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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/3/e/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 / 3 / e / Solution by

Codex  0 2026-10-31

Use the symmetric tree fold equations for the other algebraic data type:

foldZag (fZag,eZag) (fZig,eZig) Nal = eZag
foldZag (fZag,eZag) (fZig,eZig) (Cans x z) =
  fZag x (foldZig (fZig,eZig) (fZag,eZag) z)

The most general type signatures, using the same schematic variable names independently for each function, are

foldZig :: (a -> s -> r, r) -> (b -> r -> s, s)
        -> Zig a b -> r
foldZag :: (a -> s -> r, r) -> (b -> r -> s, s)
        -> Zag a b -> r

For foldZig, the first empty value has type r, so the first combining function must return r. Its label argument has type a. The recursively folded tail has some independent result type s, giving a -> s -> r. The second combining function consumes a b label and an r result and returns s, giving b -> r -> s. The second empty value has type s. Swapping the two handler pairs and the label types makes the recursive call well-typed. The identical argument applies to foldZag.

The two result types need not coincide. For example, one handler can return an integer and the other a list; forcing both results to have one type would lose parametric polymorphism allowed by the definitions.

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