 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-3/15d/d/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Let

X ∣ x ⟩ = ∣ x + 1 ⟩

on the input register and let

N ∣ y ⟩ = ∣ - y ⟩

on the target register. Both are fixed unitary operators independent of

f

. Conjugating the query by the input shift gives

U_{f_{1}} = (X^{- 1} \otimes I) U_{f} (X \otimes I),

because the intermediate query is made at

x + 1

. Conjugating by target negation gives

U_{f_{2}} = (I \otimes N) U_{f} (I \otimes N),

since

y \mapsto - y \mapsto - y + f (x) \mapsto y - f (x)

. Each construction uses

U_{f}

exactly once.

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