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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2023/digital-systems-linear-algebra-and-introduction-to-formal-proof/1/e/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2023 / Digital systems, linear algebra and introduction to formal proof / 1 / e / Solution by

Codex  0 2026-10-25

Use the same bias, then an arithmetic right shift by ten:

div1024:
    asrs r3, r0, #31    // -1 when a<0, otherwise 0
    lsrs r3, r3, #22    // b = 1023 when a<0, otherwise 0
    adds r0, r0, r3     // a+b, representable for every signed input
    asrs r0, r0, #10    // floor((a+b)/1024)
    bx lr

An arithmetic right shift of a two's complement integer computes division rounded down. For

a \geq 0

, this already equals truncation towards zero. For

a < 0

, the signed power-of-two division by biasing identity is

trunc (a /1024) = ⌊ \frac{a + 1023}{1024} ⌋ .

(1)

Indeed, writing

- a = 1024 q + r

with

0 \leq r < 1024

makes the adjusted numerator

- 1024 q + (1023 - r)

; its quotient rounded down is

- q

for every

r

, including zero. Thus the code returns the signed quotient in all cases and contains no data-dependent branch. The final bx lr simply returns to the caller.

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