CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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A real-valued optimum is not a fractional Grover iteration

Statement refuted

The real-valued optimizer $k_{*} = π / (4 θ) - 1 / 2$ is always a legal number of standard Grover-iterate queries.

Witness

Take $N = 2$ and $t = 1$ . Then $θ = \arcsin (1 / \sqrt{2}) = π / 4$ , so $k_{*} = 1 / 2$ , which is not an integer. Standard Grover circuits apply an integer number of iterates. The exact continuous formula at $k_{*}$ gives $1$ , while the legal integer choices $k = 0$ and $k = 1$ each give success probability $1 / 2$ .

Facts & Assumptions

Given: $N = 2 = 2^{1}$ , one marked item, and the standard Grover iterate.

[F1]

Exact success probability after an integer number of Grover iterates: for integer $k \geq 0$ , $p_{k} = \sin^{2} ((2 k + 1) θ)$ .

[F2]

Principal inverse sine and inverse cosine: $\arcsin$ is the inverse of sine on $[- π / 2, π / 2]$ .

[F3]

Signs, monotonicity intervals, and ranges of sine and cosine: sine and cosine have their usual signs on $(0, π / 2)$ .

[F4]

Pi as twice the smallest positive zero of cosine and Cosine has a smallest positive zero, lying strictly between zero and two: $π > 0$ .

[F5]

Quarter-turn values and shifts by pi/2 and pi: $\sin (π / 2) = 1$ and $\cos (π / 2) = 0$ .

[F6]

The addition formulas for sine and cosine: sine and cosine obey their addition formulas.

[F7]

Double-angle and quadratic power-reduction identities: $\sin (2 x) = 2 \sin x \cos x$ and $\cos (2 x) = \cos^{2} x - \sin^{2} x$ .

[F9]

Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ : nonnegative square roots are unique.

Refutation

1.1givenF2F3F4F5F7F9algebra

Put $x = π / 4$ , $s = \sin x$ , and $c = \cos x$ . Since $0 < π / 4 < π / 2$ by [F4], [F3] gives $s, c > 0$ . By [F7] and [F5], $0 = \cos (π / 2) = \cos (2 x) = c^{2} - s^{2}$ , so $(c - s) (c + s) = 0$ . Since $c + s > 0$ , we have $c = s$ . Then [F7] and [F5] give $1 = \sin (π / 2) = \sin (2 x) = 2 s c = 2 s^{2}$ , so [F9] gives $s = 1 / \sqrt{2}$ . Thus $\arcsin (1 / \sqrt{2}) = π / 4$ by [F2], and $k_{*} = π / (4 (π / 4)) - 1 / 2 = 1 / 2$ .

2.1givenF1F5F6step 1.1algebra∎

The legal standard query counts are integers, so $k_{*} = 1 / 2$ is not executable. Applying the integer formula [F1] gives $p_{0} = \sin^{2} (π / 4) = 1 / 2$ and $p_{1} = \sin^{2} (3 π / 4) = 1 / 2$ , whereas substituting the noninteger $k_{*}$ into the trigonometric expression would give $\sin^{2} (π / 2) = 1$ . The continuous expression therefore does not describe an allowed standard Grover circuit at this input.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Ronald de Wolf, Quantum Computing: Lecture Notes, §7.2 (standard reference, not scraped)
Andrew M. Childs, Lecture Notes on Quantum Algorithms, §19.1 (standard reference, not scraped)