Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 324 / 2 / b / ii

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- Solution ii

Solution

Applying the two modular-addition quantum oracles consecutively adds

h (x) + r (x) = 0 (mod M)

to the answer register. Thus

U_{r} = U_{h}^{- 1} .

Let

S_{M} ∣ y ⟩ = ∣ - y (mod M)⟩

. Applying

I \otimes S_{M}

, then

U_{h}

, then

I \otimes S_{M}

gives

∣ x, y ⟩ ⟼ ∣ x, - y ⟩ ⟼ ∣ x, - y + h (x)⟩ ⟼ ∣ x, y - h (x)⟩ .

Therefore

U_{h}^{- 1} = (I \otimes S_{M}) U_{h} (I \otimes S_{M}),

using one query and two unitary operators independent of

h

. This is modular-oracle inversion by negation.

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