Show that if $|E|\ge |V|$, $G = (V,E)$ contains a cycle.

Question

Show that if $|E|\ge |V|$, the graph $G=(V,E)$ contains a cycle.

So I am trying to prove this problem. So I am assuming this is saying that if there are more edges than vertices in G, then G contains at least 1 cycle. Intuitively, this makes perfect sense, but I am having a little trouble getting it on paper because I am not very familiar with graph theory yet so it is difficult to prove things in it. Can someone help please?

score 3 · Answer 1 · edited Feb 03 '14 at 22:37

3

I'd do this: if there are no cycles, there must be vertices with only one edge. Then we can remove a vertex and an edge without changing $|E| - |V|$ and still have a graph with no cycles. This can be repeated until we are left with just one edge and two vertices, so $|E| - |V| = -1$ and $|E| < |V|$.

edited Feb 03 '14 at 22:37

Sammy Black

28,409

answered Feb 03 '14 at 22:26

Pifagor

529

What does E-v=-1 mean? – Brian Feb 03 '14 at 22:34
I edited the answer to make the notation consistent with that in your question. – Sammy Black Feb 03 '14 at 22:37
It means number of edges minus number of vertices. so when we have just one edge left, we have two vertices. (I assumed a connected graph, but the generalization is simple I think) – Pifagor Feb 03 '14 at 22:38
@Pifagor How exactly would you do the generalization? – Saksham Sethi Jul 19 '22 at 17:15
@SakshamSethi You can do the same procedure on each partial graph. If the number of partials is n, you'll end up with |E|−|V|=−n instead, and again |E|<|V|. – Pifagor Jul 20 '22 at 13:21

score 1 · Answer 2 · answered Feb 03 '14 at 22:42

It is a very important result in graph theory that a (finite) connected graph is acyclic if and only if it has exactly one fewer edges than vertices: adding any more edges will create a cycle. (Such graph are called trees.) If the graph is not connected, consider its connected components separately.

It is also true that an acyclic graph is connected if and only if it has exactly one fewer edges than vertices: removing any one edge will make it disconnected. Pifagor's proof above is fine, but you should really look those things up if you plan to use graph theory.

Show that if $|E|\ge |V|$, $G = (V,E)$ contains a cycle.

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