ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)

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.

A one-pattern sign budget misses the best amplitude fit

Example

Let $d = 1$ , $m = 2$ , $A = (1, - 1)^{T}$ , and $b = (1, 4)$ , so $r = (1, 2)$ . In noisy sign enumeration, fix the first sign to $+ 1$ , order the remaining signs with $+ 1$ before $- 1$ , and use $P = L = 1$ . The algorithm processes only $(+ 1, + 1)$ , returns $x = - 1 / 2$ with $J (x) = \sum_{i = 1}^{2} (∣ (A x)_{i} ∣ - r_{i})^{2} = 5 / 2$ , and reports budget-exhausted. The unprocessed pattern $(+ 1, - 1)$ has least-squares candidate $x = 3 / 2$ with $J (x) = 1 / 2$ .

Example references

Pattern order, budgets, and noisy objective

Finite enumeration's complete-search guarantee

Facts & Assumptions

Given: The matrix $A = (1, - 1)^{T}$ , data $b = (1, 4)$ , noisy mode, and budgets $P = L = 1$ .

[F1]

Finite sign enumeration for real phase retrieval: for $m = 2$ , the canonical signs are ordered $(+, +)$ then $(+, -)$ ; a pattern cap or solve budget that stops before both are processed returns budget-exhausted and the noisy objective is $J (x) = \sum_{i} (∣ (A x)_{i} ∣ - r_{i})^{2}$ .

[F2]

Correctness and amplitude-residual objective of finite sign enumeration: each processed signed candidate minimizes its signed least-squares residual, while a global-minimum guarantee is made only after every canonical pattern is processed.

Verification

1.1givenF1

Since $m = 2$ , the canonical sign set is exactly $(+, +), (+, -)$ in the stated order. Both budgets permit one solve, so only $(+, +)$ is processed and the algorithm reports budget-exhausted.

1.2givenF1F2algebra

For the processed target $y_{+ +} = (1, 2)$ , the squared signed residual is $(x - 1)^{2} + (- x - 2)^{2} = 2 (x + 1 / 2)^{2} + 9 / 2$ , minimized uniquely at $x = - 1 / 2$ by [F2]. Its amplitudes are $(1 / 2, 1 / 2)$ , so $J (- 1 / 2) = (1 / 2 - 1)^{2} + (1 / 2 - 2)^{2} = 5 / 2$ .

2.1F2step 1.2algebra∎

For the unprocessed target $y_{+ -} = (1, - 2)$ , the squared signed residual is $(x - 1)^{2} + (- x + 2)^{2} = 2 (x - 3 / 2)^{2} + 1 / 2$ , minimized uniquely at $x = 3 / 2$ by [F2]. Its amplitudes are $(3 / 2, 3 / 2)$ , so $J (3 / 2) = (3 / 2 - 1)^{2} + (3 / 2 - 2)^{2} = 1 / 2 < 5 / 2$ . Thus the budget-limited incumbent is not the best candidate over all patterns.

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