Minimum Cost Arrangement

Question

I want to find an arrangement to evenly place 12 items ($a_1, a_2, ..., a_{12}$) into 4 boxes ($b_1, b_2, b_3, b_4$) such that the cost is minimal.

Let $b(x)$ be the index of the box that contains item $x$. The cost of an arrangement is

$$\sum_{i=1}^{12} \sum_{j=1}^{12} |b(a_i) - b(a_j)| M_{i,j}$$

where $M$ is a $12\times 12$ constant matrix where all values are non-negatives.

---

For this instance I could use brute force with $\frac{12!}{3!3!3!3!}$ iterations, but I hope to get a generalised solution that can solve for $N$ items and $K$ boxes in a faster way.

Is there a solution faster than brute force? I have tried some operations research library (such as Google OR tools, CVXPY) but am unable to reduce the problem into a readily acceptable representation.

My intuition tells me there is no faster solution than brute force. In that case, is there an approximation algorithm that can provide good enough solution?

Answer 1

You could try expressing this as an integer linear expression: let $v_{x,b}$ be 1 if item $x$ is placed in box $b$, and let $w_{x,x',b,b'}$ be 1 if $v_{x,b}=1$ and $v_{x',b'}=1$; then your objective function is a linear function of these zero-or-one variables. You can also enforce the relationship between the $v$'s and $w$'s, and ensure that the $v$'s represent a valid placement into boxes, using the techniques in Express boolean logic operations in zero-one integer linear programming (ILP). Then, feed the resulting ILP to an off-the-shelf ILP solver. I don't know whether this would be faster than brute force, but it is something you could try.