 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2016/imperative-programming/4/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 / 4 / b / Solution by

Codex  0 2026-10-31

Exchange symmetric pairs, stopping before the middle:

PROCEDURE reverse(i, n: INTEGER);
VAR t: INTEGER;
BEGIN
  t := 0;
  WHILE t < n DIV 2 DO
    swap(i+t, i+n-1-t);
    t := t + 1
  END
END reverse;

Writing

m_{0}

for the pre-call array, the loop invariant says that the first and last t positions of the segment already satisfy

m [i + s] = m_{0} [i + n - 1 - s]

, all positions between them retain their original values, and positions outside the segment are unchanged. Each swap fixes the next symmetric pair. The loop variant

⌊ n /2 ⌋ - t

decreases; the two indices are distinct and in bounds whenever the body executes. Exactly $⌊ n /2 ⌋$ swaps reverse the segment, with the odd-length middle element left in place and no accesses for

n \leq 1

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