Phyllotaxis
the spirals are counted, not admired
What a plant might be doing

Four accounts of one angle

The exchanged pair in a wrecked stem is about one divergence step, and about is doing twelve per cent of work. Four candidate units were written down and scored on the same seventeen rows: the cut stem's own step, the surviving family's step, the control's step, and the control's corrected.
12 min read 20 figures State the claim, state the testTwo routes, one number

Worth reading first: The damage has a period · The organ that was taken away · Counting the spirals.

The exchanged pair in a wrecked stem is one organ’s step — the two displaced chains move by 88.0° to 147.2°, against control divergences of 99.1° to 138.0°, and every row is within twelve per cent.

Twelve per cent is a lot to leave in a sentence. It is not measurement error: the azimuths sit on a grid a quarter of a degree wide, so twelve per cent of a divergence is fifty grid steps. Something is producing it and this essay is the attempt to say what.

The method is the one this site uses whenever a quantity has a rival account: write the candidates down, apply every one of them to every row, and report the scores together. It is what four accounts of a survivor were put through, and the useful part there was not the winner but the ordering of the losers.

Four accounts of one angle, scored on the same 17 rows. Each bar is how far an account of the exchanged pair's size sits from the measurement, as a share of the control's divergence, averaged over the census. A step of the wrecked stem's own divergence is the obvious candidate and the worst of the four. The surviving family's own step is not the unit at all: that angle is a fifth of the exchange. A step of the control's divergence is close, and correcting it by a fifth of the surviving hop takes the worst row from 11.8 per cent to 4.0.
Fig. 1 Four candidate units for the exchange’s size, scored on the same seventeen rows. The mean error is a share of the control’s own divergence.

The rule for scoring

Four candidates were written down, each one a unit somebody would state out loud, and each is applied to every one of the seventeen rows. The error is the difference between the candidate and the measurement, as a share of the control’s divergence, so that a Lucas row at 99° and a golden row at 138° are comparable.

Writing the candidates down before reading the table is the part that makes a score mean anything. A unit found by looking at the numbers and then quoted against them has no denominator.

The exceptional pair, measured in divergences. One row per wrecked cut whose profile has exactly one pair of exceptional classes, drawn at the size of that pair divided by the stem's own settled divergence. Every row sits between 0.882 and 1.076, so the two chains that came apart moved by one organ's step rather than by two or by half of one. The residual is not scatter: rows are grouped by the lag the stem kept, and every lag sits wholly above the line or wholly below it. Why a surviving 5 or 7 overshoots and a surviving 4 or 8 falls short is not answered here.
Fig. 2 The exchanged pairs by size, which is the quantity all four candidates are trying to be.

The first candidate: the cut stem’s own step

The obvious one. The exchange happens in the wrecked stem, and the wrecked stem does not run at its control’s divergence: it runs at the control’s plus a slip of a whole turn over the surviving lag. So the step in which the exchange is measured ought to be the wrecked stem’s own.

It is the worst of the four: wrong by up to 82.2 per cent, by 28.9 on average, and right on three rows of seventeen. On the Lucas 0.013 stem it predicts 190.0° where the measurement is 88.0°, which is not a near miss in any direction — it is past a half turn, where an angle folds back on itself.

A wreck is a whole number of extra turns. For each of the 19 stems that never repair, the slip of its settled divergence multiplied by the lag whose hop survived. Every value lands on a whole number of turns — the horizontal lines — with a largest departure of 2.97 degrees, against divergences that have moved between 0 and 103 degrees. 17 of the 19 close on exactly one turn. So a wrecked stem is the stem it was with one extra turn threaded through every period of the family that survived, which is a dislocation with a stated size rather than damage.
Fig. 3 The slip on every wrecked cut. It is what makes the cut stem’s own divergence differ from its control’s, and it is large.

Which is a real negative

It is easy to read a failed candidate as a candidate nobody believed. This one is worth believing: the two chains that changed places are chains of the wrecked stem, their organs are placed by the wrecked stem’s rule, and every angle in the wrecked stem is separated by the wrecked stem’s divergence.

That it fails by a factor of two says the exchange is not measured in the arrangement it happens in. It is measured in the arrangement it came from.

How far every organ moved, 4 places back at a rise of 0.013. One mark per organ above the hole, at the angle it sits from where the same organ sits in a control that shares its history. The collection has read two numbers out of profiles like this one — the largest displacement anywhere, and the first organ's — and never the profile. It is not a bump that decays. After about 3 organs it settles into a repeating pattern of five levels, one per residue class modulo 5, which is the lag whose hop this stem kept. three of those levels sit together and two do not.
Fig. 4 A whole profile. Every angle in it is a difference between the two runs, which is the sense in which the control is still present in the wrecked stem’s readings.

The second candidate: the surviving family’s step

Also worth believing, and worse. Every wrecked stem keeps exactly one lag rigid — the angle from an organ to the one k places above it is unchanged from the control, while every other lag moves. That surviving lag is the period the profile is folded on and the family the arrangement is holding onto.

Its own step is the natural unit if the exchange is a movement within that family. The step is 19.5° to 39.1° across the census; the exchange is 88.0° to 147.2°. Wrong by 79.3 per cent on average and right on no row at all.

One wrecked stem, lag by lag — golden, rise 0.013, organ 4 back. How much each lattice hop moves from organ to organ in a stem that never repaired after a single removal, over its last 120 organs. The lag-one hop is the divergence itself and it swings by 65 degrees. The lag-5 hop swings by 0.11 degrees and sits 0.09 degrees from where the undisturbed stem put it, so that one family of the original lattice is still standing organ by organ. Its multiples inherit the same steadiness and nothing else comes within a factor of twenty. The block of angles this stem repeats has a period of 5, which is the surviving lag and not a coincidence.
Fig. 5 The lag spectrum that identifies the surviving hop. Its own angle is a fifth of the exchange, not the whole of it.

Two candidates gone, and a shape appearing

Both failures are informative in the same direction. The exchange is not a movement of the wrecked stem measured in the wrecked stem’s units, and it is not a movement within the surviving family measured in that family’s units.

What is left is that it is a movement of the control’s arrangement: two organs of the pattern that was there before, one place out of order. That is the third candidate and it is what the direction of the exchange already said.

Two readings arriving at one conclusion from opposite ends is the pattern this collection trusts. The direction is a discrete fact with no error bar on it and the size is a continuous one with twelve per cent of slack; they agree about which arrangement the exchange belongs to before either is asked to be precise.

A period of 5, and the two classes that are not with the rest. The same wrecked stem, folded on the lag it kept: one row per residue class, each drawn at the mean displacement of its own organs against the level the rest of them share. The bar through each row is the spread inside that class, and the widest of them is 0.66° — so within a class the displacement is a constant. three classes sit at the common level. The two that do not sit at 145.9° and -148.4°, equal and opposite to within 1.7 per cent, and they are neighbouring residues. The stem's own divergence is 136.78°, so an exception is one organ's step.
Fig. 6 One row’s levels. The two exceptions are the pair whose size the four candidates are competing to be.

The third candidate: the control’s step

One step of the control’s own divergence. Wrong by at most 11.8 per cent, by 4.3 on average, and right — within a twentieth — on twelve rows of seventeen.

That is the reading the thread already had, and this scoring is what turns it from “about one step” into “better than the two rivals by a factor of six on the average and a factor of seven on the worst row”.

A period of 8, with the hole's own chain at the top. Each mark is one residue class of the displacement profile, placed round a ring at its own residue, with the chain the removed organ sat on at the top. The radius is how far that class sits from the level the rest of them share. six of the eight classes sit together at the middle ring; two do not, and on this row they are one pair, equal and opposite to within a twentieth. The forward one is chain 7 and the backward one is chain 0, one residue above it, which is the order every row of the census puts them in.
Fig. 7 One row’s chains round its period, with the two exceptions the size is read from.

Where the residual is

Not scattered. The rows whose surviving lag is 5 or 7 all overshoot one step, and the rows whose lag is 4 or 8 all fall short of it. No lag appears on both sides, on seventeen rows across two branches.

A residual that sorts by a column is a residual with something in it, and this one sorts by the same column the profile is folded on. That is the fourth candidate and it gets its own essay.

What the exchange misses one divergence step by, against the surviving hop. Each mark is one wrecked cut. Across: the angle from an organ to the one its surviving lag places above it, on the control. Down: how much the exchanged pair's size falls short of one divergence step. The relationship is a straight line through the origin with a slope of about 0.2, and its sign is right on every row of the census — the four surviving lags give hops on both sides of zero and the residual follows each of them. The honest limit is the four: the hop hardly moves inside a lag, so the line rests on 11 distinct positions and not on 17.
Fig. 8 What the size misses one step by, against the surviving hop’s own angle. The line through the origin is the correction the fourth candidate applies.

The fourth candidate: the control’s step, corrected

One step of the control’s divergence, less a fifth of the surviving hop’s own angle. Wrong by at most 4.0 per cent, by 1.0 on average, and right on all seventeen rows.

A fifth is stated rather than fitted. The least-squares coefficient over the seventeen rows is 0.188, and a sixth, a fifth and a quarter all take the worst row from 11.8 per cent to under six.

Four accounts of one angle, scored on the same 17 rows. Each bar is how far an account of the exchanged pair's size sits from the measurement, as a share of the control's divergence, on its worst row. A step of the wrecked stem's own divergence is the obvious candidate and the worst of the four. The surviving family's own step is not the unit at all: that angle is a fifth of the exchange. A step of the control's divergence is close, and correcting it by a fifth of the surviving hop takes the worst row from 11.8 per cent to 4.0.
Fig. 9 The same four accounts scored on their worst row rather than on the average, which is the harder test and the one the ordering does not change under.

Four candidates is not many

The scoring’s value depends on how many rules were available and how many were tried. Four is few, and a reader is entitled to ask what a fifth would have done.

The honest answer is that the four are the units this thread already has names for: the two divergences it measures on every cut, the surviving hop it measures on every cut, and the correction that the residual’s own structure suggested. A fifth would have to be a quantity nobody in the thread has computed.

Every stem that never repaired, and the lag it kept. The 19 offsets across six lattices at which a single removal leaves a stem that never returns to its divergence. For each one: which organ was removed, the period of the block of angles the stem settles into, the lag whose hop survived the cut unchanged, and how many whole turns the stem gains over one period of that lag. The block and the surviving lag are the same number in every row. The marked row is the one whose survivor is not a parastichy number of the lattice that was cut — a hop 6.8 times the length of a contact hop, which no census would report and which the rule held rigid all the same.
Fig. 10 Every quantity the census records for each wrecked cut. Four of these columns are the four candidates, and there are not many more to try.

The fourth is not on the same footing as the other three

It has a free parameter and they do not. A step of the control’s divergence is a number the row already carries; a step of the control’s divergence less a fifth of the hop is that number with a coefficient in it, and a coefficient bought with seventeen rows is a coefficient worth about seventeen rows.

So the four scores should be read as three predictions and one fit. The fit is a good one and it is stated in the essay that makes it that it rests on four hops rather than on seventeen.

Four accounts of one angle, scored on the same 17 rows. Each bar is how far an account of the exchanged pair's size sits from the measurement, as a share of the control's divergence, averaged over the census. A step of the wrecked stem's own divergence is the obvious candidate and the worst of the four. The surviving family's own step is not the unit at all: that angle is a fifth of the exchange. A step of the control's divergence is close, and correcting it by a fifth of the surviving hop takes the worst row from 11.8 per cent to 4.0.
Fig. 11 The four accounts again by mean error. Three of them have nothing to adjust and the best of them has one thing.

What “right on twelve of seventeen” means

The line is five per cent of the control’s divergence, which is about seven degrees. It is not a threshold anybody tuned: the four candidates score 79.3, 28.9, 4.3 and 1.0 per cent on average, so any line between about ten and twenty per cent puts them in the same order and any line at all separates the top two from the bottom two.

Where the line does work is inside the top pair, and there it is worth being explicit: the third candidate is right on twelve rows and the fourth on seventeen, and moving the line to three per cent would make it eight and fifteen.

The exceptional pair, measured in divergences. One row per wrecked cut whose profile has exactly one pair of exceptional classes, drawn at the size of that pair divided by the stem's own settled divergence. Every row sits between 0.882 and 1.076, so the two chains that came apart moved by one organ's step rather than by two or by half of one. The residual is not scatter: rows are grouped by the lag the stem kept, and every lag sits wholly above the line or wholly below it. Why a surviving 5 or 7 overshoots and a surviving 4 or 8 falls short is not answered here.
Fig. 12 The exceptional pairs by surviving lag, which is the column the third candidate’s failures sort by.

The three losers are not equally wrong

There is a difference between the two failures worth keeping. The cut stem’s own step is wrong by 28.9 per cent on average and right on three rows; the surviving family’s step is wrong by 79.3 and right on none. So one of them is a bad unit and the other is not a unit at all.

The three rows the cut stem’s step gets right are the three whose slip is smallest — the rows where the wrecked stem’s divergence is nearly its control’s, so the two candidates coincide. That is not the candidate working; it is the candidate becoming the winner, and saying so is what stops three rows out of seventeen from being read as partial support.

What the exchange misses one divergence step by, against the surviving hop. Each mark is one wrecked cut. Across: the angle from an organ to the one its surviving lag places above it, on the control. Down: how much the exchanged pair's size falls short of one divergence step. The relationship is a straight line through the origin with a slope of about 0.2, and its sign is right on every row of the census — the four surviving lags give hops on both sides of zero and the residual follows each of them. The honest limit is the four: the hop hardly moves inside a lag, so the line rests on 11 distinct positions and not on 17.
Fig. 13 The residual without the fitted line, so the rows where two candidates coincide can be read as the small ones they are.

Two of the four could have been scored years ago

Both divergences and the surviving hop have been on disk for every wrecked cut since the census was built. The exchange’s size has been on disk since the pair was found. Scoring four columns against one is arithmetic on a table.

This is the third time in the thread that the expensive thing turned out to be already computed. The profile was nine thousand numbers read for two; the position of the exchange was an index already recorded; the units were four columns of one table.

How constant the displacement is inside one residue class. One row per wrecked cut in the census, drawn at the widest spread found inside any one residue class when the profile is folded on the lag that stem kept. 25 of 30 rows sit between 0.12 and 6.09 degrees, which on a quantity whose between-class differences run past a hundred and fifty degrees is a constant. The five that do not sit from 10.3° up. There is nothing in between, so the line drawn at 10° could have been drawn anywhere in a wide interval.
Fig. 14 The census by stem. Everything four candidates need is in a table like this one and was there before any of them was written down.

The rows that carry no pair

Thirteen of the thirty wrecked cuts have three or more exceptional chains, or two that do not balance, and they are refused rather than scored. That is nearly half the census left out of every number in this essay.

They are not the ragged rows — their profiles are as periodic as the seventeen — and nobody has an account of what their exceptions are. Any candidate unit that explained them as well would be a much stronger result than the one here.

How constant the displacement is inside one residue class. One row per wrecked cut in the census, drawn at the widest spread found inside any one residue class when the profile is folded on the lag that stem kept. 25 of 30 rows sit between 0.12 and 6.09 degrees, which on a quantity whose between-class differences run past a hundred and fifty degrees is a constant. The five that do not sit from 10.3° up. There is nothing in between, so the line drawn at 10° could have been drawn anywhere in a wide interval.
Fig. 15 The whole census by how steady its classes are. The thirteen rows this essay cannot score are spread through it rather than piled at one end.

What a fifth candidate might be

One suggests itself and has not been tried: the incoming counted number’s own step. A lattice has two contact numbers and a third waiting above them, and the third is what the arrangement moves towards as the rise falls.

It is a column the census already carries. It has not been scored here because it was not written down before the table was read, and adding it now would be exactly the move this essay’s own rule forbids. It belongs in the next round’s list.

The hops of a 5/8 lattice, shortest first — golden, rise 0.010. Every lattice hop at this rise, ordered by how long its step is across the surface of the stem, with the two that a wrecked stem here leaves standing picked out. The two shortest are the contact families the counter returns, 5 and 8, and they differ in length by a factor of 1.076. The lags left standing after a removal are 5 and 8, sitting at rank 2 and 1 in this order, so the family the rule holds is a short step but not always the shortest one.
Fig. 16 The lags ordered by step length at one lattice. The incoming counted number is in this picture and is not among the four candidates.

What the scoring changes

Before it, the thread had a sentence: the exchange is about one divergence step. It was true and it had a twelve per cent shrug in it, and a shrug is where a result goes to stop being tested.

After it, the thread has an ordering of four units, a best one that is right on every row, and a residual with a stated structure and a stated denominator. The sentence is the same length and it can now be wrong in a specific way.

Four accounts of one angle, scored on the same 17 rows. Each bar is how far an account of the exchanged pair's size sits from the measurement, as a share of the control's divergence, on its worst row. A step of the wrecked stem's own divergence is the obvious candidate and the worst of the four. The surviving family's own step is not the unit at all: that angle is a fifth of the exchange. A step of the control's divergence is close, and correcting it by a fifth of the surviving hop takes the worst row from 11.8 per cent to 4.0.
Fig. 17 The four on their worst rows. The ordering survives the change of statistic, which is the least a scoring should have to do.

The instrument, once

Every size here is the mean of the two exceptional chains’ distances from the level the rest of them share, taken over a hundred and twenty organs at the top of a three-hundred-organ run, on a cut and a control sharing history to the last digit.

One row of the seventeen is a row a longer run says is not periodic, so its size is read over a window still inside a transient. It is also the row that fits worst. That is stated here and pursued elsewhere.

How steady each class is, over 300 organs and over 600. Each mark is one wrecked cut, placed across at the widest spread found inside any one of its residue classes over the shorter run and up at the same reading over the longer one. A mark on the diagonal is a row the two lengths agree about. The rules are the 10 degrees that separates a profile called periodic from one that is not: three rows fall in different quadrants at the two lengths, two of them becoming periodic and one ceasing to be. The gap between the two groups narrows from 1.69 times to 1.27.
Fig. 18 How steady each class is at two run lengths. One of the seventeen rows scored here crosses the line between them.

What a reader should not conclude

That the exchange is caused by the control. The control is a run that was never disturbed; it does not act on anything. What the scoring says is that the unit in which the exchange is measured is the divergence of the arrangement the cut stem inherited, and that is a statement about arithmetic rather than about influence.

The distinction matters because the wrecked stem does settle to its own divergence and does hold its own arrangement. Both are true, and the exchange is a feature of the difference between two runs rather than of either of them — a description of that difference is not yet a mechanism behind it.

How far every organ moved, 5 places back at a rise of 0.013. One mark per organ above the hole, at the angle it sits from where the same organ sits in a control that shares its history. The collection has read two numbers out of profiles like this one — the largest displacement anywhere, and the first organ's — and never the profile. It is not a bump that decays. After about 40 organs it settles into a repeating pattern of five levels, one per residue class modulo 5, which is the lag whose hop this stem kept. two of those levels sit together and three do not.
Fig. 19 A profile is a difference between two runs, organ by organ, which is the object every unit in this essay is a unit for.

What would make the scoring stronger

More rows, and one particular kind of row. Every row here has a surviving lag of 4, 5, 7 or 8, and the fourth candidate’s correction depends on that lag through the hop’s own angle. A census with a surviving lag of 11 or 13 in it would put the correction somewhere new on its own axis.

Those lags exist — the census’s own ranking places a survivor as deep as thirty-seventh — and no wrecked cut in this census keeps one. Finding a lattice that does is a search rather than a sweep, and it is the sharpest thing this thread could do next for the smallest number of runs.

Which lag survives, at every lattice and every offset. A row for each of twelve lattices and a column for each offset an organ is removed at, with the lag whose hop survived written in the cell. 30 cells never repair. The lag left standing is one of the two contact families at 29 of them, and at 17 it is the second shortest step on the lattice rather than the shortest, which is what refuses the reading that a rule holds its nearest neighbours. The ringed cells are the five the offset rule gets wrong. Two lattices carry no cells at all because no single removal wrecks them.
Fig. 20 Which offsets wreck at each lattice and what each keeps. A cut keeping a lag outside the four this essay has would be a new column for the correction.

The one line

Four candidate units for the exchange, scored on seventeen rows: the wrecked stem’s own divergence step is wrong by up to 82 per cent, the surviving family’s step by up to 84, the control’s divergence step by up to 12, and the control’s step corrected by a fifth of the surviving hop by 4.

The exchange is measured in the arrangement the cut came from, not in the one it produced — which is what the direction of the exchange says as well, by a different route.

Where this goes next

The exceptional pair, measured in divergences. One row per wrecked cut whose profile has exactly one pair of exceptional classes, drawn at the size of that pair divided by the stem's own settled divergence. Every row sits between 0.882 and 1.076, so the two chains that came apart moved by one organ's step rather than by two or by half of one. The residual is not scatter: rows are grouped by the lag the stem kept, and every lag sits wholly above the line or wholly below it. Why a surviving 5 or 7 overshoots and a surviving 4 or 8 falls short is not answered here. A step of one organ The balanced pair inside a wrecked stem's period measures 88.0° to 147.2° against divergences of 99.1° to 138.0° — one organ's step, to within twelve per cent on every row. The residual is not scatter: every stem keeping a 5 or a 7 overshoots and every stem keeping a 4 or an 8 falls short. What the exchange misses one divergence step by, against the surviving hop. Each mark is one wrecked cut. Across: the angle from an organ to the one its surviving lag places above it, on the control. Down: how much the exchanged pair's size falls short of one divergence step. The relationship is a straight line through the origin with a slope of about 0.2, and its sign is right on every row of the census — the four surviving lags give hops on both sides of zero and the residual follows each of them. The honest limit is the four: the hop hardly moves inside a lag, so the line rests on 11 distinct positions and not on 17. A fifth of the hop The exchanged pair misses one divergence step by up to twelve per cent, and the miss is not scatter: every row keeping a lag of 5 or 7 overshoots and every row keeping a 4 or an 8 falls short. Subtract a fifth of the surviving hop's own angle and the worst row is four per cent. A period of 8, with the hole's own chain at the top. Each mark is one residue class of the displacement profile, placed round a ring at its own residue, with the chain the removed organ sat on at the top. The radius is how far that class sits from the level the rest of them share. six of the eight classes sit together at the middle ring; two do not, and on this row they are one pair, equal and opposite to within a twentieth. The forward one is chain 7 and the backward one is chain 0, one residue above it, which is the order every row of the census puts them in. One way round, seventeen times The two chains that change places in a wrecked stem are adjacent, which is symmetric and says nothing about direction. Label them by lag from the hole and the one displaced forwards is always the lower of the two — on every row of the census, without an exception.

What links here

Computed from the collection, not written here: the essays that point at this one.

Shares its objects with

Essays that name at least two of the same things, and that neither author linked.

  • A window nobody aligned — both name ablation, census, claim testing, control, honest limits, measurement, negative result, resolution, rigid hop, summary statistic
  • The alternation is not a period — both name ablation, claim testing, control, honest limits, measurement, negative result, null model, resolution, rigid hop
  • The offsets that never change — both name ablation, census, claim testing, control, honest limits, measurement, negative result, resolution, rigid hop
  • The plateau was a prediction — both name ablation, claim testing, control, honest limits, measurement, negative result, prediction, rigid hop, summary statistic
  • Two regimes above a hole — both name ablation, claim testing, control, honest limits, measurement, negative result, resolution, rigid hop, summary statistic
  • A list that was a rounding — both name census, claim testing, divergence angle, honest limits, measurement, measurement error, negative result, resolution

Named objects

A flat tag is an object no other essay names yet.

AblationCensusClaim testingControlDivergence angleHonest limitsMeasurementMeasurement errorNegative resultNull modelPredictionResidualResolutionRigid hopSlipSummary statistic