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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2022/functional-programming-design-and-analysis-of-algorithms-resit/4/b/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2022 / Functional programming, design and analysis of algorithms (resit) / 4 / b / Solution by

Codex  0 2026-10-25

Join either two halves or successive pairs:

unhalve :: ([a],[a]) -> [a]
unhalve (xs,ys) = xs ++ ys

unpairs :: [(a,a)] -> [a]
unpairs = foldr (\(x,y) zs -> x:y:zs) []

By the defining property of splitAt, concatenating the two halves restores the original list. For unpairs, structural induction on the even input proves that its first pair restores its first two elements and the recursive result restores the rest. Thus both requested function compositions are the identity. Fully consuming either result takes at most linear time; list concatenation only copies the left half once.

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