A stem coarse enough to cut

A two-organ cut wrecks a stem at the 3/5 rung, where every single removal heals, and what the wrecked stem settles into is the lattice it was cut from wound the other way: the same divergence subtracted from a full turn, the same counted pair, the same order of shortest steps, mirrored. At the 5/8 rung nothing of the kind happens.

The account offered for that difference was reachability. A reflection is not a small change — every organ of the front has to end up somewhere it was not — so a stem falls into its mirror only if the rearrangement is short enough to be carried there. A front of five has fewer arrangements to pass through than a front of eight, so the coarse rung reaches its mirror and the fine one does not.

That predicts something testable: a 2/3 stem should reverse more readily than a 3/5 one. The rung below 3/5 is coarser than anything this collection has grown, and the obvious hazard was stated when the prediction was: a stem that coarse may not be a lattice at all, and cutting a thing that is not a lattice answers a different question.

This essay establishes what is there before anything is cut. It turns out the hazard was real, it sits in the middle of the rung rather than at its edges, and the instrument that would normally catch it cannot.

Where the coarse rungs are #

Grow a stem by the rule at a stated rise, let it settle over four hundred organs, and read two things off it: the mean divergence over the last sixty, and the pair a counter returns when it is shown the positions.

Swept from a rise of 0.040 up to 0.130 in steps of 0.005, the counted pair is 3/5 at the first two, 2/3 from 0.050 to 0.120, and 1/2 from 0.125 upwards. So there is a 2/3 rung, it is nine rises wide at this resolution, and there is a coarser rung below it again.

The settled divergence slides across the rung the way it does across every rung — and what that slide does to a front is its own question, as is what a coarse enough rung does to the mirror: 140.47° at the coarse end, down through 139.45°, 138.75°, 137.93°, 137.11° and 136.17° to 135.35°, with the counted pair constant across the whole slide. That is the ordinary picture and there is nothing wrong with it.

The band in the middle that does not settle #

From a rise of 0.090 to 0.115 it stops.

At those five rises the mean divergence is 135.0000° — to four decimal places, at every one of them — and it does not hold still. The scatter over the last sixty organs runs from 0.79° to 1.60°, where every other rise on the rung scatters between 0.0000° and 0.3356°.

135° is three eighths of a turn. The stem has not settled on a lattice divergence; it has locked on a rational one, and it wobbles about it by more than a degree.

The rung is real before the band is a problem #

Before the band is worth calling a hazard, the rung it sits in has to be shown to be an ordinary rung, or the whole ladder is suspect rather than one part of it.

Four things say it is. The counted pair agrees with the contact geometry at every rise — the counter, shown the positions and told nothing, returns the pair that the two shortest steps say it should. The rung has boundaries in the right places: 3/5 above it and 1/2 below, each replacing the pair one step along the ladder that every other rung on this site follows. The settled divergence slides monotonically across it. And the front, measured by removing one organ at a time and asking which offsets are felt, is three organs deep, which is the larger of the two counts, exactly as at every finer rung.

So the coarse rung is a rung. What sits in the middle of it is not a failure of the ladder; it is a state the rule can occupy that the ladder has no row for.

The counter cannot see it #

This is the part that makes the band a hazard rather than a curiosity.

A counter shown a locked stem returns 2/3 — the same pair as every other rise on the rung. It is not confused and it is not wrong. Three eighths of a turn is close enough to the rung’s own divergences that the two shortest chains of near neighbours are still the same two, and the wobble is large enough that the eight files three eighths would make never quite form.

So every instrument that decides what a stem is by counting it says the band is ordinary 2/3. The only measurement that separates it is the one nobody thinks to make: how much the divergence moves once it has stopped moving.

Two things a locked stem still does #

It is worth recording what does not break in the band, because “not a lattice” is a strong phrase and the object is more specific than that.

A locked stem still packs. Its organs are placed by the same rule at the same prescribed heights, they do not collide, and a counter finds two families in them. It is not disordered, and none of the disorder statistics this collection uses reports anything unusual about it.

And it still repeats, after a fashion. Three eighths of a turn brings the pattern back to itself every eight organs, so the neighbourhood an organ is placed into is nearly periodic. That is what makes the state stable at all — the rule is sitting in a landscape that repeats, rather than being pushed around at random.

What it does not do is hold a single divergence, which is the one property every claim in this thread is stated over. A cut is made against a control, the control’s settled value is the reference, and a stem with no settled value has no reference to be compared against.

The threshold, and where it sits #

A rise is admitted to the sweep if its divergence scatters by no more than half a degree and its counted pair agrees with its contact geometry.

Half a degree is not a delicate number here. The rises that settle scatter by at most 0.3356° and the rises that do not scatter by at least 0.7948°, so any threshold between those two sorts the ladder identically. The gap is four tenths of a degree wide and the threshold sits in the middle of it.

That matters because the whole point of the test is that it rejects something. A test written after the rows were chosen would have admitted everything; this one throws out five of the nine rises in the middle of the rung, which is a third of the ladder and includes the rises a sweep by round numbers would have picked first.

Why nobody would have noticed #

The band is invisible to four of the five things anybody would check, and going through them is the point of the essay.

The counted pair says 2/3 across the whole rung, band included. The front, measured by removing one organ and asking which offsets are felt, is three organs deep across the whole rung. The mean divergence at 135.0000° sits between the values at 0.085 and 0.120 and is exactly where a smooth slide would put it — it is not an outlier on the curve. And a drawing of a locked stem looks like a drawing of a coarse stem, because a wobble of a degree and a half at a spacing of a hundred and twenty degrees is not something an eye picks up.

The fifth thing does see it, and it is the one this collection has been in the habit of measuring since the collection began: how much the divergence moves once the transient is over. That is not a check anybody runs to decide whether a stem is a lattice; it is a check run to decide whether a run is long enough. It happens to be the only instrument here that separates the band from the rung.

That is the shape of a defect worth recording. Not a wrong number produced by a broken instrument, but a state that every instrument reports correctly and that nobody would think to ask about.

What the band probably is #

Not the subject of this essay, and worth a paragraph because leaving it unnamed would suggest it is unexplained.

Three eighths of a turn is a rational divergence with a small denominator, and a rational divergence is a fixed point of a different kind: the pattern repeats exactly every eight organs, so an organ placed at the argmin of the profile has a perfectly periodic landscape to sit in. The rule can hold that, and at rises where the true lattice divergence passes near it the rule apparently prefers it — but holding it exactly would require the wobble to vanish, and it does not.

Why only the coarse rung is exposed #

The account above says the rule can hold a small-denominator rational when its own divergence passes near one. That immediately raises the question the rest of the ladder should be asked: why is there no such band at 3/5, or at 5/8, where four rounds of work have found nothing of the kind?

The answer is that a rung’s exposure is set by how wide a range of divergences it covers, and the coarse rungs cover wide ranges while the fine ones do not. Across the 2/3 rung the settled divergence slides from 136.17° to 135.35° — nearly a degree — and three eighths of a turn is inside that interval. Climbing the ladder, each rung’s divergence range narrows towards the limit angle, and by 8/13 it is a few hundredths of a degree. A window that narrow contains a low-denominator rational only by accident.

And it does not contain one, for a reason stronger than accident. The limit angle is the worst approximable number there is, which is exactly the statement that no small denominator gets close to it: the nearest fraction with denominator q misses by at least 161/q degrees when the miss is weighted by q, so the fractions crowd towards the limit angle far more slowly than the rungs do. The rungs contract towards a point that the rationals are constitutionally bad at reaching.

So the hazard is a coarse-rung hazard by construction, and it gets worse downwards rather than better. The 1/2 rung below this one covers a wider range still and contains more small fractions; the 2/3 rung is the first place on the way down where a range has grown wide enough to swallow one. That is a prediction rather than a measurement, and it is checkable in the cheapest way this collection has — sweep the rise across the rung below and look for a mean divergence that is a round fraction of a turn with a scatter an order of magnitude too large.

It also says what the coarse rungs cost the ladder as a whole. Nothing above 3/5 needs the threshold this essay introduces, because nothing above 3/5 has room for a rational to sit in. A gate that applied it everywhere would be paying for a check that can only fire at one end.

Whether the wobble is the grid’s #

Before the band can be called a state of the rule, one alternative has to be closed, and it is the alternative this collection has been caught by three times: a number that belongs to the program rather than to the model. The rule places organs on a finite grid of azimuths, so every divergence is a whole number of grid steps, and a “wobble” of a degree could in principle be a stem stepping between two adjacent samples.

The table settles it, using nothing but its own figures. Every divergence in it is a whole multiple of 0.1172° — 135.0000, 135.3516, 135.4688 and 136.1719 are 1152, 1155, 1156 and 1162 steps of that size — so 0.1172° is the resolution the placement grid imposes here.

Against that, the two populations sit in different worlds. The settled rises scatter by 0.0000° to 0.3356°, which is between nought and three grid steps, and the exact zeros are stems that have locked onto a single sample. The band scatters by 0.79° to 1.60°, which is seven to fourteen grid steps.

An order of magnitude above the resolution is not a rounding artefact. A stem oscillating between adjacent samples would show a scatter of one step or two, in the way a block of three angles two steps apart did when the same question was asked of a wrecked stem’s period. This is a different object: the stem is moving across a dozen resolvable positions and returning, which is a motion the instrument can see rather than one it is manufacturing.

The check that would close it completely is the one that closes all of these, and it costs a rerun rather than a stem: refine the grid threefold and read the scatter again. An artefact’s scatter in degrees falls to a third while its scatter in steps holds; a real wobble’s scatter in degrees holds while its scatter in steps triples. The arithmetic above makes the second outcome much the likelier, and it does not make the first impossible, so the honest form is that the band survives the cheap version of the test and has not yet been given the decisive one.

Stating that matters here more than it usually would. The band’s whole significance is that no counting instrument can see it, and an essay resting on one measurement that separates two populations owes an account of whether that measurement is measuring the stem.

What the band costs the experiment #

Five rises of nine is a third of the rung, and the ones lost are in the middle, which is where a sweep would naturally have been made.

A sweep at three round rises — 0.05, 0.10 and 0.15 — would have taken one member of the rung, one member of the band and one member of the rung below. Two of those three rows would have been about something other than a 2/3 lattice, and the table would have shown it as a 2/3 lattice, because the counter says so at all three.

That is not hypothetical caution. The result the coarse rung is being cut to test is a share — how often a two-organ cut reverses a stem — and a share computed over thirty-six arrangements at each of three rises is a share over a hundred and eight cells, two thirds of which would have been cells of a stem that had no divergence to reverse.

What the test costs, then, is a factor of ten in the resolution of the rise sweep and a stated criterion. What it buys is a table whose rows are all about the same object.

What is now permitted #

Nine rises of the 2/3 rung, from 0.050 to 0.085 and at 0.120, each a settled lattice by a stated test, each counted at the pair its contact geometry gives, and each therefore a legitimate thing to remove two organs from.

That is a smaller ladder than the rung, and the difference is the essay. The prediction under test — that a shallower front reverses more readily — is worth nothing measured on stems that were never lattices, and the rows it would have been measured on are the ones a sweep at 0.09, 0.10 and 0.11 would have chosen.