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

Codex  0 2026-10-31

Replace the sole recursive call by updates to the current segment's parameters:

PROCEDURE rotate(i, n, k: INTEGER);
VAR a, b: INTEGER;
BEGIN
  WHILE (k > 0) AND (k < n) DO
    a := n-k; b := k;
    IF a = b THEN
      blockswap(i, i+a, a);
      RETURN
    ELSIF a < b THEN
      blockswap(i, i+b, a);
      n := b; k := b-a
    ELSE
      blockswap(i, i+a, b);
      i := i+b; n := a; k := b
    END
  END
END rotate;

The loop invariant is that every position outside the current unresolved segment already has its final value, while rotating the current segment by the current k completes the original array rotation. The block decompositions in part (ii) prove preservation. The positive unresolved length decreases in each continuing branch; the equal-block branch finishes immediately. The same rotation now uses $O (1)$ auxiliary space, including control storage, and leaves the rest of the array untouched.

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