 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-3/10d/b/ii/solution

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

f (b_{1} b_{2}) = b_{1} \oplus b_{2}

, then

(- 1)^{f (b_{1} b_{2})} = (- 1)^{b_{1} + b_{2}},

∣ f ⟩ = \frac{∣0 ⟩ - ∣1 ⟩}{2} \otimes \frac{∣0 ⟩ - ∣1 ⟩}{2} = H^{\otimes 2} ∣11 ⟩ .

In either promised case, construct

∣ f ⟩

by the one-query procedure in part (a), apply the Walsh-Hadamard transform

H^{\otimes 2}

00

for case (i) and certainly

11

for case (ii). Therefore

00 ⟹ constant, 11 ⟹ f (b_{1} b_{2}) = b_{1} \oplus b_{2} .

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