The hop that survived

Take a stem that has settled, remove one organ from its recent history, and let the rule place what comes next. Most of the time the pattern repairs: the divergence wobbles for a few dozen organs and comes back to where it was. Some of the time it does not. At those offsets the stem never returns, and what it does instead is oddly tidy — it settles into an exactly repeating sequence of divergence angles, the same handful of numbers over and over, to the resolution of the grid the azimuths are computed on.

That repeating sequence is the block, and its length is a number. Measured across six lattices it is 4, 5, 7 or 8, and at eighteen of nineteen wrecked offsets it is one of the two spiral counts of the lattice the stem was cut from. That much has been on the table for a while, and so has the objection to every explanation offered for it. The obvious reading is that the removal destroys one of the two contact families and the surviving one sets the period, which sounds right and is refused by the arithmetic: at the 5/8 lattice the offsets that give five are four and five, and the offsets that give eight are six and seven, and neither set is a multiple of anything.

This essay reads the same runs a different way and the explanation falls out in one measurement. The trick is to stop looking at the divergence.

A lattice has more hops than the one everybody measures #

The divergence is the angle from one organ to the next. It is the quantity a botanist can measure with a protractor and it is the quantity every claim about phyllotaxis is phrased in, which is why it is the one this collection has been plotting. But it is only the first of a family. For any whole number k there is a hop: the angle from an organ to the one k places above it. On a settled lattice each hop is the same from every organ — that is what makes the k-th family a family — and the two hops with the shortest steps across the surface are the parastichy pair.

So a lattice on a stem is not one number repeated. It is a whole spectrum of them, one per lag, and they are related: the k-hop is k times the divergence, reduced to a turn. Anything that is true of the divergence propagates to all of them, which is exactly why nobody thinks to look at them separately.

The measurement is then obvious to make and had not been made. Take a wrecked stem, take its last hundred and twenty organs, and for every lag from one to twenty-four ask how much that hop moves from organ to organ. On a settled lattice every answer would be zero. On a wandering stem every answer would be large. On a wrecked stem, the answers are not all alike.

One bar on the floor #

The lag-one bar is the divergence, and it is enormous: a spread of ninety-one degrees, which is what “never repaired” means. The lag-eight bar is 0.00°. Not small — zero to the precision the azimuths are computed at, over a hundred and twenty organs, in a stem whose divergences are swinging through more than a quarter of a circle. Its multiples inherit the same steadiness, as they must. No other lag is within a factor of twenty.

And the eight-hop has not merely stopped moving; it has stopped moving where it was. Compared with the undisturbed control — the same stem, same history, with nothing removed — it sits 0.23° from where it always was. So the eighth family of the original lattice is not a new regularity the wrecked stem has invented. It is the old one, still there, organ by organ, with everything else in the pattern rebuilt around it.

The block of this stem has period eight.

Nineteen times, on six lattices #

One case is a coincidence waiting to be found out. The measurement is cheap enough to make everywhere, so it was: every offset at which a single removal leaves a stem that never comes back, on four rises of the golden branch and two of the Lucas branch.

Nineteen wrecked offsets, and the block equals the surviving lag at every one of them. The surviving hop’s spread runs from 0.000° to 0.117°; the divergences in the same stems run from 47.9° to 91.1°. The largest distance any survivor sits from where the control put it is 2.97°, on the Lucas branch at a rise of 0.020, and every other row is under 0.71°.

That is the explanation, and it is worth stating without hedging. A wrecked stem is one family of the lattice it was cut from, still intact, with the rest rebuilt around it. The block is that family’s period.

Why it comes out a spiral count, and why that was the wrong emphasis #

The reason the block is usually a parastichy number now needs no separate account. Parastichy numbers are the lags whose steps across the surface are shortest — that is the definition — and a placement rule responds most strongly to the organs nearest it. A short hop is the one the rule has the best chance of holding rigid while everything else moves. So the block comes out a spiral count for the same reason the pattern has spirals at all, and not because of anything about counting.

This matters because it changes what the claim forbids. “The block is a parastichy number” is a statement about a census: check enough wrecked stems and see whether the periods stay inside the pair. “The block is a hop that did not move” is a statement about a single stem, testable on that stem, and it makes a prediction the census version does not — that the hop in question is unchanged rather than merely present. A stem whose block was five and whose five-hop had moved by forty degrees would satisfy the first and refute the second.

None does. That is what the third column of the census is.

The lag that is not a count #

There is one offset in the nineteen where the surviving lag is not a parastichy number of the lattice, and it is the reason this essay is about hops. At the 8/13 lattice, with the organ six places back removed, the block is four. Four is not 8, it is not 13, and a counter shown the wrecked stem’s positions returns 8 and 19.

The four-hop of an 8/13 lattice is a long step: 6.8 times the distance of a contact hop, right across the pattern. It is not a spiral anybody counts, it is not a neighbour relation, and the rule held it rigid all the same — spread 0.117°, displacement from the control 0.006°.

That single row is worth more than the eighteen that agree. It says the quantity being conserved is a lag and not a family of contacts, and it explains why every attempt to derive the block from which contacts survive had to fail. The rule is not preserving a set of neighbours. It is preserving a step.

What a counter says about a wreck #

There is a second reading of the same stems, and it is the one this collection usually trusts most: hand the positions to a counter that is told nothing — not the divergence, not the order of arrival, not what was done to the stem — and ask what pair it returns.

The answers are informative in a way that is easy to misread. At the 5/8 lattice cut four places back the counter returns 5 and 12; cut five places back, 5 and 13. At the 8/13 lattice cut four places back it returns 8 and 16, and cut seven places back, 8 and 14. In every one of those the surviving lag is the first number. The counter is finding it, because a rigid family is a family of equally spaced steps and equally spaced steps are what a counter counts.

What the counter also shows is that the other half of the lattice is gone. The conjugate number is 12, 13, 16 or 14 where the undisturbed stem gave 8 or 13, and those are not near misses — they are different numbers, because the family that was not held rigid has been rebuilt at a different pitch. A wreck is therefore half a lattice preserved exactly and half a lattice replaced, which is a much more specific object than “damaged”.

One thing about the four replacements is worth reading off, because it points the same way every time. The conjugate goes from 8 to 12 and 13 at the finer of the two lattices, and from 13 to 16 and 14 at the coarser one. All four are larger than the number they replace: the family that was not held rigid is always rebuilt at more chains than it had, never at fewer, and the survivor is the smaller of the two numbers the counter returns in every case. So a wrecked stem is not merely a different pair — it is a pair that has moved off the ladder in one direction, since 5/12, 5/13, 8/16 and 8/14 are none of them arrangements a settled stem carries at any rise. A census that recorded only counted pairs would show these stems as off-ladder without saying which half of them had stayed exactly put.

It also explains a small puzzle in the census. At the 8/13 lattice cut six places back the counter returns 8 and 19, and the block is four — so the counted pair does not contain the block at all. It does contain twice it, and it has to: if the four-hop is rigid then so is the eight-hop, and eight is the shorter of the two steps, so eight is what a census of that stem reports. The counter is not wrong and neither is the block. They are answering different questions, and only one of them is a question about lags.

What was actually being asked #

It is worth being precise about which question this answers, because it does not answer all of them.

The question it answers is what is a wrecked stem. The answer is: the original lattice, with one lag left exactly where it was and everything else rearranged around it, which is a description with no free parameters in it. That collapses the space of possible outcomes enormously. Before, a wrecked stem could have been any repeating orbit at all; now it is one of the four or five lags the rule can hold, and the whole of the outcome is a single small whole number.

The question it does not answer is which lag survives. The same lattice cut at four keeps its eight-hop and cut at six keeps its four-hop, and nothing here predicts that from the offset. The one thing that can be said is negative and useful: it is not simply the nearest contact family, because the offsets that keep five at the 5/8 lattice are four and five while the offsets that keep eight are six and seven, and the removal at offset five is not more or less adjacent to the five-family than the removal at offset six.

How a real stem would show it #

The measurement transfers, which is not true of everything in this thread. A wrecked stem’s block is a sequence of divergence angles, and reading it off a plant means measuring many consecutive divergences accurately enough to see a repeat — a demand this collection has priced elsewhere and found expensive. The hop is cheaper in one specific way: it does not need the order of the organs to be known, only the positions, because a hop is a step between two organs a fixed number of places apart and that relation is visible in the point set.

So an experiment that removed a primordium and photographed the stem two hundred organs later would not need to have watched it. It would need to identify one parastichy family and ask whether its step is the step the undisturbed side of the same plant shows. That is a comparison between two chains of contacts, which is the kind of measurement a photograph supports.

Whether a meristem does anything like this is the standing question and is not touched here. What has changed is that the prediction now has a shape a measurement can take hold of: not the pattern will be disturbed, which every account agrees on, but one named family will be exactly where it was and the others will not.

What this does not say #

It does not say the surviving lag is always the smallest rigid one. It is at eighteen of nineteen offsets. The exception is a stem whose divergence barely moved at all — the mean sits 0.02° from the original — so lag one is trivially steady and the smallest-rigid rule picks it, while the stem is nonetheless running an eight-fold orbit about that mean. The rule stated here is that the block is a rigid lag, and the qualification is that smallest is not always the right selector — the shorter of the two contact steps is not the one that survives either.

It does not say the surviving family is undamaged as a set of contacts. The hop is preserved as an angle. Whether the organs in that family are still each other’s nearest neighbours in the wrecked stem is a separate question — and one a wider sweep answers by narrowing it to the two counted families — while at several offsets the answer here is no: the conjugate count changes a great deal, and a family whose partner family has moved may no longer be a contact family at all.

It does not say anything about how the stem gets there. The measurement is made on the last hundred and twenty organs of a three-hundred-organ continuation. What happens in the first few dozen, and whether the surviving lag is decided immediately or settles out of a competition, is not measured here. That the choice looks like a coin flip when it is scored across the census and not along one rung is as close as this thread comes to an answer.

And it does not say a plant does this. These are stems grown by a rule that places each organ where the repulsion from the ones already there is least. The claim licensed is about what an ablation experiment would find if that rule is what a meristem does, which remains the only reason to run one.

The check that would refuse it #

Three assertions run while the figures are drawn, and each can fail on its own.

The first is that every wrecked stem has a lag whose hop is steady to under half a degree and sits within three degrees of the control’s, and that the block the stem repeats is that lag. Nineteen offsets are checked and one disagreement stops the collection being built, and the period that comes out of it is what makes the slip a whole number of turns a checkable claim rather than a coincidence of small numbers. The tolerance is stated rather than fitted: the quantity is bimodal, with the survivors at 0.00° to 0.12° and the next steadiest lag in any stem at 54°, so the threshold has two and a half orders of magnitude to sit in and nothing turns on where in that gap it is put.

The second is the contrast, and without it the first would be satisfied by a stem that had barely moved. The divergence of every stem in the census must itself be unsteady — a spread of at least twenty degrees — and no lag that is not a multiple of the survivor may come within a factor of ten of the survivor’s own spread. A recovery would fail the first half; a stem with two independent rigid families would fail the second, and would mean the story here is incomplete.

The third is the exception, asserted as an exception. At least one offset in the census must keep a lag that is not one of the lattice’s counted numbers, and its hop must be more than three times the length of a contact hop rather than a near miss. A census in which every survivor was a parastichy number would pass the first two checks and would have left the wrong statement standing — that this is about spiral counts — so the assertion is written to fail if the awkward case ever disappears.