579 lines
83 KiB
Markdown
579 lines
83 KiB
Markdown
# Similarity as a Step Rule: Parameter-Free Root Finding Derived from a Similarity Function
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V. Sitnikov
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---
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## Abstract
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Bisection places the next iterate at the midpoint because it uses no information about the function's value. We replace that rule with a *similarity* rule: the step is proportional to how *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 *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 *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; on the eight standard problems of §2.2, five to eight of them fail. 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.
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The derivation fixes the method 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. What results is, explicitly, a Newton step with a three-point divided-difference derivative, clipped to the bracket and projected onto the bisection budget; the similarity function is scaffolding of the derivation and does not appear in the final rule. It carries a proven worst case of twice bisection plus one step, and exactly one free constant, which we declare rather than argue away: the multiplier of that budget. On the canonical set it does not matter — the total is the same for every multiplier from 1.25 to 4 — and on the wide-bracket set it does, moving the total from 2752 to 4006. On the 154 problems of Alefeld, Potra and Shi the rule needs 2806 evaluations of $f$ against 2820 for TOMS 748 — a margin of 0.5%, which is a tie — 2889 for Brent, and 2787 for Chandrupatla's method, 0.7% fewer in total; problem by problem it beats Chandrupatla 87 to 42 (exact sign test $p=9\cdot10^{-5}$) and Brent 102 to 40. On the 45 problems of Chandrupatla it beats Brent 30 to 4 and TOMS 748 31 to 8, and loses to ITP and to Chandrupatla (3220 against 2857 and 2441). A variant whose budget is earned step by step has no free constant and a tighter guarantee, bisection plus one step, at a price of 762 evaluations on the canonical set. For every data-derived scale, the ranking of the nine similarity functions matches the derived law without exception.
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Reading the same similarity in one more place — which three points the local model is built on, selected by the shape of their configuration rather than by recency — lowers the Alefeld–Potra–Shi count to 2527, below every method measured here including Chandrupatla's 2787; reading it instead on the choice of far end gives 5511 over both sets, and reading it in the decision to trust the model or halve lowers the canonical count further, to 2523. On the canonical set these records barely feel the multiplier — 2523 is the same for every multiplier from 1.25 to 4, 2527 for every one from 1.5 to 4 and 2541 at 1.25; under the constant-free earned budget the same readings gain far less (6494 over both sets against 6588). Where the similarity is read turns out to matter more than which similarity it is. The gap that remains on the wide-bracket set is taken apart on the same evaluations: about a quarter of it is the order of the model — a convex two-point step cannot carry the negative weight of quadratic interpolation — and two thirds is not a model at all but the decision when to give the model up, which Chandrupatla takes by bisecting on up to all of his steps and which no reading measured here takes correctly for ours.
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---
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## 1. Introduction
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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]. Each of those answers carries constants: Illinois halves the stale ordinate, Brent accepts an interpolation subject to three empirical inequalities, ITP is governed by $\kappa_1,\kappa_2,n_0$.
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This paper asks a different question. Suppose the only thing available is a *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?
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The answer is sharp in both directions, and the negative half is as informative as the positive half.
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---
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## 2. The similarity step rule
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**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
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$$x' \;=\; x \;+\; \bigl(1 - S(f(x)/\sigma)\bigr)\,(x_\mathrm{far} - x). \tag{1}$$
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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 $f$.
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### 2.1 The two-point normalisation is false position
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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.
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**Proposition 1.** For the ratio similarity $S(z)=1/(1+|z|)$ with $\sigma=|f(x_\mathrm{far})|$, equivalently for the endpoint weights above,
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$$S(a)\,a + S(b)\,b \;=\; \frac{a f(b) - b f(a)}{f(b)-f(a)}.$$
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That is the false-position formula. Two things must be said about how far it reaches, because the obvious generalisation is false.
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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 first and fourth place on Chandrupatla and one of them solves 41 of 45, where false position solves 6.
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**Proposition 1'.** It is, however, specific to the *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.
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Together the two say where the boundary of novelty runs, and it is not where one would guess: it runs through the words *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 153 for bisection.
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### 2.2 The contraction law
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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, $1-S(z) = c\,z + O(z^2)$.
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**Proposition 2 (Contraction law).** For a simple root and $e_n$ small, rule (1) satisfies
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$$e_{n+1} \;=\; e_n\left(1 - \frac{c\,|f'(r)|\,D}{\sigma}\right) + O(e_n^2), \qquad \text{rate} = \left|1 - \frac{c\,|f'(r)|\,D}{\sigma}\right|. \tag{2}$$
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*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). $\square$
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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 $\sigma=2$.
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**Table 1.** The contraction law against measurement, all rows at the *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 $c=0$.
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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
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$$e_n \;\simeq\; \frac{\sigma^2}{c_2\,|f'(r)|^2\,|D|}\cdot\frac1n, \tag{2a}$$
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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.
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| $S(z)$ | $c$ | predicted | measured |
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| $1/(1+\lvert z\rvert)$ | 1 | rate 0.1716 | 0.1716 |
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| $e^{-\lvert z\rvert}$ | 1 | rate 0.1716 | 0.1716 |
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| $1/(1+z^2)$ | 0, $c_2=1$ | $n\,e_n\to0.8536$ | 0.8503 |
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| $e^{-z^2}$ | 0, $c_2=1$ | $n\,e_n\to0.8536$ | 0.8498 |
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| $1/\sqrt{1+z^2}$ | 0, $c_2=\tfrac12$ | $n\,e_n\to1.7071$ | 1.6988 |
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| $\max(0,1-\lvert z\rvert)$ | 1 | rate 0.1716 | captured, see below |
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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. **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.
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**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 $(0,2)$.
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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 $[-1,3]$.
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**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 they finish at 10497 and 11941, ahead of all three $c=0$ similarities at 14314, 15102 and 16394, and one of them heads Table 3. Overshooting is the milder failure.
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**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 *local scale of the value axis*, and that scale is not available a priori: it must be read off the evaluations.
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---
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## 3. The method
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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.
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**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 **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.
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**Scale: the finest data-derived one.** We set $\sigma = |\hat f'|\cdot|x_\mathrm{far}-x|$ with $\hat f'$ a divided difference of the **most recent evaluations** — 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 $O(h^2)$. A Newton step whose derivative carries an $O(h^2)$ error satisfies $e' = e-f/\hat f' = e-(e-ce^2+\dots)(1+O(e^2)) = ce^2+O(e^3)$, so the predicted order is **2** — not Muller's 1.84, which belongs to a method using the parabola's *root* rather than its *derivative*. Measured medians are 1.584 for the two-point scale and 2.000 for the three-point, against 1.809 for Brent and 2.654 for TOMS 748.
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**What the rule reduces to.** With $S(z)=\max(0,1-|z|)$ one has $1-S=\min(1,|z|)$, and (1) becomes
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$$x' \;=\; x + \operatorname{sign}(x_\mathrm{far}-x)\cdot\min\!\Bigl(\frac{|f(x)|}{|\hat f'|},\;|x_\mathrm{far}-x|\Bigr), \tag{4}$$
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a Newton step with a three-point divided-difference derivative, clipped to the bracket and then projected as in §3.1. The similarity function does not appear in (4): it is scaffolding of the derivation, not machinery of the algorithm.
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We state this ourselves rather than leave it to be discovered, because it locates the contribution precisely. It 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 then predicts the ranking of all nine similarity functions correctly; (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.
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**Algorithm.** Given $[a,b]$ with $f(a)f(b)<0$ and tolerance $\varepsilon_x$:
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```
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evaluate fa = f(a), fb = f(b)
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A, B <- a, b the proven bracket
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x <- the endpoint with the smaller |f|; xp <- the other; xp2 <- none
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n_max <- 2 * ceil(log2(|b-a| / xtol)); allow <- 2^n_max * xtol/2
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while |B - A| > xtol:
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far <- the end of [A,B] further from x
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f' <- derivative at x of the parabola through (x,xp,xp2)
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(two-point difference while xp2 does not yet exist)
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sigma <- |f'| * |far - x|
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if sigma is 0 or not finite: xn <- midpoint(A,B)
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else:
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xn <- x + min(1, |f(x)|/sigma) * (far - x) # the step, eq (4)
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xn <- clamp(xn, hi - allow, lo + allow) # projection, eq (3)
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if xn outside (A,B) or already evaluated: xn <- midpoint(A,B)
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evaluate fn = f(xn); allow <- allow / 2
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xp2, xp, x <- xp, x, xn
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update the bracket: whichever of A,B keeps the sign change
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return midpoint(A,B)
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```
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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 `n_max`, which §3.1 declares as the method's one free constant. The constant-free variant of §3.1 replaces the two `allow` lines by the earned credit of (3').
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### 3.1 The safeguard: project, do not replace
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An interpolation-only rule has no worst-case bound. The obvious safeguards *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 *error* need not halve the *bracket*, so a bracket-based test fires on exactly the steps that were best. With the three-point scale, replacing on that test drops the observed order from 1.94 to 1.00 and the Alefeld–Potra–Shi total from 2806 to 7260 — worse than bisection.
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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
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$$\alpha_j = 2^{\,n_\max-j}\epsilon, \qquad x \;=\; \operatorname{clamp}\bigl(x_\mathrm{sim},\; h-\alpha_j,\; \ell+\alpha_j\bigr). \tag{3}$$
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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.
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**The multiplier is the method's one free constant.** An earlier version of this section argued that it is "the factor already claimed in the bound". That is circular — the bound is twice bisection because two was chosen — and measurement settles it (`measure_budget_factor.py`, Linux):
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**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.
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| $c$ | APS-154 | CHA-45 | both |
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| 1 | 4817 | 2997 | 7814 |
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| 1.25 | 2806 | 2752 | 5558 |
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| 1.5 | 2806 | 2889 | 5695 |
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| 2 | 2806 | 3220 | 6026 |
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| 3 | 2806 | 3756 | 6562 |
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| 4 | 2806 | 4006 | 6812 |
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On the canonical set the projection never binds once $c\ge1.25$, so the total does not move. On the wide-bracket set it moves monotonically with $c$, and 2 is not even the best value. Nor is zero slack ($c=1$) the degenerate case it was claimed to be, "a worst case of exactly bisection" under which "the guard behaves as bisection": it costs 4817 and 2997, far from bisection's 7530 and 3090, 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.
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**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
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$$C \leftarrow \max(0,\;C+s_j-1),\qquad \alpha_j = w_j\,2^{\,C-1},\qquad x=\operatorname{clamp}\bigl(x_\mathrm{sim},\,h-\alpha_j,\,\ell+\alpha_j\bigr), \tag{3'}$$
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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:
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**Table 1b.** The earned projection at endowment $C_0$. Total evaluations, failures charged.
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| $C_0$ | promise | APS-154 | CHA-45 | both |
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|---|---|---|---|---|
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| 0.1 | bisection + 1 step | 5344 | 3051 | 8395 |
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| 0.5 | bisection + 1 step | 4656 | 3002 | 7658 |
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| **1** | **bisection + 1 step** | **3568** | **3020** | **6588** |
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| 2 | bisection + 2 steps | 3692 | 2712 | 6404 |
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| 5 | bisection + 5 steps | 3006 | 2615 | 5621 |
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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.
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We nonetheless propose the scheduled projection with $c=2$ and declare its multiplier. On the canonical set it is 762 evaluations cheaper than the earned variant (2806 against 3568), and there its total does not depend on the multiplier at all; on the wide-bracket set the earned variant is the cheaper (3020 against 3220). 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; what changed after the question whether it was load-bearing is only that it is now declared as a constant rather than argued to be none.
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Two remarks apply to both forms.
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- **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.
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- **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.
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---
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## 4. Experimental protocol
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**Problem sets.** The 154 problems of Alefeld, Potra and Shi [1] and the 45 of Chandrupatla [3], taken programmatically from SciPy's `scipy.optimize._tstutils` so that no bracket or root is transcribed by hand.
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**Cost model.** Evaluations of $f$. A run counts as solved when it reports convergence and lands within $10^{-6}$ of the published root; $x$-tolerance $10^{-12}$; budget 200 evaluations. *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.
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**Fairness to the competitors.** Three deliberate concessions.
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1. 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.
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2. ITP is reported both at the authors' recommended constants and as a *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.
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3. 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 $n_0\in\{0,1,2,5\}$.
|
||
|
||
4. Where SciPy ships the method, the competitor **is the library**: Chandrupatla is `scipy.optimize.elementwise.find_root`, Brent is `brentq`, TOMS 748 is `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, and Chandrupatla's totals move from 2785 to 2787 on the canonical set and from 2423 to 2441 on the wide-bracket set. For Brent the library is slightly dearer than our implementation (7004 against 6991 over both sets); 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.
|
||
|
||
**Reproducibility.** All numbers come from a single Linux run. The test functions call `libm`, whose last digit differs between platforms; the same SciPy version scores 2819 for TOMS 748 on Windows and 2820 on Linux for the same set. Every number stated in this paper is checked against the run data by a script that is part of the test suite.
|
||
|
||
---
|
||
|
||
## 5. 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.
|
||
|
||
**Table 2.** Alefeld–Potra–Shi, 154 problems, out of 264 configurations. Places by failures first, then total.
|
||
|
||
| # | configuration | tuned | solved | total evals | $>2\times$ |
|
||
|---|---|---|---|---|---|
|
||
| 1 | Chandrupatla | no | 153/154 | 2787 | 0 |
|
||
| 2 | the same rule with no safeguard at all | no | 153/154 | 2806 | 0 |
|
||
| **3** | **base rule: tent, 3-point, projection ($c=2$)** | **no** | 153/154 | **2806** | 0 |
|
||
| 4 | TOMS 748 | yes | 153/154 | 2820 | 0 |
|
||
| 5 | Brent | yes | 153/154 | 2889 | 0 |
|
||
| 12 | Dekker | no | 153/154 | 3158 | 0 |
|
||
| 13 | ITP, per-problem oracle | yes | 153/154 | 3260 | 0 |
|
||
| 20 | Ridders | no | 153/154 | 3313 | 0 |
|
||
| 29 | constant-free variant: earned projection | no | 153/154 | 3568 | 0 |
|
||
| 34 | ITP, recommended constants | yes | 153/154 | 3957 | 0 |
|
||
| 73 | Illinois | yes | 153/154 | 7045 | 0 |
|
||
| 79 | bisection | no | 153/154 | 7530 | 0 |
|
||
| 244 | false position | no | 127/154 | 10560 | — |
|
||
|
||
The base rule and the same rule without any safeguard tie at 2806: the projection never binds on this set. By total, Chandrupatla's method is first with 2787, 0.7% fewer — the same order of margin as the 0.5% by which the base rule leads TOMS 748, which we read as a tie. Problem by problem the picture is the other way round: on the problems both solve the base rule beats Chandrupatla 87/24/42 (wins/ties/losses; exact two-sided sign test $p=9\cdot10^{-5}$), TOMS 748 86/16/51, Brent 102/11/40, ITP at its recommended constants 132/3/18, the per-problem ITP oracle 121/4/28, and bisection 152/1/0. It wins twice as many problems as it loses against Chandrupatla and trails him in total because its losses are the larger ones. The constant-free variant goes 79/3/71 against Chandrupatla ($p=0.57$, a tie in problems), 77/9/67 against TOMS 748 and 78/5/70 against Brent, and loses on the total to all three.
|
||
|
||
**Table 3.** Chandrupatla, 45 problems, brackets up to $10^{12}$ wide, out of 264 configurations.
|
||
|
||
| # | configuration | tuned | solved | total evals | $>2\times$ |
|
||
|---|---|---|---|---|---|
|
||
| 1 | similarity, $1-\sqrt{\lvert z\rvert}$, bracket scale, projection | no | **41/45** | 4516 | 0 |
|
||
| 2 | Chandrupatla | no | 40/45 | 2441 | 0 |
|
||
| 3 | ITP, per-problem oracle | yes | 40/45 | 2647 | 0 |
|
||
| 4 | similarity, $e^{-\lvert z\rvert}$, bracket scale, both guards | no | 40/45 | 2736 | 0 |
|
||
| 6 | ITP, recommended constants | yes | 40/45 | 2896 | 0 |
|
||
| 10 | constant-free variant: earned projection | no | 40/45 | 3020 | 0 |
|
||
| 16 | bisection | no | 40/45 | 3090 | 0 |
|
||
| **45** | **base rule: tent, 3-point, projection ($c=2$)** | **no** | 40/45 | **3220** | 0 |
|
||
| 159 | TOMS 748 | yes | 40/45 | 4000 | 15 |
|
||
| 227 | Brent | yes | 38/45 | 4115 | 18 |
|
||
| 228 | the same rule with no safeguard at all | no | 35/45 | 4104 | 10 |
|
||
|
||
The set's nominal winner is a similarity configuration that solves one problem more than anything else, 41 of 45, at a worse total; ranking by failures first puts it on top, and we report it as such rather than by the total that would flatter our proposal. Here Chandrupatla's method is genuinely the stronger: it beats the base rule 26 problems to 9 ($p=0.006$) and by 3220 against 2441 in total. The proposed rule beats Brent 30/4/4 and TOMS 748 31/1/8 — both of which exceed the $2\times$ bound on 18 and 15 problems — and loses to ITP 17/0/23 and to the ITP oracle 16/0/24. The wide-bracket set is where the multiplier matters (Table 1a) and where the constant-free variant is the cheaper form, 3020 against 3220. Over both sets the base rule needs 6026 and the constant-free variant 6588, against 5228 for Chandrupatla, 5907 for the ITP oracle, 6820 for TOMS 748 and 7004 for Brent.
|
||
|
||
### 5.1 What the guarantee costs
|
||
|
||
It depends on which bound is bought. Against the identical rule with no safeguard at all:
|
||
|
||
| | solved | total evals | runs over the $2\times$ bound |
|
||
|---|---|---|---|
|
||
| APS, no safeguard | 153/154 | 2806 | 0 |
|
||
| APS, projection — base rule ($2\times$ bisection) | 153/154 | 2806 | 0 |
|
||
| APS, earned projection — constant-free (bisection + 1 step) | 153/154 | 3568 | 0 |
|
||
| Chandrupatla, no safeguard | 35/45 | 4104 | 10 |
|
||
| Chandrupatla, projection — base rule | 40/45 | 3220 | 0 |
|
||
| Chandrupatla, earned projection — constant-free | **40/45** | **3020** | **0** |
|
||
|
||
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 five more problems for 21% 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, 762 evaluations or 27%, 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 7260 and 7239 on Alefeld–Potra–Shi, against 2806.
|
||
|
||
### 5.2 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 **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.
|
||
|
||
Those evaluation counts are *raw*: they are what the timing loop actually spent, and a failed run contributes only the evaluations it made before the budget ran out. They are therefore not the totals of §5, where a failure is charged the full 200-evaluation budget so that no method can buy a low score by giving up — 4991 raw against 6026 charged for the base rule, 5932 against 6991 for Brent (our pure-Python implementation, which is what is timed; the tables use the library's `brentq`, 7004), 9474 against 10620 for bisection. Wall-clock time can only be compared against what was really executed, so the raw counts are the right companions to the milliseconds; every comparison in §5 uses the charged ones. Both are stated to avoid the appearance of two incompatible sets of numbers for the same methods.
|
||
|
||
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 *negative* overheads for four of six methods — an impossibility that condemned the estimator, and the reason nothing is subtracted here.
|
||
|
||
### 5.3 The law across the whole grid
|
||
|
||
Corollaries 1–3 predict an ordering of the nine similarity functions and of the four scales. Table 4 reports it. The prediction holds without exception for every data-derived scale: the three similarities with $c=1$ occupy the top three places and the three with $c=0$ the bottom three. For the chosen scale the two groups are indistinguishable (16102 against 15345), which is consistent — with a chosen scale nothing converges well and the ordering carries no information.
|
||
|
||
**Table 4.** Mean total evaluations over both sets, step guard, grouped by the slope $c$ of $1-S$ at the origin.
|
||
|
||
| scale | $c=1$ (linear approach) | $c=0$ (quadratic approach) |
|
||
|---|---|---|
|
||
| 3-point derived | **6893** | 22315 |
|
||
| 2-point derived | 6981 | 22206 |
|
||
| bracket | 10279 | 19127 |
|
||
| start (chosen) | 16102 | 15345 |
|
||
|
||
Ranked individually and best over scales, the similarities come out $1-|z|$ (6560), $e^{-|z|}$ (6895), $1/(1+|z|)$ (7118), $(1+|z|)^{-2}$ (9578), $1-\sqrt{|z|}$ (10546), $1-|z|^{1/3}$ (12002), $e^{-z^2}$ (14307), $1/(1+z^2)$ (15102), $1/\sqrt{1+z^2}$ (16394); the scales come out 3-point (6560), 2-point (6702), bracket (9107), start (14540). Both orderings are those the law predicts, and both were predicted before the grid was run.
|
||
|
||
---
|
||
|
||
## 6. 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 *approach* from the *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 *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 *quality*. Reconstructing each method's bracket from the signs it observed, the median probe removes 0.500 of the bracket for every method on Alefeld–Potra–Shi, but on Chandrupatla it removes 0.275 for the similarity rule against 0.058 for Brent and 0.068 for TOMS 748 — and that comparison already flatters them, since 69 and 141 of their probes respectively fall outside the proven bracket and are excluded, against none of ours. High-order interpolation is a wager that $f$ is locally polynomial. When the wager pays, order 1.84 or 2.65 wins the endgame; when it fails, the probes are wasted. The similarity rule makes the weakest available assumption — a single local slope — and degrades gracefully.
|
||
|
||
---
|
||
|
||
## 7. Where the similarity is read
|
||
|
||
Everything above reads the similarity in one place: the *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 `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 193 of the 199 problems.
|
||
|
||
Where a safeguard other than the projection of §3.1 appears, it is one of the grid's. *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. *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. *bracket* and *both* are the replacing safeguards of §3.1, which keep a bound at the price of the order; *step* alone and *none* carry no bound.
|
||
|
||
1. **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.
|
||
2. **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 *selected* — an argmin over what is already in hand, with no threshold.
|
||
3. **Which end is far.** Read by dissimilarity instead of distance: the end whose $|f|$ is larger is the one less similar to the target.
|
||
|
||
**Table 5.** Reading the similarity in one more place is worth more than changing $S$. Evaluations of $f$; failures charged the full budget.
|
||
|
||
| configuration (all with the guarantee, 193/199) | APS-154 | CHA-45 | both |
|
||
|---|---|---|---|
|
||
| base rule: tent, 3-point, projection $c=2$ (§3) | 2806 | 3220 | 6026 |
|
||
| constant-free variant: earned projection (§3.1) | 3568 | 3020 | 6588 |
|
||
| (1) trust before the step, projection | 2848 | 3114 | 5962 |
|
||
| (1)+(2) points chosen by shape, *safe* | **2527** | 3126 | 5653 |
|
||
| (1)+(3) far end by dissimilarity, projection | 2726 | **2799** | **5525** |
|
||
| (1)+(2)+(3), *safe* | 2884 | 2871 | 5755 |
|
||
| Chandrupatla | 2787 | 2441 | 5228 |
|
||
| TOMS 748 | 2820 | 4000 | 6820 |
|
||
| ITP, per-problem oracle | 3260 | 2647 | 5907 |
|
||
|
||
Two results here are worth stating plainly. First, the canonical set improves from 2806 to **2527**, which is below every method measured, including Chandrupatla's 2787, with the worst-case guarantee intact. It hardly feels the declared multiplier of §3.1: 2527 for every multiplier from 1.5 to 4, 2541 at 1.25 (`measure_records_vs_multiplier.py`). Second, on the two sets together the 5525 of reading (3) passes ITP's oracle (5907) but not Chandrupatla (5228): the remaining gap is entirely on the wide-bracket families, where Chandrupatla needs 2441 against our 2695 (reading (3) under *safe*).
|
||
|
||
The readings do not compose. Taken together, (2) and (3) give 2884 and 2871 — 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.
|
||
|
||
Under the constant-free earned projection the same readings gain far less. Reading (1) gives 3507 and 2997, (1)+(2) 3755 and 3021, (1)+(3) 3857 and 3002, all three together 4354 and 3037, against 3568 and 3020 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.
|
||
|
||
### 7.1 The similarity function belongs to the place, not to the method
|
||
|
||
Everything above varies *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 *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 *at order one* on an approach of order $m$: its three-century-old stall is a mismatch between the similarity function and the place it is read in. The exponent $1/m$ is not a constant — $m=1/(1-ff''/f'^2)$ comes from three evaluations already paid for, as in item 5 of §10, and is 1 whenever the power model does not apply.
|
||
|
||
Measured over the 252 cells of the grid, the correction behaves exactly as that reading predicts and not at all as a universal improvement. Paired inside each (function, safeguard) cell of the far-end reading it is cheaper in 29 of 60 cells — indistinguishable from a coin. But in the one pairing where that reading *is* exact inverse interpolation, the ratio similarity, it is cheaper under all seven safeguards:
|
||
|
||
**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 29 of 60 cells over all nine functions.
|
||
|
||
| safeguard | APS, $m=1$ | APS, measured $m$ | CHA, $m=1$ | CHA, measured $m$ |
|
||
|---|---|---|---|---|
|
||
| both | 4522 | **3675** | 3406 | **3241** |
|
||
| bracket | 4606 | **3542** | 3602 | **3359** |
|
||
| credit | 5190 | **4686** | 3013 | **2884** |
|
||
| projection | 6932 | **6364** | 4789 | **4554** |
|
||
| safe | 5987 | **5229** | 4134 | **3954** |
|
||
| step | 6435 | **5551** | 4287 | **4068** |
|
||
| none | 10676 | **10043** | 8265 | **7794** |
|
||
|
||
Two limits must be stated with it. The estimator is *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 2884, against 2695 for reading (3) above and 2441 for Chandrupatla. In the *slope* reading of §3 the same exponent is harmful, $2806\to2854$, which is the same statement read from the other side — a function that does not belong to a place does not become right by being applied there.
|
||
|
||
### 7.2 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 *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 *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 *same* $S$ in both places ("simpair"). The third is *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 3059 under the projection and 2892 under *credit*, against his 2441, while the canonical set rises from 2806 to 5432. The band is not idle — it refuses 63.5% of steps — so this is a decision that was taken and did not pay *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 *our* step.
|
||
|
||
Our own readings do not close it either (best 2897 on the wide-bracket set), but "defect" sets a new best on the canonical set: **2523** at 153/154 under the projection of the base rule, below the 2527 of reading (2) and below every method measured, with the guarantee intact — and, unlike 2527, the same for every multiplier from 1.25 to 4. Under the earned projection it gives 3545 and 2949 (6494 against 6588), 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 5733 against the 5525 of reading (3), because it gives back on the wide brackets (3210) what it gains on the canonical problems — so it is a second point on the same trade, not a replacement.
|
||
|
||
### 7.3 Six problems of the two sets are scored unsolved for every method
|
||
|
||
Chasing the remaining gap to Chandrupatla led to a property of the benchmarks themselves. On the five problems of the wide-bracket family `fun7`, and on the single canonical problem `aps.13.00`, the declared root is 0 while $f$ vanishes *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 $f(x)=0$ *exactly* — a genuine root — and every method is then scored as having failed, because the harness asks for $|x-\text{root}|\le10^{-6}$ while the returned point sits up to $2.8\times10^{-2}$ away.
|
||
|
||
The consequence is arithmetic. Each such problem charges the full budget of 200 evaluations to every method: about 1150 evaluations of every reported two-set total — ours, Chandrupatla's, TOMS 748's, ITP's, Brent's, bisection's — are this artefact, and the "153 of 154" and "40 of 45" that every method reports are statements about the scoring rather than about the methods. Re-scored with the criterion "a root is a root" ($|x-\text{root}|\le10^{-6}$ *or* $f(x)=0$), the same runs solve 199 of 199 for every method except Brent's 197, and the totals fall as follows.
|
||
|
||
**Table 7.** The same runs under the harness criterion and under "a root is a root". The ranking is unchanged; the magnitudes are not, and neither is the size of the remaining gap on the wide-bracket set: 235 evaluations, or 16%, not 10%.
|
||
|
||
| method | APS as scored | APS re-scored | CHA as scored | CHA re-scored |
|
||
|---|---|---|---|---|
|
||
| ours, (1)+(2) points by shape | **2527** | **2337** | 3108 | 2159 |
|
||
| ours, base rule (projection) | 2806 | 2634 | 3220 | 2357 |
|
||
| Chandrupatla | 2787 | 2595 | **2441** | **1505** |
|
||
| TOMS 748 | 2820 | 2636 | 4000 | 3088 |
|
||
| Brent | 2889 | 2707 | 4115 | 3238 |
|
||
| ITP | 3957 | 3767 | 2896 | 1907 |
|
||
| bisection | 7530 | 7338 | 3090 | 2136 |
|
||
|
||
### 7.4 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 `fun1` family is the same function with brackets of growing width, and our cost against Chandrupatla's across them is 7/8, 12/12, 19/15, 59/24, 63/44: we tie or win while the bracket is narrow and lose progressively as it widens. The endgame is not the cause — our observed order on those very problems is the higher one (3.63 against 2.02 on the $2\times10^4$-wide bracket). The cause is the rate of progress far from the root: 59 evaluations on a $2\times10^4$-wide bracket is what bisection itself needs ($\lceil\log_2(2\times10^{16})\rceil\approx54$), i.e. there we advance by about one halving per evaluation while Chandrupatla advances by about 2.3.
|
||
|
||
It is not the guarantee that caps us. With the safeguard removed entirely the same problems cost 46, 57, 61 and 68 against bisection's 57, 77, 45 and 45 — better on the wide brackets, *worse than bisection* on `fun3.1` and `fun5.1` — and still far from Chandrupatla's 24 and 44.
|
||
|
||
Nor is it that the step fails to be relative to the bracket. With the tent similarity $1-S=\min(1,|f|/(|\hat f'|D))$, so by (4) the step is exactly $\min(D,|f|/|\hat f'|)$: a *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 *more* often than Chandrupatla's 87.5%.
|
||
|
||
What differs on `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 `fun1` and *wrong as a statement about the whole gap*. Replaying our own trajectories on the 33 problems where we lose, and placing both models on the *same* evaluations — one extra evaluation each, from the same state — splits the 256 evaluations of the gap in two. On `fun1` (58 of them, about a quarter) 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 *negative* weight when the triple brackets the root (verified on all 428 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 256 — two thirds — sit on `fun3`–`fun6`, which are pure powers of order 3, 5, 9 and 19, and there *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 `fun4`, `fun5`, `fun6` from 573, 456, 456 to 296, 239, 228 — against his own 297, 233, 233 — and costs +164, +111, +170, +144 on `fun1`, `fun2`, `fun8`, `fun9`, where it refuses steps that were good.
|
||
|
||
So most of what remains is not the order of the model but the decision *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" (4449 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 (0.2% 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 `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.
|
||
|
||
On the re-scored wide-bracket set of Table 7 our shape reading (2159) is *worse* than bisection (2136); only the dissimilarity reading (1740) is clearly ahead, which is the same statement seen from the other end: where the step cannot march, the reading that helps least is the one that only improves the local model.
|
||
|
||
Every statement in §§7.3–7.4 is reproduced by `diag_rescore_cha.py`, `diag_rescore_aps.py`, `diag_fun7_root.py`, `diag_cha_gap.py`, `diag_model_order.py`, `diag_model_order_checks.py`, `diag_flat_detectable.py` and `diag_band_by_family.py`, whose outputs are kept beside them.
|
||
|
||
## 8. The flat regime, recognised by analogy
|
||
|
||
§7.4 left one decision open: *when to give the model up*. On the pure powers `fun3`–`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 *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.
|
||
|
||
### 8.1 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 *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 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 2806 → 3029, 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 *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).
|
||
|
||
### 8.2 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 — `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 *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 *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 *read from the analogy itself*,
|
||
|
||
$$m=\frac{\ln L_f}{\ln L_x}.\tag{6}$$
|
||
|
||
A chance resemblance at a simple root reads $m\approx1$; a pure power of order 5 reads 4.9. The regime is the one where a tangent step loses, $m\ge2$ (§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 `fun2`, `fun8`, `fun9`.
|
||
|
||
**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 *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).
|
||
|
||
**The translation limit.** The exponential is self-similar by *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 `fun9` it was built for, that the limit paid.
|
||
|
||
**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 6897; with it, 3125 — the exit is what makes a false entry cheap.
|
||
|
||
### 8.3 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 $m\to\infty$. The order read from a *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 *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.
|
||
|
||
### 8.4 Five more readings
|
||
|
||
**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 *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 `fun9.5` — this is the false promise "the root is here", and removing it alone takes the canonical set from 2806 to 2564.
|
||
|
||
**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 `fun9` well: a near-cancelling end reads the root as close while $|f|$ doubles for seventeen steps.
|
||
|
||
**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 $\tfrac12$.
|
||
|
||
**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 *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 *not* cleared by it (measured on the development corpora: 19 evaluations with the memory kept, 65 without).
|
||
|
||
**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 *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.
|
||
|
||
### 8.5 The proposed method
|
||
|
||
The method this paper proposes, **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.
|
||
|
||
**Table 8.** A′ on the protocol of §4 (published roots, failures charged 200). Per problem against each rival: wins/ties/losses, exact two-sided sign test.
|
||
|
||
| | APS-154 | CHA-45 | both | vs A′ per problem, APS | vs A′ per problem, CHA |
|
||
|---|---|---|---|---|---|
|
||
| **A′ (proposed)** | **2438** | **2005** | **4443** | — | — |
|
||
| A′, earned budget (no free constant) | 3608 | 2807 | 6415 | | |
|
||
| base rule (§3) | 2806 | 3220 | 6026 | | |
|
||
| Chandrupatla | 2787 | 2441 | 5228 | A′ 103/20/30, $p=1.5\cdot10^{-10}$ | A′ 29/4/7, $p=3\cdot10^{-4}$ |
|
||
| TOMS 748 | 2820 | 4000 | 6820 | A′ 127/6/20 | A′ 38/2/0 |
|
||
| Brent | 2889 | 4115 | 7004 | A′ 118/17/18 | A′ 33/3/2 |
|
||
| ITP, per-problem oracle | 3260 | 2647 | 5907 | A′ 126/7/20 | A′ 35/1/4 |
|
||
|
||
A′ is 12.5% below Chandrupatla on the canonical set and 17.9% below on the wide-bracket set, and beats every rival problem by problem on both. On the wide-bracket families it is ahead on seven of nine, level on `fun7` (the scoring artefact of §7.3) and behind on `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.
|
||
|
||
**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.
|
||
|
||
| component removed | APS | CHA | new-1 | Roots.jl | applied | fourth | sum |
|
||
|---|---|---|---|---|---|---|---|
|
||
| the whole regime | −7 | +1013 | +1660 | +324 | +77 | +2269 | **+5336** |
|
||
| the split (7) (midpoint instead) | −5 | +479 | +667 | +272 | +30 | +1672 | **+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 |
|
||
|
||
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.
|
||
|
||
## 9. 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 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.
|
||
|
||
1. **New corpus 1** — 206 problems: SciPy's `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 `fun9` kind.
|
||
2. **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.
|
||
3. **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.
|
||
4. **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 *before* this corpus was run, and nothing was changed after.
|
||
|
||
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.
|
||
|
||
**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 *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.
|
||
|
||
**Table 10.** A′ against the library rivals on all six corpora (uniform criterion). Totals; in brackets the runs that break the $2\times$-bisection bound. Per problem against Chandrupatla: wins/ties/losses.
|
||
|
||
| corpus | A′ | Chandrupatla | Brent | TOMS 748 | A′ vs Chandrupatla |
|
||
|---|---|---|---|---|---|
|
||
| APS (154) | **2259** (0) | 2595 (0) | 2707 (0) | 2636 (0) | 103/20/31, $p=3\cdot10^{-10}$ |
|
||
| Chandrupatla (45) | **1074** (0) | 1505 (0) | 3238 (18) | 3088 (15) | 31/4/10, $p=0.001$ |
|
||
| new corpus 1 (206) | **3290** (0) | 3716 (0) | 6395 (38) | 5746 (22) | 108/33/65, $p=0.001$ |
|
||
| Roots.jl (78) | **1371** (0) | 1566 (0) | 2455 (11) | 2339 (9) | 44/14/20, $p=0.004$ |
|
||
| applied + GSL (132) | 1382 (0) | 1382 (0) | 1503 (1) | 1756 (1) | 48/52/32, $p=0.09$ |
|
||
| **corpus 4 (240), the one test** | **3656** (0) | 5717 (0) | 8267 (50) | 8084 (37) | **177/35/28, $p=10^{-27}$** |
|
||
| all 855 | **13032** | 16481 | 24565 | 23649 | |
|
||
|
||
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. 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 ($p=3\cdot10^{-4}$).
|
||
|
||
**Table 11.** Corpus 4 by class: totals of A′ / Chandrupatla, per problem wins/ties/losses.
|
||
|
||
| class | A′ | Chandrupatla | per problem |
|
||
|---|---|---|---|
|
||
| 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 |
|
||
|
||
**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 keeps its bound on all 855 problems, which Brent breaks on 118 and TOMS 748 on 84. 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: `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.
|
||
|
||
---
|
||
|
||
## 10. Negative results
|
||
|
||
1. **The prototype-aggregation form of the method loses.** A richer construction — prototypes over *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.
|
||
|
||
2. **Every safeguard that replaces the step destroys the order.** Bracket-based and step-based replacement both drop the three-point rule from order 1.94 to 1.00 and cost 7260 and 7239 evaluations on Alefeld–Potra–Shi. Only projection preserves both. We record this because it is the trap the construction naturally falls into.
|
||
|
||
3. **Zero slack in the projection is not the degenerate case it was said to be.** An earlier version reported that a worst case of exactly bisection leaves the projection no freedom and makes the guard behave as bisection. Measured (Table 1a), multiplier 1 costs 4817 and 2997, against bisection's 7530 and 3090 — and on the wide-bracket set it beats the multiplier of 2. It remains the most expensive multiplier on the canonical set, which is all that survives of the claim.
|
||
|
||
4. **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 `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 4109 against 3220, so a switch chosen per benchmark is worse than no switch at all: 6915 against 6026 over both sets. The scale axis offers only 5.3% headroom even to an oracle handed the best scale per problem, two thirds of it inside a single 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 $10^{12}$.
|
||
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5. **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. It preserves the order (2.001) and improves the eight standard problems (82 against 84) — and loses on the full benchmark, 6304 against 6026. The estimator is the failure: on Alefeld–Potra–Shi, where almost every root is simple and the truth is $m=1$, it reports $m>1$ on 27.9% of steps with excursions to $m=34.8$, against 35.8% on the flat families. 27.9 against 35.8 is no separation. A cruder variant, correcting by the fraction of its own promise the previous step kept, fails everything: the order collapses to 0.904 and the standard problems double to 161.
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6. **One problem is scored unsolved for every method**, `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. It is a property of the scoring, not a failure of the methods; §7.3 finds five more such problems in the wide-bracket set.
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7. **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).
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8. **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 *promised bracket progress* instead of by the shape of the model, costs 4449 on the canonical set against 2527 — a difference of 1922 evaluations for one choice of criterion. A secant step routinely collapses the error while barely moving the proven bracket, so bracket progress rejects steps that were excellent; this is the same mechanism that makes the bracket guard of §3.1 cost 7260.
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9. **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 *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: 3288 on the canonical set against 2726, and 3111 on the wide-bracket set against 2695. Its apparent extra solves (194 instead of 193) are two `fun7` problems where it lands inside the flat zero region close enough to satisfy the harness — the artefact of §7.3, not a capability.
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10. **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 128/154 and 10043 — one problem and six per cent — where Illinois, at the price of one constant (halve the stale ordinate), reaches 153/154 and 7045. A sixth of the gap is not the gap. The correction is real and it is in the right place; it is simply much smaller than a constant that was chosen for exactly this failure. *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.
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11. **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, *safe*) the wide-bracket set improves from 3695 to 3541 — but the canonical set gets worse, 5822 to 6940, against the 2806 of the base rule. An earlier version compared the 3541 with 4516, which is the total of a different configuration. A march needs a *ratio* of distances while an identity on narrow brackets needs an *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.
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---
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## 11. Conclusion
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A root finder can be derived from a similarity function alone, and the derivation shows where the content of such a method actually sits. It is not in the shape of $S$ — nine shapes were tried and their ranking is dictated by a single number, the slope of $1-S$ at the origin. It is not in making the similarity change faster — that overshoots exactly as readily as flatness stagnates. It is in the *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.
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The resulting rule has one declared constant and carries a worst case of twice bisection plus one step. On the canonical Alefeld–Potra–Shi set it is level with TOMS 748 — 2806 evaluations against 2820 is a margin of 0.5% and should be read as a tie — ahead of Brent by 2.9%, and level with Chandrupatla's method in the same sense: 0.7% more in total, while winning 87 of the problems to his 42 ($p=9\cdot10^{-5}$). On the wide-bracket set Chandrupatla is genuinely stronger (2441 against 3220), and that is where the declared multiplier matters. The constant-free variant trades 762 evaluations on the canonical set for no constant, a tighter guarantee and the better wide-bracket total. The claim the measurements support is parity with the best methods on the canonical set, with a problem-by-problem edge over all of them, and a measured deficit on very wide brackets whose cause §7.4 locates — not superiority overall.
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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.
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### Future work
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The diagnosis of §7.4 leaves one question, and it is narrower than the one this section used to pose. An earlier version proposed a logarithmic change of variable to turn an additive step into a geometric march; that was measured and refuted (§10, item 11), and the step turned out to be a fraction of the bracket already.
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What is open is a decision, not a model. On two thirds of the remaining gap the right move is to halve, and Chandrupatla halves; our guarantee permits a step that is slower than halving, and nothing we have measured tells our step when it is. The condition itself is known exactly — a tangent step on an approach of local order $m$ lands at $e(1-1/m)$ and loses to a halving when $m\ge2$ — so the constant is bisection's $\tfrac12$ and not a choice. What is missing is a reading of $m$ that sees it: the three-point estimator reports $m\ge2$ on 0.2% of canonical states, which is the right silence, and on 5.8% of states of pure powers of order 3 to 19, which is blindness. Two successive residuals of the run itself carry the order more directly than a parabola through three points — on an order-$m$ approach under tangent steps their ratio settles at $(1-1/m)^m$, while at a simple root it goes to zero — and that is the reading we would test next, under the same rule as every other here: pre-registered, with the canonical set as the price it must not charge.
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---
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## References
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1. G. E. Alefeld, F. A. Potra and Y. Shi, "Algorithm 748: Enclosing zeros of continuous functions," *ACM Trans. Math. Softw.*, vol. 21, no. 3, pp. 327–344, 1995.
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2. R. P. Brent, *Algorithms for Minimization without Derivatives*. Englewood Cliffs, NJ: Prentice-Hall, 1973.
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3. T. R. Chandrupatla, "A new hybrid quadratic/bisection algorithm for finding the zero of a nonlinear function without using derivatives," *Adv. Eng. Software*, vol. 28, no. 3, pp. 145–149, 1997.
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4. I. F. D. Oliveira and R. H. C. Takahashi, "An enhancement of the bisection method average performance preserving minmax optimality," *ACM Trans. Math. Softw.*, vol. 47, no. 1, pp. 1–24, 2021.
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5. C. Ridders, "A new algorithm for computing a single root of a real continuous function," *IEEE Trans. Circuits Syst.*, vol. 26, no. 11, pp. 979–980, 1979.
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6. V. Sitnikov, "Parameter-free similarity control," preprint, 2026.
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7. M. Dowell and P. Jarratt, "A modified regula falsi method for computing the root of an equation," *BIT*, vol. 11, pp. 168–174, 1971.
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