Finding a cycle in graph theory.

Question

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 Def'n)

Originally i tried to simply pick a vertex at random say $v_0$ then move away from it at random deleting each edge as i traversed it i made the assumption that eventually i would arrive back to a vertex that i had already been at and i would have a circuit. Then if this vertex i landed on a second time was $v_0$ i am done and have a cycle. If it is not a cycle i wished to delete the edge $v_0v_1$ where $v_1$ was the first vertex i moved to and proceed in this fashion until whatever vertex i started on i also ended on then claim that this was in fact a cycle.

For some reason that i dont understand this doesn't actually work.

Any ideas how to solve this would much appreciated.

Answer 1

Your method of random walking should discover a cycle in a finite simple graph of minimum degree $2$, although it may not include the starting vertex.

Eventually, because there only a finite number of edges, you will reach a point where no further steps are possible. At that point you have arrived at a vertex that you visited previously. The portion of the route from your previous visit to that vertex and the end point is a cycle.

Restarting the process is unnecessary, though; continuously following edges (without using any edge twice) until there are no possible next steps is sufficient.

Here the initial node $1$ is not part of any cycle but the walk ends at the vertex here marked $3$, so $3{-}4{-}5{-}6$ is the cycle discovered.