Selecting Columns with Minimum Overlaps

Question

Given a matrix $A=[a_{ij}]$ of positive number $0\leq a_{ij}\leq1$ that has $m$ rows and $n$ columns. I would like to select, for each row $i$, a set of columns $S_i$ such that $$\sum_{j\in S_i} a_{ij}\geq1,$$ and the number of overlaps $\Delta$ is minimum. The number of overlaps $\Delta$ is calculated as follows: $$\Delta=\sum_{1\leq p<q\leq m}\Delta(S_p,S_q),$$ where, for any pair of columns $(S_p,S_q)$, we have $S_p\cap S_q\ne\emptyset\iff\Delta(S_p,S_q)=1$.

I would like to find an algorithm that minimizes the number of overlaps.

I tried to sort $a_{ij}$ for each $i$ such that $a_{i1}\geq a_{i2}\geq \cdots\geq a_{in}$, then make $S_i=\{1\}$ and verify the sum to be larger than $1$, if not make $S_i=\{1,2\}$, and so on. But, I found a counter-example for this to be non-optimal.

Answer 1

I don't know if there is a polynomial-time algorithm, but if your matrix is not too large, you could try approaching this with integer linear programming. Define zero-or-one variables $x_{ij}$, with the intended meaning that $x_{ij}=1$ means $j \in S_i$. Then you have the constraints

$$\sum_{j} a_{ij} x_{ij} \ge 1.$$

Also, introduce zero-or-one variables $o_{pq}$, with the intended meaning that $o_{pq}=\Delta(S_p,S_q)$, and enforce that

$$o_{pq} = (x_{p1} \land x_{q1}) \lor \dots \lor (x_{pn} \land x_{qn})$$

using the techniques in Express boolean logic operations in zero-one integer linear programming (ILP). (In other words, you introduce fresh temporary zero-or-one variables $o'_{pqj}$ and add the constraints $o'_{pqj} \ge x_{pj} + x_{qj}-1$, $o'_{pqj} \le x_{pj}$, $o'_{pqj} \le x_{qj}$, $o_{pq} \le o'_{pq1} + \dots + o'_{pqn}$, $o_{pq} \ge o'_{pqj}$.)

Finally, minimize $\sum_{p<q} o_{pq}$, subject to the constraints above. This is a set of linear inequalities over integer variables, so you can feed it to an off-the-shelf ILP solver and hope it finds you a good solution. This algorithm will have exponential running time in the worst case, but ILP solvers incorporate many heuristics, so if the matrix is not too large, it's possible this might give you good results in a reasonable amount of time.