Alphabeta Math
PropositionStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passverified 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.

Termination and operation count of the periodic wave benchmark

Physical scope

A finite exact-arithmetic property of the one-dimensional scalar benchmark; it asserts no numerical-relativity solver performance or physical observation.

Statement

For every admissible integer N≥3, time-step count nmax⁡≥1, real step k>0, and pair of real periodic input arrays, the algorithm terminates after exactly nmax⁡−1 update passes and outputs all levels through unmax⁡. In the unit-cost exact-real-arithmetic model, where each scalar arithmetic operation and each array read or write has bounded cost independent of N and nmax⁡, its operation count is O(Nnmax⁡).

Facts & Assumptions

Given: The finite periodic scalar-wave algorithm with integer inputs N≥3 and nmax⁡≥1.

[F1]

A finite periodic scalar-wave benchmark: one initialization computes u1 by one finite grid pass; the update loop runs through the fixed integer range 1,…,nmax⁡−1, and each pass updates exactly N grid entries.

Proof

technique · direct
1.1givenF1algebra

The initialization uses a finite formula at each of the N sites. The outer index takes each of the finitely many values 1 through nmax⁡−1 and has no data-dependent branch, so the procedure makes exactly nmax⁡−1 update passes. Each pass writes the next N-tuple, giving the output levels u0,u1,…,unmax⁡.

2.1givenF1step 1.1algebra∎

The initialization costs at most C0N scalar and array operations for a fixed constant C0. Each update pass costs at most C1N such operations for a fixed C1, because every site uses the same fixed stencil. Thus the total is at most C0N+C1N(nmax⁡−1)≤(C0+C1)Nnmax⁡, proving the claimed bound.

Depends on

Dependency tree · 2 levels

Sources