 Solution

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/f/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 / f / Solution by

Codex  0 2026-10-31

Let

T (n) = (F_{n}, F_{n + 1})

. The equations derived in part (e) become

T (0) = (0, 1)

and

T (n + 1) = s (T (n))

, where

s (a, b) = (b, a + b)

. Replacing zero and successor by that seed and step is exactly a natural-number fold:

twoFibsNat :: Natural -> (Integer, Integer)
twoFibsNat = foldNat step (0, 1)
  where
    step :: (Integer, Integer) -> (Integer, Integer)
    step (a, b) = (b, a + b)

fibNat :: Natural -> Integer
fibNat = fst . twoFibsNat

Consequently

fibNat = fst \circ foldNat ((a, b) \mapsto (b, a + b)) (0, 1) .

(1)

Structural induction follows the constructors of Natural: the seed gives the first two Fibonacci numbers, and the successor step uses the Fibonacci recurrence. The extra component carries exactly the information needed for the next update.

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