Suppose we have collection of n sets $S_1, S_2, \dots, S_n$. Each set has a size of at least $k$. We know for sure that $\exists k$ sets where all of them contain the same $k$ elements; $|S_1 \cap S_2 \cap,\dots,\cap S_k| \geq k$. We don't know the sets nor the intersection elements. I would like to know if these questions are NP-hard or if there is a polynomial algorithm to solve them.

  1. Can we find the $k$ intersection elements in $|S_1 \cap S_2 \cap,\dots,\cap S_k| \geq k$?
  2. Assume we have set $S_t$, we would like to know if set $S_t$ is one of the $k$ sets that have $k$ common elements among them. $S_t \in ?$ $(S_1,S_2,\dots, S_k)$. In other words, does $S_t$ intersect with any $k-1$ sets on the same $k$ elements?
  • $\begingroup$ Question 2 may be related to the well known $k$-clique problem which is NPC. $\endgroup$ Feb 25 at 20:47
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    $\begingroup$ The promise that such an intersection exists makes the problem trivially not $NP$-hard (assuming the promise complexity class), though the functional variant (returning the certificate) probably could still be $FNP$-hard?.. $\endgroup$
    – rus9384
    Feb 25 at 22:06
  • $\begingroup$ Thanks for the reference to FNP; never heard of it before $\endgroup$
    – Xfae
    Feb 26 at 23:14

1 Answer 1


This is at least as hard as the problem of detecting a $k$-clique in a graph, given a promise that there contains at least one $k$-clique in the graph. In particular, given a graph $G$, let $S_i$ be equal to the set of vertices adjacent to vertex $i$, along with $i$.

This problem is NP-hard. If you had a polynomial-time algorithm for your problem, then we could use it to solve the clique problem by running it multiple times, once for each possible value of $k$, and checking which is the largest value of $k$ for which it returns a valid $k$-clique.

See also https://en.wikipedia.org/wiki/Clique_problem and https://en.wikipedia.org/wiki/Planted_clique.


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