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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/imperative-programming-resit/3/c/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Imperative programming (resit) / 3 / c / Solution by

Codex  0 2026-10-31

Treat the live array segment as a FIFO queue of runs. Remove its first two runs, append their merged result, and repeat:

PROCEDURE Sort(a, b: INTEGER; VAR c, d: INTEGER);
VAR x, y, z: INTEGER;
BEGIN
  c := a; d := b;
  WHILE c < d DO
    x := Run(c, d);
    IF x = d THEN RETURN END;
    y := Run(x, d);
    z := y-c;
    Merge(c, x, y, d);
    c := y;
    d := d+z
  END
END Sort;

Compute both run boundaries before writing the destination, which starts at the old d. The untouched suffix

u [y .. d)

is followed by the new merged run. The endpoints returned in c,d always describe the live sequence; obsolete values before c are ignored. As allowed, assume enough destination capacity.

For example,

[3, 1, 4, 2]

becomes the live sequence

[2, 1, 3, 4]

, then

[1, 2, 3, 4]

. The first merge combines

[3]

and

[1, 4]

, and the second combines

[2]

and

[1, 3, 4]

u [c .. d) contains the sorted input and d - c = b - a .

(1)

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