TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

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Median amplitude-estimation query cost

Statement

In the repeated controlled- $Q$ access mode, the median estimator uses exactly $r (M - 1)$ controlled- $Q$ calls. With the explicit thresholds in Median confidence amplification for amplitude estimation, this is $O (ϵ^{- 1} (1 + \ln (1 / δ)))$ calls. For the conventional range $0 < δ \leq 1 / 2$ , this is $O (ϵ^{- 1} \ln (1 / δ))$ . Controlled $A$ , controlled $A^{*}$ , controlled predicate reflections, state preparation, gate counts, and classical order-statistic work are separately charged.

Facts & Assumptions

Given: The median procedure with its integer parameters $M, r$ and repeated controlled- $Q$ access.

[F1]

The procedure makes $r$ fresh runs with a common $M$ (Median confidence amplification for amplitude estimation).

[F2]

Each run uses exactly $M - 1$ repeated controlled- $Q$ calls (Amplitude-estimation query and precision complexity).

[F3]

The access model charges declared quantum query actions separately from ordinary gates and classical bits (Quantum query complexity).

[F4]

The least power-of-two threshold is $M \geq 88 / (7 ϵ)$ and the least odd sample size satisfies $2^{r} \geq (2 / δ)^{8}$ (Median confidence amplification for amplitude estimation).

[F5]

$\ln$ is strictly increasing, $\ln 2 > 0$ , and $\ln 2 < 1$ (The natural logarithm as the inverse of the exponential function, The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t, The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ ).

[F6]

The natural logarithm is strictly increasing and satisfies the product, reciprocal, and power laws on positive reals (Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm).

Proof

1.1givenF1F2algebra

By [F1] and [F2], each of the $r$ runs makes $M - 1$ calls, so the exact total is $r (M - 1)$ .

2.1givenF4F5F6step 1.1algebra

The least-power rule gives $M < 2 \cdot 88 / (7 ϵ) = 176 / (7 ϵ)$ . Put $X = 2 / δ > 2$ . Since $X^{8} > 256$ , the least positive odd $r$ with $2^{r} \geq X^{8}$ satisfies $r \geq 9$ , so $r - 2$ is a positive odd integer that fails the threshold: $2^{r - 2} < X^{8}$ . By [F6], taking logarithms gives $(r - 2) \ln 2 < 8 \ln X$ , hence $r < 2 + 8 \ln X / \ln 2$ by [F5]. Also $\ln X = \ln 2 + \ln (1 / δ)$ by [F6], so [F5] gives $r = O (1 + \ln (1 / δ))$ . When $δ \leq 1 / 2$ , [F5]–[F6] give $\ln (1 / δ) \geq \ln 2 > 0$ , so this simplifies to $O (\ln (1 / δ))$ . Multiplying these bounds proves both asymptotic counts in the Statement.

3.1givenF3step 1.1step 2.1∎

Each controlled- $Q$ call expands into the separately declared controlled constituents of the task definition; [F3] keeps their gate/query costs distinct, as well as the initial preparations and classical median work.

Depends on

Used by

Amplitude-estimation expectation estimator Algorithm
Quantum bounded-expectation guarantee and query cost Theorem

Dependency tree · two levels

38 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Brassard, Høyer, Mosca, and Tapp, Quantum Amplitude Amplification and Estimation, complete article, §§4–5, Theorems 11–18 (standard reference, not scraped)
Andrew M. Childs, Lecture Notes on Quantum Algorithms, complete lecture notes, §§19.1–19.3 (standard reference, not scraped)
Ronald de Wolf, Quantum Computing: Lecture Notes, complete 239-page notes, §7.3 (standard reference, not scraped)