I ran into an interview two days ago and came across one strange definition of safe edge.

We are given an undirected weighted Graph $G = (V,E)$ with all distinct edge weights. Assume that the graph is connected.

Safe Edge Definition: if an edge $e \in E$ is not contained in any cycle, we called it a safe edge

Following is a theorem based on the above definition.

Theorem: All the safe edges must be in Minimum Spanning Tree (MST).

Please compare the definition of a "safe edge" on this interview with the classical definition of safe edge at all courses like here: MAIN SAFE EDGE THEOREM.

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    – D.W.
    Jan 11 at 22:35

They are two different definitions. The interview definition calls a safe edge one that is not part of any cycle and therefore cannot be removed from $G$ without disconnecting it, thus changing the resulting MST. Notice that this definition depends solely on the chosen edge and on $G$. Notice also that it does not depend on the edge weights in any way.

The definition in the notes defines a safe edge $e$ as one that can safely be added to a subgraph $A$ of $G$ so that $A + e$ is a partial MST of $G$. In this case $A$ is probably thought to already be a partial MST of $G$, since otherwise no edge would be safe, although this is not formally required. Notice that this definition depends on the particular choice of $A$ and on the edge weights of $G$.

As an example, let $G$ be a cycle on $n$ nodes, where the $n$ edges have weights $1, \dots, n$. None of these edges is safe according to the interview definition. If $A$ is the empty graph then all edges except for the one of weight $n$ are safe w.r.t. the definition in the notes. If $A$ is the subgraph containing the edge of weight $n$ then no edge is safe w.r.t. the definition in the notes.

  • $\begingroup$ Yes, the interview definition is defining bridges, or cut-edges. And yes, all bridges of a graph $G$ must belong to all MSTs of $G$. $\endgroup$
    – Steven
    Jan 11 at 20:46
  • $\begingroup$ I noticed that in a previous edit you were asking for a proof. Let $e=(u,v)$ be an edge of $G$. If there is a MST $T$ of $G$ such $e \not\in T$, then let $P$ the (unique) simple path from $u$ to $v$ in $T$. $P+e$ is a cycle in $G$, showing that $e$ cannot be a cut-edge. In other words $e \not\in T \implies \mbox{$e$ is not a cut edge}$. Taking the contrapositive statement: $\mbox{$e$ is a cut edge} \implies e \in T$. $\endgroup$
    – Steven
    Jan 11 at 21:08

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