Prove or disprove if $G$ is 2-edge-connected then there exist 2 edges disjoint $u-v$ trail such that every edge of $G$ lies on one of these trails.

Question

Let $G$ be a connected graph with exactly 2 odd vertices $u$ and $v$ such that $deg(u) \geq 3$ and $deg(v) \geq 3$. Prove or disprove if $G$ is 2-edge-connected then there exist 2 edges disjoint $u-v$ trail in $G$ such that every edge of $G$ lies on one of these trails.

I know that if if $G$ is 2-edge-connected then there exist 2 edges disjoint $u-v$ path in $G$. But the rest bother me, I feel like they want to say $G$ contains an Eulerian trail. I tried to find a counter example, but got no luck.

I don't know if this is correct, but let's try it. Because $G$ is 2-edge-connected then there exist 2 edges disjoint $u-v$ paths in $G$ called $p_1$ and $p_2$. But if one of these trail , says $p_1$, contain every edge of $G$ then it also contain edges in $p_2$, so $p_1$ and $p_2$ can't be edge-disjoint paths. So this is false because it contradict itself, right?

Answer 1

I'm going to disagree with the answers so far and say that the result is false.

Not only is it false, but no graph satisfying all the conditions can possibly have two such trails.

Below is an expertly drawn 2-edge connected graph with exactly two odd vertices $u,v$. It's clearly not possible to cover all the edges with exactly two edge-disjoint trails from $u$ to $v$. You can do it with one trail or three trails, but not two trails.

But why is it never true?

Say that we have a trail from $u$ to $v$. Remove it. All the vertices will now have even degree, so any trail that uses all the edges must end up where it started and so can't be a trail from $u$ to $v$.

Answer 2

If a connected graph has exactly two odd vertices it has an eulerian trail between them. Why? Call those vertices $uv$. If $uv$ is in the graph remove it, if it is not in the graph add it. You now have a graph where all the vertices have even degree, thus an eulerian circuit, that circuit becomes a path when you take the graph back to its original form.

Answer 3

As you said, if $G$ is 2-edge-connected then there exist 2 edges disjoint $u−v$ paths in $G$.

Let's suppose that those path are maximal, that is, you can't extend those paths with other edges, without intersecting them (a pair of maximal paths always exist). Suppose also that they do not cover all the edges in $G$.
That means that there exists a node in one of those paths whose edges are not totally covered by the two paths. It's easy to see that the number of such edges uncovered is even.
If you now delete all the edges covered by the two paths, you obtain a graph where all nodes have even degrees, meaning every node is included in an Eulerian cicle in his connected component. Take now a node in a path that belongs to a non trivial cicle, and glue the cicle to your path.
You've obtained a longer path that doesn't intersect the other, against the maximality hypotesis, absurd.

So The theorem is true.

Answer 4

@Raoul , @Diane Vanderwaif https://i.stack.imgur.com/lOVPS.png I think the graph is two edge connected because we can seperate the vertex in the top from the sequare in the bottom by removing two edge (We will get a vertex with degree zero and C4)