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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2016/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 / 2016 / Imperative programming (resit) / 1 / b / Solution by

Codex  0 2026-10-31

Keep both the exponent and its power. This Oberon procedure uses only the permitted arithmetic:

PROCEDURE IntLog(b, n: INTEGER): INTEGER;
VAR i, p: INTEGER;
BEGIN
  i := 0; p := 1;
  WHILE p*b <= n DO
    p := p*b; i := i+1
  END;
  RETURN i
END IntLog;

The loop invariant is

i \geq 0

and

p = b^{i} \leq n

. It holds initially because

n > 0

. A successful guard permits the next power, and the two assignments preserve the relation between

p

and

i

. At exit,

p \leq n < p b

, which is exactly part (a). Each iteration increases

i

, and only finitely many powers of

b \geq 2

are at most

n

, proving algorithm termination. The returned value is $⌊ lo g_{b} n ⌋$ . Its running time is

Θ (1 + lo g_{b} n)

arithmetic operations; integer overflow is deferred to part (e).

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