We started our discussions on graphs, and prove Euler's theorem, the first non-trivial result in graph theory. On the way we have also introduced some basic definitions in graph theory, e.g. degree sequence, isomorphism, path, cycle, etc.
In the first step, when we choose an arbitrary vertex v, and extend the path as long as possible, and come back at vertex v and got stuck. This is where we used the assumption that every vertex is of even degree, because this assumption implies that the path could not get stuck in another vertex u (otherwise we enter u once more than we leave u, and there is no more edges unvisited, meaning that u is of odd degree). So the assumption is used here to guarantee that we can come back to the same vertex v, which is crucial in constructing the Eulerian path.
I would like to ask the proof
ReplyDelete"if every vertex is of even degree, a graph has an Eulerian path"
Where does the proof use the assumption "every vertex is of even degree"?
In the first step, when we choose an arbitrary vertex v, and extend the path as long as possible, and come back at vertex v and got stuck. This is where we used the assumption that every vertex is of even degree, because this assumption implies that the path could not get stuck in another vertex u (otherwise we enter u once more than we leave u, and there is no more edges unvisited, meaning that u is of odd degree). So the assumption is used here to guarantee that we can come back to the same vertex v, which is crucial in constructing the Eulerian path.
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