 Solution

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

Codex  0 2026-10-31

Save the last entry and shift in decreasing index order, so no source is overwritten before being used:

PROCEDURE RightOne;
VAR saved: T; i: INTEGER;
BEGIN
  saved := a[N-1]; i := N-1;
  WHILE i > 0 DO
    a[i] := a[i-1]; i := i-1
  END;
  a[0] := saved
END RightOne;

The loop invariant is saved = A[N-1],

a [j] = A [j]

for

0 \leq j \leq i

, and

a [j] = A [j - 1]

for

i < j < N

. The decreasing loop variant is

i

. At exit the untouched saved value fills index zero. The result is $a [j] = A [(j - 1) mod N]$ , again with

N

reads,

N

writes, and one extra value of type T. Applying this after part (b) restores every original entry.

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