vvsandClaude Opus 5
862feaff86
Point 1. The base step's order is Muller's 1.839, not 2: the derivative of the parabola through
the last three points errs by f'''/6 (x_n - x_{n-1})(x_n - x_{n-2}) = O(e_{n-1} e_{n-2}).
order_high_precision.py (3000 digits, to 1e-2900): 1.840 on cos x - x, e^x - 2, x^3 - x - 2,
sin x - 1/2; 2.000 only on x^2 - 2; two-point 1.618. The double-precision medians (2.000, 1.94)
are removed from the text.
Point 3. Every protocol number is re-scored under the criterion of section 9 - a sign change
within 1e-6 or f(x) == 0. rescore_uniform.py re-runs all 2779 distinct configurations of the 25
protocol sweeps on APS and CHA (the runs did not keep x) -> sweep_uniform.json, same row format;
it reproduces the APS/CHA lines of Table 10 exactly (built-in check). verify_article.py,
audit_similarity_places.py and the gap diagnostics read the re-scored data; the multiplier and
endowment measurements score by the same criterion. Table 7 (two criteria side by side) is gone;
7.3 now explains why the criterion is a sign change. Consequences stated in the text: the base
rule 2634 / 2357 against Chandrupatla 2595 / 1505 (1.5% behind on APS, no longer "level"); A'
2259 / 1074 (12.9% and 28.6% below Chandrupatla); the defect reading 2351 is second to the
shape reading 2337, not the record; the remembered analogy is level with no regime (2633 vs
2634) under this criterion; Table 9a's range is 19% to 425%.
Point 8. report_sweep.head_to_head counts every problem, failures charged (it counted only the
problems both methods solved, which is why Table 8 rows summed to 153 and 40). Family-level
comparisons added (A' vs Chandrupatla 8/1/6 of 15 and 7/0/2 of 9 families, not significant), and
an independence paragraph in section 9.
Point 11. sweep_itp.py: ITP (recommended constants and per-problem oracle) on all six corpora ->
Table 10; behind A' on every corpus, 0 bound violations.
Points 7, 9, 10 and the internals: the 2x bound is ours, not the rivals' promise; the law
predicts the group order only; "outside the proven bracket" explained (narrowest bracket of all
observed signs); 7.4's fun1 paragraph corrected (five costs under safe; the unguarded numbers
belong to fun1.4, fun1.5 and to fun3.1, fun5.1 separately); Table 1's last cell; the corpus-4
composition and odd multiplicities stated, also in the abstract.
build_tex.py: pdflatex output decoded with errors="replace" (a UTF-8 character cut by TeX's log
wrap crashed the build). Table 10 column widths rebalanced.
Checks: verify_article on the server 500 OK / 0 BAD (467 existing with the new values, 33 new);
regression x2 on the server 1232 passed, 0 failed; PDF 28 pages, 0 errors, 0 undefined, 0 overfull.
Co-Authored-By: Claude Opus 5 <noreply@anthropic.com>
47 lines
1.8 KiB
Python
47 lines
1.8 KiB
Python
"""Do the section-7 records depend on the projection's multiplier?
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The plain projection's canonical total is 2806 for every multiplier from 1.25 to 4
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(measure_budget_factor.py). The records 2527 (shape reading, safe) and 2523 (defect
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reading, projection) are quoted with the scheduled projection; whether THEY are flat in
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the multiplier has to be measured, not inferred from the plain rule.
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python -u measure_records_vs_multiplier.py
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"""
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from __future__ import annotations
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from simroot.pure import pure_root
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from stage12_validation import load, per_problem, totals
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FACTORS = (1.25, 1.5, 2.0, 3.0, 4.0)
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RECORDS = {
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"shape reading (2527)": dict(similarity="ratio", scale="derived3", guard="safe",
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prestep="slope", points="shape"),
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"defect reading (2523)": dict(similarity="ratio", scale="derived3", guard="project",
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prestep="defect"),
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"far end by dissimilarity (5511)": dict(similarity="ratio", scale="derived3",
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guard="project", prestep="slope",
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direction="dissimilar"),
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}
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def main():
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probs = load()
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out = {}
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print(f"{'configuration':<34} " + " ".join(f"{f:>13.2f}" for f in FACTORS))
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for name, cfg in RECORDS.items():
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cells = []
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out[name] = {}
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for fac in FACTORS:
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p = per_problem(lambda f, a, b, xt, mx, _c=cfg, _f=fac: pure_root(
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f, a, b, xt, mx, budget_factor=_f, **_c), probs, 200, criterion="sign")
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t, _ok = totals(p, probs)
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out[name][fac] = (t["APS"], t["CHA"])
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cells.append(f"{t['APS']:>6d}/{t['CHA']:<6d}")
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print(f"{name:<34} " + " ".join(cells))
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print("\ncells are APS/CHA totals")
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return out
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if __name__ == "__main__":
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main()
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