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There is an NP-hard problem called Minimum k-Union where we are given a set system with $n$ sets and are asked to select $k$ sets in order to minimize the size of their union.

I'm currently interested in a very similar problem, but don't know how to convert one to another:

Given a set system with $n$ sets and a bound $k$. Select as many sets as possible while their union is at most $k$.

Is this problem NP-hard? Any hint is welcome!

Updated 2020: I found a paper called "Unbalanced Graph Cuts" by Hayrapetyan et al. [ESA'05] which describes the Minimum-size bounded-capacity cut (MinSBCC) problem which is very similar to what I looked for.

Muoi Tran
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2 Answers2

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Very easy. Clique problem reduces to yours.

Each set is the 2-element edges $\{u,v\}$ of $G$.

A $k$-clique has as many edges as possible while only involves $k$ vertices. So, $K={k \choose 2}$.

Thinh D. Nguyen
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Yes, it is NP-hard, and we will show that by reducing from $2$-independent set.

Problem: $2$-independent set
Input: A graph $G = (V,E)$ and an integer $\ell \leq |V|$
Definition: Is there a set of at least $\ell$ vertices $S \subseteq V$ such that for each pair of vertices $u,v \in S$, the distance $\text{dist}_G(u, v) > 2$?

The reduction is as follows:

For an input graph $G = (V, E)$, and input $\ell$, we create an instance of your problem where the set family is $\mathcal F = \{N_G[v] \mid v \in V\}$, and we set $k=0$.

Now, a solution to your problem is a set of sets $\mathcal S \subseteq \mathcal F$ with an empty pairwise intersection. For each $F \in \mathcal S$, pick out a vertex $v$ in $V$ such that $N_G[v] = F$. This vertex set is a $2$-independent set. The original problem, $G, \ell$ is a yes instance if and only if $|\mathcal S| \geq \ell$.

Ainsley H.
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