chasing Chandrupatla found a property of the BENCHMARK: on all five fun7 problems f vanishes identically near the declared root (f(0+t)=0 up to 3.2e-2), every method stops at a point where f(x)=0 exactly and is scored a failure by the |x-root|<=1e-6 rule — so 1000 evaluations of EVERY wide-bracket total are an artefact and the universal '40 of 45' is a scoring statement. Re-scored: 45/45 for everyone but Brent, ranking unchanged, gap to Chandrupatla 253 (17%) and diffuse across 8 families. Written into the paper with its reproducing scripts; pdflatex clean, verifier green

Co-Authored-By: Claude Opus 5 <noreply@anthropic.com>

diag_cha_gap.py

@@ -0,0 +1,72 @@
+"""WHERE the remaining gap to Chandrupatla on the wide-bracket set actually is.
+
+Measure before proposing: our best guaranteed configuration reaches CHA 2683 against
+Chandrupatla's 2423. Rather than invent a mechanism, this locates the loss per problem and
+per family, and reports for each problem what the two methods spend, so the next
+pre-registration can name a mechanism that addresses the measured loss.
+
+    python -u diag_cha_gap.py
+"""
+
+from __future__ import annotations
+
+import math
+from collections import defaultdict
+
+from simroot.baselines import best_of, chandrupatla, chandrupatla_scipy
+from simroot.pure import pure_root
+from sweep_full import ACC, MAXFEV, XTOL, load_problems
+
+CHA_BASE = best_of(chandrupatla, chandrupatla_scipy)
+
+
+def ours(f, a, b):
+    st = {}
+    x, n, ok = pure_root(f, a, b, XTOL, MAXFEV, similarity="ratio", scale="derived3",
+                         guard="safe", prestep="slope", direction="dissimilar", stats=st)
+    return x, n, ok, st
+
+
+def main():
+    rows = []
+    fam_loss = defaultdict(lambda: [0, 0, 0])      # family -> [ours, theirs, count]
+    tot_o = tot_c = 0
+    print(f"{'problem':<12} {'ours':>6} {'CH':>6} {'diff':>6} {'w0/xtol decades':>16} "
+          f"{'refused%':>9} {'forced%':>8} {'ok':>4}")
+    for s, pid, f, a, b, root in load_problems():
+        if s != "CHA":
+            continue
+        x_o, n_o, ok_o, st = ours(f, a, b)
+        x_c, n_c, ok_c = CHA_BASE(f, a, b, XTOL, MAXFEV)
+        ok_o = bool(ok_o and math.isfinite(x_o) and abs(x_o - root) <= ACC)
+        ok_c = bool(ok_c and math.isfinite(float(x_c)) and abs(float(x_c) - root) <= ACC)
+        c_o = n_o if ok_o else MAXFEV
+        c_c = n_c if ok_c else MAXFEV
+        tot_o += c_o
+        tot_c += c_c
+        fam = pid.split(".")[0]
+        fam_loss[fam][0] += c_o
+        fam_loss[fam][1] += c_c
+        fam_loss[fam][2] += 1
+        rows.append((c_o - c_c, pid, c_o, c_c))
+        print(f"{pid:<12} {c_o:6d} {c_c:6d} {c_o - c_c:+6d} "
+              f"{math.log2(abs(b - a) / XTOL):16.1f} "
+              f"{100.0 * st.get('pre_refused', 0) / max(n_o, 1):8.1f}% "
+              f"{100.0 * st.get('forced', 0) / max(n_o, 1):7.1f}% "
+              f"{'' if ok_o else 'OURS FAIL':>4}")
+
+    print(f"\ntotals: ours {tot_o}, Chandrupatla {tot_c}, gap {tot_o - tot_c:+d}")
+    print("\nper family (ours / theirs / gap):")
+    for fam in sorted(fam_loss, key=lambda k: -(fam_loss[k][0] - fam_loss[k][1])):
+        o, c, n = fam_loss[fam]
+        print(f"   {fam:<8} n={n}  {o:6d} / {c:6d}  gap {o - c:+6d}")
+    print("\nthe five problems that cost us most:")
+    for d, pid, o, c in sorted(rows, reverse=True)[:5]:
+        print(f"   {pid:<10} ours {o:4d}  CH {c:4d}  gap {d:+5d}")
+    print("\nthe five where we are ahead:")
+    for d, pid, o, c in sorted(rows)[:5]:
+        print(f"   {pid:<10} ours {o:4d}  CH {c:4d}  gap {d:+5d}")
+
+
+if __name__ == "__main__":
+    main()