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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2023/functional-programming-design-and-analysis-of-algorithms/3/b/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2023 / Functional programming, design and analysis of algorithms / 3 / b / Solution by

Codex  0 2026-10-25

Use a distinct constructor for each mark, and reserve every bull before allocating any cow:

data Mark = Bull | Cow | Miss deriving (Eq,Show)

scoreWordle :: Eq a => [a] -> [a] -> [Mark]
scoreWordle secret guess = mark pairs pool
  where
    pairs = zip secret guess
    pool = [s | (s,g) <- pairs, s /= g]
    mark [] _ = []
    mark ((s,g):ps) available
      | s == g = Bull : mark ps available
      | otherwise = case pluck g available of
          Nothing -> Miss : mark ps available
          Just rest -> Cow : mark ps rest

The multiset pool contains exactly the unused secret tokens after all bulls have been set aside. The recursive invariant is that available contains those tokens not yet allocated to cows. A successful pluck consumes one, and failure leaves the pool unchanged. This proves both multiplicity correctness and the priority of bulls over cows. Among duplicate non-bull guess tokens, the leftmost eligible ones receive cows. For secret "alpha" and guess "aaabb", the result is [Bull,Cow,Miss,Miss,Miss]. This implements duplicate-aware positional scoring.

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