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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2017/imperative-programming-resit/1/b/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2017 / Imperative programming (resit) / 1 / b / Solution by

Codex  0 2026-10-31

Divide numerator and denominator by their positive greatest common divisor, then fix the sign:

PROCEDURE reduce(x: rational);
VAR g: INTEGER;
BEGIN
  ASSERT(x.den # 0);
  g := gcd(ABS(x.num),ABS(x.den));
  x.num := x.num DIV g; x.den := x.den DIV g;
  IF x.den < 0 THEN
    x.num := -x.num; x.den := -x.den
  END
END reduce;

The denominator condition ensures

g > 0

, including when the numerator is zero. Dividing both components by the same nonzero integer preserves the fraction, and the final simultaneous sign change preserves it again. The resulting numerator and denominator are coprime; for zero the denominator becomes one.

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