Ryser–Brualdi–Stein conjecture for Latin squares

Let $L$ be a Latin square of order $n$. A partial transversal is a set of cells containing no two cells in the same row, column, or with the same symbol, and a full transversal has $n$ cells. Ryser–Brualdi–Stein conjecture. Every Latin square of order $n$ has a partial transversal with at least $n-1$ cells, and has a full transversal when $n$ is odd. The conjecture combines the partial-transversal and odd-order full-transversal assertions and remains open in general.