The \(k\)th border-bishop domination problem (Q2831566)

Given an \(n \times n\) chessboard, the first border consists of squares along the edge of the board. Recursively, the \(k\)th border consists of squares that border the \((k-1)\)th border squares. The \(k\)th border bishop problem asks for the minimum number \(\mathrm{Bor}(B_{n,k})\) of bishops needed so that every square is attacked or occupied by a bishop when the placement is limited only to the \(k\)th border squares.NEWLINENEWLINENEWLINE The following results are established in this paper. When either \(n \equiv 2, 3 \pmod{4}\) or \(k < \lfloor \frac{n}{4} \rfloor + 1\), \(\mathrm{Bor}(B_{n,k}) = 2n - 4k + 2\). When \(k = \lfloor \frac{n}{4} \rfloor + 1\) and \(n \equiv 0, 1 \pmod{4}\), \(\mathrm{Bor}(B_{n,k}) = 2n - 4k + 4\) and \(2n - 4k + 3\), respectively. A dominating set is impossible to find if \(k > \lfloor \frac{n}{4} \rfloor + 1\). If domination is possible, the ways to place the minimum \(k\)th border dominating sets are given as \((n - 4k + 5)(n - 4k + 7) \times 4^{n-3k}\) for odd \(n\) and \((n - 4k + 6)^2 \times 4^{n-3k}\) for even \(n\), unless \(k = \lfloor \frac{n}{4} \rfloor + 1\) and \(n \equiv 0, 1 \pmod{4}\). In this case the numbers are \(2^{\frac{n}{2}-2}\) and \(2^{\frac{n+1}{2}}\), respectively.