 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2017/functional-programming-design-and-analysis-of-algorithms/4/c/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2017 / Functional programming, design and analysis of algorithms / 4 / c / Solution by

Codex  0 2026-10-31

Use map k (u:us) = k u : map k us, list-map fusion, and the left fold equation

foldl f e (x:ys) = foldl f (f e x) ys

For the empty list, the prefix definition immediately gives scanl f e [] = [e]. For a nonempty list, calculate:

scanl f e (x:xs)
  = map (foldl f e) ([] : map (x:) (inits xs))
  = e : map ((foldl f e) . (x:)) (inits xs)
  = e : map (foldl f (f e x)) (inits xs)
  = e : scanl f (f e x) xs

The second-to-third line uses the displayed fold equation as an equality of functions of ys. Hence the efficient left scan is:

scanl :: (b -> a -> b) -> b -> [a] -> [b]
scanl f e [] = [e]
scanl f e (x:xs) = e : scanl f (f e x) xs

On a finite input of length

n

, fully demanding the original prefix-fold results performs

0 + 1 + \dots + n = n (n + 1) /2

accumulator updates. The recurrence performs

n

. Thus the update count improves from $Θ (n^{2})$ to $Θ (n)$ , assuming constant-cost updates. Both return

n + 1

results. This does not assert constant space for arbitrary accumulators or imply that lazy accumulators are evaluated strictly.

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