How can it be proved that any minimum edge cover is a forest? As far as I understand an edge cover could be one connected component therefore a spanning tree). If it's the case then it wouldn't be a forest.


A forest is a graph in which every component is a tree. If there is only one component, and that component is a tree, the graph is a forest.

So why is every minimum edge cover a forest? Suppose you have a graph $G=(V,E)$ and an edge cover $C\subseteq E$. Suppose some component of the graph $(V,C)$ contains a cycle, say $x_1x_2\dots x_kx_1$. Then $C\setminus\{x_1x_2\}$ is a smaller edge cover than $C$: $x_1$ is covered by the edge $x_kx_1$, $x_2$ is covered by $x_2x_3$ and every other vertex in $V$ is still covered by whatever covered it in $C$. Therefore, every minimum edge cover must contain no cycles, and an undirected graph that contains no cycles is a forest.

Indeed, we can go even farther. If an edge cover $C$ contains a path $x_1x_2x_3x_4$ then $C\setminus\{x_2x_3\}$ is a smaller edge cover, by essentially the same reasoning as above. Therefore any minimal edge cover is a forest that contains no path of three or more edges. So, not only is every minimum edge cover a forest, but it's a forest where every component is a star (a complete bipartite graph $K_{1,k}$ for some $k\geq 1$; note that $K_{1,1}$ is an edge and $K_{1,2}$ is a two-edge path).

  • $\begingroup$ but according to that logic a spanning tree is a also a forest. then what is the difference between spanning tree and edge cover? $\endgroup$ – Yos Jan 25 '18 at 13:41
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    $\begingroup$ A spanning tree is a tree so it is connected by definition. An edge cover doesn't have to be connected, and a non-minimal edge cover doesn't even have to be a forest. $\endgroup$ – David Richerby Jan 25 '18 at 13:54
  • $\begingroup$ it seems like the proof is almost trivial in that case: either the edge cover is one connected component then as you said it's made of one tree so it's forest; or it's made up of multiple tree then it's also a forest $\endgroup$ – Yos Jan 25 '18 at 13:57
  • $\begingroup$ The point is that an edge cover doesn't have to be a forest at all. The nontrivial part is proving that every component of a minimal cover is a tree. But you proposed "It could be a tree" as a counterexample to the proposition that it's always a forest. I've explained why that's not a counterexample. $\endgroup$ – David Richerby Jan 25 '18 at 15:16
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    $\begingroup$ @Yos Because if it wasn't a tree, it wouldn't be minimal. Edited. $\endgroup$ – David Richerby Jan 25 '18 at 18:49

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