 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2016/imperative-programming/3/b/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2016 / Imperative programming / 3 / b / Solution by

Codex  0 2026-10-31

Use the Euclidean algorithm with floor quotient and nonnegative remainder, including for a negative numerator:

PROCEDURE rat2cf(n, d: INTEGER): fraction;
VAR q, r: INTEGER; p: fraction;
BEGIN
  q := n DIV d; r := n MOD d;
  NEW(p); p.head := q; p.tail := NIL;
  IF r # 0 THEN p.tail := rat2cf(d, r) END;
  RETURN p
END rat2cf;

For

d > 0

, Euclidean division gives

n = q d + r

with

0 \leq r < d

. If

r = 0

, the singleton [q] is exact. Otherwise

\frac{n}{d} = q + \frac{r}{d} = q + \frac{1}{d / r},

(1)

so the recursive tail represents

d / r

. Its numerator is positive and greater than its denominator, so every tail head is a positive integer. The new denominator

r

is strictly smaller than

d

, proving algorithm termination by descent through the positive integers.

The returned list represents exactly $n / d$ . For example

- 7/3 = - 3 + 1/ (1 + 1/2)

gives [-3,1,2]; truncating negative division toward zero would violate the required positivity of the tail. There are

O (lo g d)

Euclidean algorithm steps, with output allocation and recursive stack depth linear in the number of generated nodes.

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