Would appreciate any insight about the following regarding set covers:

Begin with a universe set $X = \{x_1,x_2,...,x_n\}$ and a set $S=\{s_1, s_2,...,s_p\}$ such that each $s_i \subseteq X$ and $\bigcup S = X$. Consider the task of finding all inclusion-minimal covers of $X$. (Covers are inclusion-minimal if the removal of any subset $s_i$ from the cover destroys its coverage property.)

Given the set of inclusion-minimal covers $C=\{c_1, c_2, ... c_m\}$, what is an upper bound for $\max|c_i|$, that is, the largest inclusion-minimal cover? It seems fairly straightforward to show that one upper bound is $|X|$ (that is, an inclusion-minimal cover will contain at most the same number of subsets as there are elements in $X$ itself).

But can a better bound be found?

Additional information: No restrictions are assumed for $S$. The bound would ideally be expressed in terms of one or more properties of $S$ (in whatever form those might take).

A conjecture: If $\min |s_i| = k$, then $\max |c_i|$ (the size of the largest inclusion-minimal cover) is bounded above by $|X| - k + 1$. This is based upon two observations:

(1) In constructing any inclusion-minimal cover, we start with any element from $S$ (call it $s_{1'}$). As $s_{1'}$ covers a particular subset of elements from $X$, the next added cover element (call it $s_{2'}$) must cover at least one new element from $X$ not covered by $s_{1'}$.

(2) If $\min |s_i| = k$, then we can choose the first element $s_{1'}$ in our cover to have size $k$. This leaves $|X| - k$ elements in $X$ left uncovered. By the first observation, each additional $s_i$ must cover at least one new element in $X$, so we can add at most $|X| - k$ additional elements to the cover, leading to a cover size at most $|X| - k + 1$.

As noted in the comments, this type of bound might not be significantly smaller than $|X|$ itself. However, I am hoping to use this in a combinatorial context, so any savings helps. Even smaller bounds would still be useful.

  • 1
    $\begingroup$ What parameters do you want the bound to be in terms of? Certainly it's at most $|X|$ and at most $|S|$. Is there some other parameter of particular interest? $\endgroup$
    – D.W.
    Jan 4, 2020 at 3:21
  • $\begingroup$ both bounds from @D.W. are tight. An example is $S:=\{\{x\};x \in X\}$. $\endgroup$ Jan 4, 2020 at 3:23
  • $\begingroup$ It might be an interesting question to find such a bound when the sets in $S$ have a fixed size. However, in the current formulation you can't say much about the bounds. $\endgroup$ Jan 4, 2020 at 3:28
  • $\begingroup$ @D.W. It would have to based on some property of $S$ (or the individual $s_i$) but I wouldn't know how to be more specific than that at this point. For instance, perhaps it's the case that if we have $|s_i| \geq 2$ for all $i$, (that is, there are no subsets that contain only a single element from $X$) then the bound is actually less than $|X|$. $\endgroup$ Jan 4, 2020 at 5:19
  • $\begingroup$ @narekBojikian The sets in $S$ are not necessarily constrained to have a fixed size, although that is an interesting sub-case. I'd hope that whatever computation for the bound would be based on exactly those sorts of properties about $S$ (for instance, if the size of the sets in $S$ are themselves bounded, does that allow us to say something about the bound of $\max |c_i|$. $\endgroup$ Jan 4, 2020 at 5:24

1 Answer 1


The trivial bounds are that $|c| \le |X|=n$ and $|c| \le |S|=p$.

Knowing bounds on the sizes of the sets $s_i$ doesn't help much. For instance, consider the family of sets $s_1,\dots,s_{n-k+1}$ given by $s_i=\{i,n-k+2,\dots,n-1,n\}$. These sets are all of size $k$, and yet the largest inclusion-minimal cover has size $n-k+1$, i.e., exactly equal to the trivial bound of $|S|$ and fairly close to the trivial bound of $|X|$. So, knowing only the sizes of the sets does not help to provide upper bounds that are much better than the trivial ones.

  • $\begingroup$ With the $|S|$ in this example, it appears that $|S| = n-k+1$, which is equivalent to the value of the trivial bound $|S|=p$....unless I'm reading this wrong? $\endgroup$ Jan 4, 2020 at 20:10
  • $\begingroup$ @RobertRovetti, good point, see my edited answer for clarification. Thanks for helping me improve my answer. $\endgroup$
    – D.W.
    Jan 4, 2020 at 20:55
  • $\begingroup$ @DW, Yes, and thank you for your great example, it has actually led me to formulate a conjecture which I may add to the original question above. $\endgroup$ Jan 5, 2020 at 0:39

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