A survivor has to be a neighbour

A stem that has settled into a lattice, with one organ removed from its recent history, does one of two things. Usually it repairs: the divergence wobbles for a few dozen organs and comes back to where it was. Sometimes it does not, and reading those runs by lags rather than by neighbours is what says what a wrecked stem is. Exactly one hop — the angle from an organ to the one p places above it — is unchanged from the undisturbed control, organ by organ, to within a tenth of a degree, while the divergence itself swings through eighty. The block of angles the stem repeats has period p. The stem is one family of the original lattice, still standing, with everything else rebuilt around it.

That left a question stated in one line and answered in none: which p. The same lattice cut four places back keeps its five-family and cut six places back keeps its eight, and nothing on the table predicted that.

This essay sweeps the offset across every organ of the front, at twelve lattices on two branches, and tabulates the survivor. The answer is not a formula. It is a much shorter list than anybody had a right to expect — short enough that the two obvious ways to finish it, by length and by offset, can both be put to the census and both fail.

The sweep, and what it covers #

Twelve lattices: eight golden and four Lucas, at rises from 0.005 to 0.032, carrying counted pairs of 2/3, 3/5, 4/7, 5/8, 7/11 and 8/13. At each of them every offset from the tip out to two organs past the front is cut, one organ at a time, and the resulting stem is compared with a control that shares its history to the last digit.

Thirty of those offsets never repair. Two of the twelve lattices contribute none at all — the 3/5 lattice at a rise of 0.032 and the 3/4 Lucas lattice at 0.026 recover from every single removal, at every offset — and they are in the census for that reason. A table showing only the lattices with an answer would be a picture of the selection rather than of the sweep.

The sweep also has to be shown to be the whole of the question, and that is cheap. Past the front nothing wrecks. At the 8/13 lattice the front is thirteen organs deep, and every cut made from sixteen organs back out to forty repairs without exception. So the offsets the census covers are the offsets that can produce a wrecked stem at all, and the ones it omits are omitted because there is nothing there.

What “rigid” is, and why the threshold is not doing the work #

A claim that one lag holds still while the others do not needs a number for “holds still”, and the interesting thing about this one is how little it matters what the number is.

Over the last hundred and twenty organs of a wrecked run, the surviving hop’s standard deviation is between 0.00° and 0.12°. The steadiest lag that is not a multiple of it is never under 54°. That is a gap of two and a half orders of magnitude, and it means any threshold between half a degree and fifty degrees sorts the census identically. The one in the machinery is half a degree, stated rather than fitted, because a fitted threshold on a quantity with a gap like that in it is how a claim gets made true rather than tested.

The second half of the definition is that the surviving hop has not moved: its mean sits within three degrees of where the control put it. That matters more than it looks. A stem could in principle settle into a new lattice with its own perfectly steady five-hop, and every bar in the figure would look the same. The shift measurement is what says the family is the one the stem started with, and across the census it runs from 0.01° to 2.97° against divergences that have moved by forty-five to a hundred and three.

The survivor is one of the two families a counter returns #

Rank every lag of a lattice by how long its step is across the surface of the stem. The two shortest are the contact families — that is what a contact family is, and it is what a spiral counter finds when it is shown the positions and told nothing else.

At twenty-nine of the thirty wrecked offsets the surviving lag is one of those two. Every survivor in the whole census is a 4, a 5, a 7 or an 8, and each of them is a contact number of the lattice it came from.

That is worth stating carefully, because it is easy to hear as a tautology and it is not one. The rigid-hop measurement makes no reference to counting: it compares a wrecked stem with its own control, lag by lag, and reports which lag did not move. Nothing in it knows what the contact families are. The counting is done by separate machinery on the positions. The two agree at twenty-nine of thirty, and they could have disagreed at all of them.

The exception is what makes it a claim about hops #

One offset refuses. At the 8/13 lattice with a rise of 0.005, an organ removed six places back leaves a stem whose rigid lag is four — which is not a contact number of an 8/13 lattice, is not returned by any counter shown the positions, and sits thirty-seventh in the length ranking. Its step is 6.82 times the shortest step on the lattice.

It would be tidier without it, and the tidier version would be false in a way that mattered. The rule does not hold spiral counts. It holds hops, and a hop is usually a short one because short steps are what a placement rule can keep rigid — so the period that comes out is usually a number a botanist would recognise. When it is not, it still comes out, and the arithmetic of the slip works exactly as it does everywhere else: four times the change in the divergence closes on one whole turn.

There is a form of words that covers all thirty offsets rather than twenty-nine, and it is worth having because it is a sharper assertion rather than a looser one. Four is not a contact number of an 8/13 lattice, but eight is, and four divides it. A rigid four-hop forces a rigid eight-hop — two steps of a family that does not move is a step that does not move — so the exceptional stem is not holding some unrelated family instead of a contact family. It is holding a family that contains one.

Read that way the census has no exception in it. Every survivor across twelve lattices, two branches and thirty offsets is a lag that divides a contact number, with a contact dividing itself, and the awkward row is the only one where the divisor is proper. That is a stronger statement than the one it replaces, in the way that matters: it forbids more. A survivor of six at an 8/13 lattice would satisfy “usually a contact, sometimes not” and would break the divisor form outright, since six divides neither eight nor thirteen; so would a five at a 4/7 lattice, or a three at a 5/8. The permissive version has nothing to say about any of those and the divisor version rejects all of them.

It also says what the exception costs, which is nothing. A stem holding its four-hop is holding its eight-hop as well, so it has kept more of the original lattice than the stems that agree with the simpler rule, not less. The reason no counter reports the four is that a counter reports the two shortest steps and the four-hop is thirty-seventh; the reason the block comes out four rather than eight is that a block is the shortest period the angles repeat at, and four is shorter. Two instruments disagreeing about a number is what a divisor looks like from either side.

So the honest form of the result is a strong claim with one flag on it. The surviving lag is a contact family at twenty-nine of thirty offsets, and at the thirtieth it is a long hop that no count reports. Both halves are asserted in the machinery, and a version of the census restricted to lattices where the first half holds would have quietly lost the second.

The block and the survivor agree at every offset #

There is a second measurement in each of these runs, made by different machinery and for a different reason, and it is worth reporting that the two now agree everywhere.

A wrecked stem repeats a fixed sequence of divergence angles — the block — and its length is found by looking for the shortest period the tail of the run repeats at, with no reference to hops at all. The rigid-lag measurement compares two runs organ by organ and reports which lag did not move. In the census here the block equals the surviving lag at thirty of thirty offsets, with no exceptions.

The earlier statement of this was eighteen of nineteen, and the one that disagreed was a case where the period-finder had reported a motif narrower than the azimuth grid — a constant the grid could not write down rather than an orbit. Widening the sweep did not weaken the agreement; it removed the only exception, because the wider census has more offsets whose motifs are unambiguously wide.

Why a short list of hops is a short list of destinations #

The reason this narrowing is worth having is arithmetic rather than aesthetic.

If a wrecked stem keeps the p-hop rigid, then p times the change in the divergence is a whole number of turns. The destinations available to a wreck are therefore the divergence it was cut from, plus 360° divided by p, times a small whole number. That is the shape of the short list measured directly: across hundreds of wrecked stems at one lattice, six settled divergences and no others, with the largest cuts reaching nowhere the smallest had not.

Before this sweep, “the handful of lags the rule can hold rigid” was a phrase with no number in it. Now it has one. Across twelve lattices, two branches, six counted pairs and thirty wrecked offsets, the lags the rule holds are the two contact families — and once, a long hop. The list of destinations is short because the list of hops is short, and the list of hops is short because a rule that minimises a sum of inverse powers of distance is a rule about touching neighbours.

Coprimality makes the two answers tell themselves apart #

One consequence of one of two contact families is arithmetic and is worth extracting, because it turns a two-way ambiguity into something a single measurement resolves.

A wrecked stem’s settled divergence is d + 360° · j ÷ p, where p is the family it kept. So a lattice with contact numbers m and n offers two grids of destinations: multiples of 360° ÷ m, and multiples of 360° ÷ n. At the 5/8 lattice those are steps of 72° and steps of 45°.

The two grids share no point but the origin, and the reason is a property of the pair rather than of this experiment. A parastichy pair is coprime — that is what makes the two families between them touch every organ exactly once — so 360 j ÷ m equals 360 k ÷ n only when n j = m k, which with m and n coprime forces j to be a multiple of m, and a j of m is a whole turn and no move at all. At 5/8 the available offsets are 72°, 144°, 216° and 288° for a surviving five, and 45°, 90°, 135° and the rest of the eighths for a surviving eight, and no number appears in both lists.

Which means the destination names the survivor. A stem found at d + 72° kept its five-family; one found at d + 45° kept its eight; and no measurement of hops, no control run and no knowledge of what was cut is needed to say which. That is the cheapest reading in this whole thread, and it is available to anyone who can measure one divergence before an intervention and one after it, to better than half the coarser of the two spacings.

The reading is exact in one direction and not in the other, and the divisor case is why. A rigid four-hop at the 8/13 lattice sends the stem to d + 90°·j, and quarters of a turn are a subset of eighths — so that stem lands squarely on the eight-grid and a destination measurement alone would report a surviving eight. Which is not wrong so much as incomplete: the stem is holding its eight-hop, and it is also holding something finer that the destination cannot see. A destination names the survivor up to divisors, and separating four from eight needs the block, or the hops, or both.

What the reading does exclude outright is a destination on neither grid. That is a stem with no rigid lag at all, which is what removals of several organs are already known to produce, and it is the one outcome the arithmetic here forbids for a single removal.

What the census does not say #

Three things, and they are worth separating because two of them are the subject of the next two essays and one of them is a limit.

It does not say which of the two. Both contact families survive somewhere in the census. The 5/8 lattice keeps its five at some offsets and its eight at others; the 4/7 keeps its four at some and its seven at others. Which one comes out is not settled by anything here.

It does not say the shortest wins. It would be natural to guess that the family the rule holds is the shorter of the two — the one whose step is nearest. Across the census that is true at twelve of thirty offsets and false at the rest.

And it does not extend to larger cuts. A stem with several organs removed can reach destinations with no rigid lag at all, so “a wreck keeps a hop” is a statement about single removals until somebody shows otherwise. It has been measured, and it does not extend.

Two lattices that cannot be wrecked at all #

The two empty rows are worth a section, because they are the only part of the census where the answer to “which lag survives” is that the question does not arise.

At a rise of 0.032 the golden branch carries 3/5, and every one of its seven offsets recovers. At 0.026 the Lucas branch carries 3/4, and every one of its six does. Both are cut at every offset out to two organs past the front, both are compared against controls sharing their histories to the last digit, and in every case the divergence wobbles for a few dozen organs and comes back.

That is not the same as saying a coarse stem is undamageable. Two organs removed from a 3/5 stem wrecks it, and what it settles into is the lattice it came from wound the other way — an outcome no finer rung reaches. So the coarse rung is robust to the intervention this census makes and not to a slightly larger one, which is the ordinary shape of a threshold.

It matters here for one reason. A rule about which family survives has to be a rule about the cases where a family survives, and a census assembled from the rises that produce survivors would have been a census of its own selection criterion. The two empty rows cost eleven runs and they are what says the sweep was over the offsets rather than over the answers.

The shape of the answer #

What has changed is the size of the question rather than the question. Before, “which lag survives” ranged over every whole number the rule could conceivably hold. Now it ranges over two, with a stated exception and a stated boundary — and the two are not arbitrary numbers but the pair that a counter, shown the same stem and told nothing about the cut, would write down.

There is a version of this collection’s habit that applies here in an unusual direction. A result that says “one of two” rather than “this one” is normally a disappointment. Here it is the finding: the object being described — a stem that has been damaged and has settled into something that is not what it was — turns out to have two states available to it and no more, and which of the two it takes is a separate question with a separate answer. Narrowing the range is the work. What fills it is next.