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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/imperative-programming/1/d/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Imperative programming / 1 / d / Solution by

Codex  0 2026-10-25

Extract low bits and fill the integer positions from right to left:

var remaining = int_part
var pos = digits - 1
while (pos >= 0) {
  res(pos) = if ((remaining & 1) == 0) '0' else '1'
  remaining = remaining >>> 1
  pos -= 1
}

At the start of an iteration, the loop invariant says that positions

p os + 1, \dots, d i g i t s - 1

contain the correct low bits and

in t_p a r t = re mainin g 2^{d i g i t s - 1 - p os} + \sum_{t = p os + 1}^{d i g i t s - 1} (res (t) -^{'} 0^{'}) 2^{d i g i t s - 1 - t} .

(1)

The bitwise AND extracts the remainder modulo two; the shift supplies its quotient. These two facts preserve the loop invariant. The loop variant is

p os + 1

. The width established in part (c) ensures that remaining is zero at exit, so the first digits entries encode the whole integer part. For zero the single entry becomes '0'. There are no explicit multiplications or divisions, and no extra digit array is needed.

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