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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2016/functional-programming-design-and-analysis-of-algorithms/3/d/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2016 / Functional programming, design and analysis of algorithms / 3 / d / Solution by

Codex  0 2026-10-31

Map each current first-position value with the first function, then swap the functions when descending to the other algebraic data type:

mapZig :: (a -> c) -> (b -> d) -> Zig a b -> Zig c d
mapZig _ _ Nil = Nil
mapZig f g (Cins x zs) = Cins (f x) (mapZag g f zs)

mapZag :: (a -> c) -> (b -> d) -> Zag a b -> Zag c d
mapZag _ _ Nal = Nal
mapZag f g (Cans x zs) = Cans (f x) (mapZig g f zs)

All four type variables are independent: mapping may change both element types. There is no Enum or numeric constraint on the mapping functions themselves; constraints arise only from particular arguments such as succ or (+2). The mutual recursion preserves every constructor and applies the intended function at each alternating position. The results are Cins 3 (Cans 'B' (Cins 4 Nal)) and Cans "D" (Cins True (Cans "E" Nil)), respectively.

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