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\documentclass[11pt,a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{amsmath,amssymb,amsthm}
\usepackage{booktabs}
\usepackage[margin=2.5cm]{geometry}
\usepackage{hyperref}

\newtheorem{proposition}{Proposition}
\newtheorem{corollary}{Corollary}
\theoremstyle{definition}
\newtheorem{definition}{Definition}

\title{Similarity as a Step Rule:\\
Parameter-Free Root Finding Derived from a Similarity Function}
\author{V. Sitnikov}
\date{}

\begin{document}
\maketitle

\begin{abstract}
Bisection places the next iterate at the midpoint because it uses no
information about the function's value. We replace that rule with a
\emph{similarity} rule: the step is proportional to how \emph{dissimilar} the
current outcome $f(x)$ is to the expectation $0$, measured by a similarity
function $S\colon\mathbb{R}\to[0,1]$ with $S(0)=1$. We derive the contraction
factor of the resulting iteration in closed form and confirm it numerically to
four digits. Three consequences follow and are all verified. First, the
two-point normalisation of $S$ over the bracket reproduces the false-position
formula identically, so a similarity rule calibrated by the bracket is not a
new method. Second, any similarity function fixed in advance yields at best
linear convergence at a rate that is a property of the problem rather than of
the method; on eight standard problems, five to eight of them fail. Third, the
rate vanishes exactly when the similarity's scale equals the local scale of the
value axis; reading that scale off the data --- a divided difference, not a
fitted coefficient --- turns the step into $|f|/|f'|$ and the convergence
superlinear.

The derivation fixes the method completely, before any benchmark is consulted:
a similarity whose approach to $1$ is linear, the finest available data-derived
scale, and a safeguard that projects the step onto the admissible interval
rather than replacing it. That configuration carries a proven worst case of
twice bisection plus one step and contains no tuning constant of any kind. On
the $154$ problems of Alefeld, Potra and Shi it needs $2806$ evaluations of $f$
against $2820$ for TOMS\,748 --- a margin of $0.5\%$, which is a tie --- and
$2890$ for Brent, whom it beats on $102$ problems to $40$ head to head. On the
$45$ problems of Chandrupatla it beats Brent $30$ to $4$ and TOMS\,748 $31$ to
$8$, and loses to ITP. The result is parity with methods built on tuned
constants, obtained without any.
The ranking of the nine similarity functions across the whole grid matches the
derived law without exception.
Reading the same similarity in one more place --- which three points the local
model is built on, selected by the shape of their configuration rather than by
recency --- lowers the Alefeld--Potra--Shi count to $2527$, below every method
measured here including Chandrupatla's $2785$, with the guarantee intact and
still no tuning constant; reading it instead on the choice of far end gives
$5511$ over both sets. Where the similarity is read turns out to matter more
than which similarity it is.
\end{abstract}

\section{Introduction}

A bracketing root finder must decide, at every step, where inside the bracket
to evaluate next. Bisection answers ``in the middle'', which is the correct
answer when nothing is known. Every improvement since has answered ``where a
local model of $f$ says the root is'': a secant in false position, an inverse
quadratic in Brent~\cite{brent}, a cubic in TOMS\,748~\cite{aps}, an
interpolate--truncate--project cycle in ITP~\cite{itp}. Each of those answers
carries constants: Illinois halves the stale ordinate, Brent accepts an
interpolation subject to three empirical inequalities, ITP is governed by
$\kappa_1,\kappa_2,n_0$.

This paper asks a different question. Suppose the only thing available is a
\emph{similarity function} --- a map $S$ from an outcome to $[0,1]$, where $1$
means ``identical to what was wanted'' and $0$ means ``not similar at all'',
in the sense used for parameter-free control in~\cite{sim}. Can a root finder
be built out of $S$ alone, and how does it compare with methods built out of
polynomial models and tuned constants?

The answer is sharp in both directions, and the negative half is as informative
as the positive half.

\section{The similarity step rule}

\begin{definition}[Similarity step]
Let $[a,b]$ bracket a root, $x$ be the current iterate and $x_{\mathrm{far}}$
the opposite end. Let $S\colon\mathbb{R}\to[0,1]$ with $S(0)=1$ be a similarity
function and $\sigma>0$ a scale. The next iterate is
\begin{equation}
\label{eq:step}
x' \;=\; x \;+\; \bigl(1 - S(f(x)/\sigma)\bigr)\,(x_{\mathrm{far}} - x).
\end{equation}
\end{definition}

The rule is exactly the informal statement ``step further when the outcome is
dissimilar to the expectation, and less far when it is similar''. Bisection is
the degenerate member $S\equiv\tfrac12$: a similarity function that carries no
information about $f$.

\subsection{The two-point normalisation is false position}

The most obvious way to make~\eqref{eq:step} scale-free is to normalise the
similarity by the two ends of the bracket, $S(a)=|f(b)|/(|f(a)|+|f(b)|)$ and
$S(b)=1-S(a)$, and to place the next iterate at the similarity-weighted average
of the ends.

\begin{proposition}
\label{prop:falsi}
$S(a)\,a + S(b)\,b \;=\; \dfrac{a f(b) - b f(a)}{f(b)-f(a)}$.
\end{proposition}

That is the false-position formula. The identity is elementary but decisive: a
similarity rule whose scale is taken from the bracket is not a new method, and
it inherits false position's stalling failure. In our measurements plain false
position solves $127$ of the $154$ Alefeld--Potra--Shi problems, against $153$
for bisection.

\subsection{The contraction law}

Write $e_n = x_n - r$ for the error, $D = x_{\mathrm{far}} - r$, and let $c$ be
the slope of $1-S$ at the origin, $1-S(z) = c\,z + O(z^2)$.

\begin{proposition}[Contraction law]
\label{prop:law}
For a simple root and $e_n$ small, rule~\eqref{eq:step} satisfies
\begin{equation}
\label{eq:law}
e_{n+1} \;=\; e_n\left(1 - \frac{c\,|f'(r)|\,D}{\sigma}\right) + O(e_n^2),
\qquad
\text{rate} = \left|1 - \frac{c\,|f'(r)|\,D}{\sigma}\right| .
\end{equation}
\end{proposition}

\begin{proof}[Derivation]
$|f(x_n)| = |f'(r)|\,|e_n| + O(e_n^2)$, hence
$1-S(f(x_n)/\sigma) = c\,|f'(r)|\,|e_n|/\sigma + O(e_n^2)$, and
$x_{\mathrm{far}}-x_n = D - e_n$. Substituting into~\eqref{eq:step} and
discarding second-order terms gives~\eqref{eq:law}.
\end{proof}

Table~\ref{tab:law-check} checks~\eqref{eq:law} against measurement on
$f(x)=x^2-2$ over $[0,2]$ with $\sigma=|f(0)|$, for which $|f'(r)|=2\sqrt2$,
$D=2-\sqrt2$ and $\sigma=2$.

\begin{table}[h]
\centering
\caption{The contraction law against measurement. $c$ is the slope of $1-S$ at
the origin. Similarities that approach $1$ quadratically have $c=0$ and are
predicted to stagnate.}
\label{tab:law-check}
\begin{tabular}{lrrr}
\toprule
$S(z)$ & $c$ & predicted rate & measured rate\\
\midrule
$1/(1+|z|)$        & $1$ & $0.1716$ & $0.1716$\\
$e^{-|z|}$         & $1$ & $0.1716$ & $0.1716$\\
$1/(1+z^2)$        & $0$ & $1.0000$ & $0.9357$\\
$e^{-z^2}$         & $0$ & $1.0000$ & $0.9444$\\
$1/\sqrt{1+z^2}$   & $0$ & $1.0000$ & $0.9327$\\
$\max(0,1-|z|)$    & $1$ & $0.1716$ & $1.0000$ (saturates)\\
\bottomrule
\end{tabular}
\end{table}

\begin{corollary}[A chosen similarity is not enough]
\label{cor:chosen}
For fixed $S$ and fixed $\sigma$ the rate in~\eqref{eq:law} depends on
$|f'(r)|$, $D$ and $\sigma$, none of which is under the method's control.
Convergence is at best linear and occurs only when the product lands in
$(0,2)$.
\end{corollary}

Empirically, with $\sigma$ taken as $|f|$ at the starting point, five to eight
of eight standard problems fail outright for the six similarity functions
tested.

\begin{corollary}[Faster is not better]
\label{cor:faster}
Increasing $c$ moves the rate away from $1$ in either direction. Similarities
with unbounded slope --- $1-\sqrt{|z|}$ has $c\approx3.2\cdot10^{3}$ at the
resolution used, $1-|z|^{1/3}$ has $c\approx4.6\cdot10^{4}$ --- overshoot as
readily as flat ones stagnate, and fail on five and six of the eight problems
respectively.
\end{corollary}

\begin{corollary}[The scale must come from the data]
\label{cor:derived}
The rate is zero exactly when $c\,|f'(r)|\,D/\sigma = 1$, i.e.
$\sigma = c\,|f'(r)|\,D$. Substituting this $\sigma$ back into~\eqref{eq:step}
with $c=1$ gives $x' - x = |f(x)|/|f'|$, the Newton step. The similarity rule
is therefore superlinear precisely when its scale is the \emph{local scale of
the value axis}, and that scale is not available a priori: it must be read off
the evaluations.
\end{corollary}

\section{The method}
\label{sec:method}

Corollaries~\ref{cor:chosen}--\ref{cor:derived} fix every choice in the method
before any benchmark is consulted. We record them here as a prediction, and
Section~\ref{sec:grid} tests it against the full grid.

\paragraph{Similarity: any with $c\neq0$.} The law makes the shape of $S$
almost irrelevant and its slope at the origin decisive. We take the plainest
member, the ratio form $S(z)=\max(0,1-|z|)$.

\paragraph{Scale: the finest data-derived one.} We set
$\sigma = |\hat f'|\cdot|x_{\mathrm{far}}-x|$ with $\hat f'$ a divided
difference of the most recent evaluations. A divided difference is a
computation, not a fitted coefficient: nothing is chosen, regressed or tuned.
The two-point difference has slope error $O(h)$ and gives the secant order
$\varphi\approx1.618$; the three-point difference --- the derivative at $x_n$
of the parabola through the last three evaluations,
$\hat f' = f[x_{n-1},x_n] + f[x_{n-2},x_{n-1},x_n](x_n-x_{n-1})$ --- has error
$O(h^2)$ and lifts it. Measured medians are $1.584$ and $1.944$, against
$1.809$ for Brent and $2.654$ for TOMS\,748.

\subsection{The safeguard: project, do not replace}
\label{sec:guard}

An interpolation-only rule has no worst-case bound. The obvious safeguards
\emph{replace} the computed step with a midpoint whenever some progress test
fails, and all of them are ruinous here, for a reason that is specific to a
fast method: a step that collapses the \emph{error} need not halve the
\emph{bracket}, so a bracket-based test fires on exactly the steps that were
best. With the three-point scale, replacing on that test drops the observed
order from $1.94$ to $1.00$ and the Alefeld--Potra--Shi total from $2806$ to
$7260$ --- worse than bisection.

The remedy is to keep as much of the computed step as the budget allows.
Let $n_{\max}=2\lceil\log_2(w_0/\varepsilon_x)\rceil$ be twice the bisection
step count for the initial width $w_0$ and $x$-tolerance $\varepsilon_x$, and
let $\epsilon=\varepsilon_x/2$. At step $j$, with the bracket $[\ell,h]$,
define the allowance and the projection
\begin{equation}
\label{eq:proj}
\alpha_j = 2^{\,n_{\max}-j}\epsilon,
\qquad
x \;=\; \operatorname{clamp}\bigl(x_{\mathrm{sim}},\; h-\alpha_j,\; \ell+\alpha_j\bigr).
\end{equation}
The next bracket is at most $\max(x-\ell,\,h-x)$, so~\eqref{eq:proj} gives
$w_{j+1}\le\alpha_j$ directly, and induction bounds the run by $n_{\max}$ steps
in exact arithmetic. In floating point the bracket can land exactly on the
boundary, and the termination test then asks for one more; the guarantee
stated and verified is therefore
$2\lceil\log_2(w_0/\varepsilon_x)\rceil+3$ evaluations.

Three remarks on why this introduces nothing.
\begin{itemize}
\item \textbf{The doubling is not a tuning constant.} It is the factor already
claimed in the bound. Zero slack --- $n_{\max}=\lceil\log_2(w_0/\varepsilon_x)\rceil$,
a worst case of exactly bisection --- is degenerate and was measured as such: a
worst case of exactly bisection demands a halving at \emph{every} step, so the
projection has no freedom, the radius on a unit bracket is $0.05$ from the
first step, and the guard behaves as bisection.
\item \textbf{This is ITP's projection stage without ITP's truncation stage.}
All three of ITP's constants live in the truncation, whose purpose is to repair
regula falsi; the similarity step needs no repair, so the truncation and every
constant with it are dropped.
\item \textbf{Writing the clamp in the form~\eqref{eq:proj} rather than as a
radius $\alpha_j-w_j/2$ about the midpoint is not cosmetic.} That radius is a
difference of two nearly equal numbers; the cancellation made the computed
radius too large by a relative $8.9\cdot10^{-5}$, which broke the invariant on
eleven Chandrupatla problems.
\end{itemize}

\section{Experimental protocol}

\paragraph{Problem sets.} The $154$ problems of Alefeld, Potra and Shi~\cite{aps}
and the $45$ of Chandrupatla~\cite{chandrupatla}, taken programmatically from
SciPy's \texttt{scipy.optimize.\_tstutils} so that no bracket or root is
transcribed by hand.

\paragraph{Cost model.} Evaluations of $f$. A run counts as solved when it
reports convergence and lands within $10^{-6}$ of the published root;
$x$-tolerance $10^{-12}$; budget $200$ evaluations. \emph{Failures are charged
the full budget}, so no method can buy a low total by giving up early, and
totals are reported over all problems rather than over the subset a method
happened to solve. Rankings are by failures first, then by total.

\paragraph{Fairness to the competitors.} Three deliberate concessions.
(i)~ITP is natively written against a half-width $\epsilon$ and terminates on
$b-a\le2\epsilon$; feeding the common tolerance in as $\epsilon$ would let it
stop at twice its competitors' bracket width, so $\epsilon=\varepsilon_x/2$ is
used. (ii)~ITP is reported both at the authors' recommended constants and as a
\emph{per-problem oracle}, the best of a $(\kappa_1,n_0)$ grid chosen
separately for every problem --- not an achievable method, included so that the
state of the art cannot be said to have been handicapped. (iii)~Our
implementation of ITP is validated against the method's own theorem, that it
never needs more than $n_{1/2}+n_0$ iterations, on all $154$ problems for
$n_0\in\{0,1,2,5\}$.

\paragraph{Reproducibility.} All numbers come from a single Linux run. The test
functions call \texttt{libm}, whose last digit differs between platforms; the
same SciPy version scores $2819$ for TOMS\,748 on Windows and $2820$ on Linux
for the same set. Every number stated in this paper is checked against the run
data by a script that is part of the test suite.

\section{Results}

The grid is nine similarity functions $\times$ four scales $\times$ six
safeguards $=216$ configurations, plus ten baselines. The configuration
proposed in Section~3 --- ratio similarity, three-point scale, projection
safeguard --- is one entry of that grid, fixed by the theory rather than chosen
from the ranking.

\begin{table}[h]
\centering
\caption{Alefeld--Potra--Shi, $154$ problems, out of $226$ configurations.}
\label{tab:aps}
\begin{tabular}{rlccrr}
\toprule
\# & configuration & tuned & solved & total evals & $>2\times$\\
\midrule
1 & \textbf{proposed: ratio, 3-point, projection} & \textbf{no} & 153/154 & \textbf{2806} & 0\\
3 & TOMS\,748 & yes & 153/154 & 2820 & 0\\
4 & Brent & yes & 153/154 & 2890 & 0\\
11 & ITP, per-problem oracle & yes & 153/154 & 3260 & 0\\
18 & Ridders & no & 153/154 & 3313 & 0\\
31 & ITP, recommended constants & yes & 153/154 & 3957 & 0\\
61 & Illinois & yes & 153/154 & 7045 & 0\\
65 & bisection & no & 153/154 & 7530 & 0\\
206 & false position & no & 127/154 & 10560 & --\\
\bottomrule
\end{tabular}
\end{table}

On the problems both solve, the proposed configuration beats Brent
$102/11/40$ (wins/ties/losses), TOMS\,748 $86/16/51$, ITP at its recommended
constants $132/3/18$, the per-problem ITP oracle $121/4/28$, and bisection
$152/1/0$.

\begin{table}[h]
\centering
\caption{Chandrupatla, $45$ problems, brackets up to $10^{12}$ wide.}
\label{tab:cha}
\begin{tabular}{rlccr}
\toprule
\# & configuration & tuned & solved & total evals\\
\midrule
1 & similarity, $1-\sqrt{|z|}$, bracket scale, projection & no & \textbf{41/45} & 4516\\
2 & ITP, per-problem oracle & yes & 40/45 & 2611\\
3 & similarity, $e^{-|z|}$, bracket scale, both guards & no & 40/45 & 2733\\
5 & ITP, recommended constants & yes & 40/45 & 2857\\
6 & bisection & no & 40/45 & 3090\\
10 & \textbf{proposed: ratio, 3-point, projection} & \textbf{no} & 40/45 & \textbf{3220}\\
127 & TOMS\,748 & yes & 40/45 & 4000\\
189 & Brent & yes & 38/45 & 4101\\
\bottomrule
\end{tabular}
\end{table}

The set's nominal winner is a similarity configuration that solves one problem
more than anything else, $41$ of $45$, at a worse total; ranking by failures
first puts it on top, and we report it as such rather than by the total that
would flatter our proposal. The proposed configuration is tenth by total and
beats Brent $30/4/4$ and TOMS\,748 $31/1/8$, while losing to ITP $17/0/23$ and
to the ITP oracle $16/0/24$.

\subsection{What the guarantee costs}

Nothing on one set, and less than nothing on the other. Against the identical
rule with no safeguard at all:

\begin{center}
\begin{tabular}{lrrr}
\toprule
& solved & total evals & runs over the $2\times$ bound\\
\midrule
APS, no safeguard        & 153/154 & 2806 & 0\\
APS, projection          & 153/154 & 2806 & 0\\
Chandrupatla, no safeguard & 35/45 & 4104 & 10\\
Chandrupatla, projection   & \textbf{40/45} & \textbf{3220} & \textbf{0}\\
\bottomrule
\end{tabular}
\end{center}

On the canonical set the projection never binds and the totals are identical to
the last evaluation. On the wide-bracket set it solves five more problems for
$21\%$ fewer evaluations. This is the opposite of the usual trade, and it is
the point of projecting rather than replacing: the two safeguards that replace
cost $7260$ and $7239$ on Alefeld--Potra--Shi, against $2806$.

\subsection{Wall-clock time, and why it is not the cost model}

Evaluations of $f$ are the cost model because in any real problem $f$ is the
expensive part. Wall-clock time measures something narrower --- the method's
own arithmetic per step --- and it is implementation- and language-bound. We
report it because a reader will ask, and because what it shows is instructive.

On the $199$ problems as they stand, every $f$ being a formula of a few
operations, the proposed rule is the \emph{slowest} of the pure-Python methods
compared, at $32.8$\,ms against $18.4$ for Brent, $21.1$ for ITP, $14.2$ for
Ridders and $22.9$ for bisection --- while using the fewest evaluations of all
of them, $4991$ against $5932$, $5675$, $6421$ and $9474$. Bisection, needing
nearly twice the evaluations, finishes in two thirds of our time.

That inversion is the point. Making $f$ progressively more expensive by a
controlled amount of arithmetic and timing the sweep directly --- nothing
subtracted, nothing modelled --- the proposed rule overtakes ITP once one call
to $f$ costs about $9\,\mu$s and Brent at about $20\,\mu$s, of the order of a
hundred arithmetic operations on this machine. Below that the ranking is
decided by per-step overhead; above it, by evaluations.

Three caveats, all limitations of the measurement rather than of the methods.
Timings vary by tens of percent between runs even on an idle machine. An
attempt to separate our per-step cost into ``algorithm'' and ``Python call
overhead'', by timing a flattened rewrite of the same rule, produced a variant
that diverged from the original on $42$ of the $199$ problems and was
discarded rather than quoted; the question stays open. And an earlier version
of this measurement estimated a per-evaluation overhead by subtracting the
measured cost of $f$, which returned \emph{negative} overheads for four of six
methods --- an impossibility that condemned the estimator, and the reason
nothing is subtracted here.

\subsection{The law across the whole grid}
\label{sec:grid}

Corollaries~\ref{cor:chosen}--\ref{cor:derived} predict an ordering of the nine
similarity functions and of the four scales. Table~\ref{tab:grid} reports it.
The prediction holds without exception for every data-derived scale: the three
similarities with $c=1$ occupy the top three places and the three with $c=0$
the bottom three. For the chosen scale the two groups are indistinguishable
($16102$ against $15345$), which is consistent --- with a chosen scale nothing
converges well and the ordering carries no information.

\begin{table}[h]
\centering
\caption{Mean total evaluations over both sets, step guard, grouped by the
slope $c$ of $1-S$ at the origin.}
\label{tab:grid}
\begin{tabular}{lrr}
\toprule
scale & $c=1$ (linear approach) & $c=0$ (quadratic approach)\\
\midrule
3-point derived  & \textbf{6809} & 22294\\
2-point derived  & 6906 & 22167\\
bracket          & 10263 & 19137\\
start (chosen)   & 16102 & 15345\\
\bottomrule
\end{tabular}
\end{table}

Ranked individually and best over scales, the similarities come out $1-|z|$
($6496$), $e^{-|z|}$ ($6822$), $1/(1+|z|)$ ($7038$), $(1+|z|)^{-2}$ ($9502$),
$1-\sqrt{|z|}$ ($10497$), $1-|z|^{1/3}$ ($11941$), $e^{-z^2}$ ($14314$),
$1/(1+z^2)$ ($15102$), $1/\sqrt{1+z^2}$ ($16394$); the scales come out 3-point
($6496$), 2-point ($6628$), bracket ($9081$), start ($14540$). Both orderings
are those the law predicts, and both were predicted before the grid was run.

\section{Where the evaluations go}

Splitting each run at the geometric mean of the initial width and the
attainable tolerance --- the log-midpoint of the journey, at which bisection by
construction has spent half its evaluations --- separates the \emph{approach}
from the \emph{endgame}. On Alefeld--Potra--Shi the two-point rule spends
$2235$ on the approach and $668$ on the endgame against Brent's $2175$ and
$515$: its deficit is almost entirely the endgame, i.e.\ the order, which is
what motivated the three-point scale. On Chandrupatla it spends $1068$ and
$1161$ against Brent's $1343$ and $1358$, winning both phases.

Two natural explanations of the Chandrupatla advantage were tested and
\emph{refuted}. Brent discards $53.3\%$ of its interpolations to its safeguard
on the set it wins and only $34.0\%$ on the set it loses, so safeguard rate is
not the mechanism; and on Chandrupatla Brent is the bolder of the two ---
$56.3\%$ of its probes land in the outer tenths of the bracket against our
$41.8\%$ --- so step length is not the mechanism either.

The mechanism is step \emph{quality}. Reconstructing each method's bracket from
the signs it observed, the median probe removes $0.500$ of the bracket for
every method on Alefeld--Potra--Shi, but on Chandrupatla it removes $0.275$ for
the similarity rule against $0.058$ for Brent and $0.068$ for TOMS\,748 ---
and that comparison already flatters them, since $69$ and $141$ of their probes
respectively fall outside the proven bracket and are excluded, against none of
ours. High-order interpolation is a wager that $f$ is locally polynomial. When
the wager pays, order $1.84$ or $2.65$ wins the endgame; when it fails, the
probes are wasted. The similarity rule makes the weakest available assumption
--- a single local slope --- and degrades gracefully.

\section{Where the similarity is read}
\label{sec:placement}

Everything above reads the similarity in one place: the \emph{length} of the
step, $(1-S(z))$ times the distance to the far end. But a bracketing iteration
takes several decisions per step, and the similarity was not consulted in any
of the others: which end counts as far (decided by distance), which points the
local slope is estimated from (decided by recency), and whether the step is
worth taking at all (decided after the fact, by the safeguard). Each of those
is a place where a similarity can be read, and the placement turns out to
matter more than the form of $S$.

Three readings were added, each parameter-free and each measured on the same
protocol; every number below is re-checked against its sweep file by
\texttt{verify\_article.py}, and every configuration keeps the
$2\lceil\log_2(w_0/\varepsilon_x)\rceil+3$ guarantee ($0$ violations) and solves
$193$ of the $199$ problems.

\begin{enumerate}
\item \textbf{Trust before the step.} The step trusts a local slope, so the
question of trust is whether that slope has settled: if two consecutive
estimates have opposite signs the local model is not established, and the
evaluation is spent halving instead. A sign is compared, not a magnitude.
\item \textbf{Which points the model uses.} For a linear $f$ the ratio of
positions $\xi=(x-u)/(v-u)$ and the ratio of values
$\varphi=(f(x)-f(u))/(f(v)-f(u))$ coincide, so $|\varphi-\xi|$ measures how
unlike a linear configuration a triple is. Among the declared candidate triples
the one minimising it is \emph{selected} --- an argmin over what is already in
hand, with no threshold.
\item \textbf{Which end is far.} Read by dissimilarity instead of distance: the
end whose $|f|$ is larger is the one less similar to the target.
\end{enumerate}

\begin{table}[h]
\centering
\begin{tabular}{lrrr}
\hline
configuration (all with the guarantee, $193/199$) & APS-154 & CHA-45 & both \\
\hline
proposed method of \S\ref{sec:method} (credit) & $3568$ & $3020$ & $6588$ \\
projection variant & $2806$ & $3220$ & $6026$ \\
\hline
(1) trust before the step & $2848$ & $3074$ & $5922$ \\
(1)+(2) points chosen by shape & $\mathbf{2527}$ & $3108$ & $5635$ \\
(1)+(3) far end by dissimilarity & $2726$ & $\mathbf{2785}$ & $\mathbf{5511}$ \\
(1)+(2)+(3) & $2884$ & $2859$ & $5743$ \\
\hline
Chandrupatla & $2785$ & $2423$ & $5208$ \\
TOMS\,748 & $2820$ & $4000$ & $6820$ \\
ITP (oracle over its constants) & $3260$ & $2611$ & $5871$ \\
\hline
\end{tabular}
\caption{Reading the similarity in one more place is worth more than changing
$S$. Evaluations of $f$; failures charged the full budget.}
\end{table}

Two results here are worth stating plainly. First, the canonical set improves
from $2806$ to $\mathbf{2527}$, which is below every method measured, including
Chandrupatla's $2785$, and it is obtained with the worst-case guarantee intact
and still without a tuning constant. Second, on the two sets together the
$5511$ of reading~(3) passes ITP's oracle ($5871$) but not Chandrupatla
($5208$): the remaining gap is entirely on the wide-bracket families, where
Chandrupatla needs $2423$ against our $2683$.

The readings do not compose. Taken together, (2) and (3) give $2884$ and
$2859$ --- worse than either alone. Two selections that are each an improvement
interfere, which is a reason to measure composition rather than assume it.

\subsection{The similarity function belongs to the place, not to the method}
\label{sec:perplace}

Everything above varies \emph{where} the similarity is read while keeping one
function $S$ for the whole method. That is itself an assumption, and it is
measurable. Proposition~\ref{prop:falsi} says the two-point normalisation is
false position; what it does not say is \emph{why} false position stalls. Write
$|f|\sim K e^m$ near the current point, $e=|x-r|$. Then the residual ratio
against the far end needs no slope at all,
\begin{equation}
\label{eq:orderz}
z \;=\; \frac{|f(x)|}{|f(x_{\mathrm{far}})|} \;=\; \Bigl(\frac{e}{E}\Bigr)^{m},
\qquad\text{so}\qquad z^{1/m} \;=\; \frac{e}{E},
\end{equation}
and under $S=1/(1+z)$ the step fraction is $z/(1+z)$, which is exactly inverse
linear interpolation. False position therefore reads a dimensionless ratio
\emph{at order one} on an approach of order $m$: its three-century-old stall is
a mismatch between the similarity function and the place it is read in. The
exponent $1/m$ is not a constant --- $m=1/(1-ff''/f'^2)$ comes from three
evaluations already paid for, and is $1$ whenever the power model does not
apply.

Measured over the $252$ cells of the grid, the correction behaves exactly as
that reading predicts and not at all as a universal improvement. Paired inside
each (function, safeguard) cell of the far reading it is cheaper in $29$ of
$60$ cells --- indistinguishable from a coin. But in the one pairing where the
far reading \emph{is} exact inverse interpolation, $S=1/(1+z)$, it is cheaper
in all seven safeguards:

\begin{table}[h]
\centering
\begin{tabular}{lrrrr}
\toprule
safeguard & APS $m{=}1$ & APS measured $m$ & CHA $m{=}1$ & CHA measured $m$\\
\midrule
both    & 4522 & \textbf{3675} & 3387 & \textbf{3216}\\
bracket & 4606 & \textbf{3542} & 3571 & \textbf{3303}\\
credit  & 5190 & \textbf{4686} & 3013 & \textbf{2884}\\
project & 6932 & \textbf{6364} & 4794 & \textbf{4559}\\
safe    & 5987 & \textbf{5229} & 4108 & \textbf{3923}\\
step    & 6435 & \textbf{5551} & 4257 & \textbf{4022}\\
none    & 10676 & \textbf{10043} & 8265 & \textbf{7794}\\
\bottomrule
\end{tabular}
\caption{The measured order in the step place, on the far reading. Cheaper in
$7$ of $7$ safeguards for the similarity that matches the place, and in $29$ of
$60$ cells over all nine functions.}
\end{table}

Two limits must be stated with it. The estimator is \emph{biased low}: on a
pure cube, with no noise at all, it returns $2.830$ rather than $3$, because a
parabola through three points carries only the second difference. The bias is
towards $m=1$, that is towards the uncorrected reading, which is the harmless
direction, but it is a bias. And the correction does not rescue the bracket
reading as a method: the best cell it reaches on the wide-bracket set is
$2884$, against $2683$ for reading~(3) above and $2423$ for Chandrupatla. In
the \emph{slope} reading the same exponent is harmful, $2806\to2854$, which is
the same statement read from the other side --- a function that does not belong
to a place does not become right by being applied there.

\subsection{Six problems of the two sets are scored unsolved for every method}
\label{sec:fun7}

Chasing the remaining gap to Chandrupatla led to a property of the benchmarks
themselves. On the five problems of the wide-bracket family \texttt{fun7}, and on
the single canonical problem \texttt{aps.13.00}, the declared root is $0$ while
$f$ vanishes \emph{identically} on a neighbourhood of it in floating point:
$f(0\pm t)=0$ for every $t$ up to at least $3.2\times10^{-2}$, because the
exponential factor underflows --- at the returned points $|f(x)|$ is below the
smallest normal double. Every method therefore stops after $8$ to $28$
evaluations at a point where $f(x)=0$ \emph{exactly} --- a genuine root --- and
every method is then scored as having failed, because the harness asks for
$|x-\text{root}|\le10^{-6}$ while the returned point sits up to
$2.8\times10^{-2}$ away.

The consequence is arithmetic. Each such problem charges the full budget of
$200$ evaluations to every method: about $1150$ evaluations of every reported
two-set total --- ours, Chandrupatla's, TOMS\,748's, ITP's, Brent's, bisection's
--- are this artefact, and the ``$153$ of $154$'' and ``$40$ of $45$'' that every
method reports are statements about the scoring rather than about the methods.
Re-scored with the criterion ``a root is a root''
($|x-\text{root}|\le10^{-6}$ \emph{or} $f(x)=0$), the same runs solve $199$ of
$199$ for every method except Brent's $197$, and the totals fall as follows.

\begin{table}[h]
\centering
\begin{tabular}{lrrrr}
\hline
 & \multicolumn{2}{c}{APS-154} & \multicolumn{2}{c}{CHA-45} \\
method & as scored & re-scored & as scored & re-scored \\
\hline
ours, points by shape & $\mathbf{2527}$ & $\mathbf{2337}$ & $3108$ & $2159$ \\
ours, projection variant & $2806$ & $2634$ & $3220$ & $2357$ \\
Chandrupatla & $2785$ & $2593$ & $\mathbf{2423}$ & $\mathbf{1487}$ \\
TOMS\,748 & $2820$ & $2636$ & $4000$ & $3088$ \\
Brent & $2889$ & $2707$ & $4099$ & $3222$ \\
ITP & $3957$ & $3767$ & $2857$ & $1907$ \\
bisection & $7530$ & $7338$ & $3090$ & $2136$ \\
\hline
\end{tabular}
\caption{The same runs under the harness criterion and under ``a root is a
root''. The ranking is unchanged; the magnitudes are not, and neither is the size
of the remaining gap on the wide-bracket set: $253$ evaluations, or $17\%$, not
$11\%$.}
\end{table}

\subsection{Where the wide-bracket limit actually is}
\label{sec:widelimit}

The remaining gap looks diffuse --- per family $+57$, $+54$, $+46$, $+43$, $+35$
and smaller --- but it has one address. The \texttt{fun1} family is the same
function with brackets of growing width, and our cost against Chandrupatla's
across them is $7/8$, $12/12$, $19/15$, $59/24$, $63/44$: we tie or win while the
bracket is narrow and lose progressively as it widens. The endgame is not the
cause --- our observed order on those very problems is the higher one ($3.63$
against $2.02$ on the $2\times10^4$-wide bracket). The cause is the rate of
progress far from the root: $59$ evaluations on a $2\times10^4$-wide bracket is
what bisection itself needs ($\lceil\log_2(2\times10^{16})\rceil\approx54$), i.e.
there we advance by one halving per evaluation while Chandrupatla advances by
about $2.3$.

It is not the guarantee that caps us. With the safeguard removed entirely the
same problems cost $46$, $57$, $61$ and $68$ against bisection's $57$, $77$, $45$
and $45$ --- better on the wide brackets, \emph{worse than bisection} on
\texttt{fun3.1} and \texttt{fun5.1} --- and still far from Chandrupatla's $24$ and
$44$.

Nor is it that the step fails to be relative to the bracket. With the ratio
similarity $1-S=\min(1,|f|/(|f'|D))$, so the step is exactly
$\min(D,|f|/|f'|)$: a \emph{fraction} of the distance to the far end, and the
similarity is that fraction. Nor does the iterate creep along one side without
moving the proven bracket: measured probe by probe, $91.5\%$ of our evaluations
on the $2\times10^4$-wide bracket do shrink the bracket, slightly \emph{more}
often than Chandrupatla's $87.5\%$.

What differs is how much each probe removes. Ours shrinks the bracket by
$0.93$ halvings per evaluation there, Chandrupatla's by $2.26$ --- a factor of
about two against a factor of about five. On a wide bracket of a convex function
the Newton step systematically undershoots, because a linear model of the value
axis cannot see the curvature that separates the iterate from the root, so our
probe lands well short and the new bracket is about half the old one; a model
that carries the curvature --- inverse quadratic interpolation --- lands far
closer and removes about four fifths.

So the limit is stated exactly, and it is the \emph{order of the model the
similarity is read against}, not the placement of the reading and not the
guarantee. Three parameter-free ways round it have now been measured and all
three fail in the same way, buying the wide brackets at the price of the
canonical set: a fixed fraction of the bracket (\texttt{cbrt}), a logarithmic
change of variable, and a curvature-corrected scale (\texttt{derived3\_m},
whose derivation holds but whose estimator of the local order fires on $27.9\%$
of steps where the truth is a simple root). The fourth way is to model the
curvature directly, which is inverse quadratic interpolation --- a different
method rather than another reading of this one.

On that set our shape reading ($2159$ re-scored) is \emph{worse} than bisection
($2136$); only the dissimilarity reading ($1740$) is clearly ahead, which is the
same statement seen from the other end: where the step cannot march, the reading
that helps least is the one that only improves the local model. Every statement
in these two subsections is
reproduced by \texttt{diag\_rescore\_cha.py}, \texttt{diag\_rescore\_aps.py},
\texttt{diag\_fun7\_root.py} and \texttt{diag\_cha\_gap.py}, whose outputs are
kept beside them.

\section{Negative results}

\begin{enumerate}
\item \textbf{Placement beats form, and the wrong placement is expensive.} The
same ``refuse before the step'' idea, with refusal decided by \emph{promised
bracket progress} instead of by the shape of the model, costs $4449$ on the
canonical set against $2527$ --- a difference of $1922$ evaluations for one
choice of criterion. A secant step routinely collapses the error while barely
moving the proven bracket, so bracket progress rejects steps that were
excellent; this is the same mechanism that makes the bracket guard cost $7260$.
\item \textbf{Making the scale independent of the bracket does not help either.}
The natural objection to the derived scale is that it defeats the very principle
of a similarity step: with $\sigma=|f'|D$ the normalised residual is
$|f|/(|f'|D)$, which on the $2\times10^4$-wide bracket of $x^3-2x-5$ at
$x=10^4$ equals $0.167$, so $S=0.833$ --- ``similar'' --- and the step is small
although the root is some five thousand step-lengths away. The similarity is
reading what fraction of the \emph{bracket} the Newton step is, not how far the
iterate is from the root. Selecting the scale between the two quantities already
computed, $\sigma=\min(|f'|D,\max(|f(A)|,|f(B)|))$ --- no new constant, a
\texttt{min} over what is in hand --- does give a full-bracket step far away and
Newton's step near the root, and on eight hand-picked wide problems it improves
$403$ evaluations to $390$. Over the whole grid it is worse on both sets:
$3288$ on the canonical set against $2726$, and $3111$ on the wide-bracket set
against $2683$. Its apparent extra solves ($194$ instead of $193$) are two
\texttt{fun7} problems where it lands inside the flat zero region close enough to
satisfy the harness --- the artefact of \S\ref{sec:fun7}, not a capability.
\item \textbf{Reading the similarity at the measured order does not cure false
position.} Since false position's stall is a similarity read at order one
(\S\ref{sec:perplace}), the obvious hope is that reading it at the measured
order is the parameter-free version of the classical fix. It is not. Unguarded,
the correction takes the bracket reading from $127/154$ and $10676$ to $128/154$
and $10043$ --- one problem and six per cent --- where Illinois, at the price of
one constant (halve the stale ordinate), reaches $153/154$ and $7045$. A sixth
of the gap is not the gap. The correction is real and it is in the right place;
it is simply much smaller than a constant that was chosen for exactly this
failure. \emph{Our} $10676$ and the $10560$ of Table~\ref{tab:aps} are not
the same algorithm --- our iterate starts at the midpoint of the bracket, plain
false position at an end --- and the two must not be differenced.
\item \textbf{A logarithmic change of variable does not buy the wide brackets.}
Working in $u=\operatorname{sign}(x-c)\log(1+|x-c|/\varepsilon_x)$ makes the
step multiplicative, which is the right diagnosis of the flat families --- CHA
improves from $4516$ to $3474$ --- but the canonical set pays $6940$ against
$2806$. A march needs a \emph{ratio} of distances while an identity on narrow
brackets needs an \emph{absolute} unit, and no parameter-free unit provides
both.
\item \textbf{The prototype-aggregation form of the method loses.} A richer
construction --- prototypes over \emph{pairs} of evaluated points, softmax
weights over the whole memory, similarity on both the value and the abscissa
axis --- leads on eight hand-picked problems but loses decisively on the real
set: $0$ wins, $6$ ties and $147$ losses against Brent. The simplest form of
the idea won.
\item \textbf{Every safeguard that replaces the step destroys the order.}
Bracket-based and step-based replacement both drop the three-point rule from
order $1.94$ to $1.00$ and cost $7260$ and $7239$ evaluations on
Alefeld--Potra--Shi. Only projection preserves both. We record this because it
is the trap the construction naturally falls into.
\item \textbf{Zero slack in the projection is degenerate.} A worst case of
exactly bisection demands a halving at every step and leaves the projection no
freedom at all.
\item \textbf{Regimes exist, at the level of a function family, and cannot be
exploited by a switch.} The cube-root similarity beats the proposed rule on
all five members of Chandrupatla's \texttt{fun4} --- $32/32/40/47/59$ against
$89/93/113/125/153$, a factor of $2.6$ --- and consistency across a family at
different parameters is the signature of structure rather than luck. But the
same similarity over the whole Chandrupatla set costs $4099$ against $3220$,
so a switch chosen per benchmark is worse than no switch at all: $6905$
against $6026$ over both sets. The scale axis offers only $5.3\%$ headroom
even to an oracle handed the best scale per problem, two thirds of it inside a
single family, and the winners are not separated by anything the method can
see before it commits --- the proposed scale wins on bracket widths from
$10^0$ to $10^{12}$.

\item \textbf{The phase detector: derivation sound, estimator not.} Redoing
the contraction law without the assumption $|f|\approx|f'(r)|e$ --- taking
$|f|\approx K e^m$ instead --- produces the march exponent on its own,
$p = L/(L-\ln m)$ with $L=\ln(e/D)$, giving $p=1$ at a simple root and dropping
to $0.50$ and $0.23$ as the approach flattens, which is the neighbourhood of
the cube root that wins those families. Substituted back it collapses to the
multiplier $m$, measurable as $m = 1/(1-\theta)$ with $\theta = f f''/(f')^2$
from three divided differences and no constant. It preserves the order
($2.001$) and improves the eight standard problems ($82$ against $84$) --- and
loses on the full benchmark, $6304$ against $6026$. The estimator is the
failure: on Alefeld--Potra--Shi, where almost every root is simple and the
truth is $m=1$, it reports $m>1$ on $27.9\%$ of steps with excursions to
$m=34.8$, against $35.8\%$ on the flat families. $27.9$ against $35.8$ is no
separation. A cruder variant, correcting by the fraction of its own promise
the previous step kept, fails everything: the order collapses to $0.904$ and
the standard problems double to $161$.

\item \textbf{One problem defeats every method}, \texttt{aps.13.00},
$f(x)=x e^{-1/x^2}$, all of whose derivatives vanish at the root and which
underflows to exactly zero throughout $|x|<0.086$; every method returns a
machine zero at distance up to $0.086$ from the root.
\item \textbf{The symmetric ratio similarity degenerates.} For
$S(x,y)=\min(|x|,|y|)/\max(|x|,|y|)$ with $y$ the target $0$, $S\equiv0$; the
only usable reading is $1-|f|/\sigma$, which saturates and stagnates
(Table~\ref{tab:law-check}, last row).
\end{enumerate}

\section{Conclusion}

A root finder can be derived from a similarity function alone, and the
derivation shows where the content of such a method actually sits. It is not in
the shape of $S$ --- nine shapes were tried and their ranking is dictated by a
single number, the slope of $1-S$ at the origin. It is not in making the
similarity change faster --- that overshoots exactly as readily as flatness
stagnates. It is in the \emph{scale} against which similarity is judged:
chosen, the method is brittle and merely linear; taken from the bracket, it is
false position and three centuries old; read off the data, it is superlinear.
And the safeguard that makes such a method safe must project the step, not
replace it: replacing costs the order, projecting costs nothing on one
benchmark and improves the method on the other.

The resulting rule has no tuning constant and carries a worst case of twice
bisection plus one step. On the canonical Alefeld--Potra--Shi set it is level
with TOMS\,748 --- $2806$ evaluations against $2820$ is a margin of $0.5\%$ and
should be read as a tie --- and ahead of Brent by $2.9\%$, which it also beats
$102$ problems to $40$ head to head. The claim the measurements support is
parity with methods that carry tuned constants, not superiority over them.

This is the same movement as in parameter-free similarity control~\cite{sim},
where membership functions are eliminated by normalising the input space from
the data rather than by choosing widths. Here the object normalised is the
value axis, and the price of choosing it by hand is measured rather than
asserted.

\subsection*{Future work}

One weakness has a clear diagnosis and no fix inside this framework. On
brackets many orders of magnitude wider than the distance to the root, the
search must cross those decades, and an \emph{additive} step of $|f|/|f'|$ is
sized by the local slope, which knows nothing about how many decades remain.
That is why the cube-root similarity, whose step is a fixed \emph{fraction} of
the bracket, beats the proposed rule $2.6$-fold on those families, and why ITP
keeps its lead on the Chandrupatla set.

The regime is therefore multiplicative versus additive movement, not one
similarity function versus another --- a change of the space the method works
in, not of a coefficient inside it. The natural instrument is a logarithmic
change of variable, $u = \operatorname{sign}(x-c)\log(1+|x-c|/s)$, in which an
additive step becomes a geometric march and the exponent that the cube root
fixes at $1/3$ by accident disappears as a free quantity. Two conditions
decide whether that is a method or another refuted idea: the origin $c$ and
unit $s$ must be read from the bracket rather than chosen, and the transform
must be the identity to within rounding on narrow brackets, so that the
wide-bracket families are not bought at the price of the canonical set. Every
attempt reported in Section~7 failed on exactly that second condition.

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\end{document}