the scoring artefact is in BOTH sets: aps.13.00 joins the five fun7 problems — 6 of 199 problems where f underflows to exactly zero on a ~3.2e-2 neighbourhood of the declared root, so every method returns a genuine root (|f(x)| below the smallest normal) and every method is scored a failure. About 1150 evaluations of EVERY two-set total are this artefact; re-scored, all methods solve 199/199 except Brent 197. Ranking unchanged: ours-by-shape 2337 on APS against chandrupatla 2593, chandrupatla 1487 on CHA against our 1740. Written into the paper as a table with both criteria; pdflatex clean, verifier green
Co-Authored-By: Claude Opus 5 <noreply@anthropic.com>
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APS problems: 154
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method strict solved/total honest solved/total disagreements
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ours (shape) 153/154 2527 154/154 2337 1
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ours (dissimilar) 153/154 3123 154/154 2934 1
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ours (published proj) 153/154 2806 154/154 2634 1
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chandrupatla 153/154 2785 154/154 2593 1
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toms748 153/154 2820 154/154 2636 1
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brent 153/154 2889 154/154 2707 1
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itp 153/154 3957 154/154 3767 1
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bisection 153/154 7530 154/154 7338 1
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problems where the two criteria disagree (a root found, scored a failure):
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ours (shape) aps.13.00 x=+0.008638489218 root=+0 f(x)=+0.000e+00 after 10 evals
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ours (dissimilar) aps.13.00 x=+0.0248491106 root=+0 f(x)=+0.000e+00 after 11 evals
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ours (published proj) aps.13.00 x=-0.01020453457 root=+0 f(x)=+0.000e+00 after 28 evals
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chandrupatla aps.13.00 x=-0.008282706313 root=+0 f(x)=+0.000e+00 after 8 evals
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toms748 aps.13.00 x=-0.0002236853352 root=+0 f(x)=+0.000e+00 after 16 evals
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brent aps.13.00 x=+0.0154213326 root=+0 f(x)=+0.000e+00 after 18 evals
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itp aps.13.00 x=+0.03079215104 root=+0 f(x)=+0.000e+00 after 10 evals
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bisection aps.13.00 x=+0.015625 root=+0 f(x)=+0.000e+00 after 8 evals
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problems no method solved under the strict rule:
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aps.13.00 solved by 0 of 8 methods
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aps.01.00 solved by 8 of 8 methods
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aps.02.00 solved by 8 of 8 methods
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aps.02.01 solved by 8 of 8 methods
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aps.02.02 solved by 8 of 8 methods
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aps.02.03 solved by 8 of 8 methods
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diagnosis of aps.13.00: bracket [-1, 4] declared root 0
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f(a)=-3.678794e-01 f(b)=+3.757652e+00 f(root)=+0.000000e+00
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f(root+1e-12) = +0.000000e+00 f(root-1e-12) = +0.000000e+00
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f(root+1e-09) = +0.000000e+00 f(root-1e-09) = +0.000000e+00
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f(root+1e-06) = +0.000000e+00 f(root-1e-06) = +0.000000e+00
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f(root+0.001) = +0.000000e+00 f(root-0.001) = +0.000000e+00
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ours returns x=+0.0248491106033 after 11 evals, f(x)=+0.000e+00, |x-root|=0.0248
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|f(x)| below smallest normal? True
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