I recently asked about the minimum number of black (forbidden) squares needed to guarantee a unique "loop" cycle on an $n \times n$ grid. It seems natural, then, to consider $m \times n$ rectangular grid - which might be more insightful.
What is the minimum number of black (forbidden) squares needed on an $m \times n$ grid of squares to guarantee that there is only one possible closed loop visiting every white square exactly once? Denote this minimum number $\mathcal{M}(m, n)$
More details are available on the linked (original) post. Misha Lavrov has given an upper bound that $\mathcal{M}(4k, 4k) \leq 2k$. I do not know of a linear lower bound. Experimental methods are given in the same post, and I have performed my own calculations (my code is very crude and uses exhaustive search; if I were a better coder, I would probably restrict the search space). Below are the values $\mathcal{M}(m,n)$ I have computed. (Rob Pratt has given more values, which I have added)
\begin{array}{c|ccccccccccccc} m \backslash n & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\ \hline 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \dots \\ 3 & 0 & 1 & 2 & 1 & 2 & 3 & 2 & 3 & 4 & 3 & 4 & \dots & \dots \\ 4 & 0 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & \dots & \dots & \dots & \dots \\ 5 & 0 & 1 & 2 & 3 & 4 & 3& 4 & 5 & 4 & \dots & \dots & \dots & \dots \\ 6 & 0 & 2 & 2 & 4 & 4 & 4 & 4 & 4 & \dots & \dots & \dots & \dots & \dots \\ 7 & 0 & 3 & 2 & 3 & 4 & 5 & 4 & \dots & \dots & \dots & \dots & \dots & \dots \\ 8 & 0 & 2 & 2 & 4 & 4 & 4 & 4 & 4 & \dots & \dots & \dots & \dots & \dots \\ 9 & 0 & 3 & 2 & 5 & 4 & \dots & 4 & \dots & \dots & \dots & \dots & \dots & \dots \\ 10 & 0 & 4 & 2 & 4 & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots \\ 11 & 0 & 3 & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots \\ 12 & 0 & 4 & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots \\ \end{array}
The table is far from complete. Of course, $\mathcal{M}(m,n) = \mathcal{M}(n,m)$ and $\mathcal{M}(2,n) = 0$. For larger values, the problem is more interesting. I have already found patterns that the 'optimal' arrangement of black squares must follow for $\mathcal{M}(3,n)$, namely
$$\mathcal{M}(3,n) = \left\lfloor \frac{n}{3} \right\rfloor + \begin{cases} 1 & \text{if } n \equiv 1 \pmod 3,\\[2mm] 0 & \text{otherwise.} \end{cases}$$
For $\mathcal{M}(4,n)$, I am stumped. The long sequence of $2,2,2,2, \dots$ makes me think $\mathcal{M}(4,n) = 2$ for $n > 2$, but I do not want to mistakenly extrapolate anything. I will continue to review the arrangements for $\mathcal{M}(4,n)$; perhaps I have missed something obvious. After $4$, it becomes increasingly (exponentially...) hard to fill the table.
I am looking for any insight into $\mathcal{M}(m,n)$. This includes a linear lower bound, the calculation of values, etc.
