The forgery needs a history

The comb in a divergence sequence was offered here as evidence that a plant computes its pattern rather than merely having one. It stopped being that when a kinematic lattice — organs at exact multiples of a divergence, no rule anywhere — was given errors inherited from its contact neighbours and reproduced the comb, the second comb and the pair on eight seeds of eight.

The conclusion drawn was that a comb is evidence that something is transmitted between neighbours, of which a placement rule is one instance. This essay narrows that, because the transmitted disturbance was doing two things and only one of them turns out to be necessary.

The comparison #

Two disturbances, the same size, at the same coupling, on the same eight seeds, driven into the same kinematic lattice.

The inherited one is the original: each organ takes a share of what its two contact neighbours were displaced by. The shared one takes the same share of the fresh deviates those neighbours received — the same correlation at the same two lags, with nothing handed on twice.

The result #

The inherited disturbance forges everything. Main comb 0.205 against a band of 0.073, second comb 0.242, and the pair returned as 8/13 on eight seeds of eight, with a recorded scatter of 0.73°.

The shared disturbance forges nothing. Main comb 0.099 — the band is 0.073 — second comb 0.098, and the pair returned on one seed of eight, at a recorded scatter of 0.71°.

Same lags, same coupling, same recorded scatter, opposite verdicts. The forgery is the accumulation, not the structure.

It is not a matter of size #

The obvious objection is that the shared disturbance is simply weaker — a moving average is a milder object than a recursion, and the comparison was made at a coupling where the transport barely works.

The objection is testable and it fails. Push the shared disturbance to a coupling of 0.7 and its main comb is 0.114 on two seeds of eight. Push it to 0.9 — a coupling the transport cannot even be run at, since its recursion diverges there and the stream has no variance to normalise by — and the main comb is 0.109 on two seeds of eight.

So the shared disturbance is at the noise floor at every coupling it can be given, including couplings the inherited one cannot survive. Its failure to forge is not a failure to be large enough.

Why re-transmission is what a comb needs #

The reason is in what the readout looks at, and it is arithmetic rather than empirical.

The comb is not a peak at one lag. It is a residue class: the readout finds the lags at which the divergence sequence’s autocorrelation stands above its band, and what makes a comb is that they are the multiples of one number — eight, sixteen, twenty-four, thirty-two — rather than a single tooth at eight.

A moving average over lags eight and thirteen puts correlation at eight, at thirteen and at five. It puts none at sixteen, twenty-four or twenty-one, because organ i and organ i−16 share no innovation: the terms of the first are at i−8 and i−13, and of the second at i−24 and i−29.

A recursion does. Organ i contains a share of organ i−8’s error, which contains a share of organ i−16’s, which contains a share of organ i−24’s, and so on down. The whole residue class is populated by the recursion and only by the recursion.

So the comb is not evidence of correlation at the contact offsets. It is evidence of correlation at the multiples of them, and only a process that hands a disturbance on and on produces that.

The arithmetic, worked #

It is worth doing the bookkeeping once, because the conclusion is a statement about which lags exist and that can be checked by hand.

Write e for the fresh deviates and v for what an organ ends up displaced by.

The shared disturbance is v(i) = a·(c₈·e(i−8) + c₁₃·e(i−13)) + b·e(i). Every v is a combination of exactly three innovations. Two organs’ displacements are correlated only if their sets of three overlap, and the overlaps are: lag 8 (e(i−8) against e(i−8) from the other’s own term), lag 13 likewise, and lag 5, where organ i’s e(i−13) is organ i−5’s e(i−18)… no, precisely: organ i−5 contains e(i−13) and e(i−18), and organ i contains e(i−8) and e(i−13). They share e(i−13). That is the whole list. At lag 16 organ i−16 contains e(i−24) and e(i−29) and shares nothing.

The inherited disturbance is v(i) = a·(c₈·v(i−8) + c₁₃·v(i−13)) + b·e(i), and each v on the right expands into its own three terms, which expand again. Organ i’s displacement therefore contains e(i−16) through the chain i → i−8 → i−16, and e(i−21) through i → i−8 → i−21, and so on for every sum of eights and thirteens.

So the two streams’ correlation functions are: three lags for one, and every non-negative combination 8a + 13b for the other. The readout looks for a family of lags at the multiples of a number. Only one of the two objects has one.

That also explains the measured numbers rather than merely accompanying them. The shared stream’s correlation at lags 16, 21, 24, 26, 29 and 34 is 0.000, 0.002, 0.008, 0.002, 0.004 and 0.006 — zero to the precision of forty thousand draws — and the inherited stream’s is 0.175, 0.298, 0.081, 0.176, 0.170 and 0.172.

The second comb goes too #

The main comb is the headline, and the second comb is the one that mattered longest, so it is worth its own paragraph.

The second comb is the family of lags offset from the main one by the difference of the pair — the residue class containing thirteen when the main class is the multiples of eight. Both streams are correlated at that difference: it is lag five, and the arithmetic above shows both objects populating it.

Measured: the inherited disturbance’s second comb is 0.242 and the shared one’s is 0.098, against a band of 0.073. So the shared disturbance does not manufacture a second comb either, despite carrying the correlation that a single-lag account would say produces one.

The reason is the same. A comb is a residue class rather than a tooth, and the second class needs 5, 18, 21, 26 and so on — which again requires the chain.

That is worth recording because the second comb is the observable this site spent a whole phase pricing, refining and eventually retiring. Its behaviour here is one more piece of evidence that it was always about the same property as the first: not “are the contacts correlated” but “does the correlation propagate”.

What that does to the evidential position #

The position after the forgery was found had three lines. It now has three different ones.

A comb is still not evidence of a placement rule. The transported disturbance manufactures one and contains no rule. That stands.

But a comb is evidence of something stronger than “transmission”. It requires a process in which an organ’s disturbance reaches organs many contacts away — not merely its neighbours, but its neighbours’ neighbours, down a chain of eight or sixteen or twenty-four organs. A process in which each organ is jostled by what arrived at its neighbours, and no further, makes no comb at all.

And the class of rivals is narrower than it looked. “Errors are correlated between touching organs” is a very weak hypothesis, satisfied by almost any mechanical coupling. “Errors are inherited, so that a displacement propagates along a parastichy for dozens of organs” is a specific claim with its own consequences — the most obvious being the slow drift, which is measurable in a divergence sequence without any comb at all.

One seed in eight, which is a warning about single stems #

The shared disturbance returned the pair on one seed of eight. That is not nothing, and it is worth saying what it means for an experiment rather than letting it pass as a rounding error.

The readout refuses when it cannot find a clear enough comb, and it returns a pair when it can. On an arrangement that manufactured nothing, one stem in eight still produced a clear enough comb by chance to return 8/13 — the correct pair, which is the awkward part: a false positive here does not look like nonsense, it looks like a confirmation.

That is a rate of about 12% per stem, measured on a control that is known to contain no signal at all. So a comb read off one stem is not evidence of very much, whatever it says; the strength of the original forgery result was that it reproduced on every seed, and the strength of this one is that its control does not.

The site’s own specification for a survey has always asked for tens of specimens rather than one, on separate grounds — how many are needed to pin a divergence, or to tell two accounts apart. This is a third reason with a different shape: the instrument itself has a false-positive rate, and the only way to see it is to run the control many times.

A test a plant could fail #

The narrowing is worth something because it is checkable on a stem rather than in a model.

If a plant’s disturbances are inherited, the divergence sequence carries a slow wander as well as a comb: block means over a hundred organs vary several times more than independent errors would allow. If they are merely shared between touching organs, the sequence has teeth at the contacts and no wander.

That is two statistics off one sequence, and this site has already priced what a sequence costs: about nine hundred organs at half a degree of protractor precision for a comb. The wander is cheaper — it is a variance ratio, not a spectral feature — so a stem long enough for the comb is more than long enough for both.

And the two hypotheses disagree about it. A placement rule produces neither: its errors are corrections rather than inheritances, and its divergence sequence is anticorrelated at lag one rather than drifting. So a stem with a comb and no drift is evidence for the rule against the transport, which is an observable this site did not have a week ago and lost the previous one to.

What a plant would have to be doing #

It is worth stating the surviving rival as a claim about a meristem rather than as a property of a stream, because the narrowing changes what it would take to believe it.

The weak version — organs that touch share their errors — is nearly free. Two primordia in contact are mechanically coupled; if one is displaced, the other is displaced a little. Almost any tissue satisfies it, and it is the version most people would assent to without thinking.

The version that forges a comb is not free. It requires that when organ i is displaced, the displacement it passes to organ i+8 is passed on again to organ i+16, and again to i+24, with only a slow decay. A displacement entering the apex would then be detectable dozens of organs later, along a parastichy, having travelled around the stem several times.

That is a claim about the tissue with consequences beyond phyllotaxis, and it is the kind of claim an anatomist could have an opinion about. It also has an awkward corollary: a mechanism that propagates displacements that far would tend to accumulate them, and an apex that accumulated displacements for a hundred organs would not have a lattice left — which is roughly what the measurements show, since the transported disturbance destroys a lattice at half the amplitude white noise does.

What this does not say #

It does not restore the comb as evidence of a rule. It restores part of the comb’s evidential value against one specific class of rival — the one where sharing is local and does not propagate — and leaves the propagating version exactly where it was.

It does not say plants share errors rather than inherit them. Nothing here measures a plant. If anything the physically natural version is the inherited one, which is why it was written first.

And it does not depend on the coupling being small. The result is stated at the coupling where the transport works and repeated at two couplings where it does not, precisely because a control that only wins in a narrow range is not a control.

The check #

The forgery comparison is asserted in three parts.

The inherited disturbance must return the pair on every seed, with its main comb clear of the band — so the essay cannot quietly rest on a weakened version of the result it is narrowing.

The shared one must return the pair on fewer than half the seeds and must produce a comb less than two-thirds of the inherited one’s, at the same coupling and the same scatter.

And raising the shared one’s coupling must not rescue it, checked at 0.7 and 0.9. That third assertion is the one that answers the objection this result invites, and it is checked at a coupling the rival cannot be run at.