# Two definitions of Safe Edge

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.
• Yes, the interview definition is defining bridges, or cut-edges. And yes, all bridges of a graph $G$ must belong to all MSTs of $G$. Jan 11 at 20:46
• 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$is a cut edge} \implies e \in T$. Jan 11 at 21:08