vertices (pick any one vertex - connected means there is a path, so just draw all the paths from that vertex to the others and back) the usual restriction is to limit each edge to exactly one traversal (so the order of a vertex is the total number of arrivals and departures it encounters in the continuous path), which is called an eulerian path (the seven bridges of koenigsburg apparently brought the problem to his attention) there are no finite graphs with exactly one odd order vertex it is clear that any eulerian path in a graph with two odd order vertices will necessarily start at one of them and end at the other also, it is clear that any eulerian path for a graph with no odd order vertices must end at the same vertex at which it starts i expect there is a reasonably straightforward induction proof - it is certainly an old theorem more later, cal