Maximal subsets of $n\times n$ board without cycles

Define $\mathrm{\nabla }(n)$ as the least possible number of cells that must be removed from the $n\times n$ square board in order to leave no cycles. Put $L_n=\lceil ((n-1{)}^2+1)\mathrm{/}3\rceil$. The results of Luccio, Madelaine-Stewart, and Lien-Fu-Shih imply that $\mathrm{\nabla }(n)=L_n$ in case $n\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}6)$ and $n\equiv 4(\mathrm{m}\mathrm{o}\mathrm{d}6)$, while for the remaining $4$ residue classes $\mathrm{\nabla }(n)=L_n$ or $\mathrm{\nabla }(n)=L_n+1$. In this paper, we provide optimal constructions in the odd residue classes $n\equiv 3,5(\mathrm{m}\mathrm{o}\mathrm{d}6)$, (proving $\mathrm{\nabla }(n)=L_n$) and a geometric obstruction showing that equality in lower bound is impossible when $n\equiv 0,2(\mathrm{m}\mathrm{o}\mathrm{d}6)$ (yielding $\mathrm{\nabla }(n)=L_n+1$). Starting with $n=3$, the resulting sequence is thus given by \nabla(n)=\bigg{\lceil}\frac{n^2}{3}-\frac{n}{6}+\frac{4}{3}\bigg{\rceil}-\bigg{\lceil}\frac{n}{2}\bigg{\rceil}.