TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck pass

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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 certified floating determinant filter returns only the exact sign

Statement

Whenever Outward-bounded floating determinant filter returns positive or negative, the exact determinant has that same sign. An inconclusive or nonfinite result carries no sign claim.

Facts & Assumptions

Given: A finite matrix $M$ and a completed filter evaluation satisfying the stated model and range hypotheses in Outward-bounded floating determinant filter.

[F1]

Outward-bounded floating determinant filter, step 4: the filter returns positive only when $\hat{D} - E > 0$ , and negative only when $\hat{D} + E < 0$ ; otherwise it returns inconclusive. These tests compare exact dyadic rationals.

[F2]

A left-to-right Leibniz determinant has an explicit forward error bound: under the filter's operation hypotheses, $∣ \hat{D} - \det M ∣ \leq γ_{q} S$ .

[F3]

Outward-rounded operations are inclusion-isotone and enclose DAG evaluations: every exact node value of a finite expression DAG with finite operation results belongs to its computed interval enclosure.

[F4]

Directed dyadic rounding encloses with explicit underflow and overflow bounds: directed upward rounding encloses each finite exact endpoint from above.

Proof

1.1givenF2F3F4algebra

The filter's absolute-term DAG evaluates $S$ from [F2]. Its returned upper endpoint $U$ satisfies $S \leq U$ by [F3]. The definition of $E$ and [F4] give $E \geq γ_{q} U$ , hence $E \geq γ_{q} S$ .

2.1step 1.1F2algebra

Combining [F2] with [step 1.1] yields $∣ \hat{D} - \det M ∣ \leq E$ , so $\det M \in [\hat{D} - E, \hat{D} + E]$ .

3.1F1step 2.1algebra∎

If the algorithm returns positive, [F1] says $\hat{D} - E > 0$ ; by [step 2.1], $\det M \geq \hat{D} - E > 0$ . If it returns negative, then $\hat{D} + E < 0$ and [step 2.1] gives $\det M \leq \hat{D} + E < 0$ . In all other cases it returns no sign.

Depends on

Used by

Cited to discharge well-definedness by Outward-bounded floating determinant filter.

Dependency tree · two levels

16 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

Jonathan R. Shewchuk, Adaptive Precision Floating-Point Arithmetic and Fast Robust Geometric Predicates, §§4.2–4.4 (standard reference, not scraped)
CGAL Developer Manual, Robustness Issues, §11.1 (standard reference, not scraped)