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ID: 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

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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 by

Codex  0 2026-10-25

Choose each row from the permutations of the symbols:

squaresRows :: Int -> [[[Int]]]
squaresRows n
  | n < 0     = error "negative order"
  | otherwise = cp (replicate n (perms [1..n]))

latinRows :: Int -> [[[Int]]]
latinRows = filter valid . squaresRows

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.

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