 Solution

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

Codex  0 2026-10-31

Let

a = n - k

and

b = k

be the lengths of the left and right blocks. Exchange the smaller block with the equally long block at the opposite end.

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

a < b

, split

B = B_{1} B_{2}

with

∣ B_{2} ∣ = a

. The first block swap sends

A B_{1} B_{2}

B_{2} B_{1} A

;

A

is now in its final suffix. Rotate the remaining prefix of length

b

right by

b - a

to send

B_{2} B_{1}

B_{1} B_{2}

. If

a > b

, split

A = A_{1} A_{2}

with

∣ A_{1} ∣ = b

. The swap sends

A_{1} A_{2} B

B A_{2} A_{1}

; the fixed prefix is

B

, and rotating the remaining length-

a

suffix right by

b

restores

A_{1} A_{2}

. In both cases the exchanged segments are disjoint.

The remaining length strictly decreases, so induction on

n

proves algorithm termination and correctness. This implements the block-exchange rotation. It uses only simple explicit variables, but its recursive call stack can have

Θ (n)

depth when one block has length

1

; the iterative version removes that extra storage.

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