 Solution

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

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

Codex  0 2026-10-31

The cyclic strict comparisons in the hint identify a head that cannot be the answer:

PROCEDURE common3(a, N, b, M, c, L): INTEGER;
VAR i, j, k: INTEGER;
BEGIN
  i := 0; j := 0; k := 0;
  WHILE (a[i] # b[j]) OR (b[j] # c[k]) DO
    IF a[i] < b[j] THEN i := i + 1
    ELSIF b[j] < c[k] THEN j := j + 1
    ELSE k := k + 1
    END
  END;
  RETURN a[i]
END common3;

Let

z = a [p] = b [q] = c [r]

be the smallest common value. The loop invariant is

0 \leq i \leq p < N

0 \leq j \leq q < M

0 \leq k \leq r < L

, and every discarded entry is not common to all three arrays. If

a [i] < b [j]

, the entry

a [i]

cannot occur in the unprocessed part of

b

, so discard it. The branch

b [j] < c [k]

has the analogous justification. If neither comparison holds but the heads are not all equal, the hint guarantees

c [k] < a [i]

, so discard

c [k]

A pointer at its target position cannot be advanced: for example

i = p

and

a [i] < b [j]

would imply

z < b [j] \leq z

. Thus all reads stay within array bounds. The loop variant

(p - i) + (q - j) + (r - k)

decreases. At exit all heads equal

z

. The cost is $O (N + M + L)$ time and $O (1)$ auxiliary space.

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