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

Codex  0 2026-10-31

Use the first binary relation for the outer generator, the second for the inner generator, and match their middle values. Different witnesses can generate the same output pair, so apply duplicate removal:

relcomp :: Eq a => [(a, a)] -> [(a, a)] -> [(a, a)]
relcomp r s = removedup [(a, b) | (a, c) <- r,
                                (d, b) <- s, c == d]

The guards give precisely the existential witness required for composition of relations:

(a, b) \in relcomp (R, S) ⟺ \exists c : (a, c) \in R \land (c, b) \in S .

(1)

Every output is sound by its generating witness, and every witness is visited for finite inputs. Duplicate removal converts the resulting enumeration back to the required representation by distinct pairs.

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