# Root Finding by Similarity and Analogy: A Bracketing Method with One Declared Constant V. Sitnikov --- ## Abstract Bisection places the next iterate at the midpoint because it uses no information about the function's value. We replace that rule with a *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. Third, the rate vanishes exactly when the scale equals the local scale of the value axis; reading that scale off the data — a divided difference, not a fitted coefficient — turns the step into $|f|/|f'|$ and the convergence superlinear. The derivation fixes a *base rule* before any benchmark is consulted: a similarity whose approach to 1 is linear, the finest available data-derived scale, and a safeguard that projects the step onto the admissible interval rather than replacing it. It is, explicitly, a Newton step with a three-point divided-difference derivative, clipped to the bracket and projected onto the bisection budget. The similarity sets that step, and its form is decisive: with the tent the step is Newton's, with any of the other eight functions the proposed method costs from 19% to five times more. Its order is Muller's, 1.839. It carries a proven worst case of twice bisection plus one step and one free numerical constant, the multiplier of that budget, which we declare. On the 154 problems of Alefeld, Potra and Shi it needs 2634 evaluations of $f$ against 2636 for TOMS 748, 2707 for Brent and 2595 for Chandrupatla's method; it wins more problems than it loses against Chandrupatla, 87 to 43, and needs 1.5% more in total because its losses are the larger ones. On the 45 wide-bracket problems of Chandrupatla it beats Brent and TOMS 748 and loses to ITP and to Chandrupatla (2357 against 1946 and 1505). For every data-derived scale the three similarity functions whose $1-S$ has a non-zero slope at the origin come before the three whose slope is zero, as the derived law predicts. The wide-bracket gap turns out to be not the order of the model but a decision — when to give the model up — and the decision is a similarity after all, of an unfamiliar kind: the similarity of a run to its own past. A pure power is self-similar under zoom, so the dimensionless configuration of the last three points recurs, while at a simple root it tends to the linear prototype; the analogous pair yields the local order, $m=\ln L_f/\ln L_x$, and an exponential, self-similar by translation, is its limit $m=\infty$. Entering a halving regime when that analogy holds with $m\ge2$ — the condition under which a tangent step loses to a halving — splitting at the point the order predicts, and leaving when the split looks linear, together with four smaller readings of the same kind and one borrowed element, gives the method proposed here, A′. On the two sets it needs 2259 and 1074 evaluations against Chandrupatla's 2595 and 1505 and wins more problems than it loses on both, 103 to 31 and 31 to 10. The chain of decisions that produced it was selected on those sets; tested once, after its code was frozen, on 240 problems from six new families, it needs 3656 evaluations against Chandrupatla's 5717 and wins 177 problems to 28 — a corpus the author composed knowing where the method is strong: two of its six families have multiple or fractional roots, and without them the lead is 2403 against 3569, on the simple-root families alone 737 against 747. Against simpler detectors put in the place of the analogy — Lagouanelle's multiplicity estimate with the same memory, Chandrupatla's acceptance band, or none — it wins on that corpus in total and problem by problem (3656 against 4488, 6236 and 6165); against the classical transformation $\operatorname{sign}(f)|f|^{1/m}$ with the multiplicity given in advance it loses every multiple or fractional root, by a factor of 2.4 to 6.8 — the price of estimating the order. Over 855 problems in six corpora it is ahead of Brent, TOMS 748, ITP and the recent ModAB everywhere and stays within its own bound of twice bisection on every problem, a bound that Brent, TOMS 748 and ModAB, which do not promise it, exceed on 118, 84 and 9 problems; against Chandrupatla it is ahead in total on five corpora and level on the sixth. The lead is large on multiple, fractional and nearly flat roots and a matter of per cent on simple ones. Several of the resulting mechanisms coincide with classical ones — Lagouanelle's multiplicity estimate, false position on $|f|^{1/m}$ — reached here from similarity and analogy and recognised afterwards; what we found no counterpart for is the decision layer, where every choice — when to give the model up, where to split, when to return — is a judgment of similarity or of analogy, and the similarity that sets the step, whose form is decisive (replacing it costs 19% to 425%). --- ## 1. Introduction A bracketing root finder must decide, at every step, where inside the bracket to evaluate next. Bisection answers "in the middle", which is the correct answer when nothing is known. Every improvement since has answered "where a local model of $f$ says the root is": a secant in false position, an inverse quadratic in Brent [2], a cubic in TOMS 748 [1], an interpolate–truncate–project cycle in ITP [4], an Anderson–Björck step behind a linearity check in ModAB [10, 20]. Each of those answers carries constants: Illinois halves the stale ordinate, Brent accepts an interpolation subject to three empirical inequalities, ITP is governed by $\kappa_1,\kappa_2,n_0$. This paper asks a different question. Suppose the only thing available is a *similarity function* — a map $S$ from an outcome to $[0,1]$, where 1 means "identical to what was wanted" and 0 means "not similar at all", in the sense used for parameter-free control in [6]. Can a root finder be built out of $S$ alone, and how does it compare with methods built out of polynomial models and tuned constants? The answer is sharp in both directions, and the negative half is as informative as the positive half. The paper proceeds in two stages. Sections 2–7 derive a base rule from the similarity alone, measure it, and locate what it lacks: not a better model but a decision, when to stop trusting the model it has. Section 8 supplies that decision with a similarity of the run to its own past, and Section 9 tests it on four further corpora: three of them exposed two last gaps of the base step and were used to close them, and the fourth was fixed after the method's code was frozen and run once. --- ## 2. The similarity step rule **Definition 1 (Similarity step).** Let $[a,b]$ bracket a root, $x$ be the current iterate and $x_\mathrm{far}$ the opposite end. Let $S:\mathbb{R}\to[0,1]$ with $S(0)=1$ be a similarity function and $\sigma>0$ a scale. The next iterate is $$x' \;=\; x \;+\; \bigl(1 - S(f(x)/\sigma)\bigr)\,(x_\mathrm{far} - x). \tag{1}$$ The rule is exactly the informal statement "step further when the outcome is dissimilar to the expectation, and less far when it is similar". Bisection is the degenerate member $S\equiv\tfrac12$: a similarity function that carries no information about $f$. ### 2.1 The two-point normalisation is false position The most obvious way to make (1) scale-free is to normalise the similarity by the two ends of the bracket, $S(a)=|f(b)|/(|f(a)|+|f(b)|)$ and $S(b)=1-S(a)$, and to place the next iterate at the similarity-weighted average of the ends. **Proposition 1.** For the ratio similarity $S(z)=1/(1+|z|)$ with $\sigma=|f(x_\mathrm{far})|$, equivalently for the endpoint weights above, $$S(a)\,a + S(b)\,b \;=\; \frac{a f(b) - b f(a)}{f(b)-f(a)}.$$ That is the false-position formula. Two things must be said about how far it reaches, because the obvious generalisation is false. It is specific to that similarity. On $f=x^2-2$ over $[1,4]$, where false position gives 1.200000, the same construction gives 1.206812 for $e^{-|z|}$, 1.015228 for $1/(1+z^2)$, 1.007624 for $1/\sqrt{1+z^2}$ and 1.214286 for the truncated linear form: one of five reproduces it. So "a similarity rule scaled by the bracket is not a new method" would be too broad a conclusion, and this paper's own Table 3 refutes it — bracket-scaled configurations take third, fourth and sixth place on Chandrupatla's set, where false position solves 6 of 45. **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. 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 154 for bisection. ### 2.2 The contraction law Write $e_n = x_n - r$ for the error, $D = x_\mathrm{far} - r$, and let $c$ be the slope of $1-S$ at the origin, $1-S(z) = c\,z + O(z^2)$. **Proposition 2 (Contraction law).** For a simple root and $e_n$ small, rule (1) satisfies $$e_{n+1} \;=\; e_n\left(1 - \frac{c\,|f'(r)|\,D}{\sigma}\right) + O(e_n^2), \qquad \text{rate} = \left|1 - \frac{c\,|f'(r)|\,D}{\sigma}\right|. \tag{2}$$ *Derivation.* $|f(x_n)| = |f'(r)|\,|e_n| + O(e_n^2)$, hence $1-S(f(x_n)/\sigma) = c\,|f'(r)|\,|e_n|/\sigma + O(e_n^2)$, and $x_\mathrm{far}-x_n = D - e_n$. Substituting into (1) and discarding second-order terms gives (2). $\square$ Table 1 checks (2) against measurement on $f(x)=x^2-2$ over $[0,2]$ with $\sigma=|f(0)|$, for which $|f'(r)|=2\sqrt2$, $D=2-\sqrt2$ and $\sigma=2$. **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 $c=0$. A similarity with $c=0$ does not converge at a rate at all. Expanding $1-S=c_2z^2+O(z^4)$ in the same way as (2) gives $e_{n+1}=e_n-c_2|f'(r)|^2|D|e_n^2/\sigma^2+O(e_n^3)$, hence $$e_n \;\simeq\; \frac{\sigma^2}{c_2\,|f'(r)|^2\,|D|}\cdot\frac1n, \tag{2a}$$ sublinear rather than stagnant. Reporting a "measured rate" for these rows would report the value of $e_{n+1}/e_n$ at whatever $n$ the window happened to end — it creeps to 1 monotonically (0.925 at $n=10$, 0.991 at $n=100$, 0.9998 at $n=4000$) — so the prediction tested is the coefficient in (2a) instead. | $S(z)$ | $c$ | predicted | measured | |---|---|---|---| | $1/(1+\lvert z\rvert)$ | 1 | rate 0.1716 | 0.1716 | | $e^{-\lvert z\rvert}$ | 1 | rate 0.1716 | 0.1716 | | $1/(1+z^2)$ | 0, $c_2=1$ | $n\,e_n\to0.8536$ | 0.8503 | | $e^{-z^2}$ | 0, $c_2=1$ | $n\,e_n\to0.8536$ | 0.8498 | | $1/\sqrt{1+z^2}$ | 0, $c_2=\tfrac12$ | $n\,e_n\to1.7071$ | 1.6988 | | $\max(0,1-\lvert z\rvert)$ | 1 | rate 0.1716 | fails: stays at $x_\mathrm{far}$ | 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. **Corollary 1 (A chosen similarity is not enough).** For fixed $S$ and fixed $\sigma$ the rate in (2) depends on $|f'(r)|$, $D$ and $\sigma$, none of which is under the method's control. Convergence is at best linear and occurs only when the product lands in $(0,2)$. Empirically, with $\sigma$ taken as $|f|$ at the starting point, five to eight of the eight standard problems fail outright for the six similarity functions tested. Those eight, used throughout as a quick set and never as evidence on their own, are: $x^2-2$ on $[0,2]$; $\cos x-x$ on $[0,1]$; $e^x-2$ on $[0,2]$; $x^3-x-2$ on $[1,2]$; $\sin x-\tfrac12$ on $[0,1.5]$; $x^{10}-1$ on $[0,1.3]$, which stalls false position; $e^{-x}(x-1)$ on $[-1,40]$, whose right end is flat to $10^{-16}$; and $\arctan x$ on $[-1,3]$. **Corollary 2 (Missing the calibration in either direction costs).** The rate degrades as the product $c\,|f'(r)|\,|D|/\sigma$ departs from 1, upwards or downwards. We claim no symmetry between the two directions, and the measurements do not support one: similarities with unbounded slope — $1-\sqrt{|z|}$ has $c\approx3.2\cdot10^{3}$ at the resolution used, $1-|z|^{1/3}$ has $c\approx4.6\cdot10^{4}$ — fail on five and six of the eight problems with a chosen scale, yet over the full benchmark (best over scales, Table 4) they finish at 9411 and 10874, ahead of all three $c=0$ similarities at 13187, 13990 and 15272. Overshooting is the milder failure. **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. --- ## 3. The method Corollaries 1–3 fix every choice in the method before any benchmark is consulted. We record them here as a prediction, and Section 5.3 tests it against the full grid. One number in the safeguard is not fixed by them, and §3.1 declares it as the method's one free constant. **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. **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 $\tfrac16 f'''(\eta)\,(x_n-x_{n-1})(x_n-x_{n-2})$, that is $O(e_{n-1}e_{n-2})$. The Newton step with that derivative satisfies $e_{n+1}=\tfrac{f''}{2f'}e_n^2+\tfrac{f'''}{6f'}\,e_n e_{n-1}e_{n-2}+\dots$; the second term dominates, and the order is the root of $p^3=p^2+p+1$, $p\approx1.839$ — **Muller's order**, as it should be, since the step uses the same parabola as Muller's method, through its derivative rather than its root. Only where the parabola is exact ($f'''=0$) does the order rise to Newton's 2. Carried to $10^{-2900}$ in 3000-digit arithmetic (`order_high_precision.py`), the computational order per step settles at 1.840 on $\cos x-x$, $e^x-2$, $x^3-x-2$ and $\sin x-\tfrac12$ and at 2.000 on $x^2-2$; the two-point scale settles at 1.618. In double precision a run takes three or four superlinear steps before it reaches the tolerance, too few to tell 1.84 from 2, and we report no order measured there. **What the rule reduces to.** With $S(z)=\max(0,1-|z|)$ one has $1-S=\min(1,|z|)$, and (1) becomes $$x' \;=\; x + \operatorname{sign}(x_\mathrm{far}-x)\cdot\min\!\Bigl(\frac{|f(x)|}{|\hat f'|},\;|x_\mathrm{far}-x|\Bigr), \tag{4}$$ a Newton step with a three-point divided-difference derivative, clipped to the bracket and then projected as in §3.1. That the tent turns the similarity step into Newton's is a property of the similarity chosen, not a sign that the similarity has dropped out: which $S$ stands in the step decides the step, and it is measured, not assumed — Table 4 ranks the nine functions over the whole grid, the readings of §7 pay most with the tent in the step and some of them lose with other functions (§7), and in the proposed method replacing the tent by any other function costs 19% to 425% (Table 9a). We state the reduction ourselves rather than leave it to be discovered, because it locates the contribution precisely. The base rule is not a new family of methods. It is (i) a derivation that fixes every choice — which similarity, which scale, which safeguard — before any benchmark is consulted, and predicts which similarity functions come first; (ii) a safeguard of ITP's projection type with ITP's truncation stage, and hence all three of its constants, removed; (iii) the negative results of §10; (iv) the measurements. We call it the **base rule**; the method this paper proposes builds on it in Section 8. **Algorithm.** Given $[a,b]$ with $f(a)f(b)<0$ and tolerance $\varepsilon_x$: ``` evaluate fa = f(a), fb = f(b) A, B <- a, b the proven bracket x <- the endpoint with the smaller |f|; xp <- the other; xp2 <- none n_max <- 2 * ceil(log2(|b-a| / xtol)); allow <- 2^n_max * xtol/2 while |B - A| > xtol: far <- the end of [A,B] further from x f' <- derivative at x of the parabola through (x,xp,xp2) (two-point difference while xp2 does not yet exist) sigma <- |f'| * |far - x| if sigma is 0 or not finite: xn <- midpoint(A,B) else: xn <- x + min(1, |f(x)|/sigma) * (far - x) # the step, eq (4) xn <- clamp(xn, hi - allow, lo + allow) # projection, eq (3) if xn outside (A,B) or already evaluated: xn <- midpoint(A,B) evaluate fn = f(xn); allow <- allow / 2 xp2, xp, x <- xp, x, xn update the bracket: whichever of A,B keeps the sign change return midpoint(A,B) ``` 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'). ### 3.1 The safeguard: project, do not replace 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 makes the convergence linear and takes the Alefeld–Potra–Shi total from 2634 to 7068 — nearly the cost of bisection (7338). The remedy is to keep as much of the computed step as the budget allows. Let $n_{\max}=2\lceil\log_2(w_0/\varepsilon_x)\rceil$ be twice the bisection step count for the initial width $w_0$ and $x$-tolerance $\varepsilon_x$, and let $\epsilon=\varepsilon_x/2$. At step $j$, with the bracket $[\ell,h]$, define the allowance and the projection $$\alpha_j = 2^{\,n_{\max}-j}\epsilon, \qquad x \;=\; \operatorname{clamp}\bigl(x_\mathrm{sim},\; h-\alpha_j,\; \ell+\alpha_j\bigr). \tag{3}$$ The next bracket is at most $\max(x-\ell,\,h-x)$, so (3) gives $w_{j+1}\le\alpha_j$ directly, and induction bounds the run by $n_{\max}$ steps in exact arithmetic. In floating point the bracket can land exactly on the boundary, and the termination test then asks for one more; the guarantee stated and verified is therefore $2\lceil\log_2(w_0/\varepsilon_x)\rceil+3$ evaluations. **The multiplier is the method's one free numerical constant.** It is not "the factor already claimed in the bound" — that argument is circular, the bound is twice bisection because two was chosen — and measurement settles it (`measure_budget_factor.py`, Linux): **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. | $c$ | APS-154 | CHA-45 | both | |---|---|---|---| | 1 | 4625 | 2039 | 6664 | | 1.25 | 2626 | 1822 | 4448 | | 1.5 | 2636 | 1991 | 4627 | | 2 | 2634 | 2357 | 4991 | | 3 | 2634 | 2936 | 5570 | | 4 | 2634 | 3229 | 5863 | On the canonical set the projection binds on one problem only once $c\ge1.25$ — `aps.13.00`, whose $f$ underflows to zero around the root (20, 30 and 28 evaluations at $c=1.25$, 1.5 and 2) — so the total barely moves. On the wide-bracket set it moves monotonically with $c$ from 1.25 on, and 2 is not the best value. Nor is zero slack ($c=1$) a degenerate case under which the guard behaves as bisection: it costs 4625 and 2039, far from bisection's 7338 and 2136, and on the wide-bracket set it beats $c=2$. The multiplicative family has no member that the bound singles out, so its multiplier is a genuinely free constant. **The budget earned rather than scheduled: a constant-free variant.** The constant can be removed. Let $s_j=\log_2(w_j/w_{j+1})$ be the shrink a step delivers in bits — bisection delivers exactly 1 — and keep a credit $$C \leftarrow \max(0,\;C+s_j-1),\qquad \alpha_j = w_j\,2^{\,C-1},\qquad x=\operatorname{clamp}\bigl(x_\mathrm{sim},\,h-\alpha_j,\,\ell+\alpha_j\bigr), \tag{3'}$$ leaving the step alone while $C\ge1$. A step may under-deliver only by what earlier steps banked, and since $C$ never goes negative the shrinks over $j$ steps sum to at least $j-C_0$: the run costs at most bisection plus $\lceil C_0\rceil$ steps. The endowment $C_0$ is therefore the promise, not a setting. $C_0=0$ makes the allowance exactly half the bracket on the first step, which forces the midpoint and makes the method bisection; every $C_0\in(0,1]$ promises the same single extra step; and among those, 1 is the most permissive and — measured — the cheapest: **Table 1b.** The earned projection at endowment $C_0$. Total evaluations, failures charged. | $C_0$ | promise | APS-154 | CHA-45 | both | |---|---|---|---|---| | 0.1 | bisection + 1 step | 5150 | 2097 | 7247 | | 0.5 | bisection + 1 step | 4462 | 2037 | 6499 | | **1** | **bisection + 1 step** | **3377** | **2073** | **5450** | | 2 | bisection + 2 steps | 3502 | 1757 | 5259 | | 5 | bisection + 5 steps | 2819 | 1666 | 4485 | A looser promise buys evaluations, as it must; within one promise the cheapest endowment is determined. The constant-free variant is therefore the earned projection with $C_0=1$, whose guarantee — $\lceil\log_2(w_0/\varepsilon_x)\rceil+4$ evaluations, bisection plus one step with one more for the floating-point boundary — is verified on all 199 problems. We nonetheless propose the scheduled projection with $c=2$ and declare its multiplier. On the canonical set it is 743 evaluations cheaper than the earned variant (2634 against 3377), and there its total barely depends on the multiplier; on the wide-bracket set the earned variant is the cheaper (2073 against 2357). The two are the ends of a trade stated in numbers: one declared constant and the better canonical result, or no constant, a tighter bound and the better wide-bracket result. The value 2 was fixed before any benchmark was consulted. Two remarks apply to both forms. - **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. - **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. --- ## 4. Experimental protocol **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. **Cost model.** Evaluations of $f$. A run counts as solved when it reports convergence and $f$ changes sign within $10^{-6}$ of the returned point, or vanishes there — the bracketing notion of a root, the same criterion on every corpus of this paper; $x$-tolerance $10^{-12}$; budget 200 evaluations. Scoring against the published root instead, as the benchmarks' harness does, charges every method the full budget on six problems where $f$ vanishes identically around the root in floating point (§7.3). Head-to-head counts (wins/ties/losses) are taken over every problem of a set with the same charging as the totals. *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. **Fairness to the competitors.** Three deliberate concessions. 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. 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. 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. For Brent the library is slightly dearer than our implementation; we use the library anyway, because it is the implementation a reader can check. ITP has no library implementation, which is why it is checked against its own theorem instead; bisection and the false-position family are not in SciPy either and are ours. **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 2635 for TOMS 748 on Windows and 2636 on Linux for the same set. Every number stated in this paper is checked against the run data by a script that is part of the test suite. --- ## 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 | 154/154 | 2595 | 0 | | 2 | the same rule with no safeguard at all | no | 154/154 | 2634 | 0 | | **3** | **base rule: tent, 3-point, projection ($c=2$)** | **no** | 154/154 | **2634** | 0 | | 4 | TOMS 748 | yes | 154/154 | 2636 | 0 | | 5 | Brent | yes | 154/154 | 2707 | 0 | | 11 | ITP, per-problem oracle | yes | 154/154 | 3068 | 0 | | 15 | Ridders | no | 154/154 | 3124 | 0 | | 25 | constant-free variant: earned projection | no | 154/154 | 3377 | 0 | | 29 | ITP, recommended constants | yes | 154/154 | 3767 | 0 | | 73 | bisection | no | 154/154 | 7338 | 0 | | 232 | Dekker | no | 153/154 | 3158 | 0 | | 237 | Illinois | yes | 153/154 | 7045 | 0 | | 244 | false position | no | 127/154 | 10560 | — | The base rule and the same rule without any safeguard tie at 2634: the projection never binds on this set. By total, Chandrupatla's method is first with 2595, 1.5% fewer; the base rule and TOMS 748 are level (2634 against 2636). Problem by problem the base rule wins against Chandrupatla 87/24/43 (wins/ties/losses; exact two-sided sign test $p=1.4\cdot10^{-4}$), against TOMS 748 86/16/52, Brent 102/11/41, ITP at its recommended constants 132/3/19, the per-problem ITP oracle 121/4/29, and bisection 152/1/1. It wins twice as many problems as it loses against Chandrupatla and trails that method in total because its losses are the larger ones; the problems are not independent (§9), and taken by function family the comparison is 6/1/8 of 15 families, a tie. Dekker and Illinois solve 153 of 154 and are ranked after every method that solves all. The constant-free variant goes 79/3/72 against Chandrupatla ($p=0.63$, a tie in problems), 78/9/67 against TOMS 748 and 79/5/70 against Brent, and loses on the total to all three. **Table 3.** Chandrupatla, 45 problems, brackets up to $10^{12}$ wide, out of 264 configurations. | # | configuration | tuned | solved | total evals | $>2\times$ | |---|---|---|---|---|---| | 1 | Chandrupatla | no | 45/45 | 1505 | 0 | | 2 | ITP, per-problem oracle | yes | 45/45 | 1675 | 0 | | 3 | similarity, $e^{-\lvert z\rvert}$, bracket scale, both guards | no | 45/45 | 1778 | 0 | | 5 | ITP, recommended constants | yes | 45/45 | 1946 | 0 | | 9 | constant-free variant: earned projection | no | 45/45 | 2073 | 0 | | 15 | bisection | no | 45/45 | 2136 | 0 | | **49** | **base rule: tent, 3-point, projection ($c=2$)** | **no** | 45/45 | **2357** | 0 | | 158 | TOMS 748 | yes | 45/45 | 3088 | 15 | | 227 | Brent | yes | 43/45 | 3238 | 18 | | 229 | the same rule with no safeguard at all | no | 39/45 | 3391 | 10 | Here Chandrupatla's method is genuinely the stronger: it beats the base rule 29 problems to 11 ($p=0.006$) and by 1505 against 2357 in total. The base rule beats Brent 34/4/7 and TOMS 748 33/1/11 — which exceed $2\times$ bisection, a bound they do not promise, on 18 and 15 problems — and loses to ITP 18/1/26 and to the ITP oracle 16/1/28. The wide-bracket set is where the multiplier matters (Table 1a) and where the constant-free variant is the cheaper form, 2073 against 2357. Over both sets the base rule needs 4991 and the constant-free variant 5450, against 4100 for Chandrupatla, 4743 for the ITP oracle, 5724 for TOMS 748 and 5945 for Brent. ### 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 | 154/154 | 2634 | 0 | | APS, projection — base rule ($2\times$ bisection) | 154/154 | 2634 | 0 | | APS, earned projection — constant-free (bisection + 1 step) | 154/154 | 3377 | 0 | | Chandrupatla, no safeguard | 39/45 | 3391 | 10 | | Chandrupatla, projection — base rule | 45/45 | 2357 | 0 | | Chandrupatla, earned projection — constant-free | **45/45** | **2073** | **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 six more problems for 30% fewer evaluations. The earned projection promises twice as much — a worst case of bisection plus one step instead of twice bisection — and pays for it on the canonical set, 743 evaluations or 28%, while on the wide-bracket set it is the cheapest of the three. Both are the opposite of the usual trade on at least one set, and that is the point of projecting rather than replacing: the two safeguards that replace cost 7068 and 7046 on Alefeld–Potra–Shi, against 2634. ### 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. (These are the evaluations the timing loop actually spent; for the base rule and bisection, which solve every problem, they equal the totals of §5; Brent here is our pure-Python implementation, which is what is timed.) That inversion is the point. Making $f$ progressively more expensive by a controlled amount of arithmetic and timing the sweep directly — nothing subtracted, nothing modelled — the base rule overtakes ITP once one call to $f$ costs about 9 µs and Brent at about 20 µs, which on this machine is of the order of a hundred arithmetic operations. Below that the ranking is decided by per-step overhead; above it, by evaluations. Three caveats, all of them limitations of the measurement rather than of the methods. Timings vary by tens of percent between runs even on an idle machine. An attempt to separate our per-step cost into "algorithm" and "Python call overhead", by timing a flattened rewrite of the same rule, produced a variant that diverged from the original on 42 of the 199 problems and was therefore discarded rather than quoted; the question stays open. And an earlier version of this measurement estimated a per-evaluation overhead by subtracting the measured cost of $f$, which returned *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 that the similarities with $c=1$ come before those with $c=0$ under a data-derived scale; they say nothing about the order within a group, nor about the functions with unbounded slope. Table 4 reports it. For every data-derived scale the three similarities with $c=1$ occupy the top three places and the three with $c=0$ the bottom three. That $c=0$ is slow is the unsurprising half — a step proportional to $z^2$ is damped quadratically. For the chosen scale the two groups are indistinguishable (14983 against 14228), which is consistent — with a chosen scale nothing converges well and the ordering carries no information. **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 | **5764** | 21200 | | 2-point derived | 5854 | 21089 | | bracket | 9149 | 18015 | | start (chosen) | 14983 | 14228 | Ranked individually and best over scales, the similarities come out $1-|z|$ (5422), $e^{-|z|}$ (5772), $1/(1+|z|)$ (5993), $(1+|z|)^{-2}$ (8455), $1-\sqrt{|z|}$ (9411), $1-|z|^{1/3}$ (10874), $e^{-z^2}$ (13187), $1/(1+z^2)$ (13990), $1/\sqrt{1+z^2}$ (15272); the scales come out 3-point (5422), 2-point (5564), bracket (7972), start (13423). The group order and the order of the scales are those the law predicts and were predicted before the grid was run; the order inside each group is measured, not predicted. --- ## 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. The bracket here is reconstructed from *every* sign the method has observed — the narrowest interval the evaluations prove, not the one the method keeps; Brent and TOMS 748 always probe inside their own bracket, but 69 and 141 of their probes fall outside this narrowest one and remove nothing from it, and those probes are left out of the median. High-order interpolation is a wager that $f$ is locally polynomial. When the wager pays, a high order wins the endgame; when it fails, the probe removes little. The similarity rule makes the weakest available assumption — a single local slope — and degrades gracefully. --- ## 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 all 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, 199/199) | APS-154 | CHA-45 | both | |---|---|---|---| | base rule: tent, 3-point, projection $c=2$ (§3) | 2634 | 2357 | 4991 | | constant-free variant: earned projection (§3.1) | 3377 | 2073 | 5450 | | (1) trust before the step, projection | 2663 | 2170 | 4833 | | (1)+(2) points chosen by shape, *safe* | **2337** | 2177 | 4514 | | (1)+(3) far end by dissimilarity, projection | 2539 | **1864** | **4403** | | (1)+(2)+(3), *safe* | 2695 | 1928 | 4623 | | Chandrupatla | 2595 | 1505 | 4100 | | TOMS 748 | 2636 | 3088 | 5724 | | ITP, per-problem oracle | 3068 | 1675 | 4743 | Two results here are worth stating plainly. First, the canonical set improves from 2634 to **2337**, which is below every method measured, including Chandrupatla's 2595, with the worst-case guarantee intact. It hardly feels the declared multiplier of §3.1: 2337 for every multiplier from 1.5 to 4, 2351 at 1.25 (`measure_records_vs_multiplier.py`). Second, on the two sets together the 4403 of reading (3) passes ITP's oracle (4743) but not Chandrupatla (4100): the remaining gap is entirely on the wide-bracket families, where Chandrupatla needs 1505 against our 1752 (reading (3) under *safe*). The readings do not compose. Taken together, (2) and (3) give 2695 and 1928 — worse than either alone on the canonical set, and on the wide-bracket set worse than (3) alone. Two selections that are each an improvement interfere, which is a reason to measure composition rather than assume it. **Which similarity stands in the step decides whether a reading pays.** The readings above, and the decision *defect* of §7.2, compare quantities directly — an ordinal judgment of which is more similar, in which every decreasing $S$ would decide alike. Their gains nevertheless depend on the $S$ that sets the step (`audit_similarity_places.py`, on the same sweeps): on the canonical set the triple chosen by shape gains 493 evaluations with the tent in the step and loses 250 with $1/\sqrt{1+z^2}$; *defect* gains 283 with the tent and loses 215 with $1-|z|^{1/3}$. The records of this section are the pairing of the tent step with these judgments, not the judgments alone. Under the constant-free earned projection the same readings gain little or lose. Reading (1) gives 3316 and 2050, (1)+(2) 3564 and 2074, (1)+(3) 3666 and 2052, all three together 4163 and 2087, against 3377 and 2073 for the constant-free variant itself: the earned budget binds early on the canonical problems, and what the readings buy under the projection is mostly room that the earned budget does not grant. ### 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 — $m=1/(1-ff''/f'^2)$, Lagouanelle's estimate of the multiplicity [13] with the derivatives replaced by divided differences, comes from three evaluations already paid for, as in item 5 of §10, and is 1 whenever the power model does not apply. Measured over the grid, the correction behaves as that reading predicts and not as a universal improvement. Paired inside each (function, safeguard) cell of the far-end reading it is cheaper in 33 of 63 cells — indistinguishable from a coin. But in the one pairing where that reading *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 33 of 63 cells over all nine functions. | safeguard | APS, $m=1$ | APS, measured $m$ | CHA, $m=1$ | CHA, measured $m$ | |---|---|---|---|---| | both | 4331 | **3485** | 2469 | **2306** | | bracket | 4418 | **3352** | 2671 | **2428** | | credit | 4999 | **4495** | 2064 | **1933** | | projection | 6783 | **6176** | 4057 | **3751** | | safe | 5803 | **5040** | 3218 | **3032** | | step | 6251 | **5362** | 3371 | **3146** | | none | 10676 | **9855** | 8265 | **7424** | 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 1933, against 1752 for reading (3) above and 1505 for Chandrupatla. In the *slope* reading of §3 the same exponent is harmful, $2634\to2703$, which is the same statement read from the other side — a function that does not belong to a place does not become right by being applied there. ### 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, $S(|\varphi-\xi|)\tfrac12$. 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 two sets of §4 (failures charged 200). A′'s wins/ties/losses against each rival over all problems of the set, exact two-sided sign test. | | APS-154 | CHA-45 | both | A′ per problem, APS | A′ per problem, CHA | |---|---|---|---|---|---| | **A′ (proposed)** | **2259** | **1074** | **3333** | — | — | | A′, earned budget (no free constant) | 3417 | 1852 | 5269 | | | | base rule (§3) | 2634 | 2357 | 4991 | | | | Chandrupatla | 2595 | 1505 | 4100 | 103/20/31, $p=3\cdot10^{-10}$ | 31/4/10, $p=0.001$ | | TOMS 748 | 2636 | 3088 | 5724 | 127/6/21 | 42/2/1 | | Brent | 2707 | 3238 | 5945 | 118/17/19 | 39/3/3 | | ITP, per-problem oracle | 3068 | 1675 | 4743 | 126/7/21 | 35/1/9 | | ModAB [10, 20] | 2726 | 1681 | 4407 | 118/10/26 | 34/4/7 | A′ is 12.9% below Chandrupatla on the canonical set and 28.6% below on the wide-bracket set, and wins more problems than it loses against every rival on both. By function family, the unit in which these problems are independent (§9), it is ahead of Chandrupatla on 8 of the 15 canonical families (6 behind, one level) and on 7 of the 9 wide-bracket families — too few families for either count to be significant ($p=0.79$, $0.18$). The two it is behind on are `fun7`, 69 against 64, and `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. **Table 9a.** The similarity in the step of A′: the same method with each of the nine similarity functions of §5.3 in place of the tent, everything else unchanged. Totals on the two sets of §4 and over the six corpora of §9; the guarantee holds with every function. | $S$ in the step | APS + CHA | six corpora (855) | |---|---|---| | **tent $1-|z|$ (A′)** | **3333** | **13032** | | $e^{-|z|}$ | 3965 | 14142 | | $1/(1+|z|)$ | 4424 | 14843 | | $(1+|z|)^{-2}$ | 7249 | 28500 | | $1-\sqrt{|z|}$ | 11613 | 45552 | | $1-|z|^{1/3}$ | 16094 | 63031 | | $e^{-z^2}$ | 17429 | 67701 | | $1/(1+z^2)$ | 17452 | 67394 | | $1/\sqrt{1+z^2}$ | 17487 | 67854 | The nearest other function costs 19% more on the two sets and 9% more over the six corpora, and the functions that approach 1 quadratically cost five times more; the three functions with a linear approach come first and the three with a quadratic approach last, as the contraction law of §2.2 predicts. What this measures should be said plainly: with the tent the step is the undamped Newton step clipped to the bracket, and the other functions damp it, so the table shows that the undamped step is the better one — expected, and evidence for the choice of $S$ within the rule rather than for the similarity framing as such. ### 8.6 What is known, and what is new Several mechanisms of A′ coincide with classical ones, and the coincidence is worth stating, because they were not taken from the literature but reached from the principles of the paper — similarity to a prototype and analogy as similarity of relations — and recognised afterwards: | in A′ | classical counterpart | |---|---| | the three-point order estimate of §7.1 | Lagouanelle's multiplicity estimate $m=f'^2/(f'^2-ff'')$ [13] | | the contraction $e(1-1/m)$ behind the flag of §8.1 | multiplicity estimates from successive Newton iterates [14, 15] | | the order (6) of an analogous pair | the log–log slope of a power law; the (approximate) computational order of convergence [18, 19] | | the split (7) | false position on $\operatorname{sign}(f)|f|^{1/m}$ [16, 17] | | the half the step promises | a relative of Brent's acceptance condition [2] | | the tolerance step | Dekker [9], Brent [2] — borrowed and named as such | | the exit judgment of §8.2 (the value at a split against the line) | ModAB's linearity check at the midpoint [10, 20], $|y_m-y_3|1$ on 27.9% of steps with excursions to $m=34.8$, against 35.8% on the flat families. 27.9 against 35.8 is no separation. A cruder variant, correcting by the fraction of its own promise the previous step kept, fails everything: the convergence becomes sublinear and the standard problems double to 161. 6. **The harness scores one canonical problem 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. With the five `fun7` problems of §7.3 it is why this paper scores by a sign change. 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). 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 4258 on the canonical set against 2337 — a difference of 1921 evaluations for one choice of criterion. A secant step routinely collapses the error while barely moving the proven bracket, so bracket progress rejects steps that were excellent; this is the same mechanism that makes the bracket guard of §3.1 cost 7068. 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: its best cell is 3101 on the canonical set against 2539 for the best uncapped cell of the same grid, and 1766 on the wide-bracket set against 1752. 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 129/154 and 9855 — two problems and eight per cent — where Illinois, at the price of one constant (halve the stale ordinate), reaches 153/154 and 7045. A sixth of the gap is not the gap. The correction is real and it is in the right place; it is simply much smaller than a constant that was chosen for exactly this failure. *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. 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 2759 to 2598 — but the canonical set gets worse, 5635 to 6749, against the 2634 of the base rule. 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. 12. **Reading the unkept promise one step at a time does not pay.** The flag of §8.1 sees the flat regime, but a halving in place of each flagged step takes the canonical set from 2634 to 2839 and the wide-bracket set only from 2357 to 2243: a simple root far from convergence breaks its promises too. Requiring the promise to fail twice in a row keeps the canonical set (2613) but fires one step late, before the crossing, and replaces the most useful step of the cycle (wide brackets 2388); requiring a creep in one direction as well gives 2688 and 2399. 13. **A halving that sticks once begun wins the flat families and loses everything else.** Without an exit the flat families fall to 1057 — pure bisection's level — and the canonical set rises to 6707, because a false entry now lasts to the end of the run. Splitting at the order read from the failed promise, $(1-1/m)^m=r$, changes nothing: that order is $m=\infty$ in 163 of 174 entries (6591). 14. **The model's own point cannot tell a flat root from a simple one.** Judging the evaluated point against the line through the bracket ends and against 0 — the exit judgment of §8.2 — also at *entry* gives 2820 and 1532 against 2935 and 1546 for the exit alone: the model aims at the root, so $f$ is small there by construction and looks flat at a simple root as readily as at a power. The same judgment works at a split, which the model did not choose. 15. **The decision reading "defect" does not transfer.** Combined with the regime it was the best configuration on the two development sets (§7.2's defect reading with the flat families of §8), and it passed every pre-registered criterion there. On new corpus 1 it did not: per problem 95/33/78 against Chandrupatla ($p=0.22$), losing the class of exponential flanks 9 to 26. It is therefore not part of A′. 16. **A component rejected alone can pay in combination.** Remembering the analogy for the run was rejected by its own pre-registration, under the harness scoring of the time (canonical set 2814 against 2806 without the regime; under the sign-change criterion the two are level, 2633 against 2634), and with the split of §8.3 it is the better form (with the step pattern, wide brackets 1250 against 1388 for the analogy read in one state). Composition was measured, not assumed, throughout — which is also why §7's readings (2) and (3), each an improvement, are not combined. --- ## 11. Conclusion A root finder can be derived from a similarity function alone, and the derivation shows where the content of such a method actually sits. It is not in the shape of $S$ — nine shapes were tried and which group of them comes first is dictated by a single number, the slope of $1-S$ at the origin. It is not in making the similarity change faster — that overshoots exactly as readily as flatness stagnates. It is in the *scale* against which similarity is judged: chosen, the method is brittle and merely linear; taken from the bracket, it is false position and three centuries old; read off the data, it is superlinear. And the safeguard that makes such a method safe must project the step, not replace it: replacing costs the order, projecting costs nothing on one benchmark and improves the method on the other. The projection's budget multiplier is a free constant, and we declare it: it does not move the canonical result and it does move the wide-bracket one; projecting onto a budget the steps earn instead removes it, at a measured price. The base rule has one declared numerical constant, converges with Muller's order 1.839 and carries a worst case of twice bisection plus one step. On the canonical Alefeld–Potra–Shi set it is level with TOMS 748 in total and 1.5% behind Chandrupatla's method, while winning more problems than it loses against both; on the wide-bracket set Chandrupatla is genuinely stronger (1505 against 2357). §7.4 locates the cause: not the order of the model but the decision to give it up, which Chandrupatla takes by halving and which no reading of a single state takes correctly for our step. That decision is a similarity too — of the run to its own past. A pure power repeats its configuration under zoom, an exponential under translation, a simple root tends to the line; the analogous pair reads the local order, the order decides between the model and a halving by bisection's own $\tfrac12$, and it says where to split. The method built on it, A′, is ahead of Chandrupatla on both canonical sets in total and problem by problem, and on a corpus fixed after its code was frozen it needs 36% fewer evaluations and wins 177 problems to 28. Its decision layer beats Lagouanelle's estimate, Chandrupatla's band and no detector put in its place; with the multiplicity known in advance the classical transformation is 2.4 to 6.8 times cheaper on multiple roots, which measures what estimating the order costs. The claim the measurements support is a lead over Brent, TOMS 748, ITP and ModAB everywhere, within a bound of twice bisection that Brent, TOMS 748 and ModAB do not promise and exceed, and over Chandrupatla a lead that is large where roots are multiple, fractional or nearly flat and small where they are simple — not superiority by the same margin everywhere. `fun9`, near-cancelling terms and smooth applied equations are where it is behind or level. This is the same movement as in parameter-free similarity control [6], where membership functions are eliminated by normalising the input space from the data rather than by choosing widths. Here the object normalised is the value axis, and the price of choosing it by hand is measured rather than asserted. ### Future work Three questions remain, all narrower than the one this section used to pose. The first is the price of estimating the order. With the multiplicity given, the transformation of §8.3 is 2.4 to 6.8 times cheaper than A′ on multiple and fractional roots (Table 9c): the order read from the analogous pair, or its use in the split, loses most of what the transformation could deliver. The second is the steep flank. Where the loss remains — `fun9`, the near-cancelling class, 4 − tan x at its pole — the base step approaches along a flank far steeper than its model, and the half contradiction of §8.4 removes only part of the cost. The third is the one geometric choice of §8.2, the Euclidean metric of the $(\xi,\varphi)$ plane: a metric derived from the configuration's own sensitivity would remove it. --- ## References 1. G. E. Alefeld, F. A. Potra and Y. 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