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Problem

I am considering the following maximization problem:

  • Input is a finite collection of finite sets $\mathcal{F} = \{ X_1, X_2, \ldots, X_n \}$.
  • Goal is to find a subset $G \subseteq \mathcal{F}$ that maximizes $|G| \times |\bigcap G|$ where
    • $|G|$ is the cardinality of the set $G$, and
    • $\bigcap G = \bigcap \{X_{i_1}, X_{i_2}, \ldots, X_{i_m} \} = X_{i_1} \cap X_{i_2} \cap \cdots \cap X_{i_m}$.

As an example, for the collection $$ \mathcal{F} = \{ \{a, b, c\}, \{a, b, c, x\}, \{b, c, y\}, \{a, b, c, z\} \}, $$ the maximizing subset is $G = \{ \{a, b, c\}, \{a, b, c, x\}, \{a, b, c, z\} \}$ and the score is $3 \times |\{a, b, c\}| = 9$.

Note: the score of $\mathcal{F}$ itself is $4 \times |\{b, c\}| = 8$.

Question

I am planning to use a procedure of this problem for compressing data (represented by finite collections of finite sets). However, I don't have any good idea to solve this problem efficiently. As yow know, we can solve this by enumerating all the collections of $\mathcal{F}$; but, it's too slow for practical use.

Is there a polynomial-time or some kind of efficient algorithm for this problem? Or, does this problem belong to the complexity class that cannot be solved in polynomial time?

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  • $\begingroup$ I think I understand why $\bigcap G = \bigcap \{X_{i_1}, X_{i_2}, \ldots, X_{i_m} \} = X_{i_1} \cap X_{i_2} \cap \cdots \cap X_{i_m}$ is of advantage (there may be one $ \cap G$ missing in both explications). I fail to see the value of maximising $|G|$, let alone multiplying that to the former: please provide the intuition for that goal. $\endgroup$
    – greybeard
    Commented Jul 25, 2020 at 8:24
  • $\begingroup$ @greybeard Thank you for your interesting. I try to answer your questions. 1. Is maximizing | \bigcap G | useful ? This problem equals to find one of the largest set $X_i$ from $\mathcal{F}$ and it can be solved in linear time. BTW, I've found this pdf about "Maximum k-Subset Intersection" (MSI) problem. On this, for a given $k$, we find a collection $H \subseteq \mathcal{F}$ with $|H| = k$ that maximizes $| \bigcap H |$. This is also interesting!! The paper shows MSI and Maximum-edge biclique problem are closely related. $\endgroup$
    – yuezato
    Commented Jul 26, 2020 at 16:07
  • $\begingroup$ 2. How do we use this problem? Let me explain my original motivation, compactifying finite collections of finite sets, by reusing the above instance. On the example, the collection $\mathcal{F}$ can be rephrased as follows by using $\bigcap G = \{ a, b, c \}$: $$ \mathcal{F} \approx \langle \mathcal{G} = \{a, b, c\} \&\& \{ \mathcal{G}, \mathcal{G} \cup x, \{ b, c, y \}, \mathcal{G} \cup z \} \rangle. $$ Using $\bigcap G$, we can obtain the shorter representation. To this end, I need the above problem rather than MSI. I hope these will help. $\endgroup$
    – yuezato
    Commented Jul 26, 2020 at 16:09

1 Answer 1

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This problem is NP-complete. Let's reformulate it first: we have a bipartite graph, where

  • The left side corresponds to elements
  • The right side corresponds to sets
  • The edge $(u,v)$ means that $u \in v$.

Our goal is to find the bipartite clique with the maximum number of edges. As stated in Rene Peeters, "The maximum edge biclique problem is NP-complete", the decision problem is NP-complete.

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    $\begingroup$ Thanks so much for your great answer! I'd have explored a useful variant of the (biparate) maximum matching, but I couldn't. Your answer, "the maximum edge biclique problem", quite naturally encodes my problem. Reading the mentioned and some related papers, I've also understood to develop an approximating algorithm may be hard. $\endgroup$
    – yuezato
    Commented Jul 21, 2020 at 21:55

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