0
$\begingroup$

Prove that if every vertex of a simple graph $G$ has degree at least $2$ then $G$ contains a cycle. (Our definition says no multiple edges and no loops allowed; also all simple graphs are finite by our definition.)

I have already proved this in a different way, but would like to try this by using the concept of the longest path.

Consider the longest path in $G$. I'm going to assume that I can pick one of the vertices on the end or beginning (preferably both). I know that each of these vertices on the end must have degree 1 because it's the longest path and I want to say that there is an edge that connects them to each other as if the edge leaving the end or the beginning connected to something else it wouldn't be the longest path. Lastly I'd like to claim that the path plus this edge is a cycle.

This method is supposed to be easier, but I find I'm confused almost every step of the way on whether each statement is true. Could someone show me how to prove this using the concept of longest path or if possible explain why my attempt works.

  • 1
    How does your proposed method work on a graph consisting of two disjoint triangles and a path connecting a vertex in one triangle to a vertex in the other triangle? – 2017-01-14
  • 0
    my argument fails entirely but you shown me how to fix it. – 2017-01-14
  • 0
    if i make the argument on the end point and it having degree at least 1 then say it connect to some vertex already in the path.( if it wasnt then it would be a longer path contraditing the longest path) giving me a circuit that i can jury rig into a cycle – 2017-01-14
  • 0
    I believe that correcting your argument results in the "canonical" proof: Start walking on the graph; and since the degrees are at least $2$ whenever you visit a new vertex you can leave it by a different edge; since the graph is finite you will eventually return to a previously visited vertex,.This is the "Proof in the Book"; I don't know why you are looking for another. – 2017-01-14

1 Answers 1

2

Idea:

Let $u,v_2,v_3, \ldots ,v_{n-1}, v_n,w$ be the longest simple path in $G$. Since deg$(u) \geq 2$, therefore there must be a vertex (other than $v_2$), call it $t$ such that $ut$ is an edge in $G$.

Case(1): If $t \in \{v_3, \ldots , w\}$, then we have a cycle in $G$.

Case2): If $t \not\in \{v_2, v_3, \ldots , w\}$, then consider the path $\color{red}{t}, u,v_2,v_3, \ldots ,v_{n-1}, v_n,w$. This path is longer than the longest pah, hence.......

  • 1
    How is this different from the proof where you start at a vertex $w$ and walk a path $w,v_n,v_{n-1},\dots,v_3,v_2,u$ and then find that the only exit from $u$ is to a vertex $t$ that you've visited before? – 2017-01-14
  • 0
    Strictly speaking, we are looking at *a* longest *simple* path. – 2017-01-14
  • 1
    @bof im not sure how its different and i prefer the idea of wandering aimlessly deleting edges till i arrive somewhere i have already been – 2017-01-14
  • 0
    Looking for a *longest* simple path is a needless conplication. The same argument works with any *maximal* simple path, or more generally still, a simple path which is inextendible at one end — but that is exactly the "start somewhere and walk till you've seen the same vertex twice" proof. – 2017-01-14