 Solution

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

Codex  0 2026-10-31

Let

A

denote the original array, used only in the proof. Save its first entry before overwriting it:

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

The loop invariant is saved = A[0],

a [j] = A [j + 1]

for

0 \leq j < i

, and

a [j] = A [j]

for

i \leq j < N

. Thus the next source is untouched, and the assignment extends the completed prefix. At exit, the saved entry supplies the last position. The loop variant is

N - 1 - i

. The result is $a [j] = A [(j + 1) mod N]$ . This also handles

N = 1

, and uses exactly

N

element reads and

N

element writes, counting the saved read and final write, with one extra value of type T.

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