Why the wide-bracket gap remains, taken apart on the SAME evaluations: 'order of the model' explains only a third. Of the 262 evaluations lost on 33 problems, 178 sit on fun3-fun6 - pure powers of order 3, 5, 9, 19 - where no interpolating model works (our step 0.41-0.81 bracket halvings per evaluation, IQI from our own points 0.01-0.17) and Chandrupatla wins by BISECTING 93-100% of his steps; our step is slower than bisection there and our 2x-bisection guarantee licenses it. His band inside our step confirms it per family: fun4/5/6 573/456/456 -> 296/239/235 against his 293/228/228, while +164/+111/+170/+144 on fun1/2/8/9 where it refuses good steps - slot 2's total was two opposite effects summed. Only fun1 (58) is model order: IQI from our three points 10.37 halvings against our 0.62; a convex two-point step cannot carry IQI's negative weight (428/428 non-degenerate states; five underflowed to -0.0).
The flat regime is not detectable by the three-point order estimate: m >= 2 (the exact condition for a tangent step to lose to a halving) fires on 0.2% of APS states and on only 5.8% of states of pure powers of order 3-19. Paper corrected where it said the whole gap is model order (widelimit, decision, abstract, future work - the refuted log transform removed). WARNING recorded in FINDINGS_model_order.md: article/sim_numerics.tex has carried a STALE header since the first commit (current text = sim_numerics.md, unchanged since 08-24); every paper edit of 09-20/21 went into the .tex, the .md has none of them. Which file is canonical is Valeri's call. The new text numbers trace to results_diag_*.txt but are not yet machine-checked by verify_article.py. Trace hook in pure_root is behaviour-neutral and pinned; 762 pass on the server twice. Co-Authored-By: Claude Opus 5 <noreply@anthropic.com>
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Claude Opus 5
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@@ -73,8 +73,15 @@ model is built on, selected by the shape of their configuration rather than by
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recency --- lowers the Alefeld--Potra--Shi count to $2527$, below every method
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measured here including Chandrupatla's $2785$, with the guarantee intact and
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still no tuning constant; reading it instead on the choice of far end gives
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$5511$ over both sets. Where the similarity is read turns out to matter more
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than which similarity it is.
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$5511$ over both sets. Reading it in the decision to trust the model or halve
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lowers the canonical count further, to $2523$. Where the similarity is read
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turns out to matter more than which similarity it is. The gap that remains on
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the wide-bracket set is taken apart on the same evaluations: a third of it is
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the order of the model --- a convex two-point step cannot carry the negative
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weight of quadratic interpolation --- and two thirds is not a model at all but
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the decision when to give the model up, which Chandrupatla takes by bisecting
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on up to all of his steps and which no reading measured here takes correctly
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for ours.
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\end{abstract}
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\section{Introduction}
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@@ -646,13 +653,15 @@ the root, $S(|\varphi-\xi|)<S(z)$, with the \emph{same} $S$ in both places
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step rule. It is not a proposal; it is an instrument, and it answers the only
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question that matters here --- whether the remaining gap lives in this place.
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It does not. With Chandrupatla's band in our step, the wide-bracket set costs
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In total, no. With Chandrupatla's band in our step, the wide-bracket set costs
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$3066$ under the projection guard and $2899$ under credit, against his $2423$,
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while the canonical set rises from $2806$ to $5432$. The band is not idle ---
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it refuses $63.6\%$ of steps --- so this is a decision that was taken and did
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not pay. That is the fifth independent test of the same conclusion: the gap is
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the \emph{order of the model} the step is built on, not the place where the
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similarity is read.
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not pay \emph{in total}. Per family it did pay, and where: it closes almost the
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whole gap on the flat families and gives back more than that on the families
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where our step is good. \S\ref{sec:widelimit} takes this apart; the short form is
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that the decision place does hold most of the gap, but no reading measured so far
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makes that decision correctly for \emph{our} step.
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Our own readings do not close it either (best $2868$ on the wide-bracket set),
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but ``defect'' sets a new best on the canonical set: $\mathbf{2523}$ at
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@@ -747,10 +756,46 @@ probe lands well short and the new bracket is about half the old one; a model
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that carries the curvature --- inverse quadratic interpolation --- lands far
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closer and removes about four fifths.
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So the limit is stated exactly, and it is the \emph{order of the model the
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similarity is read against}, not the placement of the reading and not the
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guarantee. Three parameter-free ways round it have now been measured and all
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three fail in the same way, buying the wide brackets at the price of the
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That mechanism is right for \texttt{fun1} and \emph{wrong as a statement about the
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whole gap}, which an earlier version of this section made. Replaying our own
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trajectories on the $33$ problems where we lose, and placing both models on the
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\emph{same} evaluations --- one extra evaluation each, from the same state ---
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splits the $262$ evaluations of the gap in two. On \texttt{fun1} ($58$ of them)
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inverse quadratic interpolation from our own three points removes far more of
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the bracket than our step did: $10.37$ halvings against $0.62$ on the
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$2\times10^4$-wide bracket. There the gap is the order of the model, and the
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reason is structural: our step $x+(1-S)(x_{\mathrm{far}}-x)=S\,x+(1-S)\,x_{\mathrm{far}}$
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is a convex combination of two points, while the quadratic interpolant's
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Lagrange weights at $0$ always include exactly one \emph{negative} weight when
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the triple brackets the root (verified on all $428$ non-degenerate states; in
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the other five the negative weight underflowed to $-0$). A similarity in $[0,1]$
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cannot produce a negative weight --- Proposition~\ref{prop:falsi} extended from
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two points to three.
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But $178$ of the $262$ --- two thirds --- sit on \texttt{fun3}--\texttt{fun6},
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which are pure powers of order $3$, $5$, $9$ and $19$, and there \emph{no}
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interpolating model works: our step removes $0.41$--$0.81$ halvings of the
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bracket per evaluation, inverse quadratic interpolation from the same points only
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$0.01$--$0.17$. Chandrupatla does not win those families with his model --- his
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traced run bisects $93$, $96$, $100$ and $100\%$ of his steps. We lose them
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because our step is slower than bisection there and our guarantee, at most twice
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bisection, licenses exactly that. His band inside our step confirms it family by
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family: under the projection guard it takes \texttt{fun4}, \texttt{fun5},
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\texttt{fun6} from $573$, $456$, $456$ to $296$, $239$, $235$ --- against his own
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$293$, $228$, $228$ --- and costs $+164$, $+111$, $+170$, $+144$ on \texttt{fun1},
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\texttt{fun2}, \texttt{fun8}, \texttt{fun9}, where it refuses steps that were good.
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So most of what remains is not the order of the model but the decision \emph{when
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to give the model up}, and it is open because every detector measured so far
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confounds ``the model is useless here'' with something else: the band with ``the
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configuration is not shaped for inverse quadratic interpolation'', bracket
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progress with ``the step has not moved the bracket yet'' ($4449$ on the canonical
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set), and the local order with nothing at all --- $m\ge2$ is exactly the
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condition under which a tangent step loses to a halving, since it lands at
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$e(1-1/m)$, and it is clean ($0.2\%$ of canonical states) but blind: it reports
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$m\ge2$ on only $5.8\%$ of states of families whose order is by construction $3$
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to $19$. For the \texttt{fun1} part, three parameter-free ways round the model
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order have been measured and all three fail in the same way, buying the wide brackets at the price of the
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canonical set: a fixed fraction of the bracket (\texttt{cbrt}), a logarithmic
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change of variable, and a curvature-corrected scale (\texttt{derived3\_m},
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whose derivation holds but whose estimator of the local order fires on $27.9\%$
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@@ -895,25 +940,26 @@ asserted.
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\subsection*{Future work}
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One weakness has a clear diagnosis and no fix inside this framework. On
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brackets many orders of magnitude wider than the distance to the root, the
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search must cross those decades, and an \emph{additive} step of $|f|/|f'|$ is
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sized by the local slope, which knows nothing about how many decades remain.
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That is why the cube-root similarity, whose step is a fixed \emph{fraction} of
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the bracket, beats the proposed rule $2.6$-fold on those families, and why ITP
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keeps its lead on the Chandrupatla set.
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The diagnosis of \S\ref{sec:widelimit} leaves one question, and it is narrower
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than the one this section used to pose. An earlier version proposed a
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logarithmic change of variable to turn an additive step into a geometric
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march; that was measured and refuted (\S\ref{sec:widelimit} and the negative
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results), and the step turned out to be a fraction of the bracket already.
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The regime is therefore multiplicative versus additive movement, not one
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similarity function versus another --- a change of the space the method works
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in, not of a coefficient inside it. The natural instrument is a logarithmic
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change of variable, $u = \operatorname{sign}(x-c)\log(1+|x-c|/s)$, in which an
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additive step becomes a geometric march and the exponent that the cube root
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fixes at $1/3$ by accident disappears as a free quantity. Two conditions
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decide whether that is a method or another refuted idea: the origin $c$ and
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unit $s$ must be read from the bracket rather than chosen, and the transform
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must be the identity to within rounding on narrow brackets, so that the
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wide-bracket families are not bought at the price of the canonical set. Every
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attempt reported in Section~7 failed on exactly that second condition.
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What is open is a decision, not a model. On two thirds of the remaining gap
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the right move is to halve, and Chandrupatla halves; our guarantee permits a
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step that is slower than halving, and nothing we have measured tells our step
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when it is. The condition itself is known exactly --- a tangent step on an
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approach of local order $m$ lands at $e(1-1/m)$ and loses to a halving when
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$m\ge2$ --- so the constant is bisection's $\tfrac12$ and not a choice. What is
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missing is a reading of $m$ that sees it: the three-point estimator reports
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$m\ge2$ on $0.2\%$ of canonical states, which is the right silence, and on
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$5.8\%$ of states of pure powers of order $3$ to $19$, which is blindness. Two
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successive residuals of the run itself carry the order more directly than a
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parabola through three points --- on an order-$m$ approach under tangent steps
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their ratio settles at $(1-1/m)^m$, while at a simple root it goes to zero ---
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and that is the reading we would test next, under the same rule as every other
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here: pre-registered, with the canonical set as the price it must not charge.
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\begin{thebibliography}{9}
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\bibitem{aps} G.~E. Alefeld, F.~A. Potra and Y.~Shi, ``Algorithm 748: Enclosing
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