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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2023/functional-programming-design-and-analysis-of-algorithms/4/g/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2023 / Functional programming, design and analysis of algorithms / 4 / g / Solution by

Codex  0 2026-10-25

A tree fold can compute insertion while retaining the unchanged children by carrying a pair. The following fold-derived implementation preserves the lazy evaluation behaviour of part (f):

insertViaFold :: Ord a => a -> Tree a -> Tree a
insertViaFold q t = snd (foldTree node base t)
  where
    base = (Empty, Tree Empty q Empty)
    node ~(l,li) x ~(r,ri) = (original,updated)
      where
        original = Tree l x r
        updated = case compare q x of
          LT -> Tree li x r
          EQ -> original
          GT -> Tree l x ri

The invariant of the folded pair is (original subtree, subtree with q inserted). The empty case establishes it. At a node, the first components reconstruct the original tree, and the second components supply whichever child insertion is needed. The irrefutable pair patterns ~(l,li) and ~(r,ri) avoid demanding an unused child's fold. Induction on finite approximations therefore proves equivalence even for partial trees, and only the selected search route needs insertion work.

There is a distinction if “written as a fold” is required to mean a direct same-result-type expression insert q = foldTree node base, with no richer result and final projection. On finite fully defined binary search trees, such an expression is possible by undoing insertion of a new extreme key in the child that should remain unchanged. For completeness, a literal direct fold on that finite, fully defined domain is:

eraseFresh :: Ord a => a -> Tree a -> Tree a
eraseFresh _ Empty = Empty
eraseFresh q (Tree l x r) = case compare q x of
  LT -> Tree (eraseFresh q l) x r
  EQ -> Empty
  GT -> Tree l x (eraseFresh q r)

insertDirectFold :: Ord a => a -> Tree a -> Tree a
insertDirectFold q = foldTree node (Tree Empty q Empty)
  where
    node left x right = case compare q x of
      LT -> Tree left x (eraseFresh q right)
      EQ -> Tree (eraseFresh q left) x (eraseFresh q right)
      GT -> Tree (eraseFresh q left) x right

Here eraseFresh is used only to undo a newly inserted extreme key, which is a leaf; it is not a general deletion operation. The binary search tree bound at each root guarantees exactly that precondition on every child being restored.

On all partial Haskell trees, however, a direct expression with result type Tree a cannot implement part (f).

For that obstruction, fix q = 0, and consider the two right subtrees

r1 = Tree Empty undefined Empty
r2 = Tree Empty undefined (Tree Empty 3 Empty)

Both can be completed to valid right subtrees of a root labelled 1. Yet insert 0 r1 and insert 0 r2 are both bottom, because insertion must compare 0 with their undefined root label. A direct fold equal to insertion would therefore receive the same folded left value, root 1, and bottom folded right value on both larger trees Tree Empty 1 r1 and Tree Empty 1 r2. It must return the same result. Part (f) instead preserves r1 or r2, which are distinguishable by inspecting their right-child constructors without inspecting the undefined labels. This proves the direct-fold obstruction. The pair-valued implementation avoids losing that information and states precisely which sense of fold implementation is being used.

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