A wreck has a short list

The obvious way to think about damage is as a matter of degree: take more away and the result should be worse, in the sense of further from where it started and less like anything in particular. Sweeping cuts of one, two, three, four and five organs at one arrangement makes it possible to check that, and it is wrong in a specific and useful way.

The share of arrangements that never repair does climb with the size of the cut: 0.25, 0.56, 0.71, 0.83, 0.98. That much is ordinary. What does not change is where the wrecked stems go.

The climb is worth one paragraph of arithmetic before moving on, because it is not the climb the obvious model gives. If each organ removed were an independent chance to wreck the stem, the share would be one minus the survival raised to the size of the cut, and the single-organ figure of 0.25 fixes that survival at 0.75. The predicted shares are then 0.44, 0.58, 0.68 and 0.76 against measured 0.56, 0.71, 0.83 and 0.98 — above the prediction at every size, and by a widening margin.

So removals are not independent: taking two organs is worse than taking one organ twice, and the excess grows with the cut. That is a small result and it points the same way as the front — organs removed from one neighbourhood are removed from each other’s neighbourhood as well, so a cut of five does not deliver five separate disturbances but one disturbance to a region the rule cannot see around. It also means the wreck rate is the wrong quantity to reason about a dose with, which is the theme of the essay that tries a share instead and finds the same thing from the other side.

Over two hundred and forty-six wrecked stems there are six distinct destinations, and the largest cut reaches nothing the smallest did not already find. The dominant one takes more than half of every row, and it is the old lattice with one family left standing.

The list #

Written out, with a representative cut for each:

The stem was cut from 136.78°, counted 5/8. Two arrangements that both settle near 208.8° are counted as one place here even though a counter reads their positions as 5/12 and 5/13 — the same slip of the same family with different conjugate structure, which the divergence cannot separate and the positions can.

What counts as one destination #

A list of six needs a rule for when two settled values are the same place, and the rule affects the count, so it is worth being explicit.

Two wrecked stems are counted as reaching one destination when their settled divergences agree to within a degree. A degree is four steps of the azimuth grid the runs are computed on, and it is far smaller than the gaps in the list: the closest pair of genuinely different destinations here are 33.7° apart, and most neighbouring pairs are 20° to 45° apart.

So the tolerance has a wide empty band to sit in and little about the count turns on it. Anything from a degree to fifteen gives the same six places, which is the property a threshold should have and the property the mirror tolerance elsewhere in this thread deliberately does not have.

The one place it bites is a pair of arrangements settling at 208.73° and 208.80°, 0.07° apart and therefore one place by the rule. A counter shown the two stems returns 5/13 and 5/12, so they are the same slip of the same family reached with different conjugate structure. That is a distinction the divergence cannot make and the positions can, and it is the reason the count of destinations is reported beside what each one turned out to be rather than on its own.

Half the list is the old lattice, slipped #

Three of the destinations have an immediate description, and it is the one the single-organ work established: the stem keeps one family of the lattice it was cut from, rigid organ by organ, and gains a whole number of turns over that family’s period.

Ask each destination whether any lag’s hop is unchanged from the control and the answer at 136.99°, 208.80° and 280.43° is the same lag: five.

• At 136.99° the five-hop is rigid and the divergence has moved by 0.21° — zero turns. The mean is where it started and the sequence never settles.
• At 208.80° the five-hop is rigid and the divergence has moved by 72.02°, against 360 divided by five, which is 72.00°. One turn.
• At 280.43° the five-hop is rigid and the divergence has moved by 143.65°, which is two turns over five organs, or 144.00°.

So one family of the arrangement — the five-family — accounts for half the list, at three different numbers of inserted turns. A counter shown any of those three returns 5 as one of its two numbers, which is the surviving family being found without anything being known about the intervention.

Why the dominant destination dominates #

More than half of every row goes to one place — 208.78°, the five-family kept with one turn inserted — and it is worth asking why that one rather than another.

Two reasons are available and this work separates them only partly.

The arithmetic reason is that one turn is the smallest non-zero slip a surviving five-family admits, so it is the nearest of the available destinations to where the stem started: 72° away, against 144° for two turns. If a wrecked stem falls to the nearest available equilibrium, that is the one.

The structural reason would be that the surviving family is the one the rule holds most tightly, which ought to be the one with the shortest step. That one is measurable and it comes out wrong.

At this rise the eight-hop is 0.111 of the circumference and the five-hop is 0.120, so the family that survives is the longer of the two contact steps. That does not refute the structural reason outright — both are far shorter than any other lag, and the rule’s grip on the two may be indistinguishable — but it removes it as an explanation of which of the two is kept, and it leaves the arithmetic reason standing alone.

Separating them properly needs a lattice where the two accounts disagree more sharply: one whose counts differ by more than a factor of two, so that a large surviving period gives a small slip while a short step belongs to the small period. None of the three arrangements swept here is of that kind, so the question is named rather than answered.

One destination that a mean would miss #

The first entry on the list is worth a paragraph of its own, because it is the one an ordinary measurement would classify wrongly.

At 136.99° the stem has settled 0.21° from the divergence it was cut from. Any survey that recorded a plant’s average divergence and compared it with an undisturbed control would call that a full recovery — the difference is a fifth of a degree, well inside what a protractor on a real apex resolves and inside the scatter this collection’s own noisy runs produce.

It is not a recovery. The individual divergences never settle: they run a repeating cycle whose members are tens of degrees apart, about a mean that happens to sit where it started. The five-family is rigid, as at the other slipped destinations, and the number of turns inserted over its period is zero rather than one.

So the list has an entry that is invisible to the measurement most likely to be made. Everything that distinguishes it lives in the sequence of divergences rather than in their average, which is the same distinction this collection has had to draw for combs, for wanders and for the second moment of a cell area, and it is the reason a mean is treated here as a summary rather than as a measurement.

And half of it is not #

The other three have no rigid lag at all. Not a long one, not a marginal one: no lag from one to twenty-four has a hop steady to half a degree and within three degrees of the control’s.

What they have instead is a different lattice. A counter shown their positions returns 2/6, 4/6 and 3/6 — every one of them containing a six, and none of them containing an eight or a thirteen. The stem cut from 5/8 is not on a slipped 5/8; it is on something else.

Those three are reached only by cuts of two organs and more, and two of the three only by cuts of four. The single-organ cut, at this arrangement, produces nothing but slips.

That is a limit on the description the single-organ work arrived at, and it is worth stating as one. Every stem that never repairs after one organ is removed keeps a rigid lag; that stops being true when more than one organ goes. The census that established it covered nineteen single removals on six lattices, and it does not extend.

The other rungs, briefly #

The list is a property of the arrangement, so it is worth saying what the other two arrangements do, even though neither was swept as thoroughly.

At the coarse 3/5 arrangement a two-organ cut wrecks two arrangements of twenty and they go to two different places: 258.75°, which a counter reads as a 3/4 lattice, and 220.31°, which is the mirror of the divergence the stem was cut from. So the coarse arrangement’s short list contains something the fine one’s does not, and it is the subject of another essay in this thread.

At the finest 8/13 arrangement a three-organ cut wrecks seventy arrangements of seventy-two and they land on nineteen distinct places, which is more than twice the fine arrangement’s list from a single size of cut. The dominant one takes twenty-four of the seventy and sits 45.0° from where the stem started, which is one turn over eight organs — the eight-family surviving, exactly as the single-organ census at that arrangement found.

So the list gets longer as the arrangement gets finer, which is what the slip account predicts: more lags are available to be held rigid, and each admits several numbers of turns. Nothing here counts them carefully enough to test that as a rate.

Why the list is short at all #

The short list is the more surprising half of this, and it now has an account for at least the slips.

A slipped destination is specified by two small whole numbers: the period of the family that survives and the number of turns inserted. The periods available are the handful of lags a placement rule can hold rigid, and the turns are zero, one or two in everything measured. Multiply the two and there are not many combinations, so there are not many places to land.

For the three that are not slips there is no such account here. What can be said is that they too are a small set rather than a scatter, that every one of them is counted at a pair containing six, and that the sixes are suspicious: six is twice three, and a pattern whose counts share a factor is a whorled pattern, which this collection has grown deliberately elsewhere and which the ordinary rule does not produce from an ordinary seed.

Whether a large cut is knocking a stem onto a jugate lattice is a question this essay raises and does not answer. It would be answered by measuring the rotational symmetry of the positions, which is a measurement this collection has and which was not run here.

What the dose does and does not decide #

Putting the two sweeps together gives a clean division of labour.

The dose decides whether. More organs removed means more arrangements that never return: a quarter at one organ, nearly all at five, monotone at every step. That is an entirely ordinary dose-response and it is what a larger intervention should do.

The dose does not decide where. The destinations reached by a five-organ cut are, without exception, destinations a smaller cut already reached. What varies with the dose is the mix — the dominant slip takes 100 per cent of the wrecked single cuts and about half of the wrecked five-organ cuts, with the rest spread over the other seven — but the set does not grow.

The one thing the dose does add is access. Three of the six destinations are unreachable by a single cut, and they are the three that are not slips. So a larger intervention does buy something — not a worse outcome, but a different kind of outcome, and the kinds are two.

What a census would see #

The transferable form is short and it is unusually cheap to check.

If a wrecked plant’s divergence is measured and it is not the one it should have, the value is predicted to be one of a short list rather than anything at all. On this arrangement the list has six entries in a range of a hundred and forty-four degrees, so a measurement good to ten degrees would place a specimen on it — and a specimen landing between entries would refute the whole picture.

That is a rare shape for a prediction in this subject: a discrete set of allowed values, with the gaps between them large compared with any plausible measurement error. Most of this collection’s predictions are about differences of a few degrees and need careful protractors. This one needs a bad protractor and a plant that has been damaged.

What this does not say #

It does not say six is the number of destinations. It is the number found by these sweeps, which cover between forty and eighty arrangements at each size of cut and do not exhaust the possibilities. A destination reached by one arrangement in five hundred would not appear here.

It does not say the three non-slips are one thing. They are three places with no rigid lag and counted pairs containing a six. Whether they are a single family of outcomes or three unrelated ones is not measured.

It does not say the single-organ description was wrong. It says its scope is single-organ cuts. Within that scope it holds at every wrecked offset in a census of nineteen across six lattices.

And it does not say a plant would land on this list. The list belongs to a placement rule at one rise. What a meristem would do after losing several primordia is a question about tissue as much as about placement, and nothing here addresses it.

The check that would refuse it #

Three assertions carry this, and the third is the one that keeps the first two honest.

The first is that the destinations really are a short list: the wrecked stems of every size of cut land, between them, on at most a dozen distinct settled divergences. The bound is stated generously because the claim is qualitative — a few rather than a continuum — and a tight bound would be a claim about the sweep rather than about the stems.

The second is the containment: the largest cut’s destinations must be places the smallest cut already reached. That is what says the dose does not decide where, and it would fail the moment a big cut invented somewhere new.

The third is the split, and it is asserted in both directions. At least two destinations must be slips — the old lattice with a rigid lag and a whole number of turns, checked to within four degrees of closure — and at least one must not be, with a counted pair the original lattice does not carry. Asserting only the first would let the single-organ description quietly extend past its evidence; asserting only the second would make the slips look like an exception rather than half the list.