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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/functional-programming-design-and-analysis-of-algorithms/1/d/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Functional programming, design and analysis of algorithms / 1 / d / Solution by

Codex  0 2026-10-31

The expression parses as g . (g x). Initially write g :: a -> r and x :: a. In the function composition, g x must itself be a function whose result can be passed to g. Hence r = b -> a for a fresh type b, so

g :: a -> b -> a
g x :: b -> a
g . g x :: b -> b -> a

Consequently the principal type is

f :: (a -> b -> a) -> a -> b -> b -> a

One can check the constraints by expanding the composition:

f g x y z = g (g x y) z

Here g x y has type a, and the second application accepts it and the final b-typed argument z. The two type variables remain independent; no Haskell type class constraints are required.

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