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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/2/c/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 / 2 / c / Solution by

Codex  0 2026-10-31

Use concat [] = [], concat (ys:yss) = ys ++ concat yss, the corresponding equations for map, associativity of list concatenation, and (r:ys) ++ zs = r:(ys ++ zs). For an empty forest,

flatten2 [] xs
= concat (map flatten []) ++ xs
= [] ++ xs
= xs

For a nonempty forest, write its first rose tree as Node r children:

flatten2 (Node r children : trees) xs
= (flatten (Node r children) ++ concat (map flatten trees)) ++ xs
= (r : concat (map flatten children))
    ++ (concat (map flatten trees) ++ xs)
= r : (concat (map flatten children) ++ flatten2 trees xs)
= r : flatten2 children (flatten2 trees xs)

Thus the recursive Haskell definition is

flatten2 :: [RTree a] -> [a] -> [a]
flatten2 [] xs = xs
flatten2 (Node r children : trees) xs =
  r : flatten2 children (flatten2 trees xs)

The nested call supplies the continuation of the preorder traversal as a suffix. Each label is emitted once, without repeated list concatenation of an already flattened subtree. For a finite forest, constructing its entire contribution takes linear time in its number of vertices, without traversing the supplied suffix xs.

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