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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/2/a/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) / 2 / a / Solution by

Codex  0 2026-10-31

Separate integer partitions according to whether they contain a part equal to

i

. Removing one such part is a bijection to partitions of

n - i

whose parts are at most

i

. The alternatives are disjoint, giving

p (i, n) = p (i - 1, n) + p (i, n - i) (i \geq 1, n \geq 1) .

Use

p (i, 0) = 1

, because the empty integer partition is counted once, and

p (0, n) = 0

for

n > 0

, with

p (i, n) = 0

for

n < 0

. This Oberon implementation avoids negative recursive arguments:

PROCEDURE P(i, n: INTEGER): INTEGER;
VAR answer: INTEGER;
BEGIN
  IF n = 0 THEN answer := 1
  ELSIF i = 0 THEN answer := 0
  ELSIF i > n THEN answer := P(i-1, n)
  ELSE answer := P(i-1, n) + P(i, n-i)
  END;
  RETURN answer
END P;

PROCEDURE Partitions(n: INTEGER): INTEGER;
BEGIN
  RETURN P(n, n)
END Partitions;

The precondition is

i, n \geq 0

. Each recursive call reduces

i + n

, proving algorithm termination. Every positive part of a partition of

n

is at most

n

, so the required one-argument count is $P (n, n)$ . Counts require a sufficiently wide integer representation.

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