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.

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

The product of ILU(0) factors can contain a fill position

Statement refuted

For every successful fixed-pattern ILU(0) factorization, every nonzero entry of the product $L U$ lies in the nonzero pattern of the input matrix $A$ .

Facts & Assumptions

Given: The fixed-pattern ILU(0) recurrence on a canonical sorted CSR input, and the usual matrix-product convention.

[F1]

Fixed-pattern ILU(0) factorization and application: the algorithm stores strict-lower and upper factor entries only at input-pattern positions and updates a work entry only when both relevant positions are in that pattern.

[F2]

Compressed sparse row matrices with stored-entry semantics: the represented matrix entry is the sum of stored values in its row with the specified column index; canonical records have no repeated columns or stored zeros.

[F3]

Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes: $(L U)_{i j} = \sum_{k} L_{i k} U_{k j}$ .

[F4]

The rationals form a field: $Q$ with its usual operations is a field.

Counterexample

1.1givenF2F4algebra

Take $A = (\begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{array}) .$ By [F4], this is a matrix over a field. Its canonical sorted CSR pattern is $P = {(1, 1), (1, 2), (2, 2), (3, 1), (3, 3)}$ , with all diagonal entries equal to $1$ . Thus the input satisfies the algorithm's domain and has no missing diagonal.

2.1F1step 1.1algebra

The first row gives $U_{11} = U_{12} = 1$ . Row 2 has no strict-lower position and gives $U_{22} = 1$ . In row 3 the only strict-lower position is $(3, 1)$ , so $L_{31} = 1$ . The update from this multiplier can reach only a column $j > 1$ present in both row 3 and row 1; no such column exists. Hence $L_{32}$ is not stored, $U_{33} = 1$ , and the procedure returns ok with $L = (\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{array}), U = (\begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}) .$

3.1F3step 2.1algebra∎

By [F3], $(L U)_{32} = L_{31} U_{12} + L_{32} U_{22} + L_{33} U_{32} = 1$ , whereas $A_{32} = 0$ . Therefore $L U$ has a nonzero outside the input pattern, so the refuted statement is false.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 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

Yousef Saad, Iterative Methods for Sparse Linear Systems, 2nd ed., §10.3.2, printed pp. 307–310 (standard reference, not scraped)
Richard Barrett et al., Templates for the Solution of Linear Systems: Building Blocks for Iterative Methods, 2nd ed., §3.4.2, printed pp. 39–41 (standard reference, not scraped)