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So , Basically assume there is a graph $G$ which has no bridges. Is it always true that there exists two edge disjoint paths between any two vertices in the Graph ?


$\text{My Attempt at the Proof}$:-

Suppose there isn't , then there must be at-least one edge common in both the paths , and if we remove it then it disconnects $G$ and thus is a bridge. As there are no Bridges , Hence Proved !


$\text{Why I feel this Proof is incorrect }$:-

Suppose the exits three paths from u to v.

Path 1 :- $u$ ----> $E_1$ -----> $E_2$ -----> $v$

Path 2 :- $u$ ----> $E_2$ -----> $E_3$ -----> $v$

Path 3 :- $u$ ----> $E_1$ -----> $E_3$ -----> $v$

(here -----> represents some path and $E_1 , E_2 , E_3$ represent Edges.

This is an example which violates the above property. It has no bridges as well as no edge disjoint paths.

Can someone either Prove or disprove this Property ? I am really confused.

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  • $\begingroup$ I'm confused. Paths 1 and 3 share edge $E_1$ so they are not edge-disjoint. And you want to prove the existence of two, not three edge-disjoint paths in your statement. $\endgroup$ – Anthony Labarre Apr 3 at 9:53
  • $\begingroup$ I was just trying to think of a counter example to my proof. My Proof is wrong for sure. I wanted to prove the statement , could you help me with the proof. I want to prove the existence of atleast two edge disjoin paths in $G$ $\endgroup$ – rajdeep dhingra Apr 3 at 11:17
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Yes, it is true that there exist two edge-disjoint paths between any two vertices in a connected bridgeless graph.

Define a closed trail as a trail whose last vertex is its first vertex. Note, a trail contains two edge-disjoint paths between any two of its vertices.

Trail combining lemma. Let $G$ be a graph. Suppose a closed trail passes vertex $v$. Another closed trail passes vertex $u$. If those two trails share one vertex, there is a closed trail that passes both $u$ and $v$.

Proof. You can convince yourself easily by drawing a few situations. Done.


Let us prove the following proposition.

A graph is connect and bridgeless if and only if any two vertices are included in a closed trail.

Proof.

"$\Leftarrow$": Obvious.

"$\Rightarrow$": Suppose $G$ is connected and bridgeless. $v$ and $w$ are two vertices.

Let $$T_v=\{u : u \text{ is a vertex of } G \text{ such that there is closed trail that passes }v\text{ and } u\}.$$ Suppose $u\in T_v$ and $s$ be a neighboring vertex of $u$, i.e., $\{u,s\}$ is an edge.

  • There is a closed trail that passes $v$ and $u$ by the definition of $T_v$.

  • There is a closed trail that passes $u$ and $s$, since edge $\{u,s\}$ is not a bridge.

Thanks to the lemma, we have a closed trail that passes $v$ and $s$. In other words, $T_v$ contains all neighborhoods of all its vertices. That means $T_v$ (with all edges between its points) is a connected component, which must be $G$ itself. In particular, there is a closed trail that passes $v$ and $w$. Proof is complete.


A very similar characterization of bridgeless connected graphs is Robbins' theorem. If $G$ is connected and bridgeless, that theorem says we can choose a direction for each edge of $G$, turning it into a directed graph that has a path from every vertex to every other vertex.

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