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>
263 lines
9.9 KiB
Python
263 lines
9.9 KiB
Python
"""Stage 12 - the validation runs added on revision.
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Five independent questions, each printed as it completes so the log can be
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read at any moment:
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1. RE-RANKING with the baselines that were missing - Chandrupatla's own
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method and Dekker - against the proposed rule.
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2. STABILITY to arithmetic noise. The margin over TOMS 748 is 0.5% and the
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paper itself reports that libm moves counts by a unit between platforms.
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Every f is perturbed by +-1 ulp at random, and the distribution of the
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total is reported for each method. Without this, "a tie" is rhetoric.
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3. THE BUDGET MULTIPLIER. The method fixes it at 2 and argues that 1 is
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degenerate. Showing 1 fails is not showing 2 is right; the sweep over
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1, 1.25, 1.5, 2, 3, 4 answers whether the result is flat in it.
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4. A LARGER EVALUATION BUDGET. On Chandrupatla the guarantee alone can
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reach 160 evaluations against a budget of 200, so some "failures" may be
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the budget rather than the method. Re-run at 400.
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5. STATISTICS. A sum of 2806 against 2820 is 14 evaluations over 154
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problems. The head-to-head count gets an exact sign test and the total
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gets a bootstrap interval, so that "tie" and "win" are separated by
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evidence rather than by wording.
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Seeded; the seed is printed. Nothing here selects a configuration.
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"""
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from __future__ import annotations
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import math
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import random
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import sys
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import time
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from scipy.optimize._tstutils import (_APS_TESTS_DICTS,
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_CHANDRUPATLA_TESTS_DICTS)
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from simroot.baselines import BRENT_BASELINE, CHANDRUPATLA_BASELINE
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from simroot.baselines import (anderson_bjorck, best_of, bisection, brent,
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brent_scipy, chandrupatla, chandrupatla_scipy,
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dekker, itp, ridders, toms748)
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from simroot.pure import pure_root
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XTOL = 1e-12
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ACC = 1e-6
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SEED = 20260824
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def ours(f, a, b, xtol, maxfev, budget_factor=2.0):
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return pure_root(f, a, b, xtol, maxfev, similarity="ratio",
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scale="derived3", guard="project",
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budget_factor=budget_factor)
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# Competitors are taken at the better of SciPy's implementation and ours,
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# problem by problem, so no result can be attributed to a weak reimplementation.
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# ITP has no library version, which is why it is validated against its own
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# theorem instead; the false-position family and bisection are not in SciPy.
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METHODS = [
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("OURS (similarity)", ours),
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("chandrupatla", CHANDRUPATLA_BASELINE),
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("toms748 (scipy)", toms748),
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("brent", BRENT_BASELINE),
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("itp", itp),
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("ridders", ridders),
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("anderson_bjorck", anderson_bjorck),
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("dekker", dekker),
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("bisection", bisection),
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]
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def load():
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out = []
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for d in _APS_TESTS_DICTS:
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a, b = d["bracket"]
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args = d.get("args", ())
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out.append(("APS", d["ID"],
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(lambda ff, aa: lambda x: ff(x, *aa))(d["f"], args),
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float(a), float(b), float(d["root"])))
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for d in _CHANDRUPATLA_TESTS_DICTS:
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a, b = d["bracket"]
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out.append(("CHA", d["ID"], (lambda ff: lambda x: ff(x))(d["f"]),
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float(a), float(b), float(d["root"])))
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return out
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def per_problem(method, probs, maxfev, criterion="root"):
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"""Charged evaluation count per problem, and whether it was solved. criterion="root":
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within ACC of the published root (this stage's original scoring); "sign": the paper's
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single criterion since 2026-09-22, a sign change within 1e-6 or f(x) == 0."""
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from corpus_roots_jl import solved
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out = {}
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for _s, pid, f, a, b, root in probs:
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try:
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x, n, conv = method(f, a, b, XTOL, maxfev)
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x = float(x)
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except Exception:
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x, n, conv = float("nan"), maxfev, False
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if criterion == "sign":
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ok = bool(conv and solved(f, x))
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else:
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ok = bool(conv and math.isfinite(x) and abs(x - root) <= ACC)
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out[pid] = (n if ok else maxfev, ok)
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return out
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def totals(per, probs):
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t = {"APS": 0, "CHA": 0}
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ok = {"APS": 0, "CHA": 0}
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for s, pid, *_ in probs:
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t[s] += per[pid][0]
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ok[s] += per[pid][1]
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return t, ok
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def section(n, title):
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print("\n" + "=" * 78, flush=True)
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print(f"{n}. {title}", flush=True)
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print("=" * 78, flush=True)
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def main():
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probs = load()
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t0 = time.time()
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print(f"seed {SEED}; {len(probs)} problems; started", flush=True)
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# ------------------------------------------------------------------
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section(1, "RE-RANKING WITH THE MISSING BASELINES")
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base = {}
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for name, m in METHODS:
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p = per_problem(m, probs, 200)
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base[name] = p
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t, ok = totals(p, probs)
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print(f" {name:22s} APS {t['APS']:>6d} {ok['APS']:>3d}/154 "
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f"CHA {t['CHA']:>6d} {ok['CHA']:>3d}/45 "
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f"both {t['APS'] + t['CHA']:>6d}", flush=True)
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# ------------------------------------------------------------------
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section(2, "STABILITY TO +-1 ULP PERTURBATION OF f")
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rng = random.Random(SEED)
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REAL = 40
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print(f" {REAL} realisations, every evaluation of f nudged one ulp "
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f"either way\n", flush=True)
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def perturb(f, r):
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# The nudge is CACHED per abscissa. Drawing it afresh on every call
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# would model a noisy function rather than one computed to within an
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# ulp, and would penalise any method that asks about a point twice.
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seen = {}
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def g(x):
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if x in seen:
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return seen[x]
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v = f(x)
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if v != 0.0 and math.isfinite(v):
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v = math.nextafter(v, math.inf if r.random() < 0.5
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else -math.inf)
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seen[x] = v
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return v
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return g
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dist = {name: [] for name, _ in METHODS}
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for k in range(REAL):
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r = random.Random(SEED + k)
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pp = [(s, pid, perturb(f, r), a, b, root)
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for s, pid, f, a, b, root in probs]
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for name, m in METHODS:
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t, _ok = totals(per_problem(m, pp, 200), pp)
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dist[name].append(t["APS"] + t["CHA"])
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if (k + 1) % 5 == 0:
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print(f" ... {k + 1}/{REAL} realisations, "
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f"{time.time() - t0:.0f}s", flush=True)
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print(f"\n {'method':22s}{'min':>8s}{'median':>9s}{'max':>8s}"
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f"{'spread':>9s}{'unperturbed':>13s}", flush=True)
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print(" " + "-" * 70, flush=True)
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for name, _m in METHODS:
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v = sorted(dist[name])
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t, _ = totals(base[name], probs)
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print(f" {name:22s}{v[0]:>8d}{v[len(v) // 2]:>9d}{v[-1]:>8d}"
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f"{v[-1] - v[0]:>9d}{t['APS'] + t['CHA']:>13d}", flush=True)
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print("\n If the spread of a method is comparable to the gap between two"
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"\n methods, that gap is not a result.", flush=True)
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# ------------------------------------------------------------------
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section(3, "IS THE BUDGET MULTIPLIER 2 LOAD-BEARING?")
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print(f" {'factor':>8s}{'APS':>9s}{'APS ok':>9s}{'CHA':>9s}"
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f"{'CHA ok':>9s}{'both':>9s}", flush=True)
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print(" " + "-" * 54, flush=True)
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for fac in (1.0, 1.25, 1.5, 2.0, 3.0, 4.0):
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p = per_problem(lambda f, a, b, xt, mx, _f=fac: ours(f, a, b, xt, mx, _f),
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probs, 200)
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t, ok = totals(p, probs)
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mark = " <- the method" if fac == 2.0 else ""
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print(f" {fac:>8.2f}{t['APS']:>9d}{ok['APS']:>6d}/154"
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f"{t['CHA']:>9d}{ok['CHA']:>6d}/45"
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f"{t['APS'] + t['CHA']:>9d}{mark}", flush=True)
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# ------------------------------------------------------------------
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section(4, "A LARGER EVALUATION BUDGET (400 INSTEAD OF 200)")
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print(f" {'method':22s}{'solved@200':>12s}{'solved@400':>12s}"
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f"{'evals@400':>12s}", flush=True)
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print(" " + "-" * 60, flush=True)
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for name, m in METHODS:
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p4 = per_problem(m, probs, 400)
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t4, ok4 = totals(p4, probs)
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_t2, ok2 = totals(base[name], probs)
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print(f" {name:22s}{ok2['APS'] + ok2['CHA']:>9d}/199"
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f"{ok4['APS'] + ok4['CHA']:>9d}/199"
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f"{t4['APS'] + t4['CHA']:>12d}", flush=True)
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# ------------------------------------------------------------------
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section(5, "STATISTICS ON THE MARGINS")
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ourp = base["OURS (similarity)"]
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def sign_test(w, l):
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n = w + l
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if n == 0:
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return 1.0
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k = min(w, l)
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p = sum(math.comb(n, i) for i in range(k + 1)) / 2.0 ** n
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return min(1.0, 2.0 * p)
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print(f" {'competitor':18s}{'W/T/L':>14s}{'sign test p':>14s}"
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f"{'total diff':>13s}{'bootstrap 95%':>22s}", flush=True)
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print(" " + "-" * 82, flush=True)
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for name, _m in METHODS:
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if name.startswith("OURS"):
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continue
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opp = base[name]
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w = t_ = l = 0
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diffs = []
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for _s, pid, *_ in probs:
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(no, oko), (nc, okc) = ourp[pid], opp[pid]
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if not (oko and okc):
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continue
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diffs.append(no - nc)
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if no < nc:
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w += 1
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elif no > nc:
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l += 1
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else:
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t_ += 1
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if not diffs:
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continue
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p = sign_test(w, l)
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tot = sum(diffs)
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r = random.Random(SEED)
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boots = []
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for _ in range(4000):
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boots.append(sum(diffs[r.randrange(len(diffs))]
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for _ in range(len(diffs))))
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boots.sort()
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lo, hi = boots[int(0.025 * len(boots))], boots[int(0.975 * len(boots))]
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print(f" {name:18s}{f'{w}/{t_}/{l}':>14s}{p:>14.2e}"
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f"{tot:>13d}{f'[{lo}, {hi}]':>22s}", flush=True)
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print("\n Negative total difference means we spend fewer evaluations.",
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flush=True)
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print(" An interval straddling zero means the totals are indistinguishable.",
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flush=True)
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print(f"\nfinished in {time.time() - t0:.0f}s", flush=True)
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if __name__ == "__main__":
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main()
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