 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2016/imperative-programming/2/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 / 2 / b / Solution by

Codex  0 2026-10-31

Test each possible positive divisor directly:

PROCEDURE divisors(n: INTEGER): INTEGER;
VAR i, count: INTEGER;
BEGIN
  i := 1; count := 0;
  WHILE i <= n DO
    IF n MOD i = 0 THEN count := count + 1 END;
    i := i + 1
  END;
  RETURN count
END divisors;

For

n \geq 1

, the loop invariant is

1 \leq i \leq n + 1

and count

= ∣ {d : 1 \leq d < i, d ∣ n} ∣

. Testing

i

adds exactly the next possible divisor. The loop variant

n + 1 - i

decreases, and the divisor in MOD is always positive. At exit the interval

1 \leq d \leq n

contains every positive divisor. This computes the divisor function $τ (n)$ using $n$ remainder operations, $Θ (n)$ time, and $O (1)$ auxiliary space.

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