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RDTvlokip PRO

RDTvlokip

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https://rdtvlokip.fr

AI & ML interests

I'm builds complete AI systems from the ground up, tokenizers, search engines, and small-scale language models to investigate, through exact mathematics and public experimentation, how language and intelligence emerge from minimal resources and pure signal.

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repliedto their post 1 day ago

I published an article about training a network to write from reward alone. Code on GitHub with it. Then someone read the code. Dipankar Sarkar commented four times in a day. Each time he had run something first. He rebuilt my statistics in numpy because he had no torch installed. He found a bound I had missed. A policy that never learned the determiner to noun dependency has a product support, so at full validity it cannot exceed the largest fully valid product in the sublanguage it entered. That is 12 on one side and 24 on the other, computable before any training. Over 70 seeds it is never crossed, and the most common outcome is the bound itself. I had published one of those numbers as an interesting coincidence. Then four of my published numbers came apart. Three were a single seed. The fourth was twenty seeds, and I had produced it while fixing the other three. And the test I built to validate his bound tested nothing. I had swapped two conditions so cleanly that the two grammars were isomorphic. Seventy seeds would have returned the mirror image by construction. The real lesson: A relabelling can permute, but it cannot change a ratio. A perfectly symmetric control is often a perfectly empty one. None of my errors were in the reasoning. They were in the plumbing, and nothing in my own process caught a single one. Code, figures, and the notebook with eight dated refutations 👇 🔗 https://huggingface.co/blog/RDTvlokip/i-published-my-rl-experiments 💻 https://github.com/RDTvlokip/RDTRL 📦 https://doi.org/10.5281/zenodo.21726216

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replied to their post 1 day ago

Main limit first: reduction, b1 = 0.9, b2 = 0.999, x = 1e-4, five starts per lr that are near clones at low lr, lr 0.1 to 0.25 (at 0.275 and 0.3 the groups merge). The length L of a burst varies by only 1.2 to 2.6 steps (standard deviation) inside a record, around 107 to 131, so every "per step" slope below is a local response to a few steps. Several lines are post hoc: the controls on the entry state, the mediation, the peak-aligned profile and the re-reading of gamma were done after my own prediction about the state failed. I predicted two of seven lines.

Your numbers

1/107 = 0.00935 and 1/131 = 0.00763: your +0.0076 to +0.0093 holds. The slopes of ln I on L I regenerated reproduce round 91 to the digit: e3 +3.8e-4, -9.4e-4, -2.4e-4 (lr 0.1, 0.125, 0.15) then -3.69e-3, -8.90e-3, -1.312e-2, -2.039e-2. Two corrections to my own wording. Up to 0.15 the slopes are within +-1e-3 as means over starts; start by start they reach -2.5e-3 and +1.3e-3, and the block-bootstrap standard deviation inside one record is 1.1e-3 to 2.0e-3. And gamma = half the gap of the two slopes is an identity, not an agreement: ln rho* = (ln I_e - ln I_l) / 2 and a least-squares slope is linear in y. It gives +1.16e-4 at 0.175 and +7.38e-4 at 0.25; the l3 / e3 slope ratio is 1.063, 1.066, 1.072, 1.073 at your four lr, and 1.041, 1.040, 1.061, 1.063, 1.066, 1.072, 1.072 over the seven, smooth. Your "the calm adds little to I" holds: the share of the calm in I, maximum over every burst of every start, is 6.8e-5 at lr 0.1 and 3 to 4e-5 at the others (0.007 percent at most).

Does ln(I / L) sit at -1/L up to lr 0.15

With the burst alone, E_b = the sum of g^2 over the group widened by 5 steps (no v(i0) in it): the slope of ln E_b on L is +7.6e-4, -1.44e-3, -4.1e-4, and ln(E_b / L) minus (-ln L) is +7.6e-4, -1.4e-3, -4.1e-4, all inside 2e-3 against a bootstrap standard deviation of 1.6 to 2.0e-3. So yes, in the sense that a fixed energy per step (slope +0.009) is rejected at about 4 sigma. But the second clause is the same measurement as the first, and it does not say the budget is fixed. The standard deviation of ln E_b is 5.7, 6.5 and 6.6 percent at lr 0.1, 0.125, 0.15. It is explained at R2 = 0.98, 0.96, 0.94 (e3; 0.99 for l3) by ln v_e(i0), the entry value of the second moment, with slope -12.7, -12.0, -11.7 (the standard deviation of ln v(i0) is 0.4 to 0.6 percent). L explains nothing of it (R2 of L on ln v(i0) is 0.00). So up to 0.15 the budget is not fixed: it varies by 6 percent, and L is not what it depends on. Across lr it is not fixed either: the mean ln I_e goes from -34.42 to -33.13 between lr 0.1 and 0.25, a factor 3.6, while L grows 22 percent.

From 0.175, is the loss in the peak or in the end

In the peak. The slope of ln(peak) on L (peak = the largest 3-step mean of g^2) over that of ln E_b is 1.20, 1.19, 1.20, 1.23 (e3) at lr 0.175 to 0.25, and 1.20, 1.17, 1.20, 1.23 (l3): the peak falls faster than the energy, and the effective width E_b / peak grows with L (+1.1e-3, +2.3e-3, +3.7e-3, +6.0e-3 per step). The fraction of E_b in the last third of the group has a slope of -8.1e-5, -7.1e-5, -5.4e-5, -2.1e-5 against -1.3e-4 to -9e-5 up to 0.15: it does not turn on. The reason is that the end carries almost nothing. With the profile of g^2 aligned on each burst's peak in 10-step bins, 0.27 to 0.33 of the energy is in the 10 steps before the peak, 0.47 to 0.50 in the 10 after, 0.12 to 0.16 in the next 10, 0.04 to 0.05 in the next, and about 0.01 beyond 40 steps after the peak, whereas the group lasts 107 to 131 steps. The shape of the high-v and low-v terciles is almost the same (share differences up to 0.03); the energy of the high tercile is 0.72 to 0.92 times that of the low one over the core, with 2 to 4 times more in the 30 to 20 steps before the peak (a share of 0.001 to 0.005).

What I got wrong in round 91

This is the main result, and it was post hoc. With ln v_e(i0) and ln v_l(i0) as controls, the slope of ln I on L is, for e3, +3.5e-4, +3.4e-4, +5.3e-4, +4.3e-4, +6e-5, +5.1e-4, -2.8e-5 (lr 0.1 to 0.25) against -3.7e-3 to -2.04e-2 without them: 0.7 percent of its value at 0.2 and 0.14 percent at 0.25. The same holds for E_b (+1.5e-4 to +7.4e-4 with the control) and for the peak (-3e-5 to +1.0e-3). L is correlated with the entry value: the slope of L on ln v_l(i0) is -0.02, +13, +8, +33, +71, +114, +222 steps per unit of ln v, R2 of L on ln v_e(i0) 0.00, 0.00, 0.00, 0.01, 0.05, 0.12, 0.31, correlation of L with ln v_l(i0) -0.00, +0.04, +0.02, +0.11, +0.23, +0.34, +0.58. I read a higher v at entry as a lower effective step lr / sqrt(v), hence a lower and longer burst; I did not test that reading. There is no jump at 0.175, and v(i0) precedes the burst, so L cannot have caused it; what I cannot exclude is a common cause outside v (the other coordinates).

Gamma: with the same controls it is -7e-6, -9e-6, -1.3e-5, -1.3e-5, -1.5e-5, -1.4e-5, -2.6e-5 (lr 0.1 to 0.25), against +1.16e-4, +2.96e-4, +4.73e-4, +7.38e-4 at 0.175 to 0.25: under 4 percent, and of opposite sign. The dependence of the target on the length of its burst, which I called in round 91 "an amplitude that moves with lr, turning on just before the step", does not survive this control. I withdraw that reading. What it does not touch: the step of M carried by the target rho* (0.0093 against 0.0090 on lr 0.15 to 0.25), which did not use L.

The turn-on at 0.175 also belongs to my cutoff. With the group threshold at 1e-4 instead of 3e-4 (L = 132 to 154), the slope of ln I on L is -1.8e-3 already at lr 0.1, then -4.0e-3, -5.4e-3, -9.5e-3, -1.44e-2, -1.74e-2: the first lr under -2e-3 is 0.125, with no kink. At 1e-3 (L = 80 to 100) it is positive, +8.8e-4, +3.4e-3, +3.2e-3, +4.4e-3, +5.1e-3, +1.9e-3, and negative (-8.1e-3) only at 0.25. L is the moment when the tail passes the threshold, and the tail holds about 1 percent of the energy; I did not test that reading apart.

Where I missed

Held: P92-1 (a slope of ln E_b within 2e-3 up to 0.15), P92-3 (the calm under 3 percent of I; it is under 0.01 percent). Missed: the E_b slope within 0.6 to 1.4 times that of ln I (it is 1.44 at 0.175; the other three lr are inside); the slope staying within 30 percent of itself under the state control (it falls to 4 percent or less); the turn-on at 0.175 +- 0.025 with the other cutoffs (0.125 at 1e-4, 0.25 at 1e-3, and the sign changes); the loss in the width or the end (it is in the peak: I bet 45 percent on the other side); the end-third fraction slope tripling (it does not).

What I have not tested

Why the entry value matters so steeply (a slope of -12 of ln E_b on ln v(i0), with the standard deviation of ln v(i0) only 0.4 to 1 percent); whether it acts through the reading lr p_z at the crossing of the level 38; a control by the other coordinates (x, m), not only v; other x, b1, b2; the full network. A residual slope of +1.5e-4 to +7e-4 per step on ln E_b under the control is not excluded as a direct effect of L (at most 0.04 to 0.07 percent per step).

One question back

Since a burst's energy is set to 94 to 99 percent (R2 of ln E_b on ln v(i0)) by the value of v at entry, with a slope near -12 at every lr, and its dependence on L disappears under that control, is the next measurement ln E_b against the value of lr p_z at the instant the level 38 is crossed before each burst (the reading of rounds 85 to 88), at fixed lr, to see whether v(i0) acts through that reading and whether this dependence changes at 0.2: yes or no?

Scripts: verifier_tour92_budget.py (modes enregistrer, analyse, etat, mediation, profil, profil_pic), lancer_tour92.sh.

Notebook: section "VRAIE CRITIQUE DE DIPANKARSARKAR, 13/11/2026 (tour 92, PAS simulée)".

replied to their post 10 days ago

Main limit first: reduction, b1 = 0.9, b2 = 0.999, x = 1e-4, five to ten starts per lr. At lr 0.275 and 0.3 the groups merge, so a "cycle" is no longer a period and I do not read those two lr. Several lines below were found after seeing the data and are marked post hoc.

Your numbers

The slope of ln(1/rho) on F is -0.0295 (lr 0.275) to -0.0578 (lr 0.15), your "-0.030 to -0.058". Between lr, d ln M / df = +0.0489 between f = 0.35 and 0.45. |S_F| is 0.0578, 0.0577, 0.0565 at lr 0.15, 0.175, 0.2, then 0.0486, 0.0411. Your sign argument holds: with ln(1/rho) = h(F) = -1/2 ln[(A_e F + a_e (1 - F)) / (A_l F + a_l (1 - F))], the slope inside a record is h'(F) and across lr M is about h(f), so d ln M / df = h'(f), the same sign. We observe opposite signs.

One correction to how S_F reads. h'(F) = -1/2 [(A_e - a_e) / v_e - (A_l - a_l) / v_l] with v_i = A_i F + a_i (1 - F). It is the difference of two relative burst excesses, each over the current mix, not the contrast between the amplitudes. S_F < 0 means the relative excess of e3 is larger than that of l3.

Does rho move inside a calm

No. Summed |change of ln rho|, calm against bursts, on regenerated records: lr 0.1: 0.00016 against 0.08138 (ratio 0.0020); lr 0.2: 0.00016 against 0.36341 (0.0004); lr 0.3: 0.00008 against 0.76665 (0.0001). The standard deviation of the change over a calm is 0.00000. So d_g is the increment of burst g.

d_g against rho_g

lr      slope beta   r        R2       R2 of quintile means   cycle (steps)   slope of the model -(1 - b2^cycle)
0.1     -0.0707      -0.188   0.035    0.919                  426             -0.347
0.15    -0.0599      -0.171   0.030    0.915                  364             -0.306
0.2     -0.0497      -0.157   0.025    0.946                  296             -0.256
0.225   -0.0351      -0.132   0.017    0.920                  256             -0.226
0.25    -0.0313      -0.125   0.016    0.912                  224             -0.201
0.275   -0.0672      -0.183   0.033    0.702                  236             (merging)
0.3     -0.1975      -0.314   0.099    0.879                  500             (merging)

The mean of d_g given rho_g lies on a line (quintile means at 0.9 or more for seven of nine lr). The cloud is not a clean line: |r| = 0.13 to 0.19 for lr <= 0.275. The simple slope falls from 0.0707 to 0.0313 over lr 0.1 to 0.25. Between lr 0.2 and 0.225 it drops 29 percent (local exponent of |beta| -2.97 against -0.70, -0.06, -1.00, -0.25 and -1.06 at the other intervals up to 0.25). The starts spread by about 15 percent (lr 0.2: -0.0583 to -0.0421; lr 0.225: -0.0408 to -0.0270), so that drop is about two sigma. A two-segment fit of ln|beta| over the nine lr puts the kink at 0.26, pulled up by the two merging points. So the slope does not clearly move at 0.2.

Adding the burst length, ln sqrt(v_l3) and the calm before to the regression takes R2 from 0.025 to 0.984 at lr 0.2 (0.035 to 0.983 at lr 0.1; 0.016 to 0.964 at lr 0.25). The map is nearly exact once these are in. The partial coefficient of ln rho is -0.3522, -0.3261, -0.3029, -0.2749, -0.2501, -0.2191, -0.1962 for lr 0.1 to 0.25, against the model -(1 - 0.999^cycle) = -0.3469, -0.3264, -0.3055, -0.2802, -0.2560, -0.2261, -0.2005: ratio 0.97 to 1.015. It is smooth in lr and follows the cycle length. That follows from the identity below; it is not an independent confirmation.

The exact map

Between two instants just before successive bursts (i0 = a_g - 1, i2 = a_(g+1) - 1, T steps), v_i(i2) = theta v_i(i0) + I_i, with theta = b2^T and I_i the net injection (cycle, calm included). Then rho_(g+1)^2 = omega_g rho_g^2 + (1 - omega_g) rho*_g^2, omega_g = theta v_l(i0) / v_l(i2), rho*^2 = I_e / I_l. This holds with no linearity assumption. I computed it on five regenerated records per lr. It holds to 7e-16 on every cycle, which is algebra since I defined I from the v's; the content is that both injections are positive in 100 percent of the cycles at every lr.

omega: mean 0.6542, 0.6735, 0.6944, 0.7199, 0.7433, 0.7741, 0.7988 for lr 0.1 to 0.25, against b2^cycle 0.6541, 0.6736, 0.6943, 0.7198, 0.7432, 0.7739, 0.7986. That is stationarity of v_l (the ratio v_l(i0) / v_l(i2) averages to 1), not a test. Its sd is 0.014 to 0.021 (0.088 at lr 0.275, 0.212 at lr 0.3). Its local exponent rises from 0.240 to 0.344 as lr passes 0.2, because the cycle shortens faster (the number of groups grows with exponent 0.77, then 1.2).
Target: the weighted mean of ln rho* equals the weighted mean of ln rho (stationarity again). Over lr 0.15 to 0.25, post hoc since the two merging points leave: exponent of 1/rho 0.02265 against 0.01367 (0.2 to 0.25), step 0.0090; of 1/rho* 0.02248 against 0.01314, step 0.0093. The step is in the target; omega does not enter. On the window I had written down (0.2 to 0.3, which contains the merging lr), the unweighted target step is 0.0041 against 0.0089 for M, so my prediction fails as written.

What moves with lr (post hoc)

The slope gamma of ln rho*_g on the length of burst g, per step:

lr        0.1       0.125    0.15      0.175     0.2       0.225     0.25
gamma    -7.8e-6   +1.9e-5  +7.5e-6   +1.16e-4  +2.96e-4  +4.73e-4  +7.38e-4
corr     -0.017    +0.029   +0.008    +0.099    +0.220    +0.337    +0.567

Up to lr 0.15 the target does not depend on the length of its burst (gamma within +-2e-5 across starts). From lr 0.175 it does, and increasingly. This is an amplitude that moves with lr, as you require, and it turns on just before the step at 0.2 to 0.225. It does not read as a contrast shrinking. Both injections depend on the length: the slope of ln I_e and ln I_l on length is -3.7e-3 and -3.9e-3 at lr 0.175, -8.9e-3 and -9.5e-3 at 0.2, -1.31e-2 and -1.41e-2 at 0.225, -2.04e-2 and -2.19e-2 at 0.25, against slopes within +-1e-3 up to lr 0.15. The longer the burst (by my cutoff), the less energy it injects into each of the two coordinates, and the two differ by 6 percent. The target keeps only that difference. So the thing the formula has to reproduce is the shared dependence, with the e3 / l3 difference second.

What I cannot claim: that this term explains the step. I tried to split the exponent of 1/rho* into a level at a fixed reference length and the length term. The result depends on the reference length: at 100 steps the exponents at the six intervals up to 0.25 are 0.019, 0.021, 0.036, 0.057, 0.062, 0.108; at 120 steps 0.017, 0.022, 0.022, 0.030, 0.031, 0.058; at 140 steps 0.014, 0.024, 0.008, 0.003, 0.001, 0.007. The sign of the contribution changes within the range of lengths I have (107 to 131 steps). I drop that split.

Where I missed

Held: rho constant in a calm (under 0.2 percent of the change); the multiple regression (R2 +0.15 or more); the injections positive; the target carries the step on lr <= 0.25. Missed: |r| of the simple line at or above 0.3 (it is 0.13 to 0.19); the simple slope within 30 percent of -(1 - b2^cycle) (it is 0.15 to 0.20 of it; the partial coefficient is within 3 percent, which I did not predict); a kink of the slope in [0.19, 0.22] (the fit puts it at 0.26); the unweighted target step in 0.005 to 0.013 on the window 0.2 to 0.3 (0.0041). The identity check (7e-16) and the omega agreement with b2^cycle are bookkeeping.

What I have not tested

x other than 1e-4, b1 other than 0.9, b2 other than 0.999 for this map. Whether the dependence on length is the same thing as the dependence on F found at round 90. Why the injections fall with length, and why from lr 0.175. The merged regime. A fit of A_e, A_l, a_e, a_l with the injections as inputs.

One question back

Since rho does not move inside a calm and the exact map has omega = b2^cycle smooth in lr, with the step in the burst's target; and since both injections of a burst start to fall with its length at lr 0.175 (slope -3.7e-3 against under 1e-3 up to 0.15) while the e3 / l3 difference is 6 percent of that, is the next measurement the energy a burst injects per step, its peak amplitude and its length taken separately (the slopes above only give the injection against length), to see why a longer burst injects less and why from lr 0.175, before any formula with fixed amplitudes: yes or no?

Scripts: verifier_tour91_carte.py (modes calme, carte, injection, injection_synthese).

Notebook: section "VRAIE CRITIQUE DE DIPANKARSARKAR, 04/11/2026 (tour 91, PAS simulée)".

replied to their post 13 days ago

Main limit first: every new configuration is x = 1e-4, five starts per lr, the reduction. The "largest drop" is searched among 10 to 14 intervals per configuration, after the data. At short memory the calm is 40 to 96 steps, just above the 30-step offset where rho is read. At b2 = 0.9995 and 0.9999 the curve is different from the base (M decreases with lr), so a "drop of the exponent" there measures something else than the base step.

The free check on b1 = 0.95

Your counts hold: 11 of the 15 lr are past f = 0.5 and the slope of ln M on ln f is 0.0072 to 0.0089 where the x curves have flattened (0.0089 at f = 0.455 for x = 1e-4, 0.0086 at 0.492 for x = 1e-3, 0.0072 at 0.488 for x = 1e-5). But the b1 = 0.95 grid does not start after the step: it starts at f = 0.316 and has 4 points below f = 0.47. I plotted b1 = 0.95 against f. 100 times the change of ln M:

f range         x 1e-4    x 1e-3    x 1e-5    b1 0.95    b1 0.95 / x curves
0.32 -> 0.50    +0.807    +0.916    +0.912    +0.023     0.029, 0.025, 0.025
0.50 -> 0.79    +0.319    +0.355    +0.258    +0.269     0.84, 0.76, 1.05

So b1 = 0.95 is not on the same M(f) curve below f = 0.5: it rises 40 times less where the x curves rise 0.8 to 0.9 percent, with slopes of ln M on ln f of 0.0002, 0.0005, 0.0009, 0.0011 (f = 0.35 to 0.49) against 0.017 to 0.044 for the x curves. Above f = 0.5 it has the same amplitude as the x curves (0.76 to 1.05 times). It has no drop between f = 0.45 and 0.5; its slope goes from 0.0011 (f = 0.486) to 0.0027 (0.515). The missing step is a missing rise before the crossover. I did this check before writing my predictions down; it is not a preregistered result.

b1 = 0.92 and 0.85: your prediction

Your prediction: the slope of ln M on ln f drops between f = 0.45 and 0.5.

b1 = 0.92 (lr* = 0.2127): the exponent of M in lr goes +0.0186 then +0.0105 across lr 0.2375; its largest drop is at lr 0.2375, f = 0.546, +0.00812 +- 0.00287 (2.8 sigma), 0.025 above lr*. The slope in ln f goes from 0.0207 (f = 0.533) to 0.0072 (f = 0.567). So the drop is at f about 0.55: 0.05 to 0.1 above your window, 0.1 above the base. I had written the window [0.40, 0.55] and it falls at the edge.
b1 = 0.85 (lr* = 0.2017, collapses from lr 0.3): the slope in ln f is 0.098, 0.117, 0.057, 0.023, 0.013, 0.010, -0.003, +0.0025 (f = 0.452), -0.0134 (0.529), -0.0055, +0.0033. M decreases for lr >= 0.175. The largest drops of the slope are at f = 0.206 (+0.060) and 0.252 (+0.034); the one near f = 0.5 (+0.016) is third. Your prediction does not hold there.

Is the memory b2, and can it change without touching b1

Yes. The module has B1, B2 = 0.9, 0.999 on one line; B2 appears only in the update of v and its bias correction. I copied the module and edited that line: b2 = 0.99, 0.995, 0.9995, 0.9999 at b1 = 0.9, x = 1e-4. The memory 1 / (1 - b2) is 100, 200, 2 000, 10 000 steps. It changes the whole regime, so calm = group length cannot be held fixed while keeping lr: lr* (calm = group) is 0.0096, 0.0144, 0.2254 (b2 = 0.999), 0.2347, 0.1619.

b2 = 0.99 and 0.995: M does not depend on lr at all (exponent between -0.00005 and +0.0007 over lr 0.001-0.015 and 0.002-0.04). Neither the rise nor the drop.
b2 = 0.999 (base): rise of M, drop at f = 0.45 to 0.49.
b2 = 0.9995: the exponent of M is +0.0288, +0.0258, +0.0153, +0.0028, then -0.0059 to -0.0251; M peaks near lr 0.2 (f about 0.3 to 0.35); the largest drops of the slope in ln f are at f = 0.205 (+0.053), 0.248 (+0.019), 0.313 (+0.010).
b2 = 0.9999: M peaks near lr 0.14 (f about 0.19); the exponent goes from +0.0173 to -0.0335 between lr 0.1375 and 0.15 (drop +0.0508 +- 0.0011 at f = 0.191); the groups merge abruptly between lr 0.15 and 0.2 (f from 0.30 to 0.93).

By the rule I wrote before the scans: the location of the drop in f changes by more than 0.1 with b2 (0.45-0.49, 0.2-0.27, 0.19), so the memory intervenes in this protocol. And below a memory of 200 steps the dependence on lr is gone. What it does not say: calm / N at the drop is 0.13-0.17, 0.16, 0.056 for N = 1 000, 2 000, 10 000, not constant; I have no rule for where the drop goes.

What rho tracks inside a record

For each burst g >= 9, I took ln(1 / rho_g) (rho read 30 steps before the start) and the exponential average F_lambda of the oscillation indicator (1 during a group) at the same instant. Correlation, five starts per lr:

lr      lambda = 0.99  0.995   0.998   0.999   0.9995  0.9998  0.9999   calm before g   previous group length
0.15    -0.836        -0.866  -0.949  -0.987  -0.869  -0.271  -0.057   +0.828          -0.010
0.2     -0.809        -0.851  -0.948  -0.988  -0.885  -0.353  -0.090   +0.789          -0.181
0.225   -0.808        -0.853  -0.947  -0.987  -0.905  -0.456  -0.168   +0.794          -0.299
0.25    -0.785        -0.838  -0.942  -0.985  -0.910  -0.516  -0.204   +0.763          -0.465
0.3     -0.528        -0.693  -0.888  -0.956  -0.919  -0.729  -0.531   +0.405          -0.498

The maximum is at lambda = 0.999 = b2 at each of the nine lr I ran (0.1 to 0.3 in steps of 0.025), with a correlation of -0.952 to -0.988, and it falls on both sides. Inside a record, more oscillation over the last 1 000 steps means a larger rho (smaller 1 / rho), slope -0.041, -0.051, -0.058, -0.058, -0.057, -0.049, -0.041, -0.030, -0.032 per unit of F for lr 0.1 to 0.3. That agrees with your picture: rho read in the calm tracks f over the memory of v. What it does not give: independence. F_lambda and rho are both functions of the gradient history (rho^2 = ratio of two averages of g^2 with memory 1 / (1 - b2), and g^2 is large during bursts), so a correlation of -0.99 at lambda = b2 is close to what that formula says; it does not separate your statement from it.

The sign is the opposite of the trend between lr: M increases with f between lr (d ln M / df = +0.049 between f = 0.35 and 0.45) while inside a record 1 / rho decreases with F. The explicit lr term at fixed F is positive and larger. I tried to split the exponent of M into S_F times df / d ln lr plus a remainder E:

0.15 -> 0.2:  M +0.02265 = f term -0.01927 + E +0.04192
0.2  -> 0.3:  M +0.01380 = f term -0.04387 + E +0.05766
step          +0.00885   = +0.02460 + (-0.01574)

The f term carries 278 percent of the step and E offsets it. My preregistered prediction (E has a step under 0.003) fails: E rises from +0.024 to +0.073 with a pause between lr 0.2 and 0.25. The step of M is a small remainder of two large terms of opposite sign, and I assumed a linear response of the mean level to the within-record fluctuations. This decomposition does not explain the step. One descriptive fact: |S_F| is flat (0.0577 to 0.0565) from lr 0.15 to 0.2, then falls (0.0486, 0.0411, 0.0295, 0.0318): the sensitivity of rho to the recent oscillation fraction starts to change at lr 0.2, where M's exponent drops.

Where I missed

Held: the correlation maximum in {0.998, 0.999, 0.9995} (nine lr out of nine, absolute value above 0.5, F_lambda the best predictor); no identifiable drop at b2 = 0.9999 (one lr in f 0.40 to 0.55); b1 = 0.92 inside my window, just. Missed: b1 = 0.85; b2 = 0.9995 drop within 0.05 of the base in f (it is at 0.2 to 0.27); a rise of M of at least 0.4 percent at b2 = 0.995 and 0.99 (it is about 0.001 percent); the within-record slope being within a factor 2 of the trend between lr (same size, opposite sign); and the decomposition.

What I have not tested

Other x for b1 = 0.92 and the b2 variants. Whether the change of regime at b2 = 0.9995 and 0.9999 (M decreasing with lr) is the same mechanism as the step at b2 = 0.999. Why |S_F| starts falling at lr 0.2. A derivation of rho from the two averages and the duty cycle; I have the correlation, not the formula. The full network.

One question back

Since the correlation of ln(1 / rho) with the exponential average of the oscillation indicator peaks at exactly lambda = b2, and the drop of M moves in f with b2 while M becomes flat below a memory of 200 steps, is the working statement "the lr dependence of M comes from rho following the duty cycle over the memory of v" the one to turn into a formula (rho^2 as a ratio of two averages of g^2 over bursts and calm, fitted for the amplitudes A and a) before any more scans: yes or no?

Scripts: verifier_tour90_memoire.py (modes libre, essai, enregistrer, config, regression, decomposer), lancer_tour90a.sh, lancer_tour90b.sh, reduction_numba_tour90_b1_085.py, _b1_092, _b2_099, _b2_0995, _b2_09995, _b2_09999.

Notebook: section "VRAIE CRITIQUE DE DIPANKARSARKAR, 02/11/2026 (tour 90, PAS simulée)".

replied to their post 13 days ago

Main limit first: everything below is the reduction, five starts per lr. Three x (1e-5, 1e-4, 1e-3) at b1 = 0.9, one x at b1 = 0.95. b1 = 0.8 collapses from lr 0.15 and was not scanned. At b1 = 0.95 the gain barely moves with lr, so that test has no leverage. The "largest drop" analysis was done after seeing the data, and it looks for the largest of about twelve intervals.

What M is in my code

Your decomposition assumes M is a time average. It is not. M is the mean, over the bursts g >= 9, of 1 / rho_g with rho_g = sqrt(v_e3) / sqrt(v_l3) taken 30 steps before the start of burst g. Each burst counts once, the reading is always in the calm, and there is no f in it. Inside a calm, rho is constant (both square roots decay by the same factor), so rho_g is also the value for the whole calm before burst g. The time average T of 1 / rho(t) over every step is another object, and it does split exactly: T = f T_osc + (1 - f) T_calme. T_calme is the calm average weighted by calm duration; M is the unweighted average over bursts. I computed T and its split from 280 new records (300 000 steps each).

Your numbers

f exponents from my three-digit table of round 88: 0.961, 1.480, 1.540 (your 0.96, 1.48, 1.54). With ten starts: 1.00, 1.44, 1.59. The median calm equals the median group length (127 against 127) at lr 0.225. Both reproduce.

What the same records add, which I had not looked at before: the change of the f exponent from one interval to the next is +0.104, +0.230, +0.133, +0.440 +- 0.056 (at lr 0.2), +0.143, +0.555 +- 0.093 (at lr 0.25), +0.299. The f exponent jumps at 0.2 and at least as much at 0.25. The exponent of M does not change at 0.25 (0.0127, 0.0138, 0.0139). So the jump of f at 0.2 is not specific.

Which piece carries the step

Base configuration (b1 = 0.9, x = 1e-4), exponent averaged over 0.15 to 0.2 against 0.2 to 0.3:

                      0.15 -> 0.2   0.2 -> 0.3   step (below - above)
T (time average)      +0.02291      +0.00811     +0.0148
  weights  df (T_osc - T_calme) / T  +0.00001   -0.00116   +0.00118   (8 %)
  conditional, osc     f dT_osc / T   +0.00829   +0.00337   +0.00492   (33 %)
  conditional, calm   (1-f) dT_calme / T +0.01461 +0.00590 +0.00871   (59 %)
M (burst average)     +0.02265      +0.01380     +0.00885
T_calme               +0.02300      +0.01448     +0.00851
M / T_calme           -0.00035      -0.00069     +0.00034

The identity T = f T_osc + (1 - f) T_calme holds to 1.6e-7. The weights carry nothing because T_osc and T_calme are almost the same number: 11.2292 against 11.2285 at lr 0.1; 11.393 against 11.3931 at lr 0.2; 11.4274 against 11.4321 at lr 0.25; 11.4265 against 11.4602 at lr 0.3. Rho barely changes inside a burst (the two second-moment averages have a memory of 1 000 steps against 120 for a burst), so the weight f multiplies a difference of 0.01 to 0.3 percent. For T, the step is in the conditional means, 92 percent. For M, which has no weights, the step of 0.00885 is the step of T_calme within 0.00034. The step is the conditional mean of 1 / rho in the calm. It is not a weighting by the oscillation fraction.

The step of T (+0.0148) is larger than that of M because T_osc stops growing at lr 0.25 to 0.3 while T_calme keeps growing.

x and b1: where the step goes

Preregistered rule, written before the data: if the kink of M (continuous two-segment fit of ln M against ln lr, kink position scanned; a "kink" means RSS ratio under 0.3 and a slope drop above 0.004) follows lr* at x = 1e-3 and at b1 = 0.95, the crossover is compatible; if there is no kink in one of the two while there is one at the base, it is not supported.

configuration     lr* (calm = group)  kink of M by the rule                          result
base x 1e-4       0.2254              0.2002, drop +0.0084, RSS ratio 0.012          kink
x 1e-3            0.2892              0.1285, drop -0.0067, ratio 0.54               no kink
x 1e-5            0.1953              0.075, drop -0.0086, ratio 0.68                no kink
b1 0.95 x 1e-4    0.0608              0.1056, drop -0.0030, ratio 0.06               no step down

By the rule as written: not supported (no kink at x = 1e-3, none at b1 = 0.95). What I see in the exponent profiles, post hoc: the largest drop of the exponent of M from one interval to the next, in every b1 = 0.9 configuration, sits just below lr*: base at lr 0.2, +0.01024 +- 0.00147, 0.025 under lr*; x = 1e-3 at lr 0.275, +0.00796 +- 0.00102 (7.8 sigma), 0.014 under lr*, followed by a rebound (0.0200, 0.0120, then 0.0176); x = 1e-5 at lr 0.1875, +0.00999 +- 0.00454 (2.2 sigma), 0.008 under lr*; a fit restricted to 0.1375-0.25 gives a kink at 0.1818, drop +0.0135, ratio 0.027. So the location moves with x: 0.1875, 0.2, 0.275 in the order x = 1e-5, 1e-4, 1e-3, as lr* does (0.195, 0.225, 0.289). My rule failed at x = 1e-3 because that drop is not followed by a lasting step; and my kink fit at x = 1e-5 on the whole grid is dominated by the early rise of the exponent from 0.013 to 0.031. At b1 = 0.95 the exponent of M grows from +0.0001 to +0.0045, 5 to 50 times smaller than at b1 = 0.9, and has no drop.

M as a function of f only (post hoc): 100 times ln M(f) minus ln M(f = 0.3), interpolated in f.

f      x 1e-4   x 1e-3   x 1e-5
0.40   0.769    0.777    0.866
0.50   1.028    1.121    1.182
0.60   1.185    1.318    1.281
0.80   1.356    1.491    1.444

The three curves agree to 0.07 to 0.15 percentage points, about 10 percent of the change, while the same change takes lr 0.1-0.2 at x = 1e-4, 0.15-0.3 at x = 1e-3 and 0.075-0.2 at x = 1e-5. The slope of ln M against ln f drops at f between 0.39 and 0.49 in all three: x = 1e-4 from 0.0241 (f = 0.39) to 0.0089 (0.455); x = 1e-3 from 0.0192 (0.44) to 0.0086 (0.49); x = 1e-5 from 0.0169 (0.441) to 0.0072 (0.488). The step is where f reaches about 0.45 to 0.5, which is by construction where calm = group length.

Where I missed

Held: the weights under 30 percent (8); the step of T same sign and within a factor 2 of M's (0.0148 against 0.0088); the step of M / T_calme under 0.003 (0.00034); the base kink at 0.2002 with ratio 0.012. Missed: the kink rule at x = 1e-5 (the whole-grid fit puts it at 0.075). The outcomes at x = 1e-3 and b1 = 0.95 were the "no kink" case I had given 40 percent. What I did not write down before looking: the f exponent also jumps at lr 0.25; and the post hoc profile at x = 1e-3 and x = 1e-5, which contradicts the plain reading of my rule.

What I have not tested

Whether M depends on lr only through f: the collapse is on three x at one b1. Another b1 with a non-flat gain (0.85 or 0.92, between the collapse at 0.8 and the flat 0.95). x = 1e-6 and 1e-2. How rho (a ratio of two second-moment averages with memory 1 000 steps) can depend on the duty cycle at f about 0.5; I have a description, not a derivation. The full network, and the readings by Jacobian at other x and b1: all levels here come from rho times the constant C' of the base configuration, which was not recomputed at x = 1e-5, 1e-3 or b1 = 0.95; the exponents of M do not use C'.

One question back

Since the weights carry 8 percent of the step of T and none of M, the step of M is the step of T_calme within 0.00034, and the largest drop of the exponent of M sits just below lr* at f about 0.45 to 0.49 for x = 1e-5, 1e-4 and 1e-3 with M(f) agreeing across x to about 10 percent, is the working statement "M depends on lr through f, by the dynamics of rho and not through weights" the one to test next, at b1 = 0.85 and 0.92 and at x = 1e-6 and 1e-2: yes or no?

Scripts: verifier_tour89_poids.py (modes f_exposants, essai, enregistrer, temps, config, universel), lancer_tour89a.sh, lancer_tour89b.sh.

Notebook: section "VRAIE CRITIQUE DE DIPANKARSARKAR, 01/11/2026 (tour 89, PAS simulée)".

replied to their post 20 days ago

Main limit first: everything below is the reduction, x = 1e-4, b1 = 0.9. The scan variants use 3 starts per lr (60 bursts each, the same bursts as the reference); the four new lr (0.125, 0.175, 0.225, 0.275) have 5 starts. I varied the scan step, the window and the position where rho is taken. I did not vary the perturbation seed, the scale of the starting moments, the fixed level 38, eps / lr, b1, x or the network. I cannot say the lr 0.2 step is not an artefact of one of those.

Your numbers

Unweighted three-point weights: low segment (0.1, 0.15, 0.2) -1.510, +0.162, +1.348; high segment (0.2, 0.25, 0.3) -2.540, +0.165, +2.376. So lr 0.2 weighs 1.348 + 2.540 = 3.89 in low - high, lr 0.3 weighs 2.38. Lowering the lr 0.2 level of round 87 by 0.10 percent: local exponents +0.0171 +0.0198 +0.0118 +0.0123 become +0.0171 +0.0163 +0.0163 +0.0123, and low - high goes from +0.0061 to +0.0023 unweighted (+0.0057 to +0.0021 weighted by my round 87 errors). All of it reproduces. The lr 0.2 level is 0.100 percent above the log-log chord from 0.15 to 0.25; its noise there is 0.100 +- 0.026, 3.9 times the noise. For +0.10 percent at one lr, the effect on low - high is -0.0015 (lr 0.1), +0.0002 (0.15), +0.0039 (0.2), -0.0002 (0.25), -0.0024 (0.3) unweighted; lr 0.2 is the one that matters, as you say.

One thing your arithmetic does not separate: a smooth curve whose exponent falls from 0.0198 to 0.0118 over a width of about 0.25 in ln lr also puts lr 0.2 about 0.1 percent above the chord. An isolated offset at lr 0.2 and a smooth bend give the same number at 0.2. They differ at 0.175 and 0.225.

Does the lr 0.2 level move when the scan step goes to 1?

Same 60 bursts per start, last crossing, 3 starts, paired differences against the reference (step 3, window 140):

shift of the level (percent)   lr 0.1     lr 0.15    lr 0.2     lr 0.25    lr 0.3
step 1 against step 3          +0.00000   +0.00001   +0.00003   +0.00059   +0.00021
step 6 against step 3          -0.00000   -0.00001   -0.00002   +0.00016   -0.00105
window 210 against 140         +0.00000   +0.00000   +0.00001   +0.00015   +0.00005

lr 0.2 moves by 0.00003 percent, 0.0006 percent at most across the five lr, and the local exponents move by under 0.0001 (step 1: +0.00000 +0.00000 +0.00002 -0.00002; low - high -0.00000). lr 0.2 against its neighbours: -0.0003 percent. At lr 0.1 to 0.2 all 180 bursts are identical to 0.01 percent in every variant; at lr 0.25 and 0.3 individual bursts differ by up to 0.05 percent (step 1: 175 and 134 of 180 identical) with a mean under 0.0006 percent. Your two reasons for lr dependence (the lead runs from 63.5 to 20 steps; the 140-step window against 305-step groups) are real for the burst-by-burst readings at lr 0.25 and 0.3, and they are not what moves a level. The step law (step 6 against step 1) cannot be tested: no shift exceeds 0.003 percent on average. I also took rho 1 step or 30 steps before the group (C' recomputed at the same offset): shift 0.00000 percent at every lr, which is what you expect when both square roots decay by the same factor in the calm; that variant says nothing.

lr 0.175 and 0.225, and the grid

Whole-record levels (C' times the mean of 1/rho, bursts 9 and later), step 0.025 from 0.1 to 0.3 (10 starts at 0.1, 0.15, 0.2, 0.25, 0.3; 5 at the others):

interval        local exponent (+- noise + spread between starts)
0.1  -> 0.125   +0.0155 +- 0.0005
0.125-> 0.15    +0.0191 +- 0.0008
0.15 -> 0.175   +0.0195 +- 0.0012
0.175-> 0.2     +0.0201 +- 0.0018
0.2  -> 0.225   +0.0124 +- 0.0028
0.225-> 0.25    +0.0111 +- 0.0038
0.25 -> 0.275   +0.0121 +- 0.0040
0.275-> 0.3     +0.0125 +- 0.0048

Level above the chord 0.15 to 0.25 (percent +- noise): 0.175, +0.049 +- 0.019; 0.2, +0.100 +- 0.026; 0.225, +0.055 +- 0.036; 0.125, -0.052 +- 0.020; 0.275, -0.040 +- 0.042; 0.1, -0.034 +- 0.030; 0.3, -0.073 +- 0.053. A smooth bend predicts +0.085 at 0.175 and +0.067 at 0.225; an isolated offset at 0.2 predicts about 0 at both. The data sit in between: 0.175 is 2.6 sigma from 0 and 0.225 is 1.5 sigma from 0; both are above the 0.04 percent I had written down. The exponent reads about 0.020 up to lr 0.2 and about 0.012 from lr 0.2 on: 0.0077 +- 0.0033 across the two intervals touching 0.2 (2.3 sigma), 0.0198 against 0.0118 for 0.15 to 0.2 against 0.2 to 0.25 (4.5 sigma).

What the grid does not settle: a degree-3 polynomial in ln lr fits the nine levels (chi2 5.8 for 5 degrees of freedom, residuals under 0.043 percent, +0.025 at lr 0.2); degree 2 does not (chi2 55 for 6). Without lr 0.2, the degree-3 curve of its neighbours passes 0.038 +- 0.019 percent below the lr 0.2 level (degree 2: 0.097). So I cannot tell a kink at 0.2 from a smooth bend of scale 0.2. I can say the levels do not need an isolated +0.10 percent offset at lr 0.2.

Where the step lives

The level is the product of C'(lr), the median of reading times rho, which is the only factor that uses the scan, and M(lr), the mean of 1/rho over the record, which uses none. C' runs from 2.893703e5 (lr 0.1) to 2.887053e5 (lr 0.3), 0.23 percent over the whole grid. Local exponents of M: +0.0175 +0.0216 +0.0221 +0.0227 | +0.0140 +0.0128 +0.0138 +0.0139; of C': -0.0020 -0.0025 -0.0026 -0.0026 | -0.0015 -0.0017 -0.0018 -0.0014. Exponent 0.15 to 0.2 against 0.2 to 0.3: M +0.02236 against +0.01360 (difference +0.00876); C' -0.00261 against -0.00159 (difference -0.00102); level +0.01975 against +0.01201 (difference +0.00774). The step is in M: 113 percent of it, with C' removing 13. The exponent of C' varies by at most 0.0011 from one interval to the next.

So the scan cannot make this step, because the part of the level that does not use the scan carries all of it. What this does not exclude: a lr-dependent artefact in the reduction itself, in the perturbation seed, or in the scale of the starting moments.

For context only, not tested: the fraction of time in oscillation grows without a knee (0.365 at lr 0.175, 0.415 at 0.2, 0.494 at 0.225, 0.581 at 0.25); the median calm drops under the median group length between lr 0.2 (170 against 123 steps) and lr 0.225 (127 against 127), where the exponent of M falls from 0.0227 to 0.0140.

Where I missed

Held: your arithmetic; lr 0.2 shift under 0.03 percent (0.00003); range over lr under 0.03 percent (0.0006); exponent shifts under 0.002 (under 0.0001); the grid (0.175 and 0.225 above the chord by 0.049 and 0.055, predicted at least 0.04); the step in M (predicted at least 80 percent, found 113). Missed: bursts identical to 0.01 percent between window 210 and 140 (94 percent at lr 0.25 and 65 at lr 0.3, predicted 99 and 90; the gaps are at most 0.05 percent and average under 0.0015). Not testable: the step-6 against step-1 law.

What I have not tested

Other b1, other x, the full network. The perturbation seed and the starting-moment scale at lr 0.175 to 0.225 (the wait was varied at lr 0.2 and 0.3 only, level shifts 0.002 and -0.013 percent). Whether the step moves with x or b1, which is what would tell a property of the reduction from an artefact of my protocol. What changes in the trajectory between lr 0.2 and 0.225.

One question back

Since the scan step, the window and the position of rho in the calm move no level by more than 0.0016 percent and no local exponent by more than 0.0001, and the step sits in the mean of 1/rho, which uses no scan, is the next test whether the step moves with x and b1 rather than with any part of my protocol: yes or no?

Scripts: verifier_tour88_protocole.py (modes arithmetique, variantes, grille, decomposition, regime, decalage_rho), lancer_tour88.sh, verifier_tour78_lr.py (TOUR88_FEN), verifier_tour86_longue.py.

Notebook: section "VRAIE CRITIQUE DE DIPANKARSARKAR, 25/10/2026 (tour 88, PAS simulée)".

replied to their post 21 days ago

Main limit first: everything below is the reduction, x = 1e-4, b1 = 0.9, ten starts per lr, 60 random bursts per start for the direct readings, whole records (575 to 1 317 bursts) for the levels through rho. The cut for "late" (lead above 100 steps) was found after I had seen the histograms. I withdraw two things I wrote in letter 86; see below.

Your numbers

ln(3.29510e6 / 3.30752e6) = -0.376 percent; 31 / 300 = 10.3 percent; 0.376 / 0.103 = 3.6 percent; 0.0127 - 0.0053 = 0.0074; 0.0074 / 0.00376 = 1.97; 0.0074 / 0.0131 = 56.5 percent. All reproduce. The 1.97 is a real weight: with the spread between starts plus a 0.1 percent floor, the 0.3 point weighs 1.976 in the all-readings fit and 2.354 in the fit without the late readings. The low segment is the same in both (0.0183) because no reading at lr 0.1 to 0.25 was flagged late.

What I could not leave standing: I had defined "late" by the reading (C / rho minus the reading above 0.3 percent), which is the circularity you point to.

Late, defined before the reading

The lead is the start of the group minus the interpolated instant where the level crosses 38. It is known from the trajectory before the group. It is not one number across lr: median lead 63.5 steps at lr 0.1, 52.9 at 0.15, 41.2 at 0.2, 26.1 at 0.25, 20.0 at 0.3. My precommitted rule, lead above 40, would flag 100 percent of the bursts at lr 0.1 and 59 percent at lr 0.2. It failed.

At lr 0.3 the lead is bimodal with an empty gap, in classes of 10 steps (starts 0 to 4 / starts 5 to 9): 0 to 49 steps: 267 / 271 bursts; 50 to 119: 1 / 0; 120 to 136: 32 / 21. Cutting at 100 steps (found after seeing this, so post hoc) gives p_late = 0 of 2 400 at lr 0.1 to 0.25 and 53 of 592 (9.0 percent) at lr 0.3 (10.7 then 7.2 percent in the two sets of five starts).

Those leads of 120 to 136 are the edge of my scan window, which starts 140 steps before the group. For all 53 the lead is longer than the calm before the group (median calm 35 steps; 0 of 539 normal bursts at lr 0.3 do that) and the previous group lasts 305 steps in the median. So the first upward crossing in the window was taken inside the previous long group. Rereading the same 2 992 bursts with the last upward crossing in the window: at lr 0.1 to 0.25, 600 of 600 readings are identical at each lr; at lr 0.3, 539 of 592 are identical and the 53 others are exactly those 53. For them the lead becomes 9 to 30 steps (all under 60) and the reading sits at -0.50 / -0.29 / -0.15 / +0.02 / +0.24 percent (5, 25, 50, 75, 95th percentile) from the median of its start; it was -2.7 to -4.3 percent. 50 of 53 are within 0.5 percent. With the last crossing, no burst has a lead above 100 at any lr: p_late = 0.

What this does to your two numbers: p_late is 0 at every lr, and the exponent on non-late readings is the exponent on all readings. At the last crossing, ten starts, all readings: low (0.1 to 0.2) +0.0183, high (0.2 to 0.3) +0.0119 +- 0.0035, difference +0.0064, 1.6 sigma with the 0.1 percent floor. Mean at lr 0.3: 3.30669e6, against 3.30714e6 for the first-crossing readings without the late ones (-0.014 percent). So the 57 percent of 0.0131 that came from late bursts was my choice of the first crossing, not something the system does. What could bias this: the cut at 100 is post hoc; both crossings come from the same instrument (the Jacobian block scanned every 3 steps in a window of 140 steps); the 53 readings at the last crossing are a little low (median -0.15 percent), which weighs -0.014 percent on the lr 0.3 mean.

The +0.4 percent population

It is not a population. It is two whole starts, lr 0.3 starts 2 and 3. The median gap C / rho minus reading, on their non-late readings, is -0.348 and -0.413 percent, with sd 0.040 and 0.031, and the same in each third of the burst index (-0.334 -0.360 -0.348 and -0.413 -0.415 -0.402). The other three starts: -0.008, -0.028, -0.031. My C is the geometric mean of reading times rho over the first eight readings of letter 85, and those two starts each contained one late reading there (-3.5 and -3.2 percent), which lowered C by 3.5 / 8 = 0.44 and 3.2 / 8 = 0.40 percent. That is the 92 readings (about two starts) I called 93 in letter 86.

So I withdraw from letter 86: "31 percent of the readings 0.3 to 0.45 percent above C / rho", "lr 0.3 is a mixture of reading populations", "the 38 crossing came about 90 steps before the group" (I deduced it from 4.5 percent and 0.05 percent per step; the measured lead is 120 to 136 steps at the window edge), the C / rho record test at 2.1 sigma, and the lr 0.3 mean by C / rho (3.30156e6, 0.18 percent too low because of those two calibrations). The direct readings never used C.

Your question, whether something before the group names it: no variable does, because there is nothing to name. Among the non-late lr 0.3 readings, ln(reading) has sd 0.26 percent and is unimodal (classes of 0.2 percent: 3 12 39 79 77 38 17 3); ln(reading times rho) has sd 0.035 percent. So rho, taken in the calm before the group, accounts for 98 percent of the variance. C' (median of reading times rho per start) is the same across ten starts to 0.001 percent at lr 0.1, 0.0025, 0.0037, 0.0079 and 0.0071 percent at lr 0.3. The best single threshold on lead, calm, previous group length, rho, sqrt(v_l3), z3 or burst index reaches at most 0.55 balanced accuracy; the lead is shorter than the median for only 53 percent of the readings I had labelled +0.4 percent.

The floor, measured instead of chosen

Whole records, blocks of 100 bursts of 1 / rho (rho taken in the calm before the group, so independent of the crossing), level = C' times the mean of 1 / rho over bursts 9 and later:

lr     noise of one record (blocks)   spread between 10 starts   9-559 against all >= 9
0.1    0.005 %                        0.003 %                    -0.002 %
0.15   0.010 %                        0.002 %                    +0.000 %
0.2    0.018 %                        0.006 %                    +0.011 %
0.25   0.028 %                        0.010 %                    -0.000 %
0.3    0.033 %                        0.011 %                    +0.001 %

Segment test on those levels: low +0.0177, high +0.0120 +- 0.0009, difference +0.0057, 6.1 sigma with the noise added to the spread (18.4 with the spread alone, which measures near-copies); with the floor of 0.1 percent chosen in round 84: 1.5 sigma. The wait before recording, which I had suspected for the first eight bursts, moves the whole-record level by +0.002 +- 0.011 and -0.000 +- 0.009 percent at lr 0.2 (waits of 30 000 and 100 000 steps against 10 000) and by +0.003 +- 0.025 and -0.013 +- 0.023 percent at lr 0.3, which moves the high exponent by 0.0000 and -0.0003.

What this contains and what it does not. It is the noise of one trajectory inside one protocol, plus the wait at two lr. It does not contain the window, the scan step (3 against 1 was tested at lr 0.15 only, under 1e-5), the definition of the crossing, b1, x or the network. The gap between 1.5 and 6.1 sigma is exactly the gap between a protocol floor chosen at 0.1 percent and a noise floor measured at 0.005 to 0.033 percent. Neither is a measured protocol floor.

Local exponents between consecutive lr, from these levels (post hoc, five points; the first value uses the noise only, the value in brackets the 0.1 percent floor): 0.1 to 0.15: +0.0171 +- 0.0003 (0.0035); 0.15 to 0.2: +0.0198 +- 0.0007 (0.0049); 0.2 to 0.25: +0.0118 +- 0.0016 (0.0064); 0.25 to 0.3: +0.0123 +- 0.0025 (0.0078). It looks like a step of 0.008 across lr 0.2 and a plateau near 0.012, not a steady decline; lr 0.225 and 0.275 are not in these data.

Where I missed

Held: your numbers; the result of the last crossing (50 of 53 within 0.5 percent, all under 60 steps); the share of late bursts (5 to 15 percent); the identical readings at lr 0.1 to 0.25; the non-late exponent (+0.0122 +- 0.0035, 1.5 sigma); the wait (under 0.05 percent, exponent under 0.002); C' spread. Missed: the rule of 40 steps (the lead depends on lr); the slope of ln(reading) against the lead on all readings (-0.031 percent per step, predicted -0.045 to -0.055); the idea that the corrected reading would be more stable (sd 0.44 percent against 0.26); the idea that a variable before the group would name the +0.4 percent readings (the readings were my calibration); the measured block noise (0.005 to 0.033 percent, predicted 0.03 to 0.08 and 0.05 to 0.15) and the test at 6.1 sigma (predicted 1.0 to 2.5). Also in letters 83 to 85: the lr 0.3 readings at -3.2 and -3.5 percent and the "long groups" there were largely this first-crossing choice, so the lr 0.3 means of those rounds depend on it.

What I have not tested

Other b1, other x, the full network. A protocol floor: the window (140 steps), the scan step, the definition of the crossing, at every lr. Whether lr 0.225 and 0.275 sit on the step at 0.2 or on the plateau. What makes the first crossing in the window fall inside a long group, and what the long groups are at lr 0.3.

One question back

Since at the last crossing p_late is 0 at every lr, and the whole-record test gives 6.1 sigma with the measured noise and 1.5 sigma with a floor I chose at 0.1 percent, is the remaining disagreement only about the protocol floor, so that the next step is to measure that floor by varying the window and the scan step at every lr rather than to choose it: yes or no?

Scripts: verifier_tour87_populations.py (modes arithmetique, analyse, dernier, plancher, attente), lancer_tour87.sh, lancer_tour87b.sh, verifier_tour78_lr.py (TOUR87_DERNIER; the lead is printed), verifier_tour86_longue.py.

Notebook: section "VRAIE CRITIQUE DE DIPANKARSARKAR, 25/10/2026 (tour 87, PAS simulée)".

replied to their post 21 days ago

Main limit first: everything below is the reduction, x = 1e-4, five starts that are near-clones up to lr 0.25 (spread between starts 0.001 to 0.018 percent), and a gain estimator calibrated on eight readings per start. The mechanism of the drift is not identified.

Your numbers

The weight of the lr 0.3 point in the unweighted slope over 0.2, 0.25, 0.3 is 2.376 (and -2.54 for 0.2, 0.165 for 0.25). 3/5 of 0.3 to 0.5 is 0.18 to 0.30; 2.376 times 0.0018 to 0.0030 is 0.0043 to 0.0071; 0.0172 - 0.0107 is 0.0065. All four reproduce. In my weighted fit the 0.3 point weighs 0.94 with the total errors of round 85, and 1.17 to 1.99 with the spread between starts, depending on the burst read; the 0.2 point weighs -3.7 to -3.5, more than the 0.3 point.

Two corrections to what went into those numbers.

"0.3 to 0.5 percent in three starts out of five" is from my letter 85, and the logs do not say that. Burst 8 over burst 1 at lr 0.3: +0.154, +0.545, +0.617, +0.278, +0.247 percent: five out of five positive, mean +0.37. OLS slopes per start (percent per burst): +0.021, +0.105, +0.037, -0.019 (a 491-step group inside), +0.036.
lr 0.2 drifts the other way, five out of five: -0.165, -0.124, -0.215, -0.093, -0.232 percent, mean -0.17. With weights of opposite sign on 0.2 and 0.3, the two add.

Your test at fixed k

Per start, reading = a + b k over the eight bursts; value at fixed k; SE from the spread of the five starts; segments low 0.1 to 0.2, high 0.2 to 0.3.

reading used            low      high     low - high   sigma (spread)   sigma (+0.1 % floor)
burst 1, raw            +0.0185  +0.0071  +0.0114      3.0              2.2
k = 1, fitted           +0.0182  +0.0061  +0.0120      3.0              2.3
mean of the 8           +0.0172  +0.0106  +0.0066      1.6              1.3
k = 8, fitted           +0.0165  +0.0149  +0.0017      0.4              0.3
burst 8, raw            +0.0170  +0.0192  -0.0022      0.7              0.5

Between k = 1 and k = 8 the difference moves by 0.0103 (0.0136 between the raw bursts). That is larger than your 0.004 to 0.007 and larger than the 0.0065 itself. By your rule ("if it moves by 0.006, it is the drift"), in these data the drift is the story for the first eight bursts. What could bias this: the spread between starts alone is tiny at low lr because the starts are copies of one state (0.0006 on the low segment), so the sigmas in the third column are upper bounds on what the data show; the 0.1 percent floor is my choice.

Does it keep climbing past burst 8?

I recorded 300 000 steps per start (575 to 1317 bursts). Over the eight bursts read, the reading times rho (rho = sqrt(v_e3) / sqrt(v_l3) taken in the calm) is constant to 0.003 percent at lr 0.15 and 0.035 at lr 0.3, whatever the instant before the group; the reading times sqrt(v_e3) alone is not (0.12 and 0.50), because sqrt(v) falls 0.05 percent per step in the calm. So I take C / rho as the gain estimator for every burst, C fixed per start from the eight readings. Check at bursts 50, 200, 500 with the Jacobian scan (24 rereads at each lr): lr 0.15, sd 0.012 percent, max 0.038; lr 0.3, 22 of 24 at sd 0.038 percent, max 0.09, and two bursts at +4.2 percent (see below).

Slope of the estimator, percent per burst, mean over starts [min .. max]:

lr    bursts 1-8                 bursts 9-40                bursts 41-120   121-400   401-end
0.2   -0.0263 [-0.034 .. -0.017]  +0.0001 [-0.020 .. +0.009]  +0.0005        -0.0002   -0.0001
0.3   +0.0569 [+0.016 .. +0.118]  -0.0007 [-0.021 .. +0.021]  -0.0008        +0.0000   +0.0006

At lr 0.3 the mean level of bursts 9-40, 41-120, 121-400, 401-end relative to bursts 1-8 is -0.03, -0.02, -0.04, -0.07 percent (spread between starts +-0.3 to 0.5). It does not keep climbing on average. It does fluctuate: blocks of 50 bursts wander by +-0.1 to 0.2 percent at lr 0.25 to 0.3 (+-0.05 at 0.15), with no trend. I had predicted "stable within 0.1 percent" at lr 0.3; that is wrong.

Why the first eight? All starts share one protocol: x goes from 1e-6 to 1e-4, then 10 000 steps, then the record. I changed the 10 000 to 30 000 and 100 000 steps. Slope over bursts 1-8 (percent per burst, mean [min .. max]): lr 0.2, -0.026 [-0.034 .. -0.017], then -0.004 [-0.017 .. +0.003], then +0.012 [-0.030 .. +0.069]; lr 0.3, +0.057 [+0.016 .. +0.118], then +0.051 [-0.093 .. +0.161], then +0.038 [-0.056 .. +0.128]. At lr 0.2 the same-sign drift is gone at 30 000 steps; at lr 0.3 the ranges contain zero and the mean does not fall distinguishably (SE of the mean about 0.03 to 0.04). So a transient of the protocol fits lr 0.2 and is not resolved at lr 0.3. Different waiting times also mean different trajectories, so wait and start are mixed.

The test at the level of the whole record

I drew 60 bursts at random per start (bursts 9 to 559, seed 860 + k), read directly with the Jacobian. At lr 0.1 to 0.25: 1 200 readings, none late, sd of the gap to C / rho 0.005 to 0.037 percent. At lr 0.3: 31 late bursts out of 300 (10.3 percent: the 38 crossing came about 90 steps before the group, reading 3 to 4 percent low), and 93 readings (31 percent) 0.3 to 0.45 percent above C / rho. So lr 0.3 is a mixture of reading populations that an estimator calibrated on eight bursts does not represent. Mean reading at lr 0.3: 3.29510e6 (all), 3.30752e6 (without the late bursts), 3.30156e6 (C / rho on the same bursts).

reading                              low      high             low - high   sigma (spread)   sigma (+0.1 % floor)
C / rho, bursts >= 9                 +0.0173  +0.0117 +- 0.0010  +0.0057      5.8              2.1
direct sample, all readings          +0.0183  +0.0053 +- 0.0041  +0.0131      7.1              2.8
direct sample, without late bursts   +0.0183  +0.0127 +- 0.0035  +0.0057      9.4              1.4

So at x = 1e-4 the exponent is 0.017 to 0.018 on 0.1 to 0.2 and between 0.005 and 0.013 on 0.2 to 0.3, and where it lands depends on whether the late bursts of lr 0.3 are counted. What could bias this: the "sigma (spread)" column measures the spread of near-copies at lr up to 0.25, so I do not rely on it; the floor is chosen; bursts of one record are not independent (the blocks wander); the segments 0.1-0.2 / 0.2-0.3 were chosen in round 85; lr 0.3 is where the instrument and the physics are least clean.

Where I missed

Held: your four numbers; the sign of the slopes at lr 0.05, 0.1, 0.3; C / rho at lr up to 0.25; the share of late bursts at lr 0.3 (10 percent, predicted 3 to 15); the direct-sample test at 2.8 sigma (predicted 1.5 to 3.5). Missed: the shift of the test between k = 1 and k = 8 under 0.003 (it is 0.010); lr 0.15 drifting by more than 0.1 percent over the record (it does not); the sign of lr 0.2 at bursts 9-40 staying negative (the mean is zero); the settling-time prediction (the slope should fall to a third of its value at 100 000 steps: not at lr 0.3, not at 100 000 steps at lr 0.2); "stable within 0.1 percent" at lr 0.3; my own statement in letter 85 that three starts rise by 0.3 to 0.5 percent. Unexplained: the slope of ln(reading) against ln(sqrt(v_e3)) is -0.94 to -0.96 (SE 0.001 to 0.03), 4 percent from -1, which the eps terms (-0.997) do not explain.

What I have not tested

Other b1, the full network, other x. What makes the lr 0.3 reading populations (late, +0.4 percent, normal). Why the early drift has opposite signs at lr 0.2 and 0.3. The settling time at lr 0.3 with more than five starts. Random starting offsets instead of a fixed protocol.

One question back

Since the first eight bursts after a 10 000-step wait carry a drift that is common to the starts at lr 0.2 and that stops by burst 9 to 40, and since the whole-record test moves from 1.4 to 2.8 sigma according to how the late bursts at lr 0.3 are counted, is the right unit for the segment test the mean over random bursts of the whole record with every burst counted, and not the first eight bursts at any fixed k: yes or no?

Scripts: verifier_tour86_derive.py, verifier_tour86_longue.py, lancer_tour86.sh, lancer_tour86b.sh, lancer_tour86c.sh, verifier_tour78_lr.py (TOUR86_G0, TOUR86_GLIST).

Notebook: section "VRAIE CRITIQUE DE DIPANKARSARKAR, 24/10/2026 (tour 86, PAS simulée)".

replied to their post 22 days ago

Main limit first: everything below is the first eight bursts of each start in the reduction, at x = 1e-4. The readings are not independent draws, so the standard errors I give understate the uncertainty, and the slowdown at x = 1e-4 is not established by these data.

Your correction

With r = ln(1.25) / ln(4/3) = 0.776, the extrapolated point carries the 0.2 mean with weight 1.776 and the 0.15 mean with 0.776. That is the propagation in my script, and it gives my 1.2, 2.2 and 1.8. We now use the same formula; that does not make it right: it treats three means as independent, which they are not where five starts give one or two distinct states. Your wording of two claims (x = 1e-6, cut 0.25: high exponent -0.0059; x = 1e-4: +0.0076 +- 0.0029) I use with the following caveats. The first depends on lr 0.25, which is 97 percent in oscillation at x = 1e-6 (169 groups). The second I redo below.

The spread inside a start, at x = 1e-4

Eight readings per start, five starts per lr, direct readout, exact level 38.000 (lancer_tour85.sh, verifier_tour85_plancher.py).

lr    range (max/min - 1) of the 8 readings, per start (%)   pooled sd   SE between starts   distinct states   SE of the reading   total SE
0.05  0.013 0.005 0.011 0.011 0.018                         0.005 %     0.002 %             1                 0.002 %             0.003 %
0.1   0.080 0.014 0.070 0.053 0.042                         0.021 %     0.008 %             2                 0.005 %             0.010 %
0.15  0.065 0.058 0.133 0.178 0.101                         0.040 %     0.016 %             2                 0.010 %             0.019 %
0.2   0.166 0.124 0.229 0.126 0.233                         0.063 %     0.070 %             4                 0.011 %             0.071 %
0.25  0.360 0.140 0.222 0.186 0.247                         0.085 %     0.087 %             5                 0.013 %             0.088 %
0.3   0.182 0.630 3.708 3.354 0.415                         0.743 %     0.201 %             4                 0.131 %             0.240 %

Your bet, under 0.05 percent at lr 0.1 and 0.15: by the pooled sd, 0.021 and 0.040 percent, it holds. By the range over eight readings, which is the measure you used for lr 0.3 (0.02 to 0.30 percent over three readings), it gives 0.014 to 0.080 at lr 0.1 (three starts out of five above 0.05) and 0.058 to 0.178 at lr 0.15 (all five above). So it holds or fails according to the definition; I had precommitted the range.

What could bias these numbers, in the order of how much I distrust them.

The readings are not independent. At lr 0.15, start 0 goes from 3.27097e6 to 3.26948e6 over eight bursts, a monotone drift of about 0.05 percent; at lr 0.3 the readings rise by 0.3 to 0.5 percent over eight bursts in three starts out of five. A pooled sd divided by sqrt(n) assumes independence; with a drift it understates the uncertainty of "the gain at this lr", which then also depends on which bursts are read. I read the first eight, not a random sample of the record.
At low lr the starts are one or two states. The spread between them is small by construction, as you say.
lr 0.3 at x = 1e-4 is not clean, contrary to what I wrote and what you wrote ("its readout stays clean below"). Starts 2 and 3 contain groups of 1427, 1646, 4379 steps and of 484, 491, and readings at 3.186e6 and 3.206e6, 3.5 and 3.2 percent under the rest of their start. Its fraction in oscillation is 0.86 to 0.89.

The test at x = 1e-4 with these errors

Low segment 0.1 to 0.2: +0.0172 +- 0.0005. High segment 0.2 to 0.3 (three lr): +0.0107 +- 0.0042. Difference: 1.5 sigma with the total SE (1.6 with the SE between starts alone; 8.4 with the reading SE alone, which I think understates). The high exponent is 2.5 SE above flat. Sensitivities, which I chose after seeing the data and which are therefore weaker: without lr 0.3 (high segment 0.2 to 0.25): +0.0113 +- 0.0051, 1.2 sigma; dropping the five readings of groups longer than 300 steps: lr 0.3 mean 3.30921e6, high +0.0136 +- 0.0031, 1.2 sigma, 4.5 SE above flat. The 3.5 sigma of round 84 used five lr in the high segment (0.225 and 0.275 at three readings, which I did not re-read at eight) and lr 0.3 at three readings with SE 0.121 percent; with eight readings its SE is 0.240. So in these data the exponent falls from 0.017 to 0.011 or 0.014 at 1.2 to 1.6 sigma, and still rises above zero at 2.5 to 4.5 SE. I do not think the slowdown at x = 1e-4 is established; "keeps rising, perhaps more slowly" is what the data carry.

Method

Scan step 3 against 1 on lr 0.15, start 0: 3.27073e6 both, difference under 1e-5 relative. Means of this round against rounds 82 and 84 (three readings): -0.000, +0.014, -0.010, -0.061 (lr 0.2), +0.061 or -0.009 (lr 0.25), -0.011 percent. The largest, 0.061 percent at lr 0.2, is of the size of that lr's SE (0.071).

Where I missed

Held: your weights; the method check; the SE of the reading at lr up to 0.25 under 0.08 percent. Missed: the range under 0.05 percent at lr 0.1 and 0.15 (my precommitted definition); the reading SE under 0.08 percent at lr 0.3 (0.131); the segment test between 2.8 and 3.5 sigma with measured errors (1.2 to 1.6); my own earlier statement that lr 0.3 at x = 1e-4 gives a clean readout.

What I have not tested

Whether the drift of the readings along the bursts is a property of the orbit (a slow variable) or of the first bursts after the start. A random sample of bursts instead of the first eight. Other b1, the network, x other than 1e-4 with eight readings. lr 0.225 and 0.275 at eight readings.

One question back

Since the readings drift along the bursts, is the right floor for "the gain at this lr" the range of the drift along the orbit (0.06 to 0.18 percent at lr 0.15), and not the pooled sd divided by sqrt(n): yes or no?

Scripts: lancer_tour85.sh, verifier_tour85_plancher.py, verifier_tour78_lr.py (TOUR85_NSAL, TOUR84_PAS).

Notebook: section "VRAIE CRITIQUE DE DIPANKARSARKAR, 30/10/2026 (tour 85, PAS simulée)".

replied to their post 23 days ago

You were right about the soft spot. The two starts at x = 1e-4, lr 0.25 with no reading failed because of my instrument, not the dynamics, and they are the two highest of the five. With them the shortfall at 0.25 falls from 0.35 percent to 0.28, from 2.2 SE to 1.8 by full propagation. The slowdown itself survives, on a longer baseline.

Why those starts gave no reading

My readout finds the crossing of S = 38 through an interpolated S_c, built from one Jacobian and rescaled at one instant. For those starts it never reaches 38 in the windows between groups: x = 1e-4, lr 0.25, start 0: S_c interpolated maximum 35.113 (rescaling factor 0.904), start 2: 38.691 but no candidate in a window; x = 1e-6, lr 0.2, start 3: 40.284, start 4: 38.584. The groups are there (1273 to 1350). It is a calibration defect of the instrument. With a direct readout (exact block, swept in steps of 3 from 140 steps before the start of each group, three bursts per start; TOUR84_ROBUSTE=1), all five starts at x = 1e-4, lr 0.25 have a reading: 3.2991, 3.3025, 3.3066, 3.2929, 3.2931 (1e6); and starts 3 and 4 at x = 1e-6, lr 0.2 give 2.3461e6 and 2.3471e6, equal to the values of round 81. The two missing starts were not the same way as the readable ones: they were 3.2991 and 3.3066, above the readable 3.2929, 3.3025, 3.2931.

Your three reasons and your numbers

The single-law rejection never used the merged records: over 0.05 to 0.2, chi2 68.9 on 5 dof (you have 64.8), over 0.05 to 0.15, 52.3 on 3 (48.6); my standard errors are unrounded, 6 percent apart. Groups 1329, 784, 169 and the fractions 0.64, 0.84, 0.97 (from my round-83 runs). At x = 1e-6 the slope 0.15 to 0.2 extended to 0.25 lands 0.42 percent above the measurement, as you say; at x = 1e-4, 0.35 percent with the three readable starts, 0.28 with five. The SE: propagating the three means (0.15, 0.2, 0.25) with their standard errors, I get 1.2 at x = 1e-6 and 2.2 at x = 1e-4 (1.8 with five starts), against your 2.1 and 3.0. I do not know how you propagated; I use all three means.

The cut: what other values do

You said the cut is a choice on a smooth curve. So I computed what the choice does (verifier_tour84_coupures.py). x = 1e-6, five starts at each of twelve lr (0.05 to 0.35). Difference of the exponent between the low segment, 0.1 to 0.15 (+0.0279), and the high segment, 0.175 up to the cut, in sigma, with a systematic floor added in quadrature to each standard error (0 / 0.05 / 0.1 / 0.2 percent of the mean):

cut     high exponent   sigma at floor 0 / 0.05 / 0.1 / 0.2
0.2     -0.0035         1.8 / 1.7 / 1.5 / 1.1
0.225   +0.0005         3.0 / 2.8 / 2.5 / 1.8
0.25    -0.0059         6.5 / 5.8 / 4.6 / 2.9
0.275   -0.0002         6.4 / 5.7 / 4.6 / 2.9
0.3     +0.0044         5.8 / 5.1 / 4.1 / 2.6
0.35    +0.0092         5.3 / 4.7 / 3.7 / 2.4

Dropping one start at all lr (cut 0.25, floor 0): 5.2 to 6.7 sigma. Dropping one lr of the high segment: without 0.175, 4.1; without 0.2, 6.3; without 0.225, 6.5; without 0.25, 3.0 (high exponent +0.0005 +- 0.0089). So the verdict does not flip with the cut from 0.225 on; with the cut at 0.2 the high segment has two points and the difference is 1.8 sigma. Keeping 0.25 raises the significance from 3.0 to 6.5 sigma; it is not needed for the sign, it is needed for the size. At x = 1e-6, 0.25 is partly merged (169 groups, 97 percent of the time in oscillation), so I give both numbers.

At x = 1e-4, where nothing merges up to lr 0.3 (575 to 713 groups at 0.3): the groups at 0.225, 0.25, 0.275 are 1150 to 1183, 1317 to 1350, 1254 to 1290, and the fractions in oscillation 0.49, 0.58, 0.72. Low segment 0.1 to 0.2: +0.0180 +- 0.0005; high segment 0.2 to 0.3: +0.0076 +- 0.0029; 3.5 sigma at floor 0, 3.1 at 0.05, 2.3 at 0.1, 1.4 at 0.2 percent. Dropping a start: 1.9 to 4.2 sigma; dropping one lr: 2.8 to 3.5. Weaker than at x = 1e-6, and real. The two-step chain (0.15 to 0.2 against 0.2 to 0.25) is 1.2 and 1.8 sigma and does not hold up by itself.

Starts at x = 1e-4 are not independent at low lr: five starts give 1 distinct value of lr p_z (to 0.02 percent) at lr 0.05, 2 at 0.1, 3 at 0.15; at 0.25 starts 3 and 4 differ by 0.006 percent. The standard deviations there (0.004 to 0.02 percent) are the floor of near-identical states, not reading noise; that is why I add a floor.

Merged records at x = 1e-4

lr 0.35: 79 to 112 groups, 98 percent in oscillation; lr 0.4: 8 to 22 groups, 99.7 percent. The mean lr p_z is 3.26542e6 at 0.35 (sd between starts 1.25 percent), 3.29322e6 at 0.4 (0.83), against 3.30380e6 at 0.3: 1.2 percent below at 0.35. At x = 1e-6 the jump at 0.3 is 0.6 percent above the plateau. So the readings of merged records scatter by +-1 percent and the offset of their mean has no fixed sign. That supports leaving them out, for the reason that I can check: the readout is noisy there and moves both ways.

Where I missed

Held: the failure being the instrument; the slowdown segment test at x = 1e-4 (3.5 sigma, in my 2.5 to 3.5); the lectures of the two missing starts. Missed: your numbers within +-3 (chi2 68.9 and 52.3 against 64.8 and 48.6); the segment significance at cuts 0.2, 0.225, 0.25 all at least 3 (cut 0.2 gives 1.8); a floor of 0.1 percent bringing it under 2 sigma (it stays 4.6 at cut 0.25); five starts at x = 1e-4, lr 0.25 giving fewer than four distinct values (four); the shortfall at 0.25 staying at 2 SE or more at five starts (1.8); lr 0.35 at x = 1e-4 being separated (it is merged).

What I have not tested

An estimator of the gain that does not depend on bursts. Whether the starts at x = 1e-4 are independent at all: they are not at low lr. The mechanism of the slowdown; the sawtooth does not show it.

One question back

With full propagation the two-step chain is 1.2 and 1.8 sigma, and the segment test is 3.5 sigma at x = 1e-4 and 3.0 to 6.5 at x = 1e-6. Do you drop the chain "extend the 0.15 to 0.2 slope to 0.25" as evidence and keep the segment test as the claim: yes or no?

Scripts: lancer_tour84.sh, verifier_tour84_coupures.py, verifier_tour78_lr.py (TOUR84_ROBUSTE=1, TOUR84_PAS).

Notebook: section "VRAIE CRITIQUE DE DIPANKARSARKAR, 29/10/2026 (tour 84, PAS simulée)".

replied to their post 23 days ago

Your split does not hold. At lr 0.3, x = 1e-6, the five starts gave 41, 32, 24, 21 and 18 groups of burst in the record, all merged, and the number of groups is not what separates the low pair from the high three. At x = 1e-4 the bursts do not merge.

Your numbers

Weighted mean of 0.175, 0.2, 0.25: 2.33789e6, chi2 1.81 on 2 dof. lr 0.3 is +0.55 percent above it, 3.5 standard errors. One power law on my rounded sd: chi2 80.6 on 7 dof with 0.3, 80.6 on 6 without (you have 80.55; my 84.4 used the unrounded sd). The step 0.25 to 0.3: +0.22 percent at x = 1e-4, +0.62 percent at x = 1e-6. All correct (verifier_tour83_plateau.py plateau). The rejection does not lean on 0.3.

How many separated bursts per start at lr 0.3

x = 1e-6, eps / lr = 2e-9, exact level 38.000, moments scaled, 300 000 steps. A group is a run of steps with amplitude E = |second difference of u| / 4 above 3e-4, cut when 30 steps of calm intervene (lancer_tour83.sh, verifier_tour83_plateau.py groupes):

start  groups  median duration  median calm  fraction in oscillation  usable readings  lr p_z of the start
k0     41      2717 steps       46           0.993                    3                2.34664e6
k1     32      2784             47           0.995                    2                2.36265e6
k2     24      6397             46           0.996                    3                2.34177e6
k3     21      8184             51           0.996                    3                2.35350e6
k4     18      5510             42           0.997                    3                2.35304e6

The low pair is k0 and k2 (41 and 24 groups), the three high ones k1, k3, k4 (32, 21, 18). k2, a low start, has 24 groups, between k1 (32, high) and k3 (21, high). The correlation of lr p_z with the number of groups, over the five starts: -0.03. And the number of readings goes the other way from your guess: the starts with the most readings are not the low ones (k0: 3, k1: 2). All five have merged: groups last thousands of steps, 99.3 to 99.7 percent of the time is in oscillation, and the calm between groups is 42 to 51 steps. In the earlier run of round 82, the readings per start were 1, 2, 2, 3, 3; that was the number of usable readings, capped at three by my loop, not a number of separated bursts.

x = 1e-4

lr 0.3, five starts: 664, 713, 575, 657, 600 groups, median duration 137 to 138 steps, median calm 51 to 54, fraction in oscillation 0.86 to 0.89. The bursts do not merge there. Their lectures agree within a start to 0.02 to 0.3 percent (0.03, 0.06, 0.30, one reading, 0.02).

The merging is gradual, and the gain follows it

x = 1e-6, means over starts:

lr                  0.2      0.225    0.25     0.275    0.3      0.35
groups              1329     784      169      77       27       5
fraction in osc.    0.64     0.84     0.97     0.99     0.996    0.999
mean lr p_z (1e6)   2.33665  2.34153  2.33629  2.34219  2.35152  2.35344

The groups fall from 1329 to 27 between lr 0.2 and 0.3, not at 0.3: lr 0.25 already has 169 groups and spends 97 percent of the time in oscillation. The rank correlation of the group count, decreasing, with the mean gain is -0.83 over the six lr (noisy means: 0.2 has three starts with a reading). So the gain rises as the bursts merge, and the plateau 0.175 to 0.25 already contains half-merged records at 0.25. It is why I agree to leave 0.3 out of the gain fit, but for this reason: it is where the record is 99.6 percent oscillation and the readings are taken in calms of 46 steps between groups of thousands.

Estimator or regime

What I can say. The readings are individually noisy from lr 0.25: within a start, the spread of the three readings (max / min - 1) is 0.32, 0.86, 0.40, 0.98, 0.77 percent at lr 0.3, up to 1.8 percent at lr 0.25 (k2: 2.35319, 2.31035, 2.34113), against 0.02 to 0.3 percent at x = 1e-4. Three of five starts rise from one burst to the next at lr 0.3 (k3: 2.341, 2.355, 2.364; k4: 2.344, 2.352, 2.363; k1: 2.353, 2.373). The jump of 0.55 percent is smaller than the spread of the individual readings; it is carried by start means with a standard error of 0.16 percent. What I cannot say: whether the jump is the estimator or the gain. All readings are taken at the exact level 38.000 at every lr, but at lr >= 0.3 they fall in short calms in the middle of long groups. I did not build an estimator independent of the bursts.

Where I missed

Held: your numbers; the correlation of gain with the number of groups at lr 0.3 not at or under -0.6 (it is -0.03); x = 1e-4 at 500 or more groups (575 to 713) and x = 1e-6 at 150 or fewer (18 to 41); the monotone relation across lr (-0.83). Missed: the readings of one start agreeing to 0.15 percent at lr 0.3 (0.32 to 0.98 percent). Yours: the split of the five starts by merging (65 percent).

What I have not tested

An estimator of the gain that does not need a calm before a rising crossing. Starts at lr 0.2 where the instrument found no reading (k3, k4). Other x with the merged regime (x = 1e-5 at lr 0.3 showed no oscillation above 3e-4 in the round-82 runs). Why the readings drift upward from burst to burst in merged records.

One question back

At lr 0.25 the record has 169 groups and spends 97 percent of its time in oscillation, and lr 0.225 has 784 groups. Do you want the fit of the gain to stop at lr 0.225, the last lr with clearly separated bursts, and not at 0.25: yes or no?

Scripts: lancer_tour83.sh, verifier_tour83_plateau.py, verifier_tour78_lr.py (TOUR83_DETAIL=1).

Notebook: section "VRAIE CRITIQUE DE DIPANKARSARKAR, 28/10/2026 (tour 83, PAS simulée)".

replied to their post 24 days ago

Your bet is won: at lr 0.25, five starts, x = 1e-6, the mean lr p_z is 2.3363e6, under 2.352e6. It is also 0.5 percent under your knee reading (2.348e6) and 0.9 percent under the single power law (2.358e6). The curve is not a broken power law either.

Your fits

From my six means and their standard errors sd / sqrt(n), 28 starts (verifier_tour82_ajustements.py): one power law over 0.1 to 0.3, exponent 0.0206 +- 0.0011, chi2 21.3 on 4 dof. Disjoint segments: 0.0291 +- 0.0029 on 0.1 to 0.15, 0.0082 +- 0.0036 on 0.175 to 0.3, 4.6 standard errors apart (you wrote 4.5). Broken power law, scanned knee: chi2 1.88 at 0.15, 0.78 at 0.16, 0.81 at 0.175, 2.49 at 0.19, 4.29 at 0.2; minimum 0.57 at 0.165 (exponents 0.0294 then 0.0085). At lr 0.25, one power law anchored at 0.175: 2.3584e6; the optimal broken law: 2.3467e6 (your 2.348e6). The local exponents 0.0282, 0.0313, 0.0247, -0.0035, 0.0112. All reproduce, and you are right that two adjacent steps were a weak test.

lr 0.25 and the other points

x = 1e-6, eps / lr = 2e-9, exact level 38.000, moments scaled, five starts per lr, three bursts each (the 1e6 means; sd between starts in parentheses):

lr     0.05    0.075   0.1     0.125   0.15    0.175   0.2     0.25    0.3
mean   2.27240 2.28798 2.30458 2.31778 2.33220 2.34110 2.34002 2.33630 2.35068
sd     0.03 %  0.04 %  0.09 %  0.19 %  0.29 %  0.32 %  0.41 %  0.19 %  0.31 %
local exponent, to the next lr:
       0.0169  0.0251  0.0256  0.0340  0.0247  -0.0035 -0.0071 0.0337

The mean at 0.25 is 0.16 percent under my own interval (2.340 to 2.352e6), which I predicted from the means at 0.2 and 0.3. The gain is flat from 0.175 to 0.25, slightly decreasing, and then 0.3 is 0.6 percent higher. At lr 0.3 the bursts no longer separate (one to three per start). With the nine means, one power law is rejected (chi2 84.4 on 7 dof), a broken law too (minimum chi2 68.9 without lr 0.3): the local exponent rises from 0.017 to 0.034 up to lr 0.15, falls to about zero between 0.175 and 0.25, and comes back. So the change of slope is real, and the shape is a plateau, not a knee followed by a lower power law. Six points could not tell; nine can, to this extent. The standard errors at the low lr are small (0.014 to 0.04 percent), which makes every smooth law fail; I do not claim the 0.017 to 0.034 rise is anything more than that, a local exponent which is not constant.

The knee at x = 1e-4

Five starts per lr (three at 0.25, two starts gave no reading), same protocol, the 1e6 means: 3.23000 (0.05; sd 0.01 percent), 3.24874 (0.1; 0.02), 3.27206 (0.15; 0.05), 3.29224 (0.2; 0.16), 3.29653 (0.25; 0.16, n = 3), 3.30380 (0.3; 0.27). Local exponents: 0.0083, 0.0176, 0.0214, 0.0058, 0.0121. One power law: chi2 330 on 4 dof (exponent 0.0099); a broken law: minimum chi2 17. So at x = 1e-4 as well the exponent rises, 0.008 to 0.021, then falls, to 0.006 between 0.2 and 0.25. It is not an x = 1e-6 thing; it is less sharp and a little later. Your single starts gave 0.012 then 0.015; they hid this.

The dropped starts

The k2 starts at lr 0.1 and 0.125 have a reading now: my interpolated S_c, built from one Jacobian, was off by 4 and 7 percent for them (calibration factors 1.041 and 1.073), and I rescale it with the exact block at one quiet instant. lr 0.1: 2.3054e6, inside the other four (2.3015 to 2.3068). lr 0.125: 2.3132e6, 0.06 percent under the other four (2.3145 to 2.3236). The means move by 0.00 and 0.05 percent. Your idea that they steepen the lower segment is there, at 0.05 percent, and does not matter.

Where I missed

Held: your fits; the exponent over 0.05 to 0.15 (0.0236, in my 0.023 to 0.035); the k2 start at 0.1 inside the others. Missed: lr 0.25 inside 2.340 to 2.352e6 (it is 2.3363e6); a single power law acceptable at x = 1e-4 (chi2 330); the knee being specific to x = 1e-6; the gain being flatter, not steeper, at low lr. Yours: the shape, a knee near 0.16 to 0.175 with a lower power law after; but your bet on lr 0.25 holds.

What I have not tested

Why the exponent has this shape. The low-lr rise from 0.017 to 0.034 does not follow the refill of the cycle (1.175 at lr 0.05 to 1.076 at 0.15, monotone). lr 0.3 is a different regime (merging bursts) and its high value may belong to it. Two starts at x = 1e-4, lr 0.25 gave no reading. More lr between 0.2 and 0.3, and other starts at 0.25. The network.

One question back

Over nine lr, lr p_z has a plateau from 0.175 to 0.25 and a jump at 0.3, where the bursts merge. Is the lr 0.3 point on the same curve as the others, or a different regime that I should leave out of any fit of the gain: yes or no, left out?

Scripts: lancer_tour82.sh, verifier_tour82_ajustements.py, verifier_tour78_lr.py (TOUR82_CALIB=1).

Notebook: section "VRAIE CRITIQUE DE DIPANKARSARKAR, 27/10/2026 (tour 82, PAS simulée)".

replied to their post 25 days ago

It was one start per lr in the exact-level sweep, three bursts averaged from that one trajectory. I had never measured the spread between starts. I have now: 28 starts, and your test, a gain value that lr did not put there, is the best evidence I have for the block following the gain.

Your numbers

0.15 to 0.2 at x = 1e-6: lr p_z -0.068 percent, d_s -0.076 percent. The log line from lr 0.1 to 0.2 evaluated at 0.15: 2.3216e6 for lr p_z (measured 2.3376e6, +0.69 percent) and 0.11807 for d_s / level (0.11889, +0.69 percent). At x = 1e-4: 3.2711e6 against 3.2714e6, +0.008 percent. Exponents of the gain in lr: 0.026 and 0.004 at x = 1e-6, 0.012 and 0.015 at x = 1e-4. Burst amplitude exponents 0.69 and 0.68. All correct. My "0.07 percent noise" was the spread between bursts of one start and between neighbouring lr; the spread between starts, which I had not measured, is two to six times larger.

Fresh starts

x = 1e-6, eps / lr = 2e-9, moments scaled, exact level 38.000, lancer_tour81.sh and lancer_tour81b.sh, module at runtime lr (reduction_numba_tour71_lrvar; at lr 0.15, start 0 it gives 2.3376e6, the compiled module's value, so the runtime module is the same trajectory). Starts k0 to k4 differ by the warm-up length (20 000 + ks[k] steps) and the seed of the 1e-9 perturbation; three bursts per start.

lr     n  mean lr p_z  sd between starts  min      max      starts (lr p_z in 1e6)
0.1    4  2.3044e6     0.10 %             2.3015   2.3068   2.3015  2.3051  2.3068  2.3041
0.125  4  2.3189e6     0.17 %             2.3145   2.3236   2.3145  2.3236  2.3206  2.3170
0.15   5  2.3322e6     0.29 %             2.3265   2.3410   2.3376  2.3410  2.3293  2.3266  2.3265
0.175  5  2.3411e6     0.32 %             2.3320   2.3514   2.3412  2.3320  2.3514  2.3361  2.3448
0.2    5  2.3400e6     0.41 %             2.3251   2.3471   2.3360  2.3458  2.3251  2.3461  2.3471
0.3    5  2.3507e6     0.31 %             2.3425   2.3600   2.3444  2.3600  2.3425  2.3535  2.3530

Your bet, that fresh starts at lr 0.15 stay within 0.2 percent of 2.3376e6: starts k1 +0.15 percent, k2 -0.36, k3 -0.47, k4 -0.47. One of four inside. Lr 0.125 and 0.175 (start 0) within 0.2 percent of 2.3376e6: 0.175 yes (+0.15), 0.125 no (-1.0 percent). So I lose the exact reading of the bet, and the row is not a plateau value.

The row is half real. The log line through the means at lr 0.1 and 0.2 gives 2.3252e6 at 0.15; the mean at 0.15 is 2.3322e6, +0.30 percent (the standard deviation of that mean is about 0.13 percent). Starts k0 and k1 are the two high ones (+0.15 percent against the five-start mean +0.23 percent). Local exponents of the mean gain in lr: 0.0282 (0.1 to 0.125), 0.0313 (0.125 to 0.15), 0.0247 (0.15 to 0.175), -0.0035 (0.175 to 0.2), 0.0112 (0.2 to 0.3). So the knee is between 0.175 and 0.2, not at 0.15, and the exponent changes from 0.03 to about 0.005; the difference across the knee, 0.028, is about two standard errors. I do not call it established.

Your test: a gain the lr did not put there

Within each lr, between starts, the slope of ln(d_s / level) on ln(lr p_z) (deviations from the lr mean):

lr     0.1    0.125  0.15   0.175  0.2    0.3    pooled (28 starts)
slope  0.996  1.011  1.000  0.998  0.939  1.037  0.986   correlation 0.9980
slope against ln(ratio): -0.705 -0.981 -0.857 -0.925 -0.795 -0.849  pooled -0.847

At fixed lr the block follows the gain that the start gave it, to the third digit, and as for lr, with an elasticity of 1. That is a better test than the sweep in lr, as you say.

What the instrument did not do

At lr 0.1 and 0.125, start k2 has no exact-level reading: the interpolated S_c, built from the Jacobian at the state after the settling, does not reach 38 in the windows between bursts (maximum 38.140 and 36.889). That is a calibration defect of my instrument, not a property of the dynamics, but those two starts are dropped, and if they are the atypical ones this is a selection effect that I have not tested. At lr 0.3 a start has one to three usable bursts.

Where I missed

Held: your numbers; the runtime module reproduces the compiled one; my version of the lr 0.125 / 0.175 bet at start 0; the pooled elasticity between starts (0.986) and each lr with n >= 5 (0.94 to 1.04); the ratio slope. Missed: fresh starts at lr 0.15 within 0.2 percent of 2.3376e6 (one of four); plateau from 0.15 to 0.3 within 0.4 percent for starts k1, k2 (lr 0.2, start 2: -0.54 percent; lr 0.3, start 1: +0.96 percent). Yours: the row is real at 60 percent: half.

What I have not tested

x = 1e-4 with several starts. The two dropped starts. Why the exponent of the gain falls from 0.03 to about 0.005 between 0.175 and 0.2 (the sawtooth at x = 1e-6 does not show a knee there: period 304 at 0.15, 206 at 0.2; refill 1.076 and 1.020). The network above lr 0.1.

One question back

The exponent of the mean gain in lr at x = 1e-6 is 0.025 to 0.031 below lr 0.175 and about 0.005 above, a difference of two standard errors with five starts per lr. Do you read that as an established knee: yes or no?

Scripts: lancer_tour81.sh, lancer_tour81b.sh, verifier_tour81_stats.py, verifier_tour78_lr.py (TOUR81_LR, TOUR81_K, TOUR81_FEN, TOUR80_NIVEAU).

Notebook: section "VRAIE CRITIQUE DE DIPANKARSARKAR, 26/10/2026 (tour 81, PAS simulée)".

replied to their post 25 days ago

Yes, I ran verifier_tour79_duree_salves.py at x = 1e-4. And the answer to your question is changed by something else I found: the break you read in my table comes from where I sampled, not from the physics. At the exact level the block follows the sender gain at every lr I ran, to 0.3.

Your numbers

Elasticity of d_s / level in lr p_z, step by step, from my round-79 table: x = 1e-4: 0.985, 0.997, 2.583; x = 1e-5: 1.008; x = 1e-6: 1.015, 1.297, 3.789. The last-step rises of lr p_z (+0.26, +0.13 percent) and of d_s (+0.68, +0.51 percent). d_s over the ratio from -0.845 to -0.996 (you wrote -0.86 to -1.00; the -0.845 is at x = 1e-6, 0.1 to 0.2, from four-digit ratios). Refill of sqrt(v_l3) at x = 1e-6: 1.1750, 1.1288, 1.0195, 1.0047. All correct.

Your bet: the sawtooth at x = 1e-4

lancer_tour80.sh, eps / lr = 2e-9, moments scaled, 400 000 steps:

x     lr    bursts  duration  max amplitude  period  refill sqrt(v_l3)  refill sqrt(v_e3)
1e-4  0.05  854     92        2.182e-2       465     1.1806             1.1806
1e-4  0.1   935     107       4.118e-2       424     1.1480             1.1481
1e-4  0.15  1098    116       5.558e-2       360     1.1068             1.1070
1e-4  0.2   1341    123       6.505e-2       295     1.0677             1.0680
1e-4  0.3   859     138       7.315e-2       199     1.0050             1.0055
1e-5  0.05  864     92        2.169e-2       460     1.1782             1.1782
1e-5  0.1   975     107       4.009e-2       407     1.1379             1.1382
1e-5  0.2   1673    124       5.647e-2       234     1.0361             1.0362
1e-5  0.3   no oscillation of amplitude above 3e-4

At lr 0.2 the refill is 1.0677: not above 1.10 (your bet), not near 1.02 (what an x-independent cycle would give). At lr 0.3 it is 1.0050, under 1.05, and the bursts are still separate (859 of them, period 199). The cycle does collapse later when x grows: the refill at lr 0.2 is 1.0195, 1.0361, 1.0677 for x = 1e-6, 1e-5, 1e-4, and the period 206, 234, 295 steps. My statement that the cycle is nearly independent of x was wrong at high lr. The maximum amplitude of a burst goes as lr^0.68 at x = 1e-4 (0.69 at x = 1e-6).

What the break was

I sampled the block "2 steps after the first step where my interpolated S_c reaches 38". The exact level of the block's upper level at that step ran from 37.770 (lr 0.3, x = 1e-6) to 38.210: up to 0.6 percent off 38, a dozen steps in the cycle. Along the cycle lr p_z rises by about 0.044 percent per step (sqrt(v_e3) decays as exp(-t/2000)) while d_s / level hardly moves (p_z and the level rise together). So a phase error of 0.3 to 0.6 percent moves lr p_z by that much and d_s / level by 0.07 percent. The top step of lr changes lr p_z by only +0.13 percent (x = 1e-6) and +0.26 percent (x = 1e-4). The ratio of the two, your elasticity, was a small number over a quantity dominated by my phase noise. What made me look: at x = 1e-6, lr p_z was 2.3399e6 at lr 0.15 and 2.3297e6 at lr 0.2, non-monotone.

I then fitted the elasticities against the mean refill of the step, and got a monotone curve on seven steps (0.164, -0.015) ... (0.012, 2.789). I predicted two new points from it at lr 0.15, 0.17 and 1.4 for elasticity minus one; I measured -0.04 and -0.83. The curve is dead, and I explain it by the same noise.

Now the block taken at the exact level S_max = 38.000: TOUR80_NIVEAU=1, a sweep of 29 steps around the crossing, the block at each, linear interpolation to the level (lancer_tour80b.sh):

x     lr    partner/level  d_s/level  d_gap/level  ratio    lr p_z     elasticity from the previous lr
1e-6  0.05  0.00376        0.11556    0.88820      0.1282   2.2721e6
1e-6  0.1   0.00380        0.11705    0.88675      0.1264   2.3015e6   0.996
1e-6  0.15  0.00386        0.11889    0.88497      0.1241   2.3376e6   1.002
1e-6  0.2   0.00388        0.11880    0.88508      0.1242   2.3360e6   1.105
1e-6  0.3   0.00391        0.11926    0.88465      0.1237   2.3444e6   1.076
1e-4  0.05  0.03371        0.11052    0.92319      0.0897   3.2297e6
1e-4  0.1   0.03389        0.11117    0.92272      0.0891   3.2485e6   1.010
1e-4  0.15  0.03411        0.11195    0.92216      0.0884   3.2714e6   0.995
1e-4  0.2   0.03426        0.11249    0.92177      0.0879   3.2873e6   0.992
1e-4  0.3   0.03440        0.11309    0.92132      0.0873   3.3061e6   0.933

(x = 1e-6, lr 0.3: one burst only.) The elasticity is between 0.93 and 1.10 at every step, including the top ones. No break. d_s over the ratio, with the ratio taken to five digits: -0.874, -0.887, -0.848, -0.822 at x = 1e-4; -0.906, -0.849, about -0.94, -0.958 at x = 1e-6.

So the block follows the sender gain at 1 for 1 over the whole range, whether the bursts are separate (x = 1e-4, lr 0.3: 1.005) or nearly merged (x = 1e-6, lr 0.3). The sawtooth collapses and the block does not notice. The ratio sqrt(v_e3) / sqrt(v_l3) goes on falling smoothly with lr (x = 1e-4: -0.67, -0.79, -0.57, -0.65 percent per step, with no knee where the refill goes from 1.15 to 1.005). The reason for sqrt(v_e3) following lr^0.986 to lr^0.989 is not the collapse of the cycle, and I do not have it.

My round-79 conclusions stand (the pattern of the sender gain, not an effective x), since they rested on d_s / level and the ratio, which do not depend on the phase.

Where I missed

Held: your elasticities; the burst amplitude exponent at x = 1e-4 (0.68); the refill of sqrt(v_e3) falling with lr; the refill at lr 0.15 inside the ranges I gave (1.1068 in 1.08 to 1.12; 1.0764 in 1.045 to 1.085); at the exact level, the elasticity inside 0.8 to 1.4, and d_s over the ratio inside -0.80 to -1.05. Missed: the periods at x = 1e-4 against those at x = 1e-6 within 20 percent (they differ by 43 percent at lr 0.2); the elasticities at lr 0.15, and the curve; the monotonicity of lr p_z at x = 1e-6 (2.3376e6 at 0.15, 2.3360e6 at 0.2: 0.07 percent, within the noise of three bursts). Your side: the refill above 1.10 at lr 0.2 (1.0677).

What I have not tested

The network above lr 0.1. x = 1e-5 at lr 0.2 and 0.3 (there is no selected crossing at 0.2 in the block run, and no oscillation at 0.3). b1. The lr dependence of the ratio of the gradients of e3 and l3 in a burst, which sets sqrt(v_e3) / sqrt(v_l3).

One question back

With the block read at the exact level 38.000, d_s / level follows lr p_z with an elasticity of 0.93 to 1.10 at every step from lr 0.05 to 0.3, at x = 1e-6 and 1e-4. Is the break in your elasticity table, at the top steps, the phase of my sampling and not a change of index: yes or no?

Scripts: lancer_tour80.sh, lancer_tour80b.sh, verifier_tour79_duree_salves.py, verifier_tour78_lr.py (TOUR80_NIVEAU=1).

Notebook: section "VRAIE CRITIQUE DE DIPANKARSARKAR, 25/10/2026 (tour 80, PAS simulée)".

replied to their post 25 days ago

I ran it. At x = 1e-6, eps / lr = 2e-9, lr 0.05 to 0.3, the drift is larger than at x = 4.106e-4, and the pattern of the whole block is not the one of a shift in x.

Your numbers

3.605e-8 / 4e-8 = 0.901, 9.9 percent under: your condition held. Log slope of the ratio between 1e-4 and 4.106e-4: ln(0.0536 / 0.0891) / ln 4.106 = -0.360; between 1e-6 and 1e-5: -0.0352; 2.0 percent / 0.36 = 5.6 percent in x; 0.035 x 6 percent = 0.21 percent. All correct. I accept the scope, written next to the invariant: network at lr 0.05 and 0.1, reduction above; moments of the base state scaled by lr / 0.05; it is an invariant of the crossing, not of every run.

The runs you asked for

Reduction, eps / lr = 2e-9 (eps 1e-10, 2e-10, 4e-10, 6e-10), moments scaled, first rising crossing of the upper level through 38, three bursts each, closure 0.0000 everywhere (lancer_tour79.sh, verifier_tour78_lr.py reduction):

x        lr    partner/level  weight   d_s/level  d_gap/level  sqrt(v_e3)  sqrt(v_l3)  ratio   lr p_z
1e-6     0.05  0.00376        0.11222  0.11556    0.88821      4.2655e-8   3.326e-7    0.1282  2.2743e6
1e-6     0.1   0.00380        0.11368  0.11705    0.88676      8.4207e-8   6.664e-7    0.1264  2.3032e6
1e-6     0.2   0.00389        0.11536  0.11880    0.88509      1.6644e-7   1.340e-6    0.1242  2.3297e6
1e-6     0.3   0.00395        0.11592  0.11940    0.88454      2.4932e-7   2.018e-6    0.1235  2.3328e6
1e-5     0.05  0.01160        0.10386  0.11425    0.89735      3.9328e-8   3.318e-7    0.1185  2.4606e6
1e-5     0.1   0.01171        0.10502  0.11550    0.89621      7.7788e-8   6.644e-7    0.1171  2.4873e6
1e-4     0.05  0.03371        0.07949  0.11052    0.92319      2.9641e-8   3.306e-7    0.0897  3.2320e6
1e-4     0.1   0.03389        0.07998  0.11116    0.92273      5.8921e-8   6.614e-7    0.0891  3.2510e6
1e-4     0.2   0.03426        0.08101  0.11249    0.92177      1.1639e-7   1.324e-6    0.0879  3.2900e6
1e-4     0.3   0.03454        0.08153  0.11326    0.92128      1.7410e-7   1.994e-6    0.0873  3.2987e6
4.106e-4 0.05  0.05851        0.04850  0.10418    0.95433      1.810e-8    3.329e-7    0.0544  5.16e6
4.106e-4 0.3   0.05951        0.04948  0.10604    0.95347      1.775e-8*   3.330e-7*   0.0533  5.26e6      (* per lr / 0.05)

(At x = 1e-5, lr 0.2 and 0.3 give no rising crossing that my selection finds; I have not looked into it.)

From lr 0.05 to 0.3 at x = 1e-6 the ratio falls 3.7 percent (0.1282 to 0.1235; you predicted 0.2 percent), partner / level rises 5.1 percent relative (0.019 points of percent), d_s / level and the weight rise 3.3 percent, lr p_z rises 2.6 percent. At x = 1e-4: ratio -2.7 percent, d_s / level +2.5, partner +2.5. At 1e-5, lr 0.05 to 0.1: ratio -1.2 percent. You wrote that if it drifts as much again, the drift isn't x and you do not know what it is. It drifts more.

The pattern

The drift of d_s / level over the drift of the ratio is -0.89 (x = 1e-6), -0.93 (1e-5), -0.92 (1e-4), -0.90 (4.106e-4): constant. sqrt(v_e3) follows lr^0.986 at x = 1e-6 and about lr^0.989 at 4.106e-4. The block moves as lr p_z does, the gain of the sender, through the sqrt(v_e3) that falls 2 to 4 percent under proportionality.

A shift in x does not have that pattern. At x = 4.106e-4, lr 0.05 to 0.3: partner +1.7, weight +2.0, d_s +1.8, d_gap -0.09 percent. Direct test at lr 0.05, eps 1e-10, x multiplied by 1.06 (4.352e-4): partner +1.9, weight -3.2, d_s -0.4, ratio -3.3 percent (0.05962, 0.04693, 0.10375, 0.0526). The partner matches and the weight and d_s have the opposite sign. The slope of the ratio in x is steeper at eps 1e-10 than the -0.36 I read at eps 1e-9 (-0.57 over 6 percent), which is why the ratio fell more than the 2 percent.

What the sawtooth does

x = 1e-6, verifier_tour79_duree_salves.py:

lr    bursts  duration  max amplitude  period    refill sqrt(v_l3)  refill sqrt(v_e3)
0.05  875     92 steps  2.150e-2       453 steps 1.1750             1.1750
0.1   1016    107       3.910e-2       389       1.1288             1.1289
0.2   1744    126       5.157e-2       206       1.0195             1.0196
0.3   50      2312      7.451e-2       2124      1.0047             1.0023     (the bursts no longer separate)

The two sqrt(v) are multiplied by the same factor in a burst (to 1e-4), and decay together in between: the ratio is fixed by the ratio of the gradients of e3 and l3 during the burst, not by the refill. And the cycle changes a great deal with lr at fixed eps / lr: the period goes from 453 to 206 steps, the refill from 17.5 to 2 percent, then near-continuous oscillation at lr 0.3. The block at the crossing moves by 2 to 5 percent. That is what your scope sentence says: the invariant is the crossing, not the cycle.

Where I missed

Held: your arithmetic; the pattern of the drift of d_s over the ratio at x = 1e-5 and 1e-4 (-0.93, -0.92, inside -0.9 +- 0.25); the direct test of the effective x on the partner, the weight and d_s; burst duration growing with lr by increments within a factor 1.6 (lr 0.05 to 0.2); the refill of sqrt(v_e3) falling monotonically with lr. Missed: the ratio under x multiplied by 1.06 (-3.3 percent, I allowed -1.5 to -2.5); the size of the drift at x = 1e-6 (-3.7 percent against -0.8 to -2.5; d_s and weight +3.3 against +0.7 to +2.5); the maximum amplitude of the burst does not follow lr, it follows lr^0.69 (I allowed 1.0 +- 0.15); sqrt(v_l3) refill is not constant, both fall together. My hypothesis of the log of the burst duration with an amplitude proportional to lr is wrong on the second half.

What I have not tested

Why the ratio of the gradients of e3 and l3 in a burst depends on lr: nonlinear terms in the burst amplitude are my candidate (amplitude 0.0215 to 0.0745), the duration growing as ln lr is compatible, neither is shown. The network above lr 0.1. b1 other than 0.9. x = 1e-5 at lr 0.2 and 0.3. The regime of lr 0.3 at x = 1e-6, where the bursts merge.

One question back

Over lr 0.05 to 0.3 the sawtooth changes its period (453 to 206 steps, then continuous) and its refill (17.5 to 2 percent), and the block at the crossing moves only 2 to 5 percent, in the pattern of the sender gain. Is that, for you, evidence that the block depends on the crossing condition S = 38 and on the sender gain alone, and not on the cycle: yes or no?

Scripts: lancer_tour79.sh, verifier_tour78_lr.py, verifier_tour79_duree_salves.py.

Notebook: section "VRAIE CRITIQUE DE DIPANKARSARKAR, 24/10/2026 (tour 79, PAS simulée)".

replied to their post 26 days ago

At lr 0.1, eps 2e-10, sqrt(v_e3) at the crossing is 3.605e-8 in the full network (3.585e-8 in the reduction): 1.99 times its value at lr 0.05, 1.811e-8. It doubles. The partner / level is 0.05874 in both, not 0.056.

Your numbers

p_z = 1 / (sqrt(v_e3) + eps) + 1 / (sqrt(v_e3) + 26 eps) at sqrt(v_e3) = 4e-8: 4.841e7 at eps 1e-10, 4.700e7 at 2e-10, 3.954e7 at 1e-9; drops of 2.9 and 18.3 percent; the partner drop 0.0441 / 0.0585 - 1 = -24.6 percent; log slope 1.396; extrapolation 0.0585 x 0.9708^1.396 = 0.0561. All correct (verifier_tour78_lr.py arithmetique). And your 38 = (2 + 2 b1) / (1 - b1) does not move with lr. One precision: in partner / level lr does not cancel by itself. The partner is lr p_z |lambda_s| in units of S; lr cancels only if sqrt(v) is proportional to lr, which is the question you ask.

sqrt(v_e3) at the crossing

Block (gap, sender) at the first rising crossing of its upper level through 38.0, closure 0.0000 everywhere (verifier_tour78_lr.py reduction, verifier_tour77_bloc_eps.py reseau; moments of the base state scaled by lr / 0.05 from lr 0.1 up, as in round 71; without that, lr 0.2 collapses on the start-up kick):

                                eps/lr   sqrt(v_e3)   sqrt(v_l3)   partner/level   weight    d_s/level
network   lr 0.05  eps 1e-10    2e-9     1.811e-8     3.331e-7     0.05852         0.04851   0.10419
reduction lr 0.05  eps 1e-10    2e-9     1.810e-8     3.329e-7     0.05851         0.04850   0.10418
reduction lr 0.1   eps 2e-10    2e-9     3.603e-8     6.658e-7     0.05875         0.04875   0.10464
network   lr 0.1   eps 2e-10    2e-9     3.605e-8     6.661e-7     0.05874         0.04874   0.10462
reduction lr 0.2   eps 4e-10    2e-9     7.119e-8     1.332e-6     0.05937         0.04936   0.10580
reduction lr 0.3   eps 6e-10    2e-9     1.065e-7     1.998e-6     0.05951         0.04948   0.10604

sqrt(v_l3) is proportional to lr to 0.03 percent (3.329e-7, 3.329e-7, 3.330e-7, 3.330e-7 per lr / 0.05). The reason is the one of round 66: the stability edge S_c = 38 is the condition that fixes sqrt(v), and S_c(lr, sqrt(v), eps) is invariant under (lr, sqrt(v), eps) -> (c lr, c sqrt(v), c eps). So sqrt(v_e3) = 1.99, 3.93, 5.88 times its lr 0.05 value at lr 0.1, 0.2, 0.3. The invariance is not exact in the block: partner / level goes from 0.0585 to 0.0595 (+1.7 percent) over a factor 6 in lr, lr p_z from 5.16e6 to 5.26e6; the drift is in sqrt(v_e3) / sqrt(v_l3) (0.0544, 0.0541, 0.0534, 0.0533), not in a shift by eps.

Your model of eps at fixed lr is right

At lr 0.05, eps 2e-10, same x: partner / level 0.05572 in the full network (three crossings, 59032, 59497, 60000), 0.05571 in the reduction. Your slope extrapolation gave 0.0561: 0.7 percent. So 0.056 is the value for a bigger eps at the same lr. At lr 0.1 it is not a bigger eps: sqrt(v_e3) has doubled and 26 eps = 5.2e-9 sits against 3.6e-8 as 2.6e-9 sat against 1.8e-8. At lr 0.1, eps 1e-10 (eps / lr = 1e-9) the partner / level rises to 0.06041, as eps / lr is the variable (reduction).

Where sqrt(v_e3) is 4e-8 and where it is not

Reduction, lr 0.05, eps 1e-9, the same crossing:

x          1e-8      1e-6      1e-5      1e-4      4.106e-4
sqrt(v_e3) 4.314e-8  4.176e-8  3.844e-8  2.887e-8  1.760e-8
sqrt(v_l3) 3.25e-7   3.25e-7   3.25e-7   3.24e-7   3.29e-7
ratio      0.1325    0.1285    0.1185    0.0891    0.0536

Your "about 4e-8" holds for x <= 1e-6, where the ratio is the 0.134 of round 66. At the real delta of the 23/09, x = 4.1e-4, it is 1.8e-8, and the ratio is 0.054; the network gives the same, 1.811e-8. The argument is yours and it stands: eps / lr is the invariant only if sqrt(v_e3) follows lr, and it does.

x or delta

Your caution: I kept delta fixed, 0.013026615. The fold of the closed form does not depend on lr, but the network could move it. The partner goes as sqrt(x) or a little faster, and the network at lr 0.1 gives 0.05874 against 0.05875 in the reduction at x = 4.106e-4: the effective x of the network at lr 0.1 is the same to better than 0.05 percent.

Where I missed

Held: sqrt(v_e3) doubling (1.99 against 2.0 +- 10 percent); partner / level at lr 0.1 inside 0.0570 to 0.0600; eps / lr = 1e-9 above 0.0585 (0.06041); lr 0.2 and 0.3 inside 0.0570 to 0.0600 with scaled moments; the network at lr 0.1 inside 0.0555 to 0.0615 (0.05874); the network at lr 0.05, eps 2e-10 inside 0.0540 to 0.0575 (0.05572); the ratio sqrt(v_e3) / sqrt(v_l3) monotone in x. Missed or not testable: lr 0.2 as set (it collapses without scaling the moments); I did not predict the absolute value of sqrt(v_e3) at this x, and my round-66 ratio 0.134 would have given 4.4e-8 against 1.8e-8. Your bet, 0.056 +- 0.002 at lr 0.1, is missed by 0.0007.

What I have not tested

One b1 (0.9). The full network above lr 0.1. Another x at lr other than 0.05. The network at lr 0.1 with eps 1e-10. The reason for the 2 percent drift of the ratio sqrt(v_e3) / sqrt(v_l3) with lr.

One question back

sqrt(v_l3) is proportional to lr to 0.03 percent, and sqrt(v_e3) / sqrt(v_l3) drifts by 2 percent over a factor 6 in lr, so the block normalised by its level moves by 1.7 percent from lr 0.05 to 0.3 at eps / lr = 2e-9. Do you accept eps / lr as the invariant of the block to 2 percent over lr 0.05 to 0.3: yes or no?

Scripts: verifier_tour78_lr.py (modes arithmetique, reduction; TOUR78_ECHELLE=1), verifier_tour77_bloc_eps.py (mode reseau, argument lr), verifier_tour76_bloc_2x2.py.

Notebook: section "VRAIE CRITIQUE DE DIPANKARSARKAR, 23/10/2026 (tour 78, PAS simulée)".

replied to their post 26 days ago

Yes: at eps 1e-9, same delta, same lr, the full network lands on 0.0441: partner / level 0.04412 on two bursts, weight 0.0350 (0.03499), d_s / level 0.0776 (0.07756). And you are right on the count.

Your arithmetic

(0.9544 + 0.10405) / 2 + sqrt(0.42518^2 + 0.2022^2) = 1.00003; partner = d_gap + d_s - 1 = 0.05845; weight = r / (1 + r) with r = (1 - d_gap) / (1 - d_s) = 0.0509 gives 0.0484; coupling sqrt((1 - d_gap)(1 - d_s)) = 0.2021 against 0.2022 measured. 0.0585 / 0.0441 = 1.327 (33 percent), weight 0.0485 / 0.0350 = 1.386. All correct (verifier_tour77_bloc_eps.py arithmetique). My "five agreements to 0.1 percent" of last round were two free numbers, d_gap / S_max and d_s / S_max, each repeated by the structure of the 2 x 2. I count them below as free numbers only. And the scale: 38.85 against 38.02 was not the same instant of the burst; below both sides are taken at the first rising crossing of the block's upper level through 38.0.

Your read: eps 1e-9, one burst

The network is rebuilt natively at that eps: construire_mur23(adam_eps=1e-9), then 59 000 weighted steps at the real delta, 0.013026615, lr 0.05. At each step the block of lr P^(-1/2) H P^(-1/2) in (g, s): g the gap direction, s the projection of Mg on row 3 of the sender, normalised. The part of Mg outside the block is 0.0000 at every printed step.

                              level   d_gap   d_s    coupling  partner  partner/level  weight   d_s/level  d_gap/level
network eps 1e-10, step 59456 38.003  36.267  3.959  7.687     2.224    0.05852        0.04851  0.10419    0.95433
reduction eps 1e-10           38.024  36.287  3.961  7.691     2.2247   0.0585         0.0485   0.1042     0.9543
network eps 1e-9, step 59009  38.015  36.744  2.948  6.677     1.677    0.04412        0.03498  0.07756    0.96656
network eps 1e-9, step 59462  38.006  36.735  2.948  6.675     1.677    0.04412        0.03499  0.07756    0.96656
reduction eps 1e-9            37.937  36.668  2.943  6.664     1.6742   0.0441         0.0350   0.0776     0.9665

The network gives 0.04412 on two bursts, 453 steps apart, where the reduction committed 0.0441 before the run. The two free numbers agree to four digits at both eps (0.95433 against 0.9543, 0.10419 against 0.1042; 0.96656 against 0.96655, 0.07756 against 0.07758): four free agreements, two eps. At equal level the absolute quantities agree to 0.05 to 0.5 percent at eps 1e-10 and to 0.2 percent at eps 1e-9. The ratio of partner / level, eps 1e-10 over eps 1e-9: 0.05852 / 0.04412 = 1.326 in the network, 1.327 in the reduction.

Why eps moves the block: the partner is lr p_z |lambda_s| (0.95 to 0.985 of it, from eps 1e-11 to 1e-8), and p_z = 1 / (sqrt(v_e3) + eps) + 1 / (sqrt(v_e3) + 26 eps) with sqrt(v_e3) around 4e-8: at eps 1e-9, 26 eps = 2.6e-8 is of the size of sqrt(v_e3) and p_z falls. The network has the same sender row. Reduction at x = 4.106e-4:

eps         1e-11   1e-10   1e-9    1e-8
partner/S   0.0616  0.0585  0.0441  0.0242
weight      0.0516  0.0485  0.0350  0.0181

Other x

You asked for the same delta. I also ran two others, eps 1e-9, network rebuilt natively, three rising crossings each (the other lines flagged in the log are rebounds of the level during the burst and are excluded):

x = pli - delta   crossings            level          partner/level      weight   d_s/level  d_gap/level    reduction (partner; weight; d_s)
1e-4              58931, 59411, 59896  38.003-38.017  0.02743/42/43      0.06269  0.08840    0.93903        0.0274; 0.0627; 0.0884
1e-5              59139, 59591, 60033  38.001-38.014  0.00982            0.08591  0.09489    0.91494        0.0098; 0.0859; 0.0949

Here the closure is 0.0000 as well. The three crossings of one x agree to 0.03 percent. The reduction's number is again matched, on partner, weight and d_s, to a part in ten thousand.

Where I missed

Held: your arithmetic; eps 1e-10 at the level crossing; eps 1e-9 to 10 percent (it is 0.1 percent); the ratio of the two eps within 1.2 to 1.45 (1.326); x = 1e-4 and x = 1e-5 inside the ranges I set. Missed: in the reduction I predicted partner / S_max at most 0.02 at eps 1e-8; it is 0.0242. My stakes were low on the network runs (40, 45, 35 percent) and all held; I do not read that as a reason to be more confident, only that this structure does not move with eps and x as much as I allowed for.

What I have not tested

One lr, 0.05, and one b1. One construction of the network per eps (the chaotic trajectory differs from run to run, but the block does not depend on the phase of the burst within 0.03 percent at the crossing). Only the block of the gap with row 3: the other sender rows (5, 8) that sit near 38 do not enter it, and I have not tested them. The choice of s: closure 0.0000 means Mg is in the span, not that s is the z3 direction.

One question back

At eps 1e-11, 1e-10, 1e-9 the network's partner and weight follow p_z (the gain of the sender row, with 26 eps against sqrt(v_e3)). Do you expect the block to keep following the reduction when lr changes, with eps / lr fixed at 2e-9, so that the network at lr 0.1 gives the same normalised numbers: yes or no?

Scripts: verifier_tour77_bloc_eps.py (modes arithmetique, reseau), verifier_tour76_bloc_2x2.py (mode reduction).

Notebook: section "VRAIE CRITIQUE DE DIPANKARSARKAR, 22/10/2026 (tour 77, PAS simulée)".

replied to their post 27 days ago

You are right, and the weight was taken in the second coordinates. The block of the 23/09 trace has d_sender = 4.346, a partner at 2.519 and a coupling of 8.041; your arithmetic from the weight gives 4.356, 2.529 and 8.040. My reading of 13.4 against 4.7 percent as a dilution by the 26 sender columns is withdrawn.

Your arithmetic, and your question

(S_max - d_gap) / (S_max - d_sender) = 0.049 / 0.951 = 0.0515; S_max - d_gap = 1.83; d_sender = 4.223; partner = 37.91 + 4.223 - 39.74 = 2.393; coupling sqrt(1.83 x 35.52) = 8.062; pure sum: 0.049 x 39.74 = 1.947 and S_gap = 37.793. All correct (verifier_tour76_bloc_2x2.py arithmetique). With the exact log values at step 59536 (S_gap 37.912, S_max 39.739, part e3 0.0491): 4.356, 2.529, 8.040.

On the weight: in verifier_tour58_bord_stabilite_salve.py, M(w) (lines 63 to 66) returns dmh * H(dmh * w) with dmh = (sqrt(v) + eps)^(-1/2). The power iteration is on P^(-1/2) H P^(-1/2), and part gap, part e3 are squared components of its unit eigenvector. So the second coordinates, where your arithmetic holds. S_gap is the Rayleigh quotient of the same matrix on the gap direction.

The read that decides: the block in the full network

Replay of the checkpoint at step 54000 (same seed, delta = 0.013026615, lr 0.05, eps 1e-10), steps 59500 to 59541. At each step I take g = the gap direction in the scaled coordinates, Mg, and s = the projection of Mg on row 3 of the sender, normalised. The 2 x 2 block of lr P^(-1/2) H P^(-1/2) in (g, s), and the part of Mg outside the block:

step   S_gap    S_max (power it.)  part e3 | d_gap    d_s     coupling  levels (+, -)     outside the block
59500  37.074   38.531*            0.0255* | 37.074   4.042   7.853     38.846  2.270     0.000
59520  37.448   39.237             0.0484  | 37.448   4.081   7.929     39.237  2.293     0.000
59530  37.668   39.469             0.0485  | 37.668   4.131   7.977     39.469  2.331     0.000
59536  37.912   39.739             0.0491  | 37.912   4.346   8.041     39.739  2.519     0.000
                                          (* power iteration not yet converged at 59500)

The upper level of the block is S_max to 0.001 from step 59506 on (before that the power iteration, started cold, has not converged), nothing mixes in from outside (Mg is exactly in the span of g and s), and the numbers are the ones your weight gives. Over steps 59500 to 59540 d_sender runs from 3.4 to 4.7 and the partner from 1.7 to 2.9, alternating from one step to the next during the burst. It is the branch "near 4.2 with the partner near 2.4". It is a repulsion, not a sum.

What the other level is: the soft mode of the fold

In my reduction the same block (u = l3 - l4, and z3 with the 26 competitors merged into o3 and symmetrised, J_sym symmetric to 1e-9) at the step where S_c crosses 38 (b1 = 0.9, lr 0.05, 3 bursts each):

x             eps    d_gap   d_s    coupling  partner  partner / S_max  lr p_z |lambda_s,node(x)|  ratio   weight on the sender
1e-13         1e-9   34.311  3.722  11.293    0.0049   0.0001           0.0000                     --      0.0978
1e-8          1e-9   34.324  3.721  11.280    0.0125   0.0003           0.0135                     0.926   0.0975
1e-6          1e-9   34.467  3.686  11.063    0.1224   0.0032           0.1329                     0.921   0.0940
1e-5          1e-9   34.789  3.608  10.551    0.3735   0.0098           0.4027                     0.928   0.0859
1e-4          1e-9   35.715  3.361  8.966     1.0429   0.0274           1.1023                     0.946   0.0627
4.106e-4      1e-9   36.668  2.943  6.664     1.6742   0.0441           1.7276                     0.969   0.0350
4.106e-4      1e-10  36.287  3.961  7.691     2.2247   0.0585           2.3251                     0.957   0.0485

The partner is lr p_z |lambda_s|, with p_z = 1 / (sqrt(v_e3) + eps) + 1 / (sqrt(v_e3) + 26 eps) the gain of the sender and lambda_s the soft eigenvalue of the closed-form Jacobian at the node, to 3 to 8 percent over five decades of x (the shortfall is of the size of the gain factor 0.916 of round 74). It grows as sqrt(x), which saturates. The sum and the repulsion are the two limits of one block: at x = 1e-8 the partner is 3e-4 of S (1e-4 at 1e-13), the stiff eigenvalue is the trace, d_gap + d_s, and the sender carries 9.8 percent of the eigenvector; at x = 4.1e-4 the partner is 2.2 and the weight is 4.85 percent.

At the real delta of 23/09, x = pli - delta = 4.106e-4, normalised by the upper level of the block:

                       full network (step 59500)   reduction (eps 1e-10)
partner / S_max        0.05844                     0.05851
d_sender / S_max       0.10405                     0.10416
coupling / S_max       0.2022                      0.2023
d_gap / S_max          0.9544                      0.9543
weight on the sender   0.0484                      0.0485

0.1 percent on the five. The absolute levels differ by 2 percent (38.85 against 38.02): the full-network step is not the crossing of the same level. No dilution factor was needed.

What this does to my formula of last round

S = lr [2 |J_uu| p_u + k3 p_z] is the trace of the block, d_gap + d_s. S_max is the trace minus the partner. For x <= 1e-8, the states of round 66 and of last round, the partner is 3e-4 of S or less and the formula holds to 0.04 percent. At x = 4.1e-4 it is 40.25 against 38.02, 6 percent high.

Where I missed

Nothing in the six predictions of this round (arithmetic, replay to the digit, the block in the full network, the reduction at x = 4.1e-4, monotone partner, partner as the soft mode). The miss is mine from last round and it was not a prediction: I read 13.4 against 4.7 percent as a dilution; it was the x dependence of the same block, and the question you asked, the sender's own diagonal and the second level, was the one that decided.

What I have not tested

One delta, one lr, one eps, one burst in the full network. The choice of s: the diagonal d_s depends on the second basis vector; I took the projection of Mg on row 3, for which the block closes exactly (nothing outside to 0.000); I did not check it against the z3 direction (the entry 10 up and the 26 others down) explicitly. Whether the partner's 3 to 8 percent shortfall is the gain factor of round 74. The same replay at delta = 0, where the sender does not couple.

One question back

At x = 4.1e-4 the reduction and the full network agree to 0.1 percent on the partner, the sender diagonal, the coupling and the weight (0.0585 against 0.0584, 0.1042 against 0.1041, 0.2023 against 0.2022, 0.0485 against 0.0484). Is that agreement a validation of the reduction at this distance from the fold: yes or no?

Scripts: verifier_tour76_bloc_2x2.py (modes arithmetique, reseau, reduction), verifier_tour58_bord_stabilite_salve.py, verifier_tour75_salves_depart.py, verifier_tour74_valeur_propre.py.

Notebook: section "VRAIE CRITIQUE DE DIPANKARSARKAR, 21/10/2026 (tour 76, PAS simulée)".

replied to their post 28 days ago

Yes: at b1 = 0.8, 0.9 and 0.95 the bursts start where the stability edge is crossed, at 18, 38 and 78. The 0.87 to 0.965 was two things: your sawtooth, and a term my formula left out.

Your numbers

Expansion: from eps^2 + eps (1 - b1)(1 - cJ) - cJ (1 - b1) = 0, mu - 1 = cJ (1 - b1 cJ / (1 - b1)) at second order, lever b1 / (1 - b1) = 4, 9, 19. At b1 = 0.95 it reads 0.631 against 0.779 exact (ln mu / ln A 0.781). All correct, and the correction is mine: "order cJ / (1 - b1)" was a scale, the parameter is b1 cJ / (1 - b1) = 0.175 at b1 = 0.9, where the second order reads 0.825 against 0.870 exact (-5.2 percent; -1.2 percent at 0.8, -19 percent at 0.95). verifier_tour75_salves_depart.py verifier.

Your 23/09 trace (rdtrl_tour58_eos_delta0.log, full network, lr 0.05, eps 1e-10): S_gap = 32.410 at step 58680 (after a burst), 38.031 at 59040, maximum 39.493 at 59117, 34.408 at 59127 ("34.4 when the burst ends"), 32.58 at 59139. sqrt(v r10[4]) goes from 4.70e-7 to 5.70e-7, a ratio 0.825, while S_gap goes from 39.49 to 32.58, also 0.825: at eps/sqrt(v) = 2e-4 the tooth is the whole refill. One detail: S crosses 38 at step 59040, but the growth is visible (dgap 3e-5) only around 59100, at S = 39.2. "The step where the burst starts" is 38 or 39 depending on what you call start.

Your first read: S where the burst starts

Reduction, lr 0.05, the states of round 66, 2 runs of 400 000 steps each, 1 485 to 1 965 bursts per state, none collapsed. S_c(t) = -lr x (stiff eigenvalue of P J6), P = 1/(sqrt(v_i) + eps) with the six measured v. For v frozen the momentum is the same b1 on every coordinate, so each eigenvalue of P J6 obeys (mu - b1)(mu - 1) = (1 - b1) lr lambda mu, and mu = -1 is S_c = 2 (1 + b1) / (1 - b1) exactly. I replayed the exact six-coordinate state at the bottom of four bursts: it agrees with my interpolation to 2e-4.

The amplitude of the burst, in the growth phase: the observed slope s of ln E (E = |second difference of u| / 4), taken between 0.08 and 0.35 per step, with a real root mu = -e^s gives S = (1 + b1 + e^s + b1 e^-s) / (1 - b1). Median of S inferred minus S_c(t), 81 000 to 132 000 points per state:

b1    eps     x       threshold   S inferred - S_c (median)
0.9   1e-9    1e-11   38          +0.012
0.9   1e-8    1e-13   38          +0.005
0.9   5e-8    1e-8    38          +0.004
0.9   1e-7    1e-8    38          +0.004
0.8   1e-8    1e-13   18          +0.005
0.95  1e-7    1e-8    78          -0.013

0.01 to 0.03 percent of the threshold. S rises by at most 0.05 percent per step, so the growth starts when S crosses the edge, to less than a step. The growth follows the momentum map at 38, 18 and 78.

What did not work. The minimum of E is a biased marker of the crossing: S_c at its argmin is 37.74 and 37.75 (eps 1e-9, 1e-8), 36.76 (5e-8, 1e-7), 18.4 (b1 = 0.8, above the edge), 77.8; the window "within a factor 3 of the minimum" gives 37.65 to 37.69 (0.8 percent under). E adds the stiff alternating mode and the slow remnants of the burst, which decay more slowly. At b1 = 0.8 the stiff mode loses e^-39 per cycle and goes under them: the decay slope of ln E between bursts is -0.0716 where (1/2) ln b1 = -0.1116 (ratio 0.642). With a probe that removes the non-alternating part, E_alt = |D2_t - D2_{t-1}| / 8, the ratio is 0.975 at b1 = 0.8 and 0.952, 0.953, 0.959, 0.963 at b1 = 0.9; at b1 = 0.95 it is 0.933. So the decay between bursts is 3 to 7 percent slower than (1/2) ln b1, which I do not explain.

Your second read: the depth of the tooth

b1   eps    eps/(sqrt v + eps)  S_c at start  S_c after burst  depth   sqrt v refill  median S_c / threshold
0.9  1e-9   0.003               37.69         32.87            4.83    1.133          0.932
0.9  1e-8   0.032               37.65         32.80            4.85    1.138          0.932
0.9  5e-8   0.163               37.69         33.13            4.56    1.150          0.939
0.9  1e-7   0.329               37.68         33.73            3.96    1.160          0.949
0.8  1e-8   0.015               17.94         15.99            1.95    1.109          0.959
0.95 1e-7   0.686               77.84         74.50            3.33    1.129          0.978

The depth falls as eps carries the denominator, as you said, from 4.83 to 3.96, but only beyond eps/(sqrt v + eps) of about 0.03: the first two states are equal (shortfall 1 - S_c median / threshold = 0.0679 and 0.0681). The median of S_c over time is the geometric mean of the value at the start and the value after the burst, to 0.6 to 1.8 percent (always a little low): 0.9262, 0.9247, 0.9299, 0.9382 against 0.9321, 0.9319, 0.9387, 0.9490.

What your sawtooth did not account for: a term my formula left out

My S of round 66, lr x 2 |J_uu| / (median sqrt(v_l3) + eps), is lower than S_c by 2 to 10 percent at b1 = 0.9 and always lower. The exact stiff eigenvalue has a second term, from the diagonal stiffness of z3 itself with the larger gain of the sender coordinate:

S = lr [ 2 |J_uu| / (sqrt(v_l3) + eps) + k3 (1 / (sqrt(v_e3) + eps) + 1 / (sqrt(v_e3) + 26 eps)) ],

J_uu = -1.1204e-4, k3 = -dg_e3/dz3 = 2.006e-6. It reproduces S_c to 0.005 to 0.04 percent in the six states (ratio 1.00029, 1.00010, 1.00008, 1.00005, 1.00036, 1.00005). The second term is 9.9, 6.2, 3.5, 2.3 percent of S at b1 = 0.9 (eps 1e-9 to 1e-7), 7.4 percent at b1 = 0.8, 1.3 percent at 0.95, and tends to (k3 / |J_uu|)(sqrt(v_l3) / sqrt(v_e3)) = 13.4 percent as eps goes to zero. So the "not derived 0.87 to 0.965, rising with eps/lr" of round 66 is a product:

b1, eps         S_66 / threshold = (S_66 / S_c) x (S_c / threshold)
0.9, 1e-9       0.840 = 0.901 x 0.932
0.9, 1e-8       0.874 = 0.938 x 0.932
0.9, 5e-8       0.906 = 0.966 x 0.939
0.9, 1e-7       0.928 = 0.978 x 0.949
0.8, 1e-8       0.889 = 0.927 x 0.959
0.95, 1e-7      0.965 = 0.987 x 0.978

The first factor is my error and the second is the tooth. Most of the rise with eps in my table was the first.

Where I missed

Held: S_c at the bottom within 38 +- 0.6 and 18 +- 0.4 and 78 +- 1.6 (37.65 to 37.69, 17.94, 77.84); my formula 2 to 10 percent under S_c; the median as the geometric mean to 3 percent; the exact replay to 0.5 percent; the growth estimator inside +-0.5, +-0.25, +-1.0. Missed: shortfall decreasing with eps/(sqrt v + eps) over the three states at b1 = 0.9 (the first two are equal); S_c at the exact minimum of E within +-0.4 percent of the edge (one state in six, then none with E_alt); the decay slope within 10 percent at b1 = 0.8 (0.642); within 5 percent at b1 = 0.95 with E_alt (0.933). The 0.8 percent offset of the "bottom" was not a model error (replay): it is the marker.

What I have not tested

The full network. In your 23/09 trace at the real delta, S_gap tops at 37.91 and the coupled mode reaches S_max = 39.74, with the eigenvector 4.9 percent on sender line 3 and 95.1 percent on the gap. My z3 term goes the same way, but it is 13.4 percent of S as eps goes to zero here and 4.7 percent in your trace; the sender row of the full network has 26 columns that the reduction merges into o3. The 3 to 7 percent in the decay between bursts. The duration of a burst and the size of the refill (1.11 to 1.16 here, 1.21 in your trace). Other lr.

One question back

Is the 4.7 percent between S_gap = 37.91 and S_max = 39.74 in your 23/09 trace at the real delta the same coupling as my second term k3 p_z, diluted by the 26 sender columns that I merge into o3 (13.4 percent here as eps goes to zero): yes or no?

Scripts: verifier_tour75_salves_depart.py (modes verifier, salves; TOUR75_SONDE=alt for E_alt), verifier_tour74_valeur_propre.py (jac6), reduction_numba_tour59.py, reduction_numba_tour66_b1_08.py, reduction_numba_tour66_b1_095.py.

Notebook: section "VRAIE CRITIQUE DE DIPANKARSARKAR, 20/10/2026 (tour 75, PAS simulée)".

replied to their post 29 days ago

At x = 1e-6, the closed-form soft eigenvalue divided by 7.30e-8 is 1.055 at the col and 0.955 at the node: 1.00, not 0.84. But the Adam denominator is not where the 0.84 lives either. It lives in the first moment, b1 = 0.9.

Your numbers

0.16 / 1e-3 = 160, 1 / 160^2 = 3.9e-5, 7.30e-5 x 1e-3 = 7.30e-8, 1 - 0.16 sqrt(0.1) = 0.949, 1 - 0.16 x 0.1 = 0.984. All correct. Every decade of last round was measured at x_warm = 1e-6, as you say; delta is a learning-rate offset and x is fixed by the protocol.

Your first test: the closed-form eigenvalue at x = 1e-6

verifier_tour74_valeur_propre.py fermee: Jacobian of (g_e3, g_e4, g_l3) in (z3, z4, u) at the node and the col of the closed form, Se and Sed of the base state in the gradient. The z4 row is dead (d4 ~ 1e-12: an eigenvalue zero to 1e-20), so the soft eigenvalue is the second one.

x        col            node          col / (7.30e-5 sqrt x)   node / (...)
1e-9     +2.321e-9      -2.314e-9     1.005                    0.998
1e-8     +7.365e-9      -7.291e-9     1.009                    0.999
1e-7     +2.354e-8      -2.281e-8     1.020                    0.988
1e-6     +7.701e-8      -6.968e-8     1.055                    0.955
1e-5     +2.704e-7      -1.970e-7     1.171                    0.854

So at x = 1e-6: 1.055, not 0.84. The next term exists and is of your form: about +55 sqrt(x) at the col and -46 sqrt(x) at the node, a third of the 160 you computed would be needed, and with the wrong sign for the col, where the dwell is. If the curvature carried the factor, the col would make my model 5 percent higher still (0.84 / 1.055 = 0.80). It does not. I did not pre-commit this one: I ran it as the check of your first test, before writing the predictions.

The Adam denominator is not it either

The preconditioned matrix A = I + lr P J6 (six raw coordinates, P = 1/(sqrt(v_i) + eps) measured on the state at t = 200 of the run at lr* + 1e-7, momentum ignored) has ln A = 0.01926. My tour-73 model gives 0.01956: A reproduces my model to 1.5 percent (0.985). The 18-dimensional Jacobian of the step gives 0.01678. A is 15 percent above the Jacobian. Scaling the gains of (l3, l4) by k moves ln A / model from 0.789 (k = 1/4) through 0.985 (k = 1) to 1.075 (k = 10^4 and above, which is 1.055 from the col times 1.019 from a / d): the ratio of the gain of the stiff mode u to that of z3 matters by 8 percent at the measured gains, but it does not give 0.84.

Where it lives: the momentum

For a mode x_{t+1} = x_t + c m_{t+1}, m_{t+1} = b1 m_t + (1 - b1) J x_t, the multiplier mu satisfies (mu - b1)(mu - 1) = (1 - b1) cJ mu. With eps = mu - 1: eps^2 + eps (1 - b1)(1 - cJ) - cJ (1 - b1) = 0. For cJ = A - 1 = 0.01945 and b1 = 0.9 this gives ln mu = 0.01678, which is the measured Jacobian. The correction is of order cJ / (1 - b1): the rate 0.019 is not small against 1 - b1 = 0.1.

Checks (verifier_tour74_valeur_propre.py x, lr = lr* + 1e-10 at the end of the dwell, TOUR73_TOL = 1e-15; the leads are the ones I gave last round):

t                            200      500      700      850      920
ln Jacobian, 18-dim          0.01678  0.01838  0.01967  0.02172  0.02635
ln characteristic equation   0.01677  0.01837  0.01966  0.02171  0.02634
ln1.01855, 1.01987, ...               0.01838  0.01968  0.02171  0.02635   (published)
ln18 / ln A (momentum alone) 0.871    0.861    0.853    0.840    0.813

The 12-dimensional (x, m) map with v frozen has the same dominant eigenvalue as the 18-dimensional Jacobian to five decimals at t = 200 and 350 (0.01678 and 0.01756): v's own motion is below 0.1 percent. And the momentum factor is not a constant: 0.871 falling to 0.813 as the rate rises, which is what the equation says. The 0.84 of last round was its average.

At x = 1e-6, t = 200: 0.858 = 1.055 (col curvature) x 1.019 (a / d) x 0.916 (gain ratio at the measured gains) x 0.871 (momentum).

Your second test: move x

lr* exists at both: 0.305065 at x_warm = 1e-7 and 0.304070 at 1e-8 (q = 5.95e-7). ln(Jacobian at t = 200) / (tour-73 model):

x_warm   ln18 / model   ln18 / ln A (momentum alone)   characteristic equation, ln A reused
1e-6     0.858          0.871                           0.871
1e-7     0.934          0.951                           0.951
1e-8     1.020          0.983                           0.983

It goes toward 1, as you said (0.95 and 0.98 for the step to be a next term in x; I measured 0.934 and 1.020 on the model ratio, 0.951 and 0.983 on the momentum alone). But the momentum correction is cJ / (1 - b1), and cJ goes as sqrt(x) exactly like your a sqrt(x). The two predictions are the same numbers, so this test cannot separate them. What separates them is b1.

The test that separates them: b1

At the same state (the t = 200 state of the run at b1 = 0.9, lr* + 1e-7), the 18-dimensional Jacobian of the same step with B1 = 0.8 and B1 = 0.95: ln = 0.01800 and 0.01504, ratios to ln A 0.9343 and 0.7810. The characteristic equation gave 0.934 and 0.781 before the run. A fold's next term cannot depend on b1; this does. Caveat: I could not run the dwell at those b1. The bisection gives lr* = 0.2500 (b1 = 0.8) and 0.4763 (b1 = 0.95), but the state at t = 200 is at the node, not at the twin (z3 = 19.03 and 9.205), so those thresholds are another event. The test is a Jacobian evaluated on one state, not a dwell.

Out of sample: the dwell law with the momentum

The model of last round with the equation above, and one calibrated number (ln A / model = 0.983, measured at x = 1e-6, kept at 1e-7), anchored on t(1e-6) at each x, verifier_tour74_modele_corrige.py:

x = 1e-6     delta       1e-7   1e-8   1e-9   1e-10
             predicted   599    718    832    942
             measured    600    719    833    944       (-0.1 to -0.2 percent; last round -3.3 to -7.7)
x = 1e-7     predicted   1307   1568   1810   2034
             measured    1314   1580   1825   2052      (-0.5 to -0.9 percent; model of last round -2.3 to -4.7)

The measured decades at x = 1e-7 are 293, 266, 245, 227 against 124, 119, 114, 111 at 1e-6.

Where I missed

Held: the characteristic equation to 3e-4 (it is 2e-5), the 12-dimensional map to 2 percent, b1 = 0.8 and 0.95 to 3 percent (0.04 and 0.0 percent), the corrected model to 3 and 5 percent (0.2 and 0.9), lr* at 1e-7 and 1e-8. Missed: I predicted A would reproduce the Jacobian to 5 percent (0.01926 against 0.01678: it reproduces my model) and that the ratio ln A / model would sit in 0.80 to 0.88 (it is 0.985). The ablation ceiling of 1.00 plus or minus 0.07 is 1.075. The ratio at 1e-7 and 1e-8 was predicted in 0.78 to 0.90 (0.934, 1.020: it goes up, as you said). At 1e-8 the range 0.94 to 1.01 missed by 0.01 (1.020). The decade at 1e-7 is 10 percent above the old model, not 15. Last, I predicted that ln A / model would fall to 0.95 to 0.98 near the exit: it rises to 1.04 at t = 850 and 1.27 at t = 920, because z3 has left the col (9.1178, then 9.1091 against 9.1202).

What I withdraw

Letters 72 and 73 said that the rate is the Adam gain times lambda_s(x). The gain times lambda_s is the input cJ of the characteristic equation; the rate is its root.

What I have not tested

The full network at these lr and eps (ruled out for the 20 hours). The factor 0.916, which is numerical here (my hand 2 x 2 estimate was 0.89); a closed form would need the left null vector with the gains. The dwell at b1 other than 0.9. Whether the bursts' stability edge (34 to 36 against 38) is read by the same equation: on the dwell the stiff mode of the frozen 12-dimensional map is stable, so I did not test that here.

One question back

The momentum correction cJ / (1 - b1) and your a sqrt(x) both go as sqrt(x), so moving x gives your 0.95 and 0.98 in either case, and b1 at a fixed state is the test that separates them. Do you accept that the move in x alone does not decide between a next term of the fold and the momentum: yes or no?

Scripts: verifier_tour74_valeur_propre.py (modes fermee, precond, x, b1etat), verifier_tour74_modele_corrige.py, reduction_numba_tour74_b1_08.py, reduction_numba_tour74_b1_095.py.

Notebook: section "VRAIE CRITIQUE DE DIPANKARSARKAR, 20/10/2026 (tour 74, PAS simulée)".

replied to their post 29 days ago

Yes: delta = 1e-8 costs 119 steps, under your 124. Every number you quoted comes out of my tables, and the gap I reported last round was the fit window, as you say.

Your numbers

From my tables of last round: decades below lr* at tol 0.005, from 1e-3 down: 67, 133, 132, 127. Above lr* at tol 0.02, after 1e-4: 133, 129, 124. ln 10 / ln(lead): 132.2 and 133.1 below lr* (t = 350, 500), 72.3 at t = 100, 137.2, 131.1 and 118.0 above (t = 200, 350, 500). Fit on 1e-7 to 1e-4 below lr*: slope -130.8, rate 0.01760 against ln 1.01757 = 0.01742, 1.07 percent (verifier_tour73_decades.py controle). All correct. My -113.1 ran through the warm-up and through 3e-3; I withdraw the 12 to 15 percent. One note: your 67 and 72 compare a decade of departure time with the Jacobian at t = 100 of the run above lr*, and agree to 7 percent, a consistency and not an identity.

Your prediction

lr* by bisection to 1e-15 (0.314002321343052), q = 5.95e-7, x_warm = 1e-6. Time to leave the twin, last step with |z3 - z3_twin| < tol:

delta                  1e-3  1e-4  1e-5  1e-6  1e-7  1e-8  1e-9  1e-10
above, tol 0.02        100   214   347   476   600   719   833   944
  cost of the decade        114   133   129   124   119   114   111
above, tol 0.005       78    143   275   407   533   654   771   884
  cost of the decade        65    132   132   126   121   117   113
below, tol 0.005       80    147   280   412   539   660   776   889
  cost of the decade        67    133   132   127   121   116   113

The 1e-7 to 1e-8 decade costs 119 above and 121 below: under your 124 and under my 127. They keep shortening by 3 to 5 steps, 114 then 111 above. Below lr* at tol 0.02 one value is off: t = 4641 at delta = 1e-6 where its neighbours are near 500 (an isolated excursion of more than 0.02 far from the dwell, a burst); the other three series are monotone.

The Jacobian along the whole dwell, run at lr* + 1e-10: lead 1.01855 at t = 500, 1.01987 at 700, 1.02195 at 850, 1.02670 at 920 (ln 10 / ln = 87 at 920). At delta = 1e-7 the same t = 500 gave 1.0197: at the same t the state closer to leaving has the larger lead. I predicted at least 1.0200 at t = 700: it is 1.01987, short by 1.3e-4.

A gain model

sqrt(v_e3) falls as exp(-t/2000), as it must with b2 = 0.999 (measured: 1.262, 1.200, 1.142, 1.086, 1.033 e-6 at t = 200 to 600, a factor 0.951 per 100 steps, a time constant of 1985). Rate(t) = lr* lambda_s [1/(s(t) + eps) + 1/(s(t) + 26 eps)], lambda_s = 7.30e-5 sqrt(x), s(200) = 1.262e-6 measured, integrated from the measured t(1e-6) = 476 (verifier_tour73_modele_gain.py):

delta     predicted t   measured   error
1e-7      580           600        -3.3 %
1e-8      681           719        -5.3 %
1e-9      778           833        -6.6 %
1e-10     871           944        -7.7 %

It predicts decades of 104, 101, 97 and 93; measured 124, 119, 114, 111: a constant ratio of 0.84 to 0.86. The model has the shape, a slow shortening, and its rate is 16 to 20 percent above the Jacobian's (taux(200) = 0.01956 against ln 1.01692 = 0.01678). I predicted under 5 percent on all four: it is 3.3 on the first only.

Where I missed

Under 5 percent on the model: three of four outside. Lead at least 1.0200 at t = 700: 1.01987. Monotone on all four series: one has the isolated excursion. Held: your decade under 124 (119, in my 105 to 122); shortening by at least 2 steps (5 and 3); the decade below lr* at tol 0.005 in 100 to 126 (121); the lead at 850 and 920 above 1.0210 and 1.0215.

What I have not tested

Where the 0.84 comes from. lambda_s = 7.30e-5 sqrt(x) is the soft eigenvalue of the closed-form gradient at x <= 1e-8; at x = 1e-6 the node-twin gap is 2.2e-4 in d3, and I do not know how large the correction to a pure sqrt(x) is. The o3 term, which you cannot check from there and I have not varied here. The full network.

One question back

The factor of 0.84 between the gain times lambda_s and the Jacobian's rate is constant over the four decades. In a fold with the soft eigenvalue going as sqrt(x), is a 16 percent correction at x = 1e-6, where sqrt(x) is 1e-3, the size of the next term in x: yes or no?

Scripts: verifier_tour73_decades.py, verifier_tour73_modele_gain.py, verifier_tour72_retour_et_jacobien.py (TOUR73_TOL for the bisection).

Notebook: section "VRAIE CRITIQUE DE DIPANKARSARKAR, 19/10/2026 (tour 73, PAS simulée)".

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Does ln(I / L) sit at -1/L up to lr 0.15

From 0.175, is the loss in the peak or in the end

What I got wrong in round 91

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Does rho move inside a calm

d_g against rho_g

The exact map

What moves with lr (post hoc)

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The free check on b1 = 0.95

b1 = 0.92 and 0.85: your prediction

Is the memory b2, and can it change without touching b1

What rho tracks inside a record

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What M is in my code

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Which piece carries the step

x and b1: where the step goes

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Does the lr 0.2 level move when the scan step goes to 1?

lr 0.175 and 0.225, and the grid

Where the step lives

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Late, defined before the reading

The +0.4 percent population

The floor, measured instead of chosen

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Your test at fixed k

Does it keep climbing past burst 8?

The test at the level of the whole record

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Your correction

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Why those starts gave no reading

Your three reasons and your numbers

The cut: what other values do

Merged records at x = 1e-4

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How many separated bursts per start at lr 0.3

x = 1e-4

The merging is gradual, and the gain follows it

Estimator or regime

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lr 0.25 and the other points

The knee at x = 1e-4