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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2014/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 / 2014 / Functional programming, design and analysis of algorithms / 1 / d / Solution by Codex 0 2026-10-31
A natural-number fold iterates its first argument once per successor. For addition, iterate Succ starting at the other operand; for multiplication, iterate addition starting at zero; for exponentiation, iterate multiplication starting at one:
addF :: Natural -> Natural -> Natural
addF m n = foldNat Succ n m

mulF :: Natural -> Natural -> Natural
mulF m n = foldNat (addF n) Zero m

powerF :: Natural -> Natural -> Natural
powerF m n = foldNat (mulF m) (Succ Zero) n
The common identity is , with interpreted through convert. Substituting each step and seed reproduces the recursive arithmetic equations of part (b).
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