THE CHINESE POSTMAN PROBLEM - HOW CAN WE TELL WHETHER THERE W...

Basically, if all vertices in a graph are of an even order, then for any path the goes into a given vertex (excluding the one from which the path started), there will exist a continuation of the path leading out of the vertex.

Highlighted by mikeem

Basically, if all vertices in a graph are of an even order, then for any path the goes into a given vertex (excluding the one from which the path started), there will exist a continuation of the path leading out of the vertex. Therefore, it will be possible to find a circuit through the network in which every edge is included, by selective choice of path.

Highlighted by mikeem

it has been proven that, if every vertex in a network is of an even degree, then at least one (though possibly a lot more) circuit through that network can be found such that each edge is traversed once and once only, arriving back at the starting vertex once all other edges have been included.

Highlighted by mikeem

Obviously, it would be possible to start at any vertex on the graph, as they are all identical in that their degrees are all even.
Now, this will only be the case if there are no odd vertices in the network. This is because, if there is an odd degree vertex in the network, although a path may have a route into the vertex, it may not necessarily be able to leave again. There will be one occasion a path is not able to leave an odd vertex if a graph contains one or more of these, and so the cycle will break down.

Highlighted by mikeem

A connected graph does not have an Euler circuit or a semi-Euler circuit if and only if it has more than two odd vertices. This is because any path constructed, even if the initial vertex is an odd one, will necessarily reach an odd vertex that it is not able to leave again without repeating the same edge twice (which is not allowed as we are aiming to achieve an Eulerian or semi-Eulerian circuit).

Highlighted by mikeem