The linear system for Sudoku and a fractional completion threshold (Q6634436)

Let \(P\) be a partial Latin square with symbols in the set \([n]:=\{1,\ldots,n\}\). This is \(\epsilon\)-dense if no row, column, or symbol is used more than \(\epsilon n\) times in the entries of \(P\). Further, if \(f_P:[n]^3\rightarrow \{0,1\}\) is defined so that, for any \(i,j,k\in [n]\), the value \(f_P(i,j,k)\) is the number of times that symbol \(k\) appears in the cell \((i,j)\) of \(P\), then a fractional completion of \(P\) is a function \(f:[n]^3\rightarrow [0,1]\) such that, for any \(i,j,k\in [n]\),\N\begin{itemize}\N\item \(f_P(i,j,k)=1\) implies that \(f(i,j,k)=1\); and\N\item \(\sum_{i\in [n]}f(i,j,k)=\sum_{j\in [n]}f(i,j,k)=\sum_{k\in [n]}f(i,j,k)=1\).\N\end{itemize}\NIn case of dealing with a partial Sudoku, it must also be \(\sum_{(i,j)\in b}f(i,j,k)=1\) for every box \(b\) within the Sudoku. Based on spectral decomposition and linear perturbation methods applied to the resulting normal system of linear equations, the authors prove that if \(\epsilon<1/101\), then every \(\epsilon\)-dense partial Sudoku with sufficiently large and sparse rectangular-boxes has a fractional completion.