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Valeri's decisions (1) competitors from the library, (2) variant a. (1) Chandrupatla is scipy.optimize.elementwise.find_root, Brent is brentq, no more per-problem best_of. Our Chandrupatla stopped on a bracket twice as wide as every other method (the factor the protocol corrects for ITP); fixed (+0.5*xtol) in baselines.py and in the two copies found by grep (stage16_blend, stage19_modern) and the traced mirror, and pinned: ours == library on every problem. Regenerating sweep_full.json changed exactly three rows - chandrupatla APS 2785->2787, CHA 2423->2441, brent CHA 4099->4115 - and nothing else (diff_sweeps.py: 0 unexpected).

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(2) The proposed method is again the scheduled projection (2806/3220/6026), its budget multiplier DECLARED as the one free constant: the canonical total is the same for every multiplier 1.25-4, the wide-bracket one is not. The earned projection is the constant-free variant (3568/3020/6588). What the old tables hid: problem by problem the proposed rule beats Chandrupatla on the canonical set 87/24/42, exact sign test p = 9e-5, while trailing him by 0.7% in total; on the wide-bracket set he is genuinely stronger (26 to 9, p = 0.006). Gap decomposition recomputed with the library: 244 (was 262), 160 of it on the flat families (two thirds), fun1 58 (a quarter, not a third as the abstract said).

Also corrected on the way: the 'subsumed by Chandrupatla' pin claimed he is better on the worst single problem - true only with the looser tolerance (84 against the credit rule's 83 once fixed); §8.3 still called zero slack degenerate, which Table 1a refutes; §5.2's Brent figure is our pure-Python implementation and now says so. verify_article.py 286/286, 802 tests pass twice.

Co-Authored-By: Claude Opus 5 <noreply@anthropic.com>
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@@ -8,9 +8,9 @@ V. Sitnikov
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.
The derivation fixes the similarity and the scale before any benchmark is consulted — one whose approach to 1 is linear, and the finest available data-derived scale — and one requirement fixes the safeguard: it must carry a bound without a free constant, which leaves a projection of the step onto the admissible interval with the bisection budget *earned* step by step rather than scheduled in advance. What results is, explicitly, a Newton step with a three-point divided-difference derivative, clipped to the bracket and projected onto an earned 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 bisection plus one step and contains no free constant. On the 154 problems of Alefeld, Potra and Shi it needs 3568 evaluations of $f$, 29th of 264 configurations, against 2785 for Chandrupatla's method, 2820 for TOMS 748 and 2889 for Brent. The variant whose budget is scheduled at twice bisection needs 2806 — level with TOMS 748 and 0.75% behind Chandrupatla — but its multiplier is a free constant, and a load-bearing one: on the wide-bracket set its value moves the total from 2752 to 4006. On the 45 problems of Chandrupatla the proposed rule needs 3020, ahead of TOMS 748 (4000) and Brent (4099) and behind Chandrupatla (2423) and ITP (2857). For every data-derived scale, the ranking of the nine similarity functions matches the derived law without exception.
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.
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 under the scheduled projection, below every method measured here including Chandrupatla's 2785; 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. Those records carry the projection's multiplier, although on the canonical set they barely feel it — 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: a third 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.
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.
---
@@ -90,7 +90,7 @@ Empirically, with $\sigma$ taken as $|f|$ at the starting point, five to eight o
## 3. The method
Corollaries 1–3 fix the similarity and the scale before any benchmark is consulted; the safeguard is fixed by one requirement, that it carry a bound without a free constant (§3.1). We record the first two as a prediction, and Section 5.3 tests it against the full grid.
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.
@@ -110,7 +110,7 @@ We state this ourselves rather than leave it to be discovered, because it locate
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
C <- 1 one bit: the promise "bisection + 1 step"
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)
@@ -119,18 +119,15 @@ while |B - A| > xtol:
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)
if C < 1: # earned projection, eq (3')
allow <- |B - A| * 2^(C - 1)
xn <- clamp(xn, hi - allow, lo + allow) # the midpoint if empty
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); w <- |B - A|
evaluate fn = f(xn); allow <- allow / 2
xp2, xp, x <- xp, x, xn
update the bracket: whichever of A,B keeps the sign change
C <- max(0, C + log2(w / |B - A|) - 1) # bank what beat a halving
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. The one number in it, the endowment of one bit, is the bound being promised rather than a setting (§3.1).
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
@@ -142,7 +139,7 @@ $$\alpha_j = 2^{\,n_\max-j}\epsilon, \qquad x \;=\; \operatorname{clamp}\bigl(x_
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 doubling is a free constant, and it is load-bearing.** 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):
**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):
**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.
@@ -157,7 +154,7 @@ The next bracket is at most $\max(x-\ell,\,h-x)$, so (3) gives $w_{j+1}\le\alpha
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.
**The budget earned rather than scheduled.** The proposed safeguard removes that constant. 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
**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'}$$
@@ -173,9 +170,9 @@ leaving the step alone while $C\ge1$. A step may under-deliver only by what earl
| 2 | bisection + 2 steps | 3692 | 2712 | 6404 |
| 5 | bisection + 5 steps | 3006 | 2615 | 5621 |
A looser promise buys evaluations, as it must; within one promise the cheapest endowment is determined. So the proposed method is 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. The scheduled projection with $c=2$ is kept as the other end of a stated trade: one free constant, and on the canonical set 762 evaluations cheaper (2806 against 3568).
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.
This is the one choice in the method that was *not* fixed before a benchmark was consulted: the earned budget replaced the scheduled one after the question whether the multiplier was load-bearing. The reason for the replacement is structural — it removes a free constant — and not its cost, which on the canonical set is higher; a replacement chosen by the numbers would have gone the other way.
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.
Two remarks apply to both forms.
@@ -196,7 +193,7 @@ Two remarks apply to both forms.
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 run at **the better of SciPy's implementation and ours, problem by problem**. This is not a formality. Our implementations turned out to be slightly *cheaper* than the library's — 6991 evaluations against `brentq`'s 7004, and 5209 against `_chandrupatla`'s 5228 — so quoting the library figure alone would have flattered us by a fifth of a percent, which on a 0.5% margin is not nothing. The two agree on 191 of 199 problems for Brent and 179 of 199 for Chandrupatla, the differences being one or two evaluations from the relative-tolerance term in the stopping rules. 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.
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.
@@ -204,45 +201,45 @@ Two remarks apply to both forms.
## 5. Results
The grid is nine similarity functions × four scales × seven safeguards = 252 configurations, plus twelve baselines: 264 in all. The configuration proposed in Section 3 — tent similarity, three-point scale, earned projection — is one entry of that grid, fixed by the theory and by the requirement of §3.1 rather than chosen from the ranking; the scheduled projection is shown beside it as the other end of the trade. Chandrupatla's own method and Dekker's are among the baselines; an earlier version of these tables was drawn up before either was added, and did not show that Chandrupatla's method is first on the canonical set.
The grid is nine similarity functions × four scales × seven safeguards = 252 configurations, plus twelve baselines: 264 in all. The configuration proposed in 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 | 2785 | 0 |
| 1 | Chandrupatla | no | 153/154 | 2787 | 0 |
| 2 | the same rule with no safeguard at all | no | 153/154 | 2806 | 0 |
| 3 | scheduled projection, $c=2$ (one free constant) | no | 153/154 | 2806 | 0 |
| **3** | **proposed: 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** | **proposed: tent, 3-point, earned projection** | **no** | 153/154 | **3568** | 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 | — |
On the problems both solve, the proposed configuration goes 79/3/71 (wins/ties/losses) against Chandrupatla, 77/9/67 against TOMS 748, 78/5/70 against Brent, 103/10/40 against ITP at its recommended constants, 86/6/61 against the per-problem ITP oracle and 122/1/30 against bisection: it wins more problems than it loses against every competitor, and still loses on the total to Chandrupatla, TOMS 748, Brent and the ITP oracle, because its losses are the larger ones. The scheduled projection goes 86/25/42 against Chandrupatla, 86/16/51 against TOMS 748, 102/11/40 against Brent, 132/3/18 against ITP, 121/4/28 against the oracle and 152/1/0 against bisection. Its total, 2806 against TOMS 748's 2820, is a margin of 0.5% and should be read as a tie; against Chandrupatla's 2785 it is 0.75% behind.
The proposed 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 proposed 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 proposed 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 | 2423 | 0 |
| 2 | Chandrupatla | no | 40/45 | 2441 | 0 |
| 3 | ITP, per-problem oracle | yes | 40/45 | 2611 | 0 |
| 4 | similarity, $e^{-\lvert z\rvert}$, bracket scale, both guards | no | 40/45 | 2733 | 0 |
| 6 | ITP, recommended constants | yes | 40/45 | 2857 | 0 |
| **10** | **proposed: tent, 3-point, earned projection** | **no** | 40/45 | **3020** | 0 |
| 10 | constant-free variant: earned projection | no | 40/45 | 3020 | 0 |
| 16 | bisection | no | 40/45 | 3090 | 0 |
| 47 | scheduled projection, $c=2$ | no | 40/45 | 3220 | 0 |
| **47** | **proposed: tent, 3-point, projection ($c=2$)** | **no** | 40/45 | **3220** | 0 |
| 164 | TOMS 748 | yes | 40/45 | 4000 | 15 |
| 226 | Brent | yes | 38/45 | 4099 | 18 |
| 226 | 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 a proposal. The proposed configuration is tenth, and here the earned budget is the cheaper form: 3020 against the scheduled projection's 3220. Problem by problem it goes 24/0/16 against TOMS 748 and 18/0/20 against Brent — both of which exceed the $2\times$ bound on 15 and 18 problems — and loses clearly to Chandrupatla (5/0/35) and to the ITP oracle (2/3/35); to ITP it loses on the total but not on problems (12/20/8, and 3020 against 2857); and problem by problem it loses even to bisection (5/0/35), whose total of 3090 it nonetheless beats. Over both sets the proposed rule needs 6588 and the scheduled projection 6026, against 5208 for Chandrupatla, 5871 for the ITP oracle, 6820 for TOMS 748 and 6988 for Brent.
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 proposed 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 proposed rule needs 6026 and the constant-free variant 6588, against 5228 for Chandrupatla, 5871 for the ITP oracle, 6820 for TOMS 748 and 7004 for Brent.
### 5.1 What the guarantee costs
@@ -251,23 +248,23 @@ It depends on which bound is bought. Against the identical rule with no safeguar
| | solved | total evals | runs over the $2\times$ bound |
|---|---|---|---|
| APS, no safeguard | 153/154 | 2806 | 0 |
| APS, scheduled projection ($2\times$ bisection) | 153/154 | 2806 | 0 |
| APS, earned projection (bisection + 1 step) | 153/154 | 3568 | 0 |
| APS, projection — proposed ($2\times$ 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, scheduled projection | 40/45 | 3220 | 0 |
| Chandrupatla, earned projection | **40/45** | **3020** | **0** |
| Chandrupatla, projection — proposed | 40/45 | 3220 | 0 |
| Chandrupatla, earned projection — constant-free | **40/45** | **3020** | **0** |
The scheduled projection 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.
The proposed projection 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 scheduled projection (the configuration timed) 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.
On the 199 problems as they stand, every $f$ being a formula of a few operations, the proposed 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 scheduled projection, 5932 against 6991 for Brent, 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.
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 proposed 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 scheduled projection 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.
That inversion is the point. Making $f$ progressively more expensive by a controlled amount of arithmetic and timing the sweep directly — nothing subtracted, nothing modelled — the proposed rule overtakes ITP once one call to $f$ costs about 9 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.
@@ -314,21 +311,21 @@ Where a safeguard other than the projection of §3.1 appears, it is one of the g
| configuration (all with the guarantee, 193/199) | APS-154 | CHA-45 | both |
|---|---|---|---|
| proposed: tent, 3-point, earned projection (§3.1) | 3568 | 3020 | 6588 |
| scheduled projection, $c=2$ (§3.1) | 2806 | 3220 | 6026 |
| proposed: 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 | 3074 | 5922 |
| (1)+(2) points chosen by shape, *safe* | **2527** | 3108 | 5635 |
| (1)+(3) far end by dissimilarity, projection | 2726 | **2785** | **5511** |
| (1)+(2)+(3), *safe* | 2884 | 2859 | 5743 |
| Chandrupatla | 2785 | 2423 | 5208 |
| Chandrupatla | 2787 | 2441 | 5228 |
| TOMS 748 | 2820 | 4000 | 6820 |
| ITP, per-problem oracle | 3260 | 2611 | 5871 |
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 2785, with the worst-case guarantee intact. It is obtained under the scheduled projection, whose multiplier §3.1 shows to be a free constant, although this total hardly feels it: 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 5511 of reading (3) passes ITP's oracle (5871) but not Chandrupatla (5208): the remaining gap is entirely on the wide-bracket families, where Chandrupatla needs 2423 against our 2683 (reading (3) under *safe*).
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 5511 of reading (3) passes ITP's oracle (5871) but not Chandrupatla (5228): the remaining gap is entirely on the wide-bracket families, where Chandrupatla needs 2441 against our 2683 (reading (3) under *safe*).
The readings do not compose. Taken together, (2) and (3) give 2884 and 2859 — worse than either alone. Two selections that are each an improvement interfere, which is a reason to measure composition rather than assume it.
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 proposed rule itself: the earned budget binds early on the canonical problems, and what the readings buy under the scheduled projection is mostly room that the earned budget does not grant.
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
@@ -360,9 +357,9 @@ Three places have now been measured: the step length, the far end, and the point
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 3066 under the projection and 2899 under *credit*, against his 2423, while the canonical set rises from 2806 to 5432. The band is not idle — it refuses 63.6% 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.
In total, no. With Chandrupatla's band in our step, the wide-bracket set costs 3066 under the projection and 2899 under *credit*, against his 2441, while the canonical set rises from 2806 to 5432. The band is not idle — it refuses 63.6% 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 2868 on the wide-bracket set), but "defect" sets a new best on the canonical set: **2523** at 153/154 under the scheduled projection, 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, the best total over both sets of any constant-free configuration measured (6494 against 6588). It refuses only 7.0% of steps, the fewest of the three. Over both sets it costs 5733 against the 5511 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.
Our own readings do not close it either (best 2868 on the wide-bracket set), but "defect" sets a new best on the canonical set: **2523** at 153/154 under the proposed projection, 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, the best total over both sets of any constant-free configuration measured (6494 against 6588). It refuses only 7.0% of steps, the fewest of the three. Over both sets it costs 5733 against the 5511 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
@@ -370,31 +367,31 @@ Chasing the remaining gap to Chandrupatla led to a property of the benchmarks th
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: 253 evaluations, or 17%, not 11%.
**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, scheduled projection | 2806 | 2634 | 3220 | 2357 |
| Chandrupatla | 2785 | 2593 | **2423** | **1487** |
| ours, proposed (projection) | 2806 | 2634 | 3220 | 2357 |
| Chandrupatla | 2787 | 2595 | **2441** | **1505** |
| TOMS 748 | 2820 | 2636 | 4000 | 3088 |
| Brent | 2889 | 2707 | 4099 | 3222 |
| Brent | 2889 | 2707 | 4115 | 3238 |
| ITP | 3957 | 3767 | 2857 | 1907 |
| bisection | 7530 | 7338 | 3090 | 2136 |
### 7.4 Where the wide-bracket limit actually is
The remaining gap looks diffuse — per family +57, +54, +46, +43, +35 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.
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 2.26. 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.
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 262 evaluations of the gap in two. On `fun1` (58 of them) 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 five the negative weight underflowed to $-0$). A similarity in $[0,1]$ cannot produce a negative weight — Proposition 1 extended from two points to three.
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 244 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 five 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 178 of the 262 — 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 0.41–0.81 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 93, 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, 235 — against his own 293, 228, 228 — and costs +164, +111, +170, +144 on `fun1`, `fun2`, `fun8`, `fun9`, where it refuses steps that were good.
But 160 of the 244 — 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 0.41–0.81 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, 235 — 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, §8), 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 only 5.8% of 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 §8 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.
@@ -410,9 +407,9 @@ Every statement in §§7.3–7.4 is reproduced by `diag_rescore_cha.py`, `diag_r
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 degenerate.** A worst case of exactly bisection demands a halving at every step and leaves the projection no freedom at all.
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 scheduled projection 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 4099 against 3220, so a switch chosen per benchmark is worse than no switch at all: 6905 against 6026 over both sets. The scale axis offers only 5.3% headroom even to an oracle handed the best scale per problem, two thirds of it inside a single problem family, and the winners are not separated by anything the method can see before it commits — the proposed scale wins on bracket widths from $10^0$ to $10^{12}$.
4. **Regimes exist, at the level of a function family, and cannot be exploited by a switch.** The cube-root similarity beats the proposed rule on all five members of Chandrupatla's `fun4` — 32/32/40/47/59 against 89/93/113/125/153, a factor of 2.6 — and consistency across a family at different parameters is the signature of structure rather than luck. But the same similarity over the whole Chandrupatla set costs 4099 against 3220, so a switch chosen per benchmark is worse than no switch at all: 6905 against 6026 over both sets. The scale axis offers only 5.3% headroom even to an oracle handed the best scale per problem, two thirds of it inside a single problem family, and the winners are not separated by anything the method can see before it commits — the proposed scale wins on bracket widths from $10^0$ to $10^{12}$.
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.
@@ -426,15 +423,15 @@ Every statement in §§7.3–7.4 is reproduced by `diag_rescore_cha.py`, `diag_r
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.
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 3643 to 3474 — but the canonical set gets worse, 5822 to 6940, against the 2806 of the scheduled projection. An earlier version compared the 3474 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.
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 3643 to 3474 — but the canonical set gets worse, 5822 to 6940, against the 2806 of the proposed rule. An earlier version compared the 3474 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.
---
## 9. Conclusion
A root finder can be derived from a similarity function alone, and the derivation shows where the content of such a method actually sits. It is not in the shape of $S$ — nine shapes were tried and their ranking is dictated by a single number, the slope of $1-S$ at the origin. It is not in making the similarity change faster — that overshoots exactly as readily as flatness stagnates. It is in the *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 onto a budget scheduled in advance costs nothing on one benchmark and improves the method on the other, but the schedule carries a free, load-bearing multiplier; projecting onto a budget the steps earn removes it, and the price of removing it is measured rather than hidden.
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.
The resulting rule has no free constant and carries a worst case of bisection plus one step. It is not level with the best methods. On the canonical Alefeld–Potra–Shi set it needs 3568 evaluations, 29th of 264, against 2785 for Chandrupatla's method, 2820 for TOMS 748 and 2889 for Brent — though it wins more problems than it loses against each of them. On the wide-bracket set it needs 3020, ahead of TOMS 748 and Brent and behind Chandrupatla and ITP. Accept one free constant, the scheduled budget of twice bisection, and the same rule needs 2806 on the canonical set, a tie with TOMS 748 (0.5%) and 0.75% behind Chandrupatla. The claim the measurements support is therefore a trade stated in numbers: parity with the tuned methods at the price of one constant, or a constant-free rule with a tighter guarantee that trails them on the canonical set — not superiority either way.
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.
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.