TheoremStatement: Literature-sourcedProof: 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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Nine-round randomized known-count search succeeds with probability at least two-thirds

Statement

For every exact known count $t > 0$ , the randomized known-count procedure returns a marked item with probability at least $2 / 3$ . If $0 < t \leq N / 2$ , each round succeeds with probability at least $1 / 8$ and nine-round failure is at most $(7 / 8)^{9} < 1 / 3$ . If $t > N / 2$ , each direct-sampling round succeeds with probability greater than $1 / 2$ , so nine-round failure is less than $1 / 512$ . For $t = 0$ the procedure returns none exactly.

Facts & Assumptions

Given: The nine-round algorithm with its exact $K$ , rejection sampling, and candidate verification.

[F1]

Known-count search by randomized Grover iteration: for $0 < t \leq N / 2$ , each round draws $b$ fair bits, accepts $j < K$ , applies $j$ Grover iterates, and verifies the measured candidate; for $t > N / 2$ it makes a direct uniform sample; for $t = 0$ it returns none.

[F2]

Average Grover success for a uniform integer iteration count: conditional on a uniform $J \in {0, \dots, K - 1}$ , average Grover success is at least $1 / 4$ .

Proof

1.1givenF1algebra

If $0 < t \leq N / 2$ , the algorithm's accepted labels are exactly $0, \dots, K - 1$ . If $K = 1$ , then $b = 0$ and acceptance has probability $1$ . If $K > 1$ , minimality of $b$ gives $2^{b - 1} < K \leq 2^{b}$ , so acceptance probability $K / 2^{b} > 1 / 2$ and, conditional on acceptance, each accepted $j$ is equally likely.

2.1step 1.1F1F2algebra

By [F2], an accepted round succeeds with probability at least $1 / 4$ . Multiplying by the acceptance probability gives per-round success at least $1 / 8$ . Fresh random bits, registers, and measurements make the nine rounds independent conditional on the fixed oracle, so total failure is at most $(7 / 8)^{9}$ . Since $3 \cdot 7^{9} = 121060821 < 134217728 = 8^{9}$ , this is less than $1 / 3$ .

3.1givenF1algebra∎

If $t > N / 2$ , a uniform address is marked with probability $t / N > 1 / 2$ . Nine fresh direct samples therefore fail with probability $(1 - t / N)^{9} < (1 / 2)^{9} = 1 / 512 < 1 / 3$ . If $t = 0$ , the algorithm's explicit branch returns none. Every returned item in the other branches is verified by a standard bit query, so no unmarked item is returned.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

22 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

Gilles Brassard, Peter Høyer, Michele Mosca, and Alain Tapp, Quantum Amplitude Amplification and Estimation, §2 (standard reference, not scraped)