In their proof of the sparsification lemma, Impagliazzo et. al describe the following operator $\pi$:

For a familiy of sets $\mathcal{F}$, let $\pi(\mathcal{F}) \subseteq \mathcal{F}$ be the family of all sets $S$ in $\mathcal{F}$, such that no other set $S'$ in $\mathcal{F}$ is a proper subset of $S$.

I am trying to understand what this operator does. So far, I thought that it could be calculated by the following algorithm:

$$ \text{While}~\exists S_i, S_j \in \mathcal{F}: S_i \subset S_j \text{do} \\ \mathcal{F} \leftarrow \mathcal{F} \setminus S_i$$

By the rest of the proof seems to indicate that actually, one has to remove the superset: $$ \text{While}~\exists S_i, S_j \in \mathcal{F}: S_i \subset S_j \text{do} \\ \mathcal{F} \leftarrow \mathcal{F} \setminus S_j$$

Which leads me to believe that I did not fully understand the description of the operator. What would be an algorithm to calculate that $\pi(F)$?

  • 2
    $\begingroup$ Writing it this way is confusing (or wrong). $\mathcal F$ is fixed, you aren't changing that. It would be clearer to say that $\pi(\mathcal F)=\mathcal G$ and you apply your algorithm to $\mathcal G$ with $\mathcal G$ initialized to $\mathcal F$. At that point the issue Tassle brings up becomes clear as we're then asking that no set in $\mathcal F$ is contained by a set in $\mathcal G$. Removing $S_i$ from $\mathcal G$ doesn't remove it from $\mathcal F$ so doesn't help at all. $\endgroup$ Commented Aug 24, 2019 at 23:58

1 Answer 1


From the definition you gave, the supersets have to be removed, as you want to keep only those sets for which there is no proper subset in $\mathcal{F}$.

For example:

enter image description here

In blue are the sets of $\mathcal{F}$ which don't have proper subsets in $\mathcal{F}$, in red are the ones who do. You want to keep the blue ones, by definition of $\pi$.


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