2
$\begingroup$

I have a simple question and probably stupid in graph theory:

Prove that if there are more edges than vertices in a connected graph then there is a cycle:

Let $G=(V,E)$ a connected graph,where $|V| \le |E|$.Prove that there is a cycle in the graph.

How can I prove that?

  • 2
    Possible duplicate of [If a graph with $n$ vertices and $n$ edges there must a cycle?](http://math.stackexchange.com/questions/414733/if-a-graph-with-n-vertices-and-n-edges-there-must-a-cycle) – 2017-02-23

1 Answers 1

4

Consider a vertex of minimal degree. If the degree was $1$, then we're done by induction if we remove that vertex. Otherwise, the minimal degree in the graph is at least $2$. Take a longest path $v_0 v_1 \dots v_k$; then $v_0$ has degree $2$ or higher, so the only way this can be a longest path is if $v_0$ is connected to some other $v_i$ than $v_1$. (If not, if $v_0$ is connected to $x$ which is not a $v_i$, then we could extend the path by adding $x$.) So we have found a cycle.

Why is there a longest path? There is certainly a path of length $1$ (pick any vertex!). There are only finitely many paths, since there are only $|V|$ vertices. So we can just list the paths, and their lengths; at least one of them will have longest length (because every finite set has a maximum).

Question for you: where have we used that the graph was connected?

  • 0
    When you assume that there is such long path? – 2017-02-23
  • 0
    @TakaTiki Added to the answer. – 2017-02-23
  • 0
    I meant,that we have used the fact that the graph was connected when you assumed that there is such a long path? – 2017-02-23
  • 0
    Oh, sorry, I misread your comment :) Note that if the minimal degree is $0$ then we can proceed by induction, removing any degree-$0$ vertex. So I don't think we need to use connectedness at all. – 2017-02-24
  • 0
    I understand that when you say "remove that vertex" we are going to a subgraph $H$ without that vertex, and then you take a longest path in $H$. But you cannot assert that the extremes of this path have degree $>1$, since some of them could be connected to some 1-degree vertex in the supergraph. What am I missing? – 2017-11-28
  • 1
    @PedroAlves Remove that vertex; then find a *cycle* inductively (not a longest path, though the inductive "find a cycle" process may involve finding a longest path in the reduced graph). – 2017-11-28
  • 0
    @PatrickStevens So, I will walk through subgraphs removing every 1-degree vertex until finding some with $\delta \geq 2$? What guarantee do I have that this will happen? – 2017-11-28
  • 0
    @PedroAlves Every graph has a vertex of degree $\geq 2$ unless it's disconnected or the tree on two vertices (which is forbidden by $|V| \leq |E|$). Note that removing a degree-$1$ vertex removes exactly one edge and exactly one vertex, so it doesn't change the fact that $|V| \leq |E|$ from the original graph. – 2017-11-28