I'd like to show that the language DISCONNECTED-SET, defined by $$ \{\langle G, k\rangle | \text{$G$ is an undirected graph that contains a disconnected set of size $k$}\},$$ is NP-complete. (A disconnected set $S$ is defined as a set of vertices such that for every pair of vertices $u,v \in G$, if there is no edge between $u$ and $v$, then at least one of the vertices is in $S$.)

I'm under the assumption that it could be done by reduction from CLIQUE. I've made the following observations (which I'm unable to utilize, though):

  • If we have a graph $G$, then the maximum disconnected set of its complement $G'$ has a size equal to the number of nodes in $G$ that belong to connected components of size at least 2 in $G$.
  • If we have isolated nodes in $G$, the sizes of the disconnected sets in $G'$ are unchanged.

May I just ask for a tiny tip that would get me going in the right direction, not a complete proof. Thank you.


1 Answer 1


This is just VERTEX-COVER in disguise. The reduction from CLIQUE is also simple – a set is disconnected iff its complement is a clique.

  • $\begingroup$ I'm sorry, I don't see how the last statement is true. Might it be that you are confusing the language INDEPENDENT-SET to the one under discussion, the language DISCONNECTED-SET (see the definition in my original post). $\endgroup$
    – stensootla
    Dec 17, 2016 at 22:38
  • $\begingroup$ Nevertheless, it is true, just as the complement of a vertex cover is an independent set and vice versa. $\endgroup$ Dec 17, 2016 at 22:41
  • $\begingroup$ Your problem is the same as VERTEX-COVER (exercise). Look up any proof of the NP-completeness of VERTEX-COVER. $\endgroup$ Dec 17, 2016 at 23:00
  • $\begingroup$ Oh, my bad, at my first reading, I assumed you were talking about the graphs G and G' that contain the disconnected sets and cliques. In any case, I was able to show a reduction from VERTEX-COVER to my problem, as you suggested. Thanks. $\endgroup$
    – stensootla
    Dec 17, 2016 at 23:28

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