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\hypersetup{
pdftitle={Root Finding by Similarity and Analogy: A Bracketing Method with One Declared Constant},
pdfauthor={V. Sitnikov},
hidelinks,
pdfcreator={LaTeX via pandoc}}
\title{Root Finding by Similarity and Analogy: A Bracketing Method with
One Declared Constant}
\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 \(S:\mathbb{R}\to[0,1]\) with
\(S(0)=1\). We derive the contraction factor of the resulting iteration
in closed form; it predicts a linear rate to four digits when \(1-S\)
has a non-zero slope at the origin, and a sublinear \(e_n\sim C/n\) with
a predicted \(C\) when it does not, both confirmed. Three consequences
follow. First, with the \emph{ratio} similarity scaled by the far end
--- and with that combination only, of the five tested --- the rule
reproduces the false-position formula identically; so does the truncated
linear similarity scaled by a divided difference over the \emph{bracket
ends}, which shows that the boundary of novelty runs through which
points the difference is taken over. Second, any similarity fixed in
advance with the scale also fixed yields at best linear convergence at a
rate that is a property of the problem rather than of the method. Third,
the rate vanishes exactly when the 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 a \emph{base rule} 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. It is,
explicitly, a Newton step with a three-point divided-difference
derivative, clipped to the bracket and projected onto the bisection
budget. The similarity sets that step, and its form is decisive: with
the tent the step is Newton's, with any of the other eight functions the
proposed method costs from 19\% to five times more. Its order is
Muller's, 1.839. It carries a proven worst case of twice bisection plus
one step and one free numerical constant, the multiplier of that budget,
which we declare. On the 154 problems of Alefeld, Potra and Shi it needs
2634 evaluations of \(f\) against 2636 for TOMS 748, 2707 for Brent and
2595 for Chandrupatla's method; it wins more problems than it loses
against Chandrupatla, 87 to 43, and needs 1.5\% more in total because
its losses are the larger ones. On the 45 wide-bracket problems of
Chandrupatla it beats Brent and TOMS 748 and loses to ITP and to
Chandrupatla (2357 against 1946 and 1505). For every data-derived scale
the three similarity functions whose \(1-S\) has a non-zero slope at the
origin come before the three whose slope is zero, as the derived law
predicts.\\
The wide-bracket gap turns out to be not the order of the model but a
decision --- when to give the model up --- and the decision is a
similarity after all, of an unfamiliar kind: the similarity of a run to
its own past. A pure power is self-similar under zoom, so the
dimensionless configuration of the last three points recurs, while at a
simple root it tends to the linear prototype; the analogous pair yields
the local order, \(m=\ln L_f/\ln L_x\), and an exponential, self-similar
by translation, is its limit \(m=\infty\). Entering a halving regime
when that analogy holds with \(m\ge2\) --- the condition under which a
tangent step loses to a halving --- splitting at the point the order
predicts, and leaving when the split looks linear, together with four
smaller readings of the same kind and one borrowed element, gives the
method proposed here, A′. On the two sets it needs 2259 and 1074
evaluations against Chandrupatla's 2595 and 1505 and wins more problems
than it loses on both, 103 to 31 and 31 to 10. The chain of decisions
that produced it was selected on those sets; tested once, after its code
was frozen, on 240 problems from six new families, it needs 3656
evaluations against Chandrupatla's 5717 and wins 177 problems to 28 ---
a corpus the author composed knowing where the method is strong: two of
its six families have multiple or fractional roots, and without them the
lead is 2403 against 3569, on the simple-root families alone 737 against
747. Against simpler detectors put in the place of the analogy ---
Lagouanelle's multiplicity estimate with the same memory, Chandrupatla's
acceptance band, or none --- it wins on that corpus in total and problem
by problem (3656 against 4488, 6236 and 6165); against the classical
transformation \(\operatorname{sign}(f)|f|^{1/m}\) with the multiplicity
given in advance it loses every multiple or fractional root, by a factor
of 2.4 to 6.8 --- the price of estimating the order. Over 855 problems
in six corpora it is ahead of Brent, TOMS 748, ITP and the recent ModAB
everywhere and stays within its own bound of twice bisection on every
problem, a bound that Brent, TOMS 748 and ModAB, which do not promise
it, exceed on 118, 84 and 9 problems; against Chandrupatla it is ahead
in total on five corpora and level on the sixth. The lead is large on
multiple, fractional and nearly flat roots and a matter of per cent on
simple ones. Several of the resulting mechanisms coincide with classical
ones --- Lagouanelle's multiplicity estimate, false position on
\(|f|^{1/m}\) --- reached here from similarity and analogy and
recognised afterwards; what we found no counterpart for is the decision
layer, where every choice --- when to give the model up, where to split,
when to return --- is a judgment of similarity or of analogy, and the
similarity that sets the step, whose form is decisive (replacing it
costs 19\% to 425\%).
\end{abstract}
\hypertarget{introduction}{%
\subsection{1. Introduction}\label{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 {[}2{]}, a cubic in TOMS
748 {[}1{]}, an interpolate--truncate--project cycle in ITP {[}4{]}, an
Anderson--Björck step behind a linearity check in ModAB {[}10, 20{]}.
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
{[}6{]}. 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.
The paper proceeds in two stages. Sections 2--7 derive a base rule from
the similarity alone, measure it, and locate what it lacks: not a better
model but a decision, when to stop trusting the model it has. Section 8
supplies that decision with a similarity of the run to its own past, and
Section 9 tests it on four further corpora: three of them exposed two
last gaps of the base step and were used to close them, and the fourth
was fixed after the method's code was frozen and run once.
\begin{center}\rule{0.5\linewidth}{0.5pt}\end{center}
\hypertarget{the-similarity-step-rule}{%
\subsection{2. The similarity step
rule}\label{the-similarity-step-rule}}
\textbf{Definition 1 (Similarity step).} Let \([a,b]\) bracket a root,
\(x\) be the current iterate and \(x_\mathrm{far}\) the opposite end.
Let \(S:\mathbb{R}\to[0,1]\) with \(S(0)=1\) be a similarity function
and \(\sigma>0\) a scale. The next iterate is
\[x' \;=\; x \;+\; \bigl(1 - S(f(x)/\sigma)\bigr)\,(x_\mathrm{far} - x). \tag{1}\]
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 \(f\).
\hypertarget{the-two-point-normalisation-is-false-position}{%
\subsubsection{2.1 The two-point normalisation is false
position}\label{the-two-point-normalisation-is-false-position}}
The most obvious way to make (1) 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.
\textbf{Proposition 1.} For the ratio similarity \(S(z)=1/(1+|z|)\) with
\(\sigma=|f(x_\mathrm{far})|\), equivalently for the endpoint weights
above, \[S(a)\,a + S(b)\,b \;=\; \frac{a f(b) - b f(a)}{f(b)-f(a)}.\]
That is the false-position formula. Two things must be said about how
far it reaches, because the obvious generalisation is false.
It is specific to that similarity. On \(f=x^2-2\) over \([1,4]\), where
false position gives 1.200000, the same construction gives 1.206812 for
\(e^{-|z|}\), 1.015228 for \(1/(1+z^2)\), 1.007624 for
\(1/\sqrt{1+z^2}\) and 1.214286 for the truncated linear form: one of
five reproduces it. So ``a similarity rule scaled by the bracket is not
a new method'' would be too broad a conclusion, and this paper's own
Table 3 refutes it --- bracket-scaled configurations take third, fourth
and sixth place on Chandrupatla's set, where false position solves 6 of
45.
\textbf{Proposition 1'.} It is, however, specific to the \emph{bracket
ends} rather than to the ratio form. Take the truncated linear
similarity \(S(z)=\max(0,1-|z|)\) and the divided difference over the
two ends, \(\sigma=|f[a,b]|\cdot|b-a|=|f(b)-f(a)|=|f(a)|+|f(b)|\). Then
\(1-S=|f(a)|/(|f(a)|+|f(b)|)\), which is false position again, and the
agreement is exact to the last bit.
Together the two say where the boundary of novelty runs, and it is not
where one would guess: it runs through the words \emph{most recent} in
the definition of \(\hat f'\) in §3. A divided difference over the
bracket ends is three centuries old; the same difference over the most
recent iterates is the secant. In our measurements plain false position
solves 127 of the 154 Alefeld--Potra--Shi problems, against 154 for
bisection.
\hypertarget{the-contraction-law}{%
\subsubsection{2.2 The contraction law}\label{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,
\(1-S(z) = c\,z + O(z^2)\).
\textbf{Proposition 2 (Contraction law).} For a simple root and \(e_n\)
small, rule (1) satisfies
\[e_{n+1} \;=\; e_n\left(1 - \frac{c\,|f'(r)|\,D}{\sigma}\right) + O(e_n^2), \qquad \text{rate} = \left|1 - \frac{c\,|f'(r)|\,D}{\sigma}\right|. \tag{2}\]
\emph{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 (1) and discarding
second-order terms gives (2). \(\square\)
Table 1 checks (2) 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 \(\sigma=2\).
\textbf{Table 1.} The contraction law against measurement, all rows at
the \emph{chosen} scale \(\sigma=|f(x_0)|\). \(c\) is the slope of
\(1-S\) at the origin and \(c_2\) the coefficient of \(z^2\) when
\(c=0\).
A similarity with \(c=0\) does not converge at a rate at all. Expanding
\(1-S=c_2z^2+O(z^4)\) in the same way as (2) gives
\(e_{n+1}=e_n-c_2|f'(r)|^2|D|e_n^2/\sigma^2+O(e_n^3)\), hence
\[e_n \;\simeq\; \frac{\sigma^2}{c_2\,|f'(r)|^2\,|D|}\cdot\frac1n, \tag{2a}\]
sublinear rather than stagnant. Reporting a ``measured rate'' for these
rows would report the value of \(e_{n+1}/e_n\) at whatever \(n\) the
window happened to end --- it creeps to 1 monotonically (0.925 at
\(n=10\), 0.991 at \(n=100\), 0.9998 at \(n=4000\)) --- so the
prediction tested is the coefficient in (2a) instead.
\begin{longtable}[]{@{}
>{\raggedright\arraybackslash}p{(\columnwidth - 6\tabcolsep) * \real{0.2500}}
>{\raggedright\arraybackslash}p{(\columnwidth - 6\tabcolsep) * \real{0.2500}}
>{\raggedright\arraybackslash}p{(\columnwidth - 6\tabcolsep) * \real{0.2500}}
>{\raggedright\arraybackslash}p{(\columnwidth - 6\tabcolsep) * \real{0.2500}}@{}}
\toprule\noalign{}
\begin{minipage}[b]{\linewidth}\raggedright
\(S(z)\)
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
\(c\)
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
predicted
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
measured
\end{minipage} \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
\(1/(1+\lvert z\rvert)\) & 1 & rate 0.1716 & 0.1716 \\
\(e^{-\lvert z\rvert}\) & 1 & rate 0.1716 & 0.1716 \\
\(1/(1+z^2)\) & 0, \(c_2=1\) & \(n\,e_n\to0.8536\) & 0.8503 \\
\(e^{-z^2}\) & 0, \(c_2=1\) & \(n\,e_n\to0.8536\) & 0.8498 \\
\(1/\sqrt{1+z^2}\) & 0, \(c_2=\tfrac12\) & \(n\,e_n\to1.7071\) &
1.6988 \\
\(\max(0,1-\lvert z\rvert)\) & 1 & rate 0.1716 & fails: stays at
\(x_\mathrm{far}\) \\
\end{longtable}
The last row is not a slow rate but an immediate failure, and it matters
because that similarity is the one this paper goes on to propose. At
\(x_0=0\) the residual is \(z=-1\), so \(S=\max(0,1-1)=0\), the step is
the whole bracket, the iterate lands exactly on \(x_\mathrm{far}\) and
never leaves: the error is 0.5858 after one step and after two hundred.
\textbf{The failure is a property of the chosen scale, not of the
similarity.} Under the derived scale of §3 the residual \(|f|/\sigma\)
stays below 1 --- on this problem the clip does not engage on a single
one of the seven steps the method takes --- the truncation never fires,
and the same function is the best of the nine (Table 4). This is the
paper's own thesis, that the content sits in the scale and not in the
shape of \(S\), showing up as a discontinuity in the behaviour of one
function.
\textbf{Corollary 1 (A chosen similarity is not enough).} For fixed
\(S\) and fixed \(\sigma\) the rate in (2) 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 \((0,2)\).
Empirically, with \(\sigma\) taken as \(|f|\) at the starting point,
five to eight of the eight standard problems fail outright for the six
similarity functions tested. Those eight, used throughout as a quick set
and never as evidence on their own, are: \(x^2-2\) on \([0,2]\);
\(\cos x-x\) on \([0,1]\); \(e^x-2\) on \([0,2]\); \(x^3-x-2\) on
\([1,2]\); \(\sin x-\tfrac12\) on \([0,1.5]\); \(x^{10}-1\) on
\([0,1.3]\), which stalls false position; \(e^{-x}(x-1)\) on
\([-1,40]\), whose right end is flat to \(10^{-16}\); and \(\arctan x\)
on \([-1,3]\).
\textbf{Corollary 2 (Missing the calibration in either direction
costs).} The rate degrades as the product \(c\,|f'(r)|\,|D|/\sigma\)
departs from 1, upwards or downwards. We claim no symmetry between the
two directions, and the measurements do not support one: similarities
with unbounded slope --- \(1-\sqrt{|z|}\) has \(c\approx3.2\cdot10^{3}\)
at the resolution used, \(1-|z|^{1/3}\) has \(c\approx4.6\cdot10^{4}\)
--- fail on five and six of the eight problems with a chosen scale, yet
over the full benchmark (best over scales, Table 4) they finish at 9411
and 10874, ahead of all three \(c=0\) similarities at 13187, 13990 and
15272. Overshooting is the milder failure.
\textbf{Corollary 3 (The scale must come from the data).} The rate is
zero exactly when \(c\,|f'(r)|\,D/\sigma = 1\),
i.e.~\(\sigma = c\,|f'(r)|\,D\). Substituting this \(\sigma\) back into
(1) 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.
\begin{center}\rule{0.5\linewidth}{0.5pt}\end{center}
\hypertarget{the-method}{%
\subsection{3. The method}\label{the-method}}
Corollaries 1--3 fix every choice in the method before any benchmark is
consulted. We record them here as a prediction, and Section 5.3 tests it
against the full grid. One number in the safeguard is not fixed by them,
and §3.1 declares it as the method's one free constant.
\textbf{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 \textbf{truncated linear} or tent form
\(S(z)=\max(0,1-|z|)\). It is not the ratio form \(1/(1+|z|)\), which is
a separate row of every table here; the two are distinguished throughout
because Proposition 1 applies to the ratio form and Table 1's last row
to the tent.
\textbf{Scale: the finest data-derived one.} We set
\(\sigma = |\hat f'|\cdot|x_\mathrm{far}-x|\) with \(\hat f'\) a divided
difference of the \textbf{most recent evaluations} --- not of the
bracket ends, which by Proposition 1' would give false position. 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 \(\tfrac16 f'''(\eta)\,(x_n-x_{n-1})(x_n-x_{n-2})\), that is
\(O(e_{n-1}e_{n-2})\). The Newton step with that derivative satisfies
\(e_{n+1}=\tfrac{f''}{2f'}e_n^2+\tfrac{f'''}{6f'}\,e_n e_{n-1}e_{n-2}+\dots\);
the second term dominates, and the order is the root of \(p^3=p^2+p+1\),
\(p\approx1.839\) --- \textbf{Muller's order}, as it should be, since
the step uses the same parabola as Muller's method, through its
derivative rather than its root. Only where the parabola is exact
(\(f'''=0\)) does the order rise to Newton's 2. Carried to
\(10^{-2900}\) in 3000-digit arithmetic
(\texttt{order\_high\_precision.py}), the computational order per step
settles at 1.840 on \(\cos x-x\), \(e^x-2\), \(x^3-x-2\) and
\(\sin x-\tfrac12\) and at 2.000 on \(x^2-2\); the two-point scale
settles at 1.618. In double precision a run takes three or four
superlinear steps before it reaches the tolerance, too few to tell 1.84
from 2, and we report no order measured there.
\textbf{What the rule reduces to.} With \(S(z)=\max(0,1-|z|)\) one has
\(1-S=\min(1,|z|)\), and (1) becomes
\[x' \;=\; x + \operatorname{sign}(x_\mathrm{far}-x)\cdot\min\!\Bigl(\frac{|f(x)|}{|\hat f'|},\;|x_\mathrm{far}-x|\Bigr), \tag{4}\]
a Newton step with a three-point divided-difference derivative, clipped
to the bracket and then projected as in §3.1. That the tent turns the
similarity step into Newton's is a property of the similarity chosen,
not a sign that the similarity has dropped out: which \(S\) stands in
the step decides the step, and it is measured, not assumed --- Table 4
ranks the nine functions over the whole grid, the readings of §7 pay
most with the tent in the step and some of them lose with other
functions (§7), and in the proposed method replacing the tent by any
other function costs 19\% to 425\% (Table 9a).
We state the reduction ourselves rather than leave it to be discovered,
because it locates the contribution precisely. The base rule is not a
new family of methods. It is (i) a derivation that fixes every choice
--- which similarity, which scale, which safeguard --- before any
benchmark is consulted, and predicts which similarity functions come
first; (ii) a safeguard of ITP's projection type with ITP's truncation
stage, and hence all three of its constants, removed; (iii) the negative
results of §10; (iv) the measurements. We call it the \textbf{base
rule}; the method this paper proposes builds on it in Section 8.
\textbf{Algorithm.} Given \([a,b]\) with \(f(a)f(b)<0\) and tolerance
\(\varepsilon_x\):
\begin{verbatim}
evaluate fa = f(a), fb = f(b)
A, B <- a, b the proven bracket
x <- the endpoint with the smaller |f|; xp <- the other; xp2 <- none
n_max <- 2 * ceil(log2(|b-a| / xtol)); allow <- 2^n_max * xtol/2
while |B - A| > xtol:
far <- the end of [A,B] further from x
f' <- derivative at x of the parabola through (x,xp,xp2)
(two-point difference while xp2 does not yet exist)
sigma <- |f'| * |far - x|
if sigma is 0 or not finite: xn <- midpoint(A,B)
else:
xn <- x + min(1, |f(x)|/sigma) * (far - x) # the step, eq (4)
xn <- clamp(xn, hi - allow, lo + allow) # projection, eq (3)
if xn outside (A,B) or already evaluated: xn <- midpoint(A,B)
evaluate fn = f(xn); allow <- allow / 2
xp2, xp, x <- xp, x, xn
update the bracket: whichever of A,B keeps the sign change
return midpoint(A,B)
\end{verbatim}
Every quantity in it is either an evaluation of \(f\), a divided
difference of evaluations, or the halving that defines bisection ---
with one exception, the factor 2 in \texttt{n\_max}, which §3.1 declares
as the method's one free constant. The constant-free variant of §3.1
replaces the two \texttt{allow} lines by the earned credit of (3').
\hypertarget{the-safeguard-project-do-not-replace}{%
\subsubsection{3.1 The safeguard: project, do not
replace}\label{the-safeguard-project-do-not-replace}}
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
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 makes the convergence linear and takes the Alefeld--Potra--Shi
total from 2634 to 7068 --- nearly the cost of bisection (7338).
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
\[\alpha_j = 2^{\,n_{\max}-j}\epsilon, \qquad x \;=\; \operatorname{clamp}\bigl(x_\mathrm{sim},\; h-\alpha_j,\; \ell+\alpha_j\bigr). \tag{3}\]
The next bracket is at most \(\max(x-\ell,\,h-x)\), so (3) 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.
\textbf{The multiplier is the method's one free numerical constant.} It
is not ``the factor already claimed in the bound'' --- that argument is
circular, the bound is twice bisection because two was chosen --- and
measurement settles it (\texttt{measure\_budget\_factor.py}, Linux):
\textbf{Table 1a.} The scheduled projection at budget multiplier \(c\),
i.e.~\(n_{\max}=\lceil c\cdot\log_2(w_0/\varepsilon_x)\rceil\). Total
evaluations, failures charged.
\begin{longtable}[]{@{}llll@{}}
\toprule\noalign{}
\(c\) & APS-154 & CHA-45 & both \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
1 & 4625 & 2039 & 6664 \\
1.25 & 2626 & 1822 & 4448 \\
1.5 & 2636 & 1991 & 4627 \\
2 & 2634 & 2357 & 4991 \\
3 & 2634 & 2936 & 5570 \\
4 & 2634 & 3229 & 5863 \\
\end{longtable}
On the canonical set the projection binds on one problem only once
\(c\ge1.25\) --- \texttt{aps.13.00}, whose \(f\) underflows to zero
around the root (20, 30 and 28 evaluations at \(c=1.25\), 1.5 and 2) ---
so the total barely moves. On the wide-bracket set it moves
monotonically with \(c\) from 1.25 on, and 2 is not the best value. Nor
is zero slack (\(c=1\)) a degenerate case under which the guard behaves
as bisection: it costs 4625 and 2039, far from bisection's 7338 and
2136, and on the wide-bracket set it beats \(c=2\). The multiplicative
family has no member that the bound singles out, so its multiplier is a
genuinely free constant.
\textbf{The budget earned rather than scheduled: a constant-free
variant.} The constant can be removed. Let \(s_j=\log_2(w_j/w_{j+1})\)
be the shrink a step delivers in bits --- bisection delivers exactly 1
--- and keep a credit
\[C \leftarrow \max(0,\;C+s_j-1),\qquad \alpha_j = w_j\,2^{\,C-1},\qquad x=\operatorname{clamp}\bigl(x_\mathrm{sim},\,h-\alpha_j,\,\ell+\alpha_j\bigr), \tag{3'}\]
leaving the step alone while \(C\ge1\). A step may under-deliver only by
what earlier steps banked, and since \(C\) never goes negative the
shrinks over \(j\) steps sum to at least \(j-C_0\): the run costs at
most bisection plus \(\lceil C_0\rceil\) steps. The endowment \(C_0\) is
therefore the promise, not a setting. \(C_0=0\) makes the allowance
exactly half the bracket on the first step, which forces the midpoint
and makes the method bisection; every \(C_0\in(0,1]\) promises the same
single extra step; and among those, 1 is the most permissive and ---
measured --- the cheapest:
\textbf{Table 1b.} The earned projection at endowment \(C_0\). Total
evaluations, failures charged.
\begin{longtable}[]{@{}lllll@{}}
\toprule\noalign{}
\(C_0\) & promise & APS-154 & CHA-45 & both \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
0.1 & bisection + 1 step & 5150 & 2097 & 7247 \\
0.5 & bisection + 1 step & 4462 & 2037 & 6499 \\
\textbf{1} & \textbf{bisection + 1 step} & \textbf{3377} & \textbf{2073}
& \textbf{5450} \\
2 & bisection + 2 steps & 3502 & 1757 & 5259 \\
5 & bisection + 5 steps & 2819 & 1666 & 4485 \\
\end{longtable}
A looser promise buys evaluations, as it must; within one promise the
cheapest endowment is determined. The constant-free variant is therefore
the earned projection with \(C_0=1\), whose guarantee ---
\(\lceil\log_2(w_0/\varepsilon_x)\rceil+4\) evaluations, bisection plus
one step with one more for the floating-point boundary --- is verified
on all 199 problems.
We nonetheless propose the scheduled projection with \(c=2\) and declare
its multiplier. On the canonical set it is 743 evaluations cheaper than
the earned variant (2634 against 3377), and there its total barely
depends on the multiplier; on the wide-bracket set the earned variant is
the cheaper (2073 against 2357). The two are the ends of a trade stated
in numbers: one declared constant and the better canonical result, or no
constant, a tighter bound and the better wide-bracket result. The value
2 was fixed before any benchmark was consulted.
Two remarks apply to both forms.
\begin{itemize}
\tightlist
\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 (3) 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}
\begin{center}\rule{0.5\linewidth}{0.5pt}\end{center}
\hypertarget{experimental-protocol}{%
\subsection{4. Experimental protocol}\label{experimental-protocol}}
\textbf{Problem sets.} The 154 problems of Alefeld, Potra and Shi
{[}1{]} and the 45 of Chandrupatla {[}3{]}, taken programmatically from
SciPy's \texttt{scipy.optimize.\_tstutils} so that no bracket or root is
transcribed by hand.
\textbf{Cost model.} Evaluations of \(f\). A run counts as solved when
it reports convergence and \(f\) changes sign within \(10^{-6}\) of the
returned point, or vanishes there --- the bracketing notion of a root,
the same criterion on every corpus of this paper; \(x\)-tolerance
\(10^{-12}\); budget 200 evaluations. Scoring against the published root
instead, as the benchmarks' harness does, charges every method the full
budget on six problems where \(f\) vanishes identically around the root
in floating point (§7.3). Head-to-head counts (wins/ties/losses) are
taken over every problem of a set with the same charging as the totals.
\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.
\textbf{Fairness to the competitors.} Three deliberate concessions.
\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
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.
\item
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.
\item
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 \(n_0\in\{0,1,2,5\}\).
\item
Where SciPy ships the method, the competitor \textbf{is the library}:
Chandrupatla is \texttt{scipy.optimize.elementwise.find\_root}, Brent
is \texttt{brentq}, TOMS 748 is \texttt{toms748}, each run with the
same stopping rule as every other method --- a bracket no wider than
\(\varepsilon_x\), the relative term at machine precision, no stopping
on \(|f|\) --- and each evaluation counted by the same wrapper around
\(f\). An earlier version gave each competitor the better of the
library's implementation and ours, problem by problem. That concession
backfired in the competitor's favour: our Chandrupatla stopped on a
bracket twice as wide as everybody else's --- the factor this protocol
corrects for ITP in item 1 and had not corrected there --- and the
per-problem minimum let the looser run win. With the tolerance
corrected our implementation and the library agree on all 199 problems
exactly. For Brent the library is slightly dearer than our
implementation; we use the library anyway, because it is the
implementation a reader can check. ITP has no library implementation,
which is why it is checked against its own theorem instead; bisection
and the false-position family are not in SciPy either and are ours.
\end{enumerate}
\textbf{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 2635 for TOMS 748 on Windows
and 2636 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.
\begin{center}\rule{0.5\linewidth}{0.5pt}\end{center}
\hypertarget{results}{%
\subsection{5. Results}\label{results}}
The grid is nine similarity functions × four scales × seven safeguards =
252 configurations, plus twelve baselines: 264 in all. The base rule of
Section 3 --- tent similarity, three-point scale, projection safeguard
with its multiplier declared --- is one entry of that grid, fixed by the
theory rather than chosen from the ranking; the constant-free earned
variant of §3.1 is shown beside it. Chandrupatla's own method and
Dekker's are among the baselines; an earlier version of these tables was
drawn up before either was added.
\textbf{Table 2.} Alefeld--Potra--Shi, 154 problems, out of 264
configurations. Places by failures first, then total.
\begin{longtable}[]{@{}
>{\raggedright\arraybackslash}p{(\columnwidth - 10\tabcolsep) * \real{0.1667}}
>{\raggedright\arraybackslash}p{(\columnwidth - 10\tabcolsep) * \real{0.1667}}
>{\raggedright\arraybackslash}p{(\columnwidth - 10\tabcolsep) * \real{0.1667}}
>{\raggedright\arraybackslash}p{(\columnwidth - 10\tabcolsep) * \real{0.1667}}
>{\raggedright\arraybackslash}p{(\columnwidth - 10\tabcolsep) * \real{0.1667}}
>{\raggedright\arraybackslash}p{(\columnwidth - 10\tabcolsep) * \real{0.1667}}@{}}
\toprule\noalign{}
\begin{minipage}[b]{\linewidth}\raggedright
\#
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
configuration
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
tuned
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
solved
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
total evals
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
\(>2\times\)
\end{minipage} \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
1 & Chandrupatla & no & 154/154 & 2595 & 0 \\
2 & the same rule with no safeguard at all & no & 154/154 & 2634 & 0 \\
\textbf{3} & \textbf{base rule: tent, 3-point, projection (\(c=2\))} &
\textbf{no} & 154/154 & \textbf{2634} & 0 \\
4 & TOMS 748 & yes & 154/154 & 2636 & 0 \\
5 & Brent & yes & 154/154 & 2707 & 0 \\
11 & ITP, per-problem oracle & yes & 154/154 & 3068 & 0 \\
15 & Ridders & no & 154/154 & 3124 & 0 \\
25 & constant-free variant: earned projection & no & 154/154 & 3377 &
0 \\
29 & ITP, recommended constants & yes & 154/154 & 3767 & 0 \\
73 & bisection & no & 154/154 & 7338 & 0 \\
232 & Dekker & no & 153/154 & 3158 & 0 \\
237 & Illinois & yes & 153/154 & 7045 & 0 \\
244 & false position & no & 127/154 & 10560 & --- \\
\end{longtable}
The base rule and the same rule without any safeguard tie at 2634: the
projection never binds on this set. By total, Chandrupatla's method is
first with 2595, 1.5\% fewer; the base rule and TOMS 748 are level (2634
against 2636). Problem by problem the base rule wins against
Chandrupatla 87/24/43 (wins/ties/losses; exact two-sided sign test
\(p=1.4\cdot10^{-4}\)), against TOMS 748 86/16/52, Brent 102/11/41, ITP
at its recommended constants 132/3/19, the per-problem ITP oracle
121/4/29, and bisection 152/1/1. It wins twice as many problems as it
loses against Chandrupatla and trails that method in total because its
losses are the larger ones; the problems are not independent (§9), and
taken by function family the comparison is 6/1/8 of 15 families, a tie.
Dekker and Illinois solve 153 of 154 and are ranked after every method
that solves all. The constant-free variant goes 79/3/72 against
Chandrupatla (\(p=0.63\), a tie in problems), 78/9/67 against TOMS 748
and 79/5/70 against Brent, and loses on the total to all three.
\textbf{Table 3.} Chandrupatla, 45 problems, brackets up to \(10^{12}\)
wide, out of 264 configurations.
\begin{longtable}[]{@{}
>{\raggedright\arraybackslash}p{(\columnwidth - 10\tabcolsep) * \real{0.1667}}
>{\raggedright\arraybackslash}p{(\columnwidth - 10\tabcolsep) * \real{0.1667}}
>{\raggedright\arraybackslash}p{(\columnwidth - 10\tabcolsep) * \real{0.1667}}
>{\raggedright\arraybackslash}p{(\columnwidth - 10\tabcolsep) * \real{0.1667}}
>{\raggedright\arraybackslash}p{(\columnwidth - 10\tabcolsep) * \real{0.1667}}
>{\raggedright\arraybackslash}p{(\columnwidth - 10\tabcolsep) * \real{0.1667}}@{}}
\toprule\noalign{}
\begin{minipage}[b]{\linewidth}\raggedright
\#
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
configuration
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
tuned
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
solved
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
total evals
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
\(>2\times\)
\end{minipage} \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
1 & Chandrupatla & no & 45/45 & 1505 & 0 \\
2 & ITP, per-problem oracle & yes & 45/45 & 1675 & 0 \\
3 & similarity, \(e^{-\lvert z\rvert}\), bracket scale, both guards & no
& 45/45 & 1778 & 0 \\
5 & ITP, recommended constants & yes & 45/45 & 1946 & 0 \\
9 & constant-free variant: earned projection & no & 45/45 & 2073 & 0 \\
15 & bisection & no & 45/45 & 2136 & 0 \\
\textbf{49} & \textbf{base rule: tent, 3-point, projection (\(c=2\))} &
\textbf{no} & 45/45 & \textbf{2357} & 0 \\
158 & TOMS 748 & yes & 45/45 & 3088 & 15 \\
227 & Brent & yes & 43/45 & 3238 & 18 \\
229 & the same rule with no safeguard at all & no & 39/45 & 3391 & 10 \\
\end{longtable}
Here Chandrupatla's method is genuinely the stronger: it beats the base
rule 29 problems to 11 (\(p=0.006\)) and by 1505 against 2357 in total.
The base rule beats Brent 34/4/7 and TOMS 748 33/1/11 --- which exceed
\(2\times\) bisection, a bound they do not promise, on 18 and 15
problems --- and loses to ITP 18/1/26 and to the ITP oracle 16/1/28. The
wide-bracket set is where the multiplier matters (Table 1a) and where
the constant-free variant is the cheaper form, 2073 against 2357. Over
both sets the base rule needs 4991 and the constant-free variant 5450,
against 4100 for Chandrupatla, 4743 for the ITP oracle, 5724 for TOMS
748 and 5945 for Brent.
\hypertarget{what-the-guarantee-costs}{%
\subsubsection{5.1 What the guarantee
costs}\label{what-the-guarantee-costs}}
It depends on which bound is bought. Against the identical rule with no
safeguard at all:
\begin{longtable}[]{@{}
>{\raggedright\arraybackslash}p{(\columnwidth - 6\tabcolsep) * \real{0.2500}}
>{\raggedright\arraybackslash}p{(\columnwidth - 6\tabcolsep) * \real{0.2500}}
>{\raggedright\arraybackslash}p{(\columnwidth - 6\tabcolsep) * \real{0.2500}}
>{\raggedright\arraybackslash}p{(\columnwidth - 6\tabcolsep) * \real{0.2500}}@{}}
\toprule\noalign{}
\begin{minipage}[b]{\linewidth}\raggedright
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
solved
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
total evals
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
runs over the \(2\times\) bound
\end{minipage} \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
APS, no safeguard & 154/154 & 2634 & 0 \\
APS, projection --- base rule (\(2\times\) bisection) & 154/154 & 2634 &
0 \\
APS, earned projection --- constant-free (bisection + 1 step) & 154/154
& 3377 & 0 \\
Chandrupatla, no safeguard & 39/45 & 3391 & 10 \\
Chandrupatla, projection --- base rule & 45/45 & 2357 & 0 \\
Chandrupatla, earned projection --- constant-free & \textbf{45/45} &
\textbf{2073} & \textbf{0} \\
\end{longtable}
The projection of the base rule never binds on the canonical set, so its
totals are identical to the last evaluation; on the wide-bracket set it
solves six more problems for 30\% fewer evaluations. The earned
projection promises twice as much --- a worst case of bisection plus one
step instead of twice bisection --- and pays for it on the canonical
set, 743 evaluations or 28\%, while on the wide-bracket set it is the
cheapest of the three. Both are the opposite of the usual trade on at
least one set, and that is the point of projecting rather than
replacing: the two safeguards that replace cost 7068 and 7046 on
Alefeld--Potra--Shi, against 2634.
\hypertarget{wall-clock-time-and-why-it-is-not-the-cost-model}{%
\subsubsection{5.2 Wall-clock time, and why it is not the cost
model}\label{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 base rule is the \textbf{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. (These are the evaluations the timing loop actually spent;
for the base rule and bisection, which solve every problem, they equal
the totals of §5; Brent here is our pure-Python implementation, which is
what is timed.)
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 base rule overtakes ITP
once one call to \(f\) costs about 9 µs and Brent at about 20 µs, which
on this machine is of the order of a hundred arithmetic operations.
Below that the ranking is decided by per-step overhead; above it, by
evaluations.
Three caveats, all of them 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 therefore 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.
\hypertarget{the-law-across-the-whole-grid}{%
\subsubsection{5.3 The law across the whole
grid}\label{the-law-across-the-whole-grid}}
Corollaries 1--3 predict that the similarities with \(c=1\) come before
those with \(c=0\) under a data-derived scale; they say nothing about
the order within a group, nor about the functions with unbounded slope.
Table 4 reports it. 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. That \(c=0\) is slow is the unsurprising half --- a step
proportional to \(z^2\) is damped quadratically. For the chosen scale
the two groups are indistinguishable (14983 against 14228), which is
consistent --- with a chosen scale nothing converges well and the
ordering carries no information.
\textbf{Table 4.} Mean total evaluations over both sets, step guard,
grouped by the slope \(c\) of \(1-S\) at the origin.
\begin{longtable}[]{@{}lll@{}}
\toprule\noalign{}
scale & \(c=1\) (linear approach) & \(c=0\) (quadratic approach) \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
3-point derived & \textbf{5764} & 21200 \\
2-point derived & 5854 & 21089 \\
bracket & 9149 & 18015 \\
start (chosen) & 14983 & 14228 \\
\end{longtable}
Ranked individually and best over scales, the similarities come out
\(1-|z|\) (5422), \(e^{-|z|}\) (5772), \(1/(1+|z|)\) (5993),
\((1+|z|)^{-2}\) (8455), \(1-\sqrt{|z|}\) (9411), \(1-|z|^{1/3}\)
(10874), \(e^{-z^2}\) (13187), \(1/(1+z^2)\) (13990), \(1/\sqrt{1+z^2}\)
(15272); the scales come out 3-point (5422), 2-point (5564), bracket
(7972), start (13423). The group order and the order of the scales are
those the law predicts and were predicted before the grid was run; the
order inside each group is measured, not predicted.
\begin{center}\rule{0.5\linewidth}{0.5pt}\end{center}
\hypertarget{where-the-evaluations-go}{%
\subsection{6. Where the evaluations
go}\label{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. The bracket here is reconstructed from \emph{every}
sign the method has observed --- the narrowest interval the evaluations
prove, not the one the method keeps; Brent and TOMS 748 always probe
inside their own bracket, but 69 and 141 of their probes fall outside
this narrowest one and remove nothing from it, and those probes are left
out of the median. High-order interpolation is a wager that \(f\) is
locally polynomial. When the wager pays, a high order wins the endgame;
when it fails, the probe removes little. The similarity rule makes the
weakest available assumption --- a single local slope --- and degrades
gracefully.
\begin{center}\rule{0.5\linewidth}{0.5pt}\end{center}
\hypertarget{where-the-similarity-is-read}{%
\subsection{7. Where the similarity is
read}\label{where-the-similarity-is-read}}
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 to the base rule of §3 (tent similarity,
three-point scale), each parameter-free and each measured on the
protocol of §4; every number below is re-checked against its own sweep
file by \texttt{verify\_article.py}, and every configuration in Table 5
keeps the \(2\lceil\log_2(w_0/\varepsilon_x)\rceil+3\) guarantee (0
violations) and solves all 199 problems.
Where a safeguard other than the projection of §3.1 appears, it is one
of the grid's. \emph{safe} is the projection preceded by a step-length
test --- a step not at most half the step taken two evaluations earlier
is replaced by a halving --- with the projection still carrying the
bound. \emph{credit} replaces the scheduled allowance of (3) by one
earned step by step: a surplus is banked whenever a step shrinks the
bracket by more than half and drawn on when it shrinks it by less, which
bounds the run by bisection plus one step. \emph{bracket} and
\emph{both} are the replacing safeguards of §3.1, which keep a bound at
the price of the order; \emph{step} alone and \emph{none} carry no
bound.
\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\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}
\textbf{Table 5.} Reading the similarity in one more place is worth more
than changing \(S\). Evaluations of \(f\); failures charged the full
budget.
\begin{longtable}[]{@{}
>{\raggedright\arraybackslash}p{(\columnwidth - 6\tabcolsep) * \real{0.2500}}
>{\raggedright\arraybackslash}p{(\columnwidth - 6\tabcolsep) * \real{0.2500}}
>{\raggedright\arraybackslash}p{(\columnwidth - 6\tabcolsep) * \real{0.2500}}
>{\raggedright\arraybackslash}p{(\columnwidth - 6\tabcolsep) * \real{0.2500}}@{}}
\toprule\noalign{}
\begin{minipage}[b]{\linewidth}\raggedright
configuration (all with the guarantee, 199/199)
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
APS-154
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
CHA-45
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
both
\end{minipage} \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
base rule: tent, 3-point, projection \(c=2\) (§3) & 2634 & 2357 &
4991 \\
constant-free variant: earned projection (§3.1) & 3377 & 2073 & 5450 \\
(1) trust before the step, projection & 2663 & 2170 & 4833 \\
(1)+(2) points chosen by shape, \emph{safe} & \textbf{2337} & 2177 &
4514 \\
(1)+(3) far end by dissimilarity, projection & 2539 & \textbf{1864} &
\textbf{4403} \\
(1)+(2)+(3), \emph{safe} & 2695 & 1928 & 4623 \\
Chandrupatla & 2595 & 1505 & 4100 \\
TOMS 748 & 2636 & 3088 & 5724 \\
ITP, per-problem oracle & 3068 & 1675 & 4743 \\
\end{longtable}
Two results here are worth stating plainly. First, the canonical set
improves from 2634 to \textbf{2337}, which is below every method
measured, including Chandrupatla's 2595, with the worst-case guarantee
intact. It hardly feels the declared multiplier of §3.1: 2337 for every
multiplier from 1.5 to 4, 2351 at 1.25
(\texttt{measure\_records\_vs\_multiplier.py}). Second, on the two sets
together the 4403 of reading (3) passes ITP's oracle (4743) but not
Chandrupatla (4100): the remaining gap is entirely on the wide-bracket
families, where Chandrupatla needs 1505 against our 1752 (reading (3)
under \emph{safe}).
The readings do not compose. Taken together, (2) and (3) give 2695 and
1928 --- worse than either alone on the canonical set, and on the
wide-bracket set worse than (3) alone. Two selections that are each an
improvement interfere, which is a reason to measure composition rather
than assume it.
\textbf{Which similarity stands in the step decides whether a reading
pays.} The readings above, and the decision \emph{defect} of §7.2,
compare quantities directly --- an ordinal judgment of which is more
similar, in which every decreasing \(S\) would decide alike. Their gains
nevertheless depend on the \(S\) that sets the step
(\texttt{audit\_similarity\_places.py}, on the same sweeps): on the
canonical set the triple chosen by shape gains 493 evaluations with the
tent in the step and loses 250 with \(1/\sqrt{1+z^2}\); \emph{defect}
gains 283 with the tent and loses 215 with \(1-|z|^{1/3}\). The records
of this section are the pairing of the tent step with these judgments,
not the judgments alone.
Under the constant-free earned projection the same readings gain little
or lose. Reading (1) gives 3316 and 2050, (1)+(2) 3564 and 2074, (1)+(3)
3666 and 2052, all three together 4163 and 2087, against 3377 and 2073
for the constant-free variant itself: the earned budget binds early on
the canonical problems, and what the readings buy under the projection
is mostly room that the earned budget does not grant.
\hypertarget{the-similarity-function-belongs-to-the-place-not-to-the-method}{%
\subsubsection{7.1 The similarity function belongs to the place, not to
the
method}\label{the-similarity-function-belongs-to-the-place-not-to-the-method}}
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 1 says that the ratio
similarity scaled by the far end 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,
\[z \;=\; \frac{|f(x)|}{|f(x_\mathrm{far})|} \;=\; \Bigl(\frac{e}{E}\Bigr)^{m}, \qquad\text{so}\qquad z^{1/m} \;=\; \frac{e}{E}, \tag{5}\]
and under the ratio similarity \(S=1/(1+|z|)\) 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 --- \(m=1/(1-ff''/f'^2)\),
Lagouanelle's estimate of the multiplicity {[}13{]} with the derivatives
replaced by divided differences, comes from three evaluations already
paid for, as in item 5 of §10, and is 1 whenever the power model does
not apply.
Measured over the grid, the correction behaves as that reading predicts
and not as a universal improvement. Paired inside each (function,
safeguard) cell of the far-end reading it is cheaper in 33 of 63 cells
--- indistinguishable from a coin. But in the one pairing where that
reading \emph{is} exact inverse interpolation, the ratio similarity, it
is cheaper under all seven safeguards:
\textbf{Table 6.} The measured order in the step place, on the far-end
reading with the ratio similarity \(1/(1+|z|)\). Cheaper in 7 of 7
safeguards; in 33 of 63 cells over all nine functions.
\begin{longtable}[]{@{}
>{\raggedright\arraybackslash}p{(\columnwidth - 8\tabcolsep) * \real{0.2000}}
>{\raggedright\arraybackslash}p{(\columnwidth - 8\tabcolsep) * \real{0.2000}}
>{\raggedright\arraybackslash}p{(\columnwidth - 8\tabcolsep) * \real{0.2000}}
>{\raggedright\arraybackslash}p{(\columnwidth - 8\tabcolsep) * \real{0.2000}}
>{\raggedright\arraybackslash}p{(\columnwidth - 8\tabcolsep) * \real{0.2000}}@{}}
\toprule\noalign{}
\begin{minipage}[b]{\linewidth}\raggedright
safeguard
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
APS, \(m=1\)
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
APS, measured \(m\)
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
CHA, \(m=1\)
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
CHA, measured \(m\)
\end{minipage} \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
both & 4331 & \textbf{3485} & 2469 & \textbf{2306} \\
bracket & 4418 & \textbf{3352} & 2671 & \textbf{2428} \\
credit & 4999 & \textbf{4495} & 2064 & \textbf{1933} \\
projection & 6783 & \textbf{6176} & 4057 & \textbf{3751} \\
safe & 5803 & \textbf{5040} & 3218 & \textbf{3032} \\
step & 6251 & \textbf{5362} & 3371 & \textbf{3146} \\
none & 10676 & \textbf{9855} & 8265 & \textbf{7424} \\
\end{longtable}
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 far-end reading as a method: the best
cell it reaches on the wide-bracket set is 1933, against 1752 for
reading (3) above and 1505 for Chandrupatla. In the \emph{slope} reading
of §3 the same exponent is harmful, \(2634\to2703\), 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.
\hypertarget{the-decision-place}{%
\subsubsection{7.2 The decision place}\label{the-decision-place}}
Three places have now been measured: the step length, the far end, and
the points the model is built on. The fourth is the \emph{decision} ---
whether the local model deserves this evaluation or halving is the
better use of it. Chandrupatla's method is strong exactly there: it
refuses its model on most steps by the acceptance band
\(1-\sqrt{1-\xi}<\varphi<\sqrt{\xi}\), with \(\xi\) and \(\varphi\) the
ratios of positions and of values defined above. The rule of §3 has no
similarity in this place at all: a sign gate and a projection that
repairs \emph{after} the step.
Three readings of that place were measured, each on quantities already
in hand and each without a constant. Two are ours: refuse when the
configuration is less like a linear one than the iterate is off-centre,
\(|\varphi-\xi|>\min(\xi,1-\xi)\) (``defect''); and refuse when the
configuration's similarity to linear falls below the iterate's
similarity to the root, \(S(|\varphi-\xi|)<S(z)\), with the \emph{same}
\(S\) in both places (``simpair''). The third is \emph{Chandrupatla's
own band}, placed inside our step rule. It is not a proposal; it is an
instrument, and it answers whether the remaining gap lives in this
place.
In total, no. With Chandrupatla's band in our step, the wide-bracket set
costs 2101 under the projection and 1942 under \emph{credit}, against
that method's own 1505, while the canonical set rises from 2634 to 5242.
The band is not idle --- it refuses 63.5\% of steps --- so this is a
decision that was taken and did not pay \emph{in total}. Per family it
did pay, and where: it closes almost the whole gap on the flat families
and gives back more than that on the families where our step is good.
§7.4 takes this apart; the short form is that the decision place does
hold most of the gap, but no reading measured so far makes that decision
correctly for \emph{our} step.
Our own readings do not close it either (best 1941 on the wide-bracket
set), but ``defect'' gives the second-best canonical total:
\textbf{2351} under the projection of the base rule, next to the 2337 of
reading (2) and below every method measured, with the guarantee intact,
and between 2343 and 2353 for every multiplier from 1.25 to 4. Under the
earned projection it gives 3354 and 2002 (5356 against 5450), the best
of the readings of this section without a constant; the flat regime of
§8 does better still. It refuses only 6.6\% of steps, the fewest of the
three. Over both sets it costs 4697 against the 4403 of reading (3),
because it gives back on the wide brackets (2346) what it gains on the
canonical problems --- so it is a second point on the same trade, not a
replacement.
\hypertarget{why-the-criterion-is-a-sign-change-six-problems-the-published-root-cannot-score}{%
\subsubsection{7.3 Why the criterion is a sign change: six problems the
published root cannot
score}\label{why-the-criterion-is-a-sign-change-six-problems-the-published-root-cannot-score}}
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, so about 1150 evaluations of every
two-set total under the harness --- ours and every competitor's --- are
this artefact, and the ``153 of 154'' and ``40 of 45'' that every method
reports under the harness are statements about the scoring rather than
about the methods. This is why every number in this paper uses the
bracketing criterion of §4, a sign change within \(10^{-6}\) of the
returned point or \(f=0\) there: under it Chandrupatla, TOMS 748, ITP,
bisection and every configuration of Table 5 solve all 199, Brent 197,
Dekker and Illinois 178 and 177.
\hypertarget{where-the-wide-bracket-limit-actually-is}{%
\subsubsection{7.4 Where the wide-bracket limit actually
is}\label{where-the-wide-bracket-limit-actually-is}}
The remaining gap looks diffuse --- per family +57, +49, +42, +39, +30
and smaller --- but part of 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; the cost is incurred 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 about one halving per evaluation while
Chandrupatla advances by about 2.3.
It is not the guarantee that caps us. (The five costs above are under
the \emph{safe} safeguard; under the projection of the base rule they
are 7, 12, 30, 46, 57.) With the safeguard removed entirely the two
widest brackets, \texttt{fun1.4} and \texttt{fun1.5}, cost 46 and 57:
better than bisection's 57 and 77, still far from Chandrupatla's 24 and
44. Unguarded, the rule is worse than bisection elsewhere --- 61 and 68
against 45 on the flat problems \texttt{fun3.1} and \texttt{fun5.1} ---
which is what the guarantee is for.
Nor is it that the step fails to be relative to the bracket. With the
tent similarity \(1-S=\min(1,|f|/(|\hat f'|D))\), so by (4) the step is
exactly \(\min(D,|f|/|\hat f'|)\): 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 on \texttt{fun1} is how much each probe removes. Ours
shrinks the bracket by 0.93 halvings per evaluation there,
Chandrupatla's by 2.30. 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; a model that carries the curvature --- inverse quadratic
interpolation --- lands far closer.
That mechanism is right for \texttt{fun1} and \emph{wrong as a statement
about the whole gap}. Replaying our own trajectories on the 35 problems
where we lose, and placing both models on the \emph{same} evaluations
--- one extra evaluation each, from the same state --- splits the 262
evaluations of the gap in two. On \texttt{fun1} (58 of them, about a
fifth) inverse quadratic interpolation from our own three points removes
far more of the bracket than our step did: 10.37 halvings against 0.62
on the \(2\times10^4\)-wide bracket. There the gap is the order of the
model, and the reason is structural: our step
\(x+(1-S)(x_\mathrm{far}-x)=S\,x+(1-S)\,x_\mathrm{far}\) is a convex
combination of two points, while the quadratic interpolant's Lagrange
weights at 0 always include exactly one \emph{negative} weight when the
triple brackets the root (verified on all 431 non-degenerate states; in
the other three the negative weight underflowed to \(-0\)). A similarity
in \([0,1]\) cannot produce a negative weight --- Proposition 1 extended
from two points to three.
But 172 of the 262 --- two thirds --- sit on
\texttt{fun3}--\texttt{fun6}, which are pure powers of order 3, 5, 9 and
19, and there \emph{no} interpolating model works: our step removes
0.41--0.75 halvings of the bracket per evaluation, inverse quadratic
interpolation from the same points only 0.01--0.17. Chandrupatla does
not win those families with his model --- his traced run bisects 92, 96,
100 and 100\% of his steps. We lose them because our step is slower than
bisection there and our guarantee, at most twice bisection, licenses
exactly that. His band inside our step confirms it family by family:
under the projection it takes \texttt{fun4}, \texttt{fun5},
\texttt{fun6} from 573, 456, 456 to 296, 239, 228 --- against his own
297, 233, 233 --- and costs +164, +111, +170, +144 on \texttt{fun1},
\texttt{fun2}, \texttt{fun8}, \texttt{fun9}, where it refuses steps that
were good.
So most of what remains is not the order of the model but the decision
\emph{when to give the model up}, and it is open because every detector
measured so far confounds ``the model is useless here'' with something
else: the band with ``the configuration is not shaped for inverse
quadratic interpolation'', bracket progress with ``the step has not
moved the bracket yet'' (4258 on the canonical set, §10), and the local
order with nothing at all --- \(m\ge2\) is exactly the condition under
which a tangent step loses to a halving, since it lands at \(e(1-1/m)\),
and it is clean (0.3\% of canonical states) but blind: it reports
\(m\ge2\) on none of the states of families whose order is by
construction 3 to 19. For the \texttt{fun1} part, three parameter-free
ways round the model order have 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 (the cube-root similarity), a logarithmic
change of variable, and the curvature-corrected scale of §10 item 5. The
fourth way is to model the curvature directly, which is inverse
quadratic interpolation --- a different method rather than another
reading of this one. Section 8 shows that the decision is reachable
without a new model: by a similarity of the run to its own past.
On the wide-bracket set our shape reading (2177) is \emph{worse} than
bisection (2136); only the dissimilarity reading (1752 under
\emph{safe}) 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 §§7.3--7.4 is reproduced by
\texttt{diag\_fun7\_root.py}, \texttt{diag\_cha\_gap.py},
\texttt{diag\_who\_caps\_us.py}, \texttt{diag\_model\_order.py},
\texttt{diag\_model\_order\_checks.py},
\texttt{diag\_flat\_detectable.py} and
\texttt{diag\_band\_by\_family.py}, whose outputs are kept beside them.
\hypertarget{the-flat-regime-recognised-by-analogy}{%
\subsection{8. The flat regime, recognised by
analogy}\label{the-flat-regime-recognised-by-analogy}}
§7.4 left one decision open: \emph{when to give the model up}. On the
pure powers \texttt{fun3}--\texttt{fun6} the right move is to halve,
Chandrupatla halves on up to all of his steps, and every reading of a
single state measured so far confuses ``the model is useless here'' with
something else. This section closes that decision, and the reading that
closes it is not a similarity of the current state to a prototype but a
similarity of the run to \emph{its own past} --- an analogy in Pólya's
sense {[}8{]}, two situations agreeing in more than one relation. Every
rule below was pre-registered before its code and its run, with a stop
rule; the rejected ones are listed in §10 with their numbers.
\hypertarget{the-flag-a-promise-that-was-not-kept}{%
\subsubsection{8.1 The flag: a promise that was not
kept}\label{the-flag-a-promise-that-was-not-kept}}
A tangent step on an approach of local order \(m\) lands at \(e(1-1/m)\)
and leaves the residual \(r=|f_\text{new}/f_\text{old}|=(1-1/m)^m\) on
the same side of the root. It beats a halving exactly when
\(|1-1/m|<\tfrac12\), that is for \(\tfrac23<m<2\). At the upper end
\(r=(1-\tfrac12)^2=\tfrac14\). So a model step that did not cross the
root and left \(r\ge\tfrac14\) is a \emph{promise not kept}: the model
is in the regime where a halving would have done better. The constant is
bisection's \(\tfrac12\), carried through the formula. The relation
between the order and the contraction, \(e(1-1/m)\), is the one behind
the classical estimates of multiplicity from successive Newton iterates
{[}14, 15{]}; here it is not used to estimate \(m\) but to set the point
at which a model step has lost to a halving.
The flag sees the regime --- on the flat families it rises on most model
steps --- but acted upon step by step it does not pay. Replacing the one
flagged step by a halving costs the canonical set 2634 → 2839, because a
simple root far from convergence also breaks its promises (the flag
rises at least once on 142 of the 154 canonical problems). Every
refinement of \emph{when} to halve that reads the moment alone --- twice
in a row, the promise together with a creep in one direction, a halving
that sticks once begun, a judgment of the evaluated point against the
line through the bracket --- trades one set for the other (§10, items
12--14).
\hypertarget{the-analogy-similarity-of-a-run-to-its-own-past}{%
\subsubsection{8.2 The analogy: similarity of a run to its own
past}\label{the-analogy-similarity-of-a-run-to-its-own-past}}
What distinguishes a pure power from a simple root is not the shape of
one configuration --- \texttt{fun1} on a wide bracket looks as
non-linear as \(x^9\) --- but what happens under zoom. A pure power
\(f=c(x-r)^m\) is similar to itself: zooming about the root by \(L\)
scales \(x-r\) by \(L\) and \(f\) by \(L^m\), so the \emph{dimensionless
configuration} of the last three points,
\[\xi=\frac{x_k-x_{k-1}}{x_{k-2}-x_{k-1}},\qquad \varphi=\frac{f_k-f_{k-1}}{f_{k-2}-f_{k-1}},\]
recurs as the iteration proceeds. At a simple root, zooming in makes
\(f\) look like its tangent, and the configuration tends to the
\emph{linear prototype} \(\varphi=\xi\) instead. Two prototypes, then:
the line, and the run's own past. The judgment is ordinal --- is the
current \((\xi,\varphi)\) nearer to some earlier configuration of this
run than to the line? --- with the plane's Euclidean distance and the
perpendicular to the line, the one geometric choice in this section,
declared as such.
Nearness of configurations is one relation. An analogy in the full sense
needs a second, and the pair supplies it: if the analogous pair shrank
\(x\) by \(L_x\) and \(f\) by \(L_f\), the power law requires
\(L_f=L_x^{\,m}\), so the order is \emph{read from the analogy itself},
\[m=\frac{\ln L_f}{\ln L_x}.\tag{6}\]
Formula (6) is the log--log slope by which any power-law exponent is
estimated, and has the shape of the computational order of convergence
and its approximation from differences of iterates {[}18, 19{]}. What is
not classical is \emph{which} two states it compares: not consecutive
iterates, but a pair the similarity of their configurations has found.
Under an exact zoom a pure power of order \(m\) reads exactly \(m\);
along a real run the analogous pairs are only nearly homothetic and the
reading scatters about the order, but the regime asks only whether it
reaches 2, the order from which a tangent step loses to a halving (§8.1)
--- no new constant. On the traces of the base rule, at the moments the
flag rises, the three conditions together hold on 3 of the 154 canonical
problems, on all 20 flat problems, and on none of \texttt{fun2},
\texttt{fun8}, \texttt{fun9}.
\textbf{The analogy belongs to the run, the flag to the moment.} In the
cycle the base rule runs on a pure power --- undershoot, undershoot,
crossing, a nearly empty step --- the flag rises in the phases whose
configuration does \emph{not} recur (the empty step grows emptier every
cycle), while the recurring phases have the flag down. Requiring both in
one state delayed the entry on the flat problems to their thirteenth
state on average. Once established, the analogy is therefore remembered
for the run, and the regime is entered at the next unkept promise
(eighth state on average).
\textbf{The translation limit.} The exponential is self-similar by
\emph{translation}, \(e^{x-1}=e^x/e\): along its flank the step does not
shrink while \(f\) does, so (6) is undefined or negative. That is the
limit \(m\to\infty\) of (6), steeper than any power, and it is read as
\(m=\infty\): when an analogous pair shows \(L_f<1\le L_x\). A high
power far from its root is locally an exponential, \(x^m=e^{m\ln x}\),
and it was there, not on the flank of \texttt{fun9} it was built for,
that the limit paid.
\textbf{Exit.} Inside the regime the step splits the bracket (§8.3). At
every split the evaluated \(f\) is judged against two predictions: the
line through the bracket ends, \(F_A(1-t)+F_Bt\) at the split fraction
\(t\), and 0. Nearer the line, the regime is left and the remembered
analogy is cleared. A sticky halving without this exit cost the
canonical set 6707; with it, 2935 --- the exit is what makes a false
entry cheap.
\hypertarget{where-to-split-the-point-from-the-order}{%
\subsubsection{8.3 Where to split: the point from the
order}\label{where-to-split-the-point-from-the-order}}
Halving is the answer when nothing is known about the shape; inside the
regime the order is known. For \(|f|\propto|x-r|^m\) the ends of the
bracket lie at distances in the ratio \(|F_A|^{1/m}:|F_B|^{1/m}\) from
the root, so the regime splits at
\[t=\frac{|F_A|^{1/m}}{|F_A|^{1/m}+|F_B|^{1/m}},\tag{7}\]
which is exact for a pure power and becomes the midpoint as
\(m\to\infty\). It is false position applied to
\(\operatorname{sign}(f)|f|^{1/m}\) --- the transformation that turns a
root of multiplicity \(m\) into a simple one, classical since Schröder
{[}16{]} and Traub {[}17{]}; what is new here is only where \(m\) comes
from. With the multiplicity given in advance the transformation is far
cheaper still (§9.1): the contribution is the estimate, and its price is
measured there. The order read from a \emph{failed promise},
\((1-1/m)^m=r\), was tried first and gave \(m=\infty\) in 163 of 174
entries, because the base step is not a pure tangent; the order read
from the \emph{analogy} is the one that works. With it the flat families
of the wide-bracket set cost 743--847 evaluations against 1044 for pure
bisection and 1068 for Chandrupatla.
\hypertarget{five-more-readings}{%
\subsubsection{8.4 Five more readings}\label{five-more-readings}}
\textbf{The half the step promises.} \(S(z)\ge\tfrac12\) means a step of
at most half the way to the far end: the model promises the root in the
half of the bracket \emph{near} \(x\). The line through the bracket ends
puts it nearer the end with the smaller \(|f|\). When the far end has
the smaller \(|f|\) the two disagree about the half, and the evaluation
halves. On a steep flank --- a slope dominated by
\(f(81)=1.5\cdot10^{35}\) on \texttt{fun9.5} --- this is the false
promise ``the root is here'', and removing it alone takes the canonical
set from 2634 to 2392. It is a relative of Brent's acceptance condition
{[}2{]}, which admits an interpolant only between the best point and
\((3a+b)/4\): both refuse a model that points towards the worse end of
the bracket.
\textbf{The sign contradiction.} The bracket says which way \(f\) must
go between its ends; the last two points say which way it goes here.
Opposite signs --- the local picture contradicts the bracket --- and the
evaluation halves. It is the cause of the \texttt{fun9} well: a
near-cancelling end reads the root as close while \(|f|\) doubles for
seventeen steps.
\textbf{The repeated step pattern.} Two successive moves the same way,
of similar length, mean the iterate creeps along a geometric series
whose remainder is \(\text{step}\cdot s/(1-s)\); the step is lengthened
by \(1/(1-s)\), with \(s\) the similarity of successive step lengths,
but only when the pattern of successive length ratios repeats with a
period read from the data (the lag of highest mean similarity) and that
similarity exceeds \(\tfrac12\).
\textbf{The overshoot branch.} §8.1's inequality has a second branch,
\(m<\tfrac23\): a root of order below one, where the tangent step
overshoots and the iterate oscillates about the root
(\(\sqrt{|x|}\operatorname{sign}x\): 84 evaluations of the base rule).
The flag for it is the unkept promise \emph{with} a sign change,
\(f_\text{new}/f_\text{old}\le-\tfrac14\), and an analogous pair reading
\(0<m\le\tfrac23\); the split (7) is again exact for a pure power. The
exit judgment of §8.2 separates a flat root from a simple one; at a root
of order below one \(f\) is far from 0 on both sides and the exit fires
falsely, so the memory of this branch is \emph{not} cleared by it;
clearing it was measured on the development corpora and was worse
(recorded in the pre-registration amendment of this branch).
\textbf{Borrowed: the tolerance step.} Once the root is reached from one
side the bracket's far end can stay far: on \(\sin\) over \([3,4]\) the
base rule reaches \(|f|\approx10^{-16}\) in 7 evaluations and then
halves the remaining \(1.3\cdot10^{-7}\) down to the tolerance in 17
more. Dekker {[}9{]} and Brent {[}2{]} step at least the tolerance; we
do the same, \(\varepsilon_x/2\) towards the far end, when the model's
step is shorter --- but only after the model has \emph{kept} its
previous promise (\(r<\tfrac14\)), because at the start of a run a slope
dominated by a huge far value promises the root at an end. This element
is not ours and is named as such.
\hypertarget{the-proposed-method}{%
\subsubsection{8.5 The proposed method}\label{the-proposed-method}}
The method this paper proposes, \textbf{A′}, is the base rule of §3 ---
tent similarity, three-point scale, projection with the declared
multiplier \(c=2\) --- with the regime of §8.2--8.3 (the analogy
remembered, its translation limit, the split (7), the exit) and the five
readings of §8.4. Its constants are bisection's \(\tfrac12\) and what
follows from it: \(\tfrac14=(1-\tfrac12)^2\) and the two branches
\(m\ge2\), \(m\le\tfrac23\) of \(|1-1/m|>\tfrac12\). Three choices are
declared rather than derived: the multiplier \(c\) of the projection
(§3.1), the Euclidean geometry of the \((\xi,\varphi)\) plane, and the
pattern test of the step lengthening (lags from 2 to half the history,
threshold \(\tfrac12\)). The projection carries the guarantee,
\(2\lceil\log_2(w_0/\varepsilon_x)\rceil+3\) evaluations; it is verified
with no violation on every corpus of this paper and under every
safeguard of the grid.
\textbf{Table 8.} A′ on the two sets of §4 (failures charged 200). A′'s
wins/ties/losses against each rival over all problems of the set, exact
two-sided sign test.
\begin{longtable}[]{@{}
>{\raggedright\arraybackslash}p{(\columnwidth - 10\tabcolsep) * \real{0.1667}}
>{\raggedright\arraybackslash}p{(\columnwidth - 10\tabcolsep) * \real{0.1667}}
>{\raggedright\arraybackslash}p{(\columnwidth - 10\tabcolsep) * \real{0.1667}}
>{\raggedright\arraybackslash}p{(\columnwidth - 10\tabcolsep) * \real{0.1667}}
>{\raggedright\arraybackslash}p{(\columnwidth - 10\tabcolsep) * \real{0.1667}}
>{\raggedright\arraybackslash}p{(\columnwidth - 10\tabcolsep) * \real{0.1667}}@{}}
\toprule\noalign{}
\begin{minipage}[b]{\linewidth}\raggedright
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
APS-154
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
CHA-45
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
both
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
A′ per problem, APS
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
A′ per problem, CHA
\end{minipage} \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
\textbf{A′ (proposed)} & \textbf{2259} & \textbf{1074} & \textbf{3333} &
--- & --- \\
A′, earned budget (no free constant) & 3417 & 1852 & 5269 & & \\
base rule (§3) & 2634 & 2357 & 4991 & & \\
Chandrupatla & 2595 & 1505 & 4100 & 103/20/31, \(p=3\cdot10^{-10}\) &
31/4/10, \(p=0.001\) \\
TOMS 748 & 2636 & 3088 & 5724 & 127/6/21 & 42/2/1 \\
Brent & 2707 & 3238 & 5945 & 118/17/19 & 39/3/3 \\
ITP, per-problem oracle & 3068 & 1675 & 4743 & 126/7/21 & 35/1/9 \\
ModAB {[}10, 20{]} & 2726 & 1681 & 4407 & 118/10/26 & 34/4/7 \\
\end{longtable}
A′ is 12.9\% below Chandrupatla on the canonical set and 28.6\% below on
the wide-bracket set, and wins more problems than it loses against every
rival on both. By function family, the unit in which these problems are
independent (§9), it is ahead of Chandrupatla on 8 of the 15 canonical
families (6 behind, one level) and on 7 of the 9 wide-bracket families
--- too few families for either count to be significant (\(p=0.79\),
\(0.18\)). The two it is behind on are \texttt{fun7}, 69 against 64, and
\texttt{fun9}, 65 against 55. The earned-budget form keeps its
constant-free guarantee of bisection plus one step and pays for it as in
§3.1.
\textbf{Table 9.} Leave-one-out ablation of A′ on all six corpora of §9
(uniform criterion, a sign change within \(10^{-6}\)). Δ = total without
the component − total with it; positive means the component earns its
place.
\begin{longtable}[]{@{}
>{\raggedright\arraybackslash}p{(\columnwidth - 14\tabcolsep) * \real{0.1250}}
>{\raggedright\arraybackslash}p{(\columnwidth - 14\tabcolsep) * \real{0.1250}}
>{\raggedright\arraybackslash}p{(\columnwidth - 14\tabcolsep) * \real{0.1250}}
>{\raggedright\arraybackslash}p{(\columnwidth - 14\tabcolsep) * \real{0.1250}}
>{\raggedright\arraybackslash}p{(\columnwidth - 14\tabcolsep) * \real{0.1250}}
>{\raggedright\arraybackslash}p{(\columnwidth - 14\tabcolsep) * \real{0.1250}}
>{\raggedright\arraybackslash}p{(\columnwidth - 14\tabcolsep) * \real{0.1250}}
>{\raggedright\arraybackslash}p{(\columnwidth - 14\tabcolsep) * \real{0.1250}}@{}}
\toprule\noalign{}
\begin{minipage}[b]{\linewidth}\raggedright
component removed
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
APS
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
CHA
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
new-1
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
Roots.jl
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
applied
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
fourth
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
sum
\end{minipage} \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
the whole regime & −7 & +1013 & +1660 & +324 & +77 & +2269 &
\textbf{+5336} \\
the split (7) (midpoint instead) & −5 & +479 & +667 & +272 & +30 & +1672
& \textbf{+3115} \\
overshoot branch & −6 & 0 & −7 & +16 & +63 & +893 & +959 \\
tolerance step (borrowed) & +130 & +35 & +97 & +78 & +135 & −2 & +473 \\
the half the step promises & +249 & +65 & +82 & −16 & −5 & +1 & +376 \\
translation limit & +31 & +143 & +100 & −21 & −2 & +87 & +338 \\
repeated step pattern & −22 & −6 & +123 & −30 & −10 & +34 & +89 \\
sign contradiction & +11 & −12 & +29 & 0 & +1 & −1 & +28 \\
\end{longtable}
Every component is positive in sum; the regime and the split carry the
method. Two are marginal and we say so: the step pattern earns on two
corpora of six and loses on four, and the sign contradiction is within
noise (+28 over 855 problems). Removing both would change A′'s total by
under one per cent (117 of 13032 evaluations); they are kept because the
pre-registered rule removes only a component that earns nowhere.
\textbf{Table 9a.} The similarity in the step of A′: the same method
with each of the nine similarity functions of §5.3 in place of the tent,
everything else unchanged. Totals on the two sets of §4 and over the six
corpora of §9; the guarantee holds with every function.
\begin{longtable}[]{@{}lll@{}}
\toprule\noalign{}
\(S\) in the step & APS + CHA & six corpora (855) \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
\textbf{tent \(1-|z|\) (A′)} & \textbf{3333} & \textbf{13032} \\
\(e^{-|z|}\) & 3965 & 14142 \\
\(1/(1+|z|)\) & 4424 & 14843 \\
\((1+|z|)^{-2}\) & 7249 & 28500 \\
\(1-\sqrt{|z|}\) & 11613 & 45552 \\
\(1-|z|^{1/3}\) & 16094 & 63031 \\
\(e^{-z^2}\) & 17429 & 67701 \\
\(1/(1+z^2)\) & 17452 & 67394 \\
\(1/\sqrt{1+z^2}\) & 17487 & 67854 \\
\end{longtable}
The nearest other function costs 19\% more on the two sets and 9\% more
over the six corpora, and the functions that approach 1 quadratically
cost five times more; the three functions with a linear approach come
first and the three with a quadratic approach last, as the contraction
law of §2.2 predicts. What this measures should be said plainly: with
the tent the step is the undamped Newton step clipped to the bracket,
and the other functions damp it, so the table shows that the undamped
step is the better one --- expected, and evidence for the choice of
\(S\) within the rule rather than for the similarity framing as such.
\hypertarget{what-is-known-and-what-is-new}{%
\subsubsection{8.6 What is known, and what is
new}\label{what-is-known-and-what-is-new}}
Several mechanisms of A′ coincide with classical ones, and the
coincidence is worth stating, because they were not taken from the
literature but reached from the principles of the paper --- similarity
to a prototype and analogy as similarity of relations --- and recognised
afterwards:
\begin{longtable}[]{@{}
>{\raggedright\arraybackslash}p{(\columnwidth - 2\tabcolsep) * \real{0.5000}}
>{\raggedright\arraybackslash}p{(\columnwidth - 2\tabcolsep) * \real{0.5000}}@{}}
\toprule\noalign{}
\begin{minipage}[b]{\linewidth}\raggedright
in A′
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
classical counterpart
\end{minipage} \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
the three-point order estimate of §7.1 & Lagouanelle's multiplicity
estimate \(m=f'^2/(f'^2-ff'')\) {[}13{]} \\
the contraction \(e(1-1/m)\) behind the flag of §8.1 & multiplicity
estimates from successive Newton iterates {[}14, 15{]} \\
the order (6) of an analogous pair & the log--log slope of a power law;
the (approximate) computational order of convergence {[}18, 19{]} \\
the split (7) & false position on \(\operatorname{sign}(f)|f|^{1/m}\)
{[}16, 17{]} \\
the half the step promises & a relative of Brent's acceptance condition
{[}2{]} \\
the tolerance step & Dekker {[}9{]}, Brent {[}2{]} --- borrowed and
named as such \\
the exit judgment of §8.2 (the value at a split against the line) &
ModAB's linearity check at the midpoint {[}10, 20{]},
\(|y_m-y_3|<k\,(|y_3|+|y_m|)\) with \(k\) from a symmetry factor; ours
compares two predictions, the line and 0, without a factor \\
\end{longtable}
What we found no counterpart for, in the literature we could reach, is
the part that carries the gain in Table 9: the \emph{decision} layer. A
bracketing method that recognises the regime in which its model has lost
by the similarity of the run to its own past --- the recurrence of a
dimensionless configuration against the linear prototype --- reads the
order from the analogous pair itself, distinguishes similarity by
scaling from similarity by translation, and leaves the regime by a
similarity judgment at a point the model did not choose; and a step
whose length is set by a similarity function whose form, measured, is
decisive. The methods for multiple roots we found are single-point
iterations without a bracket, with the multiplicity assumed, estimated
from derivatives, or removed by a transformation ({[}17{]}; Roots.jl's
\texttt{Order1B} and \texttt{Order2B} {[}12{]}); Chandrupatla takes the
decision to halve by an empirical acceptance band {[}3{]}. §9.1 puts
this claim against the detectors a reader would try first, and against
the transformation with the multiplicity known in advance.
\hypertarget{out-of-sample-evidence}{%
\subsection{9. Out-of-sample evidence}\label{out-of-sample-evidence}}
Every decision above was taken on the two sets of §4. Pre-registration
protects each decision but not the chain: the final configuration --- a
decision layer of about ten discrete rules, chosen among the many
candidates listed in §10 --- was selected on the same 199 problems. Four
further corpora were therefore used, each fixed and committed before any
method was run on it, each with its criteria and stop rule written down
first.
\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
\textbf{New corpus 1} --- 206 problems: SciPy's \texttt{original} set
(6) and five generated classes of 40 (seed 20260921): smooth simple
roots, transcendental simple roots, multiple roots \(|x-r|^p\) with
\(p\sim U(2,20)\), exponential flanks, and near-cancelling terms of
the \texttt{fun9} kind.
\item
\textbf{Roots.jl / ModAB} --- the published bracketing benchmark of
the Julia package Roots.jl {[}12{]}, taken from the ModAB benchmark of
Ganchovski and Traykov {[}10{]}: 91 problems, of which 13 coincide
with APS or Chandrupatla functions (a seven-point identity test) and
are excluded, leaving 78, many of them discontinuous or pathological.
\item
\textbf{Applied and GSL} --- the 10 bracketed problems of the GNU
Scientific Library's root-finding test suite {[}11{]} and 122
equations from physics and engineering on grids fixed beforehand:
Kepler's equation (eccentricity up to 0.999), Colebrook--White,
Lambert's W, Planck/Wien, van der Waals for CO₂ (including
near-critical), internal rate of return, the finite square well.
\item
\textbf{Corpus 4} --- 240 problems in six new families (seed
20260922): random polynomials with a simple root, polynomials with a
multiple root, fractional powers \(|x-r|^p\) with
\(p\sim U(0.15,0.95)\), sums of exponentials and sines, compositions
\(\tanh\), \(\arctan\), \(\log(1+q^2)\) of a cubic, and functions odd
about the root. The code of A′ was committed \emph{before} this corpus
was run, and nothing was changed after. The families were chosen by
the author, who knew by then where the method is strong: two of the
six (multiple roots, fractional powers) are the regime §8 was built
for.
\end{enumerate}
A root must carry a sign change to be bracketed, so the multiple roots
in these corpora have odd multiplicity (3, 5, 7, 9 in corpus 4) or the
form \(\operatorname{sign}(x-r)|x-r|^p\) (corpora 1 and 4).
ModAB, the most recent bracketing method we know of, is run from its
authors' package (\texttt{pymodab} 1.0.9, a wrapper of their C code)
with the stopping rule of every other method --- a bracket no wider than
\(\varepsilon_x\), the relative term at machine precision, no stopping
on \(|f|\) --- and counted by the same wrapper around \(f\), whose count
agrees with the package's own. Where roots are not known in closed form,
and for uniformity on all six corpora, a run is solved when \(f\)
changes sign within \(10^{-6}\) of its answer; the \(x\)-tolerance, the
budget and the charge for a failure are those of §4.
\textbf{The history is part of the result.} The variant carried to
corpora 1--3 was A′ without the tolerance step and the overshoot branch.
It transferred to corpus 1 by every pre-registered criterion (total 10\%
below Chandrupatla, per problem 105/34/67, \(p=0.005\)), did \emph{not}
transfer to Roots.jl (total 6.5\% below, but per problem 38/10/30,
\(p=0.4\)) and lost on the applied corpus (14\% above). The losses there
were traced to two gaps of the base step --- no closing of the bracket
once the root is reached, and no recognition of roots of order below one
--- which became the last two readings of §8.4; they were developed on
the seen corpora and tested once, on corpus 4.
\textbf{Table 10.} A′ against the library rivals on all six corpora
(uniform criterion): total evaluations; in brackets the runs that break
the \(2\times\)-bisection bound.
\begin{longtable}[]{@{}
>{\raggedright\arraybackslash}p{(\columnwidth - 14\tabcolsep) * \real{0.1942}}
>{\raggedright\arraybackslash}p{(\columnwidth - 14\tabcolsep) * \real{0.1068}}
>{\raggedright\arraybackslash}p{(\columnwidth - 14\tabcolsep) * \real{0.1650}}
>{\raggedright\arraybackslash}p{(\columnwidth - 14\tabcolsep) * \real{0.1068}}
>{\raggedright\arraybackslash}p{(\columnwidth - 14\tabcolsep) * \real{0.1068}}
>{\raggedright\arraybackslash}p{(\columnwidth - 14\tabcolsep) * \real{0.1068}}
>{\raggedright\arraybackslash}p{(\columnwidth - 14\tabcolsep) * \real{0.1068}}
>{\raggedright\arraybackslash}p{(\columnwidth - 14\tabcolsep) * \real{0.1068}}@{}}
\toprule\noalign{}
\begin{minipage}[b]{\linewidth}\raggedright
corpus
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
A′
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
Chandrupatla
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
ModAB
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
ITP
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
ITP oracle
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
Brent
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
TOMS 748
\end{minipage} \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
APS (154) & \textbf{2259} (0) & 2595 (0) & 2726 (0) & 3767 (0) & 3068
(0) & 2707 (0) & 2636 (0) \\
Chandrupatla (45) & \textbf{1074} (0) & 1505 (0) & 1681 (0) & 1946 (0) &
1675 (0) & 3238 (18) & 3088 (15) \\
new corpus 1 (206) & \textbf{3290} (0) & 3716 (0) & 3809 (1) & 5674 (0)
& 4707 (0) & 6395 (38) & 5746 (22) \\
Roots.jl (78) & \textbf{1371} (0) & 1566 (0) & 1446 (0) & 2161 (0) &
1629 (0) & 2455 (11) & 2339 (9) \\
applied + GSL (132) & 1382 (0) & 1382 (0) & 1460 (0) & 3245 (0) & 2386
(0) & 1503 (1) & 1756 (1) \\
\textbf{corpus 4 (240), the one test} & \textbf{3656} (0) & 5717 (0) &
5535 (8) & 5721 (0) & 4832 (0) & 8267 (50) & 8084 (37) \\
all 855 & \textbf{13032} & 16481 & 16657 & 22514 & 18297 & 24565 &
23649 \\
\end{longtable}
\textbf{Table 10b.} The same runs problem by problem: A′'s
wins/ties/losses against Chandrupatla and against ModAB (exact two-sided
sign test).
\begin{longtable}[]{@{}
>{\raggedright\arraybackslash}p{(\columnwidth - 4\tabcolsep) * \real{0.3333}}
>{\raggedright\arraybackslash}p{(\columnwidth - 4\tabcolsep) * \real{0.4545}}
>{\raggedright\arraybackslash}p{(\columnwidth - 4\tabcolsep) * \real{0.2121}}@{}}
\toprule\noalign{}
\begin{minipage}[b]{\linewidth}\raggedright
corpus
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
A′ vs Chandrupatla
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
A′ vs ModAB
\end{minipage} \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
APS (154) & 103/20/31, \(p=3\cdot10^{-10}\) & 118/10/26 \\
Chandrupatla (45) & 31/4/10, \(p=0.001\) & 34/4/7 \\
new corpus 1 (206) & 108/33/65, \(p=0.001\) & 125/26/55 \\
Roots.jl (78) & 44/14/20, \(p=0.004\) & 47/11/20 \\
applied + GSL (132) & 48/52/32, \(p=0.09\) & 76/35/21 \\
\textbf{corpus 4 (240), the one test} & \textbf{177/35/28,
\(p=10^{-27}\)} & \textbf{183/23/34} \\
\end{longtable}
Only the last line of corpora is evidence for A′ in the strict sense;
the first five were seen while it was developed and are reported as a
description. ITP, at its recommended constants and as the per-problem
oracle, keeps its own bound of bisection plus \(n_0\) everywhere and is
behind A′ on every corpus.
\textbf{Independence.} A sign test counts problems as independent draws,
and not all of them are. In the two canonical sets one function appears
with several brackets or parameters, so the independent unit is the
function family (15 in APS, 9 in CHA); at that level the counts of Table
8 are not significant, because there are too few families. In corpora 1
and 4 every problem is drawn independently from its family's generator,
so the per-problem test is valid --- for the population those generators
define, which the author chose (above). On corpus 4 A′ is 36\% below
Chandrupatla and wins 177 problems to 28. Without the two classes aimed
at by the last two readings it is 2403 against 3569, per problem
107/31/22 (\(p=10^{-14}\)); on the simple-root classes alone, 737
against 747 (−1.3\%), per problem 43/22/15 (\(p=3\cdot10^{-4}\)).
\textbf{Table 11.} Corpus 4 by class: totals of A′ / Chandrupatla, per
problem wins/ties/losses.
\begin{longtable}[]{@{}llll@{}}
\toprule\noalign{}
class & A′ & Chandrupatla & per problem \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
random polynomial, simple root & 433 & 419 & 20/8/12 \\
polynomial, multiple root & 1163 & 1805 & 39/1/0 \\
fractional power, \(p<1\) & 924 & 1784 & 37/1/2 \\
sums of exponentials and sines & 304 & 328 & 23/14/3 \\
\(\tanh\), \(\arctan\), \(\log(1+q^2)\) of a cubic & 503 & 1017 &
25/8/7 \\
odd about the root & 329 & 364 & 33/3/4 \\
\end{longtable}
\textbf{What the evidence supports, and what it does not.} Against Brent
and TOMS 748 A′ is ahead on every corpus, by 8\% to 67\% in total, and
it stays within its own bound of twice bisection on all 855 problems, a
bound that Brent exceeds on 118 and TOMS 748 on 84 --- neither promises
it. Against ModAB it is ahead on every corpus, in total and problem by
problem (\(p\le0.0013\)), on the one test of corpus 4 by 3656 to 5535
and 183 problems to 34 (\(p=7\cdot10^{-26}\)), and on ModAB's own
benchmark as well --- all 91 problems of it, 1507 against 1588, 56/13/22
(\(p=1.5\cdot10^{-4}\)); ModAB exceeds that bound, which it does not
promise either, on 9 problems. Against Chandrupatla it is ahead in total
on five corpora and level on the sixth, and ahead problem by problem on
all six, significantly on five. The size of the lead is not uniform: it
is large on multiple, fractional and nearly flat roots --- the regime §8
recognises --- and small on simple roots, where it is a matter of per
cent. Where it is behind is named: \texttt{fun9} (65 against 55), the
near-cancelling class of corpus 1 (491 against 391), random polynomials
with a simple root in total (433 against 419, though ahead 20 to 12 in
problems), and the smooth equations of the applied corpus, where the two
are level.
\hypertarget{controls-simpler-detectors-and-the-multiplicity-known-in-advance}{%
\subsubsection{9.1 Controls: simpler detectors, and the multiplicity
known in
advance}\label{controls-simpler-detectors-and-the-multiplicity-known-in-advance}}
Two controls ask whether the gain of §8 comes from the recognition by
analogy or only from the split (7) with any reasonable detector. Both
were pre-registered, with what the paper would say under every outcome,
before the run (\texttt{sweep\_controls.py}).
\textbf{Other detectors in A′.} A′ is run with only its detector
swapped; the memory, the split, the exit and the five readings of §8.4
are unchanged. (a) \emph{Lagouanelle's estimate} \(m_L=1/(1-\theta)\),
\(\theta=ff''/f'^2\) from the parabola through the last three points
(\(m_L=\infty\) for \(\theta\ge1\)), with the same memory --- once seen,
held until the exit, a hysteresis --- entering at \(m_L\ge2\) and at
\(0<m_L\le\tfrac23\) and splitting at \(m_L\). (b) \emph{Chandrupatla's
acceptance band}, entering when the band rejects the current triple and
splitting at the current \(m_L\). (c) \emph{No detector}: the regime is
entered at every unkept promise.
\textbf{Table 9b.} A′ against its detector swapped: totals on the two
canonical sets, on corpus 4 and over the six corpora of §9 (uniform
criterion), and A′'s wins/ties/losses on corpus 4 with the exact sign
test. No variant breaks the bound.
\begin{longtable}[]{@{}
>{\raggedright\arraybackslash}p{(\columnwidth - 12\tabcolsep) * \real{0.3291}}
>{\raggedright\arraybackslash}p{(\columnwidth - 12\tabcolsep) * \real{0.0759}}
>{\raggedright\arraybackslash}p{(\columnwidth - 12\tabcolsep) * \real{0.0759}}
>{\raggedright\arraybackslash}p{(\columnwidth - 12\tabcolsep) * \real{0.1013}}
>{\raggedright\arraybackslash}p{(\columnwidth - 12\tabcolsep) * \real{0.1139}}
>{\raggedright\arraybackslash}p{(\columnwidth - 12\tabcolsep) * \real{0.1392}}
>{\raggedright\arraybackslash}p{(\columnwidth - 12\tabcolsep) * \real{0.1646}}@{}}
\toprule\noalign{}
\begin{minipage}[b]{\linewidth}\raggedright
detector
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
APS
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
CHA
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
corpus 4
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
six corpora
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
A′ on corpus 4
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
\(p\)
\end{minipage} \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
\textbf{analogy (A′)} & 2259 & \textbf{1074} & \textbf{3656} &
\textbf{13032} & --- & \\
Lagouanelle, with memory & \textbf{2235} & 1437 & 4488 & 15171 &
84/121/35 & \(8\cdot10^{-6}\) \\
Chandrupatla's band & 2682 & 2526 & 6236 & 20668 & 110/96/34 &
\(1.5\cdot10^{-10}\) \\
none & 2693 & 2526 & 6165 & 20749 & 113/87/40 & \(3\cdot10^{-9}\) \\
\end{longtable}
On corpus 4, the one test, A′ beats all three in total and problem by
problem, as the pre-registered criterion required. The competent
alternative is Lagouanelle's estimate: it is slightly better on the
canonical set (2235 against 2259), and on corpus 4 A′'s lead over it
lies in the multiple-root and fractional-power families (1163 against
1739, 924 against 1207); it is behind on \(\log(1+q^2)\) of a cubic (503
against 477) and level on the rest. The band and the regime without a
detector are worse than no regime at all. The decision layer by analogy
therefore earns its place against the detectors a reader would try
first.
\textbf{The multiplicity known in advance.} The second control asks how
good the classical transformation is when its multiplicity is given.
Library Chandrupatla and Brent are run on
\(g=\operatorname{sign}(f)\,|f|^{1/m^*}\), with \(m^*\) the order read
at the known root --- the slope of \(\ln|f|\) against \(\ln|x-r|\) on
both sides at \(h=10^{-5}\), rounded to 0.01, and 1 where either side
shows no zero (a jump). It is not an achievable method but an upper
bound. On the multiple and fractional roots, \(m^*\ge2\) or
\(m^*\le\tfrac23\) --- the branches of §8.1 where a tangent step loses
to a halving:
\textbf{Table 9c.} Multiple and fractional roots with the multiplicity
given: totals, failures charged. A′ wins none of the 148 problems.
\begin{longtable}[]{@{}
>{\raggedright\arraybackslash}p{(\columnwidth - 10\tabcolsep) * \real{0.1667}}
>{\raggedright\arraybackslash}p{(\columnwidth - 10\tabcolsep) * \real{0.1667}}
>{\raggedright\arraybackslash}p{(\columnwidth - 10\tabcolsep) * \real{0.1667}}
>{\raggedright\arraybackslash}p{(\columnwidth - 10\tabcolsep) * \real{0.1667}}
>{\raggedright\arraybackslash}p{(\columnwidth - 10\tabcolsep) * \real{0.1667}}
>{\raggedright\arraybackslash}p{(\columnwidth - 10\tabcolsep) * \real{0.1667}}@{}}
\toprule\noalign{}
\begin{minipage}[b]{\linewidth}\raggedright
corpus
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
problems
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
A′
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
Chandrupatla on \(g\)
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
Brent on \(g\)
\end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright
Chandrupatla on \(f\)
\end{minipage} \\
\midrule\noalign{}
\endhead
\bottomrule\noalign{}
\endlastfoot
Chandrupatla & 20 & 654 & 96 & 66 & 1068 \\
new corpus 1 & 40 & 1229 & 358 & 337 & 1792 \\
corpus 4 & 88 & 2159 & 901 & 844 & 4151 \\
\end{longtable}
With the multiplicity known the transformation is 2.4 to 6.8 times
cheaper than A′. A′ is far ahead of Chandrupatla on \(f\) itself, which
does not know the multiplicity either. On the remaining problems the
transformed and untransformed runs differ little and A′ is close to
them: ahead on the canonical sets (2259 against 2640, 420 against 437
for Chandrupatla on \(g\)), behind on corpora 1 and 4 (2061 against
1924, 1497 against 1427). So on multiple roots the contribution of A′ is
the estimate of the order without knowing it, and the gap to Table 9c is
the price of that estimate --- large, and the clearest direction for
improving the method.
\begin{center}\rule{0.5\linewidth}{0.5pt}\end{center}
\hypertarget{negative-results}{%
\subsection{10. Negative results}\label{negative-results}}
\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\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 make the three-point
rule converge linearly and cost 7068 and 7046 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 not a degenerate case.} A
worst case of exactly bisection might be expected to leave the
projection no freedom and make the guard behave as bisection. Measured
(Table 1a), multiplier 1 costs 4625 and 2039, against bisection's 7338
and 2136 --- and on the wide-bracket set it beats the multiplier of 2.
It is the most expensive multiplier on the canonical set.
\item
\textbf{Regimes exist, at the level of a function family, and cannot
be exploited by a switch.} The cube-root similarity beats the base
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 3204 against 2357, so a switch chosen per
benchmark is worse than no switch at all: 5838 against 4991 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
problem family, and the winners are not separated by anything the
method can see before it commits --- the derived 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 --- does produce 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 --- which is Lagouanelle's
multiplicity estimate {[}13{]}. It improves the eight standard
problems (82 against 84) --- and loses on the full benchmark, 5336
against 4991. 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 convergence becomes
sublinear and the standard problems double to 161.
\item
\textbf{The harness scores one canonical problem unsolved for 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 point where
\(f(x)=0\) exactly --- a genuine root --- which the harness scores as
a failure because it lies more than \(10^{-6}\) from the declared
root. With the five \texttt{fun7} problems of §7.3 it is why this
paper scores by a sign change.
\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 1, last row).
\item
\textbf{Placement beats form, and the wrong placement is expensive.}
The same ``refuse before the step'' idea as reading (1) of §7, with
refusal decided by \emph{promised bracket progress} instead of by the
shape of the model, costs 4258 on the canonical set against 2337 --- a
difference of 1921 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 of §3.1 cost 7068.
\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 \(\sigma=|\hat f'|D\)
the normalised residual is \(|f|/(|\hat f'|D)\), 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,
\(\sigma=\min(|\hat f'|D,\max(|f(A)|,|f(B)|))\) --- no new constant, a
minimum 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: its best cell is 3101 on the canonical set against
2539 for the best uncapped cell of the same grid, and 1766 on the
wide-bracket set against 1752.
\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 (§7.1), 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 far-end reading from 127/154 and
10676 to 129/154 and 9855 --- two problems and eight 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 2 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 looked like the right diagnosis of the flat
families --- for the same similarity, scale and safeguard
(\((1+|z|)^{-2}\), two-point, \emph{safe}) the wide-bracket set
improves from 2759 to 2598 --- but the canonical set gets worse, 5635
to 6749, against the 2634 of the base rule. 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.
§7.4 shows the diagnosis itself was wrong: on the flat families the
right move is to halve, not to march.
\item
\textbf{Reading the unkept promise one step at a time does not pay.}
The flag of §8.1 sees the flat regime, but a halving in place of each
flagged step takes the canonical set from 2634 to 2839 and the
wide-bracket set only from 2357 to 2243: a simple root far from
convergence breaks its promises too. Requiring the promise to fail
twice in a row keeps the canonical set (2613) but fires one step late,
before the crossing, and replaces the most useful step of the cycle
(wide brackets 2388); requiring a creep in one direction as well gives
2688 and 2399.
\item
\textbf{A halving that sticks once begun wins the flat families and
loses everything else.} Without an exit the flat families fall to 1057
--- pure bisection's level --- and the canonical set rises to 6707,
because a false entry now lasts to the end of the run. Splitting at
the order read from the failed promise, \((1-1/m)^m=r\), changes
nothing: that order is \(m=\infty\) in 163 of 174 entries (6591).
\item
\textbf{The model's own point cannot tell a flat root from a simple
one.} Judging the evaluated point against the line through the bracket
ends and against 0 --- the exit judgment of §8.2 --- also at
\emph{entry} gives 2820 and 1532 against 2935 and 1546 for the exit
alone: the model aims at the root, so \(f\) is small there by
construction and looks flat at a simple root as readily as at a power.
The same judgment works at a split, which the model did not choose.
\item
\textbf{The decision reading ``defect'' does not transfer.} Combined
with the regime it was the best configuration on the two development
sets (§7.2's defect reading with the flat families of §8), and it
passed every pre-registered criterion there. On new corpus 1 it did
not: per problem 95/33/78 against Chandrupatla (\(p=0.22\)), losing
the class of exponential flanks 9 to 26. It is therefore not part of
A′.
\item
\textbf{A component rejected alone can pay in combination.}
Remembering the analogy for the run was rejected by its own
pre-registration, under the harness scoring of the time (canonical set
2814 against 2806 without the regime; under the sign-change criterion
the two are level, 2633 against 2634), and with the split of §8.3 it
is the better form (with the step pattern, wide brackets 1250 against
1388 for the analogy read in one state). Composition was measured, not
assumed, throughout --- which is also why §7's readings (2) and (3),
each an improvement, are not combined.
\end{enumerate}
\begin{center}\rule{0.5\linewidth}{0.5pt}\end{center}
\hypertarget{conclusion}{%
\subsection{11. Conclusion}\label{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 which group of
them comes first 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 projection's budget
multiplier is a free constant, and we declare it: it does not move the
canonical result and it does move the wide-bracket one; projecting onto
a budget the steps earn instead removes it, at a measured price.
The base rule has one declared numerical constant, converges with
Muller's order 1.839 and carries a worst case of twice bisection plus
one step. On the canonical Alefeld--Potra--Shi set it is level with TOMS
748 in total and 1.5\% behind Chandrupatla's method, while winning more
problems than it loses against both; on the wide-bracket set
Chandrupatla is genuinely stronger (1505 against 2357). §7.4 locates the
cause: not the order of the model but the decision to give it up, which
Chandrupatla takes by halving and which no reading of a single state
takes correctly for our step.
That decision is a similarity too --- of the run to its own past. A pure
power repeats its configuration under zoom, an exponential under
translation, a simple root tends to the line; the analogous pair reads
the local order, the order decides between the model and a halving by
bisection's own \(\tfrac12\), and it says where to split. The method
built on it, A′, is ahead of Chandrupatla on both canonical sets in
total and problem by problem, and on a corpus fixed after its code was
frozen it needs 36\% fewer evaluations and wins 177 problems to 28. Its
decision layer beats Lagouanelle's estimate, Chandrupatla's band and no
detector put in its place; with the multiplicity known in advance the
classical transformation is 2.4 to 6.8 times cheaper on multiple roots,
which measures what estimating the order costs. The claim the
measurements support is a lead over Brent, TOMS 748, ITP and ModAB
everywhere, within a bound of twice bisection that Brent, TOMS 748 and
ModAB do not promise and exceed, and over Chandrupatla a lead that is
large where roots are multiple, fractional or nearly flat and small
where they are simple --- not superiority by the same margin everywhere.
\texttt{fun9}, near-cancelling terms and smooth applied equations are
where it is behind or level.
This is the same movement as in parameter-free similarity control
{[}6{]}, 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.
\hypertarget{future-work}{%
\subsubsection{Future work}\label{future-work}}
Three questions remain, all narrower than the one this section used to
pose.
The first is the price of estimating the order. With the multiplicity
given, the transformation of §8.3 is 2.4 to 6.8 times cheaper than A′ on
multiple and fractional roots (Table 9c): the order read from the
analogous pair, or its use in the split, loses most of what the
transformation could deliver.
The second is the steep flank. Where the loss remains --- \texttt{fun9},
the near-cancelling class, 4 − tan x at its pole --- the base step
approaches along a flank far steeper than its model, and the half
contradiction of §8.4 removes only part of the cost. The third is the
one geometric choice of §8.2, the Euclidean metric of the
\((\xi,\varphi)\) plane: a metric derived from the configuration's own
sensitivity would remove it.
\begin{center}\rule{0.5\linewidth}{0.5pt}\end{center}
\hypertarget{references}{%
\subsection{References}\label{references}}
\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
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\end{document}