Partial \(n\)-solutions to the modular \(n\)-queen problem
Let \(Z_ n\) be the set of residue classes modulo \(n\) and let \(r,s,\dots\) be its elements. If the set \(S=\{(R_ i,S_ i)| i=1,2,\dots,m\}\subset Z_ n\times Z_ n\) \((0<m\leq n)\) satisfies the conditions that \(R_ i\neq R_ j\), \(S_ i\neq S_ j\), \(S_ i+R_ i\neq S_ j+R_ j\) and \(S_ i-R_ i\neq S_ j-R_ j\) when \(i\neq j\), then we say that \(S\) is a partial \(n\)-solution with \(m\) elements. A partial \(n\)-solution with one element exists for all \(n\). Let \(M(n)\) be the maximal \(m\) such that there exists a partial \(n\)-solution with \(m\) elements. Then \(M(n)\leq n\). The modular \(n\)-queens are those pieces on the \(n\times n\) chessboard, which can move freely along ranks, or files, or diagonal lines (either broken or unbroken ones). If we regard \(Z_ n\times Z_ n\) as an \(n\times n\) chessboard, then \(M(n)\) is the maximal number of modular \(n\)-queens in such a way that none of them can attack any other queen. The known result about the values of \(M(n)\) are as follows. Theorem A: \(M(n)=n\) if and only if \(n\equiv 1,5\pmod 6\). Theorem B: When \(n\equiv 3\pmod 6\), \(M(n)=n-2\). When \(n\equiv 2,10\pmod{12}\), \(M(n)=n-1\). In this note we give results about the exact values or boundaries of \(M(n)\) for the other cases.
scientific article; zbMATH DE number 148458