In the classical set cover problem, we are given the set $U$ of elements $\{1, \dots, n\}$ and a collection $C$ of some subsets such that their union is the whole set. Now, I will introduce the first small twist: instead of searching for one cover, we have a target number $t$ and we need to find $t$ disjoint covers. For example, the instance $$ U = \{1,2,3,4\}, C = \{\{1,2,3,4\}, \{1,2,3\}, \{1,3\}, \{2,4\}\}$$ with $t=2$ has a solution ($S_1 = \{1,2,3,4\}, S_2 = \{1,3\}, \{2,4\}$) while it has no solutions for $t=3$. Clearly, even for $t=2$ the problem may be infeasible (e.g. when $C$ contains exactly one set, $U$).
In order to be able to guarantee a solution we must impose further restrictions upon $C$. We will do it in the following manner:
- $C$ can now be multiset
- Every element of $U$ must appear in exactly $k$ subsets of $C$
. Moreover, we will have that $t = n$ so the first question is `
What is the minimum number of subsets an element from a $n$-sized universe must appear in such that the collection of subsets contains $n$ disjoint set covers?
Does this seems related with an existing problem?
Now, I will proceed to my second and main question. Let me define the "almost set cover" $SC_i$ as the set cover of $U\setminus i$. Returning to the 4-element universe of my previous example we have $SC_1 = \{2,3,4\}, SC_2 = \{1,3,4\}$ and so on. So the question now is
What is the minimum number of subsets an element from a $n$-sized universe must appear in such that the collection of subsets can be partitioned into $n$ subcollections with each subcollection containing an almost set cover?
If the answer to my question is denoted by $g(n)$ I have so far that $$ g(2) = 2, g(3) = 3\text{ and }n \le g(n) \le n^2 - 3n + 3$$ At this point, I should mention that I ask the same question here under a different name and formulation but upon realizing the set cover connection I thought this might be a more appropriate place. I think that the proofs can be easily parsed despite the slight change in notation so I will not repeat them here right now but I will gladly do if asked.
