Past exam of the year 1 of the computer science course of the University of Oxford / 2021 / Functional programming, design and analysis of algorithms (resit) / 3 / e / Solution

Choose each row from the permutations of the symbols:Every Latin square remains a candidate, because each of its rows is one of those permutations. Conversely every candidate has valid rows, so only its columns still need checking. There are $(n!{)}^n$ candidates instead of $n^{n^2}$: at order four, $331776$ rather than more than four billion. The improvement is substantial but does not make exhaustive generation practical at arbitrary orders. One can replace valid by a column-only test here, since row validity is guaranteed by construction.